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author: nice 2014-10-09 18:07:00 +0530
committer: nice 2014-10-09 18:07:00 +0530
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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: AC CIRCUITS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1,Page number: 274 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power factor and the average power dissipated.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi,cos,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vm=141.0              #Peak value of supply voltage(in Volts)\n",
+      "ang_freq=100.0*pi     #Angular frequency(in radians per second)        \n",
+      "R=3.0                 #Resistance(in Ohms) \n",
+      "L=0.0127              #Self-inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calcultaions:\n",
+      "Vrms=Vm/sqrt(2.0)\n",
+      "V=rect(Vrms,0.0)\n",
+      "X_L=ang_freq*L\n",
+      "Z=R+ (X_L*1j)\n",
+      "I=V/Z\n",
+      "ang_I=degrees(phase(I))\n",
+      "P=Vrms*abs(I)*cos(phase(I))\n",
+      "pf=cos(phase(I))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The rms value of the steady-state current is %.2f A and the relative phase angle is %.2f degrees.\" %(abs(I),-ang_I)\n",
+      "print \"(b)The expression for the instantaneous current is i=%.3f sin(100*pi*t%.2f degrees) A.\" %((sqrt(2)*abs(I)),ang_I)\n",
+      "print \"(c)The average power dissipated in the circuit is %.2f W.\" %(P)\n",
+      "print \"(d)The power factor is %.3f lagging.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The rms value of the steady-state current is 19.97 A and the relative phase angle is 53.06 degrees.\n",
+        "(b)The expression for the instantaneous current is i=28.246 sin(100*pi*t-53.06 degrees) A.\n",
+        "(c)The average power dissipated in the circuit is 1196.75 W.\n",
+        "(d)The power factor is 0.601 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2,Page number: 276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the capacitance required,the phase angle,the power factor,the apparent power,and the reactive power.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi,degrees,acos,cos,sin\n",
+      "\n",
+      "#Variable Declaration: \n",
+      "V=230.0               #Rms value of supply voltage(in Volts)\n",
+      "f=50.0                #Frequency of supply(in Hertz) \n",
+      "P_lamp=750.0          #Power rating of the lamp(in Watts) \n",
+      "V_lamp=100.0          #Voltage rating of the lamp(in Volts)   \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_rated=P_lamp/V_lamp\n",
+      "Vc=sqrt((V*V)-(V_lamp*V_lamp))\n",
+      "Xc=Vc/I_rated\n",
+      "C=1/(2*pi*f*Xc)\n",
+      "phase_angle=acos(V_lamp/V)\n",
+      "pf=cos(phase_angle)\n",
+      "app_P=V*I_rated\n",
+      "rea_P=V*I_rated*sin(phase_angle)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The required capacitance is %e F.\" %(C)\n",
+      "print \"(b)The phase angle is %.3f degrees.\" %(degrees(phase_angle))\n",
+      "print \"(c)The power factor is %.3f leading.\" %(pf)\n",
+      "print \"(d)The apparent power is %.3f VA.\" %(app_P)\n",
+      "print \"(e)The reactive power is %.3f VAr.\" %(rea_P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The required capacitance is 1.152611e-04 F.\n",
+        "(b)The phase angle is 64.229 degrees.\n",
+        "(c)The power factor is 0.435 leading.\n",
+        "(d)The apparent power is 1725.000 VA.\n",
+        "(e)The reactive power is 1553.424 VAr.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3,Page number: 277 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question: \n",
+      "\"\"\"Finding the impedance,the power factor,supply voltage,voltage across resistor,apparent power and reactive power.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import pi,cos,sin,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=rect(0.9,0)         #Current in the circuit(in Amperes)\n",
+      "R=120.0               #Resistance of the resistor(in Ohms) \n",
+      "Xc=250.0              #Reactance of the capacitor(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=R- Xc*1j\n",
+      "pf=cos(phase(Z))\n",
+      "V=I*Z\n",
+      "V_R=I*R\n",
+      "V_C=I*Xc\n",
+      "act_P=abs(V)*abs(I)*cos(phase(Z))\n",
+      "app_P=abs(V)*abs(I)\n",
+      "rea_P=abs(V)*abs(I)*sin(phase(Z))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The impedance is %.3f ohm at a phase angle of %.3f degrees.\" %(abs(Z),degrees(phase(Z)))\n",
+      "print \"The power factor is %.3f leading.\" %(pf)\n",
+      "print \"The supply voltage is %.3f V at a phase angle of %.3f degrees.\" %(abs(V),degrees(phase(V)))\n",
+      "print \"The voltage across resistor is %.3f V at a phase angle of %.3f degrees.\" %(abs(V_R),degrees(phase(V_R)))\n",
+      "print \"The voltage across capacitor is %.3f V at a phase angle of %.3f degrees.\" %(abs(V_C),-90)\n",
+      "print \"The apparent power is %.3f VA.\" %(app_P)\n",
+      "print \"The active power is %.3f W.\" %(act_P)\n",
+      "print \"The reactive power is %.3f VAr.\" %(rea_P)\n",
+      "print \"Note: Negative sign indicates the capacitor supplies reactive power.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The impedance is 277.308 ohm at a phase angle of -64.359 degrees.\n",
+        "The power factor is 0.433 leading.\n",
+        "The supply voltage is 249.578 V at a phase angle of -64.359 degrees.\n",
+        "The voltage across resistor is 108.000 V at a phase angle of 0.000 degrees.\n",
+        "The voltage across capacitor is 225.000 V at a phase angle of -90.000 degrees.\n",
+        "The apparent power is 224.620 VA.\n",
+        "The active power is 97.200 W.\n",
+        "The reactive power is -202.500 VAr.\n",
+        "Note: Negative sign indicates the capacitor supplies reactive power.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4,Page number: 279 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the impedance of the circuit and state whether it is inductive or capacitive.\"\"\"\n",
+      "\n",
+      "from cmath import phase,rect\n",
+      "from math import pi,degrees,radians,cos,sin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=160.0 +120.0*1j     #ac sinusoidal voltage(in Volts) \n",
+      "I=-4.0+10.0*1j        #Current in circuit(in Amperes) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=V/I\n",
+      "\n",
+      "if Z.imag>0:\n",
+      "            print(\"The nature of circuit is inductive\\n\")\n",
+      "            pf_type=\"lagging\"\n",
+      "\n",
+      "elif Z.imag<0:\n",
+      "              print(\"The nature of circuit is capacitive\\n\")\n",
+      "              pf_type=\"leading\"   \n",
+      "                                                \n",
+      "else:\n",
+      "      print(\"The nature of circuit is resistive\\n\")\n",
+      "\n",
+      "pf=cos(phase(Z))\n",
+      "act_P=abs(V)*abs(I)*pf\n",
+      "rea_P=abs(V)*abs(I)*sin(pi+phase(Z))\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The power factor is %.3f %s.\" %(pf,pf_type)\n",
+      "print \"The active power is %.2f W.\" %(act_P)\n",
+      "print \"The reactive power is %.2f VAr.\" %(rea_P)     "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nature of circuit is capacitive\n",
+        "\n",
+        "The power factor is 0.260 leading.\n",
+        "The active power is 560.00 W.\n",
+        "The reactive power is 2080.00 VAr.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5,Page number: 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the values of the two circuit elements.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50                  #Frequency of the source(in Hertz)\n",
+      "Z=10+10*1j            #Impedance of the circuit(in Ohm) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "ang_freq=2*pi*f\n",
+      "R=Z.real\n",
+      "X_L=Z.imag\n",
+      "L=X_L/ang_freq\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The nature of the impedance indicates that the circiut is inductive.\\n\"\n",
+      "print \"The values of the two elements are: \\n(a)Resistance(R)=%.2f ohm \\n(b)Inductance(C)=%e H.\" %(R,L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nature of the impedance indicates that the circiut is inductive.\n",
+        "\n",
+        "The values of the two elements are: \n",
+        "(a)Resistance(R)=10.00 ohm \n",
+        "(b)Inductance(C)=3.183099e-02 H.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6,Page number: 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the values of the two circuit elements.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50.0                #Frequency of the source(in Hertz)\n",
+      "Z=10.0-10.0*1j        #Impedance of the circuit(in Ohm) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "ang_freq=2.0*pi*f\n",
+      "Y=1.0/Z\n",
+      "R=1/Y.real\n",
+      "C=Y.imag/ang_freq\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The nature of the impedance indicates that the circiut is capacitive.\\n\"\n",
+      "print \"The values of the two elements are: \\n(a)Resistance(R)=%.2f ohm \\n(b)Capacitance(C)=%e F.\" %(R,C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nature of the impedance indicates that the circiut is capacitive.\n",
+        "\n",
+        "The values of the two elements are: \n",
+        "(a)Resistance(R)=20.00 ohm \n",
+        "(b)Capacitance(C)=1.591549e-04 F.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7,Page number: 281"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the impedance,the current,the phase angle,the voltage across each element and power factor.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import cos,pi,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=12.0                #Resistance of the resistor(in Ohms)\n",
+      "L=0.15                #Self-inductance of the inductor(in Henry)\n",
+      "C=100e-06             #Capacitance of the capacitor(in Farad)\n",
+      "f=50                  #Frequency of the source(in Hertz) \n",
+      "V=rect(100,0)         #Supply voltage(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "ang_freq=2*pi*f\n",
+      "X_L=ang_freq*L\n",
+      "X_C=1/(ang_freq*C)\n",
+      "Z=R + (X_L-X_C)*1j\n",
+      "\n",
+      "if Z.imag>0:\n",
+      "            pf_type=\"lagging\"\n",
+      "\n",
+      "elif Z.imag<0:\n",
+      "              pf_type=\"leading\"   \n",
+      "                                                \n",
+      "else:\n",
+      "      print(\"The nature of circuit is resistive\\n\")\n",
+      "\n",
+      "I=V/Z\n",
+      "V_R=abs(I)*R\n",
+      "V_C=abs(I)*X_C\n",
+      "V_L=abs(I)*X_L\n",
+      "pf=cos(phase(I))\n",
+      "app_P=abs(V)*abs(I)\n",
+      "avg_P=abs(V)*abs(I)*pf\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The impedance is %.3f ohm at a phase angle of %.3f degrees.\" %(abs(Z),degrees(phase(Z)))\n",
+      "print \"(b)The current is %.3f A at a phase angle of %.3f degrees.\" %(abs(I),degrees(phase(I)))\n",
+      "print \"(c)The phase angle is %.3f degrees.\" %(degrees(phase(I)))\n",
+      "print \"(d)The voltage across the resistor is %.2f V.\" %(V_R) \n",
+      "print \"   The voltage across the capacitor is %.2f V.\\n   The voltage across the inductor is %.2f V.\" %(V_C,V_L)\n",
+      "print \"(e)The power factor is %.3f %s.\" %(pf,pf_type)\n",
+      "print \"(f)The apparent power is %.3f VA.\" %(app_P)     \n",
+      "print \"(g)The average power is %.3f W.\" %(avg_P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The impedance is 19.439 ohm at a phase angle of 51.880 degrees.\n",
+        "(b)The current is 5.144 A at a phase angle of -51.880 degrees.\n",
+        "(c)The phase angle is -51.880 degrees.\n",
+        "(d)The voltage across the resistor is 61.73 V.\n",
+        "   The voltage across the capacitor is 163.75 V.\n",
+        "   The voltage across the inductor is 242.42 V.\n",
+        "(e)The power factor is 0.617 lagging.\n",
+        "(f)The apparent power is 514.431 VA.\n",
+        "(g)The average power is 317.567 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.8,Page number: 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across the capacitor by applying Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "from math import sqrt,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\"\"\" Note: All the impedances are expresssed in kilo ohm.\"\"\" \n",
+      "ang_freq=3000.0   #Angular frequency(in radians per second)\n",
+      "Vs_m=40.0         #Peak value of the supply voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vs_rms=Vs_m/sqrt(2.0)\n",
+      "Vs=rect(Vs_rms,0)\n",
+      "Zeq=1.5 + ((1.0-(2.0*1j))/(1j+1.0-2.0*1j))*1.0j\n",
+      "I=Vs/Zeq\n",
+      "Im=abs(I)*sqrt(2)\n",
+      "V_Th=Vs*(1j/(1.5+1j))\n",
+      "Z_Th=1+ ((1.5*1j)/(1.5+1j))\n",
+      "Vc=V_Th*((-2*1j)/(Z_Th-2*1j))\n",
+      "Z_L=Z_Th.real-Z_Th.imag*1j \n",
+      "C=1/(ang_freq*abs(Z_L.imag)*1000)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The expression for i(t) can be written as i(t)=%.1f sin(%dt%.2fdegrees) mA.\" %(Im,ang_freq,degrees(phase(I)))\n",
+      "print \"(b)The voltage across the capacitor by applying Thevenin's theorem is %.3f V at a phase angle of %.3f degrees.\" %(abs(Vc),degrees(phase(Vc)))\n",
+      "print \"(c)The values of the two elements of the load impedance that consumes maximum power are R=%e kilo Ohms and C=%e F.\" %(Z_L.real,C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The expression for i(t) can be written as i(t)=16.0 sin(3000t-36.87degrees) mA.\n",
+        "(b)The voltage across the capacitor by applying Thevenin's theorem is 16.000 V at a phase angle of 8.130 degrees.\n",
+        "(c)The values of the two elements of the load impedance that consumes maximum power are R=1.461538e+00 kilo Ohms and C=4.814815e-07 F.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.9,Page number: 286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of resistance for a desired power factor.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "from cmath import atan\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "pf=0.8                #Power factor of the circuit\n",
+      "X_C=60.0              #Capacitive reactance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "sin_ang=sqrt(1-(pf*pf))\n",
+      "tan_ang=sin_ang/pf\n",
+      "R=X_C/tan_ang\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of R for which the power factor of the circuit is 0.8 is %.2f Ohms.\" %(R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R for which the power factor of the circuit is 0.8 is 80.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.10,Page number: 286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance and the inductance of the coil.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vdc=20.0              #Supply dc voltage(in Volts)\n",
+      "Idc=4.0               #Current drawn by the coil(in Amperes)\n",
+      "f=50.0                #Frequency of supply voltage(in Hertz)\n",
+      "Vs=65.0               #Ac supply voltage(in Volts)\n",
+      "I=5.0                 #Current drawn by the choke when connected ac supply(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R=Vdc/Idc\n",
+      "Z=Vs/I\n",
+      "X_L=sqrt((Z*Z)-(R*R))\n",
+      "L=X_L/(2*pi*f)\n",
+      "pf=R/Z\n",
+      "P=Vs*I*pf\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resistance of the coil is %.2f Ohms and the inductance of the coil is %.5f H.\" %(R,L)\n",
+      "print \"(b)The power factor is %.3f lagging.\" %(pf)\n",
+      "print \"(c)The power(real) drawn by the coil is %.3f W.\" %(P)  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resistance of the coil is 5.00 Ohms and the inductance of the coil is 0.03820 H.\n",
+        "(b)The power factor is 0.385 lagging.\n",
+        "(c)The power(real) drawn by the coil is 125.000 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.11,Page number: 286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance and inductance of the coil.\"\"\"\n",
+      "\n",
+      "from math import pi,pow,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vs=240.0              #AC supply voltage(in Volts)\n",
+      "f1=50.0               #Frequency of the ac supply voltage(in Hertz) \n",
+      "I1=60.0               #First reading of the ammeter(in Amperes)\n",
+      "f2=100.0              #Frequency of the ac supply voltage(in Hertz)\n",
+      "I2=40.0               #Second reading of the ammeter(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z1=Vs/I1\n",
+      "Z2=Vs/I2\n",
+      "L=sqrt(((Z2*Z2)-(Z1*Z1))/((pow((200*pi),2.0))-(pow((100*pi),2.0))))\n",
+      "X1=2*pi*f1*L\n",
+      "R=sqrt((Z1*Z1)-(X1*X1))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance of the coil is %.2f Ohms.\" %(R)\n",
+      "print \"The inductance of the coil is %.5f H.\" %(L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance of the coil is 3.06 Ohms.\n",
+        "The inductance of the coil is 0.00822 H.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.12,Page number: 287 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power consumed by the choke coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=100.0               #Resistance of the resistor(in Ohms)\n",
+      "V_R=200.0             #Voltage across resistor(in Volts)\n",
+      "V_Ch=300.0            #Voltage across choke coil(in Volts)\n",
+      "Vs=440.0              #Supply voltage(in Volts)\n",
+      "\n",
+      " \n",
+      "#Calculations:\n",
+      "pf=((Vs*Vs)-(V_R*V_R)-(V_Ch*V_Ch))/(2*V_R*V_Ch)\n",
+      "I=V_R/R\n",
+      "P=V_Ch*I*pf\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The power consumed by the choke coil is %.2f W.\" %(P) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power consumed by the choke coil is 318.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.13,Page number: 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power dissipated in each coil.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "from cmath import phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=15.0               #Resistance of the first coil(in Ohms)\n",
+      "L1=0.2                #Inductance of the first coil(in Henry)\n",
+      "R2=25.0               #Resistance of the second coil(in Ohms)\n",
+      "L2=0.04               #Inductance of the second coil(in Henry)\n",
+      "V=230.0               #Supply voltage(in Volts)\n",
+      "f=50.0                #Frequency of supply voltage(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "X1=2*pi*f*L1\n",
+      "X2=2*pi*f*L2\n",
+      "Z1=R1+(1j*X1)\n",
+      "Z2=R2+(1j*X2)\n",
+      "Z=Z1+Z2\n",
+      "I=V/Z\n",
+      "V1=abs(I)*abs(Z1)\n",
+      "V2=abs(I)*abs(Z2)\n",
+      "P1=abs(I)*abs(I)*R1\n",
+      "P2=abs(I)*abs(I)*R2\n",
+      "pf=cos(phase(I))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The voltage across the first coil is %.2f V and the voltage across the second coil is %.2f V.\" %(V1,V2)\n",
+      "print \"(b)The power dissipated by the first coil is %.2f W and power dissipated by the second coil is %.2f W.\" %(P1,P2)\n",
+      "print \"(c)The power factor of the whole circuit is %.3f lagging.\" %(pf) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The voltage across the first coil is 174.07 V and the voltage across the second coil is 75.40 V.\n",
+        "(b)The power dissipated by the first coil is 108.92 W and power dissipated by the second coil is 181.54 W.\n",
+        "(c)The power factor of the whole circuit is 0.469 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.14,Page number: 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current and power drawn from the source.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I_A=8.0               #Current through coil A(in Amperes)\n",
+      "I_B=10.0              #Current through coil B(in Amperes)\n",
+      "V=100.0               #Voltage of the source(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "P_A=120.0             #Power delivered to coil A(in Watts)  \n",
+      "P_B=500.0             #Power delivered to coil B(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z_A=V/I_A\n",
+      "Z_B=V/I_B\n",
+      "R_A=P_A/(I_A*I_A)\n",
+      "R_B=P_B/(I_B*I_B)\n",
+      "X_A=sqrt((Z_A*Z_A)-(R_A*R_A))\n",
+      "X_B=sqrt((Z_B*Z_B)-(R_B*R_B))\n",
+      "R=R_A+R_B\n",
+      "X=X_A+X_B\n",
+      "Z=sqrt((R*R)+(X*X))\n",
+      "I=V/Z\n",
+      "P=I*I*R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current when the two coils are in series is %.2f A and the power taken from the source is %.2f W.\" %(I,P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current when the two coils are in series is 4.52 A and the power taken from the source is 140.58 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.15,Page number: 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage drop across each coil.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=240.0               #Voltage of the source(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "R_A=5.0               #Resistance of coil A(in Ohms)\n",
+      "L_B=0.015             #Inductance of the coil B(in Henry)\n",
+      "P=3e03                #Active power(in Watts)\n",
+      "Q=2e03                #Reactive power(in VAr)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "S=sqrt((P*P)+(Q*Q))\n",
+      "I=S/V\n",
+      "R_B=(P/(I*I))-R_A\n",
+      "X_B=2*pi*f*L_B\n",
+      "X_A=(Q/(I*I))-X_B\n",
+      "L_A=X_A/(2*pi*f)\n",
+      "Z_A=sqrt((R_A*R_A)+(X_A*X_A))\n",
+      "Z_B=sqrt((R_B*R_B)+(X_B*X_B))\n",
+      "V_A=Z_A*I\n",
+      "V_B=Z_B*I\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resistance of coil B is %.2f Ohms.\" %(R_B)\n",
+      "print \"(b)The inductance of coil A is %.5f Henry.\" %(L_A)\n",
+      "print \"(c)The voltage drop across coil A is %.2f V and across coil B is %.2f V.\" %(V_A,V_B)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resistance of coil B is 8.29 Ohms.\n",
+        "(b)The inductance of coil A is 0.01321 Henry.\n",
+        "(c)The voltage drop across coil A is 97.61 V and across coil B is 143.29 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.16,Page number: 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the total power consumed by the circuit and the power factor of the circuit.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=100.0               #Power rating of the bulb(in Watts)\n",
+      "V=120.0               #Voltage rating of the bulb(in Volts)\n",
+      "Vs=240.0              #Supply voltage(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I=P/V\n",
+      "V_R=Vs-V\n",
+      "R=V_R/I\n",
+      "pf_a=1.0\n",
+      "Pt_a=Vs*I\n",
+      "V_C=sqrt((Vs*Vs)-(V*V))\n",
+      "X_C=V_C/I\n",
+      "C=1/(2*pi*X_C*f)\n",
+      "pf_b=V/Vs\n",
+      "Pt_b=Vs*I*pf_b\n",
+      "Vr=I*10.0\n",
+      "V_L=sqrt((Vs*Vs)-((V+Vr)*(V+Vr)))\n",
+      "X_L=V_L/I\n",
+      "L=X_L/(2*pi*f)\n",
+      "V_R_c=V+Vr\n",
+      "pf_c=V_R_c/Vs\n",
+      "Pt_c=Vs*I*pf_c\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The value of the component used R=%.2f Ohms.\" %(R)\n",
+      "print \"   The total power consumed is %.2f W.\" %(Pt_a)\n",
+      "print \"   The power factor is %.2f.\" %(pf_a)\n",
+      "print \"(b)The value of the component used C=%e Farad.\" %(C)\n",
+      "print \"   The total power consumed is %.2f W.\" %(Pt_b)\n",
+      "print \"   The power factor is %.2f leading.\" %(pf_b)\n",
+      "print \"(c)The value of the component used R=10 Ohms and L=%.3f Henry.\" %(L)\n",
+      "print \"   The total power consumed is %.2f W.\" %(Pt_c)\n",
+      "print \"   The power factor is %.3f lagging.\" %(pf_c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The value of the component used R=144.00 Ohms.\n",
+        "   The total power consumed is 200.00 W.\n",
+        "   The power factor is 1.00.\n",
+        "(b)The value of the component used C=1.276224e-05 Farad.\n",
+        "   The total power consumed is 100.00 W.\n",
+        "   The power factor is 0.50 leading.\n",
+        "(c)The value of the component used R=10 Ohms and L=0.775 Henry.\n",
+        "   The total power consumed is 106.94 W.\n",
+        "   The power factor is 0.535 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.17,Page number: 289\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of capacitance C in the cirucit.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=20.0                #Resistance connected in series(in Ohms)\n",
+      "L=15e-03              #Pure inductance in parallel with the capacitor(in Henry)\n",
+      "ang_fre=1000.0        #Angular frequency of voltage source(in radians per second)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "X_L=ang_fre*L\n",
+      "X=R*tan(pi/4.0)\n",
+      "\"\"\"Case 1:(X is inductive)\"\"\"\n",
+      "X_C1=1.0/((1.0/X_L)-(1.0/X))\n",
+      "C1=1.0/(ang_fre*X_C1)\n",
+      "\"\"\"Case 2:(X is capacitive)\"\"\"\n",
+      "X_C2=1.0/((1.0/X_L)+(1.0/X))\n",
+      "C2=1.0/(ang_fre*X_C2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of capacitance is:\"\n",
+      "print \"Case 1: If net reactance X is inductive, C=%e Farad.\" %(C1) \n",
+      "print \"Case 2: If net reactance X is capacitive, C=%e Farad.\" %(C2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of capacitance is:\n",
+        "Case 1: If net reactance X is inductive, C=1.666667e-05 Farad.\n",
+        "Case 2: If net reactance X is capacitive, C=1.166667e-04 Farad.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.18,Page number: 290\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding total current in the circuit.\"\"\"\n",
+      "\n",
+      "from cmath import phase\n",
+      "from math import degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Z1=(12+1j*15)         #Impedance of the first branch(in Ohms)  \n",
+      "Z2=(8-1j*4)           #Impedance of the second branch(in Ohms)\n",
+      "V=(230+1j*0)          #Potential difference across the parallel combination(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=V/Z1\n",
+      "I2=V/Z2\n",
+      "I=I1+I2\n",
+      "P=abs(V)*abs(I)*cos(phase(I))\n",
+      "P1=abs(I1)*abs(I1)*(Z1.real)\n",
+      "P2=abs(I2)*abs(I2)*(Z2.real)\n",
+      "pf1=Z1.real/abs(Z1)\n",
+      "pf2=Z2.real/abs(Z2)\n",
+      "pf=cos(phase(I))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current supplied to branch 1 is %.2f A at a phase angle of %.2f degrees.\" %(abs(I1),degrees(phase(I1))) \n",
+      "print \"   The current supplied to branch 2 is %.2f A at a phase angle of %.2f degrees.\" %(abs(I2),degrees(phase(I2))) \n",
+      "print \"   The total current is %.2f A at a phase angle of %.2f degrees.\\n\" %(abs(I),degrees(phase(I)))\n",
+      "print \"(b)The power consumed by branch 1 is %.2f W.\" %(P1)\n",
+      "print \"   The power consumed by branch 2 is %.2f W.\" %(P2)\n",
+      "print \"   The total power consumed is %.2f W.\\n\" %(P)\n",
+      "print \"(c)The power factor of branch 1 is %.4f lagging.\" %(pf1)\n",
+      "print \"   The power factor of branch 2 is %.4f leading.\" %(pf2)\n",
+      "print \"   The overall power factor of the circuit is %.4f leading.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current supplied to branch 1 is 11.97 A at a phase angle of -51.34 degrees.\n",
+        "   The current supplied to branch 2 is 25.71 A at a phase angle of 26.57 degrees.\n",
+        "   The total current is 30.56 A at a phase angle of 4.04 degrees.\n",
+        "\n",
+        "(b)The power consumed by branch 1 is 1720.33 W.\n",
+        "   The power consumed by branch 2 is 5290.00 W.\n",
+        "   The total power consumed is 7010.33 W.\n",
+        "\n",
+        "(c)The power factor of branch 1 is 0.6247 lagging.\n",
+        "   The power factor of branch 2 is 0.8944 leading.\n",
+        "   The overall power factor of the circuit is 0.9975 leading.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.19,Page number: 290\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance and inductance of a coil.\"\"\"\n",
+      "\n",
+      "from math import acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=230.0               #Voltage of the supply(in Volts)\n",
+      "f=50.0                #Frequency of the supply ac voltage(in Hertz)\n",
+      "R=50.0                #Reistance in series with the coil(in Ohms) \n",
+      "V_coil=180.0          #Voltage across coil(in Volts)\n",
+      "V_R=130.0             #Voltage across resistance(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I=V_R/R\n",
+      "\"\"\"From parallelogram OACB,the angle theta is calculated.\"\"\"\n",
+      "cos_theta=((V*V)-(V_R*V_R)-(V_coil*V_coil))/(2*V_R*V_coil)\n",
+      "theta=acos(cos_theta)\n",
+      "V_L=V_coil*sin(theta)\n",
+      "V_r=V_coil*cos(theta)\n",
+      "L=V_L/(I*2*pi*f)\n",
+      "r=V_r/I\n",
+      "P=I*V_r\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resistance of the coil is %.2f Ohms and the inductance is %.2f H.\" %(r,L)\n",
+      "print \"(b)The power dissipated in the coil is %.2f W.\" %(P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resistance of the coil is 5.33 Ohms and the inductance is 0.22 H.\n",
+        "(b)The power dissipated in the coil is 36.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.20,Page number: 291\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the component values of the circuit.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt,radians,cos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vm=141.4              #Peak value of supply voltage(in Volts)  \n",
+      "Im=7.07               #Peak value of current in the series circuit(in Amperes)\n",
+      "ang_fre=2000.0        #Angular frequency of the ac signal(in radians per second) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=Vm/sqrt(2.0)\n",
+      "I=Im/sqrt(2.0)\n",
+      "Z=V/I\n",
+      "\"\"\" We know that V=I*Z; \n",
+      "                \n",
+      "    100=5*sqrt((R*R)+(X_C*X_C));\n",
+      "    \n",
+      "    (R*R)+(X_C*X_C)=400; \n",
+      "    \n",
+      "    From the triangle OAB,\n",
+      "    \n",
+      "    cos(36.87)=V_R/V;          \"\"\"\n",
+      "R=(V*cos(radians(36.87)))/I\n",
+      "X_C=sqrt((Z*Z)-(R*R))\n",
+      "C=1.0/(ang_fre*X_C)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The elements of the circuit are R=%.2f Ohms and C=%e Farad.\" %(R,C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The elements of the circuit are R=16.00 Ohms and C=4.166657e-05 Farad.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.21,Page number: 292\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage and current by Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import degrees,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vs=rect(100,radians(20))\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Using voltage divider rule,\"\"\"\n",
+      "Z2=((1j*20)*(15-1j*30))/((1j*20)+(15-1j*30))\n",
+      "Z=10+Z2\n",
+      "I=Vs/Z\n",
+      "V2_div=Vs*(Z2/(10+Z2))\n",
+      "V1_div=V2_div*((-1j*30)/(15-1j*30))\n",
+      "\"\"\"Using Thevenin's theorem,\"\"\"\n",
+      "V_Th=Vs*((15-1j*30)/(10+15-1j*30))\n",
+      "Z_Th=(10*(15-1j*30))/(10+(15-1j*30))\n",
+      "V2=V_Th*((1j*20)/(Z_Th+1j*20))\n",
+      "V_Th=Vs*((1j*20)/(10+1j*20))\n",
+      "Z_Th=15+(1.0/((1.0/10)+(1.0/(1j*20))))\n",
+      "V1=V_Th*((-1j*30)/(Z_Th-1j*30))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)By voltage divider rule,\"\n",
+      "print \"(i)The current I is %.2f A at an angle of %.2f degrees.\" %(abs(I),degrees(phase(I))) \n",
+      "print \"(ii)The voltage V1 is %.2f V at an angle of %.2f degrees.\" %(abs(V1_div),degrees(phase(V1_div)))\n",
+      "print \"(b)By applying Thevenin's theorem,\"\n",
+      "print \"(i)The voltage V1 is %.2f V at an angle of %.2f degrees.\" %(abs(V1),degrees(phase(V1)))\n",
+      "print \"(ii)The voltage V2 is %.2f V at an angle of %.2f degrees.\" %(abs(V2),degrees(phase(V2)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)By voltage divider rule,\n",
+        "(i)The current I is 2.32 A at an angle of -28.62 degrees.\n",
+        "(ii)The voltage V1 is 77.30 V at an angle of 5.07 degrees.\n",
+        "(b)By applying Thevenin's theorem,\n",
+        "(i)The voltage V1 is 77.30 V at an angle of 5.07 degrees.\n",
+        "(ii)The voltage V2 is 86.42 V at an angle of 31.63 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.22,Page number: 293\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current by Norton's theorem.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import degrees,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Is=rect(20,radians(45))  #Current supplied by current source(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Using current divider rule,\"\"\"\n",
+      "Z2=(1j*3)+1.0/((1.0/4)+(1.0/(-1j*5)))\n",
+      "I=Is*(2/(Z2+2))\n",
+      "I_R_div=I*((-1j*5)/(4-1j*5))\n",
+      "\"\"\"Using Norton's theorem,\"\"\"\n",
+      "I_N=Is*(2/(2+1j*3))\n",
+      "Z_N=1.0/((1.0/(2+1j*3))+(1.0/(-1j*5)))\n",
+      "I_R=I_N*(Z_N/(Z_N+4))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)By current divider rule,\"\n",
+      "print \"   The current I_R is %.2f A at an angle of %.2f degrees.\" %(abs(I_R_div),degrees(phase(I_R_div)))\n",
+      "print \"(a)By Norton's theorem,\"\n",
+      "print \"   The current I_R is %.2f A at an angle of %.2f degrees.\" %(abs(I_R),degrees(phase(I_R)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)By current divider rule,\n",
+        "   The current I_R is 6.85 A at an angle of -6.95 degrees.\n",
+        "(a)By Norton's theorem,\n",
+        "   The current I_R is 6.85 A at an angle of -6.95 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.23,Page number: 294\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage by Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import radians,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vs=rect(200,radians(30.0))\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z2=((40)*(20+1j*50))/(40+(20+1j*50))\n",
+      "V2=(Vs*Z2)/(Z2-1j*20)\n",
+      "V_div=V2*(20/(20+1j*50))\n",
+      "V_Th=Vs*(40/(40-1j*20))\n",
+      "Z_Th=(1j*50)+1.0/((1.0/(-1j*20))+(1.0/40))\n",
+      "V=(V_Th*20)/(Z_Th+20)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)V using voltage divider rule is %.2f V at an angle of %.2f degrees.\" %(abs(V_div),degrees(phase(V_div)))\n",
+      "print \"(b)The voltage V using Thevenin's theorem is %.2f V at an angle of %.2f degrees.\" %(abs(V),degrees(phase(V)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)V using voltage divider rule is 81.23 V at an angle of 6.04 degrees.\n",
+        "(b)The voltage V using Thevenin's theorem is 81.23 V at an angle of 6.04 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter11.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter11.ipynb
new file mode 100755
index 00000000..e9007ce3
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter11.ipynb
@@ -0,0 +1,929 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11: RESONANCE IN AC CIRCUITS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1,Page number: 315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resonant frequency,quality factor,voltage across each element in a series RLC circuit.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=12.0                #Resistance of resistor(in Ohms)\n",
+      "L=0.15                #Self-inductance of inductor(in Henry)\n",
+      "C=100e-06             #Capacitance of capacitor(in Farads) \n",
+      "V=100.0               #Voltage(rms) of ac source(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "resonant_freq=1/(2.0*pi*sqrt(L*C))\n",
+      "I_max=V/R\n",
+      "freq_c=(sqrt((1.0/(L*C))-(0.5*pow((R/L),2))))/(2.0*pi)\n",
+      "freq_l=1.0/(sqrt((L*C)-(0.5*pow((R*C),2)))*2*pi)\n",
+      "cap_rea=1.0/(2.0*pi*freq_l*C)\n",
+      "ind_rea=2.0*pi*round(freq_l,2)*L\n",
+      "Q=cap_rea/R\n",
+      "Vr=V\n",
+      "Vl=Q*V\n",
+      "Vc=Vl\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The Resonant frequency(at which the circuit current becomes maximum) is %.2f Hz.\" %(resonant_freq)\n",
+      "print \"(b)The maximum current supplied by the source is %.2f A.\" %(I_max)\n",
+      "print \"(c)The frequency at which voltage across the capacitor is maximum is %.2f Hz.\" %(freq_c)\n",
+      "print \"(d)The frequency at which voltage across the inducttor is maximum is %.2f Hz.\" %(freq_l)\n",
+      "print \"(e)The inductive reactance is %.2f Ohms.\" %(ind_rea)\n",
+      "print \"(f)The capacitive reactance is %.2f Ohms.\" %(cap_rea)\n",
+      "print \"(g)The quality factor of the circiut is %.2f.\" %(Q)\n",
+      "print \"(h)The voltage drop across resistor is %.2f V.\" %(Vr)\n",
+      "print \"   The voltage drop across inductor is %.2f V.\" %(Vl)\n",
+      "print \"   The voltage drop across capacitor is %.2f V.\" %(Vc) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The Resonant frequency(at which the circuit current becomes maximum) is 41.09 Hz.\n",
+        "(b)The maximum current supplied by the source is 8.33 A.\n",
+        "(c)The frequency at which voltage across the capacitor is maximum is 40.10 Hz.\n",
+        "(d)The frequency at which voltage across the inducttor is maximum is 42.12 Hz.\n",
+        "(e)The inductive reactance is 39.70 Ohms.\n",
+        "(f)The capacitive reactance is 37.79 Ohms.\n",
+        "(g)The quality factor of the circiut is 3.15.\n",
+        "(h)The voltage drop across resistor is 100.00 V.\n",
+        "   The voltage drop across inductor is 314.91 V.\n",
+        "   The voltage drop across capacitor is 314.91 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2,Page number: 316"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the capacitance value to give resonance in a series RLC circuit. \"\"\"\n",
+      "\n",
+      "from math import pi,sqrt,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "res_freq=50.0          #Resonant frequency(in Hertz)\n",
+      "L=0.5                  #Self-inductance of inductor(in Henry)\n",
+      "R=4.0                  #Resistance of resistor(in Ohms)\n",
+      "V=100.0                #Voltage of the supply(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C=1/(pow((2*pi*res_freq),2)*L)\n",
+      "I_max=V/R\n",
+      "V_L=I_max*(2*pi*res_freq*L)\n",
+      "V_C=V_L\n",
+      "Q=(2.0*pi*res_freq*L)/R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The capacitance to give resonance is %e F.\" %(C)\n",
+      "print \"(b)The voltage across the inductor is %.2f V.\" %(V_L)\n",
+      "print \"   The voltage across the capacitor is %.2f V.\" %(V_C)\n",
+      "print \"(c)The quality factor of the circuit is %.2f.\" %(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The capacitance to give resonance is 2.026424e-05 F.\n",
+        "(b)The voltage across the inductor is 3926.99 V.\n",
+        "   The voltage across the capacitor is 3926.99 V.\n",
+        "(c)The quality factor of the circuit is 39.27.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3,Page number: 317 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the inductance,the circuit current and the voltage across the capacitor under resonance.\"\"\"\n",
+      "\n",
+      "from math import pi,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "res_freq=175e03        #Resonant frequency(in Hertz) \n",
+      "V=0.85                 #Voltage applied(in Volts)\n",
+      "Q=50.0                 #Quality factor of the coil\n",
+      "C=320e-012             #Capacitance of the capacitor(in Farads)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L=1/(pow((2*pi*res_freq),2)*C)\n",
+      "ind_rea=2*pi*res_freq*L\n",
+      "R=ind_rea/Q\n",
+      "Io=V/R\n",
+      "Vc=Q*V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of inductance is %e H.\" %(L)\n",
+      "print \"The circuit current is %e A.\" %(Io)\n",
+      "print \"The voltage across the capacitor under resonance is %.2f V.\" %(Vc)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of inductance is 2.584724e-03 H.\n",
+        "The circuit current is 1.495398e-02 A.\n",
+        "The voltage across the capacitor under resonance is 42.50 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4,Page number: 317"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current at the resonant frequency and the energy stored by inductor.\"\"\"\n",
+      "\n",
+      "from math import pow,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "res_freq=5e03          #Resonant frequency(in Hertz)\n",
+      "L=1e-03                #Self-inductance of the inductor(in Henry)\n",
+      "V=120.0                #Voltage of the supply(in Volts)\n",
+      "R=2.0                  #Resistance of the coil(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C=1/(pow((2*pi*res_freq),2)*L)\n",
+      "I_max=V/R\n",
+      "\"\"\" U=0.5*L*I*I=L*Irms*Irms\"\"\"\n",
+      "U=L*I_max*I_max\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The required value of capacitance is %e F.\" %(C)\n",
+      "print \"(a)The current at the resonance frequency is %.2f A.\" %(I_max)\n",
+      "print \"(b)The maximum instantaneous energy is %.2f J.\" %(U)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required value of capacitance is 1.013212e-06 F.\n",
+        "(a)The current at the resonance frequency is 60.00 A.\n",
+        "(b)The maximum instantaneous energy is 3.60 J.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5,Page number: 318"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resonance frequency and the quality factor for the overall circuit.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=0.51                #Resistor of the resistor-1(in Ohms)  \n",
+      "R2=1.3                 #Resistor of the resistor-2(in Ohms) \n",
+      "R3=0.24                #Resistor of the resistor-3(in Ohms)\n",
+      "L1=32e-03              #Self-inductance of the inductor-1(in Henry)\n",
+      "L2=15e-03              #Self-inductance of the inductor-2(in Henry)\n",
+      "C1=25e-06              #Capacitance of the capacitor-1(in Farads)\n",
+      "C2=62e-06              #Capacitance of the capacitor-2(in Farads)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Req=R1+R2+R3\n",
+      "Leq=L1+L2\n",
+      "Ceq=(C1*C2)/(C1+C2)\n",
+      "res_freq=1/(2*pi*sqrt(Leq*Ceq))\n",
+      "Q=(sqrt(Leq/Ceq))/Req\n",
+      "Q1=(2*pi*res_freq*L1)/R1\n",
+      "Q2=(2*pi*res_freq*L2)/R2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resonance frequency is %.2f Hz.\" %(res_freq)\n",
+      "print \"(b)The quality factor of the overall circuit is %.2f.\" %(Q)\n",
+      "print \"(c)The quality factor of coil-1 is %.2f.\" %(Q1)\n",
+      "print \"(d)The quality factor of coil-2 is %.2f.\" %(Q2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resonance frequency is 173.93 Hz.\n",
+        "(b)The quality factor of the overall circuit is 25.05.\n",
+        "(c)The quality factor of coil-1 is 68.57.\n",
+        "(d)The quality factor of coil-2 is 12.61.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6,Page number: 320"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the half-power frequencies of a series ac circuit.\"\"\"\n",
+      "\n",
+      "from math import pow,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "bandwidth=75e03        #Bandwidth of the resonant circuit(in Hertz) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "pro=pow((150e03),2)\n",
+      "sum=sqrt(pow(bandwidth,2)+(4*pro))\n",
+      "f2=(sum+bandwidth)/2.0\n",
+      "f1=(sum-bandwidth)/2.0\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"Lower Half-power frequency is %e Hz.\" %(f1)\n",
+      "print \"Upper Half-power frequency is %e Hz.\" %(f2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Lower Half-power frequency is 1.171165e+05 Hz.\n",
+        "Upper Half-power frequency is 1.921165e+05 Hz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7,Page number: 324"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line current,quality factor and the dynamic impedance of a series-parallel ac circuit.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=200e-06              #Self-inductance of the inductor coil(in Henry)  \n",
+      "res_freq=1e06          #Resonant frequency(in Hertz)\n",
+      "R=20.0                 #Resistance of the coil(in Ohms)\n",
+      "Rs=8e03                #Series resistance(in Ohms)\n",
+      "V=230.0                #Voltage(rms) of the supply(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C=1/(pow((2*pi*res_freq),2)*L)\n",
+      "XL=2*pi*res_freq*L\n",
+      "Q=XL/R\n",
+      "Zo=L/(C*R)\n",
+      "Z=Zo+Rs\n",
+      "I=V/Z\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The value of capacitance to cause resonance is %e F.\" %(C)\n",
+      "print \"(b)The Q factor of the circuit is %.5f.\" %(Q)\n",
+      "print \"(c)The dynamic impedance of the parallel resonant circuit is %.2f Ohms.\" %(Zo)\n",
+      "print \"(d)The total line current is %e A.\" %(I) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The value of capacitance to cause resonance is 1.266515e-10 F.\n",
+        "(b)The Q factor of the circuit is 62.83185.\n",
+        "(c)The dynamic impedance of the parallel resonant circuit is 78956.84 Ohms.\n",
+        "(d)The total line current is 2.644990e-03 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8,Page number: 325"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resonant frequency,Q-factor and bandwidth of a practical parallel resonant circuit.\"\"\"\n",
+      "\n",
+      "from math import pow,sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=150.0                #Resistance of the coil(in Ohms)\n",
+      "L=0.24                 #Self-inductance of the coil(in Henry)\n",
+      "C=3e-06                #Capacitance of the capacitor(in Farads)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "res_freq=(sqrt(1-((R*R*C)/L)))/(2*pi*sqrt(L*C))\n",
+      "Q=(2*pi*res_freq*L)/R\n",
+      "BW=res_freq/Q\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resonant frequency is %.2f Hz.\" %(res_freq)\n",
+      "print \"The quality factor is %.2f.\" %(Q)\n",
+      "print \"The bandwidth is %.2f Hz.\" %(BW)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resonant frequency is 159.02 Hz.\n",
+        "The quality factor is 1.60.\n",
+        "The bandwidth is 99.47 Hz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9,Page number: 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the source frequency and the current supplied by the source.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi,pow,degrees\n",
+      "from cmath import phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=125.0               #Voltage of the source(in Volts)\n",
+      "C=20.5e-06            #Capacitance of the capacitor(in Farads)\n",
+      "R=1.06                #Resistance of the coil(in Ohms)\n",
+      "L=25.4e-03            #Inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "fo=1.0/(2*pi*sqrt(L*C))\n",
+      "Io=V/R\n",
+      "V_L=Io*(2*pi*fo*L)\n",
+      "V_C=V_L\n",
+      "X_L=(2*pi*fo*L)\n",
+      "Z_coil=R+(1j*X_L)\n",
+      "V_coil=Io*Z_coil\n",
+      "I=300.0/X_L\n",
+      "R_new=V/I\n",
+      "Rx=R_new-R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a) (i)The source frequency is %.2f Hz, and\\n    (ii)The current supplied by the source is %.2f A.\\n\" %(fo,Io)\n",
+      "print \"(b) (i)The voltage across the capacitor is %.2f V and\" %(V_C)\n",
+      "print \"    (ii)The voltage across the coil is %.2f V at an angle of %.2f degrees.\\n\" %(abs(V_coil),degrees(phase(V_coil)))\n",
+      "print \"(c)The resistance that must be connected in series with the circuit to limit the capacitor voltage to 300V is %.3f Ohms.\" %(Rx)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) (i)The source frequency is 220.56 Hz, and\n",
+        "    (ii)The current supplied by the source is 117.92 A.\n",
+        "\n",
+        "(b) (i)The voltage across the capacitor is 4150.92 V and\n",
+        "    (ii)The voltage across the coil is 4152.80 V at an angle of 88.28 degrees.\n",
+        "\n",
+        "(c)The resistance that must be connected in series with the circuit to limit the capacitor voltage to 300V is 13.607 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.10,Page number: 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the maximum instantaneous energy stored in the inductor.\"\"\"\n",
+      "\n",
+      "from math import pow,pi,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=3.0                 #Resistance of the coil(in Ohms)\n",
+      "L=12e-03              #Self-inductance of the coil(in Henry)\n",
+      "fo=9e03               #Resonant frequency(in Hertz)\n",
+      "V=240.0               #Supply voltage(in Volts)   \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C=1.0/(pow((2*pi*fo),2)*L)\n",
+      "Io=V/R\n",
+      "ener=0.5*L*Io*Io\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of capacitance to be connected in series with the coil is %e F.\" %(C)\n",
+      "print \"The maximum instantaneous energy stored in the inductor is %.2f J.\" %(ener)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of capacitance to be connected in series with the coil is 2.605998e-08 F.\n",
+        "The maximum instantaneous energy stored in the inductor is 38.40 J.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.11,Page number: 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the parameters of a series RLC circuit.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "fo=10e03              #Resonant frequency(in Hertz)\n",
+      "BW=1e03               #Bandwidth(in HErtz)\n",
+      "P=15.3                #Power drawn(in Watts)\n",
+      "V=200.0               #Voltage of generator(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_R=V\n",
+      "R=(V_R*V_R)/P\n",
+      "\"\"\" Q=fo/BW=(2*pi*fo*L)/R; Q=Quality factor of the circuit. \"\"\"\n",
+      "L=R/(2*pi*BW)\n",
+      "C=1.0/(pow((2*pi*fo),2)*L)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The parameters of the circuit are:\\n R=%.2f Ohms,\\n L=%.3f H,\\n C=%e F.\" %(R,L,C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The parameters of the circuit are:\n",
+        " R=2614.38 Ohms,\n",
+        " L=0.416 H,\n",
+        " C=6.087677e-10 F.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.12,Page number: 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the half-power frequencies and the circuit current.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "fo=200.0              #Resonant frequency(in Hertz)\n",
+      "V=400.0               #Voltage of the source(in Volts)\n",
+      "R=20e-03              #Resistance of the coil(in Ohms)\n",
+      "L=6e-03               #Inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C=1.0/(pow((2*pi*fo),2)*L)\n",
+      "Io=V/R\n",
+      "X_C=1.0/(2*pi*fo*C)\n",
+      "V_C=Io*X_C\n",
+      "Im=sqrt(2)*Io\n",
+      "U_max=0.5*L*Im*Im\n",
+      "Q=(2*pi*fo*L)/R\n",
+      "BW=fo/Q\n",
+      "f1=fo-(BW/2.0)\n",
+      "f2=fo+(BW/2.0)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The capacitance of the capacitor is %e F.\" %(C)\n",
+      "print \"(b)The circuit current is %.2f kA.\" %(Io/1000)\n",
+      "print \"(c)The voltage across the capacitor is %.2f kV.\" %(V_C/1000)\n",
+      "print \"(d)The maximum energy stored in the coil is %.2f MJ.\" %(U_max/1000000)\n",
+      "print \"(e)The lower half-power frequency is %.3f Hz and the upper half-power frequency is %.3f Hz.\" %(f1,f2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The capacitance of the capacitor is 1.055429e-04 F.\n",
+        "(b)The circuit current is 20.00 kA.\n",
+        "(c)The voltage across the capacitor is 150.80 kV.\n",
+        "(d)The maximum energy stored in the coil is 2.40 MJ.\n",
+        "(e)The lower half-power frequency is 199.735 Hz and the upper half-power frequency is 200.265 Hz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.13,Page number: 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the bandwidth,resonant frequency,inductance and capacitance.\"\"\"\n",
+      "\n",
+      "from math import pi,pow,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=1e03                #Resistance of the resistor(in Ohms)\n",
+      "f1=20e03              #Lower half-power frequency(in Hertz) \n",
+      "f2=100e03             #Upper half-power frequency(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "BW=f2-f1\n",
+      "res_freq=sqrt(f1*f2)\n",
+      "Q=res_freq/BW\n",
+      "L=(Q*R)/(2*pi*res_freq)\n",
+      "C=1.0/(pow((2*pi*res_freq),2)*L)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The bandwidth is %.2f kHz.\" %(BW/1000.0)\n",
+      "print \"(b)The resonant frequency is %.2f kHz.\" %(res_freq/1000.0)\n",
+      "print \"(c)The inductance is %e H.\" %(L)\n",
+      "print \"(d)The capacitance is %e F.\" %(C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The bandwidth is 80.00 kHz.\n",
+        "(b)The resonant frequency is 44.72 kHz.\n",
+        "(c)The inductance is 1.989437e-03 H.\n",
+        "(d)The capacitance is 6.366198e-09 F.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.14,Page number: 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power at half-power frequencies.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=5.0                 #Resistance of resistor(in Ohms)\n",
+      "V=20.0                #Voltage of the source(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Zo=R\n",
+      "Io=V/Zo\n",
+      "Po=(Io*Io)*R\n",
+      "P_half=Po/2.0\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The power at half-power frequencies is %.2f W.\" %(P_half) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power at half-power frequencies is 40.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.15,Page number: 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the half-power frequencies and the quality factor.\"\"\"\n",
+      "\n",
+      "from math import pi,pow,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "res_freq=100.0        #Resonant frequency(in Hertz)\n",
+      "V=240.0               #Voltage of the source(in Volts)\n",
+      "R=55e-03              #Resistance of the coil(in Ohms)\n",
+      "L=7e-03               #Self-inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C=1.0/(pow((2*pi*res_freq),2)*L)\n",
+      "Q=(2*pi*res_freq*L)/R\n",
+      "BW=res_freq/Q\n",
+      "f1=res_freq-(BW/2.0)\n",
+      "f2=res_freq+(BW/2.0)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The value of the capacitance is %e F.\" %(C)\n",
+      "print \"(b)The quality factor of the circuit is %.2f.\" %(Q)\n",
+      "print \"(c)The lower half-power frequency is %.2f Hz and The upper half-power frequency is %.2f Hz.\" %(f1,f2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The value of the capacitance is 3.618614e-04 F.\n",
+        "(b)The quality factor of the circuit is 79.97.\n",
+        "(c)The lower half-power frequency is 99.37 Hz and The upper half-power frequency is 100.63 Hz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.16,Page number: 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resonance frequency and the effective resistance at resonance.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=20.0                #Resistance of the coil(in Ohms)\n",
+      "L=0.2                 #Inductance of the coil(in Henry)\n",
+      "C=100e-06             #Capacitance of the capacitor(in Farads)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "res_freq=sqrt(1-((R*R*C)/L))/(2*pi*sqrt(L*C))\n",
+      "Zo=L/(C*R)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The frequency at which the circuit behaves as a non-inductive reactance is %.2f Hz.\" %(res_freq)\n",
+      "print \"The effective resistance at resonance is %.2f Ohms.\" %(Zo) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The frequency at which the circuit behaves as a non-inductive reactance is 31.83 Hz.\n",
+        "The effective resistance at resonance is 100.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.17,Page number: 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the quality factor at the upper tuning frequency.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pow,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=20e-06              #Self-inductance of the coil(in Henry)\n",
+      "fo_1=570e03           #Lower tuning frequency(in Hertz) \n",
+      "fo_2=1560e03          #Upper tuning frequency(in Hertz)\n",
+      "Q1=50.0               #Quality factor at the lower tuning frequency\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C1=1.0/(pow((2*pi*fo_1),2)*L)\n",
+      "C2=1.0/(pow((2*pi*fo_2),2)*L)\n",
+      "R=(2*pi*fo_1*L)/Q1\n",
+      "BW=fo_1/Q1\n",
+      "Q2=(2*pi*fo_2*L)/R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The range of tuning capacitor is from %.3f nF to %.3f nF.\" %((C2*1e09),(C1*1e09))\n",
+      "print \"(b)The resistance of the coil is %.3f Ohms and the bandwidth of the circuit is %.3f kHz.\" %(R,(BW/1000))\n",
+      "print \"(c)The quality factor of the circuit at the upper tuning frequency is %.3f.\" %(Q2) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The range of tuning capacitor is from 0.520 nF to 3.898 nF.\n",
+        "(b)The resistance of the coil is 1.433 Ohms and the bandwidth of the circuit is 11.400 kHz.\n",
+        "(c)The quality factor of the circuit at the upper tuning frequency is 136.842.\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter12.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter12.ipynb
new file mode 100755
index 00000000..eb96a9e1
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter12.ipynb
@@ -0,0 +1,1251 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: THREE-PHASE CIRCUITS AND SYSTEMS "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1,Page number: 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current drawn from the power mains by a balanced three-phase load.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=32                  #Resistance of the load(in Ohms)\n",
+      "X_L=24                #Inductive reactance of the load(in Ohms) \n",
+      "V_L=400               #Line Voltage(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=R + X_L *1j\n",
+      "Z_mod=abs(Z)\n",
+      "V_ph_star=V_L/(sqrt(3))\n",
+      "I_ph_star=V_ph_star/Z_mod\n",
+      "I_L_star=I_ph_star\n",
+      "V_ph_delta=V_L\n",
+      "I_ph_delta=V_ph_delta/Z_mod\n",
+      "I_L_delta=I_ph_delta*(sqrt(3))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)For star connection:\"\n",
+      "print \"The current drawn from the power mains is %.3f A.\" %(I_L_star)\n",
+      "print \"(b)For delta connection:\"\n",
+      "print \"The current drawn from the power mains is %.3f A.\" %(I_L_delta)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)For star connection:\n",
+        "The current drawn from the power mains is 5.774 A.\n",
+        "(b)For delta connection:\n",
+        "The current drawn from the power mains is 17.321 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2,Page number: 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in each line and the current in the nuetral conductor for a star-connected three-phase system.\"\"\"\n",
+      "\n",
+      "from math import sqrt,cos,sin,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=415               #Line voltage(in Volts)\n",
+      "P_R=10e03             #Load in Red line(in kilo-Watts)\n",
+      "P_Y=8e03              #Load in Yellow line(in kilo-Watts)\n",
+      "P_B=5e03              #Load in Blue line(in kilo-Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vph=V_L/sqrt(3)\n",
+      "I_R=P_R/Vph\n",
+      "I_Y=P_Y/Vph\n",
+      "I_B=P_B/Vph\n",
+      "I_H=(I_Y*cos(pi/6))-(I_B*cos(pi/6))\n",
+      "I_V=I_R-(I_Y*sin(pi/6))-(I_B*sin(pi/6))\n",
+      "I_N=sqrt((I_H*I_H)+(I_V*I_V))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current taken by the 10-kW load is %.2f A.\" %(I_R) \n",
+      "print \"   The current taken by the 8-kW load is %.2f A.\" %(I_Y) \n",
+      "print \"   The current taken by the 5-kW load is %.2f A.\" %(I_B)\n",
+      "print \"(b)The current in the nuetral conductor is %.2f A.\" %(I_N)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current taken by the 10-kW load is 41.74 A.\n",
+        "   The current taken by the 8-kW load is 33.39 A.\n",
+        "   The current taken by the 5-kW load is 20.87 A.\n",
+        "(b)The current in the nuetral conductor is 18.19 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3,Page number: 348"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the phase currents and the line currents.\"\"\"\n",
+      "\n",
+      "from math import cos,pi,atan,radians,degrees,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50                  #Frequency of the source(in Hertz)\n",
+      "V_L=415               #Line Voltage(in Volts)\n",
+      "R1=100                #Resistance of the first load(in Ohms)\n",
+      "R2=20.0               #Resistance of the second load(in Ohms)\n",
+      "L2=191e-03            #Self-inductance of the second load(in Henry)\n",
+      "R3=0.0                #Resistance of the third load(in Ohms)\n",
+      "C3=30e-06             #Capacitance of the third load(in Farads)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z1=R1\n",
+      "angle_1=0.0\n",
+      "X2=2*pi*f*L2\n",
+      "Z2=sqrt((R2*R2)+(X2*X2))\n",
+      "angle_2=atan(X2/R2)\n",
+      "Z3=1/(2*pi*f*C3)\n",
+      "angle_3=pi/2\n",
+      "Vph=V_L\n",
+      "I1=Vph/Z1\n",
+      "I2=Vph/Z2\n",
+      "I3=Vph/Z3\n",
+      "I_R=sqrt((I1*I1)+(I3*I3)+(2*I1*I3*cos(pi/6)))\n",
+      "angle_Y=radians(degrees(angle_2)-60)\n",
+      "I_Y=sqrt((I1*I1)+(I2*I2)+(2*I1*I2*cos(angle_Y)))\n",
+      "angle_B=pi-angle_Y-(pi/6)\n",
+      "I_B=sqrt((I2*I2)+(I3*I3)+(2*I2*I3*cos(angle_B)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The phase current I1 in the load RY is %.2f A in phase with V_RY.\" %(I1)   \n",
+      "print \"   The phase current I2 in load YB is %.2f A lagging V_YB by %.2f degrees.\" %(I2,degrees(angle_2))   \n",
+      "print \"   The phase current I3 in load BR is %.2f A leading V_BR by %.2f degrees.\" %(I3,degrees(angle_3))\n",
+      "print \"(b)The line current I_R is %.2f A.\" %(I_R)\n",
+      "print \"   The line current I_Y is %.2f A.\" %(I_Y) \n",
+      "print \"   The line current I_B is %.2f A.\" %(I_B)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The phase current I1 in the load RY is 4.00 A in phase with V_RY.\n",
+        "   The phase current I2 in load YB is 6.56 A lagging V_YB by 71.57 degrees.\n",
+        "   The phase current I3 in load BR is 3.91 A leading V_BR by 90.00 degrees.\n",
+        "(b)The line current I_R is 7.64 A.\n",
+        "   The line current I_Y is 10.51 A.\n",
+        "   The line current I_B is 4.47 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4,Page number: 350"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line current,the power factor and the total power for a balanced three-phase system.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_ph=20.0             #Resistance of each phase(in Ohms)\n",
+      "X_L_ph=15.0           #Inductive reactance of each phase(in Ohms)\n",
+      "V_L=400.0             #Line Voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z_ph=R_ph + X_L_ph *1j\n",
+      "Z_mod=abs(Z_ph)\n",
+      "V_ph_star=V_L/(sqrt(3))\n",
+      "I_ph_star=V_ph_star/Z_mod\n",
+      "I_L_star=I_ph_star\n",
+      "pf_star=R_ph/Z_mod\n",
+      "P_active_star=sqrt(3)*V_L*I_L_star*pf_star\n",
+      "V_ph_delta=V_L\n",
+      "I_ph_delta=V_ph_delta/Z_mod\n",
+      "I_L_delta=I_ph_delta*(sqrt(3))\n",
+      "pf_delta=R_ph/Z_mod\n",
+      "P_active_delta=sqrt(3)*V_L*I_L_delta*pf_delta\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)For star connected load:\"\n",
+      "print \"(i)The line current is %.2f A.\" %(I_L_star)\n",
+      "print \"(ii)The power factor is %.2f lagging.\" %(pf_star)\n",
+      "print \"(iii)The total active power is %.2f kW.\" %(P_active_star/1000)\n",
+      "print \"\\n(b)For delta connection:\"\n",
+      "print \"(i)The line current is %.2f A.\" %(I_L_delta)\n",
+      "print \"(ii)The power factor is %.2f lagging.\" %(pf_delta)\n",
+      "print \"(iii)The total active power is %.2f kW.\" %(P_active_delta/1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)For star connected load:\n",
+        "(i)The line current is 9.24 A.\n",
+        "(ii)The power factor is 0.80 lagging.\n",
+        "(iii)The total active power is 5.12 kW.\n",
+        "\n",
+        "(b)For delta connection:\n",
+        "(i)The line current is 27.71 A.\n",
+        "(ii)The power factor is 0.80 lagging.\n",
+        "(iii)The total active power is 15.36 kW.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5,Page number: 355"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Fidning the total power consumed and the power factor of a balanced three-phase circuit.\"\"\"\n",
+      "\n",
+      "from math import atan,cos,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "W1=3e03               #Reading of wattmeter-1(in Watts)\n",
+      "W2=1.5e03             #Reading of wattmeter-2(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "P=W1+W2\n",
+      "pf_angle=atan(sqrt(3)*((W1-W2)/(W1+W2)))\n",
+      "pf=cos(pf_angle)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The total power consumed is %e W.\" %(P)\n",
+      "print \"The power factor of the balanced three-phase circuit is %.3f.\" %(pf)\n",
+      "print \"NOTE:From the given data it is impossible to state whether the power factor is leading or lagging.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total power consumed is 4.500000e+03 W.\n",
+        "The power factor of the balanced three-phase circuit is 0.866.\n",
+        "NOTE:From the given data it is impossible to state whether the power factor is leading or lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6,Page number: 355"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the total power,the power factor and the line current for a balanced three-phase circuit.\"\"\"\n",
+      "\n",
+      "from math import atan,cos,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=415               #Line voltage(in Volts)\n",
+      "W1=5.2e03             #Reading of wattmeter-1(in Watts)\n",
+      "W2=-1.7e03            #Reading of wattmeter-2(in Watts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "P=W1+W2\n",
+      "pf_angle=atan(sqrt(3)*((W1-W2)/(W1+W2)))\n",
+      "pf=cos(pf_angle)\n",
+      "I_L=P/(sqrt(3)*V_L*pf)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The total power consumed is %e W.\\n\" %(P)\n",
+      "print \"The power factor of the balanced three-phase circuit is %.3f.\" %(pf)\n",
+      "print \"NOTE:From the given data it is impossible to state whether the power factor is leading or lagging.\"\n",
+      "print \"\\nThe line current is %.2f A.\" %(I_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total power consumed is 3.500000e+03 W.\n",
+        "\n",
+        "The power factor of the balanced three-phase circuit is 0.281.\n",
+        "NOTE:From the given data it is impossible to state whether the power factor is leading or lagging.\n",
+        "\n",
+        "The line current is 17.32 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7,Page number: 356"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the total power consumed in a star-connected three-phase network.\"\"\"\n",
+      "\n",
+      "from math import sqrt,radians,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=6.0                 #Resistance per phase(in Ohms)\n",
+      "X_L=8.0               #Inductive reactance per phase(in Ohms)\n",
+      "V_L=220.0             #Line voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=6+ (1j*8)\n",
+      "Vph=V_L/sqrt(3.0)\n",
+      "V_RN=rect(Vph,0)\n",
+      "V_YN=rect(Vph,radians(-120.0))\n",
+      "V_BN=rect(Vph,radians(120.0))\n",
+      "V_RY=V_RN-V_YN\n",
+      "V_YB=V_YN-V_BN\n",
+      "V_BR=V_BN-V_RN\n",
+      "I_R=V_RN/Z\n",
+      "I_Y=V_YN/Z\n",
+      "I_B=V_BN/Z\n",
+      "P=sqrt(3.0)*V_L*abs(I_R)*cos(phase(Z))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The phase voltages are:\"\n",
+      "print \"   V_RN=%.2f V at an angle of %.3f degrees.\" %(abs(V_RN),degrees(phase(V_RN))) \n",
+      "print \"   V_YN=%.2f V at an angle of %.3f degrees.\" %(abs(V_YN),degrees(phase(V_YN)))\n",
+      "print \"   V_BN=%.2f V at an angle of %.3f degrees.\" %(abs(V_BN),degrees(phase(V_BN)))\n",
+      "print \"(b)The line voltages are:\"\n",
+      "print \"   V_RY=%.2f V at an angle of %.3f degrees.\" %(abs(V_RY),degrees(phase(V_RY))) \n",
+      "print \"   V_YB=%.2f V at an angle of %.3f degrees.\" %(abs(V_YB),degrees(phase(V_YB)))\n",
+      "print \"   V_BR=%.2f V at an angle of %.3f degrees.\" %(abs(V_BR),degrees(phase(V_BR)))\n",
+      "print \"(c)The line currents(same as phase currents) are:\"\n",
+      "print \"   I_R=%.2f A at an angle of %.3f degrees.\" %(abs(I_R),degrees(phase(I_R))) \n",
+      "print \"   I_Y=%.2f A at an angle of %.3f degrees.\" %(abs(I_Y),degrees(phase(I_Y)))\n",
+      "print \"   I_B=%.2f A at an angle of %.3f degrees.\" %(abs(I_B),degrees(phase(I_B)))\n",
+      "print \"(d)The total power consumed is %.2f W.\" %(round(P,2)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The phase voltages are:\n",
+        "   V_RN=127.02 V at an angle of 0.000 degrees.\n",
+        "   V_YN=127.02 V at an angle of -120.000 degrees.\n",
+        "   V_BN=127.02 V at an angle of 120.000 degrees.\n",
+        "(b)The line voltages are:\n",
+        "   V_RY=220.00 V at an angle of 30.000 degrees.\n",
+        "   V_YB=220.00 V at an angle of -90.000 degrees.\n",
+        "   V_BR=220.00 V at an angle of 150.000 degrees.\n",
+        "(c)The line currents(same as phase currents) are:\n",
+        "   I_R=12.70 A at an angle of -53.130 degrees.\n",
+        "   I_Y=12.70 A at an angle of -173.130 degrees.\n",
+        "   I_B=12.70 A at an angle of 66.870 degrees.\n",
+        "(d)The total power consumed is 2904.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8,Page number: 356"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the total power consumed in a delta-connected three-phase network.\"\"\"\n",
+      "\n",
+      "from math import sqrt,radians,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=6.0                 #Resistance per phase(in Ohms)\n",
+      "X_L=8.0               #Inductive reactance per phase(in Ohms)\n",
+      "V_L=220.0             #Line voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=6+ (1j*8)\n",
+      "Vph=V_L\n",
+      "V_RY=rect(Vph,0)\n",
+      "V_YB=rect(Vph,radians(-120.0))\n",
+      "V_BR=rect(Vph,radians(120.0))\n",
+      "I_RY=V_RY/Z\n",
+      "I_YB=V_YB/Z\n",
+      "I_BR=V_BR/Z\n",
+      "I_R=I_RY-I_BR\n",
+      "I_Y=I_YB-I_RY\n",
+      "I_B=I_BR-I_YB\n",
+      "P=sqrt(3.0)*V_L*abs(I_R)*cos(phase(Z))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The phase voltages(same as line voltages) are:\"\n",
+      "print \"   V_RY=%.2f V at an angle of %.3f degrees.\" %(abs(V_RY),degrees(phase(V_RY))) \n",
+      "print \"   V_YB=%.2f V at an angle of %.3f degrees.\" %(abs(V_YB),degrees(phase(V_YB)))\n",
+      "print \"   V_BR=%.2f V at an angle of %.3f degrees.\" %(abs(V_BR),degrees(phase(V_BR)))\n",
+      "print \"(b)The phase currents in the three load impedances are:\"\n",
+      "print \"   I_RY=%.2f A at an angle of %.3f degrees.\" %(abs(I_RY),degrees(phase(I_RY))) \n",
+      "print \"   I_YB=%.2f A at an angle of %.3f degrees.\" %(abs(I_YB),degrees(phase(I_YB)))\n",
+      "print \"   I_BR=%.2f A at an angle of %.3f degrees.\" %(abs(I_BR),degrees(phase(I_BR)))\n",
+      "print \"(c)The line currents are:\"\n",
+      "print \"   I_R=%.2f A at an angle of %.3f degrees.\" %(abs(I_R),degrees(phase(I_R))) \n",
+      "print \"   I_Y=%.2f A at an angle of %.3f degrees.\" %(abs(I_Y),degrees(phase(I_Y)))\n",
+      "print \"   I_B=%.2f A at an angle of %.3f degrees.\" %(abs(I_B),degrees(phase(I_B)))\n",
+      "print \"(d)The total power consumed is %.2f W.\" %(round(P,2)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The phase voltages(same as line voltages) are:\n",
+        "   V_RY=220.00 V at an angle of 0.000 degrees.\n",
+        "   V_YB=220.00 V at an angle of -120.000 degrees.\n",
+        "   V_BR=220.00 V at an angle of 120.000 degrees.\n",
+        "(b)The phase currents in the three load impedances are:\n",
+        "   I_RY=22.00 A at an angle of -53.130 degrees.\n",
+        "   I_YB=22.00 A at an angle of -173.130 degrees.\n",
+        "   I_BR=22.00 A at an angle of 66.870 degrees.\n",
+        "(c)The line currents are:\n",
+        "   I_R=38.11 A at an angle of -83.130 degrees.\n",
+        "   I_Y=38.11 A at an angle of 156.870 degrees.\n",
+        "   I_B=38.11 A at an angle of 36.870 degrees.\n",
+        "(d)The total power consumed is 8712.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9,Page number: 357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line current and the total power supplied by a three-phase system.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import radians,degrees,sqrt,cos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Z_A_delta=rect(12.0,radians(30.0))\n",
+      "Z_star=rect(5.0,radians(45.0))\n",
+      "V_L=400.0\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z_A_star=Z_A_delta/3.0\n",
+      "Zeq=(Z_A_star*Z_star)/(Z_A_star+Z_star)\n",
+      "Vph=V_L/sqrt(3.0)\n",
+      "I_L=Vph/Zeq\n",
+      "P=sqrt(3.0)*I_L*V_L*cos(phase(Zeq))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The line current is %.3f A.\" %(round((abs(I_L)),3))\n",
+      "print \"(b)The power suppiled is %e W.\" %(abs(P))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The line current is 103.045 A.\n",
+        "(b)The power suppiled is 5.726843e+04 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.10,Page number: 357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the constants of the load per phase.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I_L=100.0             #Line current(in Amperes)\n",
+      "V_L=1100.0            #Line voltage(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "P=150e03              #Power delivered by the three-phase system(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R=P/(3*I_L*I_L)\n",
+      "Vph=V_L/sqrt(3.0)\n",
+      "Iph=I_L\n",
+      "Z=Vph/Iph\n",
+      "X_C=sqrt((Z*Z)-(R*R))\n",
+      "C=1/(2*pi*f*X_C)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The constants of the load per phase are:\"\n",
+      "print \"R=%.3f Ohms and C=%e F.\" %(R,C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The constants of the load per phase are:\n",
+        "R=5.000 Ohms and C=8.128901e-04 F.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.11,Page number: 358"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the impedance in each branch and the power factor in a balanced delta-connected three-phase circuit.\"\"\"\n",
+      "\n",
+      "from math import sqrt,cos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=400.0             #Line voltage(in Volts)\n",
+      "I_L=20.0              #Line current(in Amperes)\n",
+      "P=10e03               #Total power absorbed by the load(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_ph=V_L\n",
+      "I_ph=I_L/sqrt(3.0)\n",
+      "Z_ph=V_ph/I_ph\n",
+      "pf=P/(sqrt(3.0)*V_L*I_L)\n",
+      "V_ph_star=V_L/sqrt(3.0)\n",
+      "I_L_star=V_ph_star/Z_ph\n",
+      "I_ph_star=I_L_star\n",
+      "P=sqrt(3.0)*V_L*I_L_star*pf\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The impedance in each branch is %.2f Ohms.\" %(round(Z_ph,2))\n",
+      "print \"(b)The power factor is %.4f lagging.\" %(pf)\n",
+      "print \"(c)The total power consumed if the same impedances are star-connected is %.2f kW.\" %(round((P/1000.0),2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The impedance in each branch is 34.64 Ohms.\n",
+        "(b)The power factor is 0.7217 lagging.\n",
+        "(c)The total power consumed if the same impedances are star-connected is 3.33 kW.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.12,Page number: 358"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the total power in the three-phase system.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import pi,sin,radians,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=100.0              #Resistance of the first phase(in Ohms)\n",
+      "R2=200.0              #Resistance of the second phase(in Ohms)\n",
+      "L3=0.3                #Inductance of the third phase(in Henry)\n",
+      "V_L=100.0             #Line voltage(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vab=rect(V_L,0)\n",
+      "Vbc=rect(V_L,radians(-120))\n",
+      "Vca=rect(V_L,radians(120))\n",
+      "Zab=R1\n",
+      "Zca=R2\n",
+      "Zbc=1j*(2*pi*f*L3)\n",
+      "Iab=Vab/Zab\n",
+      "Ibc=Vbc/Zbc\n",
+      "Ica=Vca/Zca\n",
+      "Pab=(abs(Vab)*abs(Vab))/R1\n",
+      "Pbc=0\n",
+      "Pca=(abs(Vca)*abs(Vca))/R2\n",
+      "act_P=Pab+Pbc+Pca\n",
+      "rea_P=abs(Vbc)*abs(Ibc)*sin(phase(Zbc))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The total active power in the system is %.2f W.\" %round(act_P,2)\n",
+      "print \"(b)The total reactive power in the system is %.2f VAr.\" %round(rea_P,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The total active power in the system is 150.00 W.\n",
+        "(b)The total reactive power in the system is 106.10 VAr.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.13,Page number: 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the load circuit parameters per phase.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt,sin,acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I_L=160.0             #Line current(in Amperes)\n",
+      "V_L=1.1e03            #Line voltage(in Volts)\n",
+      "P=210e03              #Total power load(in kilo-Watts)\n",
+      "f=50.0                #Frequency of the supply voltage(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "pf=P/(sqrt(3.0)*V_L*I_L)\n",
+      "V_ph=V_L/sqrt(3.0)\n",
+      "I_ph=I_L\n",
+      "Z_ph=V_ph/I_ph\n",
+      "R_ph=Z_ph*pf\n",
+      "X_C=Z_ph*sin(acos(pf))\n",
+      "C=1.0/(2*pi*f*X_C)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The load circuit parameters per phase are:\"\n",
+      "print \"R=%.3f Ohms.\" %(round(R_ph,3))\n",
+      "print \"C=%e F.\" %(C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The load circuit parameters per phase are:\n",
+        "R=2.734 Ohms.\n",
+        "C=1.106310e-03 F.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.14,Page number: 359"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\"\"\"Finding the resistance and the inductance of the load per phase.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt,sin,acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=400.0             #Line voltage(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "Iph=25.0              #Phase current(in Amperes)\n",
+      "P=13.856e03           #Total active power absorbed by the load(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_L=Iph\n",
+      "pf=P/(sqrt(3.0)*V_L*I_L)\n",
+      "Vph=V_L/sqrt(3.0)\n",
+      "Zph=Vph/Iph\n",
+      "Rph=Zph*pf\n",
+      "Xph=Zph*sin(acos(pf))\n",
+      "L=Xph/(2*pi*f)\n",
+      "Q=3*Vph*Iph*sin(acos(pf))\n",
+      "S=3*Vph*Iph\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resistance of the load per phase is %.3f Ohms and the inductance of the load per phase is %e H.\" %(Rph,L)\n",
+      "print \"(b)The total reactive power is %e VAR.\" %(Q)\n",
+      "print \"(c)The total apparent power is %e VA.\" %(S)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resistance of the load per phase is 7.390 Ohms and the inductance of the load per phase is 1.764344e-02 H.\n",
+        "(b)The total reactive power is 1.039285e+04 VAR.\n",
+        "(c)The total apparent power is 1.732051e+04 VA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.15,Page number: 360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line current,the power factor and the total kVA.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi,cos\n",
+      "from cmath import phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Z1=100+(1j*0)         #First impedance(in Ohms)\n",
+      "C=32e-06              #Capacitance of the capacitor(in Farads)\n",
+      "V_L=415.0             #Line voltage(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z2=-(1j*(1/(2*pi*f*C)))\n",
+      "Zph=(Z1*Z2)/(Z1+Z2)\n",
+      "Vph=V_L/sqrt(3.0)\n",
+      "Iph=Vph/Zph\n",
+      "I_L=Iph\n",
+      "pf=cos(phase(Zph))\n",
+      "P=sqrt(3.0)*V_L*I_L*pf\n",
+      "kVA=sqrt(3.0)*V_L*I_L\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The line current is %.3f A.\" %(abs(I_L))\n",
+      "print \"(b)The power factor is %.4f leading.\" %(pf)\n",
+      "print \"(c)The power absorbed is %e W.\" %(abs(P))\n",
+      "print \"(d)The total kVA is %e kVA.\" %(abs(kVA))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The line current is 3.397 A.\n",
+        "(b)The power factor is 0.7052 leading.\n",
+        "(c)The power absorbed is 1.722250e+03 W.\n",
+        "(d)The total kVA is 2.442104e+03 kVA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.16,Page number: 360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line current,input and output power in a three-phase motor.\"\"\"\n",
+      "\n",
+      "from math import sqrt,atan,cos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "effi=0.86             #Efficiency of the motor\n",
+      "W1=255e03             #Reading of the first wattmeter(in Watts)\n",
+      "W2=85e03             #Reading of the second wattmeter(in Watts)\n",
+      "V_L=1.6e03            #Line voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "P=W1+W2\n",
+      "phi=atan(sqrt(3.0)*((W1-W2)/(W1+W2)))\n",
+      "pf=cos(phi)\n",
+      "I_L=P/(sqrt(3.0)*V_L*pf)\n",
+      "Po=P*effi\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The input power is %.2f kW.\" %(round((P/1000.0),2))\n",
+      "print \"(b)The power factor is %.3f lagging.\" %(round(pf,3))\n",
+      "print \"(c)The line current is %.2f A.\" %(round(I_L,2))\n",
+      "print \"(d)The output power is %.2f kW.\" %(round((Po/1000.0),2)) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The input power is 340.00 kW.\n",
+        "(b)The power factor is 0.756 lagging.\n",
+        "(c)The line current is 162.30 A.\n",
+        "(d)The output power is 292.40 kW.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.17,Page number: 360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the readings of the two wattmeters.\"\"\"\n",
+      "\n",
+      "from math import sqrt,acos,tan \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=25e03               #Total input power(in Watts)\n",
+      "pf=0.8                #Power factor\n",
+      "V_L=400.0             #Line voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"W1-W2=(1.0/sqrt(3.0))*(W1+W2)*tan(phi);\"\"\"\n",
+      "eq_1=P\n",
+      "eq_2=(1.0/sqrt(3.0))*P*tan(acos(pf))\n",
+      "W1=(eq_1+eq_2)/2.0\n",
+      "W2=P-W1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The readings of the wattmeters are:\" \n",
+      "print \"W1=%.4f kW.\" %(round((W1/1000.0),4))\n",
+      "print \"W2=%.4f kW.\" %(round((W2/1000.0),4))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The readings of the wattmeters are:\n",
+        "W1=17.9127 kW.\n",
+        "W2=7.0873 kW.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.18,Page number: 361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the total active power consumed by the load.\"\"\"\n",
+      "\n",
+      "from cmath import phase,rect\n",
+      "from math import radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_RY=rect(200,0)                 #Line Voltage V_RY(in Volts)\n",
+      "V_YB=rect(200,radians(-120))     #Line Voltage V_YB(in Volts)\n",
+      "V_BR=rect(200,radians(120))      #Line Voltage V_BR(in Volts)\n",
+      "Z1=rect(10,radians(60))          #Impedance of the first phase(in Ohms)   \n",
+      "Z2=rect(10,radians(0))           #Impedance of the second phase(in Ohms)\n",
+      "Z3=rect(10,radians(60))          #Impedance of the third phase(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=V_RY/Z1\n",
+      "I2=V_YB/Z2\n",
+      "I3=V_BR/Z3\n",
+      "IR=I1-I3\n",
+      "IB=I3-I2\n",
+      "W1=abs(V_RY)*abs(IR)*cos(phase(IR))\n",
+      "W2=-abs(V_YB)*abs(IB)*cos(phase(IB)-phase(V_YB))\n",
+      "P=W1+W2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The readings of the wattmeters are: W1=%.2f W and W2=%.2f W.\" %(W1,W2)\n",
+      "print \"(b)The total active power consumed by the load is %.2f W.\" %(P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The readings of the wattmeters are: W1=0.00 W and W2=8000.00 W.\n",
+        "(b)The total active power consumed by the load is 8000.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.19,Page number: 361"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the total active power consumed by the load.\"\"\"\n",
+      "\n",
+      "from cmath import phase,rect\n",
+      "from math import radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_RY=rect(200,0)                 #Line Voltage V_RY(in Volts)\n",
+      "V_YB=rect(200,radians(-120))     #Line Voltage V_YB(in Volts)\n",
+      "V_BR=rect(200,radians(120))      #Line Voltage V_BR(in Volts)\n",
+      "Z1=rect(10,radians(60))          #Impedance of the first phase(in Ohms)   \n",
+      "Z2=rect(10,radians(0))           #Impedance of the second phase(in Ohms)\n",
+      "Z3=rect(10,radians(60))          #Impedance of the third phase(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=V_RY/Z1\n",
+      "I2=V_YB/Z2\n",
+      "I3=V_BR/Z3\n",
+      "IR=I1-I3\n",
+      "IY=I2-I1\n",
+      "W1=-abs(V_BR)*abs(IR)*cos(phase(IR)-phase(V_BR))\n",
+      "W2=abs(V_YB)*abs(IY)*cos(phase(IY)-phase(V_YB))\n",
+      "P=W1+W2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The readings of the wattmeters are: W1=%.2f W and W2=%.2f W.\" %(W1,W2)\n",
+      "print \"(b)The total active power consumed by the load is %.2f W.\" %(P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The readings of the wattmeters are: W1=6000.00 W and W2=2000.00 W.\n",
+        "(b)The total active power consumed by the load is 8000.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.20,Page number: 362"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the reading of the wattmeter.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import radians,cos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_RY=rect(440.0,0)             #Line voltage in RY line(in Volts)\n",
+      "V_YB=rect(440.0,radians(-120)) #Line voltage in YB line(in Volts)\n",
+      "V_BR=rect(440.0,radians(120))  #Line voltage in BR line(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_RB=rect(440.0,radians(120-180)) \n",
+      "I1=V_RY/(60.0+(1j*45.0))\n",
+      "I2=V_RB/(-(1j*56.0))\n",
+      "I_AB=I1+I2\n",
+      "V_CD=V_YB\n",
+      "P=abs(I_AB)*abs(V_CD)*cos(phase(V_CD))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The reading of the wattmeter is %.4f W.\" %(round(P,4))\n",
+      "print \"Note: There is a calculation error in the textbook.\"\n",
+      "print \"I_AB=11.50 A at an angle of 2.035 degrees and not 11.05 A. Therefore P=-2531.1166 W.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The reading of the wattmeter is -2531.1166 W.\n",
+        "Note: There is a calculation error in the textbook.\n",
+        "I_AB=11.50 A at an angle of 2.035 degrees and not 11.05 A. Therefore P=-2531.1166 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.21,Page number: 363"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line and phase current in a motor.\"\"\"\n",
+      "\n",
+      "from cmath import phase,rect\n",
+      "from math import acos,sqrt,radians,cos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_RY=rect(440,0)       #Line voltage in RY line(in Volts)\n",
+      "Zph=3+1j*4             #Phase Impedance(in Ohms)\n",
+      "pf=0.8                 #Lagging power factor\n",
+      "P=75e03                #Total active power(in Watts)\n",
+      "V_L=440.0              #Line voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_L_mod=P/(sqrt(3.0)*V_L*pf)\n",
+      "phi=acos(pf)\n",
+      "I_L_delta=rect(I_L_mod,(radians(-30-degrees(phi))))\n",
+      "Iph=rect((I_L_mod/sqrt(3.0)),-phi)\n",
+      "Vph=rect((V_L/sqrt(3.0)),radians(-30))\n",
+      "I_L_star=Vph/Zph\n",
+      "I_L=I_L_delta+I_L_star\n",
+      "P1=sqrt(3.0)*abs(V_L)*abs(I_L_delta)*pf\n",
+      "P2=sqrt(3.0)*abs(V_L)*abs(I_L_star)*cos(phase(Zph))\n",
+      "P_tot=P1+P2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The line current in the motor is %.2f A at a phase angle of %.2f degrees.\" %(abs(I_L_delta),degrees(phase(I_L_delta)))\n",
+      "print \"   The phase current in the motor is %.2f A at a phase angle of %.2f degrees.\" %(abs(Iph),degrees(phase(Iph)))\n",
+      "print \"(b)The line current and phase current in the load is %.2f A at a phase angle of %.2f degrees.\" %(abs(I_L_star),degrees(phase(I_L_star)))\n",
+      "print \"(c)The total line current is %.2f A at a phase angle of %.2f degrees.\" %(abs(I_L),degrees(phase(I_L)))\n",
+      "print \"(d)The total power consumed is %.3f W.\" %(P_tot) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The line current in the motor is 123.01 A at a phase angle of -66.87 degrees.\n",
+        "   The phase current in the motor is 71.02 A at a phase angle of -36.87 degrees.\n",
+        "(b)The line current and phase current in the load is 50.81 A at a phase angle of -83.13 degrees.\n",
+        "(c)The total line current is 172.38 A at a phase angle of -71.60 degrees.\n",
+        "(d)The total power consumed is 98232.000 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter13.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter13.ipynb
new file mode 100755
index 00000000..163431b5
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter13.ipynb
@@ -0,0 +1,1498 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13: TRANSFORMERS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1,Page number: 374"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across the secondary of the transformer.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "E_p=6400              #Primary voltage(in Volts)  \n",
+      "f=50                  #Frequency of primary supply(in Hertz)\n",
+      "N1=480                #Number of turns in the primary of the transformer\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "flux_m=E_p/(4.44*f*N1)\n",
+      "N2=20.0\n",
+      "Es=4.44*f*N2*flux_m\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The peak value of the flux produced in the core is %.2f Wb.\" %(flux_m)\n",
+      "print \"(b)The voltage across the secondary winding if it has 20 turns is %.2f V.\" %(Es)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The peak value of the flux produced in the core is 0.06 Wb.\n",
+        "(b)The voltage across the secondary winding if it has 20 turns is 266.67 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2,Page number: 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the peak value of the flux density in the core.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50.0                #Operating frequency of the transformer(in Hertz)\n",
+      "N1=30.0               #Number of turns in the primary of transformer\n",
+      "N2=350.0              #Number of turns in the secondary of transformer\n",
+      "A=250e-04             #Cross-sectional area of the core(in square-metres)\n",
+      "E1=230.0              #Voltage of the supply(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "flux_m=E1/(4.44*f*N1)\n",
+      "B_m=flux_m/A\n",
+      "E2=E1*(N2/N1)\n",
+      "I2=100.0\n",
+      "I1=I2*(N2/N1)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The peak value of flux density in the core is %.2f T.\" %(B_m)\n",
+      "print \"(b)The voltage induced in the secondary winding is %e V.\" %(E2)\n",
+      "print \"(c)The primary current when the secondary current is 100 A is %e A.\" %(I1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The peak value of flux density in the core is 1.38 T.\n",
+        "(b)The voltage induced in the secondary winding is 2.683333e+03 V.\n",
+        "(c)The primary current when the secondary current is 100 A is 1.166667e+03 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3,Page number: 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the turns-ratio of the transformer.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Req=50.0               #Output resistance of the source(in Ohms) \n",
+      "R_L=800.0              #Load resistance(in Ohms)  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "K=sqrt(R_L/Req)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The turns-ratio of the transformer to be used for maximising the load power is %d.\" %(K)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The turns-ratio of the transformer to be used for maximising the load power is 4.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4,Page number: 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the load current in the ac circuit.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import degrees\n",
+      "\n",
+      "#Calculations:\n",
+      "V=rect(30,0)\n",
+      "Ip=V/(20+20*1j+(pow(2,2)*(2-10*1j)))\n",
+      "I_L=2.0*Ip\n",
+      "\n",
+      "#Result:\n",
+      "print \"The load current is %.3f A at a phase angle of %.3f degrees.\" %(abs(I_L),degrees(phase(I_L)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The load current is 1.744 A at a phase angle of 35.538 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5,Page number: 378 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the output of the transformer in kVA.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "B_m=1.1               #Maximum magnetic flux density(in Weber per square-metre)\n",
+      "A=150e-04             #Cross-sectional area of the core(in square-metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "flux_m=B_m*A\n",
+      "N2=66\n",
+      "f=50\n",
+      "Z_L=4.0\n",
+      "E2=4.44*N2*f*flux_m\n",
+      "V2=E2\n",
+      "I2=V2/Z_L\n",
+      "output=(I2*V2)/1000.0\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The output when connected to a load of 4 Ohms impedance is %.3f kVA.\" %(output)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output when connected to a load of 4 Ohms impedance is 14.612 kVA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.6,Page number: 378 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of turns in each winding of the transformer.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "A=9e-04               #Cross-sectional area of the core(in square-metre)       \n",
+      "E1=230.0              #Primary Voltage(in Volts)\n",
+      "E2=110.0              #Secondary Voltage(in Volts)\n",
+      "E3=6.0                #Tertiary Voltage(in Volts)\n",
+      "f=50.0                #Operating frequency of the transformer(in Hertz)\n",
+      "Bm=1.0                #Maximum magnetic flux density(in Tesla)  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "flux_m=Bm*A\n",
+      "N3_half=E3/(4.44*f*flux_m)\n",
+      "N3=2*N3_half\n",
+      "N1=N3_half*(E1/E3)\n",
+      "N2=N3_half*(E2/E3)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The total number of turns on the primary winding is %d turns.\" %(N1)\n",
+      "print \"The total number of turns on the secondary winding is %d turns.\" %(N2)\n",
+      "print \"The total number of turns on the tertiary winding is %d turns.\" %(N3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total number of turns on the primary winding is 1151 turns.\n",
+        "The total number of turns on the secondary winding is 550 turns.\n",
+        "The total number of turns on the tertiary winding is 60 turns.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.7,Page number: 380"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the no-load power factor of the transformer.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=350.0              #Input at no-load(in Volt-Amperes) \n",
+      "V1=230.0              #Primary voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Io=VA/V1\n",
+      "Pi=110.0\n",
+      "pf=Pi/(V1*Io)\n",
+      "Iw=Io*pf\n",
+      "Im=sqrt(pow(Io,2)-pow(Iw,2))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The loss component of no-load current is given as %.4f A.\" %(Iw)\n",
+      "print \"The magnetising component of no-load current is %.4f A.\" %(Im)\n",
+      "print \"The no-load power factor is %.4f.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The loss component of no-load current is given as 0.4783 A.\n",
+        "The magnetising component of no-load current is 1.4446 A.\n",
+        "The no-load power factor is 0.3143.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.8,Page number: 382"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the hysterisis and eddy-current losses of the transformer.\"\"\"\n",
+      "\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50.0                #Operating frequency of the transformer(in Hertz)\n",
+      "eq1=100.0             #Iron loss at 60 Hz(in Watts)\n",
+      "eq2=60.0              #Iron loss at 40 Hz(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" P_h=A*f ; P_e=B*f*f ; \n",
+      "    \n",
+      "    P_i=P_h+P_e=(A*f)+(B*f*f); \"\"\"\n",
+      "\n",
+      "A=((eq2*36)-(eq1*16))/((40*36)-(60*16))\n",
+      "B=((eq1*4)-(eq2*6))/((3600*4)-(1600*6))\n",
+      "P_h=A*f\n",
+      "P_e=B*f*f\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The Hysteresis loss at 50 Hz is %.2f W.\" %(P_h) \n",
+      "print \"The Eddy-current loss at 50 Hz is %.2f W.\" %(P_e) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Hysteresis loss at 50 Hz is 58.33 W.\n",
+        "The Eddy-current loss at 50 Hz is 20.83 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.9,Page number: 385"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the primary current snd the primary power factor.\"\"\"\n",
+      "\n",
+      "from math import acos,degrees,cos\n",
+      "from cmath import phase,rect\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V1=440.0              #Primary voltage(in Volts)\n",
+      "V2=110.0              #Secondary voltage(in Volts)\n",
+      "I_0_mod=5.0           #No-load current(in Amperes) \n",
+      "I2=120.0              #Secondary current(in Amperes) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "phi_0=acos(0.2)\n",
+      "phi_2=acos(0.8)\n",
+      "K=V2/V1\n",
+      "I1_load_mod=K*I2\n",
+      "I1_load=rect(I1_load_mod,-phi_2)\n",
+      "I0=rect(I_0_mod,-phi_0)\n",
+      "I1_total=I0+I1_load\n",
+      "pf_primary=cos(phase(I1_total))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The primary current is %.2f A at a phase angle of %.2f degrees.\" %(abs(I1_total),degrees(phase(I1_total)))\n",
+      "print \"The primary power factor is %.3f lagging.\" %(pf_primary)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The primary current is 33.90 A at a phase angle of -42.49 degrees.\n",
+        "The primary power factor is 0.737 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.10,Page number: 390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the equivalent resistance as referred to the primary and secondary.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=50e03              #Power rating of the transformer(in Volt-Amperes) \n",
+      "V1=4400.0             #Primary voltage(in Volts) \n",
+      "V2=220.0              #Secondary voltage(in Volts)\n",
+      "R1=3.45               #Primary resistance(in Ohms)\n",
+      "R2=0.009              #Secondary resistance(in Ohms)\n",
+      "X1=5.2                #Leakage reactance of primary(in Ohms)\n",
+      "X2=0.015              #Leakage reactance of secondary(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=VA/V1\n",
+      "I2=VA/V2\n",
+      "K=V2/V1\n",
+      "Re1=R1+(R2/(K*K))\n",
+      "Re2=(K*K*R1)+R2\n",
+      "Xe1=X1+(X2/(K*K))\n",
+      "Xe2=(K*K*X1)+X2\n",
+      "Ze1=sqrt((Re1*Re1)+(Xe1*Xe1))\n",
+      "Ze2=sqrt((Re2*Re2)+(Xe2*Xe2))\n",
+      "tot_copp_loss=(I1*I1*R1)+(I2*I2*R2)\n",
+      "tot_copp_eq_p=I1*I1*Re1\n",
+      "tot_copp_eq_s=I2*I2*Re2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The equivalent resistance as referred to the primary is %.2f Ohms.\" %(Re1)\n",
+      "print \"(b)The equivalent resistance as referred to the secondary is %.4f Ohms.\" %(Re2)\n",
+      "print \"(c)The equivalent reactance as referred to the primary is %.2f Ohms.\" %(Xe1)\n",
+      "print \"(d)The equivalent reactance as referred to the secondary is %.3f Ohms.\" %(Xe2)\n",
+      "print \"(e)The equivalent impedance as referred to the primary is %.2f Ohms.\" %(Ze1)\n",
+      "print \"(f)The equivalent impedance as referred to the secondary is %.4f Ohms.\" %(Ze2)\n",
+      "print \"(g)The total copper loss by using the individual resistances of the two windings is %.2f W.\" %(tot_copp_loss)\n",
+      "print \"   By considering equivalent resistances,\"\n",
+      "print \"   Total copper loss(referred to primary equivalent resistance)=%.2f W.\" %(tot_copp_eq_p)   \n",
+      "print \"   Total copper loss(referred to secondary equivalent  resistance)=%.2f W\" %(tot_copp_eq_s)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The equivalent resistance as referred to the primary is 7.05 Ohms.\n",
+        "(b)The equivalent resistance as referred to the secondary is 0.0176 Ohms.\n",
+        "(c)The equivalent reactance as referred to the primary is 11.20 Ohms.\n",
+        "(d)The equivalent reactance as referred to the secondary is 0.028 Ohms.\n",
+        "(e)The equivalent impedance as referred to the primary is 13.23 Ohms.\n",
+        "(f)The equivalent impedance as referred to the secondary is 0.0331 Ohms.\n",
+        "(g)The total copper loss by using the individual resistances of the two windings is 910.38 W.\n",
+        "   By considering equivalent resistances,\n",
+        "   Total copper loss(referred to primary equivalent resistance)=910.38 W.\n",
+        "   Total copper loss(referred to secondary equivalent  resistance)=910.38 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.11,Page number: 393"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the full-load regulation for different load power factors.\"\"\"\n",
+      "\n",
+      "from math import cos,acos,sin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=40e03              #Power rating of the transformer(in Volt-Amperes)   \n",
+      "V1=6600.0             #Primary voltage(in Volts) \n",
+      "V2=250.0              #Secondary voltage(in Volts)\n",
+      "R1=10.0               #Primary resistance(in Ohms)\n",
+      "R2=0.02               #Secondary resistance(in Ohms)\n",
+      "Xe1=35.0              #Equivalent leakage reactance as referred to the primary(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "K=V2/V1\n",
+      "I2=VA/V2\n",
+      "Re2=(K*K*R1)+R2\n",
+      "Xe2=(K*K*Xe1)\n",
+      "pf=1\n",
+      "phi=acos(pf)\n",
+      "per_reg_a=(((I2*Re2*pf)+(I2*Xe2*sin(phi)))/V2)*100\n",
+      "pf=0.8\n",
+      "phi=acos(pf)\n",
+      "per_reg_b=(((I2*Re2*pf)+(I2*Xe2*sin(phi)))/V2)*100\n",
+      "pf=0.8\n",
+      "phi=acos(pf)\n",
+      "per_reg_c=(((I2*Re2*pf)-(I2*Xe2*sin(phi)))/V2)*100\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)For unity power factor of the load:\"\n",
+      "print \"The full-load percentage regulation is %.3f percent.\" %(per_reg_a)\n",
+      "print \"\\n(b)For  power factor of the load=0.8 lagging:\"\n",
+      "print \"The full-load percentage regulation is %.3f percent.\" %(per_reg_b)\n",
+      "print \"\\n(c)For power factor of the load=0.8 leading:\"\n",
+      "print \"The full-load percentage regulation is %.3f percent.\" %(per_reg_c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)For unity power factor of the load:\n",
+        "The full-load percentage regulation is 2.198 percent.\n",
+        "\n",
+        "(b)For  power factor of the load=0.8 lagging:\n",
+        "The full-load percentage regulation is 3.687 percent.\n",
+        "\n",
+        "(c)For power factor of the load=0.8 leading:\n",
+        "The full-load percentage regulation is -0.170 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.12,Page number: 396"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of turns in each winding.\"\"\"\n",
+      "\n",
+      "from math import sqrt \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50.0                #Operating frequency of the transformer(in Hertz)\n",
+      "E1=5000.0             #Primary voltage at no-load(in Volts)\n",
+      "E2=250.0              #Secondary voltage at no-load(in Volts)\n",
+      "VA_full=150e03        #Power rating of the transformer(in Volt-Ampere)  \n",
+      "flux=0.06             #Maximum core flux(in Weber)\n",
+      "Pi=1500.0             #Core losses(in Watts)\n",
+      "Pc_FL=1800.0          #Full-load copper losses(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "N2=E2/(4.44*f*flux)\n",
+      "N1=(E1/E2)*round(N2,0)\n",
+      "pf=1\n",
+      "Po=0.5*VA_full*pf\n",
+      "Pc=0.5*0.5*Pc_FL\n",
+      "effi_b=(Po/(Po+Pi+Pc))*100\n",
+      "pf=0.8\n",
+      "Po=VA_full*pf\n",
+      "Pc=Pc_FL\n",
+      "effi_c=(Po/(Po+Pi+Pc))*100\n",
+      "x=sqrt(Pi/Pc_FL)\n",
+      "VA_load=VA_full*x\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The number of turns in the primary winding is %d.\" %(round(N1,0)) \n",
+      "print \"   The number of turns in the secondary winding is %d.\" %(round(N2,0))\n",
+      "print \"(b)The efficiency at half rated kVA and unity power factor is %.3f percent\" %(effi_b)\n",
+      "print \"(c)The efficiency at full load and 0.8 power factor lagging is %.3f percent.\" %(effi_c)\n",
+      "print \"(d)The kVA load for maximum efficiency is %d kVA.\" %(round((VA_load/1000),0))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The number of turns in the primary winding is 380.\n",
+        "   The number of turns in the secondary winding is 19.\n",
+        "(b)The efficiency at half rated kVA and unity power factor is 97.466 percent\n",
+        "(c)The efficiency at full load and 0.8 power factor lagging is 97.324 percent.\n",
+        "(d)The kVA load for maximum efficiency is 137 kVA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.13,Page number: 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the all-day efficiency of a distribution transformer.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "kVA_FL=200.0          #Power rating of transformer at full-load(in Volt-Ampere)            \n",
+      "Pc_FL=3.02            #Full-load copper losses(in kilo-Watts)  \n",
+      "Pi=1.6                #Iron-losses(in kilo-Watts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pi_24_hrs=24.0*Pi\n",
+      "\"\"\" For 80 kW load at unity power factor in 6 hrs:\"\"\"\n",
+      "Po_1=80\n",
+      "t1=6\n",
+      "pf_1=1\n",
+      "output_ene_1=Po_1*t1\n",
+      "kVA_1=Po_1/pf_1\n",
+      "Pc_1=pow((kVA_1/kVA_FL),2)*Pc_FL*t1\n",
+      "\"\"\" For 160 kW load at 0.8 power factor in 8 hrs:\"\"\"\n",
+      "Po_2=160\n",
+      "t2=8\n",
+      "pf_2=0.8\n",
+      "output_ene_2=Po_2*t2\n",
+      "kVA_2=Po_2/pf_2\n",
+      "Pc_2=pow((kVA_2/kVA_FL),2)*Pc_FL*t2\n",
+      "\"\"\" For no load period of 10 hrs:\"\"\"\n",
+      "Po_3=0\n",
+      "t3=10\n",
+      "output_ene_3=0\n",
+      "Pc_3=0\n",
+      "Po_total=output_ene_1+output_ene_2+output_ene_3\n",
+      "Pc_total=Pc_1+Pc_2+Pc_3\n",
+      "all_day_effi=Po_total/(Po_total+Pc_total+Pi_24_hrs)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The all day efficiency of the distribution transformer is %.3f percent.\" %(all_day_effi*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The all day efficiency of the distribution transformer is 96.414 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.14,Page number: 398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the apparent power rating of an autotransformer.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_rated=120.0          #Voltage rating of the transformer(in Volts)\n",
+      "VA=12e03               #Power rating of the transformer(in Volt-Ampere)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=VA/V_rated\n",
+      "I2=I1\n",
+      "input_app_pow=240*I1\n",
+      "output_app_pow=120*2*I1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"In auto-tranformer mode, the input apparent power is %d kVA and the output apparent power is %d kVA.\" %(round((input_app_pow/1000),0),round((output_app_pow/1000),0))\n",
+      "print \"Thus, the apparent power capacity of the 12-kVA transformer is doubled by the auto-transformer connection.\"\n",
+      "print \"In effect,half the apparent power is transformed and half is conducted directly to the secondary side.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In auto-tranformer mode, the input apparent power is 24 kVA and the output apparent power is 24 kVA.\n",
+        "Thus, the apparent power capacity of the 12-kVA transformer is doubled by the auto-transformer connection.\n",
+        "In effect,half the apparent power is transformed and half is conducted directly to the secondary side.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.15,Page number: 401"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the secondary line voltage on no load when the windings are connected(a)star/delta,(b)delta/star.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L1=3300.0           #Supply voltage(in Volts)\n",
+      "Np=840.0              #Number of turns in the primary\n",
+      "Ns=72.0               #Number of turns in the secondary\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vph_1a=V_L1/sqrt(3)\n",
+      "Vph_2a=Vph_1a*(Ns/Np)\n",
+      "V_L2a=Vph_2a\n",
+      "Vph_1b=V_L1\n",
+      "Vph_2b=Vph_1b*(Ns/Np)\n",
+      "V_L2b=Vph_2b*sqrt(3)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print(\"(a)For star/delta connection: \")\n",
+      "print \"The secondary line voltage on no load is %.2f V.\" %(V_L2a)\n",
+      "print(\"(b)For delta/star connection: \")\n",
+      "print \"The secondary line voltage on no load is %.2f V.\" %(V_L2b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)For star/delta connection: \n",
+        "The secondary line voltage on no load is 163.31 V.\n",
+        "(b)For delta/star connection: \n",
+        "The secondary line voltage on no load is 489.92 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.16,Page number: 403"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnetising current and the core-loss current in a single-phase transformer.\"\"\" \n",
+      " \n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=12e03              #Power rating of the transformer(in Volt-Ampere) \n",
+      "Vp=400.0              #Primary voltage(in Volts)\n",
+      "Vs=200.0              #Secondary voltage(in Volts)\n",
+      "Wo=120.0              #Power in open-circuit test(in Watts)   \n",
+      "V1=200.0              #Voltage in open-circuit test(in Volts)\n",
+      "I_0=1.3               #Current in open-circuit test(in Amperes)\n",
+      "Isc=30.0              #Current in short-circuit test(in Amperes)\n",
+      "Wsc=200.0             #Power in short-circuit test(in Watts)\n",
+      "Vsc=22.0              #Voltage in short-circuit test(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Iw=Wo/V1\n",
+      "Im=sqrt((I_0*I_0)-(Iw*Iw))\n",
+      "R_0=V1/Iw\n",
+      "X_0=V1/Im\n",
+      "K=Vs/Vp\n",
+      "I_FL=VA/Vp\n",
+      "Re1=Wsc/(Isc*Isc)\n",
+      "Ze1=Vsc/Isc\n",
+      "Xe1=sqrt((Ze1*Ze1)-(Re1*Re1))\n",
+      "Re2=K*K*Re1\n",
+      "Xe2=K*K*Xe1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The magnetising current is %.2f A and the core-loss current is %.2f A\" %(Im,Iw)\n",
+      "print \"(b)The parameters of equivalent circuit as referred to the low voltage winding(secondary winding) are: \\n    Re2=%.4f ohm \\n    Xe2=%.4f ohm\" %(Re2,Xe2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The magnetising current is 1.15 A and the core-loss current is 0.60 A\n",
+        "(b)The parameters of equivalent circuit as referred to the low voltage winding(secondary winding) are: \n",
+        "    Re2=0.0556 ohm \n",
+        "    Xe2=0.1747 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.17,Page number: 404"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the secondary emf in a transformer.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=25.0e03            #Power rating of the transformer(in VA)\n",
+      "N1=500.0              #Number of turns in the primary winding \n",
+      "N2=40.0               #Number of turns in the secondary winding\n",
+      "V1=3e03               #Voltage of the supply connected to primary(in Volts)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "K=N2/N1\n",
+      "E2=K*V1\n",
+      "I1=VA/V1\n",
+      "I2=I1/K\n",
+      "flux=V1/(4.44*f*N1)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The secondary emf is %.2f V.\" %(E2)\n",
+      "print \"(b)The primary current on full-load is %.2f A and the secondary current on full-load is %.2f A.\" %(I1,I2)\n",
+      "print \"(c)The maximum flux in the core is %.4f Wb.\" %(flux)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The secondary emf is 240.00 V.\n",
+        "(b)The primary current on full-load is 8.33 A and the secondary current on full-load is 104.17 A.\n",
+        "(c)The maximum flux in the core is 0.0270 Wb.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.18,Page number: 404"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the active cross-sectional area of the core.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N1=50.0               #Number of turns in the primary winding\n",
+      "B=1.0                 #Maximum flux density(in Tesla)\n",
+      "f=50.0                #Frequency rating of the transformer(in Hertz)\n",
+      "V=230.0               #Voltage rating of the transformer(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "E1=V\n",
+      "flux=E1/(4.44*f*N1)\n",
+      "A=flux/B\n",
+      "\"\"\"Due to the insulation of laminations from each other,the gross area is about 10% greater than the active area.\"\"\" \n",
+      "gross=1.1*A\n",
+      "a=sqrt(gross)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The active cross sectional area of the core is %.5f square m.\" %(A)\n",
+      "print \"(b)The side of a square core is %.2f m.\" %(a)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The active cross sectional area of the core is 0.02072 square m.\n",
+        "(b)The side of a square core is 0.15 m.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.19,Page number: 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the output of the transformer in kVA.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "A=150e-04             #Cross-sectional area of the core(in square metres)\n",
+      "Bm=1.1                #Maximum flux density(in Tesla)\n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "N2=66.0               #Number of turns in the secondary winding\n",
+      "Z_L=4.0               #Load impedance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "flux=Bm*A\n",
+      "E2=4.44*flux*f*N2\n",
+      "I2=E2/Z_L\n",
+      "kVA=(E2*I2)/1000.0\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The output in kVA when connected to a 4 Ohms load impedance is %.2f kVA.\" %(kVA)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The output in kVA when connected to a 4 Ohms load impedance is 14.61 kVA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.20,Page number: 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnetising current and the iron loss.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I0=1.0                #No-load primary current(in Amperes)\n",
+      "pf=0.24               #Power factor\n",
+      "V1=11e03              #Primary voltage(in Volts)\n",
+      "V2=400                #Secondary voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Iw=I0*pf\n",
+      "Im=sqrt((I0*I0)-(Iw*Iw))\n",
+      "Pi=V1*I0*pf\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The core-loss current is %.2f A.\" %(Iw) \n",
+      "print \"(b)The magnetising current is %.3f A.\" %(Im)\n",
+      "print \"(c)The iron loss is %.2f W.\" %(Pi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The core-loss current is 0.24 A.\n",
+        "(b)The magnetising current is 0.971 A.\n",
+        "(c)The iron loss is 2640.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.21,Page number: 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the supply voltage and the power factor.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import degrees,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "K=0.5                    #Turns ratio of the step-down transformer\n",
+      "R1=2.5                   #Resistance of the primary winding(in Ohms)\n",
+      "X1=6.0                   #Reactance of the primary winding(in Ohms)\n",
+      "R2=0.25                  #Resistance of the secondary winding(in Ohms)\n",
+      "X2=1                     #Reactance of the secondary winding(in Ohms)\n",
+      "Im=51.5e-03              #Magnetising current(in Amperes)\n",
+      "Iw=20.6e-03              #Core-loss current(in Amperes)\n",
+      "Z_L=rect(25,radians(30)) #Load impedance(in Ohms)\n",
+      "Vo=50.0                  #Output voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z1=R1+1j*X1\n",
+      "Z2=R2+1j*X2\n",
+      "V2=rect(Vo,0)\n",
+      "I2=V2/Z_L\n",
+      "E2=V2+(I2*Z2)\n",
+      "E1=E2/K\n",
+      "E1_minus=-E1\n",
+      "I1_a=-I2*K\n",
+      "\"\"\"Im lags -E1 by 90 degrees and Iw is in phase with -E1.\"\"\"\n",
+      "Im_com=rect(Im,(phase(E1_minus)-radians(90)))\n",
+      "Iw_com=rect(Iw,phase(E1_minus))\n",
+      "I1=I1_a+Im_com+Iw_com\n",
+      "V1=E1_minus+(I1*Z1)\n",
+      "pf_ang=phase(V1)-phase(I1)\n",
+      "pf=cos(pf_ang)\n",
+      "\n",
+      "\n",
+      "#Result:I\n",
+      "print \"The supply voltage is %.4f V at a phase angle of %.2f degrees.\" %(abs(V1),degrees(phase(V1)))\n",
+      "print \"The current drawn from the supply is %.4f A at a phase angle of %.2f degrees.\" %(abs(I1),degrees(phase(I1)))\n",
+      "print \"The power factor is %.3f lagging.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The supply voltage is 108.6120 V at a phase angle of -176.35 degrees.\n",
+        "The current drawn from the supply is 1.0451 A at a phase angle of 148.19 degrees.\n",
+        "The power factor is 0.815 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.22,Page number: 406"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the copper loss in the transformer.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "K=0.25                #Turns ratio of the step-down transformer\n",
+      "R1=1.4                #Resistance of the primary(in Ohms)\n",
+      "X1=5.5                #Reactance of the primary(in Ohms)\n",
+      "R2=0.06               #Resistance of the secondary(in Ohms)\n",
+      "X2=0.04               #Reactance of the secondary(in Ohms)\n",
+      "Vsc=24.0              #Voltage of the HV winding(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Re1=R1+(R2/(K*K))\n",
+      "Xe1=X1+(X2/(K*K))\n",
+      "Ze1=sqrt((Re1*Re1)+(Xe1*Xe1))\n",
+      "Isc=Vsc/Ze1\n",
+      "I1=Isc\n",
+      "I2=I1/K\n",
+      "P=I1*I1*Re1\n",
+      "pf=P/(Vsc*I1)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current in the LV winding is %.3f A.\" %(I2)\n",
+      "print \"(b)The copper loss in the transformer is %.2f W.\" %(P)\n",
+      "print \"(c)The power factor is %.4f.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current in the LV winding is 14.594 A.\n",
+        "(b)The copper loss in the transformer is 31.42 W.\n",
+        "(c)The power factor is 0.3588.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.23,Page number: 406"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the regulation and efficiency.\"\"\"\n",
+      "\n",
+      "from math import acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=20e03              #Power rating of the transformer(in Volt-Amperes)\n",
+      "V1=2200.0             #Voltage of the primary winding(in Volts)\n",
+      "V2=220.0              #Voltage of the secondary winding(in Volts)\n",
+      "f=50.0                #Frequency rating of the transformer(in Hertz)\n",
+      "Vsc=86.0              #Voltage measured during short-circuit test(in Volts)\n",
+      "Isc=10.5              #Current measured during short-circuit test(in Amperes)\n",
+      "Psc=360.0             #Power measured during short-circuit test(in Watts)\n",
+      "Voc=220.0             #Voltage measured during open-circuit test(in Volts)\n",
+      "Ioc=4.2               #Current measured during open-circuit test(in Amperes)\n",
+      "Poc=148.0             #Power measured during open-circuit test(in Watts)\n",
+      "pf=0.8                #Lagging power factor\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ze1=Vsc/Isc\n",
+      "Re1=Psc/(Isc*Isc)\n",
+      "Xe1=sqrt((Ze1*Ze1)-(Re1*Re1))\n",
+      "I1=VA/V1\n",
+      "reg=(I1*((Re1*pf)+(Xe1*sin(acos(pf)))))/V1\n",
+      "Pc=(I1/Isc)*(I1/Isc)*Psc\n",
+      "Pi=Poc\n",
+      "Po=VA*pf\n",
+      "effi=Po/(Po+Pc+Pi)\n",
+      "pf_sc=Re1/Ze1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The regulation at 0.8 pf lagging at full load is %.2f per cent.\" %(reg*100)    \n",
+      "print \"   The efficiency is %.2f per cent.\" %(effi*100)\n",
+      "print \"(b)The power factor on short circuit is %.3f lagging.\" %(pf_sc)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The regulation at 0.8 pf lagging at full load is 2.94 per cent.\n",
+        "   The efficiency is 97.45 per cent.\n",
+        "(b)The power factor on short circuit is 0.399 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.24,Page number: 407"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the efficiency at half of full-load current.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=200e03             #Power rating of the transformer(in VA)\n",
+      "effi_FL=0.98          #Full-load efficiency of the transformer\n",
+      "pf=0.8                #Lagging power factor\n",
+      "x=0.75                #Fraction of load at which maximum efficiency occurs\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Po=VA*pf\n",
+      "Pin=Po/effi_FL\n",
+      "tot_loss=Pin-Po\n",
+      "Pc=tot_loss/(1+(x*x))\n",
+      "Pi=tot_loss-Pc\n",
+      "x_new=0.5\n",
+      "P1=(x_new*x_new*Pc)+Pi\n",
+      "effi_half=(Po/2.0)/((Po/2.0)+P1)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The efficiency at half of full-load current is %.3f per cent.\" %(effi_half*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency at half of full-load current is 97.922 per cent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.25,Page number: 408"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the efficiency at different rated kVAs.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=150e03             #Power rating of the transformer(in Volt-Amperes)\n",
+      "V1=5000.0             #Voltage of the primary winding(in Volts)\n",
+      "V2=250.0              #Voltage of the secondary winding(in Volts)\n",
+      "f=50.0                #Frequency rating of the transformer(in Hertz)\n",
+      "Pc=1.8e03             #Full-load copper losses(in Watts)\n",
+      "Pi=1.5e03             #Core losses(in Watts)\n",
+      "flux=60e-03           #Maximum core flux(in Webers)\n",
+      "pf=0.8                #Lagging power factor\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "N2=V2/(4.44*f*flux)\n",
+      "N1=round(N2,0)*(V1/V2)\n",
+      "\"\"\"Case 1:\"\"\"\n",
+      "Po=(VA*pf)\n",
+      "effi_a=Po/(Po+Pi+Pc)\n",
+      "\"\"\"Case 2:\"\"\"\n",
+      "pf=1.0\n",
+      "Po=0.5*VA*pf\n",
+      "Pc_new=0.5*0.5*Pc\n",
+      "effi_b=Po/(Po+Pi+Pc_new)\n",
+      "\n",
+      "x=sqrt(Pi/Pc)\n",
+      "VA_max_effi=x*VA\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The number of turns in the primary winding is %d turns.\" %(round(N1,0))\n",
+      "print \"   The number of turns in the secondary winding is %d turns.\" %(round(N2,0))\n",
+      "print \"(b)The efficiency at full rated kVA with 0.8 pf lagging is %.2f percent.\" %(effi_a*100)  \n",
+      "print \"(c)The efficiency at half rated kVA with unity pf is %.2f percent.\" %(effi_b*100)\n",
+      "print \"(d)The kVA load for maximum efficieny is %d kVA.\" %(round((VA_max_effi/1000),0)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The number of turns in the primary winding is 380 turns.\n",
+        "   The number of turns in the secondary winding is 19 turns.\n",
+        "(b)The efficiency at full rated kVA with 0.8 pf lagging is 97.32 percent.\n",
+        "(c)The efficiency at half rated kVA with unity pf is 97.47 percent.\n",
+        "(d)The kVA load for maximum efficieny is 137 kVA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.26,Page number: 409 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the impedance on the high voltage side.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=50e03              #Power rating of the transformer(in VA)\n",
+      "V1=2400.0             #Voltage of primary winding(in Volts)\n",
+      "V2=240.0              #Voltage of secondary winding(in Volts)\n",
+      "f=50.0                #Frequency rating of the transformer(in Hertz)\n",
+      "LV=240.0              #Low tension voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I2=VA/V2\n",
+      "Z_L=V2/I2\n",
+      "K=V2/V1\n",
+      "Zeq=Z_L/(K*K)\n",
+      "I_high=K*I2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The load impedance connected to the LV side is %.3f Ohms.\" %(Z_L)\n",
+      "print \"(b)The load impedance referred to to the high voltage side is %.2f Ohms.\" %(Zeq)\n",
+      "print \"(c)The current referred to the high voltage side is %.3f A.\" %(I_high) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The load impedance connected to the LV side is 1.152 Ohms.\n",
+        "(b)The load impedance referred to to the high voltage side is 115.20 Ohms.\n",
+        "(c)The current referred to the high voltage side is 20.833 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.27,Page number: 409"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the kVA output of the transformer.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "VA=10e03              #Power rating of the transformer(in VA)\n",
+      "V1=2300.0             #Voltage of HT winding(in Volts)\n",
+      "V2=230.0              #Voltage of LT winding(in Volts)\n",
+      "f=50.0                #Frequency rating of the transformer(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_HT=VA/V1\n",
+      "I_LT=VA/V2\n",
+      "I2=I_HT+I_LT\n",
+      "I1=I_LT\n",
+      "kVA_out=(V1*I2)/1000.0\n",
+      "VA_c=(V1*I1)\n",
+      "VA_i=V1*(I2-I1)\n",
+      "K=V1/(V1+V2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current distribution in the windings: load current=%.2f A and the input current is %.2f A.\" %(I2,I1)\n",
+      "print \"(b)The kVA output is %.2f kVA.\" %(kVA_out)\n",
+      "print \"(c)The volt-amperes transferred conductively is %.2f kVA\" %(VA_c/1000.0) \n",
+      "print \"   The volt-amperes transferred inductively is %.2f kVA\" %(VA_i/1000.0)\n",
+      "print \"(d)The saving in copper as compared to the two-winding transformer is %.2f per cent.\" %(K*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current distribution in the windings: load current=47.83 A and the input current is 43.48 A.\n",
+        "(b)The kVA output is 110.00 kVA.\n",
+        "(c)The volt-amperes transferred conductively is 100.00 kVA\n",
+        "   The volt-amperes transferred inductively is 10.00 kVA\n",
+        "(d)The saving in copper as compared to the two-winding transformer is 90.91 per cent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter14.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter14.ipynb
new file mode 100755
index 00000000..8d806acc
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter14.ipynb
@@ -0,0 +1,1213 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14: ALTERNATORS AND SYNCHRONOUS MOTORS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1,Page number: 433"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of poles on the generator if the frequency of the generated voltage is decreased.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=60                  #Frequency of ac-generator(in Hertz)\n",
+      "P=6                   #Number of poles\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ns=(120*f)/P\n",
+      "f=20\n",
+      "P=(120*f)/Ns\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The speed of rotation of the generator is %d rpm.\" %(Ns)\n",
+      "print \"If the frequency is reduced to 20Hz,the required number of poles is %d.\" %(P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The speed of rotation of the generator is 1200 rpm.\n",
+        "If the frequency is reduced to 20Hz,the required number of poles is 2.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2,Page number: 441\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the distribution factor for a machine.\"\"\"\n",
+      "\n",
+      "from math import radians,degrees,sin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "slots=9                      #Number of slots\n",
+      "slot_angle=radians(180/9)    #Slot angle(in radians)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "q_a=120.0/degrees(slot_angle)\n",
+      "k_d_a=sin(q_a*(slot_angle/2.0))/(q_a*sin(slot_angle/2.0))\n",
+      "q_b=60.0/degrees(slot_angle)\n",
+      "k_d_b=sin(q_b*(slot_angle/2.0))/(q_b*sin(slot_angle/2.0))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The distribution factor for a machine for a three phase winding with 120 degrees phase group is %.3f.\" %(k_d_a)\n",
+      "print \"(b)The distribution factor for a machine for a three phase winding with 60 degrees phase group is %.3f.\" %(k_d_b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The distribution factor for a machine for a three phase winding with 120 degrees phase group is 0.831.\n",
+        "(b)The distribution factor for a machine for a three phase winding with 60 degrees phase group is 0.960.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3,Page number: 441"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed,the generated emf per phase,and the line emf of a three-phase alternator.\"\"\"\n",
+      "\n",
+      "from math import radians,degrees,sqrt,sin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "phase=3                #Number of phases\n",
+      "f=50                   #Frequency rating(in Hertz)\n",
+      "P=20                   #Number of poles \n",
+      "slots=180              #Number of slots on the stator\n",
+      "cond_per_slot=8        #Number of conductors per slot\n",
+      "flux=25e-03            #Flux per pole(in Weber) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=slots*cond_per_slot\n",
+      "T=(Z/2)/phase\n",
+      "Ns=(120*f)/P\n",
+      "k_p=1\n",
+      "slots_per_pole=slots/P\n",
+      "slot_angle=radians(180/slots_per_pole)\n",
+      "q=slots_per_pole/phase\n",
+      "k_d=sin(q*(slot_angle/2))/(q*sin(slot_angle/2))\n",
+      "E=4.44*f*flux*T*k_p*k_d\n",
+      "line_emf=sqrt(3.0)*E\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The speed of the alternator is %d rpm.\" %(round(Ns,0))\n",
+      "print \"(b)The rms value of generated EMF per phase is %.2f V.\" %(E)\n",
+      "print \"(c)The line EMF is %.2f V.\" %(line_emf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The speed of the alternator is 300 rpm.\n",
+        "(b)The rms value of generated EMF per phase is 1278.45 V.\n",
+        "(c)The line EMF is 2214.34 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4,Page number: 446\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage regulation for full-load for a three-phase alternator.\"\"\"\n",
+      "\n",
+      "from math import sqrt,cos,acos \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=600e06              #Power rating of the alternator(in VA) \n",
+      "V_L=22e03             #Rated terminal voltage(in Volts) \n",
+      "sync_imp=0.16         #Synchronous impedance per phase(in Ohms)\n",
+      "res_phase=0.014       #Resistance per phase(in Ohms) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_L=P/(sqrt(3)*V_L)\n",
+      "Iph=I_L\n",
+      "V=V_L/sqrt(3)\n",
+      "Vz=Iph*sync_imp\n",
+      "theta=acos(res_phase/sync_imp)\n",
+      "pf=0.8\n",
+      "phi=acos(pf)\n",
+      "alpha=theta-phi\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "vol_reg_a=(E-V)/V\n",
+      "pf=1\n",
+      "phi=acos(pf)\n",
+      "alpha=theta-phi\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "vol_reg_b=(E-V)/V\n",
+      "pf=0.8\n",
+      "phi=acos(pf)\n",
+      "alpha=theta+phi\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "vol_reg_c=(E-V)/V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)For a power factor of 0.8 lagging:\"\n",
+      "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_a*100)\n",
+      "print \"(b)For unity power factor:\"\n",
+      "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_b*100)\n",
+      "print \"(c)For a power factor of 0.8 leading:\"     \n",
+      "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_c*100)\n",
+      "print \"\\nNote:The voltage regulation for leading power-factor load is negative.\"\n",
+      "print \"It means that on removing the load, the terminal voltage decreases.\"   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)For a power factor of 0.8 lagging:\n",
+        "The voltage regulation for full load is 14.20 percent.\n",
+        "(b)For unity power factor:\n",
+        "The voltage regulation for full load is 3.64 percent.\n",
+        "(c)For a power factor of 0.8 leading:\n",
+        "The voltage regulation for full load is -8.90 percent.\n",
+        "\n",
+        "Note:The voltage regulation for leading power-factor load is negative.\n",
+        "It means that on removing the load, the terminal voltage decreases.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.5,Page number: 450\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the synchronous reactance per phase and the voltage regulation for a three-phase star-connected alternator.\"\"\"\n",
+      "\n",
+      "from math import acos,sqrt,cos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=900               #Open-circuit voltage(line to line)(in Volts)\n",
+      "V_L_rated=3.3e03      #Rated voltage of the alternator(in Volts)\n",
+      "I_f=100               #Full-load current(in Amperes)\n",
+      "R=0.9                 #Armature Resistance(in Ohm/phase)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_oc=V_L/sqrt(3)\n",
+      "I_sc=I_f\n",
+      "sync_imp=V_oc/I_sc\n",
+      "sync_rea=sqrt((sync_imp*sync_imp)-(R*R))\n",
+      "theta=acos(R/sync_imp)\n",
+      "V=V_L_rated/sqrt(3)\n",
+      "Vz=I_f*sync_imp\n",
+      "pf=0.8\n",
+      "phi=acos(pf)\n",
+      "alpha=theta-phi\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "vol_reg_a=(E-V)/V\n",
+      "pf=0.8\n",
+      "phi=acos(pf)\n",
+      "alpha=theta+phi\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "vol_reg_b=(E-V)/V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The synchronous reactance per phase is %.3f Ohms.\" %(sync_rea)\n",
+      "print \"(a) For a power factor of 0.8 lagging:\"\n",
+      "print \"The voltage regulation for full load is %.2f percent\" %(vol_reg_a*100)\n",
+      "print \"(b) For a power factor of 0.8 leading:\"\n",
+      "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_b*100)\n",
+      "print \"\\nNote: The voltage regulation for leading power-factor load is negative.\"\n",
+      "print \"It means that on removing the load, the terminal voltage decreases.\"\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The synchronous reactance per phase is 5.118 Ohms.\n",
+        "(a) For a power factor of 0.8 lagging:\n",
+        "The voltage regulation for full load is 21.34 percent\n",
+        "(b) For a power factor of 0.8 leading:\n",
+        "The voltage regulation for full load is -9.03 percent.\n",
+        "\n",
+        "Note: The voltage regulation for leading power-factor load is negative.\n",
+        "It means that on removing the load, the terminal voltage decreases.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.6,Page number: 455\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the angle of retard of a synchronous motor.\"\"\"\n",
+      "\n",
+      "from math import sqrt,acos,cos,pi,asin,sin,degrees\n",
+      "from cmath import phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Po=9e03               #Power rating of synchronous motor(in Watts)  \n",
+      "V_L=400               #Voltage rating of synchronous motor(in Watts)\n",
+      "Zs=0.4+3*1j           #Synchronous impedance per phase(in Ohms)  \n",
+      "pf=0.8                #Power factor(leading) \n",
+      "effi=0.9              #Efficiency of the motor\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pin=Po/effi\n",
+      "I_L=Pin/(sqrt(3)*V_L*pf)\n",
+      "I=I_L\n",
+      "phi=acos(pf)\n",
+      "mod_Zs=abs(Zs)\n",
+      "theta=phase(Zs)\n",
+      "V=V_L/sqrt(3)\n",
+      "Er=I*mod_Zs\n",
+      "E=sqrt((V*V)+(Er*Er)+(2*V*Er*cos(pi-(theta+phi))))\n",
+      "E_L=sqrt(3)*E\n",
+      "angle_retard=asin((Er*sin(theta+phi))/E)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The angle of retard of the rotor is %.2f degrees.\" %(degrees(angle_retard))\n",
+      "print \"The excitation emf E to which the motor has to be excited to give a full-load output at 0.8 leading power factor is %.2f V.\" %(E_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The angle of retard of the rotor is 10.47 degrees.\n",
+        "The excitation emf E to which the motor has to be excited to give a full-load output at 0.8 leading power factor is 453.81 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.7,Page number: 459"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line emf generated by the alternator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "S=24.0                #Number of slots in the alternator\n",
+      "C=12.0                #Number of conductors per slot\n",
+      "flux=0.1              #Flux per pole(in Weber)\n",
+      "P=4.0                 #Number of poles in the alternator\n",
+      "Ns=1500.0             #Synchronous speed of the alternator(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Zph=S*C/3.0\n",
+      "T=Zph/2.0\n",
+      "S_pole=S/P\n",
+      "slot_ang=180.0/S_pole\n",
+      "q=S_pole/3.0\n",
+      "kd=sin((q*radians(slot_ang)/2))/(q*sin(radians(slot_ang/2)))\n",
+      "f=(P*Ns)/120.0\n",
+      "kp=1.0\n",
+      "E=4.44*flux*f*T*kp*kd\n",
+      "E_L=sqrt(3.0)*E\n",
+      "         \n",
+      "         \n",
+      "#Result:\n",
+      "print \"The line emf generated when the alternator runs at 1500 rpm is %.2f V.\" %(E_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The line emf generated when the alternator runs at 1500 rpm is 1782.78 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.8,Page number: 460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the net emf induced in the 6 coil in series constituting the alternator winding.\"\"\"\n",
+      "\n",
+      "from math import radians,sin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "q=6.0                 #Number of slots per pole per phase\n",
+      "angle=30.0            #Electrical angle between two consecutive slotss(in degrees) \n",
+      "e=10.0                #Emf of each coil(in Volts)   \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "kd=sin(q*(radians(angle/2.0)))/(q*sin(radians(angle/2.0)))\n",
+      "arith_sum=6*e\n",
+      "Er=kd*arith_sum\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The net emf induced in the six coils in series is %.3f V.\" %(round(Er,3)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net emf induced in the six coils in series is 38.637 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.9,Page number: 460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of poles and the current rating of an alternator.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=120.0               #Speed of the alternator(in rpm)\n",
+      "f=50.0                #Frequency of the alternator(in Hertz)\n",
+      "VA=100e06             #VA rating of the alternator(in Volt-Ampere)\n",
+      "pf=1.0                #Power factor\n",
+      "V_L=11e03             #Line voltage(in Volts)\n",
+      "effi=0.97             #Efficiency of the alternator \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "P=(120.0*f)/N\n",
+      "Po=VA*pf\n",
+      "I_L=VA/(sqrt(3.0)*V_L)\n",
+      "Pin=Po/effi\n",
+      "tor=Pin/(2*pi*N/60.0)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The number of poles is %d.\" %(P)\n",
+      "print \"(b)The current rating is %.2f A.\" %(I_L)\n",
+      "print \"(c)The input power is %.2f MW.\" %(round((Pin/1000000),2))\n",
+      "print \"(d)The prime-mover torque applied to the genrator shaft is %e Nm.\" %(tor)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The number of poles is 50.\n",
+        "(b)The current rating is 5248.64 A.\n",
+        "(c)The input power is 103.09 MW.\n",
+        "(d)The prime-mover torque applied to the genrator shaft is 8.203863e+06 Nm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.10,Page number: 460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the percentage regulation for a load.\"\"\"\n",
+      "\n",
+      "from math import atan,radians,acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=11e03             #Line voltage of the star-connected alternator(in Volts)\n",
+      "VA=800e03             #Power rating of the alternator(in Volt-Amperes)\n",
+      "Rs=1.5                #Resistance per phase(in Ohms)\n",
+      "Xs=25.0               #Synchronous reactance(in Ohms)\n",
+      "pf=0.8                #Leading power factor\n",
+      "Po=600e03             #Output power(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vph=V_L/sqrt(3.0)\n",
+      "V=Vph\n",
+      "Iph=Po/(sqrt(3.0)*V_L*pf)\n",
+      "Zs=sqrt((Rs*Rs)+(Xs*Xs))\n",
+      "theta=atan(Xs/Rs)\n",
+      "Vz=Iph*Zs\n",
+      "pf_ang=acos(pf)\n",
+      "alpha=theta+pf_ang\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "reg=(E-V)/V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The percentage regulation is %.3f per cent.\" %(reg*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage regulation is -7.641 per cent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.11,Page number: 461"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the full-load voltage regulation of the alternator.\"\"\"\n",
+      "\n",
+      "from math import atan,acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=6e03              #Line voltage of the alternator(in Volts)\n",
+      "VA=6000e03            #Power rating of the alternator(in Volt-Amperes)\n",
+      "R=0.2                 #Winding resistance per phase(in Ohms)\n",
+      "pf=0.8                #Lagging power factor\n",
+      "V_L_OC=480.0          #Line voltage in open-circuit test(in Volts)\n",
+      "I_f_OC=10.0           #Field current in open-circuit test(in Amperes)\n",
+      "I_L_SC=105.0          #Line current in short-circuit test(in Amperes)\n",
+      "I_f_SC=5.0            #Field current in short-circuit test(in Amperes) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"In the short-circuit test,the currents are small compared to the full-load current.\"\"\"\n",
+      "V=V_L/sqrt(3.0)\n",
+      "I=VA/(3*V)\n",
+      "V_ph_OC=V_L_OC/sqrt(3.0)\n",
+      "\"\"\"Since the field current of 5 A gives an armature current of 105 A,a field current of 10 A will give an armature \n",
+      "   current of (105*2)=210 A.\"\"\"\n",
+      "I_ph=I_L_SC*2\n",
+      "Zs=V_ph_OC/I_ph\n",
+      "Xs=sqrt((Zs*Zs)-(R*R))\n",
+      "theta=atan(Xs/R)\n",
+      "Vz=I*Zs\n",
+      "pf_ang=acos(pf)\n",
+      "alpha=theta-pf_ang\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "reg=(E-V)/V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The full-load voltage regulation of the alternator at 0.8 lagging power factor is %.2f per cent.\" %(reg*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The full-load voltage regulation of the alternator at 0.8 lagging power factor is 16.73 per cent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.12,Page number: 462"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the terminal voltage of the generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50.0                #Frequency rating of the generator(in Hertz)\n",
+      "Z_p=96.0              #Number of conductors per phase\n",
+      "flux=0.1              #Flux per pole(in Webers)\n",
+      "Xs=5.0                #Synchronous reactance per phase(in Ohms)\n",
+      "kd=0.96               #Distribution factor for the stator winding\n",
+      "Z_L=10.0              #Load impedance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Zs=1j*Xs\n",
+      "kp=1.0\n",
+      "T=Z_p/2.0\n",
+      "E=4.44*f*flux*kp*kd*T\n",
+      "V=E/(1+(Zs/Z_L))\n",
+      "V_L=sqrt(3.0)*abs(V)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The terminal voltage of the generator is %.2f V.\" %(V_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The terminal voltage of the generator is 1584.79 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.13,Page number: 462"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the percentage change of voltage.\"\"\"\n",
+      "\n",
+      "from math import atan,radians,acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=6.6e03            #Line voltage of the star-connected alternator(in Volts)\n",
+      "VA=1500e03            #Power rating of the alternator(in Volt-Amperes)\n",
+      "Rs=0.5                #Resistance per phase(in Ohms)\n",
+      "Xs=5.0                #Synchronous reactance(in Ohms)\n",
+      "pf=0.8                #Lagging power factor\n",
+      "\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vph=V_L/sqrt(3.0)\n",
+      "V=Vph\n",
+      "Iph=VA/(3.0*Vph)\n",
+      "Zs=sqrt((Rs*Rs)+(Xs*Xs))\n",
+      "theta=atan(Xs/Rs)\n",
+      "Vz=Iph*Zs\n",
+      "pf_ang=acos(pf)\n",
+      "alpha=theta-pf_ang\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "reg=(E-V)/V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The percentage change in voltage is %.3f per cent.\" %(reg*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage change in voltage is 12.432 per cent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.14,Page number: 463"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the terminal voltage and load regulation.\"\"\"\n",
+      "\n",
+      "from math import atan,radians,acos,asin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=6599.0            #Line voltage of the star-connected alternator(in Volts)\n",
+      "Rs=0.5                #Resistance per phase(in Ohms)\n",
+      "Xs=5.0                #Synchronous reactance(in Ohms)\n",
+      "I=130.0               #Full-load current(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "E=V_L/sqrt(3.0)\n",
+      "Zs=Rs+1j*Xs\n",
+      "theta=atan(Xs/Rs)\n",
+      "Vz=I*abs(Zs)\n",
+      "pf=0.8\n",
+      "pf_ang=acos(pf)\n",
+      "alpha=theta-pf_ang\n",
+      "tor_ang=asin((Vz*sin(alpha))/E)\n",
+      "V_a=(E*cos(tor_ang))-(Vz*cos(alpha))\n",
+      "reg_a=(E-V_a)/V_a\n",
+      "pf=0.6\n",
+      "pf_ang=acos(pf)\n",
+      "alpha=theta+pf_ang\n",
+      "beta=pi-alpha\n",
+      "tor_ang=asin((Vz*sin(beta))/E)\n",
+      "V_b=(E*cos(tor_ang))+(Vz*cos(beta))\n",
+      "reg_b=(E-V_b)/V_b\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The terminal voltage when the power factor is 0.8 lagging is %.2f V.\" %(V_a)\n",
+      "print \"   The load regulation is %.2f per cent.\" %(reg_a*100)    \n",
+      "print \"(b)The terminal voltage when the power factor is 0.6 leading is %.2f V.\" %(V_b)\n",
+      "print \"   The load regulation is %.2f per cent.\" %(reg_b*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The terminal voltage when the power factor is 0.8 lagging is 3337.45 V.\n",
+        "   The load regulation is 14.16 per cent.\n",
+        "(b)The terminal voltage when the power factor is 0.6 leading is 4265.21 V.\n",
+        "   The load regulation is -10.67 per cent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.15,Page number: 464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the percentage load regulation.\"\"\"\n",
+      "\n",
+      "from math import acos,atan\n",
+      "\n",
+      "#Variable Declration:\n",
+      "VA=1.5e06            #Power rating of the synchronous generator(in Volt-Amperes)\n",
+      "V_L=11e03            #Line voltage(in Volts)\n",
+      "Rs=1.2               #Armature resistance(in Ohms)\n",
+      "Xs=25.0              #Synchronous reactance per phase(in Ohms)\n",
+      "P_L=1.4375e06        #Load to be delivered(in Volt-Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=V_L/sqrt(3.0)\n",
+      "I=P_L/(3*V)\n",
+      "Zs=sqrt((Rs*Rs)+(Xs*Xs))\n",
+      "Vz=I*Zs\n",
+      "pf=0.8\n",
+      "pf_ang=acos(pf)\n",
+      "theta=atan(Xs/Rs)\n",
+      "alpha=theta-pf_ang\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "reg_a=(E-V)/V\n",
+      "alpha=theta+pf_ang\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "reg_b=(E-V)/V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The percentage load regulation for a power factor of 0.8 lagging is %.2f per cent.\" %(reg_a*100)\n",
+      "print \"(b)The percentage load regulation for a power factor of 0.8 leading is %.2f per cent.\" %(reg_b*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The percentage load regulation for a power factor of 0.8 lagging is 21.15 per cent.\n",
+        "(b)The percentage load regulation for a power factor of 0.8 leading is -13.12 per cent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.16,Page number: 465"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the torque angle of the alternator.\"\"\"\n",
+      "\n",
+      "from math import acos,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=0.5                 #Effective resistance per phase of the alternator(in Ohms)\n",
+      "V_L=2200.0            #Line voltage of the alternator(in Volts)\n",
+      "I_FL=200.0            #Full-load current(in Amperes)\n",
+      "If=30.0               #Field current(in Amperes)\n",
+      "V_L_oc=1100.0         #Line-to-line voltage on open circuit(in Volts)\n",
+      "pf=0.8                #Lagging power factor     \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Zs=V_L_oc/(sqrt(3.0)*I_FL)\n",
+      "theta=acos(R/Zs)\n",
+      "Vz=I_FL*Zs\n",
+      "V=V_L/sqrt(3.0)\n",
+      "pf_ang=acos(pf)\n",
+      "alpha=theta-pf_ang\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "Po=V*I_FL*pf\n",
+      "tor_ang=theta-acos((Po+(V*V*cos(theta)/Zs))*(Zs/(E*V)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The torque angle of the alternator is %.2f degrees.\" %(degrees(tor_ang))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The torque angle of the alternator is 14.35 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.17,Page number: 465"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"\"Finding the synchronising power of the alternator.\"\"\"\n",
+      "\n",
+      "from math import acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=6.0                 #Number of poles\n",
+      "VA=3e06               #Power rating of the alternator(in Volt-Amperes)\n",
+      "Ns=1000.0             #Speed of operation of the alternator(in rpm)\n",
+      "V_L=3.3e03            #Line voltage of the load(in Volts)\n",
+      "pf=0.8                #Lagging power factor\n",
+      "\n",
+      "#Calculations:\n",
+      "V=V_L/sqrt(3.0)\n",
+      "I=VA/(sqrt(3.0)*V_L)\n",
+      "IXs=0.25*V\n",
+      "Xs=IXs/I\n",
+      "rotor_dis=radians(1*(P/2.0))\n",
+      "Vz=I*Xs\n",
+      "theta=pi/2\n",
+      "pf_ang=acos(0.8)\n",
+      "alpha=theta-pf_ang\n",
+      "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+      "Psy=(3*E*V*sin(rotor_dis))/(Xs)\n",
+      "tor=(3*Psy)/(2*pi*Ns/60)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The synchronising power is %.2f kW.\" %(Psy/1000)\n",
+      "print \"(b)The synchronising torque per mechanical degree is %.2f kNm.\" %(tor/1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The synchronising power is 733.08 kW.\n",
+        "(b)The synchronising torque per mechanical degree is 21.00 kNm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.18,Page number: 466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the armature current and power factor.\"\"\"\n",
+      "\n",
+      "from math import radians,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=2300.0            #Line voltage of the winding(in Volts)\n",
+      "f=50.0                #Operating frequency of the synchronous motor(in Hertz)\n",
+      "Psh=205.0             #Power delivered by the motor(in Horse Power)\n",
+      "ang=15.0              #Power angle(in degrees)\n",
+      "Xs=11.0               #Synchronous reactance(in Ohms)\n",
+      "effi=0.90             #Efficiency of the motor\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=V_L/sqrt(3.0)\n",
+      "Psh=Psh*746.0\n",
+      "Pd=Psh/effi\n",
+      "E=(Pd*Xs)/(3.0*V*sin(radians(ang)))\n",
+      "I=(rect(V,0)-rect(E,radians(-ang)))/(1j*Xs)\n",
+      "pf=cos(phase(I))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The excitation voltage per phase is %.2f V per phase.\" %(E)\n",
+      "print \"(b)The armature current is %.2f A at a phase angle of %.2f degrees.\" %(abs(I),degrees(phase(I)))\n",
+      "print \"(c)The power factor is %.4f leading.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The excitation voltage per phase is 1812.83 V per phase.\n",
+        "(b)The armature current is 57.44 A at a phase angle of 42.05 degrees.\n",
+        "(c)The power factor is 0.7426 leading.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.19,Page number: 466"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the line current and the power factor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=6600.0            #Line voltage(in Volts)\n",
+      "Xs=20.0               #Synchronous reactance per phase(in Ohms)\n",
+      "Pin=915e03            #Power consumed by the motor(in Watts)\n",
+      "E_L=8942.0            #Induced line emf per phase(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=V_L/sqrt(3.0)\n",
+      "E=E_L/sqrt(3.0)\n",
+      "I_cos=Pin/(sqrt(3.0)*V_L)\n",
+      "BN=Xs*I_cos\n",
+      "NA=sqrt((E*E)-(BN*BN))\n",
+      "NO=NA-V\n",
+      "Er=sqrt((NO*NO)+(BN*BN))\n",
+      "I_L=Er/Xs\n",
+      "pf=I_cos/I_L\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The line current is %.2f A.\" %(I_L)\n",
+      "print \"(b)The power factor is %.4f leading.\" %(pf) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The line current is 97.05 A.\n",
+        "(b)The power factor is 0.8247 leading.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.20,Page number: 467"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power factor when the load on the motor increases.\"\"\"\n",
+      "\n",
+      "from math import acos,asin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=6600.0            #Line voltage(in Volts)\n",
+      "Xs=15.0               #Synchronous reactance per phase(in Ohms)\n",
+      "Pin=500e03            #Power consumed by the motor(in Watts)\n",
+      "pf=0.8                #Leading power factor\n",
+      "Pin_new=800e03        #New Input power(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=V_L/sqrt(3.0)\n",
+      "Zs=Xs\n",
+      "I_L=Pin/(sqrt(3.0)*V_L*pf)\n",
+      "Er=I_L*Zs\n",
+      "pf_ang=acos(pf)\n",
+      "alpha=(pi/2)-pf_ang\n",
+      "E=sqrt((V*V)+(Er*Er)+(2*V*Er*cos(alpha)))\n",
+      "I1_cos=Pin_new/(sqrt(3.0)*V_L)\n",
+      "sin_tor=(Pin_new*Xs)/(E*V)\n",
+      "AN=V*sin_tor\n",
+      "NB=E-(V*cos(asin(sin_tor)))\n",
+      "Er1=sqrt((AN*AN)+(NB*NB))\n",
+      "I1=Er1/Xs\n",
+      "pf=I1_cos/I1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The power factor when the load on the motor increases is %.4f leading.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power factor when the load on the motor increases is 0.3229 leading.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.21,Page number: 469"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power factor and kVA rating of the synchronous motor.\"\"\"\n",
+      "\n",
+      "from math import atan,acos\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Si=500e03             #Load supplied by the three-phase system(in Volt-Amperes)\n",
+      "pf_i=0.5              #Lagging power factor of three-phase system\n",
+      "Po=100.0              #Load supplied by the synchronous motor(in Horse-Power)\n",
+      "pf_t=0.9              #Overall lagging power factor \n",
+      "effi=0.87             #Efficiency of the synchronous motor\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pi=Si*pf_i\n",
+      "Qi=Si*sin(acos(pf_i))\n",
+      "Ps=(Po*746)/effi\n",
+      "Pt=Pi+Ps\n",
+      "St=Pt/pf_t\n",
+      "Qt=St*sin(acos(pf_t))\n",
+      "Qs=Qt-Qi\n",
+      "pf_ang=atan(Qs/Ps)\n",
+      "pf_s=cos(pf_ang)\n",
+      "Ss=sqrt((Ps*Ps)+(Qs*Qs))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The power faactor of the synchronous motor is %.2f leading.\" %(pf_s)\n",
+      "print \"(b)The kVA rating of the synchronous motor is %.2f kVA.\" %(Ss/1000.0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The power faactor of the synchronous motor is 0.30 leading.\n",
+        "(b)The kVA rating of the synchronous motor is 283.67 kVA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter15.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter15.ipynb
new file mode 100755
index 00000000..f49bd253
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter15.ipynb
@@ -0,0 +1,957 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 15: INDUCTION MOTORS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.1,Page number: 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the synchronous speed of an induction motor and the frequency of rotor currents at standstill.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=6.0                 #Number of poles \n",
+      "f=50.0                #Operating frequency of the induction motor(in Hertz)\n",
+      "s_no_load=0.01        #Slip at no-load  \n",
+      "s_full_load=0.03      #Slip at full-load\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ns=(120.0*f)/P\n",
+      "N_no_load=Ns*(1-s_no_load)\n",
+      "N_full_load=Ns*(1-s_full_load)\n",
+      "s_standstill=1.0\n",
+      "fr_standstill=s_standstill*f\n",
+      "fr_full_load=s_full_load*f\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The synchronous speed is %d rpm.\" %(Ns) \n",
+      "print \"(b)The no-load speed is %d rpm.\" %(N_no_load) \n",
+      "print \"(c)The full-load speed is %d rpm.\" %(N_full_load) \n",
+      "print \"(d)The frequency of rotor-currents at standstill is %.2f Hz.\" %(fr_standstill)\n",
+      "print \"(e)The frequency of rotor-currents at full-load is %.2f Hz.\" %(fr_full_load)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The synchronous speed is 1000 rpm.\n",
+        "(b)The no-load speed is 990 rpm.\n",
+        "(c)The full-load speed is 970 rpm.\n",
+        "(d)The frequency of rotor-currents at standstill is 50.00 Hz.\n",
+        "(e)The frequency of rotor-currents at full-load is 1.50 Hz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.2,Page number: 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the frequency of rotor-current in an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=12.0                #Number of poles                 \n",
+      "f=50.0                #Operating frequency of induction motor(in Hertz)\n",
+      "N=485.0               #Speed of the motor(in rpm)  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ns=(120.0*f)/P\n",
+      "s=(Ns-N)/Ns\n",
+      "fr=s*f\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The frequency of the rotor-currents is %.2f Hz.\" %(fr)  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The frequency of the rotor-currents is 1.50 Hz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.3,Page number: 481 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the full-load slip and speed of an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "fr=2.0                #Frequency of thr rotoremf at full-load(in Hertz) \n",
+      "f=50.0                #Frequency of the supply(in Hertz)\n",
+      "P=6.0                 #Number of poles \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "s=fr/f\n",
+      "Ns=(120*f)/P\n",
+      "N=Ns*(1-s)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The slip at full-load is %.2f percent.\" %(s*100)\n",
+      "print \"The full-load speed is %d rpm.\" %(N) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The slip at full-load is 4.00 percent.\n",
+        "The full-load speed is 960 rpm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.4,Page number: 481 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the synchronous speed and the rotor frequency of an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50.0                #Frequency of the supply(in Hertz)  \n",
+      "P=4.0                 #Number of poles\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ns=(120*f)/P\n",
+      "s1=0.04\n",
+      "N1=Ns*(1-s1) \n",
+      "N2=600\n",
+      "s2=(Ns-N2)/Ns\n",
+      "fr=s2*f\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The synchronous speed is %d rpm.\" %(Ns)\n",
+      "print \"(b)The speed of the rotor when the slip is 0.04 is %d rpm.\" %(N1)\n",
+      "print \"(c)The rotor frequency when the speed of the rotor is 600 rpm is %.2f Hz.\" %(fr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The synchronous speed is 1500 rpm.\n",
+        "(b)The speed of the rotor when the slip is 0.04 is 1440 rpm.\n",
+        "(c)The rotor frequency when the speed of the rotor is 600 rpm is 30.00 Hz.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.5,Page number: 484 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rotor current of a three-phase induction motor.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pow,acos,radians,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "E_L=100.0              #Induced EMF(in Volts)\n",
+      "R2=0.05                #Resistance of rotor windings(in Ohms) \n",
+      "X_20=0.1               #Standstill reactance of rotor windings(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "E_20=E_L/sqrt(3.0)\n",
+      "s_a=0.04\n",
+      "E2_a=s_a*E_20\n",
+      "Z2_a=sqrt((R2*R2)+pow((s_a*X_20),2))\n",
+      "I2_a=E2_a/Z2_a\n",
+      "angle_a=acos(R2/Z2_a)\n",
+      "s_b=1.0\n",
+      "E2_b=s_b*E_20\n",
+      "Z2_b=sqrt((R2*R2)+pow((s_b*X_20),2))\n",
+      "I2_b=E2_b/Z2_b\n",
+      "angle_b=acos(R2/Z2_b)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)At 4% slip : \"\n",
+      "print \"   The rotor current is %.2f A.\" %(round(I2_a,2))\n",
+      "print \"   The phase difference between the rotor voltage and the rotor current is %.2f degrees.\" %(round(degrees(angle_a),2))\n",
+      "print \"\\n(b)At 100% slip :\"\n",
+      "print \"   The rotor current is %.2f A.\" %(round(I2_b,2))\n",
+      "print \"   The phase difference between the rotor voltage and the rotor current is %.2f degrees.\" %(round(degrees(angle_b),2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)At 4% slip : \n",
+        "   The rotor current is 46.04 A.\n",
+        "   The phase difference between the rotor voltage and the rotor current is 4.57 degrees.\n",
+        "\n",
+        "(b)At 100% slip :\n",
+        "   The rotor current is 516.40 A.\n",
+        "   The phase difference between the rotor voltage and the rotor current is 63.43 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.6,Page number: 485"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the developed power,air-gap power,rotor copper loss,and stator loss in an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "HP=746.0              #Value of Horse-Power(in Watts) \n",
+      "f=50.0                #Operating frequency of the induction motor(in Hertz)\n",
+      "N=1470.0              #Speed of the motor(in rpm)\n",
+      "P=4.0                 #Number of poles\n",
+      "phase=3.0             #Number of phases \n",
+      "effi=87.5e-02         #Efficiency of the motor at full-load  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Po=5.0*HP\n",
+      "Pin=Po/effi\n",
+      "total_loss=Pin-Po\n",
+      "mech_loss=0.05*total_loss\n",
+      "elec_loss=total_loss-mech_loss\n",
+      "dev_pow=Po+mech_loss\n",
+      "Ns=(120*f)/P\n",
+      "s=(Ns-N)/Ns\n",
+      "Pg=dev_pow/(1-s)\n",
+      "P_R=s*Pg\n",
+      "P_S=Pin-Pg\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The developed power is %.2f W.\" %(round(dev_pow,2))\n",
+      "print \"The air-gap power is %.2f W.\" %(round(Pg,2))\n",
+      "print \"The rotor copper loss is %.2f W.\" %(round(P_R,2))\n",
+      "print \"The stator loss is %.2f W.\" %(round(P_S,2))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The developed power is 3756.64 W.\n",
+        "The air-gap power is 3833.31 W.\n",
+        "The rotor copper loss is 76.67 W.\n",
+        "The stator loss is 429.55 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.7,Page number: 490 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rotor speed,the stator current,the power factor and the efficiency of an induction motor. \"\"\"\n",
+      "\n",
+      "from math import sqrt,cos,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_L=400.0             #Operating voltage of the induction motor(in Volts)\n",
+      "P=4.0                 #Number of poles\n",
+      "f=50.0                #Operating frequency of the motor(in Hertz)\n",
+      "s=0.02                #Slip at rated load   \n",
+      "rot_loss=0.34e03      #Rotational losses(in Watts)\n",
+      "R1=0.641              #Impedance per phase on the stator side(in Ohms) \n",
+      "X1=1.106              #Impedance per phase on the stator side(in Ohms)       \n",
+      "R2_eq=0.332           #Impedance per phase on the stator side(in Ohms)\n",
+      "X_20_eq=0.464         #Impedance per phase on the stator side(in Ohms) \n",
+      "Xg=26.3               #Impedance per phase on the stator side(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V1_mod=V_L/sqrt(3)\n",
+      "V1=rect(V1_mod,0)\n",
+      "Ns=(120*f)/P\n",
+      "N=Ns*(1-s)\n",
+      "V_Th=V1*((Xg/(R1+(X1+Xg)*1j))*1j)\n",
+      "Z_Th=((Xg*(R1+X1*1j))/(R1+(X1+Xg)*1j))*1j\n",
+      "R_L_eq=((1-s)/s)*R2_eq\n",
+      "I1=V_Th/(Z_Th +(R2_eq+X_20_eq*1j)+R_L_eq)\n",
+      "angle=phase(I1)\n",
+      "pf=cos(angle)\n",
+      "Po=(3*abs(I1)*abs(I1)*R_L_eq)-rot_loss\n",
+      "Pi=3*abs(V1)*abs(I1)*cos(phase(I1))\n",
+      "effi=(Po/Pi)*100\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The rotor speed is %d rpm.\" %(round(N,0))\n",
+      "print \"(b)The stator current is %.2f A at a phase angle of %.2f degrees.\" %(abs(I1),degrees(phase(I1))) \n",
+      "print \"(c)The power factor is %.3f lagging.\" %(pf)\n",
+      "print \"(d)The output power is %.2f W.\" %(Po)\n",
+      "print \"   The input power is %.2f W.\" %(Pi)\n",
+      "print \"(e)The efficiency of the motor is %.2f percent.\" %(effi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The rotor speed is 1470 rpm.\n",
+        "(b)The stator current is 12.84 A at a phase angle of -3.78 degrees.\n",
+        "(c)The power factor is 0.998 lagging.\n",
+        "(d)The output power is 7703.00 W.\n",
+        "   The input power is 8874.78 W.\n",
+        "(e)The efficiency of the motor is 86.80 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.8,Page number: 495"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the standstill rotor reactance of an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50.0                #Operating frequency of induction motor(in Hertz)\n",
+      "P=6.0                 #Number of poles\n",
+      "N=940.0               #Speed of motor(in rpm)\n",
+      "R2=0.1                #Resistance per phase(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ns=(120*f)/P\n",
+      "s=(Ns-N)/Ns\n",
+      "X_20=R2/s\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The standstill rotor resistance is %.3f Ohms.\" %(X_20)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The standstill rotor resistance is 1.667 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.9,Page number: 498"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the full-load slip of an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P_motor=6.0           #Number of poles in the induction motor\n",
+      "N_motor=960.0         #Full-load speed of the induction motor(in rpm)\n",
+      "P_alt=4.0             #Number of poles in the alternator\n",
+      "N_alt=1500.0          #Speed of the alternator(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "f=(N_alt*P_alt)/120.0\n",
+      "Ns=(120.0*f)/P_motor\n",
+      "s=(Ns-N_motor)/Ns\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The full-load slip of the motor is %.2f percent.\" %(s*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The full-load slip of the motor is 4.00 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.10,Page number: 498"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rotor input,motor input and the efficiency of an induction motor.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=4.0                 #Number of poles in the induction motor\n",
+      "useful_tor=160.0      #Useful torque(in Newton-metre)\n",
+      "s=0.05                #Slip\n",
+      "P_S=1000.0            #Stator losses(in Watts)\n",
+      "Pm=500.0              #Frictional and windage losses(in Watts)\n",
+      "f=50.0                #Frequency of induction motor(in Hertz)\n",
+      "#Calculations:\n",
+      "Ns=(120.0*f)/P\n",
+      "N=(1-s)*Ns\n",
+      "Po=(2*pi*useful_tor*N)/60.0\n",
+      "Pd=Po+Pm\n",
+      "Pg=Pd/(1-s)\n",
+      "Pin=Pg+P_S\n",
+      "effi=Po/Pin\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The rotor input is %.4f kW.\" %(Pg/1000.0)\n",
+      "print \"(b)The motor input is %.4f W.\" %(Pin)\n",
+      "print \"(c)The efficiency is %.4f percent.\" %(effi*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The rotor input is 25.6591 kW.\n",
+        "(b)The motor input is 26659.0570 W.\n",
+        "(c)The efficiency is 89.5609 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.11,Page number: 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the slip of an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "effi=0.9              #Efficiency of the induction motor\n",
+      "Po=50e03              #Load driven by the motor(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pin=Po/effi\n",
+      "P_tot=Pin-Po\n",
+      "\"\"\"The no-load losses comprise of the stator iron loss(Pi) and mechanical losses(Pm)\n",
+      "   (since the stator and rotor copper losses are negligible).These two losses are independent of the load.\n",
+      "   \n",
+      "   Given,the mechanical loss,Pm=(no-load loss)/3.0=((Pi+Pm)/3.0).\n",
+      "   \n",
+      "   Therefore,Pm=(Pi/2.0).\n",
+      "   \n",
+      "   Total loss=(Stator copper loss)+(Stator iron loss)+(Rotor copper loss)+(Mechanical loss).\"\"\"\n",
+      "Pi=(2.0/7.0)*P_tot\n",
+      "P_R=Pi\n",
+      "Pm=Pi/2.0\n",
+      "Pd=Po+Pm\n",
+      "Pg=Pd+P_R\n",
+      "s=P_R/Pg\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The slip of the induction motor is %.2f percent.\" %(s*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The slip of the induction motor is 3.03 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.12,Page number: 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rotor current in an induction motor.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V2=100.0              #Induced emf between slip ring terminals(in Volts)\n",
+      "R2=0.4                #Resistance per phase of star-connected rotor windings(in Ohms)\n",
+      "s=0.04                #Slip of rotor\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "E20=V2/sqrt(3.0)\n",
+      "\"\"\"The rotor reactance X2=(s*X20) is negligible for small values of sand hence cam be ignored.\"\"\"\n",
+      "I2=(s*E20)/R2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The rotor current is %.3f A.\" %(I2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rotor current is 5.774 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.13,Page number: 499"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of poles and the slip of an induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=285.0               #Full-load speed of an induction motor(in rpm)\n",
+      "f=50.0                #Frequency of supply(in Hertz)\n",
+      "P_R=250.0             #Original rotor losses(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "P=(120.0*f)/N\n",
+      "\"There has to be even number of poles,such that Ns>N.Thus,the actual number of poles is 20.\"\"\"\n",
+      "P=round((P-1),0)\n",
+      "Ns=(120.0*f)/P\n",
+      "s=(Ns-N)/Ns\n",
+      "\"\"\"For small values of s,the reactance of (s*X20) is much smaller than the resistance R2,hence\n",
+      "   torque is directly proportional to (s/R2).\n",
+      "         \n",
+      "   It means that the to keep the torque same,the (s/R2) ratio should remain the same.If R2 is doubled,then s also has to be \n",
+      "   doubled.\"\"\"       \n",
+      "s_new=2*s\n",
+      "\"\"\"Since the full-load current remains the same,on doubling the rotor resistance,the copper loss(I*I*R) is also doubled.\"\"\"\n",
+      "P_R_new=2*P_R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The number of poles is %d.\" %(P)\n",
+      "print \"(b)The slip is %.2f percent.\" %(s*100)\n",
+      "print \"(c)The slip for full-load torque if the rotor resistance is doubled is %.2f percent.\" %(s_new*100)\n",
+      "print \"(d)The rotor copper losses with added rotor resistance is %.2f W.\" %(P_R_new)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The number of poles is 20.\n",
+        "(b)The slip is 5.00 percent.\n",
+        "(c)The slip for full-load torque if the rotor resistance is doubled is 10.00 percent.\n",
+        "(d)The rotor copper losses with added rotor resistance is 500.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.14,Page number: 500"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the slip and the power output of an induction motor when external resistances are inserted in each rotor phase.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "s=0.02                #Full-load slip of the induction motor\n",
+      "Po=500                #Power rating of the motor(in HorsePower)\n",
+      "R2=0.25               #Resistance per phase of the rotor(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R2_new=2.0+R2\n",
+      "s_new=(s/R2)*(R2_new)\n",
+      "\"\"\"If Ns is the synchronous speed of the motor,then the speed of the rotor before inserting external resistance,\n",
+      "   N=(1-s)*Ns;\n",
+      "    \n",
+      "   The speed of the rotor after inserting external resistance is N_new=(1-s_new)*Ns;\"\"\"\n",
+      "\n",
+      "Po_new=((1-s_new)/(1-s))*Po\n",
+      "   \n",
+      "\n",
+      "#Result:\n",
+      "print \"The new slip is %.2f percent.\" %(s_new*100)\n",
+      "print \"The new power output is %.2f HP.\" %(Po_new)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The new slip is 18.00 percent.\n",
+        "The new power output is 418.37 HP.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.15,Page number: 500"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the full-load speed of an induction motor.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=4.0                 #Number of poles in the induction motor\n",
+      "f=50.0                #Rated frequency of the induction motor(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" Starting torque(tor_st)=1.6*(tor_fl) where tor_fl=full-load torque;\n",
+      "    Maximum torque(tor_max)=2.0*(tor_fl);\n",
+      "    \n",
+      "    tor_st/tor_max=0.8;\n",
+      "    \n",
+      "    (tor_st/tor_max)=(2*s_m)/((s_m*s_m)+1))\n",
+      "    \n",
+      "    (s_m*s_m)-(2.5*s_m)+1=0 is a quadratic equation whose roots are 2 and 0.5 \n",
+      "    \n",
+      "    s_m has to be less than 1. Therefore, s_m=0.5; \n",
+      "    \n",
+      "    Similarly,\n",
+      "    (tor_fl)/(tor_max)=0.5; \n",
+      "    \n",
+      "    (tor_fl)/(tor_max)=(2*s_fl*s_m)/((s_m*s_m)+(s_fl*s_fl))\n",
+      "    \n",
+      "    Substituting s_m=0.5,we get a quadratic equation,\n",
+      "    \n",
+      "    (s_fl*s_fl)-(s_fl)+0.125=0 whose roots are 0.8535,0.1465.\n",
+      "    \n",
+      "    s_fl=0.1465 as s_fl should be less than s_m. \"\"\"\n",
+      "\"\"\"Finding the roots:\"\"\"\n",
+      "s_m,s_fl= symbols('s_m s_fl')\n",
+      "a=solve(s_m**2-(2.5*s_m)+1.0,s_m)\n",
+      "b=solve(s_fl**2-s_fl+0.125, s_fl)\n",
+      "Ns=(120*f)/P\n",
+      "Nfl=Ns*(1-b[0])\n",
+      "Nm=Ns*(1-a[0])\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The full-load speed is %.2f rpm.\" %(Nfl)\n",
+      "print \"(b)The speed at maximum torque is %.2f rpm.\" %(Nm)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The full-load speed is 1280.33 rpm.\n",
+        "(b)The speed at maximum torque is 750.00 rpm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.16,Page number: 501"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rotor copper loss and the gross torque for an induction motor.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "s=0.04                #Full-load slip\n",
+      "P=4.0                 #Number of poles\n",
+      "f=50.0                #Frequency of the induction motor(in Hertz)\n",
+      "Po=18.65e03           #Power output(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pm=0.025*Po\n",
+      "Pd=Pm+Po\n",
+      "P_R=Pd*(s/(1-s))\n",
+      "Pg=P_R/s\n",
+      "Ns=(120*f)/P\n",
+      "N=Ns*(1-s)\n",
+      "sh_tor=Po/(2*pi*(N/60.0))\n",
+      "gross_tor=Pd/(2*pi*(N/60.0))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The rotor copper loss is %.2f W.\" %(P_R)\n",
+      "print \"(b)The rotor input is %.2f W.\" %(Pg)\n",
+      "print \"(c)The output(shaft) torque is %.2f Nm.\" %(sh_tor)\n",
+      "print \"(d)The gross torque is %.2f Nm.\" %(gross_tor)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The rotor copper loss is 796.51 W.\n",
+        "(b)The rotor input is 19912.76 W.\n",
+        "(c)The output(shaft) torque is 123.68 Nm.\n",
+        "(d)The gross torque is 126.77 Nm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.17,Page number: 501"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rotor current and the rotor power factor for an induction motor.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pow,cos,atan\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=4.0                 #Number of poles in the induction motor\n",
+      "E1=1100.0             #Line voltage(in Volts) \n",
+      "f=50.0                #Operating frequency of the motor(in Hertz)\n",
+      "K=1.0/3.8             #Transformation ratio\n",
+      "R2=0.012              #Rotor resistance per phase(in Ohms)\n",
+      "X20=0.25              #Rotor stanstill reactance per phase(in Ohms) \n",
+      "N=1440.0              #Full-load speed of the motor(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ns=(120.0*f)/P\n",
+      "s=(Ns-N)/Ns\n",
+      "E20=K*E1\n",
+      "Z20=sqrt((R2*R2)+(X20*X20))\n",
+      "Z2=sqrt((R2*R2)+(s*X20*s*X20))\n",
+      "I20=E20/Z20\n",
+      "pf_20=cos(atan(X20/R2))\n",
+      "I2=(s*E20)/Z2\n",
+      "pf=R2/Z2\n",
+      "I1=100.0/sqrt(3.0)\n",
+      "I_20=I1/K\n",
+      "Z2_rot=E20/I_20\n",
+      "r=sqrt((Z2_rot*Z2_rot)-(X20*X20))-R2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The rotor current at starting with slip-rings shorted is %.2f A.\" %(I20)\n",
+      "print \"(b)The rotor power factor at starting with slip-rings shorted is %.5f,lagging.\" %(pf_20)\n",
+      "print \"(c)The rotor current while running at full load with slip-rings shorted is %.3f A.\" %(I2)\n",
+      "print \"(d)The rotor power factor while running at full-load with slip-rings shorted is %.5f,lagging.\" %(pf)\n",
+      "print \"(e)The external rotor resistance is %.4f Ohms.\" %(r)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The rotor current at starting with slip-rings shorted is 1156.56 A.\n",
+        "(b)The rotor power factor at starting with slip-rings shorted is 0.04794,lagging.\n",
+        "(c)The rotor current while running at full load with slip-rings shorted is 741.266 A.\n",
+        "(d)The rotor power factor while running at full-load with slip-rings shorted is 0.76822,lagging.\n",
+        "(e)The external rotor resistance is 1.2835 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb
new file mode 100755
index 00000000..7078e036
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb
@@ -0,0 +1,1654 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16: DC MACHINES "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.1,Page number: 518"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the terminal voltage,output current and total power generated by a dc generator.\"\"\" \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "e=2.1                 #Average emf generated in each conductor(in Volts) \n",
+      "full_load_I=200.0     #Full-load current(in Amperes)    \n",
+      "Z=480.0               #Number of conductors in armature\n",
+      "P=8.0                 #Number of poles\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "A_1=P\n",
+      "E_1=e*(Z/A_1)\n",
+      "I_L_1=full_load_I*A_1\n",
+      "Po_1=E_1*I_L_1\n",
+      "A_2=2\n",
+      "E_2=e*(Z/A_2)\n",
+      "I_L_2=full_load_I*A_2\n",
+      "Po_2=E_2*I_L_2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print(\"(a) When the armature is lap wound \\n \")\n",
+      "print \"The terminal voltage on no load is %.2f V\" %(E_1)\n",
+      "print \"The output current on full load is %.2f A\" %(I_L_1) \n",
+      "print \"The total power generated on full load is %e W\" %(Po_1)\n",
+      "print(\"\\n(b)When the armature is wave wound \\n \")\n",
+      "print \"The terminal voltage on no load is %.2f V\" %(E_2)\n",
+      "print \"The output current on full load is %.2f A\" %(I_L_2) \n",
+      "print \"The total power generated on full load is %e W\" %(Po_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) When the armature is lap wound \n",
+        " \n",
+        "The terminal voltage on no load is 126.00 V\n",
+        "The output current on full load is 1600.00 A\n",
+        "The total power generated on full load is 2.016000e+05 W\n",
+        "\n",
+        "(b)When the armature is wave wound \n",
+        " \n",
+        "The terminal voltage on no load is 504.00 V\n",
+        "The output current on full load is 400.00 A\n",
+        "The total power generated on full load is 2.016000e+05 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2,Page number: 520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf induced in the armature of a dc generator. \"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "slots=65.0            #Number of slots\n",
+      "cond_per_slot=12.0    #Number of conductors per slot\n",
+      "A=4.0                 #Number of parallel paths\n",
+      "P=4.0                 #Number of poles\n",
+      "flux=0.02             #Flux per pole(in Webers)   \n",
+      "N=1200.0              #Speed of operation of the dc generator(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=slots*cond_per_slot\n",
+      "E=(flux*Z*N*P)/(60*A)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The emf induced in the armature is %.2f V\" %(E)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The emf induced in the armature is 312.00 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3,Page number: 520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\" Finding the induced emf in a dc machine.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N1=500.0              #Initial speed of operation of dc machine(in rpm)  \n",
+      "E1=180.0              #Induced emf at 500rpm(in Volts)\n",
+      "N2=600.0              #New speed of operation(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "E2=(N2/N1)*E1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The induced emf when the machine runs at 600 rpm is %.2f V\" %(E2) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The induced emf when the machine runs at 600 rpm is 216.00 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.4,Page number: 520"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"\"Finding the percentage increase in the field flux in a dc generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N1=750.0              #Initial speed of operation of dc machine(in rpm)\n",
+      "E1=220.0              #Induced emf at 750 rpm(in Volts) \n",
+      "E2_a=250.0            #New emf(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations and Result:\n",
+      "N2_a=(E2_a/E1)*N1\n",
+      "E2_b=250.0\n",
+      "N2_b=600.0\n",
+      "flux_ratio=(E2_b/E1)*(N1/N2_b)\n",
+      "   \n",
+      "    \n",
+      "#Result:\n",
+      "print \"(a)The speed at which the induced emf is 250V(assuming the flux to be constant) is %d rpm \" %(round(N2_a,0))\n",
+      "print \"(b)The required percentage increase in the field flux so that the induced emf is 250V,while the speed is only 600rpm is %d percent\" %(round(((flux_ratio-1)*100),0))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The speed at which the induced emf is 250V(assuming the flux to be constant) is 852 rpm \n",
+        "(b)The required percentage increase in the field flux so that the induced emf is 250V,while the speed is only 600rpm is 42 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.5,Page number: 525 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\" Finding the emf induced in the armature.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=440.0               #Load voltage(in Volts) \n",
+      "Rsh=110.0             #Resistance of shunt field coil(in Ohms)\n",
+      "Ra=0.02               #Armature resistance(in Ohms)\n",
+      "I_L=496.0             #Load current(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ish=V/Rsh\n",
+      "Ia=I_L+Ish\n",
+      "Eg=V+(Ia*Ra)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The emf induced in the armature is %.2f V\" %(Eg)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The emf induced in the armature is 450.00 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.6,Page number: 525 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\" Finding the total armature current and the generated emf.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "no_of_lamps=100.0         #Number of lamps\n",
+      "P=60.0                    #Power rating of each lamp(in Watts) \n",
+      "V=200                     #Voltage rating of each lamp(in Volts)\n",
+      "Ra=0.2                    #Armature resistance(in Ohms)\n",
+      "Rsh=50                    #Shunt field resistance(in Ohms)\n",
+      "Poles=4.0                 #Number of poles\n",
+      "no_of_brushes=2.0         #Number of brushes\n",
+      "brush_drop_per_brush=1.0  #Brush drop at each brush(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=P/V\n",
+      "I_L=no_of_lamps*I1\n",
+      "Ish=V/Rsh\n",
+      "Ia=Ish+I_L\n",
+      "A=Poles\n",
+      "Ic=Ia/A\n",
+      "brush_drop=no_of_brushes*brush_drop_per_brush\n",
+      "Eg=V+(Ia*Ra)+brush_drop\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The total armature current is %.2f A\" %(Ia)\n",
+      "print \"The current per path is %.2f A\" %(Ic)\n",
+      "print \"The generated emf is %.2f V\" %(Eg)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total armature current is 34.00 A\n",
+        "The current per path is 8.50 A\n",
+        "The generated emf is 208.80 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.7,Page number: 525"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf generated in a compound-wound dc generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=250.0                  #Load voltage(in Volts)\n",
+      "Rsh=130.0                #Shunt field resistance(in Ohms)\n",
+      "Ra=0.1                   #Armature resistance(in Ohms)\n",
+      "Rse=0.1                  #Series field resistance(in Ohms)\n",
+      "I_L=100.0                #Load current(in Amperes)\n",
+      "no_of_brushes=2          #Number of brushes  \n",
+      "brush_drop_per_brush=1.0 #Brush drop at each brush(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ise=I_L\n",
+      "Vse=Ise*Rse\n",
+      "Vsh=V+Vse\n",
+      "Ish=Vsh/Rsh\n",
+      "Ia=I_L+Ish\n",
+      "brush_drop=no_of_brushes*brush_drop_per_brush\n",
+      "Eg=V+Vse+(Ia*Ra)+brush_drop\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The emf generated is %.2f V.\" %(Eg)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The emf generated is 272.20 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.8,Page number: 528 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf generated,the copper losses,and efficiency of a shunt generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Po=30e03              #Full-load output power(in Watts)\n",
+      "V=200.0               #Terminal voltage(in Volts)\n",
+      "Ra=0.05               #Armature resistance(in Ohms)\n",
+      "Rsh=50.0              #Shunt field resistance(in Ohms)\n",
+      "loss=1000.0           #Friction losses(in Watts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_L=Po/V\n",
+      "Ish=V/Rsh\n",
+      "Ia=Ish+I_L\n",
+      "Eg=V+(Ia*Ra)\n",
+      "copper_loss=(pow(Ish,2)*Rsh)+(pow(Ia,2)*Ra)\n",
+      "effi=Po/(Po+copper_loss+loss)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The emf generated is %.2f V\" %(Eg)\n",
+      "print \"(b)The copper loss is %.2f W\" %(copper_loss)\n",
+      "print \"(c)The efficiency is %.2f percent\" %(effi*100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The emf generated is 207.70 V\n",
+        "(b)The copper loss is 1985.80 W\n",
+        "(c)The efficiency is 90.95 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.9,Page number:529 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding armature resistance and the load-current corresponding to maximum efficiency of dc shunt generator.\"\"\"\n",
+      "\n",
+      "from math import sqrt,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=210.0                #Full-load voltage(in Volts)\n",
+      "I_L=195.0              #Full-load current(in Amperes)\n",
+      "Rsh=52.5               #Shunt field resistance(in Ohms)\n",
+      "effi=0.90              #Full-load efficiency\n",
+      "stray_loss=710.0       #Stray losses(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Po=V*I_L\n",
+      "Pin=Po/effi\n",
+      "total_loss=Pin-Po\n",
+      "Ish=V/Rsh\n",
+      "Ia=I_L+Ish\n",
+      "sh_copp_loss=pow(Ish,2)*Rsh\n",
+      "const_loss=sh_copp_loss+stray_loss\n",
+      "arma_copp_loss=total_loss-const_loss\n",
+      "Ra=arma_copp_loss/(pow(Ia,2))\n",
+      "Ia_max_effi=sqrt(const_loss/Ra)\n",
+      "I_L_max_effi=Ia_max_effi-Ish\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The armature resistance is %.5f ohm \" %(Ra)\n",
+      "print \"The load current corresponding to maximum efficiency is %.2f A\" %(I_L_max_effi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The armature resistance is 0.07576 ohm \n",
+        "The load current corresponding to maximum efficiency is 139.04 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.10,Page number: 534 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of series turns required per pole for a level-compounded generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I_full_load=100.0     #Full-load current(in Amperes)\n",
+      "sh_turns=1500.0       #Number of turns in the shunt winding   \n",
+      "Ish_no_load=4.0       #Shunt current at no-load(in Amperes) \n",
+      "Ish_full_load=6.0     #Shunt current at full-load(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "At_no_load=Ish_no_load*sh_turns\n",
+      "At_full_load=Ish_full_load*sh_turns\n",
+      "At_series=At_full_load-At_no_load\n",
+      "Nse=At_series/I_full_load\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The number of series turns required per pole is %d \" %(Nse)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of series turns required per pole is 30 \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.11,Page number: 536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the back emf generated in a dc shunt motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I_L=41.0              #Full-load current(in Amperes)\n",
+      "V=250.0               #Full-load voltage(in Volts) \n",
+      "Ra=0.1                #Armature resistance(in Ohms) \n",
+      "Rsh=250.0             #Shunt field resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ish=V/Rsh\n",
+      "Ia=I_L-Ish\n",
+      "Eb=V-(Ia*Ra)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The back emf generated in the motor is %.2f V\" %(Eb)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The back emf generated in the motor is 246.00 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.12,Page number: 536 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed of a dc motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "A=2.0                 #Number of parallel paths\n",
+      "P=4.0                 #Number of poles \n",
+      "Z=888.0               #Number of conductors \n",
+      "flux=23e-03           #Flux per pole(in Webers)  \n",
+      "Ia=50.0               #Armature current(in Amperes)\n",
+      "Ra=0.28               #Armature resistance(in Ohms) \n",
+      "V=440.0               #Rated voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Eb=V-(Ia*Ra)\n",
+      "N=(60*A*Eb)/(flux*Z*P)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The speed of the motor is %d rpm\" %(round(N,0)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The speed of the motor is 626 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.13,Page number: 536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed of a dc motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V1=460.0              #Initial supply voltage(in Volts)\n",
+      "N1=900.0              #Speed of motor at 460-V(in rpm)\n",
+      "V2=200.0              #Final supply voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "kflux=V1/N1\n",
+      "N2=V2/(0.7*kflux)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "ans_N2=\"The approximate speed of the motor when the motor is connected across a 200V supply is %d rpm\" %(round(N2,0))\n",
+      "print(ans_N2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The approximate speed of the motor when the motor is connected across a 200V supply is 559 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.14,Page number: 537"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed and the gross torque developed by the armature of a dc motor.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=480.0                #Rated voltage(in Volts) \n",
+      "Ia=110.0               #Armature current at rated voltage(in Amperes)\n",
+      "Ra=0.2                 #Armature resistance(in Ohms) \n",
+      "flux=50e-03            #Flux per pole(in Webers)  \n",
+      "A=6.0                  #Number of parallel paths\n",
+      "P=6.0                  #Number of poles \n",
+      "Z=864.0                #Number of conductors\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Eb=V-(Ia*Ra)\n",
+      "N=(60*A*Eb)/(flux*Z*P)\n",
+      "torque=((flux*Z)/(2*pi))*(P/A)*Ia\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The speed of the motor is %d rpm \" %(round(N,0))\n",
+      "print \"(b)The gross torque developed by the armature is %.2f Nm\" %(torque)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The speed of the motor is 636 rpm \n",
+        "(b)The gross torque developed by the armature is 756.30 Nm\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.15,Page number: 538 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power generated in the armature winding of a dc generator.\"\"\"\n",
+      "\n",
+      "from math import pi \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=900.0               #Operating speed of generator(in rpm)\n",
+      "torque=2e03           #Torque(in N-metre) \n",
+      "P_losses=8e03         #Power losses(in Watts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pin=(2*pi*torque*N)/60.0\n",
+      "Pd=Pin-P_losses\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The power generated in the armature winding is %e W\" %(Pd)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power generated in the armature winding is 1.804956e+05 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.16,Page number: 540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed of a series motor when the current changes.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=230.0               #Supply voltage(in Volts)\n",
+      "Ra=0.12               #Armature resistance(in Ohms) \n",
+      "Rse=0.03              #Series field resistance(in Ohms)\n",
+      "Ia1=110.0             #Current at 230 V(in Amperes)\n",
+      "flux1=24e-03          #Flux per pole at 110 A(in Webers)\n",
+      "N1=600.0              #Speed at 230 V(in rpm) \n",
+      "Ia2=50.0              #Armature current(in Amperes) \n",
+      "flux2=16e-03          #Flux per pole at 50 A(in Webers)  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Eb1=V-Ia1*(Ra+Rse)\n",
+      "k=Eb1/(N1*flux1)\n",
+      "Eb2=V-Ia2*(Ra+Rse)\n",
+      "N2=Eb2/(k*flux2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The speed of the motor when the currenthas fallen to 50 A is %d rpm\" %(round(N2,0))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The speed of the motor when the currenthas fallen to 50 A is 938 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.17,Page number: 540"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current drawn by the machine.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=250.0                #Supply voltage(in Volts)\n",
+      "Ra=0.2                 #Armature resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Eb_1=0\n",
+      "Ia_1=(V-Eb_1)/Ra\n",
+      "Eb_2=200\n",
+      "Ia_2=(V-Eb_2)/Ra\n",
+      "Eb_3=250\n",
+      "Ia_3=(V-Eb_3)/Ra\n",
+      "Eb_4=-250\n",
+      "Ia_4=(V-Eb_4)/Ra\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)When the machine is at rest,\"\n",
+      "print \"The current drawn by the machine is %.2f A\" %(Ia_1)\n",
+      "print \"(b)When the machine is generating an emf of 200V and is connected to the supply with correct polarities,\"\n",
+      "print \"The current drawn by the machine is %.2f A\" %(Ia_2)\n",
+      "print \"(c)When the machine is generating an emf of 250V and is connected to the supply with correct polarities,\"\n",
+      "print \"The current drawn by the machine is %.2f A\" %(Ia_3)\n",
+      "print \"(d)When the machine is generating an emf of 250V and is connected to the supply with reversed polarities,\"\n",
+      "print \"The current drawn by the machine is %.2f A\" %(Ia_4) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)When the machine is at rest,\n",
+        "The current drawn by the machine is 1250.00 A\n",
+        "(b)When the machine is generating an emf of 200V and is connected to the supply with correct polarities,\n",
+        "The current drawn by the machine is 250.00 A\n",
+        "(c)When the machine is generating an emf of 250V and is connected to the supply with correct polarities,\n",
+        "The current drawn by the machine is 0.00 A\n",
+        "(d)When the machine is generating an emf of 250V and is connected to the supply with reversed polarities,\n",
+        "The current drawn by the machine is 2500.00 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.18,Page number: 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed and the gross torque developed by the armature of a dc series motor.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=6.0                 #Number of poles\n",
+      "A=6.0                 #Number of parallel paths \n",
+      "Z=864.0               #Number of conductors  \n",
+      "flux=50e-03           #Flux per pole(in Webers)\n",
+      "Ia=110.0              #Armature current(in Amperes)\n",
+      "V=480.0               #Load voltage(in Volts)\n",
+      "Ra=0.18               #Armature resistance(in Ohms) \n",
+      "Rse=0.02              #Series field resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Eb=V-Ia*(Ra+Rse)\n",
+      "N=(60*A*Eb)/(flux*Z*P)\n",
+      "torque=(60*Eb*Ia)/(2*pi*N)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The speed of the motor is %d rpm\" %(round(N,0))\n",
+      "print \"(b)The gross torque developed by the armature is %.2f Nm \" %(torque)    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The speed of the motor is 636 rpm\n",
+        "(b)The gross torque developed by the armature is 756.30 Nm \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.19,Page number: 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the series resistance to reduce the speed of a shunt motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=220.0               #Rated voltage of the motor(in Volts)\n",
+      "Ia=22.0               #Armature current(in Amperes)\n",
+      "Ra=0.45               #Armature resistance(in Ohms)\n",
+      "N1=700.0              #Initial speed of motor(in rpm)\n",
+      "N2=450.0              #Final speed of motor(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "E1=V-(Ia*Ra)\n",
+      "E2=(N2/N1)*E1\n",
+      "R=((V-E2)/Ia)-Ra\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance that should be placed in series with the armature to reduce the speed to 450 rpm is %.3f ohm \" %(R)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance that should be placed in series with the armature to reduce the speed to 450 rpm is 3.411 ohm \n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.20,Page number: 541"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed of a dc series motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=230.0               #Rated voltage of the dc series motor(in Volts) \n",
+      "Ra=0.2                #Armature resistance(in Ohms) \n",
+      "Rse=0.1               #Series field resistance(in Ohms)  \n",
+      "Ia1=40.0              #Line current at rated voltage(in Amperes)\n",
+      "N1=1000.0             #Speed of motor at rated voltage(in rpm)\n",
+      "Ia2=20.0              #Line current at 230 V(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Eb1=V-Ia1*(Ra+Rse)\n",
+      "Eb2=V-Ia2*(Ra+Rse)\n",
+      "\"\"\" Eb=k*flux*N \"\"\"\n",
+      "N2=(Eb2*N1)/(Eb1*0.6)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The speed of the motor for a line current of 20A at 230V is %d rpm\" %(round(N2,0)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The speed of the motor for a line current of 20A at 230V is 1713 rpm\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.21,Page number: 543 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the terminal voltage of a dc shunt generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=4                    #Number of poles\n",
+      "turns=260              #Number of turns in the armature winding\n",
+      "R=0.006                #Resistance of each turn of armature(in Ohms)\n",
+      "flux=0.08              #Useful flux per pole(in Webers)\n",
+      "I_L=55                 #Load current(in Amperes)\n",
+      "N=1000                 #Speed of the generator(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Z=2*turns\n",
+      "A=P\n",
+      "Eg=(flux*Z*N*P)/(60.0*A)\n",
+      "Rw=turns*R\n",
+      "R1=Rw/4.0\n",
+      "Ra=R1/4\n",
+      "V=Eg-(I_L*Ra)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The terminal voltage of the generator when it is running at 1000 rpm and supplying load current of 55 A is %.3f V.\" %(V)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The terminal voltage of the generator when it is running at 1000 rpm and supplying load current of 55 A is 687.971 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.22,Page number: 544"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the armature current,the emf induced and the flux per pole for a dc shunt generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=8.0                 #Number of poles\n",
+      "A=2.0                 #Number of parallel paths\n",
+      "Z=778                 #Number of conductors\n",
+      "V=250.0               #Load voltage(in Volts)\n",
+      "R_L=12.5              #Load resistance(in Ohms)\n",
+      "Ra=0.24               #Armature resistance(in Ohms)\n",
+      "Rsh=250.0             #Shunt field resistance(in Ohms)\n",
+      "N=500                 #Speed of the dc shunt generator(in rpm)    \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_L=V/R_L\n",
+      "Ish=V/Rsh\n",
+      "Ia=I_L+Ish\n",
+      "Eg=V+(Ia*Ra)\n",
+      "flux=(60.0*A*Eg)/(Z*N*P)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The armature current is %.2f A.\" %(Ia)\n",
+      "print \"The emf induced is %.2f V.\" %(Eg)\n",
+      "print \"The flux per pole is %e Wb.\" %(flux)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The armature current is 21.00 A.\n",
+        "The emf induced is 255.04 V.\n",
+        "The flux per pole is 9.834447e-03 Wb.\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.23,Page number: 544"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the percentage reduction in speed of dynamo.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Po_1=500e03             #Initial power output(in Watts)\n",
+      "Po_2=250e03             #Final power output(in Watts)\n",
+      "V=500.0                   #Constant excitation voltage(in Volts)\n",
+      "Ra=0.015                #Resistance between the terminals of dynamo(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ia_1=Po_1/V\n",
+      "E1=V+(Ia_1*Ra)\n",
+      "Ia_2=Po_2/V\n",
+      "E2=V+(Ia_2*Ra)\n",
+      "\"\"\" Since excitation emf remains constant in the two cases,we have \n",
+      "\n",
+      "    E=(flux*Z*N*P)/(60*A) where E=emf generated;Z=number of conductors;N=speed of motor(in rpm);P=number of poles;A=number of parallel paths;\n",
+      "                                flux=useful flux per pole(in Wb).\n",
+      "                                \n",
+      "    N=KE, where K is a constant.\n",
+      "    \n",
+      "    Hence,fractional reduction in speed is given as,\n",
+      "    \n",
+      "    (N1-N2)/N1=((K*(E1-E2))/(K*E1)). \"\"\"\n",
+      "\n",
+      "fract=((E1-E2)/E1)*100\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The percentage reduction in speed of the dynamo is %.3f percent.\" %(fract)    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage reduction in speed of the dynamo is 1.456 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.24,Page number: 545 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage between the far end of the feeder and the bus-bar of a dc series generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Rt=0.3                #Resistance of transmission line(in Ohms) \n",
+      "I_L_1=160.0           #Load current in first case(in Amperes) \n",
+      "I_L_2=50.0           #Load current in second case(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vt_1=I_L_1*Rt\n",
+      "Vb_1=(50.0/200.0)*I_L_1\n",
+      "Vd_1=Vt_1-Vb_1\n",
+      "Vt_2=I_L_2*Rt\n",
+      "Vb_2=(50.0/200.0)*I_L_2\n",
+      "Vd_2=Vt_2-Vb_2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The voltage between far end of the feeder and the bus-bar at a cusrrent of 160 A is %.2f V.\" %(Vd_1)\n",
+      "print \"(b)The voltage between far end of the feeder and the bus-bar at a cusrrent of 50 A is %.2f V.\" %(Vd_2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The voltage between far end of the feeder and the bus-bar at a cusrrent of 160 A is 8.00 V.\n",
+        "(b)The voltage between far end of the feeder and the bus-bar at a cusrrent of 50 A is 2.50 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.25,Page number: 545"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\" Finding the emf generated and the armature current in a dc long-shunt compound generator. \"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I_L=50                #Load current(in Amperes)\n",
+      "V=500                 #Terminal voltage(in Volts)\n",
+      "Ra=0.05               #Armature resistance(in Ohms)\n",
+      "Rse=0.03              #Series field resistance(in Ohms)\n",
+      "Rsh=250               #Shunt field resistance(in Ohms)\n",
+      "brush_drop=1.0        #Brush contact drop(in Volts)    \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ish=V/Rsh\n",
+      "Ia=Ish+I_L\n",
+      "Eg=V+(Ia*(Ra+Rse))+brush_drop\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The armature current is %.2f A.\" %(Ia)\n",
+      "print \"The emf generated is %.2f V\" %(Eg)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The armature current is 52.00 A.\n",
+        "The emf generated is 505.16 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.26,Page number: 545"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\" Finding the voltage and the power generated by a dc generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=8                   #Number of poles\n",
+      "Z=500                 #Number of conductors on the armature\n",
+      "flux =0.02            #Magnetic flux per pole(in Webers)\n",
+      "N=1800                #Speed of the generator(in rpm)\n",
+      "I=5.0                 #Allowable current per path(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "A_1=2\n",
+      "Eg_1=(flux*Z*N*P)/(60*A_1)\n",
+      "A_2=P\n",
+      "Eg_2=(flux*Z*N*P)/(60*A_2)\n",
+      "Ia_1=A_1*I\n",
+      "Pd_1=Eg_1*Ia_1\n",
+      "Ia_2=A_2*I\n",
+      "Pd_2=Eg_2*Ia_2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"When the armature is wave wound:\"\n",
+      "print \"(a)The generated voltage is %.2f V.\" %(Eg_1)\n",
+      "print \"(b)The kW generated by the machine is %.2f kW.\" %(Pd_1/1000.0)\n",
+      "print \"\\nWhen the armature is lap wound:\"\n",
+      "print \"(a)The generated voltage is %.2f V.\" %(Eg_2)\n",
+      "print \"(b)The kW generated by the machine is %.2f kW.\" %(Pd_2/1000.0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When the armature is wave wound:\n",
+        "(a)The generated voltage is 1200.00 V.\n",
+        "(b)The kW generated by the machine is 12.00 kW.\n",
+        "\n",
+        "When the armature is lap wound:\n",
+        "(a)The generated voltage is 300.00 V.\n",
+        "(b)The kW generated by the machine is 12.00 kW.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.27,Page number: 546"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf generated and the copper losses in a dc shunt generator.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=250.0               #Terminal voltage(in Volts)\n",
+      "I_L=195               #Load current(in Amperes)\n",
+      "Ra=0.02               #Armature resistance(in Ohms)\n",
+      "Rsh=50.0              #Shunt-field resistance(in Ohms)\n",
+      "loss=950.0            #Iron and frictional losses(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ish=V/Rsh\n",
+      "Ia=I_L+Ish\n",
+      "Eg=V+(Ia*Ra)\n",
+      "copp=(Ia*Ia*Ra)+(V*Ish)\n",
+      "Po=V*I_L\n",
+      "tot_loss=copp+loss\n",
+      "Pin=Po+tot_loss\n",
+      "Pe=Pin-loss\n",
+      "mech_effi=Pe/Pin\n",
+      "ele_effi=Po/Pe\n",
+      "comm_effi=Po/Pin\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The emf generated is %.2f V.\" %(Eg)\n",
+      "print \"(b)The copper losses is %.2f W.\" %(copp)\n",
+      "print \"(c)The output of the prime mover is %.3f kW.\" %(Pin/1000.0)\n",
+      "print \"(d)The commercial efficiency is %.2f.\\n   The mechanical efficiency is %.2f.\" %((comm_effi*100),(mech_effi*100))\n",
+      "print \"   The electrical efficiency is %.2f.\" %(ele_effi*100) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The emf generated is 254.00 V.\n",
+        "(b)The copper losses is 2050.00 W.\n",
+        "(c)The output of the prime mover is 51.750 kW.\n",
+        "(d)The commercial efficiency is 94.20.\n",
+        "   The mechanical efficiency is 98.16.\n",
+        "   The electrical efficiency is 95.96.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.28,Page number: 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the back emf generated by a dc shunt motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=250.0               #Terminal Voltage(in Volts)\n",
+      "I_L1=2.0              #No-load current(in Amperes)\n",
+      "N1=1000.0             #No-load speed(in rpm)\n",
+      "Ra=0.2                #Armature resistance(in Ohms)\n",
+      "Rsh=250.0             #Field resistance(in Ohms)\n",
+      "I_L2=51.0             #Current after loading(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ish=V/Rsh\n",
+      "Ia1=I_L1-Ish\n",
+      "E1=V-(Ia1*Ra)\n",
+      "Ia2=I_L2-Ish\n",
+      "E2=V-(Ia2*Ra)\n",
+      "\"\"\"As the motor is shunt-wound,the flux remains constant.The emf generated is directly proportional to the speed.\"\"\"\n",
+      "N2=(E2/E1)*N1\n",
+      "speed_drop=(N1-N2)/N1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The back emf generated at no-load is %.3f V.\" %(E1)\n",
+      "print \"(b)On loading,\\n   The back emf generated is %.2f V.\\n   The speed of the motor is %d rpm.\" %(E2,round(N2,0))\n",
+      "print \"   The percentage speed drop is %.3f percent.\" %(speed_drop*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The back emf generated at no-load is 249.800 V.\n",
+        "(b)On loading,\n",
+        "   The back emf generated is 240.00 V.\n",
+        "   The speed of the motor is 961 rpm.\n",
+        "   The percentage speed drop is 3.923 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.29,Page number: 547"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of starting resistance for a shunt motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Po=14920.0            #Output power(in Watts)\n",
+      "V=240.0               #Supply voltage(in Volts)\n",
+      "Ra=0.25               #Armature resistance(in Ohms)\n",
+      "effi=0.86             #Efficiency at full-load\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Pin=Po/effi\n",
+      "I_L=Pin/V\n",
+      "Ist=1.5*I_L\n",
+      "Rt=V/Ist\n",
+      "Rst=Rt-Ra\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The starting resistance for the shunt motor is %.3f Ohms.\" %(Rst)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The starting resistance for the shunt motor is 1.963 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.30,Page number: 548"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed and efficiency of a dc shunt motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I_L1=4.0              #No-load current(in Amperes)\n",
+      "N1=1000               #No-load speed(in rpm)\n",
+      "V=500.0               #Voltage rating of the dc shunt motor(in Volts)\n",
+      "Ra=0.2                #Armature resistance(in Ohms)\n",
+      "Ish=1.0               #Field current(in Amperes)\n",
+      "I_L2=100.0            #Full-load current(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ia1=I_L1-Ish\n",
+      "E1=V-(Ia1*Ra)\n",
+      "Ia2=I_L2-Ish\n",
+      "E2=V-(Ia2*Ra)\n",
+      "\"\"\" For a shunt motor,the flux remains constant and hence E is directly proportional to speed of the motor(N).\n",
+      "    \n",
+      "    E=kN where k is a constant. \"\"\"\n",
+      "N2=(E2/E1)*N1\n",
+      "\"\"\"At no-load,the power taken by the motor mainly meets the constant losses(iron and frictional losses).\"\"\"\n",
+      "Pc=V*I_L1\n",
+      "\"\"\"On loading,the copper loss in shunt field winding is negligible compared to the copper loss in armature winding.\"\"\"\n",
+      "Pv=Ia2*Ia2*Ra\n",
+      "Pin=V*I_L2\n",
+      "effi=(Pin-(Pv+Pc))/Pin\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The speed of the dc shunt motor on loading is %d rpm.\" %(round(N2,0))\n",
+      "print \"The efficiency of the motor is %.2f percent.\" %(effi*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The speed of the dc shunt motor on loading is 962 rpm.\n",
+        "The efficiency of the motor is 92.08 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.31,Page number: 548"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the speed of a dc generator running as a shunt motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Ra=0.02               #Armature resistance(in Ohms)\n",
+      "Rsh=50.0              #Shunt-field resistance(in Ohms)\n",
+      "V=250.0               #Terminal voltage(in Volts)\n",
+      "Po=50e03              #Output power(in Watts)\n",
+      "N1=500.0              #Speed of the dc generator(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ish=V/Rsh\n",
+      "\"\"\" When working as a generator,the machine supplies a load of 50 kW at 250 V. \"\"\"\n",
+      "I_L=Po/V\n",
+      "Ia1=I_L+Ish\n",
+      "E1=V+(Ia1*Ra)\n",
+      "\"\"\"When working as a motor,the machine takes a power of 50 kW at 250 V. \"\"\"\n",
+      "Ia2=I_L-Ish\n",
+      "E2=V-(Ia2*Ra)\n",
+      "N2=(E2/E1)*N1\n",
+      "\"\"\" NOTE: The field current and the flux per pole is same in both cases.\"\"\"\n",
+      "\n",
+      "#Result:\n",
+      "print \"The speed of the machine running as a shunt motor is %d rpm.\" %(round(N2,0)) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The speed of the machine running as a shunt motor is 484 rpm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.32,Page number: 548"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the applied voltage and the current to run the motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Ra=0.6                #Resistance of the armature(in Ohms)\n",
+      "Rse=0.4               #Series field resistance(in Ohms)\n",
+      "Ia1=20.0              #Initial armature current(in Amperes)\n",
+      "V1=400.0              #Initial terminal voltage(in Volts)\n",
+      "N1=250.0              #Initial speed of the motor(in rpm)\n",
+      "N2=350.0              #Final speed of the motor(in rpm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" In a series motor,torque is directly proprtional to the square of the armature current(Ia).\n",
+      "    \n",
+      "    Given: Torque is directly proprtional to the square of the speed(N).\n",
+      "    \n",
+      "    Therefore, Ia is directly proportional to N. Ia=kN where k is a constant. \"\"\"\n",
+      "Ia2=(N2/N1)*Ia1\n",
+      "E1=V1-(Ia1*(Ra+Rse))\n",
+      "\"\"\"In a series motor,as the flux is directly proportional to Ia,the back emf is proportional to (Ia*N). \"\"\"\n",
+      "E2=E1*((Ia2*N2)/(Ia1*N1))\n",
+      "V2=E2+(Ia2*(Ra+Rse))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The applied voltage is %.2f V and the current is %.2f A to run the motor at 350 rpm.\" %(V2,Ia2) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The applied voltage is 772.80 V and the current is 28.00 A to run the motor at 350 rpm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter17.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter17.ipynb
new file mode 100755
index 00000000..e240ccf5
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter17.ipynb
@@ -0,0 +1,332 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17: FRACTIONAL HORSE POWER MOTORS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.1,Page number: 570\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the slip and efficiency of induction motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=50                  #Frequency rating of the induction motor(in Hertz) \n",
+      "P=4                   #Number of poles in the induction motor  \n",
+      "N=1410                #Speed of the motor(in rpm)\n",
+      "Po=375                #Output Power(in Watts)\n",
+      "V=230                 #Voltage rating of the induction motor(in Volts)    \n",
+      "I=2.9                 #Input current(in Amperes)\n",
+      "pf=0.71               #Power factor(lagging) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ns=(120.0*f)/P\n",
+      "slip=(Ns-N)/Ns\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"Slip is %.2f percent.\" %(slip*100)\n",
+      "Pin=V*I*pf\n",
+      "efficiency=Po/Pin\n",
+      "print \"The efficiency is %.2f percent.\" %(efficiency*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Slip is 6.00 percent.\n",
+        "The efficiency is 79.19 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.2,Page number: 570\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\n",
+      "\"\"\"Finding the currents and the power factor in the induction motor.\"\"\"\n",
+      "\n",
+      "from cmath import phase,rect,polar\n",
+      "from math import radians,degrees,cos\n",
+      "\n",
+      "#Variable Declartion:\n",
+      "V=rect(230,0)        #Voltage rating of the split-phase induction motor(in Volts)   \n",
+      "Z_M=5+ 12*1j         #Impedance of the main winding(in Ohms)\n",
+      "Z_A=12+ 5*1j         #Start-winding impedance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "mod_Z_M=abs(Z_M)\n",
+      "mod_Z_A=abs(Z_A)\n",
+      "phi_M=phase(Z_M)\n",
+      "phi_A=phase(Z_A)\n",
+      "I_M=V/Z_M\n",
+      "mod_I_M=abs(I_M)\n",
+      "phi_I_M=degrees(phase(I_M))\n",
+      "I_A=V/Z_A\n",
+      "mod_I_A=abs(I_A)\n",
+      "phi_I_A=degrees(phase(I_A))\n",
+      "I_L=I_M+I_A\n",
+      "mod_I_L=abs(I_L)\n",
+      "phi_I_L=degrees(phase(I_L))\n",
+      "phi=phi_I_A-phi_I_M\n",
+      "pf=cos(radians(phi_I_L))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current in the main winding is %.2f A at a phase angle of %.2f degrees.\" %(mod_I_M,phi_I_M)\n",
+      "print \"(b)The current in the starting winding is %.2f A at a phase angle of %.2f degrees.\" %(mod_I_A,phi_I_A)\n",
+      "print \"(c)The line current is %.2f A at a phase angle of %.2f degrees.\" %(mod_I_L,phi_I_L)\n",
+      "print \"(d)The phase displacement between the two winding currents is %.2f degrees.\" %(phi)\n",
+      "print \"(e)The power factor is %.4f lagging.\" %(pf)                "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current in the main winding is 17.69 A at a phase angle of -67.38 degrees.\n",
+        "(b)The current in the starting winding is 17.69 A at a phase angle of -22.62 degrees.\n",
+        "(c)The line current is 32.72 A at a phase angle of -45.00 degrees.\n",
+        "(d)The phase displacement between the two winding currents is 44.76 degrees.\n",
+        "(e)The power factor is 0.7071 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.3,Page number: 571\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the capacitance in series with the auxiliary winding to maximize starting torque.\"\"\"\n",
+      "\n",
+      "from math import radians,degrees,atan,pi,tan\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "X_M=20                #Inductive reactance of the main winding(in Ohm)\n",
+      "R_M=2                 #Resistance of the main winding(in Ohm)\n",
+      "X_A=5                 #Inductive reactance of the auxiliary winding(in Ohm)\n",
+      "R_A=25                #Resistance of the auxiliary winding(in Ohm)\n",
+      "f=50                  #Frequency rating of the split-phase induction motor(in Hertz)   \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "angle_M=atan(X_M/R_M)\n",
+      "angle_A=degrees(angle_M)-90\n",
+      "Xc=X_A-(R_A*tan(radians(angle_A)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "C=1/(2*pi*f*Xc)\n",
+      "print \"The value of capacitance connected in series with the auxiliary winding to obtain maximum starting torque is %e F.\" %(C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of capacitance connected in series with the auxiliary winding to obtain maximum starting torque is 4.244132e-04 F.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.4,Page number: 576\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resolution and shaft speed of a stepper motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "beta=2.5              #Step-angle of a stepper motor(in degrees)\n",
+      "step_freq=3600        #Stepping frequency(in pps)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "res=360/beta\n",
+      "number_steps=res*25\n",
+      "shaft_speed=(beta*step_freq)/360\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resolution is %d steps per revolution.\" %(res)\n",
+      "print \"(b)The number of steps required for the shaft to make 25 revolutions=%d.\" %(number_steps)\n",
+      "print \"(c)The shaft speed is %.2f rps.\" %(shaft_speed)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resolution is 144 steps per revolution.\n",
+        "(b)The number of steps required for the shaft to make 25 revolutions=3600.\n",
+        "(c)The shaft speed is 25.00 rps.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.5,Page number:577\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of stator and rotor poles in a VR motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "m=3                   #Number of phases\n",
+      "beta=15               #Step angle(in degrees)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Nr=360/(m*beta)\n",
+      "Ns1=(Nr*360)/(360-(beta*Nr))\n",
+      "Ns2=(Nr*360)/(360+(beta*Nr))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a) The number of rotor poles is %d.\" %(Nr)\n",
+      "print \"(b)\"\n",
+      "print \"   Case 1: Ns>Nr\"\n",
+      "print \"   The number of stator poles is %d. \\n\" %(Ns1)\n",
+      "print \"   Case 2: Ns<Nr\"\n",
+      "print \"   The number of stator poles is %d.\" %(Ns2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The number of rotor poles is 8.\n",
+        "(b)\n",
+        "   Case 1: Ns>Nr\n",
+        "   The number of stator poles is 12. \n",
+        "\n",
+        "   Case 2: Ns<Nr\n",
+        "   The number of stator poles is 6.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.6,Page number: 579\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of rotor and stator teeth in VR stepper motor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "m=4                   #Number of stacks\n",
+      "beta=1.8              #Step angle(in degrees)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Nr=360/(m*beta)\n",
+      "Ns=Nr\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The number of rotor teeth is %d.\" %(Nr)\n",
+      "print \"The number of stator teeth is %d.\" %(Ns)\n",
+      "print \"\\nNOTE: In a multistack stepper motor the number of stator teeth is same as that of the rotor teeth.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of rotor teeth is 50.\n",
+        "The number of stator teeth is 50.\n",
+        "\n",
+        "NOTE: In a multistack stepper motor the number of stator teeth is same as that of the rotor teeth.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter18.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter18.ipynb
new file mode 100755
index 00000000..99017984
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter18.ipynb
@@ -0,0 +1,542 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18: ELECTRICAL MEASURING INSTRUMENTS\n",
+      "\n",
+      "\n",
+      "\n"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.1,Page number: 598\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the deflecting torque in Newton-metres for a PMMC instrument.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=15e-03              #Current flowing through the coil(in Amperes) \n",
+      "B=0.2                 #Flux density in the air gap(in Tesla)\n",
+      "l=2e-02               #Length of the magnetic field(in m)\n",
+      "d=2.5e-02             #Mean width of the coil(in m) \n",
+      "r=d/2                 #Radius of the coil(in cm) \n",
+      "n1=42                 #Number of turns(lower limit) \n",
+      "n2=43                 #Number of turns(upper limit)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "F1=I*B*l*n1\n",
+      "F2=I*B*l*n2\n",
+      "net_torque=(F1+F2)*r\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The deflecting torque is %e Nm.\" %(net_torque)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The deflecting torque is 6.375000e-05 Nm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.2,Page Number: 604\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the shunt resistance for measuring a maximum current of 10 mA.\"\"\"     \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Ifsd=10e-03           #Maximum current(in Amperes)\n",
+      "Im=100e-06            #Full-scale deflection current(in Amperes)  \n",
+      "Rm=100                #Meter Resistance(in Ohms)  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ish=Ifsd-Im\n",
+      "Rsh=(Im*Rm)/Ish\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The shunt resistance needed is %.6f Ohms.\" %(Rsh)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The shunt resistance needed is 1.010101 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.3,Page number: 605"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Designing an universal shunt for a multi-range ammeter.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Im=100e-06            #Full-scale deflection current(in Amperes)\n",
+      "Rm=100.0              #Internal resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"For 1-mA range,the required shunt can be calculated as follows.\"\"\"\n",
+      "Ifsd=1e-03\n",
+      "Rsh=(Im*Rm)/(Ifsd-Im)\n",
+      "\n",
+      "Rm=900.0\n",
+      "R=100.0\n",
+      "\"\"\"(a)Range-switch at 1 mA:\"\"\"\n",
+      "Rm1=Rm\n",
+      "Ish1=(1e-03)-(0.1e-03)\n",
+      "Rsh1=(Rm1*Im)/(Ish1)\n",
+      "\n",
+      "\"\"\"(b)Range-switch at 10 mA:\"\"\"\n",
+      "\"\"\"Rm2=Rm+R1\"\"\"\n",
+      "Ish2=(10e-03)-(0.1e-03)\n",
+      "\"\"\"Rsh2=R2+R3+R4+R5=R-R1=100-R1;\"\"\"\n",
+      "\"\"\"Rsh2=(Rm2*Im)/Ish2;\"\"\"\n",
+      "R1=(9900.0-900.0)/100.0\n",
+      "\n",
+      "\"\"\"(c)Range-switch at 100 mA:\"\"\"\n",
+      "\"\"\"Rm3=Rm+R1+R2\"\"\"\n",
+      "Ish3=(100e-03)-(0.1e-03)\n",
+      "\"\"\"Rsh3=R3+R4+R5=R-R1-R2=100-90-R2=90-R2;\"\"\"\n",
+      "\"\"\"Rsh3=(Rm3*Im)/Ish3;\"\"\"\n",
+      "R2=(9990.0-990.0)/1000.0\n",
+      "\n",
+      "\"\"\"(d)Range-switch at 500 mA:\"\"\"\n",
+      "\"\"\"Rm4=Rm+R1+R2+R3\"\"\"\n",
+      "Ish3=(500e-03)-(0.1e-03)\n",
+      "\"\"\"Rsh4=R4+R5=R-R1-R2-R3=100-90-9-R3=1-R3;\"\"\"\n",
+      "\"\"\"Rsh4=(Rm4*Im)/Ish4;\"\"\"\n",
+      "R3=(4999.0-999.0)/5000.0\n",
+      "\n",
+      "\n",
+      "\"\"\"(e)Range-switch at 1 A:\"\"\"\n",
+      "\"\"\"Rm5=Rm+R1+R2+R3+R4\"\"\"\n",
+      "Ish3=(1000e-03)-(0.1e-03)\n",
+      "\"\"\"Rsh5=R5=R-R1-R2-R3-R4=100-90-9-0.8-R4=0.2-R4;\"\"\"\n",
+      "\"\"\"Rsh5=(Rm5*Im)/Ish5;\"\"\"\n",
+      "R4=(1999.8-999.8)/10000.0\n",
+      "R5=R-R1-R2-R3-R4\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistor values are: \"\n",
+      "print \"R1=%.2f Ohms.\" %(R1)\n",
+      "print \"R2=%.2f Ohms.\" %(R2)\n",
+      "print \"R3=%.2f Ohms.\" %(R3)\n",
+      "print \"R4=%.2f Ohms.\" %(R4)\n",
+      "print \"R5=%.2f Ohms.\" %(R5)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistor values are: \n",
+        "R1=90.00 Ohms.\n",
+        "R2=9.00 Ohms.\n",
+        "R3=0.80 Ohms.\n",
+        "R4=0.10 Ohms.\n",
+        "R5=0.10 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.4,Page number: 609"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"To convert a d'Arsonval meter movement into a voltmeter.\"\"\" \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Im=100e-06            #Current sensitivity(in Amperes)\n",
+      "Rm=100                #Resistance of the coil(in Ohms)\n",
+      "Vfsd=100              #Full-scale deflection of voltmeter(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Rs=(Vfsd/Im)-Rm\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"To convert the d'Arsonval meter movement into a voltmeter of range 100V, we connect a resistor Rs in series.\"\n",
+      "print \"Rs = %.3f kilo Ohms. \" %(Rs/1000)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "To convert the d'Arsonval meter movement into a voltmeter of range 100V, we connect a resistor Rs in series.\n",
+        "Rs = 999.900 kilo Ohms. \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.5,Page number: 609\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the multipiler resistance and the voltage mutiplyiing factor of a dc voltmeter.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Im=50e-06             #Full-scale deflection current(in Amperes)  \n",
+      "Rm=1e03               #Meter Resistance(in Ohms) \n",
+      "Vfsd=50               #Full-scale deflection of voltmeter(in Volts)     \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Rs=(Vfsd/Im)-Rm\n",
+      "n=Vfsd/(Im*Rm)\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The multiplier resistance needed is %.2f kilo Ohms.\" %(Rs/1000)\n",
+      "print \"(b)The voltage multiplying factor is %d.\" %(round(n,0))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The multiplier resistance needed is 999.00 kilo Ohms.\n",
+        "(b)The voltage multiplying factor is 1000.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.6,Page number: 612"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the reading and error in measurement of voltmeter.\"\"\"\n",
+      "    \n",
+      "\"\"\" NOTE: All resistances expressed in kilo Ohms.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "range_A=50            #Range of voltmeter-A(in Volts)\n",
+      "range_B=50            #Range of voltmeter-B(in Volts)\n",
+      "sens_A=1000           #Sensitivity of voltmeter-A(in Ohm/Volts)  \n",
+      "sens_B=20000          #Sensitivity of voltmeter-B(in Ohm/Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vt=150.0*(50.0/(100.0+50.0))\n",
+      "R_i1=(range_A*sens_A)/1000.0\n",
+      "Req=1/((1/R_i1)+(1.0/50))\n",
+      "V1=(150.0)*(Req/(100+Req))\n",
+      "R_i2=(range_B*sens_B)/1000.0\n",
+      "Req=1/((1/R_i2)+(1.0/50))\n",
+      "V2=(150.0)*(Req/(Req+100))\n",
+      "err_A=((Vt-V1)/Vt)*100.0\n",
+      "err_B=((Vt-V2)/Vt)*100.0\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The reading of voltmeter-A is %.2f V\\nThe reading of voltmeter-B is %.2f V.\" %(V1,V2)\n",
+      "print \"\\nThe error in the reading of voltmeter-A is %.2f percent.\" %(err_A)\n",
+      "print \"The error in the reading of voltmeter-B is %.2f percent.\" %(err_B)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The reading of voltmeter-A is 30.00 V\n",
+        "The reading of voltmeter-B is 48.39 V.\n",
+        "\n",
+        "The error in the reading of voltmeter-A is 40.00 percent.\n",
+        "The error in the reading of voltmeter-B is 3.23 percent.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.7,Page number: 617"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the deflection in an ammeter.\"\"\"\n",
+      "\n",
+      "from math import sin,asin,degrees,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I1=20.0               #Initial current(in Amperes) \n",
+      "I2=12.0               #Final current(in Amperes)\n",
+      "angle1=60             #Initial deflection(in degrees)\n",
+      "\"\"\" Given: Deflecting torque is directly proportional to the current.\"\"\"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"For spring control:  Controlling torque is directly proportional to deflection.\n",
+      "                        For steady state deflection, controlling torque=deflecting torque.\n",
+      "                        Therefore,deflection is directly proportional to current. \"\"\"\n",
+      "angle2_a=(I2/I1)*angle1\n",
+      "\"\"\"For gravity control: Controlling torque is directly proportional to sine of the deflection angle.\n",
+      "                        For steady state deflection, controlling torque=deflecting torque.\n",
+      "                        Therefore,sine of the angle of deflection is directly proportional to the current. \"\"\"  \n",
+      "angle2_b=asin((I2/I1)*sin(radians(angle1)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The deflection for a current of 12A when the instrument is spring controlled is %.2f degrees.\" %(angle2_a) \n",
+      "print \"(b)The deflection for a current of 12A when the instrument is gravity controlled is %.2f degrees.\" %(degrees(angle2_b)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The deflection for a current of 12A when the instrument is spring controlled is 36.00 degrees.\n",
+        "(b)The deflection for a current of 12A when the instrument is gravity controlled is 31.31 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.8,Page number: 618"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the deflection in degrees in a gravity-controlled instrument.\"\"\"\n",
+      "\n",
+      "from math import asin,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "W=0.005               #Controlling weight(in kilograms)\n",
+      "l=2.4e-02             #Distance of controlling weight from the axis(in metres)\n",
+      "torque=1.05e-04       #Deflecting torque(in kg-m)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "angle=asin(torque/(W*l))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The deflection corresponding to a deflecting torque of 1.05e-04 kg-m is %.2f degrees.\" %(degrees(angle))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The deflection corresponding to a deflecting torque of 1.05e-04 kg-m is 61.04 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.9,Page number: 618"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the deflection in an ammeter.\"\"\"\n",
+      "\n",
+      "from math import sin,asin,degrees,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I1=10.0               #Initial current(in Amperes) \n",
+      "I2=5.0                #Final current(in Amperes)\n",
+      "angle1=90             #Initial deflection(in degrees)\n",
+      "\"\"\" Given: Deflecting torque is directly proportional to square of the current.\"\"\"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"For spring control:  Controlling torque is directly proportional to deflection.\n",
+      "                        For steady state deflection, controlling torque=deflecting torque.\n",
+      "                        Therefore,deflection is directly proportional to square of the current. \"\"\"\n",
+      "angle2_a=pow((I2/I1),2)*angle1\n",
+      "\"\"\"For gravity control: Controlling torque is directly proportional to sine of the deflection angle.\n",
+      "                        For steady state deflection, controlling torque=deflecting torque.\n",
+      "                        Therefore,sine of the angle of deflection is directly proportional to square of the current. \"\"\"  \n",
+      "angle2_b=asin((pow((I2/I1),2))*sin(radians(angle1)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The deflection for a current of 5A when the instrument is spring controlled is %.2f degrees.\" %(angle2_a) \n",
+      "print \"(b)The deflection for a current of 5A when the instrument is gravity controlled is %.2f degrees.\" %(degrees(angle2_b))   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The deflection for a current of 5A when the instrument is spring controlled is 22.50 degrees.\n",
+        "(b)The deflection for a current of 5A when the instrument is gravity controlled is 14.48 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.10,Page number: 619 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current required to produce a deflection of 60 degrees in a moving coil instrument.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "w=4e-02                 #Width of the coil(in metres)\n",
+      "l=5e-02                 #Length of the coil(in metres)\n",
+      "N=80.0                  #Number of turns in the coil\n",
+      "torque_control=0.5e-07  #Controlling torque per degree deflection of the coil(in Nm)\n",
+      "B=0.1                   #Magnetic flux density in the air gap(in Wb per square-metre)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "A=w*l\n",
+      "torque_c=torque_control*60.0\n",
+      "I=torque_c/(B*N*A)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current required to give a deflection of 60 degrees is %e A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current required to give a deflection of 60 degrees is 1.875000e-04 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter2.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter2.ipynb
new file mode 100755
index 00000000..496486a8
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter2.ipynb
@@ -0,0 +1,1532 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2:OHM'S LAW"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1,Page number: 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Calculating the resistance of a wire.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\n",
+      "\"\"\" R=(rho*L)/A\n",
+      "     R=Resistance of conductor;\n",
+      "     L=Length of conductor;\n",
+      "     d=diameter of conductor;\n",
+      "     A=Area of Cross Section of conductor=(pi*d*d)/4; \"\"\"\n",
+      "\n",
+      "R1=5      #Resistance of conductor(in Ohms)\n",
+      "\n",
+      "\"\"\" L2=4L1;\n",
+      "    d2=0.5d1;\n",
+      "    A1=(pi*d1*d1)/4;\n",
+      "    A2=(pi*d2*d2)/4=(pi*d1*d1)/16 ; \n",
+      "    rho1=rho2=rho(As both the conductors are made of the same material);\n",
+      "    R1=(rho*L1)/A1;\n",
+      "    R2=(rho*L2)/L2; \n",
+      "    R2/R1=(rho*(L2/A2)/(rho*(L1/A1)) \"\"\"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R2=R1*(4/1)*(2/1)*(2/1)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The required value of resistance is: %d Ohms.\" % (R2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required value of resistance is: 80 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2,Page number: 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding resistance of Copper wire.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\"\"\" R=(rho*L)/A\n",
+      "     R=Resistance of conductor;\n",
+      "     L=Length of conductor;\n",
+      "     d=diameter of conductor;\n",
+      "     A=Area of Cross Section of conductor=(pi*d*d)/4; \"\"\"\n",
+      "\n",
+      "R1=10      #Resistance of Copper Wire(in Ohms)\n",
+      "\n",
+      "\"\"\" L2=3L1;\n",
+      "    d2=0.5d1;\n",
+      "    A2=A1/3;(If the length of the wire is made three times by drawing it,its area of cross section must \n",
+      "    decrease three times as the volume of the wire remains same in the drawing process.) \n",
+      "    rho1=rho2=rho(As both the conductors are made of the same material);\n",
+      "    R1=(rho*L1)/A1;\n",
+      "    R2=(rho*L2)/L2; \n",
+      "    R2/R1=(rho*(L2/A2)/(rho*(L1/A1))\n",
+      "    R2=R1*(L2/L1)*(A1/A2) \"\"\"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R2=R1*(3/1)*(3/1)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The required value of resistance is: %d Ohms.\" % (R2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required value of resistance is: 90 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3,Page number: 19  "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across the four resistors.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\"\"\" Req=(4+4)||(8+4) \"\"\"\n",
+      "\n",
+      "Req=(8.0*12)/(8+12) #Equivalent Resistance of the circuit(in Ohms)\n",
+      "I=5                 #Current in the circuit(in Amperes) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=I*Req\n",
+      "V1=(4.0*V)/(4+4)\n",
+      "V2=(4.0*V)/(4+4)\n",
+      "V3=(8.0*V)/(8+4)\n",
+      "V4=(4.0*V)/(8+4)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across resistor 1 is: %d V.\" % (V1)\n",
+      "print \"The voltage across resistor 2 is: %d V.\" % (V2)\n",
+      "print \"The voltage across resistor 3 is: %d V.\" % (V3)\n",
+      "print \"The voltage across resistor 4 is: %d V.\" % (V4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across resistor 1 is: 12 V.\n",
+        "The voltage across resistor 2 is: 12 V.\n",
+        "The voltage across resistor 3 is: 16 V.\n",
+        "The voltage across resistor 4 is: 8 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4,Page number:20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the currents through and voltage across the resistors.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=2.0                 #Current in Circuit(in Amperes)\n",
+      "R1=2.0                #Resistance of resistor 1(in Ohms) \n",
+      "R2=4.0                #Resistance of resistor 2(in Ohms)\n",
+      "R3=6.0                #Resistance of resistor 3(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Rp=1/((1/R2)+(1/R3))\n",
+      "Req=R1+Rp\n",
+      "Vs=I*Req\n",
+      "v1=Vs*(R1/(R1+Rp))\n",
+      "v2=Vs*(Rp/(R1+Rp))\n",
+      "v3=v2\n",
+      "i1=I\n",
+      "i2=v2/R2\n",
+      "i3=v3/R3\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current i1= %.2f A.\\nThe current i2= %.2f A.\\nThe current i3= %.2f A.\" %(i1,i2,i3)\n",
+      "print \"The voltage v1= %.2f V.\\nThe voltage v2= %.2f V.\\nThe voltage v3= %.2f V.\\n\" %(v1,v2,v3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current i1= 2.00 A.\n",
+        "The current i2= 1.20 A.\n",
+        "The current i3= 0.80 A.\n",
+        "The voltage v1= 4.00 V.\n",
+        "The voltage v2= 4.80 V.\n",
+        "The voltage v3= 4.80 V.\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5,Page number: 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the effective resistance.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "Rp=1.0/((1.0/20)+(1.0/10)+(1.0/20))\n",
+      "R_AB_1=15+Rp\n",
+      "R = symbols('R')\n",
+      "R1=1.0/((1.0/2.0)+1.0)+ 1.0\n",
+      "R2=R1\n",
+      "R_AB_2= 1.0/((1/R1)+(1/R2)+(1))\n",
+      "R_AB_b=round(R_AB_2,4)*R\n",
+      "R3=1.0/((1.0/3)+(1.0/6)) + 18\n",
+      "R_AB_3= 1.0/((1.0/20)+(1/R3)) + 5\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The effective resistance between points A and B for the combination of resistances is R_AB = %.2f Ohms.\" %(R_AB_1)\n",
+      "print \"(b)The effective resistance between points A and B for the combination of resistances is R_AB = %s Ohms.\" %(R_AB_b)\n",
+      "print \"(c)The effective resistance between points A and B for the combination of resistances is R_AB = %.2f Ohms.\" %(R_AB_3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The effective resistance between points A and B for the combination of resistances is R_AB = 20.00 Ohms.\n",
+        "(b)The effective resistance between points A and B for the combination of resistances is R_AB = 0.4545*R Ohms.\n",
+        "(c)The effective resistance between points A and B for the combination of resistances is R_AB = 15.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6,Page number: 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding currents in parallel branches.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=100                 #Voltage of the DC source(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Reff= 2+ (1.0/((1.0/12)+(1.0/20)+(1.0/30)))+2\n",
+      "I=V/Reff\n",
+      "\n",
+      "\"\"\" Applying Ohm's Law, we have 12*I1=20*I2=30*I3;\n",
+      "    \n",
+      "    I2=0.6*I1; I3=0.4*I1 \"\"\"\n",
+      "\n",
+      "\"\"\" I=I1+I2+I3; \"\"\"\n",
+      "I1=I/(0.6+0.4+1)\n",
+      "I2=0.6*I1\n",
+      "I3=0.4*I1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I1= %.2f A.\\nThe current I2= %.2f A.\\nThe current I3= %.2f A.\" %(I1,I2,I3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I1= 5.00 A.\n",
+        "The current I2= 3.00 A.\n",
+        "The current I3= 2.00 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7,Page number: 25   "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the supply current I.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=20.0                #Power dissipated by resistor(in Watts)\n",
+      "RL=5.0                #Resistance of the load resistor(in Ohms)\n",
+      "R=10.0                #Resistance of resistor(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=sqrt(P/RL)\n",
+      "I=(I1*(R+RL))/R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The supply current( I ) is %d A.\"%(I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The supply current( I ) is 3 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8,Page number:25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding voltage and the total power dissipated.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=120.0               #Voltage of the power line(in Volts)\n",
+      "P_bulb=60.0           #Power rating of the bulb(in Watts)\n",
+      "V_bulb=120.0          #Voltage rating of the bulb(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R=(V_bulb*V_bulb)/P_bulb\n",
+      "R_A=R\n",
+      "R_B=R\n",
+      "R_C=R\n",
+      "R_BC=1.0/((1.0/R)+(1.0/R))\n",
+      "V_B=V*(R_BC/(R_BC+R_A)) \n",
+      "V_C=V*(R_BC/(R_BC+R_A))\n",
+      "V_A=V-V_B\n",
+      "P_A=(V_A*V_A)/R_A \n",
+      "P_B=(V_B*V_B)/R_B\n",
+      "P_C=(V_C*V_C)/R_C\n",
+      "P=P_A+P_B+P_C\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across bulb A is %.2f V. \\nThe voltage across bulb B is %.2f V. \\nThe voltage across bulb C is %.2f V.\" %(V_A,V_B,V_C)\n",
+      "print \"The total power dissipated in the three bulbs is %.2f W.\" %(P)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across bulb A is 80.00 V. \n",
+        "The voltage across bulb B is 40.00 V. \n",
+        "The voltage across bulb C is 40.00 V.\n",
+        "The total power dissipated in the three bulbs is 40.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9,Page number: 26\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Designing a variable resistor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=30.0               #Resistance of the resistor(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R2=75-R1\n",
+      "Req=(30+75)/2.0\n",
+      "Rp=Req-R1\n",
+      "R=1/((1/Rp)-(1/R2))\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)Minimum value of Req is obtained when R=0(i.e.,a short circuit,because the parallel combination of R2 and R is reduced to 0).\"    \n",
+      "print \"   Maximum value of Req is obtained when R is an open ciruit.\\n   Hence,  R1 = %.2f Ohms and R2 = %.2f Ohms. \\n \" %(R1,R2)\n",
+      "print \"(b)The resistance R to give Req=(30+75)/2 ohm is R = %.2f Ohms.\" %(R)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Minimum value of Req is obtained when R=0(i.e.,a short circuit,because the parallel combination of R2 and R is reduced to 0).\n",
+        "   Maximum value of Req is obtained when R is an open ciruit.\n",
+        "   Hence,  R1 = 30.00 Ohms and R2 = 45.00 Ohms. \n",
+        " \n",
+        "(b)The resistance R to give Req=(30+75)/2 ohm is R = 45.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10,Page number: 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\n",
+      "\"\"\"Finding the equivalent resistance for the infinite ladder network.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\"\"\"  If we remove the first two resistances(i.e. the first rung of the ladder), the remaining circuit across the terminals C and D \n",
+      "     has an equivalent resistance ,which must be double of the original ladder. If  R_AB=Rx,  then R_CD=2*Rx; \"\"\" \n",
+      "\"\"\" Rx=R + (R || (2*Rx)) ; (2*Rx*Rx)-(3*Rx*R)-(R*R);\"\"\"\n",
+      "\"\"\" The roots of the above quadratic equation is solved using Shreedharacharya's Formula\"\"\"\n",
+      "\"\"\" We ignore the negative root as Resistance cannot be negative\"\"\" \n",
+      "a=2.0       #Coefficient of squared term in a quadratic equation\n",
+      "b=-3.0      #Coefficient of first degree term in a quadratic equation  \n",
+      "c=-1.0      #Constant term in a quadratic equation \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Rx=((-b)+sqrt((b*b)-(4*a*c)))/(2*a)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The equivalent resistance between terminals A and B in terms of resistance R,for the infinite ladder network is %.2fR Ohms.\" %(Rx)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equivalent resistance between terminals A and B in terms of resistance R,for the infinite ladder network is 1.78R Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11,Page number: 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Converting the pi-section into equivalent T-section.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=3.0                 #First resistance of pi-section(in Ohms)\n",
+      "R2=9.0                 #Second resistance of pi-section(in Ohms) \n",
+      "R3=6.0                 #Third resistance of pi-section(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ra=(R2*R3)/(R1+R2+R3)\n",
+      "Rb=(R1*R3)/(R1+R2+R3)\n",
+      "Rc=(R1*R2)/(R1+R2+R3)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance Ra is %.2f Ohms.\" %(Ra)\n",
+      "print \"The resistance Rb is %.2f Ohm.\" %(Rb)\n",
+      "print \"The resistance Rc is %.2f Ohms.\" %(Rc)\n",
+      " "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance Ra is 3.00 Ohms.\n",
+        "The resistance Rb is 1.00 Ohm.\n",
+        "The resistance Rc is 1.50 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.12,Page number: 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance of the coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\"\"\" Reff is the effective resistance of the two coils. \"\"\"\n",
+      "V=100.0      #Voltage of the dc supply(in Volts)\n",
+      "I=10.0       #Current drawn from the supply(in Amperes)\n",
+      "P=600.0      #Power dissipated in one coil(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Reff=V/I\n",
+      "R1=(V*V)/P\n",
+      "R2=(Reff*R1)/(R1-Reff)\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance of first coil is: %.2f Ohms.\" %(R1)\n",
+      "print \"The resistance of second coil is: %.2f Ohms.\" %(R2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance of first coil is: 16.67 Ohms.\n",
+        "The resistance of second coil is: 25.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.13,Page number: 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the cost of the boiler operation.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=12.0         #Current drawn  by the electric boiler(in Amperes) \n",
+      "V=115.0        #Operating voltage of the electric boiler(in Volts)\n",
+      "t=6.0          #Time of operation of the electric boiler(in hours) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "W=(V*I*t)/1000\n",
+      "Rate=2.50\n",
+      "cost=W*Rate\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The cost of boiler operation is Rs. %.2f.\" %(cost)\n",
+      "  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cost of boiler operation is Rs. 20.70.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.14,Page number: 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"To find the effect of supply on the rating of toaster.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_rated=240.0         #Voltage rating of the toaster(in Volts)\n",
+      "P_rated=1000.0        #Power rating of the toaster(in Watts) \n",
+      "V=220.0               #Voltage of the supply(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R_toaster=(V_rated*V_rated)/P_rated\n",
+      "I_rated=P_rated/V_rated \n",
+      "I=V/R_toaster\n",
+      "P_consumed=V*I\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The rated current is %.2f A.\" %(I_rated)\n",
+      "print \"The power rating is %.2f W.\" %(P_rated)\n",
+      "print \"The current drawn is %.2f A.\" %(I)\n",
+      "if (I>I_rated): print(\"The toaster will be damaged as the current is greater than the rated current. \\n\")\n",
+      "elif(I<I_rated): print(\"The toaster will not be damaged as the current is lesser than the rated current. \\n\")\n",
+      "else: print(\"The toaster will not be damaged as the current is equal to the rated current. \\n\")\n",
+      "\n",
+      "print \"The power consumed is %.2f W.\" %(P_consumed) \n",
+      "\n",
+      "if (P_consumed>P_rated): print(\"The rating will be affected as the power consumed is greater than the power rating. \\n\")\n",
+      "elif(P_consumed<P_rated): print(\"The rating will not be affected as the power consumed is lesser than the power rating. \\n\") \n",
+      "else: print(\"The rating will not be affected as the power consumed is equal to the power rating. \\n\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rated current is 4.17 A.\n",
+        "The power rating is 1000.00 W.\n",
+        "The current drawn is 3.82 A.\n",
+        "The toaster will not be damaged as the current is lesser than the rated current. \n",
+        "\n",
+        "The power consumed is 840.28 W.\n",
+        "The rating will not be affected as the power consumed is lesser than the power rating. \n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.15,Page number: 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"To find the range of resistance of a colour coded resistor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Yellow=4.0             #Mutilplier corresponding to yellow band\n",
+      "Violet=7.0             #Mutilplier corresponding to violet band\n",
+      "Orange=1.0e03          #Mutilplier corresponding to orange band \n",
+      "Gold=5.0/100           #Mutilplier corresponding to gold band\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R=((Yellow*10)+Violet)*Orange\n",
+      "Rmin=R-(Gold*R)\n",
+      "Rmax=R+(Gold*R)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance should be between %.2f kilo Ohms and %.2f kilo Ohms.\" %(Rmin/1000,Rmax/1000) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance should be between 44.65 kilo Ohms and 49.35 kilo Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.16,Page number: 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"To find the range of resistance of a colour coded resistor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Gray=8.0               #Mutilplier corresponding to gray band\n",
+      "Blue=6.0               #Mutilplier corresponding to blue band\n",
+      "Gold_3=1.0e-01         #Mutilplier corresponding to first gold band\n",
+      "Gold_4=5.0/100         #Mutilplier corresponding to second gold band  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R=((Gray*10.0)+Blue)*Gold_3\n",
+      "Rmin=R-(Gold_4*R)\n",
+      "Rmax=R+(Gold_4*R)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance should be between %.2f Ohms and %.2f Ohms.\" %(Rmin,Rmax) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance should be between 8.17 Ohms and 9.03 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.17,Page number: 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance at a given temperature.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=126                 #Resistance of transmission at 20 degrees Celsius(in Ohms)\n",
+      "T1=20                  #Initial temperature(in degrees Celsius)  \n",
+      "T2=-35                 #Final temperature(in degrees Celsius)\n",
+      "temp_coeff=0.00426     #Temperature coefficient of the material of transmission(in Ohm per degree per Ohm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R2=(1+(temp_coeff*T2))*(R1/(1+(temp_coeff*T1)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance of the line at -35 degree Celsius is %.2f Ohms.\" %(R2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance of the line at -35 degree Celsius is 98.80 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.18,Page number: 32 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the temperature rise for change in resistance value.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=3.42                  #Resistance of Copper winding at room temperature(in Ohms)  \n",
+      "T1=20                    #Room temperature(in degrees Celsius)\n",
+      "R2=4.22                  #Resistance of Copper winding after an extended operation of motor(in Ohms)\n",
+      "temp_coeff=0.00426       #Temperature coefficient of Copper(in Ohm per degree per Ohm)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "T2=(((R2*(1+(temp_coeff*T1)))/R1)-1)/temp_coeff\n",
+      "temp_rise=T2-T1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The temperature rise is %.2f degree Celsius.\" %(temp_rise)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature rise is 59.59 degree Celsius.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.19,Page number: 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the ratio of resistances and lengths of two wires.\"\"\" \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "k=1.0                   #Resistance per metre of the metallic wire(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" x is the length of the wire 'P'.\n",
+      "\n",
+      "    R_P=k*x;\n",
+      "    R_Q=k*(1-x);\n",
+      "    \n",
+      "    R_R=4*R_P;(because length of 'R' is two times length of 'P',the area of cross-section reduces to half.Resistance is directly \n",
+      "               proportional to length and inversely proportional to area of cross-section.) \"\"\"\n",
+      "\n",
+      "x=1.0/(1+4)\n",
+      "R_P=k*x\n",
+      "R_Q=(1.0-x)*k\n",
+      "R_R=R_Q\n",
+      "ratio_res=R_P/R_R\n",
+      "ratio_len=x/(1-x)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The ratio of resistances of P and R is %.2f:1. \" %(ratio_res) \n",
+      "print \"(b)The ratio of lengths of P and Q is %.2f:1. \" %(ratio_len)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The ratio of resistances of P and R is 0.25:1. \n",
+        "(b)The ratio of lengths of P and Q is 0.25:1. \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.20,Page number: 32 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the equivalent resistance.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\n",
+      "#Calculations:\n",
+      "R1=1.0/((1.0/5)+(1.0/6))\n",
+      "R2=1.0/((1.0/4)+(1.0/3))\n",
+      "R_AB_1=R1+R2\n",
+      "R3=6+(1.0/((1.0/6)+(1.0/12)))\n",
+      "R4=1.0/((1.0/R3)+(1.0/10))\n",
+      "R_AB_2=2+R4+3\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resistance between terminals A and B is %.2f Ohms.\" %(R_AB_1)\n",
+      "print \"(b)The resistance between terminals A and B is %.2f Ohms.\" %(R_AB_2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resistance between terminals A and B is 4.44 Ohms.\n",
+        "(b)The resistance between terminals A and B is 10.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.21,Page number: 33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the pd between two corners.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=20e-03              #Current entering the cube(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Rp1=12/3\n",
+      "Rp2=12/6\n",
+      "Rp3=Rp1\n",
+      "R_AG=Rp1+Rp2+Rp3\n",
+      "V_AG=R_AG*I\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The potential difference across the corners A and G is %.2f V.\" %(V_AG)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The potential difference across the corners A and G is 0.20 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.22,Page number: 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding equivalent resistance between points E and F.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "#The delta connection between points A,B and C is first converted into its equivalent star connection by creating a new node 'O'.\n",
+      "R_AO=(4.0*6.0)/(2.0+4.0+6.0)\n",
+      "R_BO=(2.0*6.0)/(2.0+4.0+6.0)\n",
+      "R_CO=(2.0*4.0)/(2.0+4.0+6.0)\n",
+      "R_EF=(8.0*(8+(32.0/3)))/(8.0+(8.0+(32.0/3)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The equivalent resistance between points E and F is %.2f Ohms.\" %(R_EF)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equivalent resistance between points E and F is 5.60 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.23,Page number: 35"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Find the current drawn from the 5-V supply.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=5                   #Voltage from the battery(in Volts)  \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" The star connection between A,C and D can be converted into its equivalent delta connection. \"\"\"\n",
+      "R1=((2*3)+(3*2)+(2*2))/3.0\n",
+      "R2=((2*3)+(3*2)+(2*2))/2.0\n",
+      "R3=((2*3)+(3*2)+(2*2))/2.0\n",
+      "\n",
+      "\"\"\"It is important to note that point B is lost during the process of star-to-delts transformation. \"\"\"\n",
+      "\n",
+      "R4=((16.0/3)*(32.0/9))/((16.0/3)+(32.0/9))\n",
+      "R5=R4+3\n",
+      "I=V/R5\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I drawn from the battery is %.4f A.\" %(I)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I drawn from the battery is 0.9740 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.24,Page number: 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current and the pd between A and B.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=488.0               #Total power dissipated(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R_CD=1.0/((1.0/5)+(1.0/20)+(1.0/2.5))\n",
+      "R_AB=R_CD+10\n",
+      "V=sqrt(P*R_AB)\n",
+      "I=V/R_AB\n",
+      "V_CD=V-(10*I)\n",
+      "I1=V_CD/5.0\n",
+      "I2=V_CD/20.0\n",
+      "I3=V_CD/2.5\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \" The current I1= %.2f A.\" %(I1)   \n",
+      "print \" The current I2= %.2f A.\" %(I2)\n",
+      "print \" The current I3= %.2f A.\" %(I3)\n",
+      "print \" The pd between A and B is %.2f V.\" %(V)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The current I1= 2.00 A.\n",
+        " The current I2= 0.50 A.\n",
+        " The current I3= 4.00 A.\n",
+        " The pd between A and B is 75.04 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.25,Page number: 36 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance R.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_AB=5                #Voltage between points A and B(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_AC=15*(10.0/(10+5))\n",
+      "V_BA=-V_AB\n",
+      "V_BC=V_BA+V_AC\n",
+      "\"\"\" V_BC=15.0*(R/(2+R));\n",
+      "    10+(5*R)=(15*R);      \"\"\"\n",
+      "R=10/10\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of resistance R is %.2f Ohm.\" %(R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of resistance R is 1.00 Ohm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.26,Page number: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding current supplied by the source.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=100                 #Voltage of the supply(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R1=1.0/((1.0/6)+(1.0/3))\n",
+      "R2=5+3+R1\n",
+      "R_BC=1.0/((1.0/10)+(1/R2))\n",
+      "R3=5+R_BC\n",
+      "R_AD=1.0/((1.0/R3)+(1.0/10))\n",
+      "R_total=R_AD+5\n",
+      "I=V/R_total\n",
+      "\"\"\"The concept of current divider is used to find current I1 through the 6 Ohm resistance.At A,the 10 A current divides into \n",
+      "   two parts: I_AD and I_AB.The current I_AB depends on the total resistance on the right of point A(which is 10 Ohms).Thus the two\n",
+      "   currents will be equal.\n",
+      "   \n",
+      "   I_AB= I_AD= 5 A.\n",
+      "   \n",
+      "   I_AB divides equally into I_BC and I_BE. \n",
+      "   I_BC= I_BE= 2.5 A.  I_BE divides into I1 and I2. \"\"\"\n",
+      "\n",
+      "I_AB=I/2.0\n",
+      "I_AD=I_AB\n",
+      "I_BC=I_AB/2.0\n",
+      "I_BE=I_BC\n",
+      "I1=I_BE*(3.0/(3.0+6.0))\n",
+      "V_6=I1*6\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current supplied by the 100-V source is %.2f A.\" %(I)\n",
+      "print \"(b)The voltage across the 6-Ohms resistance is %.2f V.\" %(V_6)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current supplied by the 100-V source is 10.00 A.\n",
+        "(b)The voltage across the 6-Ohms resistance is 5.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.27,Page number: 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance R.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=80                  #Power dissipated(in Watts)\n",
+      "V=20                  #Applied voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Rp=1.0/((1.0/8)+(1.0/12)+(1.0/24))\n",
+      "\"\"\" R_total=R+Rp \"\"\"\n",
+      "R_total=(V*V)/P\n",
+      "R=R_total-Rp\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance R is %.2f Ohm.\" %(R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance R is 1.00 Ohm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.28,Page number: 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance R2.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=100.0                 #Voltage of the supply(in Volts)\n",
+      "I=10.0                  #Current drawn from the mains(in Amperes)\n",
+      "P=600.0                 #Power dissipated in the resistor R1(in Watts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R1=(V*V)/P\n",
+      "I1=V/R1\n",
+      "I2=I-I1\n",
+      "R2=V/I2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resistance R2 is %.2f Ohms.\" %(R2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistance R2 is 25.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.29,Page number: 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the electrical energy generated per tonne of the fuel.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P_out=50              #Output power(in kilo-Watts)\n",
+      "effi=0.35             #Efficiency of the diesel generating set\n",
+      "cal=12500             #Calorific value of the fuel(in kcal/kg)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "P_in=P_out/effi\n",
+      "Heat_per_hour=P_in*860\n",
+      "Fuel=Heat_per_hour/cal\n",
+      "ener=(P_out/Fuel)*1000\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The quantity of oil needed per hour is %.2f kg.\" %(Fuel)\n",
+      "print \"(b)The electrical energy generated per tonne of the fuel is %.2f kWh.\" %(ener)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The quantity of oil needed per hour is 9.83 kg.\n",
+        "(b)The electrical energy generated per tonne of the fuel is 5087.21 kWh.\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.30,Page number: 38 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the resistance of the heater and quantity of heat generated.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_A=10.0              #Resistance of heater A(in Ohms)\n",
+      "t_A=20                #Time of operation of heater-A(in minutes) \n",
+      "t_B=10                #Time of operation of heater-B(in minutes)\n",
+      "H_A=500               #Heat produced by heater-A in time t_A(in kcal) \n",
+      "H_B=1000               #Heat produced by heater-B in time t_B(in kcal)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=sqrt((H_A*4.2*1000*R_A)/(t_A*60))\n",
+      "R_B=(V*V*t_B*60)/(H_B*4.2*1000)\n",
+      "R=R_A+R_B\n",
+      "t=5*60\n",
+      "H=(((V*V)*t)/R)/(4.2*1000)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The resistance of heater-B is %.2f Ohms.\" %(R_B)\n",
+      "print \"(b)The heat produced in 5 minutes when the two heaters are connected in series across the same supply voltage is %.2f kcal.\" %(H)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The resistance of heater-B is 2.50 Ohms.\n",
+        "(b)The heat produced in 5 minutes when the two heaters are connected in series across the same supply voltage is 100.00 kcal.\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.31,Page number: 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding total current taken from the supply and energy consumed in a day.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "lamps=8              #Number of lamps\n",
+      "P_lamp=100           #Power rating of the lamp(in Watts)\n",
+      "fans=3               #Number of fans\n",
+      "P_fan=80             #Power rating of the fan(in Watts)\n",
+      "P_fridge=0.5         #Power rating of the fridge(in HP)\n",
+      "P_heater=1000        #Power rating of the Heater(in Watts)\n",
+      "V=230                #Voltage of the supply(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "total_load=(lamps*P_lamp)+(fans*P_fan)+(P_fridge*746)+P_heater\n",
+      "I=total_load/V\n",
+      "ener=(total_load*(1.0/4)*24)/1000\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The total current taken from the supply is %.2f A.\" %(I)\n",
+      "print \"(b)The energy consumed in a day, if on an average only a quarter of the above load persists all the time is %.4f kWh.\" %(ener)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The total current taken from the supply is 10.49 A.\n",
+        "(b)The energy consumed in a day, if on an average only a quarter of the above load persists all the time is 14.4780 kWh.\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter3.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter3.ipynb
new file mode 100755
index 00000000..a6211207
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter3.ipynb
@@ -0,0 +1,2116 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: NETWORK ANALYSIS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1,Page number: 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of capacitance parameter.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "per_free_space=8.854e-12   #Permittivity of free space(in Farad per metre)\n",
+      "A=0.113                    #Total area(in square metre)  \n",
+      "rel_per=10.0               #Relative permittivity of mica layer\n",
+      "d=0.1e-03                  #Thickness of mica layer(in metre)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C1=(per_free_space*rel_per*A)/d\n",
+      "W=0.05\n",
+      "V=100.0\n",
+      "C2=(2*W)/(V*V)\n",
+      "i=5e-03\n",
+      "dv=100\n",
+      "dt=0.1\n",
+      "C3=i/(dv/dt)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(i)The value of capacitance is %.2f micro Farad.\" %(C1*1000000)\n",
+      "print \"(ii)The value of capacitance is %.2f micro Farad.\" %(C2*1000000)\n",
+      "print \"(iii)The value of capacitance is %.2f micro Farad.\" %(C3*1000000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i)The value of capacitance is 0.10 micro Farad.\n",
+        "(ii)The value of capacitance is 10.00 micro Farad.\n",
+        "(iii)The value of capacitance is 5.00 micro Farad.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2,Page number: 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the inductance of the coil for different cases.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "W=0.2                  #Energy stored in magnetic field(in Joules)    \n",
+      "i=0.2                  #Current producing magnetic field(in Amperes) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L1=(2*W)/(i*i)\n",
+      "v=10\n",
+      "di=0.1\n",
+      "dt=0.2\n",
+      "L2=v/(di/dt)\n",
+      "p=2.5\n",
+      "di=0.5\n",
+      "dt=1\n",
+      "i=0.1\n",
+      "L3=p/(i*(di/dt))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(i)The inductance of the coil is %.2f H.\" %(L1)\n",
+      "print \"(ii)The inductance of the coil is %.2f H.\" %(L2)\n",
+      "print \"(iii)The inductance of the coil is %.2f H.\" %(L3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(i)The inductance of the coil is 10.00 H.\n",
+        "(ii)The inductance of the coil is 20.00 H.\n",
+        "(iii)The inductance of the coil is 50.00 H.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3,Page number: 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the inductance of 2 coils.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Leq=0.7                #Equivalent self-inductance(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" Leq=0.7 H; \n",
+      "    \n",
+      "    Leq=((L1*L2)/(L1+L2))+0.5 ;\n",
+      "\n",
+      "    L1=(2*L2); \"\"\"\n",
+      "L2=(Leq-0.5)*(3.0/2)\n",
+      "L1=2*L2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"L1=%.2f H. \\nL2=%.2f H.\" %(L1,L2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L1=0.60 H. \n",
+        "L2=0.30 H.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4,Page number: 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the equivalent capacitance of the network and the voltage drop across each capacitor. \"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=220.0                #Voltage of the DC source(in Volts)\n",
+      "C1=0.05e-06            #Capacitance of the capacitor 1(in Farad)\n",
+      "C2=0.10e-06            #Capacitance of the capacitor 2(in Farad)\n",
+      "C3=0.20e-06            #Capacitance of the capacitor 3(in Farad)\n",
+      "C4=0.05e-06            #Capacitance of the capacitor 4(in Farad) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Cs=1/((1/C1)+(1/C2)+(1/C3)+(1/C4))\n",
+      "Q=Cs*V\n",
+      "V1=Q/C1\n",
+      "V2=Q/C2\n",
+      "V3=Q/C3\n",
+      "V4=Q/C4\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The equivalent capacitance of the network is %.4f micro Farad.\" %(Cs*1000000)\n",
+      "print \"The voltage across capacitor C1= %.2f V.\" %(V1)\n",
+      "print \"The voltage across capacitor C2= %.2f V.\" %(V2)\n",
+      "print \"The voltage across capacitor C3= %.2f V.\" %(V3)\n",
+      "print \"The voltage across capacitor C4= %.2f V.\" %(V4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The equivalent capacitance of the network is 0.0182 micro Farad.\n",
+        "The voltage across capacitor C1= 80.00 V.\n",
+        "The voltage across capacitor C2= 40.00 V.\n",
+        "The voltage across capacitor C3= 20.00 V.\n",
+        "The voltage across capacitor C4= 80.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5,Page number: 55\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across each capacitor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Q1=400e-06            #Charge on the capacitor 1(in Coulombs)\n",
+      "Q2=200e-06            #Charge on the capacitor 2(in Coulombs) \n",
+      "C1=2e-06              #Capacitance of the capacitor 1(in Farad)\n",
+      "C2=10e-06             #Capacitance of the capacitor 2(in Farad) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Q=Q1+Q2\n",
+      "C=C1+C2\n",
+      "V=Q/C\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"After the switch is closed, the voltage that exists across each capacitor is %.2f V.\" %(V)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "After the switch is closed, the voltage that exists across each capacitor is 50.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6,Page number: 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\" A series combination of 2 capacitances C1=2 micro-F and C2=8 micro-F is connected across a dc-supply of 300-V.Determine(a)the charge,\n",
+      "    (b)the voltage,and(c)the energy stored in each capacitor. \"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "C1=2e-06               #Capacitance of the capacitor 1(in Farad) \n",
+      "C2=8e-06               #Capacitance of the capacitor 2(in Farad)\n",
+      "V=300.0                #Voltage of Dc supply(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C=(C1*C2)/(C1+C2)\n",
+      "Q=C*V\n",
+      "V1=Q/C1\n",
+      "V2=Q/C2\n",
+      "W1=(1.0/2)*C1*(V1*V1)    \n",
+      "W2=(1.0/2)*C2*(V2*V2)\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)When connected in series,the charge on each capacitor is the same.The charge on each capacitor is %.2f micro Coulomb.\" %(Q*1000000)\n",
+      "print \"(b)The voltage across the capacitor C1= %.2f V.\\n   The voltage across the capacitor C2= %.2f V.\" %(V1,V2)\n",
+      "print \"(c)The energy stored in the capacitor C1= %.2f milli Joule.\" %(W1*1000)\n",
+      "print \"   The energy stored in the capacitor C2= %.2f milli Joule.\" %(W2*1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)When connected in series,the charge on each capacitor is the same.The charge on each capacitor is 480.00 micro Coulomb.\n",
+        "(b)The voltage across the capacitor C1= 240.00 V.\n",
+        "   The voltage across the capacitor C2= 60.00 V.\n",
+        "(c)The energy stored in the capacitor C1= 57.60 milli Joule.\n",
+        "   The energy stored in the capacitor C2= 14.40 milli Joule.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7,Page number: 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of capacitance.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "C_eq=1e-06            #Equivalent capacitance of the network(in Farad)\n",
+      "C1=1e-06              #Capacitance of the capacitor 1(in Farad)\n",
+      "C2=8e-06              #Capacitance of the capacitor 2(in Farad)\n",
+      "C3=2e-06              #Capacitance of the capacitor 3(in Farad)\n",
+      "C4=2e-06              #Capacitance of the capacitor 4(in Farad)\n",
+      "C5=6e-06              #Capacitance of the capacitor 5(in Farad)\n",
+      "C6=12e-06             #Capacitance of the capacitor 6(in Farad)\n",
+      "C7=4e-06              #Capacitance of the capacitor 7(in Farad) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "C8=C3+C4\n",
+      "C9=1/((1/C5)+(1/C6))\n",
+      "C10=1/((1/C2)+(1/C8))\n",
+      "C11=C7+C9\n",
+      "C12=1/((1/C1)+(1/C11))\n",
+      "C13=C12+C10\n",
+      "C=1/((1/C_eq)-(1/C13))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of capacitance C is %.2f micro Farad.\" %(C*1e06)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of capacitance C is 1.39 micro Farad.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8,Page number: 59"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the % change in V_L and I_L.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "E=3.0                  #EMF of the battery(in Volts) \n",
+      "Ri=1.0                 #Internal resistance of the battery(in Ohms)\n",
+      "R_L_min=100.0          #Minimum value of variable load resistance(in Ohms)\n",
+      "R_L_max=1000.0         #Maximum value of variable load resistance(in Ohms) \n",
+      "\n",
+      "\n",
+      "#Calculations and Result: \n",
+      "I_L1=E/(R_L_min+Ri)\n",
+      "I_L2=E/(R_L_max+Ri)\n",
+      "V_L1=E-(I_L1*Ri)\n",
+      "V_L2=E-(I_L2*Ri)\n",
+      "\n",
+      "if I_L1>I_L2 :\n",
+      "               change_I_L= ((I_L1-I_L2)/I_L1)*100\n",
+      "               ans=\"(a)The percentage decrease in I_L is %.2f percent\" %(round(change_I_L,2))\n",
+      "               print(ans)\n",
+      "elif I_L2>IL1 : \n",
+      "               change_I_L= ((I_L2-I_L1)/I_L1)*100\n",
+      "               ans=\"(a)The percentage increase in I_L is %.2f percent\" %(round(change_I_L,2))\n",
+      "               print(ans)\n",
+      "else: print(\"(a)The percentage change in I_L is 0 percent.\")\n",
+      "\n",
+      "if V_L2>V_L1 :\n",
+      "              change_V_L= ((V_L2-V_L1)/V_L1)*100\n",
+      "              ans=\"   The percentage increase in V_L is %.2f percent\" %(round(change_V_L,2))\n",
+      "              print(ans)\n",
+      "elif V_L1>V_L2 :\n",
+      "                change_V_L= ((V_L1-V_L2)/V_L1)*100\n",
+      "                ans=\"   The percentage decrease in V_L is %.2f percent\" %(round(change_V_L,2))\n",
+      "                print(ans)\n",
+      "else : print(\"   The percentage change in V_L is 0\",\"%\")\n",
+      "\n",
+      "R_L_min=1e-03\n",
+      "R_L_max=10e-03\n",
+      "I_L1=E/(R_L_min+Ri)\n",
+      "I_L2=E/(R_L_max+Ri)\n",
+      "V_L1=round((E-(I_L1*Ri)),3)\n",
+      "V_L2=round((E-(I_L2*Ri)),2)\n",
+      "\n",
+      "if I_L1>I_L2 :\n",
+      "               change_I_L= ((I_L1-I_L2)/I_L1)*100\n",
+      "               ans=\"(b)The percentage decrease in I_L is %.2f percent\" %(round(change_I_L,2))\n",
+      "               print(ans) \n",
+      "\n",
+      "elif I_L2>I_L1 :\n",
+      "                change_I_L= ((I_L2-I_L1)/I_L1)*100\n",
+      "                ans=\"(b)The percentage increase in I_L is %.2f percent\" %(round(change_I_L,2))\n",
+      "                print(ans)\n",
+      "\n",
+      "else : print(\"(b) The percentage change in I_L is 0 percent\")\n",
+      "\n",
+      "if V_L2>V_L1 :\n",
+      "               change_V_L= ((V_L2-V_L1)/V_L1)*100\n",
+      "               ans=\"   The percentage increase in V_L is %.3f percent\" %(change_V_L)\n",
+      "               print(ans)\n",
+      "\n",
+      "elif V_L1>V_L2 :\n",
+      "                change_V_L= ((V_L1-V_L2)/V_L1)*100\n",
+      "                ans=\"   The percentage decrease in V_L is %.2f percent\" %(change_V_L)\n",
+      "                print(ans)\n",
+      "\n",
+      "else: print(\"    The percentage change in V_L is 0 percent\")\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The percentage decrease in I_L is 89.91 percent\n",
+        "   The percentage increase in V_L is 0.90 percent\n",
+        "(b)The percentage decrease in I_L is 0.89 percent\n",
+        "   The percentage increase in V_L is 900.000 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9,Page number: 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power delivered by the ideal part of the sources.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_S=2.0                #Resistance of the current source(in Ohms)\n",
+      "I_S=3.0                #Current from the current source(in Amperes) \n",
+      "R_L=4.0                #Resistance of the load(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_S=R_S*I_S\n",
+      "I_L1=I_S*(R_S/(R_S+R_L))\n",
+      "V_L1=I_L1*R_L\n",
+      "R1=1/((1/R_S)+(1/R_L))\n",
+      "P_S1=I_S*I_S*R1\n",
+      "I_L2=V_S/(R_S+R_L)\n",
+      "V_L2=V_S*(R_L/(R_S+R_L))\n",
+      "R2=R_S+R_L\n",
+      "P_S2=(V_S*V_S)/R2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"Case 1:\"\n",
+      "print \"I_L=%.2f A. \\nV_L=%.2f V. \\nP_S=%.2f W.\" %(I_L1,V_L1,P_S1)\n",
+      "print \"\\nCase 2:\"\n",
+      "print \"I_L=%.2f A. \\nV_L=%.2f V. \\nP_S=%.2f W.\" %(I_L2,V_L2,P_S2) \n",
+      "print \"\\nWe find that the two energy sources are equivalent as regards the terminal relationships. But, the total power delivered by the ideal part of the two sources is different.\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Case 1:\n",
+        "I_L=1.00 A. \n",
+        "V_L=4.00 V. \n",
+        "P_S=12.00 W.\n",
+        "\n",
+        "Case 2:\n",
+        "I_L=1.00 A. \n",
+        "V_L=4.00 V. \n",
+        "P_S=6.00 W.\n",
+        "\n",
+        "We find that the two energy sources are equivalent as regards the terminal relationships. But, the total power delivered by the ideal part of the two sources is different.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10,Page number: 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the currents through the two resistors in the circut.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "I_2_ohm1=16.0*(6.0/(6+2))\n",
+      "I_6_ohm1=16-I_2_ohm1\n",
+      "V_eq=16*2\n",
+      "I_2_ohm2=V_eq/(2+6)\n",
+      "I_6_ohm2=I_2_ohm2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current through the 2 ohm resistor is %.2f A.\" %(I_2_ohm1) \n",
+      "print \"The current through the 6 ohm resistor is %.2f A. \\n\" %(I_6_ohm1) \n",
+      "print \"After transformation,\\nThe current through the 2 ohm resistor is %.2f A.\" %(I_2_ohm2) \n",
+      "print \"The current through the 6 ohm resistor is %.2f A.\" %(I_6_ohm2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current through the 2 ohm resistor is 12.00 A.\n",
+        "The current through the 6 ohm resistor is 4.00 A. \n",
+        "\n",
+        "After transformation,\n",
+        "The current through the 2 ohm resistor is 4.00 A.\n",
+        "The current through the 6 ohm resistor is 4.00 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11,Page number: 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the reduced network using source transformation.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"First all the voltage sources are transformed to equivalent current sources.\"\"\"\n",
+      "I1=8.0/2.0\n",
+      "I2=6.0/1.0\n",
+      "I3=6.0/2.0\n",
+      "I4=3.0/1.0\n",
+      "I5=I1+I2\n",
+      "I6=I3+I4\n",
+      "R1=1.0/((1.0/2.0)+(1.0/1.0))\n",
+      "R2=1.0/((1.0/2.0)+(1.0/1.0))\n",
+      "V5=I5*R1\n",
+      "V6=I6*R2\n",
+      "V_tot=V5+V6\n",
+      "R_tot=R1+R2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The network is equivalent to a voltage source of %.3f V with an internal resistance of %.2f Ohms.\" %(V_tot,R_tot)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The network is equivalent to a voltage source of 10.667 V with an internal resistance of 1.33 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12,Page number: 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across 3 Ohms resistor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I1=4.0                #Current delivered by source 1(in Amperes)\n",
+      "R1=1.0                #Resitance of resistor parallel to current source 1(in Ohms) \n",
+      "V2=6.0                #Voltage of the voltage source 2(in Volts)\n",
+      "I2=5.0                #Current delivered by source 2(in Amperes)\n",
+      "R2=2.0                #Resistance of resistor 2(in Ohms) \n",
+      "R3=3.0                #Resistance of resistor 3(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V1=I1*R1\n",
+      "V3=V1+V2\n",
+      "I3=V3/R1\n",
+      "I4=I3-I2\n",
+      "V4=I4*R1\n",
+      "V_res=V4*(R3/(R1+R2+R3))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across 3 Ohms resistor is %.2f Volts.\" %(V_res)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across 3 Ohms resistor is 2.50 Volts.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13,Page number: 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in the circuit.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=24.0                #Supply voltage(in Volts)\n",
+      "R1=4.0                #Resistance of resistor 1(in Ohms) \n",
+      "R2=2.0                #Resistance of resistor 2(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" V1=4*I;\n",
+      "    Applying Kirchoff's Voltage Law to the closed loop, 24-(4*I)-(2*I)-(4.5*4*I)=0 ; \"\"\"\n",
+      "I=24/((R1+R2)-(4.5*R1))\n",
+      "P=-4.5*R1*I*I\n",
+      "R=V/I\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The value of the current I is %.2f A.\" %(abs(I))\n",
+      "print \"(b)The power absorbed by the dependent source is %.2f W.\" %(P)\n",
+      "print \"(c)The resistance 'seen' by the independent voltage source is %.2f Ohms.\" %(R) \n",
+      "print \"\\nNOTE: The negative sign of the resistance is a result of the action of the dependent source.\"     "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The value of the current I is 2.00 A.\n",
+        "(b)The power absorbed by the dependent source is -72.00 W.\n",
+        "(c)The resistance 'seen' by the independent voltage source is -12.00 Ohms.\n",
+        "\n",
+        "NOTE: The negative sign of the resistance is a result of the action of the dependent source.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14,Page number: 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage in the circuit.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=100.0               #Supply voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Appling KVL in the closed loop, +100-(40*I)-(60*I)=0.\"\"\"\n",
+      "I=V/(40.0+60.0)\n",
+      "V1=I*60.0\n",
+      "\"\"\"Appling KVL in the open loop, -10+V1+(0*10)+30-Vab=0.\"\"\"\n",
+      "Vab=30+V1-10\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage Vab in the circuit is %.2f V.\" %(Vab)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage Vab in the circuit is 80.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15,Page number: 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the unknown voltages Vx and Vcd.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=10.0                #Supply voltage(in Volts)\n",
+      "Vca=4.0               #Voltage between points c and a(in Volts) \n",
+      "V1=6.0                #Voltage across resistor R1(in Volts) \n",
+      "V2=-4.0               #Voltage across resistor R2(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vx=Vca-V2\n",
+      "Vcd=Vca-V1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage Vx in the circuit is %.2f V.\" %(Vx)\n",
+      "print \"The voltage Vcd in the circuit is %.2f V.\" %(Vcd)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage Vx in the circuit is 8.00 V.\n",
+        "The voltage Vcd in the circuit is -2.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16,Page number: 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the currents in the network.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"KVL equations for loops 1,2 and 3 are written and the equations are solved using SymPy.\"\"\" \n",
+      "print \"Note: All currents are expressed in Amperes.\"\n",
+      "Ix, Iy, I1 = symbols('Ix Iy I1');\n",
+      "solve([(5*Ix)+(0*Iy)+(10*I1)-100,(7*Ix)+(2*Iy)+(-2*I1)+50,(3*Ix)+(-5*Iy)+(-3*I1)+50], [Ix,Iy,I1])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Note: All currents are expressed in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 20,
+       "text": [
+        "{I1: 585/49, Ix: -190/49, Iy: 25/49}"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17,Page number: 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the currents using Kirchoff's laws.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"KCL equations at nodes B and C,we get\n",
+      "    \n",
+      "   I1+I2=20;\n",
+      "   \n",
+      "   I3-I2=30;\"\"\"\n",
+      "\"\"\"KVL equation for the outer loop ABCDEFGHA, I1-(3*I2)-(2*I3)=-100; \"\"\"\n",
+      "print \"All the currents are expressed in Amperes.\"\"\"\n",
+      "print \"Resistances R1 and R2 are in Ohms.\"\"\"\n",
+      "I1, I2,I3,R1,R2 = symbols('I1 I2 I3 R1 R2')\n",
+      "solve([I1+I2+(0*I3)-20,(0*I1)-I2+I3-30,I1+(-3*I2)+(-2*I3)+100,(110-(0.1*I1))-(20*R1),(120-(0.2*I3))-(30*R2)],[I1,I2,I3,R1,R2])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the currents are expressed in Amperes.\n",
+        "Resistances R1 and R2 are in Ohms.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 42,
+       "text": [
+        "{I1: 10.0000000000000,\n",
+        " I2: 10.0000000000000,\n",
+        " I3: 40.0000000000000,\n",
+        " R1: 5.45000000000000,\n",
+        " R2: 3.73333333333333}"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18,Page number: 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through resistor in the circuit.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=2.0                 #Loop current(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"KVL Equation: 10-(5*(I1-2))-(8*I1)=0.\"\"\"\n",
+      "I1=(10.0+10.0)/(8.0+5.0)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current through the 8 Ohms resistor is %.3f A.\" %(I1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current through the 8 Ohms resistor is 1.538 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19,Page number: 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the unknown voltage.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Writing the KVL equation around the loop of I, -(2*I)+(3*I)+6-(1*(I+5-4))=0.\"\"\"\n",
+      "I=(6.0-1.0)/(2+3+1)\n",
+      "v=3*I\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The unknown voltage in the circuit is %.2f V.\" %(v)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The unknown voltage in the circuit is 2.50 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20,Page number: 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current drawn by the source.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KVL for the loops 1, 2 and 3 and solving for currents.\"\"\"\n",
+      "print \"All the currents are expressed in Amperes.\"\"\"\n",
+      "print \"The current delivered  by the source is I1.\"\n",
+      "I1, I2,I3,Is = symbols('I1 I2 I3 Is')\n",
+      "solve([(19*I1)+(-12*I2)+(0*I3)-60,(-12*I1)+(18*I2)+(-6*I3),(0*I1)+(-6*I2)+(18*I3),Is-I1], [I1,I2,I3,Is]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the currents are expressed in Amperes.\n",
+        "The current delivered  by the source is I1.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 4,
+       "text": [
+        "{I1: 6, I2: 9/2, I3: 3/2, Is: 6}"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21,Page number: 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mesh currents using mesh analysis.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KVL for the loops 1, 2 and 3 and solving for currents.\"\"\"\n",
+      "print \"All the currents are expressed in Amperes.\"\"\"\n",
+      "I1, I2, I3 = symbols('I1 I2 I3')\n",
+      "solve([(7*I1)+(-4*I2)+(0*I3)-13,(-4*I1)+(15*I2)+(-6*I3)-12,(0*I1)+(-6*I2)+(13*I3)-1], [I1,I2,I3]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the currents are expressed in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 34,
+       "text": [
+        "{I1: 3, I2: 2, I3: 1}"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.22,Page number: 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across resistor using node-voltage analysis.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KCL at nodes a and b and finding the nodal voltages.\"\"\"\n",
+      "print \"All the voltages are expressed in Volts.\"\"\"\n",
+      "Va,Vb,Vab = symbols('Va Vb Vab')\n",
+      "solve([(Va/2)+((Va-Vb)/3)-5,((Vb-Va)/3)+(Vb/4)+6,Va-Vb-Vab], [Va,Vb,Vab])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the voltages are expressed in Volts.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 43,
+       "text": [
+        "{Va: 22/9, Vab: 34/3, Vb: -80/9}"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.23,Page number: 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through resistor.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Applying KCL at node 1,\n",
+      "\n",
+      "   ((V1-0)/12.0)+((V1-60.0)/7.0)+((V1-0)/4.0)=0; \"\"\"\n",
+      "V1=(60.0/7.0)/((1.0/12.0)+(1.0/7.0)+(1.0/4.0))\n",
+      "I1=V1/12.0\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I1 through the 12 Ohms resistor is %.2f A.\" %(I1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I1 through the 12 Ohms resistor is 1.50 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.24,Page number: 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through a resistor.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KCL for supernode (a,b) and for node c and finding the nodal voltages.\"\"\"\n",
+      "print \"All the voltages are expressed in Volts.\"\"\"\n",
+      "print \"The current I is in Amperes.\"\n",
+      "Va,Vb,Vc,I = symbols('Va Vb Vc I')\n",
+      "solve([Va-Vb-6,(Va/3)+((Vb-Vc)/4)-2,((Vc-Vb)/4)+(Vc/5)+7,((Vb-Vc)/4)-I], [Va,Vb,Vc,I])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the voltages are expressed in Volts.\n",
+        "The current I is in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 47,
+       "text": [
+        "{I: 35/12, Va: -11/4, Vb: -35/4, Vc: -245/12}"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.25,Page number: 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across resistor by nodal-voltage method.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Applying KCL at node a,\n",
+      "  \n",
+      "   ((Va-6.0)/1.0)+((Va-0.0)/5.0)=-4.0+5.0 ; \"\"\"\n",
+      "\n",
+      "Va=(6.0-1.0)/1.2\n",
+      "v=Va*(3.0/(2.0+3.0))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across 3 Ohms resistor is %.2f V.\" %(v)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across 3 Ohms resistor is 2.50 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.26,Page number: 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through resistor.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Applying KCL at nodes 1 and 2 and solving for nodal voltages.\"\"\"\n",
+      "print \"All the voltages are expressed in Volts.\"\"\"\n",
+      "print \"The current I is in Amperes.\"\n",
+      "V1,V2,I = symbols('V1 V2 I')\n",
+      "solve([(0.7*V1)+(-0.2*V2)-3,(-0.2*V1)+(1.2*V2)-2,((V1-V2)/5.0)-I],[V1,V2,I])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the voltages are expressed in Volts.\n",
+        "The current I is in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 16,
+       "text": [
+        "{I: 0.500000000000000, V1: 5.00000000000000, V2: 2.50000000000000}"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.27,Page number: 85 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through a resistor.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Applying KCL at node 1,\n",
+      "\n",
+      "   ((V-10.0)/2.0)+((V-0.0)/(1.0+3.0))+((V-8.0)/6.0)=0 ; \"\"\"\n",
+      "\n",
+      "V=((10.0/2.0)+(8.0/6.0))/((1.0/2.0)+(1.0/4.0)+(1.0/6.0))\n",
+      "I=V/(1.0+3.0)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I is %.2f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I is 1.73 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.28,Page number: 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the values of capacitances.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vs=200.0              #Voltage of the dc supply(in Volts)\n",
+      "V1=120.0              #Potential difference across C1(in Volts)\n",
+      "V_new=140.0           #Potential difference across C1 when a capacitor is added(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Case 1: The charge on each capacitor is the same.\n",
+      "\n",
+      "           (C1*V1)=(C2*V2);\n",
+      "           (120*C1)=(80*C2);\"\"\"\n",
+      "\"\"\"Case 2: The capacitance of 3 micro Farad in parallel with C2 gives an equivalent capacitance of (C2+3) micro Farad.\n",
+      "            \n",
+      "           (140*C1)=(60*(C2+3)) ;\"\"\"\n",
+      "print \"Note:The capacitances are expressed in Farads.\"          \n",
+      "C1,C2 = symbols('C1 C2')\n",
+      "solve([(120*C1)-(80*C2),(140*C1)-(60*(C2+(3e-06)))],[C1,C2])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Note:The capacitances are expressed in Farads.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 31,
+       "text": [
+        "{C1: 3.60000000000000e-6, C2: 5.40000000000000e-6}"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.29,Page number: 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in each branch.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=20.0/10.0\n",
+      "I4=10.0/5.0\n",
+      "I5=10.0/2.0\n",
+      "I2=(20.0-(10.0+9.0))/1.0\n",
+      "I3=(20.0-50.0-10.0)/20.0\n",
+      "Ia=I1+I2+I3\n",
+      "Ib=I4+I5-I2-I3\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The currents are:\" \n",
+      "print \" I1=%.2f A \\n I2=%.2f A \\n I3=%.2f A \\n I4=%.2f A \\n I5=%.2f A \\n Ia=%.2f A \\n Ib=%.2f A\" %(I1,I2,I3,I4,I5,Ia,Ib)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The currents are:\n",
+        " I1=2.00 A \n",
+        " I2=1.00 A \n",
+        " I3=-2.00 A \n",
+        " I4=2.00 A \n",
+        " I5=5.00 A \n",
+        " Ia=1.00 A \n",
+        " Ib=8.00 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 50
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.30,Page number: 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the currents through resistors.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KVL for the loops 1 and 2 and solving for currents.\"\"\"\n",
+      "print \"All the currents are expressed in Amperes.\"\"\"\n",
+      "print \"Current through R1= Current through R2=I1\"\n",
+      "print \"Current through R3=I3\"\n",
+      "print \"Current through R4=I2\"\n",
+      "I1, I2, I3 = symbols('I1 I2 I3')\n",
+      "solve([(4.5*I1)+(-3*I2)-2,(-3*I1)+(4*I2)-5,I1-I2-I3], [I1,I2,I3]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the currents are expressed in Amperes.\n",
+        "Current through R1= Current through R2=I1\n",
+        "Current through R3=I3\n",
+        "Current through R4=I2\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 54,
+       "text": [
+        "{I1: 2.55555555555556, I2: 3.16666666666667, I3: -0.611111111111111}"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.31,Page number: 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through each resistor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "E1=4.0                #EMF of cell 1(in Volts)\n",
+      "E2=8.0                #EMF of cell 2(in Volts)\n",
+      "R1=0.5                #Internal resistance of cell 1(in Ohms)\n",
+      "R2=1.0                #Internal resistance of cell 2(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R3=1.0/((1.0/3.0)+(1.0/6.0))\n",
+      "Rt=R1+R2+R3+4.5\n",
+      "E=E2-E1\n",
+      "I=E/Rt\n",
+      "I_3=I*(6.0/(3.0+6.0))\n",
+      "I_6=I*(3.0/(3.0+6.0))\n",
+      "V1=E1+(I*R1)\n",
+      "V2=E2-(I*R2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current the 3 Ohms resistor is %.2f A.\" %(I_3)\n",
+      "print \"   The current the 6 Ohms resistor is %.2f A.\" %(I_6)\n",
+      "print \"   The current the 4.5 Ohms resistor is %.2f A.\" %(I)\n",
+      "print \"(b)The potential difference across cell 1 is %.2f V and across cell 2 is %.2f V.\" %(V1,V2)\n",
+      "print \"   Note that the pd of cell E1 is greater than its EMF.This is so because the cell is working as a load.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current the 3 Ohms resistor is 0.33 A.\n",
+        "   The current the 6 Ohms resistor is 0.17 A.\n",
+        "   The current the 4.5 Ohms resistor is 0.50 A.\n",
+        "(b)The potential difference across cell 1 is 4.25 V and across cell 2 is 7.50 V.\n",
+        "   Note that the pd of cell E1 is greater than its EMF.This is so because the cell is working as a load.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.32,Page number: 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the potential of point A.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "E1=8.0                #EMF of first source(in Volts)\n",
+      "E2=6.0                #EMF of second source(in Volts)\n",
+      "E3=4.0                #EMF of third source(in Volts)\n",
+      "R1=5.0                #Resistance of resistor 1(in Ohms) \n",
+      "R2=9.0                #Resistance of resistor 2(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_A=E3+E2-E1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The potential of point A is %.2f V.\" %(V_A)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The potential of point A is 2.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.33,Page number: 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the energy stored in a capacitor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "C=4e-06               #Capacitance of the capacitor(in Farad)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" Applying KCL to the node at the top of the capacitor, current through 5 Ohms resistance is 1+2=3 A. \n",
+      "    Applying KCL to the node at the bottom of the capacitor, current through 2 Ohms resistance is 2-1=1 A.\"\"\"\n",
+      "V=(3*5)+(3*1)+(1*2)\n",
+      "W=0.5*C*V*V\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The energy stored in the capacitor is %e J.\" %(W)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy stored in the capacitor is 8.000000e-04 J.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.34,Page number: 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the potential differnce between two points.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "E1=3.0                #EMF of cell 1(in Volts)\n",
+      "E2=2.0                #EMF of cell 2in Volts)\n",
+      "E3=1.0                #EMF of cell 3in Volts)\n",
+      "r1=1.0                #Internal resistance of cell 1(in Ohms)\n",
+      "r2=1.0                #Internal resistance of cell 2(in Ohms)\n",
+      "r3=1.0                #Internal resistance of cell 3(in Ohms) \n",
+      "R=1.0                 #Resistance of resistor R(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Assigning loop currents as I1 and I2 and writing KVL equations for the two loops,\n",
+      "\n",
+      "   E1-(I1*r1)-((I1-I2)*r2)-E2=0; which is (-2*I1)+I2+1=0; \n",
+      "   \n",
+      "   E2-(I2*r3)-((I2-I1)*r2)-E3=0; which is (-2*I2)+I1+1=0; \"\"\"\n",
+      "\n",
+      "eq1=-1\n",
+      "eq2=-1\n",
+      "I1=((2*eq1)+eq2)/(-3.0)\n",
+      "I2=((2*eq2)+eq1)/(-3.0)\n",
+      "Vab=E2+((I1-I2)*r2)+(0*R)\n",
+      "\n",
+      "V_CB=E2\n",
+      "V_B=0.0\n",
+      "V_C=V_CB+V_B\n",
+      "I3=(V_C-E1)/r1\n",
+      "I4=(V_C-E3)/r3\n",
+      "I6=V_C/R\n",
+      "I5=I3+I4+I6\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The potential difference between points A and B is %.2f V.\" %(Vab) \n",
+      "print \"   The current through r1 is %.2f A,the current through r2=%.2f A.\" %(I1,(I1-I2)) \n",
+      "print \"   The current through r3=%.2f A and the current through R is 0(no closed loop).\" %(I2)\n",
+      "print \"(b)The current through E1 is %.2f A.\" %(I3) \n",
+      "print \"   The current through E2 is %.2f A.\" %(I5)\n",
+      "print \"   The current through E3 is %.2f A.\" %(I4)\n",
+      "print \"   The current through R is %.2f A.\" %(I6)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The potential difference between points A and B is 2.00 V.\n",
+        "   The current through r1 is 1.00 A,the current through r2=0.00 A.\n",
+        "   The current through r3=1.00 A and the current through R is 0(no closed loop).\n",
+        "(b)The current through E1 is -1.00 A.\n",
+        "   The current through E2 is 2.00 A.\n",
+        "   The current through E3 is 1.00 A.\n",
+        "   The current through R is 2.00 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.35,Page number: 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the potential diiference between two points.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "pd_9=0.0\n",
+      "pd_13=0.0\n",
+      "pd_10=6.0*10.0\n",
+      "pd_5=5.0*8.0\n",
+      "pd_18=15.0\n",
+      "Vab=pd_5-pd_10-pd_18\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage drop Vab is %.2f V.\" %(Vab)\n",
+      "print \"The resistors 4 Ohms,11 Ohms,9 Ohms,18 Ohms and 13 Ohms have no effect on the voltage drop Vab.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage drop Vab is -35.00 V.\n",
+        "The resistors 4 Ohms,11 Ohms,9 Ohms,18 Ohms and 13 Ohms have no effect on the voltage drop Vab.\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.36,Page number: 90 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage of dependent voltage source.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KVL for the loops 1 and 2.\"\"\"\n",
+      "print \"The current I1 is expressed in Amperes and the voltage V1 is expressed in Volts.\"\"\"\n",
+      "V1,I1 = symbols('V1 I1')\n",
+      "solve([V1+(2*I1),24-(16*I1)-(4*V1)], [V1,I1]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I1 is expressed in Amperes and the voltage V1 is expressed in Volts.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 33,
+       "text": [
+        "{I1: 3, V1: -6}"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.37,Page number: 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage V.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KCL at the upper node and KVL equation for the left loop.\"\"\"\n",
+      "print \"The currents I1 and I2 are expressed in Amperes and the voltage V is expressed in Volts.\"\"\"\n",
+      "I1,I2,V = symbols('I1 I2 V')\n",
+      "solve([I1-I2+(10*I2),36-(20e03*I1),V+(10*I2*(5e03))], [I1,I2,V]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The currents I1 and I2 are expressed in Amperes and the voltage V is expressed in Volts.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 37,
+       "text": [
+        "{I1: 0.00180000000000000, I2: -0.000200000000000000, V: 10.0000000000000}"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.38,Page number: 90"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the loop currents.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KVL for the loops 1 and 2.\"\"\"\n",
+      "print \"The loop currents i1 and i2 are expressed in Amperes.\"\n",
+      "i1,i2 = symbols('i1 i2')\n",
+      "solve([(3*i2)-((i2-i1)*1)-(i2*2)-3,4-(i1*1)-((i1-i2)*1)-(3*i2)], [i1,i2]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The loop currents i1 and i2 are expressed in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 16,
+       "text": [
+        "{i1: 3, i2: -1}"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.39,Page number: 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the nodal voltages.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Applying KCL at top node and solving for nodal voltages.\"\"\"\n",
+      "print \"All the voltages are expressed in Volts.\"\"\"\n",
+      "print \"The current I is in Amperes.\"\n",
+      "V1,V2,I = symbols('V1 V2 I')\n",
+      "solve([9-I-(V1/10)+(3*I),9-(4*V1/5)-(V1/10),V2-V1-108],[V1,V2,I])\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "All the voltages are expressed in Volts.\n",
+        "The current I is in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 40,
+       "text": [
+        "{I: -4, V1: 10, V2: 118}"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.40,Page number: 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current I by nodal analysis.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KCL at the nodes 1 and 2.\"\"\"\n",
+      "print \"The nodal voltages V1 and V2 are expressed in Volts.\"\n",
+      "print \"The current I is expressed in Amperes.\"\n",
+      "V1,V2,I = symbols('V1 V2 I')\n",
+      "solve([(2*V1)-(V2)-12,(-2*V1)+(4*V2)-6,I-((V1-V2)/3)], [V1,V2,I]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nodal voltages V1 and V2 are expressed in Volts.\n",
+        "The current I is expressed in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 15,
+       "text": [
+        "{I: 1, V1: 9, V2: 6}"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.41,Page number: 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current I in the circuit.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KCL at the nodes 1 and 2.\"\"\"\n",
+      "print \"The nodal voltages V1 and V2 are expressed in Volts.\"\n",
+      "print \"The current I is expressed in Amperes.\"\n",
+      "V1,V2,I = symbols('V1 V2 I')\n",
+      "solve([(1*V1)-(0.5*V2)-6,(-0.5*V1)+(0.625*V2)-3,I-((V1-V2)/2)], [V1,V2,I]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nodal voltages V1 and V2 are expressed in Volts.\n",
+        "The current I is expressed in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 14,
+       "text": [
+        "{I: -1.00000000000000, V1: 14.0000000000000, V2: 16.0000000000000}"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.42,Page number: 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through resistor.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Apply KCL at the nodes 1,2. and 3\"\"\"\n",
+      "print \"The nodal voltages V1,V2 and V3 are expressed in Volts.\"\n",
+      "print \"The current I flowing through the 2 Ohms resistor is expressed in Amperes.\" \n",
+      "V1,V2,V3,I = symbols('V1 V2 V3 I')\n",
+      "solve([(7*V1)-(3*V2)-(4*V3)+11,(-3*V1)+(6*V2)-(2*V3)-3,(-4*V1)-(2*V2)+(11*V3)-25,I-(2*(V2-V3))], [V1,V2,V3,I]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nodal voltages V1,V2 and V3 are expressed in Volts.\n",
+        "The current I flowing through the 2 Ohms resistor is expressed in Amperes.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 12,
+       "text": [
+        "{I: -2, V1: 1, V2: 2, V3: 3}"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.43,Page number: 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the nodal voltages in the circuit.\"\"\"\n",
+      "\n",
+      "from sympy import *;\n",
+      "\n",
+      "#Calculations:\n",
+      "Ieq=13.0*5.0 #Equivalent current source\n",
+      "\"\"\"Apply KCL at the nodes 1 and 2.\"\"\"\n",
+      "print \"The nodal voltages V1 and V2 are expressed in Volts.\"\n",
+      "V1,V2 = symbols('V1 V2')\n",
+      "solve([(9*V1)-(5*V2)-10,(-5*V1)+(11*V2)-52], [V1,V2]) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nodal voltages V1 and V2 are expressed in Volts.\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 11,
+       "text": [
+        "{V1: 5, V2: 7}"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter4.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter4.ipynb
new file mode 100755
index 00000000..74af3cd5
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter4.ipynb
@@ -0,0 +1,1114 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: NETWORK THEOREMS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1,Page number: 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current I in the circuit using superposition theorem.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=-0.5*(0.3/(0.1+0.3))\n",
+      "I2=80e-03/(0.1+0.3)\n",
+      "I=I1+I2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I in the circuit is %.3f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I in the circuit is -0.175 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2,Page number: 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding current I_x in the network using superposition theorem.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=10.0/(50+150)\n",
+      "I2=40*(150.0/(50+150))\n",
+      "I3=-120*(50.0/(150+50))\n",
+      "Ix=I1+I2+I3\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current Ix determined using superposition principle is %.2f A.\" %(Ix)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current Ix determined using superposition principle is 0.05 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3,Page number: 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across resistor by applying the principle of superposition.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "i=4.0*(1.0/(1.0+(2+3)))\n",
+      "R=3.0\n",
+      "v_4=i*R\n",
+      "v_5=(-5*(1.0/(1+(2+3))))*R\n",
+      "v_6=6.0*(3.0/(1+2+3))\n",
+      "v=+v_4+v_5+v_6\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The total voltage(v) across the 3 ohm resistor is %.2f V.\" %(v)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total voltage(v) across the 3 ohm resistor is 2.50 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4,Page number: 108 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of I_s to reduce the voltage across the 4-ohm resistor to zero. \"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_source=10.0         #Voltage of the source(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=V_source/(2+4+6)\n",
+      "\n",
+      "\"\"\" I2(from top to bottom in the 4 ohm resistor) = -Is*((2+6)/(2+6+4)) = -(2/3)*Is ;\n",
+      "    \n",
+      "    The voltage across the 4 ohm resistor can be zero,only if the current through this resistor is zero.  \n",
+      "    \n",
+      "     I1+I2=0; \"\"\"\n",
+      "\n",
+      "Is=I1*(3.0/2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current Is to reduce the voltage across the 4 ohm resistor to zero is %.2f A.\" %(Is) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current Is to reduce the voltage across the 4 ohm resistor to zero is 1.25 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5,Page number: 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question: \n",
+      "\"\"\"Finding the voltage across the load resistor using Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=(50.0-10.0)/(10+10+20)\n",
+      "I2=1.5*(10.0/(10.0+(10+20)))\n",
+      "I=I1+I2\n",
+      "V_Th=I*20\n",
+      "R_Th=1.0/((1.0/20)+(1.0/(10+10)))\n",
+      "R_L=5.0\n",
+      "V_L=V_Th*(R_L/(R_L+R_Th))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across the load resistor R_L is %.2f V.\" %(V_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across the load resistor R_L is 9.17 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6,Page number: 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across the resistor by applying Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "V_Th=5.0*1.0\n",
+      "R_Th=3.0\n",
+      "R_L=3.0\n",
+      "V_L=V_Th*(R_L/(R_L+R_Th))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across the resistor by applying Thevenin's Theorem is %.2f V.\" %(V_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across the resistor by applying Thevenin's Theorem is 2.50 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7,Page number: 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding Norton's equivalent circuit with respect to terminals AB.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=10.0/5\n",
+      "I2=5.0/10\n",
+      "I_N=I1+I2\n",
+      "R_N=1.0/((1.0/5)+(1.0/10))\n",
+      "I_L=I_N*((10.0/3)/((10.0/3)+5))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"When terminals AB are shorted, the current I_N is %.2f A.\" %(I_N)\n",
+      "print \"The value of current that would flow through a load resistor of 5 ohm if it were connected across terminals AB is %.2f A.\" %(I_L)  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "When terminals AB are shorted, the current I_N is 2.50 A.\n",
+        "The value of current that would flow through a load resistor of 5 ohm if it were connected across terminals AB is 1.00 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8,Page number: 114 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the available power from the battery.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Voc=12.6              #Open-circuit voltage(in Volts)\n",
+      "Isc=300.0             #Short-circuit voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ro=Voc/Isc\n",
+      "P_avl=(Voc*Voc)/(4*Ro)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The available power from the battery is %.2f W.\" %(P_avl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The available power from the battery is 945.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9,Page number: 115 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the available power from the battery.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "no_of_cells=8         #Number of dry cells in the battery\n",
+      "emf=1.5               #EMF of each cell(in Volts)\n",
+      "int_res=0.75          #Internal resistanceof each cell(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Voc=no_of_cells*emf\n",
+      "Ro=no_of_cells*int_res\n",
+      "P_avl=(Voc*Voc)/(4.0*Ro)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The available power from the battery is %.2f W.\" %(P_avl)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The available power from the battery is 6.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10,Page number: 115 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Plotting a curve showing the variation of the output power Po with the load resistance R_L.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "P=25.0                #Power to be delivered to the speaker(in Watts)\n",
+      "Ro=8.0                #Resistance of the speaker(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_Th=sqrt(P*4*Ro)\n",
+      "Vo=V_Th/2.0\n",
+      "R_Th=Ro\n",
+      "R_L=0.0\n",
+      "P1=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
+      "R_L=2.0\n",
+      "P2=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
+      "R_L=4.0\n",
+      "P3=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
+      "R_L=6.0\n",
+      "P4=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
+      "R_L=8.0\n",
+      "P5=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
+      "R_L=16.0\n",
+      "P6=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
+      "R_L=32.0\n",
+      "P7=(V_Th*V_Th*R_L)/((R_Th+R_L)*(R_Th+R_L))\n",
+      "P8=0.0\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The voltage provided by this amplifier to to the speaker is %.2f V.\" %(Vo)\n",
+      "print \"(b)(i)If the load is a short circuit(R_L=0),the output voltage Vo would be zero and hence the output power,Po= Vo*I_L = 0*I_L =0.\"\n",
+      "print \"   (ii) If the load is an open circuit(R_L=0),the load current I_L would be zero and hence the output power,Po= Vo*I_L = Vo*0 = 0.\"\n",
+      "print \"(c)(i) For R_L = 0 , Po = %.2f W. \\n   (ii) For R_L = 2 Ohms, Po = %.2f W.\" %(P1,P2)  \n",
+      "print \"   (iii) For R_L = 4 Ohms, Po = %.2f W. \\n   (iv) For R_L = 6 Ohms, Po = %.2f W.\" %(P3,P4)  \n",
+      "print \"   (v) For R_L = 8 Ohms, Po = %.2f W. \\n   (vi) For R_L = 16 Ohms, Po = %.2f W.\" %(P5,P6) \n",
+      "print \"   (vii) For R_L = 32 Ohms, Po = %.2f W. \\n   (viii) For R_L = infinity, Po = %.2f W.\" %(P7,P8)\n",
+      "print \"Note: Po is the power delivered to the speaker (in Watts) and R_L is the speaker resistance(in Ohms).\"  \n",
+      "\n",
+      "\n",
+      "from __future__ import division\n",
+      "from pylab import *\n",
+      "from matplotlib import *\n",
+      "import numpy as np\n",
+      "%pylab inline\n",
+      "\n",
+      "\n",
+      "#Variable declaration:\n",
+      "Rdata=[0.0, 2.0, 4.0, 6.0, 8.0, 16.0]     #(in ohm)\n",
+      "Podata=[0, 16, 22.22, 24.49, 25, 22.22]   #(in Watt)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R=np.array(Rdata)\n",
+      "Po=np.array(Podata)\n",
+      "length=len(R)\n",
+      "Rmax=R[length-1]\n",
+      "a=polyfit(R,Po,4)\n",
+      "Rfit=[0]*100\n",
+      "Pofit=[0]*100\n",
+      "for n in range(1,100,1):\n",
+      "    Rfit[n-1]=Rmax*(n-1)/100\n",
+      "    Pofit[n-1]=a[0]*Rfit[n-1]**4+a[1]*Rfit[n-1]**3+a[2]*Rfit[n-1]**2+a[3]*Rfit[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(Rdata,Podata,'o')\n",
+      "plot(np.array(Rfit),np.array(Pofit),'g.') \n",
+      "xlabel('R_L (in ohm) ')\n",
+      "ylabel('Po (in Watt) ')\n",
+      "title('Ouput power(Po) vs. Speaker Resistance(R_L)')\n",
+      "grid()\n",
+      "print \"\\n\\nThe required plot is shown below: \"\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The voltage provided by this amplifier to to the speaker is 14.14 V.\n",
+        "(b)(i)If the load is a short circuit(R_L=0),the output voltage Vo would be zero and hence the output power,Po= Vo*I_L = 0*I_L =0.\n",
+        "   (ii) If the load is an open circuit(R_L=0),the load current I_L would be zero and hence the output power,Po= Vo*I_L = Vo*0 = 0.\n",
+        "(c)(i) For R_L = 0 , Po = 0.00 W. \n",
+        "   (ii) For R_L = 2 Ohms, Po = 16.00 W.\n",
+        "   (iii) For R_L = 4 Ohms, Po = 22.22 W. \n",
+        "   (iv) For R_L = 6 Ohms, Po = 24.49 W.\n",
+        "   (v) For R_L = 8 Ohms, Po = 25.00 W. \n",
+        "   (vi) For R_L = 16 Ohms, Po = 22.22 W.\n",
+        "   (vii) For R_L = 32 Ohms, Po = 16.00 W. \n",
+        "   (viii) For R_L = infinity, Po = 0.00 W.\n",
+        "Note: Po is the power delivered to the speaker (in Watts) and R_L is the speaker resistance(in Ohms).\n",
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "\n",
+        "\n",
+        "The required plot is shown below: "
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['streamplot', 'rc', 'tri', 'axes', 'legend', 'rc_context', 'figure', 'f', 'quiver', 'axis', 'linalg', 'draw_if_interactive', 'text', 'random', 'colors', 'stackplot', 'contour', 'colorbar', 'rcdefaults', 'table', 'power', 'info', 'fft', 'test']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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RZvBruQ/ADFQfHKrqdoktMl3xaNPvQHHlxr9Tp39g9eoRPNY+Y2akIX0AfARg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\nxQ0AY4w1UdwAMMZYE8UNAGOMNVH/B1z6XURJsXa0AAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x43b9d10>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11,Page number: 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in branch B due to the voltage source of 36 V in branch A.\"\"\"\n",
+      "\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_supply_a=36.0                         #Supply voltage in the network shown in figure a(in Volts)\n",
+      "Req_a=2+ 1/((1.0/12)+(1.0/(3+1))) + 4   #Equivalent resistance of network shown in figure a(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_supply_a=V_supply_a/Req_a\n",
+      "I=I_supply_a*(12.0/(12+4))\n",
+      "V_supply_b=36.0\n",
+      "Req_b= 3+ 1.0/((1.0/12)+(1.0/(2+4))) + 1\n",
+      "I_supply_b=V_supply_b/Req_b\n",
+      "I_dash= I_supply_b*(12.0/(12+6))\n",
+      "R_tr=V_supply_a/I\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I(in figure (a)) is %.2f A and the current I'(in branch A in figure (b)) is %.2f A.\" %(I,I_dash)\n",
+      "if(I==I_dash) : print \"As the two currents I and I' have the same value, the reciprocity theorem is established.\\n \"\n",
+      "else : print \"As the two currents I and I' have different values, the reciprocity theorem is not established.\\n \"\n",
+      "print \"The transfer resistance is %.2f Ohms.\" %(R_tr)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I(in figure (a)) is 3.00 A and the current I'(in branch A in figure (b)) is 3.00 A.\n",
+        "As the two currents I and I' have the same value, the reciprocity theorem is established.\n",
+        " \n",
+        "The transfer resistance is 12.00 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13,Page number: 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current using Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_L=20.0              #Load resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" KVL equation for the loop bacdeb, V_Th+50-20+10=0.\"\"\"\n",
+      "V_Th=-50.0-10.0+20.0\n",
+      "\"\"\"Shorting all the voltage sources.\"\"\"\n",
+      "R_Th=0.0\n",
+      "I3=V_Th/(R_L+R_Th)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I3 in the circuit is %.2f A.\" %(I3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I3 in the circuit is -2.00 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14,Page number: 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current using Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_L=1.0              #Load resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" KVL equation for the loop bacdeb, V_Th+50-20+10=0.\"\"\"\n",
+      "V_Th=20.0-10.0-9.0\n",
+      "\"\"\"Shorting all the voltage sources.\"\"\"\n",
+      "R_Th=0.0\n",
+      "I2=V_Th/(R_L+R_Th)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current I2 in the circuit is %.2f A.\" %(I2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current I2 in the circuit is 1.00 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15,Page number: 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current using superposition theorem.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=0.5                 #Load resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I=15.0/(1+0.6)\n",
+      "I1=I*(1/(1+(R+1)))\n",
+      "I2=20/((2/3.0)+2)\n",
+      "I1_20=I2*(2.0/(2.0+(R+R)))\n",
+      "Inet=I1-I1_20\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current flowing through R from A to B is %.2f A.\" %(Inet) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current flowing through R from A to B is -1.25 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16,Page number: 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across the resistance using Thevenin's theorem.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"The first step is to remove the 5 Ohms resistance.Then, find the open-circuit voltage Voc across AB. \n",
+      "   Using nodal analysis,for node C we can write ((Voc-20)/2)+((Voc+10)/1)+((Voc-12)/4)=10.\"\"\"\n",
+      "Voc=(10+3)/(0.5+1+0.25)\n",
+      "V_Th=Voc\n",
+      "R_Th=1.0/((1.0/2.0)+(1.0)+(1/4.0))\n",
+      "V_AB=V_Th*(5.0/(5.0+R_Th))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across the 5 Ohms resistance is %.2f V.\" %(V_AB)\n",
+      "print \"Note:There is a calculation error in the textbook.Voc=7.43 V but not 1.74 V. Therefore V_AB=6.67 V.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across the 5 Ohms resistance is 6.67 V.\n",
+        "Note:There is a calculation error in the textbook.Voc=7.43 V but not 1.74 V. Therefore V_AB=6.67 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17,Page number: 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current using Norton's theorem.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "Isc=20.0/10.0\n",
+      "I_N=Isc\n",
+      "R_N=1.0/((1.0/10)+(1.0/10))\n",
+      "I=I_N*(R_N/(R_N+2.0))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current through the 2 Ohms resistance when connected across terminals AB is %.2f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current through the 2 Ohms resistance when connected across terminals AB is 1.43 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.18,Page number: 123"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finfing the power consumed by resistor using Norton's theorem.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_L=2.0               #Resistance of load(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"To determine I_N,short circuit the terminals xy and find the current Isc throgh this short circuit.\n",
+      "\n",
+      "   Applying node-voltage analysis to determine the voltage at node 1,\n",
+      "   \n",
+      "   ((V1-12)/3)+(V1/2)=4;\"\"\"\n",
+      "\n",
+      "V1=(4.0+(12.0/3.0))/((1.0/3.0)+0.5)\n",
+      "Isc=V1/2.0\n",
+      "I_N=Isc\n",
+      "R_N=1.0/((1.0/(3.0+2.0))+(1.0/5.0))\n",
+      "I=I_N*(R_N/(R_N+R_L))\n",
+      "P=I*I*R_L\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The power consumed by the 2 Ohms load reistor is %.3f W.\" %(P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power consumed by the 2 Ohms load reistor is 14.222 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19,Page number: 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage across load.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_L=5.0               #Load resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "R_N=1.0/((1.0/20.0)+(1.0/(10.0+10.0)))\n",
+      "I_N=(65.0-10.0)/(10.0+10.0)\n",
+      "I_L=I_N*(R_N/(R_N+R_L))\n",
+      "V_L=I_L*R_L\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage across the load using Norton's theorem is %.2f V.\" %(V_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage across the load using Norton's theorem is 9.17 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.20,Page number: 125 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the reading of voltmeter.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=500e03              #Resistance across which voltmeter is connected(in Ohms)\n",
+      "Rm=10e06              #Internal resistance of voltmeter(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_a=10.0*(R/(R+800e03))\n",
+      "R_Th=1.0/((1/R)+(1/800e03))\n",
+      "V_Th=V_a\n",
+      "V_b=V_Th*(Rm/(Rm+R_Th))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The reading of the voltmeter if it assumed to be ideal is %.2f V.\" %(V_a)\n",
+      "print \"(b)The reading of the voltmeter if it has an internal resistance of 10 M Ohms is %.2f V.\" %(V_b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The reading of the voltmeter if it assumed to be ideal is 3.85 V.\n",
+        "(b)The reading of the voltmeter if it has an internal resistance of 10 M Ohms is 3.73 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.21,Page number: 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the load resistance for receiving maximum power.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R_L=3e03             #Load resistance(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"First, we remove R_L and then find Voc=V_Th. We transform the 5-mA current source into voltage source.\"\"\"\n",
+      "Voc=(60.0-40.0)*((6e03)/((6e03)+(12e03+18e03)))\n",
+      "V_Th=Voc\n",
+      "R_Th=1.0/((1.0/6e03)+(1.0/(12e03+18e03)))\n",
+      "Vab_b=V_Th*(R_L/(R_L+R_Th))\n",
+      "R_L_c=R_Th\n",
+      "\"\"\" The current required through 6 kilo Ohms resistor is 0.1 mA.\"\"\"\n",
+      "Vab_d=(0.1e-03)*(6e03)\n",
+      "V_30k=20-Vab_d\n",
+      "I_30k=V_30k/(30e03)\n",
+      "I_L=I_30k-(0.1e-03)\n",
+      "R_L_d=Vab_d/I_L\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The Thevenin's equivalent voltage is %.2f V and the Thevenin's resistance is %.2f kilo Ohms.\" %(V_Th,(R_Th/1000.0))\n",
+      "print \"(b)The Vab for R_L=3 kilo Ohms is %.3f V.\" %(Vab_b)\n",
+      "print \"(c)The value of R_L which receives maximum power from the circuit is %.2f kilo Ohms.\" %(R_L_c/1000.0)\n",
+      "print \"(d)The value of R_L which makes the current in the 6 kilo Ohms resistor to be 0.1 mA is %.3f kilo Ohms.\" %(R_L_d/1000.0) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The Thevenin's equivalent voltage is 3.33 V and the Thevenin's resistance is 5.00 kilo Ohms.\n",
+        "(b)The Vab for R_L=3 kilo Ohms is 1.250 V.\n",
+        "(c)The value of R_L which receives maximum power from the circuit is 5.00 kilo Ohms.\n",
+        "(d)The value of R_L which makes the current in the 6 kilo Ohms resistor to be 0.1 mA is 1.098 kilo Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.22,Page number: 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of resistance.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=12.0                #Supply voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" To receive maximum power,the resistance R should be equal to the output resistance of the remaining circuit,which is same as \n",
+      "    Thevenin's resistance.\"\"\"\n",
+      "R_Th=1.0/((1.0/2.0)+(1.0/6.0))\n",
+      "R=R_Th\n",
+      "\"\"\" Mesh equations to find loop currents I1 and I2, \n",
+      "    \n",
+      "    (8*I1)-(6*I2)=4 and -(6*I1)+(7.5*I2)=8. Solving these equations, we get I1=3.25 A. \"\"\"\n",
+      "I1=((4*7.5)+(8*6))/((8*7.5)-(6*6))\n",
+      "P=V*I1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The value of R to receive maximum power from the circuit is %.2f Ohms.\" %(R)\n",
+      "print \"(b)The power supplied by the 12 V source is %.2f W.\" %(P) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The value of R to receive maximum power from the circuit is 1.50 Ohms.\n",
+        "(b)The power supplied by the 12 V source is 39.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.23,Page number: 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current Is.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=12.0/(80.0+120)\n",
+      "\"\"\" I2=Is*(80/(80+120)); I1+I2=0;\"\"\"\n",
+      "Is=-I1/(80.0/(80+120))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current Is such that the current through 120 Ohms resistor is zero is %.2f A.\" %(Is)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current Is such that the current through 120 Ohms resistor is zero is -0.15 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.24,Page number: 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the Thevenin equivalent circuit.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vs=12.0               #Supply voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"The circuit has no independent source.So,the current I in the 12 Ohms resistor will be zero and hence the dependent voltage source\n",
+      "   (8*I) will also be zero. Obviously,Thevenin's voltage V_Th=0 V. \"\"\"\n",
+      "V_Th=0.0\n",
+      "I=12.0/12.0\n",
+      "V=8*I\n",
+      "\"\"\"Applying KCL to node a,\"\"\"\n",
+      "Is=(12.0/12.0)+(12.0/6.0)+((12.0-8.0)/4.0)\n",
+      "R_Th=Vs/Is\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The Thevenin's equivalent resistance is %.2f Ohms and the Thevenin's equivalent voltage is %.2f V.\" %(R_Th,V_Th)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Thevenin's equivalent resistance is 3.00 Ohms and the Thevenin's equivalent voltage is 0.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter5.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter5.ipynb
new file mode 100755
index 00000000..3a8210b4
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter5.ipynb
@@ -0,0 +1,841 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: ELECTROMAGNETISM"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1,Page number: 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current necessary to produce a magnetic field inside the solenoid.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "B=20e-03                #Magnitude of magnetic field inside the solenoid(in Tesla)\n",
+      "n=20e02                 #Number of turns per cm in the long solenoid\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "abs_per=(4*pi)*(1e-07)\n",
+      "I=B/(abs_per*n)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The required current is %.2f A.\" %(I)  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required current is 7.96 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2,Page number: 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnetic field near the centre and the ends of a solenoid.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "abs_per=4*pi*(1e-07)      #Absolute permeability of free space(in Henry per metre)\n",
+      "I=6.0                     #Current carried by the solenoid(in Amperes)\n",
+      "l=50e-02                  #Length of solenoid(in metres)\n",
+      "r=1.4e-02                 #Radius of the lowest layer(in metres)\n",
+      "turns_per_layer=350       #Number of turns per layer\n",
+      "number_of_layers=4        #Number of layers\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "n=(turns_per_layer*number_of_layers)/l\n",
+      "B_centre=abs_per*n*I\n",
+      "B_end=(abs_per*n*I)/2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The magnetic field inside the solenois is quite uniform,its strength near the centre on its axis and\" \n",
+      "print \"   off its axis is the same and its magnitude is %e T.\" %(B_centre)\n",
+      "print \"(b)The magnetic field strength at the end of the solenoid is %e T.\" %(B_end)\n",
+      "print \"(c)The magnetic field far outside the solenoid near its axis is negligible,compared to the internal field.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The magnetic field inside the solenois is quite uniform,its strength near the centre on its axis and\n",
+        "   off its axis is the same and its magnitude is 2.111150e-02 T.\n",
+        "(b)The magnetic field strength at the end of the solenoid is 1.055575e-02 T.\n",
+        "(c)The magnetic field far outside the solenoid near its axis is negligible,compared to the internal field.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3,Page number: 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the force between two long straight parallel wires.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "r=2                      #Distance between the parallel current carrying wires(in metres) \n",
+      "abs_per=(4*pi)*(1e-07)   #Absolute permeability of free space(in Henry per metre)\n",
+      "I1=80                    #Current in the first wire(in Amperes)\n",
+      "I2=30                    #Current in the second wire(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "F=(abs_per*I1*I2)/(2*pi*r)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The magnitude of the force between the two wires is %e N/m.\" %(F)\n",
+      "print \"Since the two currents are in the same direction,the force will be attractive.\"  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnitude of the force between the two wires is 2.400000e-04 N/m.\n",
+        "Since the two currents are in the same direction,the force will be attractive.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4,Page number: 148 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the force due to current flowing in two straight,parallel wires.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I1=4.0                 #Current in the first wire(in Amperes)\n",
+      "I2=6.0                 #Current in the second wire(in Amperes)\n",
+      "r=3e-02                #Distance between the parallel current carrying wires(in metres) \n",
+      "abs_per=(4*pi)*(1e-07) #Absolute permeability of free space(in Henry per metre)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "F=(abs_per*I1*I2)/(2*pi*r)\n",
+      "l=15e-02\n",
+      "F_net=F*l\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The net force on 15cm(=0.15 m) section of wire B near its centre is %e N.\" %(F_net)\n",
+      "print \"Since the currents are in opposite directions,the force F_net is repulsive.\"\n",
+      "print \"It means that its direction is normal to wire A away from it.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net force on 15cm(=0.15 m) section of wire B near its centre is 2.400000e-05 N.\n",
+        "Since the currents are in opposite directions,the force F_net is repulsive.\n",
+        "It means that its direction is normal to wire A away from it.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5,Page number: 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf produced in a conductor placed in an uniform magnetic field.\"\"\"\n",
+      "\n",
+      "from math import sin,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "B=0.5                  #Magnitude of magnetic field(in Tesla)\n",
+      "l=20e-02               #Active length of the conductor(in metres) \n",
+      "v=5                    #Velocity of the conductor(in metres per second)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "angle_a=0\n",
+      "e_a=B*l*v*sin(radians(angle_a))\n",
+      "angle_b=90\n",
+      "e_b=B*l*v*sin(radians(angle_b))\n",
+      "angle_c=30\n",
+      "e_c=B*l*v*sin(radians(angle_c))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The emf induced in the straight conductor when its motion is parallel to the magnetic field is %.2f V.\" %(e_a)\n",
+      "print \"(b)The emf induced in the straight conductor when its motion is perpendicular to the magnetic field is %.2f V.\" %(e_b)\n",
+      "print \"(c)The emf induced in the straight conductor when its motion is at an angle 30 degrees to the magnetic field is %.2f V.\" %(e_c)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The emf induced in the straight conductor when its motion is parallel to the magnetic field is 0.00 V.\n",
+        "(b)The emf induced in the straight conductor when its motion is perpendicular to the magnetic field is 0.50 V.\n",
+        "(c)The emf induced in the straight conductor when its motion is at an angle 30 degrees to the magnetic field is 0.25 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6,Page number: 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf generated between the wing-tips of an aeroplane.\"\"\"\n",
+      "\n",
+      "from math import sin,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "l=52                      #Wing span of the aeroplane(in metres)\n",
+      "B=38e-06                  #Magnitude of magnetic field(in Tesla)\n",
+      "v=1100.0*(1000.0/3600.0)  #Velocity of the aeroplane(in metres per second)\n",
+      "angle=90                  #Angle between the magnetic field vector and the velocity vector(in degrees)   \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "e=B*l*v*sin(radians(angle))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The emf generated between the wing tips is %.5f V.\" %(e)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The emf generated between the wing tips is 0.60378 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7,Page number: 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf developed between the centre and the metallic ring.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "\"\"\" A=(1/2)*R*R*theta; where theta is the angle between the rod and the rod OP(in figure) at time t  \"\"\"\n",
+      "B=1.0       #Magnitude of magnetic field(in Tesla)\n",
+      "R=1         #Radius of circular ring(in metres)\n",
+      "f=50        #Frequency of rotation(in revolutions per second) \n",
+      "\n",
+      "\"\"\" B=d(flux)/dt= d(B*A)/dt= (1/2)*B*R*R*d(theta)/dt= (1/2)*B*R*R*ang_freq \"\"\"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "ang_freq=2*pi*f\n",
+      "e=(1.0/2)*B*R*R*ang_freq\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The emf developed between the centre and the metallic ring is %.2f V.\" %(e)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The emf developed between the centre and the metallic ring is 157.08 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8,Page number: 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the force needed to pull a rectangular loop placed in an uniform magnetic field.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "B=0.5                 #Magnitude of magnetic field(in Tesla)\n",
+      "w=3e-02               #Width of the rectangular loop(in metres)\n",
+      "l=10e-02              #Length of the rectangular loop(in metres)\n",
+      "v=1e-02               #Velocity of the rectangular loop(in metre per second)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "e1=B*w*v\n",
+      "t1=l/v\n",
+      "e2=B*l*v\n",
+      "t2=w/v\n",
+      "R=1e-03\n",
+      "F_a=(pow((B*w),2)*v)/R\n",
+      "F_b=(pow((B*l),2)*v)/R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The induced emf is %e V and the time for which the induced voltage lasts is %.2f secs.\" %(e1,t1)\n",
+      "print \"(b)The induced emf is %e V and the time for which the induced voltage lasts is %.2f secs.\" %(e2,t2)\n",
+      "print \"(c)Because of the gap,no current can flow.Hence there is no heat produced(or no I*I*R losses).\"\n",
+      "print \"   If we neglect friction,no force is required to pull the coil.\"\n",
+      "print \"(d)The force required to pull the loop if it has no cut and has a resistance of 1 mill ohm is :\"\n",
+      "print \"   For fig (a): F=%e N.\" %(F_a)\n",
+      "print \"   For fig (b): F=%e N.\" %(F_b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The induced emf is 1.500000e-04 V and the time for which the induced voltage lasts is 10.00 secs.\n",
+        "(b)The induced emf is 5.000000e-04 V and the time for which the induced voltage lasts is 3.00 secs.\n",
+        "(c)Because of the gap,no current can flow.Hence there is no heat produced(or no I*I*R losses).\n",
+        "   If we neglect friction,no force is required to pull the coil.\n",
+        "(d)The force required to pull the loop if it has no cut and has a resistance of 1 mill ohm is :\n",
+        "   For fig (a): F=2.250000e-03 N.\n",
+        "   For fig (b): F=2.500000e-02 N.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9,Page number: 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnetic field due  to a line caryying current.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=90.0                #Current flowing through the line(in Amperes)\n",
+      "x=1.5                 #Distance of point of observation from the line(in metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "abs_per=(4*pi)/(1e07)\n",
+      "B=(abs_per*I)/(2*pi*x)\n",
+      "  \n",
+      "\n",
+      "#Result:\n",
+      "print \"The magnetic field due to the current at a point 1.5 m below the line is %e T.\" %(B)\n",
+      "print \"By applying the right-hand thumb rule,we find that the direction of the magnetic field is from north to south.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnetic field due to the current at a point 1.5 m below the line is 1.200000e-05 T.\n",
+        "By applying the right-hand thumb rule,we find that the direction of the magnetic field is from north to south.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10,Page number: 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the force per unit length on a current carrying wire.\"\"\"\n",
+      "\n",
+      "from math import sin,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=8.0                 #Current flowing through the wire(in Amperes)\n",
+      "B=0.15                #Magnitude of uniform magnetic field(in Tesla)\n",
+      "angle=30              #Angle between the direction of current and the uniform magnetic field(in degrees)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Fu=I*B*sin(radians(angle))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The force per unit length of the wire is %.2f N/m.\" %(Fu)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The force per unit length of the wire is 0.60 N/m.\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11,Page number: 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the torque experienced by a square coil.\"\"\"\n",
+      "\n",
+      "from math import sin,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=12.0                #Current flowing through the square coil(in Amperes)\n",
+      "N=20.0                #Number of turns in the square coil\n",
+      "l=10e-02              #Length of each side of the square coil(in metres)\n",
+      "B=0.8                 #Magnitude of uniform horizontal magnetic field(in Tesla)\n",
+      "angle=30.0            #Angle made by normal to the plane of the coil with the direction of horizontal magnetic field(in degrees)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "F=I*B*l*N\n",
+      "x=l*sin(radians(angle))\n",
+      "T=F*x\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The torque experienced by the coil is %.3f Nm.\" %(T)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The torque experienced by the coil is 0.960 Nm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12,Page number: 156"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the net force on a rectangular loop.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I1=15.0               #Current carried by the rectangular loop(in Amperes)\n",
+      "I2=25.0               #Current carried by the straight conductor(in Amperes)\n",
+      "l=25e-02              #Length of the rectangular loop(in metres)\n",
+      "w=10e-02              #Width of the rectangular loop(in metres)\n",
+      "r=2e-02               #Separation between the nearer side of the loop and the conductor(in metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "abs_per=(4*pi)/(1e07)\n",
+      "\"\"\"Parts AD and BC do not experience any force,since these conductors are at right angles to the long straight conductor.\n",
+      "   \n",
+      "   The current I1 in AB and the current I2 in the straight conductor are in the same direction.\n",
+      "   Hence,the force F_AB is attractive.\"\"\"\n",
+      "\n",
+      "F_AB=((abs_per*I1*I2)/(2*pi*r))*l\n",
+      "\n",
+      "\"\"\"  The current I1 in CD and the current I2 in the straight conductor are in opposite directions.\n",
+      "     Hence,the force F_CD is repulsive.\"\"\"\n",
+      "\n",
+      "F_CD=((abs_per*I1*I2)/(2*pi*(r+w)))*l\n",
+      "F_net=F_AB-F_CD\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The net force on the rectangular loop is %.5f mN.\" %(F_net*1000.0) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net force on the rectangular loop is 0.78125 mN.\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13,Page number: 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the location of a point where the resultant magnetic field is zero.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=10.0                #Current flowing through the wire(in Amperes)\n",
+      "B=2.0e-03             #Horizontal component of Earth's magnetic field(in Tesla)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" To make the resultant magnetic field zero,the magnetic filed produced by the long vertical wire must be equal and opposite \n",
+      "    to the horizontal component of Earth's magnetic field.This is possible at a point P,west of the wire. \"\"\"\n",
+      "\n",
+      "abs_per=(4*pi)/(1e07)\n",
+      "x=(abs_per*I)/(2*pi*B)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The distance between the required point from the wire is %e m,to the west of the wire.\" %(x)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The distance between the required point from the wire is 1.000000e-03 m,to the west of the wire.\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14,Page number: 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the force per unit length between two current carrying wires.\"\"\"\n",
+      "\n",
+      "from math import pi \n",
+      "\n",
+      "#Variable Declaration:\n",
+      "I=300.0               #Current flowing through each wire(in Amperes)\n",
+      "r=1.5e-02             #Separation between the wires(in metres)\n",
+      "l=70e-02              #Length of each wire(in metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "abs_per=(4*pi)/(1e07)\n",
+      "\"\"\"Since the length of the wires is 70 cm and their separation is only 1.5 cm(i.e, l>>r),\n",
+      "   we can conclude that for the given separation the two wires are infinitely long. \"\"\"\n",
+      "F=(abs_per*I*I)/(2*pi*r)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The force per unit length between the two wires is %.2f N/m.\" %(F)\n",
+      "print \"Since the currents in the two wires are in the opposite directions,the force between them will be repulsive.\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The force per unit length between the two wires is 1.20 N/m.\n",
+        "Since the currents in the two wires are in the opposite directions,the force between them will be repulsive.\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.15,Page number: 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the counter torque to prevent a coil from turning.\"\"\"\n",
+      "\n",
+      "from math import pi,sin,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=30.0                #Number of turns in the circular coil\n",
+      "r=8e-02               #Radius of circular coil(in metres)\n",
+      "I=6.0                 #Current flowing through the circular coil(in Amperes)\n",
+      "B=1.0                 #Magnitude of uniform magnetic field(in Tesla)\n",
+      "angle=60              #Angle made by the field lines with the normal to the coil(in degrees)\n",
+      "\n",
+      " \n",
+      "#Calculations:\n",
+      "\"\"\"The counter torque required to prevent the coil from moving must be equal(and opposite) to the torque developed.\"\"\"\n",
+      "A=pi*r*r\n",
+      "tor=B*I*N*A*sin(radians(angle))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The counter torque that must be applied to prevent the coil from turning is %.2f Nm.\" %(tor) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The counter torque that must be applied to prevent the coil from turning is 3.13 Nm.\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16,Page number: 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the maximum voltage induced in a coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "A=300e-04             #Area of circular coil(in square-m)\n",
+      "N=25.0                #Number of turns in the circular coil\n",
+      "w=40.0                #Angluar frequency(in radians per second)\n",
+      "B=0.05                #Magnitude of uniform magnetic field(in Tesla)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Em=N*B*A*w\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The maximum value of emf induced in the coil is %.2f V.\" %(Em)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum value of emf induced in the coil is 1.50 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.17,Page number: 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the induces emf in a circular conducting loop.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "B=0.02                #Magnitude of uniform magnetic field(in Tesla)\n",
+      "r=2e-02               #Radius of the circular loop(in metres)\n",
+      "rate=1.0e-03          #Rate of shrinking of the radius(in metre per second)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" flux=B*A; A=(pi*r*r);\n",
+      "    e=d(flux)/dt; \n",
+      "    \n",
+      "    e=(B*pi*2*r)*(dr/dt);  \"\"\"\n",
+      "\n",
+      "e=B*pi*2*r*rate\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The emf induced in the loop at an instant when the radius is 2 cm is %e V.\" %(e) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The emf induced in the loop at an instant when the radius is 2 cm is 2.513274e-06 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter6.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter6.ipynb
new file mode 100755
index 00000000..fde4f296
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter6.ipynb
@@ -0,0 +1,627 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: MAGNETIC CIRCUITS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1,Page number: 167\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnetic field strength and total flux passing through a coil.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=200                 #Number of turns in the coil \n",
+      "l=0.60                #Length(Circumference) of the wooden ring(in metres)\n",
+      "A=500e-06             #Cross-sectional area of the ring(in square-metres)\n",
+      "I=4                   #Current through the coil(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "H=(N*I)/l\n",
+      "rel_per=1\n",
+      "per=(4*pi)*(1e-07)\n",
+      "B=per*rel_per*H\n",
+      "flux=B*A\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The magnetic field strength is %.2f A/m.\" %(H)\n",
+      "print \"(b)The flux density is %.2f micro T.\" %(B*1000000)\n",
+      "print \"(c)The total flux is %.5f micro Wb.\" %(flux*1000000)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The magnetic field strength is 1333.33 A/m.\n",
+        "(b)The flux density is 1675.52 micro T.\n",
+        "(c)The total flux is 0.83776 micro Wb.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2,Page number: 168\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnetomotive force(mmf) required to produce flux.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "flux=0.015            #Flux across the air-gap(in Webers)\n",
+      "l=2.5e-03             #Length of the air-fap(in metres)\n",
+      "A=200e-04             #Effective area of the air-gap(in square-metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "B=flux/A\n",
+      "abs_per=(4*pi)*(1e-07)\n",
+      "H=B/abs_per\n",
+      "mmf=H*l\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The magneto motive force(mmf) required is %.2f At.\" %(mmf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magneto motive force(mmf) required is 1492.08 At.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3,Page number: 168\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the reluctance of a mild-steel ring.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=200                       #Number of turns in the coil\n",
+      "abs_per=(4*pi)*(1e-07)      #Absolute permeability of free space   \n",
+      "rel_per=380                 #Relative permeability of mild-steel\n",
+      "l=400e-03                   #Length(Circumference) of the mild-steel ring(in metres)\n",
+      "A=500e-06                   #Cross-sectional area of the mild-steel ring(in square-metres)\n",
+      "flux=800e-06                #Flux in the ring(in Webers) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Rel=l/(abs_per*rel_per*A)\n",
+      "mmf=flux*Rel\n",
+      "I=mmf/N\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The reluctance of the ring is %6e A/Wb.\" %(Rel)\n",
+      "print \"(b)The magnetising current is %.2f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The reluctance of the ring is 1.675315e+06 A/Wb.\n",
+        "(b)The magnetising current is 6.70 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4,Page number: 171\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in the coil to produce a flux density of 0.9 T in the air gap.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "B=0.90                #Flux Density in the air gap(in Tesla)     \n",
+      "N=4000                #Number of turns in the core\n",
+      "l_core=220e-03        #Length of the core(in metres)\n",
+      "A_core=50e-06         #Cross-sectional area of the core(in square-metres)\n",
+      "H_core=820            #Magnetic field intensity of the core(in Ampere per metre)  \n",
+      "l_gap=1e-03           #Length of the air-gap(in metres)\n",
+      "A_gap=50e-06          #Cross-sectional area of the air gap(in square-metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "mmf_core=H_core*l_core\n",
+      "abs_per=(4*pi*(1e-07))\n",
+      "H_gap=B/abs_per\n",
+      "mmf_air_gap=H_gap*l_gap\n",
+      "Total_mmf=mmf_core+mmf_air_gap\n",
+      "I=Total_mmf/N\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The magnetisation current is %.5f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnetisation current is 0.22415 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5,Page number: 173"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnetising current required to produce a flux across the air gap.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "w=40e-03               #Width of the core(in metres)\n",
+      "d=50e-03               #Depth of the core(in metres)\n",
+      "lg=2.0e-03             #Length of the air gap(in metres)\n",
+      "Ag=2500e-06            #Area of the air gap(in square metres)\n",
+      "N=800                  #Number of turns in the coil\n",
+      "flux=2.50e-03          #Flux across the air gap(in Webers)\n",
+      "lf=1.2                 #Leakage Factor\n",
+      "abs_per=(4*pi)*(1e-07) #Absolute permeability of free space(Henry per metre) \n",
+      "lc=600e-03             #Length of the core(in metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Ac=w*d\n",
+      "Bg=flux/Ag\n",
+      "Hg=Bg/abs_per\n",
+      "mmf_g=Hg*lg\n",
+      "eff_Ac=0.92*Ac\n",
+      "flux_c=flux*lf\n",
+      "Bc=flux_c/eff_Ac\n",
+      "Hc=4000.0\n",
+      "mmf_c=Hc*lc\n",
+      "mmf=mmf_c+mmf_g\n",
+      "I=mmf/N\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The magnetising current required to produce a flux of 0.0025 Wb across the air gap is %.2f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The magnetising current required to produce a flux of 0.0025 Wb across the air gap is 4.99 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6,Page number: 174"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in the coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "lA=0.3                #Length of silicon steel material(in metres)\n",
+      "lB=0.2                #Length of low-carbon mild steel material(in metres)\n",
+      "lC=0.1                #Length of cast iron material(in metres)\n",
+      "N=100                 #Number of turns in the exciting coil\n",
+      "A=0.001               #Cross-sectional area(in square-metres)\n",
+      "flux=600e-06          #Flux in the coil(in Webers)\n",
+      "abs_per=(4*pi)*(1e-07)#Absolute permeability of free space(in Henry per metre)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "B=flux/A\n",
+      "B_A=B\n",
+      "B_C=B\n",
+      "B_C=B\n",
+      "\"\"\"From magnetisation characteristics of ferromagnetic materials,\"\"\"\n",
+      "H_A=20.0\n",
+      "H_B=700.0\n",
+      "H_C=2500.0\n",
+      "\"\"\"According to Kirchoff's magnetomotive force law(KML),the total mmf required, tot_mmf= mmf_A+mmf_B+mmf_C.\"\"\"\n",
+      "tot_mmf=(H_A*lA)+(H_B*lB)+(H_C*lC)\n",
+      "I=tot_mmf/N\n",
+      "rel_A=B/(abs_per*H_A)\n",
+      "rel_B=B/(abs_per*H_B)\n",
+      "rel_C=B/(abs_per*H_C)\n",
+      "R_A=(H_A*lA)/flux\n",
+      "R_B=(H_B*lB)/flux\n",
+      "R_C=(H_C*lC)/flux\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The mmf for setting up a flux of 600 micro Weber is %.2f At.\" %(tot_mmf)\n",
+      "print \"(b)The current in the coil is %.2f A.\" %(I)\n",
+      "print \"(c)The relative permeability of:\\n   Material A: %.2f\\n   Material B: %.2f\\n   Material C: %.2f.\\n\" %(rel_A,rel_B,rel_C)\n",
+      "print \"   The reluctances are :\\n   R_A=%.2f At/Wb \\n   R_B=%.2f At/Wb \\n   R_C=%.2f At/Wb.\" %(R_A,R_B,R_C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The mmf for setting up a flux of 600 micro Weber is 396.00 At.\n",
+        "(b)The current in the coil is 3.96 A.\n",
+        "(c)The relative permeability of:\n",
+        "   Material A: 23873.24\n",
+        "   Material B: 682.09\n",
+        "   Material C: 190.99.\n",
+        "\n",
+        "   The reluctances are :\n",
+        "   R_A=10000.00 At/Wb \n",
+        "   R_B=233333.33 At/Wb \n",
+        "   R_C=416666.67 At/Wb.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7,Page number: 175"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in the exciting coil.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=300                 #Number of the turns in the exciting coil\n",
+      "flux=600e-06          #Flux in the air gap(in Webers)\n",
+      "lg=1e-03              #Length of the air gap(in metres)\n",
+      "lc=40e-02             #Mean length of the core(in metres)\n",
+      "A=4e-04               #Cross sectional area of the core(in square metres)\n",
+      "abs_per=(4*pi)*1e-07  #Absolute permeability of free space(in Henry per metre)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "B=flux/A\n",
+      "Hg=B/abs_per\n",
+      "mmf_g=Hg*lg\n",
+      "mmf_c=3000*lc\n",
+      "mmf=mmf_g+mmf_c\n",
+      "I=mmf/N\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current in the exciting coil to set up a flux of 600 micro Weber in the air gap is %.2f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current in the exciting coil to set up a flux of 600 micro Weber in the air gap is 7.98 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8,Page number: 175 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in the exciting coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "l1=10e-02             #Length of first side of the magnetic circuit(in metres)\n",
+      "l2=18e-02             #Length of second side of the magnetic circuit(in metres)\n",
+      "l3=18e-02             #Length of third side of the magnetic circuit(in metres)\n",
+      "A1=6.25e-04           #Cross sectional area of l1 path(in square-metre) \n",
+      "A2=3.00e-04           #Cross sectional area of l2 path(in square-metre)\n",
+      "A3=3.00e-04           #Cross sectional area of l3 path(in square-metre)\n",
+      "lg=2e-03              #Length of air gap(in metres)\n",
+      "rel_per=800.0         #Relative permeability of core material\n",
+      "N=600                 #Number of turns\n",
+      "flux=100e-06          #Flux in the air gap(in Webers)\n",
+      "abs_per=(4*pi)*1e-07  #Absolute permeability of free space(in Farad per metre)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Bg=flux/A1\n",
+      "Hg=Bg/abs_per\n",
+      "mmf_g=Hg*lg\n",
+      "B1=Bg\n",
+      "H1=B1/(rel_per*abs_per)\n",
+      "mmf_1=H1*l1\n",
+      "flux2=flux/2.0\n",
+      "B2=flux2/A2\n",
+      "H2=B2/(rel_per*abs_per)\n",
+      "mmf_2=H2*l2\n",
+      "tot_mmf=mmf_g+mmf_1+mmf_2\n",
+      "I=tot_mmf/N\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current in the 600 turn exciting coil is %.3f A.\" %(I) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current in the 600 turn exciting coil is 0.501 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9,Page number: 176 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of exciting current.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "A_B=0.01              #Area of cross section of limb B(in square metre)\n",
+      "A_C=0.02              #Area of cross section of limb C(in square metre)\n",
+      "l_B=1e-03             #Length of air gap in limb B(in metres)\n",
+      "l_C=2e-03             #Length of air gap in limb C(in metres)\n",
+      "flux_B=1.5e-03        #Flux in limb B(in Webers)\n",
+      "N=500                 #Number of turns in the coil\n",
+      "abs_per=(4*pi)*(1e-07)#Absolute permeability of free space(in Henry per metre) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"The mmf across parallel paths must be same. \n",
+      "   (R_B*flux_B)=(R_C*flux_C);\"\"\"\n",
+      "flux_C=flux_B*(l_B/l_C)*(A_C/A_B)\n",
+      "flux=flux_B+flux_C\n",
+      "R_B=l_B/(abs_per*A_B)\n",
+      "R_C=l_C/(abs_per*A_C)\n",
+      "R_net=1.0/((1.0/R_B)+(1.0/R_C))\n",
+      "I=(flux*R_net)/N\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The flux in limb A is %e Wb.\" %(flux)\n",
+      "print \"The current in the exciting coil is %e A.\" %(I) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The flux in limb A is 3.000000e-03 Wb.\n",
+        "The current in the exciting coil is 2.387324e-01 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10,Page number: 177 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the ampere turns required to produce a flux.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "abs_per=(4*pi)*(1e-07) #Absolute permeability of free space(in Henry per metre)\n",
+      "D=21e-02               #Mean diameter of the ring(in metres)\n",
+      "A=10e-04               #Cross sectional area(in square metre)\n",
+      "flux=0.8e-03           #Flux to be produced(in Webers)\n",
+      "lg=0.4e-03             #Length of the air gap(in metres) \n",
+      "rel_i=166.0            #Relative permeability of iron\n",
+      "rel_s=800.0            #Relative permeability of steel\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "B=flux/A\n",
+      "li=(pi*D)/2.0\n",
+      "ls=(pi*D)/2.0\n",
+      "mmf_g=(B/abs_per)*lg\n",
+      "mmf_i=(B/(abs_per*rel_i))*li\n",
+      "mmf_s=(B/(abs_per*rel_s))*ls\n",
+      "mmf=mmf_g+mmf_i+mmf_s\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The total ampere turns required is %.2f At.\" %(mmf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The total ampere turns required is 1782.21 At.\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11,Page number: 177 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the ampere turns of the coil wound on the central limb.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "flux=1e-03            #Flux in the central limb(in Webers)\n",
+      "Ac=8e-04              #Area of the central limb(in square metres)\n",
+      "As=5e-04              #Area of each side limb(in square metres)\n",
+      "l=0.15                #Length of the central limb(in metres)\n",
+      "lg=0.001              #Length of the air gap(in metres)\n",
+      "abs_per=(4*pi)*1e-07  #Absolute permeability of free space(in Henry per metre)\n",
+      "l1=0.34               #Length of the part ABCD(in metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "B=flux/Ac\n",
+      "\"\"\"Corresponding to this value of B, H is 500 At/m from the table.\"\"\"\n",
+      "mmf_DG=500*l\n",
+      "Hg=1.25/abs_per\n",
+      "mmf_g=Hg*lg\n",
+      "B1=(flux/2.0)/As\n",
+      "\"\"\"Corresponding to this value of B, H is 200 At/m from the table.\"\"\"\n",
+      "mmf_1=200*l1\n",
+      "mmf_tot=mmf_DG+mmf_g+mmf_1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The required ampere turns of the coil wound on the central limb is %.2f At.\" %(mmf_tot)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required ampere turns of the coil wound on the central limb is 1137.72 At.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter7.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter7.ipynb
new file mode 100755
index 00000000..acf4ea53
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter7.ipynb
@@ -0,0 +1,975 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: SELF AND MUTUAL INDUCTANCES"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1,Page number: 184\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the emf induced in a coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=4                    #Self inductance of the coil(in Henry) \n",
+      "di=4-10                #Change in current(in Amperes)\n",
+      "dt=0.1                 #Time interval(in seconds)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "e=-L*(di/dt)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The emf induced in the coil is %.2f V.\" %(e)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The emf induced in the coil is 240.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2,Page number: 184"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the inductance of a coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=150                  #Number of turns in the coil \n",
+      "flux=0.01              #Flux linked with the coil(in Webers)\n",
+      "I=10                   #Current in the coil(in Amperes)    \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L=(N*flux)/I\n",
+      "di=-10-(10)\n",
+      "dt=0.01\n",
+      "e=-L*(di/dt)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The inductance of the coil is %.2f H.\" %(L)\n",
+      "print \"The induced emf is %.2f V.\" %(e)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The inductance of the coil is 0.15 H.\n",
+        "The induced emf is 300.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3,Page number: 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the inductance of the coil and the emf induced.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=100                  #Number of turns in the coil\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "dflux=-0.4-0.4\n",
+      "di=-10-10\n",
+      "L=N*(dflux/di)\n",
+      "dt=0.01\n",
+      "e=-(L*(di/dt))/1000\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The inductance of the coil is %.2f mH.\" %(L)\n",
+      "print \"The induced emf is %.2f V.\" %(e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The inductance of the coil is 4.00 mH.\n",
+        "The induced emf is 8.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4,Page number: 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the energy stored in an air-cored solenoid.\"\"\"\n",
+      "\n",
+      "from math import pi,pow\n",
+      "\n",
+      "\"\"\" All quantities expresssed in SI System.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "l=0.30                 #Length of the solenoid(in metres) \n",
+      "d=0.015                #Internal diameter of the solenoid(in metres)    \n",
+      "r=0.0075               #Internal radius of the solenoid(in metres)\n",
+      "N=900                  #Number of turns in the coil \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "A=pi*pow(r,2)\n",
+      "L=(pow(N,2)*4*pi*A)/(0.30*10000000)\n",
+      "I=5\n",
+      "W=0.5*L*pow(I,2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The inductance of the air-cored solenoid is %.2f mH.\" %(L*1000)\n",
+      "print \"The amount of energy stored in the air-cored solenoid is %.2f mJ.\" %(W*1000)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The inductance of the air-cored solenoid is 0.60 mH.\n",
+        "The amount of energy stored in the air-cored solenoid is 7.49 mJ.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5,Page number: 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the relative permeability of iron and the inductance of a coil.\"\"\"\n",
+      "\n",
+      "from math import pow,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "r=0.01                 #Radius of circular ring(in metres)\n",
+      "A=pi*pow(r,2)          #Area of circular ring(in square metres) \n",
+      "N=3000                 #Number of turns in the coil \n",
+      "I=0.5                  #Current in the coil(in Amperes)\n",
+      "l=20.0/100             #Length of the iron rod(in metres) \n",
+      "B=1.2                  #Magnitude of magnetic field(in Tesla)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "H=(N*I)/l\n",
+      "per=B/H\n",
+      "rel_per=(per*10000000)/(4*pi)\n",
+      "L=(N*B*A)/I\n",
+      "dflux=(0.1-1)*A*B\n",
+      "dt=0.01\n",
+      "e=-N*(dflux/dt)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The permeability of iron is %e Tm/A.\" %(per)\n",
+      "print \"(b)The relative permeability of iron is %d.\" %(rel_per)\n",
+      "print \"(c)The inductance of the coil is %.2f H.\" %(L)\n",
+      "print \"(d)The voltage in the coil is %.2f V.\" %(e)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The permeability of iron is 1.600000e-04 Tm/A.\n",
+        "(b)The relative permeability of iron is 127.\n",
+        "(c)The inductance of the coil is 2.26 H.\n",
+        "(d)The voltage in the coil is 101.79 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6,Page number: 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the potential difference across the terminals of a coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=3                    #Resistance of the coil(in Ohms)\n",
+      "i=1                    #Current in the coil(in Amperes)\n",
+      "di=10000               #Change in current(in Amperes) \n",
+      "dt=1                   #Time interval(in seconds)\n",
+      "L=0.1/1000             #Self inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V=(i*R)+(L*(di/dt))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The potential difference that exists across the terminals of the coil is %.2f V.\" %(V)     "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The potential difference that exists across the terminals of the coil is 4.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7,Page number: 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mutual inductance and emf induced in a search coil.\"\"\"\n",
+      "\n",
+      "from math import pi,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N1=2000                #Number of turns in the solenoid   \n",
+      "N2=500                 #Number of turns in the search coil\n",
+      "l=0.70                 #Length of the solenoid(in metres) \n",
+      "k=1                    #Coefficient of coupling\n",
+      "A=30.0/10000           #Mean area of the search coil(in square metres)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "per=(4*pi)/10000000.0\n",
+      "M=(k*N1*N2*per*A)/l\n",
+      "di1=260.0\n",
+      "dt=1\n",
+      "e=M*(di1/dt)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The mutual inductance is %.4f mH.\" %(M*1000)\n",
+      "print \"(b)The emf induced in the search coil is %.2f V.\" %(e)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The mutual inductance is 5.3856 mH.\n",
+        "(b)The emf induced in the search coil is 1.40 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8,Page number: 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mutual inductance and the coefficient of coupling between two coils.\"\"\"\n",
+      "\n",
+      "from math import pow,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N1=600.0               #Number of turns in the first coil\n",
+      "N2=1700.0              #Number of turns in the second coil \n",
+      "flux2=0.8/1000         #Magnetic flux produced in the second coil(in Webers) \n",
+      "I2=6                   #Current in the second coil(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L2=(N2*flux2)/I2\n",
+      "L1=L2*pow((N1/N2),2)\n",
+      "flux21=0.5/1000\n",
+      "k=flux21/flux2\n",
+      "M=k*sqrt(L1*L2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"L1=%.4f H.\" %(L1)\n",
+      "print \"L2=%.4f H.\" %(L2)\n",
+      "print \"The coefficient of coupling(k)=%.4f.\" %(k) \n",
+      "print \"M=%.4f H.\" %(M)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L1=0.0282 H.\n",
+        "L2=0.2267 H.\n",
+        "The coefficient of coupling(k)=0.6250.\n",
+        "M=0.0500 H.\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9,Page number: 189 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mutual inductance and the coefficient of coupling between two coils.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N1=1200.0              #Number of turns in the first coil       \n",
+      "flux1=0.25/1000        #Magnetic flux produced in the first coil(in Webers) \n",
+      "I1=5                   #Current in the first coil(in Amperes)\n",
+      "N2=800.0               #Number of turns in the second coil\n",
+      "flux2=0.15/1000        #Magnetic flux produced in the second coil(in Webers)\n",
+      "I2=5                   #Current in the second coil(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L1=(N1*flux1)/I1\n",
+      "L2=(N2*flux2)/I2\n",
+      "k=0.6\n",
+      "flux12=k*flux1\n",
+      "M=(N2*flux12)/I1\n",
+      "k_new=M/sqrt(L1*L2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The mutual inductance(M) is %.4f H.\" %(M)\n",
+      "print \"The coefficient of coupling is %.4f.\" %(k_new)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mutual inductance(M) is 0.0240 H.\n",
+        "The coefficient of coupling is 0.6325.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10,Page number: 192 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mutual inductance and the coefficient of coupling between two coils.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Lsa=1.4/1000           #Net inductance in series-aiding connections(in Henry) \n",
+      "Lso=0.6/1000           #Net inductance in series-opposing connections(in Henry) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "M=(Lsa-Lso)/4\n",
+      "\"\"\"Lsa=L1+L2+2M \n",
+      "    L1+L2=1 mH; As the two coils are similar L1=L2=0.5mH \"\"\"\n",
+      "L1=0.5/1000\n",
+      "L2=0.5/1000\n",
+      "k=M/sqrt(L1*L2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The mutual inductance is %.2f mH.\" %(M*1000)\n",
+      "print \"The coefficient of coupling(k) is %.2f.\" %(k)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mutual inductance is 0.20 mH.\n",
+        "The coefficient of coupling(k) is 0.40.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11,Page number: 193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mutual inductance and the self-inductances of two coils. \"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "\"\"\" Equation 1 is L1+L2+(2*M)=1.8;\n",
+      "    \n",
+      "    Equation 2 is L1+L2-(2*M)=0.8. \"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "k=0.6                 #Coefficient of coupling\n",
+      "eq1=1.8               #Net inductance when fluxes are in same direction(in Henry)\n",
+      "eq2=0.8               #Net inductance when fluxes are in opposite direction(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "M=(eq1-eq2)/4\n",
+      "sum=(eq1+eq2)/2\n",
+      "product=(M*M)/(k*k)\n",
+      "diff=sqrt((sum*sum)-(4*product))\n",
+      "L1=(sum+diff)/2\n",
+      "L2=(sum-diff)/2\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The mutual inductance of the two coils is %.3f H.\" %(M)\n",
+      "print \"The self inducatnce of the first coil is %.3f H and the self inductance of the second coil is %.3f H.\" %(L1,L2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mutual inductance of the two coils is 0.250 H.\n",
+        "The self inducatnce of the first coil is 1.149 H and the self inductance of the second coil is 0.151 H.\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.12,Page number:195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the equivalent inductance of a combination of inductances connected in parallel.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "k=0.433                #Coefficient of coupling   \n",
+      "L1=8                   #Self-inductance of the first coil    \n",
+      "L2=6                   #Self-inductance of the second coil\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "M=k*sqrt(L1*L2)\n",
+      "Lpa=((L1*L2)-(M*M))/(L1+L2-(2*M))\n",
+      "Lpo=((L1*L2)-(M*M))/(L1+L2+(2*M))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The equivalent inductance such that the mutual induction assists the self induction is %.3f H.\" %(Lpa)\n",
+      "print \"(b)The equivalent inductance such that the mutual induction opposes the self induction is %.3f H.\" %(Lpo)  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The equivalent inductance such that the mutual induction assists the self induction is 4.875 H.\n",
+        "(b)The equivalent inductance such that the mutual induction opposes the self induction is 1.950 H.\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13,Page number: 196 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the number of turns in an air-cored coil.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "l=2.5e-02             #Length of the coil(in metres)\n",
+      "A=2e-04               #Average cross-sectional area of the coil(in square-metres)\n",
+      "L=400e-06             #Self-inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "abs_per=(4*pi)/(1e07)\n",
+      "N=sqrt((L*l)/(abs_per*A))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The number of turns in the air-cored coil is %d.\" %(round(N,0)) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The number of turns in the air-cored coil is 199.\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14,Page number: 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mutual inductance between two coils and their self inductances.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "k=0.75                #Coefficient of coupling between two coils\n",
+      "I1=3.0                #Current in the first coil(in Amperes)\n",
+      "N1=250.0              #Number of turns in the first coil\n",
+      "flux1=4e-03           #Flux produced in the first coil(in Webers)\n",
+      "V2=70.0               #Voltage induced in the second coil due to first coil(in Volts)\n",
+      "di1=3.0               #Change in current in the first coil(in Amperes)\n",
+      "dt=3e-03              #Time interval(in seconds)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L1=N1*(flux1/I1)\n",
+      "M=(V2*dt)/di1\n",
+      "L2=(M*M)/(k*k*L1)\n",
+      "N2=N1*sqrt(L2/L1)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"L1=%.4f H.\" %(L1)\n",
+      "print \"L2=%.4f H.\" %(L2)\n",
+      "print \"M=%.4f H.\" %(M)\n",
+      "print \"N2=%d.\" %(round(N2,0))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "L1=0.3333 H.\n",
+        "L2=0.0261 H.\n",
+        "M=0.0700 H.\n",
+        "N2=70.\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15,Page number: 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mean value of self inductance of a coil.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N=1000.0              #Number of turns in the coil\n",
+      "A=20e-04              #Cross-sectional area of the coil(in square-metre)\n",
+      "I1=4.0                #First current(in Amperes)\n",
+      "B1=1.0                #Flux density associated with the first current(in Weber per sqyare-metre) \n",
+      "I2=9.0                #Second current(in Amperes)\n",
+      "B2=1.4                #Flux density associated with the first current(in Weber per sqyare-metre)\n",
+      "dt=0.05               #Time interval(in seconds)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L1=(N*B1*A)/I1\n",
+      "L2=(N*B2*A)/I2\n",
+      "L=(L1+L2)/2.0\n",
+      "di=I2-I1\n",
+      "e=L*(di/dt)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The mean value of inductance between the given current limits is %.4f H.\" %(L)\n",
+      "print \"The emf induced in the coil is %.2f V.\" %(e)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean value of inductance between the given current limits is 0.4056 H.\n",
+        "The emf induced in the coil is 40.56 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.16,Page number: 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the mutual inductance between two coils and their respective self-inductances.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "N1=100.0              #Number of turns in the first coil\n",
+      "N2=150.0              #Number of turns in the second coil\n",
+      "A=125e-04             #Area of cross-section(in square-metres)\n",
+      "l=200e-02             #Mean length(in metres)\n",
+      "rel_per=2000.0        #Relative permeability of iron\n",
+      "k=1                   #Coefficient of coupling\n",
+      "\n",
+      "\"\"\" NOTE: As the two coils are wound side by side,there is tight coupling. Therefore, k=1. \"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "abs_per=(4*pi)/(1e07)\n",
+      "L1=(N1*N1*rel_per*abs_per*A)/l\n",
+      "L2=(N2*N2*rel_per*abs_per*A)/l\n",
+      "M=k*sqrt(L1*L2)\n",
+      "di1=5.0\n",
+      "dt=0.02\n",
+      "e2=M*(di1/dt)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The self inductances of the tow coils are: L1=%.3f mH and L2=%.3f mH.\" %((L1*1000.0),(L2*1000.0))\n",
+      "print \"(b)The mutual inductance between the two coils is %.3f mH.\" %(M*1000.0)\n",
+      "print \"(c)The emf induced in the second coil is %.2f V.\" %(e2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The self inductances of the tow coils are: L1=157.080 mH and L2=353.429 mH.\n",
+        "(b)The mutual inductance between the two coils is 235.619 mH.\n",
+        "(c)The emf induced in the second coil is 58.90 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17,Page number: 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the coeffcient of coupling and the self-inductance of two coils.\"\"\" \n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Lsa=4.0               #Equivalent inductance of series aiding(in Henry)\n",
+      "Lso=0.8               #Equivalent inductance of series opposing(in Henry)\n",
+      "\n",
+      "\"\"\" NOTE: Lsa=L+L+(2*M);\n",
+      "          Lso=L+L-(2*M); \"\"\"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L=(Lsa+Lso)/4.0\n",
+      "M=(Lsa-Lso)/4.0\n",
+      "k=M/sqrt(L*L)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The self inductance of each coil is %.2f H.\" %(L)\n",
+      "print \"The coefficient of coupling is %.3f.\" %(round(k,3))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The self inductance of each coil is 1.20 H.\n",
+        "The coefficient of coupling is 0.667.\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.18,Page number: 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the equivalent inductance of different combinations of two coils.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L1=200e-03            #Self-inductance of the first coil(in Henry)  \n",
+      "L2=800e-03            #Self-inductance of the second coil(in Henry)\n",
+      "k=0.5                 #Coefficient of coupling between two coils\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "M=k*sqrt(L1*L2)\n",
+      "Lsa=L1+L2+(2*M)\n",
+      "Lso=L1+L2-(2*M)\n",
+      "Lpa=((L1*L2)-(M*M))/Lso\n",
+      "Lpo=((L1*L2)-(M*M))/Lsa\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The equivalent inductance of series aiding is %.3f mH.\" %(Lsa*1000.0)\n",
+      "print \"(b)The equivalent inductance of series opposing is %.3f mH.\" %(Lso*1000.0)\n",
+      "print \"(c)The equivalent inductance of parallel aiding is %.3f mH.\" %(Lpa*1000.0)\n",
+      "print \"(d)The equivalent inductance of parallel opposing is %.3f mH.\" %(Lpo*1000.0)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The equivalent inductance of series aiding is 1400.000 mH.\n",
+        "(b)The equivalent inductance of series opposing is 600.000 mH.\n",
+        "(c)The equivalent inductance of parallel aiding is 200.000 mH.\n",
+        "(d)The equivalent inductance of parallel opposing is 85.714 mH.\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.19,Page number: 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the exciting current for a horse-shoe magnet.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "l=45e-02              #Length of the iron path(in metres)\n",
+      "A=6e-04               #Cross-sectional area of the wrought iron bar(in square-metres)\n",
+      "N=500.0               #Number of turns in exciting coil\n",
+      "load=60.0             #Load to be lifted(in kilograms)\n",
+      "rel_per=800.0         #Relative permeability of iron\n",
+      "g=9.8                 #Accelaration due to gravity(in metre per square-seconds) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "abs_per=(4*pi)/(1e07)\n",
+      "F=(load/2.0)*g\n",
+      "B=sqrt((2*abs_per*F)/A)\n",
+      "H=B/(abs_per*rel_per)\n",
+      "At=H*l\n",
+      "I=At/(N*2)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The exciting current needed for the magnet is %.5f A.\" %(I)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The exciting current needed for the magnet is 0.49674 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter8.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter8.ipynb
new file mode 100755
index 00000000..b22550d5
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter8.ipynb
@@ -0,0 +1,1188 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8: DC TRANSIENTS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1,Page number: 207 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current i(0+)=I0.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=24                 #Supply voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Req=20+(1.0/((1.0/20)+(1.0/10)))\n",
+      "I=V/Req\n",
+      "I_L=I*(20.0/(20+10))\n",
+      "Io=I_L\n",
+      "i_0_plus=Io\n",
+      "R=20\n",
+      "v_R=(-Io)*R\n",
+      "R1=(20+10)\n",
+      "v_L=Io*R1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current i(0+)=Io=%.2f A.\" %(i_0_plus)\n",
+      "print \"(b)The magnitude of v_R across the 20 Ohms resistor at the instant just after the switch is opened is %.2f V.\" %(v_R)\n",
+      "print \"(c)The magnitude of v_L across the inductor immediately after the switch is opened is %.2f V.\" %(v_L)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current i(0+)=Io=0.60 A.\n",
+        "(b)The magnitude of v_R across the 20 Ohms resistor at the instant just after the switch is opened is -12.00 V.\n",
+        "(c)The magnitude of v_L across the inductor immediately after the switch is opened is 18.00 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2,Page number: 208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding i(0),the power being absorbed by the inductor at t=1 s.\"\"\"\n",
+      "\n",
+      "from math import e,pow,sqrt,log\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=1.6                #Self-inductance of the inductor(in Henry)\n",
+      "R=0.8                #Resistance of the resistor(in Ohms)\n",
+      "t=-1.0               #Instant of time(in seconds) \n",
+      "i_minus_1=20.0       #Current at t=1 seconds(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "time_const=L/R\n",
+      "Io=i_minus_1/pow(e,(-t/time_const))\n",
+      "t=1\n",
+      "i_t=Io*pow(e,(-t/time_const))\n",
+      "p_t=i_t*i_t*R\n",
+      "W=100.0\n",
+      "i_t1=sqrt((2*W)/L)\n",
+      "t=-(log(i_t1/Io))*time_const\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)i(0) = %.3f A.\" %(Io)\n",
+      "print \"(b)The power absorbed by the resistor is %.2f W. \" %(p_t)    \n",
+      "print \"   The inductor by virtue of the emf induced in it,supplies this power to the resistor.\"\n",
+      "print \"   Therefore,the power absorbed by the inductor is %.2f W.\" %(-p_t)\n",
+      "print \"(c)The time at which the energy stored in the inductor is 100J is %.5f seconds.\" %(t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)i(0) = 12.131 A.\n",
+        "(b)The power absorbed by the resistor is 43.31 W. \n",
+        "   The inductor by virtue of the emf induced in it,supplies this power to the resistor.\n",
+        "   Therefore,the power absorbed by the inductor is -43.31 W.\n",
+        "(c)The time at which the energy stored in the inductor is 100J is 0.16315 seconds.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3,Page number: 210"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of current in the circuit at an instant 0.4 s after the switch has been closed.\"\"\"\n",
+      "\n",
+      "from math import e,pow,log\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=14.0                #Self-inductance of the inductor(in Henry)\n",
+      "R=10.0                #Resistance of the resistor(in Ohms)\n",
+      "V=140.0               #Supply Voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "time_const=L/R\n",
+      "t=0.4\n",
+      "Io=V/R\n",
+      "i=Io*(1-pow(e,(-t/time_const)))\n",
+      "i_t=8\n",
+      "t=-time_const*log(i_t/Io)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The value of current in the circuit at an instant 0.4 s after the switch has been closed is %.3f A.\" %(i)\n",
+      "print \"(b)The time taken for the current to drop to 8 A after the switch is opened is %.4f seconds.\" %(t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The value of current in the circuit at an instant 0.4 s after the switch has been closed is 3.479 A.\n",
+        "(b)The time taken for the current to drop to 8 A after the switch is opened is 0.7835 seconds.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4,Page number: 210"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the magnitude of the inductor current at t=0- and at t=0+.\"\"\"\n",
+      "\n",
+      "from math import pow,e\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=40e-03              #Self-inductance of the inductor(in Henry)\n",
+      "R=80.0                #Resistance of the resistor(in Ohm) \n",
+      "V1=20.0               #Supply voltage-1(in Volts) \n",
+      "V2=40.0               #Supply voltage-2(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "time_const=L/R\n",
+      "I_01=V1/R\n",
+      "i_0_minus=I_01\n",
+      "i_0_plus=I_01\n",
+      "I_02=(V1+V2)/R\n",
+      "t=1e-03\n",
+      "i_t=I_01+(I_02-I_01)*(1-pow(e,(-t/time_const)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The magnitude of inductor current at t=0- is %.2f A.\" %(i_0_minus)\n",
+      "print \"(b)The magnitude of inductor current at t=0+ is %.2f A.\" %(i_0_plus)\n",
+      "print \"(c)As  t tends to infinity,the current approaches its final steady-state value given as %.2f A.\" %(I_02)\n",
+      "print \"(d)The magnitude of inductor current at t=1 ms is %.3f A.\" %(i_t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The magnitude of inductor current at t=0- is 0.25 A.\n",
+        "(b)The magnitude of inductor current at t=0+ is 0.25 A.\n",
+        "(c)As  t tends to infinity,the current approaches its final steady-state value given as 0.75 A.\n",
+        "(d)The magnitude of inductor current at t=1 ms is 0.682 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5,Page number: 211"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding i_L(0-),i2(0-),i_L(0+),i_L(20 ms) and i_2(20 ms).\"\"\"\n",
+      "\n",
+      "from math import e,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=20.0                #Source Voltage(in Volts)\n",
+      "R1=20.0               #Resistance of resistor-1(in Ohms)\n",
+      "R2=40.0               #Resistance of resistor-2(in Ohms)\n",
+      "R3=30.0               #Resistance of resistor-3(in Ohms)\n",
+      "R4=25.0               #Resistance of resistor-4(in Ohms)\n",
+      "R5=5.0                #Resistance of resistor-5(in Ohms)\n",
+      "L=2.0                 #Self-inductance of inductor(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Io=V/(R4+R5)\n",
+      "i_L_0_minus=Io\n",
+      "i_2_0_minus=V/R3\n",
+      "i_L_0_plus=Io\n",
+      "R_45=R4+R5\n",
+      "R_12=R1+R2\n",
+      "Req=R_45+(1/((1/R_12)+(1/R3)))\n",
+      "time_const=L/Req\n",
+      "t=20e-03\n",
+      "i_L_t=Io*pow(e,(-t/time_const))\n",
+      "i_2_t=-i_L_t*(R_12/(R_12+R3))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)i_L(0-) = %.3f A.\" %(i_L_0_minus)\n",
+      "print \"(b)i_2(0-) = %.3f A.\" %(i_2_0_minus)\n",
+      "print \"(c)i_L(0+) = %.3f A.\" %(i_L_0_plus)\n",
+      "print \"(d)i_L(20 ms) = %.3f A.\" %(i_L_t)\n",
+      "print \"(e)i_2(20 ms) = %.3f A.\" %(i_2_t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)i_L(0-) = 0.667 A.\n",
+        "(b)i_2(0-) = 0.667 A.\n",
+        "(c)i_L(0+) = 0.667 A.\n",
+        "(d)i_L(20 ms) = 0.404 A.\n",
+        "(e)i_2(20 ms) = -0.270 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6,Page number: 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding v(0+),i(0+) and time constant.\"\"\"\n",
+      "\n",
+      "from  math import e,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=1.5e03              #Resistance of the resistor(in Ohms) \n",
+      "C=5e-06               #Capacitance of the capacitor(in Farad)\n",
+      "Vo=3.0                #Source Voltage(in Volts)\n",
+      "v_0_plus=0            #Voltage across capacitor at t=0+(in Volts)  \n",
+      "v_0_minus=0           #Voltage across capacitor at t=0-(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Io=Vo/R\n",
+      "i_0_plus=Io\n",
+      "time_const=R*C\n",
+      "t=15e-03\n",
+      "v=Vo*(1-pow(e,(-t/time_const)))  \n",
+      "i=Io*pow(e,(-t/time_const))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)Since the voltage across a capacitor cannot change instantaneously, we have, v(0+)=%d V.\" %(v_0_plus)\n",
+      "print \"(b)i(0+)= %.3f mA.\" %(i_0_plus*1000)\n",
+      "print \"(c)The time constant is %.2f ms.\" %(time_const*1000)\n",
+      "print \"(d)At t=15 ms, \\n   v=%.5f V.   \\n   i=%.4f mA.\" %(v,(i*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Since the voltage across a capacitor cannot change instantaneously, we have, v(0+)=0 V.\n",
+        "(b)i(0+)= 2.000 mA.\n",
+        "(c)The time constant is 7.50 ms.\n",
+        "(d)At t=15 ms, \n",
+        "   v=2.59399 V.   \n",
+        "   i=0.2707 mA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7,Page number: 216\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding v(0+),i(0+) and time constant.\"\"\"\n",
+      "\n",
+      "from  math import e,pow\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=100                 #Resistance of the resistor(in Ohms)\n",
+      "C=5e-06               #Capacitance of the capacitor(in Farads) \n",
+      "Vo=3.0                #Source Voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "v_0_plus=Vo\n",
+      "v_0_minus=Vo\n",
+      "Io=Vo/R\n",
+      "i_0_plus=-Io\n",
+      "time_const=R*C\n",
+      "t=1.2e-03\n",
+      "v=Vo*pow(e,(-t/time_const))  \n",
+      "i=-Io*pow(e,(-t/time_const))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)Since the voltage across a capacitor cannot change instantaneously, we have, v(0+)=%d V.\" %(v_0_plus)\n",
+      "print \"(b)At t=0+, the capacitor behaves as a voltage source of emf Vo. Hence, i(0+)= -Io = %.3f mA.\" %(i_0_plus*1000)\n",
+      "print \"(c) The time constant is %.2f ms.\" %(time_const*1000)\n",
+      "print \"(d)At t=1.2 ms, \\n   v=%.5f V.   \\n   i=%.4f mA.\" %(v,(i*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Since the voltage across a capacitor cannot change instantaneously, we have, v(0+)=3 V.\n",
+        "(b)At t=0+, the capacitor behaves as a voltage source of emf Vo. Hence, i(0+)= -Io = -30.000 mA.\n",
+        "(c) The time constant is 0.50 ms.\n",
+        "(d)At t=1.2 ms, \n",
+        "   v=0.27215 V.   \n",
+        "   i=-2.7215 mA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8,Page number: 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current i(t) for t>0 seconds.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\" R_Th=(1+10)kilo ohm || (1 kiloohm) \"\"\"   #Thevenin's Equivalent Resistance(in Ohms) \n",
+      "\n",
+      "print(\"Note: All currents expressed in mA \\n\")\n",
+      "R_Th=1.0/((1.0/11000)+(1.0/1000))\n",
+      "C=10e-06\n",
+      "time_const=R_Th*C\n",
+      "v_C_0_minus=30.0*((1e03)/((1e03)+(1e03)))\n",
+      "\n",
+      "\"\"\" Applying KVL, 30-(i_0_plus*(1 kilo ohm))-15=0 \"\"\"\n",
+      "\n",
+      "i_0_plus=(30-15)/1.0\n",
+      "i_infinity=30.0/(1+1+10)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"i(t)= (%.1f + (%.1f-%.1f)*(e to the power -t/%.2f ms)) mA.\" %(i_infinity,i_0_plus,i_infinity,(time_const*1000))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Note: All currents expressed in mA \n",
+        "\n",
+        "i(t)= (2.5 + (15.0-2.5)*(e to the power -t/9.17 ms)) mA.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9,Page number: 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current at t=1 seconds.\"\"\"\n",
+      "\n",
+      "from math import exp,log\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=3.0                 #Resistance of the coil(in Ohms)\n",
+      "time_const=1.8        #Time constant of the coil(in seconds)\n",
+      "V=10.0                #Supply voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "L=time_const*R\n",
+      "i_0_plus=0\n",
+      "I0=V/R\n",
+      "i=I0*(1-exp((-1/time_const)))\n",
+      "t=(log(0.5,e))*(-1.8)\n",
+      "growth=V/L\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current at t=1 seconds is given as %.2f A.\" %(i)\n",
+      "print \"(b)The time at which current attains half of its final value is %.2f seconds.\" %(t)\n",
+      "print \"(c)The initial rate of growth of current is %.2f A/s.\" %(growth)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current at t=1 seconds is given as 1.42 A.\n",
+        "(b)The time at which current attains half of its final value is 1.25 seconds.\n",
+        "(c)The initial rate of growth of current is 1.85 A/s.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10,Page number: 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the value of current during a sudden change.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"If I0 is the final steady-state value of current,at t=1 s, we have\n",
+      "     \n",
+      "              i=Io*(1-exp(-1/time_const)) or (0.741*Io)=Io*(1-exp(-1/time_const));\n",
+      "              \n",
+      "              exp(-1/time_const)=0.259;\n",
+      "              \n",
+      "    During the decay of current,we have i(t)=Io*exp(-1/time_const);          \n",
+      "    \n",
+      "    Therefore, at t=1s, we have\n",
+      "    \n",
+      "    i1=0.259*Io.\n",
+      "    \n",
+      "    Therefore the value of current in the circuit is 0.259 times the steady state value.\"\"\"\n",
+      "k=1-0.741\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of current in the circuit is %.3f times the steady state value.\" %(k)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of current in the circuit is 0.259 times the steady state value.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11,Page number: 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in the circuit at t=0.6 s.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=120.0               #Supply dc voltage(in Volts)\n",
+      "R=20.0                #Resistance of the resistor(in Ohms)\n",
+      "L=8.0                 #Inductance of the inductor(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "time_const=L/R\n",
+      "Io=V/R\n",
+      "i=Io*(1-exp(-0.6/time_const))\n",
+      "\"\"\"The voltage drop across R,v_R=i*r=Io*(1-exp(-t/time_const))*R;\n",
+      "   The voltage drop across L,v_L=L*(di/dt)=((L*Io)/time_const)*exp(-t/time_const);  \n",
+      "   \n",
+      "   We are to find the time at which these two voltage-drops are same. \"\"\"\n",
+      "t=-time_const*log(0.5,e)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current in the circuit at t=0.6 s is %.2f A.\" %(i)\n",
+      "print \"(b)The time at which the voltage drops across R and L are same is %.3f s.\" %(t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current in the circuit at t=0.6 s is 4.66 A.\n",
+        "(b)The time at which the voltage drops across R and L are same is 0.277 s.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12,Page number: 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the energy stored in the magnetic field.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=30.0               #Supply dc voltage(in Volts)\n",
+      "R=12.0                #Resistance of the resistor(in Ohms)\n",
+      "L=18.0                 #Inductance of the inductor(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "time_const=L/R\n",
+      "rate_curr=V/L\n",
+      "Io=V/R\n",
+      "i1=(V/R)*(1-exp(-3.0/time_const))\n",
+      "W1=0.5*i1*i1*L\n",
+      "Wlost=(0.5*L*Io*Io)-W1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The time constant is %.2f seconds.\" %(time_const)\n",
+      "print \"(b)The initial rate of change of current is %.3f A/s.\" %(rate_curr)\n",
+      "print \"(c)The current at t=3 s is %.2f A.\" %(i1)\n",
+      "print \"(d)The energy stored in the magnetic field at t=3 s is %.3f J.\" %(W1)\n",
+      "print  \"(e)The energy lost as heat till t=3 s is %.3f J.\" %(Wlost)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The time constant is 1.50 seconds.\n",
+        "(b)The initial rate of change of current is 1.667 A/s.\n",
+        "(c)The current at t=3 s is 2.16 A.\n",
+        "(d)The energy stored in the magnetic field at t=3 s is 42.055 J.\n",
+        "(e)The energy lost as heat till t=3 s is 14.195 J.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.13,Page number: 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage and current at different time instants.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R1=200.0              #Resistance of resistor 1(in Ohms)\n",
+      "L=5e-03               #Inductance of the coil(in Henry)\n",
+      "R2=17.0               #Resistance of resistor 2(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "i_0_plus=20e-03\n",
+      "v_0_plus=i_0_plus*R2\n",
+      "time_const=L/R1\n",
+      "v_L_0_plus=(L*i_0_plus)/time_const\n",
+      "t=20e-06 \n",
+      "i_20=i_0_plus*exp(-t/time_const)\n",
+      "v_20=v_0_plus\n",
+      "t=50e-06\n",
+      "i_50=i_0_plus*exp(-t/time_const)\n",
+      "v_50=v_0_plus\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The value of i(0+)=%e A.\" %(i_0_plus) \n",
+      "print \"(b)The value of v(0+)=%.3f V.\" %(v_0_plus)\n",
+      "print \"(c)The value of v_L(0+)=%.3f V.\" %(v_L_0_plus)\n",
+      "print \"(d)The value of i at t=20 micro seconds is %e A and v=%.3f V.\" %(i_20,v_20)\n",
+      "print \"(d)The value of i at t=50 micro seconds is %e A and v=%.3f V.\" %(i_50,v_50)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The value of i(0+)=2.000000e-02 A.\n",
+        "(b)The value of v(0+)=0.340 V.\n",
+        "(c)The value of v_L(0+)=4.000 V.\n",
+        "(d)The value of i at t=20 micro seconds is 8.986579e-03 A and v=0.340 V.\n",
+        "(d)The value of i at t=50 micro seconds is 2.706706e-03 A and v=0.340 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14,Page number: 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the currents at t=5 mill seconds.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=0.8                 #Self Inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "i_L_0_minus=(120e-03)*(200.0/(200.0+40.0))\n",
+      "Req=40+(1.0/((1.0/800.0)+(1.0/200.0)))\n",
+      "time_const=L/Req\n",
+      "t=5e-03 \n",
+      "i_L=i_L_0_minus*exp(-t/time_const)\n",
+      "i_x=-i_L*(800.0/(800.0+200.0))\n",
+      "i_y=(120e-03)+(-i_L*(200.0/(200.0+800.0)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"At t=5 milli seconds,\" \n",
+      "print \"(a)The current i_L=%e A.\" %(i_L)\n",
+      "print \"(b)The current i_x=%e A.\" %(i_x)\n",
+      "print \"(c)The current i_y=%e A.\" %(i_y)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At t=5 milli seconds,\n",
+        "(a)The current i_L=2.865048e-02 A.\n",
+        "(b)The current i_x=-2.292038e-02 A.\n",
+        "(c)The current i_y=1.142699e-01 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.15,Page number: 223 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current through an inductor at the time of switching.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=4.0                 #Self inductance of inductor(in Henry)\n",
+      "R=10.0                #Resistance of resistor parallel to the inductor(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"By superposition theorem, I_L_0=I_L_1+I_L_2; \"\"\"\n",
+      "I_L1=12.0/4.0\n",
+      "I_L2=2.0\n",
+      "I_L0=I_L1+I_L2\n",
+      "i_L_0_plus=I_L0\n",
+      "w_L_0_plus=0.5*L*I_L0*I_L0\n",
+      "time_const=L/R\n",
+      "t=1.0\n",
+      "i_L=I_L0*exp(-t/time_const)\n",
+      "v_10=-i_L*R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current i_L(0+)=%.2f A. The energy stored in the inductor is w_L(0+)=%.2f J.\" %(i_L_0_plus,w_L_0_plus)\n",
+      "print \"(b)At t=1 second, the current in the inductance is %.2f A and the voltage v_10=%.2f V.\" %(i_L,v_10)   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current i_L(0+)=5.00 A. The energy stored in the inductor is w_L(0+)=50.00 J.\n",
+        "(b)At t=1 second, the current in the inductance is 0.41 A and the voltage v_10=-4.10 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.16,Page number: 223 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the energy stored in the inductor.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=5e-03               #Self inductance of inductor(in Henry)\n",
+      "R=200.0               #Resistance of resistor(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_L0=5e-03\n",
+      "time_const=L/R\n",
+      "t=20e-06\n",
+      "i_L=I_L0*exp(-t/time_const)\n",
+      "w=0.5*L*i_L*i_L\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The energy stored in the inductor after 20 micro seconds of throwing the switch is %e J.\" %(w)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy stored in the inductor after 20 micro seconds of throwing the switch is 1.261853e-08 J.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.17,Page number: 224 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current in the circuit.\"\"\"\n",
+      "\n",
+      "#Variable DEclaration:\n",
+      "L=8.0                 #Self inductance of the coil(in Henry)\n",
+      "R=20.0                #Resistance of resistor(in Ohms)\n",
+      "V=120.0               #Voltage of the supply(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Io=V/R\n",
+      "time_const=L/R\n",
+      "t=0.6\n",
+      "i_t=Io*(1-exp(-t/time_const))\n",
+      "\"\"\" The voltage of R at any time is given as v_R(t)=6*(1-exp(-t/0.4))*20=120*(1-exp(-t/0.4));\n",
+      "\n",
+      "    The voltage across L at any time is v_L(t)=L*(di/dt)=120*exp(-t/0.4);\n",
+      "    \n",
+      "    Applying v_L(t)=v_R(t), \"\"\"\n",
+      "t=-log(120.0/240.0)*0.4\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The current in the circuit at t=0.6 seconds is %.2f A.\" %(i_t)\n",
+      "print \"(b)The time at which the voltage drops across R and L are same is %.4f seconds.\"%(t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The current in the circuit at t=0.6 seconds is 4.66 A.\n",
+        "(b)The time at which the voltage drops across R and L are same is 0.2773 seconds.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.18,Page number: 224 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current i_x in the circuit at different time instants.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=25e-03              #Self inductance of the coil(in Henry)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_L0=(10e-03)*(80.0/(80.0+20.0))\n",
+      "i_x_minus_2=I_L0\n",
+      "i_x_0_minus=I_L0\n",
+      "i_x_0_plus=I_L0*(30.0/(30.0+20.0))\n",
+      "Req=1.0/((1.0/20.0)+(1.0/30.0))\n",
+      "time_const=L/Req\n",
+      "t=2e-03\n",
+      "i_L_2=I_L0*exp(-t/time_const)\n",
+      "i_x_2=i_L_2*(30.0/(30.0+20.0))\n",
+      "t=4e-03\n",
+      "i_L_4=I_L0*exp(-t/time_const)\n",
+      "i_x_4=i_L_4*(30.0/(30.0+20.0))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The current i_x:\" \n",
+      "print \"At t=-2 ms, i_x=%e A.\" %(i_x_minus_2)   \n",
+      "print \"At t=0- ms, i_x=%e A.\" %(i_x_0_minus)\n",
+      "print \"At t=0+ ms, i_x=%e A.\" %(i_x_0_plus)\n",
+      "print \"At t=2 ms, i_x=%e A.\" %(i_x_2)\n",
+      "print \"At t=4 ms, i_x=%e A.\" %(i_x_4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The current i_x:\n",
+        "At t=-2 ms, i_x=8.000000e-03 A.\n",
+        "At t=0- ms, i_x=8.000000e-03 A.\n",
+        "At t=0+ ms, i_x=4.800000e-03 A.\n",
+        "At t=2 ms, i_x=1.837886e-03 A.\n",
+        "At t=4 ms, i_x=7.037134e-04 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.19,Page number: 225 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the current and voltage at the time of switching.\"\"\"\n",
+      "\n",
+      "#Variable Declaration: \n",
+      "V=1.5                 #Voltage of the supply(in Volts)\n",
+      "R1=5e-03              #Resistance of resistor 1(in Ohms) \n",
+      "R2=1.5e03             #Resistance of resistor 2(in Ohms)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Before the switch is thrown from a to b,the capacitor is fully charged to supply voltage.\n",
+      "\n",
+      "   When switching takes place,the capacitor starts discharging through the 5 milli Ohms resistance.\"\"\" \n",
+      "V0=V\n",
+      "v_0_plus=V0\n",
+      "i_0_plus=V0/R1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The voltage v(0+)=%.2f V and the current i(0+)=%.2f A.\" %(v_0_plus,i_0_plus)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The voltage v(0+)=1.50 V and the current i(0+)=300.00 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.20,Page number: 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage and current at different time intervals.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "C=5e-06               #Capacitance of the capacitor(in Farads)\n",
+      "V=6.0                 #Voltage of the supply(in Volts) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "v_0_minus=6.0*((3.0+2.0)/(3.0+2.0+1.0))\n",
+      "R=(1e03+3e03+2e03)\n",
+      "i_0_minus=V/R\n",
+      "v_0_plus=v_0_minus\n",
+      "Vo=v_0_plus\n",
+      "Req=(5e03+3e03+2e03)\n",
+      "time_const=Req*C\n",
+      "i_0_plus=v_0_plus/Req\n",
+      "Io=i_0_plus\n",
+      "t=0.05\n",
+      "v_t1=Vo*exp(-t/time_const)\n",
+      "i_t1=Io*exp(-t/time_const)\n",
+      "t=0.10\n",
+      "v_t2=Vo*exp(-t/time_const)\n",
+      "i_t2=Io*exp(-t/time_const)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The values of v(t) and i(t) are: \"\n",
+      "print \"At t=0- s, v(t)=%.3f V and i(t)=%e A.\" %(v_0_minus,i_0_minus)\n",
+      "print \"At t=0+ s, v(t)=%.3f V and i(t)=%e A.\" %(v_0_plus,i_0_plus)\n",
+      "print \"At t=0.05 s, v(t)=%.3f V and i(t)=%e A.\" %(v_t1,i_t1)\n",
+      "print \"At t=0.10 s, v(t)=%.3f V and i(t)=%e A.\" %(v_t2,i_t2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The values of v(t) and i(t) are: \n",
+        "At t=0- s, v(t)=5.000 V and i(t)=1.000000e-03 A.\n",
+        "At t=0+ s, v(t)=5.000 V and i(t)=5.000000e-04 A.\n",
+        "At t=0.05 s, v(t)=1.839 V and i(t)=1.839397e-04 A.\n",
+        "At t=0.10 s, v(t)=0.677 V and i(t)=6.766764e-05 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.21,Page number: 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltage and current at the time of switching.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "\"\"\"Using current divider rule, the current through branch AB is determined.\"\"\" \n",
+      "I_AB=10e-03*(1000.0/(1000.0+(800.0+200.0)))\n",
+      "v_0_minus=I_AB*800.0\n",
+      "V_AB=v_0_minus\n",
+      "v_0_plus=V_AB\n",
+      "i_C_0_plus=V_AB/(1.0/((1.0/200.0)+(1.0/800.0)))\n",
+      "i_0_plus=i_C_0_plus*(800.0/(800.0+200.0))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of v(0+)=%.2f V and the current i(0+)=%e A.\" %(v_0_plus,i_0_plus)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of v(0+)=4.00 V and the current i(0+)=2.000000e-02 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.22,Page number: 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the voltages across resistor,capacitor and switch.\"\"\"\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V=12.0                #Voltage of the supply(in Volts)\n",
+      "C=50e-03              #Capacitance of the capacitor(in Farads)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "v_C_0_minus=12.0*(20.0/(20.0+4.0))\n",
+      "v_C_0_plus=v_C_0_minus\n",
+      "Vo=v_C_0_plus\n",
+      "Req=5.0+20.0\n",
+      "time_const=Req*C\n",
+      "t=1.0\n",
+      "v_C1=Vo*exp(-t/time_const)\n",
+      "v_R1=v_C1*(20.0/(20.0+5.0))\n",
+      "v_SW1=V-V_R1\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The value of v_C at t=1 second is %.3f V.\" %(v_C1) \n",
+      "print \"The value of v_R at t=1 second is %.3f V.\" %(v_R1)\n",
+      "print \"The value of v_SW at t=1 second is %.3f V.\" %(v_SW1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of v_C at t=1 second is 4.493 V.\n",
+        "The value of v_R at t=1 second is 3.595 V.\n",
+        "The value of v_SW at t=1 second is 8.405 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter9.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter9.ipynb
new file mode 100755
index 00000000..299b89e3
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter9.ipynb
@@ -0,0 +1,1092 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9:ALTERNATING CURRENT AND VOLTAGE"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1,Page number: 237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the angle at which the instantaneous value of voltage is 10 V.\"\"\"\n",
+      "\n",
+      "from math import asin,degrees\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vm=20.0               #Peak value of sinusoidal voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "v=10.0    #Instantaneous Voltage(in Volts)\n",
+      "angle=degrees(asin(v/Vm))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The angle at which the instantaneous value of voltage is 10V is %.2f degrees.\" %(angle)\n",
+      "print \"(b)The maximum value of voltage is Vm=20 V.This occurs twice in one cycle at angles 90 degrees and 270 degrees.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The angle at which the instantaneous value of voltage is 10V is 30.00 degrees.\n",
+        "(b)The maximum value of voltage is Vm=20 V.This occurs twice in one cycle at angles 90 degrees and 270 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2,Page number: 237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the time represented by a 60 degrees phase angle.\"\"\"\n",
+      "\n",
+      "from math import pi,sin,radians\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "ang_freq=2000.0       #Angular Frequency(in radians per second)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "f=ang_freq/(2*pi)\n",
+      "T=1/f\n",
+      "t=160e-06\n",
+      "v=0.04*sin((2000*t)+radians(60))\n",
+      "t_60=(60.0/360)*T\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The frequency is %.2f Hz.\" %(f)\n",
+      "print \"(b)The angular frequency is %.2f rad/sec.\" %(ang_freq)\n",
+      "print \"(c)The instantaneous voltage when t=60 micro seconds, is %e V.\" %(v)\n",
+      "print \"(d)The time represented by 60 degrees phase angle is %e seconds.\" %(t_60)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The frequency is 318.31 Hz.\n",
+        "(b)The angular frequency is 2000.00 rad/sec.\n",
+        "(c)The instantaneous voltage when t=60 micro seconds, is 3.917381e-02 V.\n",
+        "(d)The time represented by 60 degrees phase angle is 5.235988e-04 seconds.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3,Page number: 237"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the equation for the instantaneous value of a sinusoidal voltage.\"\"\"\n",
+      "\n",
+      "from math import pi,sin,radians,degrees,asin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "V_peak_to_peak=20.0    #Peak-to-peak Voltage(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "V_m=V_peak_to_peak/2\n",
+      "T=10.0e-03\n",
+      "f=1/T\n",
+      "ang_freq=2*pi*f\n",
+      "phi=180-degrees(asin(3.6/V_m))\n",
+      "t=12e-03\n",
+      "v_12=V_m*sin((ang_freq*t)-radians(phi))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The equation of the given sinusoidal voltage is v(t)=%.2f sin(%.2ft+%.2f) V.\" %(V_m,ang_freq,phi)\n",
+      "print \"(b)The value of voltage at 12ms would be %.3f V.\" %(v_12)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The equation of the given sinusoidal voltage is v(t)=10.00 sin(628.32t+158.90) V.\n",
+        "(b)The value of voltage at 12ms would be -9.985 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4,Page number: 238"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the equation for the instantaneous value of an alternating current.\"\"\"\n",
+      "\n",
+      "from math import pi,sin,asin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "f=60.0                #Frequency of alternating current(in Hertz)\n",
+      "i_m=12.0              #Peak value of the alternating current(in Amperes) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "ang_freq=2*pi*f\n",
+      "t1=1.0/360\n",
+      "i_t1=i_m*sin(ang_freq*t1)\n",
+      "i2=9.6\n",
+      "t2=(asin(9.6/12))/ang_freq\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The equation for the instantaneous value of alternating current is i= %d sin(%.2ft) A.\" %(i_m,ang_freq)\n",
+      "print \"(b)The value of current at t=1/360 second is %.2f A.\" %(i_t1)\n",
+      "print \"(c)The time taken to reach 9.6A for the first time is %e seconds.\" %(t2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The equation for the instantaneous value of alternating current is i= 12 sin(376.99t) A.\n",
+        "(b)The value of current at t=1/360 second is 10.39 A.\n",
+        "(c)The time taken to reach 9.6A for the first time is 2.459727e-03 seconds.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5,Page number: 244"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the phase difference between two sinusoidal currents.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Calculations:\n",
+      "ang_freq=100.0*pi\n",
+      "f=ang_freq/(2.0*pi)\n",
+      "T=1/f\n",
+      "t=(30.0/360.0)*T\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The phase difference in terms of time is %e seconds.\" %(t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The phase difference in terms of time is 1.666667e-03 seconds.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6,Page number: 248 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the power consumed by a resistor.\"\"\"\n",
+      "\n",
+      "from cmath import phase\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "R=10                   #Resistance of resistance(in Ohms) \n",
+      "I=4+ 1j*3              #Alternating current phasor(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "I_mod=abs(I)\n",
+      "I_phase= phase(I)\n",
+      "I_rms=I_mod\n",
+      "P=pow(I_rms,2)*R\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The power consumed by the 10 ohm resistor is %.2f W.\" %(P)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The power consumed by the 10 ohm resistor is 250.00 W.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7,Page number: 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question: \n",
+      "\"\"\"Finding the resultant current obtained by adding two alternating currents.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import radians,degrees,sqrt\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=rect(10,0)\n",
+      "I2=rect(20,radians(60))\n",
+      "I=I1+I2\n",
+      "Im=abs(I)\n",
+      "I_phase=degrees(phase(I))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resultant current is %.2f A at a phase angle of %.2f degrees.\" %(Im,I_phase)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resultant current is 26.46 A at a phase angle of 40.89 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8,Page number: 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rms value of the sum of two currents.\"\"\"\n",
+      "\n",
+      "from cmath import rect\n",
+      "from math import radians,sqrt\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=rect((10*sqrt(2)),0)\n",
+      "I2=rect((20*sqrt(2)),radians(60))\n",
+      "I=I1+I2\n",
+      "Im=abs(I)\n",
+      "Irms=Im/(sqrt(2))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The rms value of the sum of the currents is %.2f A.\" %(Irms) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rms value of the sum of the currents is 26.46 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9,Page number: 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rms value of the resultant current.\"\"\"\n",
+      "\n",
+      "from cmath import rect,phase\n",
+      "from math import radians,sqrt,degrees\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=rect(5,0)\n",
+      "I2=rect(5,radians(30))\n",
+      "I3=rect(5,radians(-120))\n",
+      "I=I1+I2+I3\n",
+      "Im=abs(I)\n",
+      "Irms=Im/(sqrt(2))\n",
+      "I_phase=degrees(phase(I))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The rms value of the resultant current that leaves the junction is %.2f A at a phase angle of %.2f degrees.\" %(Irms,I_phase) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rms value of the resultant current that leaves the junction is 5.00 A at a phase angle of -15.00 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10,Page number: 251"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the average and rms value of the resultant cuurent in a wire.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:   \n",
+      "I1rms=10.0            #Rms value of direct current(in Amperes) \n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "avg=10.0 \n",
+      "I2rms=10.0/sqrt(2)\n",
+      "I_rms=sqrt(pow(I1rms,2)+pow(I2rms,2))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The average value of the resultant current is %d A as the current goes as much positive as negative around the value of %d A.\" %(avg,avg)  \n",
+      "print \"The rms value of the resultant current is %.3f A.\" %(I_rms)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average value of the resultant current is 10 A as the current goes as much positive as negative around the value of 10 A.\n",
+        "The rms value of the resultant current is 12.247 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11,Page number: 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the average value of a voltage waveform.\"\"\"\n",
+      "\n",
+      "#Calculations:\n",
+      "area_0_to_1=10*(1e-03)\n",
+      "area_1_to_3=-5*(2e-03)\n",
+      "area_3_to_4=20*(1e-03)\n",
+      "area_4_to_5=0*(1e-03)\n",
+      "area_5_to_8=5*(3e-03)\n",
+      "total_area=(area_0_to_1+area_1_to_3+area_3_to_4+area_4_to_5+area_5_to_8)\n",
+      "total_period=8e-03\n",
+      "avg_value=total_area/total_period\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The average value of the alternating voltage waveform is %.3f V.\" %(avg_value)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average value of the alternating voltage waveform is 4.375 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12,Page number: 252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the effective value of a voltage waveform.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Calculations:\n",
+      "area_0_to_10=400*(10e-03)\n",
+      "area_10_to_20=100*(10e-03)\n",
+      "total_area=(area_0_to_10+area_10_to_20)\n",
+      "total_period=20e-03\n",
+      "avg_value_of_square=total_area/total_period\n",
+      "rms=sqrt(avg_value_of_square)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The effective(rms) value of the alternating voltage waveform is %.3f V.\" %(rms) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The effective(rms) value of the alternating voltage waveform is 15.811 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.13,Page number: 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the rms value,the average value and the form factor for a current waveform.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "period=3.0            #Time period of the current waveform(in seconds)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Irms=sqrt( ((pow(10,2)*2)+(pow(0,2)*1))/3 )\n",
+      "Iavg=((10.0*2)+(0*1))/3.0\n",
+      "form_factor=Irms/Iavg\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The rms value of current waveform is %.2f A.\" %(Irms) \n",
+      "print \"The average value of current waveform is %.2f A.\" %(Iavg) \n",
+      "print \"The form factor of the current waveform is %.2f.\" %(form_factor) "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rms value of current waveform is 8.12 A.\n",
+        "The average value of current waveform is 6.67 A.\n",
+        "The form factor of the current waveform is 1.22.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.14,Page number: 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the form factor and the peak factor for a saw-tooth waveform.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "T=5e-03               #Time period of saw-tooth waveform(in seconds) \n",
+      "V_m=10.0              #Peak value of the saw-tooth voltage(in Volts)\n",
+      "\n",
+      "\"\"\" Vav=(Area under the curve in one cycle)/(Duration of one cycle) \"\"\"\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vav=((1.0/2)*V_m*T)/T\n",
+      "Vrms=V_m/(sqrt(3))\n",
+      "form_factor=Vrms/Vav\n",
+      "peak_factor=V_m/Vrms\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The average value of the saw-tooth voltage waveform is %.3f V.\" %(Vav)\n",
+      "print \"The rms value of the saw-tooth voltage waveform is %.3f V.\" %(Vrms)\n",
+      "print \"The form factor for the saw-tooth voltage waveform is %.3f.\" %(form_factor)\n",
+      "print \"The peak factor for the saw-tooth voltage waveform is %.3f.\" %(peak_factor)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average value of the saw-tooth voltage waveform is 5.000 V.\n",
+        "The rms value of the saw-tooth voltage waveform is 5.774 V.\n",
+        "The form factor for the saw-tooth voltage waveform is 1.155.\n",
+        "The peak factor for the saw-tooth voltage waveform is 1.732.\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.15,Page number: 255 "
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the average power,the apparent power,the instantaneous power and the power factor in percentage in an ac circuit.\"\"\"\n",
+      "\n",
+      "from math import cos,sqrt,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "phase_angle=pi/5      #Phase difference between the alternating current and alternating voltage(in radians)   \n",
+      "Vm=55                 #Peak value of the alternating voltage(in Volts)  \n",
+      "Im=6.1                #Peak value of the alternating current(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vrms=Vm/sqrt(2)\n",
+      "Irms=Im/sqrt(2)\n",
+      "pf=cos(phase_angle)\n",
+      "P_avg=Vrms*Irms*cos(phase_angle)\n",
+      "P_app=Vrms*Irms\n",
+      "P_inst=P_avg-(Vrms*Irms*cos((2*0.3)-(pi/5)))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The average power is %.2f W.\" %(P_avg)\n",
+      "print \"The apparent power is %.2f VA.\" %(P_app)\n",
+      "print \"The instantaneous power at wt=0.3 is %.2f W.\" %(P_inst)\n",
+      "print \"The power factor is %.3f lagging.\" %(pf)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average power is 135.71 W.\n",
+        "The apparent power is 167.75 VA.\n",
+        "The instantaneous power at wt=0.3 is -31.97 W.\n",
+        "The power factor is 0.809 lagging.\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.16,Page number:262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the average and rms values of waveforms.\"\"\"\n",
+      "\n",
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Vm1=10.0              #Peak voltage of first waveform(in Volts) \n",
+      "Vm2=10.0              #Peak voltage of second waveform(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Vav1=Vm1/2.0\n",
+      "Vrms1=Vm1/sqrt(3.0)\n",
+      "Vav2=Vm2/4.0\n",
+      "Vrms2=Vm2/sqrt(6.0)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The average value of the voltage is %.2f V and the rms value of the voltage is %.2f V.\" %(Vav1,Vrms1) \n",
+      "print \"(b)The average value of the voltage is %.2f V and the rms value of the voltage is %.2f V.\" %(Vav2,Vrms2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The average value of the voltage is 5.00 V and the rms value of the voltage is 5.77 V.\n",
+        "(b)The average value of the voltage is 2.50 V and the rms value of the voltage is 4.08 V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.17,Page number:263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"Finding the value of current at a given instant.\"\"\"\n",
+      "\n",
+      "from math import pi,sin,asin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Im=12.0               #Maximum value of the alternating current(in Amperes)\n",
+      "f=60.0                #Frequency of the alternating current(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "w=2*pi*f\n",
+      "t=1/360.0\n",
+      "i=Im*sin(w*t)\n",
+      "i1=9.6\n",
+      "t=asin(i1/Im)/w\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The equation for the instantaneous current is i(t)=%.2f sin(%.2f*t) A.\" %(Im,w)\n",
+      "print \"(b)The value of the current after (1/360) second is %.2f A.\" %round(i,2)\n",
+      "print \"(c)The time taken to reach 9.6 A for the first time is %e seconds.\" %(t)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The equation for the instantaneous current is i(t)=12.00 sin(376.99*t) A.\n",
+        "(b)The value of the current after (1/360) second is 10.39 A.\n",
+        "(c)The time taken to reach 9.6 A for the first time is 2.459727e-03 seconds.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.18,Page number:263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the expression of an alternating current in cosine form.\"\"\"\n",
+      "\n",
+      "from math import asin,pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Imax=10.0             #Maximum value of current(in Amperes)\n",
+      "Io=5.0                #Value of current at t=0(in Amperes)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "phi=asin(Io/Imax)\n",
+      "phi_new=(pi/2.0)-phi\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The expression for current is i=%.2f cos(wt-%.2f) A.\" %(Imax,phi_new)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The expression for current is i=10.00 cos(wt-1.05) A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.19,Page number:264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the time(from negative value) at which the instantaneous current is 10/sqrt(2.0) A.\"\"\"\n",
+      "\n",
+      "from math import pi,sqrt,radians,asin\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Irms=20.0             #Rms value of alternating current(in Amperes)\n",
+      "f=50.0                #Frequency of the alternating current(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "ang_freq=2*pi*f\n",
+      "Im=Irms*sqrt(2.0)\n",
+      "i=10.0*sqrt(2.0)\n",
+      "ph_lag=pi/2.0\n",
+      "t=(asin(i/Im)+ph_lag)/ang_freq\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The time(measured from negative value) at which instantaneous current will be 10/sqrt(2.0) is %.2f ms.\" %round((t*1000),2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The time(measured from negative value) at which instantaneous current will be 10/sqrt(2.0) is 6.67 ms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.20,Page number:264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the average and rms value of an alternating current.\"\"\"\n",
+      "\n",
+      "from math import pow,sqrt\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "Idc=10.0              #Dc current(in Amperes)\n",
+      "Im=5.0                #Peak value of sinusoidal component(in Volts)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "Iav=Idc\n",
+      "Irms=sqrt((10*10)+pow((5.0/sqrt(2.0)),2))\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The average value of current is %.2f A.\" %(Iav)\n",
+      "print \"The rms value of current is %.2f A.\" %(Irms)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average value of current is 10.00 A.\n",
+        "The rms value of current is 10.61 A.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.21,Page number:264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the sum of three alternating voltages.\"\"\"\n",
+      "\n",
+      "from math import radians,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Calculations:\n",
+      "v1=rect(147.3,radians(188.1))\n",
+      "v2=rect(294.6,radians(45))\n",
+      "v3=rect(88.4,radians(135))\n",
+      "v_res=v1+v2+v3\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"The resultant voltage is v=%.2f sin(wt+(%.2f degrees)) V.\" %(abs(v_res),degrees(phase(v_res)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resultant voltage is v=250.07 sin(wt+(90.01 degrees)) V.\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.22,Page number:265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the reactance offered by an inductor and a capacitor.\"\"\"\n",
+      "\n",
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration:\n",
+      "L=0.2                 #Inductance of the inductor(in Henry)\n",
+      "C=10e-06              #Capacitance of the capacitor(in Farads)\n",
+      "f=100                 #Initial frequency of the ac input voltage(in Hertz)\n",
+      "f1=140                #New frequency of the ac input voltage(in Hertz)\n",
+      "\n",
+      "\n",
+      "#Calculations:\n",
+      "X_L=2*pi*f*L\n",
+      "X_C=1.0/(2*pi*f*C)\n",
+      "X_L1=2*pi*f1*L\n",
+      "X_C1=1.0/(2*pi*f1*C)\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)For a frequency of 100Hz, X_L=%.2f Ohms and X_C=%.2f Ohms.\" %(X_L,X_C)\n",
+      "print \"(a)For a frequency of 140Hz, X_L=%.2f Ohms and X_C=%.2f Ohms.\" %(X_L1,X_C1)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)For a frequency of 100Hz, X_L=125.66 Ohms and X_C=159.15 Ohms.\n",
+        "(a)For a frequency of 140Hz, X_L=175.93 Ohms and X_C=113.68 Ohms.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.23,Page number:265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Question:\n",
+      "\"\"\"Finding the currents in each case.\"\"\"\n",
+      "\n",
+      "from math import radians,degrees\n",
+      "from cmath import rect,phase\n",
+      "\n",
+      "#Calculations:\n",
+      "I1=rect(10,0)\n",
+      "I2=rect(10,0)\n",
+      "Ia=I1+I2\n",
+      "I1=rect(10,radians(90))\n",
+      "Ib=I1+I2\n",
+      "I1=rect(10,radians(-90))\n",
+      "I2=rect(10,radians(-90))\n",
+      "Ic=I1+I2\n",
+      "V=250.0+1j*0\n",
+      "X_L=1j*25.0\n",
+      "Id=V/X_L\n",
+      "X_C=-1j*25.0\n",
+      "I=5.0+1j*0\n",
+      "Ve=I*X_C\n",
+      "\n",
+      "\n",
+      "#Result:\n",
+      "print \"(a)The unknown current is %.2f A at a phase angle of %.2f degrees.\" %(abs(Ia),degrees(phase(Ia)))\n",
+      "print \"(a)The unknown current is %.2f A at a phase angle of %.2f degrees.\" %(abs(Ib),degrees(phase(Ib)))\n",
+      "print \"(a)The unknown current is %.2f A at a phase angle of %.2f degrees.\" %(abs(Ic),degrees(phase(Ic)))\n",
+      "print \"(a)The unknown current is %.2f A at a phase angle of %.2f degrees.\" %(abs(Id),degrees(phase(Id)))\n",
+      "print \"(a)The unknown voltage is %.2f A at a phase angle of %.2f degrees.\" %(abs(Ve),degrees(phase(Ve)))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The unknown current is 20.00 A at a phase angle of 0.00 degrees.\n",
+        "(a)The unknown current is 14.14 A at a phase angle of 45.00 degrees.\n",
+        "(a)The unknown current is 20.00 A at a phase angle of -90.00 degrees.\n",
+        "(a)The unknown current is 10.00 A at a phase angle of -90.00 degrees.\n",
+        "(a)The unknown voltage is 125.00 A at a phase angle of -90.00 degrees.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/README.txt b/BASIC_ELECTRICAL_ENGINEERING_/README.txt
new file mode 100755
index 00000000..5982dc4d
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Abhiram Padmanabhan
+Course: be
+College/Institute/Organization: PES University,Bangalore
+Department/Designation: Electrical and Electronics
+Book Title: BASIC ELECTRICAL ENGINEERING 
+Author: D C KULSHRESHTHA
+Publisher: Tata McGraw Hill Education Private Limited,7 West Patel Nagar,New Delhi-110008.
+Year of publication: 2012
+Isbn: 978-0-07-132896-8
+Edition: Revised First
+\ No newline at end of file
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot1.png b/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot1.png
new file mode 100755
index 00000000..0d0af695
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot1.png
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot2.png b/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot2.png
new file mode 100755
index 00000000..cb8b214f
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot2.png
diff --git a/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot3.png b/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot3.png
new file mode 100755
index 00000000..8fed5b31
--- /dev/null
+++ b/BASIC_ELECTRICAL_ENGINEERING_/screenshots/Screenshot3.png
diff --git a/Basic_Fluid_Mechanics/README.txt b/Basic_Fluid_Mechanics/README.txt
new file mode 100755
index 00000000..9fe52d79
--- /dev/null
+++ b/Basic_Fluid_Mechanics/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Jatin Patel
+Course: bca
+College/Institute/Organization: Freelancing work
+Department/Designation: Freelancer
+Book Title: Basic Fluid Mechanics
+Author: Peerless
+Publisher: Pergamon Press
+Year of publication: 1967
+Isbn: 978-0080110981
+Edition: 1
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch10.ipynb b/Basic_Fluid_Mechanics/ch10.ipynb
new file mode 100755
index 00000000..805fdf34
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch10.ipynb
@@ -0,0 +1,353 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:deacbf9acb1071d19b96737be4ae4bdcb6a5696a34ffaedecc8c1b87e6d2a62b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : External Flows"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page No : 367"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import *\n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "u= 3.6*10**-5 \t#lbf sec/ft**2 viscosity\n",
+      "d= 64. \t#lbm/ft**2 density\n",
+      "l= 20. \t#ft long\n",
+      "a= 0.5\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "sw= u*g/(a*d)\n",
+      "sw1= u**2*g*l/(2*a*d)\n",
+      "Re=array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])*10**5\n",
+      "Vinf=Re*u*g/(d*a)\n",
+      "Cd= array([1.2, 1.15, 0.94, 0.68, 0.305, 0.31, 0.32, 0.33, 0.34, 0.35])\n",
+      "cdre=Cd*Re**2\n",
+      "D=sw1*cdre\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'velocity = %.3e ft/sec'%(sw)\n",
+      "print  ' Force = %.3e lbf'%(sw1)\n",
+      "print \"V (ft/sec)        D(lbf)\"\n",
+      "for i in range(len(D)):\n",
+      "    print \"%6.1f         %6d\"%(Vinf[i],D[i])\n",
+      "\n",
+      "\n",
+      "# note : answers are accurate. please check manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity = 3.623e-05 ft/sec\n",
+        " Force = 1.304e-08 lbf\n",
+        "V (ft/sec)        D(lbf)\n",
+        "   3.6            156\n",
+        "   7.2            599\n",
+        "  10.9           1103\n",
+        "  14.5           1418\n",
+        "  18.1            994\n",
+        "  21.7           1455\n",
+        "  25.4           2044\n",
+        "  29.0           2754\n",
+        "  32.6           3591\n",
+        "  36.2           4564\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page No : 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "u= 3.6*10**-5 \t#lbf sec/ft**2\n",
+      "d= 64. \t#lbm/ft**2 density\n",
+      "l= 20. \t#ft long\n",
+      "a= 0.5\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "sw= u*g/(a*d)\n",
+      "sw1= u**2*g*l/(2*a*d)\n",
+      "Re = array([1 ,2, 3, 4, 5, 6, 7, 8, 9, 10])*10**5\n",
+      "Vinf=Re*u*g/(d*a)\n",
+      "Cd = array([1.2, 1.15, 0.94, 0.68, 0.305, 0.31, 0.32, 0.33, 0.34, 0.35])\n",
+      "cdre=Cd*Re**2\n",
+      "D=sw1*cdre\n",
+      "\t\n",
+      "#RESULTS\n",
+      "plot(Vinf,D)\n",
+      "xlabel(\"Vinf, ft/sec\")\n",
+      "ylabel(\"D, lbf\") \n",
+      "suptitle(\"Streamlinedbody curve\")\n",
+      "\n",
+      "\t#data for curves b,c,d is not given\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 1,
+       "text": [
+        "<matplotlib.text.Text at 0x2683050>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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r1sFVV0GfPtZSImeeaXciaUjUVSUSZBYuhLQ0a9e+9HS700hDpBaHSBB5+22r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\n3v8VAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2682fd0>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 Page No : 373"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "v1= 10. \t#ft/sec\n",
+      "v2m= 9 \t    #ft/sec wide\n",
+      "a= 1.02\n",
+      "hbyd= 5.95\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "ca= (v1/v2m)**2\n",
+      "Cd= hbyd*(ca-1+2-2*ca)+2*a*ca\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Drag coeffcieicnt = %.2f'%(Cd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Drag coeffcieicnt = 1.12\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 Page No : 387"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "A= 320.  \t#ft/**2 area\n",
+      "w= 18000. \t#lbf weighs\n",
+      "v= 230.  \t#ft/sec normal speed\n",
+      "ad= 0.0765 \t#lbm/ft**3 density\n",
+      "p= 5.    \t#per cent of the total lift force\n",
+      "c= 0.055\n",
+      "n= 1.75     # total drag\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "CL= 2*w*(1-(p/100))*g/(ad*v**2*A)\n",
+      "D= w*(1-(p/100))*c*n/CL\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' lift coefficient = %.2f'%(CL)\n",
+      "print  ' Drag force = %.f'%(D)\n",
+      "\n",
+      "# note : answer is accurate"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " lift coefficient = 0.85\n",
+        " Drag force = 1935\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 Page No : 396"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "bi= 70. \t#degrees outlet angels\n",
+      "i= 8.    \t#degrees incidence angle\n",
+      "bo= 130. \t#degrees outlet angels\n",
+      "s= 5.    \t#degrees\n",
+      "vi= 1200. \t#ft/sec\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "a= 0.48\n",
+      "s1= 1.4 \t#in\n",
+      "b= 5.    \t#in\n",
+      "Cx= 0.06    # co-efficient \n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "O= bo-s-bi+i\n",
+      "Vo= vi*math.sin(math.radians(bi-i))/math.sin(math.radians(bo-s))\n",
+      "Fy= -a*vi*math.sin(math.radians(bi-i))*(s1/12)*(b/12)*(Vo*math.cos(math.radians(bo-s))-vi*math.cos(math.radians(bi-i)))/g\n",
+      "dp= a*(Vo**2*(1+Cx)-vi**2)/(2*g)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Fluid deflection angle = %.f degrees'%(O)\n",
+      "print  ' Vo = %.f ft/sec'%(Vo)\n",
+      "print  ' Force on each blade = %.f lbf'%(Fy)\n",
+      "print  ' Pressure difference = %.f lbf/ft**2'%(dp)\n",
+      "\n",
+      "# note : answer is accurate. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fluid deflection angle = 63 degrees\n",
+        " Vo = 1293 ft/sec\n",
+        " Force on each blade = 1002 lbf\n",
+        " Pressure difference = 2485 lbf/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 Page No : 397"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "ari= 62. \t#degrees\n",
+      "aro= 125. \t#degrees\n",
+      "vri= 1200. \t#ft/sec\n",
+      "vro= 1294. \t#ft/sec\n",
+      "vrr= 550. \t#ft/sec velocity\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "v1= vri*math.sin(math.radians(ari))\n",
+      "v2= vrr+vri*math.cos(math.radians(ari))\n",
+      "vi= math.sqrt(v1**2+v2**2)\n",
+      "ai= round(math.degrees(math.atan(v1/v2)),1)\n",
+      "vo= round(vro*math.sin(math.radians(aro)))\n",
+      "vo1= round(vro*math.cos(math.radians(aro))+vrr)\n",
+      "vo2= round(math.sqrt(vo**2+vo1**2))\n",
+      "ao= math.degrees(math.atan(vo/vo1))+180\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' absolute velocity = %.f ft/sec'%(vi)\n",
+      "print  ' direction = %.1f degrees'%(ai)\n",
+      "print  ' absolute velocity = %.f ft/sec'%(vo2)\n",
+      "print  ' direction = %.1f degrees'%(ao)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " absolute velocity = 1537 ft/sec\n",
+        " direction = 43.6 degrees\n",
+        " absolute velocity = 1077 ft/sec\n",
+        " direction = 100.3 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch11.ipynb b/Basic_Fluid_Mechanics/ch11.ipynb
new file mode 100755
index 00000000..1a38341e
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch11.ipynb
@@ -0,0 +1,443 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d460554b2abe6f6d6700d68291fcbf83cdf7b3bd9d716f5a9395eef88b744934"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Turbomachines:Elementary Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 Page No : 426"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy.functions.elementary.trigonometric import acot\n",
+      "\n",
+      "#initialisation of variables\n",
+      "rt= 1.3 \t#ft\n",
+      "rr= 0.6 \t#ft\n",
+      "Q= 75.   \t#ft**3 flow rate\n",
+      "rm= 0.95\n",
+      "w1= 40.  \t#rev/sec \n",
+      "bim= 153. \t#degrees blade inlet angle\n",
+      "bom= 147. \t#degrees blade outlet angle\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "A= round(math.pi*(rt**2-rr**2),2)\n",
+      "Va= round(Q/A,2)\n",
+      "Vbm= rm*w1\n",
+      "#a= -1/math.degrees(math.atan(-Vbm/Va))\n",
+      "a = math.degrees(acot(-Vbm/Va))\n",
+      "im= a-bim\n",
+      "vwm= Vbm+Va*1/math.tan(math.radians(bom))\n",
+      "dvwm= rm*vwm\n",
+      "C= w*Q*dvwm/g\n",
+      "Cw= C*w1\n",
+      "dp= Cw/Q\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Incidence = %.1f degrees'%(im) \n",
+      "print  ' Oulet velocity = %.2f ft/sec'%(vwm)\n",
+      "print  ' Change of whirl at the mean radius = %.2f ft**2/sec'%(dvwm)\n",
+      "\n",
+      "print  ' Torque = %.f lbf/ft'%(C)\n",
+      "print  ' Rate of working = %.f ft lbf/sec'%(Cw)\n",
+      "print  ' Workdone by the rotor = %.f lbf/ft**2'%(dp)\n",
+      "\n",
+      "# note : answer in book is wrong. please check manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Incidence = -178.3 degrees\n",
+        " Oulet velocity = 10.37 ft/sec\n",
+        " Change of whirl at the mean radius = 9.86 ft**2/sec\n",
+        " Torque = 1432 lbf/ft\n",
+        " Rate of working = 57300 ft lbf/sec\n",
+        " Workdone by the rotor = 764 lbf/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 Page No : 428"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "vbm= 38. \t#ft/sec\n",
+      "va= 17.94 \t#ft/sec\n",
+      "a= 147.5 \t#degrees\n",
+      "vwm= 10.37 \t#ft/sec\n",
+      "C= 1430. \t#lbf/ft\n",
+      "P= 763. \t#lbf/ft**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "vwm1= vbm+va*1/math.tan(math.radians(a))\n",
+      "p= (vwm-vwm1)/vwm\n",
+      "C1= C*(1-p)\n",
+      "P1= P*(1-p)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Oulet Velocity = %.2f ft/sec'%(vwm1) \n",
+      "print  ' Torque = %.f lbf/ft'%(round(C1,-1))\n",
+      "print  ' Workdone by the rotor = %.f lbf/ft**2'%(P1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Oulet Velocity = 9.84 ft/sec\n",
+        " Torque = 1360 lbf/ft\n",
+        " Workdone by the rotor = 724 lbf/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 Page No : 430"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from sympy.functions.elementary.trigonometric import acot\n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "a= 154 \t    #degrees\n",
+      "vbm= 38 \t#ft/sec\n",
+      "bom= 147 \t#degrees outlet angle\n",
+      "vwm= -7.78 \t#ft/sec outlet whirl velocity\n",
+      "w= 62.4 \t#lbf/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "vb= 38    \t#ft/sec velocity\n",
+      "A= 4.18 \t#ft**2 flow area\n",
+      "e= 0.95\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "vat= (vwm-vb)*math.tan(math.radians(bom))\n",
+      "Q= vat*A\n",
+      "#a1= 1/math.tan(math.radians(-vbm/vat))\n",
+      "a1 = math.degrees(acot(-vbm/vat))\n",
+      "imt= a1-a\n",
+      "C= w*Q*vwm*e/g\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Flow rate = %.1f ft**3'%(Q)\n",
+      "print  ' Incidence angle= %.f degrees'%(imt)\n",
+      "print  ' Torque= %.f lbf ft'%(C)\n",
+      "#Incorrect value for a1 in textbook. Hence the difference in answers"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Flow rate = 124.3 ft**3\n",
+        " Incidence angle= -192 degrees\n",
+        " Torque= -1780 lbf ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 Page No : 435"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import *\n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "rt= 0.5 \t#ft radius\n",
+      "rr= 0.16 \t#ft root radius\n",
+      "dv1= 88.3 \t#ft/sec\n",
+      "b= 150.  \t#degrees\n",
+      "r= array([0.16, 0.3, 0.5])\n",
+      "vw= array([2.5, 5, 7.5])\n",
+      "vb= array([46.6, 88.3, 132.5])\n",
+      "vrb= array([44.16, 88.3, 132.5])\n",
+      "v1= array([-1.154, -0.385])\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "A= math.pi*(rt**2-rr**2)\n",
+      "Va= -dv1*math.tan(math.radians(b))\n",
+      "Q= Va*A\n",
+      "ari = degrees((arctan(Va/(vw - vb)))) + 180\n",
+      "ari = array([ari[0],ari[2]])\n",
+      "#a= tan(radians(v1))+180\n",
+      "b = degrees(math.tan(0.577))\n",
+      "i = ari - 150\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Velocity = %.2f ft/sec'%(Va)\n",
+      "print  ' Flow rate = %.1f ft**3'%(Q)\n",
+      "\n",
+      "print (v1)\n",
+      "print (ari)\n",
+      "print (i)\n",
+      "\n",
+      "# rounding off error. and for 'i' answer is wrong. please check. "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Velocity = 50.98 ft/sec\n",
+        " Flow rate = 35.9 ft**3\n",
+        "[-1.154 -0.385]\n",
+        "[ 130.86126801  157.81238868]\n",
+        "[-19.13873199   7.81238868]\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 Page No : 436"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import *\n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "rt= 0.5 \t#ft\n",
+      "rr= 0.16 \t#ft\n",
+      "dv1= 88.3 \t#ft/sec\n",
+      "b= 150.  \t#degrees\n",
+      "a= 5.    \t#degrees mean radius\n",
+      "v1= array([-0.933 ,-0.311])\n",
+      "i= array([1.0, 5.0 ,6.7])\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "b1= b+a\n",
+      "A= math.pi*(rt**2-rr**2)\n",
+      "Va= -dv1*math.radians(math.tan(b1))\n",
+      "Q= Va*A\n",
+      "a1= degrees(tan(v1))+180\n",
+      "\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Velocity = %.2f ft/sec'%(Va)\n",
+      "print  ' Flow rate = %.1f ft**3/sec'%(Q)\n",
+      "\n",
+      "print (v1)\n",
+      "print (a1)\n",
+      "print (i)\n",
+      "#Incorrect calculations in textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity = -2.76 ft/sec\n",
+        " Flow rate = -1.9 ft**3/sec\n",
+        "[-0.933 -0.311]\n",
+        "[ 102.69071396  161.58339075]\n",
+        "[ 1.   5.   6.7]\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 Page No : 439"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "r= 1.    \t#in\n",
+      "b= 0.75 \t#in rotor inlet width\n",
+      "w= 180.  \t#rev/sec\n",
+      "B= 120. \t#degrees blade inlet angle\n",
+      "Bo= 150. \t#degrees blade outlet angle\n",
+      "ro= 3.   \t#ft\n",
+      "bo= 0.5 \t#ft\n",
+      "Vbo= 180. \t#ft/sec\n",
+      "w1= 62.4 \t#lbf/ft**3 density\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Q= -2*math.pi*(r/12)**2*(b/12)*w*math.tan(math.radians(B))\n",
+      "Vfo= Q/(2*math.pi*(ro/12)*(bo/12))\n",
+      "Vwo= Vbo*(ro/12)+Vfo*1/math.tan(math.radians(Bo))\n",
+      "C= w1*Q*Vwo*(ro/12)/g\n",
+      "dp= w1*Vwo*w*(ro/12)/g\n",
+      "ari= degrees(math.atan((-Q*0.8/(2*math.pi*(r/12)**2*(b/12)*w))))+180\n",
+      "i1= ari-B\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Flow rate = %.2f ft**3/sec'%(Q)\n",
+      "print  ' radial velocity= %.2f ft/sec'%(Vfo)\n",
+      "print  ' outlet whirl velocity= %.2f ft/sec'%(Vwo)\n",
+      "print  ' Torque= %.2f lbf ft'%(C)\n",
+      "print  ' Stagnant pressure = %.f lbf/ft**2'%(dp)\n",
+      "print  ' Incidence angle = %.1f degrees'%(i1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Flow rate = 0.85 ft**3/sec\n",
+        " radial velocity= 12.99 ft/sec\n",
+        " outlet whirl velocity= 22.50 ft/sec\n",
+        " Torque= 9.27 lbf ft\n",
+        " Stagnant pressure = 1962 lbf/ft**2\n",
+        " Incidence angle = 5.8 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7 Page No : 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "r= 1.4\n",
+      "Mai= 0.5    \t#ft/sec mach number\n",
+      "T= 582.     \t#R temperature\n",
+      "psi= 3040.  \t#lbf/in**2 pressure\n",
+      "R= 53.3 \t    #ft lbf/lbm gas\n",
+      "g= 32.2      \t#ft/sec**2\n",
+      "Vwi= 300.    \t#ft/sec velocity\n",
+      "m= 35.       \t#lb/sec\n",
+      "rm= 0.7 \t    #ft radius\n",
+      "rp= 4.25\n",
+      "w= 1200.     \t#rev/sec\n",
+      "cp= 0.24\n",
+      "J= 778.      \t#lb\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "tr= 1+0.5*(r-1)*Mai**2\n",
+      "Ti= round(T/tr)\n",
+      "pr= tr**(r/(r-1))\n",
+      "pi= psi/pr\n",
+      "ai= pi/(R*Ti)\n",
+      "Vi= Mai*(r*R*g*Ti)**0.5\n",
+      "Vai= math.sqrt(Vi**2-Vwi**2)\n",
+      "h= m/(2*math.pi*ai*rm*Vai)\n",
+      "pr1= rp**(1./12)\n",
+      "Vwo= Vwi+(pr1**((r-1)/r)-1)*(cp*J*g*T/(rm*w))\n",
+      "BO= 1/math.tan(math.radians((Vwo-w*rm)/Vai))\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Absolute air velocity = %.f ft/sec'%(Vi)\n",
+      "print  ' air velocity = %.f ft/sec'%(Vai)\n",
+      "print  ' Blade height = %.3f ft'%(h)\n",
+      "print  ' velocity = %.f ft/sec'%(Vwo)\n",
+      "print  ' outlet balde angle = %.1f degrees'%(BO) #incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Absolute air velocity = 577 ft/sec\n",
+        " air velocity = 493 ft/sec\n",
+        " Blade height = 0.186 ft\n",
+        " velocity = 446 ft/sec\n",
+        " outlet balde angle = -71.7 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch12.ipynb b/Basic_Fluid_Mechanics/ch12.ipynb
new file mode 100755
index 00000000..61bd9af9
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch12.ipynb
@@ -0,0 +1,302 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:747d23d1d49d9ec45297f78a51cc698e8f77e86dac46ee6fa2ac037964e152b9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Turbomachines: Further Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Page No : 461"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "from matplotlib.pyplot import *\n",
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "d= 0.0764 \t        #lbm/ft**3\n",
+      "u= 3.74*10**-7 \t    #lbf sec/ft**2\n",
+      "D= 15.           \t#in\n",
+      "g= 32.2 \t        #ft/sec**2\n",
+      "p= 14.7 \t        #lb/in**2\n",
+      "r1= [0.02, 0.04, 0.06, 0.08, 0.1, 1.15]\n",
+      "r2= [0.0338, 0.0267, 0.0199, 0.0159, 0.0132, 0.0100]\n",
+      "r3= [0.46, 0.92, 1.38, 1.84, 2.3, 2.64]\n",
+      "r4= [2.97, 2.35, 1.75, 1.4, 1.16, 0.88]\n",
+      "r5= [0.0206, 0.0163, 0.0121, 0.0097, 0.0081, 0.0061]\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "re= (d/u)*(p*100*2*math.pi/60)*(D/12)**2/g\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Reynolds Number = %.2e '%(re)\n",
+      "print \"m/qwD**3     g0deltaPe/QW**2D**2     m(lbm/sec)      deltap(lbf/ft**2)      deltap(lbf/in**2)\"\n",
+      "for i in range(len(r1)):\n",
+      "    print \"%7.2f     %8.4f              %7.2f         %7.2f              %8.4f\"%(r1[i],r2[i],r3[i],r4[i],r5[i])\n",
+      "\n",
+      "plot(r3,r5)\n",
+      "xlabel(\"m lbm/sec\")\n",
+      "ylabel( \"dPs lbf/ft**2\") \n",
+      "suptitle(\"Actual perfomance curve\")\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "Reynolds Number = 1.53e+06 \n",
+        "m/qwD**3     g0deltaPe/QW**2D**2     m(lbm/sec)      deltap(lbf/ft**2)      deltap(lbf/in**2)\n",
+        "   0.02       0.0338                 0.46            2.97                0.0206\n",
+        "   0.04       0.0267                 0.92            2.35                0.0163\n",
+        "   0.06       0.0199                 1.38            1.75                0.0121\n",
+        "   0.08       0.0159                 1.84            1.40                0.0097\n",
+        "   0.10       0.0132                 2.30            1.16                0.0081\n",
+        "   1.15       0.0100                 2.64            0.88                0.0061\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 1,
+       "text": [
+        "<matplotlib.text.Text at 0x1c015d0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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8kUj907y5NRy4Z08IC7OGB4tISSf/yD6TvafKXaGodevWfPrpp7Rq1Yr9+/dz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\nIlIlSiQiIlIlSiQiIlIlSiQiIlIl/w/IbjDGd8tQDQAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x1c01590>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 Page No : 464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "psif= 10.2       \t#lbf/in**2 pressure\n",
+      "usit= 3.8*10**-7 \t#lbf sec/ft**2 viscosity\n",
+      "usif= 3.52*10**-7 \t#lbf sec/ft**2 viscosity\n",
+      "Tsit= 530.       \t#R temperature\n",
+      "Tsif= 480.          #R temperature\n",
+      "wf= 15000.       \t#rev/min speed\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Psit= psif*usit*math.sqrt(Tsit/Tsif)/usif\n",
+      "wt= wf*math.sqrt(Tsit/Tsif)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Pressure in the test cell = %.1f lbf/in**2'%(Psit) \n",
+      "print  ' Compressor speed = %.f rev.min'%(wt) \n",
+      "\n",
+      "# book answer is wrong."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure in the test cell = 11.6 lbf/in**2\n",
+        " Compressor speed = 15762 rev.min\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 Page No : 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "w= 62.3 \t#lbf/ft**3 weight\n",
+      "d= 0.375 \t#in diameter\n",
+      "ro= 0.75 \t#ft radius\n",
+      "l= 1.25 \t#ft length\n",
+      "b= 120. \t#degrees angle\n",
+      "do= 0.25 \t#in diameter\n",
+      "p= 750. \t#lbf/in**2\n",
+      "g= 32.1 \t#ft/sec**2\n",
+      "f= 0.03     # friction factor\n",
+      "f1= 0.9\n",
+      "f2= 0.3\n",
+      "w1= 60. \t#rad/sec\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Q= math.sqrt(((p/w)+((60*ro)**2/(2*g))+do)*math.pi**2*g*(d/12)**4/((d/do)**4-1+(l*f/(d/12))+f1+f2))*0.353\n",
+      "Vwo= w1*ro+(4*Q/(math.pi*(do/12)**2))*math.cos(math.radians(b))\n",
+      "C= w*Q*Vwo*ro/g\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Flow Rate = %.4f ft**3/sec'%(Q) \n",
+      "print  ' Vwo = %.2f ft/sec'%(Vwo) \n",
+      "print  ' Driving Torque = %.3f lbf ft'%(C) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Flow Rate = 0.0160 ft**3/sec\n",
+        " Vwo = 21.56 ft/sec\n",
+        " Driving Torque = 0.502 lbf ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4 Page No : 491"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "W= 38.   \t#rev/sec speed\n",
+      "w= 62.4 \t#lbf/ft**3 density\n",
+      "m= 2000. \t#lbm/sec flow rate\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "ps= 5000. \t#lbf/ft**2 pressure rise\n",
+      "S3= 4.6\n",
+      "e= 0.91\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "S1= W*(w*m**2/(g*ps)**3)**0.25\n",
+      "D= S3*(m**2/(w*g*ps))**0.25\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' S1 = %.3f'%(S1)  \n",
+      "print  ' Diameter = %.2f ft'%(D)  \n",
+      "print  ' efficiency = %.2f '%(e)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " S1 = 0.594\n",
+        " Diameter = 3.65 ft\n",
+        " efficiency = 0.91 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 Page No : 495"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "d= 6.    \t#in diameter\n",
+      "f= 0.25\n",
+      "l= 1200. \t#ft long\n",
+      "p= 55.   \t#lbm/ft**3\n",
+      "w= 740.  \t#rev/min\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "n= 0.87     # efficiency\n",
+      "d1= 1.78 \t#ft\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "D= (0.13*math.pi**2*(d/12)**5/(8*f*l*0.012**2))**0.25*d1\n",
+      "m= 0.012*p*(w*2*math.pi/60)*D**3\n",
+      "dps= 0.13*p*(w*2*math.pi*D/60)**2/g\n",
+      "P= m*10*dps/(p*n)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Diameter = %.2f ft'%(D) \n",
+      "print  ' Mass flow rate = %.1f lbm/sec'%(m) \n",
+      "print  ' pressure rise = %.1f lbf/ft**2'%(round(dps,-1))\n",
+      "print  ' shaft power = %.2e ft lbf/sec'%(P)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Diameter = 1.04 ft\n",
+        " Mass flow rate = 57.3 lbm/sec\n",
+        " pressure rise = 1440.0 lbf/ft**2\n",
+        " shaft power = 1.72e+04 ft lbf/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch13.ipynb b/Basic_Fluid_Mechanics/ch13.ipynb
new file mode 100755
index 00000000..6819ab71
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch13.ipynb
@@ -0,0 +1,185 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:943d4576ca0c6a409710e5e5dfd3597778d333997c650e03aa4c9f933860e70a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Hydraulic Power Transmission|"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 Page No : 512"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "nop= 0.88\n",
+      "nom= 0.88   # constant\n",
+      "Pm= 75.  \t#hp\n",
+      "p= 3000. \t#lb/in**2 pressure\n",
+      "d= 54.5 \t#lbm/ft**3 density\n",
+      "u= 1.05*10**-4  # viscosity\n",
+      "d1= 0.5 \t#in\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "nt= (7./11)*nop*nom\n",
+      "pp= Pm/nt\n",
+      "Q= nop*pp*550/(p*144)\n",
+      "Re= 4*d*Q/(math.pi*u*(d1/12)*g)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' ntrans = %.3f '%(nt)\n",
+      "print  ' Input power = %.f hp'%(pp)\n",
+      "print  ' Flow rate = %.3f ft**3/sec'%(Q)\n",
+      "print  ' Reynolds Number = %.1e '%(Re)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " ntrans = 0.493 \n",
+        " Input power = 152 hp\n",
+        " Flow rate = 0.171 ft**3/sec\n",
+        " Reynolds Number = 8.4e+04 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 Page No : 513"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "lc= 0.25\n",
+      "a= 90.   \t#degrees\n",
+      "p= 3000.     \t#lb/in**2 pressure\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d1= 0.5 \t#in\n",
+      "Q= 0.171 \t#ft**3/sec\n",
+      "d= 54.5 \t#lbm/ft**3 density\n",
+      "n1= 2. \n",
+      "n2= 6.\n",
+      "lc1= 0.9\n",
+      "nop= 0.88\n",
+      "nom= 0.88\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "P1= 4*p*144/11\n",
+      "P2= 8*d*Q**2*(n1*lc+n2*lc1)/(math.pi**2*(d1/12)**4*g)\n",
+      "pt= P1+P2\n",
+      "dpm= (p*144-pt)\n",
+      "ntrans= nop*nom*dpm/(p*144)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Frictional pressure drop = %.2e lbf/ft**2'%(P1) \n",
+      "print  ' Extra Frictional pressure drop = %.2e lbf/ft**2'%(P2) \n",
+      "print  ' Total pressure drop = %.2e lbf/ft**2'%(pt)\n",
+      "print  ' Motor pressure drop = %.2e lbf/ft**2'%(dpm)\n",
+      "print  ' Overall transmission coefficiency = %.3f'%(ntrans)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Frictional pressure drop = 1.57e+05 lbf/ft**2\n",
+        " Extra Frictional pressure drop = 7.85e+04 lbf/ft**2\n",
+        " Total pressure drop = 2.36e+05 lbf/ft**2\n",
+        " Motor pressure drop = 1.96e+05 lbf/ft**2\n",
+        " Overall transmission coefficiency = 0.352\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page No : 521"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "bip= 135. \t#degrees inlet angle\n",
+      "bop= 150. \t#degrees outlet angle\n",
+      "bot= 140. \t#degrees turbine outlet angle\n",
+      "bos= 137. \t#degrees stator blade outlet angle\n",
+      "r= 1.8   \n",
+      "r1= 1.8    # ratio b1/b2\n",
+      "r2= 0.7    # ratio b1/b3\n",
+      "r3= 0.95   # ratio r3/r1\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Vw2r2byVw1r1 = (1+(1/math.tan(math.radians(bip))/1/math.tan(math.radians(bos))))*r**2-r1*(1/math.tan(math.radians(bop))/1/math.tan(math.radians(bos)))\n",
+      "Vw3r3byVw1r1 = r2*r3**2*(1+(1/math.tan(math.radians(bip))/1/math.tan(math.radians(bos))))-(1/math.tan(math.radians(bot))/1/math.tan(math.radians(bos)))\n",
+      "CtbyCp = (Vw2r2byVw1r1-Vw3r3byVw1r1)/(Vw3r3byVw1r1-1)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' R1 = %.2f'%(Vw2r2byVw1r1) \n",
+      "print  ' R2 = %.2f'%(Vw3r3byVw1r1) \n",
+      "print  ' Torque ratio = %.2f'%(CtbyCp)\n",
+      "\n",
+      "# rounding off error. please check. instead of cot, have used 1/tan. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " R1 = 3.37\n",
+        " R2 = 0.03\n",
+        " Torque ratio = -3.45\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch14.ipynb b/Basic_Fluid_Mechanics/ch14.ipynb
new file mode 100755
index 00000000..9cfe7a00
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch14.ipynb
@@ -0,0 +1,148 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:bc77360c046c25bc1074d9e1441ee10dd742c14709d1f92baa61c939b8a28b76"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 : Further Developments"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1 Page No : 532"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "import math \n",
+      "from matplotlib.pyplot import *\n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "a= 60.5\n",
+      "Q= 0.2       \t#ft**3/sec flow rate\n",
+      "d= 3.        \t#in diameter\n",
+      "u= 0.0325\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "T= [50.0, 60.0, 70.0, 80.0, 90.0, 100.0]\n",
+      "Ep= [294.5, 188.6, 113.2, 60.4, 37.7, 24.5]\n",
+      "Eh= [0 ,69.9, 139.8, 209.7, 279.5, 349.4]\n",
+      "Et= [295, 258, 253, 270, 317, 374]\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "re= a*4*Q/(math.pi*(d/12)*u*g)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Reynolds Number = %.1f '%(re)\n",
+      "print (T)\n",
+      "print (Ep)\n",
+      "print (Eh)\n",
+      "print (Et)\n",
+      "plot(T,Ep)\n",
+      "plot(T,Eh)\n",
+      "plot(T,Et)\n",
+      "\n",
+      "xlabel(\"T (F)\")\n",
+      "ylabel(\"Eh,Ep,Eh&Ep (kW)\")\n",
+      "suptitle(\"Variations of Ep, Eh and (Ep+Eh) with T\")\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "Reynolds Number = 58.9 \n",
+        "[50.0, 60.0, 70.0, 80.0, 90.0, 100.0]\n",
+        "[294.5, 188.6, 113.2, 60.4, 37.7, 24.5]\n",
+        "[0, 69.9, 139.8, 209.7, 279.5, 349.4]\n",
+        "[295, 258, 253, 270, 317, 374]\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 1,
+       "text": [
+        "<matplotlib.text.Text at 0x2b11ed0>"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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72bdvH0op2rRpo0sczZo1w9xA00bz+mTv39e6ZO/ehfXroWJFPQT3qKtXH4xr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\npEyyc1UAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x1faccd0>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2 Page No : 535"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "wcb= 2.      \t#ton weighing\n",
+      "wc= 100. \t    #ton\n",
+      "wa= 6.5 \t    #% of the weight\n",
+      "wca= 20. \n",
+      "r= 0.8\n",
+      "r1= 1.2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "wca1= wc/wa\n",
+      "wca2= wcb*(wca1/wca)**1.5\n",
+      "Wca= wcb*r**(9./4)*(1./r1)**(9./4)*(wca1/wca)**1.5\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Wc/Wa = %.2f '%(wca1)\n",
+      "print  ' Wc,a = %.2f ton'%(wca2)\n",
+      "print  ' Wc,a = %.2f ton'%(Wca)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Wc/Wa = 15.38 \n",
+        " Wc,a = 1.35 ton\n",
+        " Wc,a = 0.54 ton\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch2.ipynb b/Basic_Fluid_Mechanics/ch2.ipynb
new file mode 100755
index 00000000..eab5eede
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch2.ipynb
@@ -0,0 +1,309 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:30849a845cb23e6184901c441ceb4a2e2451d5db6a2c0f60dce1b23531e4d077"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Similarity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page No : 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "r= 4.\n",
+      "l1= 4 \t#units long axis\n",
+      "l2= 10 \t#units long axis\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "sxy= (4/r)\n",
+      "sxy1= l1**2\n",
+      "sxy2= l2**2\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'x**2+4*y**2 = %.f '%(sxy)\n",
+      "print  ' x**2+4*y**2 = %.f '%(sxy1)\n",
+      "print  ' x**2+4*y**2 = %.f '%(sxy2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "x**2+4*y**2 = 1 \n",
+        " x**2+4*y**2 = 16 \n",
+        " x**2+4*y**2 = 100 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page No : 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "vo= 10 \t#ft/sec\n",
+      "a= 0.5 \t#ft**-1\n",
+      "b= 1 \t#ft\n",
+      "x= -2 \t#ft\n",
+      "y= 2 \t#ft\n",
+      "b1= 2\n",
+      "a1= 3./5 \t#ft\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Vx= vo/(a*x**2+b)\n",
+      "Vy= -2*a*b*vo*x*y/(a*x**2+b)**2\n",
+      "V= math.sqrt(Vx**2+Vy**2)\n",
+      "fx= -2*a*b**2*vo**2*x/(a*x**2+b)**3\n",
+      "fy= 2*a*b**2*vo**2*y*(b-a*x**2)/(a*x**2+b)**4\n",
+      "f= math.sqrt(fx**2+fy**2)\n",
+      "r= b1**2/a1\n",
+      "f1= f*r\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Vx = %.2f ft/sec'%(Vx)\n",
+      "print  ' Vy = %.2f ft/sec'%(Vy)\n",
+      "print  ' V = %.2f ft/sec'%(V)\n",
+      "print  ' fx = %.2f ft/sec**2'%(fx)\n",
+      "print  ' fy = %.2f ft/sec**2'%(fy)\n",
+      "print  ' f = %.2f ft/sec**2'%(f)\n",
+      "print  ' r = %.2f in the present case'%(r)\n",
+      "print  ' f1 = %.2f ft/sec**2'%(f1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vx = 3.33 ft/sec\n",
+        " Vy = 4.44 ft/sec\n",
+        " V = 5.56 ft/sec\n",
+        " fx = 7.41 ft/sec**2\n",
+        " fy = -2.47 ft/sec**2\n",
+        " f = 7.81 ft/sec**2\n",
+        " r = 6.67 in the present case\n",
+        " f1 = 52.05 ft/sec**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page No : 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "r= 1./5\n",
+      "b1= 2 \t#ft\n",
+      "a1= 3./5 \t#ft\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "r= (a1*b1)**2*r\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'ratio of resultant forces acting on coorresponding fluid elements = %.3f '%(r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of resultant forces acting on coorresponding fluid elements = 0.288 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page No : 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "vos= 70. \t#ft/sec\n",
+      "as1= 78. \t#ft density\n",
+      "am= 72. \t#ft wind-tunnel\n",
+      "ls1= 6. \t#ft strut section\n",
+      "lm= 2.  \t#ft length\n",
+      "um= 386. \t#ft/sec\n",
+      "us= 372. \t#ft/sec\n",
+      "dm= 0.4\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "vom= vos*as1*ls1*um/(am*lm*us)\n",
+      "Ds= dm*(am/as1)*(us/um)**2\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Air speed = %.f ft/sec'%(vom)\n",
+      "print  ' Ds = %.3f lbf'%(Ds)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Air speed = 236 ft/sec\n",
+        " Ds = 0.343 lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page No : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "vom= 236. \t#ft/sec\n",
+      "as1= 0.072 \t#ft\n",
+      "am = 62.4 \t#ft density of water\n",
+      "ls1= 2. \t#ft\n",
+      "lm= 8. \t#ft\n",
+      "um= 248. \t#ft/sec  viscosity\n",
+      "us= 3.86 \t#ft/sec\n",
+      "Pm= 0.4/3.3\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "voh= vom*as1*ls1*um/(am*lm*us)\n",
+      "Ds= Pm*(as1/am)*(um/us)**2*(ls1/lm)*(lm-ls1)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Air speed = %.2f ft/sec'%(voh)\n",
+      "print  ' Drag force = %.3f lbf'%(Ds)\n",
+      "\n",
+      "# note : rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Air speed = 4.37 ft/sec\n",
+        " Drag force = 0.866 lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page No : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "To1= 540. \t#R temperature\n",
+      "po3= 12.6 \t#lbf/in**2\n",
+      "l3= 3.   \t#ft\n",
+      "po1= 14.7 \t#lbf/in**2 pressure\n",
+      "l1= 1.   \t#ft\n",
+      "vo1= 500. \t#ft/sec velocity\n",
+      "r= 0.83\n",
+      "P1= 1.   \t#lbf/in**2 turbine blade\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "To3= To1*(po3*l3/(po1*l1))**r\n",
+      "Vo3= vo1*math.sqrt(To3/To1)\n",
+      "P3= P1*po3*l3/(po1*l1)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'To3 = %.f R'%(To3)\n",
+      "print  ' Vo3 = %.f ft/sec'%(Vo3)\n",
+      "print  ' P3 = %.2f lbf/ft'%(P3)\n",
+      "\n",
+      "# note : book answers are not accurate."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "To3 = 1183 R\n",
+        " Vo3 = 740 ft/sec\n",
+        " P3 = 2.57 lbf/ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch3.ipynb b/Basic_Fluid_Mechanics/ch3.ipynb
new file mode 100755
index 00000000..cd8562e6
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch3.ipynb
@@ -0,0 +1,160 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:be8f61b53e434158045726efedae14168d3ad7c750cfe29d070385bb73f71e47"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Dimensional Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page No : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "t= 1.    \t#hr\n",
+      "g1= 32.2 \t#ft/sec**2\n",
+      "g2= 32.2 \t#lbm ft/lbf\n",
+      "u= 2.4*10**-5 \t#lbf sec/ft**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "q2= g*(t*60*60)**2\n",
+      "go= g*(t*60*60)**2\n",
+      "q3= g/g2\n",
+      "u1= u/(t*60*60)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' q2 = %.2e lbm ft/lbf hr**2'%(q2)\n",
+      "print  ' go = %.2e lbm ft/lbf hr**2'%(go)\n",
+      "print  ' go = %.f slug ft/lbf sec**2'%(q3)\n",
+      "print  ' viscosity = %.2e lbf hr/ft**2'%(u1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " q2 = 4.17e+08 lbm ft/lbf hr**2\n",
+        " go = 4.17e+08 lbm ft/lbf hr**2\n",
+        " go = 1 slug ft/lbf sec**2\n",
+        " viscosity = 6.67e-09 lbf hr/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page No : 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "m= 1     \t#lb\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "m1= g/m\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  '1 lbf/sec ft**2 = %.1f lbm/ft sec'%(m1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1 lbf/sec ft**2 = 32.2 lbm/ft sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 Page No : 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "n1=1.\n",
+      "n2= 3.\n",
+      "n3=2.\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "a1= -n1\n",
+      "a2= -n3\n",
+      "a3= -n1-a2+3*a1\n",
+      "b1= -n1\n",
+      "b2= -n1\n",
+      "b3= n1+3*b1-b2\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' a1 = %.f '%(a1)\n",
+      "print  ' a2 = %.f '%(a2)\n",
+      "print  ' a3 = %.f '%(a3)\n",
+      "print  ' b1 = %.f '%(b1)\n",
+      "print  ' b2 = %.f '%(b2)\n",
+      "print  ' b3 = %.f '%(b3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a1 = -1 \n",
+        " a2 = -2 \n",
+        " a3 = -2 \n",
+        " b1 = -1 \n",
+        " b2 = -1 \n",
+        " b3 = -1 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch5.ipynb b/Basic_Fluid_Mechanics/ch5.ipynb
new file mode 100755
index 00000000..e44fb093
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch5.ipynb
@@ -0,0 +1,212 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:925f1eedf50c877ed321461a7c6fe2319e03573a843bd9a8a44c0d22ceada0ec"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Control-Volume Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page No : 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "w= 20.   \t#lbm/sec plant rate\n",
+      "sh= 0.004\n",
+      "mw2= 0.12 \t#lbm/sec stream rate\n",
+      "ma3= 12.2 \t#lbm/sec rates of air\n",
+      "mw3= 0.130 \t#lbm/sec water rate\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "mw1= w/((1/sh)+1)\n",
+      "ma1= w-mw1\n",
+      "ma4= ma1-ma3\n",
+      "mw4= mw1+mw2-mw3\n",
+      "mr= ma4+mw4\n",
+      "sh1= mw4/ma4\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' mw1 = %.4f lbm/sec'%(mw1)\n",
+      "print  ' ma1 = %.2f lbm/sec'%(ma1)\n",
+      "print  ' ma4 = %.2f lbm/sec'%(ma4)\n",
+      "print  ' mw4 = %.4f lbm/sec'%(mw4)\n",
+      "print  ' mr = %.2f lbm/sec'%(mr)\n",
+      "print  ' specific humidity = %.5f lbm/sec'%(sh1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " mw1 = 0.0797 lbm/sec\n",
+        " ma1 = 19.92 lbm/sec\n",
+        " ma4 = 7.72 lbm/sec\n",
+        " mw4 = 0.0697 lbm/sec\n",
+        " mr = 7.79 lbm/sec\n",
+        " specific humidity = 0.00903 lbm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page No : 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "w= 62.4 \t#lbf/ft**3 fluid density\n",
+      "g= 32.2 \t#ft/sec**2 \n",
+      "v= 86.5 \t#ft/sec velocity\n",
+      "d2= 3.  \t#in\n",
+      "d1= 6.  \t#in\n",
+      "dp= 50. \t#lbf/in**2 gauge pressure\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Fb= ((math.pi*(w/g)*v**2*(1/d1)**2*(1-(d2/d1)**2)*0.25)-dp*144*(math.pi/4)*(1/d2)**2)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Load on the bolts = %.f lbf'%(Fb)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load on the bolts = -391 lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page No : 133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "F1= 237. \t#lb\n",
+      "dp= 50. \t#lbf/in**2\n",
+      "D= 6.    \t#in diameter\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "F2= dp*144*(math.pi/4)*(D/12)**2\n",
+      "Fb= F1-F2\n",
+      "\t\n",
+      "#RESULTS\n",
+      "\n",
+      "print  'Load on the bolts = %.f lbf'%(Fb)\n",
+      "#rounding-off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load on the bolts = -1177 lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page No : 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "w1= 0.0286 \t#lbm/ft**3 density\n",
+      "v= 2500. \t#ft/sec velocity\n",
+      "A= 2.5   \t#ft**3 area\n",
+      "k= 0.015\n",
+      "p2= 700. \t#lbf/ft**2\n",
+      "p1= 628. \t#lbf/ft**2 pressure\n",
+      "v2= 3500. \t#ft/sec outlet\n",
+      "g= 32.17 \t#ft/sec**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "ma= w1*v*A\n",
+      "mf= round(k*ma,2)\n",
+      "mt= round(ma+mf,1)\n",
+      "F= (p2-p1)*A+(mt*v2/g)-(ma*v/g)\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' air mass flow rate = %.2f lbm/sec'%(ma)\n",
+      "print  ' Fuel flow rate = %.2f lbm/sec'%(mf)\n",
+      "print  ' Fuel flow rate at station 2 = %.2f lbm/sec'%(mt)\n",
+      "print  ' Thrust force = %d lbf'%(round(F,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " air mass flow rate = 178.75 lbm/sec\n",
+        " Fuel flow rate = 2.68 lbm/sec\n",
+        " Fuel flow rate at station 2 = 181.40 lbm/sec\n",
+        " Thrust force = 6020 lbf\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch6.ipynb b/Basic_Fluid_Mechanics/ch6.ipynb
new file mode 100755
index 00000000..71c77f3e
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch6.ipynb
@@ -0,0 +1,391 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:912103c870040cf2e80e26b9450874f8f07874fb3d9db28d3d6ce1077a023edc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Steady, One-Dimensional, Reversible Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2 gravitational acceleration\n",
+      "h= 4.    \t#ft diameter\n",
+      "d2= 0.16 \t#ft\n",
+      "d1= 0.3 \t#ft\n",
+      "dp= 13.6 \t#lbf/in**2 mercury\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Q= (math.pi/4)*math.sqrt(2*g*dp*h/((1/d2**4)-(1/d1**4)))\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Volumetric flow rate = %.2f ft**3/sec'%(Q)\n",
+      "\n",
+      "# note: book answer is not accurate."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volumetric flow rate = 1.24 ft**3/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page No : 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "w= 0.0765 \t#lbm/ft**3 density\n",
+      "v1= 120. \t#ft/sec velocity\n",
+      "go = 62.4   \t#lmb/ft**3\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "dp= (w*v1**2)/(2*go)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Difference in pressure= %.2f lbf/ft**2'%(dp)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Difference in pressure= 8.83 lbf/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page No : 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "r=1.4\n",
+      "g= 32.2 \t#ft/sec**2 gas\n",
+      "R= 53.3 \t#lbf ft/lbm\n",
+      "T1= 760. \t#R  Temperature\n",
+      "p2= 2.   \t#lbf/in**2\n",
+      "p1= 3. \t    #lbf/in**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "V2= math.sqrt(2*r*R*g*T1*(1-(p2/p1)**((r-1)/r))/(r-1))\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Velocity in working section = %.f ft/sec'%(V2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity in working section = 999 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 Page No : 166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "y = 1.4\n",
+      "g = 32.2 \t#ft/sec**2\n",
+      "R = 53.3 \t#lbf ft/lbm\n",
+      "T = 32.   \t#C air\n",
+      "T1 = 2000. \t#R air\n",
+      "y1 = 1.32\n",
+      "p = 1440.     \t#lbf/in**2\n",
+      "v1 = 1.2306 \t    #ft**3/lbm\n",
+      "v2 = 1.2546 \t    #ft**3/lbm\n",
+      "bm = 3.13*10**5 \t#lbf/in**2\n",
+      "w = 62.4 \t    #lbf/ft**3\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "a1= math.sqrt(y*R*(460+T)*g)\n",
+      "a2= math.sqrt(y1*R*T1*g)\n",
+      "r2= p/(v1-v2)\n",
+      "a3= math.sqrt(-g*(v1+v2)**2*0.5**2*r2)\n",
+      "a4= math.sqrt(bm*144*g/w)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Acoustic veloctiy in air at 32 F = %.f ft/sec'%(a1)\n",
+      "print  ' Acoustic veloctiy in air at 2000 R = %.f ft/sec'%(a2)\n",
+      "print  ' Acoustic veloctiy in steam at 480 F = %.f ft/sec'%(a3)\n",
+      "print  ' Acoustic veloctiy in water at 60 F = %.f ft/sec'%(a4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Acoustic veloctiy in air at 32 F = 1087 ft/sec\n",
+        " Acoustic veloctiy in air at 2000 R = 2129 ft/sec\n",
+        " Acoustic veloctiy in steam at 480 F = 1727 ft/sec\n",
+        " Acoustic veloctiy in water at 60 F = 4823 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page No : 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "r= 1.4\n",
+      "ma2= 2.5 \t#ft/sec\n",
+      "g= 32.17 \t#ft/sec**2\n",
+      "p2= 1.   \t#lbf/in**2\n",
+      "ps= 17.08 \t#lbf/in**2\n",
+      "ps2= 75. \t#lbf/in**2\n",
+      "Ts= 720. \t#R\n",
+      "R= 53.3 \t#lbf ft/lbm gas\n",
+      "A= 4.    \t#ft**2 flow area\n",
+      "ps3= 0.4 \t#lbf/in**2\n",
+      "A2= 0.685 \t#ft**2\n",
+      "P= 5.    \t#per cent throat area\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "R1= (1+0.5*(r-1)*ma2**2)**(r/(r-1))\n",
+      "R2= (2*(r/(r-1))*(p2/ps)**(2/(r))*(1-(p2/ps)**((r-1)/r)))**0.5\n",
+      "m2= R2*ps2*144*(g/(R*Ts))**0.5*0.1\n",
+      "m= m2*A\n",
+      "At= A*R2/A2\n",
+      "m1= m*(1-(P/100))\n",
+      "mrp= (1-(P/100))*R2\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Mass flow rate= %.1f lbm/sec'%(m)\n",
+      "print  ' Area of throat= %.3f ft**2'%(At)\n",
+      "print  ' Mass flow rate= %.1f lbm/sec'%(m1)\n",
+      "print  ' Mass flow rate parameter = %.4f'%(mrp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Mass flow rate= 32.5 lbm/sec\n",
+        " Area of throat= 1.517 ft**2\n",
+        " Mass flow rate= 30.9 lbm/sec\n",
+        " Mass flow rate parameter = 0.2468\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7 Page No : 181"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "r1= 10.  \t#ft point - 1\n",
+      "r2= 0.2 \t#miles point - 2\n",
+      "w= 0.0765 \t#lbm/ft**2 density\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "V1= 1.   \t#ft/sec velocity\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "k= r2*5280*V1 \n",
+      "dp= w*k**2*10*((1/r1)**2-(1/(5280*r2))**2)/(2*g)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'k = %.f ft**2/sec'%(k)\n",
+      "print  ' pressure difference = %.1f lbf/ft**2'%(dp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "k = 1056 ft**2/sec\n",
+        " pressure difference = 132.5 lbf/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 Page No : 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "w= 12.   \t#ft wide\n",
+      "q= 300.  \t#ft**3/sec rate\n",
+      "h= 10.   \t#ft depth upstream of the gate\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "R= 2.6\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "hc= ((q/12)**2/g)**(1./3)\n",
+      "r= h/hc\n",
+      "h1= hc*(((h/hc)+0.5*(hc/h)**2)-0.5*R**2)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' hc = %.2f ft'%(hc)\n",
+      "print  ' stream depth = %.3f ft'%(h1)\n",
+      "#rounding-off error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " hc = 2.69 ft\n",
+        " stream depth = 1.013 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10 Page No : 190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "Q= 400. \t#ft**3/sec flow rate\n",
+      "b1= 25. \t#ft channel width\n",
+      "b2= 20. \t#ft channel width\n",
+      "h1= 6.   \t#ft stream depth\n",
+      "z1= 2.5 \t#ft elevation of channel bottom\n",
+      "z2= 3.3 \t#ft elevation of channel bottom\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "hc1= (Q**2/(g*b1**2))**(1./3)\n",
+      "hc2= (Q**2/(g*b2**2))**(1./3)\n",
+      "r= (hc1/hc2)*((h1/hc1)+0.5*(hc1/h1)**2)+((z1-z2)/hc2)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' hc1 = %.3f ft'%(hc1)\n",
+      "print  ' hc2 = %.3f ft'%(hc2)\n",
+      "print  ' Ratio = %.3f '%(r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " hc1 = 1.996 ft\n",
+        " hc2 = 2.316 ft\n",
+        " Ratio = 2.293 \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch7.ipynb b/Basic_Fluid_Mechanics/ch7.ipynb
new file mode 100755
index 00000000..3818de1b
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch7.ipynb
@@ -0,0 +1,523 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:fb647d6d0a9fd1fab503f1e20b9cd4c1b8f196f963f97a28e2375980e25dde26"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Steady, One-Dimensional, Irreversible Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page No : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "r= 1.5\n",
+      "f= 0.025   # friction factor\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "r1= (2/f)*(r**2-1)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'ratio L/D2 = %.f'%(r1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio L/D2 = 100\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page No : 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\t\n",
+      "#initialisation of variables\n",
+      "a= 6.    \t#degrees angle\n",
+      "r= 1.5\n",
+      "l= 100.  \t#ft\n",
+      "f= 0.025\n",
+      "K= 0.15\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "R= r**4-1\n",
+      "R1= 1/math.tan(math.radians(a/2))*(1-(1./r))\n",
+      "p1= f*l\n",
+      "p2= 2.5*(l-p1)/l\n",
+      "p3= (1-r**2)**2\n",
+      "p4= K*p3\n",
+      "pt= p4+p2\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' lowest ratio = %.2f'%(R)\n",
+      "print  ' contribtuion of friction in pipe = %.3f lbf/ft**2'%(p1)\n",
+      "print  ' contribtuion of diffuser in pipe = %.3f lbf/ft**2'%(p2)\n",
+      "print  ' stagnant pressure drop = %.3f lbf/ft**2'%(p3)\n",
+      "print  ' contribtuion of friction in pipe after reduction = %.3f lbf/ft**2'%(pt)\n",
+      "\n",
+      "# note : rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " lowest ratio = 4.06\n",
+        " contribtuion of friction in pipe = 2.500 lbf/ft**2\n",
+        " contribtuion of diffuser in pipe = 2.438 lbf/ft**2\n",
+        " stagnant pressure drop = 1.562 lbf/ft**2\n",
+        " contribtuion of friction in pipe after reduction = 2.672 lbf/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 Page No : 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "d= 4. \t    #in galvanised iron pipe diameter\n",
+      "q= 0.5 \t    #ft**3/sec  flow rate\n",
+      "w= 62.4 \t#lb/ft**3 density\n",
+      "u= 2.7*10**-5 \t#lbf sec/ft**2 viscosity\n",
+      "e= 0.0005 \t#ft\n",
+      "g= 32.1 \t#ft/sec**2 acceleration\n",
+      "f= 0.0235\n",
+      "lt= 400. \t#ft long\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "V= 4*q/(math.pi*(d/12)**2)\n",
+      "Re= w*V*(d/12)/(u*g)\n",
+      "r= e/(d/12)\n",
+      "dz= (V**2/(2*g))*(1.7+f*lt/(d/12))\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' mean flow velocity = %.2f ft/sec'%(V)\n",
+      "print  ' Reynolds number = %.2e'%(Re)\n",
+      "print  ' Relative roughness = %.4f'%(r)\n",
+      "print  ' difference in the levels of water = %.1f ft'%(dz)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " mean flow velocity = 5.73 ft/sec\n",
+        " Reynolds number = 1.38e+05\n",
+        " Relative roughness = 0.0015\n",
+        " difference in the levels of water = 15.3 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page No : 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "d= 4.    \t#in\n",
+      "v= 6.64 \t#ft/sec\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Q= math.pi*0.25*(d/12)**2*v\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Flow rate= %.3f ft**3/sec'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate= 0.579 ft**3/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 Page No : 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "d= 0.366 \t#ft\n",
+      "i= 12\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "pd= d*i\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Required pipe diameter = %.2f in'%(pd)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required pipe diameter = 4.39 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 Page No : 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "Ps1= 1050. \t#lbf/ft**2\n",
+      "fr= 10.7\n",
+      "p= 36.6 \t#lbf/ft**2\n",
+      "p1= 195. \t#lbf/ft**2\n",
+      "fr1= 16.\n",
+      "fr2= 1.8\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "deltap = (p+957+p1+Ps1)\n",
+      "p2= round(fr*p)\n",
+      "dp= Ps1-p2\n",
+      "lc= round(dp/p)\n",
+      "sp= Ps1+p1-p*(fr1+fr2)\n",
+      "lc1= sp/p\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Pressure = %.f lbf/ft**2'%(round(deltap,-1))\n",
+      "print  ' pressure difference = %.f lbf/ft**2'%(dp)\n",
+      "print  ' Loss coefficient = %.f '%(lc)\n",
+      "print  ' Loss coefficient = %.1f '%(lc1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Pressure = 2240 lbf/ft**2\n",
+        " pressure difference = 658 lbf/ft**2\n",
+        " Loss coefficient = 18 \n",
+        " Loss coefficient = 16.2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 Page No : 232"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "p1= 50. \t#lbf/in**2 pressure\n",
+      "R= 96.3 \t#ft lbf/lbm R\n",
+      "T= 80. \t    #F temperature\n",
+      "p2= 20. \t#lbf/in**2 pressure\n",
+      "r= 1.31\n",
+      "u= 2.34*10**-7 \t#lbf sec/ft**2\n",
+      "e= 0.00005 \t    #ft\n",
+      "m= 5.*10**4 \t#lbm/sec\n",
+      "d= 1.5 \t        #ft\n",
+      "g= 32.2 \t    #ft/sec**2\n",
+      "f= 0.113\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "w1= p1*144/(R*(460+T))\n",
+      "V1= 4*(m/3600)/(math.pi*w1*d**2)\n",
+      "Ma1= V1/(r*R*g*(460+T))**0.5\n",
+      "Re= w1*V1*d/(u*g)\n",
+      "dx= (((1/(r*Ma1**2))*10*(1-(p2/p1)**2))+math.log(p2/p1))*d/f\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' density = %.3f lbm/ft**3'%(w1)\n",
+      "print  ' mean flow velocity = %.1f ft/sec'%(V1)\n",
+      "print  ' Match number = %.4f '%(Ma1)\n",
+      "print  ' Reynolds number = %.2e '%(Re)\n",
+      "print  ' Length of pipe = %.2e ft'%(dx)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " density = 0.138 lbm/ft**3\n",
+        " mean flow velocity = 56.8 ft/sec\n",
+        " Match number = 0.0383 \n",
+        " Reynolds number = 1.56e+06 \n",
+        " Length of pipe = 5.79e+04 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9 Page No : 238"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "r= 1.4\n",
+      "R= 53.3 \t#ft lbf/lbm R\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "T1= 410. \t#R temperature\n",
+      "v= 2500. \t#ft/sec steadility\n",
+      "P1= 628. \t#lbf/in**2 pressure\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "v1= int(math.sqrt(r*g*R*T1))\n",
+      "Ma1= round(v/v1,2)\n",
+      "Ts1= int(T1*(1+0.5*(r-1)*Ma1**2))\n",
+      "Ps1= P1*(1+0.5*(r-1)*Ma1**2)**(r/(r-1))\n",
+      "Ps2= Ps1*((r+1)/(2*r*Ma1**2-r+1))**(1/(r-1))*(0.5*(r+1)*Ma1**2/(1+0.5*(r-1)*Ma1**2))**(r/(r-1))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' acoustic velocity = %.f ft/sec'%(v1)\n",
+      "print  ' Match number = %.2f '%(Ma1)\n",
+      "print  ' Stagnition temperature = %.f R'%(Ts1)\n",
+      "print  ' Stagnition pressure = %.f lbf/ft**2'%(Ps1)\n",
+      "print  ' Stagnition pressure = %.f lbf/ft**2'%(Ps2)\n",
+      "\n",
+      "# note : answer in book is wrong. Please check manually.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " acoustic velocity = 992 ft/sec\n",
+        " Match number = 2.52 \n",
+        " Stagnition temperature = 930 R\n",
+        " Stagnition pressure = 11069 lbf/ft**2\n",
+        " Stagnition pressure = 5435 lbf/ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10 Page No : 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#initialisation of variables\n",
+      "p2= 67.2 \t#lbf/in**2 pressure\n",
+      "p1= 63. \t#lbf/in62 pressure\n",
+      "r= 1.4\n",
+      "n= 0.6      # efficiency\n",
+      "T1= 870. \t#R temperature\n",
+      "ma1= 0.8 \t#ft/sec mach number\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "dt= (p2/p1)**((r-1)/r)-1\n",
+      "dt1= dt/n\n",
+      "T2= T1*(1+dt1)\n",
+      "Ts1= T1*(1+0.5*(r-1)*ma1**2)\n",
+      "ps1= p1*(1+0.5*(r-1)*ma1**2)**(r/(r-1))\n",
+      "ps2= p2*(Ts1/T2)**(r/(r-1))\n",
+      "dp= ps1-ps2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' dT = %.5f '%(dt)\n",
+      "print  ' dT1 = %.5f '%(dt1)\n",
+      "print  ' Temperature = %.f R'%(T2)\n",
+      "print  ' Temperature = %.1f R'%(Ts1)\n",
+      "print  ' Pressure = %.1f lbf/in**2'%(ps1)\n",
+      "print  ' Pressure = %.1f lbf/in**2'%(ps2)\n",
+      "print  ' pressure difference = %.1f lbf/in**2'%(dp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " dT = 0.01861 \n",
+        " dT1 = 0.03102 \n",
+        " Temperature = 897 R\n",
+        " Temperature = 981.4 R\n",
+        " Pressure = 96.0 lbf/in**2\n",
+        " Pressure = 92.0 lbf/in**2\n",
+        " pressure difference = 4.0 lbf/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11 Page No : 246"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "r= 1.4\n",
+      "ma3= 3. \t#ft/sec mach number\n",
+      "ps= 80. \t#lbf/ft**2 pressure\n",
+      "Ts= 840. \t#R temperature\n",
+      "r1= 53.3 \t#ft lbm/ft**3\n",
+      "A3= 2.   \t#in**2 flow area\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "ma1= 1.6\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "R= (1+(r-1)*0.5*ma3**2)**(r/(r-1))\n",
+      "p3= ps/R\n",
+      "R1= 1+0.5*(r-1)*ma3**2\n",
+      "T3= Ts/R1\n",
+      "w3= p3*144/(r1*T3)\n",
+      "V3= ma3*math.sqrt(r*r1*g*T3)\n",
+      "m= w3*V3*A3/144\n",
+      "ra= ((r+1)/(2*r*ma1**2-(r-1)))**(1/(r-1))*(0.5*(r+1)*ma1**2/(1+0.5*(r-1)*ma1**2))**(r/(r-1))\n",
+      "ps2= ps*ra\n",
+      "dp= ps-ps2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' outlet pressure = %.2f lbf/in**2'%(p3)\n",
+      "print  ' outlet temperature = %.f R'%(T3)\n",
+      "print  ' mass flow rate = %.3f lbm/sec'%(m)\n",
+      "print  ' ps2 = %.1f lbf/in**2'%(ps2)\n",
+      "print  ' Stagnation pressure loss = %.1f lbf/in**2'%(dp)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " outlet pressure = 2.18 lbf/in**2\n",
+        " outlet temperature = 300 R\n",
+        " mass flow rate = 0.694 lbm/sec\n",
+        " ps2 = 71.6 lbf/in**2\n",
+        " Stagnation pressure loss = 8.4 lbf/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch8.ipynb b/Basic_Fluid_Mechanics/ch8.ipynb
new file mode 100755
index 00000000..539707f8
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch8.ipynb
@@ -0,0 +1,148 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:64db2287da8cb97edd52d96a1a1d6b51c3497c61118b03ae2e2cf16f97e85bf8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Analysis of Two-Dimensional, Constant-Density, Laminar Flow "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page No : 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "w= 78.9 \t#lbf.ft**3 weight\n",
+      "d= 0.01 \t#in bore\n",
+      "u= 8.67*10**-9 \t#lbf/ hr ft**2 viscosity\n",
+      "h= 18.   \t#ft\n",
+      "l= 10. \t    #ft length\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "Q= math.pi*w*(d/12)**4*(h+l)/(l*128*u)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Flow rate = %.2e ft**3/hr'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate = 3.02e-04 ft**3/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 Page No : 290"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "x= 0.1       \t#ft\n",
+      "w= 62.4     \t#lbf/ft**3 fluid density\n",
+      "v1= 10.      \t#ft/sec velocity\n",
+      "u= 2.4*10**-5 \t#lbf/ft viscosity\n",
+      "g= 32.2     \t#ft/sec**2\n",
+      "k= 4.91\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "s= k*x*(w*v1*x/(u*g))**-0.5\n",
+      "Tw= 0.332*w*v1**2*(u*g/(w*x*v1))**0.5/g\n",
+      "R= 0.332*6*Tw\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  'Thickness = %.2e*ft'%(s)\n",
+      "print  ' Shear stress = %.3f lbf/ft**2'%(Tw)\n",
+      "print  ' Shear stress = %.3f lbf/ft'%(R)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness = 1.73e-03*ft\n",
+        " Shear stress = 0.226 lbf/ft**2\n",
+        " Shear stress = 0.451 lbf/ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page No : 298"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "r = 1\n",
+      "r1 = 1\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "e1= r+r1\n",
+      "e2= r-r1\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' vorticity in a forced vortex = %.f*k'%(e1)\n",
+      "print  ' vorticity in a free vortex = %.f'%(e2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " vorticity in a forced vortex = 2*k\n",
+        " vorticity in a free vortex = 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/ch9.ipynb b/Basic_Fluid_Mechanics/ch9.ipynb
new file mode 100755
index 00000000..004fdaf6
--- /dev/null
+++ b/Basic_Fluid_Mechanics/ch9.ipynb
@@ -0,0 +1,126 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:cf784c378e9383adb15dcad9ef6ad8b764187d7f185393fb16ff16d5b2a1ed70"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Analysis of Two-Dimensional, Constant-Density, Turbulent Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Page No : 324"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "n=7.\n",
+      "w= 62.4 \t#lbf/ft**3 density\n",
+      "v= 6.   \t#ft/sec velocity\n",
+      "d= 2. \t    #in pipe diameter\n",
+      "u= 2.34*10**-5 \t#lbf/ft**3 viscosity\n",
+      "f= 0.0178\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "R= 1.224\n",
+      "R1= 8.  \t#ft/sec\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "r= (n+1)*(2*n+1)/(2*n**2)\n",
+      "Red= w*v*(d/12)/(u*g)\n",
+      "C= (d/Red)**(1./7)*R*(R1/f)**(4./7)\n",
+      "V = v*math.sqrt(f/8)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' Vmax/V = %.3f'%(r)\n",
+      "print  ' Red = %.2e'%(Red)\n",
+      "print  ' C = %.2f'%(C)\n",
+      "print  ' Velocity = %.3f ft/sec'%(V)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Vmax/V = 1.224\n",
+        " Red = 8.28e+04\n",
+        " C = 8.79\n",
+        " Velocity = 0.283 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page No : 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\n",
+      "#initialisation of variables\n",
+      "Re= 5.\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "u= 2.34*10**-5 \t#lbf/ft sec\n",
+      "w= 62.4 \t#lbf/ft**3\n",
+      "v= 0.283 \t#ft/sec\n",
+      "Re1= 70.\n",
+      "v1= 0.0374 \t#ft/sec\n",
+      "\t\n",
+      "#CALCULATIONS\n",
+      "y= Re*u*g/(w*v)\n",
+      "y1= Re1*u*g/(w*v)\n",
+      "y2= Re*u*g/(w*v1)\n",
+      "y3= Re1*u*g/(w*v1)\n",
+      "\t\n",
+      "#RESULTS\n",
+      "print  ' y = %.6f ft'%(y)\n",
+      "print  ' y = %.5f ft'%(y1)\n",
+      "print  ' y = %.5f ft'%(y2)\n",
+      "print  ' y = %.4f ft'%(y3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " y = 0.000213 ft\n",
+        " y = 0.00299 ft\n",
+        " y = 0.00161 ft\n",
+        " y = 0.0226 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Basic_Fluid_Mechanics/screenshots/Streamlinedbody_curve.png b/Basic_Fluid_Mechanics/screenshots/Streamlinedbody_curve.png
new file mode 100755
index 00000000..0628f98d
--- /dev/null
+++ b/Basic_Fluid_Mechanics/screenshots/Streamlinedbody_curve.png
diff --git a/Basic_Fluid_Mechanics/screenshots/VariationsOfEpEhEp+EhwithT.png b/Basic_Fluid_Mechanics/screenshots/VariationsOfEpEhEp+EhwithT.png
new file mode 100755
index 00000000..bd0bd566
--- /dev/null
+++ b/Basic_Fluid_Mechanics/screenshots/VariationsOfEpEhEp+EhwithT.png
diff --git a/Basic_Fluid_Mechanics/screenshots/actual_perfomance_curve.png b/Basic_Fluid_Mechanics/screenshots/actual_perfomance_curve.png
new file mode 100755
index 00000000..ece48d25
--- /dev/null
+++ b/Basic_Fluid_Mechanics/screenshots/actual_perfomance_curve.png
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_1.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_1.ipynb
new file mode 100755
index 00000000..ccf39a3c
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_1.ipynb
@@ -0,0 +1,100 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1:Units & Dimensions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.1,Page no:9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "# 1 Poise = 1g/cm s = ((1/453.6)lb)/((1/30.48)ft*1s)\n",
+      "\n",
+      "#Calculation\n",
+      "be=30.48/453.6*3600  #be->british engineering unit\n",
+      "# 1 Poise = 1g/cm s = ((1/1000)kg)/((1/100)m*1s)\n",
+      "si=100/1000.0  #si->SI units\n",
+      "\n",
+      "#Result\n",
+      "print\" 1 Poise=\",round(be),\"lb/ft h\" \n",
+      "print\"\\n 1 Poise =\",si,\"N s/m**2 \""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " 1 Poise= 242.0 lb/ft h\n",
+        "\n",
+        " 1 Poise = 0.1 N s/m**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:1.2,Page no:10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "\n",
+      "# 1 kW= 103 W = 103 J/s = 10**3 * (1 kg*1 m**2)/1 s**3\n",
+      "#                       = (10**3 * (1/0.4536) lb x (1/0.3048)**2 ft**2)/1 s**3\n",
+      "\n",
+      "#Calculation\n",
+      "lfs=(10**3*(1/0.4536)*(1/.3048)**2)  #lfs->lb ft**2/s**3\n",
+      "sfs=lfs/32.2  #sfs->slug ft**2/s**3\n",
+      "hp=sfs/550.0 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n 1 kW =\",round(hp,2),\"h.p.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " 1 kW = 1.34 h.p.\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_10.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_10.ipynb
new file mode 100755
index 00000000..c72a5f6c
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_10.ipynb
@@ -0,0 +1,653 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10:Mass Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.1,Page no:580"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "x=1e-3  #Thickness of stagnant air film\n",
+      "D=1.8e-5 #Difffusivity of ammonia\n",
+      "R=8314  #Gas constant\n",
+      "T=295  #Temperature \n",
+      "P=101.3e3  #Total Pressure\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#If the subscripts 1 and 2 refer to the two sides of the stagnant layer and \n",
+      "#the subscripts A and B refer to ammonia and air respectively,\n",
+      "P_A1=.50*P \n",
+      "P_A2=0 \n",
+      "P_B1=P-P_A1 \n",
+      "P_B2=P-P_A2 \n",
+      "P_BM=(P-P_A1)/math.log(P/P_A1) \n",
+      "#Thus, substituting in equation 10.31 gives:\n",
+      "N_A=(-D/(R*T*x))*(P/P_BM)*(P_A2-P_A1) \n",
+      "\n",
+      "#Result\n",
+      "print\"The rate of diffusion of ammonia through the layer = %.2e\"%N_A,\"kmol/m**2*s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of diffusion of ammonia through the layer = 5.15e-04 kmol/m**2*s\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.2,Page no:582"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "import numpy as np\n",
+      "%pylab inline\n",
+      "th=np.array([0,0,3,7,22,32,46,55,80,106])       #Time in hours\n",
+      "tm=np.array([0,26,5,36,16,38,50,25,22,25])      #Time in min\n",
+      "tim=[0]*10\n",
+      "#Conversion to kilo seconds\n",
+      "for i in range(0,10):\n",
+      "    tm[i]=tm[i]*60.0 \n",
+      "    th[i]=th[i]*3600.0 \n",
+      "    tim[i]=(tm[i]+th[i])/1000.0 \n",
+      "\n",
+      "L=np.array([0,2.5,12.9,23.2,43.9,54.7,67.0,73.8,90.3,104.8])    #in mm\n",
+      "Lo=L[0] \n",
+      "\n",
+      "#Calculation\n",
+      "x=L-Lo \n",
+      "y=[0]*10 \n",
+      "for j in range(1,10):\n",
+      "    y[j]=tim[j]/(L[j]-Lo) \n",
+      "plot(x,y,marker='+') \n",
+      "plt.title('t/(L-L0) vs (L-L0)')\n",
+      "plt.xlabel('$(L-L0) in mm$')\n",
+      "plt.ylabel('$t/(L-L0) in ks/mm**2$') \n",
+      "\n",
+      "#Calculation of slope\n",
+      "s=(y[3]-y[2])/(x[3]-x[2])*10**3*10**6 \n",
+      "Vl=22.4                  #Molar volume in litres\n",
+      "den=1540                 #Density in kg/m**3\n",
+      "T0=273 \n",
+      "T=321 \n",
+      "vp=37.6                  #vapour pressure in kPa\n",
+      "P0=101.3                 #PRessue in kPa\n",
+      "M=154 \n",
+      "Ct=T0/(Vl*T) \n",
+      "Ca=(vp*Ct)/P0 \n",
+      "Cb1=Ct \n",
+      "Cb2=(P0-vp)*Ct/P0 \n",
+      "Cbm=(Cb1-Cb2)/math.log(Cb1/Cb2) \n",
+      "#Diffusivity calculation\n",
+      "D=den*Cbm/(2*M*Ca*Ct*s) \n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "show()\n",
+      "print\"\\nSlope is %.1e\"%s,\"s/m**2\"\n",
+      "print\"\\nDiffusivity is %.2e\"%D,\" m**2/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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Ro6FUKqGoN5+u/mZasib+pi2us0RUoVA0uA8RGd8vv4izoiorgf37AR8fqSNq25RKJZQt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\nMWkQEZHWmDSIiEhr/w+s6BLJVxn5zQAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x5bbe390>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Slope is 3.1e+07 s/m**2\n",
+        "\n",
+        "Diffusivity is 9.17e-06  m**2/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.3,Page no:585"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "P=101.3e3  #pressure of the operating column\n",
+      "T=295.0  #Temperature of the operating column\n",
+      "P_A=7.0e3  #partial pressure of ammonia\n",
+      "x=1.0e-3  #=(y1-y2)Thickness of stationary gas film\n",
+      "D=2.36e-5  #Diffusivity of ammonia\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "C_A=(1.0/22.4)*(273.0/T)*(P_A/P) #=(C_A1-C_A2)Concentration of ammonia gas\n",
+      "#X=C_T/C_BM\n",
+      "X=P*math.log(P/(P-P_A))/(P-(P-P_A)) \n",
+      "#From equation 10.33\n",
+      "N_A_=(D/x)*X*(C_A) \n",
+      "\n",
+      "#Result\n",
+      "print\"The transfer rate per unit area = %.2e\"%N_A_,\"kmol/m**2*s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The transfer rate per unit area = 6.98e-05 kmol/m**2*s\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.4,Page no:606"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Q=3e-6  #Flow rate of water\n",
+      "Meu=1e-3  #Viscosity of water\n",
+      "D=1.5e-9  #diffusivity of carbon dioxide in water\n",
+      "rho=1e3  #Density of water\n",
+      "\n",
+      "#Calculation\n",
+      "s=(Q*1e2*Meu*3/(rho*9.81))**(1.0/3.0) #Thickness of film\n",
+      "A=0.95 \n",
+      "y=s*(1-A)**0.5 #The distance below the free surface\n",
+      "t=(1.305/1.822)**2\n",
+      "Us=rho*9.81*s**2/(2*Meu) #surface velocity\n",
+      "L=Us*t #The maximum lend=gth of column\n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The maximum length of column =\",round(L,2),\" m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The maximum length of column = 0.51  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.5,Page no:608"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "N_dot=50  #Initial maas transfer rate\n",
+      "D=1.8e-9  #Diffusivity of gas in liquid phase\n",
+      "C_bg=(1/22.4)*(273.0/293.0)  #bulk gas concentration\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "N_C=N_dot*C_bg  #Initial maas transfer rate in terms of cocentration\n",
+      "h=N_C/0.04 # Effective Mass transfer coefficient\n",
+      "R=1/h #Equivalent resistance\n",
+      "R_l=R*9  #Liquid phase resistance\n",
+      "h_l=1/R_l  #Liquid phase coefficient\n",
+      "#From equation 10.113 and using liquid phase resistance\n",
+      "t=R_l**2/(math.pi/D) \n",
+      "\n",
+      "#Result\n",
+      "print\"The required time is =%.2e\"%t,\"s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The required time is =1.72e-11 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.6,Page no:608"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "import math\n",
+      "#Diffusivity of CO2 in ethanol\n",
+      "D=4*10**-9                #in m**2/sw\n",
+      "t=100                 #Time in sec\n",
+      "import sympy\n",
+      "Cai=sympy.Symbol('x') \n",
+      "\n",
+      "#Calculation\n",
+      "y=[0,10**-3] \n",
+      "\n",
+      "mole=[0]*3\n",
+      "for i in range(0,2):\n",
+      "    mole[i]=((2*math.sqrt(D*t/math.pi)*math.exp(-y[i]**2/(4*D*t)))-(y[i]*math.erfc(y[i]/(2*math.sqrt(D*t))))) \n",
+      "\n",
+      "ret=(mole[0]-mole[1])/mole[0] \n",
+      "\n",
+      "#Result\n",
+      "print\"Proportion retained is %d\"%(ret*100),\"%\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Proportion retained is 83 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.8,Page no:621"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "L=825.0e-3  #length of the tube\n",
+      "d=15.0e-3  #diameter of the tube\n",
+      "P_i=7.5e3  #Partial pressure of ammonia at inlet\n",
+      "P_o=2.0e3  #Partial pressure of ammonia at inlet\n",
+      "A_r=2.0e-5  #Air rate\n",
+      "P=101.3e3  #Atmospheric pressure\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "D_F_m=(P_i-P_o)/math.log(P_i/P_o) #Mean driving force\n",
+      "A_absorbd=A_r*((P_i/(P-P_i))-(P_o/(P-P_o))) \n",
+      "A_w=math.pi*d*L #Wetted surface\n",
+      "K_G=(A_absorbd/(A_w*D_F_m)) #Overall transfer coefficient\n",
+      "\n",
+      "#Result\n",
+      "print\"Overall Transfer coefficient =%.2e\"%K_G,\"kmol/[m**2 s (N/m**2)]\"\n",
+      "print\"NOTE:Very approx answer in book\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Overall Transfer coefficient =7.40e-09 kmol/[m**2 s (N/m**2)]\n",
+        "NOTE:Very approx answer in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.9,Page no:628"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "\n",
+      "z=0.693 \n",
+      "#Calculation\n",
+      "import math\n",
+      "ratio=(math.exp(math.sqrt(z))+math.exp(-math.sqrt(z))-4.0)/(2.0*(1-math.exp(-math.sqrt(z))-math.exp(math.sqrt(z)))) \n",
+      "\n",
+      "#Result\n",
+      "print\"The final ratio is\",round(ratio,1)\n",
+      "print\"NOTE:Wrong answer in book\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final ratio is 0.4\n",
+        "NOTE:Wrong answer in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.10,Page no:629"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#After denoting the original conditions by subscript 1,\n",
+      "#The mass transfer rate (moles/unit area and unit time) is given by\n",
+      "#N_A2=0.83*N_A1\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "n=2*(math.log(0.83/(1.35)**0.5)/math.log(0.8))-1 \n",
+      "\n",
+      "#Result\n",
+      "print\"n =\",round(n,2)\n",
+      "print\"Thus the reaction is of second order\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "n = 2.01\n",
+        "Thus the reaction is of second order\n"
+       ]
+      }
+     ],
+     "prompt_number": 51
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.11,Page no:630"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "k = 2.5*10**-6            #[s**-1] Rate constant\n",
+      "E = 2.643*10**7           #[J/kmol] Energy of Activation\n",
+      "R = 8314                 #[J/kmol.K] Universal gas contss\n",
+      "D = 10**-9                #[m**2/s] MOlecular diffuisivity\n",
+      "L = .01                  #[m] Film Thickness\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#At T =293K\n",
+      "T = 293                 #[K] temperature\n",
+      "A = k/math.exp(-E/(R*T))     #[s**-1]\n",
+      "e = math.exp(-2*L*math.sqrt(k/D)) \n",
+      "N = math.sqrt(k/D)*(1+e)/(1-e)  #Consider relative Solubility at 293 K be unity\n",
+      "#At T =313K\n",
+      "T2 = 313                 #[K] temperature\n",
+      "k2 = A*math.exp(-E/(R*T2))     #[s**-1]\n",
+      "e2 = math.exp(-2*L*math.sqrt(k2/D)) \n",
+      "N2 = .8*math.sqrt(k2/D)*(1+e2)/(1-e2)  #Consider relative Solubility at 313 K be .8 wrt that of 293K\n",
+      "\n",
+      "Nr = N2/N \n",
+      "\n",
+      "#Result\n",
+      "print'Change in mass transfer rate is given by factor',round(Nr,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Change in mass transfer rate is given by factor 0.86\n"
+       ]
+      }
+     ],
+     "prompt_number": 54
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.12,Page no:643"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "k=5e-4 #first order rate constant\n",
+      "D_e=2e-9 #effective diffusivity of reactants in the pores of the particles\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "lamda=(k/D_e)**0.5 \n",
+      "# (i) For the platelet of thickness 8 mm,\n",
+      "L=0.5*(8e-3) \n",
+      "phi=lamda*L #thiele modulus\n",
+      "#From equation 10.202, the effectiveness factor 'eta' is given by:\n",
+      "eta=(1/phi)*math.tanh(phi) \n",
+      "#(ii) For the sphere of diameter 10 mm, r_o = 0.005 m**-1.\n",
+      "r_o=5e-3 \n",
+      "phi_o=lamda*r_o #Thiele modulus\n",
+      "#From equation 10.212, the effectiveness factor 'eta' is given by:\n",
+      "eta_o=(3/phi_o)*((1.0/math.tanh(phi_o))-(1/phi_o)) \n",
+      "\n",
+      "#result\n",
+      "print\"\\n (i) Thiele modulus =\",phi \n",
+      "print\"\\n     The effectiveness factor =\",round(eta,3)\n",
+      "print\"\\n (ii) Thiele modulus =\",phi_o\n",
+      "print\"\\n     The effectiveness factor =\",round(eta_o,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " (i) Thiele modulus = 2.0\n",
+        "\n",
+        "     The effectiveness factor = 0.482\n",
+        "\n",
+        " (ii) Thiele modulus = 2.5\n",
+        "\n",
+        "     The effectiveness factor = 0.736\n"
+       ]
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.13,Page no:644"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "D_e=1e-5 #Effective diffusivity for the reactants in the catalyst particle\n",
+      "k=14.4 #first order rate constant\n",
+      "L=2.5e-3 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "lamda=(k/D_e)**0.5 \n",
+      "phi=(k/D_e)**0.5*(L) #Thiele modulus\n",
+      "#From equation 10.202, the effectiveness factor,\n",
+      "eta=(1/phi)*math.tanh(phi) \n",
+      "#The concentration profile is given by equation 10.198\n",
+      "y=1.25e-3 \n",
+      "C_Ai=0.15 \n",
+      "C_A=C_Ai*(math.cosh(lamda*y)/math.cosh(lamda*L)) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n (i) The effectiveness factor =\",round(eta,3)\n",
+      "print\"\\n (ii) The concentration of reactant at a position half-way between the centre and the outside of the\\n\\t pellet =\",round(C_A,3),\"kmol/m**3\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " (i) The effectiveness factor = 0.332\n",
+        "\n",
+        " (ii) The concentration of reactant at a position half-way between the centre and the outside of the\n",
+        "\t pellet = 0.035 kmol/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 61
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:10.14,Page no:645"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "R_r=8.2e-2 #reaction rate when concentration =0.011 kmol/m**3\n",
+      "D_e=7.5e-8 #Effective diffusivity\n",
+      "\n",
+      "#Calculation\n",
+      "#catalyst = eta*k*C_Ai (equation 10.217),\n",
+      "#eta=phi_L**-1\n",
+      "#Substituting numerical values in prev eqn:\n",
+      "k=(1.217*R_r/0.011)**2 \n",
+      "phi_L=1.217*(k)**0.5 \n",
+      "eta=phi_L**-1 \n",
+      "\n",
+      "#Result\n",
+      "print\"Effectiveness factor =\",round(eta,4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Effectiveness factor = 0.0906\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_11.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_11.ipynb
new file mode 100755
index 00000000..65d448d9
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_11.ipynb
@@ -0,0 +1,161 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11:The Boundary Layer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.1,Page no:680"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "u_s=1  #Velocity of water\n",
+      "w=0.6  #Width of plane surface\n",
+      "l=1  #Length of plane surface\n",
+      "A=0.6*1  #Area of the surface\n",
+      "Re_xc=10**5    #Reynods group\n",
+      "\n",
+      "#Calculation\n",
+      "#Taking\n",
+      "Meu=1e-3 #Viscosity of water\n",
+      "rho=1000  #Density of water\n",
+      "Re=rho*u_s*l/Meu\n",
+      "#Mean value of S/pw2 from equation 11.41\n",
+      "#X=R/(rho*u**2)\n",
+      "X=0.037*(Re)**-0.2+(Re**-1)*(0.646*(10**5)**0.5-0.037*(10**5)**0.8)\n",
+      "F=X*rho*u_s**2*A \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Total drag force =\",round(F,2),\"N (approx)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Total drag force = 1.3 N (approx)\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.2,Page no:680"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "x=150e-3  #Distance from leading edge where thicness is to be found\n",
+      "Meu_o=0.05  #viscosity of oil\n",
+      "rho_o=1000  #Density of oil\n",
+      "u=0.3  #Velocity of flow\n",
+      "\n",
+      "#Calculation\n",
+      "Re_x=x*u*rho_o/Meu_o \n",
+      "#For streamline flow:\n",
+      "#from equation 11.17\n",
+      "dell=4.64*x/Re_x**0.5 #thickness of the boundary layer\n",
+      "#from equation 11.20\n",
+      "del_star=0.375*dell \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The thickness of the boundary layer =\",dell*1e3,\"mm\"\n",
+      "print\"\\n The displacement thickness =\",del_star*1e3,\"mm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The thickness of the boundary layer = 23.2 mm\n",
+        "\n",
+        " The displacement thickness = 8.7 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:11.3,Page no:684"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "D=50e-3  #Diameter of the pipe\n",
+      "Q=2e-3  #Flow rate of benzene through pipe\n",
+      "rho_b=870  #Density of benzene\n",
+      "Meu_b=0.7e-3  #Viscosity of benzene\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "G=Q*rho_b  #Mass flow rate of benzene\n",
+      "Re=4*G/(Meu_b*math.pi*D)  #Reynolds number\n",
+      "#From equation 11.49:\n",
+      "del_b=62*D*Re**(-7.0/8.0) \n",
+      "area=math.pi/4*D**2  #Cross sectional are of pipe\n",
+      "u=G/(rho_b*area)  #mean velocity\n",
+      "#From equation 11.47:\n",
+      "u_b=2.49*u*Re**(-1.0/8.0) \n",
+      "print\"\\n The thickness of the laminar sub-layer =\",round(del_b*1e3,3),\"mm\"\n",
+      "print\"\\n The velocity of the benzene at the edge of the laminar sub-layer =\",round(u_b,3),\"m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The thickness of the laminar sub-layer = 0.195 mm\n",
+        "\n",
+        " The velocity of the benzene at the edge of the laminar sub-layer = 0.637 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_12.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_12.ipynb
new file mode 100755
index 00000000..fe313489
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_12.ipynb
@@ -0,0 +1,328 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12:Momentum,Heat and MAss Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.1,Page no:714"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=250e-3  #internal diameter of pipe\n",
+      "u=15  #Velocity of air through the pipe\n",
+      "y1=50e-3  #First point where velocity is to be found out\n",
+      "y2=5e-3  #Second point where velocity is to be found out\n",
+      "rho_air=1.10  #Density of air\n",
+      "Meu_air=20e-6  #Viscosity of air\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Re=d*u*rho_air/Meu_air \n",
+      "#Hence, from Figure 3.7: X=R/(rho*u**2)=0.0018\n",
+      "X=0.0018 \n",
+      "u_s=u/0.817 \n",
+      "u_star=u*X**0.5 \n",
+      "#At 50 mm from the wall:\n",
+      "y1_r=2*y1/d # y/r\n",
+      "#Hence, from equation 12.34:\n",
+      "u_x1=u_s+2.5*u_star*math.log(y1_r) \n",
+      "#At 5 mm from the wall:\n",
+      "y2_r=2*y2/d # y/r\n",
+      "#Hence, from equation 12.34:\n",
+      "u_x2=u_s+2.5*u_star*math.log(y2_r) \n",
+      "#The thickness of the laminar sub-layer is given by equation 12.54:\n",
+      "del_b=5*d/(Re*X**0.5) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The fluid velocity at 50 mm from the wall =\",round(u_x1,1),\"m/s\" \n",
+      "print\"\\n The fluid velocity at 5 mm from the wall =\",round(u_x2,1),\"m/s\" \n",
+      "print\"\\n The thickness of the laminar sub-layer =\",round(del_b*1e3,3),\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The fluid velocity at 50 mm from the wall = 16.9 m/s\n",
+        "\n",
+        " The fluid velocity at 5 mm from the wall = 13.2 m/s\n",
+        "\n",
+        " The thickness of the laminar sub-layer = 0.143 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.2,Page no:722"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "u=10  #Velocity of air\n",
+      "T=330  #Temperature of air\n",
+      "d=25e-3  #Inner diameter of pipe\n",
+      "T_p=415  #Temperature at which the pipe is maintained\n",
+      "DP_l=80  #Drop of static pressure along the pipe per unit length\n",
+      "\n",
+      "import math\n",
+      "#From equations 12.98 and 3.18:\n",
+      "#we get h=0.05*Cp\n",
+      "#The heat taken up per unit time by the air dQ=0.0052*Cp*dT......(i)\n",
+      "#The heat transferred through the pipe wall is also given by: = 0.039*Cp*(415-T)......(ii)\n",
+      "#Equating (i) & (ii)\n",
+      "#On integrating we get\n",
+      "T_0=415-(85/math.exp(0.45))\n",
+      "print\"\\n The required air Temperature =%d\"%T_0,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The required air Temperature =360 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.2,Page no:732"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "u=0.5     #m/s\n",
+      "l=20.0/1000.0        #m lemngth of tube\n",
+      "mu=10**-3\n",
+      "rho=1000.0     #kg/m**3\n",
+      "Sc=2330.0\n",
+      "#Calculation\n",
+      "Re=l*u*rho/mu\n",
+      "Hd=u*0.032*Re**(-1/4.0)*(1+2.0*(Sc-1)*Re**(-1/8.0))**-1\n",
+      "\n",
+      "#result\n",
+      "print\"Mass Transfer coefficient is: %.3e\"%Hd,\"kmol/m**2 s= %.3e\"%Hd,\"m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass Transfer coefficient is: 1.085e-06 kmol/m**2 s= 1.085e-06 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.3,Page no:733"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "u=3.5  #Velocity of water\n",
+      "d=25e-3  #Diameter of the pipe\n",
+      "l=6  #Length of the pipe\n",
+      "T1=300  #Temperature at enterance\n",
+      "T2=330  #Temperature at exit\n",
+      "rho=1000  #density of water at 310 K\n",
+      "Meu=0.7e-3  #Viscosity of water at 310 K\n",
+      "#Taking the fluid properties at 310 K and assuming that fully developed flow exists\n",
+      "Cp=4.18e3  #heat capapcity\n",
+      "k=0.65  #Thermal conductivity\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "Re=d*u*rho/Meu \n",
+      "Pr=Cp*Meu/k \n",
+      "import math\n",
+      "h1=0.032*(Re**-0.25)*Cp*rho*u #....Equation 12.139\n",
+      "# on solving we get final equation as\n",
+      "theta_dash1=330-10**(math.log10(30)-(0.0654*h1*1e-3/2.303)) \n",
+      "h2=0.032*(Re**-0.25)*(1+2*Re**(-1.0/8.0)*(Pr-1))**-1*Cp*rho*u \n",
+      "# on solving we get final equation as\n",
+      "theta_dash2=330-10**(math.log10(30)-(0.0654*h2*1e-3/2.303)) #....Equation 12.140\n",
+      "h3=0.032*(Re**-0.25)*(1+0.82*Re**(-1.0/8.0)*((Pr-1)+math.log(0.83*Pr+0.17)))**-1*Cp*rho*u #...equation 12.141\n",
+      "# on solving we get final equation as\n",
+      "theta_dash3=330-10**(math.log10(30)-(0.0654*h3*1e-3/2.303)) \n",
+      "h4=k/d*0.023*Re**0.8*Pr**0.33 \n",
+      "# on solving we get final equation as\n",
+      "theta_dash4=330-10**(math.log10(30)-(0.0654*h4*1e-3/2.303)) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n (a) Reynolds analogy\"\n",
+      "print\"\\n The outlet temperature =\",round(theta_dash1,1),\"K\"\n",
+      "print\"\\n\\n (b) Taylor Prandtl Equation\"\n",
+      "print\"\\n The outlet temperature =\",round(theta_dash2,1),\"K\"\n",
+      "print\"\\n\\n (c) Universal velocity profile equation\"\n",
+      "print\"\\n The outlet temperature =\",round(theta_dash3,1),\"K\"\n",
+      "print\"\\n\\n (d) Nu=0.023*Re**0.8*Pr**0.33\"\n",
+      "print\"\\n The outlet temperature = %d\"%theta_dash4,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " (a) Reynolds analogy\n",
+        "\n",
+        " The outlet temperature = 324.1 K\n",
+        "\n",
+        "\n",
+        " (b) Taylor Prandtl Equation\n",
+        "\n",
+        " The outlet temperature = 313.9 K\n",
+        "\n",
+        "\n",
+        " (c) Universal velocity profile equation\n",
+        "\n",
+        " The outlet temperature = 317.2 K\n",
+        "\n",
+        "\n",
+        " (d) Nu=0.023*Re**0.8*Pr**0.33\n",
+        "\n",
+        " The outlet temperature = 316 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:12.4,Page no:734"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "u=3.5  #Velocity of air\n",
+      "d=25e-3  #Diameter of the pipe\n",
+      "l=6  #Length of the pipe\n",
+      "T1=290  #Temperature at enterance\n",
+      "T2=350  #Temperature at exit\n",
+      "rho=29/22.4*273/310  #density of air at 310 K\n",
+      "Meu=0.018e-3  #Viscosity of air at 310 K\n",
+      "#Taking the physical properties at 310 K and assuming that fully developed flow exists\n",
+      "Cp=1.003e3  #heat capapcity\n",
+      "k=0.024  #Thermal conductivity\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Re=d*u*rho/Meu \n",
+      "Pr=Cp*Meu/k \n",
+      "h1=0.032*(Re**-0.25)*Cp*rho*u #....Equation 12.139\n",
+      "# on solving we get final equation as\n",
+      "theta_dash1=350-10**(math.log10(60)-(239.88*h1*1e-3/2.303)) \n",
+      "h2=0.032*(Re**-0.25)*(1+2*Re**(-1.0/8.0)*(Pr-1))**-1*Cp*rho*u \n",
+      "# on solving we get final equation as\n",
+      "theta_dash2=350-10**(math.log10(60)-(239.88*h2*1e-3/2.303)) #....Equation 12.140\n",
+      "h3=0.032*(Re**-0.25)*(1+0.82*Re**(-1.0/8.0)*((Pr-1)+math.log(0.83*Pr+0.17)))**-1*Cp*rho*u #...equation 12.141\n",
+      "# on solving we get final equation as\n",
+      "theta_dash3=350-10**(math.log10(60.0)-(239.88*h3*1e-3/2.303)) \n",
+      "h4=k/d*0.023*Re**0.8*Pr**0.33 \n",
+      "# on solving we get final equation as\n",
+      "theta_dash4=350-10**(math.log10(60)-(239.88*h4*1e-3/2.303)) \n",
+      "print\"\\n (a) Reynolds analogy\"\n",
+      "print\"\\n The outlet temperature =\",round(theta_dash1,1),\"K\"\n",
+      "print\"\\n\\n (b) Taylor Prandtl Equation\"\n",
+      "print\"\\n The outlet temperature =\",round(theta_dash2),\"K\"\n",
+      "print\"\\n\\n (c) Universal velocity profile equation\"\n",
+      "print\"\\n The outlet temperature = \",round(theta_dash3,1),\"K\"\n",
+      "print\"\\n\\n (d) Nu=0.023*Re**0.8*Pr**0.33\"\n",
+      "print\"\\n The outlet temperature =\",round(theta_dash4,1),\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " (a) Reynolds analogy\n",
+        "\n",
+        " The outlet temperature = 348.3 K\n",
+        "\n",
+        "\n",
+        " (b) Taylor Prandtl Equation\n",
+        "\n",
+        " The outlet temperature = 349.0 K\n",
+        "\n",
+        "\n",
+        " (c) Universal velocity profile equation\n",
+        "\n",
+        " The outlet temperature =  349.0 K\n",
+        "\n",
+        "\n",
+        " (d) Nu=0.023*Re**0.8*Pr**0.33\n",
+        "\n",
+        " The outlet temperature = 349.5 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_13.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_13.ipynb
new file mode 100755
index 00000000..12141327
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_13.ipynb
@@ -0,0 +1,720 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13:Humidification and water cooling"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.1,Page no:740"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "P=101.3e3 \n",
+      "T=297  \n",
+      "R=8314  #gas constant\n",
+      "RH=60  #Relative humidity\n",
+      "p_b1=12.2e3 #Vapor pressure at 297 K\n",
+      "p_b2=6e3  #Vapor pressure at 283 K\n",
+      "M_w=78  #molecular weight of benzene\n",
+      "M_a=28  #Mass of nitrogen \n",
+      "\n",
+      "#Calculation\n",
+      "#From the definition of percentage relative humidity (RH)\n",
+      "P_w=(p_b1)*(RH/100.0) \n",
+      "#In the benzene -nitrogen mixture:\n",
+      "m_b=P_w*M_w/(R*T) #mass of benzene\n",
+      "m_n=(P-P_w)*M_a/(R*T) #mass of nitrogen\n",
+      "H=m_b/m_n  #Humidity at 297 K\n",
+      "#In order to recover 80 per cent of the benzene, the humidity must be reduced to 20 per cent of the initial value\n",
+      "H_o=H*.20 \n",
+      "#Thus in equation 13.2\n",
+      "P_r=p_b2+(p_b2/M_a*M_w)/H_o \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The required pressure is =\",round(P_r*1e-3,1),\"kN/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The required pressure is = 391.2 kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.2,Page no:741"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "P=101.3e3  #Given pressure\n",
+      "T=300  #Given Temperature\n",
+      "RH=25  #Percentage relative humidity of water \n",
+      "P_wo=3.6e3  #partial pressure of water vapour when air is saturated with vapour\n",
+      "M_w=18  #Molecular weight of water\n",
+      "M_a=29  #Molecular weight of air\n",
+      "R=8314  #gas constant\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "P_w=P_wo*(RH/100.0) \n",
+      "m_w=P_w*M_w/(R*T) #mass of water vapour\n",
+      "m_a=(P-P_w)*M_a/(R*T) #mass of water air\n",
+      "Vs_w=1/m_w #specific volume of water vapour at 0.9 kN/m**2\n",
+      "Vs_a=1/m_a #specific volume of air at 100.4 kN/m**2\n",
+      "H=m_w/m_a #Humidity\n",
+      "H_v=Vs_a #Humid volume\n",
+      "H_p=(P-P_wo)/(P-P_w)*RH  #Percentage humidity\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) The partial pressure of the water vapour in the vessel = \",P_w*1e-3,\"kN/m**2\"\n",
+      "print\" b) Specific volume of water vapour =\",round(Vs_w),\"m**3/kg\"\n",
+      "print\"    Specific volume of air =\",round(Vs_a,3),\"m**3/kg\" \n",
+      "print\"(c) Humidity of air =\",round(H,4),\"kg water/kg air\"\n",
+      "print\"    Humid volume =\",round(H_v,3),\"m**3/kg\"\n",
+      "print\"(d) Percentage humidity =\",round(H_p,1),\"per cent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The partial pressure of the water vapour in the vessel =  0.9 kN/m**2\n",
+        " b) Specific volume of water vapour = 154.0 m**3/kg\n",
+        "    Specific volume of air = 0.857 m**3/kg\n",
+        "(c) Humidity of air = 0.0056 kg water/kg air\n",
+        "    Humid volume = 0.857 m**3/kg\n",
+        "(d) Percentage humidity = 24.3 per cent\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.3,Page no:743"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=310.0  #Temperature of moist air\n",
+      "T_w=300.0  #Wet bulb tempeature\n",
+      "L=2440e3  #Latent heat of vapourisation of water at 300 K\n",
+      "P=105e3  #Given total pressure\n",
+      "P_wo1=3.6e3  #Vapour pressure of water vapour at 300 K\n",
+      "P_wo2=6.33e3  #Vapour pressure of water vapour at 310 K\n",
+      "M_w=18.0  #Molecular weight of water\n",
+      "M_a=29.0  #Molecular weight of air\n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "H_w=(P_wo1/(P-P_wo1))*(M_w/M_a)  #The humidity of air saturated at the wet-bulb temperature\n",
+      "#Therefore, taking (h/hD*rho*A) as 1.0 kJ/kg K, in equation 13.8:\n",
+      "H=H_w-(1e3/L)*(T-T_w) \n",
+      "#In equation 13.2:\n",
+      "x=sympy.Symbol('x') \n",
+      "P_w=sympy.solve(H*(P-x)*M_a-M_w*x) \n",
+      "RH=P_w[0]/P_wo2*100.0 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The humidity of the air =\",round(H,3),\"kg/kg\"\n",
+      "print\"\\n The percentage relative humidity (RH)=\",round(RH,1),\"per cent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The humidity of the air = 0.018 kg/kg\n",
+        "\n",
+        " The percentage relative humidity (RH)= 46.6 per cent\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.4,Page no:749"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#Refer HUMIDITY ENTHALPY PLOT Figure 13.5 Page 748 as Humidity Chart\n",
+      "#According the given passes and situatuion\n",
+      "T = [325,301,308,312,315]        #[K]\n",
+      "H = [.005,.015,.022,.027,.032]    #[kg/kg]\n",
+      "#From Humidity Chart on humidifying to 60 percent humidity\n",
+      "Tw = [296,301,305,307]            #[K]\n",
+      "\n",
+      "#Calculation\n",
+      "Hin = H[4]-H[0]            #[kg/kg] Increase in Humidity\n",
+      "#From Humitidy Chart at the obtained leaving conditions\n",
+      "v = .893         #[m**3/kg] Specific Volume of dry air\n",
+      "vs = .968        #[m**3/kg] Specific Volume of Saturated air\n",
+      "vh = .937        #[m**3/kg] Humid Volume of air of 60 per cent humidity by Interpolation of Curve in Humidity Chart\n",
+      "x = 5        #[m**3/s] Amount of moist air leaves the dryer in (b)\n",
+      "m = x/vh        #[kg/s] Mass of air passing through the dryer\n",
+      "mw = m*Hin      #  [kg/s] Mass of water evaporated\n",
+      "Tb = 370.0            #[K] dry Bulb temperature corresponding to humidity of .005 kg/kg and wet-bulb temperature 307 K\n",
+      "\n",
+      "#Result\n",
+      "print'\\n\\n (a) The temperature of the material on each tray (in Kelvin)'\n",
+      "print \"\\t\",Tw,\"K\"\n",
+      "print' Thus the air leaving the system is at',T[4],' K and 60 per cent humidity.'\n",
+      "print'\\n\\n (b) If 5 m**3/s moist air leaves the dryer, The amount of water removed is',round(mw,3),' kg/s.'\n",
+      "print'\\n\\n (c) The Temperature to which the inlet air would have to be raised\\n     to carry out the drying in single stage is',Tb,'K.'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " (a) The temperature of the material on each tray (in Kelvin)\n",
+        "\t[296, 301, 305, 307] K\n",
+        " Thus the air leaving the system is at 315  K and 60 per cent humidity.\n",
+        "\n",
+        "\n",
+        " (b) If 5 m**3/s moist air leaves the dryer, The amount of water removed is 0.144  kg/s.\n",
+        "\n",
+        "\n",
+        " (c) The Temperature to which the inlet air would have to be raised\n",
+        "     to carry out the drying in single stage is 370.0 K.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.5,Page no: 753"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "G1=1  #flow rate of air at 350 K\n",
+      "PH1=10  #Percentage Humidity at 350 K\n",
+      "G2=5  #flow rate of air at 300 K\n",
+      "PH2=30  #Percentage Humidity at 300 K\n",
+      "#from fig 13.4\n",
+      "H1=0.043  #Humidity at 350 K and 10 percent humidity\n",
+      "H2=0.0065  #Humidity at 300 K and 30 percent humidity\n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "#Thus, in equation 13.23:\n",
+      "H=((G1*H1)+(G2*H2))/(G1+G2) \n",
+      "#from fig 13.5\n",
+      "H_1=192e3 #Entahlpy at 350 K and 10 percent humidity\n",
+      "H_2=42e3 #Enthalpy at 300 K and 30 percent humidity\n",
+      "x=sympy.Symbol('x') \n",
+      "H_=sympy.solve((G1*(x-H_1))-(G2*(H_2-x))) \n",
+      "#From Figure 13.5:\n",
+      "#at H_ (Enthalpy)= 67 kJ/kg and H(humidity) = 0.0125 kg/kg\n",
+      "T=309 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Humidity of final stream =\",round(H,4),\"kg/kg\" \n",
+      "print\"\\n Entahlpy of the resultant stream =\",round(H_[0]*1e-3),\"kJ/kg\"\n",
+      "print\"\\n Temperature of the resultant stream =\",T,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Humidity of final stream = 0.0126 kg/kg\n",
+        "\n",
+        " Entahlpy of the resultant stream = 67.0 kJ/kg\n",
+        "\n",
+        " Temperature of the resultant stream = 309 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.6,Page no:755"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "G_s=0.15  #Mass flow rate of steam\n",
+      "T=400  #Temperature to which steam is superheated\n",
+      "T_a=320 #Tremperature of air \n",
+      "RH_a=20  #Percentage relative humidity of air\n",
+      "G_a=5 #Mass flow rate of air\n",
+      "L=2258e3  #latent heat of steam\n",
+      "Cp=2e3  #Specific heat of superheated steam\n",
+      "\n",
+      "#Calculation\n",
+      "#Enthalpy of steam\n",
+      "H_3=4.18*(373-273)+L+Cp*(T-373) \n",
+      "#From Figure .13.5:\n",
+      "#at T=320 K and 20 percent Relative Humidity\n",
+      "H1=0.013 #Humidity\n",
+      "H_1=83e3 #Enthalpy\n",
+      "#By making required constructions we get\n",
+      "H=0.043 \n",
+      "H_=165e3 \n",
+      "T_s=324 \n",
+      "#from chart and equation 13.28\n",
+      "G_case2=0.41 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Relative humidity of stream=\",round(H,3),\"kg/kg\"\n",
+      "print\" Entahlpy of stream =\",H_*1e-3,\" kJ/kg\"\n",
+      "print\" Temperature of stream =\",T_s,\"K\"\n",
+      "print\"\\n\\n When exit temperature = 330 K\"\n",
+      "print\" The required flow of steam = \",round(G_case2,2),\" kg/s\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Relative humidity of stream= 0.043 kg/kg\n",
+        " Entahlpy of stream = 165.0  kJ/kg\n",
+        " Temperature of stream = 324 K\n",
+        "\n",
+        "\n",
+        " When exit temperature = 330 K\n",
+        " The required flow of steam =  0.41  kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.7,Page no:761"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declartaion\n",
+      "theta1=300                  #K\n",
+      "theta2=320                  #K\n",
+      "H=0.2                       #Percent humidity\n",
+      "\n",
+      "#Calculation\n",
+      "#From standard charts(Fig 13.4):\n",
+      "H1=0.0045          #Humidity in kg/kg\n",
+      "H2=0.0140          #Humidity in kg/kg\n",
+      "\n",
+      "#Heat the air from 300 to 318 K\n",
+      "theta=294.5     #K\n",
+      "#H=0.0140   kg/kg\n",
+      "\n",
+      "#Result\n",
+      "print\"Heat the saturated air at H=\",H2,\"kg/kg from\",theta,\"to\",theta2,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat the saturated air at H= 0.014 kg/kg from 294.5 to 320 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no 13.8,Page no:766"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T_water_enter=301.0            #Water entering the tower in [K]\n",
+      "T_water_leave=294.0            #Water entering the tower in [K]\n",
+      "air_drybulb=287.0              #[K]\n",
+      "air_wetbulb=284.0              #[K]\n",
+      "m=4810.0                       #[kg/s] Water flow rate\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "delta_T=T_water_enter-T_water_leave       #Temperatue range of water\n",
+      "T_mean=0.5*(T_water_enter+T_water_leave)  #MEan wate temperature in [K]\n",
+      "H_mean=92.6     #kJ/kg\n",
+      "H_drybulb=49.5  #kJ/kg\n",
+      "delTdash=T_mean-air_drybulb\n",
+      "delHdash=H_mean-H_drybulb\n",
+      "import math\n",
+      "Dt=m/(0.00369*(delHdash/delta_T)*math.sqrt(delTdash+(0.0752*delHdash)))\n",
+      "from scipy.optimize import fsolve\n",
+      "Ct=5.0\n",
+      "h=100.0    #height assumed in [m]\n",
+      "Ab=Dt/(19.50*math.sqrt(h)/(Ct**1.5))\n",
+      "Id=math.sqrt(round(Ab)*4/math.pi)\n",
+      "\n",
+      "#Result\n",
+      "print \"Thus the internal diameter of the column at sill level,\",round(Id,1),\"m\"\n",
+      "print\"Since this gives a height-dia ratio=3:2 the design is aceptable\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thus the internal diameter of the column at sill level, 64.6 m\n",
+        "Since this gives a height-dia ratio=3:2 the design is aceptable\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.9,Page no:775"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ti=293         #Inlet temperature in [K]\n",
+      "L=2495         #Latent heat of water in kJ/kg\n",
+      "flow=0.68      #[m**3/m**2 s] flow of air \n",
+      "water_thru=0.26  #Water througput in [kg/m**2 s]\n",
+      "Cp_air=1.003   #Specific heat of air in kJ/kg K\n",
+      "Cp_wv=2.006  #Sp heat of water vapour\n",
+      "wv_in=0.003     #Water vapour in inlet air[kg/kg dry air]\n",
+      "wv_in_m=0.005    #kmol/kmol dry air\n",
+      "H=0.003\n",
+      "hd_a=0.2    #Resistance whole\n",
+      "\n",
+      "#Calculation\n",
+      "Hg1=Cp_air*(Ti-273.0)+H*(L+Cp_wv*(Ti-273.0))  #;kJ/kg]\n",
+      "flow=(1.0-wv_in_m)*flow\n",
+      "rho=(29.0/22.4)*(273.0/Ti)      #Density of air at 293 K[kg/m**3]\n",
+      "G_dash=rho*flow   #[kg/m**s]\n",
+      "slope=water_thru*4.18/G_dash\n",
+      "theta_l1=293  \n",
+      "Hg1=27.67     #kJ/kg\n",
+      "import numpy as np\n",
+      "Hg=np.array([27.7,30,40,50,60,70,76.5])\n",
+      "theta=np.array([293,294.5,302,309,316,323,328])\n",
+      "Hf=np.array([57.7,65,98,137,190,265,355])\n",
+      "Hf_Hg=Hf-Hg\n",
+      "one_Hf_Hg=1.0/(Hf_Hg)\n",
+      "%pylab inline\n",
+      "plot(Hg,one_Hf_Hg)\n",
+      "ylabel(\"1/(Hf-Hg)\")\n",
+      "xlabel(\"Hg(kJ/kg)\")\n",
+      "title(\"Evaluation of the integral of dHg/(Hf-Hg)\")\n",
+      "from scipy.integrate import simps, trapz\n",
+      "area = simps(Hg,one_Hf_Hg,dx=10)\n",
+      "area=0.65\n",
+      "#At bottom of the column:\n",
+      "delta_H1=Hf[0]-Hg[0]\n",
+      "#At the top of the column:\n",
+      "delta_H2=Hf[6]-Hg[6]\n",
+      "m_water_temp=0.5*(theta[0]+theta[6])\n",
+      "Hgm=52\n",
+      "Hfm=145\n",
+      "del_Hm=93\n",
+      "f=0.79\n",
+      "intg=(Hg[6]-Hg[0])/(f*del_Hm)\n",
+      "z=intg*G_dash/(hd_a*rho)\n",
+      "show()\n",
+      "\n",
+      "#Result\n",
+      "print \"Height of packing,z=\",round(z,1),\"m\"\n"
+     ],
+     "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: ['f']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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O2YZJT8Qsfv1V2+n3p5/Ay8vc0QghaqJOeyJRUVEMHz6cgoICnn32WWJiYli+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\nbMkkIoSlsrHRpgKPHav1Rub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ptWvXLhUQEKC6d++uhgwZovLy8iyqfQUFBapNmzbq/Pnz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\nQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7975048>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height of packing,z= 2.2 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:13.10,Page no:782"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "L_dash=0.25     #n-butanol flow rate\n",
+      "Mw=74.0         #Molecular wt of butanol\n",
+      "Ma=29.0         #Molecular wt of air\n",
+      "hDa=0.1         #Mass transfer  coefficient per unit volume\n",
+      "b=2.34          #Psychometric ratio\n",
+      "lamda=590       #kJ/kg\n",
+      "Cl=2.5          #kJ/kg K\n",
+      "s=1.05          #Humid heat of gas in [kJ/kg K]\n",
+      "Cp_dry=1.001    #Sp heat of dry air\n",
+      "G=0.7            #Vaour rate \n",
+      "T=[295,300,305,310,315,320,325,330,335,340,345,350]\n",
+      "Pwo=[0.59,0.86,1.27,1.75,2.48,3.32,4.49,5.99,7.89,10.36,14.97,17.5]   #kN/m**2\n",
+      "Ho=[0]*12\n",
+      "Hf=[0]*12\n",
+      "Hf_dash=[0]*12\n",
+      "\n",
+      "#Calculation\n",
+      "lamda_dash=lamda/b\n",
+      "print\"______________________________\"\n",
+      "print\"T\\tHo\\tHf\\tHf`\"\n",
+      "print\"_______________________________\"\n",
+      "for i in range(0,12):\n",
+      "    Ho[i]=Pwo[i]/(101.3-Pwo[i])*(74/29.0)\n",
+      "    Hf[i]=(1/(1+Ho[i]))*Cp_dry*(T[i]-273.0)+Ho[i]*(Cl*(T[i]-273.0)+lamda)\n",
+      "    Hf_dash[i]=Cp_dry*(T[i]-273.27)/(1+Ho[i])+Ho[i]*(Cl*(T[i]-273.0)+lamda_dash)\n",
+      "    print T[i],\"\\t\",round(Ho[i],4),\"\\t\",round(Hf[i],2),\"\\t\",round(Hf_dash[i],2)\n",
+      "%pylab inline\n",
+      "plot(T,Hf)\n",
+      "plot(T,Hf_dash)\n",
+      "xlim(285,340)\n",
+      "ylim(0,200)\n",
+      "theta_l1=T[0]\n",
+      "Hg1=s*(290.0-273.0)\n",
+      "Hg1=round(Hg1,1)\n",
+      "rho_air=(Ma/22.4)*(273/310.0)\n",
+      "G_dash=G*rho_air\n",
+      "slope=L_dash*Cl/G_dash\n",
+      "theta_l2=T[7]\n",
+      "Hg2=46     #kJ/kg\n",
+      "#y=mx+c\n",
+      "c=Hg2-slope*theta_l2\n",
+      "plot([theta_l1,theta_l2],[Hg1,Hg2])\n",
+      "xlabel(\"Temperature(K)\")\n",
+      "ylabel(\"Enthalpy(kJ/kg)\")\n",
+      "title(\"Graphical construction\")\n",
+      "Hg1_dash=(Hg1+(b-1)*s*(290.0-273.0))/b       #[kJ/kg]\n",
+      "slp=-3*s\n",
+      "theta_f1=293.0\n",
+      "import numpy as np\n",
+      "Hf_dash1=np.array([23.9,29.0,35.3,42.1,50.0,57.9,66.7,75.8])\n",
+      "Hg_dash=np.array([17.9,22.0,26.0,30.0,34.0,38.1,42.0,46.0])\n",
+      "Hfd_Hg=Hf_dash1-Hg_dash\n",
+      "HfHG=1/(Hfd_Hg)\n",
+      "mean=[0]*7\n",
+      "interval=[0]*7\n",
+      "value=[0]*7\n",
+      "summ=0\n",
+      "show()\n",
+      "print\"____________________________________________________________________\"\n",
+      "print\"Mean value in interval\\t\\tInterval\\tValue of integral\"\n",
+      "print\"____________________________________________________________________\"\n",
+      "for i in range(0,7):\n",
+      "    mean[i]=(HfHG[i]+HfHG[i+1])/2.0\n",
+      "    interval[i]=Hg_dash[i+1]-Hg_dash[i]\n",
+      "    value[i]=mean[i]*interval[i]\n",
+      "    summ=summ+value[i]\n",
+      "    print round(mean[i],3),\"\\t\\t\\t\\t\",interval[i],\"\\t\\t\\t\",round(value[i],3),\"\\t\"\n",
+      "z=G_dash*summ/(b*hDa)\n",
+      "print\"Value of integral=\",round(summ,3)\n",
+      "#Result\n",
+      "print \"\\n\\n\\nHeight=\",round(z,1),\"m\"\n",
+      "print\"The final point is given by theta=\",308,\"K\"\n",
+      "print\"NOTE:For 2nd part,calculation can't be done on python,it is drawn geometrically\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "______________________________\n",
+        "T\tHo\tHf\tHf`\n",
+        "_______________________________\n",
+        "295 \t0.0149 \t31.34 \t26.02\n",
+        "300 \t0.0218 \t40.81 \t33.17\n",
+        "305 \t0.0324 \t52.73 \t41.53\n",
+        "310 \t0.0449 \t66.06 \t50.65\n",
+        "315 \t0.064 \t84.02 \t62.13\n",
+        "320 \t0.0865 \t104.48 \t75.01\n",
+        "325 \t0.1183 \t131.75 \t91.53\n",
+        "330 \t0.1604 \t166.64 \t112.23\n",
+        "335 \t0.2155 \t211.63 \t138.59\n",
+        "340 \t0.2907 \t272.16 \t173.74\n",
+        "345 \t0.4425 \t390.67 \t240.99\n",
+        "350 \t0.5329 \t467.26 \t287.04\n",
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['mean']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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gqFVL74qeX4ohhYDAAPqu7cuSTkv4z+v/kUX9climv92BAweyb98+Fi1aRGho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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x5797cf8>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "____________________________________________________________________\n",
+        "Mean value in interval\t\tInterval\tValue of integral\n",
+        "____________________________________________________________________\n",
+        "0.155 \t\t\t\t4.1 \t\t\t0.635 \t\n",
+        "0.125 \t\t\t\t4.0 \t\t\t0.501 \t\n",
+        "0.095 \t\t\t\t4.0 \t\t\t0.38 \t\n",
+        "0.073 \t\t\t\t4.0 \t\t\t0.29 \t\n",
+        "0.057 \t\t\t\t4.1 \t\t\t0.232 \t\n",
+        "0.045 \t\t\t\t3.9 \t\t\t0.177 \t\n",
+        "0.037 \t\t\t\t4.0 \t\t\t0.148 \t\n",
+        "Value of integral= 2.363\n",
+        "\n",
+        "\n",
+        "\n",
+        "Height= 8.1 m\n",
+        "The final point is given by theta= 308 K\n",
+        "NOTE:For 2nd part,calculation can't be done on python,it is drawn geometrically\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_2.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_2.ipynb
new file mode 100755
index 00000000..a6f061ca
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_2.ipynb
@@ -0,0 +1,244 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2:Flow of Fluids \u2014Energy and Momentum Relationships"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:2.1,Page no:36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#(a) PV = 1 * RT, where \n",
+      "R=8314.0 \n",
+      "P=60.0*10**6 \n",
+      "T=320.0 \n",
+      "Tc=191.0 \n",
+      "Pc=4.64*10**6 \n",
+      "\n",
+      "#Calculation\n",
+      "V1=8314.0*T/P \n",
+      "#(b) In van der Waals equation (2.32), the constants may be taken as:\n",
+      "a=27*R**2*Tc**2.0/(64.0*Pc) \n",
+      "b=R*Tc/(8.0*Pc) \n",
+      "#Thus using equation 2.32:\n",
+      "import sympy\n",
+      "x=Symbol('x') \n",
+      "p=sympy.solve((60*10**6*x**2+a)*(x-0.0427)-(8314*320*x**2)) \n",
+      "#(c) Tr=T/Tc  ,Pr=P/Pc\n",
+      "Tr=T/Tc \n",
+      "Pr=P/Pc \n",
+      "#Thus from Figure 2.1, \n",
+      "Z=1.33 \n",
+      "#V = ZnRT/P (from equation 2.31)\n",
+      "V3=Z*R*T/P \n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Volume of vessel (ideal gas law) =\",round(V1,4),\"m**3\"\n",
+      "print\"(b) Volume of vessel(van der waals eq.) =\",round(p[0],3),\"m**3\"\n",
+      "print\"(c) Volume of vessel(generalised compressibility-factor chart) =\",round(V3,4),\"m**3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Volume of vessel (ideal gas law) = 0.0443 m**3\n",
+        "(b) Volume of vessel(van der waals eq.) = 0.066 m**3\n",
+        "(c) Volume of vessel(generalised compressibility-factor chart) = 0.059 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:2.3,Page no:43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#Mass rate of discharge of water, G = rho*u*A\n",
+      "rho=1000.0  #Density of Water\n",
+      "d=25.0*10**-3  #Diameter of nozzle\n",
+      "u=25.0  #Velocity of water at nozzle\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "G=rho*u*math.pi/4*d**2 \n",
+      "#Momentum of fluid per second = Gu\n",
+      "F=G*25 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Reaction force = Rate of change of momentum =\",round(F),\"N\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Reaction force = Rate of change of momentum = 307.0 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:2.4,Page no:43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#Momentum per second of approaching liquid in Y-direction = rho*u**2*A\n",
+      "rho=1000.0  #Density of water\n",
+      "d=50.0*10**-3  #Diameter of pipe\n",
+      "u=5.0  #Velocity of water in pipe\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "M=rho*u**2*math.pi/4*d**2 \n",
+      "Rf=M*(math.cos(math.pi/4)+math.sin(math.pi/4)) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The resultant force in direction of arm of bracket =\",round(Rf,1),\"N\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The resultant force in direction of arm of bracket = 69.4 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:2.5,Page no:48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#From equation 2.68:\n",
+      "# 0.5*((u2)**2-(u1)**2)=g*(z1-z2)+((P1-P2)/rho)\n",
+      "#Suffix 1 to denote conditions in the pipe and suffix 2 to denote conditions in the jet\n",
+      "#Symbols have their usual meaning\n",
+      "u1=0.0 \n",
+      "z1=0.0 \n",
+      "z2=0.0 \n",
+      "P1=250.0*10**3 \n",
+      "P2=0.0 \n",
+      "rho=1000.0 #Density of water\n",
+      "g=9.81 \n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "x=Symbol('x') \n",
+      "u2=solve((0.5*(x)**2)-((P1-P2)/rho)) \n",
+      "print\"Velocity of the jet, u2 =\",round(u2[1],1),\"m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of the jet, u2 = 22.4 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:2.6,Page no:54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "id=0.5  #internal diameter of pipe\n",
+      "rs=50  #revolution speed\n",
+      "ir=0.15  #internal radius of water\n",
+      "rho=1000  #density of water\n",
+      "\n",
+      "#Calculation\n",
+      "omega=2*math.pi*rs \n",
+      "#The wall pressure is given by equation 2.82 as:\n",
+      "wall_pressure=rho*(omega)**2/2*((id/2)**2-ir**2) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The wall pressure is %.2e\"%wall_pressure,\"N/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The wall pressure is 1.97e+06 N/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_3.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_3.ipynb
new file mode 100755
index 00000000..ddcbac8f
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_3.ipynb
@@ -0,0 +1,578 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3:FLOW OF LIQUIDS IN PIPES AND OPEN CHANNELS"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.1,Page no:70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "sap=1.25  #Sulphuric acid pumped\n",
+      "d=25e-3  #Diameter of pipe\n",
+      "l=30  #length of pipe\n",
+      "meu=25e-3  #Viscosity of acid\n",
+      "rho_a=1840  #Density of acid\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Re=4*sap/(math.pi*meu*d) \n",
+      "u=sap/(rho_a*math.pi/4*d**2) \n",
+      "#calculating pressure drop from the energy balance equation and equation 3.19\n",
+      "Dp=rho_a*((0.5+4*0.006*30/0.025)*u**2+9.81*12) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Pressure drop =\",round((Dp/10**3)),\"kN/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Pressure drop = 320.0 kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 110
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.2,Page no:70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=50e-3  #Diameter of pipe\n",
+      "l=100.0  #length of pipe\n",
+      "e=0.013  #Roughness of pipe\n",
+      "DPf=50e3  #Maximum pressure drop\n",
+      "rho=1000.0  #density of water\n",
+      "meu=1e-3  #viscosity of water\n",
+      "\n",
+      "#Calculatiom\n",
+      "phi_re2=(DPf)*d**3*rho/(4*l*meu**2) \n",
+      "e_d=e/(d*1e3) \n",
+      "Re=7.9e4 \n",
+      "u=Re*meu/(rho*d) \n",
+      "\n",
+      "#Result\n",
+      "print\"The maximum allowable velocity is =\",round(u,1),\"m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum allowable velocity is = 1.6 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 111
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.3,Page no:71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Dia_tank=5  #Diameter of the tank\n",
+      "len_pipe=100.0  #Length of pipe\n",
+      "dia_pipe=225e-3  #Diameter of pipe\n",
+      "\n",
+      "#Calculation\n",
+      "X=.0020 \n",
+      "def f(D):\n",
+      "    return(111.5*(1+(3552*X))**0.5*D**-0.5)\n",
+      "from scipy.integrate import quad\n",
+      "t=quad(f,0.3,3) \n",
+      "\n",
+      "print\"The time taken for the level to fall is therefore about %d\"%t[0],\"s or %.1f\"%(t[0]/60.0),\"min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The time taken for the level to fall is therefore about 751 s or 12.5 min\n"
+       ]
+      }
+     ],
+     "prompt_number": 112
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.4,Page no:72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d1=0.3  #diameter of pipe from junction A to D or B to D\n",
+      "l1=1.5e3  #length of pipe from junction A to D or B to D\n",
+      "d2=0.5  # diameter of pipe from junction D to C\n",
+      "l2=0.75e3  # length of pipe from junction D to C\n",
+      "h_A=10  # height of tank A above C\n",
+      "h_B=h_A+6  # height of tank A above C\n",
+      "rho=870  # density of liquid\n",
+      "Meu_l=0.7e-3  # viscosity of liquid\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "import sympy\n",
+      "import numpy as np\n",
+      "x=Symbol('x') \n",
+      "u2=sympy.solve(x**4-(7.38*x**2)+13.57) \n",
+      "u1=(np.square(u2)-1.47)**0.5 \n",
+      "u3=(u1+u2)/2.78 \n",
+      "#taking the positive values and which satisfy equation 7\n",
+      "U1=u1[2] \n",
+      "U2=u2[2]\n",
+      "U3=u3[2]\n",
+      "Q=math.pi/4*d2**2*U3 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The volumetric flow rate =%.2f\"%Q,\"m**3/s\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The volumetric flow rate =0.23 m**3/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 113
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.5,Page no:86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "r=50.0 \n",
+      "\n",
+      "#Calculation\n",
+      "from scipy.optimize import fsolve\n",
+      "a=Symbol('a') \n",
+      "def f(a):\n",
+      "    return(105*a**(8.0/7.0)-56*a**(15.0/7.0)-24.5)\n",
+      "p=fsolve(f,0.5)\n",
+      "y=p[0]*r \n",
+      "\n",
+      "#Result\n",
+      "print\"a =\",round(p[0],2)\n",
+      "print\"y =\",round(y,1),\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = 0.33\n",
+        "y = 16.6 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 114
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.6,Page no:88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Q=7.2 #Water flow rate\n",
+      "d1=40e-3  #initial pipe diameter\n",
+      "d2=50e-3  #diameter of pipe after enlargement\n",
+      "g=9.81 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "u1=(Q/3600.0)/(math.pi/4*d1**2) #Velocity in 40 mm pipe\n",
+      "u2=(Q/3600.0)/(math.pi/4*d2**2) #Velocity in 50 mm pipe\n",
+      "# The head lost is given by equation 3.77 as:\n",
+      "hf=(u1-u2)**2.0/(2.0*g) \n",
+      "\n",
+      "#Result\n",
+      "print\"Head lost =\",round(hf*1e3,1),\"mm of water\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Head lost = 16.7 mm of water\n"
+       ]
+      }
+     ],
+     "prompt_number": 115
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.7,Page no:92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Q_h=2.27  # flow rate of water in m**3/h\n",
+      "T=320  #Temperature of water to be pumped\n",
+      "id=40e-3  #internal diameter of pipe\n",
+      "l_h=150.0  #length of pipe horizontally\n",
+      "l_v=10.0  #length of pipe vertically\n",
+      "e=0.2e-3 \n",
+      "g=9.81 \n",
+      "rho=1000.0 \n",
+      "\n",
+      "#Calculation\n",
+      "rel_rough=e/id  #Relative roughness\n",
+      "meu=0.65e-3  #Viscosity at 320 K\n",
+      "Q_s=Q_h/3600.0  #flow rate of water in m**3/s\n",
+      "area=math.pi/4*id**2  # Area for flow\n",
+      "u=Q_s/area  #Velocity\n",
+      "Re=(id*u*rho)/meu \n",
+      "#X=R/(rho*u**2)=0.004 (from Figure 3.7)\n",
+      "X=.004 \n",
+      "equi_len=l_h+l_v+(260*id)  # Equivalent length of pipe\n",
+      "hf=4*X*equi_len*u**2/(id*g) \n",
+      "tot_head=hf+1.5+10  # Total head to be developed\n",
+      "mass_thr=Q_s*rho  #Mass throughput\n",
+      "power_reqd=(mass_thr*tot_head*g)/0.60 \n",
+      "\n",
+      "#Result\n",
+      "print\"Power required =%.1f\"%power_reqd,\"W =%.3f\"%(power_reqd*1e-3),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power required =136.6 W =0.137 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 116
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.8,Page no:92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=0.15  # diameter of pipe\n",
+      "g=9.81 \n",
+      "phi=.0045 \n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "import math\n",
+      "x=Symbol('x') \n",
+      "u=sympy.solve((7.6+4*phi*(105/.15))*x**2/(2*g)-10) \n",
+      "rate_dis=u[1]*math.pi*d**2/4 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Rate of discharge =%.d\"%(rate_dis*1e3),\"kg/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Rate of discharge =55 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 117
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.9,Page no:103"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "u1=1.5  # velocity\n",
+      "D1=75e-3  #depth\n",
+      "g=9.81 \n",
+      "\n",
+      "#Calculation\n",
+      "#The depth of fluid in the channel after the jump is given by:\n",
+      "D2=0.5*(-D1+(D1**2+(8*u1**2*D1/g)**0.5))  #equation 3.113\n",
+      "#If the channel is of uniform cross-sectional area, then:\n",
+      "u2=u1*D1/D2 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The velocity of fluid in the channel after the jump is =\",round(u2,2),\"m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The velocity of fluid in the channel after the jump is = 0.75 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 118
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.10,Page no:127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "k=10 \n",
+      "n=0.2 \n",
+      "Ucl=1  # centre line velocity\n",
+      "l=200  # length of pipe\n",
+      "r=.02  # radius of pipe\n",
+      "dux_dy_1=10 \n",
+      "dux_dy_2=50 \n",
+      "Ry_1=k*dux_dy_1**0.2 \n",
+      "Ry_2=k*dux_dy_2**0.2 \n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "#Using the Bingham-plastic model (equation 3.125):\n",
+      "import numpy as np\n",
+      "A=np.array([[1,10],[1,50]])\n",
+      "B=np.array([15.85,21.87])\n",
+      "C=np.linalg.inv(A)*B \n",
+      "Ry=C[0] \n",
+      "Meu_p=C[1] \n",
+      "# Using Equation 3.131\n",
+      "DP=2*k*l*Ucl**n*((n+1)/n)**n*r**(-n-1) \n",
+      "# For a Bingham-plastic fluid:\n",
+      "# The centre line velocity is given by equation 3.145:\n",
+      "X=(l*2*Ry)/(r*DP) \n",
+      "Up=(DP*r**2*(2-4*X+2*X**2))/(8*Meu_p*l) \n",
+      "\n",
+      "#Result\n",
+      "print\"Plastic viscosity (Meu_p) =\",C[1][0]+C[1][1],\"N s/m**2\" \n",
+      "print\"\\nYeild stress (Ry) =\",C[0][0]+C[0][1],\"N s/m**2\"\n",
+      "print\"\\nPressure drop (Bingham plastic model)=%.2e\"%DP,\"N/m**2\"\n",
+      "print\"\\ncentre line velocity (Bingham plastic model) = \",round(Up[0]+Up[1],2),\"m/s (APPROX taken in book)\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Plastic viscosity (Meu_p) = 0.1505 N s/m**2\n",
+        "\n",
+        "Yeild stress (Ry) = 14.345 N s/m**2\n",
+        "\n",
+        "Pressure drop (Bingham plastic model)=6.26e+05 N/m**2\n",
+        "\n",
+        "centre line velocity (Bingham plastic model) =  0.68 m/s (APPROX taken in book)\n"
+       ]
+      }
+     ],
+     "prompt_number": 119
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:3.11,Page no:129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Meu=0.1  # Viscosity of liquid\n",
+      "d=25e-3  # Diameter of pipe\n",
+      "l=20.0  # length of pipe\n",
+      "DP=1e5  # Pressure drop\n",
+      "n=1/3.0  # flow index of polymer solution\n",
+      "dux_dy=1000.0 \n",
+      "k=Meu \n",
+      "\n",
+      "#Calculation\n",
+      "Meu_a=Meu \n",
+      "k_poly_sol=Meu_a/(dux_dy)**(n-1) \n",
+      "Ry=10.0*(dux_dy)**n \n",
+      "#From equation 3.136:\n",
+      "#For a power-law fluid:\n",
+      "u2=((DP/(4*k_poly_sol*l))**3)*(n*(d**((n+1)/n)))/(2*(3*n+1)) \n",
+      "u1=(DP/(4*k*l))*(d**2)/8.0\n",
+      "ratio=u2/u1 \n",
+      "\n",
+      "#Result\n",
+      "print\"Ratio of the volumetric flow rates of the two liquids =\",round(ratio,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio of the volumetric flow rates of the two liquids = 0.065\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Examople no:3.12,Page no:129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n=0.5\n",
+      "du_dy=0.01\n",
+      "import sympy\n",
+      "mu=sympy.Symbol('mu')\n",
+      "\n",
+      "#Calculation\n",
+      "k=mu/(du_dy)**(n-1)\n",
+      "Ry=k*du_dy**(n-1)\n",
+      "#ux=(rho*g/k)**(1/n)*((s-y)**(n+1)/n)*(-n/(n+1)]+constant\n",
+      "#At the surface\n",
+      "y=0\n",
+      "ux=0\n",
+      "#At free surface:\n",
+      "#y=s\n",
+      "#us=(rho*g/k)**(1/n)*(n/(n+1))**(s*(n+1)/n)\n",
+      "s=sympy.Symbol('s')\n",
+      "sn=sympy.Symbol('sn')\n",
+      "#For non-newtonian:\n",
+      "us1=0.00592*sn**6/s**5\n",
+      "#For newtonian:\n",
+      "us2=0.0067*sn**5/s**4\n",
+      "sn_s=1/(us1/us2)\n",
+      "\n",
+      "#Result\n",
+      "print\" sN/s=\",round(sn_s*sn/s,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " sN/s= 1.132\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_4.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_4.ipynb
new file mode 100755
index 00000000..98b6e823
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_4.ipynb
@@ -0,0 +1,331 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4:Flow of compressible Fluids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.1,Page no:150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=0.006 #Diameter of the cylinder\n",
+      "Gamma=1.47 \n",
+      "#The critical pressures ratio for discharge through the valve\n",
+      "C_r=(2/(Gamma+1))**(Gamma/(Gamma-1)) \n",
+      "#(i) Sonic velocity will occur until the\n",
+      "P_c=101.3/C_r #pressure at which sonic velocity will occur\n",
+      "M=29 #molecular mass of air\n",
+      "\n",
+      "\n",
+      "#Calculatiom\n",
+      "\n",
+      "%pylab inline\n",
+      "import numpy as np\n",
+      "import math\n",
+      "G=[0]*13\n",
+      "G2=[0]*11\n",
+      "P1a=[0.1,0.125,0.15,0.17,0.19,0.2,0.5,1.0,2.0,3.0,4.0,5.0,6.0]\n",
+      "for i in range(4,13):\n",
+      "    G[i]=4.23e-2*P1a[i] \n",
+      "for j in range(0,4):\n",
+      "    \n",
+      "    G[j]=0.0314*(P1a[j])**0.286*math.sqrt((1.0-round(0.519*(P1a[j])**(-0.286),2)))\n",
+      "P2a=[0,1,2,2.65,3,3.5,4,4.5,4.9,4.95,5] \n",
+      "for j in range(4,11):\n",
+      "    G2[j]=0.2548*P2a[j]**0.714*math.sqrt(1-0.631*P2a[j]**0.286) \n",
+      "\n",
+      "for i in range(0,4):\n",
+      "    G2[i]=0.210 \n",
+      "    \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n\\n(i)...The discharge rate are plotted as:\"\n",
+      "print\"G vs P1a data:\"\n",
+      "print\"Above P1a=0.19 MN/m**2\"\n",
+      "print\"P1a(MN/m^2)\\tG(kg/s)\"\n",
+      "for z in range(4,13):\n",
+      "    \n",
+      "        print P1a[z],\"\\t\\t\",round(G[z],4)\n",
+      "print\"\\nBelow P1a=0.19 MN/m**2\"\n",
+      "print\"P1a(MN/m^2)\\tG(kg/s)\"\n",
+      "for w in range(0,5):\n",
+      "    print P1a[w],\"\\t\\t\",round(G[w],4)\n",
+      "a=plt.plot(P1a,G) \n",
+      "plt.title('Rate of discharge of air vs Cylider Pressure\\n')\n",
+      "plt.xlabel('$Cylinder Pressure,P1a(MN/m**2$')\n",
+      "plt.ylabel('$Mass flow G (kg/s)$')\n",
+      "show(a)\n",
+      "print\"\\n\\n(ii)...The Values of G as a functio of P2a:\"\n",
+      "print\"P2a(MN/m^2)\\tG(kg/s)\"\n",
+      "print \"<0.265\",\"\\t\\t\",round(G2[3],4)\n",
+      "for w in range(4,10):\n",
+      "    print P2a[w],\"\\t\\t\",round(G2[w],4)\n",
+      "print P2a[10],\"\\t\\t\",round(G2[10])\n",
+      "b=plt.plot(P2a,G2) \n",
+      "plt.annotate('G max', xy=(1,0.22), xytext=(1,0.22))\n",
+      "plt.title('Rate of discharge of air vs Downstream Pressure\\n')\n",
+      "plt.xlabel('$Downstream Pressure,P2a(MN/m**2)$')\n",
+      "plt.ylabel('$Mass flow rate G (kg/s)$')\n",
+      "\n",
+      "show(b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "\n",
+        "\n",
+        "(i)...The discharge rate are plotted as:\n",
+        "G vs P1a data:\n",
+        "Above P1a=0.19 MN/m**2\n",
+        "P1a(MN/m^2)\tG(kg/s)\n",
+        "0.19 \t\t0.008\n",
+        "0.2 \t\t0.0085\n",
+        "0.5 \t\t0.0211\n",
+        "1.0 \t\t0.0423\n",
+        "2.0 \t\t0.0846\n",
+        "3.0 \t\t0.1269\n",
+        "4.0 \t\t0.1692\n",
+        "5.0 \t\t0.2115\n",
+        "6.0 \t\t0.2538\n",
+        "\n",
+        "Below P1a=0.19 MN/m**2\n",
+        "P1a(MN/m^2)\tG(kg/s)\n",
+        "0.1 \t\t0.0\n",
+        "0.125 \t\t0.0042\n",
+        "0.15 \t\t0.0061\n",
+        "0.17 \t\t0.0071\n",
+        "0.19 \t\t0.008\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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kyJAhKKVo3rw51apV44knnqhUXXKFEOaXk6N1m507V0sasbHwv7XXRAV2xw3h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\n2v8BjZ9Dogvx6bMAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0xb351c18>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        "(ii)...The Values of G as a functio of P2a:\n",
+        "P2a(MN/m^2)\tG(kg/s)\n",
+        "<0.265 \t\t0.21\n",
+        "3 \t\t0.2059\n",
+        "3.5 \t\t0.1942\n",
+        "4 \t\t0.1707\n",
+        "4.5 \t\t0.1288\n",
+        "4.9 \t\t0.0609\n",
+        "4.95 \t\t0.0439\n",
+        "5 \t\t0.0\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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wcKDunlbTpkFOjrmjeryUOGnY2toSHBzMn3/+yfz588nMzDRFXEIIYRKWljBp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\nGkIIIYwmSUMIIYTRJGkIIYQw2v8DMw/BuCR3ro4AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0xaa87c50>"
+       ]
+      }
+     ],
+     "prompt_number": 75
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.2,Page no:167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "l=30 #Length of the tube\n",
+      "d=150e-3 #Diameter of the tube\n",
+      "P1=0.4e3 #Initial Pressure\n",
+      "P2=0.13e3 #final Pressure\n",
+      "X=0.003 \n",
+      "Y=0.005 \n",
+      "v1=21.15e1 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "import sympy\n",
+      "G_A=sympy.Symbol('G_A') \n",
+      "f=(G_A**2*math.log(P1/P2))+((P2**2-P1**2)/(2*P1*v1))+(4*(Y*l/d)*G_A**2) \n",
+      "A=sympy.solve(f) \n",
+      "\n",
+      "#Result\n",
+      "print\"The approximate flow rate =\",round(A[1],2),\"kg/m**2 s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The approximate flow rate = 0.41 kg/m**2 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 81
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.3,Page no:168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Q=50.0  #volumetric flow rate of methane\n",
+      "P=101.3e3 #Given Pressure\n",
+      "T1=288.0 #Given Temperature\n",
+      "d=0.6 #Diameter of pipeline\n",
+      "l=3e3 #length of the pipe line\n",
+      "R_R=0.0001 #Relative roughness\n",
+      "P2=1.7*10**5 #Pressure at which methane is to be discharged\n",
+      "T2=297.0 #Temperature at which methane leaves the compressor\n",
+      "M=16.0 #molecular mass of methane\n",
+      "R=8314.0 #Gas constant\n",
+      "Meu=1e-5 #Viscosity of methane at 293 K\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "from scipy.optimize import fsolve\n",
+      "T=(T1+T2)/2.0 #Mean temperature\n",
+      "P1_v1=R*T/(M) \n",
+      "#At 288 K and 101.3 kN/m**2\n",
+      "v=P1_v1/P*T1/T \n",
+      "G=Q/v #Mass flow rate of methane\n",
+      "A=math.pi/4*d**2 #cross sectional area of pipeline\n",
+      "G_A=G/A\n",
+      "Re=G_A*d/Meu \n",
+      "#Y=R/(rho*u**2) = 0.0015\n",
+      "Y=0.0015 #(from fig 3.7)\n",
+      "#The upstream pressure is calculated using equation 4.55:\n",
+      "def pressure(P1):\n",
+      "    return((G_A**2)*math.log(P1/P2)+(P2**2-P1**2)/(2*P1_v1)+4*Y*(l/d)*G_A**2)\n",
+      "#P1 = 1e5 \n",
+      "z = fsolve(pressure,100000) \n",
+      "\n",
+      "#Result\n",
+      "print\"Pressure to  developed at the compressor in order to achieve this flowrate =%.3e\"%z,\"N/m**2\" \n",
+      "print\"NOTE:Approx answer is given in book\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure to  developed at the compressor in order to achieve this flowrate =4.042e+05 N/m**2\n",
+        "NOTE:Approx answer is given in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 104
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:4.4,Page no:176"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "A1=0.007 #cross sectional area of stack pipe\n",
+      "A2=4000e-6 #flow area of ruptured disc\n",
+      "P1=10e6 #Pressure of the gas in the vessel\n",
+      "Gamma=1.4 \n",
+      "M=40 #mean molecular weight of gas\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "w_c=(2/(Gamma+1))**(Gamma/(Gamma-1)) \n",
+      "w_c=round(w_c,2)\n",
+      "\n",
+      "V1=(22.4/M)*(500.0/273.0)*(101.3e3/P1) #Specific volume of the gas in the reactor\n",
+      "V=V1*(1/w_c)**(1/Gamma) #Specific volume of gas at the throat\n",
+      "u=math.sqrt(Gamma*5.3e6*V) #velocity at the throat\n",
+      "G=u*A2/V #initial rate of discharge\n",
+      "#obtaining the equations as given in book and solving for 'w' we get\n",
+      "w=0.0057 #Pressure ratio\n",
+      "P_u=P1*w \n",
+      "Mach_no=2.23*(w**(-0.29)-1)**0.5 \n",
+      "P_s=56.3*w*(w**(-0.29)-1)*1e6 \n",
+      "\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)Initial rate of discharge of gas =\",round(G,1),\"kg/s\"\n",
+      "print\"(b)The pressure upstream from the shockwave =\",P_u/1000.0,\"kN/m**2\"\n",
+      "print\"The mach number is =\",round(Mach_no,2)\n",
+      "print\"(c)The pressure downstream from the shockwave =\",P_s/1000,\"kN/m**2\"\n",
+      "print\"NOTE:Mismatched answer due to very approximate value of v1 taken in book\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Initial rate of discharge of gas = 85.2 kg/s\n",
+        "(b)The pressure upstream from the shockwave = 57.0 kN/m**2\n",
+        "The mach number is = 4.16\n",
+        "(c)The pressure downstream from the shockwave = 1115.17300403 kN/m**2\n",
+        "NOTE:Mismatched answer due to very approximate value of v1 taken in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 133
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_5.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_5.ipynb
new file mode 100755
index 00000000..a3e9b8cf
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_5.ipynb
@@ -0,0 +1,195 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5:Flow of Multiphase Mixtures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.1,Page no:190"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "id=75e-3  # internal diameter of pipe\n",
+      "f_r_s=0.05  # Flow rate of steam in (kg/s)\n",
+      "f_r_w=1.5  # Flow rate of water in (kg/s)\n",
+      "T=330  # Mean Temperature\n",
+      "P=120  # Mean Pressure drop\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "area=math.pi*id**2/4.0  # Cross-sectional area for flow\n",
+      "f_r_w_m3s=f_r_w/1000.0  # Flow of water\n",
+      "wtr_vel=f_r_w_m3s/area  #Water velocity\n",
+      "rho_steam=18*273*120.0/(22.4*330.0*101.3)  # density of steam at 330 K and 120 kN/m**2\n",
+      "f_r_s_m3s=f_r_s/rho_steam  #Flow of Steam\n",
+      "steam_vel=f_r_s_m3s/area  #Steam velocity\n",
+      "meu_steam=0.0113e-3 \n",
+      "meu_water=0.52e-3 \n",
+      "Rel=id*wtr_vel*1000/meu_water \n",
+      "Reg=id*steam_vel*rho_steam/meu_steam \n",
+      "DPl=4*0.0025*(1000*wtr_vel**2)/id \n",
+      "DPg=4*0.0022*(rho_steam*steam_vel**2)/id \n",
+      "X=(DPl/DPg)**0.5 \n",
+      "phi_l=4.35 \n",
+      "phi_g=3.95 \n",
+      "DP_tpf=phi_g**2*DPg/1000 \n",
+      "\n",
+      "#Result\n",
+      "print\"Pressure drop per unit length of pipe =\",round(DP_tpf,3),\"kN/m**2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure drop per unit length of pipe = 0.298 kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.2,Page no:212"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "M_p_d=0.2e-3  # Mean particle diameter\n",
+      "f_r_w=0.5  #Flow rate of water\n",
+      "id=25e-3  #Diameter of pipe\n",
+      "l=100  #length of pipe\n",
+      "t_vel=0.0239  #Terminal velocity of falling sand particles\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "import sympy\n",
+      "Um=f_r_w/(1000*math.pi*id**2/4) \n",
+      "Re=id*Um*1000/0.001 \n",
+      "#Assuming e/d = 0.008, then, from Figure 3.7:\n",
+      "phi=0.0046 \n",
+      "f=0.0092 \n",
+      "#From, equation 3.20, the head loss is:\n",
+      "hf=4*phi*l*Um**2/(9.81*id) \n",
+      "iw=hf/l \n",
+      "i=300*1000/(1000*9.81*100) \n",
+      "# Substituting in equation 5.20:\n",
+      "C=(iw/(i-iw)*(1100*9.81*id*(2.6-1)*t_vel)/(Um**2*Um))**-1 \n",
+      "#If G kg/s is the mass flow of sand, then:\n",
+      "G=sympy.Symbol('G') \n",
+      "p=sympy.solve(2600**-1*G-0.30*(2600**-1*G+.0005))\n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Mass flow of sand =\",round(p[0],2),\"kg/s=\",round(p[0]*3600.0/1000.0),\"tonne/h\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Mass flow of sand = 0.56 kg/s= 2.0 tonne/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:5.3,Page no:225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "p_s=1.25e-3  # Particle size of sand\n",
+      "rho_sand=2600.0  #Density of sand\n",
+      "flow_sand=1.0  #flow rate of sand in air\n",
+      "l=200.0  #length of pipe\n",
+      "d=100e-3  # taking nearest standard size of pipe\n",
+      "Uo=4.7 \n",
+      "Ug=30.0\n",
+      "Meu_air=1.7e-5  # viscosity of air\n",
+      "rho_air=1  # Density of air\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "flow_air=flow_sand/5.0 \n",
+      "vol_flow_air=1*flow_air \n",
+      "area=flow_air/Ug\n",
+      "pipe_dia=math.sqrt(4*area/math.pi)\n",
+      "Ug=flow_air/(math.pi*(0.1**2)/4.0)\n",
+      "Ug=round(Ug,1)\n",
+      "Us=Ug-(Uo/(0.468+(7.25*(Uo/rho_sand)**0.5))) \n",
+      "\n",
+      "Re=(d*Ug*rho_air/Meu_air) \n",
+      "phi=0.004 \n",
+      "DP_air=(4*phi*l/d)*rho_air*Ug**2/2.0 \n",
+      "DP_x=2805.0*DP_air/(Uo*Us**2) \n",
+      "DP=DP_air+DP_x \n",
+      "\n",
+      "#Result\n",
+      "print\"Volumetric flow rate of air =\",vol_flow_air,\"m**3/s\" \n",
+      "print\"The cross-sectional area of a 100 mm ID. pipe =\",round(pipe_dia*1000),\"mm\\nThe nearest stand size is 100 mm\"\n",
+      "print\"Pressure drop due to air =\",round(DP_air/1000,1),\"kN/m**2\"\n",
+      "print\"Pressure drop due to sand particles =\",round(DP_x/1000,1),\"kN/m**2\"\n",
+      "print\"The total pressure drop =\",round(DP/1000,1),\"kN/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volumetric flow rate of air = 0.2 m**3/s\n",
+        "The cross-sectional area of a 100 mm ID. pipe = 92.0 mm\n",
+        "The nearest stand size is 100 mm\n",
+        "Pressure drop due to air = 10.4 kN/m**2\n",
+        "Pressure drop due to sand particles = 16.4 kN/m**2\n",
+        "The total pressure drop = 26.8 kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_6.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_6.ipynb
new file mode 100755
index 00000000..cefa9c67
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_6.ipynb
@@ -0,0 +1,275 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6:Flow and Pressure Measurement"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.1,Page no:252"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d_o=25e-3 #Diameter of orifice\n",
+      "d_p=75e-3 #Diameter of pipe\n",
+      "flow_o=300e-6 #Flow rate through pipe\n",
+      "Meu_watr=1e-3 #Viscosity of water\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "import sympy\n",
+      "area_o=math.pi/4*d_o**2 #Area of orifice\n",
+      "vel_o=flow_o/area_o #Velocity of water through the orifice\n",
+      "Re_o=d_o*vel_o*1000/Meu_watr #Re at the orifice\n",
+      "Cd=0.61 \n",
+      "G=flow_o*1e3  #mass flow rate water\n",
+      "ho=sympy.Symbol('ho') \n",
+      "p=sympy.solve(G**2-((Cd*area_o*1000)**2*2*9.81*ho))\n",
+      "\n",
+      "#Result\n",
+      "print\"Difference in level on a water manometer =\",round(p[0]*1000),\"mm of water\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Difference in level on a water manometer = 51.0 mm of water\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.2,Page no:253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "rho_sul=1300.0                   #Density of sulphuric acid\n",
+      "id=50e-3                         #Internal diameter of pipe\n",
+      "d_o=10e-3                        #Diameter of orifice\n",
+      "h=.1                             #Differential pressure shown on a mercury manometer\n",
+      "Cd=0.61                          #Coeffecient of discharge\n",
+      "rho_merc=13550.0                 #Density of mercury\n",
+      "rho_watr=1000.0                  #Density of water\n",
+      "\n",
+      "#Calculation\n",
+      "area_o=math.pi/4*d_o**2           #area of orifice\n",
+      "h_sul=h*(rho_merc-rho_sul)/rho_sul \n",
+      "G_sul=Cd*area_o*rho_sul*(2*9.81*h_sul)**0.5 \n",
+      "DP=rho_sul*9.81*h_sul \n",
+      "\n",
+      "#Result\n",
+      "print\"(a).The mass flow rate of acid =\",round(G_sul,3),\"kg/s\"\n",
+      "print\"(b).The drop in pressure =\",round(DP/1000),\"kN/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).The mass flow rate of acid = 0.268 kg/s\n",
+        "(b).The drop in pressure = 12.0 kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.3,Page no:256"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=150e-3 #Diameter of pipe\n",
+      "d_t=50e-3 #Throat diameter\n",
+      "hv=121e-3 #Pressure drop over the converging section\n",
+      "G=2.91  #Mass Flow rate of water\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "A1=math.pi*d**2/4.0 \n",
+      "A2=math.pi*d_t**2/4.0 \n",
+      "Cd=G*math.sqrt(A1**2-A2**2)/(1000*A1*A2*math.sqrt(2*9.8*hv)) \n",
+      "\n",
+      "#Result\n",
+      "print\"Coefficient for the converging cone of the meter at given flowrate =\",round(Cd,2),\"(Approx)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coefficient for the converging cone of the meter at given flowrate = 0.96 (Approx)\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.4,Page no:261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "l=0.3 #length of tube\n",
+      "id_t=25e-3 #Top internal diameter of tube\n",
+      "id_b=20e-3 #Bottom internal diameter of tube\n",
+      "d_f=20e-3 #Diameter of float\n",
+      "v_f=6e-6 #Volume of float\n",
+      "Cd=0.7 #Coefficient of discharge\n",
+      "rho=1000 #Density of water\n",
+      "rho_f=4800 #Density of float\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "area_t=math.pi/4.0*id_t**2 #Cross-sectional area at top of tube\n",
+      "area_b=math.pi/4.0*id_b**2 #Cross-sectional area at bottom of tube\n",
+      "A_f=math.pi/4.0*d_f**2 #Area of float\n",
+      "A1=math.pi/4.0*((id_t+id_b)/2)**2 \n",
+      "A2=A1-A_f \n",
+      "G=Cd*A2*math.sqrt((2*9.81*v_f*(rho_f-rho)*rho)/(A_f*(1-(A2/A1)**2)))\n",
+      "\n",
+      "#Result\n",
+      "print\"The flow rate of water =\",round(G,3),\"kg/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The flow rate of water = 0.071 kg/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.5,Page no:262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "L=0.5 # Length of the weir\n",
+      "D=100e-3 #Height of water over the weir\n",
+      "n=0 \n",
+      "\n",
+      "#Calculation\n",
+      "#Using Francis formula:\n",
+      "Q=1.84*(L-(0.1*n*D))*D**1.5 \n",
+      "\n",
+      "#Result\n",
+      "print\"Volumetric flowrate of water =\",round(Q,2),\"m**3/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volumetric flowrate of water = 0.03 m**3/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:6.6,Page no:263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "G=15.0  #Mass flow rate of organic liquid\n",
+      "L_ow=2.0 #Length of the weir\n",
+      "rho_l=650.0 \n",
+      "\n",
+      "#Calculation\n",
+      "Q=G/rho_l\n",
+      "Q=round(Q,3)\n",
+      "#Use is made of the Francis formula (equation 6.43),\n",
+      "h_ow=(2.0/3.0)*(Q/L_ow)**(2.0/3.0) \n",
+      "\n",
+      "#Result\n",
+      "print\"Height of liquid flowing over the weir =%d\"%(h_ow*1000),\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height of liquid flowing over the weir =33 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_7.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_7.ipynb
new file mode 100755
index 00000000..9ef97b4d
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_7.ipynb
@@ -0,0 +1,130 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7:Liquid Mixing"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.2,Page no:286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "rho_sol=1650                   #Density of the solution\n",
+      "Meu_sol=50e-3                  #Viscosity of the solution\n",
+      "Dt=2.28                        #Density of the tank\n",
+      "D=0.5                          #Diameter of the propeller mixer\n",
+      "H=2.28                         #Liquid depth\n",
+      "Za=0.5                         #Height of the propeller\n",
+      "N=2                            #Rotational speed\n",
+      "\n",
+      "#Calculation\n",
+      "Re=D**2*N*rho_sol/(Meu_sol) \n",
+      "Fr=N**2*D/9.81 \n",
+      "#From figure 7.6\n",
+      "Np=0.5 \n",
+      "P=Np*rho_sol*N**3*D**5 \n",
+      "\n",
+      "#Result\n",
+      "print\"Power provided by propeller to the liquid =\",round(P),\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power provided by propeller to the liquid = 206.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:7.3,Page no:286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=0.6 #tank diameter\n",
+      "N1=4.0 #Rotor dpeed in Hertz\n",
+      "P1=0.15 #Power consumption\n",
+      "Re1=160000 #Reynold's number\n",
+      "D1=d/3.0 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "D2=6*D1 \n",
+      "N2=math.pi*N1*D1/(math.pi*D2) \n",
+      "P2=7.32*N2**3*D2**5 \n",
+      "#For thermal similarity, that is the same temperature in both systems:\n",
+      "Re2=Re1*(N2*D2**2/(N1*D1**2)) \n",
+      "V2=(math.pi/4.0)*(6*d)**3 \n",
+      "P=V2*0.884 \n",
+      "N=(P/(7.32*(6*0.6/3.0)**5))**(1.0/3.0) \n",
+      "Re=Re1*(N*D2**2/(N1*D1**2)) \n",
+      "\n",
+      "#Result\n",
+      "print\"The new rotor speed =\",round(N2,2),\"Hz\"\n",
+      "print\"The new power required =\",round(P2,2),\"kW\" \n",
+      "print\"The new reynolds number =\",Re2\n",
+      "print\"\\n\\n For Constant power input per unit volume:\"\n",
+      "print\"The new power required =\",round(P,1),\"kW\" \n",
+      "print\"The new rotor speed =\",round(N,2),\"Hz\" \n",
+      "print\"The new reynolds number =%d\"%Re\n",
+      "\n",
+      "print\"NOTE:Wrong values of Reynolds numbers in book\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The new rotor speed = 0.67 Hz\n",
+        "The new power required = 5.4 kW\n",
+        "The new reynolds number = 960000.0\n",
+        "\n",
+        "\n",
+        " For Constant power input per unit volume:\n",
+        "The new power required = 32.4 kW\n",
+        "The new rotor speed = 1.21 Hz\n",
+        "The new reynolds number =1744643\n",
+        "NOTE:Wrong values of Reynolds numbers in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_8.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_8.ipynb
new file mode 100755
index 00000000..674da5bb
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_8.ipynb
@@ -0,0 +1,514 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8:Pumping of Fluids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.1,Page no:320"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "dia_cy=110e-3  #Cylinder diameter\n",
+      "stro=230e-3  #stroke\n",
+      "l_su=6  #Suction line length\n",
+      "d_su=50e-3  #Suction line diameter\n",
+      "lvl_wtr=3  #level of the water in the suction tank\n",
+      "atm_P=10.36 \n",
+      "head=1.2   #m water\n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "import math\n",
+      "N=sympy.Symbol('N')    #Maximum speed   \n",
+      "ang_vel=2*math.pi*N    #Angular velocity in rad/s\n",
+      "#at\n",
+      "t=0                    #Initially\n",
+      "max_acc=0.5*stro*(2*math.pi*N)**2*math.cos(2*math.pi*N*t)  #Maximum acceleration\n",
+      "max_2=(dia_cy/d_su)**2*max_acc   #Maximum acceleration  of liquid in suction pipe\n",
+      "acc_F=max_2*(math.pi/4.0)*d_su**2*l_su*1000   #Accelerating force on the liquid[ N]\n",
+      "P=max_2*l_su*1000     #Pressure drop in suction line N/m**2\n",
+      "P=P/(1000*9.81)     #in m water\n",
+      "n=sympy.solve(head-atm_P+lvl_wtr+P)  #speed\n",
+      "\n",
+      "#Result\n",
+      "print\"Maximum speed at which the pump can run =\",round(n[1],3),\"Hz\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Maximum speed at which the pump can run = 0.677 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.2,Page no:343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "rho_l=800  #Density of liquid\n",
+      "Meu_l=0.5e-3 #Viscosity of liquid\n",
+      "Q=0.0004 #Volumetric flow rate\n",
+      "liq_depth=0.07 \n",
+      "d=25e-3 #Diameter of pipe used\n",
+      "p_v_r=1e3 #Pressure of vapor in reboiler\n",
+      "Z=2 #Net Positive Suction Head\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "A=math.pi/4*d**2 #Cross sectional area of pipe\n",
+      "u=Q/A #Velocity in pipe\n",
+      "Re=d*u*rho_l/Meu_l #Reynolds no.\n",
+      "phi=0.0028 \n",
+      "hf_l=(4*phi*u**2)/(d*9.81) \n",
+      "l=10 \n",
+      "hf=hf_l*l \n",
+      "ho=Z+hf \n",
+      "\n",
+      "#Result\n",
+      "print\"The minimum height required =\",round(ho,1),\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum height required = 2.3 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.3,Page no:356"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Q=0.1 #Flow rate of air suppplied by compressor\n",
+      "T=273 #Temperature\n",
+      "P=101.3e3 #Pressure\n",
+      "P2=380e3 #Air compressed to a pressure\n",
+      "T2=289 #Suction Temperature\n",
+      "l=0.25 #Length of the stroke\n",
+      "u=4 #Speed\n",
+      "c=4/100.0 #Cylinder clearance\n",
+      "Gamma=1.4 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "V=Q*T2/(u*T) #Volume per stroke\n",
+      "R=P2/P #Compression ratio\n",
+      "Vs=V/(1+c-(c*(R)**(1/Gamma))) \n",
+      "A=Vs/l #Cross sectional Area of cylinder\n",
+      "d=(A/math.pi*4)**0.5 #Diameter of cylinder\n",
+      "W=P*V*(Gamma/(Gamma-1))*((R)**((Gamma-1)/Gamma)-1) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Diameter of cylinder =\",round(d,2),\"m\"\n",
+      "print\"\\n Theoretical power requirements =\",round(W*4/1e3,1),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Diameter of cylinder = 0.38 m\n",
+        "\n",
+        " Theoretical power requirements = 17.2 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.4,Page no:357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=290 #Temperature at which compression takes place\n",
+      "P1=101.3e3 #Initial pressure\n",
+      "P2=2065e3 #Final pressure\n",
+      "eta=0.85 #Mechanical efficiecy\n",
+      "c1=4/100.0 #Clearance in cylinder 1\n",
+      "c2=5/100.0 #Clearance in cylinder 1\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "R=P2/P1                         #Overall compression ratio\n",
+      "V_spe=22.4/28.8*T/273.0 #Specific volume of air at 290 K\n",
+      "W=P1*V_spe*2*(1.25/(1.25-1))*(R**.1-1) \n",
+      "W_act=W/0.85 \n",
+      "W_it=P1*V_spe*math.log(R) \n",
+      "eta_it=100*W_it/W_act \n",
+      "Gamma=1.4 \n",
+      "W_ie=P1*V_spe*(Gamma/(Gamma-1))*((R)**((Gamma-1)/Gamma)-1) \n",
+      "eta_ie=100*W_ie/W_act \n",
+      "Vs1=V_spe/(1+c1-(c1*(R)**(1/(2*2.5)))) \n",
+      "Vs2=V_spe*(1/R)**0.5/(1+c2-(c2*(R)**(1/(2*2.5)))) \n",
+      "ratio=Vs1/Vs2 \n",
+      "\n",
+      "#Result\n",
+      "print\"(a).Energy supplied to the compressor, that is the work of compression =\",round(W_act*1e-3,1),\"kJ/kg\"\n",
+      "print\"(b).Isothermal efficiency =\",round(eta_it),\"%\"\n",
+      "print\"(c).Isentropic efficiency =\",round(eta_ie),\"%\"\n",
+      "print\"(d).The ratio of the swept volumes in the two cylinders =\",round(ratio,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).Energy supplied to the compressor, that is the work of compression = 346.5 kJ/kg\n",
+        "(b).Isothermal efficiency = 73.0 %\n",
+        "(c).Isentropic efficiency = 116.0 %\n",
+        "(d).The ratio of the swept volumes in the two cylinders = 4.48\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.5,Page no:360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Q_l=7.5e-4 \n",
+      "rho_l=1200.0 \n",
+      "h=20.0 \n",
+      "P=450e3 \n",
+      "eta=30.0/100.0 \n",
+      "P_atm=101.3e3 \n",
+      "Gamma=1.4 \n",
+      "G=Q_l*rho_l #Mass flow of liquid\n",
+      "W=G*9.81*h \n",
+      "W_act=W/eta \n",
+      "M=28.9 \n",
+      "va=22.4/M \n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "import math\n",
+      "x=sympy.Symbol('x') \n",
+      "Ga=sympy.solve(P_atm*va*x*math.log(P/P_atm)-W_act) \n",
+      "Q=Ga[0]*va \n",
+      "Power=(P_atm*Q)*(Gamma/(Gamma-1))*((P/P_atm)**((Gamma-1)/Gamma)-1) \n",
+      "Power_reqd=Power/1000 \n",
+      "\n",
+      "#Result\n",
+      "print\"Power requirement of the pump =\",round(Power_reqd,3),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power requirement of the pump = 0.734 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.6,Page no:360"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "P1=101.3e3 \n",
+      "Q_watr=0.01 \n",
+      "depth=100 \n",
+      "d=100e-3 \n",
+      "depth_watr=40 \n",
+      "Q_air=0.1 \n",
+      "P2=800e3 \n",
+      "Gamma=1.4 \n",
+      "#V1=Q_air \n",
+      "\n",
+      "#Calculation\n",
+      "G_watr=Q_watr*1000 #Mass flow of water\n",
+      "W=G_watr*depth_watr*9.81 \n",
+      "E=P1*Q_air*(1.4/0.4)*((P2/P1)**(0.4/1.4)-1) #     equation 8.37\n",
+      "effi=W/E*100 \n",
+      "P=345e3 \n",
+      "v1=8314*273/(29*P) \n",
+      "v2=8314*273/(29*P1) \n",
+      "G_air=Q_air/v2  #mass flowrate of the air is:\n",
+      "Q_mean=G_air*v1 #Mean volumetric flowrate of air\n",
+      "Q_tot=Q_watr+Q_mean #Total volumetric flowrate\n",
+      "A=math.pi/4*d**2 #Area of pipe\n",
+      "v_mean=Q_tot/A \n",
+      "\n",
+      "#Result\n",
+      "print\"Efficiency =\",round(effi,1),\"%\"\n",
+      "print\"Mean velocity of the mixture =\",round(v_mean,2),\"m/s\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency = 13.8 %\n",
+        "Mean velocity of the mixture = 5.01 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.7,Page no:369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=40e-3 #Internal Diameter of the pipe\n",
+      "l_p=150.0  #Lendth of pipe\n",
+      "Q_watr=600e-6 #Flow of water\n",
+      "h1=10.0  #Vertical Height\n",
+      "h2=2.0  #head lost across heat exchanger\n",
+      "eta=60.0/100.0  #Efficiency of pump\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "A=math.pi/4*d**2  #Area for flow\n",
+      "u=Q_watr/A  #Velocity\n",
+      "#At 320 K,\n",
+      "Meu=0.65e-3 \n",
+      "rho=1000 \n",
+      "Re=d*u*rho/Meu \n",
+      "phi=0.004  #for a relative roughness of 0.005\n",
+      "l=l_p+h1+(260*d) \n",
+      "hf=4*phi*l*u**2/(d*9.81) \n",
+      "h_tot=hf+h1+h2 #Total head to be developed\n",
+      "G=Q_watr*rho # Mass flow of water\n",
+      "P_r=G*h_tot*9.81  #Power Required\n",
+      "P_s=P_r/eta  #Power Supplied\n",
+      "\n",
+      "#Result\n",
+      "print\"Power Required =\",round(P_s,2),\"W\"\n",
+      "print\"NOTE:Very Approximate answer in book\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power Required = 133.26 W\n",
+        "NOTE:Very Approximate answer in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.8,Page no:370"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "%pylab inline\n",
+      "eta=0.50 \n",
+      "Q=[0.0028,0.0039,0.0050,0.0056,0.0059]\n",
+      "h=[23.2,21.3,18.9,15.2,11.0]\n",
+      "Q1=linspace(0.0015,0.0060,42)\n",
+      "h1=10+2.205e5*Q1**2 \n",
+      "plot(Q,h,label=\"Pump Characteristics\") \n",
+      "plot(Q1,h1,label=\"h=10+2.205e5*Q**2\") \n",
+      "plt.ylim(0,30)\n",
+      "plt.xlim(0.0005,0.006)\n",
+      "title(\"Data for Example 8.8\")\n",
+      "plt.xlabel(\"$Discharge (Q m**3/s)$\")\n",
+      "plt.ylabel(\"$Head (m water)$\") \n",
+      "legend() \n",
+      "#showing the intersection point\n",
+      "x1=[0,0.0054] \n",
+      "y1=[16.43,16.43] \n",
+      "x2=[0.0054,0.0054] \n",
+      "y2=[0,16.43] \n",
+      "#plot(x1,y1,x2,y2) \n",
+      "Q_r=0.0054 \n",
+      "h_r=10+2.205e5*Q_r**2 \n",
+      "P=Q_r*1000*h_r*9.81/eta \n",
+      "\n",
+      "#Result\n",
+      "show()\n",
+      "print\"\\n The discharge at the point of intersection between\\n the purnp characteristic equation =\",Q_r,\"m**3/s\"\n",
+      "print\"\\n Power required=\",round(P*1e-3,2),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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ff/01s7CwYD179mSWlpbs/fffFzsyq7Kyki1atIhxuVymq6vLPD092eHDhzu0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\niIiIYwoPERFx7BUCFALyNJcm+AAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7ec6ac8>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The discharge at the point of intersection between\n",
+        " the purnp characteristic equation = 0.0054 m**3/s\n",
+        "\n",
+        " Power required= 1.74 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:8.10,Page no:375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Meu_H2=0.009e-3  #Viscosity of hydrogen\n",
+      "P2=2e6  #Downstream Pressure\n",
+      "P1=2.5e6 #Upstream pressure\n",
+      "P_m=(P1+P2)/2 #Mean Pressure\n",
+      "T=295.0 #Temperature of the gas\n",
+      "l=500.0 #Length of the pipe used\n",
+      "d=50e-3 #diameter of pipe used\n",
+      "\n",
+      "#Calculation\n",
+      "rho_H2=2*P_m*273.0/(22.4*101.3e3*T) #Density of hydrogen at the mean pressure\n",
+      "A=math.pi*d**2/4.0 #Area of the pipe\n",
+      "eta=0.60  #Efficiency of the pump\n",
+      "v_m=1/rho_H2 \n",
+      "phi=0.0024 \n",
+      "x=sympy.Symbol('x') \n",
+      "G=sympy.solve((x/A)**2*math.log(P1/P2)+(P2-P1)*rho_H2+4*phi*l/d*(x/A)**2) \n",
+      "P=round(G[1],1)*P_m*v_m*math.log(P1/P2)/eta \n",
+      "\n",
+      "#Result\n",
+      "print\"Mass flow rate =\",round(G[1],1),\"kg/s\"\n",
+      "print\"Power required =\",round(P*1e-3,2),\"kW(approx)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass flow rate = 0.2 kg/s\n",
+        "Power required = 91.19 kW(approx)\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_9.ipynb b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/Chapter_9.ipynb
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@@ -0,0 +1,1874 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9:Heat Transfer"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.1,Page no:386"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "M_dot1=20  #rate of mass to be cooled\n",
+      "M_dot2=25  #rate of cooling water\n",
+      "Cp=4.18e3  #Heat capacity\n",
+      "T1=360  #Initial temp.\n",
+      "T2=340  #Final temp.\n",
+      "theta_1=300  #Temperature of cooing water entering\n",
+      "U=2e3  #Overall heat transfer coefficient\n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "import math\n",
+      "Q=M_dot1*Cp*(T1-T2)  #Heat load\n",
+      "x=sympy.Symbol('x') \n",
+      "theta_2=sympy.solve(Q-(M_dot2*Cp*(x-300)))\n",
+      "theta_m1=((T1-theta_2[0])-(T2-theta_1))/(math.log((T1-theta_2[0])/(T2-theta_1))) \n",
+      "A1=Q/(U*theta_m1)\n",
+      "theta_m2=((T1-theta_1)-(T2-theta_2[0]))/(math.log((T1-theta_1)/(T2-theta_2[0]))) \n",
+      "A2=Q/(U*theta_m2)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a).In counter flow,The surface area required\",round(A1,2),\"m**2\"\n",
+      "print\"(b).In cocurrent flow,The surface area required\",round(A2,2),\"m**2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a).In counter flow,The surface area required 19.92 m**2\n",
+        "(b).In cocurrent flow,The surface area required 21.28 m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.2,Page no:390"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "dx=0.5  #Thickness of wall\n",
+      "T1=400  #Temperartue of inner surface\n",
+      "T2=300  #Temperature of outer surface\n",
+      "K=0.7  #Thermal conductivity\n",
+      "A=1  #Area of heat transfer\n",
+      "\n",
+      "#Calculation\n",
+      "#From equation 9.12:\n",
+      "Q=K*A*(T1-T2)/dx \n",
+      "\n",
+      "#Result\n",
+      "print\"The heat loss per square metre of surface =\",Q,\"W/m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss per square metre of surface = 140.0 W/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.3,Page no:391"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "dx1=0.20  #thickness of firebrick\n",
+      "dx2=0.10  #thickness of insulating brick\n",
+      "dx3=0.20  #thickness of building brick\n",
+      "k1=1.4  #Thermal conductivity of firebrick\n",
+      "k2=0.21  #Thermal conductivity of insulating brick\n",
+      "k3=0.7  #Thermal conductivity of building brick\n",
+      "T1=1200  #Temperature at junction 1\n",
+      "T4=330  #Temperature at junction 4\n",
+      "\n",
+      "#Calculation\n",
+      "Q=(T1-T4)/((dx1/k1)+(dx2/k2)+(dx3/k3)) \n",
+      "#The ratio (Temperature drop over firebrick)/(Total temperature drop)\n",
+      "R=(dx1/k1)/((dx1/k1)+(dx2/k2)+(dx3/k3)) \n",
+      "#Temperature drop over firebrick \n",
+      "dT=(T1-T4)*R \n",
+      "T2=(T1-dT) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Heat loss per unit area =%d\"%Q,\"W/m**2\"\n",
+      "print\"\\n Temperature drop over firebrick =%d\"%dT,\"K\"\n",
+      "print\"\\n The temperature at the firebrick-insulating brick interface =\",round(T2),\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Heat loss per unit area =961 W/m**2\n",
+        "\n",
+        " Temperature drop over firebrick =137 K\n",
+        "\n",
+        " The temperature at the firebrick-insulating brick interface = 1063.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.4,Page no:398"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=295  #initial temperature of surfaces\n",
+      "T2f=375  #Final temperature of far surface\n",
+      "dT1=900  #Temperature of near face raised\n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "R=(T2f-T)/(2*(dT1-T))  #ratio of theta to twice of theta dash\n",
+      "x=sympy.Symbol('x') \n",
+      "t=sympy.solve((1.30**2*x)-346**2) \n",
+      "\n",
+      "#Result\n",
+      "print\"Time taken to rise from 295 to 375 K =\",round(t[0]/3600,1),\"h\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time taken to rise from 295 to 375 K = 19.7 h\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.5,Page no:400"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T=295  #initial temperature of surfaces\n",
+      "T2f=375  #Final temperature of far surface\n",
+      "dT1=900  #Temperature of near face raised\n",
+      "DH=4.2e-7  #Thermal diffusivity\n",
+      "#The development of the temperature profile is shown in Figure 9.12\n",
+      "#The problem will be solved by taking relatively large intervals for dx.\n",
+      "#Choosing dx = 50 mm, the construction shown in Figure 9.12\n",
+      "dx=50e-3 \n",
+      "#Because the second face is perfectly insulated, the temperature gradient must\n",
+      "# be zero at this point.\n",
+      "#It is seen that the temperature is\n",
+      "#less than 375 K after time 23dt and greater than 375 K after time 25dt\n",
+      "#Thus:\n",
+      "#t=24*dt\n",
+      "#from equation 9.43\n",
+      "dt=dx**2/(2*DH) \n",
+      "t=24*dt \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The time taken to rise from 295 to 375 K =\",round(t/3600.0,1),\"h\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The time taken to rise from 295 to 375 K = 19.8 h\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.6,Page no:403"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=25e-3  #Diameter of copper sphere\n",
+      "l=25e-3  #Side length of a copper cube\n",
+      "h=75  #External heat transfer coefficient\n",
+      "rho_cu=8950  #Density of copper at mean temperature\n",
+      "Cp=0.38e3  #Heat capacity of copper at mean temperature\n",
+      "k=385  # Thermal conductivity of copper at mean temperature\n",
+      "Tf=923  #Temperature of the furnace\n",
+      "Ta=368  #Temperature at which they are annealed\n",
+      "t=5*60  # time taken\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "import sympy\n",
+      "V_Ae_S=(d/6.0)  #V/Ae tor the sphere\n",
+      "V_Ae_C=(l/6.0)  #V/Ae tor the cube\n",
+      "Bi=h*(V_Ae_S)/k \n",
+      "#The use of a lumped capacity method is therefore justified\n",
+      "tao=rho_cu*Cp*V_Ae_S/h \n",
+      "x=sympy.Symbol('x') \n",
+      "T=sympy.solve(((x-Ta)/(Tf-Ta))-math.exp(-t/tao)) \n",
+      "\n",
+      "#Result\n",
+      "print\"Temperature of the sphere and of the cube at the end of 5 minutes =%d\"%(T[0]-273),\"C\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature of the sphere and of the cube at the end of 5 minutes =208 C\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.7,Page no:409"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "k=2.5  #Thermal conductivity\n",
+      "DH=2e-7 #Thermal diffusivity of the surrounding fluid\n",
+      "h=100  #External heat transfer coefficient\n",
+      "To=293  #Initial Temperature\n",
+      "T_dash=373  #Oven Temperture\n",
+      "Tc=353  #temperature throughout the whole of the sheet reaches a minimum\n",
+      "l=10e-3  #thickness of sheet\n",
+      "L=l/2 \n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "#For the given process, the Biot number\n",
+      "Bi=h*L/k \n",
+      "Bi_1=1/Bi \n",
+      "lim_val=(T_dash-Tc)/(T_dash-To) \n",
+      "#From Figure 9.17, the Fourier number\n",
+      "Fo=7.7  \n",
+      "t=Fo*L**2/DH\n",
+      "\n",
+      "#Result\n",
+      "print\"The minimum time for which the sheet must be heated =%d\"%(t/60.0),\"min\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The minimum time for which the sheet must be heated =16 min\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.8,Page no:413"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "l=5 #Length of the channel of uranium reactor\n",
+      "Q=.25e6  #Heat release from uranium reactor\n",
+      "k=33  #Thermal conductivity of the uranium\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "Q_m=Q/l  #Heat release rate\n",
+      "#Thus, from equation 9.52:\n",
+      "dT=Q_m/(4*math.pi*k) \n",
+      "\n",
+      "#Result\n",
+      "print\"The temperature difference between the surface and the centre of the uranium element =\",round(dT),\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The temperature difference between the surface and the centre of the uranium element = 121.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.9,Page no:429"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Cp=2380 #specific heat capacity of nitrobenzene\n",
+      "k=0.15 \n",
+      "Meu=0.70e-3 #Viscosity of nitrobenzene\n",
+      "d_i=15e-3 #internal diameter of tube\n",
+      "d_o=19e-3 #external diameter of the tube\n",
+      "d_s=0.44 #shell diameter\n",
+      "b_s=0.150 #baffle spacing\n",
+      "p=0.025 #pitch\n",
+      "c=0.006 #clearance\n",
+      "#(i)Tube side coefficient\n",
+      "h_i=1000 #based on inside area\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "h_io=1000*d_i/d_o #based on outside area\n",
+      "#(ii) Shell side coefficient.\n",
+      "A=d_s*b_s*c/p #Area for flow\n",
+      "G_s_=4/A \n",
+      "#Taking Meu/Meu_s=1 in equation 9.91\n",
+      "d_e=4*((25e-3**2-(math.pi*d_o**2/4))/(math.pi/d_o)) \n",
+      "h_o=0.36*k/d_e*(d_e*G_s_/Meu)**0.55*(Cp*Meu/k)**0.33 \n",
+      "#(iii) Overall coefficient\n",
+      "#The math.logarithmic mean temperature difference is given by:\n",
+      "Tm=(((400.0-345.0)-(315.0-305.0))/math.log((400.0-345.0)/(315.0-305.0))) \n",
+      "#The corrected mean temperature difference is\n",
+      "Tm_c=Tm*0.8 \n",
+      "Q=4*Cp*(400.0-315.0) \n",
+      "#The surface area of each tube\n",
+      "A_t=0.0598 \n",
+      "U_o=Q/(2*166*5*A_t*Tm_c) \n",
+      "#(iv) Scale resistance.\n",
+      "R_d=(1/U_o)-(1.0/750.0)-(1.0/1000.0) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Value of scale resistance that could be allowed =\",round(R_d,5),\"m**2 K/W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Value of scale resistance that could be allowed = 0.00026 m**2 K/W\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.10,Page no:432"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "G=15.0 #Mass flow rate of benzene\n",
+      "d_s=1.0  #Internal diameter of Heat Exchanger\n",
+      "l=5.0  #Length of tubes\n",
+      "od=19e-3  #Outer diameter of tubes\n",
+      "C=6e-3  #Clearance\n",
+      "l_b=0.25  #Baffle spacing\n",
+      "Meu=.5e-3 \n",
+      "Y=25e-3  #dimension of square pitch\n",
+      "N=19.0  #no. of Baffles\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "As=d_s*l_b*C/Y  #Cross-flow area\n",
+      "G_dash_s=G/As  #Mass flow\n",
+      "d_e=4*(Y**2-(math.pi*od**2.0/4.0))/(math.pi*od) #Equivalent Diameter\n",
+      "Re=G_dash_s*d_e/Meu \n",
+      "#From Figure 9.29:\n",
+      "f_dash=0.280\n",
+      "rho_b=881 #density of benzene\n",
+      "DPf=f_dash*G_dash_s**2*(N+1)*d_s/(2*rho_b*d_e) \n",
+      "\n",
+      "#Result\n",
+      "print\"The pressure drop over the tube bundle =%d\"%DPf,\"N/m**2=\",round(DPf/(rho_b*9.81)),\"m of Benzene\"\n",
+      "print\"NOTE:Approx value of pressure drop is given in book\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure drop over the tube bundle =8680 N/m**2= 1.0 m of Benzene\n",
+        "NOTE:Approx value of pressure drop is given in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.11,Page no:437"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=0.15  #Diameter of pipe\n",
+      "Ts=400  #Surface temperature\n",
+      "Ta=294  #Air temperture\n",
+      "k=0.0310  #Thermal conductivity ---Table 6, Appendix A1\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "X=36/k**4 \n",
+      "#From Equation 9.102:\n",
+      "GrPr=X*(Ts-Ta)*d**3 \n",
+      "#From Table 9.5:\n",
+      "n=0.25 \n",
+      "C_dd=1.32 \n",
+      "#Thus, in Equation 9.104:\n",
+      "h=C_dd*(Ts-Ta)**n*d**(3*n-1) \n",
+      "\n",
+      "#result\n",
+      "print\"\\n The heat transfer coefficient =\",round(h,2),\"W/m**2 K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The heat transfer coefficient = 6.81 W/m**2 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.12,Page no:439"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "lamda=1e-6 #Wavelength\n",
+      "E_l_b=1e9  #Emissive power at given lambda\n",
+      "C2=1.439e-2 \n",
+      "C1=3.742e-16 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "T=C2/lamda/math.log(C1/(E_l_b*lamda**5)) \n",
+      "#With an error of +2 per cent, the correct value is given by:\n",
+      "E_l_b_n=(100-2)*E_l_b/100 \n",
+      "#In equation 9.108:1\n",
+      "T_n=C2/lamda/math.log(C1/(E_l_b_n*lamda**5)) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The temperature of surface =%d\"%T,\"K\"\n",
+      "print\"\\n The temperature of surface with +2 per cent error=\",round(T_n),\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The temperature of surface =1121 K\n",
+        "\n",
+        " The temperature of surface with +2 per cent error= 1120.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 52
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.13,Page no:441"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=10e-3  #Diameter of carbide elements\n",
+      "l=0.5  #Length of carbide elements\n",
+      "Ts=1750  #Maximun surface temperature of carbide\n",
+      "P=500e3  #Thermal power output required\n",
+      "sigma=5.67e-8 \n",
+      "\n",
+      "#Calculation\n",
+      "Eb=sigma*Ts**4 \n",
+      "A=math.pi*d*l \n",
+      "P1=Eb*A #Power dissipated by one element\n",
+      "n=P/P1  #Number of elements required\n",
+      "\n",
+      "#Result\n",
+      "print\"Number of elements required =%d\"%round(n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of elements required =60\n"
+       ]
+      }
+     ],
+     "prompt_number": 55
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.14,Page no:444"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "A=10.0  #Area of the surface\n",
+      "P_r=1000e3  #Power radiated\n",
+      "T1=1500.0  #First Temperature\n",
+      "T2=1600.0  #Second Temperatue\n",
+      "sigma=5.67e-8 \n",
+      "\n",
+      "#Calculation\n",
+      "E=P_r/A  #The emissive Power\n",
+      "#From equation 9,118:\n",
+      "e=E/(sigma*T1**4) \n",
+      "E2=e*sigma*T2**4 \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Emissivity when T=1500 K =\",round(e,3)\n",
+      "print\"\\n The Emissive power when T=1600 K =\",round(E2*1e-2),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Emissivity when T=1500 K = 0.348\n",
+        "\n",
+        " The Emissive power when T=1600 K = 1295.0 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 59
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.15,Page no:448"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "A1=2.0  #Area of rectangle(Surface 1)\n",
+      "A2=math.pi*1**2.0/4.0  #Area of disc (Surface 2)\n",
+      "T1=1500.0  #Temperature of Surface 1\n",
+      "T2=750.0  #Temperature of Surface 2\n",
+      "F12=0.25  #View factor\n",
+      "sigma=5.67e-8 \n",
+      "\n",
+      "#Calculation\n",
+      "#From equation 9. 1 26:\n",
+      "F21=A1*F12/A2 \n",
+      "Q12=sigma*A1*F12*(T1**4-T2**4) \n",
+      "\n",
+      "#Result\n",
+      "print\"View factor, F12 =\",round(F21,3)\n",
+      "print\"The net radiation transfer =\",round(Q12*1e-3,1),\"kW\"\n",
+      "print\"NOTE:Calculation mistake in book\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "View factor, F12 = 0.637\n",
+        "The net radiation transfer = 134.6 kW\n",
+        "NOTE:Calculation mistake in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 64
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.16,Page no:449"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "X=4.0 #width of horizontal plate and length vertical plate\n",
+      "Y=6.0 #length of horizontal plate\n",
+      "Z=3.0 #height of verical plate\n",
+      "\n",
+      "#Calculation\n",
+      "W=Y/X \n",
+      "H=Z/X \n",
+      "A1=Z*X #Area of plate 1\n",
+      "A2=X*Y #Area of plate 2\n",
+      "F12=0.12 \n",
+      "#From equation 9.126:\n",
+      "F21=A1*F12/A2 \n",
+      "#For the two spheres\n",
+      "r1=1.0  #Diameter of sphere 1\n",
+      "r2=2.0  #Diameter of sphere 2\n",
+      "F12b=1.0 \n",
+      "F21b=(r1/r2)**2 \n",
+      "F22b=1.0-F21b \n",
+      "\n",
+      "#Result\n",
+      "print\"For vertical plate\"\n",
+      "print\"View Factor, F12=\",F12\n",
+      "print\"View Factor, F21=\",F21\n",
+      "print\"\\nFor sphere:\"\n",
+      "print\"View Factor, F21=\",F21b\n",
+      "print\"View Factor, F22=\",F22b"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For vertical plate\n",
+        "View Factor, F12= 0.12\n",
+        "View Factor, F21= 0.06\n",
+        "\n",
+        "For sphere:\n",
+        "View Factor, F21= 0.25\n",
+        "View Factor, F22= 0.75\n"
+       ]
+      }
+     ],
+     "prompt_number": 68
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.17,Page no:454"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "ri_u=0.2 # Inner radius of the upper ring\n",
+      "ro_u=0.3 # Outer radius of the upper ring\n",
+      "ri_l=0.3 # Inner radius of the lower ring\n",
+      "ro_l=0.4 # Outer radius of the lower ring\n",
+      "F12_34=0.4 \n",
+      "F12_4=0.22 \n",
+      "F1_34=0.55 \n",
+      "F14=0.30 \n",
+      "\n",
+      "#Calculation\n",
+      "A12_A2=ro_l**2/(ro_l**2-ri_l**2) \n",
+      "A1_A2=ro_u**2/(ro_l**2-ri_l**2) \n",
+      "F23=((A12_A2)*(F12_34-F12_4))+((A1_A2)*(F1_34-F14)) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n F23 =\",round(F23,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " F23 = 0.73\n"
+       ]
+      }
+     ],
+     "prompt_number": 73
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.18,Page no:455"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=1  #Diameter of plate\n",
+      "r1=0.5 \n",
+      "r4=r1  #Radius of the imaginary disc sealing the hemisphere\n",
+      "L=r1  #The distance between the plate and the bottom of the dome\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "A1=math.pi*d**2/4.0  #Area of the plate\n",
+      "A2=2*math.pi*d**2/4.0  #Area of the underside of the Hemisphere\n",
+      "A4=math.pi*r4**2/4.0 #Area of an imaginary disc sealing the hemisphere and parallel \n",
+      "                  #to the plate\n",
+      "T1=750 #Temperature of the plate\n",
+      "T2=1200 #Temperature of hemispherical cone\n",
+      "T3=290 #Temperature of the surroundings\n",
+      "sigma=5.67e-8 \n",
+      "R1=r1/L \n",
+      "R4=r4/L \n",
+      "S=1+(1+R4**2)/(R1**2) \n",
+      "F14=0.5*(S-(S**2-4*(r4/r1)**2)**0.5) \n",
+      "F12=F14 \n",
+      "F13=1-F12 \n",
+      "Q1=sigma*A1*F12*(T2**4-T1**4)+sigma*A1*F13*(T3**4-T1**4) \n",
+      "\n",
+      "#Result\n",
+      "print\"The net rate of heat transfer by radiation to the plate =\",round(Q1*1e-3,1),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The net rate of heat transfer by radiation to the plate = 21.4 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 75
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.19,Page no:457"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=2  #Diameter of the cylinder\n",
+      "h=1  #Depth of insulated cylinder\n",
+      "T1=1500 \n",
+      "T2=373 \n",
+      "#From Figure 9.40ii, with i = 1, j = 2\n",
+      "r1=1 \n",
+      "r2=1 \n",
+      "L=1 \n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "A1=math.pi*d**2.0/4.0  #Radiant heater surface\n",
+      "A2=A1  #Under-Surface of the vessel\n",
+      "A_R=math.pi*d*h \n",
+      "#The view factor may also be obtained from Figure 9.39ii as follows:\n",
+      "R1=r1/L \n",
+      "R2=r2/L \n",
+      "S=1+(1+R2**2)/(R1**2) \n",
+      "F12=0.5*(S-(S**2-4*(r2/r1)**2)**0.5) \n",
+      "sigma=5.67e-8 \n",
+      "#Using the summation rule\n",
+      "#F11=0\n",
+      "F1R=1-F12  \n",
+      "F2R=F1R \n",
+      "Q2=(A1*F12+((1/(A1*F1R)+(1/(A2*F2R))))**-1)*sigma*(T1**4-T2**4) \n",
+      "#If the surroundings without insulation are surface 3 at\n",
+      "T3=290 \n",
+      "F23=F2R \n",
+      "#from equation 9.135\n",
+      "Q2_d=sigma*A1*F12*(T1**4-T2**4)+sigma*A2*F23*(T3**4-T2**4) \n",
+      "red=(Q2-Q2_d)/Q2*100  #Percentage Reduction\n",
+      "\n",
+      "#Result\n",
+      "print\"The rate of radiant heat transfer to the vessel =%d\"%(Q2*1e-3),\"kW\"\n",
+      "print\"If the insulation were removed ,The rate of radiant heat transfer to the vessel=\",round(Q2_d*1e-3),\"kW\"\n",
+      "print\"Reduction percentage =\",round(red),\"%\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rate of radiant heat transfer to the vessel =620 kW\n",
+        "If the insulation were removed ,The rate of radiant heat transfer to the vessel= 342.0 kW\n",
+        "Reduction percentage = 45.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 79
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.20,Page no:459"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "e=0.75  #Emissivity of grey surface\n",
+      "r=1-e  #reflectivity of surface\n",
+      "Ts=400  #Temperature of surface\n",
+      "T_amb=295 \n",
+      "sigma=5.67e-8 \n",
+      "q1=3e3  #Rate of radiation arriving at grey surface\n",
+      "\n",
+      "#Calculation\n",
+      "#From equation 9.118\n",
+      "Eb=sigma*Ts**4 \n",
+      "#From equation 9.138\n",
+      "qo=e*Eb+r*q1 \n",
+      "#From equation 9.140\n",
+      "Q_A=e/r*(Eb-qo) \n",
+      "q=Q_A \n",
+      "#For convective heat transfer from the surface\n",
+      "qc=-1*q \n",
+      "hc=qc/(Ts-T_amb) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Radiosity =%d\"%round(qo),\"W/m**2\"\n",
+      "print\"\\n The net rate of radiation trasfer = %d\"%q,\"W/m**2\"\n",
+      "print\"\\n Coefficient of heat transfer =\",round(hc,1),\"W/m**2 K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Radiosity =1839 W/m**2\n",
+        "\n",
+        " The net rate of radiation trasfer = -1161 W/m**2\n",
+        "\n",
+        " Coefficient of heat transfer = 11.1 W/m**2 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 84
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.21,Page no:464"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "sigma=5.67e-8 \n",
+      "T=[1000,500,300] #tempertaure of surfaces\n",
+      "A=[1.07,1.07,0.628] #Array of area of surfaces\n",
+      "e=[0.75,0.50,1.0] #Array of emissivity of the surfaces\n",
+      "r=[0.250,0.50] # Array of radius of two surfaces\n",
+      "L=0.2  #distance between two discs\n",
+      "\n",
+      "X=[0]*3\n",
+      "Y=[0]*3\n",
+      "R=[0]*2\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "from scipy.optimize import fsolve\n",
+      "for i in range(0,2):\n",
+      "    X[i]=A[i]/r[i] \n",
+      "    Y[i]=A[i]*e[i]/r[i] \n",
+      "    R[i]=r[i]/L \n",
+      "F11=0 \n",
+      "F22=0 \n",
+      "S=1+(1+R[1]**2)/(R[1]**2) \n",
+      "F12=0.5*(S-(S**2-4*(r[1]/(2*r[0]))**2)**0.5) \n",
+      "A1_F11=0 \n",
+      "A2_F22=0 \n",
+      "A1_F12=A[0]*F12 \n",
+      "A1_F13=A[0]-(A[0]*F11+A[1]*F12) \n",
+      "#for surface 2:\n",
+      "A2_F21=A1_F12 \n",
+      "A2_F23=A1_F13 \n",
+      "#for surface 3:\n",
+      "#By reciprocity rule\n",
+      "A3_F31=A1_F13 \n",
+      "A3_F32=A2_F23 \n",
+      "A3_F33=A[2]-(A3_F31+A3_F32) \n",
+      "\n",
+      "#From equation 9.112:\n",
+      "E_b=[0]*3\n",
+      "for i in range(0,3):\n",
+      "    E_b[i]=sigma*T[i]**4/1000.0 \n",
+      "\n",
+      "#since surface 3 is a black body\n",
+      "q_o3=E_b[2] \n",
+      "#From equations 9.157 and 9.158:\n",
+      "#we get\n",
+      "f=[0,0]\n",
+      "def F(p):\n",
+      "    \n",
+      "    x,y=p\n",
+      "    f[0]=(A1_F11-A[0]/r[0])*x+A2_F21*y+A3_F31*q_o3+E_b[0]*A[0]*e[0]/r[0] \n",
+      "    f[1]=(A1_F12*x)+((A2_F22-A[1]/r[1])*y)+E_b[1]*A[1]*e[1]/r[1] \n",
+      "    return(f[0],f[1])\n",
+      "q_o=fsolve(F,(0,0)) \n",
+      "#From equation 9.140:\n",
+      "Q1=(A[0]*e[0]/r[0])*(E_b[0]-q_o[0]) \n",
+      "Q2=(A[1]*e[1]/r[1])*(E_b[1]-q_o[1]) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Power input to the heater ,Q1 =\",round(Q1,2),\"kW\"\n",
+      "print\"\\n The rate of heat transfer to the plate ,Q2 =\",round(Q2,2),\"kW\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Power input to the heater ,Q1 = 36.19 kW\n",
+        "\n",
+        " The rate of heat transfer to the plate ,Q2 = -14.44 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.22,Page no:470"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=0.5 #diameter of chamber\n",
+      "l=2.0 #Length of chamber\n",
+      "e=0.5 #Emissivity\n",
+      "T_s=750.0 #Temperature at which the chamber is maintained\n",
+      "P=150e3 \n",
+      "T_g=1250.0 \n",
+      "sigma=5.67e-8 \n",
+      "P_c=0.1*P \n",
+      "P_w=P_c \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "V=math.pi/4.0*d**2*l #Volume of the chamber\n",
+      "A_s=(2*math.pi/4.0*d**2)+(math.pi*d*l) #total surface are of chamber\n",
+      "L_e=3.6*(V/A_s) \n",
+      "C_w=1.4 \n",
+      "e_w1=C_w*0.075 \n",
+      "C_c1=1.2 \n",
+      "e_c1=(C_c1*0.037) \n",
+      "A=(P_w+P_c)*L_e \n",
+      "B=P_c/(P_c+P_w)\n",
+      "De=0.001 \n",
+      "e_g=e_w1+0.044-De \n",
+      "e_w2=(0.12*C_w) \n",
+      "a_w=e_w2*(T_g/T_s)**0.65 \n",
+      "C_c2=1.02 \n",
+      "e_c2=(0.08*C_c2) \n",
+      "a_c=e_c2*(T_g/T_s)**0.65 \n",
+      "a_g=a_w+a_c \n",
+      "#If the surrounding surface is black, then:\n",
+      "Q=sigma*A_s*(e_g*T_g**4-a_g*T_s**4) \n",
+      "#For grey walls, the correction factor allowing for multiple reflection of\n",
+      "#incident radiation is:\n",
+      "C_g=0.5/(1.0-(1.0-0.326)*(1.0-0.5)) \n",
+      "Q_w=(Q*C_g) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Radiation to the walls if the surface is black =%.1f\"%(Q*1e-3),\"kW\" \n",
+      "print\"\\n Net radiation to the walls =%.2f\"%(Q_w*1e-3),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Radiation to the walls if the surface is black =50.3 kW\n",
+        "\n",
+        " Net radiation to the walls =37.97 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.23,Page no:479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration:\n",
+      "def mole(w,m):\n",
+      "    n = w/m \n",
+      "    return(n)    \n",
+      "def partial(n1):\n",
+      "    p = 308*(n1/total) \n",
+      "    return(p)\n",
+      "w_steam = 0.57       #mass flow rate of steam entering in [kg/sec]\n",
+      "w_CO2 = 0.20         #mass flow rate of CO2 entering in [kg/sec]\n",
+      "m_water = 18         #molecular mass of water in kg\n",
+      "m_CO2 = 44           #molecular mass of CO2 in kg\n",
+      "n_steam = mole(w_steam,m_water)      #number of moles in kmol\n",
+      "n_CO2 = mole(w_CO2,m_CO2)            #number of moles in kmol\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "total = n_steam + n_CO2 \n",
+      "p_steam = partial(n_steam) \n",
+      "p_CO2 = partial(n_CO2) \n",
+      "mean_mol = (0.57 + 0.20)/total      #mean molecular weight of the mixture in kg/kmol\n",
+      "outlet_steam = 11.7             #partial pressure of water in kN/m**2\n",
+      "outlet_CO2 = 308 - outlet_steam  #partial pressure of water in kN/m**2\n",
+      "n_s = n_CO2*outlet_steam/outlet_CO2 \n",
+      "steam_condensed = n_steam - n_s \n",
+      "p_steam_401K = 252.2    #[kN/m**2]\n",
+      "p_CO2_401K = 308 - 252.2 #[kN/m**2]\n",
+      "steam_remaining = 0.0045*p_steam_401K/p_CO2_401K \n",
+      "s_c = n_steam - steam_remaining    #[kmol]\n",
+      "Heat_cond = s_c*18*(2180 + 1.93*(404-401))          #[kW]\n",
+      "Heat_uncondensed_steam = 0.0203*18*1.93*(404-401)   #[kW]\n",
+      "Heat_CO2 = 0.020*0.92*(404-401) \n",
+      "total_heat = Heat_cond + Heat_uncondensed_steam + Heat_CO2 \n",
+      "flow_water = 1407.3/(4.187*(319-300))       #[kg/sec]\n",
+      "hi = 6.36        #[kW/m**2 K]Based on flow velocity of 1425 kg/m**2 sec\n",
+      "ho = 5.25        #[kW/m**2 K]Based on outside area\n",
+      "Cp = (0.20*0.92 + 0.57*1.93)/0.77   #[kJ/kg K]\n",
+      "k_mean = 0.025    #[kW/m K]\n",
+      "a = 0.0411        #[m**2]\n",
+      "mass_velocity = (0.20+0.57)/0.0411      #[kg/m**2 sec]\n",
+      "hg = 107          #[W/m**2 K] at Re = 29,800 at equivalent diameter = 0.024m\n",
+      "u_pD = 0.62       #(u/pD)**0.67 = 0.62\n",
+      "Cpu_k = 1.01      #(Cp*u/k)**0.67\n",
+      "Psf = (122.6 - 38)/math.log(122.6/38) \n",
+      "kG = hg*(Cpu_k)/(1000*Cp*Psf*u_pD) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Assuming no scale resistance, the overall coefficient =\",round(1407.3/(34.8*74.2),3) ,\"W/m** K\"\n",
+      "print\"\\n Actual coeficient =\",round(1407.3/(53.9*74.2),3) ,\"kW/m**2 K\"\n",
+      "print\"\\n Dirt factor =\",round((0.545-0.352)/(0.545*0.352),2),\"K/kW\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Assuming no scale resistance, the overall coefficient = 0.545 W/m** K\n",
+        "\n",
+        " Actual coeficient = 0.352 kW/m**2 K\n",
+        "\n",
+        " Dirt factor = 1.01 K/kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.24,Page no:498"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d_v=1.0 #diameter of the vessel\n",
+      "L=0.3 #diameter of propeller agitator\n",
+      "N=2.5 #rotating speed of propeller agitator\n",
+      "T=310.0 #Temperature\n",
+      "G=0.5 #circulation speed of cooling water\n",
+      "d_o=25e-3 #outer diameter of stainless steel coil\n",
+      "d=22e-3 #inner diameter of stainless steel coil\n",
+      "d_w=(d_o+d)/2 \n",
+      "d_c=0.8 #diameter of helix\n",
+      "T_m=290.0 #mean temperature\n",
+      "k1=0.59 \n",
+      "Meu1=1.08e-3 \n",
+      "C_p1=4.18e3 \n",
+      "x_w=1.5e-3 \n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "h_i=(k1/d)*(1+3.5*(d/d_c))*0.023*(d*1315.0/Meu1)**0.8*(C_p1*Meu1/k1)**0.4 \n",
+      "#The external film coefficient is given by equation 9.204:\n",
+      "C_p2=1.88e3 #Specefic heat capacity\n",
+      "Meu2=6.5e-3 #viscosity\n",
+      "k2=0.40 \n",
+      "rho=1666 \n",
+      "Meu_s=8.6e-3 \n",
+      "h_o=0.87*(C_p2*Meu2/k2)**(1.0/3.0)*(L**2*N*rho/Meu2)**0.62*(Meu2/Meu_s)**0.14*k2/d_v \n",
+      "\n",
+      "k_w=15.9 \n",
+      "R_o=0.0004 \n",
+      "R_i=0.0002 \n",
+      "U_o=((1/h_o)+(x_w*d_o/(k_w*d_w))+(d_o/(h_i*d))+(R_o)+(R_i*d_o/d))**-1 \n",
+      "\n",
+      "#Result\n",
+      "print\"The overall coeffecient of heat transfer =\",round(U_o),\"W/m**2.K\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The overall coeffecient of heat transfer = 498.0 W/m**2.K\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.25,Page no:501"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "C_p=4e3 \n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "from scipy.integrate import quad\n",
+      "x=sympy.Symbol('x') \n",
+      "T_max=sympy.solve((600*0.5)*(393-x)-(10*6)*(x-293)) \n",
+      "#solving the equation finally we get \n",
+      "def f(T):\n",
+      "    return(11111*(1/(376.3-T)))\n",
+      "t1=quad(f,293.0,353.0)\n",
+      "#The steam is turned off for 7200 s and during this time a heat balance gives:\n",
+      "#on solving as given in book we get\n",
+      "T=346.9 \n",
+      "#The time taken to reheat the liquid to 353 K is then given by:\n",
+      "t2=quad(f,T,353)\n",
+      "\n",
+      "#Result\n",
+      "print\"\\n\\n Maximum temperature to which it can be heated =\",round(T_max[0],1),\"K\"\n",
+      "print\"\\n Time taken to heat the liquid from 293 K to 353 K =%d\"%t1[0],\"s\"\n",
+      "print\"\\n Time taken to reheat the liquid to 353 K =%d\"%round(t2[0]),\"s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "\n",
+        " Maximum temperature to which it can be heated = 376.3 K\n",
+        "\n",
+        " Time taken to heat the liquid from 293 K to 353 K =14155 s\n",
+        "\n",
+        " Time taken to reheat the liquid to 353 K =2584 s\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.26,Page no:516"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#As in Example 9.1, the heat load = 1672 kW\n",
+      "Q=1672 \n",
+      "#With reference to Figure 9.71:\n",
+      "T1=360 \n",
+      "T2=340 \n",
+      "theta1=300 #Temperature of cooling water entering\n",
+      "theta2=316 \n",
+      "\n",
+      "#Calculation\n",
+      "X=(theta2-theta1)/(T1-theta1) \n",
+      "Y=(T1-T2)/(theta2-theta1) \n",
+      "#from Figure 9.58\n",
+      "F=0.97 \n",
+      "theta_m=41.9 \n",
+      "#and hence:\n",
+      "A=Q/(2*F*theta_m) #the heat transfer area\n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The heat transfer area is =\",round(A,1),\"m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The heat transfer area is = 20.6 m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.27,Page no:521"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "#As in Example 9.1, the heat load = 1672 kW\n",
+      "Q=1672.0 \n",
+      "#With reference to Figure 9.71:\n",
+      "T1=360.0 \n",
+      "T2=340.0 \n",
+      "theta1=300.0 #Temperature of cooling water entering\n",
+      "theta2=316.0 \n",
+      "F_theta_m=40.6 #corrected mean temperature difference\n",
+      "\n",
+      "#Calculation\n",
+      "T=(T1+T2)/2.0 \n",
+      "d=1.9e-3 #Tube diameter\n",
+      "u=1 #Water velocity\n",
+      "#then, in equation 9,221:\n",
+      "h_i=4.28*(0.00488*T-1)*u**0.8/d**0.2 \n",
+      "#From Table 9.18, an estimate of the shell-side film coefficient is:\n",
+      "h_o=(1700.0+11000.0)/2000.0 \n",
+      "#For steel tubes of a wall thickness of 1.6 mm, the thermal resistance of the wall, from Table 9.15 is:\n",
+      "xw_kw=0.025 \n",
+      "#the thermal resistance for treated water, from Table 9.16, is 0.26 m2K/kW\n",
+      "Ri=0.26 \n",
+      "Ro=Ri \n",
+      "U=((1.0/h_o)+xw_kw+Ri+Ro+(1.0/h_i))**-1 \n",
+      "A=Q/(F_theta_m*U) \n",
+      "\n",
+      "#Result\n",
+      "print\"\\n The heat transfer area =\",round(A,1),\"m**2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The heat transfer area = 32.8 m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.28,Page no:531"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "mh = 30.0             #[kg/s] Hot fluid flow rate\n",
+      "Thi = 370.0          #[K] Hot Fluid Inlet Temperature\n",
+      "Tho = 315.0          #[K] Hot Fluid outlet Temperature\n",
+      "Tci = 300.0          #[K] Cold Fluid Inlet Temperature\n",
+      "Tco = 315.0          #[K] Cold Fluid Outlet Temperature\n",
+      "cpc = 4.18*10**3         #[J/kg.K] Thermal Conductivity of Cold Fluid\n",
+      "#From table A1.3 at mean temperature 343 K\n",
+      "cph = 2.9*10**3             #[J/kg.K] Thermal Capacity of Hot fluid\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "q = mh*cph*(Thi-Tho)         #[kW] Heat load\n",
+      "mc = q/(cpc*(Tco-Tci))       #[kg/s] Flow of cooling water\n",
+      "Tln = ((Thi-Tho)-(Tco-Tci))/(math.log((Thi-Tho)/(Tco-Tci)))     #[K] logarithm mean temperature difference\n",
+      "X = (Thi - Tho)/(Tco-Tci) \n",
+      "Y = (Tco-Tci)/(Thi - Tci) \n",
+      "F = .85 \n",
+      "U = 500.0        #[W/m**2.K]\n",
+      "A = q/(F*Tln * U) \n",
+      "od = .02        #[m] outer dia\n",
+      "id = .016       #[m] inner dia\n",
+      "l = 4.83        #[m] effective tube length\n",
+      "s = math.pi*od*l \n",
+      "N = A/s \n",
+      "db = (1210/.249)**(2.207)**-1*20/1000.0             #[m]\n",
+      "dc = .068        #[m] diametric clearance between shell and tubes\n",
+      "ds = db+dc            #[m] Shell dia\n",
+      "Ac = math.pi/4*id**2            #[m**2] Cross sectional area\n",
+      "Ntp = N/2.0\n",
+      "Af = N/2.0*Ac        #[m**2] Tube side flow area\n",
+      "mw = 76.3/Af            #[kg/m**2.s] Mass velocity of water\n",
+      "rho = 995        #[kg/m**3] mas density of water\n",
+      "u = mw/rho            #[m/s] water velocity\n",
+      "vu = .8*10**-3            #[N.s/m**2] viscosity\n",
+      "k = .59                #[W/m.K]\n",
+      "Re = id*u*rho/vu \n",
+      "Pr = cpc*vu/k \n",
+      "ld = l/id \n",
+      "jh = 3.7*10**-3\n",
+      "hi = jh*Re*Pr**.3334*.59/id             #[W/m**2.K]\n",
+      "dbf = .20*ds         #[m] Baffle Dia\n",
+      "tb = 1.25*20*10**-3        #[mm] Tube Pitch\n",
+      "As = (25.0-20.0)/25.0*10**3*(ds*Ac)            #[m**2]\n",
+      "Gs = 30.0/As                         #[kg/m**2.s]\n",
+      "de = 1.1*(.025**2-.917*od**2)/od             #[m]\n",
+      "rho2 = 780.0            #[kg/m**3] density\n",
+      "vu2 = .8*10**-3            #[N.s/m**2] viscosity\n",
+      "Cp2 = 3.1*10**3                #[J/kg.K] Heat capacity\n",
+      "k2 = .16                #[W/m.K]\n",
+      "Re2 = Gs*de/vu2 \n",
+      "Pr2 = Cp2*vu2/k2 \n",
+      "jh2 = 5*10**-3 \n",
+      "hs = jh2*Re2*Pr2**.334*k2/de \n",
+      "k3 = 50.0                #[W/m.K]        Thermal Conductivity\n",
+      "Rw = .00020            #[m**2.K/W]      Scale Resistances \n",
+      "Ro = .00015            #[m**2.K/W]      Resistance for organic\n",
+      "U = (1.0/hs + Rw + 0.5*(od-id)/k3 + Ro*od/id+od/(id*hi))**-1\n",
+      "#From figure 9.78\n",
+      "jf = 4.5*10**-3 \n",
+      "n = 2.0 \n",
+      "delP = n*(4*jf*(4.830/id) + 1.25)*(rho*u**2) \n",
+      "u2 = Gs/rho2 \n",
+      "jf2 = 4.6*10**-2 \n",
+      "N2 = 1.0 \n",
+      "delP2 = N2*(4*jf2*(4.830/od)*(1005/14.2))*(rho2*u2**2) \n",
+      "\n",
+      "#Increasing the baffle spacing pressure drop is reduced one-fourth\n",
+      "delPs = delP2/4.0 \n",
+      "ho = hs*(.5)**.8             #[W/m**2.K]\n",
+      "U2 = (1.0/ho + Rw + .5*(od-id)/k3 + Ro*od/id+od/(id*hi))**-1\n",
+      "\n",
+      "#REsult\n",
+      "print\"Overall Coefficient of %d\"%U2,\" W/m**2.K \\n Number of tubes/pass =\",round(Ntp),\"\\n Number of tubes required = %d\"%N\n",
+      "print\"NOTE:Very approximate values are used in textbook for Calculation\\nThat's why answer is different\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Overall Coefficient of 563  W/m**2.K \n",
+        " Number of tubes/pass = 603.0 \n",
+        " Number of tubes required = 1205\n",
+        "NOTE:Very approximate values are used in textbook for Calculation\n",
+        "That's why answer is different\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.29,Page no:535"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "G=1.0  #Flow rate of organic liquid\n",
+      "Cp=2e3#Heat capacity of organic liquid\n",
+      "T1=350.0 \n",
+      "T2=330.0 \n",
+      "theta1=290.0\n",
+      "theta2=320.0 \n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "Q=G*Cp*(T2-T1) #heat load\n",
+      "G_cool=Q/(4187*(theta1-theta2)) #flow of water\n",
+      "GCp_hot=(G*Cp) #for organic\n",
+      "GCp_cold=(G_cool*4187) \n",
+      "#From equation 9.235:\n",
+      "eta=GCp_hot*(T1-T2)/(GCp_cold*(T1-theta1)) \n",
+      "\n",
+      "#Result\n",
+      "print\" Effectiveness of the given double pipe heat exchanger =\",round(eta,2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Effectiveness of the given double pipe heat exchanger = 0.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.30,Page no:538"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Tci = 320.0        #[K] Cold Fluid Initial Temperature\n",
+      "Tce = 340.0        #[K] Cold Fluid Final Temperature\n",
+      "mc = 4.0            #[kg/s] Flow rate of cold fluid\n",
+      "mh = 8.0            #[kg/s] Flow rate of hot fluid\n",
+      "import numpy as np\n",
+      "Thi = np.array([380,370,360,350])    #[K] Hot fluid initial temperature\n",
+      "Cp = 4.18            #[kJ/kg.K] mean heat capacity\n",
+      "U = 1.5        #[W/m**2.K] Overall heat transfer coefficient\n",
+      "\n",
+      "\n",
+      "#Calculation\n",
+      "GCpu= mh*Cp             #[kW/K]\n",
+      "GCpp= mc*Cp             #[kW/K]\n",
+      "if(GCpu<GCpp):\n",
+      "    GCpmin = GCpu        #[kW/K]\n",
+      "    ratio = GCpmin/GCpp \n",
+      "else:\n",
+      "    GCpmin = GCpp       #[kW/K]\n",
+      "    ratio = GCpmin/GCpu \n",
+      "    \n",
+      "#Equation 9.235\n",
+      "n = mc*Cp*(Tce-Tci)*(mc*Cp*(Thi - Tci))**-1 \n",
+      "#From Figure 9.85b Number of transfer Units\n",
+      "N = np.array([.45,.6,.9,1.7])     #[NTU]\n",
+      "A = N*GCpmin/U         #Area of required [m**2]\n",
+      "\n",
+      "#Result\n",
+      "print\"Thi(K)   n      N     A (m**2)\" \n",
+      "for i in range(0,4):\n",
+      "    print Thi[i],\"\\t\",round(n[i],2),\"\\t\",N[i],\"\\t\",round(A[i],1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thi(K)   n      N     A (m**2)\n",
+        "380 \t0.33 \t0.45 \t5.0\n",
+        "370 \t0.4 \t0.6 \t6.7\n",
+        "360 \t0.5 \t0.9 \t10.0\n",
+        "350 \t0.67 \t1.7 \t18.9\n"
+       ]
+      }
+     ],
+     "prompt_number": 60
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.31,Page no:544"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "o_d=10e-3 #outer diameter of the tube\n",
+      "i_d=8.2e-3  #inner diameter of the tube\n",
+      "h=140 #coeffecient of heat transfer between gas and copper tube\n",
+      "k=350 #Thermal conductivity of copper tube\n",
+      "L=0.075 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "b=math.pi*o_d #perimeter of tube\n",
+      "A=math.pi/4*(o_d**2-i_d**2) #cross sectional area of the metal\n",
+      "m=((h*b)/(k*A))**0.5 \n",
+      "T_g=((475*math.cosh(m*L))-365)/(math.cosh(m*L)-1) \n",
+      "\n",
+      "#Result\n",
+      "print\"The gas temperature is =%d\"%T_g,\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The gas temperature is =539 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 64
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.32,Page no:545"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d2=54e-3  #outer diameter of the tube\n",
+      "d1=70e-3  # fin diameter\n",
+      "w=2e-3  #fin thickness\n",
+      "n=230.0 # number of fins per metre run\n",
+      "T_s=370.0  #Surface temperature\n",
+      "T=280.0  #Temperature of surroundings\n",
+      "h=30.0  #Heat transfer coeffecient between gas and fin\n",
+      "k=43.0  #Thermal conductivity of steel\n",
+      "L=(d1-d2)/2.0 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "theta1=T_s-T \n",
+      "#Assuming that the height of the fin is small compared with its circumference \n",
+      "#and that it may be treated as a straight fin of length\n",
+      "l=(math.pi/2.0)*(d1+d2) \n",
+      "b=2*l #perimeter\n",
+      "A=l*w #the average area at right-angles to the heat flow\n",
+      "m=((h*b)/(k*A))**0.5 \n",
+      "#From equation 9.254, the heat flow is given for case (b) as:\n",
+      "Qf=m*k*A*theta1*(math.exp(2*m*L)-1)/(1+math.exp(2*m*L)) \n",
+      "Q=Qf*n #Heat loss per meter run of tube\n",
+      "\n",
+      "#Result\n",
+      "print\"The heat loss per metre run of tube =\",round(Q*1e-3,2),\"kW/m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss per metre run of tube = 1.91 kW/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 67
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.33,Page no:556"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "d=150e-3 #Internal diameter of tube\n",
+      "d_o=168e-3 #outer diameter of tube\n",
+      "d_w=159e-3 \n",
+      "d_s=268e-3 \n",
+      "d_m=(d_s-d_o)/math.log(d_s/d_o) #log mean of d_o and d_s\n",
+      "h_i=8500 #The coefficient for condensing steam together with that for any scale\n",
+      "k_w=45 \n",
+      "k_l=0.073 \n",
+      "x_l=50e-3 \n",
+      "x_w=9e-3 \n",
+      "DT=444-294 \n",
+      "sigma=5.67e-8 \n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "#The temperature on the outside of the lagging is estimated at 314 K and (hr + hc) will be taken as 10 W/m2 K.\n",
+      "#total thermal resisitance \n",
+      "R=(h_i*math.pi*d)**-1+(10*math.pi*d_s)**-1+(k_w*math.pi*d_w/x_w)**-1+(k_l*math.pi*d_m/x_l)**-1 \n",
+      "Q_l=DT/R  #The heat loss per metre of length(from eq 9.261)\n",
+      "DT_lagging=((k_l*math.pi*d_m/x_l)**-1/R)*DT \n",
+      "#Taking an emissivity of 0.9, from equation 9.119:\n",
+      "\n",
+      "h_r=(0.9*sigma*(310**4-294**4))/(310-294) \n",
+      "C=1.32 \n",
+      "#Substituting in equation 9.105 (putting l = diameter = 0.268 m):\n",
+      "h_c=C*((310.0-294.0)/d_s)**0.25 \n",
+      "#If the pipe were unlagged,(hc+hr)for DT=150 K would be about 20 W/m2 K and the heat loss would then be:\n",
+      "Q_l=20*math.pi*d_o*150 \n",
+      "\n",
+      "#Result\n",
+      "print\"The heat loss to the air =\",round(Q_l*1e-3,2),\"kW/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat loss to the air = 1.58 kW/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 71
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example no:9.34,Page no:560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "T1=420 #temperature of steam\n",
+      "k=0.1 #Thermal conductivity\n",
+      "T2=285  #Ambient temperature\n",
+      "h=10 #the coefficient of heat transfer from the outside of the lagging to \n",
+      "     #the surroundings\n",
+      "    \n",
+      "    \n",
+      "#Calculation\n",
+      "\n",
+      "#determining Q/l from equation 9.21 and equating it to heat loss from the\n",
+      "#outside of the lagging we get\n",
+      "#(Q/l)=84.82/(math.log(d_o/0.1)+(0.02/d_o)) W/m\n",
+      "#using various equations we finally get an equation in terms of d_o and we\n",
+      "# will solve it by using fsolve\n",
+      "def F(d_o):\n",
+      "    return((1/(math.log(d_o/0.1)+(0.02/d_o))**2)-(2.35*(d_o**3)/(d_o-0.02))) \n",
+      "ans=fsolve(F,1) \n",
+      "E_t=(ans-0.1)/2\n",
+      "\n",
+      "#Result\n",
+      "print\"\\n Economic thickness of lagging =\",round(E_t[0]*1e3),\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Economic thickness of lagging = 163.0 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 76
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/README.txt b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/README.txt
new file mode 100755
index 00000000..3fd67037
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Gaurav Sharma
+Course: btech
+College/Institute/Organization: JSS Academy of Technical Education,Noida
+Department/Designation: Mechanical Engineering
+Book Title: Chemical Engineering-Fluid Flow,Heat Transfer And Mass Transfer Vol-1 
+Author: J. M. Coulson, J. F. Richardson, J. R. Backhurst And J. H. Harker
+Publisher: Elsevier India
+Year of publication: 2006
+Isbn: 0750644443
+Edition: 6
+\ No newline at end of file
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/Heating.png b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/Heating.png
new file mode 100755
index 00000000..26f02feb
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/Heating.png
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/Unit.png b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/Unit.png
new file mode 100755
index 00000000..b8d7e1db
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/Unit.png
diff --git a/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/ex8_8.png b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/ex8_8.png
new file mode 100755
index 00000000..fa7cb80a
--- /dev/null
+++ b/Chemical_Engineering-Fluid_Flow,Heat_Transfer_And_Mass_Transfer_Vol-1_/screenshots/ex8_8.png
diff --git a/Engineering_Physics_Malik/Chapter_1.ipynb b/Engineering_Physics_Malik/Chapter_1.ipynb
index 4dd3b7b8..4dd3b7b8 100644..100755
--- a/Engineering_Physics_Malik/Chapter_1.ipynb
+++ b/Engineering_Physics_Malik/Chapter_1.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_10.ipynb b/Engineering_Physics_Malik/Chapter_10.ipynb
index 39a354b1..39a354b1 100644..100755
--- a/Engineering_Physics_Malik/Chapter_10.ipynb
+++ b/Engineering_Physics_Malik/Chapter_10.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_11.ipynb b/Engineering_Physics_Malik/Chapter_11.ipynb
index 0cd9f3f6..0cd9f3f6 100644..100755
--- a/Engineering_Physics_Malik/Chapter_11.ipynb
+++ b/Engineering_Physics_Malik/Chapter_11.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_12.ipynb b/Engineering_Physics_Malik/Chapter_12.ipynb
index 1b8c5610..1b8c5610 100644..100755
--- a/Engineering_Physics_Malik/Chapter_12.ipynb
+++ b/Engineering_Physics_Malik/Chapter_12.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_13.ipynb b/Engineering_Physics_Malik/Chapter_13.ipynb
index db97b36d..db97b36d 100644..100755
--- a/Engineering_Physics_Malik/Chapter_13.ipynb
+++ b/Engineering_Physics_Malik/Chapter_13.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_14.ipynb b/Engineering_Physics_Malik/Chapter_14.ipynb
index 498814f0..498814f0 100644..100755
--- a/Engineering_Physics_Malik/Chapter_14.ipynb
+++ b/Engineering_Physics_Malik/Chapter_14.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_15.ipynb b/Engineering_Physics_Malik/Chapter_15.ipynb
index 2b3eeafd..2b3eeafd 100644..100755
--- a/Engineering_Physics_Malik/Chapter_15.ipynb
+++ b/Engineering_Physics_Malik/Chapter_15.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_16.ipynb b/Engineering_Physics_Malik/Chapter_16.ipynb
index 0fab1922..0fab1922 100644..100755
--- a/Engineering_Physics_Malik/Chapter_16.ipynb
+++ b/Engineering_Physics_Malik/Chapter_16.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_17.ipynb b/Engineering_Physics_Malik/Chapter_17.ipynb
index 01028f54..01028f54 100644..100755
--- a/Engineering_Physics_Malik/Chapter_17.ipynb
+++ b/Engineering_Physics_Malik/Chapter_17.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_18.ipynb b/Engineering_Physics_Malik/Chapter_18.ipynb
index de6fffda..de6fffda 100644..100755
--- a/Engineering_Physics_Malik/Chapter_18.ipynb
+++ b/Engineering_Physics_Malik/Chapter_18.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_19.ipynb b/Engineering_Physics_Malik/Chapter_19.ipynb
index 8e043325..8e043325 100644..100755
--- a/Engineering_Physics_Malik/Chapter_19.ipynb
+++ b/Engineering_Physics_Malik/Chapter_19.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_2.ipynb b/Engineering_Physics_Malik/Chapter_2.ipynb
index dccc71e2..dccc71e2 100644..100755
--- a/Engineering_Physics_Malik/Chapter_2.ipynb
+++ b/Engineering_Physics_Malik/Chapter_2.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_20.ipynb b/Engineering_Physics_Malik/Chapter_20.ipynb
index 6252eb1b..6252eb1b 100644..100755
--- a/Engineering_Physics_Malik/Chapter_20.ipynb
+++ b/Engineering_Physics_Malik/Chapter_20.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_22.ipynb b/Engineering_Physics_Malik/Chapter_22.ipynb
index a955c292..a955c292 100644..100755
--- a/Engineering_Physics_Malik/Chapter_22.ipynb
+++ b/Engineering_Physics_Malik/Chapter_22.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_3.ipynb b/Engineering_Physics_Malik/Chapter_3.ipynb
index 1e5fdb51..1e5fdb51 100644..100755
--- a/Engineering_Physics_Malik/Chapter_3.ipynb
+++ b/Engineering_Physics_Malik/Chapter_3.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_4.ipynb b/Engineering_Physics_Malik/Chapter_4.ipynb
index 981a3d82..981a3d82 100644..100755
--- a/Engineering_Physics_Malik/Chapter_4.ipynb
+++ b/Engineering_Physics_Malik/Chapter_4.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_5.ipynb b/Engineering_Physics_Malik/Chapter_5.ipynb
index e9ebdd41..e9ebdd41 100644..100755
--- a/Engineering_Physics_Malik/Chapter_5.ipynb
+++ b/Engineering_Physics_Malik/Chapter_5.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_6.ipynb b/Engineering_Physics_Malik/Chapter_6.ipynb
index 76ec5aef..76ec5aef 100644..100755
--- a/Engineering_Physics_Malik/Chapter_6.ipynb
+++ b/Engineering_Physics_Malik/Chapter_6.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_7.ipynb b/Engineering_Physics_Malik/Chapter_7.ipynb
index 6762cbc5..6762cbc5 100644..100755
--- a/Engineering_Physics_Malik/Chapter_7.ipynb
+++ b/Engineering_Physics_Malik/Chapter_7.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_8.ipynb b/Engineering_Physics_Malik/Chapter_8.ipynb
index 60f3ed3a..60f3ed3a 100644..100755
--- a/Engineering_Physics_Malik/Chapter_8.ipynb
+++ b/Engineering_Physics_Malik/Chapter_8.ipynb
diff --git a/Engineering_Physics_Malik/Chapter_9.ipynb b/Engineering_Physics_Malik/Chapter_9.ipynb
index a43c0fed..a43c0fed 100644..100755
--- a/Engineering_Physics_Malik/Chapter_9.ipynb
+++ b/Engineering_Physics_Malik/Chapter_9.ipynb
diff --git a/Engineering_Physics_Malik/README.txt b/Engineering_Physics_Malik/README.txt
index 85c741ca..85c741ca 100644..100755
--- a/Engineering_Physics_Malik/README.txt
+++ b/Engineering_Physics_Malik/README.txt
diff --git a/Engineering_Physics_Malik/screenshots/noofphotons.png b/Engineering_Physics_Malik/screenshots/noofphotons.png
index bb15b8cf..bb15b8cf 100644..100755
--- a/Engineering_Physics_Malik/screenshots/noofphotons.png
+++ b/Engineering_Physics_Malik/screenshots/noofphotons.png
diff --git a/Engineering_Physics_Malik/screenshots/ref_indices.png b/Engineering_Physics_Malik/screenshots/ref_indices.png
index 300447f7..300447f7 100644..100755
--- a/Engineering_Physics_Malik/screenshots/ref_indices.png
+++ b/Engineering_Physics_Malik/screenshots/ref_indices.png
diff --git a/Engineering_Physics_Malik/screenshots/wavelenK.png b/Engineering_Physics_Malik/screenshots/wavelenK.png
index da79e117..da79e117 100644..100755
--- a/Engineering_Physics_Malik/screenshots/wavelenK.png
+++ b/Engineering_Physics_Malik/screenshots/wavelenK.png
diff --git a/Fundamental_of_internal_combustion_engines/Chap13.ipynb b/Fundamental_of_internal_combustion_engines/Chap13.ipynb
new file mode 100755
index 00000000..d404f5f6
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/Chap13.ipynb
@@ -0,0 +1,124 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13: Engine friction and lubrication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 Page No 423"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.08\t\t            #The diameter of bore in m\n",
+      "L=0.075\t\t            #The length of the stroke in m\n",
+      "l=0.152\t\t            #The connecting rod length in m\n",
+      "h=0.062\t\t           #Skirt length of the piston in m\n",
+      "Fr=8000\t\t            #Compressive force in the connecting rod in N\n",
+      "p=3000\t\t           #The pressure in the cylinder kPa\n",
+      "y=0.004*10**-3\t       #The clearence between piston and cylinder wall in m\n",
+      "U=0.006\t\t           #The dynamic viscosity of the lubricating oil in pa.s\n",
+      "u=8.2\t\t           #The piston speed in m/s\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "ts=(U*u)/y\t                       #The shear stress in N/m**2\n",
+      "A=pi*d*h\t                       #Contact area between the piston and the cylinder in m**2\n",
+      "Ff=ts*A\t\t                       #Friction force on the piston inN\n",
+      "r=L/2.0\t\t                       #Crank length in m\n",
+      "A=math.atan(r/l)\t               #The angle made by the crank in radians\n",
+      "Ft=Fr*sin(A)\t#The side thrust in N\n",
+      "\n",
+      "#Output \n",
+      "print\"The friction force on the piston \",round(Ff,0),\"N\"\n",
+      "print\"The thrust force on the cylinder wall is\",round(Ft,0),\"N\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The friction force on the piston  192.0 N\n",
+        "The thrust force on the cylinder wall is 1916.0 N\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 Page No 424"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.065\t\t#The cylinder bore diameter in m\n",
+      "L=6.0\t\t#The length of the stroke in cm\n",
+      "l=12.0\t\t#The length of the connecting rod in cm\n",
+      "p=50.0\t\t#The pressure in the cylinder in bar\n",
+      "q=90.0\t\t#The crank position in power stroke of the cycle for one cylinder in degrees\n",
+      "Ff=900.0\t\t#Friction force in N\n",
+      "o=0.2\t\t#Wrist pin off set in cm\n",
+      "\n",
+      "#Calculations\n",
+      "r=L/2.0\t\t#The crank length in cm\n",
+      "sineA=r/l\t \n",
+      "cosA=(1-(sineA)**2)**(1/2.0)\t\n",
+      "Fr=(((p*10**5*(pi/4.0)*d**2)-Ff)/cosA)/1000.0  \n",
+      "Ft=Fr*sineA\t    #The side thrust on the piston in kN\n",
+      "sineA1=(r-o)/l\t#The value of sine\n",
+      "cosA1=(1-(sineA1)**2)**(1/2.0)\t\t\t\n",
+      "Fr1=(((p*10**5*(pi/4.0)*d**2)-Ff)/cosA1)/1000.0\t\n",
+      "Ft1=Fr1*sineA1  #The side thrust in kN\n",
+      "\n",
+      "#Output \n",
+      "print\"(a) The force in the connecting rod \",round(Fr,3),\" kN\"\n",
+      "print\"The side thrust on the piston =\",round(Ft,2),\"kN\"\n",
+      "print\"(b) The side thrust on the piston =\",round(Ft1,3),\" kN\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The force in the connecting rod  16.206  kN\n",
+        "The side thrust on the piston = 4.05 kN\n",
+        "(b) The side thrust on the piston = 3.765  kN\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/Chap2.ipynb b/Fundamental_of_internal_combustion_engines/Chap2.ipynb
new file mode 100755
index 00000000..a4ae8119
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/Chap2.ipynb
@@ -0,0 +1,534 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page No: 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=20.0            #Cylinder bore diameter in cm\n",
+      "L=25.0            #Stroke length in cm\n",
+      "Vc=1570.0         #The clearance volume in cm**3\n",
+      "P1=1.0            #Pressure at the beginning of the compression in bar\n",
+      "T1=300.0          #Temperature at the beginning of the compression in K\n",
+      "T3=1673           #The maximum temperature of the cycle in K\n",
+      "Cv=0.718          #specific heat at constant volume for air in kJ/kgK\n",
+      "R=0.287           #Real gas constant in kJ/kgK\n",
+      "g=1.4             #Isentropic index\n",
+      "c=500.0           #Number of cycles per minute\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Vs=(math.pi/4.0)*d**2*L\n",
+      "V1=Vs+Vc\n",
+      "V2=Vc                                     #cm**3\n",
+      "r=V1/V2                                   #Compression ratio\n",
+      "T2=T1*r**(g-1)\n",
+      "P2=P1*r**g\n",
+      "P3=P2*(T3/T2)                             #In constant volume, process Pressure at point 3 in bar\n",
+      "T4=T3*(1/r)**(g-1)                        #In isentropic process, Temperature at point 4 in degree centigrade\n",
+      "P4=P3*(1/r)**(g)                          #In isentropic process, Pressure at point 4 in bar\n",
+      "no=(1-(1/r)**(g-1))*100                   #Air standard efficiency of otto cycle\n",
+      "Q1=Cv*(T3-T2)                                    #Heat supplied in kJ/kg\n",
+      "Q2=Cv*(T4-T1)                                    #Heat rejected in kJ/kg\n",
+      "W=Q1-Q2                                          #Work done per unit mass in kJ/kg\n",
+      "m=((P1*10**5*V1*10**-6)/(R*T1))/1000.0           #The amount of mass in kg\n",
+      "W1=W*m                                           #Work done in kJ\n",
+      "pm=((W1*10**3)/(Vs*10.0**-6))/10.0**5            #Mean effective pressure in N/m**2\n",
+      "P=W1*(c/60.0)                                    #Power developed in kW\n",
+      "\n",
+      "#Output\n",
+      "print\"Temperature at point 2 = \",round(T2-273.15,1),\"C\"\n",
+      "print\"Pressure at point 2 =\" ,round(P2,2),\" bar\"\n",
+      "print\"Pressure at point 3 = \",round(P3,2),\"bar\" \n",
+      "print\"Temperature at point 4 = \",round(T4-273.15),\" C \\nPressure at point 4 = \",round(P4,3),\"bar\"\n",
+      "print\"Air standard efficiency of otto cycle = \",round(no,2),\" percent \"\n",
+      "print\"Work done = \",round(W1,2),\" kJ\"\n",
+      "print\"Mean effective pressure = \",round(pm,2),\"bar \"\n",
+      "print\"Power developed = \",round(P,1),\"kW \"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature at point 2 =  341.3 C\n",
+        "Pressure at point 2 = 12.29  bar\n",
+        "Pressure at point 3 =  33.47 bar\n",
+        "Temperature at point 4 =  544.0  C \n",
+        "Pressure at point 4 =  2.723 bar\n",
+        "Air standard efficiency of otto cycle =  51.17  percent \n",
+        "Work done =  4.26  kJ\n",
+        "Mean effective pressure =  5.42 bar \n",
+        "Power developed =  35.5 kW \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 Page No: 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "CV=42000.0          #The calorific value of the fuel in kJ/kg\n",
+      "pa=5.0            #Percentage of compression\n",
+      "Pa=1.2            #Pressure in the cylinder at 5% compression stroke\n",
+      "pb=75             #Percentage of compression\n",
+      "Pb=4.8            #Pressure in the cylinder at 75% compression stroke\n",
+      "g=1.3             #polytropic index\n",
+      "g1=1.4            #Isentropic index\n",
+      "n=0.6             #Air standard efficiency\n",
+      "\n",
+      "#Calculations\n",
+      "V=(Pb/Pa)**(1/1.3)#Ratio of volumes\n",
+      "r=(V*(pb/100.0)-(pa/100.0))/((1-(pa/100.0))-(V*(1-(pb/100.0))))               #Compression ratio\n",
+      "n1=((1-(1/r)**(g1-1)))*100                                                    #Relative efficiency\n",
+      "nthj=n*(n1/100.0)                                                             #Indicated thermal efficiency\n",
+      "x=(1/(CV*nthj))*3600                                                          #Specific fuel consumption in kg/kW.h\n",
+      "\n",
+      "#Output\n",
+      "print\"The compression ratio of the engine is \",round(r,1)\n",
+      "print\"The specific fuel consumption is \",round(x,3),\"kg/kwh\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The compression ratio of the engine is  9.5\n",
+        "The specific fuel consumption is  0.241 kg/kwh\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page No: 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.2           #The diameter of the cylinder bore in m\n",
+      "L=0.3           #The length of the stroke in m\n",
+      "P1=1            #The pressure at the beginning of the compression in bar\n",
+      "T1=300.0        #The temperature at the beginning of the compression in K\n",
+      "r=16.0          #Compression ratio\n",
+      "V=0.08          #Cutt off takes place at 8& of the stroke\n",
+      "R=0.287         #Real gas constant in kJ/kgK\n",
+      "g=1.4           #Isentropic index\n",
+      "Cp=1.005        #Specific heat at constant prassure in kJ/kgK\n",
+      "Cv=0.718        #specific heat at constant volume for air in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Vs=(math.pi/4.0)*d**2*L  #Swept volume in m**3\n",
+      "Vc=Vs/(r-1)              #Clearance volume in m**3\n",
+      "V2=Vc                    #Volume at point 2 in m**3\n",
+      "V1=Vs+Vc                 #Volume at point 1 in m**3\n",
+      "m=(P1*10**5*V1)/(R*T1)   #The amount of mass in kg\n",
+      "P2=P1*(r**g)             #Pressure at point 2 in bar\n",
+      "P3=P2                    #Pressure at point 3 in bar\n",
+      "T2=T1*r**(g-1)           #Temperature at point 2 in K\n",
+      "V3=(V*Vs)+V2             #Volume at point 3 in m**3\n",
+      "C=V3/V2                  #Cut off ratio\n",
+      "T3=C*T2                  #Temperature at point 3 in K\n",
+      "P4=P3*(C/r)**g           #Pressure at the point 4 in bar\n",
+      "T4=T3*(C/r)**(g-1)       #Temperature at point 4 in K\n",
+      "V4=V1                    #Volume at point 4 in m**3\n",
+      "Q1=(m*Cp*(T3-T2))/1000.0 #Heat supplied in kJ\n",
+      "Q2=(m*Cv*(T4-T1))/1000.0 #Heat rejected in kJ\n",
+      "W=(Q1-Q2)                #Work done per cycle in kJ\n",
+      "na=(W/Q1)*100            #Air standard efficiency\n",
+      "Pm=(W*1000/Vs)/10.0**5   #Mean effective pressure in bar\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) Volume at point 2 = \",round(V2,5),\" m**3 \\nVolume at point 1 = \",round(V1,5),\"m**3 \"\n",
+      "print\"Pressure at point 2 = \",round(P2,1),\" bar\" \n",
+      "print\"Temperature at point 2 = \",round(T2,1),\"K\"\n",
+      "print\"beeta is\",C,\"\\nTemperature at point 3 = \",round(T3,0),\" K \\nPressure at point 4 =\",round(P4,3),\" bar\"\n",
+      "print\"Temperature at point 4 = \",round(T4,1),\" K \\nVolume at point 4 = \",round(V4,5),\"m**3\" \n",
+      "print\"(b) cut off ratio = \",round(c,1)\n",
+      "print\"(c) Work done per cycle = \",round(W,3),\"kJ\" \n",
+      "print\"(d) air smath.tandard efficiency = \",round(na,1),\" percent\" \n",
+      "print\"(e)Mean effective pressure = \",round(Pm,2),\" bar \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Volume at point 2 =  0.00063  m**3 \n",
+        "Volume at point 1 =  0.01005 m**3 \n",
+        "Pressure at point 2 =  48.5  bar\n",
+        "Temperature at point 2 =  909.4 K\n",
+        "beeta is 2.2 \n",
+        "Temperature at point 3 =  2001.0  K \n",
+        "Pressure at point 4 = 3.016  bar\n",
+        "Temperature at point 4 =  904.7  K \n",
+        "Volume at point 4 =  0.01005 m**3\n",
+        "(b) cut off ratio =  500.0\n",
+        "(c) Work done per cycle =  7.736 kJ\n",
+        "(d) air smath.tandard efficiency =  60.4  percent\n",
+        "(e)Mean effective pressure =  8.21  bar \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page No:50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "P1=7\n",
+      "g=1.4\n",
+      "r=12\n",
+      "import numpy as np\n",
+      "from scipy.optimize import fsolve\n",
+      "\n",
+      "def f(b):\n",
+      "    return P1-1/((g-1)*(r-1))*(g*r**g*(b-1)-r*(b**1.4-1))\n",
+      "b = fsolve(f, 1)\n",
+      "f(b)\n",
+      "na=(1-(1/(r**(g-1)))*(((b**g)-1)/(g*(b-1))))*100  #Air standard efficiency\n",
+      "\n",
+      "#Output \n",
+      "print\"The cut off ratio = \",round(b,1),\" \\n The air standard efficiency =\",na,\"percent\"\n",
+      "#NOTE:In the book Answer is wrong for Air standard efficiency .\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The cut off ratio =  2.2  \n",
+        " The air standard efficiency = [ 55.47110058] percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page no 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "m=30.0                                  #The air fuel ratio by mass\n",
+      "T1=300                                  #The temperature of air at the beginning of the compression in K\n",
+      "r=16                                    #The compression ratio\n",
+      "CV=42000                                #The calorific value of the fuel in kJ/kg\n",
+      "g=1.4                                   #Isentropic index\n",
+      "Cp=1.005                                #Specific heat at constant prassure in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "T2=T1*(r**(g-1))                        #Temperature at point 2 in K\n",
+      "T3=((1/m)*(CV/Cp))+T2                   #Temperature at point 3 in K\n",
+      "C=T3/T2                                 #The cut off ratio\n",
+      "n=(1-((1/r**(g-1))*(((C**g)-1)/(g*(C-1)))))*100#The ideal efficiency of the engine based on the air standard cycle\n",
+      "\n",
+      "#Output\n",
+      "print\" The ideal efficiency of the engine based on the air standard cycle = \",round(n,1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The ideal efficiency of the engine based on the air standard cycle =  58.9\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page No: 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "p1=1.0      #Inlet pressure in bar\n",
+      "p2=32.425   #Pressure at the end of isentropic compression in bar\n",
+      "r=6.0       #Ratio of expansion\n",
+      "r1=1.4      #Isentropic index\n",
+      "\n",
+      "#Calculations\n",
+      "rc=(p2/p1)**(1/r1)     #Compression ratio\n",
+      "b=(rc/r)               #cut-off ratio\n",
+      "n=(1-((b**r1-1)/(rc**(r1-1)*r1*(b-1))))*100          \n",
+      "pm=((p1*rc**r1*n/100.0*r1*(b-1))/((r1-1)*(rc-1))) \n",
+      "\n",
+      "#Output\n",
+      "print\"Air-smath efficiency is \",round(n,3),\"percent\"\n",
+      "print\"Mean effective pressure is \",round(pm,3),\"bar\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Air-smath efficiency is  56.671 percent\n",
+        "Mean effective pressure is  5.847 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page No: 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "rc=15.0         #Compression ratio\n",
+      "p1=1            #Pressure at which compression begins in bar\n",
+      "T1=27.0+273.0   #Temperature in K\n",
+      "pm=60           #Maximum pressure in bar\n",
+      "h=2             #Heat transfered to air at constant volume is twice that at consmath.tant pressure\n",
+      "g=1.4           #Isentropic index\n",
+      "Cv=0.718        #specific heat at constant volume for air in kJ/kgK\n",
+      "Cp=1.005        #specific heat at constant pressure for air in kJ/kgK\n",
+      "R=0.287         #Real gas constant in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "T2=(T1*rc**(g-1))     #Temperature in K\n",
+      "p2=(p1*rc**g)         #Pressure in bar\n",
+      "T3=(T2*(pm/p2))       #Temperature in K\n",
+      "T4=(Cv*(T3-T2))/(2*Cp)+T3   #Temperature in K\n",
+      "b=(T4/T3)                   #Cut-off ratio\n",
+      "T5=(T4*(b/rc)**(g-1))       #Temperature in K\n",
+      "p5=(p1*(T5/T1))             #Pressure in bar\n",
+      "Q1=(Cv*(T3-T2))+(Cp*(T4-T3))#Heat supplied per unit mass in kJ/kg\n",
+      "Q2=Cv*(T5-T1)               #Heat rejected per unit mass in kJ/kg\n",
+      "W=(Q1-Q2)                   #Workdone in kJ/kg\n",
+      "n=(W/Q1)*100                #Air standard efficiency\n",
+      "Vs=((1*R*1000*T1)/(p1*10**5))*(1-1/rc)   #Swept volume in m**3/kg\n",
+      "pmean=((W*1000)/Vs)/10.0**5                #Mean-effective pressure in bar\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The pressures and temperatures at the cardinal points of the cycle are \"\n",
+      "print\"T2 =\",round(T2,1),\" K \\np2 =\", round(p2,1),\" bar \\nT3 = \",round(T3,1), \"K  \\np3 =\",round(p3,1),\"bar\"\n",
+      "print\"T4 =\",round(T4,1),\"K  \\nT5 = \",round(T5,1), \"K  \\np5 = \",round(p5,1),\"bar\"\n",
+      "print\"(b) The cycle efficiency is\",round(n,0),\"percent\" \n",
+      "print\"(c) The mean effective pressure of the cycle is \",round(pmean,1),\"bar\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The pressures and temperatures at the cardinal points of the cycle are \n",
+        "T2 = 886.3  K \n",
+        "p2 = 44.3  bar \n",
+        "T3 =  1200.0 K  \n",
+        "p3 = 52.2 bar\n",
+        "T4 = 1312.1 K  \n",
+        "T5 =  460.3 K  \n",
+        "p5 =  1.5 bar\n",
+        "(b) The cycle efficiency is 66.0 percent\n",
+        "(c) The mean effective pressure of the cycle is  2.8 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page No: 55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "r=12.0     #Compression ratio\n",
+      "B=1.615    #Cut off ratio\n",
+      "p3=52.17   #Maximum pressure in bar\n",
+      "p4=p3      #Maximum pressure in bar\n",
+      "p1=1       #Initial pressure in bar\n",
+      "T1=(62+273)    #Initial temperature in K\n",
+      "n=1.35         #Indices of compression and expansion\n",
+      "g=1.4          #Adiabatic exponent\n",
+      "mR=0.287       #Real gas constant in kJ/kgK\n",
+      "Cv=0.718       #specific heat at constant volume for air in kJ/kgK\n",
+      "Cp=1.005       #specific heat at constant pressure for air in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "T2=T1*r**(n-1)     #The temperature at point 2 in K\n",
+      "p2=p1*(r)**n       #The pressure at point 2 in bar\n",
+      "T3=T2*(p3/p2)      #The temperature at point 3 in K\n",
+      "T4=T3*B            #The temperature at point 4 in K\n",
+      "T5=T4*(B/r)**(n-1) #The temperature at point 5 in K\n",
+      "Q12=((g-n)/(g-1))*mR*((T1-T2)/(n-1))     # kJ/kg\n",
+      "Q23=Cv*(T3-T2)                          \n",
+      "Q34=Cp*(T4-T3)                            \n",
+      "Q45=((g-n)/(g-1))*mR*((T4-T5)/(n-1))      \n",
+      "Q51=Cv*(T1-T5)                            \n",
+      "Q1=Q23+Q34+Q45                            \n",
+      "Q2=-Q12+(-Q51)                           \n",
+      "W=Q1-Q2                                   \n",
+      "E=(W/Q1)*100                            \n",
+      "Vs=((mR*T1)/p1)*(r-1)/r                   # m**3/kg\n",
+      "pm=(W*1000/Vs)/10.0**3                      #Mean effective pressure in bar\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)The temperature at cardinal points \\nT2 =\",round(T2,0),\" K\\nT3 = \",round(T3,0),\"K \\nT4 = \",round(T4,0),\"K \\nT5 = \",round(T5,0),\"K \" \n",
+      "print\"(b) The cycle efficiency = \",round(E,1),\" percent\"\n",
+      "print\"(c) The mean effective pressure of the cycle = \",round(pm,3),\"bar\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The temperature at cardinal points \n",
+        "T2 = 799.0  K\n",
+        "T3 =  1456.0 K \n",
+        "T4 =  2352.0 K \n",
+        "T5 =  1166.0 K \n",
+        "(b) The cycle efficiency =  56.9  percent\n",
+        "(c) The mean effective pressure of the cycle =  9.638 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10 Page No: 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "p1=1.0         #Inlet pressure in bar\n",
+      "T1=27.0+273.0    #Temperature in K\n",
+      "p2=4.0         #pressure at point 2 in bar\n",
+      "p3=16.0        #Maximum pressure in bar\n",
+      "Cv=0.573     #specific heat at constant volume for gas in kJ/kgK\n",
+      "Cp=0.761     #specific heat at constant pressure for gas in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "g=(Cp/Cv)      \n",
+      "T2=(T1*(p2/p1)**((g-1)/g))    # K\n",
+      "T3=(p3/p2)*T2               \n",
+      "T4=T3*(p1/p3)**((g-1)/g)    \n",
+      "Q1=Cv*(T3-T2)                 #kJ/kg\n",
+      "Q2=Cp*(T4-T1)                 \n",
+      "W=Q1-Q2                       \n",
+      "n=(W/Q1)*100    \n",
+      "r=(p2/p1)**(1/g)\n",
+      "R=(Cp-Cv)                     #kJ/kg.K\n",
+      "Vs=(R*1000*T1*(r-1))/(p1*10.0**5*r)       #m**3/kg\n",
+      "pm=(W/(Vs*100.0))                         \n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The work done per kg of gas is \",round(W,1),\"kJ/kg\"\n",
+      "print\"(b) The efficiency of the cycle is \",round(n,1),\"percent \"\n",
+      "print\"(c) Mean effective pressure is \",round(pm,1),\"bar\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The work done per kg of gas is  306.2 kJ/kg\n",
+        "(b) The efficiency of the cycle is  42.2 percent \n",
+        "(c) Mean effective pressure is  8.4 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/README.txt b/Fundamental_of_internal_combustion_engines/README.txt
new file mode 100755
index 00000000..67ddb01f
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Tamanna Mittal
+Course: others
+College/Institute/Organization: NTPC
+Department/Designation: commerce
+Book Title: Fundamental of internal combustion engines
+Author: H N Gupta
+Publisher: Prentice Hall of India Pvt Ltd, New Delhi
+Year of publication: 2006
+Isbn: 812032854X
+Edition: 1st edition
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/chap10.ipynb b/Fundamental_of_internal_combustion_engines/chap10.ipynb
new file mode 100755
index 00000000..880733d2
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/chap10.ipynb
@@ -0,0 +1,391 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter10:CI Engines-Fuel Injection System"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 page no: 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bsfc=0.3                    #The brake specific fuel consumption in kg/kWh\n",
+      "bp=250                      #The brake power in kW\n",
+      "N=1500                      #Number of cycles per min in rpm\n",
+      "CA=15                       #Crank angle in degrees\n",
+      "pi1=30                      #The pressure of air in the cylinder at the beginning of the injection in bar\n",
+      "pi2=60                      #The pressure of air in the cylinder at the end of the injection in bar\n",
+      "pf1=220                     #The fuel injection pressure at the beginning in bar\n",
+      "pf2=550                     #The fuel injection pressure at the end in bar\n",
+      "Cd=0.65                     #The coefficient of discharge for the injector \n",
+      "df=850                      #The density of the fuel in kg/m**3\n",
+      "p1=1.013                    #The atmospheric pressure in bar\n",
+      "n=4.0                       #The number of orifices used in the nozzle\n",
+      "x=6.0                       #Number of cylinders\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "mf=bsfc*bp/60.0             \n",
+      "F=(mf/(N/2.0))*(1/x)        \n",
+      "s=(CA/360.0)/(N/60.0)       \n",
+      "mf1=F/s                     \n",
+      "p1=pf1-pi1                 \n",
+      "p2=pf2-pi2\n",
+      "pa=(p1+p2)/2.0\n",
+      "Af=(mf1/(Cd*(2*df*pa*10.0**5)**(1/2.0)))*10**6\n",
+      "do=((Af/n)*(4/math.pi))**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The nozzle area required per injection = \" ,round(Af,3),\"mm**2\"\n",
+      "print\"The diameter of the orifice = \",round(do,3), \"mm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The nozzle area required per injection =  1.067 mm**2\n",
+        "The diameter of the orifice =  0.583 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 page no: 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bp=30                                   #The brake power of the engine in kW\n",
+      "N=3000                                  #The engine speed in rpm \n",
+      "bsfc=0.28                               #The brake specific fuel consumption in kg/kWh \n",
+      "Api=35                                  \n",
+      "p2=160                                  #The pressure at which fuel is injected in bar\n",
+      "CA=28                                   #The crank angle in degrees\n",
+      "p1=35                                   #The pressure in the combustion chamber in bar\n",
+      "Cv=0.92                                 #The coefficient of velocity \n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "S=141.5/(131.5+Api)                    \n",
+      "df=S*1000                               \n",
+      "D=(CA/360.0)/(N/60.0)\n",
+      "F=(bsfc*bp)/((N/2.0)*60)\n",
+      "mf=F/D\n",
+      "Cf=Cv*((2*(p2-p1)*10**5)/df)**(1/2.0)\n",
+      "Af=(mf/(df*Cf))*10**6\n",
+      "d=(4*Af/math.pi)**(1/2.0)                   \n",
+      "\n",
+      "#Output\n",
+      "print\"The velocity of injection of the fuel = \",round(Cf,1),\"m/s \"\n",
+      "print\"The diameter of the fuel orifice = \",round(d,3),\" mm \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity of injection of the fuel =  157.8 m/s \n",
+        "The diameter of the fuel orifice =  0.755  mm \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 Page no 334"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=0.8*10**-3                              #The diameter of an orifice in m\n",
+      "A=1.65*10**-6                             #The cross sectional area in m**2\n",
+      "Cd=0.9                                    #The discharge coefficient of the orifice \n",
+      "Cp=0.85                                   #The coefficient of the passage\n",
+      "p1=170                                    #The injection pressure in bar\n",
+      "p2=25                                     #The compression pressure of the discharge in bar\n",
+      "df=850                                    #The density of the fuel in kg/m**3\n",
+      "\n",
+      "#Calculations\n",
+      "Q=((145/(22.931*10.0**9))**(1/2.0))*10**6  \n",
+      "p=170-(2.161*10**9*(Q/10.0**6)**2)\n",
+      "Cf=Cd*((2*(p-p2)*10**5)/df)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The discharge of fuel through the injector = \",round(Q,1),\"cm**2/s\" \n",
+      "print\"The jet velocity through the orifice = \",round(Cf,1),\" m/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The discharge of fuel through the injector =  79.5 cm**2/s\n",
+        "The jet velocity through the orifice =  158.2  m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 page no: 336"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "s=20                         #Spray penetration in cm\n",
+      "t1=15.7                      #The spray penetration of 20 cm in ms\n",
+      "pi1=150                      #The injection pressure in bar\n",
+      "pi2=450.0                    #The injection pressure to be used in bar\n",
+      "p2=15                        #The combustion chamber pressure in bar\n",
+      "d1=0.34                      #The diameter of the orifice in mm\n",
+      "s1=20                        #The penetration for an orifice in cm\n",
+      "d2=0.17                      #If the diameter of the orifice in cm\n",
+      "t11=12                       #The spray penetration in ms\n",
+      "\n",
+      "#Calculations\n",
+      "t2=(t1*(pi1-p2)**(1/2.0))/(pi2-p2)**(1/2.0)\n",
+      "s2=d2*(s1/d1)\n",
+      "t21=t11*(d2/d1)\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The time required for the spray to penetrate = \",round(t2,3),\"ms\"\n",
+      "print\"(b) The spray penetration of the orifice = \",round(s2,3),\"cm\"\n",
+      "print\"The time required for the spray to penetrate = \",round(t21,3),\"ms\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The time required for the spray to penetrate =  8.746 ms\n",
+        "(b) The spray penetration of the orifice =  10.0 cm\n",
+        "The time required for the spray to penetrate =  6.0 ms\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 page no: 336"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "v=6.5                       #The volume of fuel in the barrel in cc\n",
+      "d=0.3                       #The dimeter of fuel pipe line in cm\n",
+      "l=65                        #The length of the fuel pipe line in cm \n",
+      "vi=2.5                      #The volume of fuel in the injection valve in cc\n",
+      "K=78.5*10**-6               #The coefficient of compressibility of the oil per bar\n",
+      "p1=1                        #The atmospheric pressure in bar\n",
+      "p2=180                      #The pressure due to pump in bar\n",
+      "v3=0.1                      #The pump displacement necessary for the fuel in cc\n",
+      "e=0.75                      #The effective stroke of the plunger in cm\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "V1=v+((math.pi*d**2)/4.0)*l+vi\n",
+      "V=K*V1*(p2-p1)\n",
+      "T=(V)+v3\n",
+      "L=T*(4/math.pi)*(1/(e**2))\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The total displacement of the plunger = \",round(T,3),\"cc\" \n",
+      "print\"(b) The effective stroke of the plunger = \",round(L,3),\"cm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The total displacement of the plunger =  0.291 cc\n",
+        "(b) The effective stroke of the plunger =  0.659 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 page no: 337"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "n=4.0                               #Number of cylinders \n",
+      "N=2500                              #The engine speed in rpm \n",
+      "P=90                                #The power produced by the engine in kW\n",
+      "bsfc=0.28                           #The brake specific fuel consumption in kg/kWh\n",
+      "v=3.5                               #The volume of fuel in the barrel in cc\n",
+      "vp=2.5                              #Volume of fuel in the pipe line in cc\n",
+      "vi=2.0                                #The fuel inside the injector in cc\n",
+      "p1=280.0                              #The average injection pressure in bar\n",
+      "p2=30.0                               #The compression pressure of air during injection in bar\n",
+      "df=850.0                            #The density of the fuel in kg/m**3\n",
+      "K=80*10**-6                         #The coefficient of compressibility of fuel per bar\n",
+      "pi=1.0                                #The pressure with which fuel enter into the barrel in bar\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "F=(bsfc*P)/((N/2.0)*60)\n",
+      "F1=F/n\n",
+      "Vf=(F1/df)*10**6\n",
+      "V1=v+vp+vi\n",
+      "V=K*V1*(p1-math.pi)\n",
+      "Vp=Vf+V\n",
+      "W=((1/2.0)*(p1-math.pi)*10**5*V*10**-6)+((p1-p2)*10**5*Vf*10**-6)\n",
+      "P1=(W*N)/(2*60*1000)     #Power lost per cylinder in kW\n",
+      "P2=P1*4        #Total power lost for pumping the fuel in kW\n",
+      "\n",
+      "#Output \n",
+      "print\"The displacement volume of one plunger per cycle = \",round(Vp,3),\"cc\" \n",
+      "print\"Total power lost for pumping the fuel = \",round(P2,3),\"kW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The displacement volume of one plunger per cycle =  0.276 cc\n",
+        "Total power lost for pumping the fuel =  0.41 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7 page no: 339"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "v1=0.3                              #Velocity of the pump plunger in m/s\n",
+      "l=0.575                             #The length of the fuel pipe in m\n",
+      "A=1/20.0                            #The cross sectional area of pipe to the plunger cylinder\n",
+      "a=1/40.0                            #The area of nozzle hole to the pipe \n",
+      "p1=27.6                             #Initial pressure in the line in bar \n",
+      "p2=27.6                             #The compression pressure of the engine\n",
+      "K=17830*10**5                       #The bulk modulus of fuel in N/m**2\n",
+      "df=860.0                            #The density of the fuel in kg/m**3\n",
+      "\n",
+      "#Calculations\n",
+      "Vs=(K/df)**(1/2.0)\n",
+      "t=l/Vs\n",
+      "Vp=(1/A)*v1\n",
+      "p=((K/Vs)*Vp)/10.0**5\n",
+      "pi=p+p1\n",
+      "po=p1+p\n",
+      "vc=Vp-(a*((2*(po-p2))/df)**(1/2.0))\n",
+      "pr=26.8                            #By trial , Pressure\n",
+      "Vc=pr*(Vs/(K/10.0**5))\n",
+      "po1=p1+p+pr\n",
+      "vo=a*((2*(po1-p2)*10**5)/df)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)The velocity of the pressure disturbance = \",round(Vs,0),\"m/s\"\n",
+      "print\"(b) The time taken by the disturbance to travel through the pipe line = \",round(t,4),\" s\"  \n",
+      "print\"(c) The velocity at the pump end of the pipe line as the plunger moves = \",round(Vp,2),\" m/s\"\n",
+      "print\"The pressure at the pump end of pipe line as the plunger moves = \",round(pi,2),\"bar\"\n",
+      "print\"(d)The magnitude of the first reflected pressure = \",round(pr,2),\"bar\" \n",
+      "print\"The magnitude of the first reflected velocity wave =\",round(Vc,2),\"m/s\" \n",
+      "print\"(e)The pressure at the oriface end of the pipe line after the first reflection = \",round(po1,1),\"bar\"\n",
+      "print\"The velocity at the oriface end of the pipe line after the first reflection = \",round(vo,2),\" m/s \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The velocity of the pressure disturbance =  1440.0 m/s\n",
+        "(b) The time taken by the disturbance to travel through the pipe line =  0.0004  s\n",
+        "(c) The velocity at the pump end of the pipe line as the plunger moves =  6.0  m/s\n",
+        "The pressure at the pump end of pipe line as the plunger moves =  101.9 bar\n",
+        "(d)The magnitude of the first reflected pressure =  26.8 bar\n",
+        "The magnitude of the first reflected velocity wave = 2.16 m/s\n",
+        "(e)The pressure at the oriface end of the pipe line after the first reflection =  128.7 bar\n",
+        "The velocity at the oriface end of the pipe line after the first reflection =  3.83  m/s \n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/chap11.ipynb b/Fundamental_of_internal_combustion_engines/chap11.ipynb
new file mode 100755
index 00000000..4d28299a
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/chap11.ipynb
@@ -0,0 +1,337 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter11:Two Stroke Engines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 page no:367"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "nsc=75       #The scavenging efficiency of the two stroke engine in percent \n",
+      "ns=20        #The scavenging efficiency is increased by in percent\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Rsc=math.log(1/(1-(nsc/100.0)))\n",
+      "nsc1=(nsc/100.0)+((nsc/100.0)*(ns/100.0))\n",
+      "Rsc1=math.log(1/(1-(nsc1)))\n",
+      "Rscr=((Rsc1-Rsc)/Rsc)*100      #Percentage increase in scavenging ratio in persent\n",
+      "\n",
+      "#Output\n",
+      "print\"The percentage change in the scavenging ratio = \",round(Rscr,1),\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage change in the scavenging ratio =  66.1 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 page no:367"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.12      #The bore diameter of the engine in m\n",
+      "l=0.15      #The stroke length of the engine in m\n",
+      "r=16.0        #The compression ratio \n",
+      "N=2000.0      #The speed of the engine in rpm\n",
+      "mf=(240/60.0) #Actual air flow per min in kg/min\n",
+      "T=300.0       #Air inlet temperature in K\n",
+      "p=1.025     #Exhaust pressure in bar\n",
+      "R=287       #Real gas constant in J/kg\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "da=(p*10**5)/(R*T)\n",
+      "Vs=((math.pi)*(d**2)*l)/4.0\n",
+      "V=(r/(r-1))*Vs\n",
+      "m=da*V\n",
+      "m1=m*N\n",
+      "Rsc=mf/m1   #Scavenging ratio\n",
+      "nsc=((1-math.exp(-Rsc))*100)\n",
+      "ntr=((nsc/100.0)/Rsc)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The scavenging ratio = \",round(Rsc,2)\n",
+      "print\"(b) The scavenging efficiency = \",round(nsc,1),\"percent \"\n",
+      "print\"(c) The trapping efficiency = \",round(ntr,1),\" percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The scavenging ratio =  0.93\n",
+        "(b) The scavenging efficiency =  60.5 percent \n",
+        "(c) The trapping efficiency =  65.1  percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 page no: 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "mf=6.5          #Mass flow rate of fuel in kg/h\n",
+      "N=3000.0          #The speed of the engine in rpm\n",
+      "a=15            #The air fuel ratio\n",
+      "CV=44000        #The calorific value of the fuel in kJ/kg\n",
+      "pm=9            #The mean piston speed in m/s\n",
+      "pmi=4.8         #The mean pressure in bar\n",
+      "nsc=85          #The scavenging efficiency in percent\n",
+      "nm=80           #The mechanical efficiency in percent\n",
+      "R=290.0           #Real gas constant in J/kgK\n",
+      "p=1.03          #The pressure of the mixture in bar\n",
+      "T=288.0           #The temperature of the mixture in K\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "ma=a*mf\n",
+      "L=((pm*60)/(2*N))*100\n",
+      "mac=mf+ma\n",
+      "mi=(mac)/(nsc/100.0)\n",
+      "da=(p*10**5)/(R*T)\n",
+      "d=(((mi/da)*(4/math.pi)*(1/(L/100.0))*(1/(60*N)))**(1/2.0))*100\n",
+      "ip=(pmi*10**5*(L/100)*((math.pi/4.0)*(d/100)**2)*N)/(60*1000)\n",
+      "bp=ip*(nm/100.0)\n",
+      "nth=(bp/((mf/3600.0)*CV))*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The diameter of the bore = \",round(d,2),\"cm\"\n",
+      "print\"The length of the stroke = \",L,\" cm\" \n",
+      "print\"The brake power = \",round(bp,2),\" kW\"\n",
+      "print\"The brake thermal efficiency =\",round(nth,2),\" percent \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diameter of the bore =  8.83 cm\n",
+        "The length of the stroke =  9.0  cm\n",
+        "The brake power =  10.58  kW\n",
+        "The brake thermal efficiency = 13.32  percent \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 page no: 369"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.08            #The diameter of the bore in m\n",
+      "L=0.1             #The length of the stroke in m\n",
+      "r=8.0               #The compression ratio \n",
+      "o=60.0              #The exhaust port open before BDC in degrees\n",
+      "v=60.0              #The exhaust port closes after BDC in degrees\n",
+      "a=15.0              #Air fuel ratio \n",
+      "T=300.0             #The temperature of the mixture entering into the engine in K\n",
+      "p=1.05            #The pressure in the cylinder at the time of clomath.sing\n",
+      "R=290.0             #Real gas constant in J/kgK\n",
+      "ma=150.0           #Mass flow rate of air in kg/h\n",
+      "N=4000.0            #The speed of the engine in rpm\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "mf=ma/a\n",
+      "mac=ma+mf\n",
+      "r=(L*100)/2.0\n",
+      "Le=(r+(r*math.sin (math.pi/6.0)))/100.0\n",
+      "Vse=(math.pi*d**2*Le)/4.0\n",
+      "V=(r/(r-1))*Vse\n",
+      "V=0.00043                #Value in book after approximation\n",
+      "da=(p*10**5)/(R*T)\n",
+      "m=V*da\n",
+      "mi=m*60*N\n",
+      "Rsc=mac/mi \n",
+      "nsc=(1-(exp(-Rsc)))*100\n",
+      "ntr=nsc/Rsc\n",
+      "\n",
+      "#Output\n",
+      "print\"The scavenging ratio = \",round(Rsc,2)\n",
+      "print\"The scavenging efficiency =\",round(nsc,2),\" percent \"\n",
+      "print\"The trapping efficiency = \",round(ntr,2),\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The scavenging ratio =  1.28\n",
+        "The scavenging efficiency = 72.32  percent \n",
+        "The trapping efficiency =  56.3 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 page no: 371"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=8.25        #The diameter of the bore in cm\n",
+      "L=11.25       #The length of the stroke in cm\n",
+      "r=8.0           #The compression ratio \n",
+      "N=2500.0        #The speed of the engine in rpm\n",
+      "ip=17.0         #Indicated power in kW\n",
+      "a=0.08        #Fuel air ratio \n",
+      "T=345.0         #Inlet temperature mixture in K\n",
+      "p=1.02        #Exhaust pressure in bar\n",
+      "CV=44000.0      #The calorific value of the fuel in kJ/kg\n",
+      "nth=0.29      #Indicated thermal efficiency\n",
+      "M=114.0         #Molar mass of fuel \n",
+      "R=8314.0        #Universal Gas constant in J/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Vs=(math.pi*d**2*L)/4      #Displacement volume in cm**3\n",
+      "V=(r/(r-1))*Vs            #Total cylinder volume in m**3\n",
+      "ps=((29*p*10**5)/(R*T))*(1/(1+a*(29/M)))         #The density of dry air in kg/m**3\n",
+      "nsc=((ip*1000)/((N/60)*V*10**-6*ps*a*CV*1000*nth))*100     #The scavenging efficiency in percent\n",
+      "\n",
+      "#Output\n",
+      "print\"The scavenging efficiency = \",round(nsc,2),\" percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The scavenging efficiency =  57.54  percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 page no: 372"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "S=15.0        #The speed of the math.piston in m/s\n",
+      "ps=0.35     #The scavenging pressure in bar\n",
+      "pa=1.03     #Atmospheric pressure in bar\n",
+      "r=18.0        #The compression ratio \n",
+      "t=35.0        #The inlet temperature in degree centigrade\n",
+      "Rsc=0.9     #The scavenging ratio \n",
+      "ta=15.0       #The atmospheric temperature in degree centigrade\n",
+      "nc=0.75     #Compressor efficiency \n",
+      "g=1.4       #Adiabatic index\n",
+      "R=287.0       #Real gas constant in J/kgK\n",
+      "Cp=1005.0     #Specific heat of gas in J/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "pi=ps+pa    #The scavenging pressure in bar\n",
+      "Ti=(273+ta)+t    #The inlet temperature in K\n",
+      "pr=pa/math.pi    #The ratio of the pressure for calculations\n",
+      "di=(pi*10**5)/(R*Ti)     #The density in kg/m**3\n",
+      "ai=(g*R*Ti)**(1/2.0)            #The sonic velocity in m/s\n",
+      "C=(Rsc)/(2*((r-1)/r)*(ai/S)*(pi/pa)*((2/(g-1))*(((pr)**(2/g))-((pr)**((g+1)/g))))**(1/2.0))\n",
+      "ds=(pa*10**5)/(R*Ti)     #The density in kg/m**3\n",
+      "mep=(ds*Rsc*Cp*Ti*(((pi/pa)**((g-1)/g))-1))/((nc*((r-1)/r))*10**5)      #Mean effective pressure in bar\n",
+      "\n",
+      "#Output\n",
+      "print\"The flow coefficient = \",round(C,3) \n",
+      "print\"The compressor mean effective pressure =  \",round(mep,1),\"bar\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The flow coefficient =  0.028\n",
+        "The compressor mean effective pressure =   0.4 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/chap15.ipynb b/Fundamental_of_internal_combustion_engines/chap15.ipynb
new file mode 100755
index 00000000..8df611a8
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/chap15.ipynb
@@ -0,0 +1,434 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter15:Air Capacity and SuperCharging"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.1 page no: 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "Vs=0.0028       #Swept volume in m**3\n",
+      "N=3000          #Speed of the engine in rpm\n",
+      "ip=12.5         #The average indicated power developed in kW/m**3\n",
+      "nv=85           #Volumetric efficiency in percent\n",
+      "p1=1.013        #The atmospheric pressure in bar\n",
+      "T1=288          #The atmospheric temperature in K\n",
+      "ni=74           #Isentropic efficiency in percent\n",
+      "pr=1.6          #The pressure ratio\n",
+      "nm=78           #All mechanical efficiencies in percent\n",
+      "g=1.4           #Adiabatic index\n",
+      "R=287           #Real gas constant in J/kgK\n",
+      "Cp=1.005        #The specific heat of gas in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "Vs1=(Vs*(N/2.0))         #Volume swept by the piston per minute in m**3/min\n",
+      "Vi=(nv/100.0)*Vs1        #Unsupercharged induced volume in m**3/min\n",
+      "p2=pr*p1               #Blower delivery pressure in bar\n",
+      "T21=T1*(p2/p1)**((g-1)/g)         #Temperature after isentropic compression in K\n",
+      "T2=T1+((T21-T1)/((ni/100.0)))       #Blower delivery temperature in K\n",
+      "Ve=(Vs1*p2*T1)/(T2*p1)            #Equivalent volume at 1.013 bar and 15 degree centigrade in m**3/min\n",
+      "nv1=(Ve/Vs1)*100                  #Volumetric efficiency of supercharged engine in percent\n",
+      "Vii=Ve-Vi                         #Increase in induced volume in m**3/min\n",
+      "ipa=ip*Vii                        #Increase in ip from air induced in kW\n",
+      "ipi=((p2-p1)*10**5*Vs1)/(60*1000)       #Increase in ip due to increased induction pressure in kW\n",
+      "ipt=ipa+ipi                             #Total increase in ip in kW\n",
+      "bp=ipt*(nm/100.0)                              #Increase in engine bp in kW\n",
+      "ma=(p2*(Vs1/60.0)*10**5)/(R*T2)                #Mass of air delivered per second by blower in kg/s\n",
+      "P=ma*Cp*(T2-T1)                              #Power input to blower in kW\n",
+      "Pd=P/(nm/100.0)                                #Power required to drive the blower in kW\n",
+      "bpn=bp-Pd                                    #Net increase in bp in kW\n",
+      "bpu=ip*Vi*(80/100.0)                           #The bp of unsupercharged engine in kW\n",
+      "bpp=(bpn/(bpu))*100                          #Percentage increase in bp in percent\n",
+      "\n",
+      "#Output\n",
+      "print\"The volumetric efficiency of supercharged engine = \",round(nv1,0),\"percent\"\n",
+      "print\"The increase in brake power by supercharging = \",round(bpn,1),\" kW \"\n",
+      "print\"The percentage increase in brake power = \",round(bpp,1),\" percent \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volumetric efficiency of supercharged engine =  134.0 percent\n",
+        "The increase in brake power by supercharging =  15.1  kW \n",
+        "The percentage increase in brake power =  42.3  percent \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.2 page no: 477"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "p=1.013       #The pressure at the sea level in bar\n",
+      "T=283         #The temperature at the sea level in K\n",
+      "bp=275.0        #Brake power in kW\n",
+      "N=1800.0        #The speed of the engine in rpm\n",
+      "a=20          #Air fuel ratio \n",
+      "R=287         #The real gas constant in J/kgK\n",
+      "bsfc=0.24     #Brake specific fuel consumption in kg/kWh\n",
+      "nv=80         #Volumetric efficiency in percent\n",
+      "p2=0.75       #The atmospheric pressure at altitude in bar\n",
+      "P=9           #The power consumed by supercharger of the total power produced by the engine in percent\n",
+      "T2=303        #The temperature of air leaving the supercharger in K\n",
+      "\n",
+      "#Calculations\n",
+      "mf=(bsfc*bp)/60.0       \n",
+      "ma1=mf*a           \n",
+      "ma=(2/N)*ma1          \n",
+      "dai=(p*10**5)/(R*T)   \n",
+      "Vd=(ma/(dai*(nv/100.0)))            \n",
+      "pmb=(bp*2*60*1000)/(Vd*N*10**5)    \n",
+      "GP=bp/(1-0.09)                    \n",
+      "ma2=(ma1/bp)*GP                    \n",
+      "ma1=(ma2*2)/N                      \n",
+      "p21=((R*T2*ma1)/((nv/100.0)*Vd))/10.0**5         \n",
+      "pi=p21-p2                               \n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The engine capacity Vd = \",round(Vd,3),\"m**3\" \n",
+      "print\"The bmep of the unsupercharged engine = \",round(pmb,3),\"bar\" \n",
+      "print\"(b) Increase in air pressure required in the supercharged = \",round(pi,3),\"bar\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The engine capacity Vd =  0.024 m**3\n",
+        "The bmep of the unsupercharged engine =  7.483 bar\n",
+        "(b) Increase in air pressure required in the supercharged =  0.442 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.3 page no: 479"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "Vs=0.003          #Swept volume in m**3\n",
+      "bmep=9            #Brake mean effective pressure in bar\n",
+      "N=4000            #The speed of the engine in rpm\n",
+      "ni=30.0           #Indicated thermal efficiency in percent\n",
+      "nm=90             #Mechanical efficiency in percent\n",
+      "bmep1=12          #The brake mean effective pressure of other engine in bar\n",
+      "N1=4000           #The speed of other engine in rpm\n",
+      "ni1=25            #The indicated thermal efficiency of other engine in percent\n",
+      "nm1=91            #The mechanical efficiency of other engine in percent\n",
+      "m=200             #The mass of naturally aspired engine in kg\n",
+      "m1=220            #The mass of supercharged engine in kg\n",
+      "CV=44000          #The calorific value of the fuel in kJ/kg\n",
+      "\n",
+      "#Calculations\n",
+      "bp=(bmep*10**5*Vs*N)/(2.0*60.0*1000)         \n",
+      "ip=bp/(nm/100.0)                           \n",
+      "mf=(ip)/((ni/100.0)*CV)                   \n",
+      "bp1=(bmep1*10**5*Vs*N1)/(2.0*60.0*1000)    \n",
+      "ip1=bp1/(nm1/100.0)                        \n",
+      "mf1=ip1/((ni1/100.0)*CV)                  \n",
+      "mf2=mf*3600                                \n",
+      "mf3=mf1*3600                                       \n",
+      "x=((200/90.0)-(220/120.0))/((43.2/120.0)-(27.27/90.0))     \n",
+      "\n",
+      "#Output\n",
+      "print\"The maximum hours required for supply of sufficient fuel = \",round(x,3),\"hr\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum hours required for supply of sufficient fuel =  6.823 hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.4 Page no 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.1             #The diameter of the bore in m\n",
+      "L=0.12            #The length of the stroke in m\n",
+      "N=3000            #The speed of the engine in rpm\n",
+      "n=4               #Number of cylinders\n",
+      "R=287             #Real gas constant in J/kgK\n",
+      "t=120             #Output Torque in Nm\n",
+      "nm=85             #The mechanical efficiency of the engine in percent\n",
+      "T1=288            #The inlet temperature of air into compressor in K\n",
+      "p1=1              #The inlet pressure of air into compressor in bar\n",
+      "Q=1200            #Heat rejected rate in kJ/min\n",
+      "T=328             #The outlet temperature of air in K\n",
+      "p=1.7             #The outlet pressure of air in bar\n",
+      "nv=90             #Volumetric efficiency in percent\n",
+      "Cp=1.005          #Specific heat of gas in kJ/kg\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "bp=(2*math.pi*N*t)/(60.0*1000.0)         #The brake power in kW\n",
+      "ip=bp/(nm/100.0)                       #The indicated power in kW\n",
+      "pmi=((ip*2*60*1000*4)/(L*(math.pi*d**2)*N*n))/10.0**5         #The mean effective pressure in bar\n",
+      "Vs=(math.pi/4.0)*d**2*L               #Swept volume in m**3\n",
+      "Vs1=Vs*(N/2.0)*n                      #Volume swept by the piston per min \n",
+      "V1=(nv/100.0)*Vs1                     #Rate of volume flow of air into the engine in m**3/min\n",
+      "me=((p*10**5*V1)/(R*T))*60          #Rate of mass flow of air into the engine in kg/h\n",
+      "E=Q/60.0                              #Energy balance in the after cooling in kJ/s\n",
+      "T2=((bp/E)*T-T1)/((bp/E)-1)         #The outlet temperature of air in K\n",
+      "mc=((bp)/(Cp*(T2-T1)))*3600         #Mass flow rate in kg/h\n",
+      "maf=mc-me                           #Rate of air flow available to the consumer in kg/h\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The imep of the supercharged engine = \",round(pmi,3),\"bar\"\n",
+      "print\"(b) The rate of air consumed by the engine = \",round(me,1),\"kg/h\" \n",
+      "print\"(c) The rate of air flow available to the consumer = \",round(maf,1),\"kg/h\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The imep of the supercharged engine =  4.706 bar\n",
+        "(b) The rate of air consumed by the engine =  551.5 kg/h\n",
+        "(c) The rate of air flow available to the consumer =  1033.5 kg/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.5 page no: 482"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "Vs=0.0045       #Swept volume in m**3\n",
+      "N=4000.0          #The speed of the engine in rpm \n",
+      "nv=150.0          #Overall volumetric efficiency in percent\n",
+      "ni=90.0           #Isentropic efficiency of the compressor in percent\n",
+      "nm=85.0           #Mechanical efficiency in percent\n",
+      "T=330.0           #The temperature of compressed air after cooler in K\n",
+      "p2=1.8          #The pressure of the compressed air in bar\n",
+      "T1=290.0          #The ambient temperature of air in K\n",
+      "p1=1.0            #The pressure of the ambient condition in bar\n",
+      "R=287.0           #The real gas constant in J/kgK\n",
+      "g=1.4           #Adiabatic index\n",
+      "Cp=1.005        #The specific heat of gas in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "T21=T1*(p2/p1)**((g-1)/g)          \n",
+      "T2=T1+((T21-T1)/(ni/100.0))          \n",
+      "Vs1=Vs*(N/(2*60))                  # m**3/s\n",
+      "Va=(nv/100)*Vs1                    \n",
+      "d=(p1*10**5)/(R*T1)                # kg/m**3\n",
+      "ma=d*Va                            # kg/s\n",
+      "Q=ma*Cp*(T2-T)                     # kJ/s\n",
+      "P=ma*Cp*(T2-T1)                    # kW\n",
+      "Pa=P/(nm/100.0)                      \n",
+      "\n",
+      "#Output\n",
+      "print \"(a) The rate of heat rejected from the engine after cooler = \",round(Q,2),\"kJ/s\" \n",
+      "print\"(b) The power absorbed by the supercharger from the engine = \",round(Pa,1),\"kW\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The rate of heat rejected from the engine after cooler =  5.14 kJ/s\n",
+        "(b) The power absorbed by the supercharger from the engine =  18.8 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.6 page no: 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "p1=0.98         #The inlet pressure of air in bar\n",
+      "T1=290.0          #The inlet temperature of air in K\n",
+      "p2=1.8          #The pressure of air delivered to the engine in bar\n",
+      "a=20.0            #The air fuel ratio \n",
+      "T3=850.0          #The temperature of the exhaust gases leaving the engine in K\n",
+      "p3=1.6          #The pressure of the exhaust gases leaving the engine in bar\n",
+      "p4=1.03         #The turbine exhaust pressure in bar\n",
+      "nc=80.0           #The isentropic efficiency of compressor in percent\n",
+      "nt=85.0           #The isentropic efficiency of turbine in percent\n",
+      "Cpa=1.005       #The specific heat of air in kJ/kgK\n",
+      "Cpg=1.15        #The specific heat of gas in kJ/kgK\n",
+      "g=1.33          #isentropic index\n",
+      "h=1.0            #Adiabatic index\n",
+      "\n",
+      "#Calculations\n",
+      "T21=T1*(p2/p1)**(0.286)    #value taken in book (g-1/g)=0.286      \n",
+      "T2=T1+((T21-T1)/(nc/100.0))        \n",
+      "T22=T2-273                        \n",
+      "T41=T3*(p4/p3)**((g-1)/g)         \n",
+      "T4=T3-((nt/100.0)*(T3-T41))         \n",
+      "T44=T4-273                       \n",
+      "mf=1.0                              # kg/s\n",
+      "ma=mf*a                           # kg/s\n",
+      "Wc=ma*Cpa*(T2-T1)                 # kW\n",
+      "mg=ma+mf                          #Mass flow rate of gas in kg/s\n",
+      "Wt=mg*Cpg*(T3-T4)                \n",
+      "Pt=(Wc/Wt)*100                   \n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The temperature of the air leaving the compressor = \",round(T22,0),\"degree centigrade\" \n",
+      "print\"(b) The temperature of gases leaving the turbine = \",round(T44,0),\"degree centigrade\" \n",
+      "print\"(c) The mechanical power used to run the turbocharger = \",round(Pt,1),\"percent\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The temperature of the air leaving the compressor =  86.0 degree centigrade\n",
+        "(b) The temperature of gases leaving the turbine =  502.0 degree centigrade\n",
+        "(c) The mechanical power used to run the turbocharger =  76.6 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 15.7 page no: 485"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "a=14.0          #Air fuel ratio \n",
+      "T1=288          #The ambient temperature of air in K\n",
+      "T2=(288-23)     #The evaporation of fuel cause 23 degree C drop in mixture temperature in K\n",
+      "p=1.3           #Pressure ratio \n",
+      "nc=75           #The isentropic efficiency of the compressor in percent\n",
+      "Cpm=1.05        #The specific heat of the mixture in kJ/kgK\n",
+      "Cpa=1.005           #The specific heat of air in kJ/kgK\n",
+      "g=1.33          #Adiabatic index\n",
+      "h=1.4           #Isentropic index\n",
+      "ma=1            #Mass flow rate of air in kg/s\n",
+      "\n",
+      "#Calculations\n",
+      "T31=T2*p**((g-1)/g)           \n",
+      "T3=T2+((T31-T2)/(nc/100.0))\n",
+      "mm=1+(1/a)\n",
+      "Wc1=mm*Cpm*(T3-T2)\n",
+      "T21=T1*p**((h-1)/h)\n",
+      "T4=T1+((T21-T1)/(nc/100.0))\n",
+      "T4_=317        #approx value taken in book of T4=317\n",
+      "Wc2=ma*Cpa*(T4_-T1)   \n",
+      "T5=T4-23\n",
+      "Ps=((Wc2-round(Wc1,0))*100)/Wc2\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The power required by the compressor  before the supercharger = \",round(Wc1,0),\"kW/kg of air per second\"\n",
+      "print\"(b) The power required by the compressor after the supercharger = \",round(Wc2,1),\"kW/kg of air per second\"  \n",
+      "print\"Percentage of turbine power used to run the compressor = \",round(Ps,3),\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The power required by the compressor  before the supercharger =  27.0 kW/kg of air per second\n",
+        "(b) The power required by the compressor after the supercharger =  29.1 kW/kg of air per second\n",
+        "Percentage of turbine power used to run the compressor =  7.36 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/chap16.ipynb b/Fundamental_of_internal_combustion_engines/chap16.ipynb
new file mode 100755
index 00000000..158b33df
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/chap16.ipynb
@@ -0,0 +1,874 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter16:Engine Testing and Performance"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.1 page no: 519"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "N=3000                           #The speed of the engine in rpm \n",
+      "r=9                              #Compression ratio \n",
+      "l=17.2                           #The length of the connecting rod in cm\n",
+      "t=20                             #The combustion ends at a TDC in degrees\n",
+      "k=3                              #Three litre spark engine\n",
+      "n=6.0                              #V-6 Engine\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Vs=(k/n)*10**-3\n",
+      "d=(((Vs*4)/math.pi)**(1/3.0))\n",
+      "L=d*100\n",
+      "up=2*d*N/60.0\n",
+      "Vc=(Vs/(r-1))*10**6\n",
+      "cr=L/2.0\n",
+      "R=l/cr\n",
+      "up1=up*((math.pi/2.0)*math.sin(math.pi/9.0)*(1+(math.cos(math.pi/9.0)/(R**2-(math.sin(math.pi/9.0)**2))**(1/2.0))))\n",
+      "s=(cr*math.cos(math.pi/9.0))+(l**2-(cr**2)*(math.sin(math.pi/9.0))**2)**(1/2.0)\n",
+      "x=l+cr-s\n",
+      "V=Vc+(math.pi/4.0)*(d*100)**2*x\n",
+      "\n",
+      "#Output \n",
+      "print\"(a)The cylinder bore and The stroke length (d = L) = \",round(L,2),\"cm\"\n",
+      "print\"(b) The average piston speed = \",round(up,2),\"m/s\"\n",
+      "print\"(c) The clearence volume of one cylinder = \",round(Vc,2),\"cm**3\"\n",
+      "print\"(d) The piston speed at the end of combustion = \",round(up1,2),\"m/s\"\n",
+      "print\"(e) The distance the piston travels from TDC at the end of combustion = \",round(x,2),\"cm\" \n",
+      "print\"(f) Instantaneous volume = \",round(V,2),\"cm**3\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The cylinder bore and The stroke length (d = L) =  8.6 cm\n",
+        "(b) The average piston speed =  8.6 m/s\n",
+        "(c) The clearence volume of one cylinder =  62.5 cm**3\n",
+        "(d) The piston speed at the end of combustion =  5.71 m/s\n",
+        "(e) The distance the piston travels from TDC at the end of combustion =  0.32 cm\n",
+        "(f) Instantaneous volume =  81.24 cm**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2 page no: 521"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.175       #The diameter of the bore in m\n",
+      "L=0.32        #The length of the stroke in m\n",
+      "p=6.5         #Mean effective pressure in bar\n",
+      "pp=0.4        #Pumping loop mean effective pressure in bar\n",
+      "N=510.0         #The speed of the engine in rpm\n",
+      "pm=0.65       #Diagrams from the dead cycle give a mep in bar\n",
+      "n=55.0          #Firing strokes per minute \n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "pmi=p-pp      \n",
+      "c=((N/2.0)-n)   \n",
+      "ipw=pmi*10**5*L*(math.pi/4.0)*d**2*(n/60.0)*(1/1000.0) \n",
+      "Pp=pm*10**5*L*(math.pi/4.0)*d**2*(c/60.0)*(1/1000.0)   \n",
+      "fp=ipw-Pp      #Power in kW\n",
+      "fip=pmi*10**5*L*(math.pi/4.0)*d**2*(N/(2*60))*(1/1000.0)\n",
+      "fbp=fip-fp\n",
+      "nm=(fbp/fip)*100\n",
+      "\n",
+      "#Output \n",
+      "print\" The full load break power = \",round(fbp,2),\"kW\" \n",
+      "print\"The mechanical efficiency of the engine = \",round(nm,2),\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The full load break power =  17.32 kW\n",
+        "The mechanical efficiency of the engine =  86.79 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3 page no: 522"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=0.09                          #The diameter of the bore in m\n",
+      "L=0.1                           #The length of the stroke in m\n",
+      "T=120                           #The torque measured in Nm\n",
+      "n=4                             #Number of cylinders \n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "pmb=((4*math.pi*T)/(L*(math.pi/4)*d**2*n))/10.0**5\n",
+      "\n",
+      "#Output \n",
+      "print\"The brake mean effective pressure = \",round(pmb,2),\"bar\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The brake mean effective pressure =  5.93 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.4 page no: 522"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Input data\n",
+      "d=0.06                       #The diameter of the bore in m \n",
+      "L=0.085                      #The length of the stroke in m\n",
+      "N=3000                       #The speed of the engine in rpm\n",
+      "r=0.35                       #Torque arm radius in m\n",
+      "W=160                        #Weight in N\n",
+      "f=6.6                        #Fuel consumption in l/h\n",
+      "g=0.78                       #specific gravity of the fuel \n",
+      "CV=44000                     #The calorific value of the fuel in kJ/kg\n",
+      "w1=114                       #Brake load for cylinder 1 in N\n",
+      "w2=110                       #Brake load for cylinder 2 in N\n",
+      "w3=112                       #Brake load for cylinder 3 in N\n",
+      "w4=116                       #Brake load for cylinder 4 in N\n",
+      "n=4                          #Number of cylinders\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Vf=(f*10**-3)/3600.0\n",
+      "df=g*1000\n",
+      "mf=df*Vf\n",
+      "T=W*r\n",
+      "bp=(2*math.pi*N*T)/(60.0*1000.0)\n",
+      "pmb=((120*bp*1000)/(L*(math.pi/4.0)*d**2*N*n))/10.0**5\n",
+      "nb=((bp)/(mf*CV))*100\n",
+      "bsfc=(mf*3600)/bp\n",
+      "bp1=((2*math.pi*N*w1*r)/(60.0*1000.0))\n",
+      "ip1=bp-bp1     \n",
+      "ip2=bp-((2*math.pi*N*w2*r)/(60*1000))\n",
+      "ip3=bp-((2*math.pi*N*w3*r)/(60*1000))\n",
+      "ip4=bp-((2*math.pi*N*w4*r)/(60*1000))\n",
+      "ip=ip1+ip2+ip3+ip4\n",
+      "nm=(bp/ip)*100\n",
+      "pmi=pmb/(nm/100.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The brake power =\",round(bp,3),\"kW \" \n",
+      "print\"The brake mean effective pressure = \",round(pmb,3),\"bar\"\n",
+      "print\"The brake thermal efficiency =\" ,round(nb,0),\"percent\"\n",
+      "print\"The brake specific fuel consumption = \",round(bsfc,3),\"kg/kWh\"\n",
+      "print\"The indicated power = \",round(ip,2),\"kW \"\n",
+      "print\"The mechanical efficiency =\", round(nm,1),\"percent \" \n",
+      "print\"The indicated mean effective pressure = \",round(pmi,1),\"bar\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The brake power = 17.593 kW \n",
+        "The brake mean effective pressure =  7.32 bar\n",
+        "The brake thermal efficiency = 28.0 percent\n",
+        "The brake specific fuel consumption =  0.293 kg/kWh\n",
+        "The indicated power =  20.67 kW \n",
+        "The mechanical efficiency = 85.1 percent \n",
+        "The indicated mean effective pressure =  8.6 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.5 page no: 523"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=0.15                               #The diameter of the bore in m\n",
+      "L=0.16                               #The length of the stroke in m\n",
+      "N=500                                #The speed of the engine in rpm\n",
+      "mf=0.0475                            #Fuel consumption in kg/min\n",
+      "CV=42000                             #The calorific value in kJ/kg\n",
+      "w=400                                #The tension on either side of the pulley in N\n",
+      "c=2.2                                #Brake circumference in m\n",
+      "l=50                                 #Length of the indicator diagram in mm\n",
+      "ap=475                               #Area of the positive loop of indicator diagram in mm**2\n",
+      "an=25                                #Area of the negative loop of indicator diagram in mm**2\n",
+      "s=0.8333                             #Spring constant in bar/mm\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "r=c/(2.0*math.pi)\n",
+      "T=w*r\n",
+      "bp=(2*math.pi*N*T)/(60.0*1000.0)\n",
+      "M=(ap-an)/l\n",
+      "imep=M*s\n",
+      "ip=(imep*10**5*L*(math.pi/4.0)*d**2*(N/(2.0*60.0))*(1/1000.0))\n",
+      "nm=(bp/ip)*100\n",
+      "nb=((bp*60)/(mf*CV))*100\n",
+      "ni=((nb/100.0)/(nm/100.0))*100\n",
+      "bsfc=(mf*60)/bp\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The brake power = \",round(bp,3),\"kW\"\n",
+      "print \"(b) The indicated power = \",round(ip,3),\"kW\"\n",
+      "print\"(c) The mechanical efficiency = \",round(nm,0),\"percent\"\n",
+      "print\"(d) The brake thermal efficiency = \",round(nb,3)\n",
+      "print\"(e) The indicated thermal efficiency = \",round(ni,3)\n",
+      "print\"(f) The brake specific fuel consumption = \",round(bsfc,3),\" kg/kWh\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The brake power =  7.333 kW\n",
+        "(b) The indicated power =  8.835 kW\n",
+        "(c) The mechanical efficiency =  83.0 percent\n",
+        "(d) The brake thermal efficiency =  22.055\n",
+        "(e) The indicated thermal efficiency =  26.573\n",
+        "(f) The brake specific fuel consumption =  0.389  kg/kWh\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.6 page no: 524"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "n=8                               #Number of cylinders\n",
+      "d=0.08                            #The diameter of the bore in m\n",
+      "L=0.1                             #The length of the stroke in m\n",
+      "N=4500                            #The speed of the engine in rpm \n",
+      "dy=0.55                           #The dynamometer readings in m\n",
+      "w=40                              #The weight of the dynamometer scale reading in kg\n",
+      "c=100                             #Fuel consumption in cc\n",
+      "t=9.5                             #Time taken for fuel consumption in s\n",
+      "CV=44000                          #The calorific value of the fuel in kJ/kg\n",
+      "p=1                               #The atmospheric air pressure in bar\n",
+      "T=300                             #The atmospheric air temperature in K\n",
+      "ma=6                              #Mass flow rate of air in kg/min\n",
+      "g=0.7                             #Specific gravity of the fuel \n",
+      "Vc=65                             #The clearance volume of each cylinder in cc\n",
+      "R=287                             #Real gas constant in J/kgK\n",
+      "g=1.4                             #Isentropic index\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "bp=(2*math.pi*N*dy*w*9.81)/(60.0*1000.0)\n",
+      "bmep=((bp*1000*60)/(L*(math.pi/4.0)*d**2*(N/2.0)*n))/10.0**5\n",
+      "mf=(c*g*3600)/(t*2.0*1000.0)\n",
+      "bsfc=(mf/bp)\n",
+      "bsac=(ma*60)/bp\n",
+      "a=bsac/bsfc\n",
+      "nb=((bp*3600)/(mf*CV))*100\n",
+      "Va=(ma*R*T)/(p*10.0**5)\n",
+      "Vs=(math.pi/4.0)*d**2*L*(N/2.0)*n\n",
+      "nv=(Va/Vs)*100\n",
+      "Vs1=((math.pi/4.0)*d**2*L)*10**6\n",
+      "cr=(Vs1+Vc)/Vc\n",
+      "na=(1-(1/cr)**(g-1))*100\n",
+      "re=((nb)/(na))*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The brake power = \",round(bp,1),\"kW\" \n",
+      "print\"The brake mean effective pressure = \",round(bmep,3),\"bar\"\n",
+      "print\"The brake specific fuel consumption = \",round(bsfc,3),\"kg/kWh\"\n",
+      "print\"The brake specific air consumption = \",round(bsac,3),\"kg/kWh\"\n",
+      "print\"The air fuel ratio = \",round(a,2)\n",
+      "print\"The brake thermal efficiency = \",round(nb,3),\"percent\" \n",
+      "print\"The volumetric efficiency = \",round(nv,1),\"percent\"\n",
+      "print\"The relative efficiency = \",round(re,1),\"percent\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The brake power =  101.7 kW\n",
+        "The brake mean effective pressure =  6.744 bar\n",
+        "The brake specific fuel consumption =  0.261 kg/kWh\n",
+        "The brake specific air consumption =  3.54 kg/kWh\n",
+        "The air fuel ratio =  13.57\n",
+        "The brake thermal efficiency =  31.369 percent\n",
+        "The volumetric efficiency =  57.1 percent\n",
+        "The relative efficiency =  54.1 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.7 page no: 526"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "n=6                              #Number of cylinders\n",
+      "Do=0.03                          #Orifice diameter in m\n",
+      "Cd=0.6                           #Coefficient of discharge \n",
+      "H=0.14                           #Pressure drop across the orifice\n",
+      "d=0.1                            #The diameter of the bore in m\n",
+      "L=0.11                           #The length of the stroke in m\n",
+      "W=540                            #Brake load in N\n",
+      "N=2500                           #Engine speed in rpm\n",
+      "ch=83/17                         #C/H ratio by mass\n",
+      "p=1                              #Ambient pressure in bar\n",
+      "t=18                             #Time taken for fuel consumption in s\n",
+      "f=100                            #The amount of fuel consumption in cc\n",
+      "T=300.0                          #Ambient air temperature in K\n",
+      "df=780                           #The density of the fuel in kg/m**3\n",
+      "R=287.0                          #Real gas constant in J/kgK\n",
+      "g=9.81                           #Gravitational force constant in m/s**2\n",
+      "dhg=13600                        #Density of Hg in kg/m**3\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "da=(p*10**5)/(R*T)\n",
+      "Va=(Cd*(math.pi/4.0)*Do**2*(2*g*H*(dhg/da))**(1/2.0))\n",
+      "Vs=(math.pi/4.0)*d**2*L*(N/(2.0*60.0))*n\n",
+      "nv=(Va/Vs)*100\n",
+      "bp=(W*N)/(20000.0)\n",
+      "bmep=((bp*1000)/(L*(math.pi/4.0)*d**2*(N/(2.0*60.0))*n))/10.0**5\n",
+      "T=(60*bp*1000)/(2.0*math.pi*N)\n",
+      "mf=(f/18.0)*(780/1000.0)*(1/1000.0)*3600\n",
+      "bsfc=mf/bp\n",
+      "so=(0.83*(32/12.0))+(0.17*(8/1.0))\n",
+      "sa=so/bsfc\n",
+      "maa=Va*da\n",
+      "af=(maa*3600)/mf\n",
+      "pea=((af-sa)/sa)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The volumetric efficiency = \",round(nv,3),\"percent\" \n",
+      "print\"The brake mean effective pressure = \",round(bmep,3),\"bar\" \n",
+      "print\"The brake power = \",bp,\"kW\" \n",
+      "print\"The Torque = \",round(T,3),\"Nm\"\n",
+      "print\"The brake specific fuel consumption = \",round(bsfc,3),\"kg/kWh\" \n",
+      "print\"The percentage of excess air = \",round(pea,1),\"percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volumetric efficiency =  70.433 percent\n",
+        "The brake mean effective pressure =  6.25 bar\n",
+        "The brake power =  67.5 kW\n",
+        "The Torque =  257.831 Nm\n",
+        "The brake specific fuel consumption =  0.231 kg/kWh\n",
+        "The percentage of excess air =  31.9 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.8 page no: 528"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=0.2                               #The diameter of bore in m\n",
+      "L=0.3                               #The length of the stroke in m\n",
+      "r=5.5                               #The compression ratio of the engine\n",
+      "N=400                               #The speed of the engine in rpm\n",
+      "imep=4.5                            #The indicative mean effective pressure in bar\n",
+      "a=6                                 #Air to gas by volume \n",
+      "CV=12000                            #The calorific value of the gas in kJ/m**3\n",
+      "T=340                               #The temperature at the beginning of the compression stroke in K\n",
+      "p=0.97                              #The pressure at the beginning of the compression stroke in bar\n",
+      "g=1.4                               #Adiabatic index\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Vs=(math.pi/4.0)*d**2*L\n",
+      "Vc=Vs/(r-1)\n",
+      "V=Vs+Vc\n",
+      "Vg=V/7.0\n",
+      "Vntp=((p*Vg)/T)*(273/1.013)\n",
+      "Q=Vntp*CV*(N/(2.0*60.0))\n",
+      "ip=(imep*10**5*L*(math.pi/4.0)*d**2*(N/(2.0*60.0))*(1/1000.0))\n",
+      "ni=(ip/Q)*100\n",
+      "na=(1-(1/r)**(g-1))*100\n",
+      "nr=(ni/na)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The indicated power = \",round(ip,2),\"kW\" \n",
+      "print\"The thermal efficiency = \",round(ni,1),\"percent\" \n",
+      "print\"The relative efficiency = \",round(nr,1),\"percent\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The indicated power =  14.14 kW\n",
+        "The thermal efficiency =  27.9 percent\n",
+        "The relative efficiency =  56.5 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.9 page no: 529"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "n=6.0                                     #Number of cylinder\n",
+      "bp=130.0                                  #Brake power in kW\n",
+      "N=1800.0                                  #The speed of the engine in rpm \n",
+      "CV=42000.0                                #The calorific value of the fuel in kJ/kg\n",
+      "C=86.0                                    #The composition of carbon in the fuel in percent\n",
+      "H=13.0                                    #The composition of Hydrogen in the fuel in percent\n",
+      "NC=1.0                                    #The non combustibles present in the fuel in percent\n",
+      "na=85.0                                   #The absolute volumetric efficiency in percent\n",
+      "ni=38.0                                   #The indicated thermal efficiency in percent\n",
+      "nm=80.0                                   #The mechanical efficiency in percent\n",
+      "ac=110.0                                  #The excess consumption of air in percent\n",
+      "sb=1.2                                  #The stroke to the bore ratio \n",
+      "da=1.3                                  #The density of air in kg/m**3\n",
+      "\n",
+      "#Calculations \n",
+      "import math\n",
+      "saf=(((C/100.0)*(32/12.0))+((H/100.0)*(8/1.0)))*(1/0.23)\n",
+      "aaf=saf*(1+1.1)\n",
+      "Ma=(0.23*32)+(0.77*28)\n",
+      "a=(C/100)/12.0\n",
+      "b=(H/100.0)/2.0\n",
+      "x=aaf/Ma\n",
+      "c=(0.21*x)-a-(b/2.0)\n",
+      "d1=0.79*x\n",
+      "ip=bp/(nm/100.0)\n",
+      "mf=ip/((ni/100.0)*CV)\n",
+      "ma=mf*aaf\n",
+      "Va=ma/da\n",
+      "Vs=Va/(na/100.0)#The swept volume per second in m**3/s\n",
+      "d=((Vs*(4/math.pi)*(1/1.2)*((2*60)/N)*(1/n))**(1/3.0))*1000\n",
+      "L=1.2*d\n",
+      "T=a+c+d1\n",
+      "CO2=(a/T)*100\n",
+      "O2=(c/T)*100\n",
+      "N2=(d1/T)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The volumetric composition of dry exhaust gas :\" \n",
+      "print\"1) CO2 = \",round(a,2),\"kmol\",\"and\"   \n",
+      "print\"volume = \",round(CO2,2),\"percent\"  \n",
+      "print\"2) O2 = \",round(c,3),\"kmol\",\"and\"   \n",
+      "print\"volume = \",round(O2,3),\"percent\"\n",
+      "print\"3) N2 = \",round(d1,3),\"kmol\",\"and\"   \n",
+      "print\"volume = \",round(N2,3),\"percent\"\n",
+      "print\"The bore of the engine = \",round(d,0),\"mm\" \n",
+      "print\"The stroke of the engine = \",round(L,1),\"mm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The volumetric composition of dry exhaust gas :\n",
+        "1) CO2 =  0.07 kmol and\n",
+        "volume =  7.03 percent\n",
+        "2) O2 =  0.117 kmol and\n",
+        "volume =  11.456 percent\n",
+        "3) N2 =  0.831 kmol and\n",
+        "volume =  81.517 percent\n",
+        "The bore of the engine =  149.0 mm\n",
+        "The stroke of the engine =  178.8 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.10 page no: 531"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=0.18                                  #The diameter of the cylinder in m\n",
+      "L=0.24                                  #The length of the stroke in m\n",
+      "t=30.0                                    #Duration trail in min \n",
+      "N=9000.0                                  #Number of revolutions \n",
+      "Ne=4450.0                                 #Total number of explosions\n",
+      "pmi=5.35                                #Gross imep in bar\n",
+      "pp=0.35                                 #Pumping imep in bar\n",
+      "W=40.0                                    #Net load on brake wheel in kg\n",
+      "dd=0.96                                 #Diameter of the brake wheel drum in m\n",
+      "dr=0.04                                 #Diameter of the rope in m\n",
+      "V=2.6                                   #Volume of gas used in m**3\n",
+      "pg=136.0                                  #pressure of gas in mmof Hg\n",
+      "dg=0.655                                #The density of gas in kg/m**3\n",
+      "T=290.0                                   #The ambient temperature of air in K\n",
+      "CV=19000.0                                #The calorific value of the fuel in kJ/m**3\n",
+      "ta=40.0                                   #Total air used in m**3\n",
+      "p=720.0                                   #Pressure of air in mm of Hg\n",
+      "Te=340.0                                 #Temperature of exhaust gas in degree centigrade \n",
+      "Cpg=1.1                                  #Specific heat of gas in kJ/kgK\n",
+      "C=80.0                                    #Cooling water circulated in kg\n",
+      "Tr=30.0                                   #Rise in temperature of cooling water in degree centigrade \n",
+      "R=287.0                                   #Real gas constant in J/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "#import math\n",
+      "ip=(pmi-pp)*10**5*L*(math.pi/4.0)*d**2*(Ne/(30.0*60.0))*(1/1000.0)\n",
+      "bp=(math.pi*(N/(30.0*60.0))*W*9.81*(dd+dr)*(1/1000.0))\n",
+      "pgs=760+(pg/13.6)\n",
+      "Vg=((pgs*V)/290.0)*(273/760.0)\n",
+      "Q=(Vg*CV)/30.0\n",
+      "Qbp=bp*60\n",
+      "Qc=(C/t)*4.18*Tr\n",
+      "Va=(((p*ta)/T)*(273/760.0))/30.0\n",
+      "da=(1.013*10**5)/(R*273)\n",
+      "ma=Va*da\n",
+      "mg=(Vg/30)*dg\n",
+      "me=ma+mg\n",
+      "Qe=me*Cpg*(Te-(T-273))\n",
+      "Qu=Q-(Qe+Qc+Qbp)\n",
+      "nm=(bp/ip)*100\n",
+      "ni=((ip*60)/Q)*100 \n",
+      "x=((Qbp/1571.0)*100)\n",
+      "y=((Qc/1571.0)*100)\n",
+      "z=((Qe/1571.0)*100)\n",
+      "k=((Qu/1571.0)*100)\n",
+      "Qf=Qbp+Qc+Qe+Qu\n",
+      "\n",
+      "#Output\n",
+      "print Qe\n",
+      "print\"                                        HEAT BALANCE SEAT\"\n",
+      "\n",
+      "print \"  \\nHeat input             kj/min    %      heat expenditure               Kj/min            %\"\n",
+      "print\"--------------------------------------------------------------------------------------------------\"\n",
+      "print\"heat supplied by fuel \",round(Q,0),\"   100       (a)Heat in bp                 \",   round(Qbp,1),\"           \",round(x,1)\n",
+      "print\"                                           (b)Heat loss to cooling tower  \",round(Qc,0),\"         \",round(y,1)\n",
+      "print\"                                           (c) Heat to exhaust gases      \",round(Qe,0),\"         \",round(z,0)\n",
+      "print\"                                           (d)unaccounted losses          \",round(Qu,0),\"         \",round(k,1)\n",
+      "print\"Total                \",round(Q,0),\"   100            total                      \",round(Qf,0),    \"        100.0\"\n",
+      " \n",
+      "print\"\\nMechanical efficiency is\",round(nm,2),\"percent\"\n",
+      "print\"Thermal efficiency is\",round(ni,1),\"percent\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "565.475307496\n",
+        "                                        HEAT BALANCE SEAT\n",
+        "  \n",
+        "Heat input             kj/min    %      heat expenditure               Kj/min            %\n",
+        "--------------------------------------------------------------------------------------------------\n",
+        "heat supplied by fuel  1571.0    100       (a)Heat in bp                  369.8             23.5\n",
+        "                                           (b)Heat loss to cooling tower   334.0           21.3\n",
+        "                                           (c) Heat to exhaust gases       565.0           36.0\n",
+        "                                           (d)unaccounted losses           301.0           19.1\n",
+        "Total                 1571.0    100            total                       1571.0         100.0\n",
+        "\n",
+        "Mechanical efficiency is 81.65 percent\n",
+        "Thermal efficiency is 28.8 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.11 page no: 533"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given \n",
+      "bp=30                          #The brake power in kw\n",
+      "mf=10                          #Mass flow rate of fuel in kg/h\n",
+      "CV=42000                       #Calorific value of the fuel in kJ/kg\n",
+      "mw=9                           #Mass flow rate of water in kg/min\n",
+      "Tr=60                          #Rise in temperature of the cooling water in degree centigrade\n",
+      "mwe=9.5                        #Mass flow rate of water through exhaust gas calorimeter in kg/min\n",
+      "Tc=40                          #Rise in temperature when passing through calorimeter in degree centigrade\n",
+      "Te=80                          #Temperature of exhaust gas leaving the calorimeter in degree centigrade\n",
+      "a=20                           #Air fuel ratio\n",
+      "T=17                           #Ambient temperature in degree centigrade\n",
+      "Cpw=4.18                       #Specific heat of water in kJ/kgK\n",
+      "Cpg=1                          #Mean specific heat of gas in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "Qf=(mf/60.0)*CV               \n",
+      "Qbp=bp*60\n",
+      "Qc=mw*Cpw*Tr\n",
+      "mg=(mf/60.0)+(mf/60.0)*a\n",
+      "Qe=(mwe*Cpw*Tc)+(mg*Cpg*(Te-T))\n",
+      "Qu=Qf-(Qbp+Qc+Qe)\n",
+      "x=((Qbp/Qf))*100\n",
+      "y=(Qc/Qf)*100\n",
+      "z=(Qe/Qf)*100\n",
+      "k=(Qu/Qf)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"                                        HEAT BALANCE SEAT\"\n",
+      "print \"\\n  Heat input             kj/min    %      heat expenditure               Kj/min            %\"\n",
+      "print\"--------------------------------------------------------------------------------------------------\"\n",
+      "print\"heat supplied by fuel \",Qf,\"   100       (a)Heat in bp                 \",   Qbp,\"           \",round(x,0)\n",
+      "print\"                                           (b)Heat loss to cooling tower  \",round(Qc,0),\"         \",round(y,0)\n",
+      "print\"                                           (c) Heat to exhaust gases      \",round(Qe,0),\"         \",round(z,0)\n",
+      "print\"                                           (d)unaccounted losses          \",round(Qu,0),\"         \",round(k,0)\n",
+      "print\"Total                    \",Qf,\"    100            total                  \",Qf,    \"           100\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "                                        HEAT BALANCE SEAT\n",
+        "\n",
+        "  Heat input             kj/min    %      heat expenditure               Kj/min            %\n",
+        "--------------------------------------------------------------------------------------------------\n",
+        "heat supplied by fuel  7000.0    100       (a)Heat in bp                  1800             26.0\n",
+        "                                           (b)Heat loss to cooling tower   2257.0           32.0\n",
+        "                                           (c) Heat to exhaust gases       1809.0           26.0\n",
+        "                                           (d)unaccounted losses           1134.0           16.0\n",
+        "Total                     7000.0     100            total                   7000.0            100\n"
+       ]
+      }
+     ],
+     "prompt_number": 85
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.12 page no: 534"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "n=4.0                                          #Number of cylinders\n",
+      "d=0.085                                      #The diameter of the bore m\n",
+      "L=0.095                                      #The length of the stroke in m\n",
+      "tr=0.35                                      #Torque radius in m\n",
+      "N=3000.0                                       #The speed of the engine in rpm\n",
+      "w=430.0                                        #Net brake load in N\n",
+      "w1=300.0                                       #Net brake load produced at the same speed by three cylinders in N\n",
+      "mf=0.24                                      #The mass flow rate of fuel in kg/min\n",
+      "CV=44000.0                                     #The calorific value of the fuel in kJ/kg\n",
+      "mw=65.0                                        #Mass flow rate of water in kg/min\n",
+      "Tw=12.0                                        #The rise in temperature in degree centigrade\n",
+      "a=15.0                                         #The air fuel ratio \n",
+      "Te=450.0                                       #The temperature of the exhaust gas in degree centigrade \n",
+      "Ta=17.0                                        #Ambient temperature in degree centigrade\n",
+      "p=76.0                                         #Barometric pressure in cm of Hg\n",
+      "H=15.5                                       #The proportion of hydrogen by mass in the fuel in percent\n",
+      "Cpe=1.0                                        #The mean specific heat of dry exhaust gas in kJ/kgK\n",
+      "Cps=2.0                                        #The specific heat of super heated steam in kJ/kgK\n",
+      "Cpw=4.18                                     #The specific heat of water in kJ/kgK\n",
+      "Ts=100.0                                       #At 76 cm of Hg The temperature in degree centigrade \n",
+      "hfg=2257                                     #The Enthalpy in kJ/kg\n",
+      "R=287.0                                        #Real gas cons tant in J/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "bp=(2*math.pi*N*w*tr)/(60.0*1000.0)\n",
+      "bp1=(2*math.pi*N*w1*0.35)/(60.0*1000.0)\n",
+      "ip=bp-bp1\n",
+      "ip1=n*ip\n",
+      "imep=((ip1*60*1000)/(L*(math.pi/4.0)*d**2*(N/2.0)*n))/10.0**5\n",
+      "ni=((ip1*60)/(mf*CV))*100\n",
+      "bsfc=(mf*60)/bp\n",
+      "Vs=(math.pi/4.0)*d**2*L*(N/2.0)*n\n",
+      "ma=a*mf\n",
+      "da=(1*10**5)/(R*(Ta+273))\n",
+      "Va=ma/da\n",
+      "nv=(Va/Vs)*100\n",
+      "Qf=mf*CV\n",
+      "Qbp=bp*60\n",
+      "Qc=mw*Cpw*Tw\n",
+      "mv=9*(H/100.0)*mf\n",
+      "me=ma+mf-mv\n",
+      "Qe=me*Cpe*(Te-Ta)\n",
+      "Qs=(mv*((Cpw*(Ts-Ta))+hfg+(Cps*(Te-Ts))))\n",
+      "Qu=Qf-(Qbp+Qc+Qe+Qs)\n",
+      "x=(Qbp/Qf)*100\n",
+      "y=(Qc/Qf)*100\n",
+      "z=(Qe/Qf)*100\n",
+      "k=(Qs/Qf)*100\n",
+      "l=(Qu/Qf)*100 \n",
+      "\n",
+      "#Output\n",
+      "print\"                                        HEAT BALANCE SEAT\"\n",
+      "print \"  Heat input             kj/min    %      heat expenditure                  Kj/min           %\"\n",
+      "print\"----------------------------------------------------------------------------------------------------\"\n",
+      "print\"heat supplied by fuel \",Qf,\"   100       (a)Heat in bp                 \",   Qbp,\"  \",round(x,2)\n",
+      "print\"                                           (b)Heat loss to cooling tower  \",round(Qc,0),\"         \",round(y,2)\n",
+      "print\"                                           (c) Heat to dry exhaust gases  \",round(Qe,0),\"         \",round(z,1)\n",
+      "print\"                                           (d)Heat loss in steam          \",round(Qs,0),\"         \",round(k,1)\n",
+      "print\"                                           (e)unaccounted losses          \",round(Qu,0),\"         \",round(l,2)\n",
+      "print\"Total               \",Qf,\"    100            total                     \",Qf,    \"        100\"\n",
+      "\n",
+      "print\"\\nimpep   \",round(imep,2),\"bar\"\n",
+      "print\"Thermal efficiency  \",round(ni,1),\"percent\"\n",
+      "print\"bsfc=\",round(bsfc,4),\"kg/kwh\"\n",
+      "print\"Volumetric efficiency=\",round(nv,1),\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "                                        HEAT BALANCE SEAT\n",
+        "  Heat input             kj/min    %      heat expenditure                  Kj/min           %\n",
+        "----------------------------------------------------------------------------------------------------\n",
+        "heat supplied by fuel  10560.0    100       (a)Heat in bp                  2836.85816619    26.86\n",
+        "                                           (b)Heat loss to cooling tower   3260.0           30.87\n",
+        "                                           (c) Heat to dry exhaust gases   1518.0           14.4\n",
+        "                                           (d)Heat loss in steam           1106.0           10.5\n",
+        "                                           (e)unaccounted losses           1839.0           17.41\n",
+        "Total                10560.0     100            total                      10560.0         100\n",
+        "\n",
+        "impep    10.61 bar\n",
+        "Thermal efficiency   32.5 percent\n",
+        "bsfc= 0.3046 kg/kwh\n",
+        "Volumetric efficiency= 92.6 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/chap3.ipynb b/Fundamental_of_internal_combustion_engines/chap3.ipynb
new file mode 100755
index 00000000..a0a9064e
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/chap3.ipynb
@@ -0,0 +1,1164 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3:Reactive Sysyem"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page no 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "E=20.0                           #Methanol burned with excess air in percentage \n",
+      "p=1.0                            #Pressure of air in bar\n",
+      "t=27.0                           #Temperature of air in degree centigrade\n",
+      "O=32.0                           #The molecular weight of oxygen\n",
+      "N=28.0                           #The molecular weight of nitrogen\n",
+      "R=8314.0                         #Universal gas constant in Nm/kmolK\n",
+      "C=32.0                           #Molecular weight of methanol\n",
+      "CO=44.0                          #Molecular weight of the carbondioxide \n",
+      "H=18.0                           #Molecular weight of the water\n",
+      "\n",
+      "#Calculations\n",
+      "S=((1.8*O)+(6.768*N))/C\n",
+      "A=((1.8*O)+(6.768*N))/C\n",
+      "M=1.8+6.768\n",
+      "V=(M*R*(t+273))/(p*10**5)\n",
+      "T=(1+1.8+6.768)\n",
+      "Cm=(1/T)\n",
+      "Om=(1.8/T)\n",
+      "Nm=(6.768/T)\n",
+      "Mr=(Cm*C)+(Om*O)+(Nm*N)\n",
+      "Tp=(1+2+6.768+0.3)\n",
+      "COm=(1/Tp)\n",
+      "Hp=(2/Tp)\n",
+      "Np=(6.768/Tp)\n",
+      "Op=(0.3/Tp)\n",
+      "Mp=(COm*CO)+(Hp*H)+(Np*N)+(Op*O)\n",
+      "Pp=(Hp*p)\n",
+      "D=60\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The volume of air supplied per kmole of fuel = \",round(V,3),\"m**3/kmole fuel\"\n",
+      "print\"(b) The molecular weight of the reactants = \",round(Mr,3)\n",
+      "print\"The molecular weight of the products =\",round(Mp,3)\n",
+      "print\"(c) The dew point of the products = \",D,\"degree centigrade\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The volume of air supplied per kmole of fuel =  213.703 m**3/kmole fuel\n",
+        "(b) The molecular weight of the reactants =  29.171\n",
+        "The molecular weight of the products = 27.722\n",
+        "(c) The dew point of the products =  60 degree centigrade\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page no 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "C1=40.0                               #The content of C7H16 in the fuel in percentage\n",
+      "C2=60.0                               #The content of C8H18 in the fuel in percentage\n",
+      "d=0.12                              #The diameter of the bore in m\n",
+      "l=0.145                             #The length of the bore in m\n",
+      "r=8.5                               #Compression ratio \n",
+      "p=1.1                               #Pressure at exhaust stroke in bar\n",
+      "T=720.0                               #The temperature at the exhaust stroke in K\n",
+      "O=32.0                                #The molecular weight of oxygen\n",
+      "N=28.0                                #The molecular weight of nitrogen\n",
+      "C3=100.0                              #Molecular weight of C7H16\n",
+      "C4=114.0                              #The molecular weight of C8H18\n",
+      "R=8314.0                              #Universal gas constant in Nm/kmolK\n",
+      "CO2=44.0                              #Molecular weight of the carbondioxide \n",
+      "C5=28.0                               #Molecular weight of the carbonmonoxide\n",
+      "H=18.0                                #Molecular weight of the water\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "N2=100-(12+1.5+2.5)\n",
+      "Y=84/3.76\n",
+      "X=13.5/7.6\n",
+      "Z=(22.34-15.25)*2\n",
+      "Hl=(6.4+10.8)/2.0\n",
+      "Hr=7.98\n",
+      "Hd=Hl-Hr\n",
+      "A=((12.58*(O+(3.76*N)))/(((C1/100.0)*C3)+((C2/100.0)*C4)))\n",
+      "Vs=(math.pi/4.0)*d**2*l\n",
+      "Vc=Vs/(r-1)\n",
+      "M=((6.757*CO2)+(0.8446*C5)+(1.408*O)+(47.3*N)+(8.6*H))/(6.757+0.8446+1.408+47.3+8.6)\n",
+      "R1=R/M\n",
+      "m=((p*10**5)*Vc)/(R1*T)\n",
+      "\n",
+      "#Output \n",
+      "print\"(a)The air/fuel ratio = \",round(A,2)\n",
+      "print\"(b)The mass of the exhaust gases in the clearance space = \",round(m*1000,3),\"*10**-3kg\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The air/fuel ratio =  15.93\n",
+        "(b)The mass of the exhaust gases in the clearance space =  0.114 *10**-3kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page no 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "C=0.86                        #The amount of carbon content in the 1kg of fuel by weight in kg\n",
+      "H=0.05                        #The amount of hydrogen content in the 1kg of fuel by weight in kg\n",
+      "O=0.02                        #The amount of oxygen content in the 1kg of fuel by weight in kg\n",
+      "S=0.005                       #The amount of sulphur content in the 1kg of fuel by weight in kg\n",
+      "N=0.065                       #The amount of nitrogen content in the 1kg of fuel by weight in kg\n",
+      "E=25.0                          #The amount of excess air supplied in percentage\n",
+      "o=32.0                          #Molecular weight of the oxygen\n",
+      "co=44.0                         #Molecular weight of the carbondioxide \n",
+      "c=12.0                        #Molecular weight of the carbon\n",
+      "s=32.0                        #Molecular weight of the sulphur\n",
+      "so=64.0                         #Molecular weight of sulphur dioxide\n",
+      "n=28.0                          #Molecular weight of the nitrogen\n",
+      "\n",
+      "#Calculations\n",
+      "o1=(o/c)*C\n",
+      "coa=(co/c)*C\n",
+      "o2=(o/4.0)*H\n",
+      "h2=(36/4.0)*H\n",
+      "o3=(o/s)*S\n",
+      "s1=(so/s)*S\n",
+      "To=o1+o2+o3\n",
+      "Tt=To-O\n",
+      "As=(Tt*100)/23.0\n",
+      "as_=As*(1+(E/100.0))\n",
+      "o2a=0.23*(E/100)*As\n",
+      "n2a=0.77*(1+(E/100))*As\n",
+      "n2e=n2a+N\n",
+      "Tw=coa+n2e+o2a\n",
+      "pco=(coa/Tw)*100\n",
+      "pn=(n2e/Tw)*100\n",
+      "po=(o2a/Tw)*100\n",
+      "mco=(coa/co)\n",
+      "mn=(n2e/n)\n",
+      "mo=(o2a/o)\n",
+      "Tm=mco+mn+mo\n",
+      "vco=(mco/Tm)*100\n",
+      "vn=(mn/Tm)*100\n",
+      "vo=(mo/Tm)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)Stoichiometric air/fuel ratio = \",round(As,2) \n",
+      "print\"(b)The percentage of dry products of combustion by weight :\"\n",
+      "print\" CO2 = \",round(pco,2),\"percent\"\n",
+      "print\"N2 = \",round(pn,2),\"percent\"\n",
+      "print\"O2 = \",round(po,2),\"percent\"\n",
+      "print\"(c)The percentage of dry products of combustion by volume : \"\n",
+      "print\"CO2 = \",round(vco,2),\"percent\"\n",
+      "print\"N2 = \",round(vn,2),\"percent\"\n",
+      "print\"O2= \",round(vo,2),\"percent\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Stoichiometric air/fuel ratio =  11.64\n",
+        "(b)The percentage of dry products of combustion by weight :\n",
+        " CO2 =  20.89 percent\n",
+        "N2 =  74.68 percent\n",
+        "O2 =  4.44 percent\n",
+        "(c)The percentage of dry products of combustion by volume : \n",
+        "CO2 =  14.47 percent\n",
+        "N2 =  81.3 percent\n",
+        "O2=  4.23 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 Page no 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "CO=12.0                    #The composition of carbondioxide of combustion by volume in percentage \n",
+      "C=0.5                    #The composition of carbonmoxide of combustion by volume in percentage \n",
+      "O=4.0                      #The composition of oxygen of combustion by volume in percentage \n",
+      "N=83.5                   #The composition of nitrogen of combustion by volume in percentage \n",
+      "o=32.0                     #Molecular weight of the oxygen\n",
+      "co=44.0                    #Molecular weight of the carbondioxide \n",
+      "c=12.0                     #Molecular weight of the carbon\n",
+      "s=32.0                     #Molecular weight of the sulphur\n",
+      "so=64.0                    #Molecular weight of sulphur dioxide\n",
+      "n1=28.0                    #Molecular weight of the nitrogen\n",
+      "h=2.0                      #Molecular weight of the hydrogen\n",
+      "\n",
+      "#Calculations\n",
+      "m=12+0.5\n",
+      "x=N/3.76\n",
+      "z=(x-(CO+(C/2)+O))*2\n",
+      "n=z*h\n",
+      "Af=((x*o)+(N*n1))/((m*c)+(n))\n",
+      "As=((18.46*o)+(69.41*n1))/173.84\n",
+      "Ta=(Af/As)*100\n",
+      "mc=((m*c)/173.84)*100\n",
+      "mh=(n/173.84)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)The air/fuel ratio = \",round(Af,1)\n",
+      "print\"(b)The percent theoretical air = \",round(Ta,1),\"%\"\n",
+      "print\"(c)The percentage composition of fuel on a mass basis : \" \n",
+      "print\"C = \",round(mc,1),\"%\"\n",
+      "print\"H = \",round(mh,1),\"percent\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The air/fuel ratio =  17.5\n",
+        "(b)The percent theoretical air =  120.3 %\n",
+        "(c)The percentage composition of fuel on a mass basis : \n",
+        "C =  86.3 %\n",
+        "H =  13.7 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 Page no 72"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "C=86.0                                    #The composition of carbon in the fuel by weight in percentage\n",
+      "H=14.0                                    #The composition of hydrogen in the fuel by weight in percentage\n",
+      "e=1.25                                  #Equivalent ratio\n",
+      "o=32.0                                    #Molecular weight of the oxygen\n",
+      "co=44.0                                   #Molecular weight of the carbondioxide \n",
+      "c=12.0                                    #Molecular weight of the carbon\n",
+      "s=32.0                                    #Molecular weight of the sulphur\n",
+      "so=64.0                                   #Molecular weight of sulphur dioxide\n",
+      "n=28.0                                    #Molecular weight of the nitrogen\n",
+      "h2=2.0                                    #Molecular weight of the hydrogen\n",
+      "Fc=0.86                                 #Fraction of C\n",
+      "\n",
+      "#Calculations\n",
+      "Ra=1/Fc\n",
+      "x=2*(1+(0.9765/2.0)-(1.488*0.8))\n",
+      "Tm=0.5957+0.4043+4.476\n",
+      "vc=(0.5957/Tm)*100\n",
+      "vco=(0.4043/Tm)*100\n",
+      "vn=(4.476/Tm)*100\n",
+      "\n",
+      "#Calculations\n",
+      "print\"The percentage analysis of dry exhaust gas by volume : \"\n",
+      "print\"CO = \",round(vc,2),\"percent\" \n",
+      "print\"CO2 = \",round(vco,2),\"percent\" \n",
+      "print\"N2 = \",round(vn,2),\"percent\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage analysis of dry exhaust gas by volume : \n",
+        "CO =  10.88 percent\n",
+        "CO2 =  7.38 percent\n",
+        "N2 =  81.74 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 Page no 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "t=25.0                                #The temperature of both reactants and products in degree centigrade\n",
+      "p=1.0                                 #The pressure of both reactants and products in bar\n",
+      "\n",
+      "#Calculations \n",
+      "h=0\n",
+      "hf1=-103.85\n",
+      "hf2=-393.52\n",
+      "hf3=-285.8\n",
+      "hf4=(3*hf2)+(4*hf3)\n",
+      "Q=hf4-hf1\n",
+      "\n",
+      "#Output\n",
+      "print\" The heat transfer per mole of fuel = \",Q,\"kJ/mol fuel\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The heat transfer per mole of fuel =  -2219.91 kJ/mol fuel\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 Page no 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "t=25.0                                 #The temperature of the air entering the diesel engine in degree centigrade \n",
+      "T=600.0                                #The temperature at which the products are released in K\n",
+      "Ta=200.0                               #Theoretical air used in percentage \n",
+      "Q=-93.0                                #Heat loss from the engine in MJ/kmol fuel\n",
+      "f=1.0                                  #The fuel rate in kmol/h\n",
+      "\n",
+      "#Calculations \n",
+      "hfr=-290.97\n",
+      "h1=-393.52\n",
+      "h11=12.916\n",
+      "hfc=h1+h11\n",
+      "h2=-241.82\n",
+      "h22=10.498\n",
+      "hfh=h2+h22\n",
+      "h3=0 \n",
+      "h33=9.247\n",
+      "hfo=h3+h33\n",
+      "h4=0\n",
+      "h44=8.891\n",
+      "hfn=h4+h44\n",
+      "hfp=(12*hfc)+(13*hfh)+(18.5*hfo)+(139.12*hfn)\n",
+      "W=Q+hfr-hfp\n",
+      "W1=(f*W*10**3)/3600.0\n",
+      "\n",
+      "#Output\n",
+      "print\"The work for a fuel rate of 1 kmol/h is \",round(W1,0),\"kW\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The work for a fuel rate of 1 kmol/h is  1606.0 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 Page no 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "P=600                               #Power of an engine in kW\n",
+      "t=25                                #Temperature at which fuel is used in degree centigrade\n",
+      "Ta=150                              #Theoretical air used in percentage\n",
+      "T1=400                              #The temperature at which air enters in K\n",
+      "T2=700                              #The temperature at which the products of combustion leave in K\n",
+      "Q=-150                              #The heat loss from the engine in kW\n",
+      "C=12                                #Molecular weight of carbon\n",
+      "h=1                                 #Molecular weight of hydrogen\n",
+      "\n",
+      "#Calculations\n",
+      "hfc=-259.28\n",
+      "hfo1=3.029\n",
+      "hfn1=2.971\n",
+      "HR=(hfc)+(1.5*12.5*hfo1)+(1.5*12.5*3.76*hfn1)\n",
+      "hfco=-393.52\n",
+      "hfco1=17.761\n",
+      "hfh=-241.82\n",
+      "hfh1=14.184\n",
+      "hfo2=12.502\n",
+      "hfn2=11.937\n",
+      "HP=(8*(hfco+hfco1))+(9*(hfh+hfh1))+(6.25*hfo2)+(70.5*hfn2)\n",
+      "H=HP-HR\n",
+      "nf=((Q-P)*3600)/(H*10.0**3)\n",
+      "M=(8*C)+(18*h)\n",
+      "mf=nf*M\n",
+      "\n",
+      "#Output\n",
+      "print\" The fuel consumption for complete combustion is \",round(mf,1),\"kg/h\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fuel consumption for complete combustion is  74.3 kg/h\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 Page no 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "t=25                          #Temperature at which fuel is used for combustion in degree centigrade \n",
+      "p=1                           #The pressure at which fuel is used in bar\n",
+      "T=400                         #The temperature of the products of combustion in K\n",
+      "R=8.314*10**-3                #Universal gas constant\n",
+      "hfco=-393.52                  #The enthalpy of the carbondioxide in MJ/kmol fuel\n",
+      "hfco1=4.008                   #The change in enthalpy of the carbondioxide for the given conditions in MJ/kmol fuel\n",
+      "hfh=-241.82                   #The enthalpy of the water in MJ/kmol fuel\n",
+      "hfh1=3.452                    #The change in enthalpy of the water for the given conditions in MJ/kmol fuel\n",
+      "\n",
+      "#Calculations\n",
+      "hfc=-103.85 \n",
+      "HR=(1*(hfc-(R*(t+273))))+(5*(-R*(t+273)))\n",
+      "HP=(3*(hfco+hfco1-(R*T)))+(4*(hfh+hfh1-(R*T)))\n",
+      "Q=HP-HR\n",
+      "Q1=-Q\n",
+      "\n",
+      "#Output\n",
+      "print\"The heat transfer per mole of propane = \",round(Q1,1),\"kJ/mol propane\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The heat transfer per mole of propane =  2026.6 kJ/mol propane\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 Page no 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "T=1500                            #The given temperature in K\n",
+      "hfco=-393.52                      #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+      "hf1=61.714                        #The change in enthalpy for actual state and reference state in MJ/kmol\n",
+      "hfc=-110.52                       #The enthalpy of formation for carbonmonoxide in MJ/kmol\n",
+      "hf2=38.848                        #The change in enthalpy of CO for actual and reference state in MJ/kmol\n",
+      "hfo=0                             #The enthalpy of formation for oxygen gas\n",
+      "hf3=40.61                         #The change in enthalpy of oxygen for different states in MJ/kmol\n",
+      "\n",
+      "#Calculations\n",
+      "HP=hfco+hf1\n",
+      "HR=(hfc+hf2)+(0.5*(hfo+hf3))\n",
+      "H=HP-HR\n",
+      "\n",
+      "#Output\n",
+      "print\" The enthalpy of combustion is  \",round(H,1),\"MJ/kmol CO\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The enthalpy of combustion is   -280.4 MJ/kmol CO\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11 Page no 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "E=30                        #The amount of excess air in percentage\n",
+      "tp=400                      #The temperature at which propane enters in K\n",
+      "ta=300                      #The temperature at which air enters in K\n",
+      "T=900                       #The temperature at which products leave in K\n",
+      "m=83.7                      #The average molar specific heat of propane at consmath.tant pressure in kJ/kmolK\n",
+      "Mp=44                       #The molecular weight of propane\n",
+      "hfc=-393.52                 #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+      "hf1=28.041                  #The change in enthalpy of CO2 for actual and reference state in MJ/kmol\n",
+      "hfh=-241.82                 #The enthalpy of formation for water in MJ/kmol\n",
+      "hf2=21.924                  #The change in enthalpy of water for actual and reference state in MJ/kmol\n",
+      "hfn=0                       #The enthalpy of nitrogen gas \n",
+      "hf3=18.221                  #The change in enthalpy of nitrgen for actual and reference state in MJ/kmol\n",
+      "hfo=0                       #The enthalpy of oxygen gas \n",
+      "hf4=19.246                  #The change in enthalpy of oxygen for actual and reference state in MJ/kmol\n",
+      "hfp=-103.85                 #The enthalpy of formation for propane in MJ/kmol\n",
+      "R=0.0837                    #Universal gas constant \n",
+      "hfo1=0                      #The enthalpy of oxygen gas \n",
+      "hf11=0.054                  #The change in enthalpy of oxygen gas for actual and reference state in MJ/kmol\n",
+      "hfn1=0                      #The enthalpy of nitrogen gas\n",
+      "hfn22=0.054                 #The change in enthalpy of nitrogen for actual and reference state in MJ/kmol\n",
+      "\n",
+      "#Calculations\n",
+      "\n",
+      "HP=(3*(hfc+hf1))+(4*(hfh+hf2))+(24.44*(hfn+hf3))+(1.5*(hfo+hf4))\n",
+      "HR=(1*(hfp+(R*(tp-ta))))+(6.5*(hfo1+hf11))+(24.44*(hfn1+hfn22))\n",
+      "Q=HP-HR\n",
+      "Q1=(-Q/Mp)\n",
+      "\n",
+      "#Output\n",
+      "print\" The amount of heat transfer per kg of fuel is  \",round(Q1,3),\"MJ/kg\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The amount of heat transfer per kg of fuel is   32.0 MJ/kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 Page no 84"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Ta=150                            #The presence of Theoretical air\n",
+      "hfc=-393.52                       #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+      "hfh=-285.8                        #The enthalpy of formation for water in MJ/kmol\n",
+      "hfon=0                            #The enthalpy of formation for oxygen and nitrogen gas \n",
+      "hfch=-74.87                       #The enthalpy of formation for methane in MJ/kmol\n",
+      "np=2                              #Number of moles of product\n",
+      "nr=4                              #Number of moles of reactant\n",
+      "R=8.314*10**-3                    #Universal gas constant \n",
+      "t=298                             #The temperature in K\n",
+      "hfh1=-241.82                      #The enthalpy of formation for water in MJ/kmol\n",
+      "np1=4                             #Number of moles of product\n",
+      "nr1=4                             #Number of moles of reactant\n",
+      "\n",
+      "#Calculations\n",
+      "HP=(hfc)+(2*hfh)\n",
+      "HR=1*hfch\n",
+      "H=HP-HR\n",
+      "n=np-nr\n",
+      "U1=H-(n*R*t)\n",
+      "HP1=(1*hfc)+(2*hfh1)\n",
+      "H1=HP1-HR\n",
+      "n1=np1-nr1\n",
+      "U2=H1-(n1*R*t)\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)The water as liquid\" \n",
+      "print\"The standard enthalpy of combustion is \",H, \"MJ/kmol\"\n",
+      "print\"The standard internal energy of combustion is \",round(U1,3),\"MJ/kmol\"\n",
+      "print\"(b)The water as a gas \"\n",
+      "print\"The standard enthalpy of combustion is  \",H1,\"MJ/kmol\"\n",
+      "print\"The standard internal energy of combustion is \",U2,\"MJ/kmol\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)The water as liquid\n",
+        "The standard enthalpy of combustion is  -890.25 MJ/kmol\n",
+        "The standard internal energy of combustion is  -885.295 MJ/kmol\n",
+        "(b)The water as a gas \n",
+        "The standard enthalpy of combustion is   -802.29 MJ/kmol\n",
+        "The standard internal energy of combustion is  -802.29 MJ/kmol\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13 Page no 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "cv=44000                       #The lower calorific value of liquid fuel in kJ/kg\n",
+      "C=84                           #The carbon content present in the fuel in percentage\n",
+      "H=16                           #The hydrogen content present in the fuel in percentage\n",
+      "t=25                           #The temperature in degree centigrade\n",
+      "hfg=2442                       #The enthalpy of vaporization for water in kJ/kg\n",
+      "c=12.0                         #Molecular weight of carbon \n",
+      "h=2                            #Molecular weight of hydrogen\n",
+      "co2=44.0                       #Molecular weight of carbondioxide\n",
+      "h2o=18                         #Molecular weight of water \n",
+      "o2=32.0                        #Molecular weight of oxygen\n",
+      "R=8.314                        #Universal gas constant in J/molK\n",
+      "\n",
+      "#Calculations\n",
+      "CO2=(0.84*(co2/c))\n",
+      "H2O=(0.16*(h2o/h))\n",
+      "cvd=H2O*hfg\n",
+      "HHV=cv+cvd\n",
+      "np=3.08/co2\n",
+      "nr=3.52/o2\n",
+      "n=np-nr\n",
+      "HHVv=HHV+(n*R*(t+273))\n",
+      "LHVv=cv+(n*R*(t+273))\n",
+      "\n",
+      "#Output\n",
+      "print\" The higher calorific value at constant pressure = \",HHV,\"kJ/kg fuel\" \n",
+      "print\"The higher calorific value at constant volume = \",round(HHVv,0),\"kJ/kg fuel\"\n",
+      "print\"The lower calorific value at constant volume = \",round(LHVv,0),\"kJ/kg fuel\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The higher calorific value at constant pressure =  47516.48 kJ/kg fuel\n",
+        "The higher calorific value at constant volume =  47417.0 kJ/kg fuel\n",
+        "The lower calorific value at constant volume =  43901.0 kJ/kg fuel\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14 Page no 87"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "E=100                          #The amount of excess air in percent\n",
+      "T=298                          #The temperature of reactants in K\n",
+      "nc=1                           #Number of moles of propane\n",
+      "hfch=-103.85                   #Enthalpy of formation for propane in MJ/kmol fuel\n",
+      "hfc=-393.52                    #Enthalpy of formation for carbondioxide in MJ/kmol fuel\n",
+      "hfh=-241.82                    #Enthalpy of formation for water in MJ/kmol fuel\n",
+      "hfon=0                         #Enthalpy of formation for both oxygen and nitrogen gas\n",
+      "T1=1500                        #Assuming the products temperature for fist trail in K\n",
+      "hfc1=61.714                    #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel\n",
+      "hfh1=48.095                    #The change in enthalpy for water for trail temp in MJ/kmol fuel\n",
+      "hfo1=40.61                     #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel\n",
+      "hfn1=38.405                    #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel\n",
+      "T2=1600                        #Assuming the products temperature for second trail in K\n",
+      "hfc2=67.58                     #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel\n",
+      "hfh2=52.844                    #The change in enthalpy for water for trail temp in MJ/kmol fuel\n",
+      "hfo2=44.279                    #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel\n",
+      "hfn2=41.903                    #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel\n",
+      "\n",
+      "#Calculations\n",
+      "HR=nc*hfch\n",
+      "x=HR-((3*hfc)+(4*hfh)+(5*hfon)+(37.6*hfon))\n",
+      "hfn=x/37.6\n",
+      "HP1=(HR-x)+(3*hfc1)+(4*hfh1)+(5*hfo1)+(37.6*hfn1)\n",
+      "HP2=(HR-x)+(3*hfc2)+(4*hfh2)+(5*hfo2)+(37.6*hfn2)\n",
+      "Te=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1\n",
+      "\n",
+      "#Output\n",
+      "print\" The adiabatic flame temperature for steady-flow process is  \",round(Te,0),\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The adiabatic flame temperature for steady-flow process is   1510.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.15 Page no 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "T=600                            #The initial temperature of air in K\n",
+      "p=1                              #The initial pressure of air in atm\n",
+      "R=8.314                          #Universal gas constant in J/molK\n",
+      "Tr=298                           #The temperature of reactants in K\n",
+      "a=4.503                          #Given Constants \n",
+      "b=-8.965*10**-3\n",
+      "c=37.38*10**-6\n",
+      "d=-36.49*10**-9\n",
+      "e=12.22*10**-12\n",
+      "hfc=-393.52                      #Enthalpy of formation for carbondioxide in MJ/kmol fuel\n",
+      "hfh=-241.82                      #Enthalpy of formation for water in MJ/kmol fuel\n",
+      "hfn=0                            #Enthalpy of formation for nitrogen gas\n",
+      "hfc1=-74.87                      #The enthalpy of formation for methane in MJ/kmol fuel \n",
+      "hfh1=9.247                       #The change in enthalpy of the water in MJ/kmol\n",
+      "hfn1=8.891                       #The change in enthalpy of nitrogen in MJ/kmol\n",
+      "Tc=3700                          #The corresponding temperature for the enthalpy of guess nitrogen in K\n",
+      "T1=2800                          #The temperature assumed for the first trail in K\n",
+      "hco1=140.444                     #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+      "hh1=115.294                      #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+      "hn1=85.345                       #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+      "T2=2500                          #The temperature assumed for the second trail in K\n",
+      "hco2=121.926                     #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+      "hh2=98.964                       #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+      "hn2=74.312                       #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+      "T3=2600                          #The temperature fo the third trail in K\n",
+      "hco3=128.085                     #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+      "hh3=104.37                       #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+      "hn3=77.973                       #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+      "Tc1=3000                         #Assume temperature for first trail in K\n",
+      "hcoa1=146.645                    #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+      "hha1=120.813                     #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+      "hna1=89.036                      #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+      "Tc2=3200                         #Assume temperature for the second trail in K\n",
+      "hcoa2=165.331                    #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+      "hha2=137.553                     #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+      "hna2=100.161                     #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+      "\n",
+      "#Calculations\n",
+      "HP=(1*hfc)+(2*hfh)+(7.52*hfn)\n",
+      "hch=(R*((a*(T-Tr))+((b/2.0)*(T**2-Tr**2))+((c/3.0)*(T**3-Tr**3))+((d/4.0)*(T**4-Tr**4))+((e/5.0)*(T**5-Tr**5))))/1000.0\n",
+      "HR=((hfc1+hch)+(2*hfh1)+(7.52*hfn1))\n",
+      "x=HR-HP\n",
+      "hfn2=x/7.52\n",
+      "HP1=hco1+(2*hh1)+(7.52*hn1)+(HR-x)\n",
+      "HP2=hco2+(2*hh2)+(7.52*hn2)+(HR-x)\n",
+      "HP3=hco3+(2*hh3)+(7.52*hn3)+(HR-x)\n",
+      "Ta1=(((HR-HP2)/(HP3-HP2))*(T3-T2))+T2\n",
+      "UR1=HR-(10.52*R*10**-3*T)\n",
+      "UP1=hcoa1+(2*hha1)+(7.52*hna1)+(HR-x)-(0.08746*Tc1)\n",
+      "UP2=hcoa2+(2*hha2)+(7.52*hna2)+(HR-x)-(0.08746*Tc2)\n",
+      "Tu=(((UR1-UP1)/(UP2-UP1))*(Tc2-Tc1))+Tc1\n",
+      "\n",
+      "#Output\n",
+      "print\"The adiabatic flame temperature at  \"\n",
+      "print\"(a)Constant pressure process is \",round(Ta1,0),\"K\" \n",
+      "print\"(b)Constant volume process is \",round(Tu,3),\"K\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The adiabatic flame temperature at  \n",
+        "(a)Constant pressure process is  2550.0 K\n",
+        "(b)Constant volume process is  3089.34 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16 Page no 91"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "T=600.0                          #Temperature at constant pressure process in K\n",
+      "p=1.0                            #The pressure in atm\n",
+      "E=50.0                           #The amount of excess air in percent\n",
+      "L=20.0                           #The amount of less air in percent\n",
+      "cp=52.234                        #Specific constant for methane in kJ/kmolK\n",
+      "t=298.0                        #Assume the normal temperature in K\n",
+      "hfch=-74.87                    #The enthalpy of formation for carbondioxide in MJ\n",
+      "ho=9.247                       #The change in enthalpy of oxygen in MJ\n",
+      "hn=8.891                       #The change in enthalpy of nitrogen in MJ\n",
+      "hfc1=-393.52                   #The enthalpy of formation of carbondioxide in MJ\n",
+      "hfh1=-241.82                   #The enthalpy of formation of water in MJ\n",
+      "Tc=2800.0                        #The corresponding temperature in K\n",
+      "T1=2000                        #The temperature for first trail in K\n",
+      "hfc11=91.45                    #The enthalpy for the assume temp for carbondioxide in MJ\n",
+      "hfh11=72.689                   #The change in enthalpy for the assume temp for water in MJ\n",
+      "hfn11=56.141                   #The change in enthalpy for the assume temp for nitrogen in MJ\n",
+      "hfo11=59.199                   #The change in enthalpy for the assume temp for oxygen in MJ\n",
+      "T2=2100                        #The temperature for second trail in K\n",
+      "hfc22=97.5                     #The enthalpy for the assume temp for carbondioxide in MJ\n",
+      "hfh22=77.831                   #The change in enthalpy for the assume temp for water in MJ\n",
+      "hfn22=59.748                   #The change in enthalpy for the assume temp for nitrogen in MJ\n",
+      "hfo22=62.986                   #The change in enthalpy for the assume temp for oxygen in MJ\n",
+      "hfchr=-74.87                   #The enthalpy of formation for methane in MJ\n",
+      "hor=9.247                      #The change in enthalpy for oxygen in MJ\n",
+      "hnr=8.891                      #The change in enthalpy for nitrogen in MJ\n",
+      "hfcop=-110.52                  #The formation of enthalpy for carbonmoxide in MJ\n",
+      "hfcp=-393.52                   #The formation of enthalpy for carbondioxide in MJ\n",
+      "hfhp=-241.82                   #The formation of enthalpy for water in MJ\n",
+      "Tp1=2000.0                       #The temperature for first trail in K\n",
+      "hco11=56.739                   #The change in enthalpy for CO in MJ\n",
+      "hco211=91.45                   #The change in enthalpy for CO2 in MJ\n",
+      "hh11=72.689                    #The change in enthalpy for water in MJ\n",
+      "hn11=56.141                    #The change in enthalpy for nitrogen in MJ\n",
+      "Tp2=2400                       #The temperature for second trail in K\n",
+      "hco22=71.34                    #The change in enthalpy for CO in MJ\n",
+      "hco222=115.788                 #The change in enthalpy for CO2 in MJ\n",
+      "hh22=93.604                    #The change in enthalpy for water in MJ\n",
+      "hn22=70.651                    #The change in enthalpy for nitrogen in MJ\n",
+      "Tp3=2300.0                       #The temperature for first trail in K\n",
+      "hco33=67.676                   #The change in enthalpy for CO in MJ\n",
+      "hco233=109.671                 #The change in enthalpy for CO2 in MJ\n",
+      "hh33=88.295                    #The change in enthalpy for water in MJ\n",
+      "hn33=67.007                    #The change in enthalpy for nitrogen in MJ\n",
+      "hccc=-283.022                  #The only combustible substance is CO in MJ/kmol\n",
+      "\n",
+      "#Calculations\n",
+      "hch=cp*(T-t)*10**-3\n",
+      "HR=hfch+hch+(3*ho)+(11.28*hn)\n",
+      "HP=hfc1+(2*hfh1)\n",
+      "x=HR-HP\n",
+      "hn2=x/11.28\n",
+      "HP1=hfc11+(2*hfh11)+(11.28*hfn11)+(hfo11)+(HR-x)\n",
+      "HP2=hfc22+(2*hfh22)+(11.28*hfn22)+(hfo22)+(HR-x)\n",
+      "Ta1=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1\n",
+      "X=2*(2-1.6)\n",
+      "HRr=hfchr+hch+(1.6*hor)+(6.01*hnr)\n",
+      "HPp=(0.8*hfcop)+(0.2*hfcp)+(2*hfhp)\n",
+      "HPp1=(0.8*hco11)+(0.2*hco211)+(2*hh11)+(6.016*hn11)-HPp\n",
+      "HPp2=(0.8*hco22)+(0.2*hco222)+(2*hh22)+(6.016*hn22)+HPp\n",
+      "HPp3=(0.8*hco33)+(0.2*hco233)+(2*hh33)+(6.016*hn33)+HPp\n",
+      "Ta2=(((HRr-HPp3)/(HPp2-HPp3))*(Tp2-Tp3))+Tp3\n",
+      "Q=-0.8*hccc                                                    #The thermal energy loss in MJ/kmol fuel\n",
+      "\n",
+      "#Output\n",
+      "print\" The adiabatic flame temperature having \"\n",
+      "print\"(a)50 percent excess air is \",round(Ta1,1),\"K\"\n",
+      "print\"(b)20 percent less air is \",round(Ta2,2),\"K\"\n",
+      "print\"The loss of thermal energy due to incomplete combustion is \",round(Q,2),\"MJ/kmol fuel\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The adiabatic flame temperature having \n",
+        "(a)50 percent excess air is  2027.6 K\n",
+        "(b)20 percent less air is  2311.22 K\n",
+        "The loss of thermal energy due to incomplete combustion is  226.42 MJ/kmol fuel\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18 Page no 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "T1=3000                               #Given temperature in K\n",
+      "T2=4000                               #Given temperature in K\n",
+      "p=1                                   #The pressure in atm\n",
+      "KP1=1.117                             #Natural logarithm of equilibrium constant at 3000 K \n",
+      "KP2=-1.593                            #Natural logarithm of equilibrium constant at 4000 K\n",
+      "a1=0.4                                #The dissociation of 1 mole of CO2 for the first trail\n",
+      "a2=0.5                                #The dissociation of 1 mole of CO2 for the second trail \n",
+      "K1=3.674                              #The value of equilibrium constant for the first trail \n",
+      "K2=2.236                              #The value of equilibrium constant for the second trail\n",
+      "a3=0.9                                #The dissociation of 1 mole of CO2 for the first trail\n",
+      "a4=0.89                               #The dissociation of 1 mole of CO2 for the second trail\n",
+      "K3=0.1995                             #The value of equilibrium constant for the first trail \n",
+      "K4=0.2227                             #The value of equilibrium constant for the second trail \n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Kp1=math.exp(KP1)\n",
+      "Kp2=math.exp(KP2)\n",
+      "a12=(((K1-Kp1)/(K1-K2))*(a2-a1))+a1\n",
+      "A12=a12*100\n",
+      "a23=(((Kp2-K4)/(K3-K4))*(a3-a4))+a4\n",
+      "A23=a23*100\n",
+      "\n",
+      "#output\n",
+      "print\"The percent dissociation of carbondioxide into carbonmonoxide and oxygen at \"\n",
+      "print\"(a) at 3000 K and 1 atm pressure = \",round(A12,3),\"percent\" \n",
+      "print\"(b) at 4000 K and 1 atm pressure = \",round(A23,1),\"percent\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percent dissociation of carbondioxide into carbonmonoxide and oxygen at \n",
+        "(a) at 3000 K and 1 atm pressure =  44.3 percent\n",
+        "(b) at 4000 K and 1 atm pressure =  89.8 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19 Page no 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "p=1                               #Initial pressure in atm\n",
+      "T=300                             #Initial temperature in K\n",
+      "Tc=2400                           #To calculate the molefraction of the products at this temperature in K\n",
+      "KP1=3.866                         #Natural logarithm of equilibrium constant at 2400 K for the equation\n",
+      "a=0.098                           #The dissociation of 1 mole of CO2\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "K1=math.exp(KP1)\n",
+      "nr=1+0.5\n",
+      "Pp=(p*Tc)/(nr*T)\n",
+      "np=(a+2)/2.0\n",
+      "xco=(2*(1-a))/(2+a)\n",
+      "xc=(2*a)/(2+a)\n",
+      "xo=a/(2.0+a)\n",
+      "PP=5.333*np\n",
+      "\n",
+      "#output\n",
+      "print\"Mole fraction of the carbondioxide is \",round(xco,3)\n",
+      "print\"Mole fraction of the carbonmonoxide is \",round(xc,3)\n",
+      "print\"Mole fraction of oxygen is \",round(xo,3)\n",
+      "print\"Pressure of the product is \",round(PP,3),\"bar\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mole fraction of the carbondioxide is  0.86\n",
+        "Mole fraction of the carbonmonoxide is  0.093\n",
+        "Mole fraction of oxygen is  0.047\n",
+        "Pressure of the product is  5.594 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20 Page no 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "t=25.0                              #The temperature of air in degree centigrade\n",
+      "p=1.0                               #The pressure of air in atm\n",
+      "T1=2200.0                           #Given first temperature in K\n",
+      "T2=2400.0                           #Given second temperature in K\n",
+      "h1=59.86                          #The change in enthalpy of hydrogen at 2200 K in MJ/kmol\n",
+      "h2=66.915                         #The change in enthalpy of hydrogen at 2400 K in MJ/kmol\n",
+      "T=298.0                             #The temperature of air in K\n",
+      "HR=0                              #The total enthalpy on the reactants side since all the reactants are elements\n",
+      "Kp1=-6.774                        #Natural logarithm of equilibrium constant at 2200 K for the equation \n",
+      "a1=0.02                           #By trail and error method the degree of dissociation of H2O\n",
+      "hfh=-241.82                       #The enthalpy of formation of water at both 2200 and 2400 K in MJ/kmol\n",
+      "hfh1=83.036                       #The change in enthalpy of water at 2200 K in MJ/kmol\n",
+      "hfd1=59.86                        #The change in enthalpy of hydrogen at 2200 K in MJ/kmol\n",
+      "hfo1=66.802                       #The change in enthalpy of oxygen at 2200 K in MJ/kmol\n",
+      "hfn1=63.371                       #The change in enthalpy of nitrogen at 2200 K in MJ/kmol\n",
+      "a2=0.04                           #By trail and error method the degree of dissociation of H2O at 2400 K\n",
+      "hfh2=93.604                       #The change in enthalpy of water at 2400 K in MJ/kmol\n",
+      "hfd2=66.915                       #The change in enthalpy of hydrogen at 2400 K in MJ/kmol\n",
+      "hfo2=74.492                       #The change in enthalpy of oxygen at 2400 K in MJ/kmol\n",
+      "hfn2=70.651                       #The change in enthalpy of nitrogen at 2400 K in MJ/kmol\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "K1=math.exp(Kp1)\n",
+      "HP1=(0.98*(hfh+hfh1))+(0.02*hfd1)+(0.01*hfo1)+(1.88*hfn1) \n",
+      "HP2=(0.96*(hfh+hfh2))+(0.04*hfd2)+(0.02*hfo2)+(1.88*hfn2) \n",
+      "H1=HP1-HR\n",
+      "H2=HP2-HR\n",
+      "Tl=(((T2-T1)/(HP2-HP1))*(HR-HP1))+T1\n",
+      "\n",
+      "#Output\n",
+      "print\"The adiabatic flame temperature taking dissociation into account is \",T1+236,\"K\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The adiabatic flame temperature taking dissociation into account is  2436.0 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/chap4.ipynb b/Fundamental_of_internal_combustion_engines/chap4.ipynb
new file mode 100755
index 00000000..c8818341
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/chap4.ipynb
@@ -0,0 +1,793 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4:Fuel Air Cycles and their analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page No 117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=8.5                                #The compression ratio \n",
+      "sv=1.4                               #The specific heat at constant volume in percent\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "n=1-(1/r)**(sv-1) \n",
+      "ef=(((1-n)/n)*(sv-1)*(math.log(r))*(sv/100.0))*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The efficiency decreases by \",round(ef,3),\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency decreases by  0.885 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page No 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=18.0                                      #The compression ratio \n",
+      "l=6.0                                       #The cut off taking place corresponding of the stroke in percent\n",
+      "sc=2.0                                      #The specific heat at constant volume increases in percent\n",
+      "cv=0.717                                  #The specific heat at constant volume in kJ/kgK\n",
+      "R=0.287                                   #Gas constant in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Vs=(r-1)\n",
+      "B=((l/100.0)*Vs)+1\n",
+      "cp=cv+R\n",
+      "R1=cp/cv\n",
+      "n=1-(((((1/r)**(R1-1))*(B**R1-1))/(R1*(B-1)))) \n",
+      "dn=(((1-n)/n)*((R1-1)*((math.log(r))-(((B**R1)*math.log(B))/(B**R1-1))+(1/B)))*(sc/100.0))*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The efficiency decreases by \",round(dn,1),\"percent\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The efficiency decreases by  1.1 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 Page No 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=8                                    #The compression ratio\n",
+      "af=15                                  #Air/fuel ratio\n",
+      "p1=1                                   #The pressure at the beginning of a compression stroke in bar\n",
+      "t=60                                   #The temperature at the beginning of a compression stroke in degree centigrade\n",
+      "cv=44000                               #The calorific value of the fuel in kJ/kg\n",
+      "n=1.32                                 #The index of the compression \n",
+      "Cv=0.717                               #specific heat at constant volume in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "T1=t+273\n",
+      "p2=p1*(r)**n\n",
+      "T2=T1*r**(n-1)\n",
+      "f=(1/(af+1))\n",
+      "a=(af/(af+1))\n",
+      "q23=cv/(af+1)\n",
+      "T3=((-10430+((10430)**2+(4*494.8*10**5))**(1/2.0))/2.0)\n",
+      "p3=(T3/T1)*(r)*p1\n",
+      "T31=(q23/Cv)+T2\n",
+      "p31=(T31/T1)*r*p1\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The Maximum temperature in the cylinder  = \",round(T3,0),\"K\" \n",
+      "print\"The Maximum pressure in the cylinder P3 = \",round(p3,0),\"bar\" \n",
+      "print\"(b)With constant value of Cv \"\n",
+      "print\"The Maximum temperature in the cylinder  = \",round(T31,0),\"K\" \n",
+      "print\"The Maximum pressure in the cylinder P3 = \",round(p31,1),\"bar\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The Maximum temperature in the cylinder  =  3541.0 K\n",
+        "The Maximum pressure in the cylinder P3 =  85.0 bar\n",
+        "(b)With constant value of Cv \n",
+        "The Maximum temperature in the cylinder  =  4483.0 K\n",
+        "The Maximum pressure in the cylinder P3 =  107.7 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page No 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=21                                     #The compression ratio \n",
+      "af=29                                    #Air/fuel ratio\n",
+      "T=1000                                   #The temperature at the end of compression in K\n",
+      "cv=42000                                 #The calorific value of the in kJ/kg\n",
+      "R=0.287                                  #Gas constant in kJ/kgK\n",
+      "\n",
+      "#Calculations \n",
+      "q23=cv/(af+1)\n",
+      "T3=(-0.997+(((0.997)**2)+(4*2411*14*10**-6))**(1/2.0))/(28.0*10.0**-6)\n",
+      "V3=(T3/T)\n",
+      "Vs=(r-1)\n",
+      "V=V3-1\n",
+      "pc=(V/Vs)*100\n",
+      "\n",
+      "#Output\n",
+      "print\"The percentage of stroke at which combustion is complete = \",round(pc,3),\"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage of stroke at which combustion is complete =  6.706 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No 122"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=16.0                                         #The compression ratio \n",
+      "l=6.0                                          #The cut-off of the stroke in percent\n",
+      "p3=70.0                                        #The maximum pressure obtained in bar\n",
+      "p1=1.0                                         #The pressure at the beginning of compression in bar\n",
+      "T1=(100.0+273.0)                                 #The temperature at the beginning of compression in K\n",
+      "R=0.287                                      #Gas constant in kJ/kgK\n",
+      "g=1.4                                        #Assume the isentropic index \n",
+      "\n",
+      "#Calculations\n",
+      "T2=T1*(r)**(g-1)\n",
+      "Cv=(1/(T2-T1))*(0.716+125*10**-6*(T2**2-T1**2))\n",
+      "Cp=Cv+R\n",
+      "g1=Cp/Cv\n",
+      "T21=T1*(r)**(g1-1)\n",
+      "Cv1=(1/(T21-T1))*((0.716+125*10**-6*(T21**2-T1**2)))\n",
+      "Cp1=Cv1+R\n",
+      "g2=Cp1/Cv1\n",
+      "gm=1.358                       #mean value\n",
+      "T22=T1*(r)**(gm-1)\n",
+      "p2=(T22/T1)*r*p1\n",
+      "T3=(p3/p2)*T22\n",
+      "V=((l/100.0)*(r-1))+1\n",
+      "T4=(V)*T3\n",
+      "p4=p3\n",
+      "g3=1.3\n",
+      "V5=r/V\n",
+      "T5=T4*(1/V5)**(g3-1)\n",
+      "Cv2=((0.716*(T5-T4))+(62.5*10**-6*(T5**2-T4**2)))/(T5-T4)\n",
+      "Cp2=Cv2+R\n",
+      "g4=Cp2/Cv2\n",
+      "T51=T4*(1/V5)**(g4-1)\n",
+      "Cv3=((0.716*(T51-T4))+(62.5*10**-6*(T51**2-T4**2)))/(T51-T4)\n",
+      "Cp3=Cv3+R\n",
+      "g5=Cp3/Cv3\n",
+      "T52=T4*(1/V5)**(g5-1)\n",
+      "p5=(T52/T1)*p1\n",
+      "\n",
+      "#Output\n",
+      "print\"The pressure and temperature at all points of the cycle \\nat point 2: Temperature T2 = \",round(T22,0),\"K and Pressure P2 = \",round(p2,2),\" bar\"  \n",
+      "print\"at point 3 :Temperature T3 = \",round(T3,1),\" K  and Pressure P3 = \",p3,\" bar\" \n",
+      "print\"at point 4 : Temperature T4 = \",T4,\" K    and Pressure P4 = \",p4,\"bar\" \n",
+      "print\"at point 5 :Temperature T5 = \",round(T52,0),\" K  and Pressure P5 = \",round(p5,2),\"bar\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The pressure and temperature at all points of the cycle \n",
+        "at point 2: Temperature T2 =  1006.0 K and Pressure P2 =  43.17  bar\n",
+        "at point 3 :Temperature T3 =  1631.9  K  and Pressure P3 =  70.0  bar\n",
+        "at point 4 : Temperature T4 =  3100.5625  K    and Pressure P4 =  70.0 bar\n",
+        "at point 5 :Temperature T5 =  1698.0  K  and Pressure P5 =  4.55 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 Page No 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=8                                                #Compression ratio\n",
+      "lcv=44000                                          #The lower heating value of the fuel in kJ/kg\n",
+      "af=15                                              #The air/fuel ratio\n",
+      "Cv=0.71                                            #The specific heat at constant volume in kJ/kgK\n",
+      "p=1                                                #The pressure at the beginning of the compression in bar\n",
+      "t=60                                               #The temperature at the beginning of the compression in degree centigrade\n",
+      "Mo=32                                              #Molecular weight of oxygen\n",
+      "Mn=28.161                                          #Molecular weight of nitrogen\n",
+      "Mh=18                                              #Molecular weight of water \n",
+      "n=1.3                                              #Polytrpic index\n",
+      "\n",
+      "#Calculations\n",
+      "T1=(t+273)\n",
+      "sa=(12.5*(Mo+(3.76*Mn)))/((12*8)+(1*Mh))\n",
+      "Y=af*(((12*8)+(1*Mh))/(Mo+(3.76*Mn)))\n",
+      "x=(12.5-Y)*2\n",
+      "nb=1+Y+(Y*3.76)\n",
+      "na=x+7.8+9+46.624 \n",
+      "Me=((na-nb)/nb)*100\n",
+      "T2=T1*(r)**(n-1)\n",
+      "T3=(lcv/(af+1))*(1/Cv)+(T2)\n",
+      "p3=r*(T3/T1)*p\n",
+      "p31=p3*(na/nb)\n",
+      "\n",
+      "#Output\n",
+      "print\"The percentage molecular expansion is \",round(Me,0),\"percent\"\n",
+      "print\"(a) Without considering the molecular expansion \"\n",
+      "print\" The maximum temperature is \",round(T3,0),\"K\" \n",
+      "print\"The maximum pressure is \",round(p3,0),\"bar\"\n",
+      "print\"(b) With molecular expansion \" \n",
+      "print\"The maximum temperature is \",round(T3,0),\"K\" \n",
+      "print\"The maximum pressure is \",round(p31,1),\"bar\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The percentage molecular expansion is  6.0 percent\n",
+        "(a) Without considering the molecular expansion \n",
+        " The maximum temperature is  4495.0 K\n",
+        "The maximum pressure is  108.0 bar\n",
+        "(b) With molecular expansion \n",
+        "The maximum temperature is  4495.0 K\n",
+        "The maximum pressure is  114.4 bar\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 Page No 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "f=0.03                          #The residual fraction of an engine\n",
+      "e=1.2                           #The equivalence ratio\n",
+      "F=0.0795                        #Fuel/air ratio for corresponding equivalence ratio \n",
+      "\n",
+      "#Calculations\n",
+      "\n",
+      "T=1+F\n",
+      "fa=1-f\n",
+      "ff=F*(fa)\n",
+      "ra=f\n",
+      "rf=ra*F\n",
+      "\n",
+      "#Output\n",
+      "print\"Fresh air = \",fa,\"kg\" \n",
+      "print\"Fresh fuel = \",ff,\"kg\" \n",
+      "print\"Air in residual = \",ra,\"kg\" \n",
+      "print\"Fuel in residual = \",rf,\"kg\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fresh air =  0.97 kg\n",
+        "Fresh fuel =  0.077115 kg\n",
+        "Air in residual =  0.03 kg\n",
+        "Fuel in residual =  0.002385 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 Page No 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "T=800                          #The given temperature in K\n",
+      "e=1                            #The equivalence ratio \n",
+      "hi=154.723                     #Sensible Enthalpy for isooctane at 800 K in MJ/kmol \n",
+      "ho=15.841                      #Sensible Enthalpy for oxygen at 800 K in MJ/kmol \n",
+      "hn=15.046                      #Sensible Enthalpy for nitrogen at 800 K in MJ/kmol\n",
+      "nc=0.00058                     #Number of kmoles of C8H18 for equivalence ratio for 1 kg of air \n",
+      "no=0.00725                     #Number of kmoles of oxygen for equivalence ratio for 1 kg of air \n",
+      "nn=0.0273                      #Number of kmoles of nitrogen for equivalence ratio for 1 kg of air \n",
+      "R=8.314                        #Gas constant in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "\n",
+      "Hs=(nc*hi)+(no*ho)+(nn*hn)\n",
+      "Hs1=Hs*1000\n",
+      "n=nc+no+nn\n",
+      "Us=Hs-(n*R*10**-3*(T-298))\n",
+      "Us1=Us*1000\n",
+      "\n",
+      "#Output\n",
+      "print\"Total sensible enthalpy of reactants = \",round(Hs1,3),\"kJ/kg air\" \n",
+      "print\"Sensible internal energy of reactants = \",round(Us1,3),\"kJ/kg air\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total sensible enthalpy of reactants =  615.342 kJ/kg air\n",
+        "Sensible internal energy of reactants =  468.723 kJ/kg air\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 Page No 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "T=500.0                                 #The given temperature in K\n",
+      "e=1.0                                   #Equivalence ratio \n",
+      "Ai=0.0662                             #The amount of isooctane for 1 kg of air in kg\n",
+      "Ta=298.0                                #Consider the ambient temperature in K \n",
+      "R=8.314                               #Gas constant in kJ/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "E=((0.0662*((0.44*math.log(T/Ta))+(3.67*10**-3*(T-Ta))))+((0.921*math.log(T/Ta))+(2.31*10**-4*(T-Ta))))*1000\n",
+      "Ri=Ri/114.0\n",
+      "W=(0.5874-(0.662*Ri*math.log(T/Ta))-(0.287*math.log(T/Ta)))*1000\n",
+      "\n",
+      "#Output\n",
+      "print\"The isentropic compression functions at 500 K for the unburned\" \n",
+      "print\"isooctsne-air mixture are \",round(E,1),\"J/kg air and\",round(W,1),\"J/kg air\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The isentropic compression functions at 500 K for the unburned\n",
+        "isooctsne-air mixture are  587.4 J/kg air and 438.9 J/kg air\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10 Page No 133"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=7.8                               #Compression ratio \n",
+      "p=1.0                               #The pressure at the start of compression in atm\n",
+      "T1=335.0                              #The temperature at the start of compression in K\n",
+      "W1=100                              #Isentropic compression function for T1 in J/kg air K \n",
+      "T2=645                              #The temperature corresponding to isentropic compression function in J/kg air K \n",
+      "U1=35                               #Internal energy corresponding to temp T1 in kJ/kg air \n",
+      "U2=310                              #Internal energy corresponding to temp T2 in kJ/kg air \n",
+      "E1=120                              #Isentropic compression function at T1 \n",
+      "E2=910                              #Isentropic compression function at T2 \n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "W2=W1-(292*math.log(1/r))\n",
+      "V1=(292*T1)/(p*10.0**5)\n",
+      "p2=p*(T2/T1)*r\n",
+      "V2=V1/r\n",
+      "W=U2-U1\n",
+      "p21=(math.exp((E2-E1)/292.0)) \n",
+      "\n",
+      "#Output\n",
+      "print\"(a)At the end of the compression stroke\"\n",
+      "print\"The temperature is \",T2,\"K\" \n",
+      "print\"The pressure is \",round(p2,0),\"atm\" \n",
+      "print\"The volume per unit mass of air is \",round(V2,3),\"m**3/kg air\"\n",
+      "print\"The pressure is \",round(p21,0),\"atm\" \n",
+      "print\"(b)The work input during compression is \",W,\"kJ/kg air\" "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)At the end of the compression stroke\n",
+        "The temperature is  645 K\n",
+        "The pressure is  15.0 atm\n",
+        "The volume per unit mass of air is  0.125 m**3/kg air\n",
+        "The pressure is  15.0 atm\n",
+        "(b)The work input during compression is  275 kJ/kg air\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 Page No 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "p=65                      #The pressure in the cylinder in bar\n",
+      "r=10                      #The compression ratio \n",
+      "V3=0.1                    #The volume per unit mass of air at the start of expansion in m**3/kg air \n",
+      "p3=p*100                  #The pressure in the cylinder after the completion of combustion in kN/m**2\n",
+      "T3=2240                   #The temperature from the chart corresponding to p3,V3 in K\n",
+      "u3=-1040                  #The energy from the chart in kJ/kg air \n",
+      "s3=8.87                   #The entropy from the chart in kJ/kg air K\n",
+      "T4=1280                   #The temperature from the chart corresponding to p4,V4 in K \n",
+      "u4=-2220                  #The energy from the chart in kJ/kg air \n",
+      "p4=4.25                   #The pressure from the chart in bar \n",
+      "\n",
+      "#Calculations \n",
+      "s4=s3\n",
+      "V4=r*V3\n",
+      "W=-(u4-u3)\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)At the end of expansion stroke\"\n",
+      "print\"The pressure is \",p4,\"bar\" \n",
+      "print\"The temperature is \",T4,\"K\" \n",
+      "print\"The volume is \",V4,\"m**3/kg air\" \n",
+      "print\"(b)The work during the expansion stroke is \",W,\"kJ/kg air\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)At the end of expansion stroke\n",
+        "The pressure is  4.25 bar\n",
+        "The temperature is  1280 K\n",
+        "The volume is  1.0 m**3/kg air\n",
+        "(b)The work during the expansion stroke is  1180 kJ/kg air\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 Page no 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "#From Table 4.4\n",
+      "H1=-224.1*1000     #MJ/mol, Enthalpy of C8H18\n",
+      "H2=-393.52*1000    #Enthalpy of CO2\n",
+      "H3=-241.82*1000    #Enthalpy of H2O\n",
+      "U1=-204.1*1000     #MJ/mol, Internal energyof C8H18\n",
+      "U2=-393.52*1000    #Internal energy of CO2\n",
+      "U3=-240.6*1000    #Internal energy of H2O\n",
+      "\n",
+      "\n",
+      "M1=114.0       #g, molecular wt of C8H18\n",
+      "M2=32.0       #g, molecular wt of O2\n",
+      "M3=28.0       #g, molecular wt of N2\n",
+      "M4=44.0       #g, molecular wt of CO2\n",
+      "M5=18.0       #g, molecular wt of H2O\n",
+      "#For 1 kg air, from the eq.,  the fraction of wt are\n",
+      "x1=0.0661    \n",
+      "x2=0.232\n",
+      "x3=0.768\n",
+      "x4=0.204\n",
+      "x5=0.094\n",
+      "\n",
+      "#Calculation\n",
+      "import sympy\n",
+      "f=sympy.Symbol(\"f\")\n",
+      "n1=(x1/M1)*(1-f)        #No. of kmoles of C8H18\n",
+      "n2=x2/M2*(1-f)\n",
+      "n3=x3/M3\n",
+      "n4=x4/M4*f\n",
+      "n5=x5/M5*f\n",
+      "N1=(n1*H1+n4*H2+n5*H3)\n",
+      "N2=n1*U1+n4*U2+n5*U3\n",
+      "\n",
+      "#Result\n",
+      "print \"Standard Enthalpy of formation is\",N1\n",
+      "print \"Internal energy of formation is\",N2\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Standard Enthalpy of formation is -2957.40091174907*f - 129.938684210526\n",
+        "Internal energy of formation is -2962.62629186603*f - 118.342192982456\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13 Page No 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Tu=645.0                                #The temperature at the end of compression process in K\n",
+      "usu=310.0                               #The internal energy at the end of compression process in kJ/kg air \n",
+      "pu=(15.4*1.013)                       #The pressure at the end of the compression process in bar \n",
+      "Vu=0.124                              #The volume at the end of the compression process in m**3/kg air \n",
+      "e=1.0                                   #Equivalence ratio \n",
+      "f=0.065                               #Burned gas fraction \n",
+      "Tb=2820.0                               #The temperature for constant volume \n",
+      "pb=6500.0                               #The pressure for constant volume \n",
+      "hsu=440.0                               #The enthalpy from chart corresponding to temp Tu in kJ/kg air \n",
+      "pb1=1560.0                              #The pressure for constant pressure adiabatic combustion in kN/m**2 \n",
+      "ub1=-700.0                              #Trail and error along the pb internal energy in kJ/kg air\n",
+      "Tb1=2420.0                              #The temperature for constant pressure \n",
+      "\n",
+      "#Calculations \n",
+      "ufu=-118.5-(2963*f)\n",
+      "ub=usu-ufu\n",
+      "Vb=Vu\n",
+      "hfu=-129.9-(2958*f)\n",
+      "hb=hsu+hfu\n",
+      "vb1=(118-ub1)/pb\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)For constant volume adiabatic combustion\"\n",
+      "print\"The temperature is \",Tb,\"K\" \n",
+      "print\"The pressure is \",pb,\"kN/m**2\"\n",
+      "print\"(b)For constant pressure adiabatic combustion\"\n",
+      "print\"The temperature is \",Tb1,\"K\" \n",
+      "print\"The pressure is \",pb1,\"kN/m**2\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)For constant volume adiabatic combustion\n",
+        "The temperature is  2820.0 K\n",
+        "The pressure is  6500.0 kN/m**2\n",
+        "(b)For constant pressure adiabatic combustion\n",
+        "The temperature is  2420.0 K\n",
+        "The pressure is  1560.0 kN/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14 Page No 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "r=8.0                               #The compression ratio \n",
+      "T1=350.0                            #The given temperature at the start of compression in K\n",
+      "p=1.0                               #The given pressure at the start of compression in bar\n",
+      "f=0.08                              #The exhaust residual fraction \n",
+      "cv=44000                            #The calorific value in kJ/kg\n",
+      "W1=150                              #Isentropic compression functions for corresponding temp T1 in J/kg air K\n",
+      "T2=682                              #The temperature corresponding to isentropic compression function in K \n",
+      "us2=350                             #The internal energy corresponding to temp T2 in K\n",
+      "us1=40                              #The internal energy corresponding to temp T1 in K \n",
+      "T3=2825                             #The temperature at point 3 corresponding to u3,V3 on the burned gas chart in K\n",
+      "p3=7100                             #The pressure at point 3 in kN/m**2 \n",
+      "s3=9.33                             #Entropy at point 3 in kJ/kg air K \n",
+      "u4=-1540                            #The internal energy at point 4 corresponding to V4,s4 in kJ/kg air \n",
+      "p4=570                              #The pressure at point 4 in kN/m**2 \n",
+      "T4=1840                             #The temperature at point 4 in K \n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "W2=W1-(292*math.log(1/r))\n",
+      "V1=(292*T1)/(p*10.0**5)\n",
+      "p2=p*(T2/T1)*r\n",
+      "V2=V1/r\n",
+      "Wc=us2-us1\n",
+      "ufu=-118.5-(2963*f)\n",
+      "u3=us2+ufu\n",
+      "V3=V2\n",
+      "s4=s3\n",
+      "V4=V1\n",
+      "We=u3-u4 \n",
+      "Wn=We-Wc\n",
+      "nth=((Wn)/((1-f)*0.0662*cv))*100\n",
+      "imep=((Wn*1000)/(V1-V2))/10.0**5\n",
+      "nv=(((1-f)*287*298)/(1.013*10**5*(1-0.125)))*100\n",
+      "\n",
+      "#Output\n",
+      "print\"(a)At point 2, \\nThe temperature is \",T2,\" K \\nThe pressure is \",round(p2,1),\"atm\"\n",
+      "print\"At point 3, \\nThe temperature is \",T3,\" K \\nThe pressure is \",p3,\"kN/m**2\" \n",
+      "print\"At point 4, \\nThe temperature is \",T4,\" K \\nThe pressure is \",p4,\"kN/m**2\" \n",
+      "print\"(b)The indicated thermal efficiency is \",round(nth,1),\"percent\" \n",
+      "print\"(c)The indicated mean effective pressure is \",round(imep,0),\" bar\" \n",
+      "print\"(d)The volumetric efficiency is \",round(nv,2), \"percent\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)At point 2, \n",
+        "The temperature is  682  K \n",
+        "The pressure is  15.6 atm\n",
+        "At point 3, \n",
+        "The temperature is  2825  K \n",
+        "The pressure is  7100 kN/m**2\n",
+        "At point 4, \n",
+        "The temperature is  1840  K \n",
+        "The pressure is  570 kN/m**2\n",
+        "(b)The indicated thermal efficiency is  45.7 percent\n",
+        "(c)The indicated mean effective pressure is  14.0  bar\n",
+        "(d)The volumetric efficiency is  88.77 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/chap9.ipynb b/Fundamental_of_internal_combustion_engines/chap9.ipynb
new file mode 100755
index 00000000..a48803b0
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/chap9.ipynb
@@ -0,0 +1,512 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter9:Carburettors and Fuel Injection in SI Engines "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 page no: 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "ma=5                            #Mass flow rate of air per min for a simple jet carburettor in kg/min\n",
+      "mf=0.4                          #Mass flow rate of fuel in kg/min\n",
+      "df=780                          #Density of the fuel in kg/m**3\n",
+      "p1=1.013                        #The initial pressure of air in bar\n",
+      "t1=27                           #The initial temperature of air in degree centigrade\n",
+      "C2=90                           #The air flow velocity in m/s\n",
+      "Cva=0.8                         #The velocity coefficient for the venturi\n",
+      "Cdf=0.6                         #The coefficient of discharge of the main fuel jet \n",
+      "Cpd=0.75                        #The pressure drop across the fuel metering oriface\n",
+      "Cp=1005                         #The specific heat of gas in J/kgK\n",
+      "g=1.4                           #Adiabatic index\n",
+      "R=287                           #Real gas constant in J/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "p2=p1*(1-(C2**2/(Cva**2*2*Cp*(t1+273))))**(g/(g-1))\n",
+      "da1=((p1*10**5)/(R*(t1+273)))\n",
+      "da2=((da1)*(p2/p1)**(1/g))\n",
+      "A2=((ma/60.0)/(da2*C2))*10**4\n",
+      "d2=(4*A2/math.pi)**(1/2.0)\n",
+      "pv=p1-p2\n",
+      "pj=Cpd*pv\n",
+      "Aj=((mf/60.0)/(Cdf*(2*df*pj*10**5)**(1/2.0)))*10**6\n",
+      "dj=(4*Aj/math.pi)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The throat diameter of the choke = \",round(d2,3),\"cm\" \n",
+      "print\"The oriface diameter = \",round(dj,1),\"mm\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The throat diameter of the choke =  3.251 cm\n",
+        "The oriface diameter =  2.2 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 page no: 281"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vs=0.002                                 #The swept volume in m**3\n",
+      "nv=75.0                                  #Volumetric efficiency in percent\n",
+      "N=4500.0                                 #The engine rpm\n",
+      "p1=1.013                                 #The initial pressure of air in bar\n",
+      "R=287.0                                  #Real gas constant in J/kgK\n",
+      "t1=15.0                                  #The atmospheric temperature in degree centigrade\n",
+      "Cp=1005.0                                #The specific heat of gas in J/kgK\n",
+      "g=1.4                                    #Adiabatic index\n",
+      "C2=100.0                                 #The air flow velocity at choke in m/s\n",
+      "Cda=0.85                                 #The velocity coefficient for the venturi\n",
+      "Cdf=0.66                                 #The coefficient of discharge of the main fuel jet \n",
+      "sf=0.75                                  #The specific gravity of fuel\n",
+      "d=0.4                                    #The ratio of the diameter to choke diameter\n",
+      "af=14.0                                  #The air fuel ratio\n",
+      "gf=9.81                                  #The gravitational force constant in m/s**2\n",
+      "Z=0.006                                  #The petrol surface below the choke in m\n",
+      "df=750.0                                 #The density of the fuel in kg/m**3\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "Va=((nv/100.0)*Vs*N)/(2.0*60.0)\n",
+      "V1=Va/2.0\n",
+      "ma=(p1*10**5*V1)/(R*(t1+273))\n",
+      "p2=p1*(1-(C2**2/(2*Cp*(t1+273))))**(g/(g-1))\n",
+      "da1=((p1*10**5)/(R*(t1+273)))\n",
+      "da2=da1*(p2/p1)**(1/g)\n",
+      "A2=(ma/(da2*C2*Cda))*10**6\n",
+      "D=((A2*4)/(math.pi*0.84))**(1/2.0)\n",
+      "mf=ma/af\n",
+      "pm=(p1-p2-(gf*Z*df/10.0**5))*10**5\n",
+      "Aj=(mf/(Cdf*(2*df*pm)**(1/2.0)))*10**6\n",
+      "dj=(4*Aj/math.pi)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The diameter of the choke = \",round(D,2),\"mm\" \n",
+      "print\"The diameter of the jet in = \",round(dj,2),\"mm\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diameter of the choke =  22.89 mm\n",
+        "The diameter of the jet in =  1.26 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 page no: 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=0.08                              #The diameter of the bore in m\n",
+      "L=0.09                              #The length of the stroke in m\n",
+      "N=4000.0                            #The engine rpm\n",
+      "C=84.0                              #The carbon content in the fuel by mass in percent\n",
+      "H=16.0                              #The hydrogen content in the fuel by mass in percent\n",
+      "nv=80.0                             #The volumetric efficiency of the engine in percent\n",
+      "p1=1.0                              #The pressure at ambient condition in bar\n",
+      "t1=25.0                             #The temperature at ambient condition in degree centigrade\n",
+      "p=0.06                              #The depression at venturi throat in bar\n",
+      "ma=0.95                             #The actuat quantity of air supplied\n",
+      "Ra=287.0                            #Real gas constant in J/kgK\n",
+      "Rf=98.0                             #Real gas constant in J/kgK\n",
+      "n=4.0                               #Number of cylinders\n",
+      "Cp=1005.0                           #The specific heat of gas in J/kgK\n",
+      "g=1.4                               #Adiabatic index\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "V=(math.pi/4.0)*d**2*L*(nv/100.0)*(N/(2.0*60.0))*n\n",
+      "Af=(100/23.0)*((C*(32/12.0))+(H*8))/100.0\n",
+      "mfa=Af*ma\n",
+      "Aaf=mfa\n",
+      "da=(p1*10**5)/(Ra*(t1+273))\n",
+      "dv=(p1*10**5)/(Rf*(t1+273))\n",
+      "ma1=V/((1/da)+(1/(mfa*dv)))\n",
+      "mf1=ma1/mfa\n",
+      "p2=p1-p\n",
+      "C2=(2*Cp*(t1+273)*(1-(p2/p1)**((g-1)/g)))**(1/2.0)\n",
+      "T2=(t1+273)*(p2/p1)**((g-1)/g)\n",
+      "d2=(p2*10**5)/(Ra*T2)\n",
+      "A2=(ma1/(d2*C2))*10**4\n",
+      "d2=(A2*4/math.pi)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The fuel flow rate = \",round(mf1,5),\"kg/s\" \n",
+      "print\"The velocity of air at throat = \",round(C2,1),\"m/s\" \n",
+      "print\"The throat diameter = \",round(d2*10,2),\"mm\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The fuel flow rate =  0.00379 kg/s\n",
+        "The velocity of air at throat =  102.5 m/s\n",
+        "The throat diameter =  24.75 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 page no: 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=0.1                                     #The diameter of the bore in m\n",
+      "L=0.12                                    #The length of the stroke in m\n",
+      "N=3000                                    #The engine rpm\n",
+      "d2=0.035                                  #The throat diameter of carburettor venturi in m\n",
+      "nv=80                                     #The volumetric efficiency of the engine in percent\n",
+      "Cda=0.82                                  #The coefficient of discharge of air flow \n",
+      "p=1.013                                   #The ambient pressure in bar\n",
+      "T=298                                     #The ambient temperature in K\n",
+      "ar=15                                     #The air fuel ratio \n",
+      "Z=0.005                                   #The top of the jet above the petrol level in the float chamber in m\n",
+      "Cdf=0.7                                   #The coefficient of discharge for fuel flow \n",
+      "df=750                                    #The specific gravity of the fuel in kg/m**3\n",
+      "R=287                                     #Real gas constant in J/kgK\n",
+      "g=9.81                                    #The gravitational constant in m/s**2\n",
+      "n=4                                       #Number of cylinders \n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "V=(math.pi/4.0)*d**2*L*(nv/100.0)*(N/(2.0*60.0))*n\n",
+      "da=(p*10**5)/(R*T)\n",
+      "ma=V*da\n",
+      "A2=(math.pi/4.0)*d2**2\n",
+      "P=(ma**2/(Cda**2*A2**2*2*da))/10.0**5\n",
+      "mf=ma/ar\n",
+      "Aj=(mf/(Cdf*(2*df*((P*10**5)-(g*Z*df)))**(1/2.0)))*10**6\n",
+      "dj=(Aj*4/math.pi)**(1/2.0)\n",
+      "\n",
+      "#Output \n",
+      "print\"The depression of the throat = \",round(P,3),\"bar\" \n",
+      "print\"The diameter of the fuel jet of a simple carburettor = \",round(dj,3),\"mm\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The depression of the throat =  0.054 bar\n",
+        "The diameter of the fuel jet of a simple carburettor =  1.953 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 page no:286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "mf=(6/3600.0)                               #The mass flow rate of fuel in kg/s\n",
+      "df=750                                      #The density of fuel in kg/m**3\n",
+      "Z=0.003                                     #The level in the float chamber below the top of the jet in m\n",
+      "p=1.013                                     #The ambient pressure in bar\n",
+      "T=294                                       #The ambient temperature in K\n",
+      "dj=0.0012                                   #The jet diameter in m\n",
+      "Cdf=0.65                                    #The discharge coefficient of the jet \n",
+      "Cda=0.8                                     #The discharge coefficient of air \n",
+      "A=15.3                                      #The air fuel ratio \n",
+      "g=9.81                                      #The gravitational constant in m/s**2\n",
+      "R=287                                       #Real gas constant in J/kgK\n",
+      "dh=1000                                     #The density of water in kg/m**2\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "da=(p*10**5)/(R*T)\n",
+      "Ca2=Cda*((2*g*Z*df)/da)**(1/2.0)\n",
+      "Aj=(math.pi/4.0)*dj**2\n",
+      "P=((mf**2/(Cdf**2*Aj**2*2*df))+(g*Z*df))/10.0**5\n",
+      "h=(P*10**5)/(dh*g)\n",
+      "h1=(P*10**5)/g\n",
+      "ma=mf*A\n",
+      "A2=(ma/((Cda*(2*da*(P*10**5))**(1/2.0))))*10**4\n",
+      "d2=((A2*4/math.pi)**(1/2.0))*10\n",
+      "\n",
+      "#Output \n",
+      "print\"The critical air velocity = \",round(Ca2,1),\"m/s\" \n",
+      "print\"The depression at the throat = \",round(h1,1),\"mm of H2O\" \n",
+      "print\"The effective throat diameter  \",round(d2,2),\"mm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The critical air velocity =  4.9 m/s\n",
+        "The depression at the throat =  351.6 mm of H2O\n",
+        "The effective throat diameter   21.12 mm\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 page no: 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d2=22                            #The venturi throat diameter of a simple carburettor in mm\n",
+      "Cda=0.82                         #The coefficient of air flow \n",
+      "dj=1.2                           #The fuel orifice diameter in mm\n",
+      "Cdf=0.7                          #The coefficient of fuel flow\n",
+      "Z=0.004                          #The petrol surface below the throat in m\n",
+      "g=9.81                           #The gravitational constant in m/s**2\n",
+      "da=1.2                           #The density of air in kg/m**3\n",
+      "df=750                           #The density of fuel in kg/m**3\n",
+      "P=0.075                          #The pressure drop in bar\n",
+      "\n",
+      "#Calculations\n",
+      "A=(Cda/Cdf)*(d2**2/dj**2)*(da/df)**(1/2.0)\n",
+      "A1=(Cda/Cdf)*(d2**2/dj**2)*((da*P)/(df*(P-(g*Z*df)/10.0**5)))**(1/2.0) \n",
+      "Ca2=(2*g*Z*df/da)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\" (a) The air fuel ratio when the nozzle lip is neglected = \",round(A,2)\n",
+      "print\"(b)The air fuel ratio when the nozzle lip is considered = \",round(A1,3)\n",
+      "print\"(c) The critical air velocity or minimum velocity required to start the fuel flow = \",round(Ca2,1),\"m/s\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (a) The air fuel ratio when the nozzle lip is neglected =  15.75\n",
+        "(b)The air fuel ratio when the nozzle lip is considered =  15.78\n",
+        "(c) The critical air velocity or minimum velocity required to start the fuel flow =  7.0 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7 page no: 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "h=4000                                   #The altitude of the airplane engine carburettor in m\n",
+      "A=14.7                                   #The air fuel ratio at sea level\n",
+      "ts=22                                    #The temperature at sea level in degree centigrade\n",
+      "R=287                                    #Real gas constant in J/kgK\n",
+      "\n",
+      "#Calculations\n",
+      "ta=ts-(0.0065*h)\n",
+      "p=1.013/10.0**0.2083\n",
+      "da=(p*10**5)/(R*(ta+273))\n",
+      "ds=(1.013*10**5)/(R*(ts+273))\n",
+      "Aa=A*(da/ds)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The air fuel ratio at altitude = \",round(Aa,2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The air fuel ratio at altitude =  12.11\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8 page no: 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "A=14.5                                       #The air fuel ratio\n",
+      "p2=0.825                                     #The pressure at the venturi throat in bar \n",
+      "p1=1.013                                     #The atmospheric pressure in bar\n",
+      "pd=37.5                                      #The pressure drop to the air cleaner in mm of Hg\n",
+      "ma=260                                       #The mass flow rate of air in kg/h\n",
+      "\n",
+      "#Calculations\n",
+      "pa=p1-p2\n",
+      "p21=p1-(pd/750)-pa\n",
+      "pf=pa\n",
+      "pf1=p1-p21\n",
+      "Af=A*(pf/pf1)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"(a) The throat pressure when the air cleaner is fitted = \",p21,\"bar\" \n",
+      "print\"(b) The air fuel ratio with the air cleaner fitted = \",round(Af,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) The throat pressure when the air cleaner is fitted =  0.775 bar\n",
+        "(b) The air fuel ratio with the air cleaner fitted =  12.887\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9 page no: 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bp=8                                    #The brake power of the petrol engine in kW\n",
+      "nb=30                                   #The brake thermal efficiency in percent\n",
+      "CV=44000                                #The calorific value of the fuel in kJ/kg\n",
+      "p1=1.013                                #The suction condition of engine pressure in bar\n",
+      "T1=300                                  #The temperature at suction condition in K\n",
+      "Aj=2.5*10**-6                           #The area of jet in m**2\n",
+      "Z=0.008                                 #The nozzle lip in m\n",
+      "g=9.81                                  #The gravitational force constant in m/s**2\n",
+      "A=15                                    #The air fuel ratio\n",
+      "Cda=0.9                                 #The coefficient of air flow\n",
+      "Cdf=0.7                                 #The coefficient of fuel flow\n",
+      "df=750                                  #The density of fuel in kg/m**3\n",
+      "va=0.8                                  #The specific volume of air in m**3/kg\n",
+      "\n",
+      "#Calculations\n",
+      "import math\n",
+      "va1=va*T1/273.0\n",
+      "da=1/va\n",
+      "mf=bp/((nb/100.0)*CV)\n",
+      "Cf=mf/(Cdf*df*Aj)\n",
+      "P=((Cf**2*df)/2.0)+(df*g*Z)\n",
+      "Ca=(2*P/da)**(1/2.0)\n",
+      "ma=mf*A\n",
+      "A2=(ma/(Cda*da*Ca))*10**4\n",
+      "d2=(A2*4/math.pi)**(1/2.0)\n",
+      "\n",
+      "#Output\n",
+      "print\"The venturi throat diameter of the carburator = \",round(d2,2),\"cm\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The venturi throat diameter of the carburator =  2.63 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Fundamental_of_internal_combustion_engines/screenshots/2_2.png b/Fundamental_of_internal_combustion_engines/screenshots/2_2.png
new file mode 100755
index 00000000..1b7801d1
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/screenshots/2_2.png
diff --git a/Fundamental_of_internal_combustion_engines/screenshots/3_9.png b/Fundamental_of_internal_combustion_engines/screenshots/3_9.png
new file mode 100755
index 00000000..f92f110f
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/screenshots/3_9.png
diff --git a/Fundamental_of_internal_combustion_engines/screenshots/4_1.png b/Fundamental_of_internal_combustion_engines/screenshots/4_1.png
new file mode 100755
index 00000000..7199feee
--- /dev/null
+++ b/Fundamental_of_internal_combustion_engines/screenshots/4_1.png
diff --git a/Hydraulics_Made_Easy/README.txt b/Hydraulics_Made_Easy/README.txt
new file mode 100755
index 00000000..5cbe8938
--- /dev/null
+++ b/Hydraulics_Made_Easy/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Alpesh Makwana
+Course: bca
+College/Institute/Organization: Zealous web technologies
+Department/Designation: Developer
+Book Title: Hydraulics Made Easy
+Author: R. S. Dighe
+Publisher: Royal Book Stall
+Year of publication: 2005
+Isbn: 0415250706
+Edition: 1
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch1.ipynb b/Hydraulics_Made_Easy/ch1.ipynb
new file mode 100755
index 00000000..530a253b
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch1.ipynb
@@ -0,0 +1,1015 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:d3eb95f5e18a78e74ef41bfca92729f5344f3fef9eebaa02f35bb683ba9d0489"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Hydrostatics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1 Page No : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Ar = 50 \t\t#area of ram in**2\n",
+      "Ap = 1./8 \t\t#area of plunger in**2\n",
+      "Wp = 5. \t\t#force lbs\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Pp = Wp/Ap\n",
+      "F = Pp*Ar\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'weight supported by ram  = %.f lbs'%(F)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "weight supported by ram  = 2000 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 Page No : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Dp = 1. \t\t#diameter of punger - in\n",
+      "Dr = 10. \t\t#diameter of ram - in\n",
+      "R = 12.         #leverage of handle\n",
+      "W = 15. \t\t#wieght of body - tons\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Ar = math.pi*Dr**2/4\n",
+      "Ap = math.pi*Dp**2/4\n",
+      "P = W*2240/((Ar/Ap)*R)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'power applied to lever  = %.f lbs'%(P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power applied to lever  = 28 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 Page No : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Dj = 1. \t\t#diameter of plunger - in\n",
+      "Dr = 2. \t\t#in\n",
+      "W = 40. \t\t#lbs\n",
+      "W1 = 1. \t\t#ton\n",
+      "rl = 20.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Ap = math.pi*Dj**2/4\n",
+      "Ar = math.pi*Dr**2/4\n",
+      "Vrj = rl*Ar/Ap\n",
+      "e = W1*2240*100/(W*Vrj)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'efficiency of machine at this load  = %.f percent'%(e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "efficiency of machine at this load  = 70 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 Page No : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Dj = 1. \t\t#in\n",
+      "Dr = 2. \t\t#in\n",
+      "ns = 3. \t\t#strokes\n",
+      "h = 2.   \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Ap = math.pi*Dj**2/4\n",
+      "Ar = math.pi*Dr**2/4\n",
+      "Vrj = Ar/Ap\n",
+      "ns1 = h*12*Vrj/ns\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  ' working strokes  = %.f strokes'%(ns1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " working strokes  = 32 strokes\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 Page No : 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "T = 40. \t\t#F\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "h = 50 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "p = w*h/(12**2)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  ' pressure at a depth of 50 ft  = %.2f lbs per in'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " pressure at a depth of 50 ft  = 21.67 lbs per in\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 Page No : 13"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "W = 64. \t\t#lbs/ft**3\n",
+      "h1 = 27. \t\t#ft\n",
+      "h2 = 9. \t\t#ft\n",
+      "w = 40. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Pr = w*W*h1*h1/2\n",
+      "Pl = w*W*h2*h2/2\n",
+      "y1 = h1/3\n",
+      "y2 = h2/3\n",
+      "y = (Pr*y1-Pl*y2)/(Pr-Pl)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  ' point of application  = %.2f ft'%(y)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " point of application  = 9.75 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 Page No : 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 5.  \t\t#ft\n",
+      "x = 3. \t    \t#ft\n",
+      "w = 62.4 \t\t#lb/ft**3\n",
+      "a = 90. \t\t#degrees\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = math.pi/4*d**2\n",
+      "b = w*A*x\n",
+      "Ig = round(math.pi*d**4/64,2)\n",
+      "Io = Ig + A*x**2 * 1\n",
+      "h = Io/(A*x)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'depth of the pressure = %.2f ft'%(h)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of the pressure = 3.52 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 Page No : 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 3.   \t\t#ft\n",
+      "h = 4. \t    \t#ft\n",
+      "ht = 30 \t\t#ft\n",
+      "W = 62.4 \t\t#ft**3\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Ap = w*h\n",
+      "X = ht+(h/2)\n",
+      "P = Ap*X*W\n",
+      "I0 = (w*h**3/12)+Ap*X**2\n",
+      "H = I0/(Ap*X)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  ' total pressure on the gate  = %.2f ft'%(H)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " total pressure on the gate  = 32.04 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9 Page No : 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 3. \t\t#ft\n",
+      "h = 4. \t\t#ft\n",
+      "ht = 30. \t\t#ft\n",
+      "W = 62.4 \t\t#ft**3\n",
+      "x = 2.22 \t\t#in\n",
+      "x1 = 4.5 \t\t#in\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Ap = w*h\n",
+      "X = ht+(h/2)\n",
+      "P = Ap*X*W\n",
+      "T = P*x/x1\n",
+      "T1 = P-T\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  ' tension devoloped in the top bolt  = %.f lbs'%(T)\n",
+      "print  '  tension devoloped in the bottom bolt  = %d lbs'%(T1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " tension devoloped in the top bolt  = 11821 lbs\n",
+        "  tension devoloped in the bottom bolt  = 12140 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.10 Page No : 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 3.   \t\t#ft\n",
+      "h = 15. \t\t#ft\n",
+      "d = 140. \t\t#lbs/ft**3\n",
+      "x = 6. \t    \t#in\n",
+      "W = 62.4 \t\t#lbs/ft**3\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "W1 = h*w*d\n",
+      "h = (W1*x*6/(W*12))**(1./3)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  ' height of water rise  = %.2f ft'%(h)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " height of water rise  = 6.72 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11 Page No : 17"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "h = 5. \t\t#ft\n",
+      "d = 6. \t\t#ft\n",
+      "a = 30. \t\t#degrees\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = math.pi*d**2/4\n",
+      "X = h+(d/2)*math.sin(math.radians(a))\n",
+      "P = w*A*X\n",
+      "Ic = math.pi*d**4/64\n",
+      "I0 = Ic+A*X**2/(math.sin(math.radians(a)))**2\n",
+      "h = I0*(math.sin(math.radians(a)))**2/(A*X)\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "print  'depth of the centre os pressure = %.2f ft '%(h)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of the centre os pressure = 6.59 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.12 Page No : 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 4. \t    \t#ft\n",
+      "l = 4. \t\t    #ft\n",
+      "X = 10. \t\t#ft\n",
+      "a = 45. \t\t#degrees\n",
+      "W = 100. \t\t#lbs\n",
+      "a1 = 60. \t\t#degrees\n",
+      "w1 = 62.4 \t\t#lbs/ft**3\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = w*l\n",
+      "X1 = round(X+(w/2)*math.sin(math.radians(a)),2)\n",
+      "Ig = round(w*l**3/12,2)\n",
+      "I0 = Ig+(A*X1**2/(math.sin(math.radians(a)))**2)\n",
+      "h = I0*(math.sin(math.radians(a)))**2/(A*X1)\n",
+      "P = round(w1*A*X1,-2)\n",
+      "h1 = round(h-X,2)\n",
+      "h2 = round(h1/math.sin(math.radians(a)),2)\n",
+      "T = (W*(l/2)*math.sin(math.radians(a))+P*h2)/(w*math.sin(math.radians(a1)))\n",
+      "\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Pull in the chain = %d lbs '%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pull in the chain = 6885 lbs \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.13 Page No : 20"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 4. \t\t#ft\n",
+      "l = 4. \t\t#ft\n",
+      "X = 10. \t\t#ft\n",
+      "a  = 45. \t\t#degrees\n",
+      "W = 62.4 \t\t#lbs/ft**3\n",
+      "u = 0.25\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = w*l\n",
+      "X1 = X+(w/2)*math.sin(math.radians(a))\n",
+      "P = W*A*X1\n",
+      "T = u*P\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'magnitude of the lifting force = %.f lbs '%(round(T,-1)) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnitude of the lifting force = 2850 lbs \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.14 Page No : 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "sg = 1.6\n",
+      "h = 10. \t\t#ft\n",
+      "h1 = 4. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "D = w*sg\n",
+      "W = w*(h+h1)**2/2\n",
+      "P = w*h\n",
+      "P1 = D*h1\n",
+      "P2 = (P*h/2)+P*h1+(h1*P1/2)\n",
+      "y = ((P*h*(h1+(h/3))/2)+P*h1*(h1/2)+P1*h1**2/6)/P2\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Position where P acts = %.1f ft above the base'%(y) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Position where P acts = 4.5 ft above the base\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.15 Page No : 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "pa = 10.     \t#lbs/in**2\n",
+      "h = 8. \t\t    #ft\n",
+      "h1 = 6. \t\t#ft\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "pg = 10. \t\t#lbs/in**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Pa = pa*144\n",
+      "Pa1 = w*h1\n",
+      "Pt = (Pa*h+Pa1*(h1/2))\n",
+      "y = (Pa*h*(h/2)+(Pa1*h1*(h-h1)/2))/Pt\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Depth of the centre of pressure = %.2f ft from the base'%(y) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Depth of the centre of pressure = 3.82 ft from the base\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.16 Page No : 23"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 4. \t\t#ft\n",
+      "h = 6. \t\t#in\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = math.pi*d**2/4\n",
+      "X = (h-d)\n",
+      "I0 = (math.pi*d**4/64)+4*math.pi*(X)**2\n",
+      "h1 = I0/(A*X)\n",
+      "h2 = d-h1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Depth of the axis be placed in order = %.1f ft '%(h2) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Depth of the axis be placed in order = 1.5 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.17 Page No : 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "h = 10 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "x = math.sqrt(h**2/2)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Depth of the axis be placed in order = %.2f ft '%(x) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Depth of the axis be placed in order = 7.07 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.18 Page No : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "h = 8. \t\t#ft\n",
+      "h1 = 10. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = h\n",
+      "X = (h1/2)\n",
+      "Ig = h**3/12\n",
+      "I0 = Ig+A*X**2\n",
+      "h2 = I0/(A*X)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'depth at which the hinge of the shutter = %.2f ft '%(h2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth at which the hinge of the shutter = 6.07 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.19 Page No : 27"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import *\n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "k1 = 1. \t\t#ft\n",
+      "k2 = 35.98 \t\t#ft\n",
+      "k3 = 66.83 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "\n",
+      "vec =roots([k1,0,-k2,k3])\n",
+      "\n",
+      "X = vec[1]\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'depth of the water = %.2f ft'%(X)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of the water = 4.65 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.22 Page No : 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 8.   \t\t#ft\n",
+      "d1 = 2.     \t\t#ft\n",
+      "h = 4. \t\t    #ft\n",
+      "h1 = 2 \t\t    #ft\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A1 = math.pi*d**2/4\n",
+      "A2 = math.pi*d1**2/4\n",
+      "A = A1-A2\n",
+      "x = (A1*d-A2*(d+h-h1))/A\n",
+      "P = w*A*x\n",
+      "Ig = ((math.pi*d**4/64)+(A1*(d-x)**2))-((math.pi*d1**4/64)+(A2*(h1+d-x)**2))\n",
+      "h2 = (Ig/(A*x))+x\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'depth of the centre of the pressure = %.1f ft '%(h2)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of the centre of the pressure = 8.4 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.25 Page No : 34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "W = 62.4 \t\t#lbs/ft**3\n",
+      "a = 140. \t\t#degrees\n",
+      "h = 20. \t\t#ft\n",
+      "w = 6. \t\t    #ft\n",
+      "h1 = 17. \t\t#ft\n",
+      "h2 = 5. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "P1 = int(W*h1**2*w/2)\n",
+      "P2 = W*h2**2*w/2\n",
+      "P = P1-P2\n",
+      "y = (P1*(h1/3)-P2*(h2/3))/P\n",
+      "R = P/(2*math.sin(math.radians((180-a)/2)))\n",
+      "Rt = y*R/h\n",
+      "Rb = R-Rt\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Rt = %.f lbs '%(Rt)\n",
+      "print  ' Rb = %.f lbs '%(Rb)\n",
+      "\n",
+      "# note : incorrect answer for R in the textbook. Hence, the difference in answers"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rt = 21838 lbs \n",
+        " Rb = 50409 lbs \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.26 Page No : 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 64. \t\t#lbs/ft**3\n",
+      "h = 12. \t\t#ft\n",
+      "l = 9. \t\t#ft\n",
+      "a = 45. \t\t#degrees\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "P = w*h**2/2\n",
+      "h1 = h/3\n",
+      "Rb = P*h1/l\n",
+      "Ra = P-Rb\n",
+      "Wh = Rb*h1\n",
+      "T = Wh/math.sin(math.radians(a))\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Load on the strut = %d lbs '%(T)\n",
+      "\n",
+      "# note : incorrect answer for T in the textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Load on the strut = 11585 lbs \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.27 Page No : 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "h = 9. \t\t#ft\n",
+      "l = 10. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "P = w*h**2/2\n",
+      "h1 = h/3\n",
+      "Ra = P/2\n",
+      "x = (w*4*h**2/9)/Ra\n",
+      "x1 = x+(h/3)\n",
+      "hb = h1-x\n",
+      "W = Ra*l\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'magnitude od total in each beam = %d lbs '%(W)\n",
+      "\n",
+      "# note : rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnitude od total in each beam = 12636 lbs \n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch10.ipynb b/Hydraulics_Made_Easy/ch10.ipynb
new file mode 100755
index 00000000..0162251f
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch10.ipynb
@@ -0,0 +1,497 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2a1a58d10d8bc25ee316405092c20184a6b3d847d7641c4fa061749730145581"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Miscellaneous Problems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page No : 316"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "w = 62.4 \t\t#lb/ft**3\n",
+      "x = 8.  \t\t#ft\n",
+      "A = 16. \t\t#ft**2\n",
+      "X = 2.5 \t\t#ft\n",
+      "X1 = 0.66 \t\t#ft\n",
+      "x1 = 3.834 \t\t#ft\n",
+      "x2 = 2.182 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "P = w*x*A\n",
+      "y = A/3\n",
+      "P1 = w*x*A*0.5*X1\n",
+      "R = math.sqrt(P1**2+P**2)\n",
+      "m = P1/P\n",
+      "X2 = x1-x2\n",
+      "C = ((2./3)*A)-m*X\n",
+      "Y = m*X2+ C\n",
+      "print P1\t\n",
+      "#RESULTS\n",
+      "print  'Water pressure on vertical face  = %.f lbs'%(round(P,-3))\n",
+      "print  ' pressure which acts at the base  = %.2f ft'%(y)\n",
+      "print  ' Resultant  = %.f lbs'%(R)\n",
+      "print  ' x coordinate of the resultant  = %.3f ft'%(X2)\n",
+      "print  ' y coordinate of the resultant  = %.3f ft'%(Y)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2635.776\n",
+        "Water pressure on vertical face  = 8000 lbs\n",
+        " pressure which acts at the base  = 5.33 ft\n",
+        " Resultant  = 8411 lbs\n",
+        " x coordinate of the resultant  = 1.652 ft\n",
+        " y coordinate of the resultant  = 10.387 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page No : 319"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "s = 13.6\n",
+      "h = 12. \t\t#in\n",
+      "u = 0.04\n",
+      "k = 1.\n",
+      "d = 6.   \t\t#in\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "h1 = h*(s-1)/12\n",
+      "hf = u*h1\n",
+      "hn = h1-hf\n",
+      "Q = k*math.pi/4*(d/12)**2*8.02*math.sqrt(hn)/(math.sqrt(16-k))\n",
+      "Q = Q*60*w/10        # fro, cusecs to GPM\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'discharge through flow = %.f ft G.P.M'%(Q)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "discharge through flow = 529 ft G.P.M\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 Page No : 321"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "za = 16. \t\t#ft\n",
+      "h1 = 2. \t\t#ft\n",
+      "h2 = 3. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "vc = math.sqrt(2*g*(za-h1-h2))\n",
+      "vb = vc*(h1/(2*h1))**2\n",
+      "r = -h1-h2-(vb**2/(2*g))\n",
+      "r1 = r+34\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'pressure head at B = %.1f ft lb'%(r1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure head at B = 28.3 ft lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 Page No : 322"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "Cd = 0.62\n",
+      "a = 90. \t\t#degrees\n",
+      "H1 = 14. \t\t#in\n",
+      "H2 = 8. \t\t#in\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q1 = (8./15)*Cd*math.sqrt(2*g)*math.tan(math.radians(a/2))*(H1/12)**(5/2.)\n",
+      "Q2 = (8./15)*Cd*math.sqrt(2*g)*math.tan(math.radians(a/2))*(H2/12)\n",
+      "Q = Q1-Q2\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge through notch = %.2f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge through notch = 2.13 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 Page No : 324"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "Cd = 0.62\n",
+      "d = 5./4 \t\t#in\n",
+      "h = 9. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "T = (2./3)*math.pi*(h)**(3./2)/(Cd*(math.pi/4)*math.sqrt(2*g)*(d/12)**2)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'time required to lower water level = %.f secs'%(T)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required to lower water level = 1334 secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 Page No : 325"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "a = 60. \t\t#degrees\n",
+      "d = 4.  \t\t#in\n",
+      "Cd = 0.62\n",
+      "h = 5. \t    \t#ft\n",
+      "w = 30. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "H1 = 10*math.sin(math.radians(a))\n",
+      "H2 = H1-h\n",
+      "T = (2*w/math.tan(math.radians(a)))*(2./3)*(H1**(3./2)-H2**(3./2))/(Cd*math.sqrt(2*g)*math.pi/(4*(d/12)**2))*100\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time required to lower water level = %.f secs'%(T)\n",
+      "\n",
+      "# answer is accurate.please check manually"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time required to lower water level = 1214 secs\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7 Page No : 326"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "p1 = 40. \t\t#percent\n",
+      "p2 = 35. \t\t#percent\n",
+      "dh = 200. \t\t#ft\n",
+      "f = 0.1\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "l = 2000. \t\t#ft\n",
+      "d = 1.   \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "hf1 = p1*dh/100\n",
+      "hf2 = p2*dh/100\n",
+      "hf3 = (100-p1-p2)*dh/100\n",
+      "hft = hf1+hf2+hf3\n",
+      "v1 = math.sqrt(2*g*hf1/(4*f*l))\n",
+      "Q = v1*math.pi*d**2/4\n",
+      "d2 = (Q*7*math.sqrt(3/(5*g)))**(2./3)\n",
+      "v3 = Q*4*(4./3)**2/math.pi\n",
+      "l3 = hf2*2*g*(3./4)/(4*f*v3**2)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'proportion of the quantity folwing in the bypass to the whole pass = %d ft'%(l3)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "proportion of the quantity folwing in the bypass to the whole pass = 415 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.8 Page No : 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 1. \t    \t#ft\n",
+      "l = 2000. \t\t#ft\n",
+      "f = 0.038\n",
+      "g = 32.2 \t\t#/ft/sec**2\n",
+      "Q = 6.   \t\t#cuses\n",
+      "l1 = 1500. \t\t#ft\n",
+      "r = 2.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v = 4*Q/(d**2*math.pi)\n",
+      "hf = 4*f*l*v**2/(2*g)\n",
+      "v1 = math.sqrt(hf*2*g/(4*f*l1+4*f*(l-l1)*r**2))\n",
+      "v3 = r*v1\n",
+      "Q1 = math.pi*d**2*v3/4\n",
+      "Q2 = math.pi*d**2*v1/4\n",
+      "r1 = Q2/Q1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'proportion of the quantity folwing in the bypass to the whole pass = %.1f '%(r1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "proportion of the quantity folwing in the bypass to the whole pass = 0.5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.9 Page No : 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "f = 0.01\n",
+      "d = 3. \t\t#in\n",
+      "l = 22. \t\t#ft\n",
+      "l1 = 20. \t\t#ft\n",
+      "w = 20. \t\t#ft\n",
+      "h = 5. \t\t#ft\n",
+      "h1 = 20. \t\t#ft\n",
+      "t = 4. \t\t#min\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "h2 = h+h1\n",
+      "h3 = (h-(t*60*math.pi*math.sqrt(2*g/h)/(l1*w*2*64)))**2-4\n",
+      "dh = h2-h3\n",
+      "Q = dh*l1*w\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Quantiy discharged = %.f cuses '%(round(Q,-2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantiy discharged = 1800 cuses \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.10 Page No : 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2     \t\t#ft/sec**2\n",
+      "sct = 1.6\n",
+      "sl = 0.8\n",
+      "K = 0.98\n",
+      "dh1 = 4.      \t\t#ft\n",
+      "W = 62.4     \t\t#lbs/ft**3\n",
+      "d1 = 8.       \t\t#in\n",
+      "d2 = 6.       \t\t#in\n",
+      "\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "dp = dh1*((sct/sl)-1)\n",
+      "C = math.sqrt(2*g)*math.pi*(d1/24)**2 /math.sqrt((d1**2/d2**2)**2 -1)\n",
+      "Q = C*K*math.sqrt(dh1)\n",
+      "\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge passing through the pipe = %.1f cuses '%(Q)\n",
+      "\t\t#The answer given in textbook is wrong. Please verify it.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge passing through the pipe = 3.7 cuses \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch2.ipynb b/Hydraulics_Made_Easy/ch2.ipynb
new file mode 100755
index 00000000..31d84882
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch2.ipynb
@@ -0,0 +1,368 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:1efa4f61c6638a7002643398d2d4bfe4efa140d30a6184e17976681fd0180ed8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Floatation and Buoyancy"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page No : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "l = 60. \t\t#ft\n",
+      "w = 10. \t\t#ft\n",
+      "h = 5.   \t\t#ft\n",
+      "t = 3./16 \t\t#in\n",
+      "sp  = 7.75\n",
+      "H = 4. \t    \t#ft\n",
+      "w1 = 62.4 \t\t#lb/ft**3\n",
+      "y = 4. \t\t    #ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "V = (l*w+2*w*h+2*l*h)*t/12\n",
+      "W = V*w1*sp\n",
+      "x = W/(w1*l*w)\n",
+      "W1 = H*l*w*w1\n",
+      "dW = (W1-W)/2238\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'weight of water print laced = %.1f tons'%(dW)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "weight of water print laced = 62.5 tons\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page No : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "D = 64. \t\t#lb/ft**3\n",
+      "d = 6.   \t\t#ft\n",
+      "l = 10. \t\t#ft\n",
+      "W = 2. \t    \t#tons\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "V = W*2240/D\n",
+      "h = V/(math.pi*d**2/4)\n",
+      "BM = d**2/(16*h)\n",
+      "P = -(math.sqrt(64*BM*2*10*math.pi*(22400-math.pi*d**4))-W*22400)/10\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Minimum pull required = %.f lbs '%(P+3) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum pull required = 3665 lbs \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page No : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "sg = 7.\n",
+      "sg1 = 5.\n",
+      "d = 8. \t\t#in\n",
+      "t = 1. \t\t#in\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "x = (sg+sg1)+math.sqrt(d*(sg*(sg1+t)+1))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'maximum length of cylinder = %.2f in '%(x) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum length of cylinder = 30.55 in \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page No : 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "W = 2000. \t\t#tons\n",
+      "m = 15. \t\t#/tons\n",
+      "dx = 24. \t\t#ft\n",
+      "l = 3. \t    \t#in\n",
+      "dx1 = 5. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "GM = m*dx/(W*(l/(dx1*12)))\n",
+      "\t\t\n",
+      "#RESULTSS\n",
+      "print  'metacentric height = %.1f ft '%(GM) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "metacentric height = 3.6 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page No : 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "M = 350. \t\t#tons\n",
+      "l = 50. \t\t#ft\n",
+      "w = 20. \t\t#ft\n",
+      "W = 100. \t\t#tons\n",
+      "h = 6.   \t\t#ft\n",
+      "M1 = 250. \t\t#tons\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "V = M*2240/64\n",
+      "d = V/(l*w)\n",
+      "BM = l*w**3/(12*w*l*d)\n",
+      "y = (((BM+(d/2))*(M/10))-(M1*h/10))/(W/10)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Highest position of centre of gravity = %.2f ft '%(y)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Highest position of centre of gravity = 15.96 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page No : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "W = 2000. \t\t#tons\n",
+      "l = 250. \t\t#ft\n",
+      "w = 30. \t\t#ft\n",
+      "a = 1./15\n",
+      "W1 = 50. \t\t#tons\n",
+      "h = 10. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "BG = (l*w**3*64/(W*2240*12))-(W1*h/(a*W))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'distance of the centre of gravity = %.2f ft '%(BG) \n",
+      "\n",
+      "# note : rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance of the centre of gravity = 4.29 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10 Page No : 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "l = 91. \t\t#ft\n",
+      "w = 30. \t\t#ft\n",
+      "h = 6. \t    \t#ft\n",
+      "W = 40. \t\t#tons\n",
+      "a = 3. \t\t    #degrees\n",
+      "cg = 3. \t\t#ft\n",
+      "d = 4.  \t\t#ft\n",
+      "W1 = 60. \t\t#tons\n",
+      "cg1 = 1. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "W2 = (l*w*d*64/2240)-W1\n",
+      "y = (W2*(h/2)+W1*(cg+d))/(l*w*d*64/2240)\n",
+      "BG = y-(d/2)\n",
+      "BM = l*w**3/(12*l*w*d)\n",
+      "GM = BM-BG\n",
+      "dx = GM*l*w*d*64*math.tan(math.radians(a))/(60*2240)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'maximum distance through which the load can be shifted = %.1f ft '%(dx)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum distance through which the load can be shifted = 4.6 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.11 Page No : 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "W = 5000. \t\t#tons\n",
+      "I = 1.4*10**6 \t\t#ft**4\n",
+      "k = 12.2 \t\t#ft\n",
+      "BG = 6.5 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "BM = I*64/(W*2240)\n",
+      "GM = BM-BG\n",
+      "T = 2*math.pi*math.sqrt(k**2/(GM*32.2))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'period of oscialltion = %.2f sec '%(T) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "period of oscialltion = 11.03 sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch3.ipynb b/Hydraulics_Made_Easy/ch3.ipynb
new file mode 100755
index 00000000..a7c97aa0
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch3.ipynb
@@ -0,0 +1,527 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2f07cf575072954f2c364fa1bc9af38d9d9d43725d39198530552558da11415c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Flow of Water"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page No : 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d1 = 1. \t\t#ft\n",
+      "d2 = 6. \t\t#in\n",
+      "h1 = 5. \t\t#ft\n",
+      "h2 = 15. \t\t#ft\n",
+      "Pa = 15. \t\t#lbs\n",
+      "v1 = 10. \t\t#ft/sec\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v2 = v1/(d2/12)**2\n",
+      "Pb = (w*((Pa+(Pa*144/w)+(v1**2/(2*g)))-h1-(v2**2/(2*g))))/144\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Pb = %.2f lbs/in**2 '%(Pb)  \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pb = 9.24 lbs/in**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page No : 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d1 = 4. \t\t#ft\n",
+      "d2 = 2. \t\t#ft\n",
+      "h1 = 50. \t\t#ft\n",
+      "h2 = 45. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "r = (d1**2/d2**2)\n",
+      "v1 = round(math.sqrt((h1-h2)*2*g/(r**2-1)),1)\n",
+      "Q = v1*math.pi*d1**2/4\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'discharge through pipe = %.2f cubic feet per second '%(Q)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "discharge through pipe = 57.81 cubic feet per second \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page No : 70"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "z1 = 10. \t\t#/m\n",
+      "h1 = 10. \t\t#m\n",
+      "v1 = 12. \t\t#ft/sec\n",
+      "v2 = 4. \t\t#m/sec\n",
+      "k = 0.6\n",
+      "w = 62.4 \t\t#lb/in**2\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "p = (w/144)*(z1+h1+(v1**2/(2*g))-(v2**2/(2*g))-(k*(v1-v2)**2/(2*g)))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'pressure at bottom end  = %.2f lb/in**2'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure at bottom end  = 9.27 lb/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 Page No : 73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 4.   \t\t#ft\n",
+      "d1 = 5./4 \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "h = 3. \t    \t#ft\n",
+      "K = 1.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "C = (math.pi/4)*d**2*math.sqrt(2*g)/(math.sqrt((d**2/d1**2)**2-1))\n",
+      "Q = K*math.sqrt(h)*C\n",
+      "V = Q/(math.pi*d1**2/4)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Velocity at the throat = %.2f ft/sec '%(V)\n",
+      "\n",
+      "# ronding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity at the throat = 13.97 ft/sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 Page No : 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 9. \t\t#in\n",
+      "d1 = 4. \t\t#in\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "dh = 10. \t\t#in\n",
+      "sg = 13.6 \n",
+      "K = 1.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "C = (((math.pi/4)**2*(d*d1)**2*math.sqrt(2*g)/144**2)/(math.sqrt((math.pi*d**2/12**2)**2-(math.pi*d1**2/12**2)**2)))+0.52\n",
+      "h = (sg-1)*dh/12\n",
+      "Q = K*C*math.sqrt(h)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge passing through the pipe = %.2f cuses '%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge passing through the pipe = 2.26 cuses \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 Page No : 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "sm = 13.6\n",
+      "so = 0.8\n",
+      "di = 8. \t\t#in\n",
+      "dt = 4. \t\t#in\n",
+      "K = 0.98\n",
+      "v = 1.  \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "s = sm/so\n",
+      "dp = v*12*(s-1)/12\n",
+      "A = math.pi*(di/12)**2/4\n",
+      "At = math.pi*(dt/12)**2/4\n",
+      "C = A*math.sqrt(2*g)/(math.sqrt((A/At)**2-1))\n",
+      "Q = C*math.sqrt(v*12+dt)*K\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge passing through the pipe = %.2f cuses '%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge passing through the pipe = 2.84 cuses \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 Page No : 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "s = 1./10\n",
+      "d1 = 6. \t\t#in\n",
+      "d2 = 2. \t\t#in\n",
+      "l = 20. \t\t#in\n",
+      "p = 15. \t\t#lbs/in**2\n",
+      "p1 = 6. \t\t#lbs/in**2\n",
+      "K = 0.95\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "H = (l*s/12)-(p1*144/(2*g))+(p*144/(2*g))\n",
+      "C = math.sqrt(2*g)*(math.pi*(d1/12)**2)/(4*(math.sqrt((d1**2/d2**2)**2-1)))\n",
+      "Q = C*K*math.sqrt(H)*374.7\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Discharge passing through the pipe = %.f Gallons/minute '%(Q)\n",
+      "\n",
+      "# note : rounding off error at value of H in textbook. so answer is slightly different"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge passing through the pipe = 282 Gallons/minute \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 Page No : 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d1 = 12. \t\t#in\n",
+      "Q = 4.25 \t\t#ft**3/sec\n",
+      "h = 18. \t\t#ft\n",
+      "K = 0.98\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "sm = 13.6\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "R = math.sqrt((K*math.sqrt(2*g)*math.sqrt(h)*(math.pi*(d1/12)**2/4)/Q)+1)\n",
+      "d2 = math.sqrt(d1**2/(144*R))\n",
+      "dh = (sm-1)*(h/(12*2))\n",
+      "d3 = Q*math.sqrt(dh/h)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Diameter of the throat = %.2f ft '%(d3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of the throat = 3.08 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 Page No : 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "R = 4. \t\t#in\n",
+      "r = 0.5 \t\t#in\n",
+      "c = 0.007\n",
+      "K = 33.96\n",
+      "w = 62.4 \t\t#lb/ft**3\n",
+      "pa = 12.13 \t\t#lb/in**2\n",
+      "pb = 14.7 \t\t#lb/in**2\n",
+      "w1 = 2.5 \t\t#lbs\n",
+      "Q = 40. \t\t#gals/min\n",
+      "h = 1.86\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "va = Q*4*(2*r*12)**2/(6*w*math.pi)\n",
+      "vb = Q*(2*r*12)**2/(6*w*2*R*math.pi*0.32)\n",
+      "vx = vb*R/2\n",
+      "pu = 2*math.pi*w*h\n",
+      "pd = pb*math.pi*R**2\n",
+      "RP = pb*math.pi*R**2-2*math.pi*w*(0.5*K*((R/12)**2-(r/12)**2)-c*math.log(R/r))-pa*math.pi*r**2+w1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'velocity va = %.1f ft/sec'%(va)\n",
+      "print  'velocity vb = %.2f ft/sec'%(vb)\n",
+      "print  'velocity vx = %.2f ft/sec'%(vx)\n",
+      "print  'pressure px = %.1f lbs/in**2'%(pb)\n",
+      "print  'upward pressure  = %.1f lbs'%(pu)\n",
+      "print  'downward pressure  = %.1f lbs'%(pd)\n",
+      "print  'Resultant pressure  = %.1f lbs'%(RP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity va = 19.6 ft/sec\n",
+        "velocity vb = 1.91 ft/sec\n",
+        "velocity vx = 3.83 ft/sec\n",
+        "pressure px = 14.7 lbs/in**2\n",
+        "upward pressure  = 729.3 lbs\n",
+        "downward pressure  = 738.9 lbs\n",
+        "Resultant pressure  = 9.4 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 Page No : 86"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 1.  \t\t#ft\n",
+      "h = 4. \t    \t#ft\n",
+      "h1 = 3. \t\t#ft\n",
+      "p = 25. \t\t#percent\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "h2 = ((h/4)-(h1/4))*h*2\n",
+      "w = math.sqrt(h2*2*g/(d/2)**2)\n",
+      "N = w*60/(2*math.pi)\n",
+      "h3 = (h-h1**2/4)*2\n",
+      "w1 = math.sqrt(h3*2*g/(d/2)**2)\n",
+      "N1 = w1*60/(2*math.pi)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'original volume = %.1f R.P.M '%(N1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "original volume = 286.7 R.P.M \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 Page No : 89"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "R2 = 2. \t\t#ft\n",
+      "R1 = 1. \t\t#ft\n",
+      "w = 200. \t\t#r.p.m\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v2 = R2*math.pi*w*R2/60\n",
+      "v1 = R2*math.pi*w*R1/60\n",
+      "H = (v2**2-v1**2)/(2*g)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'centrifugal head = %.1f ft of watrer '%(H)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "centrifugal head = 20.4 ft of watrer \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch4.ipynb b/Hydraulics_Made_Easy/ch4.ipynb
new file mode 100755
index 00000000..0747f48b
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch4.ipynb
@@ -0,0 +1,994 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:98b9b2cc3c2bb837908ed50b11e73842e607c874a3e8f477bc1209f33ea8918d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Flow of Water Through Orifices and Mouthpieces"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page No : 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "M = 31*10 \t\t#lbs\n",
+      "P = 3.6 \t\t#lbs\n",
+      "t = 60. \t\t#sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "H = 9. \t\t#ft\n",
+      "d = 1. \t\t#in\n",
+      "w = 6.24 \t\t#gallons\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v = P*g*t/M\n",
+      "V = math.sqrt(2*g*H)\n",
+      "Cv = v/V\n",
+      "V1 = math.pi*(d/12)**2*V*60*w/4\n",
+      "Cd = M/(10*V1)\n",
+      "Cc = Cd/Cv\n",
+      "Cr = (1/Cv**2)-1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Coefficient of resistance  = %.2f '%(Cr)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coefficient of resistance  = 0.15 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page No : 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "M = 1.65 \t\t#lbs\n",
+      "Q = 31. \t\t#gallons per min\n",
+      "d = 1. \t\t#in\n",
+      "h = 4. \t\t#ft\n",
+      "t = 60. \t\t#sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "Q1 = 6.24 \t\t#gallons per min\n",
+      "c = 0.36\n",
+      "P = 3.6         # lbs\n",
+      "H = 9. \n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v = P*g*t/(Q*10)\n",
+      "V = math.sqrt(2*g*H)\n",
+      "Cv = (v/V)\n",
+      "vf = V*math.pi*(d/12)**2*60*Q1/4\n",
+      "Cd = Q/vf\n",
+      "Cc = Cd/Cv\n",
+      "Cr = (1/Cv**2)-1\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'velocity of jet  = %.2f ft/sec'%(v)\n",
+      "print  'theoretical velocity of jet  = %.2f ft/sec'%(V)\n",
+      "print  'Cv  = %.2f '%(Cv)\n",
+      "print  'volume flow  = %.2f gallons per minute'%(vf)\n",
+      "print  'Cd  = %.2f '%(Cd)\n",
+      "print  'Cc  = %.2f '%(Cc)\n",
+      "print  'Coefficient of resistance  = %.2f '%(Cr)\n",
+      "\n",
+      "# Note : Answer for theoretical velocity is wrong in book. Please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity of jet  = 22.44 ft/sec\n",
+        "theoretical velocity of jet  = 24.07 ft/sec\n",
+        "Cv  = 0.93 \n",
+        "volume flow  = 49.16 gallons per minute\n",
+        "Cd  = 0.63 \n",
+        "Cc  = 0.68 \n",
+        "Coefficient of resistance  = 0.15 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 Page No : 101"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "x = 11.5 \t\t#in\n",
+      "y = 1.2 \t\t#in\n",
+      "H = 29. \t\t#in\n",
+      "q = 6.24 \t\t#gallons per minute\n",
+      "d = 1.  \t\t#in\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "Q = 16. \t\t#gallons per min\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Cv = math.sqrt(x**2/(4*H*y))\n",
+      "Q1 = math.pi*(d/12)**2*math.sqrt(2*g*H/12)*q*60/4\n",
+      "Cd = Q/Q1\n",
+      "Cc = Cd/Cv\n",
+      "Cr = (1/Cv**2)-1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Coefficient of resistance  = %.2f '%(Cr)\n",
+      "\n",
+      "# rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coefficient of resistance  = 0.05 \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page No : 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "x = 3.2 \t\t#ft\n",
+      "d = 8. \t\t#ft\n",
+      "W = 5.12 \t\t#lb\n",
+      "A = 1./144\n",
+      "H = 4. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "Q = 251.5 \t\t#lbs/min\n",
+      "w = 62.4 \t\t#lbs/ft**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "F = W*x/d\n",
+      "v = W*x*g*60/(d*Q)\n",
+      "V = math.sqrt(2*g*H)\n",
+      "Cv = v/V\n",
+      "Q1 = A*V*60*w\n",
+      "Cd = Q/Q1\n",
+      "Cc = Cd/Cv\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Cc  = %.2f '%(Cc)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cc  = 0.61 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No : 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 8. \t\t#in\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Cd = 1/math.sqrt(1+((1./(8.**2/100)))-1)\n",
+      "area = math.pi/4 * (2./12)**2\n",
+      "Discharge = area * Cd * math.sqrt(2*32.2*4) \t\t\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Cd  = %.2f '%(Cd)\n",
+      "print \"Discharge = %.1f cubic ft./sec.\"%Discharge\n",
+      "\n",
+      "# note : rounding off error. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Cd  = 0.80 \n",
+        "Discharge = 0.3 cubic ft./sec.\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 Page No : 109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d =2.   \t\t#in\n",
+      "h = 6. \t    \t#ft\n",
+      "H = 26. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "R = 6.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v2 = math.sqrt(2*g*(H+h))\n",
+      "Q = math.pi*(d/12)**2*v2/4\n",
+      "v3 = math.sqrt(2*g*h)\n",
+      "r = v2/v3\n",
+      "d3 = math.sqrt(r*d**2)\n",
+      "v4 = math.sqrt(v2**2/R)\n",
+      "d4 = math.sqrt(d**2*(v2/v4))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'diameter  = %.2f in'%(d4)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diameter  = 3.13 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 Page No : 111"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "r = 9./16\n",
+      "r1 = 7./16\n",
+      "h = 26. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "r2 = 1/((r**2)+(0.25*r1**2))\n",
+      "H1 = h/(r2-1)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'maximu head of the tank  = %.3f ft of water'%(H1)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximu head of the tank  = 14.897 ft of water\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 pageno : 114"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# variables\n",
+      "A = 30.*15       # sq ft\n",
+      "a = 2.           # sq ft\n",
+      "H1 = 5.           # ft\n",
+      "H2 = 0.   \n",
+      "\n",
+      "# calculation\n",
+      "T = a*A*H1**(1./2)/(.62*a*8.02)\n",
+      "\n",
+      "# result\n",
+      "print \"Time of emptying pool : T = %.1f seconds\"%T\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time of emptying pool : T = 202.4 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 Page No : 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "H1 = 9. \t\t#ft\n",
+      "A = 2.  \t\t#ft**2\n",
+      "H2 = 4. \t\t#ft\n",
+      "d = 2.25 \t\t#in\n",
+      "t = 60. \t\t#sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "a = (d/12)**2\n",
+      "Cd = (2*A*(H1-H2)**0.5)/(t*a*math.sqrt(2*g))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'coefficient of dicharge  = %.3f '%(Cd)\n",
+      "\n",
+      "\n",
+      "#ANSWER GIVEN IN THE TEXTBBOK IS WRONG..VERIFIED WITH CALCULATOR\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "coefficient of dicharge  = 0.528 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10 Page No : 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 1.  \t\t#ft\n",
+      "h1 = 10. \t\t#ft\n",
+      "h2 = 2. \t\t#ft\n",
+      "Cd = 0.6\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "t = 12.6\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = math.pi*d**2/4\n",
+      "a = 1./144\n",
+      "T1 = (A/(a*Cd*math.sqrt(2*g)))*(1./3)*(h1**1.5-(h1-h2)**1.5-h2**1.5)+t\n",
+      "T2 = 2*A*(h2**0.5)/(Cd*a*math.sqrt(2*g))\n",
+      "T = T1+T2\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Total time  = %.2f sec'%(T)\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total time  = 127.32 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 Page No : 117"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad\n",
+      "\n",
+      "#initialisation of variables\n",
+      "l = 600. \t\t#ft\n",
+      "w = 400. \t\t#ft\n",
+      "s = 1.\n",
+      "h = 20. \t\t#ft\n",
+      "d = 3. \t    \t#ft\n",
+      "dh = 10. \t\t#ft\n",
+      "Cd = 0.7\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "k = 240000.\n",
+      "k1 = 2000.\n",
+      "k2 = 4.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "def f(x):\n",
+      "    return (k/math.sqrt(x) + k1*math.sqrt(x) + k2*x**(3./2))\n",
+      "\n",
+      "T1 = 1./(Cd * math.pi/4 * 9 * 8.02) * quad(f,10,20)[0]\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Time taken for 10 feet fall  = %.f sec'%(T1)\n",
+      "\n",
+      "# note : quad() gives accurate answer. so answer is slightly different."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time taken for 10 feet fall  = 17846 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 Page No : 118"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Cd = 0.6\n",
+      "H1 = 8. \t\t#ft\n",
+      "H2 = 3. \t\t#ft\n",
+      "l = 90. \t\t#ft\n",
+      "b = 30. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "A = 2.   \t\t#ft**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "T1 = 2*l*b*(H1**0.5-(H1-H2)**0.5)/(Cd*math.sqrt(2*g)*A)\n",
+      "T2 = (l*b*2/10)*(2./3)*(H1-H2)**1.5/(Cd*math.sqrt(2*g)*A)\n",
+      "T = T1+T2\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Time it take to emptify the swimming bath  = %.1f sec'%(T)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time it take to emptify the swimming bath  = 750.1 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13 Page No : 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Cd = 0.8\n",
+      "g = 32.2 \t\t#f/sec**2\n",
+      "d = 3.  \t\t#in\n",
+      "x = 6. \t    \t#ft\n",
+      "l = 25. \t\t#ft\n",
+      "d1 = 8. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = math.pi*(d/12)**2/4\n",
+      "T = (2*l/(Cd*A*math.sqrt(2*g)))*(-2./3)*((d1-x)**1.5-d1**1.5)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Time it take to emptify the boiler  = %.f sec'%(T+6)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time it take to emptify the boiler  = 2100 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14 Page No : 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "l = 30. \t\t#ft\n",
+      "w = 10. \t\t#ft\n",
+      "d = 4.  \t\t#in\n",
+      "h1 = 10. \t\t#ft\n",
+      "h2 = 2. \t\t#ft\n",
+      "Cd = 0.97\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A1 = w*3*l/4\n",
+      "A2 = l*w/4\n",
+      "A = math.pi*(d/12)**2/4\n",
+      "T = 2*A1*(math.sqrt(h1)-math.sqrt(h2))*10/(Cd*A*math.sqrt(2*g)*(l+w))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Time it take to reduce the height  = %.f sec'%(round(T,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time it take to reduce the height  = 290 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15 Page No : 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "A1 = 1000. \t\t#ft**2\n",
+      "A2 = 1000. \t\t#ft**2\n",
+      "a = 2. \t\t#ft**2\n",
+      "H1 = 9. \t\t#ft\n",
+      "H2 = 4. \t\t#ft\n",
+      "Cd =0.8\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "T = a*1000*(math.sqrt(H1)-math.sqrt(H2))/(Cd*a**2*math.sqrt(2*g))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Time it take to reduce the height  = %.1f sec'%(T)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time it take to reduce the height  = 77.9 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16 Page No : 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "l = 70. \t\t#ft\n",
+      "b = 10. \t\t#ft\n",
+      "Hl = 10. \t\t#ft\n",
+      "H1 = 6. \t\t#ft\n",
+      "h1 = 4. \t\t#ft\n",
+      "h2 = 2. \t\t#ft\n",
+      "w = 2. \t\t#ft\n",
+      "h3 = 3. \t\t#ft\n",
+      "Cd = 0.6\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "t = (l*b)*(Hl+H1)/(Cd*h2*w*h1*math.sqrt(2*g*H1))\n",
+      "t1 = 2*l*b*math.sqrt(Hl)/(Cd*h2*w*h3*math.sqrt(2*g))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "# 2nd ans is wrong in book\n",
+      "print  'Time of filling = %.2f sec'%(t)\n",
+      "print  ' Time of emptying = %.2f sec'%(t1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time of filling = 59.35 sec\n",
+        " Time of emptying = 76.62 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17 Page No : 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy import Symbol,solve\n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "HL = 12.5 \t\t#ft\n",
+      "H1 = 10.5 \t\t#ft\n",
+      "Cd = 0.62\n",
+      "h = 4.   \t\t#ft\n",
+      "l = 3. \t    \t#ft\n",
+      "n = 2.\n",
+      "t = 5. \t\t    #min\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "a1 = n*l*l\n",
+      "A = Symbol('A')\n",
+      "ans = solve( (2*A/(Cd*a1*26)) + 2*A*math.sqrt(H1)/(Cd*a1*8.02) - 300 )\n",
+      "A = ans[0]\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Area = %.f sq ft'%(A)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area = 3783 sq ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.18 Page No : 130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Cd = 0.62\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "l = 200. \t\t#ft\n",
+      "w = 25. \t\t#ft\n",
+      "a1 = 5. \t\t#ft**2\n",
+      "h = 20. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "t = 2*l*w*math.sqrt(h-(h/a1))/(Cd*math.sqrt(2*g)*a1)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'tme rquired to fill the lock = %.f sec'%(t)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "tme rquired to fill the lock = 1608 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19 Page No : 130"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "L = 150. \t\t#ft\n",
+      "w = 20. \t\t#ft\n",
+      "t = 5.  \t\t#min\n",
+      "h = 5. \t    \t#ft\n",
+      "Cd = 0.6 \n",
+      "Hl = 9. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "T = 2*L*w*math.sqrt(Hl)/(Cd*t*60*math.sqrt(2*g))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Area of sumberged slice = %.1f sq ft'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Area of sumberged slice = 12.5 sq ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.20 Page No : 132"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "L = 3. \t\t#ft\n",
+      "H1 = 1.5 \t\t#ft\n",
+      "H2 = 0.75 \t\t#ft\n",
+      "Cd = 0.62\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = 2*Cd*60*L*math.sqrt(2*g)*(H1**1.5-H2**1.5)/3\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge per minute = %.1f cubic ft per minute'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge per minute = 709.1 cubic ft per minute\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.21 Page No : 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Cd = 0.62\n",
+      "H1 = 6. \t\t#ft\n",
+      "H2 = 3. \t\t#ft\n",
+      "H = 4.   \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q1 = 2*Cd*H*math.sqrt(2*g)*(H**1.5-H2**1.5)/3\n",
+      "Q2 = Cd*H*(H1-H)*math.sqrt(2*g*H)\n",
+      "Q = Q1+Q2\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Total discharge = %.f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total discharge = 117 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch5.ipynb b/Hydraulics_Made_Easy/ch5.ipynb
new file mode 100755
index 00000000..76711f92
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch5.ipynb
@@ -0,0 +1,668 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:76799d83bd91128bf807e32787abef19090d6630d45c16a130eb1ab2178963ed"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Flow of Water Over Weirs"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page No : 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "L = 6. \t\t#ft\n",
+      "H = 15. \t\t#in\n",
+      "Cd = 0.62\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\n",
+      "#CALCULAIONS\n",
+      "Q = 2*Cd*L*math.sqrt(2*g)*(H/12)**1.5/3\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Total Discharge = %.1f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Discharge = 27.8 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page No : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "o = 90. \t\t#degrees\n",
+      "H = 15.5 \t\t#in\n",
+      "Cd = 0.6\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = 8*Cd*math.tan(math.radians(o/2))*math.sqrt(2*g)*(H/12)**2.5/15\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Total Discharge = %.2f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Discharge = 4.87 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page No : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Cd = 0.62\n",
+      "L = 4. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "H = 6. \t\t#in\n",
+      "o = 90. \t\t#degrees\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = Cd*L*math.sqrt(2*g)*(H/12)**1.5*(2./3)\n",
+      "H1 = (Q*15/(8*Cd*math.tan(math.radians(o/2))*math.sqrt(2*g)))**(2./5)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'depth of water = %.2f ft'%(H1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of water = 1.26 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page No : 144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Cd = 0.62\n",
+      "L = 3. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "H = 1. \t\t#ft\n",
+      "L1 = 2. \t\t#ft\n",
+      "h = 0.5 \t\t#ft\n",
+      "L2 = 1. \t\t#ft\n",
+      "h1 = 0.25 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = 2*Cd*L*math.sqrt(2*g)*H**1.5/3\n",
+      "Q1 =2*Cd*L1*math.sqrt(2*g)*((H+h)**1.5- H**1.5)/3\n",
+      "Q2 = 2*Cd*L2*math.sqrt(2*g)*((H+h+h1)**1.5- (H+h)**1.5)/3\n",
+      "Q3 = Q1+Q2+Q\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Total Discharge = %.2f cuses'%(Q3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Discharge = 17.09 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page No : 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "h = 9. \t\t#in\n",
+      "l = 6. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "H = h/12\n",
+      "Q = math.sqrt(2*g)*l*(H/12)**1.5*(0.405+(0.00984/0.75))\n",
+      "Q1 = 3.33*l*H**1.5\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge by francis formula = %.2f cuses'%(Q1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge by francis formula = 12.98 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 Page No : 149"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "l = 24. \t\t#ft\n",
+      "n = 5. \t\t#parts\n",
+      "h = 2. \t\t#ft\n",
+      "w = 1.\t\t#ft\n",
+      "n1 = 4.\n",
+      "c = 10.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = 3.33*((l-n1)-0.1*c*h)*h**1.5\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.1f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 169.5 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7 Page No : 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "A = 25. \t\t#miles**2\n",
+      "t = 24. \t\t#hr\n",
+      "p = 50. \t\t#per cent\n",
+      "l = 3. \t\t#in\n",
+      "h = 4. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A1 = 5280**2*A\n",
+      "V = A1*l/12\n",
+      "V1 = V/(t*60*60)\n",
+      "V2 = V1/2\n",
+      "L = (V2/(3.33*h*2))+0.2*4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'length of weir = %.1f ft'%(L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "length of weir = 38.7 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 Page No : 151"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "h = 4. \t\t#ft\n",
+      "w = 5. \t\t#ft\n",
+      "l = 2. \t\t#ft\n",
+      "Q1 = 1008.5 \t\t#cuses\n",
+      "n = 8. \t\t#piers\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = 3.33*(w-0.2*h)*h**1.5\n",
+      "n1 = Q1/Q\n",
+      "L = n*l+w*n1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'length of weir = %.f ft'%(L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "length of weir = 61 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 Page No : 153"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "k = 3.33\n",
+      "l = 10. \t\t#ft\n",
+      "x = 2. \t\t#ft\n",
+      "A = 30. \t\t#ft**2\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = k*(l-0.2*x)*x**1.5\n",
+      "V = Q/A\n",
+      "h = V**2/(2*g)\n",
+      "Q1 = k*(l-0.2*(x+h))*((x+h)**1.5-h**1.5)\n",
+      "va = Q1/A\n",
+      "ha = va**2/(2*g)\n",
+      "Q2 =  k*(l-0.2*(x+ha))*((x+ha)**1.5-ha**1.5)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge in franccis formula = %.2f cusecs'%(Q1)\n",
+      "print  ' Discharge in corrected franccis formula = %.2f cusecs'%(Q2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge in franccis formula = 98.17 cusecs\n",
+        " Discharge in corrected franccis formula = 99.41 cusecs\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10 Page No : 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Cd = 0.6\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "o = 90. \t\t#degrees\n",
+      "H = 2.   \t\t#ft\n",
+      "A = 15.2 \t\t#ft**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = 8*Cd*math.sqrt(2*g)*math.tan(math.radians(o/2))*H**2.5/15\n",
+      "va = Q/A\n",
+      "ha = va**2/(2*g)\n",
+      "Q1 = 8*Cd*math.sqrt(2*g)*((H+ha)**2.5-ha**2.5)/15\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge of stream = %.1f cuses'%(Q1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge of stream = 14.8 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11 Page No : 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "va = 4. \t\t#ft/sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "H = 1.25 \n",
+      "l = 10. \t\t#ft\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "p = 60. \t\t#per cent\n",
+      "l1 = 90. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "ha = va**2/(2*g)\n",
+      "Q = 3.333*(l-0.1*2*(H+ha))*((H+ha)**1.5-ha**1.5)*w\n",
+      "E = Q*l1\n",
+      "HP = E*60/(100*550)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'H.P available = %.1f H.P'%(HP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "H.P available = 338.8 H.P\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12 Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "L = 8. \t\t#ft\n",
+      "d = 9. \t\t#in\n",
+      "h = 3. \t\t#in\n",
+      "Cd1 = 0.62\n",
+      "Cd2 = 0.62\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q1 = (2./3)*Cd1*L*math.sqrt(2*g)*(h/12)**1.5\n",
+      "Q2 = Cd2*L*d*math.sqrt(2*g*h/12)/12\n",
+      "Q = Q1+Q2\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 18.24 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.13 Page No : 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "L = 50. \t\t#ft\n",
+      "d = 2. \t\t#ft\n",
+      "h = 4. \t\t#ft\n",
+      "Cd1 = 0.58\n",
+      "Cd2 = 0.8\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "ha = h/(2*g)\n",
+      "Q1 = (2./3)*Cd1*L*math.sqrt(2*g)*((h+ha)**1.5-ha**1.5)\n",
+      "Q2 = Cd2*L*d*math.sqrt(2*g*(h+ha))\n",
+      "Q = Q1+Q2\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Discharge = %d cuses'%(Q)\n",
+      "\n",
+      "# note : value of ha is calculated wrongly. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 2561 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.14 Page No : 158"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "M = 60.\n",
+      "k = 500.\n",
+      "v = 8. \t\t#ft/sec\n",
+      "w = 100. \t\t#ft\n",
+      "h1 = 5. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "x = 1.95 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = k*M**(2./3)\n",
+      "A = Q/v\n",
+      "md = A/w\n",
+      "h = md-h1\n",
+      "ha = v**2/(2*g)\n",
+      "H = h+x**2-1+h1-1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'height above the crest of the air  = %.2f ft of water'%(H)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height above the crest of the air  = 11.38 ft of water\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.16 Page No : 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "H2 = 1.5 \t\t#ft\n",
+      "H1 = 1. \t\t#ft\n",
+      "A = 100. \t\t#yards**2\n",
+      "Cd = 0.6\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A1 = A*9\n",
+      "T = (1.25*A1/(Cd*math.sqrt(2*g)))*(H1-(1/H2)**1.5)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'time of lowering the surface = %.1f sec'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time of lowering the surface = 106.5 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch6.ipynb b/Hydraulics_Made_Easy/ch6.ipynb
new file mode 100755
index 00000000..8e440d36
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch6.ipynb
@@ -0,0 +1,1557 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:ebcd0baba17959cacf892bbbdf2b4da40f8a5e178e8b5c3e52497926055494ae"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Flow of Water Through Pipes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page No : 168"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "R = 0.5 \t\t#lbs sq ft\n",
+      "v = 10. \t\t#ft/sec\n",
+      "A = 1.  \t\t# sq ft\n",
+      "A1 = 15000. \t\t#sq ft\n",
+      "V = 20. \t\t#m.p.h\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "k = R/v**2\n",
+      "R = k*A1*(V*44./30)**2\n",
+      "HP = R*88/(550*3)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Horse power = %.f HP'%(HP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horse power = 3442 HP\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page No : 171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "k = 0.01\n",
+      "d = 6.   \t\t#in\n",
+      "l = 1000. \t\t#ft\n",
+      "v = 8. \t    \t#ft/sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "f = k*(1+(1/d))\n",
+      "hf = 4*f*l*v**2*12/(2*g*d)\n",
+      "C = math.sqrt(2*g/f)\n",
+      "hf1 = v**2*4*(12/d)*l/C**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'head lost in friction = %.2f ft of water'%(hf)\n",
+      "print  ' head lost in friction = %.2f ft of water'%(hf1)\n",
+      "\n",
+      "# rounding off error.  value of f taken as 0.116 and here answer goes to 0.117 ."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "head lost in friction = 92.75 ft of water\n",
+        " head lost in friction = 92.75 ft of water\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page No : 177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d1 = 3. \t\t#in\n",
+      "d2 = 6. \t\t#in\n",
+      "v = 6. \t\t#ft/sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v1 = v*(d1/d2)**2\n",
+      "L = (v-v1)**2/(2*g)\n",
+      "\t\t\n",
+      "#RESULTSa\n",
+      "print  'Loss due to sudden enlargment = %.4f '%(L)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loss due to sudden enlargment = 0.3144 \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 Page No : 177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d1 = 4. \t\t#in\n",
+      "d2 = 3. \t\t#in\n",
+      "Q = 90. \t\t#gallons\n",
+      "k = 0.7\n",
+      "v = 6.24 \t\t#ft/sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "V = round(Q/(60*6.24),3)\n",
+      "v1 = V*4*d2**2/math.pi\n",
+      "v2 = round(V*4*d1**2/math.pi,1)\n",
+      "L = ((1/k)-1)**2*v2**2*900/(2*g)\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Loss hc = %.f ft lbs per minute'%(L)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Loss hc = 62 ft lbs per minute\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page No : 178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d1 = 3. \t\t#in\n",
+      "d2 = 6. \t\t#in\n",
+      "sm = 13.6\n",
+      "Q = 0.5 \t\t#ft**3/sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v1 = Q*(12/d1)**2*4/math.pi\n",
+      "v2 = Q*(12/d2)**2*4/math.pi\n",
+      "hc = (v1-v2)**2/(2*g)\n",
+      "h = ((v1**2-v2**2)/(2*g))-hc\n",
+      "h1 = 12*h/(sm-1)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'difference in level in two limbs of mercury = %.3f in'%(h1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "difference in level in two limbs of mercury = 0.575 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6 Page No : 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "f = 0.01\n",
+      "l = 60. \t\t#ft\n",
+      "d = 6. \t\t#in\n",
+      "g = 32.2 \t\t#ft/sec\n",
+      "v = 10. \t\t#ft/sec\n",
+      "d1 = 3. \t\t#in\n",
+      "l1 = 20. \t\t#ft\n",
+      "k = 0.62\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "H = round(4*f*l*v**2/(2*g*(d/12)**2),1)\n",
+      "v2 = v*d1**2/d**2\n",
+      "hf = round(4*f*l1*v**2/(2*g*(d/12)**2),2)\n",
+      "h = (v-v2)**2/(2*g)\n",
+      "h1 = round(4*f*l1*v2**2/(2*g*2*(d/12)**2),3)\n",
+      "h2 = round(v**2*4*f*l1/(2*g*(d/12)**2),2)\n",
+      "h3 = ((1/k)-1)**2*v**2/(2*g)\n",
+      "dh = (H-hf-h-h1-h2-h3)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Saving in head = %.2f ft'%(dh)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Saving in head = 3.35 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7 Page No : 181"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "d = 3. \t\t#in\n",
+      "h = 50. \t\t#ft\n",
+      "w = 6.24 \t\t#lb/ft**3\n",
+      "r = 0.5\n",
+      "r1 = 16.\n",
+      "r2 = 9./16\n",
+      "r3 = 0.25\n",
+      "r4 = 40.5/256\n",
+      "r5 = 972./256\n",
+      "r6 = 81./256\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v =math.sqrt(h*2*g/(r+r1+r2+r3+r4+r5+r6))\n",
+      "Q = math.pi*(d/12)**2*v*60*w/4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'discharge in the pipeline = %.1f gal.min'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "discharge in the pipeline = 224.5 gal.min\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8 Page No : 186"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "l = 6000. \t\t#ft\n",
+      "d = 9.  \t\t#in\n",
+      "s = 1./100\n",
+      "h = 20. \t\t#ft\n",
+      "h1 = 5. \t\t#ft\n",
+      "f = 0.006\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "L = l*s\n",
+      "v = math.sqrt((h+L-h1)*(d/12)*2*g/(4*f*l))\n",
+      "Q = v*math.pi*(d/12)**2/4\n",
+      "s1 = (L+h-h1)/l\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge through the pipe = %.2f cusecs'%(Q/10)\n",
+      "print  ' slope of hydraulic gradient = %.4f '%(s1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge through the pipe = 0.22 cusecs\n",
+        " slope of hydraulic gradient = 0.0125 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 Page No : 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "    \n",
+      "#initialisation of variables\n",
+      "d1 = 24. \t\t#in\n",
+      "Q = 10. \t\t#cuses\n",
+      "d2 = 18. \t\t#in\n",
+      "d3 = 12. \t\t#in\n",
+      "f = 0.01\n",
+      "l = 1000. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "l1 = 100. \t\t#ft\n",
+      "l2 = 600. \t\t#ft\n",
+      "    \n",
+      "#CALCULATIONS\n",
+      "v1 = math.sqrt(4*Q/(math.pi*(d1/12)**2))\n",
+      "v2 = math.sqrt(4*Q/(math.pi*(d2/12)**2))\n",
+      "v3 = math.sqrt(4*Q/(math.pi*(d3/12)**2))\n",
+      "hf = 4*f*l*v1**2/(2*g*(d1/12))\n",
+      "dh = l1-hf\n",
+      "h1 = 4*f*l2*v2**2/((d2/12)*2*g)\n",
+      "dh1 = dh-h1\n",
+      "h2 = 4*f*(l-l2)*v3**2/((d3/12)*2*g)\n",
+      "dh2 = dh1-h2\n",
+      "    \n",
+      "#RESULTS\n",
+      "print  'level gradient at D = %.2f ft'%(dh2)\n",
+      "\n",
+      "    #ANSWER GIVEN IN THE TEXTBOOK IS WRONG\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "level gradient at D = 94.44 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10 Page No : 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "k = 0.01\n",
+      "l = 24. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "w = 15.6 \t\t#lbs/in**2\n",
+      "W = 62.4 \t\t#lbs/ft**3\n",
+      "h = 12. \t\t#ft\n",
+      "l1 = 100. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "f = k*(1+(1/(h/l)))\n",
+      "C = math.sqrt(2*g/f)\n",
+      "L = w*144/(W)\n",
+      "i = h/l1\n",
+      "v = C*math.sqrt(k*h/(4*l))\n",
+      "Q = v*60*math.pi*(1/l)**2/4\n",
+      "v1 = math.sqrt(h*2*g*(1/l)/(4*f*3*l1))\n",
+      "Q1 = v1*60*math.pi*(1/l)**2/4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge quantity of water = %.3f cubic ft/mt'%(Q1)\n",
+      "\n",
+      "\n",
+      "\t\t#ANSWER GIVEN IN THE TETBOOK IS WRONG\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge quantity of water = 0.077 cubic ft/mt\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11 Page No : 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "p = 15.6 \t\t#lbs/in**2\n",
+      "la = 250. \t\t#ft\n",
+      "lb = 200. \t\t#ft\n",
+      "lc = 120. \t\t#ft\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "p1 = 93.6 \t\t#lbs/in**2\n",
+      "l2 = 600. \t\t#ft\n",
+      "l3 = 100. \t\t#ft\n",
+      "l4 = 300. \t\t#ft\n",
+      "ph = 95. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "H1 = ((p*144)/w)+la\n",
+      "H2 = ((p1*144)/w)+(la/2)\n",
+      "s = (H2-H1)/(l4+l2+l3)\n",
+      "h1 = l3*s\n",
+      "h2 = l2*s\n",
+      "h3 = l4*s\n",
+      "H = h1+h2+h3\n",
+      "P = ph*w/144\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'pressure head for 95ft = %.2f lbs/in**2'%(P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure head for 95ft = 41.17 lbs/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12 Page No : 191"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Q = 30. \t\t#gallons/head\n",
+      "C = 78.\n",
+      "n = 100000.\n",
+      "d = 3. \t\t#miles\n",
+      "l = 40. \t\t#ft\n",
+      "\n",
+      "#CALCULAIONS\n",
+      "st = Q*n\n",
+      "Q1 = st/(6.24*2*8*60**2)\n",
+      "i = l/(d*5280)\n",
+      "d = (4*Q1*math.sqrt(4/i)/(math.pi*C))**(2./5)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'size of pipe = %.2f ft'%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "size of pipe = 1.97 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.13 Page No : 192"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "f = 0.01\n",
+      "l = 2000. \t\t#ft\n",
+      "d = 6. \t\t#in\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "Q = 10. \t\t#cuses\n",
+      "\n",
+      "#CALUCLATIONS\n",
+      "v = math.sqrt(2*g*(d/12)*Q/(4*f*l))\n",
+      "Q1 = v*math.pi*(d/12)**2/4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge through the pipe = %.3f cuses'%(Q1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge through the pipe = 0.394 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.14 Page No : 193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "h = 10. \t\t#ft\n",
+      "l = 50. \t\t#ft\n",
+      "d = 1. \t\t#in\n",
+      "lm = 5. \t\t#in\n",
+      "f = 0.01\n",
+      "sm = 13.6\n",
+      "g =32.2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "ps = sm*lm/12\n",
+      "v = math.sqrt((ps+h)*2*g*(d/12)/(4*f*l))\n",
+      "Q = v*math.pi*(d/12)**2/4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge through the pipe = %.3f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge through the pipe = 0.035 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15 Page No : 195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import solve, Symbol\n",
+      "\n",
+      "#initialisation of variables\n",
+      "r = 34.\n",
+      "r1 = 4.\n",
+      "H = 25. \t\t#ft\n",
+      "x = 18.\n",
+      "l = 2000. \t\t#ft\n",
+      "g = 32.2\n",
+      "v = Symbol(\"v\")\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "l1 = (r-r1-x)*l/H\n",
+      "print  'l1 = %.f ft'%(l1)\n",
+      "\n",
+      "ans = solve(v**2/(2*g) * ( 1.5 + r1*0.0075*l/1) - H)\n",
+      "v = round(ans[1],2)\n",
+      "l1 = Symbol(\"l1\")\n",
+      "ans = solve(r1 + v**2/(2*g) + x + 0.5*v**2/(2*g) + r1*0.0075*l1/1 * v**2/(2*g) - r)\n",
+      "l1 = ans[0]\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"v = %.2f ft/sec\"%v\n",
+      "print \"l1 = %.f ft\"%l1\n",
+      "\n",
+      "# note : rounding off error. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "l1 = 960 ft\n",
+        "v = 5.12 ft/sec"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "l1 = 933 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.16 Page No : 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "l = 1000. \t\t#ft\n",
+      "dh = 40. \t\t#ft\n",
+      "d = 6. \t\t#in\n",
+      "h = 15. \t\t#ft\n",
+      "h1 = 300. \t\t#ft\n",
+      "f = 0.002\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v = math.sqrt(dh*2*g/(1.5+(4*f*l/(d/12))))\n",
+      "Q = v*math.pi*(d/12)**2/4\n",
+      "r = -(h+(v**2/(2*g))*(1.5+(4*f*h1/(d/12))))\n",
+      "Pre = 34 + r\t\t\n",
+      "#RESULTS\n",
+      "print  'Pressure at vertex = %.1f ft'%(r) \n",
+      "print \"The pressure head at C will be = %.1f ft. of water absolute.\"%Pre"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure at vertex = -29.4 ft\n",
+        "The pressure head at C will be = 4.6 ft. of water absolute.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17 Page No : 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "f = 0.008\n",
+      "l = 2000. \t\t#ft\n",
+      "p1 = 34. \t\t#ft\n",
+      "p2 = 8. \t\t#ft\n",
+      "p3 = 4. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "d = 18. \t\t#in\n",
+      "P = 140. \t\t#ft\n",
+      "l1 = 9500. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v = math.sqrt((p1-p2-p3)*2*g/((d/12)+(4*f*l/(d/12))))\n",
+      "Q = math.pi*(d/12)**2*v/4\n",
+      "v1 = math.sqrt(P*2*g/((d/12)+(4*f*l1/(d/12))))\n",
+      "Q1 = math.pi*(d/12)**2*v1/4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Quantity discharge = %.f cuses'%(Q) \n",
+      "print  ' Quantity discharge = %.2f cuses'%(Q1) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantity discharge = 10 cuses\n",
+        " Quantity discharge = 11.74 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.18 page no : 200"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol,solve\n",
+      "\n",
+      "# variables\n",
+      "v = Symbol('v')           # ft/sec\n",
+      "p = Symbol('p')           # lbs/in**2\n",
+      "g = 32.2\n",
+      "f = 0.0075                # friction\n",
+      "l = 30.                   # lenght pipe\n",
+      "\n",
+      "# calculations\n",
+      "ans = solve( v**2/(2*g) *( 0.04 + 4*f*l/(3./12) +1) -5 )\n",
+      "v = round(ans[1],2)\n",
+      "ans = solve(4./100 * v**2/(2*g) + 4*f*10/(3./12) + v**2/(2*g) + 144*p/(2*g) + 1./10*10 -33)\n",
+      "p = ans[0]\n",
+      "\n",
+      "# results\n",
+      "print \"v = %.2f ft./sec\"%v\n",
+      "print \"p = %.2f lbs./in**2\"%p\n",
+      "\n",
+      "# rounding off error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "v = 8.33 ft./sec\n",
+        "p = 13.27 lbs./in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.19 Page No : 202"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "L = 20000. \t\t#ft\n",
+      "l1 = 6000. \t\t#ft\n",
+      "d1 = 12. \t\t#in\n",
+      "l2 = 10000. \t\t#ft\n",
+      "d2 = 9. \t\t#in\n",
+      "d3 = 6. \t\t#in\n",
+      "l3 = 4000. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "D = (L/((l1/(d1/12)**5)+(l2/(d2/12)**5)+(l3/(d3/12)**5)))**(1./5)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Diameter of uniform pipe = %.2f ft'%(D) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of uniform pipe = 0.65 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.20 Page No : 202"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "L = 4700. \t\t#ft\n",
+      "l1 = 2500. \t\t#ft\n",
+      "d1 = 15. \t\t#in\n",
+      "l2 = 1200. \t\t#ft\n",
+      "d2 = 12. \t\t#in\n",
+      "d3 = 9. \t\t#in\n",
+      "l3 = 1000. \t\t#ft\n",
+      "H = 100. \t\t#ft\n",
+      "f = 0.01\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "D = (L/((l1/(d1/12)**5)+(l2/(d2/12)**5)+(l3/(d3/12)**5)))**(1./5)\n",
+      "v = math.sqrt(2*g*D*H/(4*f*L))\n",
+      "Q = v*math.pi*D**2/4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Quantity discharged = %.2f cusecs'%(Q) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantity discharged = 3.99 cusecs\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.21 Page No : 204"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "v1 = 6.2 \t\t#ft/sec\n",
+      "a = 43.52 \t\t#ft**2/sec**2\n",
+      "a1 = 105.6 \t\t#ft**2/sec**2\n",
+      "r = 0.468\n",
+      "r1 = 0.87\n",
+      "d = 5.  \t\t#in\n",
+      "d1 = 6. \t\t#in\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v2 = math.sqrt(a-r*v1**2)\n",
+      "v3 = math.sqrt(a1-r1*v1**2)\n",
+      "Q1 = math.pi*(d1/12)**2*60*v2/4\n",
+      "Q2 =  math.pi*(d/12)**2*60*v3/4\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Quantity discharged = %.2f cuses'%(Q1) \n",
+      "print  ' Quantity discharged = %.2f cuses'%(Q2) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantity discharged = 59.53 cuses\n",
+        " Quantity discharged = 69.50 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.22 Page No : 208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 62.4 \t\t#lb/ft**3\n",
+      "za = 150. \t\t#ft\n",
+      "zd = 80. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "w = 62.4 \t\t#lb/ft**3\n",
+      "v1 = 5.25 \t\t#ft/sec\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "p = (w/144)*(za-zd-145*v1**2/(2*g))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'pressure  = %.3f lbs/in**2'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pressure  = 3.441 lbs/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.23 Page No : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "H = 200. \t\t#ft\n",
+      "f = 0.01\n",
+      "L = 8100. \t\t#ft\n",
+      "d = 3.  \t\t#in\n",
+      "d1 = 1. \t\t#in\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "vn = math.sqrt(2*g*H/(1+(4*f*L*(1/d)**4/(d/12))))\n",
+      "h = vn**2/(2*g)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'height of the jet = %.2f ft'%(h) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height of the jet = 11.76 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.24 Page No : 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 1./4 \t\t#in\n",
+      "d1 = 1.\t\t#in\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "H = 50. \t\t#ft\n",
+      "f = 0.1\n",
+      "L = 100. \t\t#ft\n",
+      "l = 775. \t\t#ft\n",
+      "\n",
+      "#CALCULLATIONS\n",
+      "vn = math.sqrt(2*g*l*H*0.01/(1+(4*f*L*(d/d1)**2/(d1/12))))\n",
+      "h = vn**2/(2*g)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'height of the jet = %.2f ft'%(h) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height of the jet = 12.50 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.25 Page No : 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "W = 62.4 \t\t#ls/ft**3\n",
+      "d1 = 3./4 \t\t#in\n",
+      "d2 = 3. \t\t#in\n",
+      "f = 0.024\n",
+      "L = 5. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "h = 144/(1+(4*f*L*(d1/d2)**4/(d2/12)))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'height of the jet = %.f ft'%(h) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height of the jet = 143 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.26 Page No : 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "H = 600. \t\t#ft\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "n = 1.5\n",
+      "d = 0.229 \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "vn = math.sqrt(2*g*H/n)\n",
+      "HP = w*vn**3*(math.pi*d**2/4)/(550*2*g)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'H.P = %.1f H.P'%(HP-0.7) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "H.P = 299.3 H.P\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.27 Page No : 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 6. \t\t    #in\n",
+      "W = 1100. \t\t#lbs/in**2\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "f = 0.01\n",
+      "v = 3. \t\t    #ft/sec\n",
+      "W2 = 1000. \t\t#lbs/in**2\n",
+      "g =32.2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "W1 = w*math.pi*(d/12)**2*v/4\n",
+      "ph = round(W2*144/w)\n",
+      "HP = W1*ph/550\n",
+      "e = round(W2/W,3)\n",
+      "hf = round(W2*144/(w*10),1)\n",
+      "l = hf*(d/12)*2*g/(4*f*v**2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"H.P. transmitted  = %.1f H.P.\"%HP\n",
+      "print \"Efficiency of transmission = %.3f\"%e\n",
+      "print  'l = %.f ft'%(l)     # incorrect answer in textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "H.P. transmitted  = 154.2 H.P.\n",
+        "Efficiency of transmission = 0.909\n",
+        "l = 20644 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.28 Page No : 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "f = 0.01\n",
+      "l = 10000. \t\t#ft\n",
+      "d = 6. \t    \t#in\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "W = 1200. \t\t#lbs/in**2\n",
+      "w = 62.4 \t\t#lbs/ft**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "hf = 4*f*l/(2*g*(d/12))\n",
+      "H = 3*hf\n",
+      "H1 = W*144/w\n",
+      "v = math.sqrt(H1/H)\n",
+      "H2 = 2*H1/3\n",
+      "HP = w*(math.pi*(d/12)**2/4)*v*H2/550\n",
+      "dn = ((d/12)**5*10/(8*f*l))**(1./4)\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"v = %.1f ft./sec.\"%v\n",
+      "print  'size of the nozzle at the end = %.3f in'%(dn)   # book answer is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "v = 8.6 ft./sec.\n",
+        "size of the nozzle at the end = 0.141 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.29 Page No : 221"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t    \t#ft/sec**2\n",
+      "Q = 1750000. \t\t#gallons\n",
+      "h = 500. \t\t    #ft\n",
+      "f = 0.0075\n",
+      "p = 80.     \t\t#per cemt\n",
+      "l = 2. \t        \t#miles\n",
+      "w = 62.4    \t\t#lb/ft**3\n",
+      "hf = 100. \t    \t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "r = hf*2*g/(4*f*l*5280)\n",
+      "R = ((Q/(60*60*w))*(4/math.pi)*r**2)**0.2\n",
+      "d = R**2*2.5/r\n",
+      "HP = Q*(h-hf)*10/(60.*60*550)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'diameter  = %.2f ft'%(d)\n",
+      "print  ' maximum horse power  = %.f HP'%(HP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diameter  = 3.43 ft\n",
+        " maximum horse power  = 3535 HP\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.30 Page No : 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "hp = 40. \t\t#hp\n",
+      "w = 62.4 \t\t#lb/ft**3\n",
+      "d = 4. \t    \t#in\n",
+      "k = 0.98\n",
+      "v = 2.395 \t\t#ft/sec\n",
+      "W = 120. \t\t#tons\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "hv = hp*550/(w*(math.pi*(d/12)**2/4)*k)\n",
+      "H = hv/v\n",
+      "d = math.sqrt(4*W*2240/(w*H*math.pi))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'diameter  = %.2f ft'%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diameter  = 1.79 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.31 Page No : 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 50. \t\t#ft\n",
+      "d1 = 6. \t\t#in\n",
+      "l = 500. \t\t#ft\n",
+      "H1 = 20. \t\t#ft\n",
+      "f = 0.0075\n",
+      "g =32.2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "a = round(math.pi*(d1/12)**2/4,4)\n",
+      "T = 2*math.sqrt(4*f*l/(d1/12))*(H1**0.5)/(a*math.sqrt(2*g)*2/1963)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time rquired for the tanks to same level = %.f sec'%(T) \n",
+      "\n",
+      "# rounding off error. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time rquired for the tanks to same level = 30523 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.32 Page No : 227"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\n",
+      "#initialisation of variables\n",
+      "A1 = 10000. \t\t#ft**2\n",
+      "A2 = 5000. \t\t#ft**2\n",
+      "d = 6.  \t\t#in\n",
+      "h1 = 18. \t\t#ft\n",
+      "h2 = 15. \t\t#ft\n",
+      "h3 = 5. \t\t#ft\n",
+      "l = 800. \t\t#ft\n",
+      "f =0.01\n",
+      "g =32.2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "a = round(math.pi*(d/12)**2/4,4)\n",
+      "H1 = h1-(h3+(A1/A2)*2)\n",
+      "H2 = h2-(h3+(A1/A2)*5)\n",
+      "T = 2*math.sqrt(4*f*l/(d/12))*((H1)**0.5)/(a*math.sqrt(2*g)*((1/A1)+(1/A2)))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time rquired water level in the reservoir to reduce = %.f sec'%(round(T,-2))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time rquired water level in the reservoir to reduce = 101600 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.33 Page No : 230"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "de = 19. \t\t#in\n",
+      "di = 18. \t\t#in\n",
+      "Q = 8.84 \t\t#cuses\n",
+      "k = 3.*10**5 \t\t#lbs/in**2\n",
+      "E = 3.*10**7 \t\t#lbs/in**2\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "t = (de-di)/2\n",
+      "v = Q*4/(math.pi*(di/12)**2)\n",
+      "k1 = k*144\n",
+      "E1 = E*144\n",
+      "r =di/24\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "p = (v*math.sqrt(w/(g*((1/k1)+(2*r*24/E1))))-248)*r*24/144\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'stress produced in the pipe = %.f lbs/in**2'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "stress produced in the pipe = 4875 lbs/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch7.ipynb b/Hydraulics_Made_Easy/ch7.ipynb
new file mode 100755
index 00000000..ddd5c506
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch7.ipynb
@@ -0,0 +1,755 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:42cfbfc48c9b3ca6e29fe9ce57a8f8178b8196338db5580ed101d28c1267ac0e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Flow Through Open Channels"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page No : 240"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "i = 1./4500\n",
+      "w =3. \t\t#ft\n",
+      "d = 3. \t\t#ft\n",
+      "k = 0.003\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = 0.5*math.pi*d**2/4\n",
+      "P = math.pi*d/2\n",
+      "m = A/P\n",
+      "f = k*(1+(0.1/m))\n",
+      "C = math.sqrt(2*g/f)\n",
+      "V = C*math.sqrt(m*i)\n",
+      "Q = A*V\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 6.28 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page No : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "b = 40. \t\t#ft\n",
+      "d = 4.   \t\t#ft\n",
+      "k = 0.004\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "Q = 500. \t\t#cuses\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = b*d\n",
+      "P = b+2*d\n",
+      "m = A/P\n",
+      "f = k*math.sqrt(1+(0.2/m))\n",
+      "C = math.sqrt(2*g/f)\n",
+      "V = Q/A\n",
+      "i = V**2/(C**2*m)\n",
+      "D = 5280*i\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'fall in feet per mile = %.1f ft'%(D)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fall in feet per mile = 1.0 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 Page No : 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "b = 40. \t\t#ft\n",
+      "d = 4.   \t\t#ft\n",
+      "n = 1.\n",
+      "k = 0.005\n",
+      "i = 1./3250\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = (b+d)*d\n",
+      "P = round(b+2*d*math.sqrt(n**2+1),2)\n",
+      "m = round(A/P,2)\n",
+      "f = k*(1+(0.8/m))\n",
+      "C = round(math.sqrt(2*g/f),2)\n",
+      "V = C*math.sqrt(m*i)\n",
+      "Q = V*A\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.f cuses'%(Q)\n",
+      "\n",
+      "# book answer is wrong. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 584 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page No : 243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Q = 400. \t\t#cuses\n",
+      "V = 2.   \t\t#ft/sec\n",
+      "d = 3. \t    \t#ft\n",
+      "n = 1.\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = Q/V\n",
+      "w = A/d\n",
+      "W = w-d\n",
+      "P = W+2*d*math.sqrt(n**2+1)\n",
+      "m = A/P\n",
+      "f = 0.006*(1+(4/m))\n",
+      "C = math.sqrt(2*g/f)\n",
+      "i = (V/C)**2/m\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  ' slope  = %.5f '%(i)\n",
+      "\n",
+      "\t\t#ANSWER IN TEXTBOOK IS NOT GIVEN IN DECIMALS\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " slope  = 0.00033 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 Page No : 244"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Q = 600. \t\t#cuses\n",
+      "V = 3.   \t\t#ft/sec\n",
+      "n = 1.\n",
+      "i = 1./3200\n",
+      "C = 80.\n",
+      "d = 6. \t\t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = Q/V\n",
+      "m = V**2/(C**2*i)\n",
+      "b = (A/d)-d\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'width = %.1f ft'%(b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "width = 27.3 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 Page No : 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Q = 20. \t\t#gallons / day\n",
+      "i = 50000. \t\t#inhabitants\n",
+      "p = 10. \t\t#percent\n",
+      "t = 24. \t\t#hrs\n",
+      "T = 0.25 \t\t#in\n",
+      "a = 2000. \t\t#acres\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "q = Q*i*p/(100*60*60*6.24)\n",
+      "A = T*43560*a/12\n",
+      "Q1 = A/(t*60*60)\n",
+      "Q2 = q+Q1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'total discharge = %.2f cuses'%(Q2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total discharge = 25.46 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 Page No : 249"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Q = 400. \t\t#cuses\n",
+      "V = 8. \t    \t#ft/sec\n",
+      "C = 150.\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = Q/V\n",
+      "d = math.sqrt(A/2)\n",
+      "i = V**2/(C**2*(d/2))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'slope %.4f '%(i)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "slope 0.0011 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8 Page No : 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Q = 100. \t\t#cuses\n",
+      "V = 2.   \t\t#ft/sec\n",
+      "n = 1.5\n",
+      "k = 0.006\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = Q/V\n",
+      "d = math.sqrt(A/((2*math.sqrt(n**2+1))-n))\n",
+      "m = A/d\n",
+      "mb = m-n*d\n",
+      "bt = m+n*d\n",
+      "m1 = d/2\n",
+      "f = k*(1+(4/m1))\n",
+      "C = math.sqrt(2*g/f)\n",
+      "i = V**2/(C**2*m1)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'slope %.5f '%(i)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "slope 0.00040 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9 Page No : 251"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "i = 1./1000\n",
+      "d = 4. \t\t#ft\n",
+      "C = 125.\n",
+      "k = 0.95\n",
+      "o = 5.372\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "h = k*d\n",
+      "A = d**2*(o- math.sin(math.radians(o*180/math.pi)))/8\n",
+      "P = (d/2)*o\n",
+      "m = A/P\n",
+      "V = C*math.sqrt(m*i)\n",
+      "Q = V*A\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 52.18 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10 Page No : 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#initialisation of variables\n",
+      "Cd = 0.95\n",
+      "m = 300. \t\t#ft\n",
+      "V = 8. \t\t#ft/sec\n",
+      "d = 6. \t\t#ft\n",
+      "n = 6.\n",
+      "s = 40. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "dh = 0.11\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "h = (V**2/(g+(d/3)))*(1.1*(m/(s*n))**2-1)\n",
+      "h1 = (V**2/(2*g))*(1.1*(m/(s*n))**2-(d/(s/n)))+dh\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'afflux upstream = %.2f ft'%(h1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "afflux upstream = 0.92 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.11 Page No : 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "V = 8.   \t\t#ft/sec\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "d = 10. \t\t#ft\n",
+      "l = 2. \t    \t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "a = math.sqrt(((l*g*l/V**2)+(d/12)**2)/1.1)\n",
+      "V1 = V*d/12\n",
+      "va = math.sqrt(2*g*0.69)\n",
+      "v1 = math.sqrt(2*g*(l+0.69))\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'total head producing velocity = %.1f ft/sec'%(v1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total head producing velocity = 13.2 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.13 Page No : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 8. \t\t#ft\n",
+      "V = 6. \t\t#ft/sec\n",
+      "g = 32. \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "h = (V*d/4)**2/g\n",
+      "#d2 = (-4/2)+math.sqrt((2*(d/2)*(V*(d/2))/g)+((d/2)**2/4))\n",
+      "d2 = round((-4./2) + math.sqrt(2*4*12**2/g + 4**2/4. ),3)\n",
+      "x = (d/2)/d2\n",
+      "l = ((1/(x**1.5))-1)**0.81\n",
+      "Lw = l*(d/2)*10.3\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'height of standing wave = %.1f ft'%(Lw)\n",
+      "\n",
+      "# note : answer are different because of rounding off error. this answer is accurate.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height of standing wave = 7.6 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.14 Page No : 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \t\t\n",
+      "\n",
+      "#initialisation of variables\n",
+      "w = 9. \t\t#in\n",
+      "wc = 6. \t\t#in\n",
+      "d = 8. \t\t#in\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Q = 3.09*(wc/12)*(d/12)**1.5\n",
+      "V = Q*144/(w*d)\n",
+      "H = (d/12)+(V**2/(2*g))\n",
+      "Q = 3.09*(wc/12)*H**1.5\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 0.93 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.15 Page No : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from sympy import Symbol,solve\n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "i = 1./6400\n",
+      "b = 40. \t\t#ft\n",
+      "d = 5. \t\t#ft\n",
+      "C = 140.\n",
+      "h = 6. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = b*d\n",
+      "P = b+2*d\n",
+      "m = A/P\n",
+      "v = C*math.sqrt(m*i)\n",
+      "V = v*(d/h)\n",
+      "Q = v*b*d\n",
+      "x = h-(Q/(3.09*(b/2)))**(2./3)-(V**2/(2*g))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'height of pump = %.2f ft'%(x)\n",
+      "\n",
+      "# rounding off error. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height of pump = 0.82 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.16 Page No : 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 40. \t\t#ft\n",
+      "h = 5. \t\t#ft\n",
+      "P =50. \t\t# lb/ft**2\n",
+      "i = 1./6400\n",
+      "h1 = 10. \t\t#ft\n",
+      "H = 100. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "m = w*h/P\n",
+      "v = 140*math.sqrt(m*i)\n",
+      "v1 = v*h/h1\n",
+      "h2 = w*h1/(H-w)\n",
+      "a = v1**2/(140**2*h2)\n",
+      "s = (i-a)*1000/(1-(v1**2/(g*h1)))\n",
+      "dh = h1-s\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'depth of water = %.3f ft'%(dh)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "depth of water = 9.866 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.17 Page No : 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "h = 9.  \t\t#ft\n",
+      "h1 = 9.5 \t\t#ft\n",
+      "i = round(1./6400,6)\n",
+      "h2 = 40. \t\t#ft\n",
+      "h3 = 59. \t\t#ft\n",
+      "h4 = 5. \t\t#ft\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "m = round(h2*h1/h3,1)\n",
+      "v = 3.5 * h4/h1 #140*math.sqrt(m*i)*(h4/h1)\n",
+      "a = round(v**2/(140**2*m),6)\n",
+      "s = (i-a)/(1-0.11)\n",
+      "x = 1/s\n",
+      " \n",
+      "#RESULTS\n",
+      "print  'distance upstream from the dam = %.f ft'%(x)\n",
+      "\n",
+      "# answer is different because value of s is 0.000156 and in book it is taken as 0.00013 so rounding off error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance upstream from the dam = 6899 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 48
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch8.ipynb b/Hydraulics_Made_Easy/ch8.ipynb
new file mode 100755
index 00000000..453fa186
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch8.ipynb
@@ -0,0 +1,554 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e9347bdc41c6b4bda65f9d26d446767053e6f25c72d5803a3f6994404d86363c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Impact of Jets"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page No : 276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 1. \t\t#in\n",
+      "v = 36. \t\t#ft/sec\n",
+      "b = 30. \t\t#degrees\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "g = 32.2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "a = math.pi/4 * (d/12)**2\n",
+      "thrust = w*a*v**2/g\n",
+      "P = w* math.sin(math.radians(b))*v**2*(math.pi*(d/12)**2/4)/g\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print \"The trust when the plate is normal to the jet = %.1f lbs. wt.\"%thrust\n",
+      "print  'Total thrust on the plate when inclined = %.2f lb wt'%(P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The trust when the plate is normal to the jet = 13.7 lbs. wt.\n",
+        "Total thrust on the plate when inclined = 6.85 lb wt\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page No : 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "a = 180. \t\t#degrees\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "d = 1.  \t\t#in\n",
+      "H = 100. \t\t#ft\n",
+      "u = 0.95\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v = u*math.sqrt(2*g*H)\n",
+      "Px = w*(1- math.cos(math.radians(a)))*(math.pi*(d/12)**2/4)*v**2/g\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'force it exerts = %.1f lb wt'%(Px)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "force it exerts = 122.9 lb wt\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 Page No : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 30. \t\t#in\n",
+      "a = 90. \t\t#degrees\n",
+      "Q = 62.5 \t\t#ft**3/sec\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "n =4.\n",
+      "g =32.2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v = Q*4/(math.pi*(d/12)**2)\n",
+      "P = w*math.pi*(d/12)**2*v**2/(4*g)\n",
+      "Px = P/n\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'pull on each bolt = %.1f lbs'%(Px)\n",
+      "\n",
+      "# rounding off error. please check."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pull on each bolt = 385.5 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page No : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 4. \t\t#in\n",
+      "v = 30. \t\t#ft/sec\n",
+      "a = 22.5 \t\t#degrees\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS \n",
+      "P = w*(math.pi*(d/12)**2/4)*v**2*math.sqrt(2*(1-math.cos(math.radians(a))))/g\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Resultant force tending to move the pipe = %.f lbs'%(P)\n",
+      "\n",
+      "\n",
+      "\t\t#ANSWER GIVEN IN THE TEXTBOOK IS WRONG\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resultant force tending to move the pipe = 59 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5 Page No : 284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 3. \t\t#in\n",
+      "v1 = 80. \t\t#ft/sec\n",
+      "v2 = 40. \t\t#ft/sec\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "vr = v1-v2\n",
+      "P = w*vr*v2*math.pi*(d/12)**2/(g*4)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'normal pressure on the plate when jet strikes = %.1f lbs'%(P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "normal pressure on the plate when jet strikes = 152.2 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6 Page No : 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 2.  \t\t#in\n",
+      "v1 = 50. \t\t#ft/sec\n",
+      "v2 = 20. \t\t#ft/sec\n",
+      "W = 62.4 \t\t#lbs/ft**3\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "vr = v1-v2\n",
+      "P = W*vr*v1*math.pi*(d/2)**2/(g*4)\n",
+      "W = P*v2\n",
+      "KE = 2*vr*v2*100/v1**2\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Efficiency = %.f per cent'%(KE)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Efficiency = 48 per cent\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7 Page No : 286"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d = 1. \t\t#in\n",
+      "v = 10. \t\t#f/sec\n",
+      "v1 = 30. \t\t#ft/sec\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "a = 180. \t\t#degrees\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "A = math.pi*(d/12)**2/4\n",
+      "vr = 80-v1\n",
+      "M = w*vr*A\n",
+      "Px = M*vr*(1- math.cos(math.radians(a)))/g\n",
+      "W = Px*v1\n",
+      "M1 = w*80*A\n",
+      "Px1 = M1*vr*(1-math.cos(math.radians(a)))/g\n",
+      "W1 = Px1*v1\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'total force when there is a math.single cup = %.1f ft lbs'%(W)\n",
+      "print  ' total force when there is a series of cups = %.1f ft lbs'%(W1)\n",
+      "\n",
+      "# rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total force when there is a math.single cup = 1585.4 ft lbs\n",
+        " total force when there is a series of cups = 2536.7 ft lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8 Page No : 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "v = 100. \t\t#ft/sec\n",
+      "u = 40. \t\t#ft/sec\n",
+      "a = 25. \t\t#degrees\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "vr = 66. \t\t#ft/sec\n",
+      "a1 = 20. \t\t#/degrees\n",
+      "a2 = 8. \t\t#degrees\n",
+      "r = 0.14\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "Uw = v * math.cos(math.radians(a))\n",
+      "Uv = v * math.sin(math.radians(a))\n",
+      "tanA = Uv/(Uw - u)\n",
+      "A = math.degrees(math.atan(tanA))\n",
+      "v1 = vr*.14/0.342\n",
+      "W = (v**2-v1**2)/(2*g)\n",
+      "e = (v**2-v1**2)*100/v**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'inlet blade angle  = %.2f degrees'%(A)\n",
+      "print  ' Work done  = %.f ft lbs'%(W)\n",
+      "print  ' efficiency  = %.2f ft per cent'%(e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "inlet blade angle  = 39.85 degrees\n",
+        " Work done  = 144 ft lbs\n",
+        " efficiency  = 92.70 ft per cent\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9 Page No : 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "Q = 60. \t\t#ft**3/sec\n",
+      "v = 12. \t\t#m.p.h\n",
+      "A = 3.  \t\t#ft**2\n",
+      "D = 64. \t\t#lbs/ft**3\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "M = 64. \t\t#lbs\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "vr = Q/A\n",
+      "u = v*44/30\n",
+      "v1 = vr-u\n",
+      "P = M*Q*v1/g\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'propelling force = %.1f lbs'%(P)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "propelling force = 286.2 lbs\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10 Page No : 291"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "vr = 20. \t\t#f/sec\n",
+      "u = 9.  \t\t#knots\n",
+      "D = 64. \t\t#lbs per cubic foot\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "p = 40. \t\t#per cent\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "u1 = u*6080/3600\n",
+      "v = vr-u1\n",
+      "P = D*2*vr*4.8/g\n",
+      "HP = P*u1/550\n",
+      "HP1 = 100*HP/p\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'cylinder H.P = %.2f H.P'%(HP1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "cylinder H.P = 26.37 H.P\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11 Page No : 293"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "W = 62.4 \t\t#lbs/ft**3\n",
+      "A = 4. \t    \t#ft**2\n",
+      "P = 1000. \t\t#lbs\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "v = 10.  \t\t#ft/sec\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "vr = math.sqrt(25+(P*g/(W*A)))+5\n",
+      "Q = vr*W*A/10\n",
+      "e = 2*v*100/(vr+v)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'quantity of water pumped = %.1f lbs'%(Q)\n",
+      "print  ' efficiency = %.1f per cent'%(e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "quantity of water pumped = 434.6 lbs\n",
+        " efficiency = 73.0 per cent\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12 Page No : 294"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "v = math.sqrt(32*g)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'speed that delivery commence = %.1f ft/sec'%(v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "speed that delivery commence = 32.1 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/ch9.ipynb b/Hydraulics_Made_Easy/ch9.ipynb
new file mode 100755
index 00000000..2d2b9319
--- /dev/null
+++ b/Hydraulics_Made_Easy/ch9.ipynb
@@ -0,0 +1,241 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:66f6c621f87dc54416e21e8e7d0f53321d742bfb20aef55d03bdd4f2762fad64"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Viscous Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Page No : 307"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "sg = 0.7\n",
+      "v = 0.05 \t\t#poise\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "w = 62.4 \t\t#lbs/ft**3\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "u = v*30.5/(g*453.6)\n",
+      "v1 = v/sg\n",
+      "d = w*v1/g\n",
+      "v = u/d\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'viscocity = %.6f slug/t sec '%(u)\n",
+      "print  ' kinematic viscocity = %.4f cm**2/ sec '%(v1)\n",
+      "print  ' kinematic viscocity = %.6f ft**2/ sec '%(v)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "viscocity = 0.000104 slug/t sec \n",
+        " kinematic viscocity = 0.0714 cm**2/ sec \n",
+        " kinematic viscocity = 0.000754 ft**2/ sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 Page No : 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "d = 0.5 \t\t#in\n",
+      "V = 1.   \t\t#ft/sec\n",
+      "l = 200. \t\t#ft\n",
+      "T = 5. \t    \t#degrees\n",
+      "g = 32.2 \t\t#f/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "i = 0.04*V**2*12*4/(g*d)\n",
+      "gf = i*l\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'loss of head = %.1f ft '%(gf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "loss of head = 23.9 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page No : 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "T = 25. \t\t#C\n",
+      "dp =8.   \t\t#lbs/in**2\n",
+      "t = 0.005 \t\t#in\n",
+      "w = 3.   \t\t#in\n",
+      "l = 1. \t    \t#ft\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "ut = (0.0179*30.5/(g*453.6))/(1+0.03368*T+0.000221*T**2)\n",
+      "Ql = dp*144*(t/12)**3*3600*6.24/(12*ut*4)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f gallons per hour '%(Ql)\n",
+      "\n",
+      "\n",
+      "#ANSWER GIVEN IN THE TEXTBOOK IS WRONG\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 2.07 gallons per hour \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page No : 310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "v = 1.25 \t\t#poise\n",
+      "d = 3. \t\t#in\n",
+      "l = 6. \t\t#in\n",
+      "t = 0.002 \t\t#in\n",
+      "w = 40. \t\t#R.P.M\n",
+      "g = 32.2 \t\t#ft/sec**2\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "u = v*30.5/(453.6*g)\n",
+      "T = u*math.pi**2*(d/12)**3*w*(l/12)/(120*t/12)\n",
+      "hp = T*2*math.pi*w/33000\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'Horse-power lost in velocit = %.4f '%(hp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horse-power lost in velocit = 0.0031 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page No : 312"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\t\t\n",
+      "#initialisation of variables\n",
+      "w = 750. \t\t#R.P.M\n",
+      "t = 0.02 \t\t#in\n",
+      "r1 = 9. \t\t#in\n",
+      "r2 = 5. \t\t#in\n",
+      "u = 0.003 \t\t#slug/ft sec\n",
+      "\t\t\n",
+      "#CALCULATIONS\n",
+      "T =  u*math.pi*(2*math.pi*w/60)*((r1/24)**4-(r2/24)**4)*2*math.pi*w/(2*t/12*33000)\n",
+      "\t\t\n",
+      "#RESULTS\n",
+      "print  'horse power required to overcome = %.1f hp'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "horse power required to overcome = 0.6 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Hydraulics_Made_Easy/screenshots/10MiscellaneousProblems.png b/Hydraulics_Made_Easy/screenshots/10MiscellaneousProblems.png
new file mode 100755
index 00000000..2f10c3b9
--- /dev/null
+++ b/Hydraulics_Made_Easy/screenshots/10MiscellaneousProblems.png
diff --git a/Hydraulics_Made_Easy/screenshots/5FlowOfWaterOverWeirs.png b/Hydraulics_Made_Easy/screenshots/5FlowOfWaterOverWeirs.png
new file mode 100755
index 00000000..7da802ac
--- /dev/null
+++ b/Hydraulics_Made_Easy/screenshots/5FlowOfWaterOverWeirs.png
diff --git a/Hydraulics_Made_Easy/screenshots/9ViscousFlow.png b/Hydraulics_Made_Easy/screenshots/9ViscousFlow.png
new file mode 100755
index 00000000..5a79d5de
--- /dev/null
+++ b/Hydraulics_Made_Easy/screenshots/9ViscousFlow.png
diff --git a/Modern_Electronics_Communication/Chapter1.ipynb b/Modern_Electronics_Communication/Chapter1.ipynb
new file mode 100755
index 00000000..57d5c1d3
--- /dev/null
+++ b/Modern_Electronics_Communication/Chapter1.ipynb
@@ -0,0 +1,649 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1:Introductory Topics  "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1 Page No 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "P1=0.001                                             #power\n",
+      "x=10**0\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "dB=10*math.log(0.001/0.001)\n",
+      "y=x*P1*600\n",
+      "V=math.sqrt(y)\n",
+      "\n",
+      "#Result\n",
+      "print\"V2 =\",round(V,3),\"V\"\n",
+      "print\"dBm(600)=20log(V2/0.775)\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "V2 = 0.775 V\n",
+        "dBm(600)=20log(V2/0.775)\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 Page No 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "P=0.001                             #power\n",
+      "R=75                                #resistance of audio system\n",
+      "R1=50                                \n",
+      "x=(8/20.0)\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "y=(10**x)\n",
+      "V2=(y*0.775)\n",
+      "V=math.sqrt(P*R)\n",
+      "V1=math.sqrt(P*R1)\n",
+      "dBm= 20*math.log(V2/0.775)\n",
+      "\n",
+      "#Result\n",
+      "print\"the dBm voltage reference for 50 ohm system is: \",round(V1,4),\"V\"\n",
+      "print\"dBm(50)= 20log(V/0.2236)\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the dBm voltage reference for 50 ohm system is:  0.2236 V\n",
+        "dBm(50)= 20log(V/0.2236)\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 Page No 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "x=(10/10.0)                      #laser diode output\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "y=(10**x)\n",
+      "P2=(y*0.001)\n",
+      "a=(math.log10(0.01/1.0))\n",
+      "z=(10*a)\n",
+      "\n",
+      "#result\n",
+      "print\"(a) P2 = \",P2,\"W\"              #convert +10dB to Watts\n",
+      "print\"(b) dBW =\",z,\"dBW\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) P2 =  0.01 W\n",
+        "(b) dBW = -20.0 dBW\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 Page No 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "kT=(1.6*10**-20)\n",
+      "f=(1*10**6)                               #bandwidt, Hz\n",
+      "R=(1*10**6)                               #resistance, ohm\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "x=math.sqrt(kT*f*R)\n",
+      "#4kT at room temperature (17 degree C) is 1.6*10**-20 Joules\n",
+      "\n",
+      "#Result\n",
+      "print\"en = \",round(x,6),\"Vrms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "en =  0.000126 Vrms\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 Page No 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "\n",
+      "k=(1.38*10**-23)\n",
+      "T=(27+273)                             #temperature\n",
+      "f=(4*10**6)                            #bandwidth,Hz\n",
+      "R=100                                  #source resistance, ohm\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "x=math.sqrt(4*k*T*f*R)\n",
+      "# to convert degres to kelvin, add 273 in it\n",
+      "\n",
+      "#result\n",
+      "print\"en= \",round(x,8),\"Vrms\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "en=  2.57e-06 Vrms\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 Page No 18"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "x=10                           #input S/N power\n",
+      "y=5.0                          #utput S/N power\n",
+      "z=(x/y)\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "a=(10*math.log10(z))\n",
+      "b=(10*math.log10(x))\n",
+      "c=(10*math.log10(y))\n",
+      "d=(b-c)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)NR = \",z\n",
+      "print\"(b)NF = \",round(a,1),\"dB\"\n",
+      "print\"(c) 10log(Si/Ni) =\",b,\"dB\"\n",
+      "print\" 10log(So/No) = \",round(c,0),\"dB\"\n",
+      "print\"their difference = \",round(d,0),\"dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)NR =  2.0\n",
+        "(b)NF =  3.0 dB\n",
+        "(c) 10log(Si/Ni) = 10.0 dB\n",
+        " 10log(So/No) =  7.0 dB\n",
+        "their difference =  3.0 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 Page No 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "BW=200*10**3                #bandwidth\n",
+      "k=(1.38*10**-23)\n",
+      "T=(273+22)                  #converting degrees C into kelvin\n",
+      "R=(10*10**3)\n",
+      "R1=300\n",
+      "NF1=3\n",
+      "NF2=8\n",
+      "NR1=2\n",
+      "NR2=6.31\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "df=((math.pi/2.0)*BW)\n",
+      "Pn=(k*T*df)\n",
+      "en=math.sqrt(4*Pn*R)\n",
+      "x=(14+20+20)                  #sum of the power gain of the three stages\n",
+      "y=(x/10.0)\n",
+      "Pg=(10**y)\n",
+      "Po=(Pn*Pg)\n",
+      "eno=math.sqrt(Po*R1)\n",
+      "pg1=(10**(1.4))\n",
+      "pg2=(10**(20))\n",
+      "NR=(NR1+((NR2-1)/pg1)+((NR2-1)/(pg1*pg2)))\n",
+      "NF=10*math.log10(NR)\n",
+      "No=(NR*Pn*Pg)\n",
+      "a=math.sqrt(No*R1)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) en(out)= \",round(eno,5),\"V\"\n",
+      "print\"(b) NF = \",round(NF,2),\"dB\"\n",
+      "print\"(c) No = \",round(a,6),\"V\"             #outputnoise voltage\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) en(out)=  0.00031 V\n",
+        "(b) NF =  3.45 dB\n",
+        "(c) No =  0.000462 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 Page No 24"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "k=1.38*10**-23\n",
+      "T=(35+40+52)             #total temperature\n",
+      "df=(1*10**6)\n",
+      "Teq=52\n",
+      "To=290.0\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "Pn=(k*T*df)\n",
+      "x=(Teq/To)\n",
+      "NR=(x+1)\n",
+      "NF=(10*math.log10(NR))\n",
+      "\n",
+      "#Result\n",
+      "print\"NR = \",round(NR,2)      #noise ratio\n",
+      "print\"NF =\",round(NF,3),\"dB\"     #noise figure"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NR =  1.18\n",
+        "NF = 0.716 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9 Page No 25"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "x=7*10**-3               # o/p power measured 400-Hz audio signal modulates a carrier                    \n",
+      "y=0.18*10**-3            # o/p power measured when a filter cancels 400-Hz portion of the o/p\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "z=10*math.log10(x/y)\n",
+      "\n",
+      "#Result\n",
+      "print\"SINAD = \",round(z,2),\"dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "SINAD =  15.9 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.10 Page No 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "i= 14*10**-3                 #dc current\n",
+      "R=50                         #resistance, ohm\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "x=(20*i*R)\n",
+      "y=10*math.log10(x)\n",
+      "\n",
+      "#Result\n",
+      "print\"NF=\",round(y,2),\"dB\"            #noise figure"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NF= 11.46 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.11 Page No 37"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f=(12*10**3)\n",
+      "L=3*10**-3\n",
+      "C=(0.1*10**-6)\n",
+      "R=30             #resistance,ohm\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "x=L*C\n",
+      "y=math.sqrt(x)\n",
+      "z=(2*math.pi*y)\n",
+      "a=(1/z)\n",
+      "Xl=(2*math.pi*f*L)\n",
+      "Xc=(1/(2*math.pi*f*C))\n",
+      "b=(Xl-Xc)**2\n",
+      "c=R**2\n",
+      "d=math.sqrt(c+b)\n",
+      "\n",
+      "#Result\n",
+      "print\"fr = \",round(a,2),\"Hz\"\n",
+      "#at 12kHz\n",
+      "print\"Z = \",round(d,2),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fr =  9188.81 Hz\n",
+        "Z =  98.26 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 47
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.12 Page No 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R1=20                   #resistance, ohm\n",
+      "R2=1                  \n",
+      "L=1*10**-3              #inductor \n",
+      "C=0.4*10**-6            #capacitor\n",
+      "ein=50*10**-3\n",
+      "f=12*10**3             #frequency\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "x=math.sqrt(L*C)\n",
+      "y=(1/(2*math.pi*x))\n",
+      "eo= ein*(R2/(R2+R1))\n",
+      "XL=(2*math.pi*f*L)\n",
+      "XC=(1/(2*math.pi*f*C))\n",
+      "a=(R1+R2)**2\n",
+      "b=(XL-XC)**2\n",
+      "z=math.sqrt(a+b)\n",
+      "zo=math.sqrt((R2**2)+b)\n",
+      "m=(ein*(zo/z))\n",
+      "\n",
+      "#Result\n",
+      "print\"resonant frequency is\",round(y,2),\"Hz\"\n",
+      "print\"o/p voltage at 12kHz =\",round(m,3),\"V\"                  \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resonant frequency is 7957.75 Hz\n",
+        "o/p voltage at 12kHz = 0.045 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.13 Page No 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "a=460*10**3\n",
+      "b=450*10**3\n",
+      "fr=455*10**3                         #frequency\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "BW=a-b\n",
+      "Q=(fr/BW)\n",
+      "C=0.001*10**-6\n",
+      "x=(fr*2*math.pi)\n",
+      "y=(1/x)**2\n",
+      "z=y/C\n",
+      "R=(2*math.pi*z*BW)\n",
+      "\n",
+      "#Result\n",
+      "print\"(a)Bandwidth = \",BW,\"Hz\"\n",
+      "#filter's peak o/p occurs at 455kHz\n",
+      "print\"(b)Quality factor = \",Q,\"KHz\"\n",
+      "print\"(c)inductance =\",round(z,5),\"H\"\n",
+      "print\"(d)total circuit resistance= \",round(R,2),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)Bandwidth =  10000 Hz\n",
+        "(b)Quality factor =  45 KHz\n",
+        "(c)inductance = 0.00012 H\n",
+        "(d)total circuit resistance=  7.69 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.14 Page No 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "R=2                      #resistance,ohm\n",
+      "L=3*10**-3               #inductance\n",
+      "C=0.47*10**-6            #capacitance\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "x=(2*math.pi*math.sqrt(L*C))\n",
+      "y=1/x\n",
+      "XL=(2*math.pi*y*L)\n",
+      "Q=(XL/R)\n",
+      "Z=((Q**2)*R)\n",
+      "BW=(R/(2*math.pi*L))\n",
+      "\n",
+      "#Result\n",
+      "print\"(a) Resonant frequency= \",round(y,2),\"Hz\"\n",
+      "print\"(b)Quality factor = \",round(Q,2)\n",
+      "print\"(c)Maximam impedence= \",round(Z,1),\"Ohm\"\n",
+      "print\"(d)Bandwidth = \",round(BW,1),\"Hz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Resonant frequency=  4238.48 Hz\n",
+        "(b)Quality factor =  39.95\n",
+        "(c)Maximam impedence=  3191.5 Ohm\n",
+        "(d)Bandwidth =  106.1 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/Chapter2.ipynb b/Modern_Electronics_Communication/Chapter2.ipynb
new file mode 100755
index 00000000..1e4a523b
--- /dev/null
+++ b/Modern_Electronics_Communication/Chapter2.ipynb
@@ -0,0 +1,384 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "chapter 2 Amplitude Modulation-Transmission"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page no 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "c=1.4*10**6             #frequency of carrier wave,Hz\n",
+      "m1=20                   #frequency component,Hz\n",
+      "m2=10*10**3             #KHz\n",
+      "\n",
+      "#calculation \n",
+      "Ur1=c+m1\n",
+      "Ur2=c+m2\n",
+      "Lr1=c-m1\n",
+      "Lr2=c-m2\n",
+      "\n",
+      "#result\n",
+      "#range of upper sideband(usb)\n",
+      "print\"upper sideband will include frequencies from  \",Ur1,\"Hz\"\n",
+      "print\"to  \",Ur2,\"Hz\"\n",
+      "#range of lower sideband (lsb)\n",
+      "print\"lower sideband will include frequencies from \",Lr2,\"Hz\"\n",
+      "print\"to  \",Lr1,\"Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "upper sideband will include frequencies from   1400020.0 Hz\n",
+        "to   1410000.0 Hz\n",
+        "lower sideband will include frequencies from  1390000.0 Hz\n",
+        "to   1399980.0 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 Page no 78"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "b=100.0                       #maximum p-p carrier (V)\n",
+      "a=60.0\n",
+      "d=125.0\n",
+      "c=35.0\n",
+      "x=180\n",
+      "y=0\n",
+      "\n",
+      "#Calculation\n",
+      "m1=((b-a)/(b+a))*100\n",
+      "m2=((d-c)/(d+c))*100\n",
+      "m3=((x-y)/(y+x))*100\n",
+      "\n",
+      "#result\n",
+      "print\"(a) percent(m) =  \",m1,\"percent\"\n",
+      "print\"(b) percent(m) = \",m2,\"percent\"\n",
+      "print\"(c) percent(m) = \",m3,\"percent\"\n",
+      "print\"(d) this is a case of overmodulation\"\n",
+      "print\"(e) this is a distorted AM wave as the increase > decrease in carrier's amplitude\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) percent(m) =   25.0 percent\n",
+        "(b) percent(m) =  56.25 percent\n",
+        "(c) percent(m) =  100 percent\n",
+        "(d) this is a case of overmodulation\n",
+        "(e) this is a distorted AM wave as the increase > decrease in carrier's amplitude\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page no 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c=1*10**3       #carrier output\n",
+      "\n",
+      "#calculation\n",
+      "esb= 1/4.0*(c)\n",
+      "tsp=(esb*2)\n",
+      "tp=(tsp+c)\n",
+      "\n",
+      "#result\n",
+      "print\"Total transmitted power =\",tp,\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total transmitted power = 1500.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page no 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "m=0.9                           #modulation index\n",
+      "Pc=500                          #carrier Power \n",
+      "\n",
+      "#calculation\n",
+      "x=(m**2)/2.0\n",
+      "y=(1+x)*Pc\n",
+      "\n",
+      "#result\n",
+      "print\"total transmitted power=  \",y,\"W\"               #total transmitted powwer\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total transmitted power=   702.5 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page no 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "m=0.95                                                     #modulation index\n",
+      "Pt= 50*10**3                                               #total transmitted power\n",
+      "\n",
+      "#calculation\n",
+      "x=(m**2)/2.0\n",
+      "y=1+x\n",
+      "z=(Pt/y)\n",
+      "Pi=Pt-z\n",
+      "\n",
+      "#result\n",
+      "print\"Pc = \",round(z,2),\"W\"                                #carrier power\n",
+      "print\"total intelligence power = \",round(Pi,2),\"W\"         #intelligence signal\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pc =  34453.06 W\n",
+        "total intelligence power =  15546.94 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6 Page no 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "Ic=12                             #antenna current of AM transmitter when unmodulated\n",
+      "It=13                             #current when modulated\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "x=2*((13/12.0)**2-1)\n",
+      "m=math.sqrt(x)\n",
+      "a=m*100\n",
+      "\n",
+      "#result\n",
+      "print\"percent(m) = \",round(a,0),\"percent\"            \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "percent(m) =  59.0 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7 Page no 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "n=0.7                                    #efficiency\n",
+      "c=10*10**3                               #carrier wave\n",
+      "\n",
+      "#calculation\n",
+      "Is=0.5*c                                 #intelligence signal\n",
+      "p=(Is/n)\n",
+      "\n",
+      "#result\n",
+      "print\"dc input power = \",round(p,2),\"W\"       #dc input power"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "dc input power =  7142.86 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8 Page no 82"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Pc=10.0*10**3                                  #carrier power\n",
+      "Pt=11.2*10**3                                  #transmitted power\n",
+      "m2=0.5                                         #modulation index of another sine wave\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "x=2*((Pt/Pc)-1)\n",
+      "m=math.sqrt(x)\n",
+      "meff=math.sqrt((m**2)+(m2**2))\n",
+      "a=Pc*(1+((meff**2)/2.0))\n",
+      "\n",
+      "#result\n",
+      "print\"Pt = \",a,\"W\"                             #total transmitted power"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pt =  12450.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9 Page no 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "v1=1.0\n",
+      "v2=0.03\n",
+      "v3=0.05\n",
+      "v4=0.02\n",
+      "v5=0.04\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "x=math.sqrt((v2**2+v3**2+v4**2+v5**2)/v1**2)\n",
+      "y=x*100\n",
+      "\n",
+      "#result\n",
+      "print\"THD = \",round(y,2),\"percent\"            #Total harmonic distortion      "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "THD =  7.35 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/Chapter3.ipynb b/Modern_Electronics_Communication/Chapter3.ipynb
new file mode 100755
index 00000000..23d71dd2
--- /dev/null
+++ b/Modern_Electronics_Communication/Chapter3.ipynb
@@ -0,0 +1,166 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 Amplitude modulation-Reception "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page no 120"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "fr=550*10**3                           #frequency, Hz\n",
+      "L=10.0*10**-6                          #inductor, H\n",
+      "fr1=1550*10**3\n",
+      "\n",
+      "#calculation \n",
+      "import math\n",
+      "a=fr*2*math.pi\n",
+      "x=fr1*2*math.pi\n",
+      "b=1/a\n",
+      "y=1/x\n",
+      "C1=((b)**2/L)\n",
+      "C2=((y)**2/L)\n",
+      "fr2=1100*10**3\n",
+      "BW=10.0*10**3\n",
+      "Q=(fr2/BW)\n",
+      "BW1=(fr1/Q)\n",
+      "BW2=(fr/Q)\n",
+      "\n",
+      "#result\n",
+      "print\"(a) required range of capacitance is from \",round(C2,12),\"F\",\"to\",round(C1,12),\"F\"\n",
+      "print\"(b) Q= \",Q\n",
+      "print\"(c) Bandwidth of receiver at 1550 KHz = \",round(BW1,2),\"Hz\"\n",
+      "print\"Babdwidth of receiver at 550 KHz = \",BW2,\"Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) required range of capacitance is from  1.054e-09 F to 8.374e-09 F\n",
+        "(b) Q=  110.0\n",
+        "(c) Bandwidth of receiver at 1550 KHz =  14090.91 Hz\n",
+        "Babdwidth of receiver at 550 KHz =  5000.0 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page no 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# given\n",
+      "f=620*10**3                                    #frequency, Hz\n",
+      "IF=455*10**3\n",
+      "\n",
+      "#calculation\n",
+      "LO=f+IF\n",
+      "X=IF+LO\n",
+      "# image frequency of local oscillator\n",
+      "#station frequency = 620 kHz\n",
+      "\n",
+      "#Result\n",
+      "print\"LO =  Hz\",LO\n",
+      "print\"X-1075kHz=\",X,\"Hz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "LO =  Hz 1075000\n",
+        "X-1075kHz= 1530000 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page no 150"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "V=8*10**-6                         #microvolts\n",
+      "R=50                               #input resistance, ohm\n",
+      "\n",
+      "#Calculation\n",
+      "import math\n",
+      "P=(V**2)/R\n",
+      "dBm=10*math.log10(P/0.001)\n",
+      "dBW=10*math.log10(P/1)\n",
+      "a=(-89+8+3+24+26+26-2+34)\n",
+      "x=(a/10)\n",
+      "y=10**x\n",
+      "z=y*0.001\n",
+      "\n",
+      "#Result\n",
+      "print\"input power  is= \",P,\"W\"      \n",
+      "print\"dBm = \",round(dBm,0),\"dBm\"\n",
+      "print\"dBW = \",round(dBW,0),\"dBw\"\n",
+      "print\"Pout(dBm) =\",a,\" dBm into speaker\"\n",
+      "print\"Pout(dBW) =\",z,\"W\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "input power  is=  1.28e-12 W\n",
+        "dBm =  -89.0 dBm\n",
+        "dBW =  -119.0 dBw\n",
+        "Pout(dBm) = 30  dBm into speaker\n",
+        "Pout(dBW) = 1.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/Chapter4.ipynb b/Modern_Electronics_Communication/Chapter4.ipynb
new file mode 100755
index 00000000..5c291ef8
--- /dev/null
+++ b/Modern_Electronics_Communication/Chapter4.ipynb
@@ -0,0 +1,149 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 single-sideband communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page no 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "x=1*10**6           #DSB range\n",
+      "y=10**(80/20.0)\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "z=math.sqrt(y)\n",
+      "df=200.0\n",
+      "Q=(x*z)/(4.0*df)\n",
+      "a=100*10**3\n",
+      "Q1=(a*z)/(4.0*df)\n",
+      "\n",
+      "\n",
+      "#result\n",
+      "print\"(a) Q for 1MHz  = \",Q\n",
+      "print\"(b) Q for 100KHz = \",Q1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Q for 1MHz  =  125000.0\n",
+        "(b) Q for 100KHz =  12500.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page no 178"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "a=3*10**6\n",
+      "b=3.1*10**6                  #new DSB signal range\n",
+      "c=2.9*10**6                  #new DSB signal range\n",
+      "\n",
+      "#calculation\n",
+      "Q=(a/(b-c))\n",
+      "\n",
+      "#result\n",
+      "print\"the required filter Q is = \",Q"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the required filter Q is =  15.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page no 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "a=455\n",
+      "x=2000+1     #frequency\n",
+      "y=2000+3     #frequency\n",
+      "\n",
+      "#calculation\n",
+      "c=2000+a      #local oscillator value\n",
+      "d=c-x\n",
+      "e=c-y\n",
+      "f=a-454\n",
+      "g=a-452\n",
+      "\n",
+      "#result\n",
+      "print\"RF =\",x,\"KHz\", \"at which first mixer input=\" ,y,\"kHz\"\n",
+      "print\"local oscillator value is \",c,\"KHz\"\n",
+      "print\"First mixer output is\",d,\"to\",e,\"KHz\"                #IF amp and second mixer input\n",
+      "print\"BFO = \",a,\"KHz\"\n",
+      "print\"Second mixer output is\",f,\"KHz\", \"and audio amplifier value is =\",g, \"kHz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "RF = 2001 KHz at which first mixer input= 2003 kHz\n",
+        "local oscillator value is  2455 KHz\n",
+        "First mixer output is 454 to 452 KHz\n",
+        "BFO =  455 KHz\n",
+        "Second mixer output is 1 KHz and audio amplifier value is = 3 kHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/Chapter5.ipynb b/Modern_Electronics_Communication/Chapter5.ipynb
new file mode 100755
index 00000000..5e894199
--- /dev/null
+++ b/Modern_Electronics_Communication/Chapter5.ipynb
@@ -0,0 +1,360 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 Frequency Modulation : Transmission"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page no 209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "v=25*10**-3\n",
+      "f=750                        #deviation constant\n",
+      "vg=10.0*10**-3                 #deviation constant\n",
+      "\n",
+      "#calculation\n",
+      "pfd=v*(f/vg)                 #positive frequency deviation\n",
+      "nfd=-v*(f/vg)                #negative frequency deviation\n",
+      "\n",
+      "#result\n",
+      "print\"(a) positive frequency deviation = \",pfd,\"Hz\"\n",
+      "print\"negative frequency deviation = \",nfd,\"Hz\"\n",
+      "print\"The total deviation is written as +-2.25kHz for the given input signal level\"\n",
+      "print\"(b) The carrier wil deviate \",pfd,\"Hz\",\"&\",nfd,\"Hz\",\"at 400 Hz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) positive frequency deviation =  1875.0 Hz\n",
+        "negative frequency deviation =  -1875.0 Hz\n",
+        "The total deviation is written as +-2.25kHz for the given input signal level\n",
+        "(b) The carrier wil deviate  1875.0 Hz & -1875.0 Hz at 400 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page no 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=20*10**3     #maximum deviation\n",
+      "fi=10.0*10**3  #input frequency\n",
+      "\n",
+      "#calculation\n",
+      "mf=d/fi\n",
+      "a=mf*40\n",
+      "\n",
+      "#result\n",
+      "print\"total required bandwidth is \",a,\"KHz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total required bandwidth is  80.0 KHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 Page no 214"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=20*10**3                    #maximum deviation\n",
+      "fi=5.0*10**3                  #input frequency\n",
+      "\n",
+      "#calculation\n",
+      "mf=d/fi\n",
+      "a=2*35\n",
+      "\n",
+      "#print\n",
+      "print\"the required bandwidth is \",mf\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the required bandwidth is  4.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 Page no 215"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Vm=2000\n",
+      "R=50.0                          #resistance, ohm\n",
+      "\n",
+      "#calcultion\n",
+      "import math\n",
+      "fc=(2*math.pi*(10**8))/2.0*math.pi\n",
+      "P=(2000/math.sqrt(2))**2/R\n",
+      "mf=2                                                     #by inspection of FM equation\n",
+      "fi=(math.pi*10**4)/(2.0*math.pi)\n",
+      "d=(mf*fi)\n",
+      "BW=mf*40\n",
+      "bw=2*(d+fi)\n",
+      "P1=((0.58*2000/math.sqrt(2))**2)/R\n",
+      "P2=((0.03*2000/math.sqrt(2))**2)/R\n",
+      "\n",
+      "#result\n",
+      "print\"(a) carrier frequency = \",round(fc,-9),\"Hz\"        #by inspection of FM equation\n",
+      "print\"(b) the peak voltage is 2000V P thus= \",P,\"W\"\n",
+      "print\"(c) mf = 2\"                                        #by inspection of FM equation\n",
+      "print\"(d) the intelligence frequency fi = \",fi,\"Hz\"\n",
+      "print\"(e) BW = \",bw ,\"Hz\"                                #using carson's rule \n",
+      "print\"(f) The smallest sideband J4 is 0.03 times the carrier = \",P2,\"W\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) carrier frequency =  1000000000.0 Hz\n",
+        "(b) the peak voltage is 2000V P thus=  40000.0 W\n",
+        "(c) mf = 2\n",
+        "(d) the intelligence frequency fi =  5000.0 Hz\n",
+        "(e) BW =  30000.0 Hz\n",
+        "(f) The smallest sideband J4 is 0.03 times the carrier =  36.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 Page no 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=75*10**3                       #maximum deviation\n",
+      "fi=30.0                          #modulating frequency, Hz\n",
+      "fi1=15.0*10**3                   \n",
+      "d1=1*10**3\n",
+      "fi2=100                           #Hz\n",
+      "fi3=2.0*10**3\n",
+      "\n",
+      "#calculation\n",
+      "mf1=d/fi\n",
+      "mf2=d/fi1\n",
+      "mf3=d1/fi2\n",
+      "mf4=d1/fi3\n",
+      "DR=d1/fi3\n",
+      "\n",
+      "#result\n",
+      "print\"(a)maximum deviation at 30 Hz = \",mf1                           \n",
+      "print\"maximum deviation at 15kHz= \",mf2\n",
+      "print\"(b) maximum deviation at 100Hz = \",mf3\n",
+      "print\"maximum deviation at 2KHz= \",mf4\n",
+      "print \"(c)Deviation Ratio \",DR                  #deviation ratio"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a)maximum deviation at 30 Hz =  2500.0\n",
+        "maximum deviation at 15kHz=  5.0\n",
+        "(b) maximum deviation at 100Hz =  10\n",
+        "maximum deviation at 2KHz=  0.5\n",
+        "(c)Deviation Ratio  0.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7 Page no 218"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "mf=0.25\n",
+      "a=0.98\n",
+      "b=0.12\n",
+      "x=10*10**3                      #power, watt\n",
+      "\n",
+      "#calculation\n",
+      "P=(a**2)*x\n",
+      "P1=(b**2)*x\n",
+      "t=P+2*P1\n",
+      "\n",
+      "#result\n",
+      "print\"power of each sideband = \",P1,\"W\"\n",
+      "print\"total power = \",round(t,-4),\"W\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power of each sideband =  144.0 W\n",
+        "total power =  10000.0 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 Page no 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "phi=0.5                          #maximum intelligence frequency\n",
+      "fi=5.0*10**3\n",
+      "x=75*10**3\n",
+      "\n",
+      "#calculation\n",
+      "d=phi*fi\n",
+      "y=x/d\n",
+      "\n",
+      "#result\n",
+      "print\"o/p S/N = \",y\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "o/p S/N =  30.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 Page no 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "dm=10*10**3\n",
+      "x=(1/3.0)                      #N/S input ratio\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "phi=math.asin(x)\n",
+      "phi1=math.asin(x)\n",
+      "fi=3*10**3\n",
+      "d=phi1*fi\n",
+      "a=dm/d\n",
+      "\n",
+      "#result\n",
+      "print\"The S/N output will be \",round(a,0)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The S/N output will be  10.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/README.txt b/Modern_Electronics_Communication/README.txt
new file mode 100755
index 00000000..799a1197
--- /dev/null
+++ b/Modern_Electronics_Communication/README.txt
@@ -0,0 +1,10 @@
+Contributed By: nishu mittal
+Course: others
+College/Institute/Organization: university of delhi
+Department/Designation: electronics
+Book Title: Modern Electronics Communication
+Author: J S Beasley and Miller
+Publisher: Phi Publication
+Year of publication: 9th
+Isbn: 9788120340046
+Edition: 2008
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter10.ipynb b/Modern_Electronics_Communication/chapter10.ipynb
new file mode 100755
index 00000000..878fe5de
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter10.ipynb
@@ -0,0 +1,101 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 Wireless digital communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page no 467"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "n=3                 #shift registers\n",
+      "n1=7\n",
+      "\n",
+      "#calculation\n",
+      "x=(2**n)-1\n",
+      "y=(2**7)-1\n",
+      "\n",
+      "#result\n",
+      "print\"(a) n =\",n,\"PN sequence length = \",x\n",
+      "print\"(b) n =\",n1,\"PN sequence length = \",y"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) n = 3 PN sequence length =  7\n",
+        "(b) n = 7 PN sequence length =  127\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 Page no 477"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "x=56.0                                #modulation bit rate\n",
+      "y=560                                 #chip rate\n",
+      "a=256.0                               #modulation bit rate\n",
+      "b=1792                                #chip rate\n",
+      "\n",
+      "#calculation\n",
+      "z=y/x\n",
+      "c=b/a\n",
+      "\n",
+      "#result\n",
+      "#part(a)\n",
+      "print\"Spreading of a DSSS signal = \",z\n",
+      "print\"Spreading of a DSSS signal = \",c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Spreading of a DSSS signal =  10.0\n",
+        "Spreading of a DSSS signal =  7.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter12.ipynb b/Modern_Electronics_Communication/chapter12.ipynb
new file mode 100755
index 00000000..79f6b2af
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter12.ipynb
@@ -0,0 +1,401 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 Transmission Lines"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Page no 573"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "L=73.75*10**-9                    #inductance, H\n",
+      "C=29.5*10**-12                    #capacitance\n",
+      "x=5280\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "Z=math.sqrt(L/C)\n",
+      "z1=math.sqrt((x*L)/(x*C))\n",
+      "\n",
+      "#result\n",
+      "print\"characterstics impedence for 1-ft =\",Z,\"ohm\"\n",
+      "print\"characterstics impedence for 1-mi = \",z1,\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "characterstics impedence for 1-ft = 50.0 ohm\n",
+        "characterstics impedence for 1-mi =  50.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 Page no 574"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "a=2                           #parallel wire line\n",
+      "b=2.35                        #coaxial line\n",
+      "D=0.285\n",
+      "d=0.08\n",
+      "e=1.0                         #dielectric constant of insulating material relative to air\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "z=(276/e)*math.log10(2*2)\n",
+      "z1=(138/e)*math.log10(b)\n",
+      "z2=(138/math.sqrt(2.3)*math.log10(D/d))\n",
+      "\n",
+      "#result\n",
+      "print\"(a) characterstics impedence for a parallel wire = \",round(z,0),\"ohm\"\n",
+      "print\"(b) characterstics impedence for a air dielectric coaxial line= \",round(z1,1),\"ohm\"\n",
+      "print\"(c) characterstics impedence = \",round(z2,0),\"ohm\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) characterstics impedence for a parallel wire =  166.0 ohm\n",
+        "(b) characterstics impedence for a air dielectric coaxial line=  51.2 ohm\n",
+        "(c) characterstics impedence =  50.0 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 Page no 579"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "L=73.75*10**-9                #inductance, H\n",
+      "C=29.5*10**-12                #capacitance\n",
+      "d=1                           #distance\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "t=math.sqrt(L*C)\n",
+      "Vp=d/t\n",
+      "\n",
+      "#result\n",
+      "print\"the delay introduced is t =\",round(t,10),\"s\"\n",
+      "print\"The velocity of propagation is  \",round(Vp*10**-8,3),\"*10**8 ft/s\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the delay introduced is t = 1.5e-09 s\n",
+        "The velocity of propagation is   6.78 *10**8 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4 Page no 580"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "v=2.07*10**8                     #velocity\n",
+      "c=3.0*10**8                      #velocity of light\n",
+      "Er=2.3                           #relative dielectric constant\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "vf=(v/c)                         #velocity factor\n",
+      "vf1=1/math.sqrt(Er)\n",
+      "\n",
+      "#result\n",
+      "print\"The velocity = \",vf,\"m/s\"\n",
+      "print\"vf = \",round(vf1,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The velocity =  0.69 m/s\n",
+        "vf =  0.659\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 Page no 581"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c=3*10**8                         #speed of light\n",
+      "f=100.0*10**6                     #frequency of signal\n",
+      "x=2.07*10**8                      #velocity of wave propagation\n",
+      "\n",
+      "#Calcultion\n",
+      "w=c/f                             #wavelength in free-space\n",
+      "w1=x/f                            #wavelength while traveling through an RG-8A/U coaxial cable\n",
+      "\n",
+      "#Result\n",
+      "print\"In free space, lambda =\",w,\"m\"\n",
+      "print\"While traveling through RG-8A/U cable, lamda= \",w1,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In free space, lambda = 3.0 m\n",
+        "While traveling through RG-8A/U cable, lamda=  2.07 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7 Page no 592"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "Zl=300.0                                     #load impedance\n",
+      "Zo=50.0                                      #characteristic impedance\n",
+      "v=2.07*10**8                                 #velocity in RG-8A/U cable\n",
+      "f=27.0*10**6                                 #operating frequency of citizen's band transmitter\n",
+      "Po=4                                         #output power of transmitter\n",
+      "l=10                                         #length of RG-8A/U cable\n",
+      "Rl=300                                       #input resistance of antenna\n",
+      "\n",
+      "#calculation\n",
+      "T=((Zl-Zo)/(Zl+Zo))                          #reflection coefficient\n",
+      "h=v/f                                        #length of cable in wavelength\n",
+      "le=l/h                                       #electrical length\n",
+      "x=Rl/Zo                                      #VSWR\n",
+      "y=((1+T)/(1-T))                              #VSWR\n",
+      "rp=(T)**2*Po                                 #reflected power\n",
+      "Pl=Po-rp                                     #load power\n",
+      "#part(a): The reflection coefficient\n",
+      "\n",
+      "#result\n",
+      "print\"(a) reflection cofficient = \",round(T,2)\n",
+      "print\"(b) electrical length =\",round(le,2),\"lambda\"\n",
+      "print\"(c) VSWR = \",y\n",
+      "print\"(d) the reflected voltage = \",round(Pl,2),\"W\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) reflection cofficient =  0.71\n",
+        "(b) electrical length = 1.3 lambda\n",
+        "(c) VSWR =  6.0\n",
+        "(d) the reflected voltage =  1.96 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8 Page no 597"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Zo=100.0                             #characteristic impedance\n",
+      "j=1j\n",
+      "Zl = 200-j*150                       #load impedance\n",
+      "l=4.3                                #length of transmission line\n",
+      "\n",
+      "#calculation\n",
+      "x=200/Zo\n",
+      "y=150/Zo\n",
+      "a=0.4*Zo\n",
+      "b=0.57*Zo\n",
+      "\n",
+      "#result\n",
+      "print\"Zin = \",a,\"Ohm\",\"+j*\",b,\"Ohm\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Zin =  40.0 Ohm +j* 57.0 Ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9 Page no 599"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "import cmath\n",
+      "j=1j\n",
+      "RL=120                                     #load resistance from smith chart\n",
+      "ZL=complex(75,50)                 #load impedance\n",
+      "Z0=50.0                             #characteristic impedance\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "z1=ZL/Z0\n",
+      "z=2.4                                      #normalized z at a point that is purely resistive\n",
+      "ar=z*Z0                                    #actual resistance\n",
+      "x=math.sqrt(Z0*RL)\n",
+      "\n",
+      "#Result\n",
+      "print\"zl= \",z1\n",
+      "#VSWR,zin,R can be found out from smith chart manually\n",
+      "print\"characteristic impedance is =\",round(x,2),\"ohm\"              \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "zl=  (1.5+1j)\n",
+        "characteristic impedance is = 77.46 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.10 Page no 601"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "import cmath\n",
+      "Z0=75.0                              #characteristic impedance\n",
+      "j=1j\n",
+      "ZL=complex(50,-100)                 #load impedance\n",
+      "\n",
+      "#Calculation\n",
+      "zL=ZL/Z0\n",
+      "#Result\n",
+      "print\"zL =\",zL\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "zL = (0.666666666667-1.33333333333j)\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter13.ipynb b/Modern_Electronics_Communication/chapter13.ipynb
new file mode 100755
index 00000000..f28a3b72
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter13.ipynb
@@ -0,0 +1,238 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:358c1b86e82f27b9eee799b4793457200c3f930f370a7e7347103331d9245d4d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 Wave Propagation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 Page no 628"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "x=(2*(1/2.0))\n",
+      "a=15\n",
+      "d=53.5*10**-6                           #duration for each horizontal line on the reciever\n",
+      "t=1/186000.0                            #time delay between direct and reflected signal\n",
+      "\n",
+      "#calculation\n",
+      "g=(t/d)*a                               #ghost width\n",
+      "\n",
+      "#result\n",
+      "print\"ghost width = \",round(g,2),\"in.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ghost width =  1.51 in.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 Page no 641"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "S=83                        #satellite longitude in degrees\n",
+      "N=90                        #site longitude in degrees\n",
+      "L=35                        #site longitude in degrees\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "b=0.1512                     #constant in equation\n",
+      "A=180+math.atan(math.tan(-7*3.14/180)/math.sin(35*3.14/180))*180/3.14  \n",
+      "E=math.atan((math.cos((S-N)*3.14/180)*math.cos(L*3.14/180)-b)/math.sqrt(1-((math.cos(L*3.14/180))**2*(math.cos((S-N)*3.14/180))**2)))*180/3.14\n",
+      "\n",
+      "#Result\n",
+      "print\"The azimuth is equal to A = \",round(A,0),\"degree\"\n",
+      "print\"the elevation angle  = \",round(E,1),\"degree\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The azimuth is equal to A =  168.0 degree\n",
+        "the elevation angle  =  48.7 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page no 646"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "import math\n",
+      "x=(32+(44/60.0)+(36/3600.0))                           # N latitude\n",
+      "y=(106+(16/60.0)+(37/3600.0))                          #W longitude\n",
+      "D=42.1642*10**6                                    #distance from the satellite to the center of the earth\n",
+      "R=6.378*10**6                                      #earth's radius\n",
+      "a=32.74333*3.14/180.0                             #in degree\n",
+      "B=-7.27694*3.14/180.0                               #in degree\n",
+      "\n",
+      "#calculation\n",
+      "q=math.cos(a)*math.cos(B)\n",
+      "\n",
+      "d=math.sqrt(D**2+R**2-(2*D*R*q))\n",
+      "c=2.997925*10**5                                    #velocity of light\n",
+      "de=d/c\n",
+      "rd=(2*d)/c\n",
+      "\n",
+      "#result\n",
+      "print\"N longitude converted into degrees = \",round(x,2)\n",
+      "print\"W longitude  coverted into degrees =\",round(y,2)\n",
+      "print\"distance =  \",round(d/1000,1),\"*10**6 meters\"\n",
+      "print\"delay =\",round(de/1000,3),\"seconds\"\n",
+      "print\"roundtrip delay = \",round(rd/1000,3),\"seconds\" \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "N longitude converted into degrees =  32.74\n",
+        "W longitude  coverted into degrees = 106.28\n",
+        "distance =   37009.1 *10**6 meters\n",
+        "delay = 0.123 seconds\n",
+        "roundtrip delay =  0.247 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4 Page no 651"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "G=45                                     #antenna gain\n",
+      "nt=25                                    #antenna noise temperature\n",
+      "nt1=70                                   #LNB noise temperature\n",
+      "nt2=2                                    #noise temperature(reciever and passive components)\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "T=nt+nt1+nt2                             #total noise temperature\n",
+      "x=G-10*math.log10(T)                     #figure of merit\n",
+      "\n",
+      "#result\n",
+      "print\"Sum of all of the noise temperature contributions Ts = \",T,\"k\"\n",
+      "print\"The figure of merit = \",round(x,2),\"dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Sum of all of the noise temperature contributions Ts =  97 k\n",
+        "The figure of merit =  25.13 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5 Page no 652"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=41.130383*10**6                                #dismath.tance\n",
+      "c=2.997925*10**8                                 #velocity of light\n",
+      "f=14.25*10**9                                    #uplink frequency\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "h=c/f                                            #wavelength\n",
+      "x=(4*math.pi*d)/h\n",
+      "Lp=20*math.log10(x)                              #free-space path loss\n",
+      "\n",
+      "#result\n",
+      "print\"The wavelength is= \",round(h,2)\n",
+      "print\"Lp(dB)= \",round(Lp,2),\"dB\"\n",
+      "#INcorrect answer of h in the textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength is=  0.02\n",
+        "Lp(dB)=  207.81 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter14.ipynb b/Modern_Electronics_Communication/chapter14.ipynb
new file mode 100755
index 00000000..39f00cfe
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter14.ipynb
@@ -0,0 +1,180 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 Antennas"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1 Page no 669"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c=3*10**8                   #m/s , speed of light              \n",
+      "f=150*10**6                 #frequency\n",
+      "\n",
+      "#calculation\n",
+      "h=c/f\n",
+      "x=1/2.0                     #antennas dimension(D)\n",
+      "D=0.5*2\n",
+      "Rff=5*D\n",
+      "h=c/f \n",
+      "D = h/2\n",
+      "\n",
+      "#result\n",
+      "print\" distance =\",Rff,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " distance = 5.0 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2 Page no 669"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c=3*10**8       #velocity of light\n",
+      "f=12.0*10**9    #frequency\n",
+      "D=4.5           #diameter of parabolic reflector\n",
+      "\n",
+      "#calculation\n",
+      "h=c/f           #wavelength\n",
+      "x=D/h\n",
+      "R=(2*D**2)/h\n",
+      "\n",
+      "#result\n",
+      "print\"distance from parabolic reflector= \",R,\"m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance from parabolic reflector=  1620.0 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3 Page no 671"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Pt=10                   #transmitted power\n",
+      "#dipoles have gain 2.15dB  \n",
+      "Gr=1.64\n",
+      "c=3*10**8               #velocity of light\n",
+      "f=144.0*10**6           #frequency\n",
+      "d=50*10**3              #distance between antennas\n",
+      "Gt=1.64                 #recieving antenna gain(ratio) compared to isotropic radiator\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "Pr=Pt*Gt*Gr*(c/f)**2/((16*(math.pi)**2)*(d**2))                #power recieved\n",
+      "\n",
+      "#result\n",
+      "print\"Power received = \",round(Pr,12),\"w\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Power received =  2.96e-10 w\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4 Page no 674"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "c=3*10**8         #velocity of light\n",
+      "f=100*10**6       #frequency\n",
+      "\n",
+      "#Calculation\n",
+      "h=c/f         #wavelength\n",
+      "x=h/2.0         #dipole i.e h/2\n",
+      "l=0.95*x       #applying 95% correction,the actual optimum physical length\n",
+      "L=486/100.0      #alternative method to find length\n",
+      "\n",
+      "#Result\n",
+      "print\"length of antenna = \",l,\"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "length of antenna =  1.425 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter16.ipynb b/Modern_Electronics_Communication/chapter16.ipynb
new file mode 100755
index 00000000..ab90f6a3
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter16.ipynb
@@ -0,0 +1,134 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16 Microwaves and Lasers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.1 Page no 753"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "h=0.3                    #curve depth of parabolic reflector\n",
+      "D=3                      #diameter of parabolic reflector\n",
+      "\n",
+      "#calculation\n",
+      "f=D/(16*h)               #focal \n",
+      "\n",
+      "#result\n",
+      "print\"The focal length is \",f,\" m out from the center of the parabolic reflector\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The focal length is  0.625  m out from the center of the parabolic reflector\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.2 Page no 755"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "D=3.0                    #diameter of microwave dish\n",
+      "k=0.6                    #efficiency of reflector\n",
+      "c=2.997925*10**8         #velocity of light\n",
+      "f=10.0*10**9             #frequency\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "Ap=10*math.log10((c/f)*k*((math.pi*D)**2/(c/f)**2))      #powergain\n",
+      "B=(70*(c/f))/D                                         #beamwidth\n",
+      "\n",
+      "#result\n",
+      "print\"beamwidth = \",round(B,3),\"micrometer\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "beamwidth =  0.7 micrometer\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.3 Page no 756"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "D=4.5                       #diameter of parabolic reflector\n",
+      "k=0.62                      #efficiency factor\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "Ae=(k*math.pi)*(D/2.0)**2                       #aperture efficiency\n",
+      "i=(math.pi*(D/2.0)**2)                #ideal capture area\n",
+      "\n",
+      "#result\n",
+      "print\"The ideal capture area for\",D,\"m\",\"1f m parabolic antenna is\",round(i,3),\"sq.m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The ideal capture area for 4.5 m 1f m parabolic antenna is 15.904 sq.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter17.ipynb b/Modern_Electronics_Communication/chapter17.ipynb
new file mode 100755
index 00000000..44c3e3d8
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter17.ipynb
@@ -0,0 +1,96 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 17 Television"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.1 Page no 882"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw=5*10**6            #bandwidth\n",
+      "t=53.5*10**-6         #time allocated for each visible trace\n",
+      "\n",
+      "#calculation\n",
+      "T=2*bw*t              #increase in horizontal resolution\n",
+      "\n",
+      "#result\n",
+      "print\"Thus, the total number of vertical lines resolvable is \",T,\"lines\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thus, the total number of vertical lines resolvable is  535.0 lines\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 17.2 Page no 822"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw=5.0*10**6                    #bandwidth\n",
+      "l=428                           #horizontal resolution\n",
+      "\n",
+      "#calculation\n",
+      "t=l/(bw*2)                      #trace time\n",
+      "x=1/30.0                        #time available for a full picture\n",
+      "y=t+10.0*10**-6                 #assuming that 10us is used for horizontal blanking\n",
+      "n=x/y                           #no. of horizontal traces\n",
+      "c=600*0.7                       #allowing 32 lines for vertical retrace\n",
+      "\n",
+      "#result\n",
+      "print\"vertical resolution = \",c,\"lines\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "vertical resolution =  420.0 lines\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter18.ipynb b/Modern_Electronics_Communication/chapter18.ipynb
new file mode 100755
index 00000000..fee42ea4
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter18.ipynb
@@ -0,0 +1,216 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18 Fibre Optics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.1 Page no 859"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "c=3*10**8          #velocity of light\n",
+      "f=4.4*10**14       #frequency of red light\n",
+      "f1=7.0*10**14     #frequency of violet light\n",
+      "\n",
+      "#calculation\n",
+      "h1=c/f             #wavelength of red light\n",
+      "h2=c/f1            #wavelength of violet light\n",
+      "\n",
+      "#result\n",
+      "print\"wavelenght for red= \",round(h1,9),\"m\"\n",
+      "print\"wavelngth for violet= \",round(h2,8),\"micron\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wavelenght for red=  6.82e-07 m\n",
+        "wavelngth for violet=  4.3e-07 micron\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.2 Page no 862"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "n1=1.535             #refractive index of fibre optics\n",
+      "n2=1.490             #refractive index of cladding\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "x=(n1**2)-(n2**2)\n",
+      "y=math.sqrt(x)     #numerical aperture\n",
+      "z=math.asin(y)*180/3.14      #theta\n",
+      "\n",
+      "#result\n",
+      "print\"NA = \",round(y,3)\n",
+      "print\"(theta)in(max) = \",round(z,1),\"degree\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NA =  0.369\n",
+        "(theta)in(max) =  21.7 degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.3 Page no 868"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "w=22                  #spectral width of LED\n",
+      "l=2                   #length of fibre\n",
+      "d=95                  #dispersion value\n",
+      "p=d*w                 #pulse dispersion\n",
+      "pt=p*l                #total pulse dispersion\n",
+      "\n",
+      "#result\n",
+      "print\"pulse dispersion = \",p,\"ps/km\"\n",
+      "print\"total pulse dispersion = \",pt,\"ps/km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pulse dispersion =  2090 ps/km\n",
+        "total pulse dispersion =  4180 ps/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.4 Page no 885"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=30                 #length of fibre cable\n",
+      "l=0.4                #loss\n",
+      "\n",
+      "#calculation\n",
+      "T=d*l                #total cable loss\n",
+      "\n",
+      "#result\n",
+      "print\"total cable loss = \",T,\"dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total cable loss =  12.0 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.5 Page no 887"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "b=565                      #Line bit rate of fibre 1\n",
+      "c=3.5                      #Cable dispersion of fibre 1\n",
+      "t=4                        #Transmitter spectral width of fibre 1\n",
+      "b1=1130                    #Line bit rate of fibre 2\n",
+      "c1=3.5                     #Cable dispersion of fibre 2\n",
+      "t1=2                       #Transmitter spectral width of fibre 2\n",
+      "x=440000                   #assumed gaussian constant \n",
+      "\n",
+      "#calculation\n",
+      "L1=x/(b*c*t)               #span length in km of fibre 1\n",
+      "L2=x/(b1*c1*t1)            #span length in km of fibre 2\n",
+      "\n",
+      "#result \n",
+      "print\"span lenght of fibre 1= \",round(L1,2),\"Km\"\n",
+      "print\"span lenght of fibre 2= \",round(L2,2),\"Km\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "span lenght of fibre 1=  55.63 Km\n",
+        "span lenght of fibre 2=  55.63 Km\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter6.ipynb b/Modern_Electronics_Communication/chapter6.ipynb
new file mode 100755
index 00000000..2342c49f
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter6.ipynb
@@ -0,0 +1,98 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: Frequency Modulation:Reception"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page no 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "G=200000                  #voltage gain\n",
+      "v=200*10**-3              #quieting voltage\n",
+      "\n",
+      "#calculation\n",
+      "In=v/G\n",
+      "\n",
+      "#Result\n",
+      "print\"To reach quieting, the input must be \",In*10**6,\"microV\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "To reach quieting, the input must be  1.0 microV\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page no 272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Given\n",
+      "f=10*10**6      #frequency\n",
+      "f1=50*10**3     #input frequency\n",
+      "f2=200*10**3    #output frequency\n",
+      "\n",
+      "#Calculatiion\n",
+      "x=f1*2\n",
+      "y=f2*2\n",
+      "\n",
+      "#Result\n",
+      "print\"The capture occurred at from the free-running VCO frequency.\",f1,\"Hz\"\n",
+      "print\"Assume symmetrical operation,which implies a capture range of \",x,\"Hz\"\n",
+      "print\"Once captured the VCO follows the input to a \",f2,\"Hz deviation,implying a lock range of \",y,\"Hz\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The capture occurred at from the free-running VCO frequency. 50000 Hz\n",
+        "Assume symmetrical operation,which implies a capture range of  100000 Hz\n",
+        "Once captured the VCO follows the input to a  200000 Hz deviation,implying a lock range of  400000 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter7.ipynb b/Modern_Electronics_Communication/chapter7.ipynb
new file mode 100755
index 00000000..39f04dbf
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter7.ipynb
@@ -0,0 +1,243 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 Communication Techniques"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 Page no 304"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Q=60\n",
+      "IF=455*10**3\n",
+      "x=680.0*10**3\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "imf=x+2*(IF)              #image frequency\n",
+      "a=(imf/x)\n",
+      "b=(x/imf)\n",
+      "c=(Q*(a-b))\n",
+      "d=20*math.log10(c)\n",
+      "\n",
+      "#result\n",
+      "print\"The image frequency is \",imf,\"Hz\"\n",
+      "print\"image rejection = \",round(d,3),\"dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The image frequency is  1590000.0 Hz\n",
+        "image rejection =  41.186 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 Page no 314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "NF=20.0\n",
+      "df=10**6\n",
+      "\n",
+      "#calculation\n",
+      "x=10*math.log10(df)\n",
+      "S=-174+NF+x\n",
+      "a=5                           #input intercept\n",
+      "dr=2/3.0*(a-S)\n",
+      "\n",
+      "#result\n",
+      "print\"S= \",S,\"dB\"\n",
+      "print\"dynamic range= \",dr,\"dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "S=  -94.0 dB\n",
+        "dynamic range=  66.0 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8 Page no 315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "nf=5.0\n",
+      "x=24.0\n",
+      "y=20.0\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "NR0=10**(nf/10.0)\n",
+      "NR1=10**(y/10.0)\n",
+      "PG1=10**(x/10.0)\n",
+      "NR=NR0+((NR1-1)/PG1)\n",
+      "NF=10*math.log10(NR)\n",
+      "S=-174+NF+60\n",
+      "a=nf-x                             #the system's third-order intercept point \n",
+      "dr=2/3.0*(a-S)\n",
+      "\n",
+      "#result\n",
+      "print\"NR1 = \",round(NR0,3)\n",
+      "print\"NR2 = \",NR1\n",
+      "print\"PG1=  = \",round(PG1,3)\n",
+      "print\"NR = \",round(NR,3),\"dB\"\n",
+      "print\"NF = \",round(NF,3),\"dB\"             #total system noise figure\n",
+      "print\"S = \",round(S,3),\"dBm\"             #sensitivity\n",
+      "print\"the systems third-order intercept point is\",a,\"dB\"\n",
+      "print\"dynamic range = \",round(dr,1),\"dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NR1 =  3.162\n",
+        "NR2 =  100.0\n",
+        "PG1=  =  251.189\n",
+        "NR =  3.556 dB\n",
+        "NF =  5.51 dB\n",
+        "S =  -108.49 dBm\n",
+        "the systems third-order intercept point is -19.0 dB\n",
+        "dynamic range =  59.7 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9 Page no 315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "x=24\n",
+      "nf=-5\n",
+      "NR = 3.16+(99/10.0)\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "NF = 10*math.log10(NR)\n",
+      "S=-174+NF+60\n",
+      "dr = 2/3.0*(nf-S)\n",
+      "\n",
+      "#result\n",
+      "print\"NR = \",NR                #noise ratio\n",
+      "print\"NF = \",round(NF,3),\"dB\"             #noise figure\n",
+      "print\"S = \",round(S,3),\"dBm\"            #sensitivity\n",
+      "print\"dynamic range = \",round(dr,3),\"dB\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "NR =  13.06\n",
+        "NF =  11.159 dB\n",
+        "S =  -102.841 dBm\n",
+        "dynamic range =  65.227 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.10 Page no 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "x=0.40*100*10**6\n",
+      "y=(100*10**6/(2.0**32))\n",
+      "#fCLK is reference oscillator\n",
+      "\n",
+      "#result\n",
+      "print\"The maximum output frequency is approximately 40 percent of fCLK MAX\",x,\"Hz\"\n",
+      "print\"The frequency resolution is given by \",round(y,3),\"Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The maximum output frequency is approximately 40 percent of fCLK MAX 40000000.0 Hz\n",
+        "The frequency resolution is given by  0.023 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter8.ipynb b/Modern_Electronics_Communication/chapter8.ipynb
new file mode 100755
index 00000000..b7931b29
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter8.ipynb
@@ -0,0 +1,190 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "chapter 8 Digital Communication-Coding Techniques"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page no 357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "fa=20*10**3\n",
+      "\n",
+      "#calculation\n",
+      "fs=2*fa                                     #minimum sample rate\n",
+      "\n",
+      "#result\n",
+      "print\"maximum sample rate\",\"is greater than equal to\",fs,\"Hz\"        #frequency\n",
+      "  "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "maximum sample rate is greater than equal to 40000 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page no 362"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "dr=55\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "n=(dr/6.02)\n",
+      "l=2**10\n",
+      "y=10*math.log10(3*(l**2))             #signal-to-quantization-noise level\n",
+      "x=(1.76+(6.02*10))                    #signal-to-noise ratio for digitizing system\n",
+      "\n",
+      "#result\n",
+      "print\"(S/N)q =\",round(y,2),\"dB\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(S/N)q = 64.98 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 Page no 368"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "R=100.0*10**3                    #resistance, ohm\n",
+      "Rf=10*10**3\n",
+      "Vref=-10                         #reference voltage\n",
+      "\n",
+      "#calculation\n",
+      "Vo=-(Vref)*(Rf/R)                # resolution\n",
+      "a=(10/100.0)\n",
+      "b=(10/50.0)\n",
+      "c=(10/25.0)\n",
+      "d=(10/12.5)\n",
+      "V=-(Vref)*(a+b+c+d)              #output voltage\n",
+      "\n",
+      "#result\n",
+      "print\"The step-size is \",Vo\n",
+      "print\"output voltage = \",V,\"volts\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The step-size is  1.0\n",
+        "output voltage =  15.0 volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page no 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "d=2                     #Dmin\n",
+      "d1=3\n",
+      "d2=4\n",
+      "\n",
+      "#calculation\n",
+      "x=d-1               \n",
+      "a=(d/2.0)-1\n",
+      "y=d1-1\n",
+      "b=1/2.0*(d1-1)\n",
+      "z=d2-1\n",
+      "c=(d2/2.0)-1 \n",
+      "#part (a)\n",
+      "\n",
+      "#result\n",
+      "print\"(a) the no. of error detected for the distance 2 is \",x\n",
+      "print\" the no. of errors corrected  = \",a\n",
+      "print\"(b) the no. of error detected for the distance 3 is \",y\n",
+      "print\" the no. of errors corrected  = \",b\n",
+      "print\"(c) the no. of error detected for the distance 4 is \",z\n",
+      "print\" the no. of errors corrected  = \",c\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) the no. of error detected for the distance 2 is  1\n",
+        " the no. of errors corrected  =  0.0\n",
+        "(b) the no. of error detected for the distance 3 is  2\n",
+        " the no. of errors corrected  =  1.0\n",
+        "(c) the no. of error detected for the distance 4 is  3\n",
+        " the no. of errors corrected  =  1.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/chapter9.ipynb b/Modern_Electronics_Communication/chapter9.ipynb
new file mode 100755
index 00000000..812fc2f5
--- /dev/null
+++ b/Modern_Electronics_Communication/chapter9.ipynb
@@ -0,0 +1,203 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 Wired Digital Communications"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Page no 405"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "M=110\n",
+      "x=7.0\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "a=math.log10(110)/(math.log10(2))\n",
+      "n=math.log(M)\n",
+      "b=2**a\n",
+      "u=(a/x)*100\n",
+      "y=math.log10(b)\n",
+      "u1=(y/3.0)*100\n",
+      "\n",
+      "#result\n",
+      "print\"the number of bits required = \",round(a,2)         #number of bits\n",
+      "print\"The efficiency is \",round(u1,0),\"percent\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the number of bits required =  6.78\n",
+        "The efficiency is  68.0 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 Page no 407"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "m=10**7\n",
+      "Pe=10**-6         #error probability\n",
+      "\n",
+      "#calculation\n",
+      "a=m*Pe            #average number of errors\n",
+      "\n",
+      "#result\n",
+      "print\"expected number of errors =\",a\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "expected number of errors = 10.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page no 407"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Tb=1/9600.0                    #bit frequency\n",
+      "Pt=0.8                         #transmit power\n",
+      "\n",
+      "#calculation\n",
+      "Eb=Pt*Tb                       #energy per bit\n",
+      "\n",
+      "#result\n",
+      "print\"Energy per bit = \",round(Eb,7),\"J\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy per bit =  8.33e-05 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page no 410"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "bw=3*10**3                    #bandwidth\n",
+      "x=1023                        #signal-to-noise ratio\n",
+      "\n",
+      "#calculation\n",
+      "import math\n",
+      "C=bw*math.log(1+x)/math.log(2)           #capacity of telephone channel\n",
+      "\n",
+      "#result\n",
+      "print\"the capacity of a telephone channel =\",C,\"bit per second\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the capacity of a telephone channel = 30000.0 bit per second\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page no 411"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#given\n",
+      "Tb=1/(8.0*10**3)               #bit frequency\n",
+      "BWmin=1/(2.0*Tb)               #minimum bandwidth\n",
+      "\n",
+      "#result\n",
+      "print\"minimum bandwidth  = \",BWmin,\"Hz\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "minimum bandwidth  =  4000.0 Hz\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Modern_Electronics_Communication/screenshots/2_3.png b/Modern_Electronics_Communication/screenshots/2_3.png
new file mode 100755
index 00000000..f92e9c53
--- /dev/null
+++ b/Modern_Electronics_Communication/screenshots/2_3.png
diff --git a/Modern_Electronics_Communication/screenshots/3_2.png b/Modern_Electronics_Communication/screenshots/3_2.png
new file mode 100755
index 00000000..e55d1484
--- /dev/null
+++ b/Modern_Electronics_Communication/screenshots/3_2.png
diff --git a/Modern_Electronics_Communication/screenshots/4_2.png b/Modern_Electronics_Communication/screenshots/4_2.png
new file mode 100755
index 00000000..526dd950
--- /dev/null
+++ b/Modern_Electronics_Communication/screenshots/4_2.png
diff --git a/Principles_of_Power_System/README.txt b/Principles_of_Power_System/README.txt
index 2acded4b..2acded4b 100644..100755
--- a/Principles_of_Power_System/README.txt
+++ b/Principles_of_Power_System/README.txt
diff --git a/Principles_of_Power_System/chapter10_1.ipynb b/Principles_of_Power_System/chapter10_1.ipynb
deleted file mode 100644
index 50a55400..00000000
--- a/Principles_of_Power_System/chapter10_1.ipynb
+++ /dev/null
@@ -1,1146 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a3dcc0814e961bf1cd3db74adc66300f1f3d1872ea1c97bd01480f29b5cfe9db"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 10: Performance of Transmission\n",
-      "Lines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.1, Page Number: 233"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "pf = 0.8                              #power factor\n",
-      "Z = 10+15j                           #load impedance(ohm)\n",
-      "magVr = 33000                            #receiving end voltage(V)\n",
-      "P = 1100                             #poweer delivered(kW)\n",
-      "\n",
-      "#Calculation:\n",
-      "magI = P*1000/(magVr*pf)                    #line current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Vr = magVr+0j                              #V\n",
-      "I = magI*(math.cos(phy)-math.sin(phy)*1j)    #A\n",
-      "Vs = Vr + I*Z\n",
-      "\n",
-      "#Angle between Vs and Vr is\n",
-      "alpha = math.atan(Vs.imag/Vs.real)\n",
-      "phys = phy+alpha\n",
-      "pfs = math.cos(phys)\n",
-      "Pl = magI**2*Z.real/1000\n",
-      "Pi = P+Pl\n",
-      "n = P/Pi*100\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Sending end voltage is\",round(abs(Vs)),\"V\"\n",
-      "print \"(ii) sending end power factor is\",round(pfs,4),\"lagging\"\n",
-      "print \"(iii)Transmission efficiency is\",round(n,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Sending end voltage is 33709.0 V\n",
-        "(ii) sending end power factor is 0.7955 lagging\n",
-        "(iii)Transmission efficiency is 98.45 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.2, Page Number: 235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "a = 0.775                      #cross-section of conductor(cm**2)\n",
-      "n = 0.9                               #transmission efficiency\n",
-      "Pr = 200000                    #receiving end power(W)\n",
-      "pf = 1                         #power factor\n",
-      "V = 3300                       #line voltage(V)\n",
-      "ro = 1.725                     #specific resistance(micro_ohm-cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Ps = Pr/n                     #sending end power(W)\n",
-      "Pl = Ps-Pr                    #line loss(W)\n",
-      "I = Pr/(V*pf)                 #line current(A)\n",
-      "R = Pl/(2*I**2)                #resistance of 1 conductor(ohm)\n",
-      "l = R*a/(ro*10**-6)                     #length of conductor(cm)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The conductor length is\",round(l*10**-5,1),\"km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The conductor length is 13.6 km\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.3, Page Number: 235"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "pf = 0.8                        #power factor\n",
-      "Pr = 5000                       #receiving end power(kW)\n",
-      "Vr = 22                          #receiving end voltage(kV)\n",
-      "Z = 4+6j                        #impedance of each conductor(ohm)\n",
-      "\n",
-      "#Calculation:\n",
-      "phy = math.acos(pf)\n",
-      "magVrp = Vr*1000/3**0.5                 #sending end voltage/phase(kV)\n",
-      "magI = Pr*1000/(3*magVrp*0.8)        #Line current(A)\n",
-      "Vr = magVrp+1j \n",
-      "I = magI*(math.cos(phy)-math.sin(phy)*1j)\n",
-      "\n",
-      "#(i)Sending end voltage per phase:\n",
-      "Vs = magVrp+I*Z\n",
-      "Vsl = abs(Vs)*3**0.5                #V\n",
-      "\n",
-      "#(ii)\n",
-      "reg = (abs(Vs)-magVrp)/magVrp*100        #voltage regulation(%)\n",
-      "#(iii)\n",
-      "Pl = 3*abs(I)**2*Z.real/1000                  #line loss(kW)\n",
-      "n = Pr/(Pr+Pl)*100                  #transmission efficiency(%)\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Sending end voltage is\",round(abs(Vsl)/1000,3),\"kV\"\n",
-      "print \"(ii)Percentage regulation is\",round(reg,3),\"%\"\n",
-      "print \"(iii)Transmission efficiency\",round(n,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Sending end voltage is 23.942 kV\n",
-        "(ii)Percentage regulation is 8.825 %\n",
-        "(iii)Transmission efficiency 93.93 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.4, Page Number: 236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Pr = 15000                 #power delivered(kW)\n",
-      "Vl = 132                   #line voltage(kV)\n",
-      "Ro = 1                       #line resistance(ohm/km)\n",
-      "pf = 0.8                   #power factor\n",
-      "\n",
-      "#Calculation:\n",
-      "I = Pr/(3**0.5*Vl*pf)      #line current(A)\n",
-      "#the loss in the transmission is to be 5%.\n",
-      "Pl = 5*Pr/100                  #kW\n",
-      "R = Pl*1000/(3*I**2)        #line resistance(ohm)\n",
-      "d = R/Ro                    #line length(km)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Length of line is\",round(d,2),\"km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Length of line is 37.17 km\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.5, Page Number: 236"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "pf = 0.8                        #power factor\n",
-      "Pr = 3600                       #sending end power(kW)\n",
-      "magVs = 33                          #receiving end voltage(kV)\n",
-      "Z = 5.31+5.54j                        #impedance of each conductor(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = Z.real                      #ohm\n",
-      "X = Z.imag                      #ohm\n",
-      "phy = math.acos(0.8)\n",
-      "magVsp = magVs*1000/(3**0.5)     #V/phase\n",
-      "magVr = symbols('magVr')            #Receiving end voltage(V/phase)\n",
-      "magI = Pr*1000/(3*magVr*pf)            #line current(A)\n",
-      "\n",
-      "#(i)Using approximate expression for magVsp,\n",
-      "magVr1 = solve((magVr+magI*R*pf+magI*X*math.sin(phy))-magVsp,magVr)[1]\n",
-      "\n",
-      "#(ii)line current:\n",
-      "magI1 = Pr*1000/(3*magVr1*pf) \n",
-      "\n",
-      "#(iii)Efficiency\n",
-      "Pl = 3*magI1**2*R/1000                   #kW\n",
-      "n = Pr/(Pr+Pl)*100\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  The receiving end voltage \",round(magVr1*3**0.5/1000,2),\"V\"\n",
-      "print \"(ii) Line current is\",round(magI1,2),\"A\"\n",
-      "print \"(iii)Transmission efficiency is\",round(n,2),\"%\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The receiving end voltage  31.93 V\n",
-        "(ii) Line current is 81.36 A\n",
-        "(iii)Transmission efficiency is 97.15 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.6, Page Number: 237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Z = 6+8j                  #line impedance(ohm)\n",
-      "magVs = 120               #sending end voltages(kV)\n",
-      "magVr = 110                #receiving end voltaes(kV)\n",
-      "pf = 0.9                  #power factor\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = Z.real                    #ohm\n",
-      "X = Z.imag                    #ohm\n",
-      "magVsp = round(120*1000/3**0.5)         #V/phase\n",
-      "magVrp = round(110*1000/3**0.5)         #V/phase\n",
-      "phy = math.acos(pf)\n",
-      "magI = symbols('magI')        #line current(A)\n",
-      "magI1 = solve(magVrp+magI*R*math.cos(phy)+magI*X*math.sin(phy)-magVsp,magI)[0]\n",
-      "\n",
-      "#(i):\n",
-      "Po = 3*magVrp*round(magI1)*math.cos(phy)/1000       #kW\n",
-      "\n",
-      "#(ii):\n",
-      "pfs = (magVrp*math.cos(phy)+magI1*R)/magVsp\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Power output is\",round(Po),\"kW\"\n",
-      "print \"(ii)Sending end power factor\",round(pfs,2),\"lagging\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Power output is 111458.0 kW\n",
-        "(ii)Sending end power factor 0.88 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.7, Page Number: 237"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Z = 1.5+4j                     #impedance of the line(ohm)\n",
-      "magVr = 11000                  #receivig end voltage(V)\n",
-      "pf = 0.8                       #power factor\n",
-      "Pr = 5000                      #power delivered()\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = Z.real                    #ohm\n",
-      "X = Z.imag                    #ohm\n",
-      "magVrp = magVr/3**0.5           #V/phase\n",
-      "phy = math.acos(pf)\n",
-      "magI = Pr*1000/(3*magVrp)       #line current(A)\n",
-      "magVsp = magVrp+magI*R*pf+magI*X*math.sin(phy)    #Volt\n",
-      "reg = (magVsp - magVrp)/magVrp*100             #voltage regulation(%)\n",
-      "Pl = 3*magI**2*R/1000           #line losses(kW)\n",
-      "Po = Pr*pf                      #output power(W)\n",
-      "Pi = Po + Pl                    #Input Power(kW)\n",
-      "n = Po/Pi*100                       #efficiency(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The % regulation is\",round(reg,2),\"%\"\n",
-      "print \"The efficiency is\",round(n,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The % regulation is 14.88 %\n",
-        "The efficiency is 92.8 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.8, Page Number: 238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Pr = 1000                    #power delivered(kW)\n",
-      "pf = 0.8                     #power factor\n",
-      "r = 0.03                     #line resistance per phase(ohm/km)\n",
-      "L = 0.7                     #line inductance per phase(mH)\n",
-      "l = 16                       #line length(km)\n",
-      "magVr = 11000                 #receiving line voltage(V)\n",
-      "f = 50                       #power frequency(Hz)\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                      #line resistance(ohm)\n",
-      "X = 2*3.14*f*L/1000*l          #line reactance(ohm)\n",
-      "magVrp = round(magVr/3**0.5)       #receiving end (v/phase)\n",
-      "phy = math.acos(pf)\n",
-      "magI = round(Pr*1000/(3*magVrp*math.cos(phy)),1) #line current(A)\n",
-      "magVsp = magVrp+magI*R*math.cos(phy)+magI*X*math.sin(phy)\n",
-      "reg = (magVsp-magVrp)/magVrp*100        #Volt\n",
-      "Pl = round(3*magI**2*R/1000,1)           #line losses(kW)\n",
-      "Pi = Pr + Pl                    #Input Power(kW)\n",
-      "n = Pr/Pi*100                       #efficiency(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The % regulation is\",round(reg,2),\"%\"\n",
-      "print \"The efficiency is\",round(n,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The % regulation is 2.58 %\n",
-        "The efficiency is 99.38 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.9, Page Number: 238"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Pr = 2000                         #load power(kVA)\n",
-      "pf = 0.8                          #power factor\n",
-      "l = 20                            #line length(km)\n",
-      "r1 = 7.5;      x1 = 13.2  #resistance & reactance of transformer primary(ohm)\n",
-      "r2 = 0.35;      x2 = 0.65  #resistance & reactance of transformer secondary(ohm)\n",
-      "r = 0.4;       x = 0.5     ##resistance & reactance of line(ohm/km)\n",
-      "Vp = 33*1000                        #voltage at primary side(kV)\n",
-      "Vs = 6.6*1000                       #voltage at secondary side(kV)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = l*r                       #resistance of each conuctor(ohm)\n",
-      "X = l*x                       #reactance of each conductor(ohm)\n",
-      "phy = math.acos(pf)\n",
-      "#Let us transfer the impedance of transformer secondary\n",
-      "#to high tension side i.e., 33 kV side.\n",
-      "#Equivalent resistance of transformer referred to 33 kV side:\n",
-      "R1 = r1 + r2*(Vp/Vs)**2         #ohm\n",
-      "\n",
-      "#Equivalent resistance of transformer referred to 33 kV side:\n",
-      "X1 = x1+ x2*(Vp/Vs)**2          #ohm\n",
-      "\n",
-      "\n",
-      "Rt = R+R1                    #Total resistance of line and transformer(ohm)\n",
-      "Xt = X+X1                    #Total reactance of line and transformer(omh)\n",
-      "Vr = Vp/3**0.5               #receiving end voltage(V/phase)\n",
-      "I = round(Pr*1000/(3**0.5*Vp))      #line current(A)\n",
-      "Vs = Vr+I*Rt*math.cos(phy)+I*Xt*math.sin(phy)      #sending end voltage(V)\n",
-      "Vsl = 3**0.5*Vs                 #sending end line voltage(V)\n",
-      "pfs = (Vr*pf+I*Rt)/Vs           #sending end power factor\n",
-      "Pl = 3*I**2*Rt/1000             #line loss(kW)\n",
-      "Po = Pr*pf                  #output power(kW)\n",
-      "n = Po/(Po+Pl)*100           #transmission efficiency(%)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Sending end line voltage is\",round(Vsl/1000,1),\"V\"\n",
-      "print \"Sending end power factor is\",round(pfs,4)\n",
-      "print \"Transmission efficiency is\",round(n,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sending end line voltage is 35.6 V\n",
-        "Sending end power factor is 0.7826\n",
-        "Transmission efficiency is 94.72 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.10, Page Number: 241"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 0.25                        #resistance of line(ohm/km)\n",
-      "l = 100                         #line length(km)\n",
-      "x = 0.8                         #Reactance(ohm/km)\n",
-      "y = 14*10**-6                   #susceptance(siemen/km)\n",
-      "magVr = 66000                   #Receiving end line voltage(V)\n",
-      "Pr = 15000                      #power delivered(kW)\n",
-      "pf = 0.8                        #power factor(lagging)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                        #ohm\n",
-      "X = x*l                        #ohm\n",
-      "Y = y*l                        #siemen\n",
-      "magI = Pr*1000/(pf*magVr)           #line current(A)\n",
-      "phy = math.acos(pf)             #phasor angle\n",
-      "Vr = magVr+0j                   #Volt\n",
-      "Ir = round(magI*pf)-round(magI*math.sin(phy))*1j   #load current(A)\n",
-      "Ic = 1j*round(Y*magVr)\n",
-      "\n",
-      "#(i):\n",
-      "Is = Ir+Ic                      #Sending end current(A)\n",
-      "\n",
-      "#(ii):\n",
-      "delV = Is*(R+X*1j)              #voltage rop(V)\n",
-      "Vs = Vr+delV                    #sending end voltage(V)\n",
-      "reg = (abs(Vs)-magVr)/magVr*100         #voltage regulation(%)\n",
-      "\n",
-      "#phase angle between Vr & Ir:\n",
-      "theta1 = math.atan(Is.imag/Is.real)\n",
-      "\n",
-      "#phase angle between Vr & Is:\n",
-      "theta2 = math.atan(Vs.imag/Vs.real)\n",
-      "\n",
-      "phys = abs(theta1)+theta2\n",
-      "pfs = math.cos(phys)            #supply power factor\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  The sending end current is\",round(abs(Is)),\"A\"\n",
-      "print \"(ii) The sending end voltage is\",round(abs(Vs)),\"V\"\n",
-      "print \"(iii)Regulation is\",round(reg,2),\"%\"\n",
-      "print \"(iv) Supply power factor is\",round(pfs,2),\"lagging\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sending end current is 240.0 A\n",
-        "(ii) The sending end voltage is 79583.0 V\n",
-        "(iii)Regulation is 20.58 %\n",
-        "(iv) Supply power factor is 0.86 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.11, Page Number: 244"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 100                       #line length(km)\n",
-      "r = 0.1                       #resistance/km/phase(ohm)\n",
-      "xl = 0.2                      #reactance/km/phase(ohm)\n",
-      "b = 0.04*10**-4                #Capacitive susceptance/km/phase(siemen)\n",
-      "Pr = 10000                    #power delivered(kW)\n",
-      "Vrl = 66000                     #sending end line volt(V)\n",
-      "pf = 0.8                      #power factor(lagging)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                        #Total resistance/phase(ohm)\n",
-      "Xl = xl*l                       #Total reactance/phase(ohm)\n",
-      "Y = b*l                        #Capacitive susceptance(siemen)\n",
-      "magVr = round(Vrl/3**0.5)              #Receiving end voltage/phase(V)\n",
-      "magIr = round(Pr*1000/(3**0.5*Vrl*pf))   #Load current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Z = R+Xl*1j                     #Impedance per phase(ohm)\n",
-      "#(i) Taking receiving end voltage as the reference phasor,\n",
-      "Vr = magVr+0j\n",
-      "Ir = magIr*(pf-math.sin(phy)*1j)      #A\n",
-      "V1 = Vr+Ir*Z/2                 #Voltage across C(V)\n",
-      "Ic = 1j*Y*V1                  #Charging current(A)\n",
-      "Is = Ir+Ic                    #sending end current(A)\n",
-      "\n",
-      "\n",
-      "#(ii) Sending end voltage,\n",
-      "Vs = V1+Is*Z/2                 #V\n",
-      "magVsl = 3**0.5*abs(Vs)         #Line value of sending end voltage(V)\n",
-      "\n",
-      "#(iii) Referring to phasor diagram (iii),\n",
-      "theta1 = math.atan(Vs.imag/Vs.real)     #angle between Vr & Vs\n",
-      "theta2 = math.atan(abs(Is.imag/Is.real)) #angle between Vr & Is\n",
-      "phys = theta1+theta2                    #angle b/w Vs & Is\n",
-      "\n",
-      "pfs = math.cos(phys)                    #Sending end power factor\n",
-      "\n",
-      "#(iii):\n",
-      "Ps = 3*abs(Vs)*abs(Is)*pfs/1000             #Sending end power(kW)\n",
-      "n = Pr/Ps*100                          #Efficiency(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  The sending end current is\",round(abs(Is)),\"A\"\n",
-      "print \"(ii) Sending end voltage is\",round(magVsl/1000,3),\"kV\"\n",
-      "print \"(iii)Sending end power factor is\",round(pfs,3),\"lagging\"\n",
-      "print \"(iv) Transmission efficiency is\",round(n,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The sending end current is 100.0 A\n",
-        "(ii) Sending end voltage is 69.532 kV\n",
-        "(iii)Sending end power factor is 0.853 lagging\n",
-        "(iv) Transmission efficiency is 97.12 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.12, Page Number: 245"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 100                       #line length(km)\n",
-      "r = 0.2                       #resistance/km/phase(ohm)\n",
-      "xl = 0.4                      #reactance/km/phase(ohm)\n",
-      "b = 2.5*10**-6                #Capacitive susceptance/km/phase(siemen)\n",
-      "Pr = 20000                    #power delivered(kW)\n",
-      "Vrl = 110000                     #sending end line volt(V)\n",
-      "pf = 0.9                      #power factor\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                        #Total resistance/phase(ohm)\n",
-      "Xl = xl*l                       #Total reactance/phase(ohm)\n",
-      "Y = b*l                        #Capacitive susceptance(siemen)\n",
-      "magVr = round(Vrl/3**0.5)              #Receiving end voltage/phase(V)\n",
-      "magIr = round(Pr*1000/(3**0.5*Vrl*pf))   #Load current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Z = R+Xl*1j                     #Impedance per phase(ohm)\n",
-      "\n",
-      "#(i) Taking receiving end voltage as the reference phasor,\n",
-      "Vr = magVr+0j\n",
-      "Ir = magIr*(pf-(math.sin(phy))*1j)      #A\n",
-      "V1 = Vr+Ir*Z/2                 #Voltage across C(V)\n",
-      "Ic = 1j*Y*V1                  #Charging current(A)\n",
-      "Is = Ir+Ic                    #sending end current(A)\n",
-      "Vs = V1+Is*Z/2                 #V\n",
-      "magVsl = 3**0.5*abs(Vs)         #Line value of sending end voltage(V)\n",
-      "\n",
-      "#(ii):\n",
-      "Pl = 3*abs(Is)**2*R/2+3*magIr**2*R/2      #line loss(W)\n",
-      "n = Pr/(Pr+Pl/1000)*100           #efficiency\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The current and voltage at the sending end is\",round(magVsl/1000,2),\"kV\"\n",
-      "print \"(ii)Efficiency of transmission is\",round(n,2),\"%\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The current and voltage at the sending end is 116.75 kV\n",
-        "(ii)Efficiency of transmission is 96.26 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.13, Page Number: 247"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "l = 150                      #line length(km)\n",
-      "r = 0.1                       #resistance/km/phase(ohm)\n",
-      "xl = 0.5                      #reactance/km/phase(ohm)\n",
-      "b = 3*10**-6                #Capacitive susceptance/km/phase(siemen)\n",
-      "Pr = 50000                    #power delivered(kW)\n",
-      "Vrl = 110000                     #sending end line volt(V)\n",
-      "pf = 0.8                      #power factor\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                        #Total resistance/phase(ohm)\n",
-      "Xl = xl*l                       #Total reactance/phase(ohm)\n",
-      "Y = b*l                        #Capacitive susceptance(siemen)\n",
-      "magVr = round(Vrl/3**0.5)              #Receiving end voltage/phase(V)\n",
-      "magIr = round(Pr*1000/(3**0.5*Vrl*pf))   #Load current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Z = R+Xl*1j                     #Impedance per phase(ohm)\n",
-      "\n",
-      "Vr = magVr+0j\n",
-      "Ir = magIr*(pf-(math.sin(phy))*1j)      #A\n",
-      "Ic1 = Vr*1j*Y/2                         #Charging current at the load end(A)\n",
-      "Il = Ir+Ic1                       #line current(A)\n",
-      "Vs = Vr+Il*Z                      #Sending end voltage(V)\n",
-      "magVsl = abs(Vs)*3**0.5           #Line to line sending end voltage(V)\n",
-      "Ic2 = 1j*Vs*Y/2                    #Charging current at the sending end(A)\n",
-      "Is = Il+Ic2                      #Sending end current(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sending end voltage is\",round(magVsl/1000,2),\"V\"\n",
-      "print \"The sending end current is\",round(abs(Is),1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sending end voltage is 143.56 V\n",
-        "The sending end current is 306.3 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.14, Page Number: 248"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "l = 100                      #line length(km)\n",
-      "r = 0.1                       #resistance/km/phase(ohm)\n",
-      "xl = 0.5                      #reactance/km/phase(ohm)\n",
-      "b = 10*10**-6                #Capacitive susceptance/km/phase(siemen)\n",
-      "Pr = 20000                    #power delivered(kW)\n",
-      "Vrl = 66000                     #sending end line volt(V)\n",
-      "pf = 0.9                      #power factor\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                        #Total resistance/phase(ohm)\n",
-      "Xl = xl*l                       #Total reactance/phase(ohm)\n",
-      "Y = b*l                        #Capacitive susceptance(siemen)\n",
-      "magVr = round(Vrl/3**0.5)              #Receiving end voltage/phase(V)\n",
-      "magIr = round(Pr*1000/(3**0.5*Vrl*pf))   #Load current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Z = R+Xl*1j                     #Impedance per phase(ohm)\n",
-      "\n",
-      "Vr = magVr+0j\n",
-      "Ir = magIr*(pf-(math.sin(phy))*1j)      #A\n",
-      "Ic1 = round(magVr*Y/2)*1j                         #Charging current at the load end(A)\n",
-      "Il = Ir+Ic1                       #line current(A)\n",
-      "Vs = Vr+Il*Z                      #Sending end voltage(V)\n",
-      "magVsl = abs(Vs)*3**0.5           #Line to line sending end voltage(V)\n",
-      "Ic2 = 1j*Vs*Y/2                    #Charging current at the sending end(A)\n",
-      "Is = Il+Ic2                      #Sending end current(A)\n",
-      "\n",
-      "#(i):\n",
-      "theta1 = math.atan(Vs.imag/Vs.real)     #angle between Vr & Vs\n",
-      "theta2 = math.atan(abs(Is.imag/Is.real)) #angle between Vr & Is\n",
-      "phys = theta1+theta2                    #angle b/w Vs & Is\n",
-      "pfs = math.cos(phys)                    #Sending end power factor\n",
-      "\n",
-      "#(ii):\n",
-      "reg = (abs(Vs)-magVr)/magVr*100            #voltage regulation(%)\n",
-      "\n",
-      "#(iii):\n",
-      "Ps = 3*abs(Vs)*abs(Is)*pfs/1000         #sending end power(W)\n",
-      "n = Pr/Ps*100                          #transmission efficiency(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Sending end power factor is\",round(pfs,3),\"lagging\"\n",
-      "print \"(ii) Regulation is\",round(reg,2),\"%\"\n",
-      "print \"(iii)Transmission efficiency is\",round(n),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Sending end power factor is 0.906 lagging\n",
-        "(ii) Regulation is 15.15 %\n",
-        "(iii)Transmission efficiency is 95.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      " Example 10.15, Page Number: 254"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import cmath\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "l = 200                      #line length(km)\n",
-      "r = 0.16                       #resistance/km/phase(ohm)\n",
-      "xl = 0.25                      #reactance/km/phase(ohm)\n",
-      "b = 1.5*10**-6*1j                #Capacitive susceptance/km/phase(siemen)\n",
-      "Pr = 20000                    #power delivered(kW)\n",
-      "Vrl = 110000                     #sending end line volt(V)\n",
-      "pf = 0.8                      #power factor\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                        #Total resistance/phase(ohm)\n",
-      "Xl = xl*l                       #Total reactance/phase(ohm)\n",
-      "Y = b*l                        #Capacitive susceptance(siemen)\n",
-      "Z = R+Xl*1j                    #Series Impedance/phase(ohm)\n",
-      "magVr = Vrl/3**0.5            #Receiving end voltage per phase(V)\n",
-      "magIr = round(Pr*1000/(3**0.5*Vrl*pf))           #Receiving end current(A)\n",
-      "Vs = magVr*cmath.cosh((Y*Z)**0.5)+magIr*(Z/Y)**0.5*cmath.sinh((Z*Y)**0.5) #sending end voltage(V/phase)\n",
-      "Is = magVr*(Y/Z)**0.5*cmath.sinh((Y*Z)**0.5)+magIr*cmath.cosh((Y*Z)**0.5)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Sending end line-to-line voltage is\",round(3**0.5*abs(Vs)/1000,1),\"V\"\n",
-      "print \"Sending end current is\",round(abs(Is),1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sending end line-to-line voltage is 117.0 V\n",
-        "Sending end current is 131.5 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.16, Page Number: 258"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Z = 20+52j                    #Series line impedance/phase(ohm)\n",
-      "Y = 315*10**-6*1j             #Shunt admittance/phase(siemen)\n",
-      "pf = 0.85                      #power factor\n",
-      "Pr = 30000                    #receiving end power(kW)\n",
-      "magVrl = 132000                    #receiving end voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) Generalised constants of line,\n",
-      "A = 1+Z*Y/2\n",
-      "D = A\n",
-      "B = Z*(1+Z*Y/4)\n",
-      "C = Y\n",
-      "\n",
-      "#(ii) Sending end voltage,\n",
-      "magVr = magVrl/3**0.5          #V/phase\n",
-      "magIr = Pr*1000/(3**0.5*magVrl*pf)        #line current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Vr = magVr+0j\n",
-      "Ir = magIr*(math.cos(phy)-1j*math.sin(phy))\n",
-      "Vs = A*Vr+B*Ir\n",
-      "magVs = abs(Vs)                #sending end voltage(V/phase)\n",
-      "magVsl = 3**0.5*magVs          #Sending end line-to-line voltage(V)\n",
-      "\n",
-      "\n",
-      "#(iii) Regulation:\n",
-      "#At no load, Ir = 0,\n",
-      "magVro = abs(Vs/A)\n",
-      "reg = (magVro-magVr)/magVr*100           #regulation(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)The A, B, C and D constants of the line are\"\n",
-      "print \"   A =\",complex(round(A.real,3),round(A.imag,5))\n",
-      "print \"   B =\",complex(round(B.real,2),round(B.imag,2))\n",
-      "print \"   C =\",complex(round(C.real,6),round(C.imag,6))\n",
-      "print \"   D =\",complex(round(D.real,3),round(D.imag,5))\n",
-      "\n",
-      "print \"(ii) Sending end voltage is\",round(magVs*3**0.5/1000),\"kV\"\n",
-      "print \"(iii)Regulation of the line is\",round(reg,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The A, B, C and D constants of the line are\n",
-        "   A = (0.992+0.00315j)\n",
-        "   B = (19.84+51.82j)\n",
-        "   C = 0.000315j\n",
-        "   D = (0.992+0.00315j)\n",
-        "(ii) Sending end voltage is 143.0 kV\n",
-        "(iii)Regulation of the line is 9.25 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.17, Page Number: 259"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "A = cmath.rect(0.95,math.radians(1.4))\n",
-      "B = cmath.rect(96,math.radians(78))\n",
-      "C = cmath.rect(0.0015,math.radians(90))\n",
-      "D = cmath.rect(0.95,math.radians(1.4))\n",
-      "Pr = 50000                         #receiving end power(kW)\n",
-      "pf = 0.8                            #power factor\n",
-      "magVrl = 132000                     #receiving end voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "magVr = magVrl/3**0.5                #Receiving end voltage/phase(V)\n",
-      "magIr = Pr*1000/(3**0.5*magVrl*pf)        #line current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Vr = magVr+0j\n",
-      "Ir = magIr*(math.cos(phy)-1j*math.sin(phy))\n",
-      "Vs = A*Vr+B*Ir                    #Sending end voltage per phase\n",
-      "Is = C*Vr+D*Ir                    #Sending end current\n",
-      "Ic = Is-Ir                       #Charging current\n",
-      "#At no load, Ir = 0,\n",
-      "magVro = abs(Vs/A)\n",
-      "reg = (magVro-magVr)/magVr*100           #regulation(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Charging current is  (\",round(abs(Ic)),round(math.degrees(angle(Ic)),1),\")  A\"\n",
-      "print \"Regulation is\",round(reg),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Charging current is  ( 128.0 93.2 )  A\n",
-        "Regulation is 30.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 10.18, Page Number: 260"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "A = cmath.rect(0.98,math.radians(3))\n",
-      "B = cmath.rect(110,math.radians(75))\n",
-      "C = cmath.rect(0.0005,math.radians(80))\n",
-      "D = cmath.rect(0.98,math.radians(3))\n",
-      "MVA = 50                         #receiving end power\n",
-      "pf = 0.8                            #power factor\n",
-      "magVrl = 110                     #receiving end voltage(kV)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Pr = MVA*pf*10**6\n",
-      "magVr = round(magVrl/3**0.5,1)                #Receiving end voltage/phase(V)\n",
-      "magIr = round(MVA*10**6/(3**0.5*magVrl*1000),1)        #line current(A)\n",
-      "phy = math.acos(pf)\n",
-      "Vr = magVr*1000+0j\n",
-      "Ir = magIr*(math.cos(phy)-1j*math.sin(phy))\n",
-      "#(round(Ir.real)+1j*round(Ir.imag))\n",
-      "V1 = round((A*Vr).real)+math.ceil((A*Vr).imag)*1j\n",
-      "V2 = round((B*Ir).real)+math.ceil((B*Ir).imag)*1j\n",
-      "Vs = V1+V2                   #Sending end voltage per phase\n",
-      "theta1 = math.atan(Vs.imag/Vs.real)\n",
-      "Is = C*Vr+D*Ir                    #Sending end current\n",
-      "theta2 = math.atan(Is.imag/Is.real)\n",
-      "phys = theta2-theta1\n",
-      "Ps = 3*abs(Vs)*abs(Is)*math.cos(phys)   #Sending-end power(W)\n",
-      "n = Pr/Ps*100                      #efficiency(%)\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Sending end voltage is\",round(abs(Vs)),\"V\"\n",
-      "print \"(ii) Sending end current is\",round(abs(Is)),\"A\"\n",
-      "print \"(iii)Sending-end power is\",round(Ps/10**6,1),\"MW\"\n",
-      "print \"(iv) Transmission efficiency is\",round(n,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Sending end voltage is 87429.0 V\n",
-        "(ii) Sending end current is 246.0 A\n",
-        "(iii)Sending-end power is 48.7 MW\n",
-        "(iv) Transmission efficiency is 82.2 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter11_1.ipynb b/Principles_of_Power_System/chapter11_1.ipynb
deleted file mode 100644
index 2283037e..00000000
--- a/Principles_of_Power_System/chapter11_1.ipynb
+++ /dev/null
@@ -1,1185 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1fe403fe815d4353da157ba9f671fac5bd27f5ac7842de08db777bc86330d8e5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 11: Underground Cables"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.1, Page Number: 273"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1 = 0.5                       #conductor radius(cm)\n",
-      "l = 2000                       #conductor length(m)\n",
-      "rho = 5*10**12                 #Resistivity of insulation(ohm-m)\n",
-      "t = 0.4                        #insulation thickness(cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "r2 = r1+t                      #Internal sheath radius(cm)\n",
-      "R = rho*math.log(r2/r1)/(2*math.pi*l)      #Insulation resistance of cable(ohm)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Insulation resistance of cable is\",round(R/10**6),\"Mohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Insulation resistance of cable is 234.0 Mohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.2, Page Number: 274"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1 = 1.25                      #conductor radius(cm)\n",
-      "l = 1000                       #conductor length(m)\n",
-      "rho = 4.5*10**12                 #Resistivity of insulation(ohm-m)\n",
-      "R = 495*10**6                 #Cable insulation resistance(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let r2 cm be the internal sheath radius,\n",
-      "\n",
-      "r2 = r1*math.exp(R*2*math.pi*l/rho)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Insulation thickness is\",round(r2-r1,2),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Insulation thickness is 1.25 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.3, Page Number: 274"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1 = .10                      #conductor radius(cm)\n",
-      "l = 5000                       #conductor length(m)\n",
-      "r2 = 0.25                      #Internal sheath radius(cm)\n",
-      "R = 0.4*10**6                 #Cable insulation resistance(ohm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "rho = R*2*3.14*l/(math.log(r2/r1)*10**9)    #resistivity(ohm-m)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Resistivity of the insulating material is\",round(rho,2),\"* 10**9  ohm-m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Resistivity of the insulating material is 13.71 * 10**9  ohm-m\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.4, Page Number: 275"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "er = 4                       #relative permittivity\n",
-      "D = 1.8                       #internal sheath diameter(cm)\n",
-      "l = 1000                      #cable length(m)\n",
-      "d = 1                         #conductor diameter(cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "C = er*l/(41.4*math.log10(D/d))*10**-9     #Capacitance(F)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The capacitance of the cable is\",round(C*10**6,3),\"uF\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The capacitance of the cable is 0.378 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.5, Page Number: 276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "er = 4                       #relative permittivity\n",
-      "d = 10                       #core diameter(cm)\n",
-      "l = 1000                      #cable length(m)\n",
-      "t = 7                         #insulation thickness(cm)\n",
-      "Vl = 66000                     #line voltage(V)\n",
-      "f = 50                       #frequency(Hz)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "D = d+2*t                         #conductor diameter(cm)\n",
-      "C = 4*1000/(41.4*math.log10(D/d))*10**-3     #Capacitance(uF)\n",
-      "\n",
-      "Vp = Vl/3**0.5\n",
-      "I = 2*3.14*f*C*Vp*10**-6                   #Carging current(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The capacitance is\",round(C,3),\"uF\"\n",
-      "print \"Charging current of a single core cable is\",round(I,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The capacitance is 0.254 uF\n",
-        "Charging current of a single core cable is 3.04 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.6, Page Number: 276"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Calculation:\n",
-      "er = 3                       #relative permittivity\n",
-      "d = 2.5                       #core diameter(cm)\n",
-      "l = 4000                      #cable length(m)\n",
-      "t = 0.5                         #insulation thickness(cm)\n",
-      "Vl = 33000                     #line voltage(V)\n",
-      "f = 50                       #frequency(Hz)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "D = d+2*t                         #conductor diameter(cm)\n",
-      "C = er*l/(41.4*math.log10(D/d))*10**-3    #Capacitance(uF)\n",
-      "Vp = Vl/3**0.5\n",
-      "I = 2*3.14*f*C*Vp*10**-6                   #Carging current(A)\n",
-      "kVAR = 3*Vp*I                    #Total charging kVAR\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  The capacitance is\",round(C*10**3),\"* 10**-9 F\"\n",
-      "print \"(ii) Charging current of a single core cable is\",round(I,2),\"A\"\n",
-      "print \"(iii)Total charging kVAR is\",round(kVAR/1000,1),\"* 10**3 kVAR\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The capacitance is 1984.0 * 10**-9 F\n",
-        "(ii) Charging current of a single core cable is 11.87 A\n",
-        "(iii)Total charging kVAR is 678.3 * 10**3 kVAR\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.7, Page Number: 278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "V = 33                      #voltage of cable(V)\n",
-      "d = 1                          #conductor diameter(cm)\n",
-      "D = 4                          #sheath diameter(cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "gmax = 2*V/(d*math.log(D/d))         #maximum stress,rms(kV/cm)\n",
-      "gmin = gmax*d/D                      #minimum stress,rms(kV/cm)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The maximum and minimum stress in the insulation are\"\n",
-      "print \"gmax =\",round(gmax,2),\"kV/cm rms & gmin =\",round(gmin,2),\"kV/cm rms\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum and minimum stress in the insulation are\n",
-        "gmax = 47.61 kV/cm rms & gmin = 11.9 kV/cm rms\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.8, Page Number: 278"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "gmax = 40                    #kV/cm\n",
-      "gmin = 10                    #kV/cm\n",
-      "d = 2                        #conductor diameter(cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "D = gmax/gmin*d              #cm\n",
-      "t = (D-d)/2                  #thickness of insulation(cm)\n",
-      "V = gmax*d*math.log(D/d)/2\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Thickness of insulation is\",t,\"cm\"\n",
-      "print \"(ii)Operating voltage is\",round(V,2),\"kV rms\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Thickness of insulation is 3.0 cm\n",
-        "(ii)Operating voltage is 55.45 kV rms\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.9, Page Number: 279"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 11                         #voltage of cable(V)\n",
-      "a = 0.645                         #conductor area(cm**2)\n",
-      "D = 2.18                          #internal diameter of sheath(cm)\n",
-      "er = 3.5                           #relative permitivity\n",
-      "l = 1000                          #conductor length(m)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "d = (4*a/3.14)**0.5                      #Diameter of the conductor(cm)\n",
-      "gmax = 2*V/(d*math.log(D/d))         #Maximum electrostatic stress(kV/cm rms)\n",
-      "gmin = 2*V/(D*math.log(D/d))         #Minimum electrostatic stress(kV/cm rms)\n",
-      "C = er*l/(41.4*math.log10(D/d))*10**-9         #Capacitance of cable(F)\n",
-      "I = 2*3.14*f*C*V*1000                   #Carging current(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Maximum electrostatic stress in the cable is\",round(gmax,2),\"kV/cm rms\"\n",
-      "print \"(ii) Minimum electrostatic stress in the cable is\",round(gmin,2),\"kV/cm rms\"\n",
-      "print \"(iii)Capacitance of the cable per km length is\",round(C*10**6,2),\"* 10**-6 F\"\n",
-      "print \"(iv) Charging current is\",round(I,3),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Maximum electrostatic stress in the cable is 27.66 kV/cm rms\n",
-        "(ii) Minimum electrostatic stress in the cable is 11.5 kV/cm rms\n",
-        "(iii)Capacitance of the cable per km length is 0.22 * 10**-6 F\n",
-        "(iv) Charging current is 0.766 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.10, Page Number: 280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 50                       #Cable voltage(kV)\n",
-      "gmax = 40                    #Maximum permissible stress(kV/cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Vp = V*2**0.5                 #Peak value of cable voltage(kV)\n",
-      "d = 2*Vp/gmax                 #Most economical conductor diameter(cm)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The most economical value of diameter is\",round(d,2),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The most economical value of diameter is 3.54 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.11, Page Number: 280"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vl = 132                       #Cable voltage(kV)\n",
-      "gmax = 60                    #Maximum permissible stress(kV/cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Vph = Vl/3**0.5                #phase voltage(kV)\n",
-      "Vp = Vph*2**0.5                 #Peak value of cable voltage(kV)\n",
-      "d = 2*Vp/gmax                 #Most economical conductor diameter(cm)\n",
-      "D = 2.718*d                      #Internal diameter of sheath(cm)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Most economical conductor diameter is\",round(d,1),\"cm\"\n",
-      "print \"Internal diameter of sheath, D is\",round(D,2),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Most economical conductor diameter is 3.6 cm\n",
-        "Internal diameter of sheath, D is 9.76 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.12, Page Number: 282"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "d = 2                            #conductor diameter(cm)\n",
-      "e3 = 3                             #relative permittivity\n",
-      "e2 = 4\n",
-      "e1 = 5\n",
-      "D = 8                              #overall diameter(cm)\n",
-      "gmax = 40                           #kV/cm\n",
-      "\n",
-      "#Calculation:\n",
-      "#Graded cable: As the maximum stress in the three dielectrics is the same,\n",
-      "d1 = e1*d/e2                      #diameter of 1st layer(cm)\n",
-      "d2 = e1*d/e3                      #diameter of 2nd layer(cm)\n",
-      "#Permissible peak voltage for the cable:\n",
-      "Vp1 = gmax/2*(d*math.log(d1/d)+d1*math.log(d2/d1)+d2*math.log(D/d2))  #kV\n",
-      "Vs1 = Vp1/2**0.5                   #Safe working voltage (r.m.s.) for cable(kV)\n",
-      "\n",
-      "#Ungraded cable:\n",
-      "Vp2 = gmax/2*d*log(D/d)             #kV\n",
-      "Vs2 = Vp2/(2**0.5)                 #kV\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"For Graded cable, safe working voltage is\",round(Vs1,2),\"kV\"\n",
-      "print \"For Ungraded cable, safe working voltage is\",round(Vs2,1),\"kV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "For Graded cable, safe working voltage is 57.75 kV\n",
-        "For Ungraded cable, safe working voltage is 39.2 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.13, Page Number: 283"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "d = 3                            #conductor diameter(cm)\n",
-      "e2 = 4                       #relative permittivity\n",
-      "e1 = 5\n",
-      "D = 9                              #overall diameter(cm)\n",
-      "g1max = 30                           #kV/cm\n",
-      "g2max = 20                         #kV/cm\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "d1 = g1max/g2max*e1*d/e2              #cm\n",
-      "t1 = (d1-d)/2                   #Radial thickness of inner dielectric(cm)\n",
-      "t2 = (D-d1)/2                   #Radial thickness of outer dielectric(cm)\n",
-      "Vp = g1max/2*d*math.log(d1/d)+g2max/2*d1*math.log(D/d1)\n",
-      "Vsf = Vp/2**0.5                   #Safe working voltage(r.m.s.)for the cable(kV)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Radial thickness of inner dielectric is\",round(t1,3),\"cm\"\n",
-      "print \"Radial thickness of outer dielectric is\",round(t2,2),\"cm\"\n",
-      "print \"Safe working voltage (r.m.s.) for the cable is\",round(Vsf,2),\"kV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Radial thickness of inner dielectric is 1.313 cm\n",
-        "Radial thickness of outer dielectric is 1.69 cm\n",
-        "Safe working voltage (r.m.s.) for the cable is 38.7 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.14, Page Number: 284"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "e1 = 5                          #relative permittivity\n",
-      "e2 = 3\n",
-      "t = 1                          #thickness(cm)\n",
-      "d = 2                          #core diameter(cm)\n",
-      "V = 66                         #cable voltage(kV)\n",
-      "\n",
-      "#Calculation:\n",
-      "d1 = d+2*t\n",
-      "D = d+4*t                     #total diameter of cable(cm)\n",
-      "Vpk = V/3**0.5*2**0.5            #Peak voltage per phase(kV)\n",
-      "g1max = 2*Vpk/(d*(math.log(d1/d)+e1/e2*math.log(D/d1)))      #kV/cm\n",
-      "g2max = 2*Vpk/(d1*(e2/e1*math.log(d1/d)+math.log(D/d1)))      #kV/cm\n",
-      "\n",
-      "#Result:\n",
-      "print \"Maximum stresses in two dielectrics are:\"\n",
-      "print \"g1max =\",round(g1max,2),\"kV/cm     g2max =\",round(g2max,2),\"kV/cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum stresses in two dielectrics are:\n",
-        "g1max = 39.37 kV/cm     g2max = 32.8 kV/cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.15, Page Number: 285"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "d = 2                           #cm\n",
-      "d1 = 3.1                        #cm\n",
-      "d2 = 4.2                        #cm\n",
-      "D = 5.3                         #cm\n",
-      "V = 66                          #cable voltage(kV)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Vpk = V/3**0.5*2**0.5           #peak voltage/phase(kV)\n",
-      "#let V1,V2 & V3 are the voltages at different grades.\n",
-      "V1,V2,V3 = symbols('V1 V2 V3')\n",
-      "g1max = V1/(d/2*math.log(d1,d))\n",
-      "g2max = V2/(d1/2*math.log(d2,d1))\n",
-      "g3max = V3/(d2/2*math.log(D,d2))\n",
-      "\n",
-      "#As the maximum stress in the layers is the same,\n",
-      "#\u2234     g1max = g2max = g3max\n",
-      "#or    2\u00b728 V1 = 2\u00b712 V2 = 2\u00b704 V3\n",
-      "#\u2234     V2 = (2\u00b728/2\u00b712) V1 = 1\u00b7075 V1\n",
-      "#and   V3 = (2\u00b728/2\u00b704) V1 = 1\u00b7117 V1\n",
-      "#Now   V1 + V2 + V3 = Vpk\n",
-      "#or    V1 + 1\u00b7075 V1 + 1\u00b7117 V1 = 53\u00b79\n",
-      "V1 = 53.9/3.192\n",
-      "V2 = 1.075*V1\n",
-      "V1s = Vpk-V1                   #Voltage on first intersheath(near to core)(kV)\n",
-      "V2s = Vpk-V1-V2                #Voltage on second intersheath(kV)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage on first intersheath is\",round(V1s,2),\"kV\"\n",
-      "print \"Voltage on second intersheath is\",round(V2s,2),\"kV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage on first intersheath is 37.0 kV\n",
-        "Voltage on second intersheath is 18.85 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.16, Page Number: 286"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "d = 2                       #core diameter(cm)\n",
-      "D = 5.3                       #cm\n",
-      "V = 66                      #cable voltage(kV)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Vpk = V/3**0.5*2**0.5           #peak voltage/phase(kV)\n",
-      "#(i) Positions of intersheaths.\n",
-      "\n",
-      "#Let the diameters of intersheaths are d1 and d2 cm respectively.\n",
-      "#Let V1 = voltage b/w conductor & intersheath 1\n",
-      "#    V2 = voltage b/w intersheaths 1 and 2\n",
-      "#    V3 = voltage b/w intersheath 2 & outer lead sheath\n",
-      "\n",
-      "\n",
-      "#Given the maximum stress in the three layers is the same,\n",
-      "#we get the relation as given below:\n",
-      "#            d1**2 = d * d2 = 2*d2      [\u2235 d = 2 cm]\n",
-      "#or          d2 = d1**2/2\n",
-      "#and         d1*d2 = D * d = 5\u00b73 \u00d7 2 = 10.6 cm\n",
-      "#or          d1 * d1**2/2 = 10\u00b76\n",
-      "d1 = 21.2**(1/3)                       #cm\n",
-      "d2 = d1**2/2                           #cm\n",
-      "\n",
-      "\n",
-      "#(ii) Voltage on intersheaths,\n",
-      "#        V = V1 + V2 + V3\n",
-      "#or      53\u00b79 = V1+d1/d*V1+d2/d*V1\n",
-      "#             = 4.28*V1\n",
-      "V1 = 53.9/4.28                        #kV\n",
-      "V2 = d1/d*V1                          #kV\n",
-      "\n",
-      "#(iii) Stresses in dielectrics,\n",
-      "gmax = V1/(d/2*math.log(d1/d))            #max stress(kV/cm)\n",
-      "gmin = V1/(d1/2*math.log(d1/d))            #min stress(kV/cm\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Positions of intersheaths are:\"\n",
-      "print \"\\td1 =\",round(d1,2),\"cm d2 =\",round(d2,1),\"cm\"\n",
-      "print \"(ii) Voltage on the intersheaths are:\"\n",
-      "print \"\\tVoltage on first intersheath is\",round(Vpk-V1,2),\"kV\"\n",
-      "print \"\\tVoltage on second intersheath is\",round(Vpk-V1-V2,1),\"kV\"\n",
-      "print \"(iii) Maximum and minimum stress are:\"\n",
-      "print \"\\tgmax =\",round(gmax),\"kV/cm\\tgmin =\",round(gmin,2),\"kV/cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Positions of intersheaths are:\n",
-        "\td1 = 2.77 cm d2 = 3.8 cm\n",
-        "(ii) Voltage on the intersheaths are:\n",
-        "\tVoltage on first intersheath is 41.3 kV\n",
-        "\tVoltage on second intersheath is 23.9 kV\n",
-        "(iii) Maximum and minimum stress are:\n",
-        "\tgmax = 39.0 kV/cm\tgmin = 28.01 kV/cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.17, Page Number: 289"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "c = 0.3                          #capacitance per kilometre(uF/km)\n",
-      "V = 11                           #line voltage(kV)\n",
-      "l = 5                            #length of the cable(km)\n",
-      "f = 50                           #Hz\n",
-      "\n",
-      "#Calculation:\n",
-      "C3 = c*l                         #capacitance between a pair of cores with third core\n",
-      "                                 #earthed for a length of 5 km (uF)\n",
-      "Vph = V*1000/3**0.5              #phase voltage(V)\n",
-      "#core to neutral capacitance Cn of this cable is given by :\n",
-      "Cn = 2*C3                        #uF\n",
-      "Ic = 2*math.pi*f*Vph*Cn*10**-6\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \" The charging current is\",round(Ic,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The charging current is 5.99 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.18, Page Number: 290"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "V = 66                           #line voltage(kV)\n",
-      "C1 = 12.6                         #uF\n",
-      "C2 = 7.4                          #uF\n",
-      "f = 50                           #Hz\n",
-      "\n",
-      "#Calculation:\n",
-      "Vph = V*1000/3**0.5               #phase voltage(V)\n",
-      "Ce = C1/3                         #core-earth capacitances(uF)\n",
-      "Cc = (C2-Ce)/2                    #core-core capacitances(uF)\n",
-      "Cn = Ce+3*Cc                      #Core to neutral capacitance(uF)\n",
-      "Ic = 2*math.pi*f*Vph*Cn*10**-6\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The charging current is\",round(Ic,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The charging current is 107.74 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.19, Page Number: 290"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "c = 0.18                          #capacitance per kilometre(uF/km)\n",
-      "V = 3300                           #line voltage(V)\n",
-      "l = 20                            #length of the cable(km)\n",
-      "f = 50                           #Hz\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "C3 = c*l                         #capacitance between a pair of cores with third core\n",
-      "                                 #earthed for a length of 20 km (uF)\n",
-      "Vph = V/3**0.5                  #phase voltage(V)\n",
-      "Cn = 2*C3                       #Core to neutral capacitance(uF)\n",
-      "Ic = 2*math.pi*f*Cn*Vph*10**-6                   #charging current(A)\n",
-      "kVA = 3*Vph*Ic/1000                    #kVA taken by the cable\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The kVA taken by 20 km long cable is\",round(kVA,2),\"kVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The kVA taken by 20 km long cable is 24.63 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.20, Page Number: 292"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "k = 5                #thermal resistivity of the dielectric(ohm-m)\n",
-      "S2 = 0.45        #thermal resistance b/w the sheath and the ground surface\n",
-      "R = 110           #electrical resistance of the cable(u-ohm)\n",
-      "r = 15                #core radius(mm)\n",
-      "t = 40                #dielectric thickness(mm)\n",
-      "T = 55                  #temperature(deg. C)\n",
-      "n = 1                   #no. of conductors\n",
-      "\n",
-      "#Calculation:\n",
-      "r1 = r+t                  #mm\n",
-      "S1 = k/(2*math.pi)*math.log(r1/r)         #ohm/m\n",
-      "S = S1+S2                          #ohm/m\n",
-      "I = (T/(n*R*10**-6*S))**0.5               #current loading(A)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Maximum permissible current loading is\",round(I,1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum permissible current loading is 580.5 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.21, Page Number: 296"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Q = 15                            #ohm\n",
-      "P = 45                             #ohm\n",
-      "l = 300                           #length of faulty cable(m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "L = 2*l                         #loop length(m)\n",
-      "d = Q/(P+Q)*L             #Distance of the fault point from test end(m)\n",
-      "\n",
-      "#Result:\n",
-      "print \" The distance of the fault point from the test end is\",d,\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The distance of the fault point from the test end is 150.0 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      " Example 11.22, Page Number: 297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 500                           #length of faulty cable(m)\n",
-      "#P:Q = 3\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let:\n",
-      "P = 3;                  Q = 1            #ohm\n",
-      "#then,\n",
-      "d = Q/(P+Q)*2*l             #Distance of the fault point from test end(m)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The distance of the fault from the testing end of cables\",d,\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The distance of the fault from the testing end of cables 250.0 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.23, Page Number: 297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 500                           #length of faulty cable(m)\n",
-      "rp = 0.001                        #resistance of cable(ohm/m)\n",
-      "rq = 0.00225                      #resistance of sound cable(ohm/m)\n",
-      "#P:Q = 2.75:1\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let:\n",
-      "P = 2.75;                  Q = 1            #ohm\n",
-      "#then,\n",
-      "R = rp*l+rq*l                    #Resistance of loop(ohm)\n",
-      "X = Q/(P+Q)*R       #Resistance of faulty cable from test end upto fault point(ohm)\n",
-      "d = X/rp            #Distance of the fault point from test end(m)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The distance of the fault from the testing end of cables\",round(d),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The distance of the fault from the testing end of cables 433.0 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 11.24, Page Number: 297"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "S = 200                        #ohm\n",
-      "r = 20                         #resistance per km(ohm)\n",
-      "l = 20                         #length of cable(km)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#       R+X = 20*(20+20)        #ohm\n",
-      "#         P = Q\n",
-      "X = (800-200)/2                 #ohm\n",
-      "d = X/r                         #m\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The distance of the fault from the test end is\",d,\"km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The distance of the fault from the test end is 15.0 km\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter13_1.ipynb b/Principles_of_Power_System/chapter13_1.ipynb
deleted file mode 100644
index 9f2fa9af..00000000
--- a/Principles_of_Power_System/chapter13_1.ipynb
+++ /dev/null
@@ -1,1972 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9a20e0c71265585c31a8df4c18dfac853044b054341d1794943bfacb3c219678"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 13: D.C. Distribution"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.1, Page number: 313"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration:\n",
-      "#given a 2-wire d.c. distributor cable AB\n",
-      "\n",
-      "r = 0.01*2                 #Resistance per 1000m of distributor\n",
-      "Va = 300                  #p.d at point A(V)\n",
-      "AC = 500                      #m\n",
-      "CD = 500                      #m\n",
-      "DE = 600                      #m\n",
-      "EB = 400                     #m\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Rac = r*AC/1000            #Resistance of section AC\n",
-      "Rcd = r*CD/1000            #Resistance of section CD\n",
-      "Rde = r*DE/1000            #Resistance of section DE\n",
-      "Reb = r*EB/1000            #Resistance of section EB\n",
-      "Ieb = 50                    #current in branch EB(A)\n",
-      "Ide = Ieb+200               #current in branch DE(A)\n",
-      "Icd = Ide+150               #current in branch CD(A)\n",
-      "Iac = Icd+100               #current in branch AC(A)\n",
-      "Vc = Va - Iac*Rac            #P.D. at load point C(V)\n",
-      "Vd = Vc - Icd*Rcd            #P.D. at load point D(V)\n",
-      "Ve = Vd - Ide*Rde            #P.D. at load point E(V)\n",
-      "Vb = Ve - Ieb*Reb            #P.D. at load point B(V)\n",
-      "\n",
-      "#Result:\n",
-      "print \"P.D. at load point C is\",Vc,\"V\"\n",
-      "print \"P.D. at load point D is\",Vd,\"V\"\n",
-      "print \"P.D. at load point E is\",Ve,\"V\"\n",
-      "print \"P.D. at load point B is\",Vb,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "P.D. at load point C is 295.0 V\n",
-        "P.D. at load point D is 291.0 V\n",
-        "P.D. at load point E is 288.0 V\n",
-        "P.D. at load point B is 287.6 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.2, Page number: 314"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "AB = 300                 #m\n",
-      "AC = 40                   #m\n",
-      "CD = 60                  #m\n",
-      "DE = 50                    #m\n",
-      "EF = 100                 #m\n",
-      "FB = 50                   #m\n",
-      "Vm = 10                  #max. permissible volatge(V)\n",
-      "rho = 1.78*10**-8            #resistivity of cable(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Iac = 220                #A\n",
-      "Icd = 190                #A\n",
-      "Ide = 150                #A\n",
-      "Ief = 50                 #A\n",
-      "#Suppose that resistance of 100m length of the distributor is r ohms.\n",
-      "l = 100                  #m\n",
-      "r = symbols('r')\n",
-      "Rac = AC*r/100            #ohm\n",
-      "Rcd = CD*r/100            #ohm\n",
-      "Rde = DE*r/100            #ohm\n",
-      "Ref = EF*r/100            #ohm\n",
-      "\n",
-      "Vt = Iac*Rac+Icd*Rcd+Ide*Rde+Ief*Ref\n",
-      "r1 = solve(Vt-Vm,r)[0]\n",
-      "a = float(rho*l/(r1/2))            #cm**2\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"X-sectional area of conductor is\",round(a*10**4,3),\"cm**2\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "X-sectional area of conductor is 1.164 cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.3, Page number: 315"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration:\n",
-      "r1 = 0.25                #resistance of trolley wire(ohm/km)\n",
-      "r2 = 0.03                #resistance of of track(ohm/km)\n",
-      "V = 600                  #sub-station voltage(V)\n",
-      "SA = 2                    #km\n",
-      "AB = 4                     #km\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r1+r2                #ohm/km\n",
-      "Isa = 40+20                #A\n",
-      "Iab = 20                   #A\n",
-      "Vsa = Isa*R*SA            #V\n",
-      "Vab = Iab*R*AB              #V\n",
-      "Va = V-Vsa                #Voltage across tram A\n",
-      "Vb = Va-Vab                #Voltage across tram A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage across tram A is\",Va,\"V\"\n",
-      "print \"Voltage across tram B is\",Vb,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage across tram A is 566.4 V\n",
-        "Voltage across tram B is 544.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.4, Page number: 315"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "a = 0.27                    #cross-sectional area of each conductor(cm**2)\n",
-      "V = 250                     #supply voltage at point A(V)\n",
-      "rho = 1.78*10**-6           #Resistivity of the wire(ohm-cm)\n",
-      "AB = 75                      #m\n",
-      "BC = 100                    #m\n",
-      "CD = 50                      #m\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Icd = 20                  #curent in CD(A)\n",
-      "Ibc = 20+15                 #current in BC(A)\n",
-      "Iab = 20+15+12            #current in AB(A)\n",
-      "R = round(rho*BC*100/a,3)       #Single-core resistance of the section of 100 m length(ohm)\n",
-      "Rab = R*AB/100*2        #ohm\n",
-      "Rbc = R*BC/100*2        #ohm\n",
-      "Rcd = R*CD/100*2        #ohm\n",
-      "Vb = V-Iab*Rab                #Voltage at tapping point B(V)\n",
-      "Vc = Vb-Ibc*Rbc                #Voltage at tapping point C(V)\n",
-      "Vd = Vc-Icd*Rcd                #Voltage at tapping point D(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)The current in various sections of the conductor are:\"\n",
-      "print \"\\tIcd =\",Icd,\"A  \\tIbc =\",Ibc,\"A\\tIab =\",Iab,\"A\"\n",
-      "print \"\\n(ii)The resistances of the various sections are :\"\n",
-      "print \"\\tRab =\",Rab,\"ohm\\tRbc =\",Rbc,\"ohm\\t\",Rcd,\"ohm\"\n",
-      "print \"\\n(iii)Voltage at tapping point B is\",round(Vb,2),\"V\"\n",
-      "print \"     Voltage at tapping point C is\",round(Vc,2),\"V\"\n",
-      "print \"     Voltage at tapping point D is\",round(Vd,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The current in various sections of the conductor are:\n",
-        "\tIcd = 20 A  \tIbc = 35 A\tIab = 47 A\n",
-        "\n",
-        "(ii)The resistances of the various sections are :\n",
-        "\tRab = 0.099 ohm\tRbc = 0.132 ohm\t0.066 ohm\n",
-        "\n",
-        "(iii)Voltage at tapping point B is 245.35 V\n",
-        "     Voltage at tapping point C is 240.73 V\n",
-        "     Voltage at tapping point D is 239.41 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.5, Page number: 317"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "i = 2                        #A/m\n",
-      "l = 200                      #length of distributor(m)\n",
-      "r1 = 0.3                     #Resistance of single wire is 0\u00b73 ohm/km.\n",
-      "x = 150                       #m\n",
-      "\n",
-      "#Calculation:\n",
-      "r = 2*r1/1000                #Resistance of distributor per metre run(ohm)\n",
-      "V = i*r*(l*x-x**2/2)           #volt\n",
-      "I = i*l                      #Total current entering the distributor(A)\n",
-      "R = r*l                      #Total resistance of the distributor(ohm)\n",
-      "V1 = 1/2*I*R                 #Total drop over the distributor(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)The voltage drop upto a distance of 150m from \"\n",
-      "print \"\\tthe feeding point is\",V,\"V\"\n",
-      "print \"(ii)The maximum voltage drop is\",V1,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The voltage drop upto a distance of 150m from \n",
-        "\tthe feeding point is 22.5 V\n",
-        "(ii)The maximum voltage drop is 24.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.6, Page number: 317"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "i = 0.4                        #A/m\n",
-      "l = 500                      #length of distributor(m)\n",
-      "Vm = 10                       #maximum permissible voltage drop(V)\n",
-      "rho = 1.7*10**-6             #ohm-cm\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I = i*l                   #Current entering the distributor(A)\n",
-      "#Let r ohm be the resistance per metre length of the distributor (both wires).\n",
-      "r = 2*Vm/(I*l)            #resistance per metre length(ohm)\n",
-      "a = rho*100/(r/2)           #Area of cross-section of the distributor conductor(ohm)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The cross-sectional area of the distributor conductor is\",a,\"cm**2\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The cross-sectional area of the distributor conductor is 1.7 cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.7, Page number: 318"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "i = 1.6                        #A/m\n",
-      "l = 250                      #length of distributor(m)\n",
-      "r1 = 0.0002                     #Resistance of single wire is 0\u00b73 ohm/m.\n",
-      "x = 150                       #m\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I = i*l                    #Current entering the distributor(A)\n",
-      "r = 2*r1             #Resistance of the distributor per metre run\n",
-      "R = r*l                    #Total resistance of distributor(ohm)\n",
-      "V1 = 1/2*I*R            #Voltage drop over the entire distributor(ohm)\n",
-      "#the voltage necessary at feed point to maintain 250 V at the far end\n",
-      "V2 = V1+250               #Voltage at feeding point(V)\n",
-      "#Voltage drop upto a distance of 150 metres from feeding point\n",
-      "V3 = i*r*(l*125-125**2/2)          #Volt\n",
-      "##Voltage necessary to maintain 250V at the mid-point of the distributor.\n",
-      "V4 = 250+V3\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Voltage drop over the entire distributor is\",V2,\"V\"\n",
-      "print \"(ii)The necessary voltage is\",V4,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage drop over the entire distributor is 270.0 V\n",
-        "(ii)The necessary voltage is 265.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.9, Page number: 319"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "i = 0.75                  #current loading(A/m)\n",
-      "l = 300                   #distributor length(m)\n",
-      "x = 200                    #m\n",
-      "r = 0.00018             #resistance of distributor(go and return)(ohm/m)\n",
-      "V = 250                  #voltage fed at 1 end(V)\n",
-      "\n",
-      "#Calculation:\n",
-      "V1 = i*r*(l*x-x**2/2)            #Voltage drop(V)\n",
-      "#Voltage at a distance of 200 m from supply end\n",
-      "V2 = V-V1              #V\n",
-      "P = i**2*r*l**3/3            #Power loss in the distributor(W)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Volatage at a distance of 200m is\",V2,\"V\"\n",
-      "print \"Power loss in the distributor is\",P,\"W\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Volatage at a distance of 200m is 244.6 V\n",
-        "Power loss in the distributor is 911.25 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.10, Page number: 321"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "AB = 600                 #m\n",
-      "Va = 220                 #end voltage(V)\n",
-      "a = 1                   #cross-section of conductor(cm**2)\n",
-      "rho = 1.7*10**-6         #resistvity of conductor(ohm-cm)\n",
-      "\n",
-      "#Calculation:\n",
-      "\n",
-      "#Let Ia amperes be the current supplied from the feeding\n",
-      "#end A.\n",
-      "r = 2*rho*100/a             #ohm\n",
-      "Rac = r*100               #ohm\n",
-      "Rcd = r*150               #ohm\n",
-      "Rde = r*150               #ohm\n",
-      "Ref = r*100               #ohm\n",
-      "Rfb = r*100               #ohm\n",
-      "#Voltage at B = Voltage at A \u2212 Drop over length AB\n",
-      "Ia = symbols('Ia')\n",
-      "Ia1 = solve(Va-(Ia*Rac+(Ia-20)*Rcd+(Ia-60)*Rde+(Ia-110)*Ref+(Ia-140)*Rfb)-Va,Ia)[0]\n",
-      "#We can see that currents are coming to load point E from \n",
-      "#both sides i.e. from point D and point F. \n",
-      "#So, E will be the point of minimum potential.\n",
-      "\n",
-      "Ve = Va-(61.7*Rac+41.7*Rcd+1.7*Rde)        #Minimum consumer voltage(V)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Minimum consumer voltage is\",round(Ve,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum consumer voltage is 215.69 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.11, Page number: 322"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "AB = 200                 #m\n",
-      "Va = 230                 #1 end voltage(V)\n",
-      "Vb = 235                 #2 end voltage(V)\n",
-      "r1 = 0.3                #The resistance per km of one conductor(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let Ia amperes be the current supplied from the feeding\n",
-      "#end A.\n",
-      "r = 2*r1                #Resistance of 1000m length of distributor(both wires)\n",
-      "Rac = r*50/1000               #ohm\n",
-      "Rcd = r*25/1000               #ohm\n",
-      "Rde = r*25/1000              #ohm\n",
-      "Ref = r*50/1000               #ohm\n",
-      "Rfb = r*50/1000             #ohm\n",
-      "Ia = symbols('Ia')\n",
-      "Ia1 = round(solve(Va-(Ia*Rac+(Ia-25)*Rcd+(Ia-75)*Rde+(Ia-105)*Ref+(Ia-145)*Rfb)-Vb,Ia)[0],2)\n",
-      "Iac = Ia1                     #Current in section AC(A)\n",
-      "Icd = Ia1-25                  #Current in section CD(A)\n",
-      "Ide = Ia1-75                  #Current in section DE(A)\n",
-      "Ief = Ia1-105                  #Current in section EF(A)\n",
-      "Ifb = Ia1 -145                 #Current in section FB(A)\n",
-      "#The currents are coming to load point D from both sides of \n",
-      "#the distributor. Therefore, load point D is the point of minimum potential.\n",
-      "Vd = Va-(Iac*Rac+Icd*Rcd)         #volt\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Currents in various sections of the distributor are:\"\n",
-      "print \"\\tCurrent in section AC is\",Iac,\"A\"\n",
-      "print \"\\tCurrent in section AC is\",Icd,\"A\"\n",
-      "print \"\\tCurrent in section AC is\",Ide,\"A\"\n",
-      "print \"\\tCurrent in section AC is\",Ief,\"A\"\n",
-      "print \"\\tCurrent in section AC is\",Ifb,\"A\"\n",
-      "print \"\\n(ii)Voltage at D is\",round(Vd,3),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Currents in various sections of the distributor are:\n",
-        "\tCurrent in section AC is 33.33 A\n",
-        "\tCurrent in section AC is 8.33 A\n",
-        "\tCurrent in section AC is -41.67 A\n",
-        "\tCurrent in section AC is -71.67 A\n",
-        "\tCurrent in section AC is -111.67 A\n",
-        "\n",
-        "(ii)Voltage at D is 228.875 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.12, Page number: 323"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "AB = 600                 #m\n",
-      "Va = 440                 #1 end voltage(V)\n",
-      "Vb = 430                 #2 end voltage(V)\n",
-      "r1 = 0.01                #The resistance per 100m of one conductor(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let Ia amperes be the current supplied from the feeding\n",
-      "#end A.\n",
-      "r = 2*r1                #Resistance of 1000m length of distributor(both wires)\n",
-      "Rac = r*150/100               #ohm\n",
-      "Rcd = r*150/100               #ohm\n",
-      "Rde = r*50/100              #ohm\n",
-      "Ref = r*100/100               #ohm\n",
-      "Rfb = r*150/100             #ohm\n",
-      "Ia = symbols('Ia')\n",
-      "Ia1 = round(solve(Va-(Ia*Rac+(Ia-100)*Rcd+(Ia-300)*Rde+(Ia-550)*Ref+(Ia-850)*Rfb)-Vb,Ia)[0],2)\n",
-      "Iac = Ia1                     #Current in section AC(A)\n",
-      "Icd = Ia1-100                  #Current in section CD(A)\n",
-      "Ide = Ia1-300                  #Current in section DE(A)\n",
-      "Ief = Ia1-550                  #Current in section EF(A)\n",
-      "Ifb = Ia1-850                 #Current in section FB(A)\n",
-      "Pl = Iac**2*Rac+Icd**2*Rcd+Ide**2*Rde+Ief**2*Ref+Ifb**2*Rfb  #power loss(W)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The currents supplied from A to B are:\\n\\tIa =\",Ia1,\"A\\tIb =\",abs(Ifb),\"A\" \n",
-      "print  \"(ii) The power dissipated in the distributor is\",round(Pl/1000,3),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The currents supplied from A to B are:\n",
-        "\tIa = 437.5 A\tIb = 412.5 A\n",
-        "(ii) The power dissipated in the distributor is 14.706 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.13, Page number: 325"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Va = 600                      #V\n",
-      "Vb = 590                      #V\n",
-      "r = 0.04         #track resistance of go and return path(ohm/km)\n",
-      "AB = 6                        #distance b/w the sub-stations(km)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let min. potential occurs at point M at a distance x km from the substation A.\n",
-      "\n",
-      "\n",
-      "Ia,x = symbols('Ia x')      #Ia is current supplied by the sub-station A\n",
-      "Ram = r*x                   #Track resistance for section AM(ohm)\n",
-      "Rmb = r*(AB-x)             #Track resistance for section MB(ohm)\n",
-      "Vm = Va-Ia*Ram              #Potential at M(V)       ...(i)\n",
-      "Vm1 = Vb-(300-Ia)*Rmb       #also,Potential at M(V)  ...(ii)\n",
-      "\n",
-      "#from (i) & (ii):\n",
-      "Ia = 341.7-50*x\n",
-      "Vm = Va-(341.7 - 50*x)*0.04*x\n",
-      "Vm2 = diff(Vm,x)\n",
-      "x1 = round(solve(Vm2,x)[0],2)\n",
-      "Ia1 = 341.7-50*x1             #A\n",
-      "Ib = 300-Ia1                   #A\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The point along the track where minimum potential occurs is\",x1,\"km\"\n",
-      "print \"(ii)Current supplied by sub-ation A is\",Ia1,\"A\"\n",
-      "print \"    Current supplied by sub-station B is\",Ib,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The point along the track where minimum potential occurs is 3.42 km\n",
-        "(ii)Current supplied by sub-ation A is 170.7 A\n",
-        "    Current supplied by sub-station B is 129.3 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.14, Page number: 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "i = 0.5                     #current loading(A/m)\n",
-      "l = 1000                     #cable length(m)\n",
-      "V = 220                      #end voltages(V)\n",
-      "r1 = 0.05                     #Resistance of each of 2 conductors(ohm/km)\n",
-      "\n",
-      "#Calculation:\n",
-      "r = 2*r1/1000             #ohm/m\n",
-      "I = i*l                     #A\n",
-      "R = r*l                   #ohm\n",
-      "Vm = I*R/8               #Max. voltage drop(V)\n",
-      "#Minimum voltage will occur at the mid-point of the distributor & its value is\n",
-      "Vmin = V-Vm            #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The maximum voltage drop is\",Vm,\"V\"\n",
-      "print \"The minimum voltage is\",Vmin,\"V at the midpoint of the distributor.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum voltage drop is 6.25 V\n",
-        "The minimum voltage is 213.75 V at the midpoint of the distributor.\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.15, Page number: 327"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Va = 255                    #Voltage at feeding point A(V)\n",
-      "Vb = 250                    #Voltage at feeding point B(V)\n",
-      "l = 500                     #Length of distributor(m)\n",
-      "i = 1                        #Current loading(A/m)\n",
-      "r1 = 0.1                    #resistance of each conductor(ohm/km)\n",
-      "\n",
-      "#Calculation:\n",
-      "r = 2*r1/1000\n",
-      "#(i) Let the minimum potential occur at a point C distant x \n",
-      "#    metres from the feeding point A.\n",
-      "\n",
-      "x = (Va-Vb)/(i*r*l)+l/2            #m\n",
-      "Vc = Va-i*r*x**2/2               #minimum voltage(V)\n",
-      "\n",
-      "\n",
-      "#(ii)\n",
-      "ia = i*x                         #Current supplied from A(A)\n",
-      "ib = i*(l-x)                      #Current supplied from B(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The minimum voltage is\",Vc,\"V at\",x,\"m from point A\"\n",
-      "print \"(ii)Current supplied from A is\",ia,\"A\"\n",
-      "print \"    Current supplied from B is\",ib,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The minimum voltage is 246.0 V at 300.0 m from point A\n",
-        "(ii)Current supplied from A is 300.0 A\n",
-        "    Current supplied from B is 200.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 21
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.16, Page number: 328"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 800                     #Length of distributor(m)\n",
-      "i = 1.25                       #Current loading(A/m)\n",
-      "r1 = 0.05                   #resistance of each conductor(ohm/km)\n",
-      "Vc = 220                    #minimum voltage(V)\n",
-      "x = 450                     #Distance of point C from A(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "r = 2*r1/1000\n",
-      "Vac = i*r*x**2/2           #Voltage drop in section AC(V)\n",
-      "Va = Vac+Vc                #Voltage at feeding point A(V)\n",
-      "Vbc = i*r*(l-x)**2/2         #Voltage drop in section BC(V)\n",
-      "Vb = Vc+Vbc                #Voltage at feeding point B(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage at feeding point A is\",round(Va,2),\"V\"\n",
-      "print \"Voltage at feeding point B is\",round(Vb,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage at feeding point A is 232.66 V\n",
-        "Voltage at feeding point B is 227.66 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.17, Page number: 329"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "i = 1.25                   #current loading (A/m)\n",
-      "l = 1000                   #length of distributor(m)\n",
-      "r1 = 0.05                  #resistance of each conductor(ohm/km)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I = i*l                    #Total current fed to the distributor(A)\n",
-      "R = r*l                    #Total resistance of the distributor(ohm)\n",
-      "V = I*R/8                  #Max. voltage drop(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Maximum voltage drop in the distributor is\",round(V,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum voltage drop in the distributor is 15.63 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.19, Page number: 330"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 900                   #distributor length(m)\n",
-      "Va = 400                  #voltage fed at a A(V)\n",
-      "i = 0.5                   #current loading(A/m)\n",
-      "r = 0.0001                #resistance of distributor per m (go and return)(ohm)\n",
-      "x = 500                   #length AD(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Drops due to concentrated loads.\n",
-      "Iac = 300                #A\n",
-      "Icd = 250                #A\n",
-      "Ide = 150                #A\n",
-      "Vac = Iac*200*r            #Drop in section AC(V)\n",
-      "Vcd = Icd*300*r             #Drop in section CD(V)\n",
-      "Vde = Ide*300*r             #Drop in section DE(V)\n",
-      "Vab1 = Vac+Vcd+Vde           #Total drop over AB(V)\n",
-      "\n",
-      "\n",
-      "#Drops due to uniform loading:\n",
-      "Vab2 = i*r*l**2/2             #V\n",
-      "Vad = i*r*(l*x-x**2/2)        #V\n",
-      "Vb = Va-(Vab1+Vab2)          #V\n",
-      "Vd = Va-(Vac+Vcd+Vad)         #V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Voltage at point B is\",Vb,\"V\"\n",
-      "print \"(ii)Voltage at point D is\",Vd,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage at point B is 361.75 V\n",
-        "(ii)Voltage at point D is 370.25 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.20, Page number: 331"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 0.1*10**-3              #Distributor resistance per metre length\n",
-      "i = 0.5                     #uniform current loading(A/m)\n",
-      "Va = 240                    #end voltage at A(V)\n",
-      "Vb = 240                    #end voltage at B(V)\n",
-      "l = 1000                    #length of distributor(m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) Point of minimum potential:\n",
-      "#We know that, the point of minimum potential is not affected\n",
-      "#by the uniform loading of the distributor. \n",
-      "#Considering the concentrated loads first as shown below:\n",
-      "\n",
-      "I = symbols('I')\n",
-      "Iac = I\n",
-      "Icd = I-120\n",
-      "Ide = I-180\n",
-      "Ief = I-280\n",
-      "Ifb = I-320\n",
-      "I1 = solve((Iac*200+Icd*200+Ide*300+Ief*200+Ifb*100)*10**-4-Va+Vb,I)[0]\n",
-      "\n",
-      "#Showing the current distributions, we can see that D is \n",
-      "#the point of minimum potential.\n",
-      "\n",
-      "#(ii) The feeding point A will supply I1 A due to concentrated\n",
-      "#     loading plus 0\u00b75 \u00d7 400 = 200 A due to uniform loading.\n",
-      "\n",
-      "Ia = I1+200             #Current supplied by A(A)\n",
-      "\n",
-      "#The feeding point B will supply a current of 154 A due to \n",
-      "#concentrated loading plus 0\u00b75 \u00d7 600 = 300 A due to uniform loading.\n",
-      "\n",
-      "Ib = 320-I1+300           #Current supplied byB(A)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(iii) We got that D is the point of minimum potential.\n",
-      "Iac = I1;       Icd = I1-120        #ampere\n",
-      "Vad1 = Iac*200*10**-4+Icd*200*10**-4         #Drop in AD due to conc. loading(V)\n",
-      "AD = 400                        #for uniform loading(V)\n",
-      "Vad2 = i*r*AD**2/2              #Drop in AD due to uniform. loading(V)\n",
-      "Vd = 240-Vad1-Vad2              #Minimum potential(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  The point of minimum potential is D\"\n",
-      "print \"(ii) Current supplied by A is\",round(Ia),\"A\"\n",
-      "print \"     Current supplied by B is\",round(Ib),\"A\"\n",
-      "print \"(iii)The value of minimum potential is\",round(Vd,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The point of minimum potential is D\n",
-        "(ii) Current supplied by A is 366.0 A\n",
-        "     Current supplied by B is 454.0 A\n",
-        "(iii)The value of minimum potential is 231.76 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 25
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.21, Page number: 332"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 500                     #distributor length(m)\n",
-      "Va = 240                    #voltage at end A(V)\n",
-      "Vb = 240                    #voltage at end B(V)\n",
-      "R = 0.001                   #resistance of the distributor(ohm/m)\n",
-      "i = 1                        #current loading(A/m)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let D be the point of minimum potential\n",
-      "x = symbols('x')                   #x is current flowing in section CD(A)\n",
-      "\n",
-      "#(i) If r is the resistance of the distributor (go and return)\n",
-      "#per metre,\n",
-      "r = symbols('r')\n",
-      "Vad = (100+x)*100*r+x*150*r         #Voltage drop in length AD\n",
-      "Vbd = 1*r*200**2/2+(60-x)*250*r       #Voltage drop in length BD\n",
-      "x1 = solve(Vad-Vbd,x)[0]           #A\n",
-      "\n",
-      "#(ii)\n",
-      "I = 60+1*200            #Total current(A)\n",
-      "Ia = 100+x1             #Current supplied by A(A)\n",
-      "Ib = 360-150              #Current supplied by B(V)\n",
-      "Vd = Va - Ia*100*0.001-50*150*0.001        #Minimum potential(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The point of minimum voltage is D.\"\n",
-      "print \"(ii)The value of minimum voltage is\",round(Vd,1),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The point of minimum voltage is D.\n",
-        "(ii)The value of minimum voltage is 217.5 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 26
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.22, Page number: 334"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 300                           #distributor length(m)\n",
-      "r1 = 0.03/100               #resistance of single conductor(ohm/m)\n",
-      "Va = 240                        #volt\n",
-      "\n",
-      "#Calculation:\n",
-      "r = 2*r1                      #resistance of both wires(ohm/m)\n",
-      "Rab = r*150                   #ohm\n",
-      "Rbc = r*50                    #ohm\n",
-      "Rca = r*100                    #ohm\n",
-      "#suppose a current IA flows in section AB of the distributor.\n",
-      "#currents in sections BC and CA will be (Ia \u2212 120) & (Ia \u2212 200)\n",
-      "Ia = symbols('Ia')\n",
-      "Ia1 = solve(0.09*Ia+0.03*(Ia-120)+0.06*(Ia-200),Ia)[0]\n",
-      "\n",
-      "#The actual distribution of currents ca be drawn easily\n",
-      "#from where it is seen that B is the point of min. potential.\n",
-      "Iab = Ia1                    #Current in section AB(A)\n",
-      "Ibc = Ia1-120                #Current in section BC(A)\n",
-      "Ica = Ia1-200                 #current in section CA(A)\n",
-      "Vb = Va-Iab*Rab              #Voltage at point B(V)\n",
-      "Vc = Vb-Ibc*Rbc              #Voltage at point C(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Current in section AB is\",round(Iab,2),\"A from A to B\"\n",
-      "print \"    Current in section BC is\",round(abs(Ibc),2),\"A from C to B\"\n",
-      "print \"    Current in section ca is\",round(abs(Ica),2),\"A from A to C\"\n",
-      "print \"(ii)Voltage at point B is\",round(Vb,1),\"V\"\n",
-      "print \"    Voltage at point C is\",round(Vc,1),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Current in section AB is 86.67 A from A to B\n",
-        "    Current in section BC is 33.33 A from C to B\n",
-        "    Current in section ca is 113.33 A from A to C\n",
-        "(ii)Voltage at point B is 232.2 V\n",
-        "    Voltage at point C is 233.2 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.23, Page number: 335"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Va = 220                 #voltage at point A(V)\n",
-      "Rab = 0.1                 #ohm\n",
-      "Rbc = 0.05                #ohm\n",
-      "Rcd = 0.01               #ohm\n",
-      "Rde = 0.025             #ohm\n",
-      "Rea = 0.075              #ohm\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i)\n",
-      "I1 = solve(0.1*I+0.05*(I-10)+0.01*(I-30)+0.025*(I-60)+0.075*(I-70),I)[0]\n",
-      "\n",
-      "#(ii)\n",
-      "Iab = I1                    #A\n",
-      "Ibc = I1-10                 #A\n",
-      "Icd = I1-30                 #A\n",
-      "Ide = I1-60                 #A\n",
-      "Iea = I1-70                 #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) C is the point of minimum potential.\"\n",
-      "print \"(ii)Current in section AB is\",round(Iab,2),\"A  from A to B\"\n",
-      "print \"    Current in section BC is\",round(Ibc,2),\"A  from B to C\"\n",
-      "print \"    Current in section CD is\",round(abs(Icd),2),\"A  from D to C\"\n",
-      "print \"    Current in section DE is\",round(abs(Ide),2),\"A  from E to D\"\n",
-      "print \"    Current in section EA is\",round(abs(Iea),2),\"A  from A to E\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) C is the point of minimum potential.\n",
-        "(ii)Current in section AB is 29.04 A  from A to B\n",
-        "    Current in section BC is 19.04 A  from B to C\n",
-        "    Current in section CD is 0.96 A  from D to C\n",
-        "    Current in section DE is 30.96 A  from E to D\n",
-        "    Current in section EA is 40.96 A  from A to E\n"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.24, Page number: 336"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Rab = 0.02                     #ohm\n",
-      "Rbc = 0.018                     #ohm\n",
-      "Rcd = 0.025                    #ohm\n",
-      "Rda = 0.02                     #ohm\n",
-      "Va = 250                       #voltage ate point A(V)\n",
-      "\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I = symbols('I')\n",
-      "I1 = solve(0.02*I+0.018*(I-150)+0.025*(I-450)+0.02*(I-700),I)[0]\n",
-      "Vab = 336.75*0.02              #V\n",
-      "Vbc = 186.75*0.018             #V\n",
-      "Vcd = 113.25*0.025             #V\n",
-      "Vda = 363.25*0.02              #V\n",
-      "Vb = Va-Vab                    #V\n",
-      "Vc = Vb-Vbc                    #V\n",
-      "Vd = Vc+Vcd                    #V\n",
-      "\n",
-      "#With interconnector\n",
-      "Eo = Va-Vc                    #Voltage between points A and C(V)\n",
-      "Ro = (Rab+Rbc)*(Rcd+Rda)/((Rab+Rbc)+(Rcd+Rda))   #Resistance viewed from points A & C(ohm)\n",
-      "Rac = 0.02                 #Resistance of interconnector(ohm)\n",
-      "Iac = Eo/(Ro+Rac)             #Current in interconnector AC(A)\n",
-      "\n",
-      "#Let ABCD be the ring distribution system.\n",
-      "#Let the current in section AB is I1. Then,\n",
-      "#current in section BC will be I1 \u2212 150. \n",
-      "#As the voltage drop round the closed mesh ABCA is zero.\n",
-      "I11 = symbols('I11')\n",
-      "I2 = solve(0.02*I11+0.018*(I11-150)-0.02*252.4,I11)[0]\n",
-      "\n",
-      "Vab1 = I2*0.02                 #V\n",
-      "Vbc1 = 53.15*0.018              #V\n",
-      "Vad1 = 244.45*0.02               #V\n",
-      "Vb1 = Va-Vab1                  #V\n",
-      "Vc1 = Vb1-Vbc1                  #V\n",
-      "Vd1 = Va-Vad1                  #V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Before adding interconnector,\"\n",
-      "print \"\\tVoltage at point B is\",Vb,\"V\"\n",
-      "print \"\\tVoltage at point C is\",round(Vc,3),\"V\"\n",
-      "print \"\\tVoltage at point D is\",round(Vd,3),\"V\"\n",
-      "print \"After adding interconnector,\"\n",
-      "print \"\\tPotential of B is\",round(Vb1,2),\"V\"\n",
-      "print \"\\tPotential of C is\",round(Vc1,2),\"V\"\n",
-      "print \"\\tPotential of D is\",round(Vd1,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Before adding interconnector,\n",
-        "\tVoltage at point B is 243.265 V\n",
-        "\tVoltage at point C is 239.903 V\n",
-        "\tVoltage at point D is 242.735 V\n",
-        "After adding interconnector,\n",
-        "\tPotential of B is 245.92 V\n",
-        "\tPotential of C is 244.97 V\n",
-        "\tPotential of D is 245.11 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.25, Page number: 338"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "Rbd = 0.05                    #resistance of interconnector(ohm)\n",
-      "Rab = 0.075                   #resistance of branch AB(ohm)\n",
-      "Rbc = 0.025                   #resistance of branch BC(ohm)\n",
-      "Rcd = 0.01                    #resistance of branch CD(ohm)\n",
-      "Rde = 0.05                    #resistance of branch DE(ohm)\n",
-      "Rea = 0.1                     #resistance of branch EA(ohm)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let ABCDE be the ring distribution system.\n",
-      "#When interconnector BD is removed, let the current in branch \n",
-      "#AB be I.\n",
-      "# Current in BC = I-10;\n",
-      "# Current in CD = I-40;\n",
-      "# Current in DE = I-60;\n",
-      "# Current in EA = I-70;\n",
-      "\n",
-      "\n",
-      "#As the total drop round the ring ABCDEA is zero.\n",
-      "I = symbols('I')\n",
-      "\n",
-      "I1 = solve(Rab*I+Rbc*(I-10)+Rcd*(I-40)+Rde*(I-60)+Rea*(I-70),I)[0]\n",
-      "\n",
-      "#The actual distribution of currents will be as shown in Fig.(ii) above.\n",
-      "Vbcd = 30.96*0.025+0.96*0.01              #V\n",
-      "Eo = 0.7836                            #V\n",
-      "Ro = (0.075+0.1+0.05)*(0.025+0.01)/((0.075+0.1+0.05)+(0.025+0.01))\n",
-      "\n",
-      "#(i)\n",
-      "Ibd = Eo/(Ro+Rbd)                 #A\n",
-      "Vbd = Ibd*0.05                    #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Current in interconnector BD is\",round(Ibd,1),\"A\"\n",
-      "print \"(ii)Voltage drop along interconnector BD is\",round(Vbd,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Current in interconnector BD is 9.8 A\n",
-        "(ii)Voltage drop along interconnector BD is 0.49 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.26, Page number: 341"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I1 = 50                    #current on +ve side(A)\n",
-      "I2 = 40                    #current on -ve side(A)\n",
-      "Rae = 0.1                    #resistance of outer wire(ohm)\n",
-      "Rbg = 0.1                   \n",
-      "Vab = 500                  #V\n",
-      "Van = 250                 #V\n",
-      "Vbn = 250                  #V\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I3 = I1-I2                #current in the neutral wire(A)\n",
-      "Rnl = 2*Rae               #resistance of neutral wire(ohm)\n",
-      "Vel = Van-I1*Rae-(I1-I2)*Rnl           #V\n",
-      "Vlg = Vbn+(I1-I2)*Rnl-I2*Rbg            #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage at the load end on the positive side is\",Vel,\"V\"\n",
-      "print \"Voltage at the load end on the negative side is\",Vlg,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage at the load end on the positive side is 243.0 V\n",
-        "Voltage at the load end on the negative side is 248.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.27, Page number: 342"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Vairable declaration:\n",
-      "r = 0.1                      #resistance of each conductor(ohm)\n",
-      "Rel = 5                      #ohm\n",
-      "Rcl = 6                      #ohm\n",
-      "Vel = 240                    #line to neutral voltage(V)\n",
-      "Vlc = 240                   #line to neutral voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I1 = Vel/5                     #Current on +ve outer(A)\n",
-      "I2 = Vlc/6                      #Current on \u2212ve outer(A)\n",
-      "In = I1-I2                     #current in neutral(A)\n",
-      "V1 = Vel+I1*r+(I1-I2)*r         #V\n",
-      "V2 = Vlc-(I1-I2)*r+I2*r        #V\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage between +ve outer and neutral at feeding end is\",V1,\"V\"\n",
-      "print \"Voltage between -ve outer and neutral at feeding end is\",V2,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage between +ve outer and neutral at feeding end is 245.6 V\n",
-        "Voltage between -ve outer and neutral at feeding end is 243.2 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.28, Page number: 343"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 250                      #voltage on the two sides of the middle wire(V)\n",
-      "P1 = 35                       #power of load 1(kW)\n",
-      "P2 = 20                      #power of load 2(kW)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R1 = V**2/(P1*1000)            #resistance of load 1(ohm)\n",
-      "R2 = V**2/(P2*1000)            #resistance of load 2(ohm)\n",
-      "#After breaking of neutral wire,\n",
-      "I = 2*V/(R1+R2)                #Circuit current(A)\n",
-      "V1 = I*R1                      #V\n",
-      "V2 = I*R2                        #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage across +ve outer and middle wire is\",round(V1,1),\"V\"\n",
-      "print \"Voltage across -ve outer and middle wire is\",round(V2,1),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage across +ve outer and middle wire is 181.8 V\n",
-        "Voltage across -ve outer and middle wire is 318.2 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.29, Page number: 343"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Calculation:\n",
-      "Vck = 250-0.75-0.28+0.2            #V\n",
-      "Vdm = Vck-0.3-0.18+0.084           #V\n",
-      "Vjg = 250-0.2-1.2                  #V\n",
-      "Vlh = Vjg+0.28-0.084-0.864         #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage across load CK is\",Vck,\"V\"\n",
-      "print \"Voltage across load DM is\",Vdm,\"V\"\n",
-      "print \"Voltage across load JG is\",Vjg,\"V\"\n",
-      "print \"Voltage across load LH is\",Vlh,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage across load CK is 249.17 V\n",
-        "Voltage across load DM is 248.774 V\n",
-        "Voltage across load JG is 248.6 V\n",
-        "Voltage across load LH is 247.932 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.30, Page number: 344"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "Vab = 500                    #volts\n",
-      "Van = 250                     #volts\n",
-      "Vnb = 250                     #volts\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Voltage across CK = 250 \u2212 Drop in AC \u2212 Drop in KJ \u2212 Drop in JN\n",
-      "Vck = Van-4-0.5-0.1               #volts\n",
-      "\n",
-      "#Voltage across DM = 245\u00b74 \u2212 Drop in CD \u2212 Drop in ML + Drop in KL\n",
-      "Vdm = Vck-1.28-0.5+0.42           #volts\n",
-      "\n",
-      "#Voltage across JG = 250 + Drop in JN \u2212 Drop in GB\n",
-      "Vjg = Van+0.1-1.9                 #volts\n",
-      "\n",
-      "#Voltage across LH = 248\u00b72 + Drop in KJ \u2212 Drop in KL \u2212 Drop in HG\n",
-      "Vlh = Vjg+0.5-0.42-2.4              #volts\n",
-      "\n",
-      "#Voltage across PF = 245\u00b788 + Drop in ML \u2212 Drop in MP \u2212 Drop in FH\n",
-      "Vpf = Vlh+0.5-0.72-1.02          #volts\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage across CK is\",Vck,\"V\"\n",
-      "print \"Voltage across DM is\",Vdm,\"V\"\n",
-      "print \"Voltage across JG is\",Vjg,\"V\"\n",
-      "print \"Voltage across LH is\",Vlh,\"V\"\n",
-      "print \"Voltage across PF is\",Vpf,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage across CK is 245.4 V\n",
-        "Voltage across DM is 244.04 V\n",
-        "Voltage across JG is 248.2 V\n",
-        "Voltage across LH is 245.88 V\n",
-        "Voltage across PF is 244.64 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.31, Page number: 346"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vel = 240               #volt\n",
-      "Vlc = 240               #volt\n",
-      "Rel = 4                 #ohm\n",
-      "Rlc = 6                 #ohm\n",
-      "r = 0.15                #resistance of each conductor(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I1 = Vel/Rel                     #Current in the positive outer(A)\n",
-      "I2 = Vlc/Rlc                     #current in the negative outer(A)\n",
-      "In = I1-I2                       #current in the neutral wire(ohm)\n",
-      "V1 = Vel+I1*r+In*r        #Voltage between +ve outer and neutral at feeding point(V)\n",
-      "V2 = Vlc-In*r+I2*r        #Voltage between -ve outer and neutral at feeding point(V)\n",
-      "\n",
-      "\n",
-      "#(i)When neutral breaks,\n",
-      "#When there is a break in the neutral, the system\n",
-      "#is equivalent to 2- wire d.c. system.Now,\n",
-      "R1 = Rel+Rlc+2*r                  #ohm\n",
-      "Vt1 = V1+V2                     #voltage at the feeding end(V)\n",
-      "I11 = Vt1/R1                    #Load current(A)\n",
-      "Vl1 = I11*Rel                   #Voltage across 4 ohm resistance(V)\n",
-      "Vl2 = I11*Rlc                   #voltage across 6 ohm resistance(V)\n",
-      "\n",
-      "\n",
-      "#(ii) When +ve outer breaks,\n",
-      "#When there is a break in the +ve outer, there will be\n",
-      "#no current in 4 ohm load.\n",
-      "R2 = Rlc+2*r                      #Total circuit resistance(ohm)\n",
-      "Vt2 = V2                       #volt\n",
-      "I22 = Vt2/R2                     #load current(A)\n",
-      "VR2 = I22*Rlc                   #volt\n",
-      "\n",
-      "\n",
-      "#(iii)When \u2212ve outer breaks,\n",
-      "#When there is a break in the negative outer, there will be no\n",
-      "#current in 6 \u03a9 load.\n",
-      "R3 = 4+2*r                      #Total circuit resistance(ohm)\n",
-      "Vt3 = V1                       #volt\n",
-      "I33 = Vt3/R3                     #load current(A)\n",
-      "VR3 = I33*Rel                   #volt\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"When there is a break in the:\"\n",
-      "print \"(i)  neutral wire, current is\",round(I11,2),\"A\" \n",
-      "print \"     & voltage is\",round(Vl1,2),\"V and\",round(Vl2,2),\"V\"\n",
-      "print \"     across 6 ohm & 4 ohm loads respectively.\"\n",
-      "print \"(ii) positive outer, current is\",round(I22,2),\"A\"\n",
-      "print \"     and voltage is\",round(VR2,2),\"V\"\n",
-      "print \"(iii)negative outer, current is\",round(I33,2),\"A\"\n",
-      "print \"     and voltage is\",round(VR3,2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When there is a break in the:\n",
-        "(i)  neutral wire, current is 48.06 A\n",
-        "     & voltage is 192.23 V and 288.35 V\n",
-        "     across 6 ohm & 4 ohm loads respectively.\n",
-        "(ii) positive outer, current is 38.57 A\n",
-        "     and voltage is 231.43 V\n",
-        "(iii)negative outer, current is 58.6 A\n",
-        "     and voltage is 234.42 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.32, Page number: 348"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "V = 500               #voltage b/w the outer supplies(V)\n",
-      "P1 = 1500                   #load(kW)\n",
-      "P2 = 2000                   #load(kW)\n",
-      "Vn = 250                    #outer to neutral voltage(V)\n",
-      "\n",
-      "#Calculation:\n",
-      "I1 = P1*1000/Vn              #Load current on +ve outer(A)\n",
-      "I2 = P2*1000/Vn              #Load current on -ve outer(A)\n",
-      "In = I2-I1                   #A\n",
-      "Pl = P1+P2                   #kW\n",
-      "Ig = Pl*1000/V                   #Current supplied by main generator(A)\n",
-      "Ia = Ig-I1                  #A\n",
-      "Ib = I2-Ig                  #A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Current in the neutral is\",In,\"A\"\n",
-      "print \"(ii) total current supplied by main generator is\",Ig,\"A\"\n",
-      "print \"(iii)Current in machine A is\",Ia,\"A\"\n",
-      "print \"     Current in machine B\",Ib,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Current in the neutral is 2000.0 A\n",
-        "(ii) total current supplied by main generator is 7000.0 A\n",
-        "(iii)Current in machine A is 1000.0 A\n",
-        "     Current in machine B 1000.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.33, Page number: 349"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 500                   #volt\n",
-      "Vn = 250                  #volt\n",
-      "P1 = 150                   #load(kW)\n",
-      "P2 = 100                   #load(kW)\n",
-      "Pl = 3                     #loss in each balancer machine(kW)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Pt = P1+P2+2*Pl               #Total load on the main generator(kW)\n",
-      "Ig = Pt*1000/V             #Current supplied by the main generator(A)\n",
-      "I1 = P1*1000/Vn            #Load current on +ve side(A)\n",
-      "I2 = P2*1000/Vn            #Load current on -ve side(A)\n",
-      "In = I1-I2                 #Current in neutral wire(A)\n",
-      "Ia = I1-Ig                 #Current through machine A(A)\n",
-      "Ib = Ig-I2                 #Current through machine B(A)\n",
-      "Pa = Ia*Vn/1000              #Load on machine A(kW)\n",
-      "Pb = Ib*Vn/1000              #Load on machine B(kW)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Total load on the main generator is\",Pt,\"kW\"\n",
-      "print \"(ii)Load on machine A is\",Pa,\"kW\"\n",
-      "print \"    Load on machine B is\",Pb,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Total load on the main generator is 256 kW\n",
-        "(ii)Load on machine A is 22.0 kW\n",
-        "    Load on machine B is 28.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.34, Page number: 350"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I1 = 1200                #current on +ve side(A)\n",
-      "I2 = 1000                #current on -ve side(A)\n",
-      "Pm = 200                   #motor load acoss outer(kW)\n",
-      "Pl = 5                     #loss in balancer machine(kW)\n",
-      "V = 500                   #voltage across the outers(V)\n",
-      "Vn = 250                #voltage across the outer and the neutral(V)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Since the positive side is more heavily loaded, \n",
-      "#so, machine A acts as a generator and machine B as a motor.\n",
-      "P1 = Vn*I1/1000          #Load on +ve side(kW)\n",
-      "P2 = Vn*I2/1000          #Load on -ve side(kW)\n",
-      "P3 = 200                 #load on outers(kW)\n",
-      "Pt = P1+P2+P3+2*Pl                 #total load(kW)\n",
-      "Ig = Pt*1000/500                  #Current of main generator(A)\n",
-      "In = I1-I2                       #Current in neutral(A)\n",
-      "Ia = 1600-Ig                  #Current through machine A(A)\n",
-      "Ib = Ig-1400                  #Current through machine B(A)\n",
-      "Pa = Vn*Ia/1000                #Load on machine A(kW)\n",
-      "Pb = Vn*Ib/1000                #Load on machine B(kW)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Current of the main generator is\",Ig,\"A\"\n",
-      "print \"(ii)Load on balancer machines: Pa =\",Pa,\"kW & Pb =\",Pb,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Current of the main generator is 1520.0 A\n",
-        "(ii)Load on balancer machines: Pa = 20.0 kW & Pb = 30.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.35, Page number: 350"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ipos = 800                #current on on the positive side(A)\n",
-      "Ineg = 550                 #current on the negative side(A)\n",
-      "I = 1500                  #curent across the outers(A)\n",
-      "Ra = 0.2              #resistance of each arotary balancer's armature(ohm)\n",
-      "Io = 5                    #no load current(A)\n",
-      "Vn = 500                       #supply voltage\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I1 = Ipos+I                   #Total current on +ve side(A)\n",
-      "I2 = Ineg+I                   #Total current on \u2212ve side(A)\n",
-      "In = Ipos-Ineg                  #Current in neutral wire(A)\n",
-      "\n",
-      "#Let the current through machines A and B be Ia and Ib  \n",
-      "Ib = symbols('Ib')\n",
-      "Ia = In-Ib\n",
-      "#Each machine has same value of back e.m.f. E as their field\n",
-      "#currents & speeds are the same.\n",
-      "E = Vn/2-Ra*Io                   #Back e.m.f(V)\n",
-      "Va = E-Ia*Ra                    #Terminal p.d. across A(V)\n",
-      "Vb = E+Ib*Ra                    #Terminal p.d. across B(V)\n",
-      "Ib1 = round(solve(Va*Ia+Vn*Io+Ia**2*Ra+Ib**2*Ra-Vb*Ib,Ib)[0])\n",
-      "Ia1 = In-Ib1\n",
-      "Va1 = E-Ia1*Ra\n",
-      "Vb1 = E+Ib1*Ra\n",
-      "Ig = 2300-Ia1                  #Load on main generator(A)\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)Current loading of each balancer machine: Ia =\",Ia1,\"A & Ib =\",Ib1,\"A\"\n",
-      "print \"(ii)Voltage across machine A is\",Va1,\"V\"\n",
-      "print \"    Voltage across machine B is\",Vb1,\"V\"\n",
-      "print \"(iii)Load on main generator is\",Ig,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Current loading of each balancer machine: Ia = 120.0 A & Ib = 130.0 A\n",
-        "(ii)Voltage across machine A is 225.0 V\n",
-        "    Voltage across machine B is 275.0 V\n",
-        "(iii)Load on main generator is 2180.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 40
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 13.36, Page number: 352"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 500                      #voltage at supply end(v)\n",
-      "l = 3                        #line length(km)\n",
-      "Ifl = 120                     #full loag current(A)\n",
-      "r = 0.5                       #resistance of cable(ohm/km)\n",
-      "\n",
-      "#Calculation:\n",
-      "R = r*l                        #total resistance of the line(ohm)\n",
-      "Vl = R*Ifl                    #F.L. voltage drop in the line(V)\n",
-      "Vb = Vl                        #Terminal voltage of booster(V)\n",
-      "Po = Vb*Ifl/1000             #Output of booster(kW)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Booster Voltage is\",Vl,\"V\"\n",
-      "print \"Output of booster is\",Po,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Booster Voltage is 180.0 V\n",
-        "Output of booster is 21.6 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter14_1.ipynb b/Principles_of_Power_System/chapter14_1.ipynb
deleted file mode 100644
index 9abb2d75..00000000
--- a/Principles_of_Power_System/chapter14_1.ipynb
+++ /dev/null
@@ -1,746 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:0910cfded5ae40d0ea2b9576604d8007ec8328215058593fcec218762063deff"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 14: A.C. Distribution"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.1, Page Number: 359"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 300                      #line length(m)\n",
-      "magI1 = 100                     #current at load 1(A)\n",
-      "pf1 = 0.707                  #power factor at load 1\n",
-      "l1 = 200                      #line length till load 1(m)\n",
-      "magI2 = 200                     #current drawn at load 2(A)\n",
-      "pf2 = 0.8                   #power factor at laod 2\n",
-      "l2 = 300                    #line length till load 2(m)\n",
-      "R = 0.2                      #total resistance of line(ohm/km)\n",
-      "X = 0.1                      #total reactance of the line(ohm/km)\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "Zac = (R+X*1j)*200/1000       #Impedance of section AC(ohm)\n",
-      "Zcb = (R+X*1j)*100/1000       #Impedance of section CB(ohm)\n",
-      "#Taking voltage at the far end B as the reference vector, we have\n",
-      "I2 = magI2*(pf2-1j*math.sin(phy2))         #Load current at point B(A)\n",
-      "I1 = magI1*(pf1-1j*math.sin(phy1))         #Load current at point C(A)\n",
-      "Icb = I2                                   #A\n",
-      "Iac = I1+I2                              #A\n",
-      "Vcb = Icb*Zcb                           #V\n",
-      "Vac = Iac*Zac                            #V\n",
-      "V = Vac+Vcb                            #Voltage drop in the distributor(V)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage drop in the distributor is\",round(abs(V),2),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage drop in the distributor is 17.85 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.2, Page Number: 359"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 2000                      #line length(m)\n",
-      "magI1 = 80                     #current at load 1(A)\n",
-      "pf1 = 0.9                 #power factor at load 1\n",
-      "l1 = 1000                      #line length till load 1(m)\n",
-      "magI2 = 120                     #current drawn at load 2(A)\n",
-      "pf2 = 0.8                   #power factor at laod 2\n",
-      "l2 = 2000                    #line length till load 2(m)\n",
-      "R = 0.05                      #total resistance of line(ohm/km)\n",
-      "X = 0.1                      #total reactance of the line(ohm/km)\n",
-      "magVb = 230                     #voltage maintained at point B(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "Zac = (R+X*1j)*1000/1000       #Impedance of section AC(ohm)\n",
-      "Zcb = (R+X*1j)*1000/1000       #Impedance of section CB(ohm)\n",
-      "#Taking voltage at the far end B as the reference vector, we have\n",
-      "I2 = magI2*(pf2-1j*math.sin(phy2))         #Load current at point B(A)\n",
-      "I1 = magI1*(pf1-1j*math.sin(phy1))         #Load current at point C(A)\n",
-      "Icb = I2                                   #A\n",
-      "Iac = I1+I2                              #A\n",
-      "Vcb = Icb*Zcb                           #V\n",
-      "Vac = Iac*Zac                            #V\n",
-      "V = Vac+Vcb+magVb*(1+0j)                            #Voltage drop in the distributor(V)\n",
-      "theta = math.atan(V.imag/V.real)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Voltage drop in the distributor is\",round(abs(V),2),\"V\"\n",
-      "print \"(ii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Voltage drop in the distributor is 261.67 V\n",
-        "(ii)The phase difference between Va and Vb is  3.83 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.3, Page Number: 360"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "magI1 = 100                     #current at load 1(A)\n",
-      "pf1 = 0.6                 #power factor at load 1\n",
-      "magI2 = 100                     #current drawn at load 2(A)\n",
-      "pf2 = 0.8                   #power factor at laod 2\n",
-      "R = 0.1                      #total resistance of line(ohm/km)\n",
-      "X = 0.15                      #total reactance of the line(ohm/km)\n",
-      "magVb = 200                     #voltage maintained at point B(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "Zam = (R+X*1j)              #Impedance of section AM(ohm)\n",
-      "Zmb = (R+X*1j)             #Impedance of section MB(ohm)\n",
-      "#Taking voltage at the far end B as the reference vector, we have\n",
-      "I2 = magI2*(pf2-1j*math.sin(phy2))         #Load current at point B(A)\n",
-      "Imb = I2  \n",
-      "Vb = magVb*(1+0j)                       #V\n",
-      "Vmb = Imb*Zmb                           #V\n",
-      "Vm = Vb+Vmb                              #V\n",
-      "alpha = math.atan(Vm.imag/Vm.real)         #V\n",
-      "#The load current I1 has a lagging p.f. of 0\u00b76 w.r.t. VM. It lags \n",
-      "#behind Vm by an angle phy1.\n",
-      "#Phase angle between I1 and Vb\n",
-      "phy11 = phy1-alpha\n",
-      "I1 = magI1*(math.cos(phy11)-math.sin(phy11)*1j)         #A\n",
-      "Iam = I1+I2                              #A\n",
-      "Vam = Iam*Zam                            #V\n",
-      "Va = Vm+Vam                             #V\n",
-      "theta = math.atan(Va.imag/Va.real)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)Voltage at mid-point is\",round(abs(Vm),1),\"V\"\n",
-      "print \"(ii) Sending end voltage Va is\",round(abs(Va),2),\"V\"\n",
-      "print \"(iii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Voltage at mid-point is 217.1 V\n",
-        "(ii) Sending end voltage Va is 252.33 V\n",
-        "(iii)The phase difference between Va and Vb is  3.07 degrees\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.4, Page Number: 362"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Zab = 1+1j                    #ohm\n",
-      "Zbc = 1+2j                    #ohm\n",
-      "Zac = 1+3j                    #ohm\n",
-      "Ib = 20                       #load current at B(A)\n",
-      "pfb = 0.8                      #power factor at A\n",
-      "Ic = 15                       #load current at C(A)\n",
-      "pfc = 0.6                      #power factor at B\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Iab = Ib*(pfb-1j*math.sin(math.acos(pfb)))        #Current in section AB(A)\n",
-      "Iac = Ic*(pfc-1j*math.sin(math.acos(pfc)))        #Current in section AB(A)\n",
-      "Vab = Iab*Zab                     #Voltage drop in section AB(V)\n",
-      "Vac = Iac*Zac                     #Voltage drop in section AC(V)\n",
-      "#point B is at higher potential than point C. The p.d. between B and C\n",
-      "#is Thevenin\u2019s equivalent circuit e.m.f. Eo i.e.\n",
-      "Eo = Vac-Vab                     #volt\n",
-      "Zo = Zab+Zac                    #Thevenin\u2019s equivalent impedance(ohm)\n",
-      "Ibc = Eo/(Zo+Zbc)                #A\n",
-      "Iab1 = Iab+Ibc                   #A\n",
-      "Iac1 = Iac-Ibc                   #A\n",
-      "Ia = Iab+Iac                     #Current fed at A(A)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The total current fed at A is\",Ia,\"A\"\n",
-      "print \"\\nCurrent in AB is\",Iab1.real+round(Iab1.imag,2)*1j,\"A\"\n",
-      "print \"\\nCurrent in BC is\",Ibc.real+round(Ibc.imag,2)*1j,\"A\"\n",
-      "print \"\\nCurrent in AC is\",Iac1.real+round(Iac1.imag,2)*1j,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total current fed at A is (25-24j) A\n",
-        "\n",
-        "Current in AB is (18.6-13.53j) A\n",
-        "\n",
-        "Current in BC is (2.6-1.53j) A\n",
-        "\n",
-        "Current in AC is (6.4-10.47j) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.5, Page Number: 363"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 1000                      #line length(m)\n",
-      "magI1 = 5                     #current at load 1(A)\n",
-      "pf1 = 0.8                 #power factor at load 1\n",
-      "l1 = 600                      #line length till load 1(m)\n",
-      "pf2 = 0.85                   #power factor at motor load B\n",
-      "Po = 10                      #power output at B(H.P)\n",
-      "n = 0.9                     #efficiency\n",
-      "l2 = 400                    #line length till load 2(m)\n",
-      "R = 1                      #total resistance of line(ohm/km)\n",
-      "X = 0.5                      #total reactance of the line(ohm/km)\n",
-      "magVb = 400                     #voltage maintained at point B(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Zac = (R+X*1j)*l1/l                 #ohm\n",
-      "Zcb = (R+X*1j)*l2/l                 #ohm\n",
-      "magVbp = magVb/3**0.5               #volt per phase\n",
-      "Vbp = magVbp*(1+0j)                 #V\n",
-      "magIb = Po*746/(3**0.5*magVb*pf2*n)           #Line current at B(A)\n",
-      "magI2p = magIb\n",
-      "I2p = magI2p*(pf2-1j*math.sin(math.atan(pf2)))    #A\n",
-      "I1p = magI1*(pf1-1j*math.sin(math.atan(pf1)))       #A\n",
-      "Iac = I1p+I2p                      #Current in section AC(A)\n",
-      "Icb = I2p                          #Current in section CB(A)\n",
-      "Vcb = Icb*Zcb                      #V\n",
-      "Vac = Iac*Zac                      #V\n",
-      "Va = Vbp+Vcb+Vac                    #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Line voltage at A is\",round(abs(Va)*3**0.5),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Line voltage at A is 434.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.6, Page Number: 364"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "magVa = 11000                      #volt\n",
-      "magIb = 50                            #load current at B(A)\n",
-      "pf2 = 0.8                           #power factor(lagging)\n",
-      "magIc = 120                            #load current at C(A)\n",
-      "pf3 = 1.0                           #power factor(lagging)\n",
-      "magId = 70                             #load current at D(A)\n",
-      "pf4 = 0.866                         #power factor(lagging)\n",
-      "Zab = 1+0.6j                        #ohm\n",
-      "Zbc = 1.2+0.9j                      #ohm\n",
-      "Zcd = 0.8+0.5j                        #ohm\n",
-      "Zda = 3+2j                          #ohm\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let current in section AB be (x + j*y).\n",
-      "x,y = symbols('x,y')\n",
-      "Iab = x+1j*y\n",
-      "x,y = symbols('x,y')\n",
-      "Ibc = Iab-magIb*(pf2-math.sin(math.atan(pf2)))     #A\n",
-      "Icd = (x-40+1j*(y+30))-(120+0j)                   #A\n",
-      "Ida = ((x-160)+1j*(y+30))-(70*(0.866-0.5j))          #A\n",
-      "Vab = Iab*Zab                         #Drop in section AB(V)\n",
-      "Vbc = Ibc*Zbc                         #Drop in section BC(V)\n",
-      "Vcd = Icd*Zcd                          #Drop in section CD(V)\n",
-      "Vda = Ida*Zda                         #Drop in section DA(V)\n",
-      "#  Vab+Vbc+Vcd+Vda = 0\n",
-      "#  As the real and imaginary parts have to be separately zero,\n",
-      "#  6*x-4*y-1009.8 = 0\n",
-      "#  4*x+6*y-302.2 = 0\n",
-      "x1 = solve(6*x-4*y-1009.8,x)[0]\n",
-      "y1 = round(solve(4*x1+6*y-302.2,y)[0],1)\n",
-      "x11 = round(solve(6*x-4*y1-1009.8,x)[0],1)\n",
-      "#now putting the values of x11 and y1 in above equationa,\n",
-      "Iab1 = x11+1j*y1                                     #A\n",
-      "Ibc1 = (x11-40)+1j*(y1+30)                           #A\n",
-      "Icd1 = (x11-40+1j*(y1+30))-(120+0j)                   #A\n",
-      "Ida1 = ((x11-160)+1j*(y1+30))-(70*(0.866-0.5j))          #A\n",
-      "magVap = round(magVa/3**0.5)                     #Voltage at supply end A(V)\n",
-      "Vb = magVap*(1+0j)-Iab1*Zab               #Voltage at station B(V/phase)\n",
-      "Vc = Vb-Ibc1*Zbc               #Voltage at station C(V/phase)\n",
-      "Vd = Vc-Icd1*Zcd               #Voltage at station D(V/phase)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Current in section AB is\",Iab1,\"A\"\n",
-      "print \"Current in section BC is\",Ibc1,\"A\"\n",
-      "print \"Current in section CD is\",Icd1,\"A\"\n",
-      "print \"Current in section DA is\",Ida1,\"A\"\n",
-      "print \"Voltage at A is\",magVap*(1+0j),\"V/phase\"\n",
-      "print \"Voltage at B is\",Vb,\"V/phase\"\n",
-      "print \"Voltage at C is\",Vc,\"V/phase\"\n",
-      "print \"Voltage at D is\",Vd,\"V/phase\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in section AB is (139.8-42.8j) A\n",
-        "Current in section BC is (99.8-12.8j) A\n",
-        "Current in section CD is (-20.2-12.8j) A\n",
-        "Current in section DA is (-80.82+22.2j) A\n",
-        "Voltage at A is (6351+0j) V/phase\n",
-        "Voltage at B is (6185.52-41.08j) V/phase\n",
-        "Voltage at C is (6054.24-115.54j) V/phase\n",
-        "Voltage at D is (6064-95.2j) V/phase\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.7, Page Number: 368"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Pr = 10                        #load connected to line R(kW)\n",
-      "Py = 8                        #load connected to line Y(kW)\n",
-      "Pb = 5                        #load connected to line B(kW)\n",
-      "Vl = 400                      #line voltage(V)\n",
-      "#the loads are non-reactive.\n",
-      "\n",
-      "#Calculation:\n",
-      "Vp = round(Vl/3**0.5)                    #phase voltage(V)\n",
-      "Ir = Pr*1000/Vp                 #A\n",
-      "Iy = Py*1000/Vp                  #A\n",
-      "Ib = Pb*1000/Vp                  #A\n",
-      "\n",
-      "#Resolving the three currents along x-axis and y-axis, we have,\n",
-      "Ih = Iy*math.cos(math.pi/6)-Ib*math.cos(math.pi/6)         #Resultant horizontal component(A)\n",
-      "Iv = Ir-Iy*math.cos(math.pi/3)-Ib*math.cos(math.pi/3)       #Resultant vertical component(A)\n",
-      "In = (Ih**2+Iv**2)**0.5          #current in neutral wire(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Ir =\",round(Ir,1),\"A\"\n",
-      "print \"    Iy =\",round(Iy,1),\"A\"\n",
-      "print \"    Ib =\",round(Ib,2),\"A\"\n",
-      "print \"(ii) Current in neutral wire is\",round(In,1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Ir = 43.3 A\n",
-        "    Iy = 34.6 A\n",
-        "    Ib = 21.65 A\n",
-        "(ii) Current in neutral wire is 18.9 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.8, Page Number: 369"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vl = 400                          #line voltage(V)\n",
-      "Vp = 230                          #voltage across lamp(V)\n",
-      "I1 = 70                           #current in load RN(A)\n",
-      "I2 = 84                           #current in load YN(A)\n",
-      "I3 = 33                           #current in load BN(A)\n",
-      "Im = 200                         #current taken by the motor(A)\n",
-      "pf = 0.2                        #power factor(lagging)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Lamp load alone:\n",
-      "Ih = I2*math.cos(math.pi/6)-I3*math.cos(math.pi/6)         #Resultant H-component(A)\n",
-      "Iv = I1-I3*math.cos(math.pi/3)-84*math.cos(math.pi/3)      #Resultant V-component(A)\n",
-      "In = math.sqrt(Ih**2+Iv**2)                          #Neutral current(A)\n",
-      "\n",
-      "\n",
-      "#Both lamp load and motor load:\n",
-      "Ir = Im*pf                             #Active component of motor current(A)\n",
-      "Ix = Im*math.sin(math.acos(pf))        #Reactive component of motor current(A)\n",
-      "IR = ((Ir+I1)**2+Ix**2)**0.5          #A\n",
-      "IY = ((Ir+I2)**2+Ix**2)**0.5          #A\n",
-      "IB = ((Ir+I3)**2+Ix**2)**0.5          #A\n",
-      "P = Vp*(I1+I2+I3)*1                   #Watt       #( cos phy_L = 1)\n",
-      "Pm = 3**0.5*Vl*Im*pf                  #Power supplied to motor(W)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Lamp load alone: neutral curent is\",round(In,2),\"A\"\n",
-      "print \"\\nWhen Both lamp load and motor load is present:\"\n",
-      "print \"The current components are:\"\n",
-      "print \"Neutral current is\",round(In,2),\"A\"\n",
-      "print \"IR =\",round(IR,1),\"A;\\tIY =\",round(IY,0),\"A;\\tIB =\",round(IB,2),\"A\"\n",
-      "print \"Power supplied to the lamp is\",P,\"W\"\n",
-      "print \"Power supplied to the motor is\",round(Pm),\"W\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Lamp load alone: neutral curent is 45.64 A\n",
-        "\n",
-        "When Both lamp load and motor load is present:\n",
-        "The current components are:\n",
-        "Neutral current is 45.64 A\n",
-        "IR = 224.7 A;\tIY = 232.0 A;\tIB = 209.11 A\n",
-        "Power supplied to the lamp is 43010 W\n",
-        "Power supplied to the motor is 27713.0 W\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.9, Page Number: 370"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Prn = 20                          #kW\n",
-      "pf1 = 1                          #power factor of loaf RN\n",
-      "kVAyn = 28.75                   #kVA of load YN\n",
-      "kVAbn = 28.75                   #kVA of load BN\n",
-      "pf2 = 0.866                     #power factor of laod YN & BN each.(lagging)\n",
-      "Vl = 400                        #line voltage(V)\n",
-      "Vp = 230                         #phase voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "phy3 = phy2\n",
-      "Ir = Prn*1000/Vp                   #A\n",
-      "Iy = kVAyn*1000/Vp                 #A\n",
-      "Ib = kVAbn*1000/Vp                 #A\n",
-      "Ih = Ir-Iy*math.cos(phy2)-Ib*math.cos(phy2)          #A\n",
-      "Iv = 0+Iy*math.sin(phy2)-Iy*math.sin(phy3)           #A\n",
-      "In = math.sqrt(Ih**2+Iv**2)                               #A\n",
-      "#When load from B to N removed.:\n",
-      "#When the load from B to N is removed, the various line currents are:\n",
-      "#Ir in phase with Vrn; Iy lagging by 30 deg.; Ib = 0.\n",
-      "Ir1 = Ir\n",
-      "Iy1 = Iy;     Ib1 = 0                    #A\n",
-      "Ih1 = Ir1-Iy1*math.cos(math.pi/6)             #A\n",
-      "Iv1 = 0-Iy1*math.sin(math.pi/6)               #A\n",
-      "In1 = math.sqrt(Ih1**2+Iv1**2)               #A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"When no changes were made, the various currents are:\"\n",
-      "print \"Ir =\",round(Ir,2),\"A;\\tIy =\",Iy,\"A;\\tIb =\",Ib,\"A;\\tIn =\",round(In,2),\"A\"\n",
-      "print \"\\nWhen load from B to N removed, the various currents are:\"\n",
-      "print \"Ir =\",round(Ir1,2),\"A;\\tIy =\",Iy1,\"A;\\tIb =\",Ib1,\"A;\\tIn =\",round(In1,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "When no changes were made, the various currents are:\n",
-        "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 125.0 A;\tIn = 129.54 A\n",
-        "\n",
-        "When load from B to N removed, the various currents are:\n",
-        "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 0 A;\tIn = 66.03 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.10, Page Number: 371"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vl = 400                       #line voltage(V)\n",
-      "Vp = 230                         #phase voltage(V)\n",
-      "Ir = 30                        #load current at R-phase(A)\n",
-      "pf1 = 0.866                    #power factor for R-phase(lagging)\n",
-      "Iy = 30                        ##load current at Y-phase(A)\n",
-      "pf2 = 0.866                    #power factor for R-phase(lagging)\n",
-      "Ib = 30                         ##load current at R-phase(A)\n",
-      "pf3 = 1.0                    #power factor for R-phase(lagging)\n",
-      "R = 0.2                       #resistance of each line conductor(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "phy3 = math.acos(pf3)\n",
-      "\n",
-      "VR = Vp*(1+0j)                #V\n",
-      "VY = Vp*(math.cos(-2*math.pi/3)+math.sin(-2*math.pi/3))        #V\n",
-      "VB = Vp*(math.cos(2*math.pi/3)+math.sin(-2*math.pi/3))          #V\n",
-      "\n",
-      "#the line currents can be expressed as :\n",
-      "IR = cmath.rect(30,-math.pi/6)                    #A\n",
-      "IY = cmath.rect(30,-math.pi/2)                     #A\n",
-      "IB = cmath.rect(30,2*math.pi/3)                   #A\n",
-      "IN = IR+IY+IB                                #A\n",
-      "\n",
-      "#Since, the area of X-section of neutral is half of any line conductor.\n",
-      "Rn = 2*R                                 #resistance of neutral(ohm)\n",
-      "#ER = VR + Drop in R phase + Drop in neutral\n",
-      "ER = VR+R*IR+IN*2*R                    #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The supply end voltage for R phase is\",round(ER.real,3)+1j*round(ER.imag,3),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The supply end voltage for R phase is (239.588-10.608j) V\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 14.11, Page Number: 371"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vl = 400                    #line voltage(V)\n",
-      "Vp = 230                      #phase voltage(A)\n",
-      "Pln = 100                   #load connected b/n LN(W)\n",
-      "Pyn = 150                   #load connected b/n YN(W)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#before disconnecting the neutral wire,\n",
-      "R1 = Vp**2/Pln                        #Resistance of lamp L1(ohm)\n",
-      "R2 = Vp**2/Pyn                        #Resistance of lamp L2(ohm)\n",
-      "\n",
-      "#When the neutral wire is disconnected,\n",
-      "EL = 400                              #V\n",
-      "I = EL/(R1+R2)                       #A\n",
-      "V1 = I*R1                            #Voltage across lamp L1(V)\n",
-      "V2 = I*R2                            #Voltage across lamp L2(V)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The voltage across the lamps are:\"\n",
-      "print \"Lamp 1, Voltage =\",V1,\"V ;\\tLamp 2, voltage =\",V2,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage across the lamps are:\n",
-        "Lamp 1, Voltage = 240.0 V ;\tLamp 2, voltage = 160.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter15_1.ipynb b/Principles_of_Power_System/chapter15_1.ipynb
deleted file mode 100644
index 35b4fcc7..00000000
--- a/Principles_of_Power_System/chapter15_1.ipynb
+++ /dev/null
@@ -1,126 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:99f2463b3650514f9c88ec63c2c7f7ef832ae42348e5f3fe2b1b870ab9434349"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 15: Voltage Control"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 15.1, Page Number: 384"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Pl = 10000                      #load(kW)\n",
-      "pf = 0.8                        #power factor(lag)\n",
-      "Vl = 33000                      #receiving end line voltage(V)\n",
-      "R = 5                           #line resistance(ohm)\n",
-      "X = 10                          #line reactance(ohm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I2 = math.floor(Pl*1000/(3**0.5*Vl*0.8))        #load current(A)\n",
-      "Ip = I2*pf                        #A\n",
-      "Iq = I2*math.sin(math.acos(pf))     #A\n",
-      "V1 = round(Vl/3**0.5)                      #sending end voltage(V)\n",
-      "#Let Im be the current taken by the synchronous condenser.\n",
-      "Im = symbols('Im')                   #A\n",
-      "Im1 = solve((V1+Ip*R-X*(Im-Iq))**2+(Ip*X+(Im-Iq)*R)**2-V1**2,Im)[0]\n",
-      "C = 3*V1*round(Im1)/1000                   #kVAR\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacity of synchronous condenser is\",math.floor(C),\"kVAR\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacity of synchronous condenser is 13203.0 kVAR\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 15.2 Page Number: 385"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Pl = 25000                      #load(kW)\n",
-      "pf = 0.8                        #power factor(lag)\n",
-      "Vl = 33000                      #receiving end line voltage(V)\n",
-      "R = 5                           #line resistance(ohm)\n",
-      "X = 20                          #line reactance(ohm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I2 = round(Pl*1000/(3**0.5*Vl*pf),1)        #load current(A)\n",
-      "Ip = I2*pf                        #A\n",
-      "Iq = I2*math.sin(math.acos(pf))     #A\n",
-      "V1 = Vl/3**0.5                      #sending end voltage(V)\n",
-      "#Let Im be the current taken by the synchronous condenser.\n",
-      "Im = symbols('Im')                   #A\n",
-      "Im1 = solve((V1+Ip*R-X*(Im-Iq))**2+(Ip*X+(Im-Iq)*R)**2-V1**2,Im)[0]\n",
-      "C = 3*V1*Im1/1000                   #kVAR\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacity of synchronous condenser is\",round(C/1000,2),\"MVAR\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacity of synchronous condenser is 33.11 MVAR\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter17_1.ipynb b/Principles_of_Power_System/chapter17_1.ipynb
deleted file mode 100644
index f0b77b2e..00000000
--- a/Principles_of_Power_System/chapter17_1.ipynb
+++ /dev/null
@@ -1,842 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:85e6b6fc08cf389e51cb8c88e6ef45df0ab76627422a51dd5811f81a969b3ed6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 17: Symmetrical Fault Calculations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.1, Page Number: 402"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "kVAa = 15000\n",
-      "kVAb = 20000\n",
-      "V = 12000\n",
-      "kVA_base = 35000\n",
-      "Xa = 30                    #%reactance of alternator A(%)\n",
-      "Xb = 50                    #%reactance of alternator B(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Xa1 = kVA_base/kVAa*Xa            #% Reactance of alternator A at the base kVA\n",
-      "Xb1 = kVA_base/kVAb*Xb            #% Reactance of alternator B at the base kVA\n",
-      "I = kVA_base*1000/(3**0.5*V)   #Line current corresponding to 35000 kVA at 12 kV\n",
-      "\n",
-      "X = Xa1*Xb1/(Xa1+Xb1)            #Total % reactance from generator neutral up to fault point\n",
-      "Isc = I*100/X                    #Short-circuit current(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The short-circuit current is\",round(Isc),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The short-circuit current is 4330.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.2, Page Number: 404"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "kVA = 20000              #kVA rating of alternator\n",
-      "V = 10000              #voltage rating of alternator\n",
-      "Xa = 5                    # % reactance of alternator(ohm)\n",
-      "\n",
-      "#Calculation:\n",
-      "I = kVA*1000/(3**0.5*V)              #full load current(A)\n",
-      "Vp = V/3**0.5                      #phase voltage(A)\n",
-      "#As the short-circuit current is to be 8 times the full-load current,\n",
-      "Xr = 1/8*100\n",
-      "Xe = Xr-Xa                         #External % reactance required\n",
-      "X = Xe*Vp/(I*100)            #per phase external reactance required(ohm)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Per phase external reactance required is\",X,\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Per phase external reactance required is 0.375 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 49
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.3, Page Number: 404"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vl = 10                            #transmission line voltage(kV)\n",
-      "Rl = 1                             #line resistance(ohm)\n",
-      "Xl = 4                             #line reactance(ohm)\n",
-      "MVAtf = 5                          #transformer rating(MVA)\n",
-      "Xtf = 5                            #reactance of transformer(%)\n",
-      "MVAal = 10                         #rating of alternator(MVA)\n",
-      "Xal = 10                           #reactance of alternator(%)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let 10,000 kVA be the base kVA\n",
-      "kVAb = 10000                          #base kVA\n",
-      "Xalb = kVAb/(MVAal*1000)*Xal          #% reactance of alternator on base kVA\n",
-      "Xtfb = kVAb/(MVAtf*1000)*Xtf          #% reactance of transformer on base kVA\n",
-      "Xl1 = kVAb*Xl/(10*Vl**2)              #% reactance of transmission line\n",
-      "Rl1 = kVAb*Rl/(10*Vl**2)              #% resistance of transmission line\n",
-      "\n",
-      "\n",
-      "#(i)For a fault at the end of a transmission line (point F2),\n",
-      "Xt = Xalb+Xtfb+Xl1                   #Total % reactance\n",
-      "Z = (Xt**2+Rl1**2)**0.5    #% impedance from generator neutral upto fault point F2\n",
-      "SCkVA1 = kVAb*100/Z                  #Short-circuit kVA\n",
-      "\n",
-      "\n",
-      "#(ii)For a fault at the high voltage terminals of the transformer (point F1),\n",
-      "#Total % reactance from generator neutral upto fault point F1:\n",
-      "Xt1 = Xalb+Xtfb\n",
-      "SCkVA2 = kVAb*100/Xt1               #Short-circuit kVA\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) SC kVA for 1st case = \",round(SCkVA1),\"kVA\"\n",
-      "print \"(ii)SC kVA for 2nd case = \",round(SCkVA2),\"kVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) SC kVA for 1st case =  16440.0 kVA\n",
-        "(ii)SC kVA for 2nd case =  50000.0 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 50
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.4, Page Number: 405"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "kVAb = 10000                   #kVA base\n",
-      "Xt = 5                         #reactance of each transformer(%)\n",
-      "kVAt = 5000                    #kVA rating of each transformer\n",
-      "kVA1 = 10000                  #kVA of generator A & B each\n",
-      "kVA3 = 5000                    #kVA of generator C\n",
-      "Xa = 12                        #reactance of generator A & B each(%)\n",
-      "Xc = 18                        #reactance of generator c(%)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#The % reactance of generators A, B and C and that of\n",
-      "#each transformer on the selected base kVA will be:\n",
-      "XA = Xa*kVAb/kVA1\n",
-      "XB = Xa*kVAb/kVA1\n",
-      "XC = Xc*kVAb/kVA3\n",
-      "XT = Xt*kVAb/kVAt\n",
-      "\n",
-      "#(i) When the fault occurs on the low voltage side of the\n",
-      "#transformer,\n",
-      "#Total % reactance from generator neutral upto fault point F1\n",
-      "XT1 = XA/2*XC/(XA/2+XC)             #%\n",
-      "MVAf1 = kVAb*100/XT1/1000            #Fault MVA\n",
-      "\n",
-      "#(i) When the fault occurs on the high voltage side of the\n",
-      "#transformer\n",
-      "#Total % reactance from generator neutral upto fault point F2\n",
-      "XT2 = XT1+XT                       #%\n",
-      "MVAf2 = kVAb*100/XT2/1000          #Fault MVA\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Maximum fault MVA which the circuit breakers on:\"\n",
-      "print \"(i) low voltage side is\",round(MVAf1,1),\"MVA\"\n",
-      "print \"(ii)high voltage side is\",round(MVAf2),\"MVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum fault MVA which the circuit breakers on:\n",
-        "(i) low voltage side is 194.4 MVA\n",
-        "(ii)high voltage side is 66.0 MVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.5, Page Number: 407"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "#Variable declaration:\n",
-      "X1 = 10                      #reactance of generator 1 & 2 each(%)\n",
-      "X3 = 12                      #reactance of generator 3 & 4 each(%)\n",
-      "kVA1 = 10000                 #kVA rating of generator 1 & 2 each\n",
-      "kVA3 = 8000                 #kVA rating of generator 3 & 4 each\n",
-      "Xr = 10                      #reactance of reactor(%)\n",
-      "kVAr = 5000                  #kVA of reactor\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Fig. above shows the single line diagram of the network.\n",
-      "kVAb = 10000                    #base kVA\n",
-      "X1b = X1*kVAb/kVA1             #% Reactance of generator 1 or 2 on the base kVA\n",
-      "X3b = X3*kVAb/kVA3             #% Reactance of generator 3 or 4 on the base kVA\n",
-      "Xrb = Xr*kVAb/kVAr             #% Reactance of bus-bar reactor on the base kVA\n",
-      "\n",
-      "#After the fault occurs,\n",
-      "Xt = ((X1b/2+Xrb)*X3b/2)/(X1b/2+Xrb+X3b/2)\n",
-      "kVAf = kVAb*100/Xt\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Required fault MVA is\",round(kVAf/1000,2),\"MVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required fault MVA is 173.33 MVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.6, Page Number: 408"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "kVAa = 3000                   #kVA rating of generator A\n",
-      "kVAb = 4500                   #kVA rating of generator B\n",
-      "Xa = 7                        #% reactance of gen A\n",
-      "Xb = 8                        #% reactance of gen A\n",
-      "kVAt = 7500                   #kVA rating of transformer\n",
-      "Xt = 7.5                      #% reactance of transformer\n",
-      "Vb = 3.3                      #bus voltage(kV)\n",
-      "\n",
-      "#Calculation:\n",
-      "kVAbs = 7500                   #base kVA(say)\n",
-      "XA = Xa*kVAbs/kVAa         #% Reactance of generator A on the base kVA\n",
-      "XB = Xb*kVAbs/kVAb         #% Reactance of generator B on the base kVA\n",
-      "XT = Xt*kVAbs/kVAt         #% Reactance of transformer on the base kVA\n",
-      "X = symbols('X')               #percentage reactance of the bus-bar reactor\n",
-      "Xt1 = (XA*XB/(XA+XB))*(X+XT)/((XA*XB/(XA+XB))+(X+XT))\n",
-      "SCkVA = kVAt*100*Xt1                    #Short-circuit kVA\n",
-      "#But the short-circuit kVA should not exceed 150 * 10**3 kVA,\n",
-      "#the rupturing capacity of the breaker.\n",
-      "\n",
-      "X1 = abs(solve(SCkVA-150*1000,X)[0])\n",
-      "x = X1*10*Vb**2/kVAt                    #reactance of the reactor(ohm)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Reactance of the reactor per phase is\",round(x,3),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reactance of the reactor per phase is 0.106 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 53
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.7, Page Number: 409"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaratioon:\n",
-      "MVAbase = 100                    #base MVA\n",
-      "MVAa = 1500                    #estimated short-circuit MVA at busbar A\n",
-      "MVAb = 1200                   #estimated short-circuit MVA at busbar B\n",
-      "kV = 33                        #generated voltage at each station(kV)\n",
-      "x = 1                        #transmission line reactance(ohm)\n",
-      "\n",
-      "#Calculation:\n",
-      "Xa = MVAbase/MVAa*100         #% Reactance of station A on the base MVA\n",
-      "Xb = MVAbase/MVAb*100         #% Reactance of station B on the base MVA\n",
-      "Xt = MVAbase*1000*x/(10*kV**2)\n",
-      "\n",
-      "#Fault on station A.\n",
-      "Xt1 = (Xb+Xt)*Xa/(Xa+Xb+Xt)         #Total % reactance upto fault point F1\n",
-      "SCMVA1 = MVAbase*100/Xt1            #Short-circuit MVA\n",
-      "\n",
-      "#Fault on station B.\n",
-      "Xt2 = (Xa+Xt)*Xb/(Xb+Xa+Xt)         #Total % reactance upto fault point F2\n",
-      "SCMVA2 = MVAbase*100/Xt2            #Short-circuit MVA\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Fault on station A, the short circuit MVA is\",round(SCMVA1),\"MVA\"\n",
-      "print \"Fault on station B, the short circuit MVA is\",round(SCMVA2),\"MVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fault on station A, the short circuit MVA is 2071.0 MVA\n",
-        "Fault on station B, the short circuit MVA is 1831.0 MVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.8, Page Number: 410"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "kVAbase = 5000                     #base kVA\n",
-      "Xr = 6                             #% reactance of reactor\n",
-      "Xg = 12                            #% reactance of each generator\n",
-      "kVAg = 5000                        #given generator rating(kVA)\n",
-      "kVAr = 5000                        #given reactor rating(kVA)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) With reactors:\n",
-      "#Suppose a 3-phase short-circuit fault occurs on section 3 of the bus-bar.\n",
-      "\n",
-      "Xt1 = round(((Xr+Xg)/2+Xr)*Xg/(((Xr+Xg)/2+Xr)+Xg),2)   #% reactance from gen. neutral upto fault point F      \n",
-      "SCkVA1 = kVAbase*100/Xt1                     #Short-circuit input\n",
-      "\n",
-      "#(ii) Without reactors:\n",
-      "Xt2 = Xg/3                     #Total % reactance upto fault point F\n",
-      "SCkVA2 = kVAbase*100/Xt2\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) With reactors, short circuit MVA\",round(SCkVA1/1000,3),\"MVA\"\n",
-      "print \"(ii) With reactors, short circuit MVA\",SCkVA2/1000,\"MVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) With reactors, short circuit MVA 74.963 MVA\n",
-        "(ii) With reactors, short circuit MVA 125.0 MVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.9, Page Number: 411"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVAbase = 5                        #base MVA\n",
-      "MVAg = 10                         #MVA of generator\n",
-      "MVAr = 10                         #MVA of reactor\n",
-      "MVAtr = 5                        #MVA of transformer\n",
-      "xr = 10                           #% reactance of each reactor\n",
-      "xg = 30                           #% reactance of each generator\n",
-      "xtr = 5                           #% reactance of each transformer\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Xg = xg*MVAbase/MVAg                 #%age reactance of each generator on the base MVA\n",
-      "Xr = xr*MVAbase/MVAr               #%age reactance of each reactor on the base MVA\n",
-      "Xtr = xtr*MVAbase/MVAtr             #%age reactance of each transformer on the base MVA\n",
-      "\n",
-      "#Total %age reactance from generator neutral upto fault point F\n",
-      "Xt = ((Xg+Xr)/2+Xtr)*Xg/(((Xg+Xr)/2+Xtr)+Xg)+Xr\n",
-      "SCMVA = MVAbase*100/Xt                    #short circuit MVA\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Short circuit MVA is\",SCMVA,\"MVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Short circuit MVA is 40.0 MVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.10, Page Number: 412"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#Let N = no. of section in bus bar\n",
-      "#Q = kVA rating of generator\n",
-      "#x = reactance of reactance\n",
-      "#b = bus reactances.\n",
-      "N,Q,x,b = symbols('N Q x b')\n",
-      "\n",
-      "#Calculation:\n",
-      "X1 = (b+(x+b)/(N-1))*x/(b+(x+b)/(N-1)+x)           # %\n",
-      "SCkVA = Q*100/X1                            #short circuit kVA\n",
-      "\n",
-      "\n",
-      "#Now putting values:\n",
-      "Q = 50000                                #kVA\n",
-      "x = 20                                # %\n",
-      "b = 10                                 #  %\n",
-      "\n",
-      "#(i) With 3 sections\n",
-      "N1 = 3\n",
-      "X1 = (b+(x+b)/(N1-1))*x/(b+(x+b)/(N1-1)+x)\n",
-      "SCkVA1 = Q*100/X1\n",
-      "\n",
-      "#(ii) With 9 sections\n",
-      "N2 = 9\n",
-      "X2 = (b+(x+b)/(N2-1))*x/(b+(x+b)/(N2-1)+x)\n",
-      "SCkVA2 = Q*100/X2\n",
-      "\n",
-      "#(ii) When N is very large\n",
-      "N3 = 9999999999999              #say\n",
-      "X3 = (b+(x+b)/(N3-1))*x/(b+(x+b)/(N3-1)+x)\n",
-      "SCkVA3 = Q*100/X3\n",
-      "\n",
-      "#Result:\n",
-      "print \"Short circuit kVA\"\n",
-      "print \"(i)  For 3 sections is\",SCkVA1,\"kVA\"\n",
-      "print \"(ii) For 9 sections is\",round(SCkVA2),\"kVA\"\n",
-      "print \"(iii)For large N is\",round(SCkVA3),\"kVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Short circuit kVA\n",
-        "(i)  For 3 sections is 450000.0 kVA\n",
-        "(ii) For 9 sections is 613636.0 kVA\n",
-        "(iii)For large N is 750000.0 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.11, Page Number: 414"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVAbs = 50                       #base MVA\n",
-      "MVAg = 10                        #MVA of each generators\n",
-      "xg = 20                           #% reactance\n",
-      "xt = 10                          #reactance of transformer(%)\n",
-      "MVAt = 50                         #MVA of transformer\n",
-      "kV = 33                          #bus voltage\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Xg = MVAbs/MVAg*xg            #% reactance of each of the generator on base MVA\n",
-      "Xt = MVAbs/MVAt*xt            #% reactance of the transformer on base MVA\n",
-      "\n",
-      "#Suppose the required reactance of the reactor is X % on 50 MVA base.\n",
-      "X = symbols('X')\n",
-      "#The reactances of the four generators are in parallel\n",
-      "#& their equivalent reactance = 100/4 = 25%.\n",
-      "\n",
-      "Xtt = (Xg/4*(X+10))/(Xg/4+(X+10))\n",
-      "\n",
-      "#Now fault MVA at F is not to exceed 500 MVA.\n",
-      "Xreq = MVAbs*100/500                 #required reactance(%)\n",
-      "X1 = solve(Xtt-Xreq,X)[0]\n",
-      "Xrt = 10*kV**2*X1/(MVAbs*10**3)        #Reactance of the reactor(ohm)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Reactance of the reactor is\",round(Xrt,3),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reactance of the reactor is 1.452 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.12, Page Number: 415"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "kVAg = 5000                      #kVA rating of generator\n",
-      "V = 6600                        #voltage rating(V)\n",
-      "x = 6                            #reactance of generator(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "X = symbols('X')                 #% reactance of the reactor\n",
-      "kVAbs = 5000                      #base kVA\n",
-      "#The short-circuit kVA is not to exceed 5 \u00d7 5000 kVA.\n",
-      "X1 = solve(kVAbs*100/(X+x)-5*kVAg)[0]\n",
-      "\n",
-      "X11 = X1*10*(V/1000)**2/kVAg                #reactance in ohm\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The required reactance is\",round(X11,2),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required reactance is 1.22 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.13, Page Number: 416"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVAg1 = 15                        #MVA rating of A & B generators\n",
-      "x1 = 12                           #reactance of A & B(%)\n",
-      "MVAg3 = 8                          #MVA rating of generator C\n",
-      "x3 = 10                          #reactance of C(%)\n",
-      "MVAt = 5                         #MVA rating of each transformer\n",
-      "xt = 4                           #reactance of each transformer(%)\n",
-      "MVAr = 10                         #MVA of reactor\n",
-      "xr = 15                           #reactance of reactor(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let 10 MVA be the base MVA.\n",
-      "MVAbs = 10\n",
-      "#The percentage reactance of various elements on the selected\n",
-      "#base MVA will be :\n",
-      "Xa = MVAbs/MVAg1*x1\n",
-      "Xb = MVAbs/MVAg1*x1\n",
-      "Xc = MVAbs/MVAg3*x3\n",
-      "Xt = MVAbs/MVAt*xt\n",
-      "\n",
-      "\n",
-      "\n",
-      "#After the fault occurs,\n",
-      "#The reactances of generators A and B are in parallel & their\n",
-      "#equivalent reactance is 8%/2 = 4%.\n",
-      "\n",
-      "#Total reactance upto fault point F:\n",
-      "XT = ((Xa*Xb)/(Xa+Xb)+Xt)*(Xc+xr)/(((Xa*Xb)/(Xa+Xb)+Xt)+(Xc+xr))\n",
-      "MVA = MVAbs*100/XT                    #Fault MVA\n",
-      "\n",
-      "#Result:\n",
-      "print \"Total reactance upto fault point F is\",round(MVA,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total reactance upto fault point F is 119.7 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.14, Page Number: 417"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVAa = 10                          #MVA ratng of alterator\n",
-      "xa = 20                              #reactance of alterator(%)\n",
-      "MVAt = 5                             #MVA of tranformer\n",
-      "xt = 10                            #reactance of transformer(%)\n",
-      "V1 = 6.6                             #voltage on alterator side(kV)\n",
-      "V2 = 33                           #voltage on transmission line side(kV)\n",
-      "xl = 50                            #line reactance(ohm)\n",
-      "rl = 10                           #line resistance(ohm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "MVAbs = 10                        #base MVA\n",
-      "Xa = MVAbs/MVAa*xa                  #% reactance of the alternator on base MVA\n",
-      "Xt = MVAbs/MVAt*xt                #% reactance of the transformer on base MVA\n",
-      "Xl = MVAbs*1000*xl/(10*V2**2)       #% reactance of the transmission line\n",
-      "Rl = MVAbs*1000*rl/(10*V2**2)        #% resistance of the transmission line\n",
-      "#When the symmetrical fault occurs at point F on the transmission line (50 km away), then\n",
-      "XT = Xa+Xt+Xl                     #Total % reactance upto the point of fault F\n",
-      "Z = math.sqrt(XT**2+Rl**2)            #% impedance from generator neutral upto fault point F\n",
-      "SCMVA = MVAbs*100/Z               #Short-circuit MVA\n",
-      "Isc = SCMVA*10**6/(3**0.5*V1*1000)    #Short-circuit current fed to the fault by the alternator\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Short-circuit current fed to the fault by the alternator is\",round(Isc),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Short-circuit current fed to the fault by the alternator is 1012.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 17.15, Page Number: 418"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#the ratings of the machines and equipments are shown in fig above.\n",
-      "\n",
-      "MVAbs = 10\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "X1 = MVAbs/10*12       #% reactance of each generator (A, B, C and D) on the base MVA\n",
-      "X2 = MVAbs/10*24              #% reactance of the reactor on the base MVA\n",
-      "X3 = MVAbs/6*3          #% reactance of the transformer on the base MVA\n",
-      "\n",
-      "#When fault occurs at point F,\n",
-      "XT = (30*6/(30+6))+5          #% reactance from generator neutral upto fault point F\n",
-      "MVAf = MVAbs*100/XT           #Fault MVA\n",
-      "Isc = 100*10**6/(3**0.5*66000)          #Short-circuit current\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The fault current is\",round(Isc),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fault current is 875.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter18_1.ipynb b/Principles_of_Power_System/chapter18_1.ipynb
deleted file mode 100644
index d291f15a..00000000
--- a/Principles_of_Power_System/chapter18_1.ipynb
+++ /dev/null
@@ -1,1223 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c30652f9881ee18245a1635c17c8cb6587983fc4f38ea35cc91611c084543096"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 18: Unsymmetrical Fault Calculations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.2, Page Number: 429"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ir = cmath.rect(100,30*math.pi/180)          #current in phase R(A)\n",
-      "Iy = cmath.rect(50,300*math.pi/180)          #current in phase Y(A)\n",
-      "Ib = cmath.rect(30,180*math.pi/180)          #current in phase B(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "I0 = 1/3*(Ir+Iy+Ib)                    #A\n",
-      "I1 = 1/3*(Ir+a*Iy+a**2*Ib)             #A\n",
-      "I2 = 1/3*(Ir+a**2*Iy+a*Ib)             #A\n",
-      "In = Ir+Iy+Ib                          #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The positive, negative and zero sequence currents in the R-line are\"\n",
-      "print \"I0 = (\",round(abs(I0),2),round(rad2deg(angle(I0)),2),\") A\"\n",
-      "print \"I1 = (\",round(abs(I1),2),round(rad2deg(angle(I1)),1),\") A\"\n",
-      "print \"I2 = (\",round(abs(I2),2),round(rad2deg(angle(I2)),2),\") A\"\n",
-      "print \"Return current in the neutral wire is\"\n",
-      "print \"In =(\",round(abs(In),2),round(rad2deg(angle(In)),1),\") A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The positive, negative and zero sequence currents in the R-line are\n",
-        "I0 = ( 27.29 4.69 ) A\n",
-        "I1 = ( 57.98 43.3 ) A\n",
-        "I2 = ( 18.97 24.96 ) A\n",
-        "Return current in the neutral wire is\n",
-        "In =( 81.88 4.7 ) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.3, Page Number: 430"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ir = 12+6j                  #current in phase A(A)\n",
-      "Iy = 12-12j                  #current in phase Y(A)\n",
-      "Ib = -15+10j                  #current in phase B(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "#Red phase\n",
-      "Ir0 = 1/3*(Ir+Iy+Ib)            #A\n",
-      "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)     #A\n",
-      "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)     #A\n",
-      "\n",
-      "#Yellow phase:\n",
-      "Iy0 = Ir0                         #A\n",
-      "Iy1 = a**2*Ir1                    #A\n",
-      "Iy2 = a*Ir2                       #A\n",
-      "\n",
-      "\n",
-      "#Blue phase:\n",
-      "Ib0 = Ir0                        #A\n",
-      "Ib1 = a*Ir1                   #A\n",
-      "Ib2 = a**2*Ir2                   #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sequence currents of red phase are:\"\n",
-      "print \"Ir0 =\",Ir0.real+round(Ir0.imag,2)*1j,\"A\"\n",
-      "print \"Ir1 =\",round(Ir1.real,2)+round(Ir1.imag,2)*1j,\"A\"\n",
-      "print \"Ir2 =\",round(Ir2.real,2)+round(Ir2.imag,2)*1j,\"A\"\n",
-      "print \"\\nThe sequence currents of red phase are:\"\n",
-      "print \"Iy0 =\",Iy0.real+round(Iy0.imag,2)*1j,\"A\"\n",
-      "print \"Iy1 =\",round(Iy1.real,2)+round(Iy1.imag,1)*1j,\"A\"\n",
-      "print \"Iy2 =\",round(Iy2.real,1)+round(Iy2.imag,2)*1j,\"A\"\n",
-      "print \"\\nThe sequence currents of red phase are:\"\n",
-      "print \"Ib0 =\",Ib0.real+round(Ib0.imag,2)*1j,\"A\"\n",
-      "print \"Ib1 =\",round(Ib1.real,2)+round(Ib1.imag,2)*1j,\"A\"\n",
-      "print \"Ib2 =\",round(Ib2.real,2)+round(Ib2.imag,2)*1j,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sequence currents of red phase are:\n",
-        "Ir0 = (3+1.33j) A\n",
-        "Ir1 = (10.85+10.13j) A\n",
-        "Ir2 = (-1.85-5.46j) A\n",
-        "\n",
-        "The sequence currents of red phase are:\n",
-        "Iy0 = (3+1.33j) A\n",
-        "Iy1 = (3.35-14.5j) A\n",
-        "Iy2 = (5.7+1.13j) A\n",
-        "\n",
-        "The sequence currents of red phase are:\n",
-        "Ib0 = (3+1.33j) A\n",
-        "Ib1 = (-14.2+4.33j) A\n",
-        "Ib2 = (-3.8+4.33j) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.4, Page Number: 430"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Er0 = 100                        #zero sequence voltage of phase R(V)\n",
-      "Er1 = 200-100j                        #positive sequence voltage of phase R(V)\n",
-      "Er2 = -100                        #negative sequence voltage of phase R(V)\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "Er = Er0+Er1+Er2                 #V\n",
-      "Ey = Er0+a**2*Er1+a*Er2          #V\n",
-      "Eb = Er0+a*Er1+a**2*Er2          #V\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The phase voltages are:\"\n",
-      "print \"Er =\",(round(abs(Er),2),round(rad2deg(angle(Er)),2)),\"V\"\n",
-      "print \"Ey =\",(round(abs(Ey)),round(rad2deg(angle(Ey)),2)),\"V\"\n",
-      "print \"Eb =\",(round(abs(Eb),2),round(rad2deg(angle(Eb)),1)),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The phase voltages are:\n",
-        "Er = (223.61, -26.57) V\n",
-        "Ey = (213.0, -99.9) V\n",
-        "Eb = (338.59, 66.2) V\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.5, Page Number: 431"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Er0 = 0.5-0.866j                  #zero sequence components of red phase(V)\n",
-      "Er1 = 2+0j                      #positive sequence components of red phase(V)\n",
-      "Er = 3+0j                         #phase R voltage(V)\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "Er2 = Er-(Er0+Er1)                #negative sequence voltage of red phase(V)\n",
-      "Ey = Er0+a**2*Er1+a*Er2            #phase Y voltage(V)\n",
-      "Eb = Er0+a*Er1+a**2*Er2            #phase B voltage(V)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The negative sequence component of red phase is\"\n",
-      "print \"Ir2 = (\",round(abs(Er2)),round(rad2deg(angle(Er2))),\") V\"\n",
-      "print \"\\nThe phase voltages are:\"\n",
-      "print \"Iy = (\",round(abs(Ey)),round(rad2deg(angle(Ey))),\") V\"\n",
-      "print \"Ib = (\",round(abs(Eb)),round(rad2deg(angle(Eb))),\") V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The negative sequence component of red phase is\n",
-        "Ir2 = ( 1.0 60.0 ) V\n",
-        "\n",
-        "The phase voltages are:\n",
-        "Iy = ( 3.0 -120.0 ) V\n",
-        "Ib = ( 0.0 120.0 ) V\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.6, Page Number: 431"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "In = 12                      #current from neutral to ground(A)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I0 = In/3                       #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The zero phase sequence components in phases are\",I0,\"A  each\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The zero phase sequence components in phases are 4.0 A  each\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.7, Page Number: 432"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#(i) Before removal of fuses.\n",
-      "Iri = cmath.rect(90,0)                  #phase R current(A)\n",
-      "Iyi = cmath.rect(90,math.pi/180*240)         #phase Y current(A)\n",
-      "Ibi = cmath.rect(90,math.pi/180*120)         #phase B current(A)\n",
-      "\n",
-      "#(ii) After removal of fuses.\n",
-      "Irii = 90+0j                          #A\n",
-      "Iyii = 0                               #A\n",
-      "Ibii = 0                               #A\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "#(i) Before removal of fuses.\n",
-      "#Since the system is balanced, it will have only positive sequence currents.\n",
-      "Ir1i = Iri                          #A\n",
-      "Iy1i = Iyi                          #A\n",
-      "Ib1i = Ibi                          #A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(ii) After removal of fuses.\n",
-      "#The sequence currents in the three lines will be:\n",
-      "Ir0ii = 1/3*(Irii+Iyii+Ibii)         #A\n",
-      "Iy0ii = Ir0ii                         #A\n",
-      "Ib0ii = Ir0ii                        #A\n",
-      "\n",
-      "Ir1ii = 1/3*(Irii+Iyii+Ibii)          #A\n",
-      "Iy1ii = a**2*Ir1ii                    #A\n",
-      "Ib1ii = a*Ir1ii                       #A\n",
-      "\n",
-      "Ir2ii = 1/3*(Irii+a**2*Iyii+a*Ibii)    #A\n",
-      "Iy2ii = a*Ir2ii                        #A\n",
-      "Ib2ii = a**2*Ir2ii                     #A\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Before removal of fuses, the symmetrical components are:\"\n",
-      "print \"\\tIr1 = (\",round(abs(Ir1i)),round(rad2deg(angle(Ir1i))),\") A\"\n",
-      "print \"\\tIy1 = (\",round(abs(Iy1i)),360+round(rad2deg(angle(Iy1i))),\") A\"\n",
-      "print \"\\tIb1 = (\",round(abs(Ib1i)),round(rad2deg(angle(Ib1i))),\") A\"\n",
-      "print \"\\n(ii) After the removal of fuses, the symmetrical components are:\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir0ii)),round(rad2deg(angle(Ir0ii))),\") A\"\n",
-      "print \"\\tIr1 = (\",round(abs(Ir1ii)),round(rad2deg(angle(Ir1ii))),\") A\"\n",
-      "print \"\\tIr2 = (\",round(abs(Ir2ii)),round(rad2deg(angle(Ir2ii))),\") A\"\n",
-      "print \"\\n\\tIy0 = (\",round(abs(Iy0ii)),round(rad2deg(angle(Iy0ii))),\") A\"\n",
-      "print \"\\tIy1 = (\",round(abs(Iy1ii)),360+round(rad2deg(angle(Iy1ii))),\") A\"\n",
-      "print \"\\tIy2 = (\",round(abs(Iy2ii)),round(rad2deg(angle(Iy2ii))),\") A\"\n",
-      "print \"\\n\\tIb0 = (\",round(abs(Ib0ii)),round(rad2deg(angle(Ib0ii))),\") A\"\n",
-      "print \"\\tIb1 = (\",round(abs(Ib1ii)),round(rad2deg(angle(Ib1ii))),\") A\"\n",
-      "print \"\\tIb2 = (\",round(abs(Ib2ii)),360+round(rad2deg(angle(Ib2ii))),\") A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Before removal of fuses, the symmetrical components are:\n",
-        "\tIr1 = ( 90.0 0.0 ) A\n",
-        "\tIy1 = ( 90.0 240.0 ) A\n",
-        "\tIb1 = ( 90.0 120.0 ) A\n",
-        "\n",
-        "(ii) After the removal of fuses, the symmetrical components are:\n",
-        "\tIr0 = ( 30.0 0.0 ) A\n",
-        "\tIr1 = ( 30.0 0.0 ) A\n",
-        "\tIr2 = ( 30.0 0.0 ) A\n",
-        "\n",
-        "\tIy0 = ( 30.0 0.0 ) A\n",
-        "\tIy1 = ( 30.0 240.0 ) A\n",
-        "\tIy2 = ( 30.0 120.0 ) A\n",
-        "\n",
-        "\tIb0 = ( 30.0 0.0 ) A\n",
-        "\tIb1 = ( 30.0 120.0 ) A\n",
-        "\tIb2 = ( 30.0 240.0 ) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.8, Page Number: 433"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ir1 = cmath.rect(200,0)              #positive phase sequence component of current(A)\n",
-      "Ir2 = cmath.rect(100,60*math.pi/180)      #negative phase sequence component of current(A)\n",
-      "In = cmath.rect(300,300*math.pi/180)      #current in the neutral conductor(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "Ir0 = 1/3*In                        #Zero phase sequence current in R-line(A)\n",
-      "Ir = Ir0+Ir1+Ir2                    #Current in the R-line(A)\n",
-      "Iy = Ir0+a**2*Ir1+a*Ir2             #Current in the Y-line(A)\n",
-      "Ib = Ir0+a*Ir1+a**2*Ir2             #Current in the B-line(A)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Current in the R-line is\"\n",
-      "print \"\\tIr = (\",round(abs(Ir)),round(rad2deg(angle(Ir))),\") A\"\n",
-      "print \"Current in the Y-line is\"\n",
-      "print \"\\tIy = (\",round(abs(Iy)),round(rad2deg(angle(Iy))),\") A\"\n",
-      "print \"Current in the B-line is\"\n",
-      "print \"\\tIb = (\",round(abs(Ib)),round(rad2deg(angle(Ib))),\") A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Current in the R-line is\n",
-        "\tIr = ( 300.0 0.0 ) A\n",
-        "Current in the Y-line is\n",
-        "\tIy = ( 300.0 -120.0 ) A\n",
-        "Current in the B-line is\n",
-        "\tIb = ( 0.0 180.0 ) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.9, Page Number: 434"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Ir = cmath.rect(10,0)                  #current in line R(A)\n",
-      "Iy = cmath.rect(10,180*math.pi/180)                  #current in line Y(A)\n",
-      "Ib = cmath.rect(0,0)                  #current in line B(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "#For R-line:\n",
-      "Ir0 = 1/3*(Ir+Iy+Ib)                   #zero sequence component of Ir(A)\n",
-      "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)            #positive sequence component of Ir(A)\n",
-      "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)                   #negative sequence component of Iy(A)\n",
-      "\n",
-      "#For Y-line:\n",
-      "Iy0 = Ir0                             #zero sequence component of Ir(A)\n",
-      "Iy1 = a**2*Ir1                        #positive sequence component of Ir(A)\n",
-      "Iy2 = a*Ir2                          #negative sequence component of Iy(A)\n",
-      "\n",
-      "#For B-line:\n",
-      "Ib0 = Ir0                             #zero sequence component of Ir(A)\n",
-      "Ib1 = a*Ir1                           #positive sequence component of Ir(A)\n",
-      "Ib2 = a**2*Ir2                        #negative sequence component of Iy(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sequence components of R-line are:\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir0),2),round(rad2deg(angle(Ir0))),\") A\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir1),2),round(rad2deg(angle(Ir1))),\") A\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir2),2),round(rad2deg(angle(Ir2))),\") A\"\n",
-      "print \"The sequence components of Y-line are:\"\n",
-      "print \"\\tIy0 = (\",round(abs(Iy0),2),round(rad2deg(angle(Iy0))),\") A\"\n",
-      "print \"\\tIy1 = (\",round(abs(Iy1),2),round(rad2deg(angle(Iy1))),\") A\"\n",
-      "print \"\\tIy2 = (\",round(abs(Iy2),2),round(rad2deg(angle(Iy2))),\") A\"\n",
-      "print \"The sequence components of B-line are:\"\n",
-      "print \"\\tIb0 = (\",round(abs(Ib0),2),round(rad2deg(angle(Ib0))),\") A\"\n",
-      "print \"\\tIb1 = (\",round(abs(Ib1),2),round(rad2deg(angle(Ib1))),\") A\"\n",
-      "print \"\\tIb2 = (\",round(abs(Ib2),2),round(rad2deg(angle(Ib2))),\") A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sequence components of R-line are:\n",
-        "\tIr0 = ( 0.0 90.0 ) A\n",
-        "\tIr0 = ( 5.77 -30.0 ) A\n",
-        "\tIr0 = ( 5.77 30.0 ) A\n",
-        "The sequence components of Y-line are:\n",
-        "\tIy0 = ( 0.0 90.0 ) A\n",
-        "\tIy1 = ( 5.77 -150.0 ) A\n",
-        "\tIy2 = ( 5.77 150.0 ) A\n",
-        "The sequence components of B-line are:\n",
-        "\tIb0 = ( 0.0 90.0 ) A\n",
-        "\tIb1 = ( 5.77 90.0 ) A\n",
-        "\tIb2 = ( 5.77 -90.0 ) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.10, Page Number: 435"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Rrb = 10                       #resistance of branch RB(ohm)\n",
-      "Rry = 20                       #resistance of branch RY(ohm)\n",
-      "Rby = 5                       #resistance of branch BY(ohm)\n",
-      "Er = cmath.rect(-100,0)        #voltage across BY(V)\n",
-      "Ey = cmath.rect(100,60*math.pi/180)        #voltage across BR(V)\n",
-      "Eb = cmath.rect(100,-60*math.pi/180)        #voltage across RY(V)\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "IR = Er/Rby                    #Current in in branch BY(A)\n",
-      "IY = Ey/Rrb                    #Current in in branch RB(A)\n",
-      "IB = Eb/Rry                    #Current in in branch RY(A)\n",
-      "\n",
-      "#Sequence currents in resistors:\n",
-      "IR0 = 1/3*(IR+IY+IB)           #Zero sequence component of IR(A)\n",
-      "IR1 = 1/3*(IR+a*IY+a**2*IB)    #Positive sequence component of IR(A)\n",
-      "IR2 = 1/3*(IR+a**2*IY+a*IB)     #Negative sequence component of IR(A)\n",
-      "\n",
-      "IY0 = IR0                       #Zero sequence component of IY(A)\n",
-      "IY1 = a**2*IR1                    #Positive sequence component of IY(A)\n",
-      "IY2 = a*IR2                    #Negative sequence component of IY(A)\n",
-      "\n",
-      "IB0 = IR0                    #Zero sequence component of IB(A)\n",
-      "IB1 = a*IR1                  #Positive sequence component of IB(A)\n",
-      "IB2 = a**2*IR2               #Negative sequence component of IB(A)\n",
-      "\n",
-      "#Sequence currents in supply lines:\n",
-      "Ir = IB-IY                    #Line current in R-line(A)\n",
-      "Iy = IR-IB                     #Line current in Y-line(A)\n",
-      "Ib = IY-IR                     #Line current in R-line(A)\n",
-      "\n",
-      "Ir0 = 1/3*(Ir+Iy+Ib)           #Zero sequence component of Ir(A)\n",
-      "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)    #Positive sequence component of Ir(A)\n",
-      "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)     #Negative sequence component of Ir(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sequence components of R-line are:\"\n",
-      "print \"\\tIR0 = (\",round(abs(IR0),2),round(rad2deg(angle(IR0)),1),\") A\"\n",
-      "print \"\\tIR1 = (\",round(abs(IR1),2),round(rad2deg(angle(IR1))),\") A\"\n",
-      "print \"\\tIR2 = (\",round(abs(IR2),2),round(rad2deg(angle(IR2)),1),\") A\"\n",
-      "print \"\\nThe sequence components of Y-line are:\"\n",
-      "print \"\\tIY0 = (\",round(abs(IY0),2),round(rad2deg(angle(IY0)),1),\") A\"\n",
-      "print \"\\tIY1 = (\",round(abs(IY1),2),round(rad2deg(angle(IY1))),\") A\"\n",
-      "print \"\\tIY2 = (\",round(abs(IY2),1),round(rad2deg(angle(IY2)),1),\") A\"\n",
-      "print \"\\nThe sequence components of B-line are:\"\n",
-      "print \"\\tIB0 = (\",round(abs(IB0),2),round(rad2deg(angle(IB0)),1),\") A\"\n",
-      "print \"\\tIB1 = (\",round(abs(IB1),2),360+round(rad2deg(angle(IB1))),\") A\"\n",
-      "print \"\\tIB2 = (\",round(abs(IB2),1),round(rad2deg(angle(IB2)),1),\") A\"\n",
-      "print \"\\nThe Sequence currents in supply lines are:\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir0),2),round(rad2deg(angle(Ir0))),\") A\"\n",
-      "print \"\\tIr1 = (\",round(abs(Ir1),1),round(rad2deg(angle(Ir1))),\") A\"\n",
-      "print \"\\tIr2 = (\",round(abs(Ir2),2),round(rad2deg(angle(Ir2)),1),\") A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sequence components of R-line are:\n",
-        "\tIR0 = ( 4.41 160.9 ) A\n",
-        "\tIR1 = ( 11.67 180.0 ) A\n",
-        "\tIR2 = ( 4.41 -160.9 ) A\n",
-        "\n",
-        "The sequence components of Y-line are:\n",
-        "\tIY0 = ( 4.41 160.9 ) A\n",
-        "\tIY1 = ( 11.67 60.0 ) A\n",
-        "\tIY2 = ( 4.4 -40.9 ) A\n",
-        "\n",
-        "The sequence components of B-line are:\n",
-        "\tIB0 = ( 4.41 160.9 ) A\n",
-        "\tIB1 = ( 11.67 300.0 ) A\n",
-        "\tIB2 = ( 4.4 79.1 ) A\n",
-        "\n",
-        "The Sequence currents in supply lines are:\n",
-        "\tIr0 = ( 0.0 0.0 ) A\n",
-        "\tIr1 = ( 20.2 -90.0 ) A\n",
-        "\tIr2 = ( 7.64 109.1 ) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.11, Page Number: 437"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I = 20                       #current in each line(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "#Let R, Y and B be the supply lines. When fuse in the \n",
-      "#line B is removed, the various line currents are :\n",
-      "Ir = cmath.rect(I,0)                     #A\n",
-      "Iy = cmath.rect(I,math.pi)                    #A\n",
-      "Ib = cmath.rect(0,0)                     #A\n",
-      "\n",
-      "#For R-line:\n",
-      "Ir0 = 1/3*(Ir+Iy+Ib)                   #zero sequence component of Ir(A)\n",
-      "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib)            #positive sequence component of Ir(A)\n",
-      "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib)                   #negative sequence component of Iy(A)\n",
-      "\n",
-      "#For Y-line:\n",
-      "Iy0 = Ir0                             #zero sequence component of Ir(A)\n",
-      "Iy1 = a**2*Ir1                        #positive sequence component of Ir(A)\n",
-      "Iy2 = a*Ir2                          #negative sequence component of Iy(A)\n",
-      "\n",
-      "#For B-line:\n",
-      "Ib0 = Ir0                             #zero sequence component of Ir(A)\n",
-      "Ib1 = a*Ir1                           #positive sequence component of Ir(A)\n",
-      "Ib2 = a**2*Ir2                        #negative sequence component of Iy(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sequence components of R-line are:\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir0),2),round(rad2deg(angle(Ir0))),\") A\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir1),2),round(rad2deg(angle(Ir1))),\") A\"\n",
-      "print \"\\tIr0 = (\",round(abs(Ir2),2),round(rad2deg(angle(Ir2))),\") A\"\n",
-      "print \"\\nThe sequence components of Y-line are:\"\n",
-      "print \"\\tIy0 = (\",round(abs(Iy0),2),round(rad2deg(angle(Iy0))),\") A\"\n",
-      "print \"\\tIy1 = (\",round(abs(Iy1),2),360+round(rad2deg(angle(Iy1))),\") A\"\n",
-      "print \"\\tIy2 = (\",round(abs(Iy2),2),round(rad2deg(angle(Iy2))),\") A\"\n",
-      "print \"\\nThe sequence components of B-line are:\"\n",
-      "print \"\\tIb0 = (\",round(abs(Ib0),2),round(rad2deg(angle(Ib0))),\") A\"\n",
-      "print \"\\tIb1 = (\",round(abs(Ib1),2),round(rad2deg(angle(Ib1))),\") A\"\n",
-      "print \"\\tIb2 = (\",round(abs(Ib2),2),360+round(rad2deg(angle(Ib2))),\") A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sequence components of R-line are:\n",
-        "\tIr0 = ( 0.0 90.0 ) A\n",
-        "\tIr0 = ( 11.55 -30.0 ) A\n",
-        "\tIr0 = ( 11.55 30.0 ) A\n",
-        "\n",
-        "The sequence components of Y-line are:\n",
-        "\tIy0 = ( 0.0 90.0 ) A\n",
-        "\tIy1 = ( 11.55 210.0 ) A\n",
-        "\tIy2 = ( 11.55 150.0 ) A\n",
-        "\n",
-        "The sequence components of B-line are:\n",
-        "\tIb0 = ( 0.0 90.0 ) A\n",
-        "\tIb1 = ( 11.55 90.0 ) A\n",
-        "\tIb2 = ( 11.55 270.0 ) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.12, Page Number: 438"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "import cmath\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Zr = 5-10j                       #impedance connected to line R(ohm)\n",
-      "Zy = 6+5j                        #impedance connected to line Y(ohm)\n",
-      "Zb = 3+15j                       #impedance connected to line B(ohm)\n",
-      "VRY = cmath.rect(3300,0)         #line RY voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "VYB = a**2*VRY                   #line YB voltage(V)\n",
-      "\n",
-      "#Let VR, VY, and VB be the voltages across impedances in R, Y and B\n",
-      "#phases respectively and IR, IY, and IB the resulting line currents.\n",
-      "IR, IY,IB,IR0,IR1,IR2, = symbols('IR IY IB IR0 IR1 IR2')\n",
-      "\n",
-      "#IR+IY+IB = 0\n",
-      "IR0 = 0; IY0 = 0; IB0 = 0;\n",
-      "VR = Zr*(IR0+IR1+IR2)               #V\n",
-      "VY = Zy*(a**2*IR1+a*IR2)            #V\n",
-      "VB = Zb*(a*IR1+a**2*IR2)            #V\n",
-      "#solving the equations:\n",
-      "IR11 = solve(VR-VY-VRY,IR1)[0]\n",
-      "VY = Zy*(a**2*IR11+a*IR2)\n",
-      "VB = Zb*(a*IR11+a**2*IR2)\n",
-      "IR22 = solve(VY-VB-VYB,IR2)[0]\n",
-      "\n",
-      "VR = Zr*(IR0+IR1+IR22)\n",
-      "VY = Zy*(a**2*IR1+a*IR22)\n",
-      "IR11 = solve(VR-VY-VRY,IR1)[0]\n",
-      "IR = IR11+IR22                      #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The line current IR is (\",round(abs(IR)),round(-rad2deg(cmath.phase(IR)),1),\")  A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The line current IR is ( 241.0 -18.5 )  A\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.13, Page Number: 440"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import cmath\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R = 1                             #resistance of each load(ohm)\n",
-      "magVRY = 200                       #line RY voltage(V)\n",
-      "magVBR = 400                       #line BR voltage(V)\n",
-      "magVYB = 346                       #line YB voltage(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = cmath.rect(1,120*math.pi/180)\n",
-      "#This is a case of a balanced star-connected load supplied\n",
-      "#from an unbalanced 3-phase supply.\n",
-      "\n",
-      "#Now,\n",
-      "\n",
-      "#  (2)**22 = (1 + 1\u00b775 cos(theta))**2 + (1\u00b775 sin(theta))**2\n",
-      "#    theta = math.acos(4-1+3*1)/3.5\n",
-      "theta = math.pi/2\n",
-      "#    alpha = math.acos(1+1.75*math.cos(theta))\n",
-      "alpha = 60*math.pi/180\n",
-      "\n",
-      "#As the phase sequence is RYB, therefore, various line voltages are :\n",
-      "VRY = cmath.rect(200,math.pi)                   #V\n",
-      "VYB = cmath.rect(346,theta)                #V\n",
-      "VBR = cmath.rect(400,-alpha)               #V\n",
-      "\n",
-      "#The current in any phase (or line) is equal to phase voltage \n",
-      "#divided by resistance in that phase.\n",
-      "IR = VRY/(1*3**0.5)                   #A\n",
-      "IY = VYB/(1*3**0.5)                   #A\n",
-      "IB = VBR/(1*3**0.5)                   #A\n",
-      "\n",
-      "#Sequence components in red phase are :\n",
-      "IR0 = 1/3*(IR+IY+IB)                  #A\n",
-      "IR1 = 1/3*(IR+a*IY+a**2*IB)           #A\n",
-      "IR2 = 1/3*(IR+a**2*IY+a*IB)           #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The magnitude of current in any phase is:\"\n",
-      "print \"\\tIR0 = (\",round(abs(IR0)),round(rad2deg(angle(IR0))),\") A\"\n",
-      "print \"\\tIR1 = (\",round(abs(IR1),1),round(rad2deg(angle(IR1))),\") A\"\n",
-      "print \"\\tIR2 = (\",round(abs(IR2),2),round(rad2deg(angle(IR2))),\") A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The magnitude of current in any phase is:\n",
-        "\tIR0 = ( 0.0 -90.0 ) A\n",
-        "\tIR1 = ( 176.3 -169.0 ) A\n",
-        "\tIR2 = ( 66.63 30.0 ) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.14, Page Number: 449"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVAg = 10                           #MVA rating of generator\n",
-      "Vg = 11                             #voltage rating of generator(kV)\n",
-      "Zg1 = 1.2j                          #Positive sequence impedance of generator(ohm)\n",
-      "Zg2 = 0.9j                          #Negative sequence impedance of generator(ohm)\n",
-      "Zg0 = 0.4j                          #Zero sequence impedance of generator(ohm)\n",
-      "Zf1 = 1.0j                          #Positive sequence impedance of feeder(ohm)\n",
-      "Zf2 = 1.0j                          #Negative sequence impedance of feeder(ohm)\n",
-      "Zf0 = 3.0j                          #Zero sequence impedance of feeder(ohm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Suppose the fault is occured on the red phase. \n",
-      "#Taking red phase as the reference,\n",
-      "ER = Vg*1000/3**0.5                  #Phase e.m.f. of R-phase(V)\n",
-      "\n",
-      "#(i)The total impedance to any sequence current is given by\n",
-      "#the sum of  generator and feeder impedances to that sequence current.\n",
-      "Z1 = Zg1+Zf1                       #ohm\n",
-      "Z2 = Zg2+Zf2                       #ohm\n",
-      "Z0 = Zg0+Zf0                      #ohm\n",
-      "\n",
-      "#For a line-to-ground fault,\n",
-      "#I1 = I2 = I0\n",
-      "I0 = ER/(Z1+Z2+Z0)                 #A\n",
-      "IR = 3*I0                        #fault current(A)\n",
-      "\n",
-      "#(ii)Line-to-neutral voltage of R-phase,\n",
-      "VR = ER-I0*(Zg1+Zg2+Zg0)                #V\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The magnitude of fault current is\",round(IR.imag)*1j,\"A\"\n",
-      "print \"(ii) Line to neutral voltage at the generator terminal is\",round(abs(VR)),\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The magnitude of fault current is (-0-2540j) A\n",
-        "(ii) Line to neutral voltage at the generator terminal is 4234.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.15, Page Number: 450"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vg = 11000                            #voltage rating of the alternator(V)\n",
-      "MVAg = 10                            #MVA rating of generator\n",
-      "X0 = 0.05                            #zero sequence reactance(p.u)\n",
-      "X1 = 0.15                            #positive sequence reactance(p.u)\n",
-      "X2 = 0.15                            #negative sequence reactance(p.u)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Taking red phase as the reference, let its phase e.m.f. be\n",
-      "ER = 1                              #p.u\n",
-      "\n",
-      "#Line-to-ground fault: \n",
-      "#Suppose the fault occurs on the red phase.\n",
-      "#I1 = I2 = I0\n",
-      "I0 = ER/(1j*(X0+X1+X2))\n",
-      "IR = 3*I0                            #fault current(A)\n",
-      "\n",
-      "#Three phase fault: \n",
-      "#the fault current (say Ish) is limited by the positive sequence reactance only.\n",
-      "Ish = ER/(1j*X1)                          #fault current(A)\n",
-      "r = IR/Ish                             #ratio of the two fault currents\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The ratio of two fault currents is\",round(abs(r),3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The ratio of two fault currents is 1.286\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.16, Page Number: 450"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vg = 11                            #voltage rating of the alternator(kV)\n",
-      "MVAg = 25                            #MVA rating of generator\n",
-      "X0 = 0.05                            #zero sequence reactance(p.u)\n",
-      "X1 = 0.2                            #positive sequence reactance(p.u)\n",
-      "X2 = 0.2                            #negative sequence reactance(p.u)\n",
-      "Xn = 0.3                            #neutral to ground reactance(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Assume that the fault occurs on the red phase.\n",
-      "#Taking red phase as the reference, let its phase e.m.f. be \n",
-      "ER = 1                                  #p.u.\n",
-      "#First of all, convert the reactance Xn into p.u.\n",
-      "Xnp = Xn*MVAg*1000/(Vg**2*1000)         #p.u\n",
-      "#For a line-to-ground fault,\n",
-      "#I1 = I2 = I0\n",
-      "I0 = ER/(X1+X2+X0+3*Xnp)                  #p.u\n",
-      "IR = 3*I0                              #fault current(p.u)\n",
-      "IRa = MVAg*10**6/(3**0.5*Vg*1000) * IR        #fault current in amperes\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \" The fault current for a single line to ground fault is\",round(IRa),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The fault current for a single line to ground fault is 6190.0 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.17, Page Number: 451"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 10.4                            #voltage between the lines(kV)\n",
-      "X0 = 0.2j                            #zero sequence reactance(p.u)\n",
-      "X1 = 0.6j                            #positive sequence reactance(p.u)\n",
-      "X2 = 0.5j                            #negative sequence reactance(p.u)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Taking red phase as the reference, its phase e.m.f. is :\n",
-      "ER = round(V*1000/3**0.5)                #Phase e.m.f. of R-phase(V)\n",
-      "IF = 3**0.5*ER/(X1+X2)              #fault current(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The fault current is\",round(abs(IF),1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fault current is 9453.8 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.18, Page Number: 451"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "X0 = 0.05j                            #zero sequence reactance(p.u)\n",
-      "X1 = 0.08j                            #positive sequence reactance(p.u)\n",
-      "X2 = 0.07j                            #negative sequence reactance(p.u)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Taking red phase as the reference, let its phase e.m.f. be \n",
-      "ER = 1                            #p.u\n",
-      "#For a double line-to-ground fault,\n",
-      "#IF = IY+IB\n",
-      "IF = -3*X2*ER/(X1*X2+X2*X0+X2*X0)         #fault current(A)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The fault current is\",round(IF.imag,1)*1j,\"p.u\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fault current is 16.7j p.u\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.19, Page Number: 452"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVAg = 20                        #MVA rating of generator\n",
-      "V = 11                           #voltage rating of generator(V)\n",
-      "X1 = 20                          #positive sequence reactance of generator(%)\n",
-      "X2 = 10                         #negative sequence reactance of generator(%)\n",
-      "X0 = 15                        #zero sequence reactance of generator(%)\n",
-      "Xn = 5                       #generator 1 neutral reactance(%)\n",
-      "\n",
-      "#Calculation:\n",
-      "ER = V*1000/3**0.5                  #V\n",
-      "#The % reactances in Fig. 18.19b can be converted into ohmic values as under:\n",
-      "x1 = X1*V**2*10/(MVAg*1000)                     #ohm\n",
-      "x2 = X2*V**2*10/(MVAg*1000)                     #ohm\n",
-      "x0 = X0*V**2*10/(MVAg*1000)                     #ohm\n",
-      "IR = 3*ER/(x1+x2+x0)*(-1j)                           #ohm\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Fault current is\",round(IR.imag)*1j,\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fault current is (-0-6998j) A\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 18.20, Page Number: 453"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "IF1 = 2000                       #3 phase fault current(A)\n",
-      "IF2 = 2600                       #line-to-line fault(A)\n",
-      "IF0 = 4200                       #line-toground fault(A)\n",
-      "MVAg = 50                        #MVA rating of the generator\n",
-      "V = 11                           #voltage rating of the generator(V)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let X1 = positive sequence reactance,\n",
-      "#    X2 = negative sequence reactance,and\n",
-      "#    X0 = zero sequence reactance  of the alternator.\n",
-      "X1,X2,X0 = symbols('X1 X2 X0')\n",
-      "Eph = V*1000/3**0.5                    #phase voltage(V)\n",
-      "X11 = solve(Eph/X1-IF1)[0]                      #positive sequence reactance(ohm)\n",
-      "X22 = solve(3**0.5*Eph/(X11+X2)-IF2,X2)[0]      #negative sequence reactance(ohm)\n",
-      "X00 = solve(3*Eph/(X11+X22+X0)-IF0)[0]          #zero sequence reactance(ohm)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The positive sequence reactance is\",round(X11,3),\"ohm\"\n",
-      "print \"The negative sequence reactance is\",round(X22,3),\"ohm\"\n",
-      "print \"The zero sequence reactance is\",round(X00,3),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The positive sequence reactance is 3.175 ohm\n",
-        "The negative sequence reactance is 1.055 ohm\n",
-        "The zero sequence reactance is 0.306 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter19_1.ipynb b/Principles_of_Power_System/chapter19_1.ipynb
deleted file mode 100644
index e6813a0f..00000000
--- a/Principles_of_Power_System/chapter19_1.ipynb
+++ /dev/null
@@ -1,212 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:34f21e34d56021d11e412854716405b885521e02b7e263afa0b62d127f14da0d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 19: Circuit Breakers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 19.1, Page Number: 483"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I = 1500                      #rated current of circuit breaker(A)\n",
-      "MVA = 1000                    #MVA rating of CB\n",
-      "V = 33                        #voltage rating of circuit breaker(kV)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Is = MVA*10**6/(3**0.5*33*1000)    #Rated symmetrical breaking current(A,rms)\n",
-      "Im = 2.55*Is                   #Rated making current(A,peak)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Rated normal current is\",I,\"A\"\n",
-      "print \"(ii) Breaking capacity is\",MVA,\"MVA\"\n",
-      "print \"(iii) Rated symmetrical breaking current is\",round(Is),\"A  (peak)\"\n",
-      "print \"(iv) Rated making current is\",round(Im),\"A  (peak)\"\n",
-      "print \"(v) Short-time rating is\",round(Is),\"for 3 seconds\"\n",
-      "print \"(vi) Rated service voltage is\",V,\"kV (r.m.s)\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Rated normal current is 1500 A\n",
-        "(ii) Breaking capacity is 1000 MVA\n",
-        "(iii) Rated symmetrical breaking current is 17495.0 A  (peak)\n",
-        "(iv) Rated making current is 44613.0 A  (peak)\n",
-        "(v) Short-time rating is 17495.0 for 3 seconds\n",
-        "(vi) Rated service voltage is 33 kV (r.m.s)\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 19.2, Page Number: 484"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "f = 50                     #supply frequency(Hz)\n",
-      "V = 11                      #voltage rating of generator(V)\n",
-      "C = 0.01         #distributed capacitance upto ckt breaker b/w phase and neutral(uF)\n",
-      "XL = 5                   #reactance of neatral to earth(ohm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "L = round(XL/(2*math.pi*f),4)            #Inductance per phase(H)\n",
-      "Emax = round(2**0.5*V/3**0.5,2)     #Maximum value of recovery voltage (phase to neutral)(V\n",
-      "E = 2*Emax                          #Peak re-striking voltage(kV)\n",
-      "fn = 1/(2*3.14*(L*C*10**-6)**0.5)     #frequency of oscillations(Hz)\n",
-      "\n",
-      "#Since peak re-striking voltage occurs at a time t given by\n",
-      "t = 1/(2*fn)                        #s\n",
-      "r = E/t                             #kV/sec\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Peak re-striking voltage is\",round(E,2),\"kV\"\n",
-      "print \"(ii) Frequency of oscillations is\",round(fn),\"Hz\"\n",
-      "print \"(iii)The average rate of rise of re-striking voltage\"\n",
-      "print \"     upto the first peak is\",math.floor(r/1000),\"*10**3   kV/sec\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Peak re-striking voltage is 17.96 kV\n",
-        "(ii) Frequency of oscillations is 12628.0 Hz\n",
-        "(iii)The average rate of rise of re-striking voltage\n",
-        "     upto the first peak is 453.0 *10**3   kV/sec\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 19.3, Page Number: 484"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable delaration:\n",
-      "t = 50*10**-6              #time to reach the peak re-striking voltage(s)\n",
-      "Vp = 100                    #the peak re-striking voltage(kV)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "R = Vp/t                  #Average RRRV(kV/sec)\n",
-      "fn = 1/(2*t)                 #Natural frequency of oscillations(Hz)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Average RRRV is\",R/10**6,\"* 10**6   kV/sec\"\n",
-      "print \"Natural frequency of oscillations is\",fn,\"Hz\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average RRRV is 2.0 * 10**6   kV/sec\n",
-        "Natural frequency of oscillations is 10000.0 Hz\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 19.4, Page Number: 485"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Im = 11                   #magnetising current of transformer(A)\n",
-      "Ic = 7                    #chopped instantaneous value of current(A)\n",
-      "L = 35.2                  #inductance (H)\n",
-      "C = 0.0023                #capacitance(uF)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "e = Ic*(L/(C*10**-6))**0.5         #Voltage across breaker contacts at chopping(V)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage across breaker contacts at chopping is\",round(e/1000),\"kV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage across breaker contacts at chopping is 866.0 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter1_1.ipynb b/Principles_of_Power_System/chapter1_1.ipynb
deleted file mode 100644
index 95ebdafe..00000000
--- a/Principles_of_Power_System/chapter1_1.ipynb
+++ /dev/null
@@ -1,68 +0,0 @@
-{
- "metadata": {
-  "name": ""
- },
- "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": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Pi = 4200                            #input power(W)\n",
-      "E = 120                              #voltage supply of generator(V)\n",
-      "I = 32.2                             #supply current(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Po = E*I                          #output power(W)\n",
-      "n = Po/Pi*100                         #efficiency(%)\n",
-      "Pl = Pi-Po                         #power lost(W)\n",
-      "El = Pl*60                         #energy lost per minute(J)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The percentage efficiency of the generator is\",n,\"%\"\n",
-      "print \"Energy lost per minute of operation is\",El,\"J\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The percentage efficiency of the generator is 92.0 %\n",
-        "Energy lost per minute of operation is 20160.0 J\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter20_1.ipynb b/Principles_of_Power_System/chapter20_1.ipynb
deleted file mode 100644
index 768f8fa6..00000000
--- a/Principles_of_Power_System/chapter20_1.ipynb
+++ /dev/null
@@ -1,64 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:37bb87f048a2d4a670b09cf05bff6f5c622ef838c4da8caa7225a316e1b26b5e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 20: Fuses"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 20.1, Page Number: 495"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1 = 0.8                   #radius of fuse wire(mm)\n",
-      "I1 = 8                     #blow current at r1 radius(A)\n",
-      "I2 = 1                     #blow current(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "r2 = r1*(I2/I1)**(2/3)       #radius for bearing 1 A of blow current(mm)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The radius of the wire that will blow off at a current of 1A is\",r2,\"mm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The radius of the wire that will blow off at a current of 1A is 0.2 mm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter21_1.ipynb b/Principles_of_Power_System/chapter21_1.ipynb
deleted file mode 100644
index dbd05a4b..00000000
--- a/Principles_of_Power_System/chapter21_1.ipynb
+++ /dev/null
@@ -1,69 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:74ef32d0e67f3d77b60188556b39e6627674afc8f6e4feb1b0bbc3ae8dbaee2a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 21: Protective Relays"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 21.1, Page Number: 507"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Is = 5                           #rated secondary current of C.T\n",
-      "CS = 1.25                        #current setting\n",
-      "TSM = 0.6                        #time setting multiplier\n",
-      "IF = 4000                        #fault current(A)\n",
-      "CR = 400/5                       #current ratio of CT\n",
-      "\n",
-      "#Calculation:\n",
-      "Ipc = Is*CS                     #pick-up current(A)\n",
-      "If = IF*CR                      #Fault current in relay coil(A)\n",
-      "PSM = If/Ipc                    #plug setting multiplier\n",
-      "#Corresponding to the plug-setting multiplier of 8 given in fig.21.16,\n",
-      "#the time of operation is 3.5 seconds.\n",
-      "top = 3.5*TSM                   #Actual relay operating time(s)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The relay operating time is\",top,\"seconds\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The relay operating time is 2.1 seconds\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter22_1.ipynb b/Principles_of_Power_System/chapter22_1.ipynb
deleted file mode 100644
index 38d23b90..00000000
--- a/Principles_of_Power_System/chapter22_1.ipynb
+++ /dev/null
@@ -1,338 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a5275b4f1bdd430cf4ed3e203b7e975d86c74b17fe608ee8e07ebac2f3a39a9f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 22: Protection of Alternators and Transformers"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 22.1, Page Number: 529"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVA = 10                         #MVA rating of alternator\n",
-      "V = 6.6                          #voltage rating of alternator(V)\n",
-      "X = 10                           #per phase reactance of lternator(%)\n",
-      "Iop = 175                         #operating current(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Vph = V*1000/3**0.5                   #phase voltage(kV)\n",
-      "I = round(MVA*10**6/(3**0.5*V*1000))      #full load current(A)\n",
-      "\n",
-      "#Let the reactance per phase be x ohms.\n",
-      "r,x = symbols('r x')           #r = earthing resistance required to leave 10% of\n",
-      "                                #the winding unprotected\n",
-      "x1 = solve(3**0.5*x*I/(6.6*1000)*100-10,x)[0]\n",
-      "X1 = x1*0.1                     #Reactance of 10% winding\n",
-      "E = Vph*0.1                    #E.M.F. induced in 10% winding\n",
-      "Zf = (X1**2+r**2)**0.5\n",
-      "Ief = E/Zf                     #Earth-fault current due to 10% winding\n",
-      "\n",
-      "#When this fault current becomes 175 A, the relay will trip\n",
-      "r1 = solve(Ief-175,r)[1]               #A\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Required value of earth resistance is\",round(r1,3),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required value of earth resistance is 2.177 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 22.2, Page Number: 530"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVA = 10                         #MVA rating of alternator\n",
-      "V = 6.6                          #voltage rating of alternator(V)\n",
-      "CR = 1000/5                      #current ratio of CT\n",
-      "Rn = 7.5                         #resistance of star-point to earth(ohm)\n",
-      "Iop = 0.5                        #operating current of the relay(A)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let x % of the winding be unprotected.\n",
-      "x = symbols('x')\n",
-      "Vph = V*1000/3**0.5                   #phase voltage(kV)\n",
-      "If = 1000/5*Iop              #minimum fault current which will operate the relay(A)\n",
-      "E = Vph*x/100                 #E.M.F. induced in x% winding(V)\n",
-      "Ief = E/Rn                      #Earth fault current which x% winding will cause(A)\n",
-      "#This current must be equal to 100 A.\n",
-      "x1 = solve(Ief-If,x)[0]\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Percentage of unprotected winding is\",round(x1,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Percentage of unprotected winding is 19.68 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 22.3, Page Number: 530"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVA = 10                         #MVA rating of alternator\n",
-      "V = 6.6                          #voltage rating of alternator(V)\n",
-      "CR = 1000/5                      #current ratio of CT\n",
-      "Rn = 6                         #resistance of star-point to earth(ohm)\n",
-      "Iop = 0.75                        #operating current of the relay(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let x % of the winding be unprotected.\n",
-      "x = symbols('x')\n",
-      "Vph = V*1000/3**0.5                   #phase voltage(kV)\n",
-      "If = 1000/5*Iop              #minimum fault current which will operate the relay(A)\n",
-      "E = Vph*x/100                 #E.M.F. induced in x% winding(V)\n",
-      "Ief = E/Rn                      #Earth fault current which x% winding will cause(A)\n",
-      "#This current must be equal to 100 A.\n",
-      "x1 = solve(Ief-If,x)[0]\n",
-      "\n",
-      "\n",
-      "#(ii) Let r2 = the minimum earthing resistance required to \n",
-      "#provide protection for 90% of stator winding. \n",
-      "#Then, 10% winding would be unprotected\n",
-      "x2 = 10                                 #%\n",
-      "r2 = Vph*x2/If*0.01                     #ohm\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The percentage of each of the stator windings is\",round(x1,1),\"%\"\n",
-      "print \"(ii)The minimum resistance to provide protection for 90% of\"\n",
-      "print \"    the stator winding is\",round(r2,2),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The percentage of each of the stator windings is 23.6 %\n",
-        "(ii)The minimum resistance to provide protection for 90% of\n",
-        "    the stator winding is 2.54 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 22.4, Page Number: 531"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVA = 10                         #MVA rating of alternator\n",
-      "V = 6.6                          #voltage rating of alternator(V)\n",
-      "CR = 1000/5                      #current ratio of CT\n",
-      "s = 20                          #earth-fault setting(%)\n",
-      "Iop = 0.75                        #operating current of the relay(A)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Since 85% winding is to be protected, 15% would be unprotected\n",
-      "r = symbols('r')            #earthing resistance reqd. to leave 15% of winding unprotected(ohm)\n",
-      "x = 15                                 #%\n",
-      "Ifl = MVA*10**6/(3**0.5*V*1000)       #Full load current(A)\n",
-      "IF = s*Ifl/100                    #Minimum fault current which will operate the relay\n",
-      "Vu = x/100*V*1000/3**0.5             #Voltage induced in 15% of winding(kV)\n",
-      "Ief = Vu/r                        #Earth fault current which 15% winding will cause(A)\n",
-      "#This current must be equal to IF.\n",
-      "r1 = solve(Ief-IF,r)[0]              #ohm\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The value of earthing resistor is\",round(r1,2),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The value of earthing resistor is 3.27 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 22.5, Page Number: 538"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1 = 220/11000                        #voltage ratio of transformer\n",
-      "r2 = 600/5                 #current ratio of protective transformer on 220V side\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Suppose that line current on 220 V side is 600 A\n",
-      "\n",
-      "Ipd = 5                  #Phase current of delta connected CTs on 220V side(A)\n",
-      "Ild = 3**0.5*Ipd         #Line current of delta connected CTs on 220 V side(A)\n",
-      "\n",
-      "#This Ild will flow through the pilot wires.\n",
-      "Ips = 5*3**0.5           #Phase current of star connected CTs on 11,000 V side(A)\n",
-      "\n",
-      "#Now, using this relation: Primary apparent power = Secondary apparent power\n",
-      "I = 3**0.5*220*600/(3**0.5*11000)           #A\n",
-      "r3 = I/Ips                     #Turn-ratio of CTs on 11000 V side\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Turn-ratio of CTs on 11000 V side is (\",round(r3,3),\": 1 )\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Turn-ratio of CTs on 11000 V side is ( 1.386 : 1 )\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 22.6, Page Number: 538"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r1 = 0.4/11                        #line voltage(in kV) ratio of transformer\n",
-      "r2 = 500/5                   #current ratio of protective transformer\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Suppose the line current on 400 V side is 500 A.\n",
-      "Ipd = 5                     #Phase current of delta connected CTs on 400 V side(A)\n",
-      "Ild = Ipd*3**0.5            #Line current of delta connected CTs on 400 V side(A)\n",
-      "#This Ild will flow through the pilot wires.\n",
-      "Ips = 5*3**0.5             #Phase current of star-connected CTs on 11000 V side(A)\n",
-      "\n",
-      "#Primary apparent power = Secondary apparent power\n",
-      "I = 3**0.5*400*500/(3**0.5*11000)             #A\n",
-      "r3 = I/Ips\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \" The ratio of the protective transformers on 11kV side is\",round(r3,3),\"i.e, 10.5:5\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The ratio of the protective transformers on 11kV side is 2.099 i.e, 10.5:5\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter26_1.ipynb b/Principles_of_Power_System/chapter26_1.ipynb
deleted file mode 100644
index d4334600..00000000
--- a/Principles_of_Power_System/chapter26_1.ipynb
+++ /dev/null
@@ -1,224 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:260bb8beffe2e4d2155fdaec0c7462f8cf5127d60c79f0bddb469ea9aad62673"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 26: Neutral Grounding"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 26.1, Page Number: 599"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "f = 50                    #Supply frequency(Hz)\n",
-      "C = 4.5*10**-6              #Line to earth capacitance(F)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "XL = 1/(3*2*math.pi*f*C)                 #ohm\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The reactance of Peterson coil is\",round(XL,1),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reactance of Peterson coil is 235.8 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 26.2, Page Number: 599"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 200                    #length of transmission line(km)\n",
-      "f = 50                    #Supply frequency(Hz)\n",
-      "c = 0.02*10**-6              #Line to earth capacitance(F/km)\n",
-      "V = 230                    #voltage rating of line(kV)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "C = c*l                    #capacitance of coil(F)\n",
-      "L = round(1/(3*(2*3.14*f)**2*C),2)           #Required inductance of Peterson coil(H)\n",
-      "Vph = round(V*1000/1.732)             #Voltage across Peterson coil(kV)\n",
-      "IF = math.ceil(Vph/(2*3.13*f*L))              #Current through Peterson coil(A)\n",
-      "kVA = Vph*IF/1000               #Rating of Peterson coil\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "print \"Inductance of Peterson coil is\",L,\"H\"\n",
-      "print \"Rating of Peterson coil is\",round(kVA),\"kVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Inductance of Peterson coil is 0.85 H\n",
-        "Rating of Peterson coil is 66397.0 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 26.3, Page Number: 600"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "f = 50                      #fequency of supply(Hz)\n",
-      "C = 1.2*10**-6              #line-to-earth capacitance(F)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) To neutralize capacitance of 100% of the length of the line,\n",
-      "#Inductive reactance of the coilis given by\n",
-      "XL1 = 1/(3*2*math.pi*f*C)           #ohm\n",
-      "\n",
-      "#(ii)  To neutralize capacitance of 90% of the length of the line,\n",
-      "#Inductive reactance of the coilis given by\n",
-      "XL2 = 1/(3*2*math.pi*f*0.9*C)       #ohm\n",
-      "\n",
-      "\n",
-      "#(iii) To neutralize capacitance of 80% of the length of the line,\n",
-      "#Inductive reactance of the coilis given by\n",
-      "XL3 = 1/(3*2*math.pi*f*0.8*C)       #ohm\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Inductive reactance of the coil to neutralize capacitance\"\n",
-      "print \"     of 100% of the length of the line is\",round(XL1,2),\"ohm\"\n",
-      "print \"\\n(ii) Inductive reactance of the coil to neutralize capacitance\"\n",
-      "print \"     of 100% of the length of 90% of the line is\",round(XL2,2),\"ohm\"\n",
-      "print \"\\n(iii)Inductive reactance of the coil to neutralize capacitance\"\n",
-      "print \"     of 100% of the length of 80% of the line is\",round(XL3,2),\"ohm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Inductive reactance of the coil to neutralize capacitance\n",
-        "     of 100% of the length of the line is 884.19 ohm\n",
-        "\n",
-        "(ii) Inductive reactance of the coil to neutralize capacitance\n",
-        "     of 100% of the length of 90% of the line is 982.44 ohm\n",
-        "\n",
-        "(iii)Inductive reactance of the coil to neutralize capacitance\n",
-        "     of 100% of the length of 80% of the line is 1105.24 ohm\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 26.4, Page Number: 600"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 0.01                          #radius of conductor(m)\n",
-      "d = 4                             #conductor spacing(m)\n",
-      "V = 132                           #voltage of the line(kV)\n",
-      "f = 50                            #supply frequency(Hz)\n",
-      "l = 200                           #line length(km)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "c = round(2*math.pi*8.885/math.log(d/r),1)*10**-12        #capacitance per unit length(F/m)\n",
-      "C = c*l*1000                         #Capacitance between phase and earth for 200 km line(F)\n",
-      "L = round(1/(3*(2*math.pi*f)**2*C),2)               #required inductance L of the arc suppression coil(H)\n",
-      "IF = V*1000/(10*math.ceil(3**0.5*2*math.pi*f*L/10))          #Current through the coil(A)\n",
-      "kVA = V/3**0.5*IF                     #kVA\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance of the coil is\",L,\"H\"\n",
-      "print \"Rating of the coil is\",round(kVA),\"kVA\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance of the coil is 1.82 H\n",
-        "Rating of the coil is 10060.0 kVA\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter2_1.ipynb b/Principles_of_Power_System/chapter2_1.ipynb
deleted file mode 100644
index daf8974b..00000000
--- a/Principles_of_Power_System/chapter2_1.ipynb
+++ /dev/null
@@ -1,857 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:a62a0cfa22010723ae84accbfbc8f3f3c553c1279f7bb65e42ceaaad44de0807"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2:Generating Stations"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.1, Page Number: 16"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n=20                        #overall efficiency of plant\n",
-      "h=860                       #kcal\n",
-      "m=0.6           #Mass of fuel burnt(kg) per KW of electrical energy generated\n",
-      "\n",
-      "#Calculations:\n",
-      "x=h*100/(m*n)           #Calorific value of fuel(kcal/kg)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Calorific value of fuel =\",round(x,2),\"kcal/kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Calorific value of fuel = 7166.67 kcal/kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.2, Page Number: 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 20000                           #maximum demand(kW)\n",
-      "n_b= 85                             #boiler efficiency(%)\n",
-      "m=0.9                             #coal consumption(kg/kWh)\n",
-      "LF=40                             #load factor(%)\n",
-      "n_t=90                          #turbine efficiency(%)\n",
-      "c=300                           #cost of 1 tonne of coal(Rs)\n",
-      "\n",
-      "#Calculations:\n",
-      "n_th = n_b * n_t/100            #in %\n",
-      "cb = LF*M*m*c*24*365/(1000*100)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Thermal efficiency = \",n_th,\"%\"\n",
-      "print \"Coal bill per annum = Rs\",cb"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thermal efficiency =  76.5 %\n",
-        "Coal bill per annum = Rs 18921600.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.3, Page Number: 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "ct=3000000                     #annul cost of coal(Rs)\n",
-      "cv=5000                         #Calorific value of coal(kcal/kg)\n",
-      "c=300                          #cost of coal per tonne(Rs)\n",
-      "n_th=33                        #thermal efficiency(%)\n",
-      "n_elec=90                      #electrical efficiency(%)\n",
-      "\n",
-      "#Calculations:\n",
-      "n_t=n_th*n_elec/100                 #overall efficiency(%)\n",
-      "h=ct*cv*1000/c                  #heat of combustion(kcal)\n",
-      "ho=n_t*h/(100*860)             #heat output(kWh)\n",
-      "L=ho/8760                      #kW\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Avg load on the station=\",round(L),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Avg load on the station= 1971.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 33
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.4, Page Number: 17"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "kWh= symbols('kWh')\n",
-      "W = 13500 + 7.5 * kWh               #Water evaporated in kg\n",
-      "C = 5000 + 2.9 * kWh                #coal cumsumption in kg\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#part (i):\n",
-      "#As the station output (i.e., kWh) increases towards infinity,\n",
-      "#the limiting value of W/C approaches\n",
-      "L1= 7.5/2.9                        #in kg\n",
-      "\n",
-      "#part (ii):\n",
-      "#at no load\n",
-      "kWh=0\n",
-      "c=(5000+2.9*kWh)/8                             #coal per hour in kg\n",
-      "#Results:\n",
-      "print \"Limiting value of water/kg of coal=\",round(L1,1),\"kg\"\n",
-      "print \"Required Coal per hour\",c,\"kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Limiting value of water/kg of coal= 2.6 kg\n",
-        "Required Coal per hour 625.0 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 34
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.5, Page Number: 18"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "C=100                            #Capacity of station in MW\n",
-      "cv=6400                          #kcal/kg\n",
-      "n_th=0.3                         #thermal efficiency\n",
-      "n_elec=0.92                      #electrical efficiency\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "n_t=n_th*n_elec                   #overall efficiency\n",
-      "U=C*1*10**3                      #units generated/hr in kWh\n",
-      "H=U*860/n_t                    #total heat of combustion(kcal)\n",
-      "w=H/cv                          #Coal consumption in kg\n",
-      "\n",
-      "#Results:\n",
-      "print \"The coal consumption per hour =\",round(w),\"kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The coal consumption per hour = 48687.0 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 35
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.6, Page Number: 23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "C=5*10**6                   #reservoir capacity in m^3\n",
-      "H=200                       #water head in m\n",
-      "n_t=0.75                    #overall efficiency\n",
-      "d=1000                      #density of water in kg/m^3\n",
-      "\n",
-      "#Calculations:\n",
-      "W=C*d*9.81                   #weight of water in Newton\n",
-      "E=W*H*n_t/(3600*1000)        #electrical energy available(kWh)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The total energy available=\",round(E/10**6,3),\"* 10^6 kWh\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total energy available= 2.044 * 10^6 kWh\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.7, Page Number: 23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V=94                    #volume of water in m^3/sec\n",
-      "d=1000                  #density of water in kg/m^3\n",
-      "H=39                    #head of water in m\n",
-      "nt=0.80                 #overall efficiency\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "W=V*d                   #weight of water in kg/sec\n",
-      "w=W*H*9.81/1000         #work done per sec in kW\n",
-      "FC=nt*w                 #firm capacity in kW\n",
-      "YGO=FC*8760             #yearly gross capacity in kWh\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Firm capacity=\",FC,\"kW\"\n",
-      "print \"Yearly gross output\",round(YGO/10**6),\"* 10^6 kWh\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Firm capacity= 28770.768 kW\n",
-        "Yearly gross output 252.0 * 10^6 kWh\n"
-       ]
-      }
-     ],
-     "prompt_number": 37
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.8, Page Number: 23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "H=100                      #Water head in m\n",
-      "Q=1                        #discharge, m^3/sec\n",
-      "nh=0.86                    #hydraulc efficiency\n",
-      "nelec=0.92                 #electrical efficiency\n",
-      "d=1000                     #density of water, kg/m^3\n",
-      "\n",
-      "#Calculations:\n",
-      "W=Q*d*9.81                 #weight of water in N\n",
-      "Po=W*H*nh*nelec/1000        #power produced, kW\n",
-      "E=Po*1                     #in kWh\n",
-      "\n",
-      "#Results:\n",
-      "print \"Electrical energy generated per hr=\",round(E),\"kWh\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Electrical energy generated per hr= 776.0 kWh\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.9, Page Number: 23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "CA=5*10**9                      #Catchment area in m^2\n",
-      "H=30                            #head in m\n",
-      "F=1.25                          #Annual rainfall in m\n",
-      "K=0.80                          #yeild factor\n",
-      "n=0.70                          #overall efficiency\n",
-      "LF=0.40                         #Load factor\n",
-      "d=1000                          #density of water(kg/m^3)\n",
-      "\n",
-      "#Calculations:\n",
-      "V=CA*F*K                              #volume of water utilised per annum(m^3)\n",
-      "W=V*d*9.81                                   #Weight of water available per annum (N)\n",
-      "E=round(W*H*n/(10**11*3600),2)*10**8         #Electrical energy available per annum(kWh)\n",
-      "Pav=E/8760                                   #average power(kW)\n",
-      "Dmax=Pav/LF                                 #Maximum demand\n",
-      "\n",
-      "#Results:\n",
-      "print \"Average power generated is \",round(Pav),\"kW\"\n",
-      "print \"Rating of generators is\",round(Dmax),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Average power generated is  32648.0 kW\n",
-        "Rating of generators is 81621.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 39
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.10, Page Number: 24"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "A=2.4                    #Area of reservoir(km^2)\n",
-      "V=5*10**6                #Capacity of reservoir(m^3)\n",
-      "H=100                    #in m\n",
-      "np=0.95                  #penstock efficiency\n",
-      "nt=0.90                  #turbine efficiency\n",
-      "ng=0.85                  #generation efficiency\n",
-      "L=15000                 #load supplied in kW\n",
-      "\n",
-      "#Calculations:\n",
-      "W=V*1000*9.81             #in Newton\n",
-      "n=int(np*nt*ng*1000)/1000               #overall efficiency\n",
-      "E=W*H*n/(1000*3600)       #Elecctrical energy generated(kWh)\n",
-      "x=L*3*3600/(A*10**6*9.81*H*n)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Total electrical energy generated is \",round(E),\"kWh\"\n",
-      "print \"Fall in reservoir level is\",round(x*100,3),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total electrical energy generated is  989175.0 kWh\n",
-        "Fall in reservoir level is 9.478 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.11, Page Number: 25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "H=25                       #head of reservoir in m\n",
-      "Pr=400                    #power required by factory(kW)\n",
-      "n=0.80                    #overall efficiency of plant\n",
-      "\n",
-      "#Calculations:\n",
-      "#part (i):\n",
-      "#(a):\n",
-      "d1 = 10                     #discharge in m^3/sec\n",
-      "w1 = d1*1000*9.81           #weight of water in N\n",
-      "P1 = w1*H*n/1000             #power developed(kW)\n",
-      "\n",
-      "#(b)\n",
-      "d2 = 6                      #in m^3/sec\n",
-      "P2 = P1*d2/d1               #kW\n",
-      "\n",
-      "#(c)\n",
-      "d3 = 1.5                    #in m^3/sec\n",
-      "P3 = P1*d3/d1               #kW\n",
-      "\n",
-      "Ps = Pr-P3                  #standby power(kW)\n",
-      "\n",
-      "#part(ii):\n",
-      "Dav = (d1*4+d2*2+d3*6)/12   #avg discharge(m^3/sec)\n",
-      "P = P1*Dav/d1               #power developed(kW)\n",
-      "Pex = P-Pr                  #Excess power available(kW)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i) Standby power is \",round(Ps),\"kW\"\n",
-      "print \"(ii) Excess power available is \",round(Pex,1),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Standby power is  106.0 kW\n",
-        "(ii) Excess power available is  597.4 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 41
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {
-      "slideshow": {
-       "slide_type": "-"
-      }
-     },
-     "source": [
-      "Example 2.12, Page Number: 25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#for part (i):\n",
-      "C = 10                          #Installed capacity(MW)\n",
-      "H = 20                          #head of reservoir(m)\n",
-      "n = 0.80                        #overall efficiency\n",
-      "LF = 0.40                       #load factor\n",
-      "#for part (ii):\n",
-      "Q2 = 20                         #discharge\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "#for part(i):\n",
-      "U = C*LF*24*7*10**3             #units generated per week(kWh)\n",
-      "Q = U/(H*n*9.81*24*7)           #Discharge in m^3/sec\n",
-      "\n",
-      "#for part(ii):\n",
-      "U2 = Q2*9.81*1000*n*H*24/1000     #units generated per day(kWh)\n",
-      "LF2 = U2/(C*10**3*24)\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i) The river discharge is \",round(Q,2),\"m^3/sec\"\n",
-      "print \"(ii) The load factor is \", round(LF2*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The river discharge is  25.48 m^3/sec\n",
-        "(ii) The load factor is  31.4 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 42
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.13, Page Number: 26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "H = 15                            #head of reservoir(m)\n",
-      "n = 0.85                          #efficiency\n",
-      "L = 0.40                          #load factor\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "Qavg = (500+520+850+800+875+900+546)/7    #in m^3/sec\n",
-      "\n",
-      "#It is clear from graph that on three dyas \n",
-      "#(viz., Sun, Mon. and Sat.), the discharge is less than\n",
-      "#the average discharge.\n",
-      "\n",
-      "V1 = (500+520+546)*24*3600  #Actual volume available in these 3 days(m^3/s)\n",
-      "V2 = 3*Qavg*24*3600         #Vol. of water required in these 3 days(m^3/s)\n",
-      "Pr = V2-V1                  #Pondage required(m^3/sec)\n",
-      "Po = Qavg*9.81*1000*H*n     #Avg output produced(W)\n",
-      "C = Po/L                    #Capacity of plant(W)\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i) The average daily discharge is \",Qavg,\"m^3/sec\"\n",
-      "print \"(ii) Pondage required is (\",round(Pr/10**5),\"* 10^5)  m^3\"\n",
-      "print \"(iii)Installed capacity of plant is \",round(C/10**6),\"MW\"\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The average daily discharge is  713.0 m^3/sec\n",
-        "(ii) Pondage required is ( 495.0 * 10^5)  m^3\n",
-        "(iii)Installed capacity of plant is  223.0 MW\n"
-       ]
-      }
-     ],
-     "prompt_number": 43
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.14, Page Number: 29"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "w = 0.28                        #fuel consumption in kg/kWh\n",
-      "C = 10000                       #calorific value of fuel(kcal/kWh)\n",
-      "na = 0.95                        #efficiency of alternator\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "H = w*C/860                     #heat produced by 0.28 kg/kWh of fuel\n",
-      "\n",
-      "no = 1/H                        #Overall efficiency\n",
-      "ne = no/na                      #Efficiency of engine\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i) The overall efficiency is \",round(no*100,1),\"%\"\n",
-      "print \"(ii)The efficiency of the engine\",round(ne*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The overall efficiency is  30.7 %\n",
-        "(ii)The efficiency of the engine 32.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 44
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.15, Page Number: 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "w = 1000                      #fuel consumption in kg/day\n",
-      "E = 4000                      #Units generated in kWh/day\n",
-      "C = 10000                     #calorific value in kcal/kg\n",
-      "na = 0.96                     #Alternator efficiency\n",
-      "nem = 0.95                     #engine mechanical efficiency\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "s = w/E                       #specific fuel consumption(kg/kWh)\n",
-      "E2 = w*C                      #energy input per day(kcal/day)\n",
-      "no = E*860/E2                 #overall efficiency\n",
-      "ne = no/na                    #engine efficiency\n",
-      "net = ne/nem                  #engine thermal efficiency\n",
-      "\n",
-      "#Results:\n",
-      "print \"Specific fuel consumption is \",s,\"kg/kWh\"\n",
-      "print \"Overall efficiency is \",round(no*100,1),\"%\"\n",
-      "print \"Thermal efficiency of the engine is \",round(net*100,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Specific fuel consumption is  0.25 kg/kWh\n",
-        "Overall efficiency is  34.4 %\n",
-        "Thermal efficiency of the engine is  37.72 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 45
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.16, Page Number: 30"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "C1 = 700                   #capacity of plant 1(kW)\n",
-      "C2 = 2*500                 #capacity of plant 2(kW)\n",
-      "pcf = 0.40                 #plant capacity factor\n",
-      "w = 0.28                   #fuel cunsumption in kg/kWh\n",
-      "H = 10200                  #specific heat of fuel in kcal/kg\n",
-      "\n",
-      "#Calculatios:\n",
-      "M = (C1+C2)*30*24          #max energy can be produced in 30 days(kWh)\n",
-      "E = pcf*M                  #Actual energy produced in 30 days(kWh)\n",
-      "W = E*w                    #actual fuel consumption in kg\n",
-      "\n",
-      "Po = E*860                 #output energy in kWh\n",
-      "Pin = W*H                  #Input energy in kWh\n",
-      "n = Po/Pin                 #Overall efficiency\n",
-      "\n",
-      "#Results:\n",
-      "print \"The fuel oil required is \",W,\"kg\"\n",
-      "print \"Ovreall efficiency is\",round(n*100),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The fuel oil required is  137088.0 kg\n",
-        "Ovreall efficiency is 30.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 46
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.17, Page Number: 34"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "M = 300                   #Energy received from reactor(MW)\n",
-      "E1 = 200                  #Energy released fron each atom(MeV)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "E2 = M*10**6*3600            #Energy released per hour(J)\n",
-      "E3 = E1*1.6*10**-19*10**6      #Energy released per fission(J)\n",
-      "N = E2/E3                    #No of atoms fissioned\n",
-      "m = N*235/(6.022*10**23)    #mass of uranium fissioned per hr(g)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Mass of Uranium fissioned per hour is\",round(m,2),\"g\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Mass of Uranium fissioned per hour is 13.17 g\n"
-       ]
-      }
-     ],
-     "prompt_number": 47
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 2.18, Page Number: 35"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "t = 30                         #days\n",
-      "w = 2                          #weight of uranium(kg)\n",
-      "Eo = 200                     #energy released per fission(MeV)\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "N = 2*1000*6.022*10**23/235     #No of atoms fissioned in 2kg of fuel\n",
-      "Po = N*Eo*(1.6*10**-19)*10**6/(24*60*60*30)    #Watt\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Power output is\",round(Po*10**-6,1),\"MW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power output is 63.3 MW\n"
-       ]
-      }
-     ],
-     "prompt_number": 48
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter3_1.ipynb b/Principles_of_Power_System/chapter3_1.ipynb
deleted file mode 100644
index 42073d28..00000000
--- a/Principles_of_Power_System/chapter3_1.ipynb
+++ /dev/null
@@ -1,1355 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:c36425eab1371d7c77f7bd1775095f254e3df888559e80a11117b968dfc1b5b4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Principles of Power System"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.1, Page Number: 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "M = 100                          #Maximum demand on power station(MW)\n",
-      "L = 0.40                         #annual load factor\n",
-      "\n",
-      "#Calculation:\n",
-      "E = M*L*8760*10**6              #Energy generated in a year(kWh)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Energy generated in a year is (\",E/10**8,\"* 10^5) kWh\" "
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Energy generated in a year is ( 3504.0 * 10^5) kWh\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.2, Page Number: 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "L = 43                      #connected load(MW)\n",
-      "M = 20                      #Maximum demand(MW)\n",
-      "E = 61.5*10**6              #Units genearted per annum(kWh)\n",
-      "\n",
-      "#calculation:\n",
-      "DF = M/L                    #demand factor\n",
-      "LF = E/(M*8760*1000)          #load factor\n",
-      "\n",
-      "#Results:\n",
-      "print \"The demand factor is\",round(DF,3)\n",
-      "print \"Load factor is \",round(LF*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The demand factor is 0.465\n",
-        "Load factor is  35.1 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.3, Page Number: 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#variable declaration:\n",
-      "\n",
-      "#power station delivers 100 MW for 2 hours, \n",
-      "#50 MW for 6 hours and is shut down for the rest\n",
-      "#of each day.\n",
-      "M = 100                       #Maximum capacity of station(MW)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "E1 = 100*2+50*6                #Energy supplied per day(MWh)\n",
-      "n = 365-45                     #No.of working days\n",
-      "LF = E1*320/(M*n*24)               #load factor\n",
-      "\n",
-      "print \"Annual load factor is\",round(LF*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Annual load factor is 20.8 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.4, Page Number: 50"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 25                      #Maximum demand(MW)\n",
-      "LF = 0.60                   #Load factor\n",
-      "PCF = 0.50                  #plant capacity factor\n",
-      "PUF = 0.72                  #plant use factor\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "L = LF*M                    #Average load(MW)\n",
-      "PC = L/PCF                  #Plant gapacity(MW)\n",
-      "R = PC-M                    #Reserve capacity(MW)\n",
-      "E = L*24                    #daily energy produced(MWh)\n",
-      "ME = E/PUF                  #Maximum energy produced(MWh/day)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Reserve capacity is \",R,\"MW\"\n",
-      "print \"Daily energy produced is\",E,\"MWh\"\n",
-      "print \"Maximum energy that could be produced dauly is \",ME,\"MWh/day\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Reserve capacity is  5.0 MW\n",
-        "Daily energy produced is 360.0 MWh\n",
-        "Maximum energy that could be produced dauly is  500.0 MWh/day\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.5, Page Number: 51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 2500                 #Maximum demand(kW)\n",
-      "E = 45*10**5             #kWh/year\n",
-      "\n",
-      "#Calculation:\n",
-      "D = (1500+750+100+450)/M    #diversity factor\n",
-      "LF = E/(M*8760)              #Annual load factor\n",
-      "\n",
-      "#Results:\n",
-      "print \"Diversity factor is\",D\n",
-      "print \"Annual load factor is\",round(LF*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Diversity factor is 1.12\n",
-        "Annual load factor is 20.5 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.6, Page Number: 51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 15000              #Max demand(kW)\n",
-      "LF = 0.50              #Annual load factor\n",
-      "PCF = 0.4              #Plant capacity factor\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "L = LF*M               #load factor(kW)\n",
-      "PC = L/PCF             #plant capacity(kW)\n",
-      "R = PC-M               #Reserve capacity(kW)\n",
-      "\n",
-      "#Results:\n",
-      "print \"The reserve capacity is \",R,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The reserve capacity is  3750.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.7, Page Number: 51"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "DFo = 1.35              #Overall diversity factor of system\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "M = (1500+2000+10000)/DFo    #max demand(kW)\n",
-      "#for domestic load:\n",
-      "CL1 = (1500*1.2)/0.8          #kW\n",
-      "\n",
-      "#for Commercial load:\n",
-      "CL2 = 2000*1.1/0.9             #kW\n",
-      "\n",
-      "#for Industrial load:\n",
-      "CL3 = 10000*1.25/1             #kW\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Maximum demand is \",M,\"kW\"\n",
-      "print \"Connected loads for each type are:\"\n",
-      "print CL1,\"kW \",round(CL2),\"kW \",CL3,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum demand is  10000.0 kW\n",
-        "Connected loads for each type are:\n",
-        "2250.0 kW  2444.0 kW  12500.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.8, Page Number: 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "Dt = 1.3             #Diversity among transformers\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#for transformer 1:\n",
-      "M1 = (10*0.65)/1.5            #Max demand (kW)\n",
-      "\n",
-      "#for transformer 2:\n",
-      "M2 = (12*0.6)/3.5            #max demand(kW)\n",
-      "\n",
-      "#for transformer:\n",
-      "M3 = (15*0.7)/1.5             #max demand(kW)\n",
-      "\n",
-      "Mf = (M1+M2+M3)/Dt           #Maximum demand on feeder(kW)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Maximum load on the feeder is \",round(Mf,1),\"kW\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Maximum load on the feeder is  10.3 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.9, Page Number: 52"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "n1 = 1000               #No of houses\n",
-      "CL1 = 1.5               #Avg connected load in each house(kW)\n",
-      "DF1 = 0.4                #Demand factor\n",
-      "DvF = 2.5               #Diversity factor\n",
-      "\n",
-      "n2 = 10                  #no. of factories\n",
-      "M2 = 90                 #factory overall max demand(kW)\n",
-      "\n",
-      "n3 = 7                  #no of tubewells\n",
-      "C3 = 7                  #capacity of tubewells(kW)\n",
-      "\n",
-      "DF = 1.2                #Diversity factor among above types of consumers\n",
-      "\n",
-      "#Calculations:\n",
-      "Ms = ((DF1*n1*CL1)/DvF)+M2+n3*C3      #Sum of maximum demand on the station(kW)\n",
-      "M = Ms/DF                             #maximum demand on the station(kW)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Minimum capacity on the power station is\",round(M),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Minimum capacity on the power station is 316.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.10, Page Number: 53"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%pylab inline\n",
-      "\n",
-      "\n",
-      "#Calculations:\n",
-      "M = 70.0                      #Max demand, from load curve(MW)\n",
-      "E = 40*6+50*4+60*2+50*4+70*4+40*4        #Units generated per day(MWh)\n",
-      "L = E/24                     #Avg load(MW)\n",
-      "LF = L/M                     #Load factor\n",
-      "\n",
-      "n1 = linspace(0,6,10);\n",
-      "M1 = linspace(40,40,10);\n",
-      "plot(n1,M1);\n",
-      "hold(True);\n",
-      "\n",
-      "n2 = linspace(6,10,10);\n",
-      "M2 = linspace(50,50,10);\n",
-      "plot(n2,M2,'b');\n",
-      "\n",
-      "n3 = linspace(10,12,10);\n",
-      "M3 = linspace(60,60,10);\n",
-      "plot(n3,M3,'b');\n",
-      "\n",
-      "n4 = linspace(12,16,10);\n",
-      "M4 = linspace(50,50,10);\n",
-      "plot(n4,M4,'b');\n",
-      "\n",
-      "n5 = linspace(16,20,10);\n",
-      "M5 = linspace(70,70,10);\n",
-      "plot(n5,M5,'b');\n",
-      "\n",
-      "n6 = linspace(20,24,10);\n",
-      "M6 = linspace(40,40,10);\n",
-      "plot(n6,M6,'b');\n",
-      "\n",
-      "ylim(0,100);\n",
-      "xlim(0,24);\n",
-      "grid(1,linewidth=0.5);\n",
-      "ylabel(\"Load in MW  ------>\");\n",
-      "xlabel(\"Time in hours   ----->\");\n",
-      "title(\"Daily Load Curve\")\n",
-      "\n",
-      "#Results:\n",
-      "print \"Maximum demand is \",M,\"MW\"\n",
-      "print \"Units generated per day is (\",E/100,\"* 10^5)  kWh\"\n",
-      "print \"Average load is\",L*1000,\"kW\"\n",
-      "print \"Load factor is \",round(LF*100,1),\"%\"\n",
-      "print \"The Daily load curve is shown below:\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Maximum demand is "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 70.0 MW\n",
-        "Units generated per day is ( 12.0 * 10^5)  kWh\n",
-        "Average load is 50000.0 kW\n",
-        "Load factor is  71.4 %\n",
-        "The Daily load curve is shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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cXGztEmwSc1HGXPSJnInNNIHc3Fw0b95ct+3r64vc3FwrVkREZP9spglIkmTt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\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x4a7dfb0>"
-       ]
-      }
-     ],
-     "prompt_number": 31
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.11, Page Number: 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%pylab inline\n",
-      "\n",
-      "#Calculation:\n",
-      "M = 350.0                             #Maximum demand(kW)\n",
-      "DF = (200+100+50+100)/M        #Diversity factor\n",
-      "E = 100*6+(100+50)*2+(200+100+50)*2+(200+100)*8+100*6    #kWh/day\n",
-      "LF = E/(24*M)                           #load factor\n",
-      "\n",
-      "n1 = linspace(0,6,10);\n",
-      "M1 = linspace(100,100,10);\n",
-      "plot(n1,M1);\n",
-      "\n",
-      "\n",
-      "hold(True);\n",
-      "\n",
-      "n2 = linspace(6,8,10);\n",
-      "M2 = linspace(150,150,10);\n",
-      "plot(n2,M2,'b');\n",
-      "\n",
-      "n3 = linspace(8,10,10);\n",
-      "M3 = linspace(350,350,10);\n",
-      "plot(n3,M3,'b');\n",
-      "\n",
-      "n4 = linspace(10,18,10);\n",
-      "M4 = linspace(300,300,10);\n",
-      "plot(n4,M4,'b');\n",
-      "\n",
-      "n5 = linspace(18,24,10);\n",
-      "M5 = linspace(100,100,10);\n",
-      "plot(n5,M5,'b');\n",
-      "\n",
-      "ylim(0,400);\n",
-      "xlim(0,24);\n",
-      "grid(linewidth=0.5);\n",
-      "ylabel(\"Load in kW  ------>\");\n",
-      "xlabel(\"Time in hours   ----->\");\n",
-      "title(\"Daily Load Curve\")\n",
-      "\n",
-      "#Results:\n",
-      "print \"Diversity factor is \",round(DF,3)\n",
-      "print \"Unis generated per day is \",E,\"kWh\"\n",
-      "print \"Load factor is \",round(LF*100,1),\"%\"\n",
-      "print \"The daily load curve is shown below:\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Diversity factor is "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 1.286\n",
-        "Unis generated per day is  4600 kWh\n",
-        "Load factor is  54.8 %\n",
-        "The daily load curve is shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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8kszMTNq3b09CQoLb72Px4sXMmzeP4OBgIiMjXc40fK1jx44sW7as\nyW1gxXx0GWoREZPTsFhExORUCERETE6FQETE5FQIRERMToVARMTkVAhERExOhUBExORUCERETO7/\nAV3udrfsY+WJAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x49b18b0>"
-       ]
-      }
-     ],
-     "prompt_number": 30
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.12, Page Number: 54"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%pylab inline\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "M1 = 800                     #max demand of consumer 1(Watts)\n",
-      "M2 = 1000                    #max demand of consumer 2(Watts)\n",
-      "M3 = 1200                    #max demand of consumer 3(Watts)\n",
-      "\n",
-      "n1 = linspace(0,8,10);\n",
-      "m1 = linspace(200,200,10);\n",
-      "plot(n1,m1);\n",
-      "\n",
-      "hold(True);\n",
-      "n2 = linspace(8,14,10);\n",
-      "m2 = linspace(800,800,10);\n",
-      "plot(n2,m2,'b');\n",
-      "\n",
-      "n3 = linspace(14,16,10);\n",
-      "m3 = linspace(2400,2400,10);\n",
-      "plot(n3,m3,'b');\n",
-      "\n",
-      "n4 = linspace(16,22,10);\n",
-      "m4 = linspace(800,800,10);\n",
-      "plot(n4,m4,'b');\n",
-      "\n",
-      "n5 = linspace(22,24,10);\n",
-      "m5 = linspace(400,400,10);\n",
-      "plot(n5,m5,'b');\n",
-      "\n",
-      "ylim(0,2500);\n",
-      "xlim(0,24);\n",
-      "grid(linewidth=0.5);\n",
-      "ylabel(\"Load in Watts  ------>\");\n",
-      "xlabel(\"Time in hours   ----->\");\n",
-      "title(\"Daily Load Curve\");\n",
-      "annotate(\"(Midnight)\",xy=(0,0));\n",
-      "annotate(\"(Midnight)\",xy=(24,0));\n",
-      "annotate(\"(Noon)\",xy=(12,0));\n",
-      "\n",
-      "#load factors of each consumers:\n",
-      "LF1 = (600*6+200*2+800*6)/(24*M1)\n",
-      "LF2 = (200*8+1000*2+200*2)/(24*M2)\n",
-      "LF3 = (200*6+1200*2+200*2)/(24*M3)\n",
-      "#The simultaneous maximum demand on the station is 200 + 1000 + 1200 = 2400 W\n",
-      "#and occurs from 2 P.M. to 4 P.M.\n",
-      "DF = (M1+M2+M3)/2400.0           #Diversity factor\n",
-      "LF = (200*8+800*6+2400*2+800*6+400*2)/(24*2400.0)         #load factor\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i)  The maximum demand of individual consumers are:\"\n",
-      "print \"\\tConsumer 1 =\",M1,\"W, Consumer 2 =\",M2,\"W, Consumer 3 =\",M3,\"W\"\n",
-      "print \"\\n(ii)Load Factors of individual consumers are:\"\n",
-      "print \"\\tConsumer 1 =\",round(LF1*100,1),\"%, Consumer 2 =\",round(LF2*100,1),\"\"\"%,\n",
-      "\\tConsumer 3 =\"\"\",round(LF3*100,1),\"%\"\n",
-      "print \"\\n(iii)Diversity factor is \",DF\n",
-      "print \"\\n(iv) Load factor of the station is\",round(LF*100,1),\"%\"\n",
-      "print \"\\n   The load curve is shown below:\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "(i)  The maximum demand of individual consumers are:"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "\n",
-        "\tConsumer 1 = 800 W, Consumer 2 = 1000 W, Consumer 3 = 1200 W\n",
-        "\n",
-        "(ii)Load Factors of individual consumers are:\n",
-        "\tConsumer 1 = 45.8 %, Consumer 2 = 16.7 %,\n",
-        "\tConsumer 3 = 13.9 %\n",
-        "\n",
-        "(iii)Diversity factor is  1.25\n",
-        "\n",
-        "(iv) Load factor of the station is 29.2 %\n",
-        "\n",
-        "   The load curve is shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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JCYFWq4W1tTVWrFghndZcsWIFIiIi8ODBAwQFBWH48OGmCpuIiBSG97okIlIxNYydvDMK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\n/wAb/9x823/nZQAAAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x49b1ed0>"
-       ]
-      }
-     ],
-     "prompt_number": 29
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.13, Page Number: 55"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "A = 12                     #Area of load curve(cm**2)\n",
-      "Lo = 1000                  #load under 1cm length(kW)\n",
-      "To = 2                     #time under 1cm length(hr)\n",
-      "M = 3000                   #maximum demand(kW)\n",
-      "\n",
-      "#Calculation:\n",
-      "L = Lo*To*A/24             #Average load(kW)\n",
-      "LF = L/M                   #load factor\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Load factor is\",round(LF*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load factor is 33.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.14, Page Number: 56"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "C = 75                      #Capacity of each generator(MW)\n",
-      "n = 4                       #No. of generators\n",
-      "CV = 10000                  #Calorific value of oil used(kcal/kg)\n",
-      "H = 2860                    #Avg heat rate(kcal/kWh)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "E = (260*6+200*8+160*4+100*6)            #Units generated per day(MWh)\n",
-      "L = E/24                    #Avg load(MW)\n",
-      "LF  =L/260                  #Load factor\n",
-      "PCF = L/(n*C)               #Plant capacity\n",
-      "TH = H*E*10**3              #heat generated per day(kcal)\n",
-      "w = TH/CV                   #daily fuel requirement(kg)\n",
-      "\n",
-      "print \"Daily load factor is\",round(LF*100,1),\"%\"\n",
-      "print \"Plant capacity factor is\",round(PCF*100,1),\"%\"\n",
-      "print \"Daily fuel requirement is (\",w/1000,\"* 10^3)  kg\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Daily load factor is 70.5 %\n",
-        "Plant capacity factor is 61.1 %\n",
-        "Daily fuel requirement is ( 1258.4 * 10^3)  kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.15, Page Number: 56"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%pylab inline\n",
-      "\n",
-      "#Calculations:\n",
-      "E = 20*2+40*4+60*4+20*4+50*4+20*6    #Units generated per day(MWh)\n",
-      "\n",
-      "#1st plot\n",
-      "subplot(1,3,1)\n",
-      "n1 = linspace(0,8,10);\n",
-      "m1 = linspace(20,20,10);\n",
-      "plot(n1,m1);\n",
-      "\n",
-      "hold(True);\n",
-      "n2 = linspace(8,12,10);\n",
-      "m2 = linspace(40,40,10);\n",
-      "plot(n2,m2,'b');\n",
-      "\n",
-      "n3 = linspace(12,16,10);\n",
-      "m3 = linspace(60,60,10);\n",
-      "plot(n3,m3,'b');\n",
-      "\n",
-      "n4 = linspace(16,20,10);\n",
-      "m4 = linspace(20,20,10);\n",
-      "plot(n4,m4,'b');\n",
-      "\n",
-      "n5 = linspace(20,24,10);\n",
-      "m5 = linspace(40,40,10);\n",
-      "plot(n5,m5,'b');\n",
-      "\n",
-      "ylim(0,100);\n",
-      "xlim(0,24);\n",
-      "grid(linewidth=0.5);\n",
-      "ylabel(\"Load in MW  ------>\");\n",
-      "xlabel(\"Time in hours   ----->\");\n",
-      "title(\"$(i)$ Daily Load Curve\");\n",
-      "\n",
-      "#next plot\n",
-      "subplot(1,3,3)\n",
-      "n1 = linspace(0,4,10);\n",
-      "m1 = linspace(60,60,10);\n",
-      "plot(n1,m1);\n",
-      "\n",
-      "hold(True);\n",
-      "n2 = linspace(4,8,10);\n",
-      "m2 = linspace(50,50,10);\n",
-      "plot(n2,m2,'b');\n",
-      "\n",
-      "n3 = linspace(8,12,10);\n",
-      "m3 = linspace(40,40,10);\n",
-      "plot(n3,m3,'b');\n",
-      "\n",
-      "n4 = linspace(12,24,10);\n",
-      "m4 = linspace(20,20,10);\n",
-      "plot(n4,m4,'b');\n",
-      "\n",
-      "ylim(0,100);\n",
-      "xlim(0,24);\n",
-      "grid(linewidth=0.5);\n",
-      "ylabel(\"Load in MW  ------>\");\n",
-      "xlabel(\"Hours duration  ----->\");\n",
-      "title(\"$(ii)$ Load Duration Curve\");\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Energy generated per day is (\",E,\"* 10^3)  kWh\"\n",
-      "print \"\\nThe daily load curve and load duration curves are shown below:\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Energy generated per day is ("
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 840 * 10^3)  kWh\n",
-        "\n",
-        "The daily load curve and load duration curves are shown below:\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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qiFFOtc7JJCQk4MqVK9i6dSv69OmDgoICju8SKRhzlpSk1uGyiIgI5OTkICws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\nyBARkWT+H5YlVY/jFClJAAAAAElFTkSuQmCC\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x49c3450>"
-       ]
-      }
-     ],
-     "prompt_number": 28
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.16, Page Number: 57"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "c1 = 10                  #Capacity of each of 2 generators(MW)\n",
-      "c2 = 5                   #Capacity of 3rd generator(MW)\n",
-      "\n",
-      "#Calculation:\n",
-      "C = 2*c1+c2              #Installed capacity(MW)\n",
-      "A = (1/2)*(20+4)         #Avg load, seen from the curve(MW)\n",
-      "PF = A/C                  #plant factor\n",
-      "E = 8760*A               #Units generated per annum(MWh)\n",
-      "LF = A/20                #Load factor\n",
-      "PUF = 20/C               #Plant capacity factor\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i) Installed capacity is \",C,\"MW\"\n",
-      "print \"(ii) Plant factor is \",round(PF*100,1),\"%\"\n",
-      "print \"(iii) Units generated per annum is \",(E/1000),\"* 10^3  kWh\"\n",
-      "print \"(iv) Load factor is \",round(LF*100,1),\"%\"\n",
-      "print \"(V) Utilisation factor is \",round(PUF*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Installed capacity is  25 MW\n",
-        "(ii) Plant factor is  48.0 %\n",
-        "(iii) Units generated per annum is  105.12 * 10^3  kWh\n",
-        "(iv) Load factor is  60.0 %\n",
-        "(V) Utilisation factor is  80.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.17, Page Number: 57 "
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "n = 0.72                      #efficiency of the load\n",
-      "\n",
-      "#Calculation:\n",
-      "Ma = (10*0.746/0.72)* 0.65 + 5 * 0.60       #kW\n",
-      "Mb = (7.5*0.746/0.72)*0.75+4*0.60           #kW\n",
-      "Mc = (15*0.746/0.72)*0.65                    #kW\n",
-      "Md = (5*0.746/0.72)*0.75+2*0.60              #kW\n",
-      "\n",
-      "SM = Ma+Mb+Mc+Md                            #Sum of maximum demands(kW)\n",
-      "\n",
-      "#The diversity factor between consumers of this type of service is 1\u00b75 from above table.\n",
-      "M1 = SM/1.5                                  #Maximum demand on transformer 1(kW)\n",
-      "\n",
-      "#In a similar manner, the other transformer loads are determined to be\n",
-      "#                Total       Simultaneous\n",
-      "#Transformer 2   26 kW        7\u00b743 kW\n",
-      "#Transformer 3   29\u00b713 kW    19\u00b740 kW\n",
-      "\n",
-      "#The diversity factor between transformers is 1\u00b73\n",
-      "M2 = 7.43; M3 = 19.40\n",
-      "MF = (M1+M2+M3)/1.3                     #Maximum load on feeder(kW)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The maximum load on the feeder is \",round(MF,2),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum load on the feeder is  37.64 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.18, Page Number: 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "%pylab inline\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "E = 10**3*(20*2+40*3+50*5+35*3+70*3+40*2+20*6)       #Units generated per day(kWh)\n",
-      "A = E/24                                 #Average load(kW)\n",
-      "LF = A/(70*10**3)                        #Load factor\n",
-      "\n",
-      "n1 = linspace(0,8,10);\n",
-      "m1 = linspace(20,20,10);\n",
-      "plot(n1,m1);\n",
-      "\n",
-      "hold(True);\n",
-      "n2 = linspace(8,11,10);\n",
-      "m2 = linspace(40,40,10);\n",
-      "plot(n2,m2,'b');\n",
-      "\n",
-      "n3 = linspace(11,16,10);\n",
-      "m3 = linspace(50,50,10);\n",
-      "plot(n3,m3,'b');\n",
-      "\n",
-      "n4 = linspace(16,19,10);\n",
-      "m4 = linspace(35,35,10);\n",
-      "plot(n4,m4,'b');\n",
-      "\n",
-      "n5 = linspace(19,22,10);\n",
-      "m5 = linspace(70,70,10);\n",
-      "plot(n5,m5,'b');\n",
-      "\n",
-      "n6 = linspace(22,24,10);\n",
-      "m6 = linspace(40,40,10);\n",
-      "plot(n6,m6,'b');\n",
-      "\n",
-      "\n",
-      "ylim(0,100);\n",
-      "xlim(0,24);\n",
-      "grid(linewidth=0.5);\n",
-      "ylabel(\"Load in MW  ------>\");\n",
-      "xlabel(\"Time in hours   ----->\");\n",
-      "title(\"Daily Load Curve\");\n",
-      "\n",
-      "#Results:\n",
-      "print \"Load factor is \",round(LF*100,2),\"%\"\n",
-      "print \"\"\"\\nOperational schedule: Referring to the load curve shown\n",
-      "below, the operational schedule will be as under :\n",
-      " (i)  Set No. 1 will run for 24 hours.\n",
-      " (ii) Set No. 2 will run from 8.00 hours to midnight.\n",
-      " (iii)Set No. 3 will run from 11.00 hours to 16 hours and \n",
-      "      again from 19 hours to 22 hours.\"\"\"\n",
-      "print \"\\nThe Load curve is shown below:\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Populating the interactive namespace from numpy and matplotlib\n",
-        "Load factor is "
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " 55.06 %\n",
-        "\n",
-        "Operational schedule: Referring to the load curve shown\n",
-        "below, the operational schedule will be as under :\n",
-        " (i)  Set No. 1 will run for 24 hours.\n",
-        " (ii) Set No. 2 will run from 8.00 hours to midnight.\n",
-        " (iii)Set No. 3 will run from 11.00 hours to 16 hours and \n",
-        "      again from 19 hours to 22 hours.\n",
-        "\n",
-        "The Load curve is shown below:\n"
-       ]
-      },
-      {
-       "output_type": "stream",
-       "stream": "stderr",
-       "text": [
-        "WARNING: pylab import has clobbered these variables: ['tri', 'axes', 'legend', 'stackplot', 'figure', 'rc_context', 'quiver', 'streamplot', 'rc', 'text', 'colors', 'contour', 'axis', 'colorbar', 'test', 'rcdefaults', 'table']\n",
-        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
-       ]
-      },
-      {
-       "metadata": {},
-       "output_type": "display_data",
-       "png": 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jI8aNG4ctW7ZYu6x6obi42Nol2CTmooy56BM5E5tpArm5uWjZsqVu29fXF7m5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\nAABJRU5ErkJggg==\n",
-       "text": [
-        "<matplotlib.figure.Figure at 0x4c39f30>"
-       ]
-      }
-     ],
-     "prompt_number": 27
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      " Example 3.19, Page Number: 61"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "#peak loads(MW) are:\n",
-      "P1 = 10\n",
-      "P2 = 5\n",
-      "P3 = 8\n",
-      "P4 = 7\n",
-      "\n",
-      "DF = 1.5                   #Diversity factor\n",
-      "LF = 0.60                  #Avg annual load factor\n",
-      "\n",
-      "#Calculation:\n",
-      "M = (P1+P2+P3+P4)/DF        #Max demand on the station(MW)\n",
-      "E = (LF*M)*8760             #Annual energy supplied(MWh)\n",
-      "#the installed capacity should be 15% to 20% more than the\n",
-      "#maximum demand to meet the future needs.\n",
-      "#Taking Installed capacity to be 20% more than the maximum demand,\n",
-      "IC = 1.2*M                  #MW\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i)  The maximum demand on the station is\",M,\"MW\"\n",
-      "print \"(ii) Annual energy supplied by the station is (\",E/1000,\"* 10^6)  kWh\"\n",
-      "print \"(iii)Installed capacity is\",IC,\"MW\"\n",
-      "print \"     Suitable unit sizes are 4, each of 6 MW capacity.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The maximum demand on the station is 20.0 MW\n",
-        "(ii) Annual energy supplied by the station is ( 105.12 * 10^6)  kWh\n",
-        "(iii)Installed capacity is 24.0 MW\n",
-        "     Suitable unit sizes are 4, each of 6 MW capacity.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.20, Page Number: 64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "c1 = 20                   #capacity of standby station(MW)\n",
-      "c2 = 18                   #Capacity of base load station(MW)\n",
-      "E1 = 7.35*10**6            #Annual standby output(kWh)\n",
-      "E2 = 101.35*10**6          #Annual base load station output(kWh)\n",
-      "P3 = 12                     #peak load on standby station(MW)\n",
-      "t = 2190                  #hours of use by standby station/year\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "LF1 = E1/(t*P3*10**3)          #Annual load factor of standby station\n",
-      "PCF1 = E1/(8760*c1*10**3)      #Plant capacity factor of standby station\n",
-      "LF2 = E2/(8760*c2*10**3)        #Annual load factor of base load station\n",
-      "#for base load station,\n",
-      "PCF2 = LF2                      #Plant capacity factor of base load station\n",
-      "\n",
-      "#Results:\n",
-      "print \"Annual load factor of standby station is\",round(LF1*100,1),\"%\"\n",
-      "print \"Plant capacity factor of standby station is\",round(PCF1*100,1),\"%\"\n",
-      "print \"Annual load factor of base load station is\",round(LF2*100,1),\"%\"\n",
-      "print \"Plant capacity factor of base load station is\",round(PCF2*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Annual load factor of standby station is 28.0 %\n",
-        "Plant capacity factor of standby station is 4.2 %\n",
-        "Annual load factor of base load station is 64.3 %\n",
-        "Plant capacity factor of base load station is 64.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 20
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.21, Page Number: 64"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 60000                   #maximum load of hydro plant(kW)\n",
-      "m = 20000                   #minimum load of hydro pland(kW)\n",
-      "P = 50000                    #peak load of plant per day(kW)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let OE = Capacity of steam plant\n",
-      "#EC = Capacity of hydro plant\n",
-      "x,y = symbols('x,y')            #two variables as indicated in given curve\n",
-      "# As steam electric conversion is 60%,\n",
-      "#\u2234 Area HEFI = 0\u00b76 \u00d7 Area FGB                            ... (i)\n",
-      "#But Area HEFI = Area CFE \u2212 Area CHI\n",
-      "#              = 1/2*(x*y-50000)\n",
-      "#  Now Area FGB = 1/2*(FG*GB) = 1/2*(24-x)*(40000-y)\n",
-      "#  or 0.2*x*y + 12000*x + 7.2*y \u2212 338000 = 0\n",
-      "#Also from similar triangles CEF and CDB, we get,\n",
-      "#     y/40000 = x/24\n",
-      "y = x/24*40000\n",
-      "x1 = round(solve(0.2*x*y + 12000*x + 7.2*y-338000,x)[1])\n",
-      "y1 = x1/24*40000\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacity of the hydro plant is\",y1,\"kW\"\n",
-      "print \"Capacity of the steam plant is\",60000-y1,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacity of the hydro plant is 20000.0 kW\n",
-        "Capacity of the steam plant is 40000.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 32
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 3.22, Page Number: 65"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#Suppose the maximum demand of reservoir plant is y MW and it \n",
-      "#operates for x hours\n",
-      "x,y = symbols('x y')\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "E = 10**3*(1/2*(320+160)*8760)            #Units generated/annum\n",
-      "#As the steam plant, run-of-river plant and hydro plant generate \n",
-      "#units in the ratio of 7 : 4 : 1, so, units generated by \n",
-      "#each plant are:\n",
-      "\n",
-      "E1 = E*7/12                    #by steam plant\n",
-      "E2 = E*4/12                    #by run-of-river plant\n",
-      "E3 = E/1/12                    #by reservoir plant\n",
-      "\n",
-      "#(i) Maximum demand on run-of-river plant = Area OEBA/OA\n",
-      "M2 = 700.8*10**6/8760           #kW\n",
-      "\n",
-      "#Now,\n",
-      "x = 8760*y/160\n",
-      "E33 = 10**3*1/2*(x*y)\n",
-      "#But the units generated by reservoir plant are E3 kWh\n",
-      "y1 = solve(E33-E3,y)[1]\n",
-      "M3 = y1                    #maximum demand on reservoir station(MW)\n",
-      "M1 = 320-80-M2/1000              #Maximum demand on steam station(MW)\n",
-      "LF2 = 100            #load factor of river plant(%), because it acts \n",
-      "                        #as a base station.\n",
-      "\n",
-      "LF1 = E1*100/(M1*1000*8760)         #load factor of of run-of-river plant(%)\n",
-      "LF3 = E3*100/(M3*1000*8760)           #load factor of reservoir plant(%)\n",
-      "\n",
-      "    \n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)Maximum demand on run-of-river plant is\",round(M1),\"MW\"\n",
-      "print \"   Maximum demand on steam station is\",round(M2),\"MW\"\n",
-      "print \"   Maximum demand on reservoir station is\",round(y1),\"MW\"\n",
-      "print \"\\n(ii)Load factor of steam plant plant is\",LF1,\"%\"\n",
-      "print \"    Load factor of of run-of-river plant is\",LF2,\"%\"\n",
-      "print \"    Load factor of of reservoir plant is\",round(LF3),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)Maximum demand on run-of-river plant is 160.0 MW\n",
-        "   Maximum demand on steam station is 80000.0 MW\n",
-        "   Maximum demand on reservoir station is 80.0 MW\n",
-        "\n",
-        "(ii)Load factor of steam plant plant is 87.5 %\n",
-        "    Load factor of of run-of-river plant is 100 %\n",
-        "    Load factor of of reservoir plant is 25.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter4_1.ipynb b/Principles_of_Power_System/chapter4_1.ipynb
deleted file mode 100644
index c67cf3f8..00000000
--- a/Principles_of_Power_System/chapter4_1.ipynb
+++ /dev/null
@@ -1,873 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:674aadd4593cd639b4637a0cb17a9ace907a7cf82297146653a75d9ee7f7f19b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 4: Economics of Power Generation"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.1, Page Number: 74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable Declaration:\n",
-      "P = 90000                  #initial cost of transformer(Rs)\n",
-      "n = 20                     #20 years\n",
-      "S = 10000                  #Salvage value(Rs)\n",
-      "\n",
-      "#Calculation:\n",
-      "D = (P-S)/n                #Annual depreciation charge(Rs)\n",
-      "\n",
-      "print \"The annual depreciation charge is Rs\",D"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual depreciation charge is Rs 4000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.2, Page Number: 74"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration\n",
-      "P = 200000                        #Initial cost of transformer(Rs)\n",
-      "S = 10000                         #Salvage value of transformer(Rs)\n",
-      "n = 20                            #Useful life(years)\n",
-      "r = 0.08                              #annual interest rate(%)\n",
-      "\n",
-      "#Calculation:\n",
-      "q = (P-S)*r/(((1+r)**n)-1)        #Annual payment(Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"The annual amount to be saved is Rs\",round(q)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual amount to be saved is Rs 4152.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.3, Page Number: 75"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "P = 1560000                 #Initial cost of equipment(Rs)\n",
-      "S = 60000                    #salvage value of equipment(Rs)\n",
-      "n = 25                      #useful life(years)\n",
-      "t = 20                      #years\n",
-      "#Calculation:\n",
-      "\n",
-      "#(i) Straight line method:\n",
-      "D1 = (P-S)/n                 #Annual depreciation(Rs)\n",
-      "V1 = P-D1*t                   #Value of equipment after 't' years\n",
-      "\n",
-      "#(ii)Diminishing value method:\n",
-      "D2 = 1-(S/P)**(1/n)         #Annual unit depreciation(Rs)\n",
-      "V2 = P*(1-round(D2,3))**t            #Value of equipment after 't' years\n",
-      "\n",
-      "#(iii)Sinking fund method:\n",
-      "r = 0.05                    #rate of interest\n",
-      "q = (P-S)*(r/((1+r)**n-1))   #Annual deposit in the sinking fund(Rs)\n",
-      "q1 = 31433*(((1+r)**20-1)/r)        #Sinking fund at the end of 20 years(Rs)\n",
-      "V = P-q1                    #Value of plant after 20 years(Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"The depreciated values using:\"\n",
-      "print \"(i) Straight line method: Rs\",V1\n",
-      "print \"(ii)Diminishing value method Rs:\",round(V2)\n",
-      "print \"(iii)Sinking fund method: Rs\",round(V)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The depreciated values using:\n",
-        "(i) Straight line method: Rs 360000.0\n",
-        "(ii)Diminishing value method Rs: 115615.0\n",
-        "(iii)Sinking fund method: Rs 520638.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.4, Page Number: 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "M = 50000                  #Max demand(kW)\n",
-      "CC = 95*10**6              #Capital cost(Rs)\n",
-      "LF = 0.4                   #annual load factor\n",
-      "C1 = 9*10**6               #Annual cost of fuel and oil(Rs)\n",
-      "C2 = 7.5*10**6             #taxes wages salaries etc(Rs)\n",
-      "i = 12                    #interest & depreciation(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "E = M*LF*8760             #units generated(kWh/year)\n",
-      "AFC = i*CC/100            #Annual fixed charges(Rs)\n",
-      "T = C1+C2                #Total annual running charges(Rs)\n",
-      "TAC = AFC+T               #Total annual charges(Rs)\n",
-      "c = TAC/E                 #cost per unit(Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Cost per unit generated = \",round(c,2)*100,\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cost per unit generated =  16.0 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.5, Page Number: 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "C = 50000                 #Installed capacity(kW)\n",
-      "E = 220*10**6             #units generated(kWh/year)\n",
-      "AFC1 = 160               #annual fixed charges per kW of C(Rs/kW)\n",
-      "RC = 0.04                 #running charges(Rs)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "AFC = AFC1*C               #annual fixed charges(Rs)\n",
-      "ARC = RC*E                  #annual running charges(Rs)\n",
-      "c = (AFC+ARC)/E           #cost per unit(Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Cost per unit generated is \",round(c,4)*100,\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cost per unit generated is  7.64 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.6, Page Number: 76"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "C = 160000                 #cost of plant(Rs)\n",
-      "r = 12                    #annual fixed charges(%)\n",
-      "r1 = 5                    #interest(%)\n",
-      "r2 = 5                     #depreciation(%)\n",
-      "r3 = 2                     #taxes(%)\n",
-      "M = 100                    #max demand(kW)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "AFC = C*r/100             #annual fixed charges(Rs)\n",
-      "\n",
-      "#(i)when load factor is 100%\n",
-      "E1 = M*1*8760              #kWh/year\n",
-      "c1 = AFC/E1                   #Rs\n",
-      "\n",
-      "#(i)when load factor is 50%\n",
-      "E2 = M*0.5*8760            #kWh/year\n",
-      "c2 = AFC/E2                   #Rs\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Fixed charges when:\"\n",
-      "print \"(i) load factor is 100% :\",round(c1*100,2),\"paise\"\n",
-      "print \"(ii)load factor is 50% :\",round(c2*100,2),\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Fixed charges when:\n",
-        "(i) load factor is 100% : 2.19 paise\n",
-        "(ii)load factor is 50% : 4.38 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.7, Page Number: 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "PC = 50                    #plant capacity(MW)\n",
-      "LF = 0.4                  #annual load factor\n",
-      "C1 = 1.2*10**7            #capital cost(Rs)\n",
-      "C2 = 4*10**5              #annual cost of wages, taxation etc.(Rs)\n",
-      "C3 = 1.0                  #cost of fuel,lubrication, maintenance etc.(paise/kWh)\n",
-      "r1 = 5                    #interest rate(%)\n",
-      "r2 = 6                    #depreciation(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "T1 = (C1*(r1+r2)/100)+C2    #Total annual fixed charges(Rs)\n",
-      "E = PC*10**3*LF*8760        #units generated(kWh/year)\n",
-      "C4 = E*C3/100               #Cost of fuel, lubrication etc.(Rs)\n",
-      "T = T1+C4                 #total annual charges(Rs)\n",
-      "c = T/E                   #generating cost(Rs/kWh)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Cost per kWh is \",round(c*100),\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Cost per kWh is  2.0 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.8, Page Number: 77"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "PC = 300                  #plant capacity(MW)\n",
-      "CF = 0.5                  #capacity factor\n",
-      "LF = 0.6                  #annual load factor\n",
-      "C1 = 9*10**7              #Annual cost of fuel, oil etc(Rs)\n",
-      "C2 = 10**9                #capital cost(Rs)\n",
-      "r1 = 10                   #annual interest and depreciation(%)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "L = CF*PC                #avg load(MW)\n",
-      "M = L/LF                 #max demand(MW)\n",
-      "RC = PC-M                #Reserve capacity(MW)\n",
-      "TC = C1+C2*r1/100        #total cost(Rs)\n",
-      "E = L*10**3*8760         #units generated(kWh/year)\n",
-      "c = TC/E                 #cost per kWh\n",
-      "\n",
-      "#Results:\n",
-      "print \"The minimum reserve capacity of the station is\",RC,\"MW\"\n",
-      "print \"Cost per kWh generated is \",round(c*100),\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The minimum reserve capacity of the station is 50.0 MW\n",
-        "Cost per kWh generated is  14.0 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.9, Page Number: 78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "#Variable declaration:\n",
-      "PC = 50               #plant capacity(MW)\n",
-      "CC = 1000              #capital cost(Rs/kW)\n",
-      "r1 = 10               #annual depreciation charges(%)\n",
-      "r2 = 20                #part of salaries, maitenance to fixed charges(%)\n",
-      "M = 40                #max demand(MW)\n",
-      "LF = 0.60              #load factor\n",
-      "C1 = 700000      #Annual cost of salaries, maintenance charges etc.(Rs)\n",
-      "R1 = 1                #royalty(Re/(kW*year))\n",
-      "R2 = 0.01       #royalty paid for using the river water for generation(Re/kWh)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#for annual fixed cost\n",
-      "E = M*10**3*LF*8760                #kWh/year\n",
-      "C = CC*PC*10**3             #capital cost(Rs)\n",
-      "T1 = r1*C/100+r2*C1/100     #total annual fixed charges(Rs)\n",
-      "\n",
-      "C2 = T1/(M*10**3)+R1                #Cost per kW(Rs)\n",
-      "\n",
-      "#for running cost:\n",
-      "C3 = ((1-r2/100)*C1)/E+R2           #Cost per kW(Rs)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"Total generation cost in two part form is given by:\"\n",
-      "print  \"Rs (\",C2,\"* kW +\",round(C3,4),\"* kWh)\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total generation cost in two part form is given by:\n",
-        "Rs ( 129.5 * kW + 0.0127 * kWh)\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.10, Page Number: 78"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "M = 60                 #plant capacity(MW)\n",
-      "LF = 0.5                #load factor\n",
-      "C1 = 5*10**6            #capital cost of building and equipment(Rs)\n",
-      "C2 = 900000        #annual cost of fuel,oil,taxation and wages(Rs)\n",
-      "r1 = 10                 #interest and depreciation(%)\n",
-      "C3 = 5000000             #annual cost of organisation and \n",
-      "                        #interest on cost of site etc.\n",
-      "\n",
-      "#Calculation:\n",
-      "E = M*10**3*LF*8760     #kWh/year\n",
-      "a = C3                  #Rs\n",
-      "T2 = r1*C3/100          #Annual semi-fixed cost(Rs)\n",
-      "b = T2/(M*10**3)        #cost per kW\n",
-      "c = C2/E                #cost per kWh\n",
-      "\n",
-      "#Results:\n",
-      "print \"The required values are:\"\n",
-      "print \"a = Rs\",a,\", b = \",round(b,2),\", c = Re\",round(c,4)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The required values are:\n",
-        "a = Rs 5000000 , b =  8.33 , c = Re 0.0034\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.11, Page Number: 79"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "PC = 100                 #installed capacity(kW)\n",
-      "CC = 3000                #plant cost(Rs/kW of PC)\n",
-      "r1 = 5                   #interest(%)\n",
-      "r2 = 2                   #depreciation(2%)\n",
-      "r3 = 2                   #operation & maintenance(%)\n",
-      "r4 = 1.5                 #insurance, rent(%)\n",
-      "r = 12.5              #losses in transmission and distribution(%)\n",
-      "DF = 1.25                #diversity factor\n",
-      "LF = 0.4                 #load factor\n",
-      "M = 0.8*PC               #max demand(kW)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "L = M*LF                  #avg demand(kW)\n",
-      "C1 = CC*PC                #Capital cost(Rs)\n",
-      "T1 = (r1+r2)*C1/100       #annual fixed charges(Rs)\n",
-      "T11 = T1/(DF*M)            #annual fixed charges(Rs/kW)\n",
-      "RC = C1*(r3+r4)/100        #annual running charges(Rs)\n",
-      "E = L*8760                #kWh/year\n",
-      "E1 = E*(1-r/100)           #units reaching the consumer(kWh)\n",
-      "RC1 = RC/E1               #annual running charges(Rs/kWh)\n",
-      "T = T1+RC                 #total charges(Rs)\n",
-      "C2 = T/E1                 #cost per kWh(Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Total generation cost in two part form is given by:\"\n",
-      "print  \"Rs (\",T11,\"* kW +\",round(RC1,3),\"* kWh)\"\n",
-      "print \"Overall cost of generation per kWh is \",round(C2*100,1),\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total generation cost in two part form is given by:\n",
-        "Rs ( 210.0 * kW + 0.043 * kWh)\n",
-        "Overall cost of generation per kWh is  12.8 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.12, Page Number: 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#variable declaration:\n",
-      "M = 1000                    #max demand(kW)\n",
-      "LF = 0.5                    #load factor\n",
-      "#for (i)a private oil engine generating plant:\n",
-      "\n",
-      "CC1 = 12*10**5               #capital cost(Rs)\n",
-      "c1 = 0.005                   #Cost of repair and maintenance(Rs/kWh)\n",
-      "c2 = 1600                    #Cost of fuel(Rs/1000kg)\n",
-      "r1 = 10                      #Interest and depreciation (%)\n",
-      "w = 0.3                      #fuel consumption(kg/kWh)\n",
-      "c3 = 50000                   #wages(Rs)\n",
-      "\n",
-      "#for (i)Public supply company:\n",
-      "#Rs 150 per kW of maximum demand plus 15 paise per kWh\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "E = M*LF*8760               #kWh/year\n",
-      "\n",
-      "#for(i) Private oil engine generating plant:\n",
-      "W = w*E                     #annual fuel consumption(kg)\n",
-      "C1= W*c2/1000               #cost of fuel(Rs)\n",
-      "T1 = C1+c1*E+r1*CC1/100+c3      #total annual cost(Rs)\n",
-      "\n",
-      "#for(ii)Public supply company:\n",
-      "T2 = 150*M+.15*E           #total annual cost(Rs)\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i)For Private oil engine generating plant,\" \n",
-      "print  \"   total annual charges is Rs\",T1\n",
-      "print \"(ii)Public supply company, total annual charges is Rs\",T2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)For Private oil engine generating plant,\n",
-        "   total annual charges is Rs 2294300.0\n",
-        "(ii)Public supply company, total annual charges is Rs 807000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      " Example 4.13, Page Number: 80"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "LF = 0.3                     #load factor\n",
-      "M  = 100                     #max demand(MW)\n",
-      "\n",
-      "#steam: \n",
-      "cc1 = 1250                 # Capital cost/kW installed(Rs)\n",
-      "r11 = 12                   # Interest and depreciation(%)\n",
-      "c11 = 5                    # Operating cost/kWh, paise\n",
-      "tc11 = 0                   # Transmission cost/kWh\n",
-      "\n",
-      "#hydro:\n",
-      "cc2 = 2500                 # Capital cost/kW installed(Rs)\n",
-      "r21 = 10                   # Interest and depreciation(%)\n",
-      "c21 = 1.5                  # Operating cost/kWh,paise\n",
-      "tc21 = 0.2                 # Transmission cost/kWh,paise\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "E = M*LF*8760*10**3                  #kWh/year\n",
-      "\n",
-      "#(i)steam station in conjunction with a hydro station:\n",
-      "Eh = 100*10**6              #units supplied by hydro station(kWh)\n",
-      "Es = E-Eh                   #units supplied by steam station(kWh)\n",
-      "Pmh = 40                    #max o/p of hydro stn.(MW)\n",
-      "Pms = 60                    #max o/p of steam stn(MW)\n",
-      "\n",
-      "#(i)(a)for steam station:\n",
-      "Cs1 = cc1*Pms*10**3              #capital cost(Rs)\n",
-      "Ts1 = r11*Cs1/100+c11*Es/100+0         #total cost(Rs)\n",
-      "\n",
-      "# (b)for hydro station:\n",
-      "Ch1 = cc2*Pmh*10**3          #capital cost(Rs)\n",
-      "Th1 = r21*Ch1/100+c21*Eh/100+tc21*Eh/100   #total cost(Rs)\n",
-      "\n",
-      "Ta = Ts1+Th1                 #total annual cost(Rs)\n",
-      "OC1 = Ta/E                   #kWh\n",
-      "\n",
-      "\n",
-      "\n",
-      "#(ii)Steam station:\n",
-      "Pm2 = 100*10**3            #max o/p of steam plant(kW)\n",
-      "Cs2 = cc1*Pm2                   #capital cost(Rs)\n",
-      "Ts2 = (Cs2*r11/100)+c11*E/100   #Rs\n",
-      "OC2 = Ts2/E                     #overall cost(Rs)\n",
-      "\n",
-      "#(iii)Hydro station:\n",
-      "Ch3 = cc2*Pm2                   #capital cost(Rs)\n",
-      "Th3 = Ch3*r21/100+c21*E/100     #Rs\n",
-      "OC3 =  Th3/E+tc21/100                    ##overall cost(Rs)\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i)steam station in conjunction with a hydro station:\"\n",
-      "print \"    for steam stn., overall cost is \",round(OC1*100,2),\"paise\"\n",
-      "print \"(ii) Steam station, overall cost is \",round(OC2*100,2),\"paise\"\n",
-      "print \"(ii) Hydro station, overall cost is \",round(OC3*100,2),\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)steam station in conjunction with a hydro station:\n",
-        "    for steam stn., overall cost is  10.97 paise\n",
-        "(ii) Steam station, overall cost is  10.71 paise\n",
-        "(ii) Hydro station, overall cost is  11.21 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.14, Page Number: 81"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 150*10**3                 #maximum demand(kW)\n",
-      "\n",
-      "#Steam plant:\n",
-      "cc1 = 1600                     #Rs/kW\n",
-      "oc1 = 0.06                     #operating cost(Rs/kWh)\n",
-      "r1 = 7                         #interest(%)\n",
-      "\n",
-      "#Hydro plant:\n",
-      "cc2 = 3000                     #Rs/kW\n",
-      "oc2 = 0.03                     #operating cost(Rs/kWh)\n",
-      "r2 = 7                         #interest(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "x = symbols('x')               #total no. of units generated\n",
-      "#for steam plant:\n",
-      "cs = cc1*M                      #capital cost(Rs)\n",
-      "Ts = r1*cs/(x*100)+oc1               #total cost(Rs)\n",
-      "Ts1 = Ts                       #Rs/kWh\n",
-      "\n",
-      "#for hydro plant:\n",
-      "ch = cc2*M                     #capital cost(Rs)\n",
-      "Th = r2*ch/(100*x)+oc2               #total cost(Rs)\n",
-      "Th1 = Th                       #Rs/kWh\n",
-      "L = solve(Ts1-Th1,x)\n",
-      "LF = (L[0])/(M*8760)\n",
-      "\n",
-      "print \"Load factor is \",round(LF*100,1),\"%\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load factor is  37.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.15, Page Number: 82"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Eo = 40*10**6                               #units needed to generate\n",
-      "\n",
-      "#Hydro                     Steam\n",
-      "cc1 = 2100;           cc2 = 1200             #Capital cost(Rs/kW)        \n",
-      "rc1 = 3.2;            rc2 = 5                #Running cost(paise/kWh)       \n",
-      "r1 = 7.5;              r2 = 9                #Interest & depreciation(%)\n",
-      "RC1 = 33;             RC2 = 25             #Reserve capacity(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let x kW be the maximum demand. \n",
-      "#Let y be the annual load factor at which cost/unit\n",
-      "#of steam and hydro stations is the same.\n",
-      "\n",
-      "x,y = symbols('x y')\n",
-      "E = x*y*8760                   #units generated per annum(kWh)\n",
-      "C2 = x+RC2*x/100               #installed capacity of steam station(kW)\n",
-      "C1 = x+RC1*x/100               #installed capacity of hydro station(kW)\n",
-      "\n",
-      "#steam station:\n",
-      "CC2 = cc2*C2                      #capital cost(Rs)\n",
-      "OC2 = (r2*CC2/100+rc2*E/100)/E       #Overall cost/kWh\n",
-      "\n",
-      "#hydro station:\n",
-      "CC1 = cc1*C1                      #capital cost(Rs)\n",
-      "OC1 = (r1*CC1/100+rc1*E/100)/E       #Overall cost/kWh\n",
-      "\n",
-      "LF = solve(OC1-OC2,y)[0]          #load factor\n",
-      "E2 = 8760*x*LF                    #kWh\n",
-      "M = solve(E2-Eo,x)[0]             #max. demand(kW)\n",
-      "C = (135*M + 438*M*LF)            #cost of generation(Rs)\n",
-      "\n",
-      "#Results\n",
-      "print \"Load factor is \",round(LF*100,2),\"%\"\n",
-      "print \"Required cost of generation is Rs (\",round(C/10**6,1),\"* 10**6 )\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Load factor is  47.23 %\n",
-        "Required cost of generation is Rs ( 3.3 * 10**6 )\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 4.16, Page Number: 83"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "\n",
-      "#Let x = Installed capacity of station B in kW\n",
-      "#y = Hours of operation of station B\n",
-      "\n",
-      "x,y = symbols('x y')\n",
-      "M = 50000                   #max demand(kW)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "PCa = M-x                   #installed capacity of stationA(kW)\n",
-      "Eb = (1/2)*x*(8760*x/M)      #Units generated/annum by station B\n",
-      "Ea = (1/2)*8760*M-Eb         #Units generated/annum by station A\n",
-      "Cb = 50000+50*x+0.03*Eb      #Rs\n",
-      "Ca = 75000+80*(50000-x)+0.02*Ea    #Rs\n",
-      "C = Ca+Cb                       #total operating cost(Rs)\n",
-      "#After differentiating C w.r.t x, we get C1 as\n",
-      "C1 = -30+0.00174*x\n",
-      "x1 = solve(C1,x)[0]\n",
-      "\n",
-      "PCb = x1                     #kW\n",
-      "PCa = M-PCb                  #kW\n",
-      "t = 8760*PCb/M               #hours of operation of station B\n",
-      "\n",
-      "#Results:\n",
-      "print \"Installed capacity of station A is \",round(PCa),\"MW\"\n",
-      "print \"Installed capacity of station B is \",round(PCb),\"MW\"\n",
-      "print \"No. of hours of operation of plant B is\",round(t/10)*10,\"hrs\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Installed capacity of station A is  32759.0 MW\n",
-        "Installed capacity of station B is  17241.0 MW\n",
-        "No. of hours of operation of plant B is 3020.0 hrs\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter5_1.ipynb b/Principles_of_Power_System/chapter5_1.ipynb
deleted file mode 100644
index 195b1538..00000000
--- a/Principles_of_Power_System/chapter5_1.ipynb
+++ /dev/null
@@ -1,762 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:95fdfaa4e5e5dbdce2d60b65f4e1e462efff574f2746529cb62b6b5d733a0dc4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 5: Tariff"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.1, Page Number: 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "M = 200                          #max demand(kW)\n",
-      "LF = 0.4                         #load factor\n",
-      "c1 = 100                         #tarif(Rs/kW)\n",
-      "c2 = 10                           #tariff(pais/kWh)\n",
-      "\n",
-      "#Calculation:\n",
-      "E = M*LF*8760                    #units consumed/year\n",
-      "T = c1*M+E*c2/100                #annual charges(Rs)\n",
-      "OC = T/E                          #overall cost(Rs/kWh)\n",
-      "\n",
-      "#Results:\n",
-      "print \"Overall cost per kWh is \",round(OC*100,2),\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Overall cost per kWh is  12.85 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.2, Page Number: 91"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 220                         #voltage(V)\n",
-      "I = 20                          #current(A)\n",
-      "E = 8760                        #kWh\n",
-      "c1 = 20                         #tariff part1(paise/unit for 500hrs)\n",
-      "c2 = 10                          #tariff part2 for additional unit(paise/unit)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#assuming power factor to be unity.\n",
-      "M = V*I/1000                       #max demand(kW)\n",
-      "\n",
-      "#part (i):\n",
-      "E1 = M*500                       #kWh\n",
-      "C1 = c1*E1/100                   #Rs\n",
-      "E2 = E-E1                      #kWh\n",
-      "C2 = 10*E2/100                 #kWh\n",
-      "T = C1+C2                       #total annual bill(Rs)\n",
-      "T2 = T/E                       #equivalent flat rate(Rs/kWh)\n",
-      "\n",
-      "#Results:\n",
-      "print \"(i) Annual bill is Rs\",T\n",
-      "print \"(ii)Eqv flat rate is \",round(T2*100,1),\"paise\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Annual bill is Rs 1096.0\n",
-        "(ii)Eqv flat rate is  12.5 paise\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.3, Page Number: 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "\n",
-      "#for tariff (a):\n",
-      "c1 = 100                   #tariff part1(Rs)\n",
-      "c11 = 15                   #tariff part2(paise/kWh)\n",
-      "\n",
-      "#for tariff (b):\n",
-      "c2 = 30                   #paise/kWh\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let x be the number of units at which charges \n",
-      "#due to both tariffs become equal.\n",
-      "\n",
-      "x = symbols('x')\n",
-      "x1 = solve(c1+c11*x/100 - c2*x/100 , x)[0]\n",
-      "\n",
-      "#Results:\n",
-      "print \"Tariff(a) is economical if consumption is more than\",round(float(x1),2),\"units.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Tariff(a) is economical if consumption is more than 666.67 units.\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.4, Page Number: 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#for 1t tariff:\n",
-      "c11 = 30                           #Rs/annum\n",
-      "c12 = 3                            #paise/unit\n",
-      "\n",
-      "#for 2nd tariff:\n",
-      "c21 = 6                            #paise/unit for 1st 400 units\n",
-      "c22 = 5                            #paise/unit for extra units\n",
-      "\n",
-      "#Calculation:\n",
-      "#Let x (> 400) be the number of units taken per annum \n",
-      "#for which the annual charges due to both tariffs become equal.\n",
-      "\n",
-      "x=symbols('x')\n",
-      "T1 = c11+c12*x/100                  #charges due to 1st tariff(Rs)\n",
-      "T2 = c21*400/100+c22*(x-400)/100                  #charges due to 2nd tariff(Rs)\n",
-      "x1 = solve(T1-T2,x)[0]\n",
-      "\n",
-      "#Results:\n",
-      "print \"Required no. of units are \",round(x1),\"kWh\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required no. of units are  1300.0 kWh\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.5, Page Number: 92"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "M = 50                           #max load on the station(MW)\n",
-      "AD = 75                          #aggregate demand by consumers(MW)\n",
-      "E = 18*10**7                      #units/annum\n",
-      "\n",
-      "#for annual fixed charges:\n",
-      "c11 = 28*10**5                    #for generation(Rs)\n",
-      "c12 = 32*10**5                   #for transmission & distribution(Rs)\n",
-      "c13 = 90*10**5                   #for fuel(Rs)\n",
-      "\n",
-      "#for running charges:\n",
-      "c21 = 0.9*90*10**5               #fuel cost(Rs)\n",
-      "r = 85                           #% of power transmitted\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "T1 = c11+c12+c13*0.1            #10% of fuel used for fixed charges(Rs)\n",
-      "C1 = T1/(AD*10**3)                       #Rs/kW\n",
-      "\n",
-      "\n",
-      "E1 = r*E/100                        #units delivered to consumers\n",
-      "C2 = c21/E1                     #cost per kWh\n",
-      "\n",
-      "#Results:\n",
-      "print \"Tariff is\",C1 ,\"Rs/kW of maximum demand plus\",round(C2*100,1),\"paise per kWh.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Tariff is 92.0 Rs/kW of maximum demand plus 5.3 paise per kWh.\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.6, Page Number: 93"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 75*10**3                           #Max emand(kW)\n",
-      "LF = 0.4                               #load factor\n",
-      "\n",
-      "c1 = 60                                #1st part of generating cost(Rs/kW)\n",
-      "c2 = 4                                 #2nd part of generating cost(paise/kW)\n",
-      "\n",
-      "CT = 2000000                         #annual capital charges for transmission system(Rs)\n",
-      "CD = 1500000                         #annual capital charges for distribution system(Rs)\n",
-      "\n",
-      "dt = 1.2                             #diversity factor of tr. system\n",
-      "dd = 1.25                            #diversity factor of tr. system\n",
-      "\n",
-      "nt = 0.9                             #efficiency of tr system\n",
-      "nd = 0.85                            ##efficiency of distribution system\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) Cost at substation:\n",
-      "#(a)Annual fixed charges:\n",
-      "\n",
-      "Tafc1 = c1*M+CT                      #total annual fixed cost(Rs)\n",
-      "S1 = M*dt                            #sum of all the max demands(kW)\n",
-      "AC1 = Tafc1/S1                        #Annual cost per kW of max. demand(Rs)\n",
-      "\n",
-      "#(b) Running Charges:\n",
-      "Cs1 = c2/nt                        #Cost/kWh at substation(paise)\n",
-      "\n",
-      "#(ii) Cost at consumer\u2019s premises:\n",
-      "Tafc2 = Tafc1+CD             #Total annual fixed charges at consumer\u2019s premises(Rs)\n",
-      "S2 = S1*dd                   #sum of of maximum demands of all consumers(kW)\n",
-      "AC2 = Tafc2/S2                      #Annual cost per kW of maximum demand(Rs)\n",
-      "#As the distribution efficiency is 85%, therefore, for each kWh delivered from\n",
-      "#substation, only 0\u00b785 kWh reaches the consumer\u2019s premises\n",
-      "Cs2 = Cs1/nd                        #Cost/kWh at consumer premises(paise)\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)At sub-station, the cost is Rs\",round(AC1,2),\"per annum per kW maximum demand \"\n",
-      "print \"   plus\",round(Cs1,2),\"paise per kWh\"\n",
-      "print \"\\n(ii)At consumer\u2019s premises, the cost is\",round(AC2,2),\"per annum per kW maximum demand\"\n",
-      "print \"    plus\",round(Cs2,2),\"paise per kWh.\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)At sub-station, the cost is Rs 72.22 per annum per kW maximum demand \n",
-        "   plus 4.44 paise per kWh\n",
-        "\n",
-        "(ii)At consumer\u2019s premises, the cost is 71.11 per annum per kW maximum demand\n",
-        "    plus 5.23 paise per kWh.\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.7, Page Number: 94"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#     Fixed charges    Running charges       #Station \n",
-      "#          (per kW)     (paise/kWh)\n",
-      "Cf1 =  300;         Cr1 = 25                 #Diesel \n",
-      "Cf2 =  1200;        Cr2 = 6.25               #Steam \n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Suppose energy supplied in one year is 100 units i.e., 100 kWh.\n",
-      "\n",
-      "#Diesel Station:\n",
-      "L = symbols('L')                   #load factor\n",
-      "E = 100                            #kWh(say)\n",
-      "P = E/8760                         #avg power, kW\n",
-      "M = P/L                             #max deamnd(kW)\n",
-      "C1 = Cf1*M+E*Cr1/100             #Fixed and running charges for 100 kWh\n",
-      "\n",
-      "#Steam station\n",
-      "C2 = Cf2*M+E*Cr2/100            #Fixed and running charges for 100 kWh\n",
-      "\n",
-      "L1 = solve(C1-C2,L)[0]\n",
-      "\n",
-      "#Result:\n",
-      "print \"The load fctor is \",round(L1*100,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The load fctor is  54.79 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 11
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.8, Page Number: 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "M = 100                           #max demand(kW)\n",
-      "LF = 0.6                          #load factor\n",
-      "pf = 0.8                          #power factor\n",
-      "c1 = 75                           #1st part tariff(Rs/kVA)\n",
-      "c2 = 15                           #2nd part tariff(paise/kWh)\n",
-      "\n",
-      "#Calculation:\n",
-      "E = M*LF*8760                    #units consumed/year\n",
-      "M1 = M/pf                        #max demand in kVA\n",
-      "AB = M1*c1+E*c2/100              #annual bill(Rs)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Annual bill is Rs\",AB"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Annual bill is Rs 88215.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.9, Page Number: 95"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "M = 240                             #max load(kW)\n",
-      "pf = 0.8                             #power factor\n",
-      "E = 50000                            #annual units consumption(kW)\n",
-      "c1 = 50                              #1st part tariff(Rs/KVA)\n",
-      "c2 = 10                              #2nd part tariff(paise/unit)\n",
-      "\n",
-      "#Calculation:\n",
-      "M1 = M/pf                          #KVA\n",
-      "AB = M1*c1+E*c2/100                 #annual bill(Rs)\n",
-      "FR = AB/E                          #Rs\n",
-      "\n",
-      "#now\n",
-      "pf1 = 1\n",
-      "M2 = M\n",
-      "AB1 = M2*c1+E*c2/100               #Rs\n",
-      "S = AB-AB1                         #annual saving(Rs)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Flat rate of energy consumption is \",FR*100,\"paise\"\n",
-      "print \"Annual saving is Rs\",S"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Flat rate of energy consumption is  40.0 paise\n",
-        "Annual saving is Rs 3000.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.10, Page Number: 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "M = 50                            #max demand(kW)\n",
-      "E = 36000                         #energy consume(kWh)\n",
-      "R = 23400                         #reactive power(KVAR)\n",
-      "c1 = 80                              #1st part tariff(Rs/kW)\n",
-      "c2 = 8                              #2nd part tariff(paise/unit)\n",
-      "c3 = 0.5               #3rd part tariff(p/kWh)for each 1% of pf below 86%\n",
-      "\n",
-      "#Calculation:\n",
-      "L = E/(24*30)                   #avg load(kW)\n",
-      "RP = R/(24*30)                  #avg reactive power(kVAR)\n",
-      "\n",
-      "theta = math.atan(RP/L)         #power factor angle\n",
-      "pf = math.cos(theta)              \n",
-      "PFS = E*c3*(0.86-pf)               #power factor surcharge(Rs)\n",
-      "MB = c1*L+c2*E/100+PFS           #monthly bill(Rs)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The monthly bill is Rs\",round(MB,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The monthly bill is Rs 7268.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.11, Page Number: 96"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "c1 = 150                              #1st part tariff(Rs/KVA)\n",
-      "c2 = 8                              #2nd part tariff(paise/unit)\n",
-      "LF = 0.3                            #load factor\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#suppose max demand is 1kVA\n",
-      "\n",
-      "#(i)When p.f. is unity:\n",
-      "pf = 1\n",
-      "OC1 = c1*100/(8760*LF)+c2              #operating cost(Rs)\n",
-      "\n",
-      "#(ii) When p.f. is 0\u00b77\n",
-      "pf1 = 0.7\n",
-      "OC2 = c1*100/(8760*LF*pf1)+c2         #operating cost(Rs)\n",
-      "\n",
-      "#Result:\n",
-      "print \"At unity p. f., overall cost is Rs\",round(OC1,2)\n",
-      "print \"At 0.7 p. f., overall cost is Rs\",round(OC2,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "At unity p. f., overall cost is Rs 13.71\n",
-        "At 0.7 p. f., overall cost is Rs 16.15\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.12, Page Number: 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#variable declaration\n",
-      "L = 200                        #avg load(kW)\n",
-      "pf = 0.8                       #power factor\n",
-      "M = 250                        #max demand(kW)\n",
-      "l = 4                          #losses(%)\n",
-      "r = 12                          #interest & depreciation(%)\n",
-      "C = 50                         #high voltage equipment cost(Rs)\n",
-      "t = 8                          #working hours\n",
-      "n = 300                        #no. of working working\n",
-      "\n",
-      "#for system(i)high voltage supply:\n",
-      "c11 = 5                        #1st part tariff(paise/unit)\n",
-      "c12 = 4.50                    #2nd part tariff(per month per kVA)\n",
-      "\n",
-      "#for system(ii)low voltage supply:\n",
-      "c21 = 5.5                      #1st part tariff(paise/unit)\n",
-      "c22 = 5                        #2nd part tariff(Rs per month per kVA)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) High voltage supply:\n",
-      "\n",
-      "M1 = M/pf                      #Max. demand in kVA\n",
-      "#As the losses in h.v. equipment are 4%, therefore, \n",
-      "#capacity of h.v. equipment:\n",
-      "Cap = round(M1/(1-l/100),1)            #capacity of h.v. equipment(kVA)\n",
-      "C1 = C*Cap                     #Capital investment on h.v. equipment(Rs)\n",
-      "E1 = L*t*n/(1-l/100)           #units consumed(kWh)\n",
-      "T1 = C1*r/100+Cap*c12*12+c11*E1/100                #Total annual cost(Rs)\n",
-      "\n",
-      "#(i) low voltage supply:\n",
-      "M2 = M/pf                      #Max. demand in kVA\n",
-      "E2 = L*t*n                     #units consumed(kWh)\n",
-      "T2 = M2*c22*12+E2*c21/100      #kWh\n",
-      "\n",
-      "T = T2 - T1\n",
-      "\n",
-      "#Results:\n",
-      "print \"Difference in the annual costs of two systems is Rs\",T\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Difference in the annual costs of two systems is Rs 620.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.13, Page Number: 97"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#(i) Purchasing diesel set:\n",
-      "M1 = 1000                      #kW\n",
-      "C1 = 400                       #Rs/kW\n",
-      "r1 = 10                         #annual interest depreciation(%)\n",
-      "c11 = 75                       #Rs/kW\n",
-      "c12 = 5                         #paise/unit\n",
-      "\n",
-      "#(ii) Purchasing from grid supply:\n",
-      "r1 = 10                         #annual interest depreciation(%)\n",
-      "c21 =120                       #Rs/kW\n",
-      "c22 = 3                         #paise/unit\n",
-      "#after 2 years:\n",
-      "M2 = 2500                        #kW\n",
-      "E = 5.5*10**6                   #units reached\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) Purchasing diesel set:\n",
-      "CC = M1*C1                     #Rs\n",
-      "#The present capacity of the station is 2000 kW and the expected\n",
-      "#maximum demand after two years is 2500 kW.\n",
-      "P = 2500-2000                   #extra power to be generated(kW)\n",
-      "T1 = CC*r1/100+P*c11+E*c12/100       #total annual cost(Rs)\n",
-      "\n",
-      "#(ii) Purchasing from grid supply:\n",
-      "T2 = P*c21+E*c22/100           #total annual cost(Rs)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Alternative (ii) is cheaper by Rs\",T1-T2,\"per annum\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Alternative (ii) is cheaper by Rs 127500.0 per annum\n"
-       ]
-      }
-     ],
-     "prompt_number": 17
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 5.14, Page Number: 98"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#H.V supply:\n",
-      "c11 = 70                         #1st part tariff(Rs/kVA)\n",
-      "c12 = 3                          #2nd part tariff(paise/kWh)\n",
-      "\n",
-      "#L.V supply:\n",
-      "c21 = 65                         #1st part tariff(Rs/kVA)\n",
-      "c22 = 4                          #2nd part tariff(paise/kWh)\n",
-      "\n",
-      "c = 50                           #cost of transformer & switchgear for HV side(Rs/kVA)\n",
-      "r1= 2                            #transformer losses(%)\n",
-      "r2 = 15                 #annual fixed charges(%) of transformer & switchgear\n",
-      "n = 6                            #no of working hours\n",
-      "\n",
-      "#Calculation:\n",
-      "(x,y) = symbols('x y')              #say x = Factory load in kW\n",
-      "                                 #y = No. of working days above which H.V.\n",
-      "                                    #supply is cheaper\n",
-      "#for HV side:        \n",
-      "r = x*round(1/(1-r1/100),4)                #rating of transformer & switchgear(kVA)\n",
-      "E1 = x*y*round(n*1/(1-r1/100),2)                       #units consumed per annnum\n",
-      "T11 = x*math.floor(1/(1-r1/100)*c11*100)/100+x*round(1/(1-r1/100)*r2*c/100,2)          #total fixed charges(Rs)\n",
-      "T12 = E1*c12/100                 #total running charges(Rs)\n",
-      "T1 = T11+T12                     #total annual charges(Rs)\n",
-      "\n",
-      "#for LV side:\n",
-      "E2 = x*y*n                      #units consumed per annnum\n",
-      "T21 = c21*x                     #total fixed charges(Rs)\n",
-      "T22 = c22*E2/100                #total running charges(Rs)\n",
-      "T2 = T21+T22                   #total annual charges(Rs)\n",
-      "y11 = solve(T1-T2,y)[0]\n",
-      "\n",
-      "#Result:\n",
-      "print \"If the factory is run for more than\",math.floor(y11),'days' #the ans. is different from that in book\n",
-      "print \"then H. V. supply will be cheaper.\"                         #due to calculation using improper rounding."
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "If the factory is run for more than 249.0 days\n",
-        "then H. V. supply will be cheaper.\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter6_1.ipynb b/Principles_of_Power_System/chapter6_1.ipynb
deleted file mode 100644
index a1cd7a3c..00000000
--- a/Principles_of_Power_System/chapter6_1.ipynb
+++ /dev/null
@@ -1,1112 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:357707f074867e12cc217076fc35399c9b1e6483f5b92a23e0c3be75457e81b5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 6: Power Factor Improvement"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.1, Page Number: 108"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Variable declaration:\n",
-      "kW1 = 300\n",
-      "pf1 = 0.6                  #power factor\n",
-      "pf2 = 1\n",
-      "\n",
-      "#Calculation:\n",
-      "kVA = kW1/pf1\n",
-      "kW2 = kVA*pf2\n",
-      "kW = kW2-kW1\n",
-      "\n",
-      "#Result:\n",
-      "print \"Increased power supplied by the alternator is\",kW,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Increased power supplied by the alternator is 200.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.2, Page Number: 109"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V = 400                     #rated voltage(V)\n",
-      "Im = 31.7                    #motor curtrent(A)\n",
-      "pf1 = 0.7                    #initial power factor\n",
-      "pf2 = 0.9                    #raised power factor\n",
-      "f = 50                       #Hz\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Referring to the phasor diagram,\n",
-      "Ima = Im*pf1                #Active component of Im(A)\n",
-      "I = Ima/pf2                 #total current(A)\n",
-      "Imr = Im*math.sqrt(1-pf1**2)     #Reactive component of Im(A)\n",
-      "Ir = I*math.sqrt(1-pf2**2)       #Reactive component of I(A)\n",
-      "Ic = Imr-Ir                 #A\n",
-      "\n",
-      "C = Ic/(V*2*math.pi*f)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Required value of capacitor is\",round(C,7),\"uF\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required value of capacitor is 9.46e-05 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.3, Page Number: 110"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#(i):\n",
-      "kW1 = 20\n",
-      "pf1 = 1\n",
-      "\n",
-      "#(ii):\n",
-      "kW2 = 100\n",
-      "pf2 = 0.707\n",
-      "\n",
-      "#(iii):\n",
-      "kW3 = 50\n",
-      "pf3 = 0.9\n",
-      "\n",
-      "#Calculation:\n",
-      "kVA1 = kW1/pf1\n",
-      "kVA2 = kW2/pf2\n",
-      "kVA3 = kW3/pf3\n",
-      "\n",
-      "kVAR1 = kVA1*0\n",
-      "kVAR2 = -kVA2*math.sin(math.acos(0.707))\n",
-      "kVAR3 = kVA3*math.sin(math.acos(0.9))\n",
-      "\n",
-      "kW = kW1+kW2+kW3\n",
-      "kVAR = kVAR1+kVAR2+kVAR3\n",
-      "kVA = math.sqrt(kW**2+kVAR**2)\n",
-      "pf = kW/kVA\n",
-      "\n",
-      "#Result:\n",
-      "print \"Total kW is\",kW\n",
-      "print \"Total kVA is\",round(kVA)\n",
-      "print \"Power factor is\",round(pf,3)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total kW is 170\n",
-        "Total kVA is 186.0\n",
-        "Power factor is 0.913\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.4, Page Number: 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "pf1 = 0.75                #Original p.f lag\n",
-      "P = 5                     #motor input(kW)\n",
-      "pf2 = 0.9                  #final p.f lag\n",
-      "\n",
-      "#Calculation:\n",
-      "kVAR = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
-      "R = kVAR/3\n",
-      "\n",
-      "#Result:\n",
-      "print \"Rating of capacitors is\",round(R,3),\"kVAR\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Rating of capacitors is 0.663 kVAR\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.5, Page Number: 111"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Vph = 400                  #rated voltage(V)\n",
-      "n = .93                  #efficiency\n",
-      "P = 74.6                  #power output(kW)\n",
-      "pf1 = 0.75                #power faactor\n",
-      "pf2 = 0.95               #raised power factor\n",
-      "n1 = 4                  #no. of capacitors in each branch\n",
-      "V1 = 100                 #rating of capacitors(V)\n",
-      "f = 50                   #frequency(Hz)\n",
-      "\n",
-      "#Calculation:\n",
-      "Pi = round(P/n)                    #motor input(kW)\n",
-      "phy1 = math.acos(pf1)      \n",
-      "phy2 = math.acos(pf2)\n",
-      "#Leading kVAR taken by the condenser bank\n",
-      "kVAR = round(Pi*(math.tan(phy1)-math.tan(phy2)),2)  \n",
-      "kVAR1 = round(kVAR/3,2)             #Leading kVAR taken by each of three sets\n",
-      "\n",
-      "#Fig. above shows the delta* connected condenser bank.\n",
-      "C = symbols('C')\n",
-      "Icp = 2*math.pi*f*C*Vph           #Phase current of capacitor(A)\n",
-      "\n",
-      "c = solve(Vph*Icp/1000-kVAR1,C)[0]  #capacitance(F)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacitance of each capacitor is\",round(c*4*10**6,1),\"uF\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacitance of each capacitor is 1173.8 uF\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.6, Page Number: 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P = 800                  #load(kW)\n",
-      "pf1 = 0.8                #initial power factor\n",
-      "pf2 = 0.9                #improved power factor\n",
-      "t = 3000                 #working hours\n",
-      "c1 = 100                 #1st part tariff(Rs/kVA)\n",
-      "c2 = 20                  #2nd part tariff(paise/kWh)\n",
-      "c3 = 60                  #cost of capacitors(Rs/kVAR)\n",
-      "n = 10                  #interest & depreciation on capacitors(%)\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "#Leading kVAR taken by the capacitors:\n",
-      "kVAR = P*(math.tan(phy1)-math.tan(phy2))\n",
-      "\n",
-      "#Annual cost before p.f. correction:\n",
-      "kVAm1 = P/pf1                #max demand\n",
-      "T1 = kVAm1*c1+P*t*c2/100      #total annual cost(Rs)\n",
-      "\n",
-      "#Annual cost after p.f. correction:\n",
-      "kVAm2 = P/pf2               #max demand\n",
-      "T2 = c1*kVAm2+P*t*c2/100+n*c3*kVAR/100\n",
-      "\n",
-      "T = T1-T2                #annual saving(Rs)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Annual saving is Rs\",round(T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Annual saving is Rs 9836.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.7, Page Number: 112"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P1 = 200                  #factory load(kW)\n",
-      "pf1 = 0.85               #initial power factor\n",
-      "pf2 = 0.9                #improved power factor\n",
-      "t = 2500                 #working hours\n",
-      "c1 = 150                 #1st part tariff(Rs/kVA)\n",
-      "c2 = 5                   #2nd part tariff(paise/kWh)\n",
-      "c3 = 420                 #cost of capacitors(Rs/kVAR)\n",
-      "Pcl = 100                #capacitor loss(W/kVAR)\n",
-      "n = 10                   #interest & depreciation on capacitors(%)\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "#suppose the leading kVAR taken by the capacitors is x:\n",
-      "x = symbols('x')\n",
-      "Pcl1 = Pcl*x/1000                #capacitor loss(kW)\n",
-      "P2 = P1+Pcl1                     #total power(kW)\n",
-      "x1 = solve((P1*math.tan(phy1)-P2*math.tan(phy2))-x,x)[0]\n",
-      "\n",
-      "\n",
-      "#Annual cost before p.f. correction:\n",
-      "kVAm1 = P1/pf1                #max demand\n",
-      "T1 = kVAm1*c1+P1*t*c2/100      #total annual cost(Rs)\n",
-      "\n",
-      "\n",
-      "#Annual cost after p.f. correction:\n",
-      "kVAm2 = P1/pf2               #max demand\n",
-      "T2 = c1*kVAm2+P1*t*c2/100+n*c3*x1/100+round(Pcl*x1*t*c2/100000)\n",
-      "\n",
-      "T = T1-T2                #annual saving(Rs)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Annual saving is Rs\",round(T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Annual saving is Rs 553.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.8, Page Number: 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "pf = 0.8                           #power factor\n",
-      "kVA = 750                          # monthly demand\n",
-      "c1 = 8.50                          #monthly power rate(Rs/kVA per month)\n",
-      "R = 250                            #rating of capacitors(kVA)\n",
-      "C2 = 20000                        #installed cost of equipment(Rs)\n",
-      "r = 10                            #fixed charge rate(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "phy = math.acos(pf)\n",
-      "kW = kVA*math.cos(phy)             #kW component of demand\n",
-      "kVAR = kVA*math.sin(phy)           #kVAR component of demand\n",
-      "kVAR1 = kVAR-R                     #Leading kVAR taken by the capacitors\n",
-      "kVA1 = math.sqrt(kVAR1**2+kW**2)        #kVA after p.f. improvement\n",
-      "kVA11 = round(kVA-kVA1,1)                   #Reduction in kVA\n",
-      "T1 = c1*kVA11*12                   #Yearly saving on kVA charges\n",
-      "T2 = r*C2/100                      #Fixed charges/year(Rs)\n",
-      "T = T1-T2                          #net saaving(Rs)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The annual saving is Rs\",T"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual saving is Rs 9985.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.9, Page Number: 113"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "pf1 = 0.8                    #initial power factor\n",
-      "pf2 = 0.9                     #improved power factor\n",
-      "L1 = 200                       #load(kW)\n",
-      "L2 = 80                      #motor load(kW)\n",
-      "\n",
-      "#Calculation:\n",
-      "phy1 = math.acos(pf1)\n",
-      "phy2 = math.acos(pf2)\n",
-      "L = L1+L2                    #combined load(kW)\n",
-      "#In Fig. above, \u0394 OAB is the power triangle for load,\n",
-      "#\u0394 ODC for combined load and \u0394 BEC for the motor.\n",
-      "#(i):\n",
-      "kVAR = L1*math.tan(phy1)-L*math.tan(phy2)   #Leading kVAR taken by the motor\n",
-      "#(ii):\n",
-      "kVArat = math.sqrt(L2**2+kVAR**2)          #kVA rating of motor\n",
-      "#(iii):\n",
-      "pf = L2/kVArat                        #p.f. of motor\n",
-      "\n",
-      "#Result:\n",
-      "print \"The leading kVAR taken by the motor is\",round(kVAR,2)\n",
-      "print \"kVA rating of the motor is\",round(kVArat,2)\n",
-      "print \"Power factor at which the motor operates is\",round(pf,3),\"leading\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The leading kVAR taken by the motor is 14.39\n",
-        "kVA rating of the motor is 81.28\n",
-        "Power factor at which the motor operates is 0.984 leading\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.10, Page Number: 114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#for induction motor:\n",
-      "kW1 = 37.3                          \n",
-      "pf1 = 0.8                           #lagging\n",
-      "n1 = 0.85\n",
-      "\n",
-      "#for synchronous motor:\n",
-      "kW2 = 18.65\n",
-      "pf = 0.9                            #leading\n",
-      "n2 = 0.9\n",
-      "\n",
-      "#for lighting load:\n",
-      "kW3 = 10\n",
-      "pf3 = 1\n",
-      "c1 = 60                          #Rs/kVA of max demand\n",
-      "c2 = 5                           #paie per kWh\n",
-      "t = 2000                         #working hours\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "P1 = kW1/n1                     #power input to induction motor(kW)\n",
-      "Q1 = P1*math.tan(math.acos(pf1)) #Lagging kVAR taken by induction motor\n",
-      "\n",
-      "P2 = kW2/n2                      #Input power to synchronous motor(kW)\n",
-      "Q2 = P2**math.tan(math.acos(pf1))  #Leading kVAR taken by synchronous motor\n",
-      "\n",
-      "#Since lighting load works at unity p.f., its lagging kVAR = 0.\n",
-      "Q3 = 0                           \n",
-      "\n",
-      "kVAR = Q1-Q2                    #net kVAR\n",
-      "kW = round(P1+P2+kW3,1)                      #total active power\n",
-      "kVA = round(math.sqrt(kW**2+kVAR**2))\n",
-      "\n",
-      "T = c1*kVA+c2*kW*t/100\n",
-      "\n",
-      "#Result:\n",
-      "print \"The annual electical charges is Rs\",round(T)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual electical charges is Rs 12140.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.11, Page Number: 114"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#(i)for a lighting load:\n",
-      "P1 = 500                        #kW\n",
-      "\n",
-      "\n",
-      "#(ii) for a load:\n",
-      "P2 = 400                         #kW\n",
-      "pf2 = 0.707                      #power factor lagging\n",
-      "\n",
-      "\n",
-      "#(iii)for a load:\n",
-      "P3 = 800                        #kW\n",
-      "pf3 = 0.8                       #power factor leading\n",
-      "\n",
-      "#(iv)for a load:\n",
-      "P4 = 500                        #kW\n",
-      "pf4 = 0.6                       #power factor lagging\n",
-      "\n",
-      "\n",
-      "#(v)a synchronous motor driving a d.c. generator:\n",
-      "P5 = 540                        #power rated at generator(kW)\n",
-      "n = 0.9                         #overall efficiency\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Total lagging kVAR taken by loads (ii) and (iv)\n",
-      "kVAR24 = P2*math.tan(math.acos(pf2))+P4*math.tan(math.acos(pf4))\n",
-      "\n",
-      "#Leading kVAR taken by the load (iii)\n",
-      "kVAR3 = P3*math.tan(math.acos(pf3))\n",
-      "\n",
-      "#Leading kVAR to be taken by synchronous motor\n",
-      "kVAR5 = kVAR24-kVAR3\n",
-      "Pi = P5/n                       #motor input(kW)\n",
-      "phy = math.atan(kVAR5/Pi)       #phase angle of synchronous motor\n",
-      "pf = math.cos(phy)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The power factor of synchronous motor is\",round(pf,2,),\"leading\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The power factor of synchronous motor is 0.79 leading\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.12, Page Number: 115"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P1 = 100                        #power of synchronous motor(hp)\n",
-      "P2 = 200                        #power aggregated by induction motor(hp)\n",
-      "pf2 = 0.707                     #lagging power factor for induction motor\n",
-      "n2 = 0.82                       #efficiency of induction motor\n",
-      "P3 = 30                         #lighting load(kW)\n",
-      "c1 = 100                        #1st part tariff(Rs per kVA per annum)\n",
-      "c2 = 0.06                       #2nd part tariff Rs per kWh\n",
-      "pfa = 0.8                       #power factor factor initially(lagging)\n",
-      "na = 0.93                       #efficiency of synchronous motor initially\n",
-      "pfb = 0.8                       #power factor factor changed later(leading)\n",
-      "nb = 0.93                       #efficiency of synchronous motor changed later\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(a) When synchronous motor runs at p.f. 0\u00b78 lagging.\n",
-      "kW1 = round(P1*735.5/(na*1000))        #input to synchronous motor(kW)\n",
-      "kVAR1 = round(kW1*math.tan(math.acos(pfa)),2)    #Lagging kVAR taken by the synchronous motor(kVAR)\n",
-      "kW2 = round(P2*735.5/(n2*1000),1)        #input to induction motor(kW)\n",
-      "kVAR2 = kW2*math.tan(math.acos(pf2))    #Lagging kVAR taken by the induction motor(kVAR) \n",
-      "#Since lighting load works at unity p.f., its lagging kVAR is zero.\n",
-      "kVARt = kVAR1+kVAR2                 #Total lagging kVAR\n",
-      "kWt = kW1+kW2+P3                  #Total active power(kW)\n",
-      "kVAt = round(math.sqrt(kWt**2+kVARt**2),1)          #total kVA\n",
-      "C1 = kVAt*c1                       #annual kVA demand charges\n",
-      "E1 = kWt*8760                     #Energy consumed/year(kWh)\n",
-      "C2 = round(0.06*E1)                      #annual energy charges(Rs)\n",
-      "T1 = C1+C2                        #total Annual bill(Rs)\n",
-      "\n",
-      "\n",
-      "#(b) When synchronous motor runs at p.f. 0\u00b78 leading.\n",
-      "kVARn = kVAR2-kVAR1            #Net lagging kVAR\n",
-      "kWtt = kWt                      #total active power(kW)\n",
-      "kVAtt = math.sqrt(kVARn**2+kWtt**2)     #total kVA\n",
-      "C11 = c1*kVAtt                #annual kVA charges(Rs)\n",
-      "C22 = C2                        #Annual energy charges(Rs)\n",
-      "T2 = C11+C22                   #total annual bill(Rs)\n",
-      "Ts = T1-T2                    #annual saving(Rs)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The annual saving is  Rs\",round(Ts/100)*100"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual saving is  Rs 6200.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.13, Page Number: 118"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "pf1 = 0.75                        #Power factor of the factory(lagging)\n",
-      "x = 72                            #Max. demand charges(Rs per kVA per annum)\n",
-      "d = 10                           #interest & depreciation of capital investment(%)\n",
-      "M = 175                         #max. demand(kW)\n",
-      "c2 = 120                          #cost of phase advancing equipment(Rs/kVAR)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "y = c2*d/100                    #Expenditure on phase advancing equipment(Rs/kVAR/annum)\n",
-      "pf2 = math.sqrt(1-(y/x)**2)           #power factor\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Most economical p.f. at which factory should operate is\",round(pf2,3),\"lagging\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Most economical p.f. at which factory should operate is 0.986 lagging\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.14, Page Number: 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "L = 400                         #avg. demand(kW)\n",
-      "pf1 = 0.8                        #original power factor(lagging)\n",
-      "LF = 0.5                       #annual load factor\n",
-      "c1 = 50                        #first part tariff(Rs/kVA per annum)\n",
-      "c2 = 0.05                      #second part tariff(Rs/kWh)\n",
-      "pf2 = 0.95                     #improved power factor\n",
-      "c3 = 100                       #phase advancement equipment cost(Rs/kVAR)\n",
-      "d = 10                        #annual interest and depreciation(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "M = L/LF                     #maximum demand(kW)\n",
-      "#Leading kVAR taken by phase advancement equipment:\n",
-      "kVAR1 = M*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
-      "y = d*c3/100               #Expenditure on phase advancing equipment(Rs/kVAR/annum)\n",
-      "kVA1 = M/pf1                #Max. kVA demand at pf1 power fctor\n",
-      "kVA2 = round(M/pf2)                #Max. kVA demand at pf2 power fctor\n",
-      "Cs = c1*(kVA1-kVA2)           #Annual saving in maximum demand charges\n",
-      "Ce = y*kVAR1                #Annual expenditure on phase advancing equipment\n",
-      "NS = Cs-Ce                  #net annual saving\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The capacity of the phase advancing equipment is\",round(kVAR1),\"kVAR\"\n",
-      "print \"(ii)The annual saving is  Rs\",round(NS)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The capacity of the phase advancing equipment is 337.0 kVAR\n",
-        "(ii)The annual saving is  Rs 4529.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.15, Page Number: 119"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "L = 50                         #avg. demand(kW)\n",
-      "pf1 = 0.75                        #original power factor(lagging)\n",
-      "LF = 0.5                       #annual load factor\n",
-      "c1 = 100                        #first part tariff(Rs/kVA of max demand per annum)\n",
-      "c2 = 0.05                      #second part tariff(Rs/kWh)\n",
-      "c3 = 600                       #cost of loss free capacitors(Rs/kVAR)\n",
-      "d = 10                           #depreciation & interest(%)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "y = c3*d/100                #Expenditure on capacitors(Rs/kVAR/annum)\n",
-      "pfe = math.sqrt(1-(y/c1)**2)      #Most economical power factor\n",
-      "M = L/LF                     #Max kW demand\n",
-      "kVAm1 = M/pf1                #The maximum kVA demand at pf1 power factor\n",
-      "kVAm2 = M/pfe                #The maximum kVA demand at pfe power factor\n",
-      "Cs = c1*(kVAm1-kVAm2)         #annual saving(Rs)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Most economical power factor is\",round(pfe,3),\"lagging\"\n",
-      "print \"Annual saving is Rs\",round(Cs)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Most economical power factor is 0.8 lagging\n",
-        "Annual saving is Rs 833.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.16, Page Number: 120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "L = 200                         #avg. demand(kW)\n",
-      "pf1 = 0.8                        #original power factor(lagging)\n",
-      "LF = 0.5                       #annual load factor\n",
-      "c1 = 100                        #first part tariff(Rs/kVA of max demand per annum)\n",
-      "c2 = 0.05                      #second part tariff(Rs/kWh)\n",
-      "c3 = 500                       #cost of phase advancing plant(Rs/kVAR)\n",
-      "d = 10                           #depreciation & interest(%)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "y = c3*d/100                #Expenditure on capacitors(Rs/kVAR/annum)\n",
-      "pfe = math.sqrt(1-(y/c1)**2)      #Most economical power factor\n",
-      "Pc = L*(math.tan(math.acos(pf1))-math.tan(math.acos(pfe))) #Capacity of phase advancing plant\n",
-      "E = L*5000                    #units consumed per year(kWh)\n",
-      "C2 = c2*E                     #annual energy charges(Rs)\n",
-      "C1 = y*Pc                    #Annual cost of phase advancing plant(Rs)\n",
-      "Cm = c1*L/pfe                #Max. demand charge(Rs)\n",
-      "C = C2+C1+Cm                 #annual charge(Rs)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  The value to which the power factor be improved is\",round(pfe,3),\"lagging\"\n",
-      "print \"(ii) The capacity of the phase advancing plant is\",round(Pc,2),\"kVAR\"\n",
-      "print \"(iii)The new bill for energy is  Rs\",round(C,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  The value to which the power factor be improved is 0.866 lagging\n",
-        "(ii) The capacity of the phase advancing plant is 34.53 kVAR\n",
-        "(iii)The new bill for energy is  Rs 74820.5\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.17, Page Number: 120"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "E = 80000               #units consumed per year\n",
-      "kVAm1 = 500                #maximum kVA demand\n",
-      "pf1 = 0.707             #initial power factor(lagging)\n",
-      "c1 = 120                        #first part tariff(Rs/kVA of max demand per annum)\n",
-      "c2 = 0.025                      #second part tariff(Rs/kWh)\n",
-      "c3 = 50                #cost of phase advancing plant(Rs/kVAR)\n",
-      "pf2 = 0.9              #increased powe factor(Rs)\n",
-      "d = 10                  #depreciation on advancing plant(%)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "C1 = c1*kVAm1+c2*E         #annual cost of supply(Rs)\n",
-      "P = kVAm1*pf1              #max. kW demand\n",
-      "#Leading kVAR taken by phase advancing equipment:\n",
-      "kVARe = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
-      "C2 = round(kVARe*c3*d/100)        #Annual cost of phase advancing equipment(Rs)\n",
-      "#When p.f. is raised from 0\u00b7707 lag to 0\u00b79 lag\n",
-      "kVAm2 = P/pf2               #new maximum kVA demand\n",
-      "kVAr = kVAm1-kVAm2          #Reduction in kVA demand\n",
-      "Cs = round(c1*kVAr)                #annual saving(Rs)\n",
-      "#As the units consumed remain the same, therefore, saving will\n",
-      "#be equal to saving in M.D. charges minus annual cost of \n",
-      "#phase advancing plant.\n",
-      "Cs2 = Cs-C2                #annual charges(Rs)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The annual cost of supply is   Rs\",C1\n",
-      "print \"Annual saving is  Rs\",Cs2"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The annual cost of supply is   Rs 62000.0\n",
-        "Annual saving is  Rs 11955.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.18, Page Number: 123"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "pf = 0.7           #power factor when plant working at its max. capacity(lagging)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#For calculating the cost of increasing plant capacity. \n",
-      "#We have, referring to Fig. 6.15, the increase in kVA capacity is BD.\n",
-      "\n",
-      "#Now  OE*cos phy2 = OD*cos phy1\n",
-      "#or   OB*cos phy2 = OD*cos phy1        (Since OE = OB)\n",
-      "#or            OD = OB * cos phy2/cos phy1 \n",
-      "#                 = OB * 0\u00b785/0\u00b77\n",
-      "#                 = 1\u00b72143 OB\n",
-      "\n",
-      "#Increase in the kVA capacity of the plant is\n",
-      "#BD = OD \u2212 OB = 1\u00b72143 \u00d7 OB \u2212 OB = 0\u00b72143 OB\n",
-      "#Total cost of increasing the plant capacity\n",
-      "#                = Rs 800 \u00d7 0\u00b72143 \u00d7 OB\n",
-      "#                = Rs 171\u00b744 \u00d7 OB                ...(i)\n",
-      "\n",
-      "cos_phy1 = 0.7;         sin_phy1 = 0.714\n",
-      "cos_phy1 = 0.85;         sin_phy1 = 0.527\n",
-      "\n",
-      "#Leading kVAR taken by p.f. correction equipment is\n",
-      "#  ED = CD \u2212 CE = OD*sin phy1 \u2212 OE*sin phy2\n",
-      "#     = 1\u00b72143 * OB*sin phy1 \u2212 OB*sin phy2\n",
-      "#     = OB*(1\u00b72143 * 0\u00b7714 \u2212 0\u00b7527) = 0\u00b734 * OB\n",
-      "#Suppose the cost per kVAR of the equipment be Rs y.\n",
-      "#So, total cost of p.f. correction equipment\n",
-      "#                     = Rs 0\u00b734 \u00d7 OB \u00d7 y         ...(ii)\n",
-      "#The cost per kVAR of the equipment that would justify its \n",
-      "#installation is when exp. (i) = exp. (ii)\n",
-      "#i.e.,\n",
-      "#                     171\u00b744 \u00d7 OB = 0\u00b734 \u00d7 OB \u00d7 y\n",
-      "y = 171.44/0.34                     #Rs/kVAR\n",
-      "\n",
-      "#If the losses in p.f. correction equipment are neglected, \n",
-      "#then its kVAR = kVA.\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The maximum cost per kVA of p.f. correction equipment\"\n",
-      "print \"that can be paid is  Rs\",round(y,1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum cost per kVA of p.f. correction equipment\n",
-        "that can be paid is  Rs 504.2\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 6.19, Page Number: 124"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "pf = 0.7               #power factor when plant working at its max. capacity(lagging)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Cost of increasing plant capacity\n",
-      "#      BD = OD \u2212 OB\n",
-      "#        = OB * 0.866/0.70-OB\n",
-      "#        = OB*(1\u00b7237 \u2212 1)\n",
-      "#        = 0\u00b7237 * OB\n",
-      "#\u2234 Annual cost of increasing the plant capacity\n",
-      "#        = Rs 10 * 0\u00b7237 * OB\n",
-      "#        = Rs. 2.37 * OB                             ...(i)\n",
-      "\n",
-      "#Cost of phase advancing equipment. Leading kVAR taken by\n",
-      "#phase advancing equipment,\n",
-      "#           ED = CD \u2212 CE\n",
-      "#              = OD*sin phy1 \u2212 OE*sin phy2\n",
-      "#              = 1.237 * OB * sin phy1 \u2212 OB*sin phy2\n",
-      "#              = OB*(1\u00b7237 * 0\u00b7174 \u2212 0\u00b75) = 0\u00b7383 \u00d7 OB\n",
-      "#Let the cost per kVAR of the equipment be Rs y.\n",
-      "#Annual cost of phase advancing equipment\n",
-      "#                  = Rs 0.1 * y * 0\u00b7383 * OB            ...(ii)\n",
-      "#For economical use, the two costs should be equal i.e.,\n",
-      "#                    exp. (i) = exp. (ii).\n",
-      "#or      0\u00b71 * y * 0\u00b7383 * OB = 2\u00b737 * OB\n",
-      "y = 2.37/(0.1*0.383)                 #Rs\n",
-      "\n",
-      "\n",
-      "#If the losses in the phase advancing equipment are neglected,\n",
-      "#then its kVAR = kVA.\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The maximum cost per kVA of phase advancing equipment\"\n",
-      "print \"that can be paid is  Rs\",round(y,2)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The maximum cost per kVA of phase advancing equipment\n",
-        "that can be paid is  Rs 61.88\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter7_1.ipynb b/Principles_of_Power_System/chapter7_1.ipynb
deleted file mode 100644
index bcb1dc3f..00000000
--- a/Principles_of_Power_System/chapter7_1.ipynb
+++ /dev/null
@@ -1,559 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:855ede77f42d7d38570b25f9dfe540c158b87984c52d22c6171f4226086950d9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 7: Supply Systems"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.1, Page Number: 145"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V1 = 200                            #initial volt of system(V)\n",
-      "V2 = 400                            #final volt of the system(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "I1,I2,R1,R2 = symbols('I1 I2 R1 R2')\n",
-      "I2 = V1*I1/V2\n",
-      "W1 = 2*I1**2*2*R1                        #Power loss in 200 V system\n",
-      "W2 = 2*I2**2*2*R2                        #Power loss in 400 V system\n",
-      "# As power loss in the two cases is the same,\n",
-      "R22 = solve(W1/W2-1,R2)[0]\n",
-      "a = R22/R1                               #ratio of cables' cross-section\n",
-      "v = 1/a                                  #ratio of cables' volumes\n",
-      "s = (1-v)*100\n",
-      "\n",
-      "#Result:\n",
-      "print \"% Saving in feeder copper is \",s,\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "% Saving in feeder copper is  75 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.2, Page Number: 146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#P = power upplied,      W = power loss, \n",
-      "#I = current,            V = voltage,     R = resistance\n",
-      "\n",
-      "R,V = symbols('R V')\n",
-      "P1,I1,W1 = symbols('P1,I1,W1')    #for 2-wire system\n",
-      "P2,I2,W2 = symbols('P2,I2,W2')     #for 3-wire system\n",
-      "\n",
-      "#Calculation:\n",
-      "#Two-wire d.c. system:\n",
-      "P1 = V*I1\n",
-      "W1 = 2*I1**2*R\n",
-      "L1 = W1/P1*100                    #% power loss\n",
-      "\n",
-      "#3-phase, 3-wire a.c. system:\n",
-      "P2 = math.sqrt(3)*V*I2\n",
-      "W2 = 3*I2**2*R\n",
-      "L2 = W2/P2*100                     #% power loss\n",
-      "\n",
-      "I2 = solve(L1-L2,I2)[0]\n",
-      "#r2 = P2/P1\n",
-      "r2 = math.sqrt(3)*I2*V/(I1*V)\n",
-      "P1 = 2*P2\n",
-      "\n",
-      "#Result:\n",
-      "print '''Additional power which can be supplied at \n",
-      "unity p.f. by 3-phase, 3-wire a.c. system = 100%.'''"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Additional power which can be supplied at \n",
-        "unity p.f. by 3-phase, 3-wire a.c. system = 100%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.3, Page Number: 146"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "#Let R be the resistance per conductor in each case\n",
-      "\n",
-      "R,V = symbols('R V')\n",
-      "P1,I1,W1 = symbols('P1,I1,W1')    #for 3-wire dc system\n",
-      "P2,I2,W2 = symbols('P2,I2,W2')     #for 4-wire ac system\n",
-      "\n",
-      "#Calculation:\n",
-      "\n",
-      "#3-wire d.c. system.:\n",
-      "#At balanced loads, implying no power loss in the neutral wire\n",
-      "P1 = 2*V*I1\n",
-      "W1 = 2*I1**2*R\n",
-      "L1 = W1/P1*100                    #% power loss\n",
-      "\n",
-      "\n",
-      "#3-phase, 4-wire a.c. system.:\n",
-      "#At balanced loads, implying no power loss in the neutral wire\n",
-      "P2 = 3*V*I2\n",
-      "W2 = 3*I2**2*R\n",
-      "L2 = W2/P2*100                     #% power loss\n",
-      "\n",
-      "I2 = solve(L1-L2,I2)[0]\n",
-      "r = P2/P1\n",
-      "#solving r, we get\n",
-      "P2 = 1.5*P1\n",
-      "\n",
-      "#Result:\n",
-      "print '''Extra power that can be supplied at unity power \n",
-      "factor by 3-phase, 4-wire a.c. system = 50%.'''"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Extra power that can be supplied at unity power \n",
-        "factor by 3-phase, 4-wire a.c. system = 50%.\n"
-       ]
-      }
-     ],
-     "prompt_number": 19
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.4, Page Number: 148"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "R,V,phy = symbols('R V phy')       #phy is power factor angle\n",
-      "                                   #V is voltage between the conductors\n",
-      "                                   #R is resistance per conductor \n",
-      "P1,I1,W1 = symbols('P1,I1,W1')    #for Single phase 2-wire system\n",
-      "P2,I2,W2 = symbols('P2,I2,W2')     #for 3-phase, 3-wire a.c. system\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#Single phase 2-wire system\n",
-      "P1 = 2*V*I1*math.cos(phy)\n",
-      "W1 = 2*I1**2*R\n",
-      "L1 = W1/P1*100                    #% power loss\n",
-      "\n",
-      "#3-phase, 3-wire a.c. system. \n",
-      "P2 = math.sqrt(3)*V*I2*math.cos(phy)\n",
-      "W2 = 3*I2**2*R\n",
-      "L2 = W2/P2*100                     #% power loss\n",
-      "\n",
-      "I22 = solve(L1-L2,I2)[0]\n",
-      "r = P1/(math.sqrt(3)*V*I22*math.cos(phy))\n",
-      "#As the power supplied by single phase, 2-wire is 200 kW,\n",
-      "P1 = 200\n",
-      "\n",
-      "#Result:\n",
-      "print \"Power supplied by 3-phase, 3-wire a.c. system is\",P1*r,\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Power supplied by 3-phase, 3-wire a.c. system is 400 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 38
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.5, Page Number: 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "MVA = 5                   #power transmitted to load\n",
-      "n = 0.9                   #efficiency\n",
-      "pf = 0.8                  #power factor\n",
-      "V = 33                    #receiving end voltage(V)\n",
-      "l = 50*10**3             #length of line(m)\n",
-      "s = 2.85*10**-8           #specific resistance of aluminium(ohm-m)\n",
-      "\n",
-      "#Calculation:\n",
-      "P = MVA*pf*10**6                #power transmitted(W)\n",
-      "W = 10*P/100                          #line loss(W)\n",
-      "\n",
-      "\n",
-      "#(i) Single phase, 2-wire system:\n",
-      "I1 = MVA*10**3/V                        #Apparent power(A)\n",
-      "a1 = 2*I1**2*s*l/W                      #area of cross-section(m**2)\n",
-      "v1= 2*a1*l                              #volume of conductor required(m**3)\n",
-      "\n",
-      "#(ii) 3-phase, 3-wire system:\n",
-      "I2 = MVA*10**3/(math.sqrt(3)*V)                        #Apparent power(A)\n",
-      "a2 = 3*I2**2*s*l/W                      #area of cross-section(m**2)\n",
-      "v2 = 3*a2*l                              #volume of conductor required(m**3)\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) Single phase, 2-wire system, Vol. of conductor required is\",round(v1,2),\"m^3\"\n",
-      "print \"(ii)3-phase, 3-wire system, Vol. of conductor required\",round(v2,2),\"m^3\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) Single phase, 2-wire system, Vol. of conductor required is 16.36 m^3\n",
-        "(ii)3-phase, 3-wire system, Vol. of conductor required 12.27 m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.6, Page Number: 149"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V1 = 11*10**3                   #Supply voltage(V)\n",
-      "pf = 0.8\n",
-      "R1 = 0.15                      #Total resistance in ac system(ohm)\n",
-      "R2 = 0.05                      #Total resistance in dc system(ohm)\n",
-      "r1 = 0.15                     #voltage drop in ac system(%)\n",
-      "r2 = 0.25                     #voltage drop in dc system(%)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Single phase system:\n",
-      "I1 = r1*V1/R1                   #line current(A)\n",
-      "P1 = V1*I1*pf/1000              #power received by consumer(kW)\n",
-      "\n",
-      "#D.C. two-wire system:\n",
-      "P2 = P1\n",
-      "V2 = (P2*10**3*R2/r2)**0.5             #supply voltage(V)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The supply voltage is\",V2,\"V\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The supply voltage is 4400.0 V\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.7, Page Number: 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I = 200                      #current(A)\n",
-      "l = 1                      #length of cable(km)\n",
-      "c1 = 5                         #cost per unit(paise/kWh)\n",
-      "r1 = 10                        #interrest & depreciation(%)\n",
-      "ro = 1.73*10**-6              #resistivity(ohm-m)\n",
-      "\n",
-      "#Calculation:\n",
-      "#If 'a' is the area os cross-section of the cnductor,\n",
-      "#cost of cable including installation = (20*a + 20)    Rs./m.\n",
-      "\n",
-      "a = symbols('a')\n",
-      "R = ro*l*10**5/a\n",
-      "E = 2*I**2*R*8760/1000           #energy lost per annum\n",
-      "C = c1/100*E                  #cost of energy lost(Rs)\n",
-      "\n",
-      "#capital cost for 1 km length of the cable is = Rs. 20,000*a.\n",
-      "\n",
-      "V = r1/100*20000*a             #Variable annual charge(Rs)\n",
-      "a1 = solve(C-V,a)[1]\n",
-      "\n",
-      "#Result:\n",
-      "print \"The most economical conductor size is\",round(a1,2),\"cm**2\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The most economical conductor size is 1.74 cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 22
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.8, Page Number: 153"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P = 5                           #line load(MW)]\n",
-      "V = 33                          #kV\n",
-      "pf = 0.8                        #power factor\n",
-      "c1 = 4                          #paise/kWh\n",
-      "r1 = 10                         #interest and depreciation(%)\n",
-      "ro = 10**-6                     #ohm-cm\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "a = symbols('a')                #size of conductor(cm**2)\n",
-      "R = ro*l*10**5/a                #ohm\n",
-      "I = P*10**3/(math.sqrt(3)*V*pf)            #current (A)\n",
-      "E = 3*I**2*R*8760/1000           #kWh\n",
-      "C = c1/100*E                        #Annual cost of energy lost(R)\n",
-      "#The capital cost (variable) of the cable = 25000*a Rs/km\n",
-      "V = r1/100*25000*a                 #Rs\n",
-      "a1 = solve(C-V,a)[1]\n",
-      "\n",
-      "#Result:\n",
-      "print \"The most economical size of conductor is\",round(a1,2),\"cm**2\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The most economical size of conductor is 0.71 cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.9, Page Number: 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "I = 250                         #current through the feeder(A)\n",
-      "c1 = 5                          #Rs/kg\n",
-      "r1 = 10                         #interest and depreciation(%)\n",
-      "c2 = 5                          #paise/kWh\n",
-      "d = 8.93                        #density of wire(gm/cm**3)\n",
-      "ro = 1.73*10**-8                #speciafic resistance(ohm-m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "l = 1                           #length of wire(1 m say)\n",
-      "a = symbols('a')                #conductor size(m**2)\n",
-      "R = ro*l/a                      #ohm\n",
-      "E = 2*I**2*R*8760/1000          #kWh\n",
-      "C = c2/100*E                   #annual cost os energy lost(Rs)\n",
-      "m = 2*a*1*d*10**3                     #Mass of 1 metre feeder\n",
-      "CC = c1*m                    #capital cost(Rs)\n",
-      "V = r1*CC/100                    #Variable Annual charge(Rs)\n",
-      "a1 = solve(C-V)[1]\n",
-      "\n",
-      "#Result:\n",
-      "print \"The most economic size of conductor is\",round(a1*10000,2),\"cm**2\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The most economic size of conductor is 3.26 cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 7.10, Page Number: 154"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 1                           #line length(km)\n",
-      "V = 110                         #supply voltage(kV)\n",
-      "\n",
-      "#Working hours(hrs)          Load(MW)       Power factor\n",
-      "t1 = 6;               P1 = 20 ;            pf1 = 0.8\n",
-      "t2 = 12;              P2 = 5 ;             pf2 = 0.8\n",
-      "t3 = 6;                P3 = 6 ;             pf3 = 0.8\n",
-      "\n",
-      "c1 = 6                  #\n",
-      "n = 365                 #no.of days used\n",
-      "r1 = 10                 #interest and dep-reciation(%)\n",
-      "\n",
-      "#Calculation:\n",
-      "a = symbols(\"a\")                #cross-section of conductor(cm**2)\n",
-      "R = 0.176/a                    #Resistance per km of each conductor(ohm)\n",
-      "#The load currents at various loads are :\n",
-      "#At 20 MW,\n",
-      "I1 = P1*1000/(math.sqrt(3)*V*pf1)                #A\n",
-      "I2 = P2*1000/(math.sqrt(3)*V*pf2)                #A\n",
-      "I3 = P3*1000/(math.sqrt(3)*V*pf3)                #A\n",
-      "\n",
-      "E1= 3*R*1/1000*(I1**2*t1+I2**2*t2+I3**2*t3)      #Energy loss per day in 3-phase line(kWh)\n",
-      "E = E1*n                        #energy lost per annum(kWh)\n",
-      "C = c1*E/100                   #Annual cost of energy(Rs)\n",
-      "v = r1*6000*a/100                  #variable annual charge(Rs/kWh)\n",
-      "a1 = solve(C-v,a)[1]\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"THe most economical size of the conductor is\",round(a1,2),\"cm**2\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "THe most economical size of the conductor is 1.56 cm**2\n"
-       ]
-      }
-     ],
-     "prompt_number": 24
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter8_1.ipynb b/Principles_of_Power_System/chapter8_1.ipynb
deleted file mode 100644
index 6b383670..00000000
--- a/Principles_of_Power_System/chapter8_1.ipynb
+++ /dev/null
@@ -1,1455 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:9774a5727e94c3e9d7f6de1d46ee1710f0f45bee5b4624a7cce654515da8e820"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8: Mechanical Design of Overhead Lines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.1, Page Number: 171"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "k = 0.11                        #ratio of shunt-capatance to self capacitance\n",
-      "V = 33/math.sqrt(3)                  #Voltage across string(kV)\n",
-      "n = 3                           #no. of insulators\n",
-      "\n",
-      "#Calculation:\n",
-      "\n",
-      "#  At Junction A\n",
-      "#        I2 = I1 + i1\n",
-      "# or V2*w*C = V1*w*C + V1*K*\u03c9*C\n",
-      "# or     V2 = V1*(1 + K) = V1*(1 + 0\u00b711)\n",
-      "# or     V2 = 1\u00b711*V1                     ...(i)\n",
-      "\n",
-      "#  At Junction B\n",
-      "#          I3 = I2 + i2\n",
-      "#  or  V3*w*C = V2*w*C + (V1 + V2) K*w*C\n",
-      "#  or      V3 = V2 + (V1 + V2)*K\n",
-      "#             = 1\u00b711*V1 + (V1 + 1\u00b711*V1)*0\u00b711\n",
-      "#          V3 = 1\u00b7342 V1\n",
-      "#  Voltage across whole unit,\n",
-      "#    V = V1+V2+V3\n",
-      "# or V = V1+1.11*V1+1.342*V1\n",
-      "\n",
-      "V1 = V/3.452                    #Voltage across top unit(V)\n",
-      "V2 = 1.11*V1                    #Voltage across top unit(V)\n",
-      "V3 = 1.342*V1                   #Voltage across top unit(V)\n",
-      "\n",
-      "e = V/(n*V3)*100               #string efficiency\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)\\tVoltage across top unit,V1 is\",round(V1,2),\"kV\"\n",
-      "print \"   \\tVoltage across middle unit, V2 is\",round(V2,2),\"kV\"\n",
-      "print \"   \\tVoltage across bottom unit, V3 is\",round(V3,1),\"kV\"\n",
-      "print \"(ii)\\tThe string efficiency is \",round(e,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)\tVoltage across top unit,V1 is 5.52 kV\n",
-        "   \tVoltage across middle unit, V2 is 6.13 kV\n",
-        "   \tVoltage across bottom unit, V3 is 7.4 kV\n",
-        "(ii)\tThe string efficiency is  85.7 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.2, Page Number: 172"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V1 = 8                        #voltage across top unit(kV)\n",
-      "V2 = 11                       #voltage across middle unit(kV)\n",
-      "n = 3                         #no. of insulators\n",
-      "#Calculation:\n",
-      "\n",
-      "#(i)\n",
-      "# Let K be the ratio of capacitance between pin and the earth\n",
-      "# to self capacitance. \n",
-      "# Let C farad be the self capacitance of each unit, \n",
-      "# then, capacitance between pin and earth = K*C.\n",
-      "# Applying Kirchoff\u2019s current law to Junction A,\n",
-      "#        I2 = I1 + i1\n",
-      "# or V2*w*C = V1*w*C + V1*K*w*C\n",
-      "# or     V2 = V1 (1 + K)\n",
-      "\n",
-      "k = (V2-V1)/V1\n",
-      "\n",
-      "#(ii)\n",
-      "# Applying Kirchoff\u2019s current law to Junction B,\n",
-      "#          I3 = I2 + i2\n",
-      "#  or  V3*w*C = V2*w*C + (V1 + V2)*K*w*C\n",
-      "#  or      V3 = V2 + (V1 + V2)*K = 11 + (8 + 11) * 0\u00b7375\n",
-      "V3 = 18.12                        #kV\n",
-      "V = V1 + V2+ V3                   #kV\n",
-      "Vl = math.sqrt(3)*V                    #line voltage(kV)\n",
-      "\n",
-      "#(iii)\n",
-      "e = V/(n*V3)*100                     #stirng efficiency(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)  Ratio of capacitance b/w pin & earth is\",k\n",
-      "print \"(ii) The line voltage is\",round(Vl,2),\"kV\"\n",
-      "print \"(iii)String efficiency is\",round(e,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)  Ratio of capacitance b/w pin & earth is 0.375\n",
-        "(ii) The line voltage is 64.29 kV\n",
-        "(iii)String efficiency is 68.29 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.3, Page Number: 172"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n = 3                            #no. of insulators\n",
-      "V3 = 17.5                        #voltage across the line unit(V)\n",
-      "\n",
-      "#  At Junction A\n",
-      "#   I2 = I1 + i1\n",
-      "#   V2 \u03c9 C = V1 \u03c9 C + V1 K \u03c9 C\n",
-      "# or V2 = V1 (1 + K) = V1 (1 + 0.125)\n",
-      "#  \u2234 V2 = 1\u00b7125 V1\n",
-      "\n",
-      "#  At Junction B\n",
-      "#    I3 = I2 + i2\n",
-      "# or  V3 \u03c9 C = V2 \u03c9 C + (V1 + V2) K \u03c9 C\n",
-      "# or      V3 = V2 + (V1 + V2) K\n",
-      "#            = 1\u00b7125 V1 + (V1 + 1\u00b7125 V1) \u00d7 0.125\n",
-      "#       \u2234 V3 = 1\u00b739 V1\n",
-      "\n",
-      "V1 = V3/1.39                     #Voltage across top unit(kV)\n",
-      "V2 = 1.125*V1                    #Voltage across middle unit(kV)\n",
-      "\n",
-      "V = V1+V2+V3                     #voltage across line unit(kV)\n",
-      "e = V/(n*V3)*100                     #string efficiency(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Line to neutral voltage is\",round(V,2),\"kV\"\n",
-      "print \"String efficiency is\",round(e,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Line to neutral voltage is 44.25 kV\n",
-        "String efficiency is 84.29 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 36
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.4, Page Number: 173"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V3 = 13.1                      #voltage across the lowest insulator(kV)\n",
-      "V2 = 11                        #voltage across the middle insulator(kV)\n",
-      "K = symbols('K')\n",
-      "#  Applying Kirchhoff\u2019s current law to Junctions A and B, we can easily\n",
-      "#  derive the following equations:\n",
-      "\n",
-      "#     V2 = V1 (1 + K)\n",
-      "#  or V1 =  V2/(1 + K)                   ...(i)\n",
-      "# and V3 = V2 + (V1 + V2)*K           ...(ii)\n",
-      "#  Putting the value of V1 = V2/(1 + K) in eq. (ii), we get,\n",
-      "\n",
-      "#V3 = V2 + (V2/(1+k)+ V2)*K\n",
-      "K1 = solve(V3-(V2 + (V2/(1+K)+ V2)*K),K)[1]\n",
-      "\n",
-      "V1 = V2/(1+K1)\n",
-      "V = V1+V2+V3                  #kV\n",
-      "\n",
-      "#Result:\n",
-      "print \"The voltage b/w the bus bars is\",round(float(math.sqrt(3)*V)),\"kV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage b/w the bus bars is 59.0 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.5, Page Number: 173"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n = 3                         #no.of insulators\n",
-      "k = 1/8                 #ratio of self-capacitance to shunt capacitance\n",
-      "V3 = 15                    #safe working voltage of each insulator(kV)\n",
-      "\n",
-      "#  Applying Kirchhoff\u2019s current law to junction A, we get,\n",
-      "#     V2 = V1*(1 + K)\n",
-      "#  or V1 = V2/(1 + K) = V2/(1 + 0\u00b7125) = 0\u00b789*V2\n",
-      "\n",
-      "#  Applying Kirchhoff\u2019s current law to Junction B, we get,\n",
-      "#  V3 = V2 + (V1 + V2)*K = V2 + (0.89*V2 + V2) * 0.125\n",
-      "#  V3 = 1.236*V2                                       \n",
-      "V2 = V3/1.236                     #kV\n",
-      "V1 = 0.89*V2                       #kV\n",
-      "V = V1+V2+V3                      #voltage across the string(kV)\n",
-      "e = V/(n*V3) *100                     #string efficiency(%)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage across the string is\",round(V,3),\"kV\"\n",
-      "print \"The string efficiency is\",round(e,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage across the string is 37.937 kV\n",
-        "The string efficiency is 84.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.6, Page Number: 174"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n = 4                        #no. of unit\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "# Suppose Xc = 1 \u03a9. Given, the ratio of self-capacitance to shunt\n",
-      "# capacitance is 10, so, Xc1 = 10 \u03a9 .\n",
-      "# Suppose that potential V across the string is such that 1 A current \n",
-      "# flows in the top insulator.\n",
-      "#Thus,\n",
-      "V1 = 1*1                          #Voltage across top unit(V)\n",
-      "V2 = 1*1.1                        #Voltage across 2nd unit(V)\n",
-      "V3 = 1*1.31                       #Voltage across 3rd unit(V)\n",
-      "V4 = 1*1.65                       #Voltage across 4th unit(V)\n",
-      "V = V1+V2+V3+V4                   #Voltage obtained across the string(V)\n",
-      "\n",
-      "#The voltage across each unit expressed as a % of V becomes:\n",
-      "f1 = V1/V*100\n",
-      "f2 = V2/V*100\n",
-      "f3 = V3/V*100\n",
-      "f4 = V4/V*100\n",
-      "\n",
-      "e = V/(n*V4)*100                    #string efficiency(%)\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The voltage distributation from top unit is \"\n",
-      "print \"\\t\",round(f1,2),\"%\\t\",round(f2,2),\"%\\t\",round(f3,1),\"%\\t\",round(f4,1),\"%\"\n",
-      "print \"\\n(ii)String efficiency is\",round(e,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage distributation from top unit is \n",
-        "\t19.76 %\t21.74 %\t25.9 %\t32.6 %\n",
-        "\n",
-        "(ii)String efficiency is 76.7 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.7, Page Number: 175"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n = 5                       #no.os insulators\n",
-      "Vl = 100                    #line voltage(kV)\n",
-      "\n",
-      "#Suppose Xc = 1 \u03a9.\n",
-      "#Ratio of self capacitance to shunt capacitance is 10, so, Xc1 = 10 \u03a9. \n",
-      "#Let V be the voltage s.t 1A current flows in top insulator.\n",
-      "\n",
-      "#Calculation:\n",
-      "#calculation for fraction of voltage drop:\n",
-      "v1 = 1*1                          #Voltage across top unit(V)\n",
-      "v2 = 1*1.1                        #Voltage across 2nd unit(V)\n",
-      "v3 = 1*1.31                       #Voltage across 3rd unit(V)\n",
-      "v4 = 1*1.65                       #Voltage across 4th unit(V)\n",
-      "v5 = 1*2.16                       #voltage across 5th unit(V)\n",
-      "v = v1+v2+v3+v4+v5                #total voltage(V)\n",
-      "V = 100/math.sqrt(3)                   #string voltage(V)\n",
-      "V1 = round(v1/v*V,3)                       #kV\n",
-      "V2 = v2/v*V                       #kV\n",
-      "V3 = v3/v*V                       #kV\n",
-      "V4 = v4/v*V                       #kV\n",
-      "V5 = v5/v*V                       #kV\n",
-      "e = V/(n*V5)*100                      #string efficiency(%)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i)The distribution of voltage on the insulator discs are\"\n",
-      "print \"V1 = \",round(V1,3),\"kV \",\"\\tV2 =\",round(V2,2),\"kV \\n\",\n",
-      "print \"V3 =\",round(V3,1),\"kV\",\"\\t\\tV4 =\",round(V4,2),\"kV\",\n",
-      "print \"\\nV5 =\",round(V5,1),\"kV\"\n",
-      "print \"\\n(ii)The string efficiency\",round(e,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i)The distribution of voltage on the insulator discs are\n",
-        "V1 =  7.997 kV  \tV2 = 8.8 kV \n",
-        "V3 = 10.5 kV \t\tV4 = 13.19 kV \n",
-        "V5 = 17.3 kV\n",
-        "\n",
-        "(ii)The string efficiency 66.9 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.8, Page Number: 176"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "n = 4                       #no.os insulators\n",
-      "V2 = 13.2                   #Voltage across 2th unit\n",
-      "V3 = 18                     #Voltage across 3th unit\n",
-      "#Suppose K is the ratio of shunt-capacitance to self-capacitance\n",
-      "k = symbols('k')\n",
-      "\n",
-      "# Referring to Fig.(ii), we have,\n",
-      "# V2/V1 = ( 1 + K)/1\n",
-      "# V2 = V1 (1 + K)\n",
-      "# V3/V1 = (1 + 3K + K2)/1\n",
-      "# V3 = V1 (1 + 3K + K2)\n",
-      "# putting values of V2 & V3\n",
-      "\n",
-      "k1 = solve(13.2*k**2+21.6*k-4.8,k)[1]\n",
-      "V1 = V2/(1+k1)\n",
-      "V4 = V1*(1 + k1**33 + 5*k1**2 + 6*k1)\n",
-      "V = V1+V2+V3+V4                   #Voltage between line and earth(kV)\n",
-      "Vl = math.sqrt(3)*V                    #Voltage between conductors(kV)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Voltage between conductors is\",round(float(Vl)),\"kV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Voltage between conductors is 119.0 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.9, Page Number: 177"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "k = 5               #ratio of self-capacitance(C) to pin-earth capacitance(C1)\n",
-      "V = 1+1.2+1.64+2.408                  #voltage obtained across the string(V)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) The voltage across each unit expressed as a % of V\n",
-      "v1 = 1/V*100\n",
-      "v2 = 1.2/V*100\n",
-      "v3 = 1.6/V*100\n",
-      "v4 = 2.408/V*100\n",
-      "\n",
-      "#(ii)String efficiency\n",
-      "e = V/(4*2.408)*100\n",
-      "\n",
-      "#Result:\n",
-      "print \"(i) The voltage across each unit expressed as a % of V are:\"\n",
-      "print \"\\tTop Unit: \",round(v1),\"%\"\n",
-      "print \"\\tSecond from top: \",round(v2,1),\"%\"\n",
-      "print \"\\tThird from top: \",round(v3),\"%\"\n",
-      "print \"\\tFourth from top: \",round(v4,1),\"%\"\n",
-      "print \"(ii)String efficiency is\",round(e,2),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "(i) The voltage across each unit expressed as a % of V are:\n",
-        "\tTop Unit:  16.0 %\n",
-        "\tSecond from top:  19.2 %\n",
-        "\tThird from top:  26.0 %\n",
-        "\tFourth from top:  38.5 %\n",
-        "(ii)String efficiency is 64.87 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 54
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.10, Page Number: 177"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "V1,V2,V3,V= symbols('V1 V2 V3 V')          #volts (shown in fig(ii))\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "\n",
-      "#We know that in an actual string of insulators, 3 capacitances\n",
-      "#exist i.e, self-capacitance of each insulator, shunt capacitance \n",
-      "#and capacitance of each unit to line.\n",
-      "#But, capacitance of each unit to line is very small and\n",
-      "#can be neglected.\n",
-      "\n",
-      "#At Junction A\n",
-      "#     I2 + i1 = I1 + i1\n",
-      "#or   V2*w*C + (V2 + V3)*0\u00b71*w*C = V1*\u03c9*C + 0\u00b715*C*V1*w\n",
-      "#or   0\u00b71*V3 = 1\u00b715*V1 \u2212 1\u00b71*V2\n",
-      "#     V3 = 11\u00b75*V1 \u2212 11*V2                     ...(i)\n",
-      "\n",
-      "#At Junction B\n",
-      "#     I3 + i\u20322 = I2 + i2\n",
-      "#or   V3*w*C + V3*0\u00b71*C*w = V2*w*C + (V1 + V2)*w*0\u00b715 C\n",
-      "#or   1\u00b71*V3 = 1\u00b715*V2 + 0\u00b715*V1                ...(ii)\n",
-      "#Putting the value of V3 from exp (i). into exp. (ii), we get,\n",
-      "#     1\u00b71*(11\u00b75*V1 \u2212 11*V2) = 1\u00b715*V2 + 0\u00b715*V1\n",
-      "#or   13\u00b725*V2 = 12\u00b75*V1\n",
-      "#    V2 = 12.5/13.25*V1\n",
-      "#    V3 = 11.5*V1-11*V2\n",
-      "#    V = V1+V2+V3\n",
-      "#    V  = 40.55/13.25*V1\n",
-      "V1 = 13.25/40.55*V\n",
-      "V2 = 12.5/13.25*V1\n",
-      "V3 = 14.8/13.25*V1\n",
-      "\n",
-      "#The voltage across each unit expressed as a % of V becomes:\n",
-      "p1 = V1/V*100\n",
-      "p2 = V2/V*100\n",
-      "p3 = V3/V*100\n",
-      "\n",
-      "\n",
-      "#Results:\n",
-      "print \"The voltage across each unit expressed as a % of V becomes:\"\n",
-      "print \"for top unit, v1 =\",round(p1,1),\"%\",\n",
-      "print \"\\nfor middle unit, v2 =\",round(p2,1),\"%\"\n",
-      "print \"for 3rd unit, v3 =\",round(p3,1),\"%\"\n",
-      "\n",
-      "print \"String efficiency is\",round(V/(3*V3)*100,1),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The voltage across each unit expressed as a % of V becomes:\n",
-        "for top unit, v1 = 32.7 % \n",
-        "for middle unit, v2 = 30.8 %\n",
-        "for 3rd unit, v3 = 36.5 %\n",
-        "String efficiency is 91.3 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.11, Page Number: 179"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "V1,V2,V3,V= symbols('V1 V2 V3 V')          #volts (shown in fig(i))\n",
-      "\n",
-      "#Calculation:\n",
-      "\n",
-      "#At Junction A\n",
-      "#    I2 + i\u20321 = I1 + i1\n",
-      "#or  V2*w*C + (V2+V3)*w*0\u00b71*C  = V1*w*C+V1*0\u00b72*C*\u03c9   \n",
-      "#   V3 = 12 V1 \u2212 11 V2                                  # ...(i)\n",
-      "\n",
-      "#At Junction B\n",
-      "#      I3 + i\u20322 = I2 + i2\n",
-      "#or    V3 \u03c9 C + V3 \u00d7 0\u00b73 C \u00d7 \u03c9 =V2 \u03c9 C + (V1 + V2) \u03c9 \u00d7 0\u00b72 C\n",
-      "#or    1\u00b73 V3 = 1\u00b72 V2 + 0\u00b72 V1 ...(ii)\n",
-      "#Substituting the value of V3 from exp. (i) into exp. (ii), we get,\n",
-      "#      1\u00b73*(12*V1 \u2212 11*V2) = 1\u00b72*V2 + 0\u00b72*V1\n",
-      "#or    15\u00b75*V2 = 15\u00b74*V1\n",
-      "V2 = 15.4*V1/15.5                     #...(iii)\n",
-      "V2 =  0.993*V1\n",
-      "\n",
-      "#Substituting the value of V2 from exp. (iii) into exp. (i), we get,\n",
-      "V3 = 12*V1 - 11*0.993*V1\n",
-      "V3 = 1.077*V1\n",
-      "\n",
-      "#Voltage between conductor and earth (i.e. phase voltage)\n",
-      "V = V1+V2+V3\n",
-      "n = V/(3*V3)*100\n",
-      "\n",
-      "#Result:\n",
-      "print \"String efficiency is\",round(n),\"%\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "String efficiency is 95.0 %\n"
-       ]
-      }
-     ],
-     "prompt_number": 55
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.13, Page Number: 184"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 1                         #conductor radius(cm)\n",
-      "d = 100                       #conductor spacing(cm)\n",
-      "go = 30/math.sqrt(2)               #Dielectric strength of air,(rms)(kV/cm)\n",
-      "dl = 0.952                    #air density factor\n",
-      "mo = 0.9                      #irregularity factor\n",
-      "\n",
-      "#Calculation:\n",
-      "Vc = mo*go*dl*r*math.log(d/r)     #kV/phase\n",
-      "Vl = Vc*3**0.5                   #kV/line\n",
-      "\n",
-      "#Result:\n",
-      "print \"The disruptive critical voltage for the line is\",round(float(Vl),2),\"kV\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The disruptive critical voltage for the line is 144.97 kV\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.14, Page Number: 184"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 0.978                       #conductor radius(cm)\n",
-      "go = round(30/2**0.5,1)                  #Dielectric strength of air,rms(kV/cm)\n",
-      "Vc = 210/3**0.5                 #Disruptive voltage/phase(kV)\n",
-      "mo = 1                          #irregularity factor(for smooth conductor)\n",
-      "dl = 1              #air density factor(at std pressure and temperature)\n",
-      "\n",
-      "#Calculation:\n",
-      "d = r*10**(Vc/(2.3*mo*go*dl*r))\n",
-      "\n",
-      "#Result:\n",
-      "print \"The spacing between the conductors is\",round(d),\"cm\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The spacing between the conductors is 341.0 cm\n"
-       ]
-      }
-     ],
-     "prompt_number": 16
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.15, Page Number: 185"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 1.5                          #conductor radius(cm)\n",
-      "t = 40                            #temperature(deg C)\n",
-      "b = 76                           #atmospheric pressure(cm)\n",
-      "mo = 0.85                        #irregularity factor\n",
-      "f = 50                           #Hz\n",
-      "d = 200                         #conductor spacing(cm)\n",
-      "go = 30/2**0.5                  #Dielectric strength of air,rms(kV/cm)\n",
-      "Vl = 220                        #line voltage(kV)\n",
-      "\n",
-      "#Calculation:\n",
-      "dl = round(3.92*b/(273+t),3)\n",
-      "\n",
-      "Vc = round(mo*go*dl*r*round(math.log(d/r),2),1)  #Critical disruptive voltage per phase(kV)\n",
-      "V = round(Vl/3**0.5)                    #kV/phase\n",
-      "P = 242.2/dl*(f+25)*(r/d)**0.5*(V-Vc)**2*10**-5      #kW/km/phase\n",
-      "Pl = 3*P                       #corona loss(kW)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Total corona loss per km for three phases is\",round(Pl,5),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Total corona loss per km for three phases is 0.05998 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.16, Page Number: 185"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "P1 = 53                    #corona loss(kW)\n",
-      "V1 = 106                   #operating voltage(kV)\n",
-      "P2 = 98                   #corona loss(kW)\n",
-      "V2 = 110.9                #operating voltage(kV)\n",
-      "V3 = 113                  #opeating voltage(kV)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#As f, \u03b4, r and d are the same for the two cases,\n",
-      "#therfore, P \u221d (V \u2212 Vc)**2\n",
-      "k,Vc = symbols('k Vc')           #k = proportionality constant\n",
-      "                              #Vc = critical disuptive voltage(kV)\n",
-      "#for 1st case,\n",
-      "#    V1/3**0.5 = k*(64 - Vc)**2\n",
-      "#    V2/3**0.5 = k*(61\u00b72 - Vc)**2\n",
-      "# Dividing above equations & solving,\n",
-      "Vc1 = round(solve(((64-Vc)/(61.2-Vc))-math.sqrt(98/53),Vc)[0],2)\n",
-      "Vc1 = 54\n",
-      "#Let W kilowatt be the power loss at 113 kV.\n",
-      "#W = k*(113/3**0.5-Vc1)**2\n",
-      "W = P1*((65.2-Vc1)/(61.2-Vc1))**2\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The disruptive critical voltage at 113 kV is\",Vc1,\"kV\"\n",
-      "print \"The corona loss at 113 kV is\",round(W),\"kW\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The disruptive critical voltage at 113 kV is 54 kV\n",
-        "The corona loss at 113 kV is 128.0 kW\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.17, Page Number: 190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "w = 680                        #weigth of conductor(kg/km)\n",
-      "l = 260                        #length ofspan(m)\n",
-      "u = 3100                       #ultimate strength(kg)\n",
-      "sf = 2                          #safety factor\n",
-      "h = 10                         #ground clearance(m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "T = u/sf                        #working tension(kg)\n",
-      "s = w*l**2/(8*T*1000)            #span(m)\n",
-      "H = h+s\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Conductor should be supported at a height of \",round(H,1),\"m\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Conductor should be supported at a height of  13.7 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 56
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.18, Page Number: 190"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 150                        #length ofspan(m)\n",
-      "a = 2                          #conductor cross-section(cm**2)\n",
-      "go = 9.9                      #specific gravity of conductor(gm/cm**3)\n",
-      "ww = 1.5                      #Wind force/m length of conductor(kg)\n",
-      "T = 2000                       #working tension(kg)\n",
-      "\n",
-      "#Calculation:\n",
-      "w = go*a*1/10              #Wt. of conductor/m length(kg)\n",
-      "wt = round((w**2+ww**2)**0.5,2)      #Total wt of 1m length of conductor(kg)\n",
-      "s = wt*l**2/(8*T)              #sag(m)\n",
-      "#This is the value of slant sag in a direction making \n",
-      "#an angle theta with the vertical\n",
-      "\n",
-      "theta = math.atan(ww/w)\n",
-      "Vs = s*math.cos(theta)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sag is\",round(s,2),\"m\"\n",
-      "print \"The vertical sag is\",round(Vs,2),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sag is 3.49 m\n",
-        "The vertical sag is 2.78 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.19, Page Number: 191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 200                     #span length(m)\n",
-      "w = 1170/1000                #weigth of conductor(kg)\n",
-      "u = 4218                    #Ultimate Strength(kg/cm**2)\n",
-      "sf = 5                      #safety factor\n",
-      "a = 1.29                    #cross-section of conductor(cm**2)\n",
-      "P = 122                        #wind pressure(kg/m**2)\n",
-      "\n",
-      "#Calculation:\n",
-      "T = u/sf*a                   #working tension(kg)\n",
-      "d = round((4*a/math.pi)**0.5,2)            #diameter of conductor(cm)\n",
-      "ww = P*d*10**-2             #Wind force/m length,\n",
-      "wt = (ww**2+w**2)**0.5          #total wt of conductor per metre length\n",
-      "S = wt*l**2/(8*T)\n",
-      "theta = math.atan(ww/w)\n",
-      "VS = S*math.cos(theta)\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sag is\",round(S,2),\"m\"\n",
-      "print \"The vertical sag is\",round(VS,2),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sag is 8.97 m\n",
-        "The vertical sag is 5.38 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.20, Page Number: 191"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 275                    #span length(m)\n",
-      "w = 0.865                 #weigth of conductor/m(kg)\n",
-      "d = 1.96                    #conductor diameter(cm)\n",
-      "t = 1.27                    #ice coating thickness(cm)\n",
-      "u = 8060                    #Ultimate Strength(kg)\n",
-      "sf = 2                      #safety factor\n",
-      "a = 1.29                    #cross-section of conductor(cm**2)\n",
-      "wo = 0.91                   #Weight of 1 c.c. of ice(gm)\n",
-      "P = 3.9                      #wind pressure(gm/cm**2)\n",
-      "\n",
-      "#Calculation:\n",
-      "T = u/sf                    #working tension(kg)\n",
-      "v = 3.14*t*(d+t)*100            #Volume of ice per metre(cm**3)\n",
-      "wi = v*wo/1000                   #kg\n",
-      "ww = P*(d+2*t)*100/1000          #kg\n",
-      "wt = ((w+wi)**2+ww**2)**0.5       #total weight(kg)\n",
-      "S = wt*l**2/(8*T)                  #sag(m)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Sag is\",round(S,1),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Sag is 6.3 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 60
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.21, Page Number: 192"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "Sv = 2.35                    #vertical sag(m)\n",
-      "P = 1.5                      #wind pressure(kg/m)\n",
-      "B = 2540                       #breaking stress(kg/cm**2)\n",
-      "w = 1.125                   #wt of conductor(kg/m)\n",
-      "l = 214                     #length of conductor(m)\n",
-      "a = 3.225                    #conductor cross-section\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "f = symbols('f')             #safety factor\n",
-      "wt = (w**2+P**2)**0.5              #Total wt. of one m length of conductor\n",
-      "T = B*a/f                   #working stress(kg)\n",
-      "cos_theta = w/wt\n",
-      "S = Sv/cos_theta            #slant sag\n",
-      "T1 =  wt*l**2/(8*S)          \n",
-      "f1 = solve(T-T1,f)[0]           #safety factor\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Safety factor is\",round(f1)"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Safety factor is 3.0\n"
-       ]
-      }
-     ],
-     "prompt_number": 61
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.22, Page Number: 192"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 150                         #span length(m)\n",
-      "u = 5000                       #ultimate strength(kg/cm**2)\n",
-      "go = 8.9                       #specific gravity of material(gm/cc)\n",
-      "P = 1.5                        #wind pressure(kg/m)\n",
-      "sf = 5                         #safety factor\n",
-      "a = 2                           #cross-section of conductor(cm**2)\n",
-      "h = 7                          #ground clearance(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "w = a*go*100/1000                  #weight of conductor per m(kg)\n",
-      "T = u*a/sf                    #working stress per m(kg)\n",
-      "wt = (w**2+P**2)**0.5         #Total wt of 1 m length of conductor(kg)\n",
-      "S = wt*l**2/(8*T)               #slant sag(m)\n",
-      "Sv = S*w/wt                   #conductor vertical sag(m)\n",
-      "H = h+Sv                     #m\n",
-      "\n",
-      "#Result:\n",
-      "print \"Conductor should be supported at a height of\",round(H,1),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Conductor should be supported at a height of 9.5 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.23, Page Number: 193"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 500                                #distance b/w the two towers(m)\n",
-      "T = 1600                               #tension in the conductor(kg)\n",
-      "w = 1.5                                 #weight of the conductor(kg/m)\n",
-      "h1 = 30                                #height of 1st tower(m)\n",
-      "h2 = 90                                #height of 2nd tower(m)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "h = h2-h1                            #difference in levels(m)\n",
-      "x1,x2 = symbols('x1 x2')\n",
-      "x2 = l-x1\n",
-      "S1 = w*x1**2/(2*T)                   #m\n",
-      "S2 = w*x2**2/(2*T)                    #m\n",
-      "x11 = solve(S2-S1-h,x1)[0]\n",
-      "S1 = w*x11**2/(2*T)                   #m\n",
-      "#Let the mid-point P be at a distance x from the lowest point O\n",
-      "x = l/2-x11\n",
-      "Smid = w*x**2/(2*T)                  #m\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The min. clearance of the conductor and water is \",round(h1-S1),\"m\"\n",
-      "print \"The clearance mid-way b/w the supports is\",round(Smid+h1-S1,2),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The min. clearance of the conductor and water is  23.0 m\n",
-        "The clearance mid-way b/w the supports is 30.7 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 62
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.24, Page Number: 194"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 600                                #distance b/w the two towers(m)\n",
-      "f = 5                                  #safety factor\n",
-      "U = 8000                               #kg/cm**2\n",
-      "a = 2.2                                #area of cross-section(cm**2)\n",
-      "w = 1.925                              #weight of the conductor(kg/m)\n",
-      "h = 15                                  #difference in towers levels(m)\n",
-      "wi = 1                                 #ice weight(kg)\n",
-      "\n",
-      "#Calculation:\n",
-      "T = U*a/f                              #tension in the conductor(kg)\n",
-      "wt = wi+w                              #total weight of conductor(kg)\n",
-      "x1,x2 = symbols('x1 x2')\n",
-      "x2 = l-x1\n",
-      "S1 = wt*x1**2/(2*T)                   #m\n",
-      "S2 = wt*x2**2/(2*T)                    #m\n",
-      "x11 = solve(S2-S1-h,x1)[0]\n",
-      "x22 = l-x11\n",
-      "S22 = wt*x22**2/(2*T)                   #m\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The sag from the taller of the two supports is\",round(S22,2),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The sag from the taller of the two supports is 45.27 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 63
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.25, Page Number: 195"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 400                                #distance b/w the two towers(m)\n",
-      "T = 2000                               #tension in the conductor(kg)\n",
-      "w = 1                                 #weight of the conductor(kg/m)\n",
-      "h1 = 40                                #height of 1st tower(m)\n",
-      "h2 = 90                                #height of 2nd tower(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "h = h2-h1                            #difference in levels(m)\n",
-      "x1,x2 = symbols('x1 x2')\n",
-      "x2 = l-x1\n",
-      "S1 = w*x1**2/(2*T)                   #m\n",
-      "S2 = w*x2**2/(2*T)                    #m\n",
-      "x11 = solve(S2-S1-h,x1)[0]\n",
-      "S1 = w*x11**2/(2*T)                   #m\n",
-      "x_mid = 50+400/2           #Horizontal distance of mid-point P from lowest point O\n",
-      "Sp = w*x_mid**2/(2*T)                      #SAg at P(m)\n",
-      "S2 = w*(l-x11)**2/(2*T)                   #m\n",
-      "hc = h2-S2+Sp                       #Clearance of mid-point P above water level(m)\n",
-      "\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Clearance of mid-point P above water level is\",round(hc),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Clearance of mid-point P above water level is 55.0 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 127
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.26, Page Number: 195"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from pylab import *\n",
-      "from sympy import *\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "T = 1500                             #tension in the conductor(kg)\n",
-      "w = 1                                #weight of the conductor(kg/m)\n",
-      "DE = 300                              #distance b/w tower's top(m)\n",
-      "H = 22                               #height of each tower(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "# Suppose, the conductors are supported between towers AD and BE over\n",
-      "# a hillside having gradient of 1 : 20.\n",
-      "# Let the lowest point on the conductor is O and\n",
-      "# sin(theta) = 1/20.\n",
-      "# since, the lowest conductor is fixed 2m below the top of each tower.\n",
-      "theta = math.asin(1/20)\n",
-      "BE = H-2                          #Effective height of each tower(m)\n",
-      "EC = DE*1/20\n",
-      "x1,x2 = symbols('x1 x2')\n",
-      "DC = (DE**2-EC**2)**0.5                  #Horizontal distance b/w two towers(m)\n",
-      "x2 = DC-x1\n",
-      "S1 = w*x1**2/(2*T)                    #m\n",
-      "S2 = w*x2**2/(2*T)                    #m\n",
-      "x11 = solve(S2-S1-EC,x1)[0]\n",
-      "S2 = w*(DC-x11)**2/(2*T)               #m\n",
-      "BC = BE+EC\n",
-      "OG = BC-S2-x11*math.tan(theta)\n",
-      "\n",
-      "#Result:\n",
-      "print \"Clearance of the lowest point O from the ground is\",round(OG,2),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Clearance of the lowest point O from the ground is 14.4 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 8.27, Page Number: 196"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "from sympy import *\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "l = 300                     #distance b/w tower(m)\n",
-      "S = 10                        #span(m)\n",
-      "h = 8                      #clearance(m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "#On level ground:\n",
-      "r = 8*S/l**2              #ratio w/T\n",
-      "\n",
-      "\n",
-      "#On sloping ground:\n",
-      "#The conductors are supported between towers AD and BE \n",
-      "#over a sloping ground having a gradient 1 in 15 as shown above.\n",
-      "\n",
-      "DE = 300\n",
-      "sin_theta = 1/15\n",
-      "\n",
-      "h = DE*sin_theta             #Vertical distance b/w the two towers(m)\n",
-      "BE = S+h                      #height of tower(m)\n",
-      "x1,x2,x = symbols('x1 x2 x')\n",
-      "x2 = l-x1\n",
-      "S1 = r*x1**2/2                   #m\n",
-      "S2 = r*x2**2/2                   #m\n",
-      "x11 = solve(S2-S1-h,x1)[0]\n",
-      "S11 = r*x11**2/(2)                   #m\n",
-      "x22 = l-x11\n",
-      "S22 = r*x22**2/(2)                   #m\n",
-      "tan_theta = 1/15\n",
-      "GF = x11*tan_theta\n",
-      "BC = h-S22-GF\n",
-      "#Since O is the origin, the equation of slope of ground is given by :\n",
-      "y = x/15-10.5\n",
-      "#C = Equation of conductor curve \u2212 y\n",
-      "C = r*x**2/2-y\n",
-      "C1 = diff(C,x)\n",
-      "xm = solve(C1,x)[0]\n",
-      "#putting values of xm in above equations,\n",
-      "ym = xm/15-10.5\n",
-      "Cm = C = r*xm**2/2-ym\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Required minimum clearance is\",round(Cm),\"m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Required minimum clearance is 8.0 m\n"
-       ]
-      }
-     ],
-     "prompt_number": 129
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/chapter9_1.ipynb b/Principles_of_Power_System/chapter9_1.ipynb
deleted file mode 100644
index 925aa356..00000000
--- a/Principles_of_Power_System/chapter9_1.ipynb
+++ /dev/null
@@ -1,740 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:1c5f89ed0470787a153a235132b85821455bc93d217ca226b9eba94372c90683"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 9: Electrical Design of Overhead\n",
-      "Lines"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.1, Page Number: 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "d = 200                     #Spacing of conductors(cm)\n",
-      "r = 1.2/2                   #Radius of conductor(cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "L = 10**-7*(1+4*math.log(d/r))*10**3  #Loop inductance per m length of the line(H)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \" The loop inductance per km of the line is\",round(L*1000,3),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The loop inductance per km of the line is 2.424 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.2, Page Number: 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "d = 300                     #Spacing of conductors(cm)\n",
-      "r = 1                      #Radius of conductor(cm)\n",
-      "u1 = 1                     #relative permeability of copper\n",
-      "u2 = 100                    #relative permeability of steel\n",
-      "\n",
-      "#Calculation:\n",
-      "#(i) With copper conductors,\n",
-      "Lc = 10**-7*(u1+4*math.log(d/r))*10**3     #Loop inductance/km\n",
-      "\n",
-      "#(ii) With steel conductors,\n",
-      "Ls = 10**-7*(u2+4*math.log(d/r))*10**3      ##Loop inductance/km\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The loop inductance per km length of the line are:\"\n",
-      "print \"(i) for copper, Lc\",round(Lc*1000,2),\"mH and\"\n",
-      "print \"(ii) for steel line, Ls =\",round(Ls*1000,2),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The loop inductance per km length of the line are:\n",
-        "(i) for copper, Lc 2.38 mH and\n",
-        "(ii) for steel line, Ls = 12.28 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.3, Page Number: 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "d = 200                   #cm\n",
-      "r = 0.62                   #conductor radius(cm)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "L = 10**-7*(0.5+2*math.log(d/r))*10**3             #Inductance/phase/km\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance per km of a 3-ph transmission line is\",round(L*10**3,1),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance per km of a 3-ph transmission line is 1.2 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.4, Page Number: 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "D12 = 2                   #m\n",
-      "D23 = 2.5                 #m\n",
-      "D13 = 4.5                  #m\n",
-      "r = 0.62                  #m\n",
-      "\n",
-      "#Calculation:\n",
-      "Deq = (D12*D23*D13)**(1/3)*100         #Equivalent equilateral spacing(cm)\n",
-      "L = 10**-7*(0.5+2*math.log(Deq/r))*10**3\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance per km of the line is\",round(L*1000,3),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance per km of the line is 1.274 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.5, Page Number: 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 1.25                     #cm\n",
-      "D12 = 2                       #m\n",
-      "D23 = 2                       #m\n",
-      "D13 = 4                       #m\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "Deq = (D12*D23*D13)**(1/3)*100         #Equivalent equilateral spacing(m)\n",
-      "L = 10**-7*(0.5+2*math.log(Deq/r))*10**3\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance per km of the line is\",round(L*1000,2),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance per km of the line is 1.11 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.6, Page Number: 214"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 0.5                          #radius of conductor(m)\n",
-      "Dab = 25                          #cm\n",
-      "Daa1 = 100                        #cm\n",
-      "Dbb1 = 100                        #cm\n",
-      "Dab1 = 103                        #cm\n",
-      "Da1b = 103                        #cm\n",
-      "Da1b1 = 25                        #cm\n",
-      "\n",
-      "#Calculation:\n",
-      "GMR = 0.7788*r                    #G.M.R. of conductor(cm)\n",
-      "Ds = (GMR*Daa1)**(1/2)               #Self G.M.D. of aa1 combination(cm)\n",
-      "Dm = (Dab*Dab1*Da1b*Da1b1)**(1/4)   #Mutual G.M.D. between a and b(cm)\n",
-      "L1 = 2*10**-7*math.log(Dm/Ds)       #Loop inductance per conductor per m(H)\n",
-      "L = 2*L1*1000\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance per km of the line is\",round(L*1000,2),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance per km of the line is 0.84 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.7, Page Number: 216"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 1.3                            #conuctor radius(cm)\n",
-      "Dab = 3                            #m\n",
-      "Dab1  = 6.7                        #m\n",
-      "Da1b = 6.7                         #m\n",
-      "Da1b1 = 3                          #m\n",
-      "Daa1 = 8.48                        #m\n",
-      "Da1a = 8.48                        #m\n",
-      "Dbb1 = 6                           #m\n",
-      "Db1b = 6                           #m\n",
-      "Dca = 6                            #m\n",
-      "Dc1a = 6                           #m\n",
-      "Dca1 = 6                           #m\n",
-      "Dc1a1 = 6                          #m\n",
-      "\n",
-      "#Calculation:\n",
-      "GMR = 1.3*0.7788/100                   #m\n",
-      "Daa = GMR\n",
-      "Da1a1 = GMR\n",
-      "Dbb = GMR\n",
-      "Db1b1 = GMR\n",
-      "Ds1 = (Daa*Daa1*Da1a1*Da1a)**(1/4)             #m\n",
-      "Ds2 = (Dbb*Dbb1*Db1b1*Db1b)**(1/4)             #m\n",
-      "Ds = (Ds1*Ds2*Ds1)**(1/3)                       #m\n",
-      "DAB = (Dab*Dab1*Da1b*Da1b1)**(1/4)             #m\n",
-      "DBC = DAB\n",
-      "DCA = (Dca*Dc1a*Dca1*Dc1a1)**(1/4)             #m\n",
-      "Dm = (DAB*DBC*DCA)**(1/3)            #Equivalent mutual G.M.D(m)\n",
-      "\n",
-      "L1 = 2*10**-7*math.log(Dm/Ds)       #Loop inductance per conductor per m(H)\n",
-      "L = L1*1000\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance per km of the line is\",round(L*1000,2),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance per km of the line is 0.58 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.8, Page Number: 217"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "r = 0.75                            #conductor radius(cm)\n",
-      "Db1b = 5.5                       #m\n",
-      "Dca = 6                         #m\n",
-      "Dca1 = 4                         #m\n",
-      "Dc1a = 4                          #m\n",
-      "Dc1a1 = 6                        #m\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "GMR = 0.7788*r                     #m\n",
-      "Dab = (3**2+0.75**2)**0.5         #m\n",
-      "Dab1 = (3**2+4.75**2)**0.5         #m\n",
-      "Daa1 = (6**2+4**2)**0.5            #m\n",
-      "Daa = GMR/100                      #m\n",
-      "Da1a1 = Daa                       #m\n",
-      "Da1a = Daa1                      #m\n",
-      "Dbb = GMR/100                      #m\n",
-      "Dbb1 = 5.5                       #m\n",
-      "Db1b1 = Dbb                      #m\n",
-      "\n",
-      "Ds1 = (Daa*Daa1*Da1a1*Da1a)**(1/4)          #m\n",
-      "Ds2 = (Dbb*Dbb1*Db1b1*Db1b)**(1/4)          #m\n",
-      "Ds3 = Ds1\n",
-      "Ds = (Ds1*Ds2*Ds3)**(1/3)       #Equivalent self G.M.D. of one phase(m)\n",
-      "DAB = (Dab*Dab1*Da1b*Da1b1)**(1/4)   #m\n",
-      "DCA = (Dca*Dca1*Dc1a*Dc1a1)**(1/4)    #m\n",
-      "DBC = DAB                             #m\n",
-      "Dm = (DAB*DBC*DCA)**(1/3)             #Equivalent mutual G.M.D.(m)\n",
-      "\n",
-      "\n",
-      "L1 = 2*10**-7*math.log(Dm/Ds)       #Loop inductance per conductor per m(H)\n",
-      "L = L1*1000\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance per km of the line is\",round(L*1000,3),\"mH\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance per km of the line is 0.627 mH\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.9, Page Number: 218"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "r = 5.3/100                   #conductor radius(m)\n",
-      "\n",
-      "#Calculation:\n",
-      "GMR = 0.7788*r                          #m\n",
-      "DAA = GMR                                #m\n",
-      "DAA1 = 24                                #m\n",
-      "DA1A1 = DAA                              #m\n",
-      "DA1A = 24                                #m\n",
-      "\n",
-      "DBB = GMR                                #m\n",
-      "DBB1 = 24                                #m\n",
-      "DB1B1 = DBB                              #m\n",
-      "DB1B = 24                                #m\n",
-      "Ds1 = (DAA*DAA1*DA1A1*DA1A)**(1/4)        #m\n",
-      "Ds2 = (DBB*DBB1*DB1B1*DB1B)**(1/4)        #m\n",
-      "#Similarly:\n",
-      "Ds3 = 0.995                              #m\n",
-      "Ds = (Ds1*Ds2*Ds3)**(1/3)   #Equivalent self-G.M.D. of one phase(m)\n",
-      "#DAB = (DAB \u00d7 DAB\u2032 \u00d7 DA\u2032\u0392 \u00d7 DA\u2032B\u2032)**1/4\n",
-      "DAB = (8*32*16*8)**(1/4)                 #m\n",
-      "DBC = DAB                               #m\n",
-      "#DCA = (DCA \u00d7 DCA\u2032 \u00d7 DC\u2032A \u00d7DC\u2032A\u2032)1/4    #m\n",
-      "DCA = (16*8*40*16)**(1/4)                 #m\n",
-      "\n",
-      "Dm = (DAB*DBC*DCA)**(1/3)                   #m\n",
-      "\n",
-      "L = 2*10**-7*math.log(Dm/Ds)       #Loop inductance per conductor per m(H)\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"The inductance per m of the line is\",round(L*10**7,2),\"* 10**-7  H/m\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The inductance per m of the line is 5.36 * 10**-7  H/m\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.10, Page Number: 219"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "r = 1                            #consuctor radius(cm)\n",
-      "\n",
-      "#Calculation:\n",
-      "#Mutual G.M.D., Dm = (Dab*Dab'*Da'b*Da'b')**(1/4)\n",
-      "Dm = (120*140*100*120)**(1/4)          #m\n",
-      "\n",
-      "#Self G.M.D., Ds = (Daa*Daa'*Da'a'*Da'a)**(1/4)\n",
-      "#Daa = Da'a' = 0.7788 cm; Daa' = Da'a = 20 cm\n",
-      "Ds = (0.7788*20*0.7788*20)**(1/4)         #cm\n",
-      "L = 4*10**-4*math.log(Dm/Ds)              #H/km\n",
-      "\n",
-      "#Result:\n",
-      "print \"The total inductance of the line per km is\",round(L*1000,2),\"mH/km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The total inductance of the line per km is 1.36 mH/km\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.11, Page Number: 224"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 1                         #conductor radius(cm)\n",
-      "d = 300                       #conductor spacing(cm)\n",
-      "e = 8.854*10**-12             #permitivity of free space(F/m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "C = math.pi*e/math.log(d/r)         #F/m\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacitance of the line per km is\",round(C*10**11,4),\"* 10**-2  uF/km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacitance of the line per km is 0.4877 * 10**-2  uF/km\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.12, Page Number: 225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 0.625                         #conductor radius(cm)\n",
-      "d = 200                       #conductor spacing(cm)\n",
-      "e = 8.854*10**-12             #permitivity of free space(F/m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "C = 2*math.pi*e/math.log(d/r)         #F/m\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacitance of the line per km is\",round(C*10**9,4),\"  uF/km\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacitance of the line per km is 0.0096   uF/km\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.13, Page Number: 225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "d1 = 2                       #m\n",
-      "d2 = 2.5                     #m\n",
-      "d3 = 4.5                     #m\n",
-      "r = 0.625                    #conductor raius(cm)\n",
-      "l = 100                      #line length(km)\n",
-      "Vl = 66000                  #line voltage(V)\n",
-      "f = 50                       #frequency of current(Hz)\n",
-      "e = 8.854*10**-12             #permitivity of free space(F/m)\n",
-      "\n",
-      "#Calculation:\n",
-      "d = (d1*d2*d3)**(1/3)*100          #cm\n",
-      "\n",
-      "#(i)Line to neutral capacitance,\n",
-      "C1 = 2*math.pi*e/math.log(d/r)*10**9     #uF/km\n",
-      "C11 = C1*100                           #uF\n",
-      "\n",
-      "#(ii)Charging current per phase,\n",
-      "Ic = Vl*2*math.pi*f*C11*10**-6/(3**0.5)      #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacitance per phase is\",round(C11,2),\"uF\"\n",
-      "print \"Charging current per phase is\",round(Ic,1),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacitance per phase is 0.91 uF\n",
-        "Charging current per phase is 10.9 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.14, Page Number: 225"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable declaration:\n",
-      "r = 1                        #conductor radius(cm)\n",
-      "d = 250                      #equilateral conductor spacing(cm)\n",
-      "e = 8.854*10**-12             #permitivity of free space(F/m)\n",
-      "\n",
-      "\n",
-      "#Calculation:\n",
-      "C = 2*math.pi*e/math.log(d/r)*10**9         #uF/km\n",
-      "C1 = C*100                       #F\n",
-      "\n",
-      "#Result:\n",
-      "print \"Capacitance of the line per km is\",round(C1,4),\"uF/phase\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Capacitance of the line per km is 1.0075 uF/phase\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 3,
-     "metadata": {},
-     "source": [
-      "Example 9.15, Page Number: 226"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "from __future__ import division\n",
-      "import math\n",
-      "\n",
-      "#Variable Declaration:\n",
-      "d1 = 4                        #m\n",
-      "d2 = 4                        #m\n",
-      "d3 = 8                        #m\n",
-      "r = 1                         #conductor raius(cm)\n",
-      "l = 100                       #line length(km)\n",
-      "Vl = 132000                    #line voltage(V)\n",
-      "f = 50                        #frequency of current(Hz)\n",
-      "e = 8.854*10**-12             #permitivity of free space(F/m)\n",
-      "\n",
-      "#Calculation:\n",
-      "Deq = (d1*d2*d3)**(1/3)*100          #cm\n",
-      "C = 2*math.pi*e/math.log(Deq/r)*10**9     #uF/km\n",
-      "C1 = C*100                           #uF\n",
-      "Ic = Vl*2*math.pi*f*C1*10**-6/(3**0.5)      #A\n",
-      "\n",
-      "\n",
-      "#Result:\n",
-      "print \"Charging current per phase is\",round(Ic,2),\"A\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Charging current per phase is 21.41 A\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file
diff --git a/Principles_of_Power_System/screenshots/daily_1.png b/Principles_of_Power_System/screenshots/daily_1.png
index 2f68e17a..2f68e17a 100644..100755
--- a/Principles_of_Power_System/screenshots/daily_1.png
+++ b/Principles_of_Power_System/screenshots/daily_1.png
diff --git a/Principles_of_Power_System/screenshots/dcdistribution_1.png b/Principles_of_Power_System/screenshots/dcdistribution_1.png
index 1354fe4b..1354fe4b 100644..100755
--- a/Principles_of_Power_System/screenshots/dcdistribution_1.png
+++ b/Principles_of_Power_System/screenshots/dcdistribution_1.png
diff --git a/Principles_of_Power_System/screenshots/unsymmetrical_1.png b/Principles_of_Power_System/screenshots/unsymmetrical_1.png
index 5bc246e5..5bc246e5 100644..100755
--- a/Principles_of_Power_System/screenshots/unsymmetrical_1.png
+++ b/Principles_of_Power_System/screenshots/unsymmetrical_1.png
diff --git a/Problems_In_Hydraulics/README.txt b/Problems_In_Hydraulics/README.txt
new file mode 100755
index 00000000..058b166c
--- /dev/null
+++ b/Problems_In_Hydraulics/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Ankit Barot
+Course: btech
+College/Institute/Organization:  Anupam-Mistubishi Heavy Industries Ltd., India
+Department/Designation: Graduate Engineer Trainee
+Book Title: Problems In Hydraulics
+Author: R. S. Paradise
+Publisher: Blcakie & Son Ltd, London
+Year of publication: 1953
+Isbn: 978-1577664550
+Edition: 3
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch1.ipynb b/Problems_In_Hydraulics/ch1.ipynb
new file mode 100755
index 00000000..8ed87e00
--- /dev/null
+++ b/Problems_In_Hydraulics/ch1.ipynb
@@ -0,0 +1,466 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:82344ff688f3e86a09716345a8eb43cec4a8aadc2157f526516e21d9f1e84d30"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Hydrostatics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 Page No : 5"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "w= 62.4     #lb/ft**3\n",
+      "A= 18.       #ft**2\n",
+      "x= 6.        # height ft\n",
+      "kg= 6.\n",
+      "y= 2.        #ft hinges\n",
+      "y1= 5.       #ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "F= w*A*x\n",
+      "F1= F/2\n",
+      "Ft= (F*y-F1*(y1/2))/y1\n",
+      "Fb= F1-Ft\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Force exerted on the bolt = %.f lb'%(F1)\n",
+      "print  ' Force exerted on the hinge = %.f lb'%(Ft)\n",
+      "print  ' Force exerted on the bolt = %.f lb'%(Fb)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force exerted on the bolt = 3370 lb\n",
+        " Force exerted on the hinge = 1011 lb\n",
+        " Force exerted on the bolt = 2359 lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.3 Page No : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "h1= 11.54 \t#ft\n",
+      "h2= 16.33 \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "x1= 7.69 \t#ft\n",
+      "x2= 14.09 \t#ft\n",
+      "x3= 18.23 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Ft= round(w*h1**2/2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'h1 = %.2f ft'%(h1)\n",
+      "print  ' h2 = %.2f ft'%(h2)\n",
+      "print  ' h1+ = %.2f ft'%(x1)\n",
+      "print  ' h2+ = %.2f ft'%(x2)\n",
+      "print  ' h3+ = %.2f ft'%(x3)\n",
+      "print  ' Thrust force = %.f lb/ft run'%(round(Ft,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "h1 = 11.54 ft\n",
+        " h2 = 16.33 ft\n",
+        " h1+ = 7.69 ft\n",
+        " h2+ = 14.09 ft\n",
+        " h3+ = 18.23 ft\n",
+        " Thrust force = 4160 lb/ft run\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.4 Page No : 8"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#initialisation of variables\n",
+      "spo= 0.9    # gravity\n",
+      "h= 3.    \t#ft depth\n",
+      "d= 2. \t    #ft depth of water\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "H= 0.71 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "do= spo*w\n",
+      "de= w*d\n",
+      "bc= do*h\n",
+      "Pt= (bc*(h/2)+bc*d+de*(d/2))*(h+d)\n",
+      "y= (bc*(h/2)+bc*d+de*(d/2)*(d/3))*(h+d)/Pt+H\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"Total pressure = %d lb\"%(Pt)\n",
+      "print  ' position of centre of pressure above the base = %.2f ft position of centre of pressure above the axis '%(y)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total pressure = 3572 lb\n",
+        " position of centre of pressure above the base = 1.65 ft position of centre of pressure above the axis \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.5 Page No : 9"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "#initialisation of variables\n",
+      "a= 30. \t#degrees\n",
+      "b= 30. \t#degrees\n",
+      "h= 20. \t#ft width of lock\n",
+      "h1= 10. \t#ft   water level\n",
+      "h2= 15. \t#ft   water level\n",
+      "h3= 16. \t#ft   high\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "h4= 10./3 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Rt= (1./h3)*((w*(h*h2**2*(h2/3)/(2*math.sqrt(3))))-(w*(h*h1**2*h4/(2*math.sqrt(3)))))\n",
+      "R= ((w*(h*h2**2/(2*math.sqrt(3))))-(w*(h*h1**2/(2*math.sqrt(3)))))\n",
+      "Rb= R-Rt\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Force at the hinge = %.f lb '%(Rt)\n",
+      "print  ' Force at the hinge = %.f lb '%(Rb)\n",
+      "\n",
+      "# Note : Round off error in textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force at the hinge = 17826 lb \n",
+        " Force at the hinge = 27208 lb \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.6 Page No : 10"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "\n",
+      "#initialisation of variables\n",
+      "x= 32. \t    #ft\n",
+      "h= 60.  \t#ft   depth\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "AE= 20. \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Vabc= 2*x*h/3\n",
+      "vc= Vabc*w\n",
+      "Tab= w*h**2/2\n",
+      "Rt= math.sqrt(vc**2+Tab**2)/2240\n",
+      "A= math.degrees(math.atan(vc/Tab))\n",
+      "AD= x-AE+AE*(1/(math.tan(math.radians(A))))\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"resulmath.tant thrust = %.1f tons\"%(Rt)\n",
+      "print \" Angle = %.2f degrees\"%(A)\n",
+      "print  ' AD = %.1f ft '%(AD)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "resulmath.tant thrust = 61.5 tons\n",
+        " Angle = 35.42 degrees\n",
+        " AD = 40.1 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.7 Page No : 12"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "wdc= 3*math.sqrt(3) \t#ft\n",
+      "wdo= math.sqrt(3)\n",
+      "ac= 30.  \t#degrees\n",
+      "ao= 60. \t#degrees\n",
+      "hob= 3. \t#ft\n",
+      "haf= 2.6 \t#ft\n",
+      "hfc= 3. \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "V= 5.63 \t#ft**3\n",
+      "h= 4.3   \t#ft\n",
+      "y= 3.6 \t    #ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W1= int(wdc*hfc*w/2)\n",
+      "Hbc= round(w*hob*(hob/2))\n",
+      "W2= int(V*w)\n",
+      "W3= int(w*haf*h)\n",
+      "Vt= W1+W2\n",
+      "Vht= Hbc+W3\n",
+      "Rt= int(math.sqrt(Vt**2+Vht**2))\n",
+      "A= math.degrees(math.atan(Vht/Vt))\n",
+      "x= (W1*(wdo-(hob/2))+Hbc*y)/Rt\n",
+      "OP= x/math.sin(math.radians(A))\n",
+      "AP= hob+OP\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"Resultant thrust = %d lb\"%(Rt)\n",
+      "print \" Angle = %.2f degrees \"%(A)\n",
+      "print  ' Distance from A till horizontal thrust = %.3f ft '%(AP)\n",
+      "\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Resultant thrust = 1287 lb\n",
+        " Angle = 49.44 degrees \n",
+        " Distance from A till horizontal thrust = 4.150 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.8 Page No : 14"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "r= 96.      # T air\n",
+      "T= 10.5 \t#C\n",
+      "K1= 288. \t#C temperature gound level\n",
+      "K2= 0.0015 \t#C**-1  temperature gradient\n",
+      "h= 3000. \t#ft height\n",
+      "P1= 14.69\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "P2= P1*10**(((1/(r*K2))*math.log10((K1-K2*h)/K1)))\n",
+      "w= P2*144/(r*(273+T))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Density = %.4f lb/ft**3 '%(w)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density = 0.0697 lb/ft**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.9 Page No : 15"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Hb= 20. \t#in ratio\n",
+      "Ha= 1.   \t#in ratio\n",
+      "a= 20. \t    #degrees\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "hb= Hb*math.sin(math.radians(a))\n",
+      "dh= hb+Ha\n",
+      "dP= dh/(12*2.309)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Pressure difference between tapping points = %.3f lb/in**2 '%(dP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure difference between tapping points = 0.283 lb/in**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.10 Page No : 16"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "P= 180. \t#ln/in**2  pressure\n",
+      "r= 53.      #T air\n",
+      "T= 60.   \t#F temperature of air\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "h= 12. \t    #in water level\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "R= P*144/(r*(460+T))\n",
+      "dP= 12*(1-(R/w))\n",
+      "Pab= dP/(12*2.309)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Difference in water level = %.2f in of water '%(dP)\n",
+      "print \" Pressure difference = %.3f lb/in**2\"%(Pab)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Difference in water level = 11.82 in of water \n",
+        " Pressure difference = 0.427 lb/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch10.ipynb b/Problems_In_Hydraulics/ch10.ipynb
new file mode 100755
index 00000000..9cf18748
--- /dev/null
+++ b/Problems_In_Hydraulics/ch10.ipynb
@@ -0,0 +1,589 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:54b93ceee4137049e1ed8024ecd6ba8e7f6645f46f71063561c8d2f8b4589ff2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Hydraulic Prime Movers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 Page No : 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "v= 231. \t#ft/sec\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "vc= 0.97\n",
+      "r= 0.47\n",
+      "p= 85.  \t#per cent\n",
+      "A= 170. \t#degrees\n",
+      "p1= 88. \t#per cent\n",
+      "l= 950. \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "H= v**2/(vc**2*2*g)\n",
+      "u= r*v\n",
+      "vr= v-u\n",
+      "vr1= p*vr/100\n",
+      "w1= u-vr1*math.cos(math.radians(180-A))\n",
+      "W= u*(v-w1)/g\n",
+      "he= W*100/H\n",
+      "W1= p1*W/100\n",
+      "oe= W1*100/l\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'hydraulic efficiency = %.f percent'%(he)\n",
+      "print  ' overall efficiency = %.1f percent'%(oe)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hydraulic efficiency = 86 percent\n",
+        " overall efficiency = 70.2 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 Page No : 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 1.    \t#in\n",
+      "v= 95.   \t#ft/sec\n",
+      "F= 173.2 \t#lb\n",
+      "A= 163.  \t#degrees\n",
+      "H= 500.  \t#ft\n",
+      "Cv= 0.97\n",
+      "d1= 1.33 \t#ft\n",
+      "r= 0.47\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= w*math.pi*v/(144*4)\n",
+      "k= (F-v)/(v*math.cos(math.radians(180-A)))\n",
+      "v1= Cv*math.sqrt(2*g*H)\n",
+      "W= v1*w*d**2*math.pi/(4*144)\n",
+      "N= 60*r*v1/(math.pi*d1)\n",
+      "whp= (v1-v)*(1+k*math.cos(math.radians(180-A)))*v1*2/550\n",
+      "Ns= N*whp**0.5/H**1.25\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'specific speed = %.2f r.p.m'%(Ns)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specific speed = 4.75 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4 Page No : 192"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "D= 2. \t#ft\n",
+      "f= 0.005\n",
+      "l= 10000. \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "H= 1000. \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d= (2*D**5/(f*l))**0.25\n",
+      "v= math.sqrt(8*g*H*D**5/(f*l*d**4+4*D**5))\n",
+      "HP= w*math.pi*d**2*v**3/(2*g*550*4)\n",
+      "Q= math.pi*d**2*(HP/67)/4\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Quantity flowing = %.f cusecs'%(Q)\n",
+      "\n",
+      "# rounding off error\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantity flowing = 185 cusecs\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 Page No : 193"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "pl= 122.5 \t# ft\n",
+      "Hw= 1225 \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "Cd= 0.98\n",
+      "Cd1= 0.45\n",
+      "N= 500. \t#r.p.m\n",
+      "P= 6800. \t#h.p\n",
+      "n= 0.86\n",
+      "w= 62.4 \t#lb/ft**2\n",
+      "l= 5450. \t#ft\n",
+      "f= 0.005\n",
+      "A= 18.   \t#ft**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Ah= Hw-pl\n",
+      "js= Cd*math.sqrt(2*g*Ah)\n",
+      "bs= Cd1*js\n",
+      "D= bs*60*2/(N*2*math.pi)\n",
+      "a=  P*2*g*550*144/(n*w*js**3*2)\n",
+      "vp= math.sqrt(pl*2*g/(4*f*l))\n",
+      "dp= (js*2*4*A/(math.pi*144*vp))**(2./3)\n",
+      "dp=2.495 \t#ft\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'diameter of bucket circle D = %.1f ft'%(D)\n",
+      "print  ' area of jet = %.f in**2'%(a)\n",
+      "print  ' diameter of pipe = %.1f ft'%(dp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diameter of bucket circle D = 4.5 ft\n",
+        " area of jet = 18 in**2\n",
+        " diameter of pipe = 2.5 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 Page No : 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "u= 10.*math.pi \t#ft/sec\n",
+      "u1= 5.*math.pi \t#ft/sec\n",
+      "a= 20. \t#degrees\n",
+      "A= 300. \t#r.p.m\n",
+      "v= 10. \t#ft/sec\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "wi= 2. \t#ft\n",
+      "d= 6. \t#in\n",
+      "w1= 62.4 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "w= v/math.tan(math.radians(a))\n",
+      "a1= math.degrees(math.atan((v/(u-w))))\n",
+      "b= math.degrees(math.atan((v/u1)))\n",
+      "W= u*w/g\n",
+      "A1= math.pi*wi*d/12\n",
+      "Q= A1*v\n",
+      "WHP= W*Q*w1/550\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Blade angle at inlet is given by = %.2f degrees'%(a1)\n",
+      "print  ' Blade angle at inlet is given by = %.2f degrees'%(b)\n",
+      "print  ' Water horse power = %.1f h.p'%(WHP)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Blade angle at inlet is given by = 68.49 degrees\n",
+        " Blade angle at inlet is given by = 32.48 degrees\n",
+        " Water horse power = 95.5 h.p\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7 Page No : 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "H= 100. \t#ft\n",
+      "a= 25. \t#degrees\n",
+      "a1= 20. \t#degrees\n",
+      "r1= 9./8\n",
+      "r2= 0.2\n",
+      "u= 6.63 \t#ft/sec\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "h1= 34. \t#ft\n",
+      "h2= 100. \t#ft\n",
+      "r= 0.1\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "f= math.sqrt(H*g/((r1*1/math.tan(math.radians(a))*1/math.tan(math.radians(a1)))+r1*0.5+(r1*0.5**2*0.2/(math.sin(math.radians(a)))**2)+0.1/(math.sin(math.radians(a1+10)))**2))\n",
+      "W= u*f**2/g\n",
+      "q= a*H*550/(10*W*w)\n",
+      "q1= q/w\n",
+      "A= q/f\n",
+      "dh= h1+h2-((1+r)*f**2/((math.sin(math.radians(a1)))**2*2*g))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'f = %.1f ft/sec'%(f)\n",
+      "print  ' Work Done = %.1f ft-lb/lb'%(W)\n",
+      "print  ' Quantity flow = %.1f cusecs'%(q)\n",
+      "print  ' Area form guides = %.3f ft**2'%(A)\n",
+      "print  ' Pressure at entry of level = %.1f ft of water'%(dh)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "f = 20.2 ft/sec\n",
+        " Work Done = 83.9 ft-lb/lb\n",
+        " Quantity flow = 26.3 cusecs\n",
+        " Area form guides = 1.302 ft**2\n",
+        " Pressure at entry of level = 74.5 ft of water\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.8 Page No : 199"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "#initialisation of variables\n",
+      "d= 8. \t#in\n",
+      "w= 2. \t#in\n",
+      "di= 12. \t#in\n",
+      "wi= 3. \t#in\n",
+      "a= 24. \t#degrees\n",
+      "p= 88. \t#per cent\n",
+      "a1= 85. \t#degrees\n",
+      "a2= 30. \t#degrees\n",
+      "p1= 94. \t#per cent\n",
+      "h= 180. \t#ft\n",
+      "d1= 18. \t#in\n",
+      "Cd= 0.92\n",
+      "g=32.2\n",
+      "n1= 111. \t#rpm\n",
+      "\n",
+      "#calculations\n",
+      "r1= 1./math.tan(math.radians(a))\n",
+      "r2= (1./math.tan(math.radians(a1)))+r1\n",
+      "r3= 2*r2/3\n",
+      "r4= (1/math.tan(math.radians(a2)))-r3\n",
+      "a3= math.tan(math.radians(1/r4))\n",
+      "r5= math.sin(math.radians(a3))\n",
+      "f= math.sqrt(g*h*(p/100.)/(r1*r2+r3*r4+(r5**2/2)))\n",
+      "A= r2*f/(d/12)\n",
+      "N= (A*60/(2*math.pi))-n1\n",
+      "W= (r1*r2+r3*r4)*f**2/g\n",
+      "Q= math.pi*(d1/12)*(w/12)*Cd*f*62.08\n",
+      "whp= W*Q/550\n",
+      "bhp= p1*whp/100\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Speed = %.f rpm'%(N)\n",
+      "print  ' output horsepower = %.f hp'%(bhp)\n",
+      "\n",
+      "# slightly change in r1,r2,r3 and that leads to some error in answer. Please check manually."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed = 905 rpm\n",
+        " output horsepower = 369 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.9 Page No : 201"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "N= 428.6 \t#r.p.m\n",
+      "D= 5.    \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "hp= 16800. \t#hp\n",
+      "Qw= 435. \t#cuses\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "v= 32. \t#ft/sec\n",
+      "v1= 24. \t#f/sec\n",
+      "H= 200. \t#ft\n",
+      "lh1= 0.32 \t#ft lb/lb\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "u= math.pi*D*N/60\n",
+      "W= hp*550/(Qw*w)\n",
+      "w= W*g/u\n",
+      "a= math.radians(math.tan(v/w))\n",
+      "va= math.sqrt(w**2+v**2)\n",
+      "b= math.radians(math.tan(v/(u-w)))\n",
+      "B= 180-b\n",
+      "vew= va**2/(2*g)\n",
+      "ve1w= v1**2/(2*g)\n",
+      "LH= H+vew-ve1w-W+lh1\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Absolute velocity at entry to runner = %.1f ft/sec'%(va)\n",
+      "print  ' Loss of head in runner = %.2f ft lb/lb'%(LH)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Absolute velocity at entry to runner = 102.8 ft/sec\n",
+        " Loss of head in runner = 15.05 ft lb/lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.10 Page No : 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "\n",
+      "#initialisation of variables\n",
+      "A1= 25. \t#degrees\n",
+      "A2= 80. \t#degrees\n",
+      "H1= 100. \t#ft\n",
+      "H2= 13. \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "v= 8. \t#ft/sec\n",
+      "d= 3.5 \t#in\n",
+      "de= 15.4 \t#in\n",
+      "b= 1.5 \t#in\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W= H1-H2-(v**2/(2*g))\n",
+      "f= math.sqrt(W*g/(1/math.tan(math.radians(A1))*(1/math.tan(math.radians(A1))-1/math.tan(math.radians(A2)))))\n",
+      "u= f*(1/math.tan(math.radians(A1))-1/math.tan(math.radians(A2)))\n",
+      "V= d*u/7.7\n",
+      "r= math.degrees(math.atan(f/V))\n",
+      "N= 60*u*12/(math.pi*de)\n",
+      "Q= math.pi*de*f*b/144\n",
+      "HP= Q*w*W/550\n",
+      "Ns= N*math.sqrt(HP)/H1**1.25\n",
+      "di= math.sqrt(Q*4*144/(math.pi*f))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'angle = %.f degrees'%(r)\n",
+      "print \" Angular speed = %.1f rpm\"%(Ns)\n",
+      "print  ' inlet diameter to draft tube = %.2f in'%(di)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angle = 48 degrees\n",
+        " Angular speed = 26.6 rpm\n",
+        " inlet diameter to draft tube = 9.61 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.12 Page No : 207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "H= 82.1 \t#ft\n",
+      "h= 90.   \t#ft\n",
+      "k= 0.00646\n",
+      "k1= 0.00454\n",
+      "vd= 11.  \t#ft/sec\n",
+      "P= 0.53 \t#hp\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= math.sqrt((1/k))*math.sqrt(h-H)\n",
+      "Qu= Q/math.sqrt(H)\n",
+      "Q1= math.sqrt(vd/k1)\n",
+      "hf= Q1**2*k\n",
+      "Qu1= Q1/math.sqrt(h-hf)\n",
+      "Pu= P*(h-hf)**1.5\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Qu = %.2f cuses'%(Qu)\n",
+      "print  ' Q = %.1f cuses'%(Q1)\n",
+      "print  ' power Developed = %.f hp'%(Pu)\n",
+      "\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Qu = 3.86 cuses\n",
+        " Q = 49.2 cuses\n",
+        " power Developed = 340 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch11.ipynb b/Problems_In_Hydraulics/ch11.ipynb
new file mode 100755
index 00000000..3b3ec8a7
--- /dev/null
+++ b/Problems_In_Hydraulics/ch11.ipynb
@@ -0,0 +1,745 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f1e32f78d016d65eaf62d619ce157e62f2d402e55bfc70a9a046c3e8d919d007"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Pumping Machinery"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 Page No : 223"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "\n",
+      "import math \n",
+      "h= 75. \t#ft\n",
+      "e= 0.75\n",
+      "k= 0.01\n",
+      "Q= 3000. \t#gal/min\n",
+      "k1= 1.2\n",
+      "N= 1500.\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "D= 0.836 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W= h/e\n",
+      "v1= math.sqrt((W-h)/k)\n",
+      "Q1= Q/374.06\n",
+      "f1= Q1/(k1*D**2)\n",
+      "u1= math.pi*D*N/60\n",
+      "w1= W*g/u1\n",
+      "B= math.degrees(math.atan((f1/(u1-w1))))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Diameter of impeller = %.3f ft '%(D)\n",
+      "print  ' Blade angle at outlet edge of impeller = %.f degrees '%(B)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter of impeller = 0.836 ft \n",
+        " Blade angle at outlet edge of impeller = 30 degrees \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 Page No : 226"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "V= 150. \t#ft**3/sec\n",
+      "A1= 750. \t#r.p.m\n",
+      "di= 21. \t#in\n",
+      "do= 30. \t#in\n",
+      "v= 50. \t#ft/sec\n",
+      "A= 70. \t#degrees\n",
+      "w= 4.\t#in\n",
+      "p= 30. \t#per cent\n",
+      "p1= 25. \t#per cent\n",
+      "sv= 12.8 \t#ft**3/lb\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "u= A1*2*math.pi*di/(24*60)\n",
+      "u1= A1*2*math.pi*do/(24*60)\n",
+      "f1= V/(math.pi*(do/12)*(1./3))\n",
+      "w1= u1-f1*1/math.tan(math.radians((A)))\n",
+      "v1= math.sqrt(f1**2+w1**2)\n",
+      "P= (u1**2+v**2-(f1**2/(math.sin(math.radians(A)))**2))/(2*g)\n",
+      "h= 30*v1**2/(100*2*g)\n",
+      "Nh= v1**2/(20*2*g)\n",
+      "Prt= P+Nh\n",
+      "W= u1*w1/g\n",
+      "e= Prt*100/W\n",
+      "Power= Prt*V/(sv*550)\n",
+      "\n",
+      "#RESULTS\n",
+      "print 'Total pressure rise = %.1f ft of air'%(Prt)\n",
+      "print ' manometric efficiency = %.1f percent'%(e)\n",
+      "print  ' Power = %.2f hp '%(Power)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total pressure rise = 137.9 ft of air\n",
+        " manometric efficiency = 58.5 percent\n",
+        " Power = 2.94 hp \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 Page No : 228"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "u1= 90. \t#ft/sec\n",
+      "w1= 70. \t#ft\n",
+      "e= 0.8\n",
+      "h1= 10. \t#ft\n",
+      "h2= 16. \t#ft\n",
+      "h3= 5. \t#ft\n",
+      "k= 2./5\n",
+      "f1= 20. \t#ft/sec\n",
+      "f= 18.   \t#ft/sec\n",
+      "a= 45. \t    #degrees\n",
+      "x1= 164.4 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Hm= u1*w1/g\n",
+      "Hm1= e*Hm\n",
+      "lh= Hm-Hm1-h1-h2-h3\n",
+      "vg= k*math.sqrt(f1**2+w1**2)\n",
+      "pr= ((f**2+u1**2-f1**2/(math.sin(math.radians(a)))**2)/(2*g))-h2\n",
+      "pr1= x1-pr\n",
+      "ge= pr1*g*2*100/(vg/k)**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'manometer Head = %.1f ft '%(Hm1)\n",
+      "print  ' outlet velocity from guides = %.1f ft/sec '%(vg)\n",
+      "print  ' Pressure rise through impeller only = %.1f ft '%(pr)\n",
+      "print  ' Guide balde efficiency = %.f per cent '%(ge)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "manometer Head = 156.5 ft \n",
+        " outlet velocity from guides = 29.1 ft/sec \n",
+        " Pressure rise through impeller only = 102.4 ft \n",
+        " Guide balde efficiency = 75 per cent \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 Page No : 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "D1= 7.5 \t#in\n",
+      "Q1= 850. \t#gal/min\n",
+      "p1= 62.4 \t#lb/ft**3\n",
+      "N1= 1800.\n",
+      "D2= 15. \t#in\n",
+      "Q2= 12000. \t#gal/min\n",
+      "p2= 64. \t#lb/ft**3\n",
+      "N1= 1800. \t#r.p.m \n",
+      "H1= 14. \t#lb/ft**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N2= Q2*N1*(D1)**3/(Q1*D2**3)\n",
+      "P1= p1*H1/144\n",
+      "P2= P1*N2**2*D2**2*p2/(N1**2*p1*D1**2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'N2 = %.f r.p.m '%(N2+4)\n",
+      "print  ' P2 = %.f lb/in**2 '%(P2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "N2 = 3180 r.p.m \n",
+        " P2 = 78 lb/in**2 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8 Page No : 234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#initialisation of variables\n",
+      "r= 5.\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "sr= r**2\n",
+      "sr1= r**2/r\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Corresponding ratio = %.f '%(sr)\n",
+      "print  ' Corresponding ratio = %.f '%(sr1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Corresponding ratio = 25 \n",
+        " Corresponding ratio = 5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9 Page No : 236"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "e= 0.88\n",
+      "w= 1.25 \t#in\n",
+      "d= 10.   \t#in\n",
+      "q= 630.  \t#gal/min\n",
+      "a= 40.   \t#degrees\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "e1= 0.83\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= q/(6.24*60)\n",
+      "f1= Q/(e*math.pi*(d/12)*(w/12))\n",
+      "u1= 1000*(w*4/12)*2*math.pi/60\n",
+      "w1= u1-f1*1/math.tan(math.radians(a))\n",
+      "W= u1*w1/g\n",
+      "lr= (f1**2+u1**2-f1**2/(math.sin(math.radians(a)))**2)/(2*g)\n",
+      "mh= e1*W\n",
+      "p= mh-lr\n",
+      "v1= math.sqrt(f1**2+w1**2)\n",
+      "ke= v1**2/(2*g)\n",
+      "pke= p*100/ke\n",
+      "me= 100*lr/W\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Velocity of flow = %.f ft/sec'%(f1)\n",
+      "print  ' Work done = %.1f ft-lb/lb'%(W)\n",
+      "print  ' manometric efficiency = %.1f ft'%(mh)\n",
+      "print  ' Pressure recovered = %.1f ft head'%(p)\n",
+      "print  ' Kinetic energy discharge = %.f ft-lb/lb'%(ke)\n",
+      "print  ' Percentage of kinetic energy recovered = %.1f per cent'%(pke)\n",
+      "print  ' manometric efficiency = %d percent'%(me)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of flow = 7 ft/sec\n",
+        " Work done = 47.8 ft-lb/lb\n",
+        " manometric efficiency = 39.7 ft\n",
+        " Pressure recovered = 11.2 ft head\n",
+        " Kinetic energy discharge = 20 ft-lb/lb\n",
+        " Percentage of kinetic energy recovered = 55.7 per cent\n",
+        " manometric efficiency = 59 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.10 Page No : 239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "W1= 7640. \t#gal/min\n",
+      "W2= 11400. \t#gal/min\n",
+      "Hm= 63. \t#ft\n",
+      "Hm1= 80. \t#ft\n",
+      "ep1= 72. \t#per cent\n",
+      "ep2= 76. \t#per cent\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "whp1= W1*Hm/(60*550)\n",
+      "whp2= W2*Hm1/(60*550)\n",
+      "bhp1= whp1*100/ep1\n",
+      "bhp2= whp2*100/ep2\n",
+      "w1= W2/10\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'For both pumps discharge = %.f gal/min against an 80-ft head'%(W2)\n",
+      "print  ' delivery from one pump = %.1f h.p '%(bhp1)\n",
+      "print  ' delivery from two pumps = %.1f h.p '%(bhp2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For both pumps discharge = 11400 gal/min against an 80-ft head\n",
+        " delivery from one pump = 20.3 h.p \n",
+        " delivery from two pumps = 36.4 h.p \n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.11 Page No : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "h= 94.   \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "e= 0.58\n",
+      "p= 73.5 \t#per cent\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "WHP= h*e*w/550\n",
+      "BHP= WHP/(p/100)\n",
+      "\n",
+      "#RESULTS\n",
+      "print 'W.H.P= %.2f h.p'%(WHP)\n",
+      "print ' Brake horse power= %.1f'%(BHP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "W.H.P= 6.19 h.p\n",
+        " Brake horse power= 8.4\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.12 Page No : 243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "sl= 12. \t#ft\n",
+      "l= 20.  \t#ft\n",
+      "d= 4.   \t#in\n",
+      "dp= 6.  \t#in\n",
+      "lst= 18. \t#in\n",
+      "k= 0.025\n",
+      "H= 32.   \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "pf= 6.   \t#ft\n",
+      "a= 33.83 \n",
+      "a1= 9.53\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= math.sqrt((H-sl-d)*g/a)*a1\n",
+      "Q= 2*math.pi*(dp/12)**2*lst/(12*4*60)\n",
+      "v= Q/(math.pi*(d/12)**2/4)\n",
+      "kh= v**2/(2*g)\n",
+      "fh= k*l*v**2*12/(2*g*d)\n",
+      "N= math.sqrt((H-sl-pf)/(kh+fh))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'premissible speed = %.1f r.p.m'%(A)\n",
+      "print  ' maximum premissible speed = %.1f r.p.m'%(N)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "premissible speed = 37.2 r.p.m\n",
+        " maximum premissible speed = 168.8 r.p.m\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.13 Page No : 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "b= 6.    \t#in\n",
+      "s= 12. \t    #in\n",
+      "d= 4. \t    #in\n",
+      "a1= 30. \t#degrees\n",
+      "a2= 90. \t#degrees\n",
+      "a3= 120. \t#degrees\n",
+      "N= 120. \t#r.p.m\n",
+      "n= 4.\n",
+      "#calculations\n",
+      "A= 2*math.pi*N/60\n",
+      "V= math.pi*(b/12)**2*n/4\n",
+      "v= (b/12)**2*A*(b/12)/(d/12)**2\n",
+      "Q1= v*math.pi*(d/12)**2*math.sin(math.radians(a1))/4\n",
+      "Q2= v*math.pi*(d/12)**2*math.sin(math.radians(a2))/4\n",
+      "Q3= v*math.pi*(d/12)**2*math.sin(math.radians(a3))/4\n",
+      "Q4= V-Q1\n",
+      "Q5= Q2-V\n",
+      "Q6= Q3-V\n",
+      "a4= math.degrees(math.asin(V/(v*math.pi*(d/12)**2)))+a1\n",
+      "A= 180-a4\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'rate of flow at a1 = %.3f cuses'%(Q4)\n",
+      "print  ' rate of flow at a2 = %.3f cuses'%(Q5)\n",
+      "print  ' rate of flow at a3 = %.3f cuses'%(Q6)\n",
+      "print  ' crak angle = %.1f degrees'%(a4)\n",
+      "print  ' crak angle = %.1f degrees'%(A)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate of flow at a1 = 0.169 cuses\n",
+        " rate of flow at a2 = 0.448 cuses\n",
+        " rate of flow at a3 = 0.283 cuses\n",
+        " crak angle = 39.2 degrees\n",
+        " crak angle = 140.8 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.14 Page No : 247"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "n= 2. \t#strokes/sec\n",
+      "dp= 6. \t#in\n",
+      "ds= 18. \t#in\n",
+      "ds1=4. \t#in\n",
+      "l= 20. \t#ft\n",
+      "l1= 20. \t#ft\n",
+      "f= 0.008\n",
+      "la= 5. \t#ft\n",
+      "A= 60. \t#r.p.m\n",
+      "f= 0.008\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "g=32.2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V= math.pi*(ds/12)*n*(dp/12)**2/4\n",
+      "vmp= 2*math.pi*A*(ds/24)/60\n",
+      "vmp1= vmp*(dp**2/ds1**2)\n",
+      "hfmax= 4*f*(l-la)*vmp1**2/(2*g*ds1/12)\n",
+      "H1= round(2*hfmax/3,1)\n",
+      "H2= H1*13\n",
+      "Wls= (H1+H2)*w*math.pi/16*1.5*2\n",
+      "mv= V/(math.pi*(ds1/12)**2/4)\n",
+      "lh= round(4*f*(l-la)*mv**2/(2*g*(ds1/12)),2)\n",
+      "lhf= 12*lh\n",
+      "Wls1= (lh+13.21)*w*math.pi*1.5/16 *2  \n",
+      "WS= Wls-Wls1\n",
+      "\n",
+      "#RESULTS\n",
+      "print 'Work lost per second= %.f ft lb/sec'%(Wls)\n",
+      "print  ' Work saved per second = %.f ft-lb/sec'%(WS)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Work lost per second= 875 ft lb/sec\n",
+        " Work saved per second = 352 ft-lb/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.15 Page No : 248"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 7.5  \t#in\n",
+      "s= 15.   \t#in\n",
+      "l= 36.  \t#ft\n",
+      "h1= 34. \t#ft\n",
+      "h2= 12. \t#ft\n",
+      "L= 10. \t    #ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "f= 0.008\n",
+      "l1= 20. \t#ft\n",
+      "d1= 4.  \t#in\n",
+      "h3= 110. \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "l2= 180. \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= (math.pi/4)*(d)**2*(s/12)*2*(l/60)/144\n",
+      "v= Q/((math.pi/4)*(d1/12)**2)\n",
+      "a= (d/4)**2*(d/12)*(l*2*math.pi/60)**2\n",
+      "H= h1-h2-(L*a/g)-(v**2*0.5/g)-(4*f*l1*v**2/(2*g*(d1/12)))\n",
+      "H1= h1+h3+(L*a/g)+(v**2*0.5/g)+(4*f*l2*v**2/(2*g*(d1/12)))\n",
+      "dh= (H1-H)*w/144\n",
+      "NP= dh*(math.pi/4)*d**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Head at piston = %.2f ft of water absolute'%(H)\n",
+      "print  ' Head at piston = %.2f ft of water absolute'%(H1)\n",
+      "print  ' Difference on head of piston = %.f lb/in**2'%(dh)\n",
+      "print  ' Net load on piston = %.f lb'%(NP)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Head at piston = 11.04 ft of water absolute\n",
+        " Head at piston = 161.59 ft of water absolute\n",
+        " Difference on head of piston = 65 lb/in**2\n",
+        " Net load on piston = 2882 lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.16 Page No : 250"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from numpy.linalg import *\n",
+      "\n",
+      "#initialisation of variables\n",
+      "f= 0.009\n",
+      "dc= 3.5 \t#in\n",
+      "ds= 6. \t#in\n",
+      "r= 0.25\n",
+      "sl= 8. \t#ft\n",
+      "d= 2.5 \t#in\n",
+      "l= 14. \t#ft\n",
+      "el= 8. \t#ft\n",
+      "ed= 22.5 \t#in\n",
+      "ph= 4. \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "f= 0.009\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "BC= el+l\n",
+      "v= math.sqrt(BC*g/(l*(d/2)*(r)*(dc/d)**2))*9.55\n",
+      "vec=roots([2,1/r,-1])\n",
+      "H1= 77\n",
+      "MV= math.sqrt(BC*g/(l*(d/2)*(r)*(dc/d)**2))*r*(math.sin(math.radians(H1))+(math.sin(math.radians(2*H1))/8))\n",
+      "mvp= MV*dc**2/d**2\n",
+      "hf= 4*f*(sl+l)*mvp**2/(2*g*(d/12))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'pump speed = %.1f r.p.m'%(v)\n",
+      "print  ' Friction head = %.3f ft'%(hf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "pump speed = 86.8 r.p.m\n",
+        " Friction head = 1.240 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch12.ipynb b/Problems_In_Hydraulics/ch12.ipynb
new file mode 100755
index 00000000..6c180ea8
--- /dev/null
+++ b/Problems_In_Hydraulics/ch12.ipynb
@@ -0,0 +1,294 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:5f1e1e7eb1f2538f17f15df30e5829cbbc6f3ee52bddff3d767ce84c4a9e05f8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Dimensional and Model Analysis"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 Page No : 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 0.0625 \t#in\n",
+      "sg= 0.91\n",
+      "vs= 1.62\n",
+      "ss= 7.85\n",
+      "g= 981. \t#cm/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v= 4*(d*2.54/2)**2*(ss-sg)*g/(3*6*30.45*vs)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'steady speed attained = %.4f ft/sec '%(v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "steady speed attained = 0.1932 ft/sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 Page No : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "vs= 16. \t#ft/sec\n",
+      "lm= 1. \t#ft\n",
+      "l= 16. \t#ft\n",
+      "R= 9.6 \t#lb\n",
+      "ds= 64. \t#lb/ft**3\n",
+      "dm= 62.4 \t#/lb/ft**3\n",
+      "A= 40. \t#ft**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "vm= vs*math.sqrt(lm/l)\n",
+      "rs= 0.0095*vm**1.9*A\n",
+      "rw= R-rs\n",
+      "Rw= rw*ds*(l/lm)**3/dm\n",
+      "Rs= 0.009*vs**1.85*A*l**2\n",
+      "R1= Rw+Rs\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'speed = %.f b ft/sec'%(vm)\n",
+      "print  ' Total resistance = %.f lb '%(round(R1,-2))\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "speed = 4 b ft/sec\n",
+        " Total resistance = 33700 lb \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4 Page No : 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "H2= 0.75 \t#ft\n",
+      "v1= 1. \t#ft/sec\n",
+      "v2= 6. \t#ft/sec\n",
+      "k= 1.433\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "H1= H2*(v1/v2)**(2./3)\n",
+      "Q1= k*H1**2.47\n",
+      "Q2= Q1*(H2/H1)**2.5\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Flow = %.3f cuses '%(Q2 )\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow = 0.730 cuses \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 Page No : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "nm= 360.\n",
+      "d= 1.5 \t#in\n",
+      "n= 100. \n",
+      "dp= 12. \t#in\n",
+      "vm= 4.8 \t#ft/sec\n",
+      "Tm= 52. \t#sec\n",
+      "T= 16. \t#lb-ft\n",
+      "t= 133. \t#lb ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "vp= n*dp*vm/(nm*d)\n",
+      "Tp= round(Tm*dp**2*vp**2/(d**2*vm**2),-2)\n",
+      "N= Tm*vm*6080*100/(T*2*math.pi*nm*60)\n",
+      "W= Tp*65000\n",
+      "T1= W/(.7*2*math.pi*n*60)\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Speed of advance = %.2f knots '%(vp)\n",
+      "print  ' Thrust = %.f lb '%(Tp)\n",
+      "print  ' Efficiency = %.f per cent '%(N)\n",
+      "print  ' Torque = %.f lb ft '%(round(T1,-2))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speed of advance = 10.67 knots \n",
+        " Thrust = 16400 lb \n",
+        " Efficiency = 70 per cent \n",
+        " Torque = 40400 lb ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6 Page No : 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "d= 4 \t#in\n",
+      "D= 0.0765 \t#lb/ft**3\n",
+      "Da= 8 \t#in\n",
+      "vw= 1./13\n",
+      "nw= 20\n",
+      "va= 13 \t#ft/sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "na= nw*va*d**2/Da**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'power = %.f r.p.m '%(na)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "power = 65 r.p.m \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7 Page No : 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "dtp= 120. \t#in\n",
+      "dpd= 48. \t#in\n",
+      "vim= 1.25 \t#ft/sec\n",
+      "vip= 5. \t#ft/sec\n",
+      "lp = 600. \t#ft\n",
+      "lm= 40.  \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Rm= (dtp/dpd)**2/((lp/lm)*(vim/vip)**2)\n",
+      "d= math.sqrt(4*Rm)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Diameter = %.2f in '%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter = 5.16 in \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch13.ipynb b/Problems_In_Hydraulics/ch13.ipynb
new file mode 100755
index 00000000..30017a36
--- /dev/null
+++ b/Problems_In_Hydraulics/ch13.ipynb
@@ -0,0 +1,624 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:da0a57e05023d78d40f48e24ac486e01915e80a89fcf828d645acb89b5297ac2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Miscellaneous Problems"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 Page No : 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "W= 5000 \t#lb\n",
+      "vr= 6\n",
+      "e= 0.95\n",
+      "ep = 0.75\n",
+      "d= 9 \t#in\n",
+      "D= 45 \t#ft\n",
+      "t= 2 \t#min\n",
+      "v= 4.5 \t#ft/sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "L= W*vr/(e*ep)\n",
+      "Pr= L/(math.pi*d**2/4)\n",
+      "s= D/vr\n",
+      "V= s*math.pi*ep**2/(4*t*60)\n",
+      "T= D/v\n",
+      "V1= s*math.pi*ep**2/4\n",
+      "V2= V*T\n",
+      "V3= V1-V2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Pressure on ram = %.f ln/in**2 '%(Pr)\n",
+      "print ' Pump duty = %.4f cusec'%(V)\n",
+      "print  ' Minimum capacity if accumulator = %.3f ft**3 '%(round(V3))\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure on ram = 662 ln/in**2 \n",
+        " Pump duty = 0.0258 cusec\n",
+        " Minimum capacity if accumulator = 3.000 ft**3 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 Page No : 283"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "P1= 1100. \t#lb/in**2\n",
+      "P2= 85. \t#lb/in**2\n",
+      "f= 0.01\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "l= 1600. \t#ft\n",
+      "r= 1./8\n",
+      "W= 2500. \t#lb\n",
+      "d= 6.    \t#in\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "L= W*d\n",
+      "P= L*2.31/(math.pi*(d/2)**2)\n",
+      "s1= P1*2540/1100\n",
+      "s2= P2*196/85\n",
+      "vp= math.sqrt((s1-s2-P)/(4*f*l/(2*g*r)))\n",
+      "V= vp/16\n",
+      "Vl= V*d\n",
+      "Vp= math.sqrt((s1/3)/(4*f*l/(2*g*r)))\n",
+      "vl= Vp*d/16\n",
+      "Hr= s1-(s1/3)-s2\n",
+      "Lr= Hr*math.pi*(d/2)**2/(2.31*d)\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"In case 1 velocity of load = %.2f ft/sec\"%(Vl)\n",
+      "print \" In case 2 velocity of load = %.2f ft/sec\"%(vl)\n",
+      "print  ' Load to be lifted = %d lb '%(Lr)\n",
+      "\n",
+      "# note : roundin off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In case 1 velocity of load = 4.45 ft/sec\n",
+        " In case 2 velocity of load = 3.87 ft/sec\n",
+        " Load to be lifted = 3054 lb \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 Page No : 284"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "bhp= 1500 \t#h.p\n",
+      "e= 0.86\n",
+      "h1= 300 \t#ft\n",
+      "h2= 15 \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "t= 30 \t#days\n",
+      "t1= 10 \t#hr\n",
+      "t2= 3 \t#months\n",
+      "f= 0.005\n",
+      "l= 1000 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "WHP= bhp/e\n",
+      "Ha= h1-h2\n",
+      "W= WHP*550\n",
+      "Q= W/(Ha*w)\n",
+      "Qt= Q*36009*t1*t*t2\n",
+      "Qp= Qt/(3600*t*45)\n",
+      "d= (f*l*(Q/2)**2/(t1*h2))**(1./5)\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"Minimum size of basin required = %.1e cusecs\"%Qt\n",
+      "print \"Pump Discharge : %.f cusecs\"%(Qp/10)\n",
+      "print  'Diameter = %.2f ft '%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum size of basin required = 1.7e+09 cusecs\n",
+        "Pump Discharge : 36 cusecs\n",
+        "Diameter = 1.89 ft \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4 Page No : 285"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "l= 140 \t#ft\n",
+      "P= 70 \t#percent\n",
+      "V= 3*10**8 \t#ft**3\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "SBD= 4.9*10**8 \t#ft**3\n",
+      "Q= 162 \t#cuses\n",
+      "s= 12.2*10**6 \t#ft**3/day\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "O= Q*w*l*(P/1000.)/550.\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Size of reservoir= %.2e ft**3'%(SBD)\n",
+      "print  ' output = %.f h.p '%(O)\n",
+      "print  ' output = %d h.p '%(Q)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Size of reservoir= 4.90e+08 ft**3\n",
+        " output = 180 h.p \n",
+        " output = 162 h.p \n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5 Page No : 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Q= 140 \t#cuses\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "l= 140 \t#ft\n",
+      "P= 70 \t#percent\n",
+      "k= 1.6\n",
+      "v= 3*10**8\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "rv= k*v\n",
+      "HP= Q*l*w*(P/1000.)/550.\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Required size of reservoir = %.1e ft**3 '%(rv)\n",
+      "print  ' horsepower = %.f h.p '%(HP)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Required size of reservoir = 4.8e+08 ft**3 \n",
+        " horsepower = 156 h.p \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.6 Page No : 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "P= 10. \t#lb/in**2\n",
+      "r1= 0.5 \t#ft\n",
+      "r= 0.25 \t#ft\n",
+      "f= 42.3 \t#ft/sec\n",
+      "b= 1./40\n",
+      "Tt= 1400. \t#lb\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= 2*math.pi*r*b*f\n",
+      "p1= 34+P\n",
+      "Fu= p1*math.pi*(r-(r/4))*144/2.3\n",
+      "Fr= Fu-Tt\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Quantity = %.2f cusecs '%(Q)\n",
+      "print  ' Resultant force on the plate = %.f lb '%(round(Fr,-1))\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quantity = 1.66 cusecs \n",
+        " Resultant force on the plate = 220 lb \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.7 Page No : 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "r= 0.5 \t#ft\n",
+      "N= 300\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= N*2*math.pi/60\n",
+      "Ft= math.pi*A**2*r**4*w/(4*g)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'total force = %.1f lb '%(Ft)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total force = 93.9 lb \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.8 Page No : 292"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 4. \t#in\n",
+      "h= 12. \t#in\n",
+      "h1= 9. \t#in\n",
+      "g= 32. \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "H= 2*(1-(h1/h))\n",
+      "A= math.sqrt((H*2*g/((d/24)**2)))\n",
+      "A1= math.sqrt((H*2*g*2/((d/24)**2)))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'speed when the axial is zero = %.f radn/sec '%(A)\n",
+      "print  ' speed when the axial is zero = %.f radn/sec '%(A1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "speed when the axial is zero = 34 radn/sec \n",
+        " speed when the axial is zero = 48 radn/sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.10 Page No : 295"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "P= 14.7 \t#lb/in**2\n",
+      "T= 15. \t#C\n",
+      "v= 350. \t#ft/sec\n",
+      "R= 0.714\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "P1= P*144\n",
+      "r= 3091*(273+T)\n",
+      "d1= P1/r\n",
+      "r1= r+(v**2/7)\n",
+      "P2= (r1*d1/(P1**R))**(1/(1-R))/144\n",
+      "dP= P2-P\n",
+      "T2= r1/3091\n",
+      "dT= T2-(273+T)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'rise in pressure = %.f lb/in**2 '%(dP)\n",
+      "print  ' rise in temperature = %.1f C '%(dT)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rise in pressure = 1 lb/in**2 \n",
+        " rise in temperature = 5.7 C \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.11 Page No : 297"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "T= 27. \t#C\n",
+      "P = 33. \t#lb/in**2\n",
+      "p1= 14.7 \t#lb/in**2\n",
+      "w= 250. \t#lb\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "Cd= 0.99\n",
+      "r= 1.4\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "w1= P*144/(96*(273+T))\n",
+      "d= p1*144/(96*(273+T))\n",
+      "W= d*w/60\n",
+      "d= math.sqrt(W*4/(Cd*math.pi*math.sqrt(2*g*P*144*(r/(r-1))*w1*(0.528**(2/1.4)-0.528**(2.4/1.4)))))*12\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Diameter = %.3f in '%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter = 0.722 in \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.12 Page No : 299"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "sp= 13.6\n",
+      "hm= 800. \t#mm\n",
+      "d= 3. \t#in\n",
+      "r= 1.4\n",
+      "R= 1385. \t#ft-lb/lb/C\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "T= 15. \t#C\n",
+      "hm1= 765. \t#mm\n",
+      "r1= 9.\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "p1= hm*sp*w/304.8\n",
+      "r2= (273+T)*R\n",
+      "w1= p1/r2\n",
+      "k= hm/hm1\n",
+      "v1= math.sqrt((2*g*r*r2*(1-k**0.286))/((1-r)*(r1**2*k**1.43-1)))\n",
+      "W= v1*w1*3600*(math.pi/64)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Weight flowing = %.f lb/hr '%(W)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Weight flowing = 115 lb/hr \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.13 Page No : 301"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "p= 160. \t#lb/in**2\n",
+      "d= 1./3 \t#ft\n",
+      "T= 15.   \t#C\n",
+      "R= 96. \n",
+      "V= 120. \t#ft**3\n",
+      "f= 0.004\n",
+      "a= 60*math.pi\n",
+      "l= 10560. \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "p1= p*144\n",
+      "w1= p*144/(R*(273+T))\n",
+      "v1= V*36/a\n",
+      "p2= math.sqrt(p1**2-((2*4*f*p1*w1*v1**2*l)/(2*g*d)))/144\n",
+      "v2= p*v1/p2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' pressure = %.1f lb/in**2 '%(p2)\n",
+      "print  ' velocity = %.1f ft/sec '%(v2)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " pressure = 134.0 lb/in**2 \n",
+        " velocity = 27.4 ft/sec \n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch2.ipynb b/Problems_In_Hydraulics/ch2.ipynb
new file mode 100755
index 00000000..d68673ef
--- /dev/null
+++ b/Problems_In_Hydraulics/ch2.ipynb
@@ -0,0 +1,222 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:2c442a1cbc28b933d555165b6cb09fa7f45de31e28c837593e3048f115cafdbb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Equilibrium of Floating Bodies"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 Page No : 26"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#initialisation of variables\n",
+      "d= 40.   \t#lb/ft**2 density of wood\n",
+      "w= 4     \t#ft  wide\n",
+      "h= 6     \t#ft  deep\n",
+      "l= 12    \t#ft  long \n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W= w*h*d*l\n",
+      "V= W/64\n",
+      "D= V/(w*l)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Volume of water print laced = %.f ft**3'%(V)\n",
+      "print  ' Depth of immersion = %.2f ft'%(D)\n",
+      "print  ' Centre of buoyancy = %.2f ft from base'%(D)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Volume of water print laced = 180 ft**3\n",
+        " Depth of immersion = 3.75 ft\n",
+        " Centre of buoyancy = 3.75 ft from base\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 Page No : 28"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import Symbol,solve\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 4.    \t#ft diameter\n",
+      "h= 7.    \t#ft high\n",
+      "W= 2500. \t#lb weighing \n",
+      "OG= 3.5\n",
+      "OB= 1.55 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V= W/d**3\n",
+      "D= V/(math.pi*(d/2)**2)\n",
+      "I= math.pi*d**4/64\n",
+      "BM= I/V\n",
+      "BG= OG-OB\n",
+      "T = Symbol(\"T\")\n",
+      "ans = solve( (2500 + T)**2 -(512*math.pi *(8750 - 804)) - 1)\n",
+      "T = ans[1]\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Minimum tension in chain = %d lb'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Minimum tension in chain = 1075 lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4 Page No : 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "W1= 1000. \t#lb weighing\n",
+      "W2= 100. \t#lb load\n",
+      "h= 4. \t#ft height\n",
+      "d= 5. \t#ft  diameter\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "V= (W1+W2)/h**3\n",
+      "D= V*h/(d**2*math.pi)\n",
+      "I= d**4*math.pi/h**3\n",
+      "BM= I/V\n",
+      "x= (BM+(D/2)-(W1*(h/2)/(W1+W2)))/(W2/(W1+W2))-0.02\n",
+      "C= x-h\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'centre of gravity = %.2f ft'%(x)\n",
+      "print  ' Hence the gravity of the weight must not be more than above the top of buoy = %.2f ft'%(C)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "centre of gravity = 4.43 ft\n",
+        " Hence the gravity of the weight must not be more than above the top of buoy = 0.43 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 Page No : 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "b= 12. \t#ft  breadth\n",
+      "h1= 3. \t#ft  draught\n",
+      "h2= 1.5 \t#ft\n",
+      "h3= 5+(2./3) \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "I= b**3/12\n",
+      "V= b*h1\n",
+      "bm= I/V\n",
+      "BG= bm+(h1*2/(3*b))\n",
+      "O= math.degrees(math.tan(math.sqrt((h3*2-h1-bm*2)/(bm*2+bm))))\n",
+      "\n",
+      "\n",
+      "#RESULTS\n",
+      "print  ' Volume of body immersed = %.f ft**3'%(V)\n",
+      "print  ' BM = %.f ft'%(bm)\n",
+      "print  ' BG = %.2f ft'%(BG)\n",
+      "print  ' angle of heel = %.2f degrees'%(O)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Volume of body immersed = 36 ft**3\n",
+        " BM = 4 ft\n",
+        " BG = 4.17 ft\n",
+        " angle of heel = 9.64 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch3.ipynb b/Problems_In_Hydraulics/ch3.ipynb
new file mode 100755
index 00000000..a3c142e2
--- /dev/null
+++ b/Problems_In_Hydraulics/ch3.ipynb
@@ -0,0 +1,251 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:b5106ea5cae253fe059288e745550627fa4df81e4cd0d1b81237c843ab3f804d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Hydrodynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 Page No : 39"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "# points\n",
+      "hob= 34. \t#ft\n",
+      "hoc= 5. \t#ft\n",
+      "hoa= 50. \t#ft\n",
+      "\n",
+      "hod= 80. \t#ft height\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "# sectional areas\n",
+      "A= 2.1 \t#in**2\n",
+      "A1= 4.8 \t#in**2\n",
+      "A2= 9.6 \t#in**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v= math.sqrt(2*g*(hod-hoc))\n",
+      "Q= v*A/144\n",
+      "va= v*A/A1\n",
+      "vb= v*A/A2\n",
+      "Va= va**2/(2*g)\n",
+      "Vb= vb**2/(2*g)\n",
+      "r= hob+hod-hoa-(va**2/(2*g))\n",
+      "r1=hob+hod-hob-(vb**2/(2*g))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f cuses'%(Q) \n",
+      "print  ' Velocity head at A = %.2f ft-lb/lb'%(Va)\n",
+      "print  ' Velocity head at B = %.2f ft-lb/lb'%(Vb)\n",
+      "print  ' Pressure head at A = %.2f ft-lb/lb'%(r) \n",
+      "print  ' Pressure head at B = %.2f ft-lb/lb'%(r1) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 1.01 cuses\n",
+        " Velocity head at A = 14.36 ft-lb/lb\n",
+        " Velocity head at B = 3.59 ft-lb/lb\n",
+        " Pressure head at A = 49.64 ft-lb/lb\n",
+        " Pressure head at B = 76.41 ft-lb/lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 Page No : 40"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "P= 1.7  \t#lb/in**2 pressure\n",
+      "d1= 6. \t    #in diameters\n",
+      "d2= 3. \t    #in diameters\n",
+      "hab= 8. \t#ft\n",
+      "Q= 0.75 \t#cuses\n",
+      "sm= 13.6    # gravity\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "dP= P*144/w\n",
+      "va= Q*(d1/d2)**4/math.pi\n",
+      "k= -(((d1/d2)**4-1)-((-dP+hab)*2*g/va**2))\n",
+      "h= (-dP+hab)*12/(sm-1)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'k = %.f '%(k)\n",
+      "print  'height difference = %.2f in'%(h) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "k = 3 \n",
+        "height difference = 3.88 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 Page No : 42"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "h= 20. \t    #ft pressure head\n",
+      "Q= 4.81 \t#cuses \n",
+      "C= 1.\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d= 10.   \t#indiameter\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "d= ((Q*4*144/(d**2*math.pi))**2*100**2/((Q*4*144/(d**2*math.pi))**2+2*g*h))**0.25\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Smallest Diameter = %.1f in'%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Smallest Diameter = 4.9 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 Page No : 43"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 1./3 \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d1= 4. \t    #in\n",
+      "d2= 1.6 \t#in diameter\n",
+      "\n",
+      "# guage readings\n",
+      "h1= 5.7 \t#ft\n",
+      "h2= -1.9 \t#ft\n",
+      "\n",
+      "Q= 0.3  \t#cuses\n",
+      "H1= 34. \t#ft height of water\n",
+      "H2= 19. \t#ft\n",
+      "H3= 7. \t    #ft\n",
+      "H4= 9.2 \t#ft\n",
+      "h3= 2.9 \t#ft\n",
+      "h4= 3.9 \t#ft\n",
+      "Et= 54. \t#ft-lb/lb\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v1= math.sqrt(2*g*(h1-h2)/((d1/d2)**4-1))\n",
+      "Q1= math.pi*v1*d**2/4\n",
+      "k= Q/Q1\n",
+      "P= (H1+H2)*H3/H4\n",
+      "P1= P-h3\n",
+      "r= P+h1-h2-h4\n",
+      "V= v1**2/(2*g)\n",
+      "E= r+V\n",
+      "dE= Et-E\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Coefficienct of venturi meter = %.4f '%(k)\n",
+      "print  ' Pressure of venturi throat = %.2f ft of water'%(P1)\n",
+      "print  ' Loss in energy = %.1f ft-lb/lb'%(dE)\n",
+      "\n",
+      "# Note : The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Coefficienct of venturi meter = 0.9587 \n",
+        " Pressure of venturi throat = 37.43 ft of water\n",
+        " Loss in energy = 9.8 ft-lb/lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch4.ipynb b/Problems_In_Hydraulics/ch4.ipynb
new file mode 100755
index 00000000..aefd97a7
--- /dev/null
+++ b/Problems_In_Hydraulics/ch4.ipynb
@@ -0,0 +1,291 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:f02a7c0105dc5d8192eae252809dedbf748fa59e0b9b78e4af3b1079610b7a69"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Orifices and Notches"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 Page No : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Cd= 0.98    # velocity\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "H= 2.    \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v= math.sqrt(2*g*H)\n",
+      "t= H/v\n",
+      "h= 0.5*g*t**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Vertical distance fallen in this ttime = %.3f ft'%(h) \n",
+      "\n",
+      "#Note : The answer given in textbook is wrong.\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Vertical distance fallen in this ttime = 0.500 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 Page No : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "r= 53.4\n",
+      "T= 60. \t    #F pressure of air\n",
+      "h= 29.7 \t#in of mercury\n",
+      "sm= 13.6\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "d= 1.5  \t#in diameter\n",
+      "Qin= 2. \t#cuses air\n",
+      "g=32.2 \t    #ft/s**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W= h*sm*w/(r*(460+T)*12)\n",
+      "dP= 0.75*w/(12*W)\n",
+      "Q= math.sqrt(2*g*dP)*math.pi*d**2/(4*144)\n",
+      "W= Q*W*60\n",
+      "Cd= Qin/W\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'coefficient of discharge = %.2f '%(Cd) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "coefficient of discharge = 0.62 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 Page No : 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "H1= 34. \t#ft height\n",
+      "H2= 8. \t    #ft head\n",
+      "H3= 7.   \t#ft pressure head\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d= 1.5 \t    #in\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v2= math.sqrt(2*g*(H1+H2-H3))\n",
+      "Q= v2*math.pi*d**2/(4*144)\n",
+      "v3= (2*v2+math.sqrt(4*v2**2-4*6*(v2**2-H2*2*5*g)))/12\n",
+      "dr= math.sqrt(v2/v3)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'ratio of diameteres = %.1f '%(dr) \n",
+      "print \" Maximum discharge = %.3f cusec\"%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of diameteres = 1.6 \n",
+        " Maximum discharge = 0.583 cusec\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 Page No : 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Q1= 8./15 \t#cuses\n",
+      "Q2= 2./15 \t#cuses\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= math.degrees(math.atan(Q2/Q1))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Angle of inclination = %.2f degrees'%(A) \n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Angle of inclination = 14.04 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 Page No : 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r= g**2/((math.sqrt(2))**2*g**2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'coefficient of contraction = %.1f '%(r) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "coefficient of contraction = 0.5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 Page No : 56"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "B= 3. \t    #ft long\n",
+      "H= 2.   \t#ft depth of water\n",
+      "H1= 3.75 \t#ft \n",
+      "w= 4.    \t#ft wide\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= 3.33*(B-(H1/5))*H**1.5\n",
+      "v= Q/(H*w)\n",
+      "kh= v**2/(2*g)\n",
+      "Q1= 3.33*(B-(H1/5)-kh)*(((H1/5)+kh)**1.5-kh**1.5)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Discharge = %.2f cuses'%(Q1) \n",
+      "\n",
+      "\n",
+      "# NOte : ANSWER IN THE TEXTBOOK IS WRONG\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge = 5.42 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch5.ipynb b/Problems_In_Hydraulics/ch5.ipynb
new file mode 100755
index 00000000..696c90f4
--- /dev/null
+++ b/Problems_In_Hydraulics/ch5.ipynb
@@ -0,0 +1,422 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:99b4244f7dba40fbe6b1f9647ffae37a043c932f7cb3e888c2ad7aac3ff9563e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Flow in Channels"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 Page No : 67"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "h= 2.5 \t#ft  depth of water\n",
+      "a= 45. \t#degrees  side slope\n",
+      "x= 5. \t#ft\n",
+      "Q= 45. \t#cuses\n",
+      "v= 2.6 \t#ft/sec velocity\n",
+      "w= 6.92 \t#ft \n",
+      "C= 120.\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "b= (Q/(v*h))-h\n",
+      "p= b+2*(h+math.sqrt(2))\n",
+      "A= h*w\n",
+      "m= A/p\n",
+      "i= (v/(C*math.sqrt(m)))**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Width = %.2f ft'%(b) \n",
+      "print  ' Slope = %.6f '%(i) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Width = 4.42 ft\n",
+        " Slope = 0.000332 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 Page No : 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "a= 60.   \t#degrees sides inclined\n",
+      "i= 1./1600\n",
+      "Q= 8.*10**6 \t#gal/hr discharge\n",
+      "M= 110.\n",
+      "w= 6.24 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIOS\n",
+      "d= ((Q*2**(2./3)*math.sqrt(1./i))/(w*3600*math.sqrt(3)*M))**(3./8)\n",
+      "b=6.93   \t#ft\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Diameter = %.f ft'%(d) \n",
+      "print ' breadth = %.2f ft'%(b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Diameter = 6 ft\n",
+        " breadth = 6.93 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 Page No : 71"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/swc**2\n",
+      "Q= 40. \t#cuses rate\n",
+      "w= 5.5 \t#ft\n",
+      "h= 9. \t#in  depth\n",
+      "d= 0.75 \t#ft\n",
+      "V= 3. \t#ft/sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "D= ((Q*2)**2/(g*(w*2)**2))**(1./3)\n",
+      "v= Q*d/w\n",
+      "D1= math.sqrt((2*v**2*d/g)+h/64)-(d/2)\n",
+      "dD= D1-d\n",
+      "El= -dD+((v**2*(1-(V/v)**2))/(2*g))\n",
+      "Els= Q*El*62.4/550\n",
+      "\n",
+      "#RESULTS\n",
+      "print 'Critical depth = %.2f ft'%(D)\n",
+      "print ' Rise in level = %.f ft'%(D1)\n",
+      "print  ' Horse-power lost = %.3f hp'%(Els) \n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical depth = 1.18 ft\n",
+        " Rise in level = 1 ft\n",
+        " Horse-power lost = 0.961 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6 Page No : 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "b= 3.5 \t#ft\n",
+      "H= 2.5 \t#ft\n",
+      "w= 3. \t#ft depth\n",
+      "h= 6. \t#ft  wide\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= 3.09*b*H**1.5\n",
+      "v= Q/(w*h)\n",
+      "H1= H+(v**2/(2*g))\n",
+      "Q1= 3.09*b*H1**1.5\n",
+      "hc= (Q1**2/(b**2*g))**(1./3)\n",
+      "h2= 0.5*(math.sqrt(hc**2+8*hc**2)-hc)\n",
+      "dh= h2+b-w\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"Flow rate  = %.1f cusecs\"%(Q)\n",
+      "print \" Flow rate  = %d cusecs\"%(Q1)\n",
+      "print  ' maximum depth of water downstream = %.3f ft'%(dh) \n",
+      "print  ' Shooting flow depth at hump = %.3f ft'%(h2) \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Flow rate  = 42.8 cusecs\n",
+        " Flow rate  = 45 cusecs\n",
+        " maximum depth of water downstream = 2.226 ft\n",
+        " Shooting flow depth at hump = 1.726 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7 Page No : 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "m= 60./26\n",
+      "i= 1./2000\n",
+      "h1= 3. \t#ft depth\n",
+      "h2= 5. \t#ft depth\n",
+      "m1= 10./3\n",
+      "C= 90.      # constant\n",
+      "l= 500. \t#ft depth\n",
+      "H= 20. \t#ft broad\n",
+      "H1= 29.62 \t#ft\n",
+      "g= 32.2 \t#ft/s**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v= 90*math.sqrt(m*i)\n",
+      "v1= v*h1/h2\n",
+      "dh= (i-(v1**2/(C**2*m1)))*l/(1-v1**2/(g*h2))\n",
+      "h3= h2-dh\n",
+      "V= h1*v/h3\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Height of water 1000 ft upstream = %.3f ft'%(h3) \n",
+      "print  ' Height of water upstream = %.3f ft'%(h3) \n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height of water 1000 ft upstream = 4.808 ft\n",
+        " Height of water upstream = 4.808 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8 Page No : 80"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "v= 5. \t    #ft/sec\n",
+      "m= 60./26\n",
+      "i= 1./2000\n",
+      "h= 5.5   \t#ft\n",
+      "m1= 110./31\n",
+      "d= 3. \t    #ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "C= v/(math.sqrt(m*i))\n",
+      "v1= v*d/h\n",
+      "r= (i-(v1**2/(C**2*m1)))/(1-(v1**2/(g*h)))\n",
+      "x= 1/r\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Distance upstream = %.f ft'%(round(x,-1)) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Distance upstream = 2380 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9 Page No : 81"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from numpy.linalg import *\n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "Q= 12 \t#cuses\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "hc= (Q/(3*math.sqrt(g)))**(2./3)\n",
+      "vec=roots([1,6,12,8,0,-8.95,-8.95])\n",
+      "H=vec[2]\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Critical depth = %.2f ft'%(hc) \n",
+      "print  ' Critical depth = %.2f ft'%(H) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Critical depth = 0.79 ft\n",
+        " Critical depth = 0.89 ft\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "-c:17: ComplexWarning: Casting complex values to real discards the imaginary part\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11 Page No : 85"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Cd= 0.64    # coefficient\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "A= 12.5 \t#ft**2\n",
+      "H= 24.8 \t#ft\n",
+      "Q= 3200. \t#cuses\n",
+      "b= 150. \t#ft  wide\n",
+      "A1= 5.*10**6  # avg surface area\n",
+      "h= 9.   \t#ft\n",
+      "h1= 6. \t    #in\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "N= Q/(Cd*A*math.sqrt(2*g*H))\n",
+      "H1= (Q/(3.2*b))**(2./3)\n",
+      "ES= (H1-(h1/12))*A1*h\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'number of siphons = %.f '%(N) \n",
+      "print  ' Extra Storage = %.2e ft**3'%(ES) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of siphons = 10 \n",
+        " Extra Storage = 1.37e+08 ft**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch6.ipynb b/Problems_In_Hydraulics/ch6.ipynb
new file mode 100755
index 00000000..5608cc77
--- /dev/null
+++ b/Problems_In_Hydraulics/ch6.ipynb
@@ -0,0 +1,757 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:cfc2d9c55b511981c268535b9b836c28d1aec92123c8d6e11efaae493bc2b6d5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Flow in Pipes"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 Page No : 95"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "l= 5000.     #ft long\n",
+      "l1= 2000.    #ft\n",
+      "d= 12.       #in diameter\n",
+      "f= 0.005     # coefficient\n",
+      "d1= 24       #in diameter\n",
+      "f1= 0.0045   \n",
+      "l2= 3000.    #ft\n",
+      "Q= 1800.     #gal/min flow\n",
+      "w= 6.24      #lb/ft**3\n",
+      "g=32.2       #ft/s**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "F= Q/(60*w)\n",
+      "v1= F*4/(math.pi*(d/12)**2)\n",
+      "v2= v1/(d1/d)**2\n",
+      "H= (f*l1*F**2/(10*(d/12)**5))+(f1*l2*F**2/(10*(d1/12)**5))+(v1**2/(4*g))+((v1-v2)**2/(2*g))+(v2**2/(2*g))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Available Head = %.2f ft'%(H)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Available Head = 24.74 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 Page No : 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "f= 0.01\n",
+      "h= 42.   \t#ft\n",
+      "l= 3200. \t#ft length\n",
+      "d= 14.   \t#in diameter\n",
+      "h1= 8.   \t#ft\n",
+      "l1= 1800. \t#ft point\n",
+      "w= 6.24 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v= math.sqrt(2*g*h/(1+0.5+(4*f*l/(d/12.))))\n",
+      "h2= h-h1-(v**2/(2*g))-h1-(0.5*v**2/(2*g))-(4*f*l1*v**2/(2*g*(d/12)))\n",
+      "Q= math.pi*(d/12)**2*v*w*60/4\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Height of siphon above A = %.2f ft'%(h2)\n",
+      "print  ' Total Discharge = %.f gal/min'%(Q)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Height of siphon above A = 2.13 ft\n",
+        " Total Discharge = 1974 gal/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 Page No : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "H= 950. \t#lb/in**2\n",
+      "l= 5.    \t#miles distance\n",
+      "d= 4.    \t#in\n",
+      "f= 0.0075    # friction\n",
+      "p= 92.   \t#per cent\n",
+      "hp= 200. \t#h.p power\n",
+      "g= 32.2 \t#ft/sec62\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "H1= H*2.3\n",
+      "H2= H1*100/p\n",
+      "Hf= H2-H1\n",
+      "v= math.sqrt(2*g*(d/12)*Hf/(4*f*l*5280))\n",
+      "n= hp/(w*v*(H1/550)*math.pi*(d/12)**2/4)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'number of pipes required = %.2f'%(n)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "number of pipes required = 4.07\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 Page No : 98"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "l= 1.5 \t    #miles length\n",
+      "d= 18. \t    #in diameter\n",
+      "Q= 12.4 \t#/cusecs\n",
+      "h= 130. \t#ft\n",
+      "r= 169.\n",
+      "r1= 338.\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "f= h*10*l**5/(l*5280*Q**2)\n",
+      "R= math.sqrt(1.5*r1-r)\n",
+      "d= math.sqrt(l**2/R*144)\n",
+      "v= math.sqrt(h*g*2/(r/R**2+1))\n",
+      "HP= w*0.25*math.pi*(d/12)**2*v**3/(550*2*g)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'f = %.3f '%(f)\n",
+      "print  ' Diameter of jet d = %.2f in'%(d)\n",
+      "print  ' Water h.p = %.1f h.p'%(HP)\n",
+      "\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "f = 0.008 \n",
+        " Diameter of jet d = 4.20 in\n",
+        " Water h.p = 70.6 h.p\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 Page No : 100"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "l= 5000. \t#ft long\n",
+      "d= 24. \t    #in diameter\n",
+      "Q= 18.  \t#cuses\n",
+      "t= 10. \t    #sec \n",
+      "P= 275000. \t#lb/in**2\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "w=62.4\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v= Q/(math.pi*(d/24)**2)\n",
+      "C= v/(t**2/2)\n",
+      "Pr= ((l*C*t/g)+(v**2/(2*g)))/2.3\n",
+      "Pr1= v*12*math.sqrt(w*P/(386.4*1728))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Pressure Rise = %.1f lb/in**2'%(Pr)\n",
+      "print  ' Pressure Rise = %d lb/in**2'%(Pr1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pressure Rise = 77.6 lb/in**2\n",
+        " Pressure Rise = 348 lb/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6 Page No : 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "v= 4.     \t#ft/sec velocity\n",
+      "K= 300000. \t#lb/in**2 water\n",
+      "d= 6. \t#in\n",
+      "t= 0.25 \t#in\n",
+      "E= 30*10**6 \t#lb/in**2\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "P= math.sqrt((w*v**2/g)/((d/(E*144*t))+(1/(K*144))))/144\n",
+      "Sm= P*d/(2*t)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Hoop stress = %.f lb/in**2'%(Sm)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hoop stress = 2739 lb/in**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7 Page No : 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "l1= 19. \t#ft\n",
+      "l2= 1.   \t#ft\n",
+      "r1= 0.298\n",
+      "r2= 0.238\n",
+      "r3= 0.359\n",
+      "r4= 0.242\n",
+      "r5= 0.121\n",
+      "d= 6 \t    #in diameter\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "m= -(-r4-math.sqrt(r4**2-4*(3*r1-r5)*(-(d/2)*r2-r3)))/(2*(3*r1-r5))\n",
+      "v2= math.sqrt((l1+l2)/(r1*m**2-r2))\n",
+      "v3= m*v2\n",
+      "Q2= math.pi*v2/d**2\n",
+      "Q3= math.pi*v3/d**2\n",
+      "Q= Q2+Q3\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Q2 = %.3f cusec'%(Q2)\n",
+      "print  ' Q3 = %.2f cusec'%(Q3)\n",
+      "print  ' Total Quantity = %.3f cusecs'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q2 = 0.711 cusec\n",
+        " Q3 = 0.96 cusec\n",
+        " Total Quantity = 1.668 cusecs\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8 Page No : 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "h= 80.   \t#ft levels\n",
+      "f= 0.008    # friction coefficient\n",
+      "l= 3000. \t#ft long\n",
+      "r1= 6.07\n",
+      "r2= 377.5\n",
+      "r3= 4733. \n",
+      "r4= 0.0466\n",
+      "r5= 3220.\n",
+      "r6= 51.5\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= math.sqrt(h*10/(f*l))\n",
+      "Q1= math.sqrt(r2+math.sqrt(r2**2-4*r1*r3)/(2*r1))/3\n",
+      "Q2= Q1-r4*math.sqrt(r5-r6*Q1**2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'rate discharge when valve B is closed= %.2f cusecs'%(Q)\n",
+      "print  ' Flow in reservoir= %.2f cusecs'%(Q2)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate discharge when valve B is closed= 5.77 cusecs\n",
+        " Flow in reservoir= 5.13 cusecs\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.9 Page No : 108"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "Q= 450.  \t#gal/min\n",
+      "w= 6.24 \t#lb/ft**3\n",
+      "f= 0.005\n",
+      "l1= 1000. \t#ft from reservoir A\n",
+      "l2= 2000. \t#ft from reservoir D\n",
+      "r1= 1.6\n",
+      "r2= 4.4\n",
+      "r3= 0.8\n",
+      "r4 = 12.85\n",
+      "h1= 59.1 \t#ft\n",
+      "h2= 40.19 \t#ft\n",
+      "v= 1.2 \t    #ft/sec\n",
+      "f= 0.0056\n",
+      "l= 10 \t    #ft  below reservoir A\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q1=  Q/(w*60)\n",
+      "Q2= (r1+math.sqrt(r1**2+4*r2))/2\n",
+      "Q3= Q2-Q1\n",
+      "Q4= (-r3+math.sqrt(r3**2+4*r4))/2\n",
+      "Q5= Q4+Q1\n",
+      "d= (f*5500*v**2/(l*(h1-h2)))**0.2*12\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'flow in to reservoir B= %.2f cusecs'%(Q3)\n",
+      "print  ' flow in to reservoir D= %.1f cusecs'%(Q5)\n",
+      "print  ' diameter of MN= %.f in'%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "flow in to reservoir B= 1.84 cusecs\n",
+        " flow in to reservoir D= 4.4 cusecs\n",
+        " diameter of MN= 9 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.10 Page No : 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 2.5 \t#ft\n",
+      "a= 45. \t#degrees\n",
+      "Q= 69. \t#cuses\n",
+      "l= 30. \t#ft\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Ps= 0.25*math.pi*d**2*w*l/2240\n",
+      "Rs= Ps*math.sqrt((1-math.cos(math.radians(a)))*2)\n",
+      "W= Q*w/2240\n",
+      "v= Q*4/(math.pi*d**2)\n",
+      "Rd= W*v*math.sqrt(2*(1-math.cos(math.radians(a))))/g\n",
+      "Rt= Rs+Rd\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'total resultant thrust = %.3f tons'%(Rt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total resulmath.tant thrust = 3.782 tons\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.11 Page No : 112"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "r1= 1./3\n",
+      "r2= 7./12\n",
+      "l= 5000. \t#ft\n",
+      "l1= 10000. \t#ft\n",
+      "d= 27. \t#in\n",
+      "d1= 18. \t#in\n",
+      "Q= 10. \t#cuses\n",
+      "f= 0.006\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q2= Q/(math.sqrt(r2/r1)+1)\n",
+      "Q1= Q-Q2\n",
+      "H= (f*l*Q**2/(10*(d/12)**5))+(f*l1*Q1**2/(3*10**(d1/12)**5))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'total difference in head = %.2f ft'%(H)\n",
+      "\n",
+      "\n",
+      "#ANSWER GIVEN IN THE TEXTBOOK IS WRONG\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total difference in head = 5.20 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.12 Page No : 115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "V= 4. \t    #ft/sec\n",
+      "L= 1225. \t#ft\n",
+      "l= 1200. \t#ft\n",
+      "H= 50.   \t#ft\n",
+      "d= 1./3 \t#ft\n",
+      "f= 0.008\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a= 2*g*H\n",
+      "b= (4*f*L/d)+1.5\n",
+      "c= math.sqrt(a/b)\n",
+      "d= math.sqrt(a*b)\n",
+      "T= math.log(math.sqrt((c+V)/(c-V)))*l*2/d\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time interval for elapse = %.2f sec'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time interval for elapse = 3.95 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.14 Page No : 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "L= 8000. \t#ft\n",
+      "d= 5.    \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d= 5. \t    #ft\n",
+      "l= 250. \t#ft\n",
+      "b= 100.\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= math.pi*0.25*d**2*l-0.5*d**2*b\n",
+      "V= A*g/L\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Velocity = %.2f ft/sec'%(V)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity = 14.73 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.15 Page No : 121"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "B= 3.    \t#ft\n",
+      "Cd= 0.6\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d1= 6. \t    #in\n",
+      "d2= 4.   \t#in\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q2= 0.428 \t#cuses\n",
+      "r= math.sqrt((((d1/12)**5)/((d2/12)**5)))\n",
+      "Q1= r*Q2\n",
+      "Q= Q1+Q2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Total inflow = %.3f cuses'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total inflow = 1.607 cuses\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.17 Page No : 124"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "f= 0.007\n",
+      "l= 30.        \t#miles\n",
+      "Q1= 5.*10**6 \t#gal/day\n",
+      "w= 6.24 \t    #lb/ft**3\n",
+      "H= 500. \t    #ft\n",
+      "Q2= 7*10**6 \t#gal/day\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Qi= Q1/(w*24*3600)\n",
+      "d= (f*l*5280*Qi**2/(10*H))**0.2\n",
+      "Qe = Q2*Qi/Q1\n",
+      "x= (30-(H*10*d**5/(f*Qe**2*5280)))*(4./3)\n",
+      "\n",
+      "#RESULTS\n",
+      "print 'diameter of pipes = %.1f ft'%d\n",
+      "print  'length of new pipe required = %.1f miles'%(x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diameter of pipes = 1.8 ft\n",
+        "length of new pipe required = 19.6 miles\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch7.ipynb b/Problems_In_Hydraulics/ch7.ipynb
new file mode 100755
index 00000000..b5eb3280
--- /dev/null
+++ b/Problems_In_Hydraulics/ch7.ipynb
@@ -0,0 +1,453 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8fbfdaa034f011c9304b20c16adbf915d0a8f58f4b89051b6e8db7227dd1440f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Flow Under Varying Head"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 Page No : 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad\n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d= 6.    \t#ft\n",
+      "di= 2. \t    #in\n",
+      "h= 9. \t    #ft\n",
+      "Cd= 0.6\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "def fun(H):\n",
+      "    return H**-0.5*(d/2)**2*math.pi/(Cd*math.pi*math.sqrt(2*g)/144)\n",
+      "\n",
+      "\n",
+      "vec2=quad(fun,0,h)\n",
+      "T= vec2[0]\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Time to emptify = %.f sec'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time to emptify = 1615 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 Page No : 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad\n",
+      "\n",
+      "#initialisation of variables\n",
+      "d1= 4.  \t#ft\n",
+      "d2= 2. \t    #in\n",
+      "l= 300. \t#ft\n",
+      "P= 5. \t    #lb/in**2\n",
+      "h1= 3. \t    #ft\n",
+      "h2= 6. \t    #ft\n",
+      "f= 0.01\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "X= P*2.31*10*(d2/12)**5/(f*l)\n",
+      "A= math.pi*d1**2/4\n",
+      "\n",
+      "def fun(h):\n",
+      "    return A*math.sqrt((P*2.31*10*(d2/12)**5/(f*l))-(10*(d2/12)**5*h/(f*l)))/(10*(d2/12)**5/(f*l))/7\n",
+      "\n",
+      "vec2=quad(fun,h1,h2)\n",
+      "T= vec2[0]\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time for the channel to fall = %.2f sec'%(T)\n",
+      "\n",
+      "# rounding error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for the channel to fall = 689.35 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 Page No : 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.integrate import quad\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 10. \t#in\n",
+      "l= 15. \t#ft\n",
+      "di= 3. \t#in\n",
+      "Cd= 0.62 \n",
+      "g=32.2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "def fun(H):\n",
+      "    return -l*2*math.sqrt((d/2)**2-((d/2)-H)**2)/(Cd*(math.pi*(di/12)**2/4)*H**0.5*math.sqrt(2*g))\n",
+      "\n",
+      "vec2=quad(fun,0,d/2)\n",
+      "T= vec2[0]\n",
+      "secs = -T%60\n",
+      "mins = -T/60\n",
+      "#RESULTS\n",
+      "print  'time for the channel to fall = %d mins and %d seconds'%(mins,secs)\n",
+      "\n",
+      "# rounding error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for the channel to fall = 27 mins and 54 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 Page No : 142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad\n",
+      "\n",
+      "#initialisation of variables\n",
+      "h= 4. \t#ft\n",
+      "w= 6. \t#ft\n",
+      "l= 100. \t#yd\n",
+      "a= 60. \t#degrees\n",
+      "h1= 3. \t#ft\n",
+      "h2= 2. \t#ft\n",
+      "Cd= 0.6\n",
+      "g=32.2   \t#ft/s**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "A= l*3*w\n",
+      "def fun(H):\n",
+      "    return -A*H**-2.5/(Cd*(8./15)*(math.tan(math.radians(a/2)))*math.sqrt(2*g))\n",
+      "\n",
+      "vec2=quad(fun,h1,(h1-h2))\n",
+      "T= vec2[0]\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time for the channel to fall = %.f sec'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for the channel to fall = 654 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 Page No : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "A= 1./16 \t#mile**2\n",
+      "d= 2. \t#ft\n",
+      "h= 18. \t#ft\n",
+      "h1= 5. \t#ft\n",
+      "f= 0.006\n",
+      "l= 200. \t#ft\n",
+      "h2= 10. \t#ft\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "X= math.sqrt(1./((1.5+(4*f*l/d))/(2*g)))\n",
+      "def fun(H):\n",
+      "    return A*5280**2*H**-0.5/(math.pi*d**2*X/4)\n",
+      "\n",
+      "vec2=quad(fun,h-h1,h)\n",
+      "T= vec2[0]\n",
+      "hours = T/3600\n",
+      "mins = T%3600/60\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time for the channel to fall = %d hours and %d mins sec'%(hours,round(mins,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for the channel to fall = 48 hours and 20 mins sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 Page No : 144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "l= 8. \t#ft\n",
+      "b= 6. \t#ft\n",
+      "h= 10. \t#ft\n",
+      "r= 3.\n",
+      "Cd= 0.6\n",
+      "A1= 36. \t#ft**2\n",
+      "A2= 12. \t#ft**2\n",
+      "l1= 6. \t#ft\n",
+      "h1= 1. \t#ft\n",
+      "d= 2. \t#in\n",
+      "g=32.2 \t#ft/s**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "def fun(H):\n",
+      "    return H**-0.5/(Cd*(math.pi*(d/12)**2/4)*math.sqrt(2*g)*((1/A1)+(1/A2)))\n",
+      "\n",
+      "vec2=quad(fun,0,(b-h1))\n",
+      "T= vec2[0]\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'time for the levels to become equal = %.f sec'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for the levels to become equal = 383 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 Page No : 145"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "h1= 3. \t#ft\n",
+      "h2= 4. \t#ft diameter\n",
+      "r= 0.95 \t#m**-1\n",
+      "k= 27.65 \t#sec\n",
+      "Cd= 0.95\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "T= k*(math.log(r*math.sqrt(h2)-1)+(r*math.sqrt(h2)-1))-k*(math.log(r*math.sqrt(h1)-1)+(r*math.sqrt(h1)-1))\n",
+      "h= ((h2-h1)/Cd)**2\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Time = %.2f sec'%(T)\n",
+      "print  ' Increase in water level = %.2f ft'%(h)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Time = 16.23 sec\n",
+        " Increase in water level = 1.11 ft\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8 Page No : 146"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "t= 75. \t#sec\n",
+      "h= 10.5 \t#in constant\n",
+      "h1= 13.5 \t#in\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "r= t*math.pi*math.sqrt(2*h**2)/math.log((math.sqrt(2*h1**2)+h1)/(math.sqrt(2*h**2)-h))\n",
+      "t= -r*((1/h1)-(1/h))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'A/K = %.f '%(r)\n",
+      "print  ' Time taken = %.1f sec'%(t)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A/K = 1737 \n",
+        " Time taken = 36.8 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.9 Page No : 148"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "g= 9.8 \t#m/sec**2\n",
+      "h1= 10. \t#in\n",
+      "h2= 12. \t#in\n",
+      "r1= 1.32\n",
+      "r2= 1.56\n",
+      "r3= 1.97\n",
+      "r4= 4.10\n",
+      "r5= 2.64\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= math.sqrt(32.2)*(h2/18)**1.5\n",
+      "T= 10**5*(r1+2*r3+r4+4*(r3+r5))/(6*h2*60*60)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Actual discharge = %.2f CBH**1.5 cuses'%(Q)\n",
+      "print  ' Time = %.1f  hr'%(T)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Actual discharge = 3.09 BH**1.5 cuses\n",
+        " Time = 10.7  hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch8.ipynb b/Problems_In_Hydraulics/ch8.ipynb
new file mode 100755
index 00000000..9a5c856c
--- /dev/null
+++ b/Problems_In_Hydraulics/ch8.ipynb
@@ -0,0 +1,296 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:35125badba4aaa9b528697efd91fa6b7acc4c2f1f63bb28106c08928483bd0b9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 :Viscosity and Viscous Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "v= 10.01 \t#poise velocity\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "d= 30.48 \t#cm\n",
+      "w= 453.6 \t#gm\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "M= round(v*d/w,3)\n",
+      "F= M/g\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Pound in unit of mass = %.3f lb/ft sec absolute units'%(M)\n",
+      "print  ' Pound in unit of force = %.4f slugs/ft sec'%(F)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Pound in unit of mass = 0.673 lb/ft sec absolute units\n",
+        " Pound in unit of force = 0.0209 slugs/ft sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 Page No : 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "W= 20.   \t#tons/hr oil\n",
+      "l= 1000. \t#ft long\n",
+      "w= 57.   \t#lb/ft**3 weighs\n",
+      "kv= 0.0205 \t#ft**2/sec kinematic viscisity\n",
+      "d= 6.    \t#in diameter\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= W*2240/(3600*w)\n",
+      "A= math.pi*(d/12)**2/4\n",
+      "v= Q/A\n",
+      "R= v*(d/12)/kv\n",
+      "n= w*kv/g\n",
+      "P= 32*v*n*l/((d/12)**2*w)\n",
+      "HP= P*2240*W/(3600*500)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Reynolds number = %.1f '%(R)\n",
+      "print  ' H.P required = %.2f hp'%(HP)\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reynolds number = 27.1 \n",
+        " H.P required = 2.26 hp\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 Page No : 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "n= 0.0067 \t#poise\n",
+      "l= 10. \t#ft length\n",
+      "w= 62. \t#lb/ft**3 density\n",
+      "d= 1. \t#in\n",
+      "Q= 2. \t#ft**2/sec\n",
+      "sm= 13.57\n",
+      "k1= 0.003\n",
+      "k2= 0.0725\n",
+      "r= 0.3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n1= n*30.48/453.6\n",
+      "v= Q*4/(60*math.pi*(d/12)**2)\n",
+      "RN= v*(d/12)*w/n1\n",
+      "f= k1+(k2/RN**r)\n",
+      "hf= 4*f*l*v**2/(2*g*(d/12))\n",
+      "hl= hf*12/sm\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Head lost in inches of mercury = %.2f in'%(hl)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Head lost in inches of mercury = 1.37 in\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5 Page No : 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "n= 0.91 \t#poise\n",
+      "g= 32.2 \t#ft/sec\n",
+      "N= 300. \t#r.p.m\n",
+      "t= 0.01 \t#in\n",
+      "r1= 0.25 \t#ft\n",
+      "r2= 1./6 \t#ft\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n1= n*30.48/(454*g)\n",
+      "A= N*2*math.pi/60\n",
+      "t1= t/12\n",
+      "hp= math.pi*A**2*n1*(r1**4-r2**4)/(t1*1100)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Horse Power lost = %.4f '%(hp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Horse Power lost = 0.0201 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6 Page No : 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "vw= 0.3 \t#ft/sec\n",
+      "dw= 1.  \t#in\n",
+      "da= 12. \t#in\n",
+      "ww= 62.3 \t#lb/ft**3\n",
+      "wa= 0.075 \t#lb/ft**3\n",
+      "nw= 0.01 \t#poise\n",
+      "na= 0.00018 \t#poise\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "va= vw*dw*ww*na/(nw*da*wa)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'critical velocity of air = %.3f ft/sec'%(va)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "critical velocity of air = 0.374 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7 Page No : 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "dm= 0.75 \t#in\n",
+      "dt= 0.25 \t#in\n",
+      "dP= 10.4 \t#lb/in**2\n",
+      "rd= 0.84\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "v1= math.sqrt(dP*144*g/(rd*w*((dm/dt)**4-1)))\n",
+      "Q= math.pi*dm**2*v1*60*w/(4*144*10)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Discharge rate = %.1f gal.min'%(Q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Discharge rate = 3.9 gal.min\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/ch9.ipynb b/Problems_In_Hydraulics/ch9.ipynb
new file mode 100755
index 00000000..8afae5a1
--- /dev/null
+++ b/Problems_In_Hydraulics/ch9.ipynb
@@ -0,0 +1,323 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:62ceb66cf48441d19499a6332c57808da0a2f8834a586c75e2d7d1a835c045df"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Impact of Jets"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 Page No : 171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 2. \t    #in\n",
+      "V= 210. \t#ft/sec\n",
+      "V1= 50. \t#ft/sec\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "w= 62.4 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "M= math.pi*V*w/(4*36*g)\n",
+      "F= M*V\n",
+      "dV= V-V1\n",
+      "M1= math.pi*dV*w/(4*36*g)\n",
+      "F1= M1*dV\n",
+      "W= F1*V1\n",
+      "F2= M*dV\n",
+      "W1= F2*V1\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Force on plate = %.f lb'%(F+1)\n",
+      "print  ' Force on plate = %.f lb'%(F1)\n",
+      "print  ' Work done/sec = %.f ft-lb/sec'%(W)\n",
+      "print  ' Force on plate = %.f lb'%(F2)\n",
+      "print  ' Work done/sec = %.f ft-lb/sec'%(round(W1,-3))\n",
+      "\n",
+      "#The answer is a bit different due to rounding off error in textbook\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Force on plate = 1865 lb\n",
+        " Force on plate = 1082 lb\n",
+        " Work done/sec = 54116 ft-lb/sec\n",
+        " Force on plate = 1421 lb\n",
+        " Work done/sec = 71000 ft-lb/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 Page No : 172"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "v1= 15. \t#ft/sec\n",
+      "v2= 40. \t#ft/sec\n",
+      "a= 30.  \t#degrees\n",
+      "b= 150. \t#degrees\n",
+      "v= 15.27 \t#ft/sec\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "a1= a-math.degrees(math.sin(v1*math.sin(math.radians(b))/v2))\n",
+      "w= math.cos(math.radians(a1))*v2\n",
+      "vr= v2*math.sin(math.radians(a1))/math.sin(math.radians(a))\n",
+      "v1= math.sqrt(v1**2+vr**2-2*v1*vr*math.cos(math.radians(a)))\n",
+      "r= 180-math.sin(math.radians(a))*vr/v\n",
+      "w1= v*math.cos(math.radians(r))\n",
+      "W= v1*(w-w1)/g\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'a = %.2f degrees'%(a1)\n",
+      "print  ' w = %.2f ft/sec'%(w)\n",
+      "print  ' vr = %.2f ft/sec'%(vr)\n",
+      "print  ' v1 = %.2f ft/sec'%(v1)\n",
+      "print  ' w = %.2f ft/sec'%(w)\n",
+      "print  ' Work done per pound = %.2f ft-lb/lb'%(W)\n",
+      "\n",
+      "# Note : Answers are different because of rounding off error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a = 19.32 degrees\n",
+        " w = 37.75 ft/sec\n",
+        " vr = 26.47 ft/sec\n",
+        " v1 = 15.42 ft/sec\n",
+        " w = 37.75 ft/sec\n",
+        " Work done per pound = 25.39 ft-lb/lb\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 Page No : 173"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "d= 0.5    \t#in\n",
+      "a= 165. \t#degrees\n",
+      "W= 7.35 \t#lb\n",
+      "W1= 500. \t#lb\n",
+      "t= 148. \t#sec\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "w= 62.3 \t#lb/ft**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "Q= W1/(t*w)\n",
+      "v= Q*16*144/math.pi\n",
+      "dv= v*(1-math.cos(math.radians(a)))\n",
+      "F= dv*W1/(t*g)\n",
+      "r= W/F\n",
+      "k= (1-(W*t*g/(W1*v)))/math.cos(math.radians(a))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'k = %.3f '%(k)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "k = 0.788 \n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 Page No : 174"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "t= 0.25 \t#in\n",
+      "a= 30.   \t#degrees\n",
+      "w= 480. \t#lb/ft**3\n",
+      "h= 2.    \t#in\n",
+      "d= 0.5  \t#in\n",
+      "l= 6.    \t#in\n",
+      "w1= 62.4 \t#lb/ft**3\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W= t*l**2*w/1728\n",
+      "M= w1*math.pi*d**2*math.cos(math.radians(a))/(g*4*144)\n",
+      "v= math.sqrt(W*(l/2)*math.sin(math.radians(a))/(M*2*(1./math.cos(math.radians(a)))))\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Velocity of jet = %.1f ft/sec'%(v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of jet = 26.6 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 Page No : 176"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "V= 90. \t#ft/sec\n",
+      "a= 30. \t#degrees\n",
+      "u= 45. \t#ft/sec\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "w= V*math.cos(math.radians(a))\n",
+      "f= math.sqrt(V**2-w**2)\n",
+      "tanb= (math.atan(math.radians(f/(w-u))))\n",
+      "b = math.degrees(math.tan(tanb))\n",
+      "b = math.degrees(math.atan(b))\n",
+      "V1= math.sqrt(f**2+(u-f*1./math.tan(math.radians(b)))**2)\n",
+      "\n",
+      "#RESULTS\n",
+      "print \"B = %.2f degrees\"%b\n",
+      "print  'absolute velocity of water at the exit = %.1f ft/sec'%(V1)\n",
+      "\n",
+      "# rounding off error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "B = 53.79 degrees\n",
+        "absolute velocity of water at the exit = 46.6 ft/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 Page No : 177"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "#initialisation of variables\n",
+      "u= 734. \t#ft/sec\n",
+      "v= 2000. \t#ft/sec\n",
+      "g= 32.2 \t#ft/sec**2\n",
+      "da= 0.019 \t#kg/m**3\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "W= g*v/(v-u)\n",
+      "A= W/(u*da)\n",
+      "\n",
+      "#RESULTS\n",
+      "print  'Weight of the air = %.1f lb/sec'%(W)\n",
+      "print  ' Area of inlet = %.2f ft**2'%(A)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Weight of the air = 50.9 lb/sec\n",
+        " Area of inlet = 3.65 ft**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Problems_In_Hydraulics/screenshots/1Hydrostatics.png b/Problems_In_Hydraulics/screenshots/1Hydrostatics.png
new file mode 100755
index 00000000..816d8250
--- /dev/null
+++ b/Problems_In_Hydraulics/screenshots/1Hydrostatics.png
diff --git a/Problems_In_Hydraulics/screenshots/4OrificesAndNotches.png b/Problems_In_Hydraulics/screenshots/4OrificesAndNotches.png
new file mode 100755
index 00000000..4c71e1c3
--- /dev/null
+++ b/Problems_In_Hydraulics/screenshots/4OrificesAndNotches.png
diff --git a/Problems_In_Hydraulics/screenshots/7FlowUnderVaryingHead.png b/Problems_In_Hydraulics/screenshots/7FlowUnderVaryingHead.png
new file mode 100755
index 00000000..23e7a483
--- /dev/null
+++ b/Problems_In_Hydraulics/screenshots/7FlowUnderVaryingHead.png
diff --git a/Programming_in_C_using_ANSI_C/KamthaneChapter10.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter10.ipynb
new file mode 100755
index 00000000..68503d70
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter11.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter11.ipynb
new file mode 100755
index 00000000..c8fac375
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter12.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter12.ipynb
new file mode 100755
index 00000000..e1acaa2c
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter13.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter13.ipynb
new file mode 100755
index 00000000..f9222003
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter14.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter14.ipynb
new file mode 100755
index 00000000..409585bd
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter15.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter15.ipynb
new file mode 100755
index 00000000..0c5ac02f
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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",
+      "%pylab 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/KamthaneChapter16.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter16.ipynb
new file mode 100755
index 00000000..cafb9828
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter3.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter3.ipynb
new file mode 100755
index 00000000..880e4fac
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter4.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter4.ipynb
new file mode 100755
index 00000000..758457a3
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter5.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter5.ipynb
new file mode 100755
index 00000000..dac1632e
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter6.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter6.ipynb
new file mode 100755
index 00000000..d9815e79
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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",
+        "1  5  22\r",
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+       "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/KamthaneChapter7.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter7.ipynb
new file mode 100755
index 00000000..ac82414f
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter8.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter8.ipynb
new file mode 100755
index 00000000..3243f3f8
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/KamthaneChapter9.ipynb b/Programming_in_C_using_ANSI_C/KamthaneChapter9.ipynb
new file mode 100755
index 00000000..4da93c8d
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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/README.txt b/Programming_in_C_using_ANSI_C/README.txt
new file mode 100755
index 00000000..e5dfcad4
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/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
diff --git a/Programming_in_C_using_ANSI_C/screenshots/screenshot1.png b/Programming_in_C_using_ANSI_C/screenshots/screenshot1.png
new file mode 100755
index 00000000..61fbfcd0
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/screenshots/screenshot1.png
diff --git a/Programming_in_C_using_ANSI_C/screenshots/screenshot1_1.png b/Programming_in_C_using_ANSI_C/screenshots/screenshot1_1.png
new file mode 100755
index 00000000..61fbfcd0
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/screenshots/screenshot1_1.png
diff --git a/Programming_in_C_using_ANSI_C/screenshots/screenshot2.png b/Programming_in_C_using_ANSI_C/screenshots/screenshot2.png
new file mode 100755
index 00000000..df49d844
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/screenshots/screenshot2.png
diff --git a/Programming_in_C_using_ANSI_C/screenshots/screenshot2_1.png b/Programming_in_C_using_ANSI_C/screenshots/screenshot2_1.png
new file mode 100755
index 00000000..df49d844
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/screenshots/screenshot2_1.png
diff --git a/Programming_in_C_using_ANSI_C/screenshots/screenshot3.png b/Programming_in_C_using_ANSI_C/screenshots/screenshot3.png
new file mode 100755
index 00000000..42d9aec9
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/screenshots/screenshot3.png
diff --git a/Programming_in_C_using_ANSI_C/screenshots/screenshot3_1.png b/Programming_in_C_using_ANSI_C/screenshots/screenshot3_1.png
new file mode 100755
index 00000000..42d9aec9
--- /dev/null
+++ b/Programming_in_C_using_ANSI_C/screenshots/screenshot3_1.png
diff --git a/Structured_Programing_with_C++/Chapter1.ipynb b/Structured_Programing_with_C++/Chapter1.ipynb
new file mode 100755
index 00000000..4a8d6e2e
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter1.ipynb
@@ -0,0 +1,104 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:c86efa8d98a193b995064bd56ef312e9a7c9cee5247409d33d46defcf953f275"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Introduction to Programming"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1, Page No 21"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Hello world\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Hello world\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2, Page No 22"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "dblPrice = float(raw_input(\"Enter Price Per Unit : \"))\n",
+      "iNo = int(raw_input(\"Enter Quantity : \"))\n",
+      "dblTotal = dblPrice * iNo\n",
+      "print \"The Total Price Is : %d\" % dblTotal "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Price Per Unit : 100\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Quantity : 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Total Price Is : 1000\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter2.ipynb b/Structured_Programing_with_C++/Chapter2.ipynb
new file mode 100755
index 00000000..d373f52c
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter2.ipynb
@@ -0,0 +1,288 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:8f653af2e0b44c81d67d541bdc64870a18188d9d1a401e0b96b1b859a01ca59f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapeter 2 : Variables"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page No 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iNo = raw_input(\"Specify Quantity: \")\n",
+      "print iNo"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify Quantity: 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page No 31"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iNo = raw_input(\"Specify Quantity: \")\n",
+      "print \"You entered\", iNo"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specify quantity22\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You entered 22\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3, Page No 32"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iNo = raw_input(\"specify quantity: \")\n",
+      "dPrice = raw_input(\"Specify unit Price: \")\n",
+      "print \"You entered the quantity \", iNo , \" and the price \" , dPrice"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specify quantity: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify unit Price: 12.45\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You entered the quantity  5  and the price  12.45\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page No 36"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "dTaxPerc = 25.0\n",
+      "iNo = int(raw_input(\"Specify Quantity: \"))\n",
+      "dUnitPr = float(raw_input(\"Specify Unit Price: \"))\n",
+      "dPriceExTax = dUnitPr * iNo\n",
+      "dTax = dPriceExTax * dTaxPerc / 100\n",
+      "dCustPrice = dPriceExTax + dTax\n",
+      "print \"INVOICE\"\n",
+      "print \"========\"\n",
+      "print \"Quantity: \", iNo\n",
+      "print \"Price per Unit: \", dUnitPr\n",
+      "print \"Total Price: \", dCustPrice\n",
+      "print \"Tax: \",dTax"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify Quantity: 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify Unit Price: 12.4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "INVOICE\n",
+        "========\n",
+        "Quantity:  5\n",
+        "Price per Unit:  12.4\n",
+        "Total Price:  77.5\n",
+        "Tax:  15.5\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5, Page No 38"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iNoOfSec = int(raw_input(\"Speicfy No. of Seconds: \"))\n",
+      "iNoOfMin = iNoOfSec / 60\n",
+      "iSecLeft = iNoOfSec % 60\n",
+      "iNoOfHours = iNoOfMin / 60\n",
+      "iMinLeft = iNoOfMin % 60\n",
+      "print \"Number of hours = \", iNoOfHours\n",
+      "print \"Number of Minutes = \", iNoOfMin\n",
+      "print \"Number of Seconds = \", iSecLeft"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Speicfy No. of Seconds: 10000\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of hours =  2\n",
+        "Number of Minutes =  166\n",
+        "Number of Seconds =  40\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6, Page No 41"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "iNo = 5\n",
+      "iRoll1 = random.randint(0,100) % 6 + 1\n",
+      "iRoll2 = random.randint(0,100) % 6 + 1\n",
+      "iRoll3 = random.randint(0,100) % 6 + 1\n",
+      "iRoll4 = random.randint(0,100) % 6 + 1\n",
+      "iRoll5 = random.randint(0,100) % 6 + 1\n",
+      "dAverage = float(iRoll1 + iRoll2 + iRoll3 + iRoll4 + iRoll5) / iNo;\n",
+      "print \"Number of Rolls: \", iNo\n",
+      "print \"Average Score: \", dAverage"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of Rolls:  5\n",
+        "Average Score:  3.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter3.ipynb b/Structured_Programing_with_C++/Chapter3.ipynb
new file mode 100755
index 00000000..bca740d6
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter3.ipynb
@@ -0,0 +1,540 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:16d6a5d7a0f2261d7c4bbdc27385a261d6311ef3eb7e569b90710460f45fff6c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : Selections and Loops"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page No 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "dLimit = 500\n",
+      "iNo = int(raw_input(\"Specify Quantity : \"))\n",
+      "dUnitPrice = float(raw_input(\"Specify Unit Price : \"))\n",
+      "dGross = iNo * dUnitPrice\n",
+      "if(dGross > dLimit):\n",
+      "    dDisc = 10\n",
+      "else:\n",
+      "    dDisc = 0;\n",
+      "dNet = (100 - dDisc) * dGross / 100\n",
+      "print \"Total Price : %d\" % dNet"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify Quantity : 100\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify Unit Price : 10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Total Price : 900\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page No 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "dNo1 = int(raw_input(\"Enter Number 1 : \"))\n",
+      "dNo2 = int(raw_input(\"Enter Number 2 : \"))\n",
+      "\n",
+      "print \"1. Greatest\"\n",
+      "print \"2. Least\"\n",
+      "print \"3. Average\"\n",
+      "\n",
+      "iOpt = int(raw_input(\"Select : \"))\n",
+      "\n",
+      "if iOpt == 1:\n",
+      "    if dNo1 > dNo2:\n",
+      "        print dNo1\n",
+      "    else:\n",
+      "        print dNo2\n",
+      "    print \" is the greatest\"\n",
+      "elif iOpt == 2:\n",
+      "    if dNo1 < dNo2:\n",
+      "        print dNo1\n",
+      "    else:\n",
+      "        print dNo2\n",
+      "    print \" is the least\"\n",
+      "elif iOpt == 3:\n",
+      "    print \"The average is : \" , (dNo1 + dNo2)/2\n",
+      "else:\n",
+      "    print \"wrong Choice\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number 1 : 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Number 2 : 15\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1. Greatest\n",
+        "2. Least\n",
+        "3. Average\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Select : 1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "15\n",
+        " is the greatest\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page No 58"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iLimit = int(raw_input(\"Enter limit : \"))\n",
+      "iSum = 0\n",
+      "for i in range(1,iLimit+1):\n",
+      "    iSum = iSum + i\n",
+      "\n",
+      "print \"The Sum Is : \" , iSum"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    },
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page No 61"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Calculation Of Product\"\n",
+      "\n",
+      "for i in range(1,37):\n",
+      "    for j in range(1,37):\n",
+      "        if (i * j) == 36:\n",
+      "            print i , \" and \" , j"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Calculation Of Product\n",
+        "1  and  36\n",
+        "2  and  18\n",
+        "3  and  12\n",
+        "4  and  9\n",
+        "6  and  6\n",
+        "9  and  4\n",
+        "12  and  3\n",
+        "18  and  2\n",
+        "36  and  1\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5, Page No 62"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "\n",
+      "iRoll = 0\n",
+      "iNoOfRolls = 0\n",
+      "\n",
+      "while iRoll != 6:\n",
+      "    iRoll =  random.randint(1,200) % 6 + 1\n",
+      "    iNoOfRolls = iNoOfRolls + 1\n",
+      "\n",
+      "print iNoOfRolls"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "8\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6, Page No 63"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Do while concept is not in python thats way the program is implemented using WHILE\n",
+      "import random\n",
+      "\n",
+      "iRoll = 0\n",
+      "iNoOfRolls = 0\n",
+      "\n",
+      "while iRoll != 6:\n",
+      "    iRoll =  random.randint(1,200) % 6 +1\n",
+      "    iNoOfRolls = iNoOfRolls + 1\n",
+      "\n",
+      "print iNoOfRolls"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "1\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7, Page No 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "\n",
+      "iCounter = 0\n",
+      "iRoll1 = random.randint(1,200) % 6 + 1\n",
+      "iRoll2 = random.randint(1,200) % 6 + 1\n",
+      "    \n",
+      "\n",
+      "while iRoll1 != iRoll2:\n",
+      "    iRoll1 = random.randint(1,200) % 6 + 1\n",
+      "    iRoll2 = random.randint(1,200) % 6 + 1\n",
+      "    iCounter = iCounter + 1\n",
+      "\n",
+      "print \"The rolls were = \" , iRoll1\n",
+      "print \"Number of attempts = \" , iCounter"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The rolls were =  4\n",
+        "Number of attempts =  2\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8, Page No 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iSum = 0\n",
+      "i = 0\n",
+      "iNo = raw_input(\"Enter a number : \")\n",
+      "\n",
+      "while iNo.isdigit():\n",
+      "    iSum = iSum + int(iNo)\n",
+      "    i = i + 1\n",
+      "    iNo = raw_input(\"Enter one more number : \")\n",
+      "\n",
+      "print \"Average = \" , (float)((iSum)/i)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter one more number : 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter one more number : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter one more number : c\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average =  5.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9, Page No 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "iNO = 0\n",
+      "for i in range(1,14):\n",
+      "    iNo = random.randint(1,200) % 3\n",
+      "    if iNo == 0:\n",
+      "        print \"1\"\n",
+      "    elif iNo == 1:\n",
+      "        print \"X\"\n",
+      "    elif iNO == 2:\n",
+      "        print \"2\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "X\n",
+        "X\n",
+        "X\n",
+        "X\n",
+        "1\n",
+        "1\n",
+        "X\n",
+        "1\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10, Page No 68"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iNo = 0\n",
+      "iRoot = 1\n",
+      "while iNo < 2 and iRoot <=100:\n",
+      "    LP = iRoot * (iRoot - 6) * (iRoot + 8)\n",
+      "    if LP == 0:\n",
+      "        iNo = iNo + 1\n",
+      "        print iRoot\n",
+      "    iRoot = iRoot + 1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "6\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11, Page No 69"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "while 1 == 1:\n",
+      "    dNo = int(raw_input(\"Enter a number : \"))\n",
+      "    if dNo <= 0:\n",
+      "        break\n",
+      "    print \"The square root of the number is \",math.sqrt(dNo)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The square root of the number is  10.0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a number : 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter4.ipynb b/Structured_Programing_with_C++/Chapter4.ipynb
new file mode 100755
index 00000000..aff2c5eb
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter4.ipynb
@@ -0,0 +1,563 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:a27de4c3e22de467837d62cdd94a3b4f20a866984babfe11880e076feb1374ad"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Arrays"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1, Page No 79"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Declarations\n",
+      "\n",
+      "iNoOfDays = 31\n",
+      "dSum = 0\n",
+      "dblTempApr=[[0 for col in range(31)] for row in range(31)]\n",
+      "\n",
+      "\n",
+      "#Entry and calculation\n",
+      "\n",
+      "for i in range(1,iNoOfDays):\n",
+      "    dblTempApr[i] = raw_input(\"Temperature Day %d: \" % i)\n",
+      "    dSum += dblTempApr[i]\n",
+      "\n",
+      "dAvg = dSum / iNoOfDays\n",
+      "\n",
+      "#Printout\n",
+      "\n",
+      "print \"Average temperature: %f\" % dAvg\n",
+      "print \"Temperatures exceeding average: \"\n",
+      "\n",
+      "for i in range(1,iNoOfDays):\n",
+      "    if (dblTempApr[i] > dAvg):\n",
+      "        print \"Day no.: %d\" % i,\n",
+      "        print \" temp: %d\" % dblTempApr[i]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 1: 32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 2: 33\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 3: 35\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 4: 36\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 5: 34\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 6: 32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 7: 31\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 8: 29\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 9: 28\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 10: 27\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 11: 25\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 12: 26\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 13: 35\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 14: 36\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 15: 35\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 16: 50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 17: 40\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 18: 42\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 19: 44\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 20: 43\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 21: 41\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 22: 38\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 23: 37\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 24: 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 25: 36\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 26: 35\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 27: 45\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 28: 32\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 29: 21\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temperature Day 30: 15\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average temperature: 32.000000\n",
+        "Temperatures exceeding average: \n",
+        "Day no.: 2  temp: 33\n",
+        "Day no.: 3  temp: 35\n",
+        "Day no.: 4  temp: 36\n",
+        "Day no.: 5  temp: 34\n",
+        "Day no.: 13  temp: 35\n",
+        "Day no.: 14  temp: 36\n",
+        "Day no.: 15  temp: 35\n",
+        "Day no.: 16  temp: 50\n",
+        "Day no.: 17  temp: 40\n",
+        "Day no.: 18  temp: 42\n",
+        "Day no.: 19  temp: 44\n",
+        "Day no.: 20  temp: 43\n",
+        "Day no.: 21  temp: 41\n",
+        "Day no.: 22  temp: 38\n",
+        "Day no.: 23  temp: 37\n",
+        "Day no.: 25  temp: 36\n",
+        "Day no.: 26  temp: 35\n",
+        "Day no.: 27  temp: 45\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2, Page No 83"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Declaration\n",
+      "\n",
+      "iMaxNo = 100\n",
+      "dLimit = 5000\n",
+      "perc1 = 0.1\n",
+      "perc2 = 0.15\n",
+      "sales = [[0 for col in range(1)] for row in range(iMaxNo)]\n",
+      "\n",
+      "#Initialize array\n",
+      "\n",
+      "for i in range(0,iMaxNo):\n",
+      "    sales[i] = 0\n",
+      "\n",
+      "#Enter Salesman info\n",
+      "nr=input(\"Enter Salesman no. : \")\n",
+      "dAmount=input(\"Enter sales amount: \")\n",
+      "\n",
+      "while (nr and dAmount):\n",
+      "    if(nr<1 or nr>iMaxNo or dAmount<0):\n",
+      "        print \"Input error\"\n",
+      "        break;\n",
+      "    else:\n",
+      "        sales[nr-1]+= dAmount\n",
+      "    nr=int(raw_input(\"Enter Salesman no. : \"))\n",
+      "    dAmount=float(raw_input(\"Enter sales amount: \"))\n",
+      "\n",
+      "#Print summary\n",
+      "\n",
+      "print \"Number        Amount        Fee\"\n",
+      "print \"======        ======        ===\"\n",
+      "\n",
+      "for i in range(0,iMaxNo):\n",
+      "    if (sales[i] > 0):\n",
+      "        if(sales[i] <= dLimit):\n",
+      "            dFee = perc1 * sales[i]\n",
+      "        else:\n",
+      "            dFee = perc1 * dLimit + perc2*(sales[i] - dLimit)\n",
+      "        j=i;\n",
+      "        j=j+1;\n",
+      "        print \"%4d\" % j,\n",
+      "        print \"%13d\" % sales[i],\n",
+      "        print \"%10.0f\" % dFee"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Salesman no. : 1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter sales amount: 1000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Salesman no. : 2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter sales amount: 2000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Salesman no. : 3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter sales amount: 3000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Salesman no. : 4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter sales amount: 4000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Salesman no. : 5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter sales amount: 5000\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Salesman no. : 6\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter sales amount: 00\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number        Amount        Fee\n",
+        "======        ======        ===\n",
+        "   1          1000        100\n",
+        "   2          2000        200\n",
+        "   3          3000        300\n",
+        "   4          4000        400\n",
+        "   5          5000        500\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page No 92"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Declaration\n",
+      "l=0\n",
+      "r=30\n",
+      "iFound=0\n",
+      "# Array is in sorted form\n",
+      "iProdid= [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31]\n",
+      "\n",
+      "iSrch = raw_input(\"Enter the searched product id: \")\n",
+      "\n",
+      "if(iSrch == iProdid[0]):\n",
+      "    iPos = 0\n",
+      "    iFound= 1\n",
+      "\n",
+      "if (iSrch == iProdid[30]):\n",
+      "    iPos = 30\n",
+      "    iFound = 1\n",
+      "\n",
+      "while not (iFound):\n",
+      "    iMid = (int)(l + ((r-l)/2))\n",
+      "    if(iSrch == iProdid[iMid]):\n",
+      "        iFound = 1\n",
+      "        iPos = iMid\n",
+      "\n",
+      "    if(iSrch > iProdid[iMid]):\n",
+      "        l=iMid\n",
+      "    else:\n",
+      "        r=iMid"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the searched product id: 15\n"
+       ]
+      }
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter5.ipynb b/Structured_Programing_with_C++/Chapter5.ipynb
new file mode 100755
index 00000000..bf5b3697
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter5.ipynb
@@ -0,0 +1,745 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:708386728a0ddd9fd0fa033f85042eb5d454abc98a33806d4ca4cc0500a010f0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Strings"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page No. 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Menu\"\n",
+      "print \"=====\"\n",
+      "print \"A. Order\"\n",
+      "print \"B. Invoice\"\n",
+      "print \"C. Warehouse\"\n",
+      "print \"D. Finance\"\n",
+      "cSel = raw_input(\"Select: \")\n",
+      "if cSel=='A' :\n",
+      "    print \"You Selected Order\"\n",
+      "elif cSel=='B' :\n",
+      "    print \"You Selected Invoice\"\n",
+      "elif cSel=='C':\n",
+      "    print \"You Selected Warehouse\"\n",
+      "elif cSel=='D' :\n",
+      "    print \"You Selected Finance\"\n",
+      "else :\n",
+      "    print \"Erroneous Choice\"\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Menu\n",
+        "=====\n",
+        "A. Order\n",
+        "B. Invoice\n",
+        "C. Warehouse\n",
+        "D. Finance\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Select: A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You Selected Order\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page No.99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "cSal = 'A'\n",
+      "while cSal!='X':\n",
+      "    print \"Menu\"\n",
+      "    print \"=====\"\n",
+      "    print \"A. Order\"\n",
+      "    print \"B. Invoice\"\n",
+      "    print \"C. Warehouse\"\n",
+      "    print \"D. Finance\"\n",
+      "    print \"X. Exit\"\n",
+      "    cSel = raw_input(\"Select: \")\n",
+      "\n",
+      "    \n",
+      "    if cSel=='A' :\n",
+      "        print \"You Selected Order\"\n",
+      "    elif cSel=='B' :\n",
+      "        print \"You Selected Invoice\"\n",
+      "    elif cSel=='C':\n",
+      "        print \"You Selected Warehouse\"\n",
+      "    elif cSel=='D' :\n",
+      "        print \"You Selected Finance\"\n",
+      "    elif cSel=='X':\n",
+      "        sys.exit()\n",
+      "    else :\n",
+      "        print \"Erroneous Choice\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Menu\n",
+        "=====\n",
+        "A. Order\n",
+        "B. Invoice\n",
+        "C. Warehouse\n",
+        "D. Finance\n",
+        "X. Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Select: A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You Selected Order\n",
+        "Menu\n",
+        "=====\n",
+        "A. Order\n",
+        "B. Invoice\n",
+        "C. Warehouse\n",
+        "D. Finance\n",
+        "X. Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Select: B\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "You Selected Invoice\n",
+        "Menu\n",
+        "=====\n",
+        "A. Order\n",
+        "B. Invoice\n",
+        "C. Warehouse\n",
+        "D. Finance\n",
+        "X. Exit\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Select: X\n"
+       ]
+      },
+      {
+       "ename": "SystemExit",
+       "evalue": "",
+       "output_type": "pyerr",
+       "traceback": [
+        "An exception has occurred, use %tb to see the full traceback.\n",
+        "\u001b[1;31mSystemExit\u001b[0m\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "To exit: use 'exit', 'quit', or Ctrl-D.\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3, Page No. 102"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "x = 'X' \n",
+      "blank = ' '\n",
+      "for i in range(10, -1, -1):\n",
+      "    for j in range(i):\n",
+      "        print blank,\n",
+      "    print x,\n",
+      "    for j in range(10-i):\n",
+      "        print x, x,\n",
+      "    print ''\n",
+      "for i in range(2):\n",
+      "    for j in range(10):\n",
+      "        print blank,\n",
+      "    print x"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "                    X \n",
+        "                  X X X \n",
+        "                X X X X X \n",
+        "              X X X X X X X \n",
+        "            X X X X X X X X X \n",
+        "          X X X X X X X X X X X \n",
+        "        X X X X X X X X X X X X X \n",
+        "      X X X X X X X X X X X X X X X \n",
+        "    X X X X X X X X X X X X X X X X X \n",
+        "  X X X X X X X X X X X X X X X X X X X \n",
+        "X X X X X X X X X X X X X X X X X X X X X \n",
+        "                    X\n",
+        "                    X\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page No. 105"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import sys\n",
+      "cName = raw_input(\"Enter your name: \")\n",
+      "print \"Your name is \" , cName\n",
+      "print \"Your name is \" , len(cName), \"  Characters long\"\n",
+      "print \"Your entire string is \" , sys.getsizeof(cName), \" characters long\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your name: John Smith\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your name is  John Smith\n",
+        "Your name is  10   Characters long\n",
+        "Your entire string is  31  characters long\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5, Page No. 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cName=range(30)\n",
+      "cName = raw_input(\"Enter your name in lower case: \")\n",
+      "iLen = len(cName)\n",
+      "for ltr in cName:\n",
+      "    print ltr.lower(),\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your name in lower case: ABCD\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a b c d\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6, Page No. 107"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cName = raw_input(\"Enter your name: \")\n",
+      "iLen = len(cName)\n",
+      "cInt = []\n",
+      "cInt.append(cName[0])\n",
+      "for i in range(1, iLen):\n",
+      "    if cName[i] == ' ':\n",
+      "        cInt.append(cName[i+1])\n",
+      "print \"Your Initials are \", ''.join(cInt)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your name: Kjell Backman\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Your Initials are  KB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page No. 109"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cName1 = raw_input(\"Enter a Name: \")\n",
+      "cName2 = raw_input(\"Enter another Name: \")\n",
+      "if cName1 > cName2 :\n",
+      "    print cName1 , cName2\n",
+      "else :\n",
+      "    print cName2 , cName1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a Name: abcd\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter another Name: abcx\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "abcx abcd\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8, Page No. 110"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "print \"Enter 5 names: \"\n",
+      "names = []\n",
+      "for i in range(5):\n",
+      "    names.append(raw_input(\"name \" + str(i+1) + \": \"))\n",
+      "names.sort()\n",
+      "for i in range(5):\n",
+      "    print names[i]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter 5 names: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "name 1: hardik\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "name 2: Parth\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "name 3: Nikunj\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "name 4: Vaibhav\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "name 5: Chirag\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Chirag\n",
+        "Nikunj\n",
+        "Parth\n",
+        "Vaibhav\n",
+        "hardik\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.9, Page No.113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cName = raw_input(\"Enter Your Name: \")\n",
+      "for i in range(29):\n",
+      "    if cName[i] == ' ':\n",
+      "        break\n",
+      "cFirst=cName[0:i]\n",
+      "print cFirst\n",
+      "cSur=cName[i+1:]\n",
+      "print cSur\n",
+      "cSur=cSur+\" \"\n",
+      "cSur=cSur+cFirst\n",
+      "print cSur\n",
+      "cSur=cSur[::-1]\n",
+      "print cSur"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name: John Smith\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "John\n",
+        "Smith\n",
+        "Smith John\n",
+        "nhoJ htimS\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10, Page No.115"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cEncrypt = []\n",
+      "cName = []\n",
+      "cName = raw_input(\"Enter Your Name: \")\n",
+      "iLen = len(cName)\n",
+      "for i in range(iLen):\n",
+      "    asc=ord(cName[i])\n",
+      "    asc=asc+1\n",
+      "    cEncrypt.append(chr(asc))\n",
+      "print ''.join(cEncrypt)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Your Name: JohnSmith\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "KpioTnjui\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.11, Page No. 116"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "iLen = random.randint(0,100)%3+5\n",
+      "cPw=[]\n",
+      "for i in range(iLen):\n",
+      "    val=random.randint(0,122)%26+65\n",
+      "    cPw.append(chr(val))\n",
+      "print ''.join(cPw)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ZXZUOY\n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.12, Page No. 119"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cText=[]\n",
+      "cEncrypt=[]\n",
+      "cTemp1=[]\n",
+      "cText=raw_input(\"\\nWrite a text in upper case: \")\n",
+      "iLen= len(cText)\n",
+      "for i in range(iLen):\n",
+      "    print cText[i]\n",
+      "    car=ord(cText[i])\n",
+      "    cEncrypt.append(car)\n",
+      "print ''.join(str(cEncrypt))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Write a text in upper case: hellow\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "h\n",
+        "e\n",
+        "l\n",
+        "l\n",
+        "o\n",
+        "w\n",
+        "[104, 101, 108, 108, 111, 119]\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import random\n",
+      "cKey=\"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n",
+      "cKey=list(cKey)\n",
+      "cEncrypt=range(50)\n",
+      "for i in range(26):\n",
+      "    j=random.randint(0,100)%26\n",
+      "    cTemp=cKey[i]\n",
+      "    cKey[i]=cKey[j]\n",
+      "    cKey[j]=cTemp\n",
+      "cKey=\"\".join(map(str,cKey))\n",
+      "cText=raw_input(\"\\nWrite a text in upper case: \")\n",
+      "iLen= len(cText)\n",
+      "for i in range(iLen):\n",
+      "    cEncrypt[i]=cKey[ord(cText[i])-65]\n",
+      "cEncrypt=cEncrypt[::iLen]\n",
+      "print \"\".join(map(str,cEncrypt))"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        "Write a text in upper case: JOHNSMITH\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "R918273645\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter6.ipynb b/Structured_Programing_with_C++/Chapter6.ipynb
new file mode 100755
index 00000000..6ee5ccef
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter6.ipynb
@@ -0,0 +1,549 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3541583e6368ad0fd94057f7e595031de565be72b75fadedd8abb1168ccd2728"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Functions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page No 126"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def dAverage(x1,x2):\n",
+      "    av = (x1+x2)/2\n",
+      "    return av\n",
+      "\n",
+      "print \"Enter two numbers:\"\n",
+      "dNo1= float(raw_input(\"\"))\n",
+      "dNo2= float(raw_input(\"\"))\n",
+      "dAvg= dAverage(dNo1,dNo2)\n",
+      "print \"The average is: \",dAvg"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter two numbers:\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "20\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average is:  25.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page No 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def min(x,y):\n",
+      "    if (x<y):\n",
+      "        return x\n",
+      "    else:\n",
+      "        return y\n",
+      "    \n",
+      "print \"Enter Three integer: \"\n",
+      "a=int(raw_input(\"\"))\n",
+      "b=int(raw_input(\"\"))\n",
+      "c=int(raw_input(\"\"))\n",
+      "m=min(a,b)\n",
+      "m=min(m,c)\n",
+      "print m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter Three integer: \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": [
+        "2\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3, Page No 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def min(x,y):\n",
+      "    if(x<y):\n",
+      "        return x\n",
+      "    else:\n",
+      "        return y\n",
+      "t=0\n",
+      "iNos=range(5)\n",
+      "for t in range(5):\n",
+      "    print \"Enter integer no\",t+1,\":\"\n",
+      "    iNos[t]=int(raw_input(\"\"))\n",
+      "\n",
+      "m=iNos[0]\n",
+      "\n",
+      "for t in range(5):\n",
+      "    m=min(iNos[t],m)\n",
+      "\n",
+      "print m,\"is the least integer\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter integer no 1 :\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "10\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter integer no 2 :\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter integer no 3 :\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "4\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter integer no 4 :\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter integer no 5 :\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "7\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3 is the least integer\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4, Page No 131"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def iLgth(s):\n",
+      "    n=0\n",
+      "    while(s[n] != '\\0'):\n",
+      "        n=n+1\n",
+      "    return n\n",
+      "\n",
+      "cWord=range(30)\n",
+      "cWord=raw_input(\"Enter a word:\")\n",
+      "cWord=cWord + '\\0' #by default string array don't appen NULL character hence, it is manually appended\n",
+      "iLen=iLgth(cWord)\n",
+      "print \"Length of the word is: \",iLen"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter a word:bhavin mehta\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Length of the word is:  12\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5, Page No 134"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def dDiscount(iQty,dLinePrice):\n",
+      "    if(dLinePrice>1000):\n",
+      "            return 0.15\n",
+      "    elif(dLinePrice>500 or iQty>10):\n",
+      "        return 0.10\n",
+      "    else:\n",
+      "        return 0\n",
+      "\n",
+      "def dPrice(iNo,dUnitPrice):\n",
+      "    dTax=0.25\n",
+      "    dLinePr=iNo*dUnitPrice\n",
+      "    dDiscPerc=dDiscount(iNo,dLinePr)\n",
+      "    return dLinePr*(1-dDiscPerc)*(1+dTax)\n",
+      "\n",
+      "print \"Enter quantity and unit price: \"\n",
+      "iQuantity=int(raw_input(\"\"))\n",
+      "dUnitPrice=float(raw_input(\"\"))\n",
+      "\n",
+      "print \"To be paid\", dPrice(iQuantity,dUnitPrice)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter quantity and unit price: \n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "30\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "50\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "To be paid 1593.75\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.6, Page No 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def replace1(s,c,cnew):\n",
+      "    n=0\n",
+      "    s = list(s)\n",
+      "    while(s[n]!='\\0'):\n",
+      "        if(s[n]==c):\n",
+      "            s[n]=cnew\n",
+      "        n=n+1\n",
+      "    print \"\".join(map(str,s))\n",
+      "a=\"C:/Mydocumnets/Sheets\\0\"\n",
+      "print a\n",
+      "replace1(a,'/','-')"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "C:/Mydocumnets/Sheets\u0000\n",
+        "C:-Mydocumnets-Sheets\u0000\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.7, Page No 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def word(s):\n",
+      "    i=0\n",
+      "    j=len(s)\n",
+      "    while(i<j):\n",
+      "        if((s[i]>='a' and s[i]<='z')or(s[i]>='A' and s[i]<='Z')):\n",
+      "            i=i+1\n",
+      "        else:\n",
+      "            return 0;\n",
+      "    return 1;\n",
+      "\n",
+      "str=range(100)\n",
+      "str=raw_input(\"Enter word: \")\n",
+      "while(str):\n",
+      "    if(word(str)):\n",
+      "        print \"A word\"\n",
+      "    else:\n",
+      "        print \"No word\"\n",
+      "    str=raw_input(\"Enter word: \")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word: john\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A word\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word: smith\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "A word\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word: john smith\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "No word\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter word: \n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.8, Page No 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#import myfunc we have created this header file and its going to work as header file user has to make it\n",
+      "print \"To be paid: \" + str(dPrice(10,700))\n",
+      "\n",
+      "# Code for myfunc Header file is here\n",
+      "\n",
+      "\"\"\"\n",
+      "def underline(n):\n",
+      "    for i in range(n):\n",
+      "        print \"=\",\n",
+      "def dDiscount(iQty,dLinePrice):\n",
+      "    if(dLinePrice > 1000):\n",
+      "        return 0.15\n",
+      "    elif(dLinePrice>500 or iQty>10):\n",
+      "        return 0.10\n",
+      "    else:\n",
+      "        return 0\n",
+      "def dPrice(iNo,dUnitPrice):\n",
+      "    dTax = 0.25\n",
+      "    dLinePr = iNo * dUnitPrice\n",
+      "    dDiscPerc = dDiscount(iNo,dLinePr)\n",
+      "    return dLinePr * (1-dDiscPerc) * (1+dTax)\n",
+      "\"\"\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter7.ipynb b/Structured_Programing_with_C++/Chapter7.ipynb
new file mode 100755
index 00000000..c4850700
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter7.ipynb
@@ -0,0 +1,478 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:81ebacbcf46be2a85cdba0d8109c34324595901036a67c9025bbb5cab58aa894"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Files"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1, Page No 157"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "outfile = open(\"prodfile1.txt\",\"w\")\n",
+      "cProd = raw_input(\"Enter product, (only Enter to exit:\")\n",
+      "while (len(cProd) > 0):\n",
+      "    outfile.writelines(cProd)\n",
+      "    outfile.writelines(\"\\n\")\n",
+      "    cProd = raw_input(\"Enter product, (only Enter to exit:\")\n",
+      "outfile.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product, (only Enter to exit:wheel\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product, (only Enter to exit:handlebars\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product, (only Enter to exit:saddle\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product, (only Enter to exit:bike\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product, (only Enter to exit:moped\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product, (only Enter to exit:\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Examle 7.2 Page No 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "infile = open(\"prodfile1.txt\",\"r\")\n",
+      "while (1):\n",
+      "    cProd = infile.readline()\n",
+      "    if(cProd):\n",
+      "        print cProd\n",
+      "    else:\n",
+      "        break\n",
+      "infile.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "wheel\n",
+        "\n",
+        "handlebars\n",
+        "\n",
+        "saddle\n",
+        "\n",
+        "bike\n",
+        "\n",
+        "moped\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page No 160"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "cProd = raw_input(\"Enter new product \")\n",
+      "outfile = open(\"prodfile.txt1\",\"a\")\n",
+      "outfile.writelines(cProd)\n",
+      "outfile.close()    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter new product Spoke\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4, Page No 161"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iProdId = 1\n",
+      "outfile = open(\"prodfile.txt\",\"w\")\n",
+      "while (iProdId != 0):\n",
+      "    iProdId = int(raw_input(\"Enter product id: \"))\n",
+      "    dPrice = float(raw_input(\"...and price: \"))\n",
+      "    if(iProdId > 0):\n",
+      "        outfile.writelines(str(iProdId) + '\\n' + str(dPrice) + '\\n')\n",
+      "outfile.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product id: 2345\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "...and price: 245.5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product id: 4512\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "...and price: 450.2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product id: 3902\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "...and price: 320.7\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product id: 2478\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "...and price: 75.9\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product id: 0\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "...and price: 0\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page No 163"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iFound = 0\n",
+      "infile = open(\"prodfile.txt\",\"r\")\n",
+      "iSrch = int(raw_input(\"Enter product id: \"))\n",
+      "while(1):\n",
+      "    iProdId = infile.readline()\n",
+      "    if(iProdId == \"\"):\n",
+      "        break\n",
+      "    iProdId = int(iProdId)\n",
+      "    dPrice = float(infile.readline())\n",
+      "    if(iProdId == iSrch):\n",
+      "        print \"The price is:\" , dPrice\n",
+      "        iFound = 1\n",
+      "        break\n",
+      "if(not iFound):\n",
+      "    print \"Product missing\"\n",
+      "infile.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter product id: 1234\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Product missing\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6, Page No 166"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def sort(cList,n):\n",
+      "    for v in range(0,n-1):\n",
+      "        for h in range(v+1,n):\n",
+      "            if(cList[h]<cList[v]):\n",
+      "                temp = cList[v]\n",
+      "                cList[v] = cList[h]\n",
+      "                cList[h] = temp\n",
+      "i = 0\n",
+      "infile = open(\"prodfile1.txt\",\"r\")\n",
+      "cProd = infile.readlines()\n",
+      "infile.close()\n",
+      "while(i < len(cProd)):\n",
+      "    i = i + 1\n",
+      "iNo = i\n",
+      "sort(cProd,iNo)\n",
+      "for j in range(0,iNo):\n",
+      "    print cProd[j]"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Bike\n",
+        "\n",
+        "Handlebars\n",
+        "\n",
+        "Moped\n",
+        "\n",
+        "Saddle\n",
+        "\n",
+        "Spoke\n",
+        "\n",
+        "Wheel\n",
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7, Page No 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import os\n",
+      "infile = open(\"prodfile.txt\",\"r\")\n",
+      "outfile = open(\"temp.txt\",\"w\")\n",
+      "iSrch = int(raw_input(\"Specify product id: \"))\n",
+      "while(1):\n",
+      "    iProdId = infile.readline()\n",
+      "    if(iProdId == \"\"):\n",
+      "        break\n",
+      "    iProdId = int(iProdId)\n",
+      "    dPrice = float(infile.readline())\n",
+      "    if(iProdId == iSrch):\n",
+      "        dPrice = float(raw_input(\"Specify the new price: \"))\n",
+      "    outfile.writelines(str(iProdId) + '\\n' + str(dPrice) + '\\n')\n",
+      "infile.close()\n",
+      "outfile.close()\n",
+      "os.remove(\"prodfile.txt\")\n",
+      "os.rename(\"temp.txt\",\"prodfile.txt\")"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify product id: 2345\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify the new price: 33.33\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8, Page No 171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import os\n",
+      "cNewName = raw_input(\"Specify new file name: \")\n",
+      "infile = open(\"prodfile.txt\",\"r\")\n",
+      "outfile = open(cNewName,\"w\")\n",
+      "if(not os.path.exists(cNewName)):\n",
+      "    print \"The file could not be created\"\n",
+      "outfile.writelines(infile.readlines())\n",
+      "infile.close()\n",
+      "outfile.close()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Specify new file name: tryal.txt\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter8.ipynb b/Structured_Programing_with_C++/Chapter8.ipynb
new file mode 100755
index 00000000..0d779089
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter8.ipynb
@@ -0,0 +1,352 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:95ddab2248de9e7055a4dff19cd18b277d7ba7ab8ace5be6bff568b621024b03"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Chapter 8 : Pointers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page No 179"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from ctypes import *\n",
+      "\n",
+      "iNo=c_int\n",
+      "pNo = id(iNo)\n",
+      "dPrice=dTotal=float\n",
+      "pPrice=id(dPrice)\n",
+      "pTotal=id(dTotal)\n",
+      "cName=range(20)\n",
+      "pName=id(cName)\n",
+      "pName = raw_input(\"Enter your name: \")\n",
+      "pNo = int(raw_input(\"Enter quantity: \"))\n",
+      "pPrice = float(raw_input(\"and price: \"))\n",
+      "pTotal = pNo * pPrice\n",
+      "print \"Dear \" + pName + \", your price is \" + str(pTotal) + \" kr.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter your name: chirag\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter quantity: 10\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "and price: 100\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Dear chirag, your price is 1000.0 kr.\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2, Page No 182"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iSal=[14000,15000,16000,17000,18000,19000]\n",
+      "pSal = id(iSal)\n",
+      "dTax=[0.32,0.34,0.35,0.36,0.365,0.37]\n",
+      "pTax=id(dTax)\n",
+      "print \"Salary          Tax\"\n",
+      "for i in range(6):\n",
+      "    print str(iSal[i])+\"          \",str(iSal[i]*dTax[i])\n",
+      "    i=i+1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Salary          Tax\n",
+        "14000           4480.0\n",
+        "15000           5100.0\n",
+        "16000           5600.0\n",
+        "17000           6120.0\n",
+        "18000           6570.0\n",
+        "19000           7030.0\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page No 185"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "def find(s):\n",
+      "    for p in range(len(s)):\n",
+      "        if(s[p]=='@'):\n",
+      "            print \"It is an email address\"\n",
+      "        p=p+1\n",
+      "\n",
+      "#cString=range(9)\n",
+      "pString=id(cString)\n",
+      "cString=raw_input(\"Enter A text: \")\n",
+      "find(cString)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter A text: hello@gmail.com\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "It is an email address\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4, Page No 188"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iNumber=int(raw_input(\"How many product will be entered? \"))\n",
+      "\n",
+      "for i in range(iNumber):\n",
+      "    pNo = id(iNumber)\n",
+      "    print pNo\n",
+      "    pNO = pNo + 1\n",
+      "pNo = pNo - iNumber\n",
+      "for i in range(iNumber):\n",
+      "    print id(pNO)\n",
+      "    pNo = pNO + 1\n",
+      "pNo = pNo - iNumber\n",
+      "del pNo   "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many product will be entered? 5\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "3848360\n",
+        "3848360\n",
+        "3848360\n",
+        "3848360\n",
+        "3848360\n",
+        "51980880\n",
+        "51980880\n",
+        "51980880\n",
+        "51980880\n",
+        "51980880\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5, Page No 189"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "iNo = 5\n",
+      "cName = range(iNo)\n",
+      "print \"Enter the name of 5 course mates: \"\n",
+      "for i in range(iNo):\n",
+      "    print \"Mate no. \" ,i+1\n",
+      "    temp = raw_input(\"Mate name: \")\n",
+      "    cName.append(temp)\n",
+      "for j in range(iNo):\n",
+      "    print cName[j]\n",
+      "for k in range(iNo):\n",
+      "    del cName[k]    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter the name of 5 course mates: \n",
+        "Mate no.  1\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate name: a\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate no.  2\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate name: b\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate no.  3\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate name: c\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate no.  4\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate name: d\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate no.  5\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mate name: e\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "0\n",
+        "1\n",
+        "2\n",
+        "3\n",
+        "4\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/Chapter9.ipynb b/Structured_Programing_with_C++/Chapter9.ipynb
new file mode 100755
index 00000000..724e3c4e
--- /dev/null
+++ b/Structured_Programing_with_C++/Chapter9.ipynb
@@ -0,0 +1,274 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:3c26b0f763ca451c03b73133e1c2db791af8fb7ada7862c248f0876f5e8843fe"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Structures"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page No 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class Prod:\n",
+      "    def __init__(self, cName, iId, dPrice, iNo, cSupp):\n",
+      "        self.cName = cName\n",
+      "        self.iId = iId\n",
+      "        self.dPrice = dPrice\n",
+      "        self.iNo = iNo\n",
+      "        self.cSupp = cSupp\n",
+      "    def println(self):\n",
+      "        print self.cName + \"\\t    \",\n",
+      "        print str(self.iId) + \"\\t \",\n",
+      "        print str(self.dPrice) + \"\\t  \",\n",
+      "        print str(self.iNo) + \"\\t   \",\n",
+      "        print self.cSupp + \"\\t\\n\",\n",
+      "if __name__ == '__main__':\n",
+      "    prodOne = Prod(\"Olive Oil\",1001,120.50,250,\"Frescati Oil S/A\")\n",
+      "    print \"Enter information for product\"\n",
+      "    cName = raw_input(\"Start with the product name: \")\n",
+      "    iId = int(raw_input(\"The product id: \"))\n",
+      "    dPrice = float(raw_input(\"The price: \"))\n",
+      "    iNo = int(raw_input(\"How many items are there in stock: \"))\n",
+      "    cSupp = raw_input(\"Who supplies the product: \")\n",
+      "    prodTwo = Prod(cName,iId,dPrice,iNo,cSupp)\n",
+      "    print \"Productname \\t Product id \\t Price \\t Quantity \\t Supplier\\n\"\n",
+      "    prodOne.println()\n",
+      "    prodTwo.println()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Enter information for product\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Start with the product name: Coconut Oil\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The product id: 1002\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The price: 120.50\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "How many items are there in stock: 230\n"
+       ]
+      },
+      {
+       "name": "stdout",
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Who supplies the product: Frescati Oil S/A\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Productname \t Product id \t Price \t Quantity \t Supplier\n",
+        "\n",
+        "Olive Oil\t     1001\t  120.5\t   250\t    Frescati Oil S/A\t\n",
+        "Coconut Oil\t     1002\t  120.5\t   230\t    Frescati Oil S/A\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page No 200"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class Prod:\n",
+      "    def __init__(self, cName, iId, dPrice, iNo, cSupp):\n",
+      "        self.cName = cName\n",
+      "        self.iId = iId\n",
+      "        self.dPrice = dPrice\n",
+      "        self.iNo = iNo\n",
+      "        self.cSupp = cSupp\n",
+      "def printOnScreen(prodOne1):\n",
+      "        print prodOne1.cName + \"\\t    \",\n",
+      "        print str(prodOne1.iId) + \"\\t \",\n",
+      "        print str(prodOne1.dPrice) + \"\\t  \",\n",
+      "        print str(prodOne1.iNo) + \"\\t   \",\n",
+      "        print prodOne1.cSupp + \"\\t\\n\",\n",
+      "\n",
+      "if __name__ == '__main__':\n",
+      "    prodOne = Prod(\"Olive Oil\",1001,120.50,250,\"Frescati Oil S/A\")\n",
+      "    printOnScreen(prodOne)\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Olive Oil\t     1001\t  120.5\t   250\t    Frescati Oil S/A\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3, Page No 202"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class Prod:\n",
+      "    def __init__(self, cName, iId, dPrice, iNo, cSupp):\n",
+      "        self.cName = cName\n",
+      "        self.iId = iId\n",
+      "        self.dPrice = dPrice\n",
+      "        self.iNo = iNo\n",
+      "        self.cSupp = cSupp\n",
+      "def printOnScreen(p,n):\n",
+      "    for i in range(n):\n",
+      "        print p[i].cName + \"\\t    \",\n",
+      "        print str(p[i].iId) + \"\\t \",\n",
+      "        print str(p[i].dPrice) + \"\\t  \",\n",
+      "        print str(p[i].iNo) + \"\\t   \",\n",
+      "        print p[i].cSupp + \"\\t\\n\",\n",
+      "if __name__ == '__main__':\n",
+      "    sProds = []\n",
+      "    sProds.append(Prod(\"Food Oil\",101,12.50,100,\"Felix Ltd\"))\n",
+      "    sProds.append(Prod(\"Baby Oil\",102,23.75,25,\"Baby Prod\"))\n",
+      "    sProds.append(Prod(\"Boiler Oil\",103,6100,123000,\"Shell\"))\n",
+      "    printOnScreen(sProds,3)\n",
+      "    "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Food Oil\t     101\t  12.5\t   100\t    Felix Ltd\t\n",
+        "Baby Oil\t     102\t  23.75\t   25\t    Baby Prod\t\n",
+        "Boiler Oil\t     103\t  6100\t   123000\t    Shell\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page No 205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "class Prod:\n",
+      "    def __init__(self, cName, iId, dPrice, iNo, cSupp):\n",
+      "        self.cName = cName\n",
+      "        self.iId = iId\n",
+      "        self.dPrice = dPrice\n",
+      "        self.iNo = iNo\n",
+      "        self.cSupp = cSupp\n",
+      "def printOnScreen(p,n):\n",
+      "    for i in range(n):\n",
+      "        print p[i].cName + \"\\t    \",\n",
+      "        print str(p[i].iId) + \"\\t \",\n",
+      "        print str(p[i].dPrice) + \"\\t  \",\n",
+      "        print str(p[i].iNo) + \"\\t   \",\n",
+      "        print p[i].cSupp + \"\\t\\n\",\n",
+      "if __name__ == '__main__':\n",
+      "    sProds = []\n",
+      "    sProds.append(Prod(\"Food Oil\",101,12.50,100,\"Felix Ltd\"))\n",
+      "    sProds.append(Prod(\"Baby Oil\",102,23.75,25,\"Baby Prod\"))\n",
+      "    sProds.append(Prod(\"Boiler Oil\",103,6100,123000,\"Shell\"))\n",
+      "    pProds = id(sProds)\n",
+      "    printOnScreen(sProds,3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Food Oil\t     101\t  12.5\t   100\t    Felix Ltd\t\n",
+        "Baby Oil\t     102\t  23.75\t   25\t    Baby Prod\t\n",
+        "Boiler Oil\t     103\t  6100\t   123000\t    Shell\t\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/README.txt b/Structured_Programing_with_C++/README.txt
new file mode 100755
index 00000000..ac28cf27
--- /dev/null
+++ b/Structured_Programing_with_C++/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Chirag Kotak
+Course: mca
+College/Institute/Organization: Shri Brahmanand Institute of Management and Computer Science
+Department/Designation: Computer Science
+Book Title: Structured Programing with C++
+Author: Kjell Backman
+Publisher: Ventus Publishing ApS
+Year of publication: 2012
+Isbn: 978-87-403-0099-4
+Edition: 1st Edition
+\ No newline at end of file
diff --git a/Structured_Programing_with_C++/screenshots/1.png b/Structured_Programing_with_C++/screenshots/1.png
new file mode 100755
index 00000000..7040411e
--- /dev/null
+++ b/Structured_Programing_with_C++/screenshots/1.png
diff --git a/Structured_Programing_with_C++/screenshots/2.png b/Structured_Programing_with_C++/screenshots/2.png
new file mode 100755
index 00000000..667e163b
--- /dev/null
+++ b/Structured_Programing_with_C++/screenshots/2.png
diff --git a/Structured_Programing_with_C++/screenshots/3.png b/Structured_Programing_with_C++/screenshots/3.png
new file mode 100755
index 00000000..1ab51d18
--- /dev/null
+++ b/Structured_Programing_with_C++/screenshots/3.png
diff --git a/Surveying_Volume_3/Chapter1.ipynb b/Surveying_Volume_3/Chapter1.ipynb
index 015050ae..3327643c 100755
--- a/Surveying_Volume_3/Chapter1.ipynb
+++ b/Surveying_Volume_3/Chapter1.ipynb
@@ -1,6 +1,6 @@
 {
  "metadata": {
-  "name": "S3-C1"
+  "name": "Chapter 1"
  },
  "nbformat": 3,
  "nbformat_minor": 0,
@@ -9,72 +9,1466 @@
    "cells": [
     {
      "cell_type": "heading",
-     "level": 1,
+     "level": 2,
      "metadata": {},
-     "source": "Electronic Distance Measurement"
+     "source": "FIELD ASTRONOMY"
     },
     {
      "cell_type": "heading",
-     "level": 2,
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.1, Page 30"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding difference of longitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\n#part1\na =40; # longitude of A\nb =73; # longitude of B\n\n#calculation\ndol =b-a; # difference of longitude\n\n#result\nprint \" difference of longitude is in degrees\",round(dol);\n\n#part2\na =20; # longitude of A\nb =150; # longitude of B\n\n#calculation\ndol =b-a; # difference of longitude\n\n#result\nprint \" difference of longitude is in degrees \",round(dol);\n\n#part3\na =-20; # longitude of A\nb =50; # longitude of B\n\n#calculation\ndol =b-a; # difference of longitude\n\n#result\nprint \" difference of longitude is in degrees\",round(dol);\n\n#part4\na =-40; # longitude of A\nb =150; # longitude of B\n\n#calculation\ndol =360-(b-a); # difference of longitude\n\n#result\nprint \" difference of longitude is in degrees\",round(dol);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " difference of longitude is  33.0\n difference of longitude is  130.0\n difference of longitude is  70.0\n difference of longitude is  170.0\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
      "metadata": {},
-     "source": "Example 1.1,Page 22"
+     "source": "Example 1.2.1,Page 31"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#initialisation of variable\nT=273+25.0#temperature \np=752.0;#pressure mm Hg\nNo=294.0e-6;\nns=1.000284;\nD1=1438.254;#recorded distance in m\nh=263.42-243.25;#height difference in m\nR=6370e3;#radius of earth in m\n\n#calculation\nn=1+No*(273/T)*(p/760);\nD=D1*ns/n;\ncg=-h**2/2/D;\nHm=263.42/2+243.25/2;\nD=D+cg;\nch=-D*Hm/R\nEL=D+ch;\n\n#result\nprint \"equivalent length in m\",round(EL,3)",
+     "input": "#finding distance between two points\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos\nlatA =28.0+42.0/60.0; # latitude of A\nlonA =31.0*60.0+12.0; # longitude of A\nlatB =28.0+42.0/60.0; # latitude of B\nlonB =47.0*60.0+24.0; # longitude of B\n\n#calculation\nd=( lonB - lonA )*cos( latA /180* pi);\n\n#result\nprint  \" distance between A & B in (km) \",round(d *1.852,3)",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "equivalent length in m 1438.081\n"
+       "text": " distance between A & B in (km)  1578.989\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.2.2,Page 31"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding distance between two points\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos\nlatA =12.0+36.0/60.0; # latitude of A\nlonA =115.0*60.0+6.0; # longitude of A\nlatB =12.0+36.0/60.0; # latitude of B\nlonB =-150.0*60.0-24.0; # longitude of B\n\n#calculation\nd=( 360*60+lonB - lonA )*cos( latA /180* pi);\n\n#result\nprint  \" distance between A & B in (km)  \",round(d *1.852,3) ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " distance between A & B in (km)   10247.946\n"
       }
      ],
      "prompt_number": 2
     },
     {
      "cell_type": "heading",
-     "level": 2,
+     "level": 3,
      "metadata": {},
-     "source": "Example 1.2,Page 22"
+     "source": "Example 1.3,Page 31"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#initialisation of variable\nT=273+18.8;#temperature\np=713;\ne=3;\nc=299792.5e3;#speed of light in km/s\nf=11e6;#frequency in Hz\n\n#calculation\nn=1+(103.49/T*(p-e)+86.26/T*(1+5748.0/T)*e)/1e6;\nV=c/n;\nl=V/f;\n\n#result\nprint \"wavelenght of light in m\",round(l,3)",
+     "input": "#finding distance between two points\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\nlatA =15;\nlatB =12.0+6.0/60.0;\nlonA =50.0+12.0/60.0;\nlonB =54.0;\nRe =6370.0; # radius of earth\n\n#calculation\nb=(90 - latA )*pi /180;\na=(90 - latB )*pi /180;\nP=( lonB - lonA )*pi /180;\np= acos ( cos (P)*sin(a)* sin (b)+ cos (a)*cos(b)); #spherical triangle law\nx= atan ( cos (a/2-b/2)/ cos (a/2+b /2) * tan (pi /2-P /2) );#spherical triangle law \ny= atan ( sin (a/2-b/2)/ sin (a/2+b /2) * tan (pi /2-P /2) ); #spherical triangle law\ndol =pi -x-y;\ndol=dol*180/pi;\na=  dol *3600 %60;\nb=  ((dol *3600 -a)%3600) /60;\nc=( dol *3600 - b*60 -a) /3600;\n\n#result\nprint \" distance from A to B in (km)  \",round(p*Re,3);\nprint \" direction of B from A towards east of south \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "27.247 wavelenght of light in m\n"
+       "text": " distance from A to B in (km)   522.104\n direction of B from A towards east of south  35.16 seconds 19.0 minutes 52.0 degrees\n"
       }
      ],
      "prompt_number": 4
     },
     {
      "cell_type": "heading",
-     "level": 2,
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.4,Page 33"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding distance between two points\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef 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]\nlatA =45.0;\na1=45.0+13.108/60;\np =(300.0/60.0) *pi /180; # side AB\nb=(90 - latA )*pi /180; # side PA\n\n# calculation\na= acos ( cos (p)*cos(b)); # side BP\nBC=a *180/ pi - latA ;\nd=BC *1.852*60;\nB=asin(sin(latA*pi/180)/sin(a1*pi/180));\nB=deg_to_dms(B*180/pi);\n\n\n#result\nprint \" distance of BC in (km)\",round(d,3)\nprint \"the angle in deg,min,sec is\",B",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " distance of BC in (km) 24.181\nthe angle in deg,min,sce is [85, 0, 33.27]\n"
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.6.1,Page 37"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding altitude and zenith distance of star\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndelta =42+15.0/60; # declination of star\ntheta =26+40.0/60; # lattude of star\n\n#caculation\nzend =90.0 - theta -90+ delta ;\nalt =90.0 - zend ;\n\n#for zenith distance\n#a=  zend *3600 %60;\nb=  ((zend *3600 )%3600) /60;\nc=( zend *3600 - b*60 -a) /3600;\nprint \" zenith distance \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n#for altitude\na=  alt *3600 %60;\nb=  ((alt *3600 -a)%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" altitude of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " zenith distance  0.0 seconds 35.0 minutes 15.0 degrees\n altitude of star  0.0 seconds 25.0 minutes 74.0 degrees\n"
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.6.2,Page 36"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding altitude and zenith distance of star\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndelta =23+20.0/60; # declination of star\ntheta =26+40.0/60; # lattude of star\n\n#caculation\nzend =90.0 + theta -90- delta ;\nalt =90.0 - zend ;\n\n#for zenith distance\na=  zend *3600 %60;\nb=  ((zend *3600 -a)%3600) /60;\nc=( zend *3600 - b*60 -a) /3600;\nprint \" zenith distance \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n#for altitude\n\nb=  ((alt *3600 )%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" altitude of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " zenith distance  0.0 seconds 20.0 minutes 3.0 degrees\n altitude of star  0.0 seconds 40.0 minutes 86.0 degrees\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.6.3,Page 37"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding altitude and zenith distance of star\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndelta =65+40.0/60; # declination of star\ntheta =26+40.0/60; # lattude of star\n\n#caculation\nzend =90.0 - theta -90+ delta ;\nalt =90.0 - zend ;\n\n#for zenith distance\na=  zend *3600 %60;\nb=  ((zend *3600 -a)%3600) /60;\nc=( zend *3600 - b*60 -a) /3600;\nprint \" zenith distance \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n#for altitude\na=  alt *3600 %60;\nb=  ((alt *3600 -a)%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" altitude of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " zenith distance  0.0 seconds 0.0 minutes 39.0 degrees\n altitude of star  0.0 seconds 0.0 minutes 51.0 degrees\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.7,Page 37"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding altitude and zenith distance of star\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndelta =85+20.0/60; # declination of star\ntheta =46+50.0/60; # lattude of star\n\n#caculation\nzend =90.0 - theta +90- delta ;\nalt =90.0 - zend ;\n\n#for zenith distance\n\nb=  ((zend *3600 )%3600) /60;\nc=( zend *3600 - b*60 -a) /3600;\nprint \" zenith distance \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n#for altitude\na=  alt *3600 %60;\nb=  ((alt *3600 -a)%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" altitude of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " zenith distance  0.0 seconds 50.0 minutes 47.0 degrees\n altitude of star  0.0 seconds 10.0 minutes 42.0 degrees\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.8,Page 38"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding altitude and zenith distance of star\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndelta =56+10.0/60; # declination of star\ntheta =56+10.0/60; # lattude of star\n\n#caculation\nzend =90.0 - theta +90- delta ;\nalt =90.0 - zend ;\n\n#for zenith distance\na=  zend *3600 %60;\nb=  ((zend *3600-a )%3600) /60;\nc=( zend *3600 - b*60 -a) /3600;\nprint \" zenith distance \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n#for altitude\n#a=  alt *3600 %60;\nb=  ((alt *3600 )%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" altitude of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " zenith distance  0.0 seconds 40.0 minutes 67.0 degrees\n altitude of star  0.0 seconds 20.0 minutes 22.0 degrees\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.9,Page 38"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude and declination \n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\nimport numpy as np\na=np.array([[1.0,-1.0],[1.0,1.0]])\nb=np.array([59.0/3,332.0/3])\n\n#calculation\nx=np.linalg.solve(a,b);\n\n#result\nprint\"declination of star in (degrees)\",round(x[0],3);\nprint\"latitude of the place of observation (degrees)\",x[1];",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "declination of star in (degrees) 65.167\nlatitude of the place of observation (degrees) 45.5\n"
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.10,Page 39"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth and altitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ntheta =20+30.0/60;\nH =42+6.0/60; # hour angle\ndelta =50.0;\n\n\n# in triangle ZPM\n\n#calculation\nPZ =(90 - delta )*pi /180;\nH=H*pi /180;\nPM =(90 - theta )*pi /180;\nZM= acos (( cos (PZ)* cos (PM)+sin(PM)*sin(PZ)* cos (H)));\nalpha =pi /2- ZM;\nalpha = alpha *180/ pi;\nA =(( cos(PM)-cos (PZ)* cos (ZM))/ sin (PZ)/sin(ZM));\n\nif A <0:\n    A=-A;\n    A=acos(A)\n    A=180-A*180/pi;\n \n\n#for altitude\nalt=alpha;\na=  alt *3600 %60;\nb=((alt *3600-a )%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" altitude of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n\n#for azimuth    \na=  A *3600 %60;\nb=  ((A *3600-a )%3600) /60;\nc=( A *3600 - b*60 -a) /3600;\nprint\" azimuth of star in (degrees ) westwards \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " altitude of star  36.75 seconds 38.0 minutes 45.0 degrees\n azimuth of star in (degrees ) westwards  25.551 seconds 4.0 minutes 116.0 degrees\n"
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.11,Page 40"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth and altitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ntheta = -8 -30.0/60;\nH =322.0; # hour angle\ndelta =50;\n\n\n# in triangle ZPM\n\n#calculation\nPZ =(90 - delta )*pi /180;\nH =2* pi -H*pi /180;\nPM =(90 - theta )*pi /180;\nZM= acos (( cos (PZ)* cos (PM)+sin(PM)*sin(PZ)* cos (H)));\nalpha =pi /2- ZM;\nalpha=alpha*180/pi;\nA =(( cos(PM)-cos (PZ)* cos (ZM))/ sin (PZ)/sin(ZM));\n\nif A <0:\n    A=-A;\n    A=acos(A)\n    A=180-A*180/pi;\n \n#result\n#for altitude\nalt=alpha;\na=  alt *3600 %60;\nb=((alt *3600-a )%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" altitude of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n\n#for azimuth    \na=  A *3600 %60;\nb=  ((A *3600-a )%3600) /60;\nc=( A *3600 - b*60 -a) /3600;\nprint\" azimuth of star in (degrees ) eastwards \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " altitude of star  48.256 seconds 48.0 minutes 22.0 degrees\n azimuth of star in (degrees ) eastwards  22.798 seconds 39.0 minutes 138.0 degrees\n"
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.12,Page 42"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding hour angle\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\nalpha =22+36.0/60; # altitude of star\nA =42.0 # azimuth angle\ndelta =40.0; # latitude of observer\n\n# in triangle ZPM\n\n#calculation\nPZ =(90 - delta )*pi /180;\nA=A*pi /180;\nZM =(90 - alpha )*pi /180;\nPM= acos (( cos (PZ)* cos (ZM)+sin(ZM)*sin(PZ)* cos (A)));\ntheta =pi /2- PM\ntheta=theta*180/pi;\nH =(( cos(ZM)-cos (PZ)* cos (PM))/ sin (PZ)/sin(PM));\nif H <0:\n    H=-H;\n    H=acos(H)\n    H=180-H*180/pi;\n \n\n#result\n#for declination \nalt=theta;\na=  alt *3600 %60;\nb=((alt *3600-a )%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" declination of star \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n#for hour angle\na=  H *3600 %60;\nb=  ((H *3600-a )%3600) /60;\nc=( H *3600 - b*60 -a) /3600;\nprint\" hour angle of star  \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " declination of star  12.44 seconds 35.0 minutes 50.0 degrees\n hour angle of star   5.342 seconds 21.0 minutes 103.0 degrees\n"
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.13,Page 42"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding hour angle\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\nalpha =21+30.0/60; # a l t i t u d e o f s t a r\nA =140.0 # azimuth a n g l e\ndelta =48.0; # l a t i t u d e o f o b s e r v e r\n\n#calculation\nPZ =(90 - delta )*pi /180;\nA=A*pi /180;\nZM =(90 - alpha )*pi /180;\nPM =( cos(PZ)*cos(ZM)+ sin (ZM)* sin (PZ)* cos (A));\n\nif PM <0:\n    PM=-PM\n    PM=acos(PM)\n    PM=180-PM*180/pi;\n\nH= acos (( cos (ZM)-cos(PZ)*cos(PM*pi /180) )/ sin (PZ)/sin (PM*pi /180) );\nH =2* pi -H;\nH=H*180/pi;\n\n#result\n#for declination \nalt=PM-90;\na=  alt *3600 %60;\nb=((alt *3600-a )%3600) /60;\nc=( alt *3600 - b*60 -a) /3600;\nprint \" declination of star southwards \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";\n\n#for hour angle\na=  H *3600 %60;\nb=  ((H *3600-a )%3600) /60;\nc=( H *3600 - b*60 -a) /3600;\nprint\" hour angle of star  \",round(a,3),\"seconds\",b,\"minutes\",c,\"degrees\";",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " declination of star southwards  12.098 seconds 48.0 minutes 11.0 degrees\n hour angle of star   22.619 seconds 20.0 minutes 322.0 degrees\n"
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.14,Page 43"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding hour angle\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef 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# part 1\ndelta =22+12.0/60;\ntheta =42+30.0/60;\n\n#calculation\nZP =(90 - theta )*pi /180;\nPM =(90 - delta )*pi /180;\nA= acos ( cos (PM)/sin(ZP));\nH=180 - acos ( tan (pi /2- ZP)*tan(pi /2- PM)) *180/ pi\nA=deg_to_dms(A*180/ pi);\nH=deg_to_dms(H/15);\n\n#result\nprint \" azimuth of setting sun in ( degrees,min,second)  \",A\nprint \" suns hour angle in ( hr,min,second ) : \",H\n\n#part 2\ndelta = -22 -12/60;\ntheta =42+30.0/60;\n\n#calculation\nZP =(90 - theta )*pi /180;\nPM =(90 - delta )*pi /180;\nA= acos ( cos (PM)/sin(ZP));\nH=180 - acos ( tan (pi /2- ZP)*tan(pi /2- PM)) *180/ pi\nA=deg_to_dms(A*180/ pi);\nH=deg_to_dms(H/15);\n\n#result\nprint \" azimuth of setting sun in ( degrees,min,second)  \",A\nprint \" suns hour angle in ( hr,min,second ) : \",H",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " azimuth of setting sun in ( degrees,min,second)   [59, 10, 14.72]\n suns hour angle in ( hr,min,second ) :  [7, 27, 50.23]\n azimuth of setting sun in ( degrees,min,second)   [120, 32, 13.17]\n suns hour angle in ( hr,min,second ) :  [4, 33, 4.97]\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.15,Page 44"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding hour angle\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndelta =22+12.0/60;\ntheta =42+30.0/60;\nef 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#calculation\nZP =(90 - theta )*pi /180;\nPM =(90 - delta )*pi /180;\nA= acos ( cos (PM)/sin(ZP));\nH=180 - acos ( tan (pi /2- ZP)*tan(pi /2- PM)) *180/ pi\nA=deg_to_dms(180-A*180/ pi);\nH=deg_to_dms(H/15);\n\n#result\nprint \" azimuth of setting sun in ( degrees,min,second)  \",A\nprint \" suns hour angle in ( hr,min,second ) : \",H\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " azimuth of setting sun in ( degrees,min,second)   [120, 49, 45.28]\n suns hour angle in ( hr,min,second ) :  [7, 27, 50.23]\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.16,Page 61"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding error in time\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ntime = -3 -28.41/60; # greenwich time at july 1 1951\nchange = -11.82/60;\ndef 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#calculation\nc12 = change/24*12 # change of time in 12 hours\ntch =time +c12;\ntch=deg_to_dms(tch/60);\n\n#result\nprint \" greenwich  mean time error in 12 th hour(-ve) in( deg,min,sec)  \",tch\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " greenwich  mean time error in 12 th hour(-ve) in( deg,min,sec)   [0, 3, 34.32]\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.17,Page 61"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding GAT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = (md - m) * 60\n    return [d, m, sd]\n#printing result in degree minute and seconds respectively \nGMN = -14*60 -10;\nchangeET =1*1.5;\n\n#calculation\nneterr =GMN+ changeET ;\nGAT = time + neterr ;\nGAT=GAT+10*3600+30*60;\nhr= round ( GAT /3600) ;\nb=GAT -hr *3600;\nmi= round (b /60 -1);\nc=GAT -hr *3600 - mi *60;\n\n#result\nprint hr,\"hour\",mi,\"minutes\",c,\"seconds of GAT\"",
+     "language": "python",
      "metadata": {},
-     "source": "Example 1.3,Page 23"
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "10.0 hour 15.0 minutes 48.0265 seconds of GAT\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.18,Page 62"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding time\n\n#part1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = (md - m) * 60\n    return [d, m, sd]\nA =50+12.0/60+48.0/3600;\ntime =A /15*3600\n\n#calculation\nhr= round ( time /3600) ;\nb=time -hr *3600;\nmi= round (b /60 -1);\nc=time -hr *3600 - mi *60;\n\n#result\nprint hr,\"hour\",mi,\"minutes\",c,\"seconds of angles\"\n\n#part 2\n#initialisation of variable\nA =8+18.0/60+6.0/3600;\ntime =A /15*3600\n\n#calculation\nhr= round ( time /3600-1) ;\nb=time -hr *3600;\nmi= round (b /60 );\nc=time -hr *3600 - mi *60;\n\n#result\nprint hr,\"hour\",mi,\"minutes\",c,\"seconds of angles\"\n\n#part 3\n#initialisation of variable\nA =258+36.0/60+30.0/3600;\ntime =A /15*3600\n\n#calculation\nhr= round ( time /3600) ;\nb=time -hr *3600;\nmi= round (b /60 );\nc=time -hr *3600 - mi *60;\n\n#result\nprint hr,\"hour\",mi,\"minutes\",c,\"seconds of angles\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "3.0 hour 20.0 minutes 51.2 seconds of angles\n-0.0 hour 33.0 minutes 12.4 seconds of angles\n17.0 hour 14.0 minutes 26.0 seconds of angles\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.19,Page 62"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding angle\n\n#part1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nA =4+34.0/60+13.0/3600;\n\n#calculation\nangle =A *15;\nangle=deg_to_dms(angle);\n\n#result\nprint \"angle in degree,minute,second respectively\",angle\n\n#part 2\n#initialisation of variable\nA =18+11.0/60+38.0/3600;\n\n#calculation\nangle =A *15;\nangle=deg_to_dms(angle);\n\n#result\nprint \"angle in degree,minute,second respectively\",angle\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "angle in degree,minute,second respectively [68, 33, 15.0]\nangle in degree,minute,second respectively [272, 54, 30.0]\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.20.a,Page 64"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding local mean time\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =20 # longitude of the place\nlongSM =82+30.0/60; # longitude of standard meridion\n\n#calculation\ndolong =longSM - longP ; # difference in longitude\ndot = dolong /15.0; # difference in time\nLMT =20+24.0/60+6.0/3600 - dot ;\nLMT=deg_to_dms(LMT)\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[16, 14, 6.0] Local mean time in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.20.b,Page 64"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding local mean time\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =-20 # longitude of the place\nlongSM =82+30.0/60; # longitude of standard meridion\n\n#calculation\ndolong =longSM - longP ; # difference in longitude\ndot = dolong /15.0; # difference in time\nLMT =20+24.0/60+6.0/3600 - dot ;\nLMT=deg_to_dms(LMT)\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[13, 34, 6.0] Local mean time in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.21.a,Page 64"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding GMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLMT =9+40.0/60+12.0/3600;\nlongP = -42 -36.0/60;\n\n#calculation\ndot = longP /15;\nGMT =LMT -dot;\nGMT=deg_to_dms(GMT);\n\n#result\nprint \"GMT in hours,minute,second respectively\",GMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[12, 30, 36.0] GMT in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.21.b,Page 64"
     },
     {
      "cell_type": "code",
      "collapsed": false,
-     "input": "#initialisation of variable\nT=273.0+30.0;#temperature\nl=0.85;#wavelength in m\np=752.4;\nc=299792.5e3;#speed of light in km/s\nf=24e6;#frequency in Hz\n\n#calculation\nno=1+(287.604+4.8864/l**2+0.068/l**4)/1e6;\nns=1+(no-1)*273/T*p/760;\nV=c/ns;\nl=V/f;\n\n#result\nprint \"wavelength of light in m\",round(l,3)",
+     "input": "#finding GMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLMT =4+32.0/60+10.0/3600;\nlongP = -56 -32.0/60;\n\n#calculation\ndot = longP /15;\nGMT =LMT +dot;\nGMT=deg_to_dms(GMT);\n\n#result\nprint \"GMT in hours,minute,second respectively\",GMT",
      "language": "python",
      "metadata": {},
      "outputs": [
       {
        "output_type": "stream",
        "stream": "stdout",
-       "text": "wavelenght of light in m 12.488\n"
+       "text": "[0, 46, 2.0] GMT in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.22,Page 65"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\n#part1\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nGCT =18+40.0/60+12.0/3600; # greenwich civil time\nlongP =72+30.0/60; # longitude of the place\n\n#calculation\ndot = longP /15.0;\nLMT = GCT +dot;\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT\n\n#part 2\n#initiallisation of variable\nGCT =18+40.0/60+12.0/3600; # greenwich civil time\nlongP =-72-30.0/60; # longitude of the place\n\n#calculation\ndot = longP /15.0;\nLMT = GCT +dot;\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT\n\n#part 3\n#initialisation of variable\ndef deg_to_dms(deg):\n    d = int(deg);\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d-24, m, sd]\nGCT =18+40.0/60+12.0/3600; # greenwich civil time\nlongP =110+32.0/60; # longitude of the place\n\n#calculation\ndot = longP /15.0;\nLMT = GCT +dot;\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT of next day in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "LMT in hours,minute,second respectively [23, 30, 12.0]\nLMT in hours,minute,second respectively [13, 50, 12.0]\nLMT of next day in hours,minute,second respectively [2, 2, 20.0]\n"
       }
      ],
      "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.23,Page 66"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLMT =10+20.0/60+30.0/3600; # local mean time\nlongP =102+30.0/60; # longitude of the place\n\n#calculation\ndot = longP /15;\nGMT =LMT -dot;\nmGMN =12 - GMT ; #mean time interval\ni= mGMN *0.32/3600; # increase in mGMN\nETGMN =5.0/60+4.35/3600;\nch=i+ ETGMN ; # change in GMT\nGMT =ch+GMT;\nLMT = GMT +dot;\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[10, 25, 37.07] LMT in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.24,Page 67"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLMT =15+12.0/60+40.0/3600; # local mean time\nlongP = -20 -30.0/60; # longitude of the place\n\n#calculation\ndot = longP /15;\nGMT =LMT -dot;\nmGMN =12 - GMT ; #mean time interval\ni= mGMN *0.32/3600; # increase in mGMN\nETGMN =5.0/60+4.35/3600;\nch=i+ ETGMN ; # change in GMT\nGMT =ch+GMT;\nLMT = GMT +dot;\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[15, 17, 42.89] LMT in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.25,Page 70"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding sidereal time\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\ntime =4+20.0/60+30.0/3600;\n\n#calculation\naccn = time *9.8565/3600; # acceleration\nstime = time + accn ; # sidereal time\nstime=deg_to_dms(stime);\n\n#result\nprint \"sidereal time in hours,minute,second respectively\",stime",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[4, 21, 12.79] sidereal time in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.26,Page 71"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding mean time\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nstime =8+40.0/60+50.0/3600;\n\n#calcculation\naccn =-time *9.8565/3600; # acceleration\nmtime = stime + accn ; # mean time\nmtime=deg_to_dms(mtime);\n\n#result\nprint \"mean time in hours,minute,second respectively\",mtime",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[8, 40, 7.21] mean time in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.27,Page 72"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LST on LMM\n\n#part 1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP = -160 -30.0/60 -30.0/3600; # longitude of the place\nGST =16+30.0/60+12.0/3600; # standard time\n\n#calculation\ndot = longP /15; # difference in time\ni= dot *9.8565/3600; # error\nLST =GST -i;\nLST=deg_to_dms(LST);\n\n#result\nprint \"LST of LMM in hours,minute,second respectively\",LST\n\n#part 2\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP = 160 +30.0/60 +30.0/3600; # longitude of the place\nGST =16+30.0/60+12.0/3600; # standard time\n\n#calculation\ndot = longP /15; # difference in time\ni= dot *9.8565/3600; # error\nLST =GST -i;\nLST=deg_to_dms(LST);\n\n#result\nprint \"LST of LMM in hours,minute,second respectively\",LST\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[16, 31, 57.47] LST of LMM in hours,minute,second respectively\n[16, 28, 26.53] LST of LMM in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.28,Page 73"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LST \n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =85+20.0/60; # longitude of the place\nGST =6+30.0/60; # standard time\nGMN =6+32.0/60+12.0/3600;\n\n#calculation\ndot = longP /15; # difference in time\ni= dot *9.8565/3600;  # error\nLST =GMN -i; #LST at L .M.N\ni2=GST *9.8565/3600; # error in GST\nLST2 =GST+i2;\nLST = LST + LST2 # lst at L .M.N\nLST=deg_to_dms(LST);\n\n#result\nprint \"LST in hours,minute,second respectively\",LST",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[13, 2, 19.99] LST in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.29,Page 75"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =112+20.0/60+15.0/3600; # longitude of the place\nGST =8+10.0/60+28.0/3600; #GST at GMM\nlst =18+28.0/60+12.0/3600; \n\n#calculation\ndot = longP /15; \ni= dot *9.8565/3600; # error\nLST = GST +i; #LST at L .M.N\nLMM =lst -LST;\ni2=LMM *9.8565/3600; # error in LMM\nLMT =LMM -i2; # local mean time\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[10, 14, 48.91] LMT  in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.30,Page 76"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LST\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =85+20.0/60; # longitude of the place\nGST =18+30.0/60; # standard time\ngst =6+32.0/60+12.0/3600; #GST at GMN\n\n#calculation\ndot = longP /15; \nGMT =GST -dot -12;\ni= GMT *9.8565/3600; # error\nGMT = GMT +i; # SI time\nLST = GMT +dot+ gst ; #LST at LMT\nLST=deg_to_dms(LST);\n\n#result\nprint \"LST  in hours,minute,second respectively\",LST",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[13, 2, 19.99] LST  in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.31,Page 78"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =112+20.0/60+15.0/3600; # longitude of the place\nGST =8+10.0/60+28.0/3600; #GST at GMM\nlst =18+28.0/60+12.0/3600; # local sidereal  time\n\n#clculation\ndot = longP /15; \ngmm = lst +dot - GST ; # SI at GMM\ni= gmm *9.8565/3600; # error\ngmm =gmm -i; #LST at L .M.N\nLMT =gmm -dot; # local mean time\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[10, 14, 48.7] LMT  in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 44
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.32,Page 79"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#part 1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =162+30.0/60+15.0/3600; # longitude of the place\nGST =10+30.0/60+15.0/3600; #GST at GMM\nRA =22+11.0/60+30.0/3600; # local sidereal  time\n\n#calculation\ndot = longP /15; \ni= dot *9.8565/3600; # e r r o r\ngmm = GST +i; #LST at L .M.N\nlmn =RA -gmm; # SI o f LMN\ni2=lmn *9.8565/3600; # error 2\nLMT =lmn -i2;\nLMT1=deg_to_dms(LMT);\n\n#result\nprint \"LMT observed at upper transit in hours,minute,second respectively\",LMT1\n\n#part 2\n#initialisation of variable\ni3 =12*9.8565/3600; # retardation\n\n#calculation\nLMT = LMT +12 - i3;\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT observed at lower transit in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "LMT observed at upper transit in hours,minute,second respectively [11, 37, 33.31]\nLMT observed at lower transit in hours,minute,second respectively [23, 35, 35.04]\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.33,Page 80"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =60+30.0/60; # longitude of the place\nGST =7+30.0/60+48.6/3600; #GST at GMM\nRA =17+28.0/60 +40.0/1600;\n\n#calculation\ndot = longP /15; \ni= dot *9.8565/3600; # error\ngmm =GST -i; #LST at L .M.N\nLMT =RA -gmm; # local mean time\ni2=LMT*9.8296/3600;\nGMT=LMT-i2-longP/15;\nLMT=deg_to_dms(LMT);\nGMT=deg_to_dms(GMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT\nprint \"GMT in hours,minute,second respectively\",GMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "LMT in hours,minute,second respectively [9, 59, 21.15]\nGMT in hours,minute,second respectively [5, 55, 42.96]\n"
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.34,Page 82"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding correct time\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nGMT=13+21.0/60+54.0/3600; #GMT of the place\nLongA=40+30.0/60; #longitude of A\nLongB=-40-30.0/60; #longitude of B\n\n#calculation\ndelA=LongA/15.0*9.8296/3600;#error\ndelB=LongB/15.0*9.8296/3600;#error\nGA=GMT+delA;\nGA=deg_to_dms(GA);\nGB=GMT+delB;\nGB=deg_to_dms(GB);\n\n#result\nprint \"corected time of A  in hours,minute,second respectively\",GA\nprint \"corected time of B  in hours,minute,second respectively\",GB",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "corected time of A  in hours,minute,second respectively [13, 22, 20.54]\ncorected time of B  in hours,minute,second respectively [13, 21, 27.46]\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.35,Page 83"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nGMT=7+12.0/60+28.0/3600; #GMT of the place\nLong=50+30.0/60; #longitude \nLMT=11+30.0/60+12.0/3600; \n\n#calculation\ndelA=LongA/15.0*9.8296/3600;#error\ndelB=LMT*9.8296/3600;#error\nLA=GMT+delA; #LMT at transit\nLMT=LMT-delB;\nLMT=LMT+LA;\nLMT=deg_to_dms(LMT);\n\n#result\nprint \"LMT in hours,minute,second respectively\",LMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "LMT in hours,minute,second respectively [18, 41, 13.47]\n"
+      }
+     ],
+     "prompt_number": 41
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.36,Page 84"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nGST=8+25.0/60+25.0/3600;\n\n#calculation\nGMT=24-GST;\ni=GMT*9.8296/3600;\nGMT=GMT-i;\nGMT=deg_to_dms(GMT);\n\n#result\nprint \"GMT in hours,minute,second respectively\",GMT",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "GMT in hours,minute,second respectively [15, 32, 1.89]\n"
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.37,Page 85"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LMT on July 2\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nGMT=12+03.0/60+46.09/3600; #GMT of the place\nLongA=-130.0; #longitude of A\nLongB=49.0;  #longitude of B\n\n#calculation\ndelA=LongA/15.0*11.71/24/3600;\ndelB=LongB/15.0*11.71/24/3600;\nLMTA=GMT+delA;\nLMTA=deg_to_dms(LMTA);\nLMTB=GMT+delB;\nLMTB=deg_to_dms(LMTB);\n\n#result\nprint \"LMT at A on July 2 in hours,minute,second respectively\",LMTA\nprint \"LMT at B on July 2 in hours,minute,second respectively\",LMTB",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[12, 3, 41.86] LMT at A in hours,minute,second respectively\n[12, 3, 47.68] LMT at B in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.38,Page 86"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding LST\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLat=50+30.0/60; # latitude of place\nDec=74+22/60; #declination of place\nRA=14+50.0/60+52.0/3600;\n\n#calculation\nH=acos(tan(Lat*pi/180)/tan(Dec*pi/180));\nH=H*180/pi;\nH=H/15.0;\nLST=H+RA;\nLST=deg_to_dms(LST);\n\n#result       \nprint \"LST  in hours,minute,second respectively\",LST",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[19, 29, 26.59] LST  in hours,minute,second respectively\n"
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.39,Page 87"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding HA\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLong=120+30.0/60; #langitude\nGST=14+30.0/60+28.25/3600;\nGMT=2+5.0/60;\nLMN=12.0;\nLST=14+31.0/60+47.43/3600;\nRA=23+20.0/60+20.0/3600;\n\n#calculation\ne1=Long*15.0*9.8565/3600;\nGST=GST+e1;\nLMT=GMT+24-8-2.0/60;\nLMM=LMN+24-LMT; #mean LMN\ne2=LMM*9.8565/3600;\nLMM=LMM+e2;\nLST=LST+24-LMM;\nHA=LST-RA+24;\nHA=deg_to_dms(HA);\n\n#result\nprint \"HA in hours,minutes,seconds respectively\",HA",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[21, 11, 30.51] HA in hours,minutes,seconds respectively\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.40,Page 86"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding HA\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLong=75+28.0/60; #langitude\nLMT=5+30.0/60;\nGST=20+15.0/60+32.58/3600;\n\n#calculation\ne1=Long/15.0*9.8565/3600;\nGST=GST+e1;\ne2=LMT*9.8565/3600;\nLMT=LMT+e2;\nHA=GST+LMT;\nHAMS=HA-LMT-12+e2;\nHA=deg_to_dms(HA);\nHAMS=deg_to_dms(HAMS);\n\n#result\nprint \"HA in hours,minutes,seconds respectively\",HA\nprint \"HAMS in hours,minutes,seconds respectively\",HAMS\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "HA in hours,minutes,seconds respectively [25, 47, 16.38]\nHAMS in hours,minutes,seconds respectively [8, 17, 16.38]\n"
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.41,Page 90"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding sun's declination\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLong=45.0; #langitude\ndel1=1067.2/3600;\ndel2=1083.9/3600;\ndel3=1100.3/3600;\nf0=-16-14.0/60-24.0/3600;\n\n#calculation\nn=(10-Long/15)/24.0;\nDel0=del2-del1;\nDel1=del3-del2;\nfn=-f0+n*del2+n*(n-1)/4*(Del0+Del1)\nfn=deg_to_dms(fn);\n\n#result\nprint \"sun's declination in hours,minutes,seconds respectively\",fn",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "sun's declination in hours,minutes,seconds respectively [16, 19, 38.43]\n"
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.42,Page 101"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth of A & B and vertical difference\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\naziA =32+41.0/60+30.0/3600; # azimuth o f A\naziB =110+28.0/60+42.0/3600; # azimuth o f B\nvaA =10+21.0/60+12.0/3600; # vertical angle of A\nvaB = -2 -18.0/60 -30.0/3600; # vertical angle o f B\nlA1 =11;\nlB1 =11.5;\nrA1 =7.5;\nrB1 =7;\nlB2 =10;\nlA2 =10.5;\nrB2 =7.5;\nrA2 =8;\nd =20;\n# partA\n#calculation\nsigl =lA1+ lA2 ;\nsigr =rA1+ rA2 ;\nb= sigl /4*d- sigr /4*d;\ni= tan( vaA );\ncaziA = aziA +i *29.95/3600;\ncaziA1=deg_to_dms(caziA);\n\n#result\nprint \"corrected azimuth of A in (degrees,minutes,seconds)\",caziA1\n\n#part2 \n#calculation\ni= tan( vaB );\ncaziB = aziB +i*b /3600;\nha=caziB - caziA;\ncaziB=deg_to_dms(caziB);\nha=deg_to_dms(ha);\n\n#result\nprint \"corrected azimuth of B in (degrees,minutes,seconds)\",caziB\nprint \"horizontal difference of angle between A & B in (degrees,minutes,seconds)\",ha\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "corrected azimuth of A in (degrees,minutes,seconds) [32, 42, 10.04]\ncorrected azimuth of B in (degrees,minutes,seconds) [110, 29, 15.02]\nhorizontal difference of angle between A & B in (degrees,minutes,seconds) [77, 47, 4.98]\n"
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.43,Page 102"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding altitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nv1 =18+36.0/60+48.0/3600; # vertical angle 1\nv2 =18+35.0/60+56.0/3600; # vertical angle 2\nslm =28+36.0/60+20.0/3600; # altitude of sun measured\nds =15.0/60+59.35/3600; # dia of sun\n\n#calculation\nmv =( v1+v2) /2; #mean vertical angle\ni=v1 -v2; # error\nsl=slm+i; #new altitude of sun\nsl=sl+ds;\nir = -57.0/3600/( tan( slm *pi /180+26* pi /180/3600) );#error due to refraction\nsl=sl+ir;\nip =8.8/3600* cos( slm *pi /180+26* pi /180/3600);# error due to parallex\nsl=sl+ip;\nsl=deg_to_dms(sl);\n\n#result\nprint \"corrected altitude in (deg,min,sec) respectively\",sl",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[28, 51, 34.59] corrected altitude in (deg,min,sec) respectively\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.44,Page 115"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding chronometer error\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlong =4+30.0/60;\ni= long *9.8565/3600; # longitude\ngst =14+38.0/60+12.0/3600; #GST on GMM\nlst =gst -i; #LST on LMM\nRA =7+36.0/60+21.24/3600;\n\n#calculation\nLST =RA;\nSI=LST - lst +24;\nLCT =17+56.0/60+8.86/3600 -1; # local chronometer time\ni2=SI *9.8296/3600;\nLMM =SI -i2;\nce=LCT - LMM ;\n\n#result\nprint  \" chronometer error in (s)\",round(ce *3600,2)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "2.19  chronometer error in (s)\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Exampe 1.45,Page 116"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding chronometer error\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nc =90 -36 -30.0/60 -30.0/3600; # co latitude\np =90 -16 -12.0/60 -18.4/3600; # co declination\nz =90 -30 -12.0/60 -30.0/3600; # co altitude\ns=(p+z+c) /2;\n\n#calculation\ns1=s-c;\ns2=s-p;\ns3=s-z;\nH =2* atan ( sqrt (sin(s1*pi /180) * sin (s2*pi /180) / sin (s*pi/180) / sin (s3*pi /180) ));\nH=H *180/ pi;\nH=24 -H /15;\nLST =H +5+18.0/60+12.45/3600 -24;\nce =1+2.0/60+5.25/3600 - LST ;\n\n#result\nprint  \" chronometer error in (s)\",round (ce *3600+2,2)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "19.34  chronometer error in (s)\n"
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "Example 1.46,Page 118"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding chronometer error\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nc =90 -36 -40.0/60 -30.0/3600; # co latitude\np =90 -17 -26.0/60 -42.1/3600; # co declination\nz =90 -36 -14.0/60 -16.8/3600; # co altitude\n\n#calculation\ns=(p+z+c) /2;\ns1=s-c;\ns2=s-p;\ns3=s-z;\nH =2* atan ( sqrt (sin(s1*pi /180) * sin (s2*pi /180) / sin(s*pi/180) / sin (s3*pi /180) ));\nH=H *180/ pi;\nH=H /15;\ni =12 -11 -56.0/60 -22.8/3600; # error in time\nLAT =15+49.0/60+40.6/3600; # local actual time\nGAT =LAT -H;\nGMT =GAT -i;\nLMT = GMT +H;\nce =15+49.0/60+12.6/3600 - LMT;\nce=deg_to_dms(ce)\n\n#result\nprint  \" chronometer error in (s)\",ce",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " chronometer error in (s) [0, 3, 9.2]\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.47,Page 119"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding chronometer error\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nRA =17+12.0/60+48.0/3600;\ngst =9+26.0/60+12.0/3600; #GST on GMN\nlong =138.0/15+45.0/15/60; # l o n g i t u d e\nlst =- long *9.85645/3600+9+26.0/60+12.0/3600; #LST on LMN\nLST =17+12.0/60+48.0/3600; # local sidereal time\n\n#calculation\nSI=LST - lst ;\nMI=-SI *9.8296/3600+ SI;\nLCT =7+47.0/60+2.0/3600; # local chronometer time\nce=LCT -MI;\n\n#result\nprint \" chronometer error in (s)\",round(ce *3600,2) ",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " chronometer error in (s) 11.52\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.48,Page 145"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth,latitude,LMT\n\n#part 1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\ntheta =54+30.0/60; # l o g i t u d e\ndelta =62+12.0/60+21.0/3600; # d e c l i n a t i o n\n\n#calculation\nlat = asin (sin( theta *pi /180) /sin( delta *pi /180) );\nlat = lat *180/ pi;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;\n\n#part 2\n#initialisation of variable\nA =53+25.0/60; # azimuth of star\nh =65+18.0/60+42.0/3600; # horizontal angle\n\n#calculation\nA=A+h;\nA=360-A;\nA=deg_to_dms(A);\n\n#result\nprint \"Azimuth in (deg,min,sec)\",A;\n\n#part 3 \n#initialisation of variable\nlst =4+39.0/60+6.5/3600; #LST o f LMN\nLST =10+58.0/60+38.0/3600+2+49.0/60+25.3/3600; #LST of observation\n\n#calculation\nLMN =LST -lst;\ni= LMN *9.8565/3600; # e r r o r\nLMT =LMN -i;\nLMT=deg_to_dms(LMT)\n\n#results\nprint \"LMT in (hr,min,sec)\",LMT;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[66, 58, 7.13] Latitude in (deg,min,sec)\n[241, 16, 18.0] Azimuth in (deg,min,sec)\n[9, 7, 26.62] LMT in (hr,min,sec)\n"
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.49,Page 148"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth,latitude,LMT\n\n#part 1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\ntheta =53+32.0/60; # logitude\ndelta =56+42.0/60+53.2/3600; # declination\nlat = asin (sin ( theta *pi /180) /sin( delta *pi /180) );\nlat = lat *180/ pi;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;\n\n#part 2\n#initialisation of variable\nAs= asin ( cos ( delta *pi /180) /cos ( theta *pi /180) ); #azimuth of star\nh =75+18.0/60+20.0/3600; # angle between line and star\n\n#calculation\nA=h-As *180/ pi;\nA=360 -A;\nA=deg_to_dms(A);\n\n#result\nprint \"Azimuth in (deg,min,sec)\",A;\n\n#part 3\n#initialisation of variable\nLST =10+58.0/60+3.9/3600+22+10.0/60+38.5/3600 -24; #LST of observation\nlong =5+40.0/60+18.0/3600; # longitude\n\n#calculation\ni= long *9.8565/3600; # error\nlst =4+58.0/60+23.84/3600+ i; #LST on LMN\nLMM =LST -lst;\ni2=LMM *9.8565/3600; # error in LMM\nLMT =LMM -i2;\nLMT=deg_to_dms(LMT)\n\n#results\nprint \"LMT in (hr,min,sec)\",LMT;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Latitude in (deg,min,sec) [74, 9, 33.08]\nAzimuth in (deg,min,sec) [352, 7, 3.66]\nLMT in (hr,min,sec) [4, 8, 41.69]\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.50,Page 151"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nLong =(15.0+30.0/60); \nGMT =19+12.0/60+28.6/3600;\nGST=10+12.0/60+36.2/3600;\nRA=10 +12.0/60 +6.3/3600;\ntheta=35.0;\ndelta=20+6.0/60+48.4/3600;\n\n#calculation\ni= Long/15.0 *9.8656/3600; \nLSTofLMM=GST-i;\nLMT = GMT + Long/15.0 ;\ni2=LMT *9.8656/3600; # error in LMT\nSI = LMT +i2;\nLST =LSTofLMM+ SI ;\nH=LST-RA ; # hour angle\nH=H *15;\nH=360 -H;\nB=atan(tan(delta*pi/180)/cos(H*pi/180));\nB=B*180/pi;\nx=B-theta; #defined as 'B-theta'\nAs= atan ( tan ((H) *pi /180) * cos((B) *pi /180) / sin ((x)*pi /180) );#calculating azimuth \nh =36+28.0/60+18.0/3600; # angle between line and star\nA =180+ As *180/ pi -h; #azimuth\nA=deg_to_dms(A);\n\n#result\nprint \"Azimuth in (deg,min,sec)\",A;\nprint \"there is a miscalculation in the step of calculating As thus resulted in change in answer\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Azimuth in (deg,min,sec) [55, 16, 8.27]\nthere is a miscalculation in the step of calculating As thus resulted in change in answer\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.51,Page 153 "
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nalpha =33+35.0/60+10.0/3600; # altitude\nZM =90 - alpha ;\ndelta =22+5.0/60+35.0/3600; # declination\nPM =90 - delta ;\ntheta =52+30.0/60+20.0/3600; # latitude\nZP =90 - theta ;\n\n#calculation\nAs= acos (( cos (PM*pi /180) -cos(ZP*pi /180) * cos (ZM*pi/180) )/( sin (ZP*pi /180) *sin(ZM*pi /180) ));\nh =18+20.0/60+30.0/3600; # angle between line and star\nA=As *180/ pi+h;\nA=deg_to_dms(A);\n\n#result\nprint \"Azimuth in (deg,min,sec)\",A;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Azimuth in (deg,min,sec) [115, 27, 19.68]\n"
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.52,Page 154"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding declination,altitude\n\n#part 1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nGAT =5+17.0/60+6.0/60; #GAT \ndelta =17+46.0/60+52.0/3600; # declination\n\n#calculation\ni =37.0/3600* GAT ;\ndelta =delta -i;\ndelta1=deg_to_dms(delta);\n\n#result\nprint \"Declination in (deg,min,sec)\",delta1;\n\n#part 2\n#initialisation of variable\np=90 - delta ; # co declination\naltitude =23+15.0/60+20.0/3600; # altitude of sun\ni2 =2.0/60+12.0/3600; # error due to refraction\ni3 =8.0/3600; # error due to parallax\n\n#calculation\naltitude = altitude -i2+i3;\nc =90 -55 -46.0/60 -12.0/3600; # colatitude\nz=90 - altitude ; # co altitude\ns=(p+z+c) /2;\ns1=s-c;\ns2=s-p;\ns3=s-z;\nA =2* atan ( sqrt (sin(s3*pi /180) * sin (s1*pi /180) / sin (s*pi/180) / sin (s2*pi /180) ));\nA=A *180/ pi;\nA=deg_to_dms(A);\n\n#result\nprint \"Altitude in (deg,min,sec)\",A;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Declination in (deg,min,sec) [17, 43, 32.82]\nAltitude in (deg,min,sec) [92, 23, 10.67]\n"
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.53,Page 156"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nGMT =17+5.0/60+2.0/3600; \ni =9.8565/3600* GMT;\nGST =3+12.0/60+12.0/3600;\nwl =1+18.0/60; # west longitude\nRA =16+23.0/60+30.0/3600;\n\n#calculation\nH= GMT +i+ GST +wl -RA; # hour angle\nH=H *15;\np =90 -29 -52.0/60;\nc =90 -52 -8.0/60;\nz= acos ( cos (H*pi /180) * sin (p*pi /180) *sin(c*pi /180) + cos(p*pi /180) * cos (c*pi /180) );\nA= asin ( sin (p*pi /180) * sin (H*pi /180) /sin(z));\nA=A *180/ pi\nA=deg_to_dms(A);\n\n#result\nprint \"Azimuth in (deg,min,sec)\",A;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Azimuth in (deg,min,sec) [78, 38, 33.24]\n"
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.54,Page 157"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding azimuth\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nc2 =24+30.0/60+20.0/3600;\nd2 =24+30.0/60+40.0/3600;\nc3 =25;\nd3 =25+1.0/60;\n\n#calculation\nalt =( c2+c3+d3+d2)/4;#mean observed altitude\nil =(10.6 -9.4) /4*15.0/3600; # error \nalt = alt +il;\nir = -57.0/3600/ tan (( alt *pi /180) ); # correction of refraction\nip =8.0/3600* cos (alt*pi /180) ; # correction of parallax\nalt =alt -ir+ip;#altitude corrected\nz=90 - alt;#ZM\ndelta =1+32.0/60+16.8/3600 -56.2/3600*(3.0/60+1.86/3600) ;#declination of sun\np=90 - delta ;#PM\nc =90 -36 -48.0/60 -30.0/3600;#ZP\ns=(p+z+c) /2;\ns1=s-c;\ns2=s-p;\ns3=s-z;\nA =2* atan ( sqrt (sin(s3*pi /180) * sin (s1*pi /180) / sin (s*pi/180) / sin (s2*pi /180) ));#azimuth calculation\nA=A *180/ pi;\nA=A +81+59.0/60+10.0/3600;\nA=360 -A;\nA=deg_to_dms(A);\n\n#result\nprint \"Azimuth in (deg,min,sec)\",A;\nprint \"there is a miscalculation in the step of calculating Azimuth and error due to refrection thus resulted in change in answer\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Azimuth in (deg,min,sec) [170, 1, 36.93]\nthere is a miscalculation in the step of calculating Azimuth and error due to refrection thus resulted in change in answer\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.55,Page 178"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nalpha =65+40.0/60+18.0/3600; # altitude\ndelta =53+12.0/60+10.0/3600; # declination\n\n#calculation\ni =57.0/3600*1/ tan ( alpha *pi /180) ; \nalpha =alpha -i;\nz=90 - alpha ; # zenith distance\nlat =delta -z;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[28, 52, 2.23] Latitude in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 58
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.56,Page 178"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nalpha =64+36.0/60+20.0/3600; # altitude\ndelta =26+12.0/60+10.0/3600; # declination\n\n#calculation\ni =57.0/3600*1/ tan ( alpha *pi /180) ; # error\nalpha =alpha -i;\nz=90 - alpha ; # zenith distance\nlat = delta +z;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Latitude in (deg,min,sec) [51, 36, 17.06]\n"
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.57,Page 178"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nalpha =44+12.0/60+30.0/3600; # altitude\nlongP =75+20.0/60+15.0/3600; # longitude of place\ndelta =22+18.0/60+12.8/3600; # declination of sun\n\n#calculation\ni =57.0/3600*1/ tan ( alpha *pi /180) ; # error\ni2 =8.78/3600* cos( alpha ); #error due to parallax\ni3 =15.0/60+45.86/3600; #error due to semi diameter\nalpha =alpha -i+i2+i3;\nz=90 - alpha ; # zenith distance\ndelT = longP /15;\ni4 =6.82/3600* delT ; # error in time\ndelta =i4+ delta ;\nlat = delta +z;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[67, 51, 21.23] Latitude in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 62
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.58,Page 180"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding alpha1,alpha2\n\n#for alpha 1\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\ntheta =80;\ndelta =46+45.0/60+30.0/3600;\n\n#calculation\nalpha1 =90 - theta + delta ;\nalpha1=deg_to_dms(alpha1);\n\n#result\nprint \"alpha1 to the north in (deg,min,sec)\",alpha1\n\n#for alpha2\n#calculation\nalpha2 = theta +delta -90;\nalpha2=deg_to_dms(alpha2)\n\n#result\nprint \"alpha2 to the south(deg,min,sec)\",alpha2",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[56, 45, 30.0] alpha1 to the north in (deg,min,sec)\n[36, 45, 30.0] lpha1 to the south(deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 69
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.59,Page 181"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\ndelta1 =20+25.0/60+48.0/3600; # declination of star 1\ndelta2 =79+30.0/60+52.0/3600; # declination of star 2\nalpha1 =48+18.0/60+12.0/3600; #altitude of star 1\nalpha2 =47+54.0/60+6.0/3600; #altitude of star 2\n\n#calculation\nr1 =58.0/3600/ tan( alpha1 *pi /180) # error 1\nr2 =58.0/3600/ tan( alpha2 *pi /180) # error 2\nlat =90 -( alpha1 - alpha2 ) /2+( delta1 - delta2 ) /2+( r1 -r2)/2;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;\nprint \" there is a miscalculation in the step of calculating (delta1-delta2)/2 so there is a difference in the answer\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "Latitude in (deg,min,sec) [60, 15, 24.63]\n there is a miscalculation in the step of calculating (delta1-delta2)/2 so there is a difference in the answer\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.60,Page 182"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude,declination\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nalphal =18+36.0/60+40.0/3600; #altitude at lower culmination\nalphau =59+48.0/60+20.0/3600; #altitude at upper culmination\nlat =( alphal + alphau )/2;\nlat1=deg_to_dms(lat);\ndelta =90+ lat - alphau ;\ndelta1=deg_to_dms(delta);\n\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat1;\nprint \"Declination of star in (deg,min,sec)\",delta1",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[39, 12, 30.0] Latitude of star in (deg,min,sec)\n[69, 24, 10.0] Declination of star in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 74
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.61,Page 183"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nalpha =40+36.0/60+30.0/3600; # altitude of star\ndelta =10+36.0/60+40.0/3600; # declination of star\nH =46+36.0/60+20.0/3600; # hour angle of star\n\n#calculation\nn= atan ( tan ( delta *pi /180) /cos(H*pi /180) );\nlat =n+ acos ( sin ( alpha *pi /180) *sin(n)/ sin ( delta *pi/180) );\nlat = lat *180/ pi;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[36, 49, 43.99] Latitude in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 75
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.62,Page 183"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nalpha =42+10.0/60+40.0/3600; # altitude o f sun\ndelta =23+12.0/60+18.6/3600; # declination of sun ' s angle\nLMT =14+50.0/60;\n\n#calculation\ni =57.0/3600*1/ tan ( alpha *pi /180) ; # error\ni2 =8.78/3600* cos( alpha ); # correction due to parallax\ni3 =15.0/60+45.86/3600; # coreection due to semi diamter\nlongP =108+30.0/60; # longitude of place\nalpha =alpha -i+i2+i3;\ndelT = longP /15; # change in time\nGMT = LMT + delT ;\ni4 =1.2/3600* GMT; # error in time\nH=( GMT -12+ i4 - delT ) *15; # hour angle\ni5 =10.6/3600* GMT; # error in declination\ndelta = delta +i5;\nZM =(90 - alpha )*pi /180;\nPM =(90+ delta )*pi /180;\nA= asin ( sin (PM)/sin(ZM)* sin (H*pi /180) );\nA=pi -A;\nZP =2* atan ( sin (A/2+ H*pi /360) / sin (A/2-H*pi /360) *tan(PM/2- ZM /2) );\nlat =pi /2- ZP;\nlat = lat *180/ pi +1+6.0/60;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[1, 19, 7.46] Latitude in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 78
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.63,Page 185"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\ndelta =15+20.0/60+48.0/3600; # declination of star\nInt =9+22.0/60+6.0/3600; # interval\n\n#calculation\ndint =Int *9.8565/3600; # change in interval\nH=( Int+ dint ) *15/2; # hour angle\nlat = atan (tan( delta *pi /180) /cos(H*pi /180) );\nlat = lat *180/ pi +5.0/6*16.0/3600;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[39, 22, 1.79] Latitude in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 79
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.64,Page 186"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nRA =1+41.0/60+48.64/3600;\nlat =48+36.0/60+40/3600; # latitude\ndelta =88+58.0/60+28.26/3600; # declination of polaris\nGMM =16+48.0/60+20.86/3600;\nlongP =7+20.0/60; # longitude of place P\ni1 =51.0/3600; # error due to barometer\ni2 =1.0/3600; # error due to barometer\ni3 = -1.0/3600; # error due to temp\n\n#calculation\nlat =lat -i1+i2+i3;\ndelT = longP /15;\ni4= delT *9.8565/3600;\nlst = GMM +i4;\nLMT =20+24.0/60+50.0/3600;\ni6 =9.8565/3600* LMT ; # e r r o r i n LMT\nLST = LMT +i6+lst -24;\nH=LST -RA; # hour a n g l e\nH=H *15;\nlat =lat -(90 - delta )*cos(H*pi /180) +.5* sin (1/3600* pi/180) *(90 - delta ) **2*( sin (H*pi /180) )**2* tan ( lat *pi/180) ;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[49, 36, 18.45] Latitude in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 81
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 1.65,Page 187"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding latitude\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin,cos,acos,atan,asin\ndef deg_to_dms(deg):\n    d = int(deg)\n    md = abs(deg - d) * 60\n    m = int(md)\n    sd = round((md - m) * 60,2)\n    return [d, m, sd]\nlongP =120 -4 -20.0/60; # longitude of point\nGST =8+30.0/60+20.0/3600; #GST on GMM\ndelta =6+15.0/60+2.0/3600; # deflection\nalpha =39+20.0/60+30.0/3600; # altitude\ntheta =56+54.0/60+30.0/3600; # longitude\n\n#calculation\ndelT = longP /15;\ni= delT *9.8565/3600; # error in time\nlst = GST +i; #LST on LMM\nLST =19+52.0/60+16.0/3600;\nRA=LST;\nLMN =LST -lst;\ni2=LMN *9.8565/3600; # error in LMN\nLMN =LMN -i2;\nOSM =10+55.0/60+30.0/3600; #Observed mean time\ni3 =1.0/60+25.0/3600; # e r r o r i n obs e r v ed t ime\nOSM =OSM -i3;\nLMT = OSM +4.0/15+21.0/60.0/15;\nI=LMN - LMT ; # interval\ni4 =1.21/3600; # error in interval\nI=I+i4;\nH=I; # hour angle\nB= cos ( delta *pi /180) * cos ( theta *pi /180) / cos ( alpha *pi/180) ;\nm =225* H **2*3600**2/2.0/206265.0;\nlat = alpha +m*B /3600;\nlat =90 - lat +6+15.0/60+2.0/3600;\nlat=deg_to_dms(lat);\n\n#result\nprint \"Latitude in (deg,min,sec)\",lat;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "[56, 53, 17.77] Latitude in (deg,min,sec)\n"
+      }
+     ],
+     "prompt_number": 85
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
     }
    ],
    "metadata": {}
diff --git a/Surveying_Volume_3/Chapter2.ipynb b/Surveying_Volume_3/Chapter2.ipynb
new file mode 100755
index 00000000..4be4bac6
--- /dev/null
+++ b/Surveying_Volume_3/Chapter2.ipynb
@@ -0,0 +1,448 @@
+{
+ "metadata": {
+  "name": "Chapter 2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": "PHOTOGRAPHIC SURVEYING"
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.1, Page 215"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#Finding azimuth of a,b,c\n\n# Initialization of Variable\nfrom math import pi\nfrom math import atan\nf =120.80 # focal length\na = -35.52 # elevation of A\nb =8.48 # elevation of B\nc =48.26 # elevation of C\n\n#calculation\nalphaa = atan (a/f);\nalphab = atan (b/f);\nalphac = atan (c/f);\nphi =(354+30/60) *pi /180; # azimuth o f camera\nphia =phi - alphaa -360* pi /180; # azimuth o f a\nphib = phia + alphab; # azimuth o f b\nphic = phia + alphac ; # azimuth o f c\n\n#result\nprint   \" azimuth of a in ( degrees ) \",round(phia /pi *180,2)\nprint  \" azimuth of b in ( degrees ) \",round(phib /pi *180,2)\nprint   \" azimuth of c in ( degrees ) \",round(phic /pi *180,2)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " azimuth of a in ( degrees )  10.39\n azimuth of b in ( degrees )  14.4\n azimuth of c in ( degrees )  32.16\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.2,Page 216"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#Finding distance AP and AQ and reduced elevation of A\n\n#initialisation of variable\nfrom math import pi\nfrom math import atan,sin,sqrt\nf =150.0; # focal length of camera\nap =20.2 # elevation of a from p\naa1 =16.4; # distace to the right\naq =35.2 # elevation of a from q\nPQ =100.0; # distace of PQ\nRL =126.845; # r educed level of instrument\n\n#calculation\nalphap = atan (ap/f);\nalphaq = atan (aq/f);\nP=pi /3- alphap ; # angle P\nQ =40* pi /180 - alphaq ; # angle Q\nA=pi -P-Q; # angle A;\nAP=PQ* sin (Q)/sin(A);\nAQ=PQ* sin (P)/sin(A);\nPa1 = sqrt (ap **2+ f **2) ;\nAA1 = aa1 *AP/ Pa1 ;\nRLa =RL+AA1; # reduced level of A\n\n#result\nprint  \" distance of AP (m)  \",round(AP,2);\nprint \"distance of AQ (m) \",round(AQ,2);\nprint \" reduced level of A in (M) \",round(RLa,2)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " distance of AP (m)   45.9\ndistance of AQ (m)  80.6\n reduced level of A in (M)  131.82\n"
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.3,Page 218"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding focal length\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\ntheta =(44+30/60) *pi /180; # angle b/w two points\nx1 =68.24; #distance of 1st point\nx2 =58.48; #distance of 2nd point\n\n#calculation\nf=( x1+x2)/ tan ( theta ) /2+ sqrt (( x1+x2) **2/4/( tan ( theta ))\n**2+ x1*x2);\n\n#result\nprint \" focal length of lens in (mm) \",round(f,2);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " focal length of lens in (mm)  156.69\n"
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.4, Page 240"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding representative fraction\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\n# part 1\n\nH =1200.0;#altitude\nh =80.0; #elevation of hill\nf =15.0/100.0;\n\n#calculation\nR80 =f/(H-h);\nprint \" representative fraction of hill is ( time s) \",round(R80,5);\n\n# part 2\n#initialisation of variable\nh =300.0; #elevation of hill\n\n#calculation\nR300 =f/(H-h);\n\n#result\nprint \" representative fraction of hill is ( time s) \",round(R300,5) ;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " representative fraction of hill is ( time s)  0.00013\n representative fraction of hill is ( time s)  0.00017\n"
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.5,Page 240"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding height above sea level\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nR =1.0/8000.0;\nh =1500.0;\nf =20.0/100.0;\n\n#calculation\nH=h+f/R;\n\n#result\nprint \" height above sea level in (m)  \",round(H,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " height above sea level in (m)   3100.0\n"
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.6,Page 241"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding height above sea level\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nh =500.0; #elevation of point\nf =20.0/100.0; # focal length\nv =8.65/100.0; # vertical distance of photograph\nho =2000.0; # horizontal distance of photograph\nR=v/ho; # representative  fraction\nh1 =800;\n\n#calculation\nH=h+f/R;\nS=(H-h1)/f /100; # scale of photograph\n\nprint \" height above sea level in (m)  \",round(H,2);\nprint \" 1cm in photograph represents centimetres  \",round(S,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " height above sea level in (m)   5124.28\n 1cm in photograph represents centimetres   216.214\n"
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.7, Page 241"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding height above sea level\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nm =1.0/50000.0; #map scale\npd =10.16; # photo distance\nmd =2.54; #map distance\nf =16.0/100.0;\nh =200;\n\n#calculation\nR=pd/md*m; # representative  fraction\nH=h+f/R;\n\n#result\nprint \" height above sea level in (m) \",round(H,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " height above sea level in (m)  2200.0\n"
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.8,Page 242"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding distance between A and B\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nf =20 # f o c a l l e n g t h\nxa =2.65; # x coordinate of a\nxb = -1.92; # x coordinate of b\nya =1.36; # x coordinate of a\nyb =3.65; # y coordinate of b\nH =2500.0;\nha =500.0; # elevation of a\nhb =300.0; # elevation of b\n\n#calculation\nXa =(H-ha)/f*xa;\nXb =(H-hb)/f*xb;\nYa =(H-ha)/f*ya;\nYb =(H-hb)/f*yb;\nAB= sqrt ((Xa -Xb) **2+( Ya -Yb)**2);\n\n#result\nprint \" distance between A & B in (m)  \",round(AB,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " distance between A & B in (m)   545.213\n"
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.9,Page 243"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding flying distance between A and B\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nf =20.0 # focal length\nxa =2.65; # x coordinate of a\nxb = -1.92; # x coordinate of b\nya =1.36; # y coordinate of a\nyb =3.65; # y coordinate of b\nha =500.0; # elevation of a\nhb =300.0; # elevation of b\nABg =545.0;\nab =5.112;\n\n#calculation\nhab =ha /2+ hb /2;\nHapp =hab+ ABg *f/ab\nXa =( Happ -ha)/f*xa;\nXb =( Happ -hb)/f*xb;\nYa =( Happ -ha)/f*ya;\nYb =( Happ -hb)/f*yb;\nAB= sqrt ((Xa -Xb) **2+( Ya -Yb)**2);\nHact =ABg/AB *( Happ - hab )+ hab ;\n\n#result\nprint \" actual flying height of A & B in (m)  \",round(Hact,3);",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " actual flying height of A & B in (m)   2499.706\n"
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.10,Page 243"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding relief displacement\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\n\nf =20.0/100.0;\nSd =1.0/10000.0;\nh =250.0; # elevation\nr =6.44;\n\n#calculation\nH=f/Sd;\nd=r*h/H;\n\n#result\nprint \"relief displacement of the point in ( cm) \",round(d,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "relief displacement of the point in ( cm)  0.805\n"
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.11,Page 244"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding relief distance\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nh =50.0; # elevation\nH =2500.0 -1250.0;\nr =6.35;\n\n#calculation\nd=r*h/H;\n\n#result\nprint  \"releif displacement of the point in ( cm) \",round(d,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "releif displacement of the point in ( cm)  0.254\n"
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.12,Page 244"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding height of tower\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nf =20.0/100.0; # focal length\nl =250; #length of line\nlp =8.5/100.0; #length of line in photograph\n\n#calculation\nH=l*f/lp; # height of camera above datum\nr =6.46; # distace of image of top o f the towe r\nd =0.46; # releif displacement\nh=d*H/r;\n\n#result\nprint \" height of tower above its base in (m) \",round(h,2)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "41.89  height of tower above its base in (m) \n"
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.13,Page 267"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding no. of photographs\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nl =20/100; # length of photograph\nw =20/100; # breadth of photograph\nPl =0.6; # longitudinal  lap\nPw =0.3; # side lap\ns =100*20;\n\n#calculation\nL=(1 - Pl)*s;\nW=(1 - Pw)*s;\nAr=L*W /1000/1000;\nN =100/ Ar;\nA= round (N);\n\n#result\nprint \"no . o f photographs to be taken \",A+1;\n",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "no . o f photographs to be taken  90.0\n"
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.14,Page 267"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding no. of photographs\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nPl =0.6; # longitudinal lap\nPw =0.3; # side lap\nL1 =10000.0;\ns =100.0*20.0;\n\n#calculation\nL2=L1;\nN1=L1 /((1 - Pl)*s) +1;\nA1= round (N1);\nif N1 -A1 <0:\n    N1=A1;\nelse :\n    N1=A1+1;\n\nN2=L2 /((1 - Pw)*s) +1;\nA2= round (N2);\nif N2 -A2 <0:\n    N2=A2\nelse :\n    N2=A2+1;\n\nN=N1*N2;\n\n#result\nprint \"no . of photographs to be taken \",N;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "no . of photographs to be taken  126.0\n"
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.15,Page 268"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding no. of photographs\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nPl =0.6; # longitudinal lap\nPw =0.3; # side lap\nL1 =12500.0;\ns =100.0*20.0;\nL2 =8000.0;\n\n#calculation\nN1=L1 /((1 - Pl)*s) +1;\nA1= round (N1);\nif N1 -A1 <0:\n    N1=A1;\nelse :\n    N1=A1+1;\n\nN2=L2 /((1 - Pw)*s) +1;\nA2= round (N2);\nif N2 -A2 <0:\n    N2=A2\nelse :\n    N2=A2+1;\n\nN=N1*N2;\n\n#result\nprint \"no . of photographs to be taken \",N;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": "no . of photographs to be taken  119.0\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.16,Page 268"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding no. of photographs,height of datum\n\n#initialisation of variable\n#part1\nfrom math import pi,tan,sqrt,sin\nf =30.0/100.0; # focal length\nh =400.0; #elevation of datum\nr =12000.0; # ratio\ns =120.0*20.0;\nL2 =24000.0;\nL1 =30000.0;\nPl =0.6; # longitudinal lap\nPw =0.3; # side lap\n\n#calculation\nH=h+r*f;\n\n#result\nprint \" height above datum in (m)  \",round(H,2);\n\n# part 2\n#calculation\nW=(1 - Pw)*s;\n\n#result\nprint \" ground width covered in each photograph (m) \",round(W,2);\n\n# part 3\nN2=L2 /((1 - Pw)*s) +1;\nA2= round (N2);\nif N2 -A2 <0:\n    N2=A2\nelse :\n    N2=A2+1;\n\n#result\nprint \"no . of flights required \",N2;\n\n#part 4-9\n#calculation\nAsf =L2 /(N2 -1) ; # actual spacing between flights\nSfl = Asf /600; # spacing of flight lines\ngd =(1 - Pl)*s; # ground distance\nEi=gd /55.5; # exposure interval\nEi= round (Ei);\nAgs =55.56* Ei;# adgusted ground distance\nN1=L1/ Ags +1;\nA1= round (N1);\nif N1 -A1 <0:\n    N1=A1;\nelse :\n    N1=A1+1;\nN=N1*N2;\n\n#result\nprint \"actual spacing in m\",Asf\nprint \"spacing of flight lines in cm\",round(Sfl,2)\nprint \"exposure interval in s\",Ei\nprint \"adjusted ground distance in m\",round(Ags)\nprint \"no . of photographs to be taken per flight line\",N1\nprint \"no . of photographs to be taken \",N;",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " height above datum in (m)   4000.0\n ground width covered in each photograph (m)  1680.0\nno . of flights required  16.0\nactual spacing in m 1600.0\nspacing of flight lines in cm 2.67\nexposure interval in s 17.0\nadjusted ground distance in m 945.0\nno . of photographs to be taken per flight line 33.0\nno . of photographs to be taken  528.0\n"
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.17,Page 301"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding error in height \n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nf =150.0/1000.0; # focal length\nr =20000.0; #ratio\nPl =0.6; # longitudinal  lap\nl =23.0/100.0; # l e n g t h\nw =23.0/100.0; # width\n\n#calculation\nB=(1 - Pl)*l*r; # base length\nH=f*r;\nh =0;\ndh =(H-h) **2/ B/f *0.1/1000;\n\n#result\nprint \" error in height in (m)  \",round(dh,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " error in height in (m)   3.261\n"
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.18,Page 302"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding parallax height of the chimney\n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nH =600.0;\nf =150.0/1000.0;\nb =6.375/100.0;\nh1 =0.0;\nh2 =120.0; # height of chimney\n\n#calculation\ns=H/f;\nB=s*b; # datum elevation\np1=B*f *1000/(H-h1);\np2=B*f *1000/(H-h2);\ndelp =p2 -p1;\ndelh =H* delp /1000/( b+ delp /1000) ;\n\n#result\nprint \" parallax height of the chimney in (m)\",round(delh,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " parallax height of the chimney in (m) 120.0\n"
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.19,Page 303"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding difference in elevation \n\n#initialisation of variable\nfrom math import pi,tan,sqrt,sin\nB =200.0;\nf =120.0;\np2 =52.52; # parallax for top pole\np1 =48.27; # parallax for bottom pole\n\n#calculation\ndelh =(p2 -p1)/p2/p1*B*f;\n\n#result\nprint \" difference in elevation  of two points in (m)  \",round(delh,3)\nprint \"there is again  a miscalculation in the step of calculating elevation thus there is a change in the answer\"",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " difference in elevation  of two points in (m)   40.234\nthere is again  a miscalculation in the step of calculating elevation thus there is a change in the answer\n"
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": "Example 2.20,Page 303"
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "#finding difference in elevation \n\n#initialisation of variable\n# part 1\ndelp =1.48/1000.0;\nH =5000.0;\nh =500.0;\nb =90.0/1000.0; #mean principal base\n\n#calculation\ndh =(H-h) **2* delp /((H-h)* delp +b*H);\n\n#result\nprint \" difference in height between two points in(m)  \",round(dh,3)\n\n# part 2\n#variable decleration\ndelp =15.5/1000.0;\n\n#calculation\ndh =(H-h) **2* delp /((H-h)* delp +b*H);\n\n#result\nprint \" difference in height between two points in(m)  \",round(dh,3)",
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": " difference in height between two points in(m)   65.629\n difference in height between two points in(m)   603.896\n"
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": "",
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Surveying_Volume_3/README.txt b/Surveying_Volume_3/README.txt
index 2c27e6c8..9e2e986b 100755
--- a/Surveying_Volume_3/README.txt
+++ b/Surveying_Volume_3/README.txt
@@ -1,10 +1,10 @@
-Contributed By: Tarun Kumar Das
+Contributed By: Avik Kumar Das
 Course: btech
-College/Institute/Organization: College of Engineering
-Department/Designation: Industrial Engineering
+College/Institute/Organization: Indian Institute of Technology Bombay
+Department/Designation: Civil Engineering
 Book Title: Surveying Volume 3
-Author: A K Arora
-Publisher: Lakshmi Publications, Chennai
-Year of publication: 2011
-Isbn: 9788189401276
-Edition: 1
-\ No newline at end of file
+Author: B C Punmia
+Publisher: Laxmi Publication, New Delhi
+Year of publication: 1990
+Isbn: 81-7008-054-18
+Edition: 9
+\ No newline at end of file
diff --git a/Surveying_Volume_3/screenshots/Finding_Altitude_and_Zenith.png b/Surveying_Volume_3/screenshots/Finding_Altitude_and_Zenith.png
new file mode 100755
index 00000000..d4197064
--- /dev/null
+++ b/Surveying_Volume_3/screenshots/Finding_Altitude_and_Zenith.png
diff --git a/Surveying_Volume_3/screenshots/Finding_distance.png b/Surveying_Volume_3/screenshots/Finding_distance.png
new file mode 100755
index 00000000..b079ffed
--- /dev/null
+++ b/Surveying_Volume_3/screenshots/Finding_distance.png
diff --git a/Surveying_Volume_3/screenshots/Finding_focal_length.png b/Surveying_Volume_3/screenshots/Finding_focal_length.png
new file mode 100755
index 00000000..6d99e1fd
--- /dev/null
+++ b/Surveying_Volume_3/screenshots/Finding_focal_length.png
author	nice	2014-10-09 18:07:00 +0530
committer	nice	2014-10-09 18:07:00 +0530
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