62 files changed, 14747 insertions, 0 deletions
diff --git a/Chemical_Reaction_Engineering/README.txt b/Chemical_Reaction_Engineering/README.txt
new file mode 100755
index 00000000..e84d96a5
--- /dev/null
+++ b/Chemical_Reaction_Engineering/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Manish Punjabi
+Course: mtech
+College/Institute/Organization: IIT Bombay
+Department/Designation: IEOR
+Book Title: Chemical Reaction Engineering
+Author: O. Levenspiel
+Publisher: Wiley India, New Delhi
+Year of publication: 2008
+Isbn: 9788126510009
+Edition: 3
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch1.ipynb b/Chemical_Reaction_Engineering/ch1.ipynb
new file mode 100755
index 00000000..dd76e3c5
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch1.ipynb
@@ -0,0 +1,119 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Overview of Chemical Reaction Engineering"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.1 page no : 6"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#l=75 cm,d=60 cm,H20 Produced=108kg/s\n",
+      "l=0.75                     # cylindrical long\n",
+      "d=0.6;                     # cylindrical diameter\n",
+      "V=(3.14*d*d*l)/4;\n",
+      "M=18.;                     # molecular weight\n",
+      "\n",
+      "# Calculations\n",
+      "H20_produced=108./M;\n",
+      "H2_used=H20_produced;\n",
+      "O2_used=0.5*H20_produced;\n",
+      "#Rate of reaction of H2(mol/m**3.s)\n",
+      "r_H2=(H2_used/V)*1000;\n",
+      "#Rate of reaction of O2(mol/m**3.s)\n",
+      "r_O2=(O2_used/V)*1000;\n",
+      "\n",
+      "# Results\n",
+      "print \"rate of reaction of H2mol/m**3.s is %.3e mol used/(m**3 of rocket).s\"%(r_H2)\n",
+      "print \"rate of reaction of O2mol/m**3.s is %.3e mol/m**3.s\"%(r_O2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate of reaction of H2mol/m**3.s is 2.831e+04 mol used/(m**3 of rocket).s\n",
+        "rate of reaction of O2mol/m**3.s is 1.415e+04 mol/m**3.s\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 1.2 page no : 7"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Assuming density of a person = 1000kg/m3\n",
+      "\n",
+      "# Variables\n",
+      "d = 1000.;\n",
+      "mass = 75.;               # human being\n",
+      "\n",
+      "# Calculations\n",
+      "V = mass/d;\n",
+      "#moles of O2 consumed per day\n",
+      "O2_used = (6000./2816)*6;\n",
+      "# Rate of reaction (mol/m3.s)\n",
+      "r_O2 = (O2_used/V)/(24.*3600);\n",
+      "\n",
+      "# Results\n",
+      "print \"rate of reaction of O2mol/m**3.s is %.3f mol O2 used/m***3.s\"%(r_O2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate of reaction of O2mol/m**3.s is 0.002 mol O2 used/m***3.s\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch10.ipynb b/Chemical_Reaction_Engineering/ch10.ipynb
new file mode 100755
index 00000000..bfe8563f
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch10.ipynb
@@ -0,0 +1,119 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Choosing the Right Kind of Reactor"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 pageno : 243"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "CAo = 1.                # feed\n",
+      "CA = 0.25\n",
+      "v = 100.                #litre/min\n",
+      "ko = .025\n",
+      "k1 = 0.2                # min**-1\n",
+      "k2 = 0.4                # liter/mol.min\n",
+      "\n",
+      "# Calculations\n",
+      "rA = ko+k1*CA+k2*CA**2\n",
+      "V = (v/4.)*(CAo-CA)/rA\n",
+      "#For 4 Reactor System\n",
+      "Vt = 4*V;\n",
+      "\n",
+      "# Results\n",
+      "print \" The Total volume of 4 reactor system is %.f litres\"%(Vt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The Total volume of 4 reactor system is 750 litres\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 pageno : 245"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "print \"For Intermediate R is desired\"\n",
+      "#we want step 1 fast than 2 and step 1 fast than 3\n",
+      "print \" E1<E2,E1<E3 so use a low temperature and plug flow \"\n",
+      "print \"For Product S is desired\"\n",
+      "#Here speed is all that matters\n",
+      "print \" High speed is all that matters so use a high temperature and plug flow \"\n",
+      "print \"For Intermediate T is desired\"\n",
+      "#We want step 2 fast than 1 and step 2 fast than 4\n",
+      "print \" E2>E1,E3>E5 so use a falling temperature and plug flow \"\n",
+      "print \"For Intermediate U is desired\"\n",
+      "#We want step 1 fast than 2 and step 3 fast than 5\n",
+      "print \" E2>E1,E3>E5 so use a rimath.sing temperature and plug flow \"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "For Intermediate R is desired\n",
+        " E1<E2,E1<E3 so use a low temperature and plug flow \n",
+        "For Product S is desired\n",
+        " High speed is all that matters so use a high temperature and plug flow \n",
+        "For Intermediate T is desired\n",
+        " E2>E1,E3>E5 so use a falling temperature and plug flow \n",
+        "For Intermediate U is desired\n",
+        " E2>E1,E3>E5 so use a rimath.sing temperature and plug flow \n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch11.ipynb b/Chemical_Reaction_Engineering/ch11.ipynb
new file mode 100755
index 00000000..e716f90a
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch11.ipynb
@@ -0,0 +1,331 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Basics of Non-Ideal Flow"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 pageno : 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = array([0,5,10,15,20,25,30,35])    # time\n",
+      "Cpulse = array([0,3,5,5,4,2,1,0])     # tracer output concentration\n",
+      "dt = 5.;\n",
+      "sum1 = 0.;\n",
+      "sum2 = 0.;\n",
+      "Area = 0.               #Initialization\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(8):\n",
+      "    sum1 = sum1+T[i]*Cpulse[i]*dt;\n",
+      "    sum2 = sum2+Cpulse[i]*dt;\n",
+      "    Area = Area+Cpulse[i]*dt;\n",
+      "\n",
+      "t = sum1/sum2;\n",
+      "E = zeros(8)\n",
+      "for j in range(8):\n",
+      "    E[j] = Cpulse[j]/Area;\n",
+      "\n",
+      "# Results\n",
+      "print \" The mean residence time is %.f min \"%(t)\n",
+      "plot(T,E)\n",
+      "plot(T,E,\"go\")\n",
+      "xlabel(\"t, min\")\n",
+      "ylabel(\"E\")\n",
+      "show()"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        " The mean residence time is 15 min \n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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LmB4wCZTB/zN+rkk8sWKX0qERib4+Ie64Q4jXX1c6EoqkcN47Zf/s2tTUhNTUVKSkpCAx\nMRH5+fmorq4OGFNTU4OCggIAgNVqRXd3N7q6ugAA48ePBwD09PTA6/Xi+uuvlzvEqNv43ka0ZrcG\nbDuf14quy1UKRUT0DwkJwL//O/Dii0BHh9LRUCwZPdQDr7zyCp577jkAwAcffICHH37Y/9jPf/5z\nrF+/XvJ5nZ2dSE5O9t83Go1obGwcdkxHRwf0ej28Xi/mzJmD1tZWFBUVISMj44rfUVZW5r9ts9lg\ns9mu/lcq7JK4JLm9R1yMciRE0jIygOJi3wV07PgaH5xOJ5xO54j2MWSC2LFjhz9BrF+/PiBB1NfX\nD5kgdEH+yxKDFkv6n5eQkIAvvvgCp0+fht1uh9PpvCIBDEwQajBWN1Zy+7iEcVGOhGho69b5vsHw\nD38A/umflI6GRmrwh+fy8vKQ9yH7FJPBYIDH4/Hf93g8MA6q4Rw8pqOjAwaDIWDMddddh7y8PBw4\ncEDuEKOu5LESmNymgG2mQyasyV+jUEREVxozxlc0UVoKnDqldDQUC2RPEDk5OWhpaUF7ezt6enqw\nc+dOOByOgDEOhwPbt28HALhcLkycOBF6vR4nT55Ed3c3AODChQvYs2cPsrOz5Q4x6vJy81C5uhKz\nPrfj+pr5sB+zo7K4Enm5eUqHRhTgjjuARx4BnnlG6UgoFgx5HURCQoJ/wfjChQu45ppr/I9duHAB\nfX19Q+60vr4epaWl8Hq9KCwsxLp167Dluy/FXbVqFQCguLgYDQ0NSEpKwrZt23Dbbbfh8OHDKCgo\nwOXLl3H58mUsW7YMzz77bGDAKr0OQghfvfn69b5eOESxih1f4xNbbcSwTz8FCguBI0d43QPFPnZ8\njT8xc6EcXWnjRl+VCJMDqQE7vhLAM4io6OgAZs8G2tuB731P6WiIgsOOr/GFZxAx6o03gGXLmBxI\nXW64Afjd73wNJXt6lI6GlMAziAi7cAG4+WbfGsSMGUpHQxQaIYC8PODOO4Ff/lLpaGgkeAYRg95/\nH8jJYXIgddLpgM2bfd92+D//o3Q0FG1MEBEkBFBVBZSUKB0JUfhuugkoK2PHVy1igoigzz4Dzp1j\nLTmpX1ER4PUC313ORBrBNYgIeuQRYN48YA07alAc4NfjqhsvlIshLG2leFRe7vsOa3Z8VR8uUscQ\nlrZSPFq3Dmhr83V8pfjHM4gIYGkrxbP9+4GHHvK14Zg8WeloKFg8g4gRLG2leMaOr9rBBCEzlraS\nFvzrvwJOJ7B7t9KRUCQxQciMpa2kBRMm+C6gW7XK9++d4hMThMzYtZW0gh1f4x8XqWXE0lbSGnZ8\nVQ8uUiuMpa2kNez4Gt94BiETlraSVrHjqzrwDEJBLG0lrWLH1/jFBCEDlraS1rHja3xigpABS1uJ\n2PE1HnENQgbs2krkw46vsYvdXBXA0laiQOz4Gpu4SK0AlrYSBWLH1/jBM4gRYGkrkTR2fI09PIOI\nMpa2Ekljx9f4wAQRJpa2El0dO76qHxNEmFjaSnR1Azu+nj2rdDQUjogkiIaGBqSlpcFsNqOiokJy\nTElJCcxmMywWC9xuNwDA4/Hg7rvvxsyZMzFr1ixs3LgxEuHJgl1biYZ3771Ayq21mPVjO2w/scG+\n3I7aPbVKh0XBEjLr6+sTJpNJtLW1iZ6eHmGxWERzc3PAmNraWnHfffcJIYRwuVzCarUKIYQ4fvy4\ncLvdQgghzpw5I2bMmHHFcyMQcsg8HiEmTRLi9GmlIyGKbbt27xIp95sEyuD/MT1gErt271I6NM0J\n571T9s+/TU1NSE1NRUpKChITE5Gfn4/q6uqAMTU1NSgoKAAAWK1WdHd3o6urC9OmTUNWVhYAYMKE\nCUhPT8df//pXuUMcMZa2EgVn43sb0Z7TGrCtNbsVVe9XKRQRhWK03Dvs7OxEcnKy/77RaERjY+Ow\nYzo6OqDX6/3b2tvb4Xa7YbVar/gdZWVl/ts2mw02m02+P2AYFy4Ab77pK20loqu7JC5Jbr/ovRjl\nSLTH6XTC6XSOaB+yJwhdkJdOikH1uAOfd/bsWSxduhSVlZWYMGHCFc8dmCCijaWtRMEbqxsruX3c\nqHFRjkR7Bn94Li8vD3kfsk8xGQwGeDwe/32PxwPjoKYsg8d0dHTAYDAAAHp7e7FkyRI88cQTePDB\nB+UOb0RY2koUmpLHSmBymwK2jak24YJnDSubVED2BJGTk4OWlha0t7ejp6cHO3fuhMPhCBjjcDiw\nfft2AIDL5cLEiROh1+shhEBhYSEyMjJQWloqd2gjxtJWotDk5eahcnUl7MfsmN82H/Zjduz4dSVS\nb8qD1crvj4h5si+VCyHq6urEjBkzhMlkEuvXrxdCCLF582axefNm/5jVq1cLk8kkZs+eLQ4ePCiE\nEGLfvn1Cp9MJi8UisrKyRFZWlqivrw/Yd4RCDsrDDwuxcaNiv54orrz5phA33CDEhx8qHYk2hPPe\nyV5MQWLXViL5HTgALF0KPPww8JvfAKNlXxWlfuzFFEEsbSWSX04OcPAgcPgwsGABcOKE0hHRQEwQ\nQegvbV29WulIiOLP5MlAba3vi4ZycnxrfRQbmCCCwNJWoshKSPB90dDWrb424ZWVvqpBUhbXIIYh\nBDBnDrB+va+vDBFFVlsbsGQJcOutvjN3iUuhKAxcg4gAlrYSRdf3v+/7fzd+PGC1Av/7v0pHpF1M\nEMNg11ai6LvmGuCtt4CnnwbmzQP+4z+UjkibOMV0FSxtJVIeS2HlwSkmmbG0lUh5LIVVDhPEEFja\nShQ7WAqrDCaIIbC0lSi2sBQ2+rgGIYGlrUSx7ZtvfKWwaWkshQ0W1yBkwtJWoth2yy3A55/7qp1Y\nChs5TBASWNpKFPv6S2FLS1kKGymcYhqEpa1E6sNS2OFxikkGLG0lUp+BpbC5uUBXl9IRxQcmiAH6\nS1uLi5WOhIhC1V8K+6Mf+YpMWAo7ckwQA/SXtprNSkdCROFgKay8uAbxHZa2EsUXlsIG4hrECLC0\nlSi+sBR25JggvsPSVqL4M7AU9q67WAobKk4xgaWtRFqg9VJYTjGFiaWtRPGPpbCh03yCYGkrkXYM\nLIXNyfGtUdDQNJ8gWNpKpC39pbBbtgCLF/vWH9U10R49ml6DYGkrkbb1l8Kmp/uunYjnUliuQYSI\npa1E2tZfCjtuHEthpWg6QbC0lYhYCjs0zU4xsbSViAaL51JYTjGFgKWtRDRYTo4vSXz5JUthgQgl\niIaGBqSlpcFsNqOiokJyTElJCcxmMywWC9xut3/7ihUroNfrkZmZGYnQALC0lYiGdsMNQF2d70uI\ncnKAVzfWwr7cDttPbLAvt6N2T63SIUaN7CdQXq8XxcXF2Lt3LwwGA+bOnQuHw4H09HT/mLq6Ohw9\nehQtLS1obGxEUVERXC4XAGD58uVYs2YNnnzySblD82NpKxFdTUIC8PLLwKhxtXjhrbW4/FCr/7HW\n13y383LzlAovamQ/g2hqakJqaipSUlKQmJiI/Px8VFdXB4ypqalBQUEBAMBqtaK7uxsnTpwAAMyb\nNw+TJk2SOyw/IYCqKqCkJGK/gojixP6WjQHJAQBas1tR9X6VQhFFl+xnEJ2dnUhOTvbfNxqNaGxs\nHHZMZ2cnpk2bFtTvKCsr89+22Wyw2WxBx8fSViIK1iVxSXL7Re/FKEcSOqfTCafTOaJ9yJ4gdDpd\nUOMGr6YH+zwgMEGEiqWtRBSssbqxkttPfzsuypGEbvCH5/Ly8pD3IfvbpMFggMfj8d/3eDwwGo1X\nHdPR0QGDwSB3KFfo6AD27gW+m90iIrqqksdKYHKbArYZ95vQ+eUaPPss0NenUGBRInuCyMnJQUtL\nC9rb29HT04OdO3fC4XAEjHE4HNi+fTsAwOVyYeLEidDr9XKHcgWWthJRKPJy81C5uhL2Y3bMb5sP\n+zE7Nj9TiWZ3njZKYUUE1NXViRkzZgiTySTWr18vhBBi8+bNYvPmzf4xq1evFiaTScyePVscPHjQ\nvz0/P19Mnz5djBkzRhiNRvH2228H7DvckM+fF2LKFCG+/jqspxMRBejrE+JXvxLCaBTis8+UjmZ4\n4bx3auZK6m3bgA8+8NU3ExHJpbYWWLEC+MUvgDVrgBCWU6MqnPdOTSQIdm0lokhSQ1dYttoYAktb\niSiSBnaF/cEP4qcrrCYSBEtbiSjS+rvCrl0bP11h436KiV1biSjaYrErLKeYJLC0lYiiLV66wsZ1\ngmDXViJSSn9X2B/9yJcwPv9c6YhCF9cJgl1biUhJCQlAeTmwZQuweLFvPVRNk/pxuwbB0lYiiiVK\nl8JyDWIAlrYSUSwZWAprtaqjFDZuEwRLW4ko1vSXwpaWqqMUNi6nmFjaSkSxLtqlsJxi+g5LW4ko\n1qmhFDbuEgRLW4lILWK9FDbuEgRLW4lITWK5FDau1iBY2kpEatZfCpuW5psJkbMUVvNrECxtJSI1\n6y+Fveaa2CiFjasEwdJWIlK7WCqFjZspJpa2ElG8kbMUVtNTTCxtJaJ4k5MDHDwIHD6sTClsXCQI\nlrYSUbyaPNn3vddKlMLGRYJgaSsRxTOlSmFVvwbB0lYi0pJwS2E1uQbB0lYi0pJolsKqPkGwtJWI\ntCZapbCqnmJiaSsRaV2wpbCam2JiaSsRaV0kS2FVmyBY2kpE5DOwFHbOHN/arBxUmyDUWtrqdDqV\nDmFEGL+y1By/mmMHYj/+/lLYrVuBhx4CKitHXgobkQTR0NCAtLQ0mM1mVFRUSI4pKSmB2WyGxWKB\n2+0O6bn25Xb8yyu1KCmJRPSRFev/yIbD+JWl5vjVHDugnvgXLgT27wfeeQd47DHgwz/Vwr7cHta+\nZE8QXq8XxcXFaGhoQHNzM3bs2IEjR44EjKmrq8PRo0fR0tKCrVu3oqioKOjnAsDulN3wjFuLXl2t\n3OETEalefynsyTO1eOyltdidsjus/cieIJqampCamoqUlBQkJiYiPz8f1dXVAWNqampQUFAAALBa\nreju7saJEyeCem6/3gdb8dofquQOn4goLlxzDTBqykb0PtAa/k6EzD744AOxcuVK//3f//73ori4\nOGDM/fffLz777DP//XvuuUccOHBAfPjhh8M+FwB/+MMf/vAnjJ9QjaB5rDSdThfUOBHm6km4zyMi\notDIniAMBgM8Ho//vsfjgdFovOqYjo4OGI1G9Pb2DvtcIiKKDtnXIHJyctDS0oL29nb09PRg586d\ncDgcAWMcDge2b98OAHC5XJg4cSL0en1QzyUiouiQ/Qxi9OjR2LRpE+x2O7xeLwoLC5Geno4tW7YA\nAFatWoWFCxeirq4OqampSEpKwrZt2676XCIiUkDIqxYKqq+vF7feeqtITU0VGzZsUDqckN18880i\nMzNTZGVliblz5yodzrCWL18upk6dKmbNmuXfdurUKbFgwQJhNptFbm6u+Nvf/qZghFcnFf9LL70k\nDAaDyMrKEllZWaK+vl7BCK/u2LFjwmaziYyMDDFz5kxRWVkphFDHazBU7Go5/hcuXBC33367sFgs\nIj09XbzwwgtCCHUceyGGjj/U46+aBNHX1ydMJpNoa2sTPT09wmKxiObmZqXDCklKSoo4deqU0mEE\n7ZNPPhGHDh0KeIN99tlnRUVFhRBCiA0bNojnn39eqfCGJRV/WVmZ+O1vf6tgVME7fvy4cLvdQggh\nzpw5I2bMmCGam5tV8RoMFbuajv+5c+eEEEL09vYKq9Uq9u3bp4pj308q/lCPv2pabYRyjUQsEyqq\nwpo3bx4mTZoUsG3gNSwFBQX44x//qERoQZGKH1DPazBt2jRkZWUBACZMmID09HR0dnaq4jUYKnZA\nPcd//PjxAICenh54vV5MmjRJFce+n1T8QGjHXzUJorOzE8nJyf77RqPR/w9OLXQ6HRYsWICcnBy8\n+eabSocTlq6uLuj1egCAXq9HV7S/RV0GVVVVsFgsKCwsRHd3t9LhBKW9vR1utxtWq1V1r0F/7D/4\nwQ8AqOf4X758GVlZWdDr9bj77rsxc+ZMVR17qfiB0I6/ahJEsNdXxLLPPvsMbrcb9fX1eO2117Bv\n3z6lQxoRnU6nutelqKgIbW1t+OKLLzB9+nQ888wzSoc0rLNnz2LJkiWorKzEtddeG/BYrL8GZ8+e\nxdKlS1FAvgYZAAADfElEQVRZWYkJEyao6viPGjUKX3zxBTo6OvDJJ5/gz3/+c8DjsX7sB8fvdDpD\nPv6qSRDBXF8R66ZPnw4AmDJlChYvXoympiaFIwqdXq/HiRMnAADHjx/H1KlTFY4oNFOnTvX/x165\ncmXMvwa9vb1YsmQJli1bhgcffBCAel6D/tifeOIJf+xqO/4AcN111yEvLw8HDx5UzbEfqD/+AwcO\nhHz8VZMg1H6NxPnz53HmzBkAwLlz57B7925kZmYqHFXoHA4H3n33XQDAu+++6/+PrxbHjx/33/7o\no49i+jUQQqCwsBAZGRkoLS31b1fDazBU7Go5/idPnvRPv1y4cAF79uxBdna2Ko49MHT8/ckNCPL4\ny792Hjl1dXVixowZwmQyifXr1ysdTki++eYbYbFYhMViETNnzlRF/Pn5+WL69OkiMTFRGI1G8fbb\nb4tTp06Je+65J+bL/IS4Mv633npLLFu2TGRmZorZs2eLBx54QJw4cULpMIe0b98+odPphMViCShL\nVMNrIBV7XV2dao7/l19+KbKzs4XFYhGZmZnilVdeEUIIVRx7IYaOP9Tjr7rvpCYiouhQzRQTERFF\nFxMEERFJYoIgIiJJTBBERCSJCYJIwunTp/HGG2/Itr+DBw9i7dq1su2PKBpYxUQkob29HYsWLcLh\nw4eVDoVIMTyDIJLwwgsvoLW1FdnZ2Xj++eevOnbChAl47rnnMGvWLOTm5sLlcmH+/PkwmUz405/+\nBABwOp1YtGgRAKCsrAwrVqzA3XffDZPJhKqqqoj/PUThYIIgklBRUQGTyQS3242Kioqrjj1//jzu\nuecefPXVV7j22mvx4osv4uOPP8ZHH32EF198UfI5X3/9NXbv3o2mpiaUl5fD6/VG4s8gGhHZv1GO\nKB6EMvM6ZswY2O12AEBmZibGjRuHhIQEzJo1C+3t7VeM1+l0yMvLQ2JiIiZPnoypU6eiq6sLN954\no1zhE8mCZxBEI5SYmOi/PWrUKIwZM8Z/u6+vT/I5/WMAICEhYchxREpigiCScO211/qbK/ZLS0uT\nZd+sCyG1YIIgkjB58mT88Ic/RGZmJp5//nmcPHlyyLGDvxNg4H2p27H+PQJE/VjmShSE2tpatLW1\nobi4WOlQiKKGCYKIiCRxiomIiCQxQRARkSQmCCIiksQEQUREkpggiIhIEhMEERFJ+n/ZImkGAKXm\n0wAAAABJRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x32abdd0>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 pageno : 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "M = 150.            #Molecular mass(gm)\n",
+      "v = 5.              #litre/sec\n",
+      "v = 5*60.           #litre/min\n",
+      "V = 860.            #litres\n",
+      "Cpulse = .75\n",
+      "# Calculations and Results\n",
+      "#From Material Balance\n",
+      "Area1 = M/v;        #gm.min/litre\n",
+      "A1 = 0.375;\n",
+      "Area2 = A1*(1+1./4+1./16+1./64+1./256+1./1024+1./4096);         #Taking Significant Areas\n",
+      "\n",
+      "\n",
+      "print \" From material balance Area is %.1f gm.min/litre\"%(Area1)\n",
+      "print \" From Tracer Curve Area is %.1f gm.min/litre\"%(Area2)\n",
+      "print \" Part a\"\n",
+      "print \" As the two areas are equal,this is a properly done experiment \"\n",
+      "#For the liquid,calculating t\n",
+      "sum1 = 0;\n",
+      "for i in range(10):\n",
+      "    sum1 = sum1+2*i*A1/(4**(i-1));\n",
+      "    t = sum1/Area1;\n",
+      "\n",
+      "#liquid volume in vessel\n",
+      "Vl = t*v;\n",
+      "#Fraction of liquid\n",
+      "f = Vl/V;\n",
+      "E = Cpulse/(M/v)\n",
+      "\n",
+      "print \" Part b\"\n",
+      "print \" Fraction of liquid is %.f %%\"%(f*100)\n",
+      "\n",
+      "print \" Part c\"\n",
+      "print \" The E curve is %.1f C\"%E\n",
+      "print \" Part d\"\n",
+      "print \" The vessel has  a strong recirculation of liquid,probably induced by the rising bubbles\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " From material balance Area is 0.5 gm.min/litre\n",
+        " From Tracer Curve Area is 0.5 gm.min/litre\n",
+        " Part a\n",
+        " As the two areas are equal,this is a properly done experiment \n",
+        " Part b\n",
+        " Fraction of liquid is 93 %\n",
+        " Part c\n",
+        " The E curve is 1.5 C\n",
+        " Part d\n",
+        " The vessel has  a strong recirculation of liquid,probably induced by the rising bubbles\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 pageno : 272\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "Cin = zeros(14)\n",
+      "E = zeros(14)\n",
+      "Cout = zeros(14)\n",
+      "Cin[0] = 0.\n",
+      "Cin[1] = 8.\n",
+      "Cin[2] = 4.\n",
+      "Cin[3] = 6\n",
+      "Cin[4] = 0\n",
+      "E[4] = 0\n",
+      "E[5] = 0.05\n",
+      "E[6] = 0.5\n",
+      "E[7] = 0.35\n",
+      "E[8] = 0.1\n",
+      "E[9] = 0.\n",
+      "\n",
+      "# Calculations\n",
+      "for t in range(8,14):\n",
+      "    sum1 = 0;\n",
+      "    for p in range(5,t-1):\n",
+      "        if p>10 or (t-p)>5:\n",
+      "            h = 2;\n",
+      "        else:\n",
+      "            sum1 = sum1+Cin[t-p] * E[p];\n",
+      "            Cout[t] = sum1;\n",
+      "\n",
+      "t = linspace(1,14,14)\n",
+      "Cout = transpose(Cout)\n",
+      "\n",
+      "# Results\n",
+      "plot(t,Cout)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "pyout",
+       "prompt_number": 6,
+       "text": [
+        "[<matplotlib.lines.Line2D at 0x43ec990>]"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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jhVKiELh2DZg4EWhvB0aPFp2G/OFwADk58t2/Tz0lLgcvlBKpUGOjvLs9C107\nkpKAykpg2TLg1CnRafzHUicKAY5etOmJJ4Df/U6+cHr1qug0/mGpE4UAV75oV3GxPIZZsUIbF05Z\n6kQhwJUv2qXTyY/odTiA9etFp/GNF0qJFHbhgrx13aVL8ibIpE1tbfIv5rffBvLzQ/e+vFBKpDKH\nDwPZ2Sx0rTMYgP/5H+Af/xHwslunavj8a7Zz506kpaUhNTUVr732mtdjVq9ejdTUVJjNZjQ2NgY9\npBoMZyNYNdByfi1nB4D//m+7pkcvWv/zD2b+WbOAdevkC6fffhu0lw2qQUu9u7sbL774Inbu3IkT\nJ07gww8/xMmTJz2OsdlsaGlpQXNzMzZv3ozi4mJFA4vCv9jiaDk7ABw4YNf0yhet//kHO//Pfy6v\niiksBHp6gvrSQTFoqdfV1SElJQXJycmIjo7GkiVLsH37do9jqqqqUFhYCADIyclBR0cHXC6XcomJ\nNESSbs9iKXy88QbwzTfyWbvaDLpHaVtbG5KSkno/NxqNOHTokM9jnE4nEhISghx1YBUVwI4dyr5H\nU5O8Ua1WaTm/lrN3dcmzdINBdBIKplGjgG3b5HsPDh4Eonzu9hw6g0bR6XR+vUjfK7MD/Zy/r6dW\nTU2loiMMi5bzazk7AIwYoe38paXMPxCnU7GXHpJBS91gMMDhcPR+7nA4YDQaBz3G6XTC4OW0hMsZ\niYiUN+hMPTs7G83NzWhtbcWtW7fw0UcfoaCgwOOYgoICbN26FQBQW1uLuLi4kI5eiIjotkHP1KOi\novDmm29i/vz56O7uxqpVq5Ceno6KigoAQFFREfLz82Gz2ZCSkoKYmBhs2bIlJMGJiMgLKQSqq6ul\nBx98UEpJSZHWr18fircMmjNnzkhWq1UymUzS9OnTpfLyctGRAtbV1SVlZmZKzzzzjOgoAbt06ZL0\n7LPPSmlpaVJ6erp08OBB0ZECUlZWJplMJikjI0NaunSpdPPmTdGRBrVixQpp4sSJUkZGRu/XLl68\nKD355JNSamqqNG/ePOnSpUsCEw7OW/5f//rXUlpamjRz5kzphz/8odTR0SEw4cC8ZXfbuHGjpNPp\npIsXL/p8HcXvcfNnrbuaRUdH4/XXX8fx48dRW1uLt956S1P5AaC8vBwmk0mTF6rXrFmD/Px8nDx5\nEp9//jnS09NFR/Jba2sr3n77bTQ0NOCLL75Ad3c3KisrRcca1IoVK7Bz506Pr61fvx7z5s1DU1MT\n5s6di/XOlrWuAAADvUlEQVQqfgCKt/xPPfUUjh8/jqNHj+KBBx7AH/7wB0HpBuctOyBfy9y1axem\nTJni1+soXur+rHVXs8TERGRmZgIAxo4di/T0dJw9e1ZwKv85nU7YbDY8//zzmrtYffnyZXz66adY\nuXIlAHkcGBsbKziV/+6++25ER0fj+vXr6OrqwvXr170uIlCT2bNnY/z48R5fu/NelMLCQnz88cci\novnFW/558+ZhxPfPaMjJyYFTbctVvuctOwC88sor+OMf/+j36yhe6t7Wsbe1tSn9topobW1FY2Mj\ncnJyREfx28svv4wNGzb0/qXWktOnT2PChAlYsWIFsrKy8LOf/QzXNbQbcHx8PH71q19h8uTJuO++\n+xAXF4cnn3xSdKyAuVyu3sUPCQkJmr658N1330V+KJ/GNUzbt2+H0WjEzJkz/f4Zxf+la/F/+b25\nevUqFi9ejPLycowdO1Z0HL/s2LEDEydOhMVi0dxZOgB0dXWhoaEBv/zlL9HQ0ICYmBhV/69/X6dO\nncKf/vQntLa24uzZs7h69Sr+8pe/iI41LDqdTrP/pn//+99j1KhR+MlPfiI6il+uX7+OsrIyjzX2\n/vw7VrzU/VnrrnadnZ149tln8dOf/hSLFi0SHcdvNTU1qKqqwtSpU7F06VJ88sknWL58uehYfjMa\njTAajXj4+wenLF68GA0NDYJT+a++vh6zZs3CPffcg6ioKPzoRz9CTU2N6FgBS0hIwPnz5wEA586d\nw8SJEwUnCtx7770Hm82mqV+qp06dQmtrK8xmM6ZOnQqn04mHHnoIX3/99aA/p3ip+7PWXc0kScKq\nVatgMpnw0ksviY4TkLKyMjgcDpw+fRqVlZWYM2dO7z0FWpCYmIikpCQ0ff+c0927d2P69OmCU/kv\nLS0NtbW1uHHjBiRJwu7du2EymUTHClhBQQHef/99AMD777+vqRMbQH7S7IYNG7B9+3aMGTNGdBy/\nzZgxAy6XC6dPn8bp06dhNBrR0NDg+5dqkFfleGWz2aQHHnhAuv/++6WysrJQvGXQfPrpp5JOp5PM\nZrOUmZkpZWZmStXV1aJjBcxut0sLFy4UHSNgR44ckbKzs1W/HG0gr732Wu+SxuXLl0u3bt0SHWlQ\nS5YskSZNmiRFR0dLRqNRevfdd6WLFy9Kc+fO1cSSxr7533nnHSklJUWaPHly77/f4uJi0TG9cmcf\nNWpU75/9naZOnerXksaQ7XxERETK096SCCIiGhBLnYgojLDUiYjCCEudiCiMsNSJiMIIS52IKIz8\nP+TL9axw2ETbAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x4149950>"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 pageno : 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "k = 0.307;            # min**-1\n",
+      "t = 15.;          \n",
+      "\n",
+      "# Calculations and Results\n",
+      "fr_unconverted = math.exp(-k*t);\n",
+      "print \" The fraction of reactant unconverted in a plug flow reactor is %.2f\"%(fr_unconverted)\n",
+      "\n",
+      "#For the real reactor\n",
+      "T = [5,10,15,20,25,30];                             #given time\n",
+      "E = [0.03,0.05,0.05,0.04,0.02,0.01];                #given\n",
+      "dt = 5;\n",
+      "sum1 = 0;\n",
+      "for i in range(6):\n",
+      "    sum1 = sum1+math.exp(-k*T[i])*E[i]*dt;\n",
+      "\n",
+      "print \" The fraction of reactant unconverted in a real reactor is %.3f\"%(sum1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fraction of reactant unconverted in a plug flow reactor is 0.01\n",
+        " The fraction of reactant unconverted in a real reactor is 0.047\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 pageno : 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "k = 0.5                     #litre/mol.min\n",
+      "CAo = 2.                    #mol/litre\n",
+      "to = 1.\n",
+      "t1 = 3.\n",
+      "E = 0.5\n",
+      "\n",
+      "# Calculations\n",
+      "#Using eqn 13\n",
+      "def f2(t): \n",
+      "\t return 1./(1+t)\n",
+      "\n",
+      "XA_avg = 1-(E* quad(f2,to,t1)[0])\n",
+      "# Results\n",
+      "print \" Average concentration of A remaining in the droplet is %.3f\"%(XA_avg)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Average concentration of A remaining in the droplet is 0.653\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch12.ipynb b/Chemical_Reaction_Engineering/ch12.ipynb
new file mode 100755
index 00000000..f90bf55b
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch12.ipynb
@@ -0,0 +1,116 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Compartment Models"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 page no : 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "tg = (8.*(9.-6)*(0.5)+11.*(15.-9)*(0.5))/((15.-6)*0.5)              #sec\n",
+      "tl = 40.                                                            #sec\n",
+      "vg = 0.5\n",
+      "vl = 0.1;\n",
+      "\n",
+      "# Calculations\n",
+      "Vg = tg*vg;\n",
+      "Vl = tl*vl;\n",
+      "G = Vg*10\n",
+      "L = Vl*10\n",
+      "Stagnant = (100-G-L)\n",
+      "\n",
+      "# Results\n",
+      "print \" fraction of gas is %.f %%\"%(G)\n",
+      "print \" fraction of liquid is %.f %%\"%(L)\n",
+      "print \" fraction of Stangnant liquid is %.f %%\"%(Stagnant)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " fraction of gas is 50 %\n",
+        " fraction of liquid is 40 %\n",
+        " fraction of Stangnant liquid is 10 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 page no : 290"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "CAo = 1.;\n",
+      "XA = 0.75               #present\n",
+      "CA = 1-XA;\n",
+      "\n",
+      "# Calculations\n",
+      "kt1 = (CAo-CA)/CA;\n",
+      "kt2 = 3*kt1;            #volume is reduced by 1/3\n",
+      "CA_unconverted = 1./(kt2+1);\n",
+      "XA = 1-CA_unconverted   #New XA after replacing the stirrer\n",
+      "\n",
+      "# Results\n",
+      "print \" New Conversion expected is %.1f %%\"%(XA*100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " New Conversion expected is 90.0 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch13.ipynb b/Chemical_Reaction_Engineering/ch13.ipynb
new file mode 100755
index 00000000..205d7e7f
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch13.ipynb
@@ -0,0 +1,174 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : The Dispersion ,Wodel"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 page no : 305"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = [0,5,10,15,20,25,30,35];       # time\n",
+      "Cpulse = [0,3,5,5,4,2,1,0];        # gm/liter \n",
+      "sum1 = 0;\n",
+      "sum2 = 0\n",
+      "sum3 = 0;\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(8):\n",
+      "    sum1 = sum1+Cpulse[i];\n",
+      "    sum2 = sum2+Cpulse[i]*T[i];\n",
+      "    sum3 = sum3+Cpulse[i]*T[i]*T[i];\n",
+      "\n",
+      "t = sum2/sum1;\n",
+      "sigma_sqr = (sum3/sum1)-((sum2/sum1))**2;\n",
+      "sigmatheta_sqr = sigma_sqr/t**2;\n",
+      "m = 0.1\n",
+      "\n",
+      "while m <= 0.2:\n",
+      "    sigmat_sqr = 2*m-2*(m**2)*(1-math.exp(-(1./m)));\n",
+      "    if sigmat_sqr >=  sigmatheta_sqr:\n",
+      "        break;\n",
+      "    m += 0.001\n",
+      "\n",
+      "# Results\n",
+      "print \" The vessel print ersion number is %.3f\"%(m)\n",
+      "\n",
+      "# answer may vary because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The vessel print ersion number is 0.100\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 13.2 pageno : 306"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "l = 1219.;             # long diameter\n",
+      "u = 0.0067;            #Velocity(mm/s)  \n",
+      "\n",
+      "#Using the probability graph\n",
+      "t1 = 178550.\n",
+      "#84th percentile point fall at\n",
+      "t2 = 187750.;\n",
+      "\n",
+      "# Calculations\n",
+      "sigma = (t2-t1)/2;\n",
+      "t = l/u;\n",
+      "sigma_theta = sigma/t;\n",
+      "#Vessel print ersion number\n",
+      "d = sigma_theta**2/2;\n",
+      "\n",
+      "# Results\n",
+      "print \" The vessel print ersion number is %.5f\"%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The vessel print ersion number is 0.00032\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 page no : 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "v = 0.4;                # bed voidage\n",
+      "u = 1.2;                # velocity of fluid\n",
+      "l = 90.                 #length(cm)\n",
+      "sigma1_sqr = 39.        # output signals\n",
+      "sigma2_sqr = 64.        # output signals    \n",
+      "\n",
+      "# Calculations\n",
+      "dsigma_sqr = sigma2_sqr-sigma1_sqr;\n",
+      "t = l*v/u;\n",
+      "sigmatheta_sqr = dsigma_sqr/t**2;\n",
+      "d = sigmatheta_sqr/2;\n",
+      "\n",
+      "# Results\n",
+      "print \" The vessel print ersion number is %.4f\"%(d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The vessel print ersion number is 0.0139\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch14.ipynb b/Chemical_Reaction_Engineering/ch14.ipynb
new file mode 100755
index 00000000..5eb6f664
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch14.ipynb
@@ -0,0 +1,232 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14 The Tanks-in-Series Model"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1 pageno : 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Original and new length(m)\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "L1 = 32.            # diameter pipe\n",
+      "L2 = 50.            # pipeline length \n",
+      "sigma1 = 8.         # bottles \n",
+      "\n",
+      "# Calculations\n",
+      "# For small deviaqtion from plug flow,sigma_sqr is directly proportional to L\n",
+      "sigma2 = sigma1*math.sqrt(L2/L1);\n",
+      "\n",
+      "# Results\n",
+      "print \" No of bottles of rose expected is %.f\"%(sigma2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " No of bottles of rose expected is 10\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2 pageno : 330"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "sigma1 = 14.        # hours\n",
+      "sigma2 = 10.5;      # hours\n",
+      "L1 = 119.           # ohio miles apart\n",
+      "\n",
+      "# Calculations\n",
+      "#spread of curve is directly proportional to sqrt of distance from origin\n",
+      "L = sigma1**2*L1/(sigma1**2-sigma2**2);\n",
+      "\n",
+      "# Results\n",
+      "print \" The dumping of toxic phenol must have occured within %.f miles upstream of cincinnati\"%(L)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The dumping of toxic phenol must have occured within 272 miles upstream of cincinnati\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3 pageno : 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "# from figure\n",
+      "vo = 1.;          \n",
+      "t1 = 1./6;\n",
+      "t2 = 1.;\n",
+      "t3 = 11./6;\n",
+      "w = 1./10;\n",
+      "\n",
+      "# Calculations\n",
+      "A2_by_A1 = 0.5;\n",
+      "R = A2_by_A1/(1-A2_by_A1);\n",
+      "#From the location of 1st peak\n",
+      "V1 = (R+1)*vo*t1;\n",
+      "#From the time between peaks\n",
+      "V2 = (R*vo)*((t2-t1)-(t1));\n",
+      "#From fig 14.3\n",
+      "N = 1+(2*(t1/w))**2;\n",
+      "\n",
+      "# Results\n",
+      "print \" The reflux ratio is %.f\"%(R)\n",
+      "print \" The volume of 1st tank is %.3f\"%( V1)\n",
+      "print \" The volume of 2nd tank is %.3f\"%(V2)\n",
+      "print \" The number of tanks are %.f \"%(N)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The reflux ratio is 1\n",
+        " The volume of 1st tank is 0.333\n",
+        " The volume of 2nd tank is 0.667\n",
+        " The number of tanks are 12 \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4 page no  :333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "import math \n",
+      "from numpy import *\n",
+      "\n",
+      "# Variables\n",
+      "# from figure E14.4a\n",
+      "t2 = 280.\n",
+      "t1 = 220.\n",
+      "sigma1_sqr = 100.\n",
+      "sigma2_sqr = 1000.\n",
+      "\n",
+      "# Calculations\n",
+      "dt = t2-t1;\n",
+      "dsigma_sqr = sigma2_sqr-sigma1_sqr;\n",
+      "N = dt**2/dsigma_sqr;\n",
+      "\n",
+      "E = zeros(200)\n",
+      "\n",
+      "for t in range(200):\n",
+      "    E[t] = ((t**(N-1))*(N**N)*math.exp(-t*N/dt))/((math.factorial(N-1))*(dt**N));\n",
+      "\n",
+      "t = zeros(200)    \n",
+      "for i in range(200):\n",
+      "    t[i] = i;\n",
+      "\n",
+      "# Results\n",
+      "plot(t,E)\n",
+      "xlabel(\"t, s\")\n",
+      "ylabel(\"E, s**-1\")\n",
+      "show()\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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BBx8EnnlG7UqIqDUME5Xt2CEXw6K2PfYYsH498N13aldCRC1RLExMJhNiY2Oh0+mQm5vb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\nRK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2625a50>"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch16.ipynb b/Chemical_Reaction_Engineering/ch16.ipynb
new file mode 100755
index 00000000..f90f61a4
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch16.ipynb
@@ -0,0 +1,98 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 16 Earliness of Mixing, Segregation, and RTD"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 16.1 pageno : 358"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "Co = 1.\n",
+      "k = 1.\n",
+      "t = 1.      #given\n",
+      "\n",
+      "# Calculations and Results\n",
+      "C1 = (-1+math.sqrt(1-4*t*(-Co)))/2*t;\n",
+      "#For the plug flow reactor\n",
+      "#t = 1/k(1/C2-1/C1)\n",
+      "C2 = C1/(1+k*t*C1);\n",
+      "print \" Conversion for flow scheme A is %.3f\"%(C2)\n",
+      "\n",
+      "#For plug flow\n",
+      "C3 = Co/(1+k*t*Co);\n",
+      "#For mixed flow reactor\n",
+      "C4 = (-1+math.sqrt(1-4*t*(-C3)))/2*t;\n",
+      "print \" Conversion for flow scheme B is %.3f\"%(C4)\n",
+      "\n",
+      "#Using exit age distribution fn for 2 equal plug-mixed flow reactor system,umath.sing fig 12.1\n",
+      "t_bar = 2.\n",
+      "in_ = 1000.\n",
+      "\n",
+      "def f3(t): \n",
+      "\t return (2/t_bar)*(math.exp(1-2*t/t_bar))/(1+Co*k*t)\n",
+      "\n",
+      "C5 =  quad(f3,t_bar/2,in_)[0]\n",
+      "\n",
+      "print \" Conversion for flow scheme C,D,E is %.3f\"%(C5)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Conversion for flow scheme A is 0.382\n",
+        " Conversion for flow scheme B is 0.366\n",
+        " Conversion for flow scheme C,D,E is 0.361"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch18.ipynb b/Chemical_Reaction_Engineering/ch18.ipynb
new file mode 100755
index 00000000..9efd8f19
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch18.ipynb
@@ -0,0 +1,465 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 18 Solid Catalyzed Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.1 pageno : 407"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "dp = 2.4*(10**-3);L = dp/6;\n",
+      "De = 5*10**-5;\n",
+      "Keff = 1.6;\n",
+      "h = 160.                #heat transfer coefficient(KJ/hr.m2cat.K)\n",
+      "kg = 300.               #mass transfer coefficient(m3/hr.m2cat)\n",
+      "Hr = -160.              #KJ/molA\n",
+      "CAg = 20.               #mol/m3\n",
+      "rA_obs = 10.**5         #mol/hr.m3cat\n",
+      "\n",
+      "# Calculations and Results\n",
+      "kobs = rA_obs/CAg;\n",
+      "Vp = 3.14*(dp**3)/6;\n",
+      "S = 3.14*(dp**2);\n",
+      "ratio = kobs*Vp/(kg*S);\n",
+      "\n",
+      "print \" Part a\"\n",
+      "if ratio<0.01 :\n",
+      "    print \" Resistance to mass transport to film should not influence rate of reaction\"\n",
+      "else:\n",
+      "    print \" Resistance to mass transport to film should influence rate of reaction\"\n",
+      "\n",
+      "\n",
+      "print \" Part b\"\n",
+      "\n",
+      "Mw = rA_obs*(L**2)/(De*CAg);\n",
+      "print \" Mw =  %.f\"%(Mw)\n",
+      "if Mw>4:\n",
+      "    print \" Pore diffusion is influencing and hence strong pore diffusion\"\n",
+      "else:\n",
+      "    print \" Pore diffusion is  not influencing and hence weak pore diffuusion\"\n",
+      "\n",
+      "dt_max_pellet = De*(CAg-0)*(-Hr)/Keff;\n",
+      "#Temp variation Across the gas film\n",
+      "dt_max_film = L*rA_obs*(-Hr)/h;\n",
+      "\n",
+      "print \" Part c\"\n",
+      "print \" dTmax,pellet is %.1f C\"%(dt_max_pellet)\n",
+      "print \" degree C  dTmax,film is %.2f C\"%(dt_max_film)\n",
+      "print \" degree C\"\n",
+      "if dt_max_pellet<1:\n",
+      "    print \" Pellet is close to uniform in temperature\"\n",
+      "else:\n",
+      "    print \" There is a variation in temp within pellet\"\n",
+      "if dt_max_film<1:\n",
+      "    print \" Film is close to uniform in temperature\"\n",
+      "else:\n",
+      "    print \" There is a variation in temp within Film\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " Resistance to mass transport to film should not influence rate of reaction\n",
+        " Part b\n",
+        " Mw =  16\n",
+        " Pore diffusion is influencing and hence strong pore diffusion\n",
+        " Part c\n",
+        " dTmax,pellet is 0.1 C\n",
+        " degree C  dTmax,film is 40.00 C\n",
+        " degree C\n",
+        " Pellet is close to uniform in temperature\n",
+        " There is a variation in temp within Film\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.2 page no : 408"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "from numpy import *\n",
+      "import math \n",
+      "from scipy import stats\n",
+      "\n",
+      "# Variables\n",
+      "PAo = 3.2;\n",
+      "R = 0.082           #litre.atm/mol.k\n",
+      "T = 390.            #k\n",
+      "v = 20.             #litre/hr\n",
+      "W = 0.01            #/kg\n",
+      "CA_in = [0.1,0.08,0.06,0.04];\n",
+      "CA_out = [0.084,0.07,0.055,0.038];\n",
+      "\n",
+      "# Calculations\n",
+      "CAo = PAo/(R*T);\n",
+      "FAo = CAo*v;\n",
+      "eA = 3;\n",
+      "\n",
+      "XA_in = zeros(4)\n",
+      "XA_out = zeros(4)\n",
+      "dXA = zeros(4)\n",
+      "rA = zeros(4)\n",
+      "CA_avg = zeros(4)\n",
+      "\n",
+      "for i in range(4):\n",
+      "    XA_in[i] = (1-CA_in[i]/CAo)/(1+eA*CA_in[i]/CAo);\n",
+      "    XA_out[i] = (1-CA_out[i]/CAo)/(1+eA*CA_out[i]/CAo);\n",
+      "    dXA[i] = XA_out[i]-XA_in[i];\n",
+      "    rA[i] = dXA[i]/(W/FAo);\n",
+      "    CA_avg[i] = (CA_in[i]+CA_out[i])/2;\n",
+      "\n",
+      "# Results\n",
+      "plot(CA_avg,rA)\n",
+      "xlabel(\"CA, mol/liter\")\n",
+      "ylabel(\"-rA, mol/hr. kg cat\")\n",
+      "coeff1 = stats.linregress(CA_avg,rA)\n",
+      "k = coeff1[0]\n",
+      "print \" The rate of reaction is %.3f mol/hr.kg\"%(k),\n",
+      "print \"CA\"\n",
+      "print ('The answer slightly differs from those given in book as regress fn is\\\n",
+      " used for calculating slope and intercept')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        " The rate of reaction is 109.395 mol/hr.kg"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " CA\n",
+        "The answer slightly differs from those given in book as regress fn is used for calculating slope and intercept\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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TyNlB+csK24vCr7/+SsuWLWnatCmRkZE899xzdkfwq0B/YSm/cwI5Oyh/WWH7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\nEBERj/8P/TvRn68kedQAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x2f0a1d0>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.3 pageno : 410"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "from numpy import zeros\n",
+      "from scipy import stats\n",
+      "\n",
+      "\n",
+      "# Variables\n",
+      "CAo = 0.1           #mol/litre\n",
+      "FAo = 2.            #mol/hr\n",
+      "eA = 3.\n",
+      "CA = [0.074,0.06,0.044,0.029]           #mol/litre\n",
+      "W = [0.02,0.04,0.08,0.16]               #kg\n",
+      "\n",
+      "XA = zeros(4)\n",
+      "y = zeros(4)\n",
+      "x = zeros(4)\n",
+      "W_by_FAo = zeros(4)\n",
+      "CAout_by_CAo = zeros(4)\n",
+      "XA1 = zeros(4)\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(4):\n",
+      "    XA[i] = (CAo-CA[i])/(CAo+eA*CA[i]);\n",
+      "    y[i]  = (1+eA)*math.log(1/(1-XA[i]))-eA*XA[i];\n",
+      "    x[i] = CAo*W[i]/FAo;\n",
+      "    W_by_FAo[i] = W[i]/FAo;\n",
+      "    CAout_by_CAo[i] = CA[i]/CAo;\n",
+      "    XA1[i] = (1-CAout_by_CAo[i])/(1+eA*CAout_by_CAo[i]);\n",
+      "\n",
+      "# Results\n",
+      "plot(x,y)\n",
+      "xlabel(\"CA0W/FAO = W/20, hr.kg cat/liter\")\n",
+      "ylabel(\"4 ln 1/(1-Xa) -3Xa\")\n",
+      "coeff3 = stats.linregress(x,y);\n",
+      "k = coeff3[0];\n",
+      "print \" Part a, using integral method of analysis\"\n",
+      "print \" The rate of reactionmol/litre is %.3f\"%(k),\n",
+      "print \"CA\"\n",
+      "\n",
+      "#Part b\n",
+      "#plotting W_by_FAo vs XA1,the calculating rA = dXA/d(W/FAo) for last 3 points,we get\n",
+      "rA = [5.62,4.13,2.715];\n",
+      "coeff2 = stats.linregress(CA[1:],rA);\n",
+      "\n",
+      "print \" Part b, using differential method of analysis\"\n",
+      "print \" The rate of reactionmol/litre) is %.3f\"%(coeff2[0]),\n",
+      "print \"CA\"\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        " Part a, using integral method of analysis"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The rate of reactionmol/litre is 96.804 CA\n",
+        " Part b, using differential method of analysis\n",
+        " The rate of reactionmol/litre) is 93.703 CA\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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IWEczvT3YTz+ZzU2pqfDEE+YigZpLISKlNNNbHBPvQkLMEU9bt2rinYhYTxP3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\nSUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7ffe3c06fe90>"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.4 pageno : 413"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "# from example 18.2\n",
+      "XA = 0.35;\n",
+      "FAo = 2000.         #mol/hr\n",
+      "eA = 3.\n",
+      "k = 96.\n",
+      "CAo = 0.1;\n",
+      "\n",
+      "# Calculations\n",
+      "W = ((1+eA)*math.log(1./(1-XA))-eA*XA)*(FAo/(k*CAo));\n",
+      "\n",
+      "# Results\n",
+      "print \" The amount of catalyst kg needed in a packed bed reactor is %.f kg\"%(W)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The amount of catalyst kg needed in a packed bed reactor is 140 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.5 pageno : 414"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from matplotlib.pyplot import *\n",
+      "from numpy import *\n",
+      "from scipy import *\n",
+      "\n",
+      "# Variables\n",
+      "CAo = 0.1;\n",
+      "eA = 3;\n",
+      "rA = [3.4,5.4,7.6,9.1];            # mol A/hr. kg cat\n",
+      "CA = [0.039,0.0575,0.075,0.092];   # mol/liter\n",
+      "\n",
+      "# Calculations\n",
+      "XA = zeros(5);\n",
+      "inv_rA = zeros(5);\n",
+      "for i in range(4):\n",
+      "    XA[i] = (1-CA[i]/CAo)/(1+eA*CA[i]/CAo);\n",
+      "    inv_rA[i] = 1/rA[i];\n",
+      "\n",
+      "for i in range(4):\n",
+      "    small = XA[i];\n",
+      "    for j in range(i,4):\n",
+      "        next_ = XA[j];\n",
+      "        if small>next_:\n",
+      "            XA[i],XA[j] = XA[j],XA[i];\n",
+      "            inv_rA[i],inv_rA[j] = inv_rA[j], inv_rA[i]\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "plot(XA,inv_rA)\n",
+      "xlabel(\"Xa, dimensionless\")\n",
+      "ylabel(\"-1/rA, hr.kg cat/mol\")\n",
+      "show()\n",
+      "XA[4] = 0.35\n",
+      "inv_rA[4] = 0.34;\n",
+      "Area = trapz(XA,inv_rA);\n",
+      "W = Area*2000.;\n",
+      "print \"Amount of catalyst needed is %.f kg cat\"%(W)\n",
+      "print ('The answer differs from those given in book as trapezoidal rule is used for calculating area')\n",
+      "\n",
+      "# Note : Answer differs because python trepz function works differnt than we do manually trepezoidal method."
+     ],
+     "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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hQ73Fe7t29o5ICMdVareVUoo2bdqwZ88eW8ZkNum2sp/8fFi0SO9sC3o8IyIC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\nRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7fa93d80a710>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amount of catalyst needed is 91 kg cat\n",
+        "The answer differs from those given in book as trapezoidal rule is used for calculating area\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 18.6 pageno : 416"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "XA = 0.35;\n",
+      "FAo = 2000.             #mol/hr\n",
+      "CAo = 0.1               #mol/litre\n",
+      "eA = 3.\n",
+      "k = 96.                 # mol/kg cat\n",
+      "\n",
+      "# Calculations\n",
+      "CA = CAo*((1-XA)/(1+eA*XA))\n",
+      "rA = k*CA;\n",
+      "W = FAo*XA/rA;\n",
+      "\n",
+      "# Results\n",
+      "print \"The amount of catalyst neededkg, is %.f kg\"%(W)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The amount of catalyst neededkg, is 230 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch19.ipynb b/Chemical_Reaction_Engineering/ch19.ipynb
new file mode 100755
index 00000000..419b424a
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch19.ipynb
@@ -0,0 +1,204 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 19 : The Packed Bed Catalytic Reactor"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.1 page no : 438"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# Variables\n",
+      "Cp = 40.                    #J/mol.k\n",
+      "Hr = 80000.                 #J/mol.k\n",
+      "FAo = 100.                  #mol/s\n",
+      "nA = 1.\n",
+      "nB = 7.\n",
+      "n = nA + nB\n",
+      "T1 = 300.                   #k\n",
+      "T2 = 600.                   #k\n",
+      "T3 = 800.                   #k\n",
+      "\n",
+      "# Calculations\n",
+      "m = Cp/Hr;\n",
+      "XA = [0.8,0.78,0.7,0.66,0.5,0.26,0.1,0];\n",
+      "inv_rA = [20,10,5,4.4,5,10,20,33];\n",
+      "\n",
+      "\n",
+      "plot(XA,inv_rA)\n",
+      "xlabel(\"Xa\")\n",
+      "ylabel(\"1/8 X 1/-rA\")\n",
+      "\n",
+      "print ('From the plot we can say that a recycle reactor should be used')\n",
+      "W = FAo*38.4\n",
+      "R = 1.;\n",
+      "Q1 = n*FAo*Cp*(T2-T1);\n",
+      "Q2 = n*FAo*Cp*(T1-T3);\n",
+      "\n",
+      "# Results\n",
+      "print \" The weight of catalyst needed is %.f kg\"%(W)\n",
+      "print \" The Recycle Ratio is %.f\"%(R)\n",
+      "print \" The heat exchange for feed is %.f MW\"%(Q1/10**6)\n",
+      "print \" The heat excahnge for the product is %.f MW\"%(Q2/10**6)\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "From the plot we can say that a recycle reactor should be used\n",
+        " The weight of catalyst needed is 3840 kg\n",
+        " The Recycle Ratio is 1\n",
+        " The heat exchange for feed is 10 MW\n",
+        " The heat excahnge for the product is -16 MW\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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cHFXUFIa8PPNSVc1v/AkJ0LGj3X/LwOJXheBSVAhE/F9FhTnHUVxsPtq3N1dO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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7ff34ba76ed0>"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 19.2 page no : 440"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# Variables\n",
+      "Cp = 40.\n",
+      "Hr = 80000.\n",
+      "m = Cp/Hr;\n",
+      "FAo = 100.              #mol/s\n",
+      "print ('We should use a mixed flow reactor operating at optimum')\n",
+      "\n",
+      "# Calculations and Results\n",
+      "XA = [0.85,0.785,0.715,0.66,0.58,0.46];\n",
+      "inv_rAopt = [20,10,5,3.6,2,1];\n",
+      "plot(XA,inv_rAopt)\n",
+      "\n",
+      "area1 = 0.66*3.6;\n",
+      "area2 = (0.85-0.66)*20;\n",
+      "W1 = FAo*area1;\n",
+      "W2 = FAo*area2;\n",
+      "print \" The weight of catalyst needed for 1st bed is %.1f kg\"%(W1)\n",
+      "print \" The weight of catalyst needed for 2ndbed is %.1f kg\"%(W2)\n",
+      "\n",
+      "#Heat exchange\n",
+      "#For the first reactor\n",
+      "Q = (820-300)*Cp+0.66*(-Hr);\n",
+      "\n",
+      "#For 100 mol/s\n",
+      "Q1 =  FAo*Q/10**6;#MW\n",
+      "print \" The amount of heat exchanged for 1st reactor is %.2f MW\"%(Q1)\n",
+      "\n",
+      "#For 2nd reactor\n",
+      "#To go from XA = 0.66 at 820 k to XA = 0.85 at 750 k\n",
+      "Q2 = FAo*((750-820)*Cp+(0.85-0.66)*(-Hr));\n",
+      "Q2 = Q2/10**6;\n",
+      "print \" The amount of heat exchanged for 2nd reactor is %.2f MW\"%(Q2)\n",
+      "\n",
+      "#For the exchanger needed to cool the exit stream from 750 k to 300 k\n",
+      "Q3 = FAo*Cp*(300. - 750);\n",
+      "Q3 = Q3/10**6;#MW\n",
+      "print \" The amount of heat exchanged for exchanger is %.2f MW\"%(Q3)\n",
+      "\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "We should use a mixed flow reactor operating at optimum\n",
+        " The weight of catalyst needed for 1st bed is 237.6 kg"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The weight of catalyst needed for 2ndbed is 380.0 kg\n",
+        " The amount of heat exchanged for 1st reactor is -3.20 MW\n",
+        " The amount of heat exchanged for 2nd reactor is -1.80 MW\n",
+        " The amount of heat exchanged for exchanger is -1.80 MW\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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jTcATe0JCgi655BLVqlVLtWvXVm5urp25ANTAoUNSRobUvLn17Be2sosuAf91ezweeb1e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\n7AAQYSh2AIgwFDsARBiKHQAizP8DWZlwLI/+Rv8AAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x3959650>"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch2.ipynb b/Chemical_Reaction_Engineering/ch2.ipynb
new file mode 100755
index 00000000..70ba8482
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch2.ipynb
@@ -0,0 +1,76 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Kinetics of Homogeneous Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 page no : 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "# Given\n",
+      "#t1 = 30 min ;T1 = 336 k;\n",
+      "#t2 = 15 sec ;T2 = 347 k;\n",
+      "# Converting t2 in min\n",
+      "t1 = 30.         # milk heated (mins)\n",
+      "T1 = 336.        # K based on t1\n",
+      "t2 = 0.25        # seconds        \n",
+      "T2 = 347.        # K based on t2\n",
+      "R = 8.314\n",
+      "\n",
+      "# Calculations\n",
+      "#math.log(t1/t2) = E(1/T1-1/T2)/R\n",
+      "E = (math.log(t1/t2)*R)/(1/T1-1/T2);\n",
+      "\n",
+      "# Results\n",
+      "print \"E is %d J/mol\"%(round(E,-3))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "E is 422000 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch20.ipynb b/Chemical_Reaction_Engineering/ch20.ipynb
new file mode 100755
index 00000000..9a87e997
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch20.ipynb
@@ -0,0 +1,112 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 20 :\n",
+      "\n",
+      "Reactors with Suspended Solid \n",
+      "\n",
+      "Catalyst, Fluidized Reactors of \n",
+      "\n",
+      "Various Types"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 20.1 pageno : 460"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import linalg\n",
+      "\n",
+      "# Variables\n",
+      "uo = 0.3\n",
+      "umf = 0.03                      #m/s\n",
+      "vo = 0.3*3.14159                #m3/s\n",
+      "d = 2.                          #m\n",
+      "db = 0.32;                      #dia of bubble(m)\n",
+      "emf = 0.5;\n",
+      "W = 7000.                       #kg\n",
+      "CAo = 100.                      #mol/m3\n",
+      "D = 20*10**-6;                  #m2/s\n",
+      "density = 2000.                 #kg/m3\n",
+      "k = 0.8;\n",
+      "alpha = 0.33;\n",
+      "g = 9.8\n",
+      "\n",
+      "# Calculations\n",
+      "#Using bubbling bed model\n",
+      "#Rise velocity of bubbles\n",
+      "ubr = 0.711*math.sqrt(g*db);\n",
+      "ub = uo-umf+ubr;\n",
+      "delta = uo/ub;\n",
+      "ef = 1-(1-emf)*(1-delta);\n",
+      "Kbc = 4.5*(umf/db)+5.85*(D**0.5)*(g**0.25)/(db**1.25);\n",
+      "Kce = 6.77*math.sqrt(emf*D*ubr/db**3);\n",
+      "fb = 0.001;\n",
+      "fc = delta*(1-emf)*((3*umf/emf)/(ubr-umf/emf)+alpha);\n",
+      "fe = (1-emf)*(1-delta)-fc-fb;\n",
+      "ft = fb+fe+fc;\n",
+      "A = 3.14*d*d/4;\n",
+      "Hbfb = W/((density*A)*(1-ef));\n",
+      "XA = 1- 0.6856948 #linalg.inv(math.exp(fb*k+(1/((1/(delta*Kbc))+1/((fc*k)+(1/((1/(delta*Kce))+(1/(fe*k)))))))*(Hbfb*ft/uo)/ft));\n",
+      "XA1 = 100*XA;                       #in percentage\n",
+      "\n",
+      "# Results\n",
+      "print \" Part a\"\n",
+      "print \" Conversion of reactant is %f \"%(XA1)\n",
+      "CA_avg = CAo*XA*vo*density/(k*W);\n",
+      "print \" Part b\"\n",
+      "print \" The proper mean concentratio (nmol/m3) of A seen by solid is %.f\"%(round(CA_avg))\n",
+      "XA1 = 1-    0.0511804  #linalg.inv(math.exp(k*ft*Hbfb/uo));\n",
+      "print \" Part c\"\n",
+      "print \" Conversion of reactant for packed bad is %f\"%(XA1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " Conversion of reactant is 31.430520 \n",
+        " Part b\n",
+        " The proper mean concentratio (nmol/m3) of A seen by solid is 11\n",
+        " Part c\n",
+        " Conversion of reactant for packed bad is 0.948820\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch21.ipynb b/Chemical_Reaction_Engineering/ch21.ipynb
new file mode 100755
index 00000000..2126aaf8
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch21.ipynb
@@ -0,0 +1,230 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 21 : The Rate and Performance Equations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 21.1 page no : 486"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "from scipy import stats\n",
+      "\n",
+      "# Variables\n",
+      "t = [0,2,4,6];                 # time\n",
+      "XA = [0.75,0.64,0.52,0.39];    # XA\n",
+      "t1 = 4000.                      #kg.s/m3\n",
+      "density_s = 1500.               #kg/m3\n",
+      "De = 5.*10**-10;\n",
+      "d = 2.4*10**-3;\n",
+      "\n",
+      "# Calculations\n",
+      "y = zeros(4)\n",
+      "for i in range(4):\n",
+      "    y[i] = math.log((1./(1-XA[i]))-1);\n",
+      "\n",
+      "plot(y,t)\n",
+      "ylabel(\"ln(CAO/CA -1)\")\n",
+      "\n",
+      "coeff  = stats.linregress(t,y);\n",
+      "kd = coeff[0];\n",
+      "k = math.exp(coeff[1])/t1;\n",
+      "L = d/6;\n",
+      "Mt = L*math.sqrt(k*density_s/De);\n",
+      "#Assuming Runs were made in regime of strong resismath.tance to  pore diffusion\n",
+      "\n",
+      "k1 = ((math.exp(coeff[1]))**2)*(L**2)*density_s/(t1*t1*De);\n",
+      "kd1 = -2*coeff[0];\n",
+      "Mt = L*math.sqrt(k1*density_s/De);\n",
+      "\n",
+      "# Results\n",
+      "print \" Rate equation in diffusion free regime with deactivation is %.2f  m**3/kg.s\"%(k1)\n",
+      "print \" CA*a with  -da/dthr-1  is %.2f a,hr**-1\"%(kd1)\n",
+      "\n",
+      "#In strong pore diffusion\n",
+      "k2 = k1*math.sqrt(De/(k1*density_s));\n",
+      "print \" CA*a**0.5/L with  -da/dthr-1 is %.2f a,hr**-1\"%(kd1),\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        " Rate equation in diffusion free regime with deactivation is 0.27  m**3/kg.s\n",
+        " CA*a with  -da/dthr-1  is 0.51 a,hr**-1\n",
+        " CA*a**0.5/L with  -da/dthr-1 is 0.51 a,hr**-1\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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44gu7liMlJcW27y1Xptfc+/Sae19RXvPYWNi2DbKznRsKZltRT0tLo1atWkRF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\nQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7f7102416750>"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 21.2 pageno : 492"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "from matplotlib.pyplot import *\n",
+      "from numpy import *\n",
+      "\n",
+      "# Variables\n",
+      "PAo = 3.                #atm\n",
+      "R = 82.06*10**-6        #m3.atm/mol.k\n",
+      "T = 730.                #k\n",
+      "W = 1000.               #kg\n",
+      "FAo = 5000.             #mol/hr\n",
+      "\n",
+      "# Calculations\n",
+      "CAo = PAo/(R*T);\n",
+      "tau = W*CAo/FAo;\n",
+      "i = 0;\n",
+      "\n",
+      "a = zeros(25)\n",
+      "XA = zeros(25)\n",
+      "a1 = zeros(25)\n",
+      "XA1 = zeros(25)\n",
+      "a2 = zeros(25)\n",
+      "XA2 = zeros(25)\n",
+      "a3 = zeros(25)\n",
+      "XA3 = zeros(25)\n",
+      "a = linspace(0,120,25)\n",
+      "for t in xrange(0,120,5):\n",
+      "    i = i+1;\n",
+      "    #Part a\n",
+      "    a[i] = 1-(8.3125*10**-3)*t;\n",
+      "    XA[i] = (tau**2)*a[i]/(1+(tau**2)*a[i]);\n",
+      "    #Part b\n",
+      "    a1[i] = math.exp(-0.05*t);\n",
+      "    XA1[i] = (tau**2)*a1[i]/(1+(tau**2)*a1[i]);\n",
+      "    #Part c\n",
+      "    a2[i] = 1/(1+3.325*t);\n",
+      "    XA2[i] = (tau**2)*a2[i]/(1+(tau**2)*a2[i]);\n",
+      "    #Part d\n",
+      "    a3[i] = 1/(math.sqrt(1+1333*t));\n",
+      "    XA3[i] = (tau**2)*a3[i]/(1+(tau**2)*a3[i]);\n",
+      "\n",
+      "t = linspace(0,120,25)\n",
+      "plot(t,XA,t,XA1,t,XA2,t,XA3)\n",
+      "suptitle(\"Decrease in conversion as a function of time for various deactivation orders.\")\n",
+      "xlabel(\"Time, days\")\n",
+      "ylabel(\"Xa\")\n",
+      "\n",
+      "def f13(t): \n",
+      "\t return ((100*(1-(8.3125*10**-3)*t))/(1+100*(1-(8.3125*10**-3)*t)))\n",
+      "\n",
+      "XA_avg = (1./120) * quad(f13,0,120)[0]\n",
+      "\n",
+      "\n",
+      "def f14(t): \n",
+      "\t return (100.*math.exp(-0.05*t))/(1+100*math.exp(-0.05*t))\n",
+      "\n",
+      "XA1_avg = (1./120)* quad(f14,0,120)[0]\n",
+      "\n",
+      "\n",
+      "def f15(t): \n",
+      "\t return ((100*(1./(1+3.325*t)))/(1+100*(1/(1+3.325*t))))\n",
+      "\n",
+      "XA2_avg = (1./120)* quad(f15,0,120)[0]\n",
+      "\n",
+      "\n",
+      "def f16(t): \n",
+      "\t return ((100*1./(math.sqrt(1+1333*t)))/(1+100*(1/math.sqrt(1+1333*t))))\n",
+      "\n",
+      "XA3_avg = (1./120)* quad(f16,0,120)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \" for d = 0,the mean conversion is % f\"%(XA_avg)\n",
+      "print \" for d = 1,the mean conversion is % f\"%(XA1_avg)\n",
+      "print \" for d = 2,the mean conversion is % f\"%(XA2_avg)\n",
+      "print \" for d = 3,the mean conversion is % f\"%(XA3_avg)\n",
+      "\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " for d = 0,the mean conversion is  0.955970\n",
+        " for d = 1,the mean conversion is  0.732280\n",
+        " for d = 2,the mean conversion is  0.400874\n",
+        " for d = 3,the mean conversion is  0.298840\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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UUzdnrZaL+cs9QKmInZur1K2TahxnB2fmdJ3D38/9zeYRmwHot6Yf7kvceXfv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\nAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x4215310>"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch22.ipynb b/Chemical_Reaction_Engineering/ch22.ipynb
new file mode 100755
index 00000000..06928688
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch22.ipynb
@@ -0,0 +1,194 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 22  :\n",
+      "\n",
+      "G/L Reactions on Solid \n",
+      "\n",
+      "Catalysts: Trickle Beds, Slurry \n",
+      "\n",
+      "Reactors, and Three-Phase \n",
+      "\n",
+      "Fluidized Beds"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.1 pageno : 511"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "PA = 101325.            #Pa\n",
+      "HA = 36845.             #PA.m3.l/mol\n",
+      "CBo = 1000.             #mol/m3\n",
+      "v = 10**-4              #m3*l/s\n",
+      "h = 5.                  #m\n",
+      "A = 0.1                 #m2\n",
+      "\n",
+      "# Calculations\n",
+      "CA = PA/HA;\n",
+      "FBo = v*CBo;\n",
+      "Vr = A*h;\n",
+      "dp = 5*10**-3;          #mcat\n",
+      "d_solid = 4500.         #kg/m3cat\n",
+      "De = 8*10**-10;         #m3l/mcat.s\n",
+      "n = 0.5;\n",
+      "b = 1.;\n",
+      "k = 2.35*10**-3;\n",
+      "L = dp/6.;\n",
+      "kai_overall = 0.02;\n",
+      "kac_ac = 0.05;\n",
+      "f = 0.6;\n",
+      "#For a half-order reaction\n",
+      "Mt = L*math.sqrt((n+1)*(k*d_solid*(CA)**(n-1))/(2*De));\n",
+      "E = 1/Mt;\n",
+      "rA = (1/((1/(kai_overall))+(1/(kac_ac))+(1/(k*b*(CA**(n-1))*E*f*d_solid))))*(PA/HA);\n",
+      "#From Material Balance\n",
+      "XB = b*rA*Vr/FBo;\n",
+      "\n",
+      "# Results\n",
+      "print \" The conversion of acetone is %.3f\"%(XB)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The conversion of acetone is 0.158\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 22.2 pageno : 513"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from matplotlib.pyplot import *\n",
+      "from numpy import *\n",
+      "\n",
+      "# Variables\n",
+      "PA = 14.6*101325;           #Pa\n",
+      "HA = 148000.;               # liquid\n",
+      "Vr = 2.;\n",
+      "Vl = Vr;\n",
+      "b = 1.;\n",
+      "fs = 0.0055;\n",
+      "\n",
+      "# Calculations\n",
+      "k = 5.*10**-5;              #m6l/kg.molcat.s\n",
+      "dp = 5*10**-5;              #mcat\n",
+      "kac = 4.4*10**-4;kai = 0.277;   #m3l/m3.r.s\n",
+      "density = 1450.;                #kg/m3\n",
+      "De = 5*10**-10;                 #m3l/mcat.s\n",
+      "L = dp/6;                       #for spherical particle\n",
+      "CA = PA/HA;\n",
+      "X = 0.9;\n",
+      "CBo = 2500.\n",
+      "CB = CBo*(1-X);\n",
+      "ac = 6*fs/dp;\n",
+      "K = kac*ac;\n",
+      "\n",
+      "CB = [2500,1000,250];\n",
+      "e = [0.19,0.29,0.5];\n",
+      "Mt = zeros(3)\n",
+      "rA = zeros(3)\n",
+      "inv_rA = zeros(3)\n",
+      "for i in range(3):\n",
+      "    Mt[i] = L*math.sqrt(k*CB[i]*density/De);\n",
+      "    rA[i] = CA/((1./kai)+(1./K)+(1./(k*density*e[i]*fs*CB[i])))\n",
+      "    inv_rA[i] = 1/rA[i];\n",
+      "\n",
+      "# Results\n",
+      "plot(CB,inv_rA)\n",
+      "ylabel(\"1/-rA\")\n",
+      "Area = 3460.\n",
+      "t = Vl*Area/(b*Vr);\n",
+      "t = t/60.                         #min\n",
+      "print \" The time required for 90 percentage conversion of reactant is %.f\"%t,\n",
+      "print \"min\"\n",
+      "show()\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        " The time required for 90 percentage conversion of reactant is 58"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " min\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive', 'e']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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JkyksLKRt27YsWLCAzMxMz3Cak/C56mrz2ok334Rbb7U7jYi0hFe/O42zOHHihPH88897\n3Hfy5EnD5XKd7c+MiooKo6SkxDAMw6iqqjKuvfZaY+fOnR7bfPLJJ8bAgQMNwzCM9evXG1lZWY2e\n5xzxxCIff2wYCQmGUVNjdxIRaQlvfnee9TqJ1q1bs2jRIo/7WrVqxc0333zWwhMdHU3GT8ufORwO\nkpKS2L9/v8c2H330EXfffTcAWVlZHD58WNdf+IlBg8zV6x5/3O4kImK3Nufa4KabbmLSpEn87ne/\no127dhiGQVhYGNddd915vUBpaSklJSVkZWV53L9v3z7iTlvQIDY2lr179xIVFeWxXUFBQf3PTqcT\np9N5Xq8rF2fuXEhLM+coUlPtTiMiZ+NyuXC5XJY89zmLRElJCWFhYTzyyCMe969evfqcT15dXc3w\n4cOZM2cODoej0ePGGWNmYU2coH96kRDf6dwZHnvMbNlRVASt1MBFxG+d+Q/omTNneu25z1kkWlqd\namtrycnJYdSoUQwZMqTR4zExMZSXl9f/vnfvXmJiYlr0WmKNvDx45x147TW47z6704iIHS7o34eD\nBg06r+0Mw2D8+PEkJyczderUJrcZPHgwb7/9NgDr16/nsssuazTUJPZq1cq8duKRR+CMKSURCREX\n1Co8MzOTkpKSc25XVFRE3759SUtLqx9CmjVrFmVlZQDk5+cDMGnSJFasWEG7du2YP39+o3kOnQLr\nH2bMgK+/hr/+1e4kInI+bFtPIjc3l/nz53vlhc+HioR/+M9/zEns556D7Gy704jIufikSOTl5TFw\n4EBuu+022rdv75UXu1AqEv5j1SoYO9Zc7tSmj4OInCefFIn169dTWFjIqlWrCA8P59e//jUDBgwg\nPT3dKy98XuFUJPzK2LFw2WXw4ot2JxGRs/H5cJPb7ebTTz9lxYoVbNu2jeuuu44BAwZwxx13eCVE\ns+FUJPyK221eM/Hxx3D99XanEZHm2LrGtWEYPPvss9TW1vLf//3fXgnRHBUJ//POO/D887Bxo9k5\nVkT8j61FAiAuLs7jGgerqEj4H8OAX/0Kfv1rdYsV8Vc+KRLdu3dv9o927drF8ePHvRLgbFQk/NO3\n30JWlnk0cfXVdqcRkTP5pEhERUWxYsUKIiIiGj32y1/+slHDPiuoSPivJ5+Ezz+H5cu13KmIv/Hm\nd2ezo8q//e1vqa6ubrTGA3DOLrAS/KZNg0WL4L334M477U4jIlZp0ZyEr+hIwr+tXw9Dh5rXTlx+\nud1pRKSO7RPXvqIi4f8mTYJjx2DePLuTiEgdFQnxG0eOQHKyOfTUt6/daUQEvPvdqVUC5KJceqm5\nQFFennlEISLBRUVCLtrQodCtGzz1lN1JRMTbNNwkXlFeDpmZ5ip23brZnUYktGm4SfxOXBw8+ijk\n58OpU3anERFvUZEQr7n/fnPtCR8uOSIiFtNwk3jV1q3Qvz9s3w5ajVbEHjoFVvza9OnmHMWiRXYn\nEQlNKhLi12pqzHUnXnkFBgywO41I6PH7ietx48YRFRXVbCdZt9vNgAEDyMjIIDU1lQULFlgRQ2zS\nti28+ircdx/8+KPdaUTkYlhyJLF27VocDgdjxoxh+/btjR4vKCjg2LFjPPnkk7jdbhITEzlw4ABt\nzljFRkcSge2uuyAmBp55xu4kIqHF748k+vTp02SL8TqdO3fmyJEjABw5coSOHTs2KhAS+F54ARYs\ngC1b7E4iIi1lyzfzhAkTuOWWW7jyyiupqqri/fffb3bbgoKC+p+dTidOp9P6gOIVkZHmVdgTJpgd\nY1u3tjuRSHByuVy4XC5LntuyievS0lKys7ObHG56/PHHcbvdvPjii3z77bf079+frVu30r59e89w\nGm4KeIYB/fqZrTumTLE7jUho8PvhpnNZt24dI0aMACA+Pp6rr76aXbt22RFFLBYWBq+9Bo89Zp4W\nKyKBxZYi0a1bNz777DMADhw4wK5du+jatasdUcQHEhNh8mSYONE8shCRwGHJcNPIkSNZs2YNbreb\nqKgoZs6cSW1tLQD5+fm43W5yc3MpKyvj1KlTPPzww/z+979vHE7DTUHj2DHIyIDHH4ecHLvTiAQ3\nXUwnAWntWhg50lzutEMHu9OIBC8VCQlYeXnQpo15NbaIWENFQgLWoUOQkgIffgi9etmdRiQ4BfzZ\nTRK6IiLMi+zy8uD4cbvTiMi5qEiIz91xh7lI0ezZdicRkXPRcJPYorQUevaEL76Aa66xO41IcNFw\nkwS8Ll3g4Yfh3nt17YSIP1ORENtMmWJOZL/zjt1JRKQ5Gm4SW23aBIMGwVdfQadOdqcRCQ46BVaC\nygMPwA8/wFtv2Z1EJDioSEhQqa42r51480249Va704gEPk1cS1BxOODllyE/H44etTuNiJxORUL8\nwqBBkJlpNgAUEf+h4SbxGxUVkJYGq1dDaqrdaUQCl4abJCh17mwuTpSXB6dO2Z1GREBFQvxMXl7D\nanYiYj8NN4nf2bEDnE7YuhWuvNLuNCKBR6fAStCbMQO+/hr++le7k4gEHs1JSNCbMQO2bYOPPrI7\niUho05GE+K1Vq2DsWHP4qX17u9OIBA6/P5IYN24cUVFRdO/evdltXC4XmZmZpKam4nQ6rYghAe6W\nW8wrsP/0J7uTiIQuS44k1q5di8PhYMyYMWzfvr3R44cPH6Z3796sXLmS2NhY3G43nZro7qYjCTl4\n0GzZ8fHHcP31dqcRCQx+fyTRp08fIiIimn180aJF5OTkEBsbC9BkgRAB6NgRnn3WPDX2xAm704iE\nnjZ2vOju3bupra2lX79+VFVVMWXKFEaPHt3ktgUFBfU/O51ODU2FoFGj4O234cUXYdo0u9OI+B+X\ny4XL5bLkuS2buC4tLSU7O7vJ4aZJkyaxefNm/vnPf1JTU0OvXr345JNPuOaMdSw13CR1vv0WsrJg\n40a4+mq704j4N78fbjqXuLg4fvWrX3HJJZfQsWNH+vbty9atW+2IIgEiPh4efBDuv1/LnYr4ki1F\n4vbbb6eoqIiTJ09SU1PDhg0bSE5OtiOKBJBp02DvXvjLX+xOIhI6LJmTGDlyJGvWrMHtdhMXF8fM\nmTOpra0FID8/n27dujFgwADS0tJo1aoVEyZMUJGQcwoPh3nzYOhQc0K7Vy9dPyFiNV1MJwHn9ddh\n4ULYvBkSE+Gmm8xb797q9SQC6t0kAsCxY2ahKCpquHXo0FA0broJunWDVmo+IyFGRUKkCadOwa5d\nnkXj8GHzCKOuaPToAT/7md1JRaylIiFynioqoLi4oWh8/bW5TGpd0fjlL+Es132KBCQVCZEWqqqC\nDRsaisaGDdCli+cQ1VVXmQsfiQQqFQkRL6mtNRc3On2IKjzcczK8e3do3drupCLnT0VCxCKGAd99\n51k09u83T7etKxw33ABt29qdVKR5KhIiPvR//wfr1jUUjW3bzKOL0482rrjC7pQiDVQkRGx09KjZ\nQ6quaKxbB1FRnvMaCQma1xD7qEiI+JGTJ83V8+qKxtq1cPy4Z9HIyDDnOkR8QUVCxM+VlTUUjeJi\nc57jhhsahqduvBEuvdTulBKsVCREAszhw/DFFw2F43//F6691vNoQy1FxFtUJEQC3JktRYqLzSML\ntRQRb1CREAkyhtG4pcihQ+YV4XVFo2dPtRSR86MiIRIC1FJEWkpFQiQEnd5SpLjY/PkXv2iYDL/p\nJvN3nXorKhIiwokTjVuKtG7tOa+hliKhSUVCRBpRSxGpoyIhIufF7fZsKbJ1K6SmerYUiYy0O6V4\nm4qEiLSIWoqEBr8vEuPGjeOTTz4hMjKS7du3N7vdxo0b6dWrF++//z7Dhg1rHE5FQsRSp7cUqTuT\n6j//8ZwMz8xUS5FA4/dFYu3atTgcDsaMGdNskTh58iT9+/enbdu25ObmkpOT0zicioSIz5WVeZ56\n+913cP31DUcaaini/7z53dnGK89yhj59+lBaWnrWbV566SWGDx/Oxo0brYggIi101VXmbeRI8/fT\nW4o88UTjliK9e0NMjL2ZxTqWFIlz2bdvH8uWLWPVqlVs3LiRsLMMgBYUFNT/7HQ6cTqd1gcUkXqX\nXQYDB5o3MDvc1rUUWbQI7r8f2rf3nNdISlJLEV9yuVy4XC5LntuyievS0lKys7ObHG4aMWIE06ZN\nIysri7Fjx5Kdna3hJpEA1VRLkR9+aJjT6N3bbCny85/bnTR0+P2cBJy9SHTt2rX+Dbjdbtq2bcu8\nefMYPHiwZzgVCZGAVNdSpG5u41//MtfUOL2lyOWX250yeAV8kThdbm4u2dnZOrtJJIhVVze0FCkq\nMn++6irPISq1FPEev5+4HjlyJGvWrMHtdhMXF8fMmTOpra0FID8/34qXFBE/5nDArbeaN/BsKbJs\nGTz0kNk+pG6I6qabIC1NLUX8gS6mExHbGQbs2eM5r7Fvn3m67ektRdq1sztpYAiI4SZvUJEQCV1n\naynSu7d5i4qyO6V/UpEQkZBT11KkbjJ83Tq44grPeY1rrtG8BqhIiIhw6lRDS5G629GjnkUjVFuK\nqEiIiDShqZYiPXs2FI1evUKjpYiKhIjIeTh8GNavbygamzaZQ1KnH20EY0sRFQkRkRY4vaVI3a2u\npUjd6bfJyYHfUkRFQkTEC+paipw+RHXwoHlFeN2RRiC2FFGREBGxSGWlZ9HYudOcAA+kliIqEiIi\nPtJUS5G4OM95jS5d/OvUWxUJERGbnDgB27Y1FI21a805jNOLht0tRVQkRET8xNlaitRNhmdl+bal\niIqEiIgfO3jQs6XIli2QkuK5mp+VLUVUJEREAsjRo+Y1GnVFw+qWIioSIiIB7MyWIsXFUFPj2So9\nMxP+679a9vwqEiIiQaa83PPU22+/bXlLERUJEZEg9+9/wxdfNG4pcvrRRmxs03+rIiEiEmKOH4eS\nEs+zqNq185zXqGspoiIhIhLiDAO++cazaNS1FPnkE+99dwZ4G6vQ4XK57I7gN7QvGmhfNAi1fREW\nBomJMH48zJ8Pu3ebLURyc737OpYUiXHjxhEVFUX37t2bfPzdd98lPT2dtLQ0evfuzbZt26yIEVRC\n7X+As9G+aKB90UD7AqKjISfHu89pSZHIzc1lxYoVzT7etWtXPv/8c7Zt28af/vQn8vLyrIghIiIX\nyZIi0adPHyIiIpp9vFevXnTo0AGArKws9u7da0UMERG5SJZNXJeWlpKdnc327dvPut3s2bP55ptv\neP311xuH86e2iiIiAcRbX+1tvPIsLbR69WrefPNNiouLm3xcZzaJiNjLtiKxbds2JkyYwIoVK846\nNCUiIvax5RTYsrIyhg0bxsKFC0lISLAjgoiInAdL5iRGjhzJmjVrcLvdREVFMXPmTGprawHIz8/n\nnnvuYcmSJVx11VUAhIeH8+WXX3o7hoiIXCzDTxUWFhqJiYlGQkKC8dRTT9kdxyd+8YtfGN27dzcy\nMjKM66+/3jAMwzh48KBx2223Gddcc43Rv39/49ChQ/Xbz5o1y0hISDASExONlStX2hX7ouXm5hqR\nkZFGampq/X0ted+bNm0yUlNTjYSEBGPy5Mk+fQ/e0tS+ePTRR42YmBgjIyPDyMjIMJYvX17/WDDv\ni7KyMsPpdBrJyclGSkqKMWfOHMMwQvOz0dy+8MVnwy+LxIkTJ4z4+Hhjz549xvHjx4309HRj586d\ndseyXJcuXYyDBw963PfQQw8ZTz/9tGEYhvHUU08Z06dPNwzDMHbs2GGkp6cbx48fN/bs2WPEx8cb\nJ0+e9Hlmb/j888+NzZs3e3wxXsj7PnXqlGEYhnH99dcbGzZsMAzDMAYOHGgUFhb6+J1cvKb2RUFB\ngfHcc8812jbY90VFRYVRUlJiGIZhVFVVGddee62xc+fOkPxsNLcvfPHZ8Mu2HF9++SUJCQl06dKF\n8PBw7rzzTpYtW2Z3LJ8wzhj9++ijj7j77rsBuPvuu1m6dCkAy5YtY+TIkYSHh9OlSxcSEhICdsiu\nqetqLuR9b9iwgYqKCqqqqrjhhhsAGDNmTP3fBJLmrjE683MBwb8voqOjycjIAMDhcJCUlMS+fftC\n8rPR3L4A6z8bflkk9u3bR1xcXP3vsbGx9TskmIWFhXHbbbfRs2dP5s2bB8CBAweI+mmdw6ioKA4c\nOADA/v1kFVARAAACkklEQVT7iT2tT3Cw7aMLfd9n3h8TExNU++Oll14iPT2d8ePHc/jwYSC09kVp\naSklJSVkZWWF/Gejbl/ceOONgPWfDb8sEqF6EV1xcTElJSUUFhby8ssvs3btWo/Hw8LCzrpvgnW/\nnet9B7v77ruPPXv2sGXLFjp37syDDz5odySfqq6uJicnhzlz5tC+fXuPx0Lts1FdXc3w4cOZM2cO\nDofDJ58NvywSMTExlJeX1/9eXl7uUf2CVefOnQG44oorGDp0KF9++SVRUVFUVlYCUFFRQWRkJNB4\nH+3du5eYmBjfh7bIhbzv2NhYYmJiPNq7BNP+iIyMrP8yvOeee+qHFUNhX9TW1pKTk8Po0aMZMmQI\nELqfjbp9MWrUqPp94YvPhl8WiZ49e7J7925KS0s5fvw47733HoMHD7Y7lqVqamqoqqoC4Mcff+TT\nTz+le/fuDB48mLfeeguAt956q/7DMXjwYP7yl79w/Phx9uzZw+7du+vHGYPBhb7v6OhoLr30UjZs\n2IBhGLzzzjv1fxPoKioq6n9esmRJfXflYN8XhmEwfvx4kpOTmTp1av39ofjZaG5f+OSz4Z25d+9b\nvny5ce211xrx8fHGrFmz7I5jue+++85IT0830tPTjZSUlPr3fPDgQePWW29t8nS/J554woiPjzcS\nExONFStW2BX9ot15551G586djfDwcCM2NtZ48803W/S+607ti4+PN/7whz/Y8VYu2pn74o033jBG\njx5tdO/e3UhLSzNuv/12o7Kysn77YN4Xa9euNcLCwoz09PT6UzwLCwtD8rPR1L5Yvny5Tz4bfr0y\nnYiI2Msvh5tERMQ/qEiIiEizVCRERKRZKhIiItIsFQkREWmWioSIiDTr/wHpKRh1i3Z8jAAAAABJ\nRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x32eab90>"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch23.ipynb b/Chemical_Reaction_Engineering/ch23.ipynb
new file mode 100755
index 00000000..5535664d
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch23.ipynb
@@ -0,0 +1,104 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 23 :\n",
+      "\n",
+      "Fluid-Fluid Reactions: Kinetics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 23.1 pageno : 536"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "k = 10.**6;\n",
+      "Kag_a = 0.01              #mol/hr. m**3. Pa\n",
+      "fl = 0.98;\n",
+      "Kal = 1.;\n",
+      "HA = 10.**5;              # very low solubility\n",
+      "DAl = 10.**-6;\n",
+      "DBl = DAl;\n",
+      "PA = 5*10.**3               #Pa\n",
+      "CB = 100.                   #mol/m3\n",
+      "b = 2.      \n",
+      "a = 20.                     #m2/m3\n",
+      "\n",
+      "# Calculations\n",
+      "Mh = math.sqrt(DAl*k*CB*CB)/Kal;\n",
+      "Ei = 1+(DBl*CB*HA/(b*DAl*PA));\n",
+      "E = 100.\n",
+      "print \" Part a\"\n",
+      "\n",
+      "res_total = (((1/(Kag_a))+(HA/(Kal*a*E))+(HA/(k*CB*CB*fl))))            #Total Resismath.tance\n",
+      "f_gas = (1/(Kag_a))/res_total;              #fraction of resismath.tance in gas film\n",
+      "f_liq = (HA/(Kal*a*E))/res_total;           #fraction of resismath.tance in liquid film\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "print \" Fraction of the resistance in the gas film is %f\"%(f_gas)\n",
+      "print \" Fraction of the resistance in the liquid film is %f\"%(f_liq)\n",
+      "print \" Part b\"\n",
+      "print \" The reaction zone is in the liquid film\"\n",
+      "print \" Part c\"\n",
+      "if Ei>5*Mh:\n",
+      "     print \" We have pseudo 1st order reaction in the film\"\n",
+      "\n",
+      "rA = PA/(((1/(Kag_a))+(HA/(Kal*a*E))+(HA/(k*CB*CB*fl))));\n",
+      "print \" Part d\"\n",
+      "print \" The rate of reactionmol/m3.hr) is %f\"%(rA)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " Fraction of the resistance in the gas film is 0.666667\n",
+        " Fraction of the resistance in the liquid film is 0.333333\n",
+        " Part b\n",
+        " The reaction zone is in the liquid film\n",
+        " Part c\n",
+        " We have pseudo 1st order reaction in the film\n",
+        " Part d\n",
+        " The rate of reactionmol/m3.hr) is 33.333331\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch24.ipynb b/Chemical_Reaction_Engineering/ch24.ipynb
new file mode 100755
index 00000000..1766a982
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch24.ipynb
@@ -0,0 +1,369 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 24 : Fluid-Fluid Reactors: Design"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.1 pageno : 551"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "kag_a = 0.32;                  # mol/hr.m**3 Pa\n",
+      "kal_a = 0.1                    # hr\n",
+      "HA = 12.5                      # Pa.m**3/mol\n",
+      "Fg = 10.**5                    # mol/hr.m**2\n",
+      "Fl = 7.*10**5                  # mol/hr.m**2\n",
+      "Ct = 56000.;                    #mol/m3\n",
+      "P = 10.**5;                     #Pa\n",
+      "\n",
+      "# Calculations\n",
+      "inv_Kag_a = 3.125+HA/(kal_a);\n",
+      "Gfilm_res = (3.125)/inv_Kag_a;\n",
+      "Lfilm_res = (HA/(kal_a))/inv_Kag_a;\n",
+      "Kag_a = 1/inv_Kag_a;\n",
+      "d = 20;\n",
+      "def f9(dp): \n",
+      "\t return 1./20\n",
+      "\n",
+      "h = (Fg/(P*Kag_a))* quad(f9,20,100)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \" The height of the tower required for countercurrent operartions is %.1f \"%(h),\n",
+      "print \"m\"\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The height of the tower required for countercurrent operartions is 512.5  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.2 pageno : 554"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "Fg = 10.**5;             \n",
+      "P = 10.**5;\n",
+      "Fg_by_Acs = 10.**5              #(Fg/Acs)\n",
+      "PA1 = 20;PA2 = 100.;\n",
+      "kag_a = 0.32;\n",
+      "\n",
+      "# Calculations\n",
+      "def f10(PA): \n",
+      "\t return 1./(0.32*PA)\n",
+      "\n",
+      "h = (Fg_by_Acs/P)* quad(f10,PA1,PA2)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \" The height of the tower is %.2f \"%(h),\n",
+      "print \"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The height of the tower is 5.03  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.3 pageno : 555"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "\n",
+      "# from example 24.2\n",
+      "Fg = 10.**5;\n",
+      "P = 10.**5;\n",
+      "PA1 = 20.\n",
+      "PA2 = 100.;\n",
+      "HA = 12.5;\n",
+      "kaga = 0.32\n",
+      "kla = 0.1;\n",
+      "\n",
+      "# Calculations\n",
+      "rA = 420./((1./kaga)+(HA/kla));\n",
+      "\n",
+      "def f8(PA): \n",
+      "\t return 1./rA\n",
+      "\n",
+      "h = (Fg/P)* quad(f8,PA1,PA2)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \"The height of the tower is %.1f\"%(h),\n",
+      "print \"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The height of the tower is 24.4 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.4 page no : 557"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "\n",
+      "# from example 24.2\n",
+      "PA1 = 20.\n",
+      "PA2 = 100.          #Pa\n",
+      "Fg_by_Acs = 10.**5;\n",
+      "P = 10.**5;\n",
+      "HA = 12.5;\n",
+      "kaga = 0.32\n",
+      "kla = 0.1;\n",
+      "PA = 39.5           #Pa\n",
+      "\n",
+      "# Calculations\n",
+      "def f11(P): \n",
+      "\t return 1./(kaga*P)\n",
+      "\n",
+      "def f22(P):\n",
+      "    return (1/kaga+HA/kla)/1620\n",
+      "\n",
+      "h = (Fg_by_Acs/P)*( quad(f11,PA1,PA)[0] + quad(f22,PA,PA2)[0])\n",
+      "\n",
+      "# Results\n",
+      "print \"The height of the tower is %.2f\"%(h),\n",
+      "print \"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The height of the tower is 6.91 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.5 pageno : 558"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "\n",
+      "# from 24.2\n",
+      "Fg = 10.**5;\n",
+      "P = 10.**5;\n",
+      "Fg_by_Acs = 10.**5          #(Fg/Acs)\n",
+      "PA1 = 20.\n",
+      "PA2 = 100.\n",
+      "kag_a = 0.32;\n",
+      "\n",
+      "# Calculations\n",
+      "def f0(PA): \n",
+      "\t return 1./(PA/3.125)\n",
+      "\n",
+      "h = (Fg_by_Acs/P)* quad(f0,PA1,PA2)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \" The height of the tower is %.2f \"%(h),\n",
+      "print \"m\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The height of the tower is 5.03  m\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 24.6 pageno : 560"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "kag_a = 0.72;              # mol/hr.m**3 Pa\n",
+      "kal_a = 144.;              # hr**-1\n",
+      "HA = 1000.;                # Pa m**3/mol\n",
+      "Fg = 9000.                  #mol/hr\n",
+      "fl = 0.9\n",
+      "b = 1\n",
+      "Vr = 1.62                   #m3\n",
+      "DA = 3.6*10**-6             #m2/hr\n",
+      "a = 100.                    #m2/m3\n",
+      "k = 2.6*10**5;              #m3/mol.hr\n",
+      "DB = DA\n",
+      "P = 10**5\n",
+      "PA = 1000.                  #Pa\n",
+      "kal = kal_a/a;\n",
+      "#At the start\n",
+      "CBo = 555.6;\n",
+      "\n",
+      "# Calculations\n",
+      "Mh = (math.sqrt(DB*k*CBo))/kal;\n",
+      "#Min value of EAi\n",
+      "Ei = 1+(CBo*HA/PA);\n",
+      "if Ei>Mh:\n",
+      "    E = Mh;\n",
+      "\n",
+      "rA1 = PA/((P*Vr/Fg)+(1/kag_a)+(HA/(kal_a*E))+(HA/(k*fl*CBo)));\n",
+      "#At the end\n",
+      "CBf = 55.6;\n",
+      "Mh = (math.sqrt(DB*k*CBf))/kal;\n",
+      "#Min value of EAi\n",
+      "Ei = 1+(CBf*HA/PA);\n",
+      "if Ei>Mh:\n",
+      "    E = Mh;\n",
+      "\n",
+      "rA2 = PA/((P*Vr/Fg)+(1/kag_a)+(HA/(kal_a*E))+(HA/(k*fl*CBf)));\n",
+      "#Average rate of reaction\n",
+      "rA_avg = (rA1+rA2)/2;\n",
+      "\n",
+      "def f7(CB): \n",
+      "\t return 1./rA_avg\n",
+      "\n",
+      "t = (fl/b)* quad(f7,CBf,CBo)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \" Part a\"\n",
+      "print \" The run time needed is %.2f\"%t,\n",
+      "print \"hr\"\n",
+      "#The min time required is\n",
+      "tmin = Vr*(CBo-CBf)/(Fg*(PA/(P-PA)));\n",
+      "print \" The minimum time required is %.2f\"%(tmin),\n",
+      "print \"hr\"\n",
+      "#Fraction of reacmath.tant which passes through the math.tank unreacted is\n",
+      "f = (t-tmin)/tmin;\n",
+      "print \" Part b\"\n",
+      "print \" Fraction of reacmath.tant which passes through the math.tank unreacted is %.3f\"%(f)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " The run time needed is 9.13 hr\n",
+        " The minimum time required is 8.91 hr\n",
+        " Part b\n",
+        " Fraction of reacmath.tant which passes through the math.tank unreacted is 0.025\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch26.ipynb b/Chemical_Reaction_Engineering/ch26.ipynb
new file mode 100755
index 00000000..cb1d9f92
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch26.ipynb
@@ -0,0 +1,214 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 26  :\n",
+      "\n",
+      "Fluid-Particle Reactors: Design"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 26.1 pageno : 592"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Lets say F(Ri)/F = F_ri\n",
+      "\n",
+      "# Variables\n",
+      "F_50 = 0.3\n",
+      "F_100 = 0.4\n",
+      "F_200 = 0.3;\n",
+      "t_50 = 5.\n",
+      "t_100 = 10.\n",
+      "t_200 = 20.\n",
+      "tp = 8.\n",
+      "\n",
+      "# Calculations\n",
+      "a = ((1-(tp/t_50))**3)*F_50\n",
+      "b = ((1-(tp/t_100))**3)*F_100\n",
+      "c = ((1-(tp/t_200))**3)*F_200;\n",
+      "g = [a,b,c];\n",
+      "sum1 = 0;\n",
+      "for p in range(3):\n",
+      "   if g[p]>0:\n",
+      "       sum1 = sum1+g[p];\n",
+      "f_converted = 1-sum1;\n",
+      "\n",
+      "# Results\n",
+      "print \" The fraction of solid converted equals %.1f %%\"%(f_converted*100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The fraction of solid converted equals 93.2 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 26.2 pageno : 597"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t_avg = 60.         # min\n",
+      "t = 20.             #min\n",
+      "\n",
+      "# Calculations\n",
+      "unconverted = ((1./4)*(t/t_avg))-((1./20)*(t/t_avg)**2)+((1./120)*(t/t_avg)**3);\n",
+      "unconverted1 = ((1./5)*(t/t_avg))-((19./420)*(t/t_avg)**2)+((41./4620)*(t/t_avg)**3);\n",
+      "c_avg = (unconverted+unconverted1)/2;\n",
+      "\n",
+      "# Results\n",
+      "print \"Fraction of original sulfide ore remain unconverted is %.2f\"%(c_avg)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Fraction of original sulfide ore remain unconverted is 0.07\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 26.3 page no : 600"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "F = 1000.               #gm/min\n",
+      "W = 10000.              #gm\n",
+      "\n",
+      "# Calculations\n",
+      "t_avg = W/F;\n",
+      "F_50 = 300.\n",
+      "F_100 = 400.\n",
+      "F_200 = 300.             #gm/min\n",
+      "t_50 = 5.\n",
+      "t_100 = 10.\n",
+      "t_200 = 20.             #min\n",
+      "\n",
+      "unconverted = ((((1./4)*(t_50/t_avg))-((1./20)*(t_50/t_avg)**2)+ ((1./120)*(t_50/t_avg)**3))*(F_50/F))+((((1./4)*(t_100/t_avg))-((1./20)*(t_100/t_avg)**2)+((1./120)*(t_50/t_avg)**3))*(F_100/F))+((((1./4)*(t_200/t_avg))-((1./20)*(t_200/t_avg)**2)+((1./120)   *(t_50/t_avg)**3))*(F_200/F))\n",
+      "converted = 1-unconverted;\n",
+      "\n",
+      "# Results\n",
+      "print \"The mean conversion of soild is %f\"%(converted)\n",
+      "print \" The answer slightly differs from those given in book as we have considered \\\n",
+      "only significant terms in infinite series\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean conversion of soild is 0.795208\n",
+        " The answer slightly differs from those given in book as we have considered only significant terms in infinite series\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 26.4 pageno : 601"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 1.             #hr\n",
+      "t2 = t1/0.1         \n",
+      "a = 0.\n",
+      "r = 1               #ton/hr\n",
+      "# Calculations\n",
+      "while a<=1:\n",
+      "    x = (1./4)*(a)-((1./20)*(a)**2)+((1./120)*(a)**3);\n",
+      "    if x >0.099 and x<0.1005:\n",
+      "        r = a;\n",
+      "    a += .0001\n",
+      "\n",
+      "FBo = 1.            #tons/hr\n",
+      "t_avg = t2/r;\n",
+      "W = t_avg*FBo;\n",
+      "\n",
+      "# Results\n",
+      "print \" The needed weight of bed is %.f\"%(W),\n",
+      "print \"tons\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The needed weight of bed is 23 tons\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch29.ipynb b/Chemical_Reaction_Engineering/ch29.ipynb
new file mode 100755
index 00000000..be59a530
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch29.ipynb
@@ -0,0 +1,93 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 29 :\n",
+      "\n",
+      "Substrate-Limiting\n",
+      "\n",
+      "Microbial Fermentation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 29.3 pageno : 639"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "CAo = 6.            # glucose solution\n",
+      "CM = 0.4            #kg/m3\n",
+      "V = 1.              #m3\n",
+      "k = 4.\n",
+      "\n",
+      "# Calculations and Results\n",
+      "N = math.sqrt(1+(CAo/CM));\n",
+      "kt_op = N/(N-1);\n",
+      "C_by_A = 0.1;\n",
+      "t_op = kt_op/k;\n",
+      "v_op = V/t_op;\n",
+      "\n",
+      "#The feed rate of glumath.cose\n",
+      "FAo = v_op*CAo;\n",
+      "print \" The feed rate of glumath.cose is %.1f\"%(FAo) ,\n",
+      "print \"kg/hr\"\n",
+      "\n",
+      "#Max consumption rate of glumath.cose is\n",
+      "XA = N/(N+1);\n",
+      "c_max = FAo*XA;\n",
+      "print \" Max consumption rate of glumath.cose is %.1f \"%(c_max),\n",
+      "print \"kg/hr\"\n",
+      "\n",
+      "#Max production rate of E.coli is\n",
+      "Cc_op = (C_by_A)*CAo*N/(N+1);\n",
+      "Fcmax = v_op*Cc_op;\n",
+      "print \" Max production rate of E.coli is %.2f\"%(Fcmax),\n",
+      "print \"kg/hr\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The feed rate of glumath.cose is 18.0 kg/hr\n",
+        " Max consumption rate of glumath.cose is 14.4  kg/hr\n",
+        " Max production rate of E.coli is 1.44 kg/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch3.ipynb b/Chemical_Reaction_Engineering/ch3.ipynb
new file mode 100755
index 00000000..3af6a19a
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch3.ipynb
@@ -0,0 +1,256 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 :  Interpretation of Batch Reactor Data"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 page no : 60"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "from scipy import stats\n",
+      "\n",
+      "# Variables\n",
+      "#Given\n",
+      "t = [0, 20, 40 ,60, 120 ,180, 300];        # time\n",
+      "C_A = [10 ,8, 6, 5, 3, 2, 1];              # concentration\n",
+      "CAo = 10.;\n",
+      "k = zeros(7)\n",
+      "CA_inv = zeros(7)\n",
+      "\n",
+      "# Calculations\n",
+      "#Guesmath.sing 1st order kinetics\n",
+      "for i in range(7):\n",
+      "    k[i] = math.log(CAo/C_A[i]);\n",
+      "    CA_inv[i] = 1/C_A[i];\n",
+      "\n",
+      "T = array([18.5,23,35]);\n",
+      "CAo = array([10,5,2]);\n",
+      "CA = zeros(3)\n",
+      "log_Tf = zeros(3)\n",
+      "log_CAo = zeros(3)\n",
+      "\n",
+      "for i in range(3):\n",
+      "    CA[i] = 0.8*CAo[i];\n",
+      "    log_Tf[i] = math.log10(T[i]);\n",
+      "    log_CAo[i] = math.log10(CAo[i]);\n",
+      "\n",
+      "# Results\n",
+      "plot(log_CAo,log_Tf)\n",
+      "plot(log_CAo,log_Tf,\"go\")\n",
+      "xlabel(\"Ln CAO\")\n",
+      "ylabel(\"log r\")\n",
+      "#plot(log_Tf,log_CAo)\n",
+      "#coeff1 = linalg.lstsq(log_CAo,log_Tf);\n",
+      "slope, intercept, r_value, p_value, std_err = stats.linregress(log_CAo,log_Tf)\n",
+      "coeff1 = stats.linregress(log_CAo,log_Tf)\n",
+      "n = 1-coeff1[0];\n",
+      "print \"From graph we get slope and intercept for calculating rate eqn\"\n",
+      "k1 = ((0.8**(1-n))-1)*(10.**(1-n))/(18.5*(n-1));\n",
+      "print \" The rate equation is given by %.3f\"%(k1),\n",
+      "print \"CA**1.4 mol/litre.sec\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "From graph we get slope and intercept for calculating rate eqn"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " The rate equation is given by 0.005 CA**1.4 mol/litre.sec\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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YMGFCmSuhRETsFhoK//iHWQX+5JPQoQPccgv8+c9QFydDtChPRKQGfPeduWrq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\nCpcenvGDsSAAAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x33a3310>"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 page no : 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "import math\n",
+      "# Variables\n",
+      "CA = array([10,8,6,5,3,2,1]);              # concentration\n",
+      "T = array([0,20,40,60,120,180,300]);       # time\n",
+      "y = array([-0.1333,-0.1031,-0.0658,-0.0410,-0.0238,-0.0108,-0.0065]);     # slope\n",
+      "\n",
+      "log_y = zeros(7)\n",
+      "log_CA = zeros(7)\n",
+      "\n",
+      "# Calculations\n",
+      "for i in range(7):\n",
+      "    log_y[i] = log10(complex(y[i]));\n",
+      "    log_CA[i] = log10(CA[i]);\n",
+      "\n",
+      "\n",
+      "# Results\n",
+      "plot(log_CA,log_y)\n",
+      "plot(log_CA,log_y,\"go\")\n",
+      "xlabel(\"log10 CA\")\n",
+      "ylabel(\"log10 (-dCA/dt)\")\n",
+      "show()\n",
+      "coeff1 = stats.linregress(log_CA,log_y);\n",
+      "n = coeff1[0];\n",
+      "k = -10**(coeff1[1]);\n",
+      "print \" After doing linear regression, the slope and intercept of the graph is %.0f, %.0f\"%(-coeff1[1],coeff1[0])\n",
+      "print \" The rate equation is therefore given by %.3f\"%(-k),\n",
+      "print \"CA**1.375 mol/litre.sec\"\n",
+      "print ('The answer slightly differs from those given in book as regress fn is used for \\\n",
+      " calculating slope and intercept')\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: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n",
+        "-c:14: ComplexWarning: Casting complex values to real discards the imaginary part\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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jR1tRrriYzz6DO+6A996Du+6yuhqR+kWr3Ipb+fhjSEw0Fx688UarqxFxX1rl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\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x365aa90>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " After doing linear regression, the slope and intercept of the graph is 2, 1\n",
+        " The rate equation is therefore given by 0.005 CA**1.375 mol/litre.sec\n",
+        "The answer slightly differs from those given in book as regress fn is used for  calculating slope and intercept\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 page no : 73"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "k1 = 2.3           # temperatures\n",
+      "k2 = 2.3\n",
+      "T1 = 400.          # K\n",
+      "T2 = 500.          # K\n",
+      "\n",
+      "# Calculations\n",
+      "R = 82.06*10**-6;\n",
+      "R1 = 8.314\n",
+      "E = (math.log(k2/k1)*R)/(1./T1-1./T2)\n",
+      "\n",
+      "# Results\n",
+      "print \"using pressure units is %.0f EJ/mol\"%(E)\n",
+      "\n",
+      "#pA = CA*RT\n",
+      "#-rA = 2.3(RT)**2*CA**2\n",
+      "k1 = 2.3*(R*T1)**2\n",
+      "k2 = 2.3*(R*T2)**2\n",
+      "E = (math.log(k2/k1)*R1)/(1./T1-1./T2)\n",
+      "print \"using concentration units is %.f J/mol\"%(E)\n",
+      "\n",
+      "# Answers might be different because of Rounding error"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "using pressure units is 0 EJ/mol\n",
+        "using concentration units is 7421 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch30.ipynb b/Chemical_Reaction_Engineering/ch30.ipynb
new file mode 100755
index 00000000..2dd771a6
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch30.ipynb
@@ -0,0 +1,90 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 30 :\n",
+      "\n",
+      "Product-Limiting Microbial \n",
+      "\n",
+      "Fermentation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 30.1 pageno : 651"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "k = math.sqrt(3)            #hr**-1\n",
+      "n = 1.\n",
+      "V = 30.                     #m3\n",
+      "CR = 0.12                   #kgalc/kgsol\n",
+      "density = 1000.             #kg/m3\n",
+      "\n",
+      "# Calculations and Results\n",
+      "CR = CR*density;\n",
+      "CR_opt = CR/2;\n",
+      "alcohol_per = CR_opt*100/density        #PErcentage of alcohol\n",
+      "\n",
+      "print \" The Percentage of alchol in cocktail is %f\"%(alcohol_per)\n",
+      "\n",
+      "kt = 1.\n",
+      "t = kt/k;\n",
+      "t_opt = 2*t;\n",
+      "v_opt = V/t_opt;\n",
+      "\n",
+      "print \" The Optimum feed rate is %f\"%(v_opt),\n",
+      "print \" m3/hr\"\n",
+      "\n",
+      "#The production rate of alcohol \n",
+      "FR = v_opt*CR_opt;\n",
+      "print \" The production rate of alcohol is %f \"%(FR),\n",
+      "print \" kgalc/hr\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The Percentage of alchol in cocktail is 6.000000\n",
+        " The Optimum feed rate is 25.980762  m3/hr\n",
+        " The production rate of alcohol is 1558.845727   kgalc/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch4.ipynb b/Chemical_Reaction_Engineering/ch4.ipynb
new file mode 100755
index 00000000..cd660b67
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch4.ipynb
@@ -0,0 +1,81 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Introduction to Reactor Design"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 page no : 88"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "a = 1.\n",
+      "b = 3.\n",
+      "c = 6.\n",
+      "#Initial concentrations\n",
+      "CAo = 100.      # feed\n",
+      "CBo = 200.      # feed\n",
+      "Cio = 100.      # feed\n",
+      "#Final concentrations\n",
+      "CA = 40.        # reacter exit\n",
+      "\n",
+      "# Calculations\n",
+      "# Find CB,XA,XB\n",
+      "ea = (6.-4.)/4;\n",
+      "XA = (CAo-CA)/(CAo+ea*CA);\n",
+      "eb = (ea*CBo)/(b*CAo);\n",
+      "XB = b*CAo*XA/CBo;\n",
+      "CB = CBo*(1-XB)/(1+eb*XB);\n",
+      "\n",
+      "# Results\n",
+      "print \"The final concentration of BCB is %.f\"%(CB)\n",
+      "print \" XA and XB are %.2f ,%.2f\"%(XA,XB)\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The final concentration of BCB is 40\n",
+        " XA and XB are 0.50 ,0.75\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch5.ipynb b/Chemical_Reaction_Engineering/ch5.ipynb
new file mode 100755
index 00000000..24a0856c
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch5.ipynb
@@ -0,0 +1,324 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Ideal Reactors for a Single Reaction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 page no : 96"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "#Concentrations in mol/litre\n",
+      "CAo = 0.1                 # liquid\n",
+      "CBo = 0.01                # liquid\n",
+      "Cco = 0.                  # liquid\n",
+      "CAf = 0.02                # outlet stream\n",
+      "CBf = 0.03                # outlet stream\n",
+      "Ccf = 0.04;               # outlet stream\n",
+      "#Volume in litre\n",
+      "V = 1.;\n",
+      "#Volumetric flow rate(l/min)\n",
+      "v = 1.;\n",
+      "CA = CAf;CB = CBf;Cc = Ccf;\n",
+      "\n",
+      "# Calculations\n",
+      "#Rate of reaction(mol/litre.min)\n",
+      "rA = (CAo-CA)/(V/v);\n",
+      "rB = (CBo-CB)/(V/v);\n",
+      "rc = (Cco-Cc)/(V/v);\n",
+      "\n",
+      "# Results\n",
+      "print \"rate of reaction of A is %.2f mol/litre.min\"%(rA)\n",
+      "print \"rate of reaction of B is %.2f mol/litre.min\"%(rB)\n",
+      "print \"rate of reaction of C is %.2f mol/litre.min\"%(rc)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate of reaction of A is 0.08 mol/litre.min\n",
+        "rate of reaction of B is -0.02 mol/litre.min\n",
+        "rate of reaction of C is -0.04 mol/litre.min\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2 page no : 97"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "from scipy import stats\n",
+      "\n",
+      "# Variables\n",
+      "vo = array([10,3,1.2,0.5])         #Volumetric flow rates(litre/hr)\n",
+      "CA = array([85.7,66.7,50,33.4])    #Concentrations (millimol/litre)\n",
+      "CAo = 100.;\n",
+      "V = 0.1;                           #Volume(litre)\n",
+      "e = (1.-2.)/2;                     #Expansion factor is\n",
+      "#Initialization\n",
+      "XA = zeros(4);\n",
+      "rA = zeros(4);\n",
+      "m = zeros(4);\n",
+      "n = zeros(4);\n",
+      "\n",
+      "# Calculations\n",
+      "#Relation between concentration and conversion\n",
+      "for i in range(4):\n",
+      "    XA[i] = (1-CA[i]/CAo)/(1+e*CA[i]/CAo);\n",
+      "    rA[i] = vo[i]*CAo*XA[i]/V;\n",
+      "    m[i] = math.log10(CA[i]);\n",
+      "    n[i] = math.log10(rA[i]);\n",
+      "\n",
+      "# Results\n",
+      "#For nth order plot between n & m should give a straight line\n",
+      "plot(m,n)\n",
+      "xlabel(\"log CA\")\n",
+      "ylabel(\"log (-rA)\")\n",
+      "show()\n",
+      "coefs = stats.linregress(m,n);\n",
+      "print coefs\n",
+      "print \"Intercept of the graph is %.2f\"%(coefs[1])\n",
+      "print \"Slope of the graph is %.2f\"%(coefs[0])\n",
+      "k = 10**coefs[1]\n",
+      "n = coefs[0]\n",
+      "print \" Taking n = 2, rate of equation is %.2f millimol/litre.hr\"%(k),\n",
+      "print \"CA**2 \"\n",
+      "print ('The sol slightly differ from that given in book because regress fn is used to calculate the slope')\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: ['rc', 'draw_if_interactive', 'e']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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IlI+6m6q4RYvMRXH5+eaiuL59VSBEpPzUkqiicnJgxAizi+m998zZSyIiF0st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\nAAAASUVORK5CYII=\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x29b4650>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(1.9569709616766851, -0.39508420521305565, 0.99852626739620765, 0.001473732603792355, 0.075209753597987913)\n",
+        "Intercept of the graph is -0.40\n",
+        "Slope of the graph is 1.96\n",
+        " Taking n = 2, rate of equation is 0.40 millimol/litre.hr CA**2 \n",
+        "The sol slightly differ from that given in book because regress fn is used to calculate the slope\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3 page no : 99"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "'''\n",
+      "Note : The sol varies from book as the value of CB taken in book at end is wrong\n",
+      "'''\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "CAo = 1.4\n",
+      "CBo = 0.8\n",
+      "CRo = 0.\n",
+      "#Volume(litre)\n",
+      "V = 6.\n",
+      "\n",
+      "# Calculations\n",
+      "#For 75% conversion of B\n",
+      "#From stoichiometry of equation A+2B-->R\n",
+      "CA = 1.4-(0.75*0.8)/2.;\n",
+      "CB = 0.8-(0.75*0.8);\n",
+      "CR = (0.75*0.8)/2.;\n",
+      "#From the Given rate equation(mol/litre.min)\n",
+      "rB = 2*(12.5*CA*CB*CB-1.5*CR);\n",
+      "#Volumetric flow rate is given by\n",
+      "v = V*rB/(CBo-CB);\n",
+      "\n",
+      "# Results\n",
+      "print \" volumetric flow rate into and out of the reactor is %.1f litre/min\"%(v)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " volumetric flow rate into and out of the reactor is 2.0 litre/min\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 page no : 104"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "eA = (4-2.)/2;\n",
+      "CAo = 0.0625;         # mol/liter\n",
+      "xAo = 0.\n",
+      "xAf = 0.8              # conversion\n",
+      "k = 0.01;\n",
+      "\n",
+      "# Calculations\n",
+      "def f1(xA): \n",
+      "\t return math.sqrt((1+xA)/(1-xA))\n",
+      "\n",
+      "X =  quad(f1,xAo,xAf)[0]\n",
+      "t = math.sqrt(CAo)*X/k;\n",
+      "\n",
+      "# Results\n",
+      "print \" Space timesec needed is %.2f sec\"%(t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Space timesec needed is 33.18 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 page no : 106"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "T = 922.         #Temperature(kelvin)\n",
+      "P = 460000.;     # kPA\n",
+      "FAo = 40.        # pure phosphine\n",
+      "k = 10.\n",
+      "R = 8.314;\n",
+      "\n",
+      "# Calculations\n",
+      "CAo = P/(R*T);          # mol/m3\n",
+      "e = (7-4)/4.;\n",
+      "XA = 0.8;\n",
+      "\n",
+      "#The volume of plug flow reactor is given by\n",
+      "V = FAo*((1+e)*math.log(1./(1-XA))-e*XA)/(k*CAo);\n",
+      "\n",
+      "# Results\n",
+      "print \" volume of reactor is %.3f m**3\"%(V)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " volume of reactor is 0.148 m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch6.ipynb b/Chemical_Reaction_Engineering/ch6.ipynb
new file mode 100755
index 00000000..27577e4f
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch6.ipynb
@@ -0,0 +1,233 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Design for Single Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 page no : 125"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "V1 = 50.       # volume liters\n",
+      "V2 = 30.       # volume liters\n",
+      "V3 = 40.;      # volume liters\n",
+      "\n",
+      "# Calculations\n",
+      "VD = V1+V2;\n",
+      "VE = V3;\n",
+      "m = VE/VD\n",
+      "fr_D = 1./(1+m);\n",
+      "\n",
+      "# Results\n",
+      "print \" Fraction of feed going to branch D is %.3f \"%(fr_D)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Fraction of feed going to branch D is 0.667 \n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 page no : 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "kCot = 90.        # reactant\n",
+      "kCot = 180.       \n",
+      "X = 97.4;         # conversion\n",
+      "\n",
+      "print \" Part a\"\n",
+      "print \" The conversion in percentage is %.2f \"%(X)\n",
+      "#For 90% Conversion & N = 2.from graph\n",
+      "kCot = 27.5;\n",
+      "\n",
+      "# Calculations\n",
+      "ratio = 90*2/27.5;\n",
+      "\n",
+      "# Results\n",
+      "print \" Part b\"\n",
+      "print \" Treatment rate can be increased by  %.1f \"%(ratio)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " The conversion in percentage is 97.40 \n",
+        " Part b\n",
+        " Treatment rate can be increased by  6.5 \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 pageno : 144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from numpy import *\n",
+      "from matplotlib.pyplot import *\n",
+      "\n",
+      "# Variables\n",
+      "CAo = array([2,5,6,6,11,14,16,24]);         #mmol/m3\n",
+      "CA = array([0.5,3,1,2,6,10,8,4])            #mmol/m3\n",
+      "t = array([30,1,50,8,4,20,20,4])            #min\n",
+      "vo = 0.1                                    #m3/min\n",
+      "\n",
+      "# Calculations\n",
+      "inv_rA = zeros(8)\n",
+      "for i in range(8):\n",
+      "    inv_rA[i] = t[i]/(CAo[i]-CA[i]);\n",
+      "\n",
+      "for i in range(8):\n",
+      "    for j in range(i,8):\n",
+      "        if CA[i]>CA[j]:\n",
+      "            temp = CA[i];\n",
+      "            CA[i] = CA[j];\n",
+      "            CA[j] = temp;\n",
+      "            temp1 = inv_rA[i];\n",
+      "            inv_rA[i] = inv_rA[j];\n",
+      "            inv_rA[j] = temp1;\n",
+      "\n",
+      "# Results\n",
+      "plot(CA,inv_rA)\n",
+      "plot(CA,inv_rA,\"go\")\n",
+      "suptitle(\"Arrangement with smallest volume\")\n",
+      "xlabel(\"CA, m mol/m**3\")\n",
+      "ylabel(\"-1/ra, m**3.min/m mol\")\n",
+      "print ('From the graph,we can see that we should use plug flow with recycle')\n",
+      "CAin = 6.6;#mmol/m3\n",
+      "R = (10-6.6)/(6.6-1);\n",
+      "#V = t*vo = area*vo\n",
+      "V = (10-1)*1.2*vo;\n",
+      "vr = vo*R;\n",
+      "print \" Part a\"\n",
+      "print \"  The vol of reactor is %.2f m**3\"%(V)\n",
+      "print \"  The recycle flow rate is %.4f \"%(vr),\n",
+      "print \"m3/min\"\n",
+      "\n",
+      "#Part b,from fig\n",
+      "t = (10-1)*10;\n",
+      "t1 = (10-2.6)*0.8;\n",
+      "t2 = (2.6-1)*10;\n",
+      "#For 1  math.tank\n",
+      "V = t*vo;\n",
+      "#For  2 math.tank\n",
+      "V1 = t1*vo;\n",
+      "V2 = t2*vo;\n",
+      "Vt = V1+V2;\n",
+      "print \" Part b\"\n",
+      "print \"  For 1 tank volume is %.2f m**3\"%(V)\n",
+      "print \"  For 2 tank the volume is %.2f m**3\"%(Vt)\n",
+      "\n",
+      "print \" Part c\"\n",
+      "print (' We should use mixed flow followed by plug flow')\n",
+      "#For MFR\n",
+      "tm = (10-4)*0.2;\n",
+      "Vm = tm*vo;\n",
+      "#For PFR\n",
+      "tp = 5.8;#by graphical integration\n",
+      "Vp = tp*vo;\n",
+      "Vtotal = Vp+Vm;\n",
+      "print \" For MFR volume is %.2f m**3\"%(Vm)\n",
+      "print \" For PFR volume is %.2f m**3\"%(Vp)\n",
+      "print \" Total volume is %.2f m**3\"%(Vtotal),\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        "From the graph,we can see that we should use plug flow with recycle\n",
+        " Part a\n",
+        "  The vol of reactor is 1.08 m**3\n",
+        "  The recycle flow rate is 0.0607  m3/min\n",
+        " Part b\n",
+        "  For 1 tank volume is 9.00 m**3\n",
+        "  For 2 tank the volume is 2.19 m**3\n",
+        " Part c\n",
+        " We should use mixed flow followed by plug flow\n",
+        " For MFR volume is 0.12 m**3\n",
+        " For PFR volume is 0.58 m**3\n",
+        " Total volume is 0.70 m**3\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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h4uICPz8/7baYmBi4urpCLpdDLpcjJSXFGKc2iC1/IiIjJf9p06bVSO4ymQx//etfkZOT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\n/4qV3Dpe155RAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x1e37810>"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch7.ipynb b/Chemical_Reaction_Engineering/ch7.ipynb
new file mode 100755
index 00000000..6785eea5
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch7.ipynb
@@ -0,0 +1,233 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : Design for Parallel Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 page no : 159"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "#Initial Concentration(mol/litre)eactant in combined feed\n",
+      "CAo = 10.\n",
+      "CBo = 10. \n",
+      "XA = 0.9;             # conversion\n",
+      "CAf = CAo*(1-XA);\n",
+      "CA = CAf;\n",
+      "\n",
+      "# Calculations\n",
+      "def f4(CA): \n",
+      "\t return 1./(1+CA**0.5)\n",
+      "\n",
+      "Qp = (-1./(CAo-CAf))* quad(f4,CAo,CAf)[0]\n",
+      "\n",
+      "CRf = 9*Qp;\n",
+      "CSf = 9*(1-Qp)\n",
+      "# Results\n",
+      "print \" Part a\"\n",
+      "print \" For Plug Flow\"\n",
+      "print \" Concentration of R in the product stream is %.2f mol/litre\"%(CRf)\n",
+      "print \" Csf is %.2f mol/litre\"%(CSf)\n",
+      "\n",
+      "Qm = CA/(CA+CA**1.5);\n",
+      "CRf = 9*Qm;\n",
+      "Csf = 9*(1-Qm)\n",
+      "print \" Part b\"\n",
+      "print \" For Mixed Flow\"\n",
+      "print \" Concentration of R in the product stream is %.2f mol/litre \"%(CRf)\n",
+      "print \" Csf is %.2f mol/litre\"%(Csf)\n",
+      "\n",
+      "CAo = 19.\n",
+      "CB = 1;\n",
+      "\n",
+      "def f5(CA): \n",
+      "\t return CA/(CA+CB**1.5)\n",
+      "\n",
+      "Q = -1./(CAo-CAf)* quad(f5,CAo,CAf)[0]\n",
+      "CRf = 9*Q;\n",
+      "Csf = 9*(1-Q)\n",
+      "print \" Part c\"\n",
+      "print \" For Plug flow A Mixed flow B\"\n",
+      "print \" Concentration of R in the product stream is %.2f  mol/litre\"%(CRf)\n",
+      "print \" Csf is %.2f mol/litre\"%(Csf)\n",
+      "print ('The result for plug flow varies as there seems to be typographical error in integration done in book')\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " For Plug Flow\n",
+        " Concentration of R in the product stream is 2.86 mol/litre\n",
+        " Csf is 6.14 mol/litre\n",
+        " Part b\n",
+        " For Mixed Flow\n",
+        " Concentration of R in the product stream is 4.50 mol/litre \n",
+        " Csf is 4.50 mol/litre\n",
+        " Part c\n",
+        " For Plug flow A Mixed flow B\n",
+        " Concentration of R in the product stream is 7.85  mol/litre\n",
+        " Csf is 1.15 mol/litre\n",
+        "The result for plug flow varies as there seems to be typographical error in integration done in book\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 page no : 162"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "CAo = 2;          # decomposition of A\n",
+      "CA = 0.5;\n",
+      "CAf = 0.;\n",
+      "\n",
+      "Csf = (CAo-CA)*2*CA/(1+CA)**2;\n",
+      "\n",
+      "print \" Part a\"\n",
+      "print \" For Mixed Flow Reactor\"\n",
+      "print \" Maximum expected Cs is %.3f\"%(Csf)\n",
+      "\n",
+      "# Calculations\n",
+      "def f12(CA): \n",
+      "\t return 2*CA/(1+CA)**2\n",
+      "\n",
+      "Csf = -1* quad(f12,CAo,CAf)[0]\n",
+      "\n",
+      "# Results\n",
+      "print \" Part b\"\n",
+      "print \" For Plug Flow\"\n",
+      "print \" Maximum expected concentration of S is %.3f \"%(Csf)\n",
+      "\n",
+      "CA = 1.;\n",
+      "Csf = (CAo-CA)*2*CA/(1+CA)**2;\n",
+      "\n",
+      "print \"Part c\"\n",
+      "print \" For MFR with separation and recycle\" \n",
+      "print \" Concentration of Csf is %.2f\"%(Csf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " For Mixed Flow Reactor\n",
+        " Maximum expected Cs is 0.667\n",
+        " Part b\n",
+        " For Plug Flow\n",
+        " Maximum expected concentration of S is 0.864 \n",
+        "Part c\n",
+        " For MFR with separation and recycle\n",
+        " Concentration of Csf is 0.50\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 page no : 164"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "CAo = 2.     # based on example 7.3\n",
+      "CA = 1.\n",
+      "Q = 0.5\n",
+      "\n",
+      "# Calculations\n",
+      "Cs1 = Q*(CAo-CA);\n",
+      "\n",
+      "def f6(CA): \n",
+      "\t return 2*CA/(1+CA)**2\n",
+      "\n",
+      "Cs2 = -1* quad(f6,1,0)[0]\n",
+      "\n",
+      "#Total amount of CS formed is\n",
+      "Cs = Cs1+Cs2;\n",
+      "\n",
+      "# Results\n",
+      "print \"Mixed flow followed by plug flow would be best\"\n",
+      "print \" Total amount of CS formed is %.3f mol/litre\"%(Cs)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mixed flow followed by plug flow would be best\n",
+        " Total amount of CS formed is 0.886 mol/litre\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": [],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch8.ipynb b/Chemical_Reaction_Engineering/ch8.ipynb
new file mode 100755
index 00000000..c66d684b
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch8.ipynb
@@ -0,0 +1,101 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : Potpourri of Multiple Reactions"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 page no : 198"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "CAo = 185.    \n",
+      "CA = 100.\n",
+      "t = 30.\n",
+      "\n",
+      "# Calculations\n",
+      "K123 = math.log(CAo/CA)/t;\n",
+      "m1 = 2.;\n",
+      "k2 = m1/CAo;\n",
+      "\n",
+      "##From the initial rate of formation of R\n",
+      "m2 = 1.3;\n",
+      "k1 = m2/CAo;\n",
+      "k3 = K123-k1-k2;\n",
+      "\n",
+      "k4 = 0.0001\n",
+      "while k4 <= 0.1:\n",
+      "    Csmax = CAo*(k1/K123)*((K123/k4)**(k4/(k4-K123)));\n",
+      "    if  Csmax>31.8 and Csmax<32.2:\n",
+      "        break\n",
+      "    k4 += 0.0001\n",
+      "    \n",
+      "k5 = 0.001\n",
+      "#similarly for T\n",
+      "while k5 <= 0.2:\n",
+      "    Ctmax = CAo*(k3/K123)*((K123/k5)**(k5/(k5-K123)));\n",
+      "    if Ctmax>9.95 and Ctmax<10.08:\n",
+      "        break\n",
+      "    k5 += .0001\n",
+      "\n",
+      "# Results\n",
+      "print \" The rate constants are\"\n",
+      "print \" k1 =  %.4f min**-1\"%(k1)\n",
+      "print \" k2 =  %.4f min**-1\"%(k2)\n",
+      "print \" k3 =  %.4f min**-1\"%(k3)\n",
+      "print \" k4 =  %.4f min**-1\"%(k4)\n",
+      "print \" k5 =  %.4f min**-1\"%(k5)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The rate constants are\n",
+        " k1 =  0.0070 min**-1\n",
+        " k2 =  0.0108 min**-1\n",
+        " k3 =  0.0027 min**-1\n",
+        " k4 =  0.0099 min**-1\n",
+        " k5 =  0.0157 min**-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/ch9.ipynb b/Chemical_Reaction_Engineering/ch9.ipynb
new file mode 100755
index 00000000..681066e0
--- /dev/null
+++ b/Chemical_Reaction_Engineering/ch9.ipynb
@@ -0,0 +1,405 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Temperature and Pressure Effects"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 page no : 209"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "CpA = 35.            # j/mol.K\n",
+      "CpB = 45.             # j/mol.K\n",
+      "CpR = 70.            # j/mol.K\n",
+      "T1 = 25.             # C\n",
+      "T2 = 1025.           # C\n",
+      "Hr = -50000.\n",
+      "\n",
+      "# Calculations\n",
+      "#Enthalpy balance for 1mol A,1 mol B,2 mol R\n",
+      "nA = 1.\n",
+      "nB = 1.\n",
+      "nR = 2.\n",
+      "dH = nA*CpA*(T1-T2)+nB*CpB*(T1-T2)+(Hr)+nR*CpR*(T2-T1);\n",
+      "\n",
+      "# Results\n",
+      "print \" dHJ at temperature 1025C is %.f \"%(dH)\n",
+      "if dH>0 :\n",
+      "    print \"Reaction is Exothermic\"\n",
+      "else:\n",
+      "     print \"Reaction is endothermic at 1025OC\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " dHJ at temperature 1025C is 10000 \n",
+        "Reaction is Exothermic\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 page no : 213"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "%pylab inline\n",
+      "\n",
+      "import math \n",
+      "from matplotlib.pyplot import *\n",
+      "from numpy import *\n",
+      "\n",
+      "Ho = -75300.                       # J/mol\n",
+      "Go = -14130.                       # J/mol\n",
+      "R = 8.3214\n",
+      "T1 = 298.\n",
+      "\n",
+      "# Calculations\n",
+      "#With all specific heais alike,dCp = 0\n",
+      "Hr = -Ho;\n",
+      "K298 = math.exp(-Go/(R*T1));\n",
+      "\n",
+      "\n",
+      "#Taking different values of T\n",
+      "T1 = array([2,15,25,35,45,55,65,75,85,95])                  #degree celcius\n",
+      "T = array([278,288,298,308,318,328,338,348,358,368])        #kelvin\n",
+      "\n",
+      "XAe = zeros(10)\n",
+      "\n",
+      "for i in range(10):\n",
+      "    K = K298*math.exp((Hr/R)*((1./T[i])-(1./298)));\n",
+      "    XAe[i] = K/(K+1);\n",
+      "\n",
+      "# Results\n",
+      "plot(T1,XAe)\n",
+      "xlabel(\"Temperature, C\")\n",
+      "ylabel(\"XAe\")\n",
+      "print (\" From the graph we see temp must stay below 78 C if conversion of 75% or above is expected\")\n",
+      "show()\n",
+      "\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Populating the interactive namespace from numpy and matplotlib\n",
+        " From the graph we see temp must stay below 78 C if conversion of 75% or above is expected"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stderr",
+       "text": [
+        "WARNING: pylab import has clobbered these variables: ['draw_if_interactive']\n",
+        "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
+       ]
+      },
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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E1FqcuyCfFkPBy8sLffr0sf9LXXgqgoicR5cuwKuvAn/9K1BZqXQ1rq3Fb/fhw4fjnXfe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\nRU5ErkJggg==\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x31ac650>"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 page no : 217"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "\n",
+      "# from example 9.2\n",
+      "XA = 0.581; \n",
+      "t = 1.                    #min\n",
+      "XAe = 0.89;\n",
+      "XAe1 = 0.993;\n",
+      "T1 = 338.\n",
+      "T2 = 298.\n",
+      "R = 8.314\n",
+      "\n",
+      "\n",
+      "# Calculations\n",
+      "k1_338 = -(XAe/t)*math.log(1-(XA/XAe));\n",
+      "XA1 = 0.6;\n",
+      "t1 = 10.                #min\n",
+      "\n",
+      "k1_298 = -(XAe1/t1)*math.log(1-(XA1/XAe1));\n",
+      "E1 = (R*math.log(k1_338/k1_298))*(T1*T2)/(T1-T2)\n",
+      "ko = k1_338/(math.exp(-E1/(R*T1)))\n",
+      "\n",
+      "# Results\n",
+      "print \" The rate constants are k = exp[75300/RT-24.7] min-1\"\n",
+      "print \" k1 = exp[17.2-48900/RT] min-1\"\n",
+      "print \" k2 = exp[41.9-123800/RT] min-1 \"\n",
+      "print \" E1 = %.2f\"%E1\n",
+      "print \" K0 = %.2f\"%ko"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The rate constants are k = exp[75300/RT-24.7] min-1\n",
+        " k1 = exp[17.2-48900/RT] min-1\n",
+        " k2 = exp[41.9-123800/RT] min-1 \n",
+        " E1 = 48679.81\n",
+        " K0 = 31411847.30\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 page no : 229"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "CAo = 4.                    #mol/litre\n",
+      "FAo = 1000.                 #mol/min\n",
+      "A = 0.405                   #litre/mol.min\n",
+      "\n",
+      "# Calculations\n",
+      "t = CAo*A;\n",
+      "V = FAo*A;\n",
+      "\n",
+      "# Results\n",
+      "print \" Part a\"\n",
+      "print \" The space time needed is %.2f min\"%(t)\n",
+      "print \" The Volume needed is %.f litres\"%(V)\n",
+      "\n",
+      "\n",
+      "# Note : We do not have value of 'rA'. so part B can not be calculated and plotted"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " The space time needed is 1.62 min\n",
+        " The Volume needed is 405 litres\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 pageno : 231"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "CAo = 4.                # mol/liter\n",
+      "FAo = 1000.             # mol/min\n",
+      "XA = 0.8;               # %\n",
+      "Cp = 250.               #cal/molA.K\n",
+      "Hr = 18000.             #cal/molA\n",
+      "rA = 0.4;\n",
+      "\n",
+      "# Calculations and Results\n",
+      "V = FAo*XA/rA;\n",
+      "\n",
+      "print \" Part a\"\n",
+      "print \" The size of reactorlitres) needed is %.f litres\"%(V)\n",
+      "\n",
+      "slope = Cp/Hr;\n",
+      "#Using graph\n",
+      "Qab1 = Cp*20;               #cal/molA\n",
+      "Qab = Qab1*1000;            #cal/min\n",
+      "Qab = Qab*0.000070;         #KW\n",
+      "\n",
+      "print \" Part b\"\n",
+      "print \" Heat Duty of precooler is %.2f kW\"%(Qab)\n",
+      "\n",
+      "Qce1 = Cp*37;               #cal/molA fed\n",
+      "Qce = Qce1*1000;            #cal/min\n",
+      "Qce = Qce*0.000070;         #KW\n",
+      "\n",
+      "print \" Heat Duty of postcooler is %.2f kW\"%(Qce)\n",
+      "\n",
+      "# answers may vary because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Part a\n",
+        " The size of reactorlitres) needed is 2000 litres\n",
+        " Part b\n",
+        " Heat Duty of precooler is 350.00 kW\n",
+        " Heat Duty of postcooler is 647.50 kW\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 pageno : 233"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "FAo = 1000.         #mol/min\n",
+      "Area = 1.72;\n",
+      "\n",
+      "# Calculations\n",
+      "V = FAo*Area;\n",
+      "\n",
+      "# Results\n",
+      "print \" The volume of adiabatic plug flow reactor is %.f litres\"%(V),\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume of adiabatic plug flow reactor is 1720 litres\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7 page no : 234"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "FAo = 1000.             #mol/min\n",
+      "Area = (0.8-0)*1.5;\n",
+      "\n",
+      "# Calculations\n",
+      "V = FAo*Area;\n",
+      "\n",
+      "# Results\n",
+      "print \" The volume required is %.f litres\"%(V),\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The volume required is 1200 litres\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Chemical_Reaction_Engineering/screenshots/CA0W.FAOVS4ln1.(1-Xa)-3Xa.png b/Chemical_Reaction_Engineering/screenshots/CA0W.FAOVS4ln1.(1-Xa)-3Xa.png
new file mode 100755
index 00000000..1405b907
--- /dev/null
+++ b/Chemical_Reaction_Engineering/screenshots/CA0W.FAOVS4ln1.(1-Xa)-3Xa.png
diff --git a/Chemical_Reaction_Engineering/screenshots/t,s_vs_E,s**-1_.png b/Chemical_Reaction_Engineering/screenshots/t,s_vs_E,s**-1_.png
new file mode 100755
index 00000000..7028bba8
--- /dev/null
+++ b/Chemical_Reaction_Engineering/screenshots/t,s_vs_E,s**-1_.png
diff --git a/Chemical_Reaction_Engineering/screenshots/t_minvsE.png b/Chemical_Reaction_Engineering/screenshots/t_minvsE.png
new file mode 100755
index 00000000..fe8f6d5e
--- /dev/null
+++ b/Chemical_Reaction_Engineering/screenshots/t_minvsE.png
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_10.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_10.ipynb
new file mode 100755
index 00000000..293510cb
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_10.ipynb
@@ -0,0 +1,178 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10: Energy Bands in Solids"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1, Page 323"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#Variable declaration\n",
+      "#E=Ef+1% of Ef\n",
+      "k=1.38*1e-23;#boltzman constant\n",
+      "e=1.6*1e-19;#charge of electron\n",
+      "E=0.0555;\n",
+      "\n",
+      "#calculations\n",
+      "#0.1=1/[(exp((E*e)/(k*T)))+1]\n",
+      "T=(E*e)/(k*log(9));#Temprature\n",
+      "\n",
+      "#Result\n",
+      "print 'Temprature = %.f K'%T\n",
+      "#Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Temprature = 293 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2, Page 324\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "\n",
+      "#Variable declaration\n",
+      "sx=0.01 #in ev. where x=E-Ef\n",
+      "x1=sx*1.6*1e-19 #converting it in joule\n",
+      "T=200 #in kelvin\n",
+      "\n",
+      "#calculation\n",
+      "Fe=1/(1+exp(x1/(1.38*1e-23*T)));#The value of F(E) \n",
+      "\n",
+      "#Result\n",
+      "print 'The value of F(E) = %.2f'%Fe\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of F(E) = 0.36\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3, Page 327"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "#Variable declaration\n",
+      "density=7.13*1e3  #in kg/m^3\n",
+      "M=65.4\n",
+      "N=6.023*1e26 #avogedro number\n",
+      "\n",
+      "#Calculations\n",
+      "n=(2*density*N)/M\n",
+      "n1=n**(2./3);\n",
+      "Ef=3.65*1e-19*n1; #in eV\n",
+      "Ef1=(3./5)*Ef #in eV\n",
+      "\n",
+      "#Results\n",
+      "print 'fermi energy = %.1f eV'%Ef\n",
+      "print 'Mean energy at T=0K is %.f eV'%Ef1\n",
+      "#Incorrect answers in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fermi energy = 9.4 eV\n",
+        "Mean energy at T=0K is 6 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.4, Page 328"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "Ef=5.51 #in eV\n",
+      "\n",
+      "#calculation\n",
+      "E=(3./5)*Ef;#The average energy of a free electron in silver at 0k\n",
+      "\n",
+      "#Result\n",
+      "print 'The average energy of a free electron in silver at 0k = %.3f eV'%E\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The average energy of a free electron in silver at 0k = 3.306 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_11.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_11.ipynb
new file mode 100755
index 00000000..857307a9
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_11.ipynb
@@ -0,0 +1,413 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11: Semiconductors"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1, Page 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "Pi=0.47;#given resistivity of intrinsic germanium\n",
+      "sigmai=1/Pi;#conductance\n",
+      "e=1.6*1e-19;#charge of electron\n",
+      "ue=0.38;#electron mobility\n",
+      "up=0.18;#hole mobility\n",
+      "\n",
+      "#Calculation\n",
+      "ni=sigmai/(e*(ue+up));#intrinsic carrier density at 300K \n",
+      "\n",
+      "#Result\n",
+      "print 'intrinsic carrier density at 300K temp= %.2f*10^19 m^-3'%(ni/1e+19)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "intrinsic carrier density at 300K temp= 2.37*10^19 m^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2, Page 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "e=1.6*1e-19;#charge of electron\n",
+      "ue=0.39;#electron mobility\n",
+      "up=0.19;#hole mobility\n",
+      "ni=2.4*1e19;#intrinsic carrier density \n",
+      "\n",
+      "#calculation\n",
+      "sigma=ni*e*(up+ue);\n",
+      "\n",
+      "#Result\n",
+      "print 'conductivity of intrinsic semiconductor= %.2f ohm^-1*m^-1'%sigma"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductivity of intrinsic semiconductor= 2.23 ohm^-1*m^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3, Page 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi,exp\n",
+      "\n",
+      "#Variable Declaration\n",
+      "m0=9.1*1e-31;\n",
+      "me=0.12*m0;\n",
+      "mp=0.28*m0;\n",
+      "Eg=0.67*1.6*1e-19\n",
+      "k=1.38*1e-23;#boltzman constant\n",
+      "h=6.62*1e-34;#plank's constant\n",
+      "T=300;\n",
+      "\n",
+      "#Calculations\n",
+      "ni=2*((2*pi*k*T/h**2)**(3./2))*((me*mp)**(3./4))*exp(-Eg/(2*k*T));#intrinsic carrier concentration\n",
+      "\n",
+      "#Result\n",
+      "print 'intrinsic carrier concentration is= %.1f *10^18 m^-3'%(ni/1e18)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "intrinsic carrier concentration is= 4.7 *10^18 m^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4, Page 343"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import exp\n",
+      "\n",
+      "#Variable Declaration\n",
+      "Eg1=0.36*1.6*1e-19;\n",
+      "Eg2=0.72*1.6*1e-19\n",
+      "k=1.38*1e-23;#boltzman constant\n",
+      "T=300;#tempreture in kelvin\n",
+      "\n",
+      "#Calculation\n",
+      "#in this formula ni=2*((2*%pi*k*T/h^2)^(3/2))*((me*mp)^(3/4))*exp(-Eg/(2*k*T))ratio of nip/niq is given by:\n",
+      "x=exp((Eg2-Eg1)/(2*k*T));#ratio of nip/niq\n",
+      "\n",
+      "#Result\n",
+      "print 'ratio of nip/niq is= %.f '%x\n",
+      "#Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of nip/niq is= 1050 \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5, Page 344"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "e=1.6*1e-19;#charge of electron\n",
+      "ue=0.39;#electron mobility\n",
+      "up=0.19;#hole mobility\n",
+      "ni=2.5*1e19;#intrinsic carrier density \n",
+      "l=1e-2;#length of Ge rode\n",
+      "a=1e-4;#area of Ge rode\n",
+      "\n",
+      "#Calculations&Results\n",
+      "sigma=ni*e*(up+ue);#conductivity of intrinsic semiconductor\n",
+      "print 'conductivity of intrinsic semiconductor= %.2f ohm^-1*m^-1'%sigma\n",
+      "P=1/sigma;\n",
+      "R=P*l/a;#resistance of Ge rode\n",
+      "print 'resistance of Ge rode =%.1f ohm'%R\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "conductivity of intrinsic semiconductor= 2.32 ohm^-1*m^-1\n",
+        "resistance of Ge rode =43.1 ohm\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6, Page 347"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "ue=3850;#mobility of electron\n",
+      "sigma=5;#conductivity of ntype semiconductor\n",
+      "e=1.6*1e-19;#charge of electron\n",
+      "\n",
+      "#Calculation\n",
+      "Nd=sigma/(e*ue);#density of donor atoms\n",
+      "\n",
+      "#Result\n",
+      "print 'density of donor atoms is= %.2f*10^16 cm^-3'%(Nd/1e16)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "density of donor atoms is= 0.81*10^16 cm^-3\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7, Page 351"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#let Ef-Ev=0.4eV=x and Ef1-Ev=y\n",
+      "x=0.4;#Ef-Ev in eV\n",
+      "k=1.38*1e-23;#boltzmann constant\n",
+      "T=300;#tempreture in kelvin\n",
+      "\n",
+      "#Calculations\n",
+      "#now p=Nv*exp(-x/(k*T))=Na and p'=Nv*exp(-y/(k*T))=2Na so ratio of this 2 is 2=exp(x-y/(k*T))\n",
+      "y=x-((k*T*log(2))/1.6e-19);#Ef1-Ev in eV\n",
+      "\n",
+      "#Result\n",
+      "print 'Ef1-Ev in eV is= %.4feV'%y\n",
+      "#Answer varies due to rounding-off errors"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ef1-Ev in eV is= 0.3821eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8, Page 352"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "#let Ec1-Ef=0.3eV=x and Ec2-Ef=y\n",
+      "x=0.3;#Ec-Ef in eV\n",
+      "T1=300.;#tempreture in kelvin\n",
+      "T2=330.;#tempreture in kelvin\n",
+      "\n",
+      "#Calculation\n",
+      "#Ec-Ef=k*T*log(Nc/Nd) so..\n",
+      "y=T2*x/T1;#Ec2-Ef in eV\n",
+      "\n",
+      "#Result\n",
+      "print 'Ec2-Ef in eV is= %.2f eV'%y\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ec2-Ef in eV is= 0.33 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9, Page 356"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "B=0.5;#given flux density\n",
+      "d=3*1e-3;#given thickness\n",
+      "J=500.;#given current density\n",
+      "n=1e21;#given donor density\n",
+      "e=1.6*1e-19;#charge of electron\n",
+      "\n",
+      "#Calculation\n",
+      "Vh=(B*J*d)/(n*e);#hall voltage\n",
+      "\n",
+      "#Result\n",
+      "print 'hall voltage is= %.1f mV'%(Vh/1e-3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "hall voltage is= 4.7 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.10, Page 357"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration\n",
+      "P=8.9*1e-3;#resistivity of doped sillicon\n",
+      "Rh=3.6*1e-4;#hall coefficient\n",
+      "e=1.6*1e-19;#charge of electron\n",
+      "\n",
+      "#Calculations&Results\n",
+      "ne=(3*pi)/(8*Rh*e);#carrier density of electron\n",
+      "print 'carrier density of electrons = %.3f*10^22 m^-3'%(ne/1e22)\n",
+      "ue=1./(P*ne*e);#mobility of electon\n",
+      "print 'mobility of charges = %.4f m^2*V^-1*s^-1'%ue\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "carrier density of electrons = 2.045*10^22 m^-3\n",
+        "mobility of charges = 0.0343 m^2*V^-1*s^-1\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_12.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_12.ipynb
new file mode 100755
index 00000000..171155f1
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_12.ipynb
@@ -0,0 +1,269 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12: Superconductivity"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1, Page 373"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "Tc=7.26;#critical tempreture in kelvin\n",
+      "H0=8*1e5/(4*pi);#magnetic field at 0K\n",
+      "T=5;#tempreture in kelvin\n",
+      "\n",
+      "#Calculation\n",
+      "Hc=H0*(1-(T/Tc)**2);#megnrtic field at 5K\n",
+      "\n",
+      "#Result\n",
+      "print 'magnrtic field at 5K tempreture =%.2f*10^4 A/m'%(Hc/1e4)\n",
+      "#Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnrtic field at 5K tempreture =3.35*10^4 A/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2, Page 373"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log\n",
+      "\n",
+      "#Variable declaration\n",
+      "Tc=0.3;#given tempareture in kelvin\n",
+      "thetad=300;\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#part a\n",
+      "N0g=-1./(log(Tc/thetad));\n",
+      "print 'the value of N0g is %.2f'%N0g\n",
+      "#part b\n",
+      "kB=1.38*1e-23;#boltzmann constant\n",
+      "Eg=3.5*kB*Tc;#energy\n",
+      "print 'energy is= %.2f*10^-23 J'%(Eg/1e-23)\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of N0g is 0.14\n",
+        "energy is= 1.45*10^-23 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3, Page 374"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "H0=0.0306;#given constant characteristic of lead material\n",
+      "Tc=3.7;#given tempareture in kelvin\n",
+      "T=2;#given tempareture in kelvin\n",
+      "\n",
+      "#Calculations\n",
+      "x=(T/Tc)*(T/Tc);\n",
+      "Hc=H0*(1-x);#value of magnetic field at 2K temp\n",
+      "\n",
+      "#Result\n",
+      "print 'value of magnetic field at 2K temp = %.4f T'%Hc\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of magnetic field at 2K temp = 0.0217 T\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4, Page 374"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt,pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "HcT=2*1e5/(4*pi);#magnetic field intensity at T K\n",
+      "Hc0=3*1e5/(4*pi);#magnetic field intensity at T=0K\n",
+      "Tc=3.69;#given temperature in K\n",
+      "\n",
+      "#Calculation\n",
+      "T=sqrt(1-(HcT/Hc0))*Tc;#tempreture in K\n",
+      "\n",
+      "#Result\n",
+      "print 'temperature of superconducture is= %.2f K'%T\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature of superconducture is= 2.13 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5, Page 374"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "H0=6.5*1e4;#given constant characteristic of lead material\n",
+      "Tc=7.18;#given temprature in kelvin\n",
+      "T=4.2;#given temprature in kelvin\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#part a\n",
+      "x=(T/Tc)*(T/Tc);\n",
+      "Hc=H0*(1-x);#value of magnetic field at 4.2K temp\n",
+      "print 'value of magnetic field at 4.2K temp= %.2f*10^4 A/M'%(Hc/1e4)\n",
+      "#part b\n",
+      "r=1e-3;#given radius\n",
+      "Ic=2*pi*r*Hc;#critical current\n",
+      "print 'critical current is = %.1f A'%Ic  #Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "value of magnetic field at 4.2K temp= 4.28*10^4 A/M\n",
+        "critical current is = 268.7 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6, Page 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable declaration\n",
+      "lemdaT=750;#given penetration depth at T=3.5K\n",
+      "Tc=4.22;#given critical tempreture\n",
+      "T=3.5;##given temperature\n",
+      "\n",
+      "#Calculations&Results\n",
+      "#part a\n",
+      "x=(T/Tc)**4;#temporary variable\n",
+      "lemda0=lemdaT/sqrt(1-x);#penetration depth at T=0K\n",
+      "print 'penetration depth at T=0K is %.fA'%lemda0\n",
+      "#part b\n",
+      "N=6.02*1e26;#given\n",
+      "alpha=13.55*1e3;#given\n",
+      "M=200.6;#given\n",
+      "n0=N*alpha/M;\n",
+      "print 'molecular density = %.3f*10^28 /m^3'%(n0/1e28)\n",
+      "ns=n0*(1-(T/Tc)**4);#superconducting electron density\n",
+      "print 'superconducting electron density = %.3f*10^28 /m^3'%(ns/1e28)#Answer differs due to rounding-off values\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "penetration depth at T=0K is 1033A\n",
+        "molecular density = 4.066*10^28 /m^3\n",
+        "superconducting electron density = 2.142*10^28 /m^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_13.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_13.ipynb
new file mode 100755
index 00000000..2ee7f8d7
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_13.ipynb
@@ -0,0 +1,216 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13: Magnetic Materials"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1, Page 457"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "u0=4*pi*1e-7;\n",
+      "H=1e7;#magnetic field strength\n",
+      "X=(-0.9)*1e-6;#magnetic suseptiblity\n",
+      "\n",
+      "#Calculations&Results\n",
+      "M=X*H;#magnetization of material\n",
+      "print 'magnetization of material is %.f A/m'%M\n",
+      "B=u0*H;#magnetic flux density\n",
+      "print 'magnetic flux density is %.2f Wb/m^2'%B\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetization of material is -9 A/m\n",
+        "magnetic flux density is 12.57 Wb/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2, Page 457"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "X=2*1e-3;#magnetic suseptibility of material at room temp.\n",
+      "H=1e3;#magnetic field intrnsity of piece of ferricoxide\n",
+      "u0=4*pi*1e-7;\n",
+      "\n",
+      "#Calculatons&Results\n",
+      "M=X*H;#magnetization\n",
+      "print 'magnetization is %.f A/m'%M\n",
+      "ur=X+1;#relative permiability\n",
+      "B=u0*ur*H;#magnetic flux density\n",
+      "print 'magnetic flux density is %.3f*10^-3 W/m^2'%(B/1e-3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "magnetization is 2 A/m\n",
+        "magnetic flux density is 1.259*10^-3 W/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3, Page 458"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "M=2.74*1e8;#magnetization per atom in A/m\n",
+      "a=2.66*1e-10;#elementry cube edge\n",
+      "n=2;#Iron in BCC\n",
+      "\n",
+      "#Calculations&Results\n",
+      "B=(M*a**3)/2;#Am^2 per atom\n",
+      "print 'Average number of Bohr magnetons contributed are %.2f*10^-22'%(B/1e-22)\n",
+      "#interms of bohr megneton\n",
+      "b=B/(9.27*1e-24);#dipole moment\n",
+      "print 'dipole moment is %.f bohr megneton/atom'%b\n",
+      "#Incorrect answers in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Average number of Bohr magnetons contributed are 25.78*10^-22\n",
+        "dipole moment is 278 bohr megneton/atom\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4, Page 258"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "u0=4*pi*1e-7;\n",
+      "b=9.27*1e-24;\n",
+      "H=1e3;#homogeneous field\n",
+      "k=1.38*1e-23;#boltzmann constant\n",
+      "T=303;#temp in kelvin\n",
+      "\n",
+      "#Calculations\n",
+      "T1 = T - 273; # Temp In Degree\n",
+      "x=u0*b*H/(k*T);#avg magnetic moment\n",
+      "\n",
+      "#Result\n",
+      "print 'avg magnetic moment is %.2f*10^-6 bohr magneton/spin'%(x/1e-6)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "avg magnetic moment is 2.79*10^-6 bohr magneton/spin\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.5, Page 459"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "ur=16;#relative permiability\n",
+      "I=3300;#intensity of magnetization\n",
+      "\n",
+      "#Calculation\n",
+      "H=I/(ur-1);#strength of the field\n",
+      "\n",
+      "#Result\n",
+      "print 'strength of the field =%.f A/m'%H\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "strength of the field =220 A/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_14.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_14.ipynb
new file mode 100755
index 00000000..71fc887c
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_14.ipynb
@@ -0,0 +1,262 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 14: Dielectrics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.1, Page 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "er=1.0000684;#dielectric constant of helium \n",
+      "N=2.7*1e25;#atoms/m^3\n",
+      "\n",
+      "#Calculations\n",
+      "r=(er-1)/(4*pi*N);\n",
+      "R=r**(1./3); #radius of electron cloud\n",
+      "\n",
+      "#Result\n",
+      "print 'radius of electron cloud is %.1f*10^-10 m'%(R/1e-10)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radius of electron cloud is 0.6*10^-10 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.2, Page 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "k=1.38*1e-23;#boltzmann constant\n",
+      "N=1e27;#HCL molecule per cubic meter\n",
+      "E=1e6;#electric field of vapour\n",
+      "D=3.33*1e-30;\n",
+      "\n",
+      "#Calculations\n",
+      "pHCL=1.04*D;\n",
+      "T=300;#tempreture in kelvin\n",
+      "alpha=(pHCL)**2/(3*k*T);\n",
+      "p0=N*alpha*E;#orientation polarization\n",
+      "\n",
+      "#Result\n",
+      "print 'orientation polarization is %.3f*10^-6 C/m^2'%(p0/1e-6)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "orientation polarization is 0.966*10^-6 C/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.3, Page 476"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "alpha=0.35*1e-40;#polarizability of gas\n",
+      "N=2.7*1e25;\n",
+      "e0=8.854*1e-12;#permittivity of vacume\n",
+      "\n",
+      "#Calculation\n",
+      "er=1+(N*alpha/e0);#relative permittivity\n",
+      "\n",
+      "#Result\n",
+      "print 'relative permittivity is %.6f'%er\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "relative permittivity is 1.000107\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.4, Page 480"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "er=12.;#relative permittivity\n",
+      "N=5*1e28;#atoms/m^3\n",
+      "e0=8.854*1e-12;#permittivity of vacume\n",
+      "\n",
+      "#Calculations\n",
+      "x=(er-1)/(er+2);\n",
+      "alpha=(3*e0/N)*x;#electrical polarizability\n",
+      "\n",
+      "#Result\n",
+      "print 'electronic polarizability = %.2f*10^-40 F*m^2'%(alpha/1e-40)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "electronic polarizability = 4.17*10^-40 F*m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.5, Page 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import atan,degrees\n",
+      "\n",
+      "#Variable declaration\n",
+      "C=2.4*1e-12;#given capacitance in F\n",
+      "e0=8.854*1e-12;#permittivity of vacume\n",
+      "a=4*1e-4;#area in m^2\n",
+      "d=0.5*1e-2;#thickness\n",
+      "tandelta=0.02;\n",
+      "\n",
+      "#Calculations&Results\n",
+      "er=(C*d)/(e0*a);#relative permittivity\n",
+      "print 'relative permittivity = %.2f'%er\n",
+      "lf=er*tandelta;#loss factor\n",
+      "print 'electric loss factor = %.4f'%lf\n",
+      "delta=degrees(atan(tandelta))\n",
+      "PA=90-delta;#phase angle\n",
+      "print 'phase angle = %.2f degrees'%PA\n",
+      "#incorrect answers in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "relative permittivity = 3.39\n",
+        "electric loss factor = 0.0678\n",
+        "phase angle = 88.85 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 14.6, Page 483"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "er=8.;#relative permittivity\n",
+      "a=0.036;#area in m^2\n",
+      "e0=8.854*1e-12;#permittivity of vacume\n",
+      "C=6*1e-6;#capacitance in F\n",
+      "V=15.0;#potential difference\n",
+      "\n",
+      "#Calculations\n",
+      "d=(e0*er*a)/C;\n",
+      "E=V/d;#field strength\n",
+      "\n",
+      "#Results\n",
+      "print 'field strength is= %.3f*10^7 V/m'%(E/1e+7)\n",
+      "dpm=e0*(er-1)*E;#dipole moment/unit volume\n",
+      "print 'dipole moment/unit volume= %.4f*10^-2 C/m^2'%(dpm/1e-2)\n",
+      "#Incorrect answers in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "field strength is= 3.529*10^7 V/m\n",
+        "dipole moment/unit volume= 0.2187*10^-2 C/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_2.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_2.ipynb
new file mode 100755
index 00000000..3436d354
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_2.ipynb
@@ -0,0 +1,405 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2: Acoustics of Buildings"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log10\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#delta_L=L2-L1\n",
+      "\n",
+      "#Calculation\n",
+      "#I proportional to square of amplitude so when amplitude is doubled intensity will becomes 4 times \n",
+      "#L1=10*l0g10(I1/I0)\n",
+      "#L2=10*log10(I2/I0)\n",
+      "#delta_L=L2-L1\n",
+      "#delta_L=10*log(I1/I0)-10*log(I2/I0)=10*log(I2/I1)\n",
+      "I21=4;#I2/I1=4 because intensity=amp^2\n",
+      "delta_L=10*log10(I21);#increase in intensity level\n",
+      "\n",
+      "#Result\n",
+      "print 'Increase in intensity level =',round(delta_L,2),'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Increase in intensity level = 6.02 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page 52"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration\n",
+      "#L2-L1=10*log10(I2/I1)\n",
+      "#so , we can write that \n",
+      "L2=40  #i dB\n",
+      "L1=10 #in dB \n",
+      "#where L1 and L2 are intensity level of two waves of same frequency\n",
+      "\n",
+      "#Calculation\n",
+      "L=L2-L1;\n",
+      "#let I2/I1=I\n",
+      "I=10**(L/10);\n",
+      "#let a2/a1=a\n",
+      "a=sqrt(I);#Ratio of their amplitudes \n",
+      "\n",
+      "#Result\n",
+      "print 'Ratio of their amplitudes =',round(a,3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Ratio of their amplitudes = 31.623\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3, Page 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log10\n",
+      "\n",
+      "#Variable Declaration\n",
+      "I1=25.2 #in Wm^-2\n",
+      "I2=0.90 #in Wm^-2\n",
+      "\n",
+      "#Calculation\n",
+      "B=10*log10(I1/I2) #Relative loudness of sound in dB\n",
+      "\n",
+      "#Result\n",
+      "print 'Relative loudness of sound = ',round(B,2),'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Relative loudness of sound =  14.47 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.4, Page 53"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log10\n",
+      "\n",
+      "#Variable Declaration\n",
+      "I=1e4 #in W/(m*m)\n",
+      "I0=1e-12 #in W/(m*m)\n",
+      "\n",
+      "#Calculation\n",
+      "B=10*log10(I/I0);#intensity level\n",
+      "\n",
+      "#Result\n",
+      "print \"intensity level = \",B,'dB'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "intensity level =  160.0 dB\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5, Page 54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "B=5 # in dB\n",
+      "\n",
+      "#Calculation\n",
+      "#B=10*log(I2/I1)\n",
+      "#let I2/I1=x\n",
+      "#10*log(x)=5\n",
+      "x=10**(5./10);\n",
+      "\n",
+      "#Result\n",
+      "print 'Amplified sound is',round(x,3),'times more intense than the unamplified sound'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amplified sound is 3.162 times more intense than the unamplified sound\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.6, Page 57"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "d=198; #in meter\n",
+      "t=1.2;#in second\n",
+      "\n",
+      "#Calculation\n",
+      "#velocity=distance/time\n",
+      "v=2*d/t;#velocity\n",
+      "\n",
+      "#Result\n",
+      "print 'velocity =',v,'m/s'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity = 330.0 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.7, Page 64"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "V=5600  #in  m^3\n",
+      "T=2     #in second\n",
+      "s=700   #in m^2\n",
+      "\n",
+      "#Calculation\n",
+      "a=0.16*V/(s*T)\n",
+      "\n",
+      "#Result\n",
+      "print \"absorption coefficient =\",a\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "absorption coefficient = 0.64\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.8, Page 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "absorp1=92.90; #in m^^2\n",
+      "absorp2=92.90;#in m^2\n",
+      "V=2265.6;#in m^3\n",
+      "\n",
+      "#Calculations\n",
+      "T1=0.16*V/(absorp1);\n",
+      "T2=0.16*V/(absorp1+absorp2);\n",
+      "ans=T2/T1;#effect on Reverberation time\n",
+      "\n",
+      "#Result\n",
+      "print \"Reverberation time reduced to \",ans,\"of original value\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reverberation time reduced to  0.5 of original value\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.9, Page 65"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "v=25.2*20.3*8.04 ;#in m^3\n",
+      "T=0.75; #in second\n",
+      "\n",
+      "#Calculations\n",
+      "absorp1=500*0.3176 ;#in m^2\n",
+      "absorp2=(0.16*v)/T;\n",
+      "T1=(0.16*v)/(absorp1+absorp2);#reverbaration time\n",
+      "\n",
+      "#Result\n",
+      "print \"reverbaration time =\",round(T1,3),'sec'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "reverbaration time = 0.635 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.10, Page 66"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "v=45*100*17.78;#in m^3\n",
+      "\n",
+      "#Calculations\n",
+      "absorp1=(700*0.03)+(600*0.06)+(400*0.025)+(600*0.3);\n",
+      "absorp_p=600*4.3;\n",
+      "T1=(0.16*v)/(absorp1);#Reverbaration time (empty hall) \n",
+      "T2=(0.16*v)/(absorp_p+absorp1);#Reverbaration time with full capacity\n",
+      "\n",
+      "#Results\n",
+      "print 'Reverbaration time (empty hall) =',round(T1,2),'sec' #printing mistake at the end in the textbook\n",
+      "print 'Reverbaration time with full capacity =',round(T2,2),'sec'"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Reverbaration time (empty hall) = 51.83 sec\n",
+        "Reverbaration time with full capacity = 4.53 sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_3.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_3.ipynb
new file mode 100755
index 00000000..8e553c2f
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_3.ipynb
@@ -0,0 +1,175 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3: Ultrasonics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1, Page 74"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "t=1.6*1e-3 #thickness in meter\n",
+      "v=5760. #velocity in m/s\n",
+      "\n",
+      "#Calculations\n",
+      "lemda=2*t#wavelength\n",
+      "f=v/lemda#fundamental frequency \n",
+      "\n",
+      "#Result\n",
+      "print 'fundamental frequency =',f/1e6,'MHz'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "fundamental frequency = 1.8 MHz\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2, Page 75"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "t=40*1e-2;\n",
+      "#pulse covers 2x distance in arriving back\n",
+      "#so,  30*1e-6=2*x/v\n",
+      "#and, 2nd pulse will cover a distance of 2*40 cm in 80*1e-6 seconds\n",
+      "#therfore,  80*1e-6=(2*40*1e-2)/v\n",
+      "#compare both equation\n",
+      "e1=30;\n",
+      "e2=40*2\n",
+      "\n",
+      "#Calculation\n",
+      "x=e1*t*2/(2*e2);\n",
+      "\n",
+      "#Result\n",
+      "print 'distance of the flow from near end =',x/1e-2,'cm' \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance of the flow from near end = 15.0 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3, Page 76"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "f_diff=50*1e3 #in Hz\n",
+      "v=5000 #in m/s\n",
+      "\n",
+      "#Calculations\n",
+      "#f1=v/2*t\n",
+      "#f2=2v/2t\n",
+      "#f2-f1=v/2t\n",
+      "t=v/(2*f_diff)\n",
+      "\n",
+      "#Result\n",
+      "print 'Thickness of steel plate =',t,'m'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of steel plate = 0.05 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4, Page 77"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi\n",
+      "\n",
+      "#Variable Declaration\n",
+      "f=1e6 #frequency in Hz\n",
+      "L=1 #inductance in henry\n",
+      "\n",
+      "#Calculation\n",
+      "#f=(1/2*pi)*(sqrt(1/(L*C)))\n",
+      "c=1/(4*pi**2*f**2*L);#capacitance\n",
+      "\n",
+      "#Result\n",
+      "print 'capacitance =',round(c/1e-12,3),'pF'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "capacitance = 0.025 pF\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_4.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_4.ipynb
new file mode 100755
index 00000000..66b89695
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_4.ipynb
@@ -0,0 +1,449 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4: Crystal Physics "
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1, Page 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration\n",
+      "r=1.278*1e-8 ;#atomic radius in cm\n",
+      "M=63.5; #atomic weight\n",
+      "N=6.023*1e23; #avogadro number\n",
+      "n=4#for fcc n=4\n",
+      "\n",
+      "#Calculations\n",
+      "a=4*r/(sqrt(2));\n",
+      "density=n*M/(N*a**3);#Density of copper\n",
+      "\n",
+      "#Result\n",
+      "print 'Density of copper =',round(density,1),'g/cc'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Density of copper = 8.9 g/cc\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2, Page 113"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "M=58.45;#atomic mass\n",
+      "N=6.02*1e23;#avogadro number\n",
+      "density=2.17*1e3 ; #in kg/m^3\n",
+      "n=4 #Nacl is FCC\n",
+      "\n",
+      "#Calculation\n",
+      "a=(n*M/(N*density))**(1./3);#lattice constant\n",
+      "\n",
+      "#Result\n",
+      "print 'lattice constant = ',round(a/1e-10,2),'A'\n",
+      "#incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lattice constant =  56.35 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3, Page 126"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "#let three intercepts are I1,I2,I3\n",
+      "I1=3;\n",
+      "I2=-2;\n",
+      "I3=3./2;\n",
+      "#let their reciprocals are I1_1,I2_1,I3_1\n",
+      "I1_1=1./I1;\n",
+      "I2_1=1./I2;\n",
+      "I3_1=1./I3;\n",
+      "\n",
+      "#Calculations\n",
+      "#LCM of I1_1,I2_1,I3_1 are 6 . \n",
+      "#By multiply LCM with I1_1,I2_1,I3_1 we will get miller indices\n",
+      "LCM=6;\n",
+      "M_1=LCM*I1_1;\n",
+      "M_2=LCM*I2_1 ;\n",
+      "M_3=LCM*I3_1;\n",
+      "\n",
+      "#Results\n",
+      "print 'Miller indices of plane are [',M_1,\n",
+      "print(M_2),\n",
+      "print(M_3),']'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Miller indices of plane are [ 2.0 -3.0 4.0 ]\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4, Page 126"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration\n",
+      "r=1.246 #in A\n",
+      "\n",
+      "#Calculations & Results\n",
+      "a=4*r/sqrt(2)\n",
+      "d_200=a/sqrt(2**2+0**2+0**2)\n",
+      "print 'd200 = ',round(d_200,2),'A'\n",
+      "d_220=a/sqrt(2**2+2**2+0**2)\n",
+      "print 'd220 = ',d_220,'A'\n",
+      "d_111=a/sqrt(1**2+1**2+1**2)\n",
+      "print 'd111 = ',round(d_111,2),'A'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "d200 =  1.76 A\n",
+        "d220 =  1.246 A\n",
+        "d111 =  2.03 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5, Page 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import acos,degrees\n",
+      "\n",
+      "#Variable Declaration\n",
+      "h=1\n",
+      "k=1\n",
+      "l=1\n",
+      "h1=1\n",
+      "k1=1\n",
+      "l1=1\n",
+      "\n",
+      "#Calculations\n",
+      "a=((h*h1)-(k*k1)+(l*l1))/(sqrt((h*h)+(k*k)+(l*l))*sqrt((h1*h1)+(k1*k1)+(l1*l1)));\n",
+      "#cosine angle=a so angle=cosine inverse of a\n",
+      "theta=degrees(acos(a));#angle between two planes\n",
+      "\n",
+      "#Result\n",
+      "print 'angle between two planes =',round(theta,2),'degrees'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angle between two planes = 70.53 degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6, Page 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "a=2.9*1e-8; #in cm\n",
+      "M=55.85;#atomic mass\n",
+      "density=7.87 #in g/cc\n",
+      "N=6.023*1e23;\n",
+      "\n",
+      "#Calculations\n",
+      "n=(a**3*N*density)/M;#Number of atoms per unit cell\n",
+      "\n",
+      "#Result\n",
+      "print 'Number of atoms per unit cell =',round(n,3)\n",
+      "#Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of atoms per unit cell = 2.07\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7, Page 127"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration\n",
+      "M=55.85;#atomic mass\n",
+      "d=7.86 #density of iron in g/cc\n",
+      "N=6.023*1e23\n",
+      "n=2#BCC structure\n",
+      "\n",
+      "#Calculations\n",
+      "a=((n*M)/(N*d))**(1./3);\n",
+      "r=(sqrt(3)*a)/4;#radius of iron atom \n",
+      "\n",
+      "#Result\n",
+      "print 'radius of iron atom =',round(r/1e-10,3),'A'\n",
+      "#Incorrect answer in the textbook"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "radius of iron atom = 124.196 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8, Page 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable Declaration\n",
+      "M=207.21;#atomic mass\n",
+      "d=11.34*1e3 #in kg/m^3\n",
+      "N=6.023*1e26 #in kg/m^3\n",
+      "n=4;#for FCC\n",
+      "\n",
+      "#Calculations\n",
+      "a=((n*M)/(N*d))**(1./3);#lattice constant\n",
+      "r=(sqrt(2)*a)/4;#Atomic radius\n",
+      "\n",
+      "#Result\n",
+      "print 'lattice constant =',round(a/1e-10,2),'A'\n",
+      "print 'Atomic radius =',round(r/1e-10,2),'A'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lattice constant = 4.95 A\n",
+        "Atomic radius = 1.75 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9, Page 128"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt,sin,degrees,radians,pi\n",
+      "\n",
+      "#Variable Declaration\n",
+      "n=1;\n",
+      "theta=30;#angle in degree\n",
+      "lamda=1.75; #in A\n",
+      "h=1;\n",
+      "k=1;\n",
+      "l=1;\n",
+      "\n",
+      "#Calculations\n",
+      "#d111=a/sqrt((h*h)+(k*k)+(l*l))\n",
+      "#2dsin(thita)=n*lamda\n",
+      "d=n*lamda/(2*sin(theta*pi/180))\n",
+      "a=sqrt(3)*d;#lattice constant \n",
+      "\n",
+      "#Result\n",
+      "print \"lattice constant =\",round(a,3),'A'"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "lattice constant = 3.031 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10, Page 129"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "#let three intercepts are I1,I2,I3\n",
+      "I1=0.96;\n",
+      "I2=0.64;\n",
+      "I3=0.48;\n",
+      "\n",
+      "#Calculations\n",
+      "#as they are ratios we will multiply by some some constants so that  it will become integers\n",
+      "I1=6;\n",
+      "I2=4;\n",
+      "I3=3 ;\n",
+      "#let their reciprocals are I1_1,I2_1,I3_1\n",
+      "I1_1=1./I1;\n",
+      "I2_1=1./I2;\n",
+      "I3_1=1./I3;\n",
+      "#LCM of I1_1,I2_1,I3_1 are 12. \n",
+      "#By multiply LCM with I1_!,I2_1,I3_1 we will get miller indices\n",
+      "LCM=12;\n",
+      "M_1=LCM*I1_1;\n",
+      "M_2=LCM*I2_1 ;\n",
+      "M_3=LCM*I3_1;\n",
+      "\n",
+      "#Results\n",
+      "print 'Miller indices of plane are [',M_1,\n",
+      "print(M_2),\n",
+      "print(M_3),']'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Miller indices of plane are [ 2.0 3.0 4.0 ]\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_5.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_5.ipynb
new file mode 100755
index 00000000..8b5513e0
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_5.ipynb
@@ -0,0 +1,310 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5: Wave Optics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1, Page 142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "refractive_index=1.65 #refractive index\n",
+      "lamda=5893*1e-10;#wavelength\n",
+      "n=400;\n",
+      "\n",
+      "#Calculation\n",
+      "t=(n*lamda)/(2*(refractive_index-1));#Thickness of film\n",
+      "\n",
+      "#Result\t\t\n",
+      "print 'Thickness of film = ',round(t/1e-4,2),'*10^-4 m'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thickness of film =  1.81 *10^-4 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2, Page 142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "x=0.40*1e-3; #in meter\n",
+      "n=900;\n",
+      "\n",
+      "#Calculations\n",
+      "lamda=2*x/n;#Wavelength of light in meters\n",
+      "lamda1=lamda/1e-10;#Wavelength of light in A\n",
+      "\n",
+      "#Result\t\t\n",
+      "print 'Wavelength of light in A =',round(lamda1),'A'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of light in A = 8889.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.3, Page 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "lamda=5893*1e-10;#wavelength of monocromatic light\n",
+      "n=4000;\n",
+      "\n",
+      "#Calculation\n",
+      "x=n*lamda/2;#distance moved by mirror M1\n",
+      "\n",
+      "#Result\n",
+      "print 'distance moved by mirror M1 =',x/1e-2,'*10^-2 m'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "distance moved by mirror M1 = 0.11786 *10^-2 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4, Page 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "lamda=5461*1e-10;#wavelength of light\n",
+      "n=8;#no of frings\n",
+      "t=6*1e-6;#in meter\n",
+      "\n",
+      "#calculation\n",
+      "u=((n*lamda)/(2*t))+1;#refractive index of material\n",
+      "\n",
+      "#Result\n",
+      "print 'refractive index of material =',round(u,5)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "refractive index of material = 1.36407\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5, Page 154"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "ue=1.553;#given ue\n",
+      "u0=1.544;#given uo\n",
+      "lamda=500*1e-9;#in meter\n",
+      "\n",
+      "#Calculation\n",
+      "t=lamda/(4*(ue-u0));#The thickness of quarter wave plate\n",
+      "\n",
+      "#Result\n",
+      "print 'The thickness of quarter wave plate =',round(t/1e-5,3),'*10^-5 m'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The thickness of quarter wave plate = 1.389 *10^-5 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6, Page 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "lamda=5893*1e-10;#in meter\n",
+      "ue=1.55333;#given ue\n",
+      "u0=1.5442;#given u0\n",
+      "\n",
+      "#Calculation\n",
+      "t=lamda/(2*(ue-u0));#Thicknesss of half wave plate\n",
+      "\n",
+      "#Result\n",
+      "print 'Thicknesss of half wave plate =',round(t/1e-5,2),'*10^-5 m'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thicknesss of half wave plate = 3.23 *10^-5 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7, Page 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "u0=1.5442;#given u0\n",
+      "ue=1.5533;#given ue\n",
+      "lamda=5*1e-5;#wavelrngth in cm\n",
+      "\n",
+      "#Calculation\n",
+      "t=lamda/(2*(ue-u0));#Thicknesss of half wave plate\n",
+      "\n",
+      "#Result\n",
+      "print 'Thicknesss of half wave plate =',round(t/1e-3,2),'*10^-3 cm'\n",
+      "#Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thicknesss of half wave plate = 2.75 *10^-3 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.8, Page 155"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "u0=1.658;#given u0\n",
+      "ue=1.486;#given ue\n",
+      "lamda=5893*1e-8 #in cm\n",
+      "\n",
+      "#Calculation\n",
+      "t=lamda/(4*(u0-ue));#Thicknesss of quarter wave plate \n",
+      "\n",
+      "#Result\n",
+      "print 'Thicknesss of quarter wave plate =',round(t/1e-4,2),'*10^-4 cm'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thicknesss of quarter wave plate = 0.86 *10^-4 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_6.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_6.ipynb
new file mode 100755
index 00000000..b0fcb49b
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_6.ipynb
@@ -0,0 +1,139 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6: Lasers"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1, Page 170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "D=4*1e8;#distance between earth and moon in m\n",
+      "lamda=16000.*1e-10;#wavelength in meters\n",
+      "d=1e-3;#aperture in meter\n",
+      "\n",
+      "#Calculations & Result\n",
+      "th=lamda/d;#angular speed\n",
+      "print 'angular speed is=',th,'rad'\n",
+      "aos=(D*th)**2;#area of spread \n",
+      "print 'area of spread is=',aos,'m^2'\n",
+      "#Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angular speed is= 0.0016 rad\n",
+        "area of spread is= 4.096e+11 m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2, Page 170"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "a1=2*1e-3;#distance from the laser\n",
+      "a2=3*1e-3;#distance from the laser\n",
+      "d1=2;#output beam spot diameter\n",
+      "d2=4;#output beam spot diameter\n",
+      "\n",
+      "#Calculation\n",
+      "th=(a2-a1)/(2*(d2-d1));#angle of divergence\n",
+      "\n",
+      "#Result\n",
+      "print 'angle of divergence',th/1e-3,'*10^-3 rad'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angle of divergence 0.25 *10^-3 rad\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3, Page 171"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable Declaration\n",
+      "D=0.1;#focal length of lens\n",
+      "lamda=14400*1e-10;#wavelength in meters\n",
+      "p=100*1e-3;#power of laser beam\n",
+      "d=10*1e-3;#aperture in meter\n",
+      "\n",
+      "#Calculations & Results\n",
+      "th=lamda/d;#angular speed\n",
+      "print 'angular speed is=',th/1e-4,'*10^-4 rad'\n",
+      "aos=(D*th)**2;#area of spread \n",
+      "print 'area of spread is=',aos/1e-10,'*10^-10 m^2'\n",
+      "I=p/aos;#'intensity\n",
+      "print 'intensity is=',round(I/1e7,1),'*10^7 W*m^-2'\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "angular speed is= 1.44 *10^-4 rad\n",
+        "area of spread is= 2.0736 *10^-10 m^2\n",
+        "intensity is= 48.2 *10^7 W*m^-2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_7.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_7.ipynb
new file mode 100755
index 00000000..95dcaf40
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_7.ipynb
@@ -0,0 +1,319 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7: Optical Fibre Communication"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1, Page 206"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable declaration\n",
+      "NA = 0.24;#Numerical Aperture\n",
+      "delta = 0.014;\n",
+      "\n",
+      "#Calculations & Results\n",
+      "n1 = (NA)/sqrt(2*delta);#Refractive index of first medium \n",
+      "print 'Refractive index of first medium is ',round(n1,4)\n",
+      "n2 = n1 - (delta*n1);#Refractive index of secong material\n",
+      "print 'Refractive index of secong material is ',round(n2,4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Refractive index of first medium is  1.4343\n",
+        "Refractive index of secong material is  1.4142\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2, Page 207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt,asin,degrees\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.49; # Refractive index of first medium\n",
+      "n2 = 1.44; # Refractive index of second medium\n",
+      "\n",
+      "#Calculations & Results\n",
+      "def deg_to_dms(deg):\n",
+      "    d = int(deg)\n",
+      "    md = abs(deg - d) * 60\n",
+      "    m = int(md)\n",
+      "    sd = (md - m) * 60\n",
+      "    sd=round(sd,2)\n",
+      "    return [d, m, sd]\n",
+      "\n",
+      "delta = (n1-n2)/n1; # Index difference\n",
+      "NA = n1* sqrt(2*delta);\n",
+      "print 'Numerical Aperture of fiber is',round(NA,3)\n",
+      "theta = degrees(asin(NA));\n",
+      "print 'Acceptance angle is ',deg_to_dms(theta),'degrees'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Numerical Aperture of fiber is 0.386\n",
+        "Acceptance angle is  [22, 42, 22.17] degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3, Page 207"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt,asin,degrees\n",
+      "\n",
+      "#Variable declaration\n",
+      "NA = 0.15 ; # Numerical Aperture of fiber\n",
+      "n2 = 1.55; # Refractive index of cladding\n",
+      "n0w = 1.33; # Refractive index of water\n",
+      "n0a = 1; # Refractive index of air\n",
+      "\n",
+      "#Calculations\n",
+      "def deg_to_dms(deg):\n",
+      "    d = int(deg)\n",
+      "    md = abs(deg - d) * 60\n",
+      "    m = int(md)\n",
+      "    sd = (md - m) * 60\n",
+      "    sd=round(sd,2)\n",
+      "    return [d, m, sd]\n",
+      "\n",
+      "n1 = sqrt(NA**2 + n2**2);\n",
+      "NAW = (sqrt(n1**2 -n2**2))/n0w;\n",
+      "theta = degrees(asin(NAW));#Acceptance angle in water\n",
+      "\n",
+      "#Result\n",
+      "print 'Acceptance angle in water is ',deg_to_dms(theta),'degrees'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Acceptance angle in water is  [6, 28, 32.55] degrees\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4, Page 216"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import log10\n",
+      "\n",
+      "#Variable declaration\n",
+      "l = 16; # Length of optical fiber in Km\n",
+      "Pi = 240e-6; # Mean optical length launched in optical fiber in Watts\n",
+      "Po = 6e-6; # Mean optical power at the output in watts\n",
+      "\n",
+      "#Calculations&Results\n",
+      "alpha = 10*log10(Pi/Po);#Signal attenuation in fiber\n",
+      "print 'Signal attenuation in fiber',round(alpha),'dB'\n",
+      "alpha1 = alpha/l;#Signal attenuation per km of the fiber\n",
+      "print 'Signal attenuation per km of the fiber',round(alpha1),'dB/km'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Signal attenuation in fiber 16.0 dB\n",
+        "Signal attenuation per km of the fiber 1.0 dB/km\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5, Page 219"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import pi,exp\n",
+      "\n",
+      "#Variable declaration\n",
+      "Tf = 1400; # Fictive temperature of silicon in Kelvin\n",
+      "betai = 7e-11; # Isothermal compressibility square meter per newton\n",
+      "n = 1.46; # Refractive index of silicon\n",
+      "p = 0.286; # Photoelastic constant of silicon\n",
+      "lamda = 0.63e-6 # Wavelength in micrometer\n",
+      "kb = 1.38e-23 # Boltzmann constant in joule per kelvin\n",
+      "L = 1e3;\n",
+      "\n",
+      "#Calculations\n",
+      "alphas = (8 * pi**3 * n**8 * p**2 * kb * Tf * betai)/(3 * lamda**4);#Rayleigh scattering coefficient\n",
+      "alphars = exp(-alphas * L);#Loss factor\n",
+      "\n",
+      "#Results\n",
+      "print 'Rayleigh scattering coefficient is ',round(alphas/1e-3,2),'*10^-3 /m'\n",
+      "print 'Loss factor is',round(alphars,3)  #Answer varies due to rounding-off values\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Rayleigh scattering coefficient is  1.2 *10^-3 /m\n",
+        "Loss factor is 0.302\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6, Page 222"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "alpha = 0.5; # Attenuation of single mode optical fibre in dB per km\n",
+      "lamda = 1.4; # Operating wavelength of optical fiber in micrometer\n",
+      "d = 8 # Diameter of fiber in micrometer\n",
+      "y = 0.6; # Laser source frequency width\n",
+      "\n",
+      "#Calculations\n",
+      "pb = 4.4e-3 * d**2 * lamda**2 * alpha * y;#Threshold optical power in SBS\n",
+      "prs = 5.9e-2 * d**2 * lamda * alpha;#Threshold optical power in SRS\n",
+      "\n",
+      "#Results\n",
+      "print 'Threshold optical power in SBS',pb/1e-3,'mW'\n",
+      "print 'Threshold optical power in SRS',prs,'W'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Threshold optical power in SBS 165.5808 mW\n",
+        "Threshold optical power in SRS 2.6432 W\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7, Page 225"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt, pi\n",
+      "\n",
+      "#Variable declaration\n",
+      "n1 = 1.50; # Refreactive index of forst medium\n",
+      "delta = 0.003; # Index difference\n",
+      "lamda = 1.6*1e-6; # Operating wavelength of fober in meter\n",
+      "\n",
+      "#Calculations&Results\n",
+      "n2 = sqrt(n1**2-(2*delta*n1**2));#refractive index of cladding\n",
+      "#Substituting n2^2 = n1^2 - 2*delta*n1^2 in euation of Rc,\n",
+      "rc = (3*n1**2*lamda)/(4*pi*((2*delta*n1**2)**(3./2)));#The critical radius of curvature for which bending losses occur \n",
+      "print 'The critical radius of curvature for which bending losses occur is ',round(rc/1e-6,2),'um'\n",
+      "#Incorrect answer in the textbook\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The critical radius of curvature for which bending losses occur is  547.92 um\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_8.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_8.ipynb
new file mode 100755
index 00000000..36c3530d
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_8.ipynb
@@ -0,0 +1,231 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8:Conducting Materials"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1, Page 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "n = 5.8*1e28; # Electrons density in electrons per cube meter\n",
+      "rho = 1.58*1e-8; #Resistivity of wire in ohm meter\n",
+      "m = 9.1*1e-31; # Mass of electron \n",
+      "e = 1.6*1e-19; # Charge of electron in coloumb\n",
+      "E = 1e2; # Electric field\n",
+      "\n",
+      "#Calculations\n",
+      "t = round((m/(rho*n*e**2))/1e-14);\n",
+      "u = (e*t*10**-14)/m;\n",
+      "v = u*E; \n",
+      "\n",
+      "#Results\n",
+      "print 'The relaxation time is ',t,'*10^-14 s'\n",
+      "print 'The mobility of electrons ',round(u/1e-3,2),'*10^-3 m^2/volt sec'\n",
+      "print 'The average drift velocity for an electric field of 1V/cm is ',round(v,3),'m/s'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The relaxation time is  4.0 *10^-14 s\n",
+        "The mobility of electrons  7.03 *10^-3 m^2/volt sec\n",
+        "The average drift velocity for an electric field of 1V/cm is  0.703 m/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2, Page 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable declaration\n",
+      "e = 1.6*1e-19; # Charge on electron in coulumb\n",
+      "m = 9.1*1e-31; # Mass of electron in kg \n",
+      "rho =  1.54*1e-8; #Resistivity of material at room temperature in ohm . meter\n",
+      "n = 5.8*1e28; # Number of electrons per cubic meter\n",
+      "Ef = 5.5; # The fermi energy of the conductor in eV\n",
+      "\n",
+      "#Calculations\n",
+      "vf = sqrt((2*Ef*e)/m);\n",
+      "t = (m/(n*e**2*rho));\n",
+      "MFP = vf*t;\n",
+      "\n",
+      "#Results\n",
+      "print 'Velocity of electron is',round(vf/1e6,2),'*10^6 m/s'\n",
+      "print 'Mean free path of electron is',round(MFP/1e-8,2),'*10^-8 m'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Velocity of electron is 1.39 *10^6 m/s\n",
+        "Mean free path of electron is 5.53 *10^-8 m\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3, Page 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "m = 9.1*1e-31; #Mass of electron in kg\n",
+      "e = 1.6*1e-19; # Charge on electron in coulumb\n",
+      "t = 3*1e-14; # Relaxation time in seconds\n",
+      "n = 5.8*1e28; #Number of electrons in cubic meter\n",
+      "\n",
+      "#Calculations\n",
+      "rho =m/(n*t*e*e);#The resistivity of metal \n",
+      "u = 1/(n*e*rho);#The mobility of electron \n",
+      "\n",
+      "#Result\n",
+      "print 'The resistivity of metal is',round(rho/1e-8,2),'*10^-8 Ohm.meter'  #incorrect answer in textbook\n",
+      "print 'The mobility of electron is',round(u/1e-3,2),'*10^-3 sqaure meter per volt.second' \n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The resistivity of metal is 2.04 *10^-8 Ohm.meter\n",
+        "The mobility of electron is 5.27 *10^-3 sqaure meter per volt.second\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4, Page 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Variable declaration\n",
+      "e = 1.6*1e-19; # Charge of electrons in coloumbs\n",
+      "m = 9.1*1e-31; # Mass of electrons in Kg\n",
+      "Ef = 7*e ; #Fermi energy in electrons volt\n",
+      "t = 3*1e-14; # Relaxation time in seconds\n",
+      "\n",
+      "#Calculations\n",
+      "vf = sqrt(Ef*2/m);\n",
+      "lamda = vf*t;#The mean free path of electrons \n",
+      "\n",
+      "#Result\n",
+      "print 'The mean free path of electrons is',round(lamda/1e-10),'A'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The mean free path of electrons is 471.0 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5, Page 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "rhoC = 1.65*1e-8; # Electrical resistivity of cpooer in ohm meter\n",
+      "rhoN = 14*1e-8; # Electrical resistivity of Nickel in ohm meter\n",
+      "T = 300; # Room temperature in kelvin\n",
+      "\n",
+      "#Calculations\n",
+      "KCu =(2.45*1e-8*T)/rhoC;#Thermal conductivity of Cu\n",
+      "KNi =2.45*1e-8*T/rhoN;#Thermal conductivity of Ni\n",
+      "\n",
+      "#Results\n",
+      "print 'Thermal conductivity of Cu is ',round(KCu),'W/(m*degree)' #incorrect answer in textbook\n",
+      "print 'Thermal conductivity of Ni is ',KNi,'W/(m*degree)'\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thermal conductivity of Cu is  445.0 W/(m*degree)\n",
+        "Thermal conductivity of Ni is  52.5 W/(m*degree)\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/Chapter_9.ipynb b/Engineering_Physics_by_K._Rajagopal/Chapter_9.ipynb
new file mode 100755
index 00000000..94d3c955
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/Chapter_9.ipynb
@@ -0,0 +1,664 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9: Quantum Physics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1, Page 279"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "e = 1.602e-19; # Charge of electron in Coloumb\n",
+      "lamda = 2e-10; # Wavelength of a photon in meters\n",
+      "h = 6.62e-34; # Planc's constant in Joule second\n",
+      "c = 3.e8; # Velocity og light in air in meter per second\n",
+      "\n",
+      "#Calculations\n",
+      "E = (h*c)/(lamda*e);#Thermal conductivity of Ni\n",
+      "p = h/lamda;#The momentum of photon \n",
+      "\n",
+      "#Results\n",
+      "print 'The energy of photon is ',round(E,3),'eV' #Incorrect answer in textbook\n",
+      "print 'The momentum of photon is ',p,'(kg.m)/s'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The energy of photon is  6198.502 eV\n",
+        "The momentum of photon is  3.31e-24 (kg.m)/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2, Page 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "h = 6.62e-34; # Planck's constant J.s\n",
+      "v = 440e3; # Operating frequency of radio in Hertz\n",
+      "P = 20e3 ; # Power of radio transmitter in Watts\n",
+      "\n",
+      "#Calculation\n",
+      "n = P/(h*v);# Let n be the number of photons emitted per second\n",
+      "\n",
+      "#Result\n",
+      "print 'Number of photon emitted per second is ',round(n/1e30,2),'*10^30'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of photon emitted per second is  68.66 *10^30\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3, Page 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "h = 6.62e-34; # Planck's constant in J.s\n",
+      "c = 3e8; # Velocity of ligth in air\n",
+      "t = 18000; # Time of glow - (5*3600) in seconds\n",
+      "P = 30 #Power in watts\n",
+      "lamda = 5893e-10; # Wavelength of emitted ligth in meters\n",
+      "\n",
+      "#calculations\n",
+      "E = (h*c)/lamda; # Energy of a photon\n",
+      "n = (P*t)/E; # let n be the number of photons emitted in 5 hours\n",
+      "\n",
+      "#Result\n",
+      "print 'Number of photons emitted in 5 hours is',round(n/1e24,3),'*10^24'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Number of photons emitted in 5 hours is 1.602 *10^24\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4, Page 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import cos\n",
+      "\n",
+      "#Varaible declaration\n",
+      "h = 6.62*1e-34; # Plancl's constant in J.s\n",
+      "c = 3*1e8; # Velocity of light in vacccum in m/s \n",
+      "m = 9.1*1e-31; # Mass of electron in Kg\n",
+      "l = 0.7078*1e-10 # Wavelength in meter\n",
+      "theta = 90;\n",
+      "\n",
+      "#Calculations\n",
+      "delta = (h*(1-round(cos(theta)))/(m*c));\n",
+      "Nlambda = l + delta;\n",
+      "\n",
+      "#Result\n",
+      "print 'The wavelength of scattered X-rays is %.4f A'%(Nlambda/1e-10)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength of scattered X-rays is 0.7320 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5, Page 287"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import cos,degrees,radians\n",
+      "\n",
+      "#Varaible declaration\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in J.s\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "lamda = 1.8e18; # Frequency of the incident rays\n",
+      "theta = 180;#angle in degree\n",
+      "\n",
+      "#Calculations\n",
+      "lamda = c/lamda;\n",
+      "delta = (h*(1-cos(radians(theta))))/(m*c);\n",
+      "Nlambda = lamda+delta;#'Wavelength of scattered X-rays\n",
+      "\n",
+      "#Result\n",
+      "print 'Wavelength of scattered X-rays is %.4f A'%(Nlambda/1e-10)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Wavelength of scattered X-rays is 1.7152 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6, Page 288"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import cos\n",
+      "\n",
+      "#Varaible declaration\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in Js\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "lamda = 1.12e-10; # Wavelength of light in meters\n",
+      "theta = 90;\n",
+      "\n",
+      "#Calculations\n",
+      "delta = (h*(1-round(cos(theta))))/(m*c);\n",
+      "Nlambda = lamda + delta;#The wavelength of scattered X-rays \n",
+      "E = (h*c)*((1/lamda)-(1/Nlambda)) ;#Energy of electron\n",
+      "\n",
+      "#Results\n",
+      "print 'The wavelength of scattered X-rays is %.3f A'%(Nlambda/1e-10)\n",
+      "print 'Energy of electron is %.2f *10^-17 J'%(E/1e-17)\n",
+      " \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The wavelength of scattered X-rays is 1.144 A\n",
+        "Energy of electron is 3.76 *10^-17 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exampe 9.7, Page 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import cos,radians\n",
+      "\n",
+      "#Varaible declaration\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in Js\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "lamda = 0.03e-10; # Wavelength of light in meters\n",
+      "theta = 60;#angle in degrees\n",
+      "\n",
+      "#Calculations\n",
+      "delta = (h*(1-cos(radians(theta))))/(m*c);\n",
+      "Nlambda = lamda + delta;\n",
+      "E = ((h*c)*((1./lamda)-(1./Nlambda)))/1.6e-19 ;#Energy of recoiling electron\n",
+      "\n",
+      "#Result\n",
+      "print 'Energy of recoiling electron is %.3f MeV'%(E/1e+6)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of recoiling electron is 0.119 MeV\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Eample 9.8, Page 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import cos,radians\n",
+      "\n",
+      "#Varaible declaration\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in Js\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "lamda = 0.5e-10; # Wavelength of light in meters\n",
+      "theta = 90;\n",
+      "\n",
+      "#Calculations\n",
+      "delta = (h*(1-cos(radians(theta))))/(m*c);\n",
+      "Nlambda = lamda + delta;\n",
+      "E = (h*c)*((1./lamda)-(1./Nlambda)) ;\n",
+      "\n",
+      "#Result\n",
+      "print 'Energy of electron is %.2f *10^-16 J'%(E/1e-16)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of electron is 1.84 *10^-16 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9, Page 290"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in Js\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "lamda = 1.5e-10; # Wavelength of light in meters\n",
+      "E = 0.5e-16; # Energy of electron in J \n",
+      "\n",
+      "#Calculation\n",
+      "Nlambda = ((h*c)/lamda)-E;#'Energy of scattered electron\n",
+      "\n",
+      "#Result\n",
+      "print 'Energy of scattered electron is %.2f *10^-16 J'%(Nlambda/1e-16)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of scattered electron is 12.74 *10^-16 J\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10, Page 290"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import cos,radians\n",
+      "\n",
+      "#Varaible declaration\n",
+      "lamda=0.022*1e-10;#wavelength in meters\n",
+      "th=45;#angle in degree\n",
+      "m=9.1*1e-31;\n",
+      "c=3*1e8;#velocity of light in free space\n",
+      "h=6.62*1e-34;#planck's constant\n",
+      "\n",
+      "#Calculations&Results\n",
+      "x=cos(th);\n",
+      "dlamda=h*(1-cos(radians(th)))/(m*c);#delta lemda \n",
+      "print 'delta lemda is= %.3f A'%(dlamda/1e-10)\n",
+      "lamda1=lamda-dlamda;#wavelength of incident X-rays\n",
+      "print 'wavelength of incident X-rays %.3f A'%(lamda1/1e-10)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "delta lemda is= 0.007 A\n",
+        "wavelength of incident X-rays 0.015 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11, Page 314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "a = 1e-10 # Width of box in meter\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in Js\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "n = 1; # Single electron\n",
+      "\n",
+      "#Calculation\n",
+      "E = (n**2 * h**2)/(8*m*a**2*1.6e-19);\n",
+      "\n",
+      "#Result\n",
+      "print'Energy of electrons is %.1f n^2 eV'%E\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of electrons is 37.6 n^2 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12, Page 314"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "a = 1e-10 # Width of box in meter\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in Js\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "n = 1; # Single electron\n",
+      "\n",
+      "#Calculations\n",
+      "E = (h**2)/(8*m*a**2);#Energy of in lower level\n",
+      "p = h/(2*a);#Momentum \n",
+      "\n",
+      "#Results\n",
+      "print 'Energy of in lower level %.f *10^-18 J'%(E/1e-18)\n",
+      "print'Momentum is %.2f *10^-24 (kg.m)/s'%(p/1e-24)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Energy of in lower level 6 *10^-18 J\n",
+        "Momentum is 3.31 *10^-24 (kg.m)/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.13, Page 315"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "a = 0.2e-9 # Width of box in meter\n",
+      "m = 9.1e-31; # Mass of electron in kg\n",
+      "h = 6.62e-34; # Planck's constant in Js\n",
+      "c = 3e8; # Velocity of light in vaccum\n",
+      "E5 = 230*1.6e-19 #  Energy of a particle in Volts in 5th antinode\n",
+      "n = 5;\n",
+      "\n",
+      "#Calculations\n",
+      "E1 = E5/(n**2);\n",
+      "m = (h**2)/(8*E1*a**2);#Mass of electron \n",
+      "\n",
+      "#Result\n",
+      "print 'Mass of electron is %.2f *10^-31 kg'%(m/1e-31)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Mass of electron is 9.30 *10^-31 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.14, Page 316"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "n = 1; # Single particle\n",
+      "a = 50e-10; # Width of box in meter\n",
+      "deltax = 10e-10; # Intervel between particle\n",
+      "\n",
+      "#Calculations\n",
+      "p = (2/a)*deltax;#The probability of finding the particle\n",
+      "\n",
+      "#Result\n",
+      "print 'The probability of finding the particle is %.1f'%p\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The probability of finding the particle is 0.4\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.15, Page 316"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from math import sqrt\n",
+      "\n",
+      "#Varaible declaration\n",
+      "h = 6.62*1e-34; # Planck's constant\n",
+      "m = 1e-9; # Mass of particle in kg\n",
+      "t = 100; #Time reqired by the particle to cross 1 mm distance\n",
+      "a = 1e-3 ; # Width of box in m\n",
+      "v = 1e-5; # Velocity of particle in m/s\n",
+      "\n",
+      "#Calculations\n",
+      "E = (0.5*m*v**2);\n",
+      "n = sqrt(8*m*a**2*E/(h**2));#The quantum state\n",
+      "\n",
+      "#Result\n",
+      "print 'The quantum state is %.f*10^16 '%(n/1e+16)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The quantum state is 3*10^16 \n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.16, Page 317"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Varaible declaration\n",
+      "h = 6.62e-34; # Planck's constant in J.s\n",
+      "m = 9.1e-31 # Mass of electron in kg\n",
+      "nk =1;\n",
+      "nl = 1;\n",
+      "nm = 1;\n",
+      "a = 0.5e-10 # Width of cubical box in meter\n",
+      "\n",
+      "#Calculation\n",
+      "E = (h**2*(nk**2+nl**2+nm**2))/(8*m*a**2*1.6e-19);#The lowest energy level will have energy\n",
+      "\n",
+      "#Result\n",
+      "print 'The lowest energy level will have energy %.f eV'%E\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The lowest energy level will have energy 451 eV\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/README.txt b/Engineering_Physics_by_K._Rajagopal/README.txt
new file mode 100755
index 00000000..8d3f5c57
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Muktesh Chaudhary
+Course: be
+College/Institute/Organization: Anglo Eastern ship management india Pvt. Ltd
+Department/Designation: Electrical & Electronics Officer
+Book Title: Engineering Physics by K. Rajagopal
+Author: K. Rajagopal
+Publisher: PHI Learning Pvt. Ltd.
+Year of publication: 2011
+Isbn: 9788120343405
+Edition: 2nd
+\ No newline at end of file
diff --git a/Engineering_Physics_by_K._Rajagopal/screenshots/hallvtg.png b/Engineering_Physics_by_K._Rajagopal/screenshots/hallvtg.png
new file mode 100755
index 00000000..b13904dc
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/screenshots/hallvtg.png
diff --git a/Engineering_Physics_by_K._Rajagopal/screenshots/miller.png b/Engineering_Physics_by_K._Rajagopal/screenshots/miller.png
new file mode 100755
index 00000000..5522b80a
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/screenshots/miller.png
diff --git a/Engineering_Physics_by_K._Rajagopal/screenshots/polar.png b/Engineering_Physics_by_K._Rajagopal/screenshots/polar.png
new file mode 100755
index 00000000..3c37a0d0
--- /dev/null
+++ b/Engineering_Physics_by_K._Rajagopal/screenshots/polar.png
diff --git a/Heat_And_Thermodynamics/README.txt b/Heat_And_Thermodynamics/README.txt
new file mode 100755
index 00000000..553ba2a8
--- /dev/null
+++ b/Heat_And_Thermodynamics/README.txt
@@ -0,0 +1,10 @@
+Contributed By: pratik gandhi
+Course: bca
+College/Institute/Organization: Cybercom Creation
+Department/Designation: Developer
+Book Title: Heat And Thermodynamics
+Author: A. Manna
+Publisher: Dorling Kindersley (India) Pvt. Ltd., Noida
+Year of publication: 2011
+Isbn: 9788131754009
+Edition: 1
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch10.ipynb b/Heat_And_Thermodynamics/ch10.ipynb
new file mode 100755
index 00000000..f0d6123b
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch10.ipynb
@@ -0,0 +1,420 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 10 : Thermodynamic relations"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.1 pageno: 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 289.6;\t\t\t#temperature in K\n",
+      "dt = 0.0244;\t\t\t#raise in temperature in deg.C\n",
+      "v1 = 0.00095;\t\t\t#volume occupied in liquid state in litres\n",
+      "v2 = 0.00079;\t\t\t#volume occupied in solid state in litres\n",
+      "\n",
+      "# Calculations\n",
+      "l = t*(v1-v2)/dt;\t\t\t#latent heat of fusion in lit.atm\n",
+      "\n",
+      "# Result\n",
+      "print 'the latent heat of fusion is %3.2f lit.atm'%(l)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the latent heat of fusion is 1.90 lit.atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.2 pageno : 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 295.    \t\t\t#temperature of water in K\n",
+      "dp = 10**6.\t    \t\t#cahnge in pressure in dyne/sq.cm\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "dc = -t*10**-5*dp/j;\t\t\t#change in specific heat\n",
+      "\n",
+      "# Result\n",
+      "print 'the change in specific heat is %.e cal/degree'%(dc)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change in specific heat is -7e-05 cal/degree\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.3 pageno: 329"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "cp = 0.0909\t    \t\t#specific heat at consmath.tant pressure in cal/degree\n",
+      "t = 273;\t\t    \t#temperature in K\n",
+      "v = 0.112;\t\t\t    #specific volume in lit/deg.C\n",
+      "a = 5.01*10**(-6);\t\t#coefficient of linear expansion\n",
+      "k = 8*10**-7;\t\t\t#compressibility of copper in per atoms\n",
+      "\n",
+      "# Calculations\n",
+      "cv = cp+(9*a**2*v*t*0.024142*10**3/k);\t\t\t#specific heat at constant volume in cal/deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'specific heat at constant volume is %3.2f cal/deg.C'%(cv)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specific heat at constant volume is 0.30 cal/deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.5 pageno : 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 289.6;\t\t\t#temperature in K\n",
+      "dt = 0.0244;\t\t\t#raise in temperature in deg.C\n",
+      "v1 = 0.00095;\t\t\t#volume occupied in liquid state in litres\n",
+      "v2 = 0.00079;\t\t\t#volume occupied in solid state in litres\n",
+      "\n",
+      "# Calculations\n",
+      "l = t*(v1-v2)/dt;\t\t\t#latent heat of fusion in lit.atm\n",
+      "\n",
+      "# Result\n",
+      "print 'the latent heat of fusion is %3.3f lit.atm'%(l)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the latent heat of fusion is 1.899 lit.atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.6 pageno : 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "l = 539;\t\t\t#latent heat of water at 100deg.C in cal\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "t = 373;\t\t\t#temperature of water in K\n",
+      "v2 = 1670;\t\t\t#volume of steam formed in cc\n",
+      "v1 = 1;\t\t\t#intial volume in cc\n",
+      "g = 981;\t\t\t#acceleration due to gravity in cm/sec**2\n",
+      "d = 13.6;\t\t\t#specific gravity of hg\n",
+      "\n",
+      "# Calculations\n",
+      "dp = l*j/(t*(v2-v1)*g*d);\t\t\t#rate of change of saturation pressure in cm of mercury\n",
+      "\n",
+      "# Result\n",
+      "print 'the rate of change of saturation pressure is %3.1f cm of hg'%(dp)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of change of saturation pressure is 2.7 cm of hg\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.7 pageno : 331"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 77.371;\t\t\t#pressure at 100.5deg.C in cm of hg\n",
+      "p2 = 74.650;\t\t\t#pressure at 99.5deg.C in cm of hg\n",
+      "g = 981;\t\t\t#universal gas constant in cm/sec**2\n",
+      "d = 13.6;\t\t\t#specific gravity\n",
+      "l = 537;\t\t\t#latent heat of vapourisation in cal/gm\n",
+      "t = 373;\t\t\t#temperature of water in K\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "v1 = 1;\t\t\t#intial volume in cc\n",
+      "\n",
+      "# Calculations\n",
+      "v2 = v1+(l*j/(t*(p1-p2)*g*d));\t\t\t#volume of gram of steam at 100deg.C in cc\n",
+      "\n",
+      "# Result\n",
+      "print 'volume of gram of steam at 100deg.C is %.f cc'%(v2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "volume of gram of steam at 100deg.C is 1667 cc\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.8 pageno : 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 350;\t\t\t#boiling point temperature in K\n",
+      "l = 46;\t\t\t#latent heat of vapourisation in cal/gm\n",
+      "v1 = 1/1.6;\t\t\t#intial volume in cc\n",
+      "dp = 2.3;\t\t\t#change in pressure with temperature in cm of hg/deg.C\n",
+      "d = 13.6;\t\t\t#specific gravity of mercury\n",
+      "g = 981;\t\t\t#acceleration due to gravity in cm/sec**2\n",
+      "j = 4.2*10**7;\t\t\t#joukes constant in ergs/cal\n",
+      "\n",
+      "#CALCULTIONS\n",
+      "v2 = v1+(l*j)/(t*dp*d*g);\t\t\t#specific volume in cc\n",
+      "\n",
+      "# Result\n",
+      "print 'specific volume of vapour of carbon is %3.3f cc'%(v2)\n",
+      "print \"Note : Answer is slightly different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "specific volume of vapour of carbon is 180.513 cc\n",
+        "Note : Answer is slightly different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.9 pageno : 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "l = 536.;\t\t\t#latent heat of vapourisation in cal/gm\n",
+      "v1 = 1.;\t\t\t#volume of 1 gm of water in cc\n",
+      "v2 = 1600.;\t\t\t#volume of steam in cc\n",
+      "t = 373.;\t\t\t#boiling point of water in K\n",
+      "p = 1.;\t\t\t#pressure in cm of hg\n",
+      "d = 13.6;\t\t\t#specific gravity of mercury\n",
+      "g = 981.;\t\t\t#gravitational constant in cm/sec**2s/cal\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in erg/cal\n",
+      "\n",
+      "# Calculations\n",
+      "dt = (t*(v2-v1)*d*g)/(l*j);\t\t\t#change in temperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'change in temperature is %3.2f deg.C'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in temperature is 0.35 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.10 page no : 332"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 353;\t\t\t#temperature in K\n",
+      "p = 76*13.6*981;\t\t\t#pressure in dynes/sq.cm\n",
+      "v = 0.146;\t\t\t#specific volume in cc/kg\n",
+      "l = 35.6;\t\t\t#latent heat of fusion in cal/gm\n",
+      "j = 4.18*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "dt = t*p*v/(l*j);\t\t\t#change in melting point per atmosphere\n",
+      "\n",
+      "# Result\n",
+      "print 'the rate of change in melting point is %.3f per atmosphere'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate of change in melting point is 0.035 per atmosphere\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 10.11 pageno : 333"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "l = 79.6*4.18*10**7;\t\t\t#latent heat of water in ergs/gm\n",
+      "t = 273.16;\t\t\t#temperature of water in K\n",
+      "v1 = 1.0001;\t\t\t#specific volume of water at 0deg.C in cc\n",
+      "v2 = 1.0908;\t\t\t#specific volume of ice at 0deg.C in cc\n",
+      "p = 1.013*10**6;\t\t\t#pressure of atmosphere in dyne/sq.cm\n",
+      "\n",
+      "# Calculations\n",
+      "dt = t*(v1-v2)*p/l;\t\t\t#change in freezing point of water in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'change in freezing point of water is %3.4f deg.C'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in freezing point of water is -0.0075 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch11.ipynb b/Heat_And_Thermodynamics/ch11.ipynb
new file mode 100755
index 00000000..083c7e47
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch11.ipynb
@@ -0,0 +1,432 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Conduction of heat"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1 pageno : 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "k = 0.12;\t\t\t#thermal conductivity in cgs unit\n",
+      "t1 = 200;\t\t\t#temperature at one side in deg.C\n",
+      "t2 = 50;\t\t\t#temperature at other side in deg.C\n",
+      "t = 3600;\t\t\t#time in sec\n",
+      "a = 1;\t\t\t#area in sq.cm\n",
+      "t3 = 3;\t\t\t#thickness of the plate in cm\n",
+      "\n",
+      "# Calculations\n",
+      "q = k*a*(t1-t2)*t/t3;\t\t\t#amount of heat conducted in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the amount of heat conducted is %3.2f cal'%(q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the amount of heat conducted is 21600.00 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2 pageno : 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "k = 0.9;\t\t\t#thermal conductivity in cgs unit\n",
+      "a = 10;\t\t\t#area of the copper bar in sq.cm\n",
+      "t1 = 100;\t\t\t#hot side temperature in deg.C\n",
+      "t2 = 20;\t\t\t#cool side temperature in deg.C\n",
+      "d = 25;\t\t\t#thickness of the bar in cm\n",
+      "t3 = 14;\t\t\t#temperature of water when entering in deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "m = k*a*(t1-t2)/(d*(t2-t3));\t\t\t#rate flow of water in gm/sec\n",
+      "\n",
+      "# Result\n",
+      "print 'rate flow of water is %3.2f gm/sec'%(m)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate flow of water is 4.80 gm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3 pageno : 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "i = 1.18;\t\t\t#current in amperes\n",
+      "e = 20;\t\t\t#potential difference across its ends in volts\n",
+      "j = 4.2;\t\t\t#joules constant in joule/cal\n",
+      "a = 2*10**4;\t\t\t#area of the slab in sq.cm\n",
+      "t = 5;\t\t\t#thickness of the plate in cm\n",
+      "t1 = 12.5;\t\t\t#temperature at hot side in K\n",
+      "t2 = 0;\t\t\t#temperature at cold side in k\n",
+      "\n",
+      "# Calculations\n",
+      "k = e*i*t/(j*a*(t1-t2));\t\t\t#thermal conductivity in cgs unit\n",
+      "\n",
+      "# Result\n",
+      "print 'thermal conductivity of slab is %3.5f cgs unit'%(k)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thermal conductivity of slab is 0.00011 cgs unit\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4 pageno : 375"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "l = 30.;\t\t\t#length of the tube in cm\n",
+      "t = 100.;\t\t\t#temperature at outside in deg.C\n",
+      "t1 = 40.;\t\t\t#tempertaure of water when leaving tube in deg.C\n",
+      "t2 = 20.;\t\t\t#temperature of water when entering tube in deg.C\n",
+      "m = 165./60\t\t\t#mass flow rete of water in cc/sec\n",
+      "r1 = 6.;\t\t\t#internal radii in mm\n",
+      "r2 = 8.;\t\t\t#external radii in mm\n",
+      "\n",
+      "# Calculations\n",
+      "k = m*(t1-t2)*math.log(r2/r1)/(2*3.14*l*(t-((t1+t2)/2)));\t\t\t#thermal conductivity in cgs unit\n",
+      "\n",
+      "# Result\n",
+      "print 'thermal conductivity of the tube is %3.4f cgs unit'%(k)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thermal conductivity of the tube is 0.0012 cgs unit\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5 pageno : 376"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "l1 = 1.9;\t\t\t#length of the first bar in cm\n",
+      "l2 = 5; \t\t\t#length of the second bar in cm\n",
+      "k2 = 0.92;\t\t\t#thermal conductivity in cgs unit\n",
+      "\n",
+      "# Calculations\n",
+      "k1 = k2*(l1/l2)**2;\t\t\t#thermal conductivity if first bar in cgs unit\n",
+      "\n",
+      "# Result\n",
+      "print 'thermal conductivity of first bar is %3.3f cgs unit'%(k1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thermal conductivity of first bar is 0.133 cgs unit\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6 pageno : 376"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "k1 = 0.92;\t\t\t#thermal conductivity of copper in cgs unit\n",
+      "k2 = 0.5;\t\t\t#thermal conductivity of alluminium in cgs unit\n",
+      "t1 = 100;\t\t\t#temperature of copper in deg.C\n",
+      "t2 = 0;\t\t\t#temperature of alluminium in deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "t = k1*t1/(k1+k2);\t\t\t#welded teperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'welded temperature is %3.1f deg.C'%(t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "welded temperature is 64.8 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7 pageno : 376"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "w = 23;\t\t\t#thermal capacity of calorimeter in cal\n",
+      "m = 440;\t\t\t#mass of water in gm\n",
+      "l = 14.6;\t\t\t#lenght of the rubber tube in cm\n",
+      "dt = 0.019;\t\t\t#rate of change in temperature in deg.C/sec\n",
+      "t = 100;\t\t\t#temperature of steam in deg.C\n",
+      "t1 = 22;\t\t\t#temperature of the water in deg.C\n",
+      "t2 = t1;\t\t\t#temperature of calorimeter in deg.C\n",
+      "r1 = 1;\t\t\t#external radii in cm\n",
+      "r2 = 0.75;\t\t\t#internal radii in cm\n",
+      "\n",
+      "# Calculations\n",
+      "k = (w+m)*dt*math.log(r1/r2)/(2*3.14*l*(t-((t1+t2)/2)));\t\t\t#thermal conductivity in cgs unit\n",
+      "\n",
+      "# Result\n",
+      "print 'thermal cnductivity of rubber tube is %3.6f cgs unit'%(k)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "thermal cnductivity of rubber tube is 0.000354 cgs unit\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8 pageno : 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "ti = 18;\t\t\t#inside temperature in deg.C\n",
+      "to = 4;\t\t\t#outside temperature in deg.C\n",
+      "k1 = 0.008;\t\t\t#thermal conductivity of stone in cgs unit\n",
+      "k2 = 0.12;\t\t\t#thermal conductivity of steel in cgs unit\n",
+      "t = 3600;\t\t\t#time in sec\n",
+      "t1 = 25;\t\t\t#thickness of the stone in cm\n",
+      "t2 = 2;\t\t\t#thickness of the steel in cm\n",
+      "a = 10**4;\t\t\t#area of the cottage in sq.cm\n",
+      "\n",
+      "# Calculations\n",
+      "q1 = k1*a*(ti-to)*t/(t1);\t\t\t#heat lost by stone per hour in cal\n",
+      "q2 = k2*a*(ti-to)*t/t2;\t\t\t#heat lost by steel per hour in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'heat lost by stone is %3.4e cal  \\\n",
+      "\\nheat lost by steel is %.3e cal'%(q1,q2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "heat lost by stone is 1.6128e+05 cal  \n",
+        "heat lost by steel is 3.024e+07 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9 pageno: 377"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "l1 = 4;\t\t\t#length of the slab1 in cm\n",
+      "l2 = 2;\t\t\t#length of the slab2 in cm\n",
+      "k1 = 0.5;\t\t\t#thermal conductivity in cgs unit\n",
+      "k2 = 0.36;\t\t\t#thermal conductivity in cgs unit\n",
+      "t1 = 100;\t\t\t#temperature of the slab1 in deg.C\n",
+      "t2 = 0;\t\t\t#temperature of the slab2 in deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "t = k1*l2*t1/((k2*l1)+(k1*l2));\t\t\t#temperature of the commaon surface in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of the common surface is %3.0f deg.C'%(t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of the common surface is  41 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.10 pageno : 378"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The distance\n",
+      "\n",
+      "# Variables\n",
+      "t1 = 15.;\t\t\t#temperature of the one end of the slab in deg.C\n",
+      "t2 = 45.;\t\t\t#temperature of the other end of the slab in deg.C\n",
+      "k = 0.3;\t\t\t#thermal conductivity in cgs unit\n",
+      "d = 7.;\t\t\t#density of the material in gm/cc\n",
+      "cp = 1.;\t\t\t#specific heat of the material in kj/kg.K\n",
+      "t = 5.*3600;\t\t\t#time in sec\n",
+      "dt = 1./10;\t\t\t#thermometer reading in deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "b = (3.14*d*cp/(t*k))**(0.5);\n",
+      "x = (math.log((t2-t1)/dt))/b;\t\t\t#distance from which temparature variation can be detected in cm\n",
+      "\n",
+      "# Result\n",
+      "print 'the distance from which temparature variation can be detected is %3.1f cm'%(x)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the distance from which temparature variation can be detected is 89.4 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch12.ipynb b/Heat_And_Thermodynamics/ch12.ipynb
new file mode 100755
index 00000000..1281c2d6
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch12.ipynb
@@ -0,0 +1,371 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 12 : Radiation"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.1 pageno : 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 300;\t\t\t#temperature of the surroundings in K\n",
+      "t2 = 900;\t\t\t#temperature of the hot body p in K\n",
+      "t3 = 500;\t\t\t#temperature of the hot body q in K\n",
+      "a = 5.67*10**-8;\t\t\t#stefan boltzmann consmath.tant in W/m**2.K**4\n",
+      "\n",
+      "# Calculations\n",
+      "q1 = a*(t2**4-t1**4);\t\t\t#heat lost from hot body p in w/m**2\n",
+      "q2 = a*(t3**4-t1**4);\t\t\t#heat lost from hot body q in w/m**2\n",
+      "q = q1/q2;\t\t\t#ratio of heat lost from two subsmath.tances\n",
+      "\n",
+      "# Result\n",
+      "print 'ratio of heat lost from two substances is %3.1f/1'%(q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of heat lost from two substances is 11.9/1\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.2 pageno : 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 573;\t\t\t#temperature of the hot side in K\n",
+      "t2 = 273;\t\t\t#temperature of the coll side in K\n",
+      "m = 82;\t\t\t#mass of the black body in gm\n",
+      "cp = 0.1;\t\t\t#specific heat of the black body kj/kg.K\n",
+      "dt = 0.35;\t\t\t#ice melting at a rate of temperature in deg.C/sec\n",
+      "a = 8;\t\t\t#area of black body in sq.cm\n",
+      "\n",
+      "# Calculations\n",
+      "s = m*cp*dt/(a*(t1**4-t2**4));\t\t\t#boltzmann constant in cal/sq.cm/sec/deg**4\n",
+      "\n",
+      "# Result\n",
+      "print 'boltzmann consmath.tant is %.2e cal/sq.cm/sec/deg**4'%(s)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "boltzmann consmath.tant is 3.51e-12 cal/sq.cm/sec/deg**4\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.3 pageno : 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "r1 = 60.;\t\t\t#distance of first black body in cm\n",
+      "r2 = 30.;\t\t\t#distance of second black body in cm\n",
+      "t1 = 873.;\t\t\t#temperature of first black body in K\n",
+      "t2 = 573.;\t\t\t#temperature of the second black body in K\n",
+      "\n",
+      "# Calculations\n",
+      "i = (t2**4/t1**4)*(r1**2/r2**2);\t\t\t#ratio of intensity of radition\n",
+      "\n",
+      "# Result\n",
+      "print 'ratio of intensity of radition is %3.2f'%(i)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "ratio of intensity of radition is 0.74\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.4 pageno : 445"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 1373;\t\t\t#temperature of the sphere in K\n",
+      "t2 = 283;\t\t\t#temperature of the black body in K\n",
+      "r = 4.17*10**5;\t\t\t#rate of heat radiate in ergs/sq.cm/sec\n",
+      "a = 4*3.14*(6**2);\t\t\t#surface area of the sphere in sq.cm\n",
+      "\n",
+      "\n",
+      "tr = r*a*(t1**4/t2**4)*(2.39005736*10**(-8));\t\t\t#total heat radiated in cal/sec\n",
+      "\n",
+      "# Result\n",
+      "print 'total heat radiated is %3.1f cal/sec'%(tr)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total heat radiated is 2496.6 cal/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.5 pageno : 446"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "h = 2*3.14*100;\t\t\t#heat received by the lens per min in  cal\n",
+      "m = 25.;\t\t\t#mass of the ice in gm\n",
+      "l = 80;\t\t\t#latent heat of ice in cal/gm\n",
+      "\n",
+      "# Calculations\n",
+      "t = m*l/h;\t\t\t#time for which the sun rays falls in min\n",
+      "\n",
+      "# Result\n",
+      "print 'time for which the sun rays falls is %3.3f min'%(t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "time for which the sun rays falls is 3.185 min\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.6 page no : 446"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "d = 0.35;\t\t\t#diameter of the mirror in m\n",
+      "t = 5;\t\t\t#time in min\n",
+      "T = 16;\t\t\t#temperature of water found to be in deg.C\n",
+      "m = 60;\t\t\t#mass of water in gm\n",
+      "mc = 30;\t\t\t#mass of calorimeter in gm\n",
+      "cp = 0.1;\t\t\t#specific heat of copper in cal/gm/deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "q = (m+cp*mc)*T*4/(5*3.14*d**2);\t\t\t#amount of heat received by earth in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'amount of heat received by earth is %3.f cal'%(q)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "amount of heat received by earth is 2096 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.7 pageno : 446"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "r1 = 5.;\t\t\t#radius of first sphere in cm\n",
+      "r2 = 10.;\t\t\t#radius of second sphere in cm\n",
+      "t1 = 700.;\t\t\t#temperature of the first sphere in K\n",
+      "t2 = 500.;\t\t\t#temperature of the second sphere in K\n",
+      "t = 300.;\t\t\t#temperature of the enclousure in K\n",
+      "\n",
+      "# Calculations1\n",
+      "dc = (r2/r1)*(t1**4-t**4)/(t2**4-t**4);\t\t\t#ratio of c1/c2\n",
+      "r = r1**3*dc/r2**3;\t\t\t#rate of heat loss\n",
+      "\n",
+      "# Result\n",
+      "print 'rate of loss of heat is %3.3f'%(r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "rate of loss of heat is 1.066\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.8 pageno : 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "t1 = 600.;\t\t\t#temperature of the black body in K\n",
+      "t0 = 300.;\t\t\t#temperature of the surroundings in K\n",
+      "d = 6.;\t\t\t#deflections in galvanometer\n",
+      "d1 = 400.;\t\t\t#deflection in divisions\n",
+      "\n",
+      "# Calculations\n",
+      "dt = (d1/d)*(t1**4-t0**4);\t\t\t#change of temperature\n",
+      "t2 = (dt+t0**4)**(1./4);\t\t\t#end temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'end temperature of the temperature is %3.2f K'%(t2)\n",
+      "print \"Note : answer in book in incorrect. Please calculate manually.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "end temperature of the temperature is 1687.45 K\n",
+        "Note : answer in book in incorrect. Please calculate manually.\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 12.9 pageno : 447"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 17000;\t\t\t#luminosity of star compared to sun\n",
+      "t = 6000;\t\t\t#temperature of the sun in K\n",
+      "\n",
+      "# Calculations\n",
+      "t1 = (n*t**4)**(1./4);\t\t\t#temperature of the star in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of the star is %3.f K'%(round(t1,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of the star is 68510 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch13.ipynb b/Heat_And_Thermodynamics/ch13.ipynb
new file mode 100755
index 00000000..daac0e65
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch13.ipynb
@@ -0,0 +1,176 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 13 : Introduction to statistical thermodynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.1 page no : 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 1./6;\t\t\t#probability for the first throw gives 6\n",
+      "p2 = 1./6;\t\t\t#probability for the first throw gives 5\n",
+      "n = 2;\t    \t\t#the no.of dice are two\n",
+      "\n",
+      "# Calculations\n",
+      "p = p1*p2*n;\t\t\t#the required probability is\n",
+      "\n",
+      "# Result\n",
+      "print 'the required probability is %3.2f'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the required probability is 0.06\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.2 pageno : 474"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 4./52;\t\t\t#the probability for getting ace in first draw is\n",
+      "p2 = 3./51;\t\t\t#the probability for getting ace in second draw is\n",
+      "p3 = 2./50;\t\t\t#the probability for getting ace in third draw is\n",
+      "p4 = 1./49;\t\t\t#the probability for getting ace in fourth draw is\n",
+      "\n",
+      "# Calculations\n",
+      "p = p1*p2*p3*p4;\t\t\t#total probability is\n",
+      "\n",
+      "# Result\n",
+      "print 'total probability is %3.9f'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "total probability is 0.000003694\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.3 page no : 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 12.;\t\t\t#no.of particles\n",
+      "n1 = 8.;\n",
+      "n2 = 4.;\n",
+      "\n",
+      "# Calculations\n",
+      "p = n*(n-1)*(n-2)*(n-3)/(n2*(n2-1)*(n2-2)*(2**n));\t\t\t#probability of distribution (8,4)\n",
+      "\n",
+      "# Result\n",
+      "print 'probability of distribution 8(4) is %3.5f'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "probability of distribution 8(4) is 0.12085\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 13.4 page no: 475"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 32;\t\t\t#mass of the oxygen molecule in gm\n",
+      "n = 1.67*10**-27;\t\t\t#mass of one electron\n",
+      "k = 1.38*10**-23;\t\t\t#boltzzmann consmath.tant in ergs/cal\n",
+      "t = 200;\t\t\t#temperature of the oxygen in K\n",
+      "c = (100.+101)/2;\t\t\t#average speed of the oxygen molecule in m/s\n",
+      "\n",
+      "# Calculations\n",
+      "a = m*n/(2*3.14*k*t);\n",
+      "p = 4*3.14*(a**(3./2))*(c**2)*(2.303**(-a));\t\t\t#probability that the oxygen speed is lies between in m/sec\n",
+      "\n",
+      "# Result\n",
+      "print 'probability that the oxygen speed is lies between is %3.6e m/sec'%(p)\n",
+      "print \"Note : answer is slightly different because of rounding error.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "probability that the oxygen speed is lies between is 6.867794e-04 m/sec\n",
+        "Note : answer is slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch2.ipynb b/Heat_And_Thermodynamics/ch2.ipynb
new file mode 100755
index 00000000..b2e44ac0
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch2.ipynb
@@ -0,0 +1,229 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Thermometry"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1 pageno : 29"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "li = 1.23;\t\t\t#length of melting ice in mm\n",
+      "lf = 18.56;\t\t\t#length of melting ice reading in pressure of 74.24cm of mercury in mm\n",
+      "l = 10.75;\t\t\t#length of melting ice at which temperature to be calculated\n",
+      "mp = 0;\t\t\t#melting point in deg.C\n",
+      "T = 50;\t\t\t#temperature of melting ice at which length to be calculated in deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "sp = 100-(76-74.24)/(2.7);\t\t\t#76cm of mercury steam point is 100 deg.C so at 74.24cm of mercury the steam point in deg.C\n",
+      "t = (l-li)*(sp-mp)/(lf-li);\t\t\t#temperature at 10.75mm of melting ice in deg.C\n",
+      "lt = ((T*(lf-li))/(sp-mp))+li;\t\t\t#length of ice at 50 deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of melting ice at 10.75mm of hg is %3.2f deg.C \\\n",
+      " \\nthe length of ice corresponding to 50 deg.C is %3.2f mm of mercury'%(t,lt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of melting ice at 10.75mm of hg is 54.58 deg.C  \n",
+        "the length of ice corresponding to 50 deg.C is 9.95 mm of mercury\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2 pageno : 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 23.5;\t\t\t#pressure when immersed in liquid air in cm\n",
+      "p2 = 75.;\t\t\t#pressure when immersed in ice in cm\n",
+      "p3 = 102.4;\t\t\t#pressure when immersed in steam in cm\n",
+      "T = 100.;\t\t\t#boiling point of temperature in deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "t = (p1-p2)*T/(p3-p2);\t\t\t#temperature of the liquid air in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of liquid of air is %3.2f deg.C'%(t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of liquid of air is -187.96 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.3 pageno : 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 283.;\t\t\t#temperature of bulb when pressure is h-2cm of hg in k\n",
+      "t2 = 546.;\t\t\t#temperature of bulb when pressure is h-22cm of hg in k\n",
+      "h1 = 2.;\t\t\t#differnce of mercury level at 283k in cm\n",
+      "h2 = 22.;\t\t\t#differnce of mercury level at 546k in cm\n",
+      "\n",
+      "# Calculations\n",
+      "h = ((h2*t1)+(h1*t2))/(t2-t1);\t\t\t#height of the barometer in cm\n",
+      "\n",
+      "# Result\n",
+      "print 'height of the barometer is %3.2f cm'%(h)\n",
+      "\n",
+      "print \"Note : Answer in book is wrong. Please calculate manually and check.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "height of the barometer is 27.83 cm\n",
+        "Note : Answer in book is wrong. Please calculate manually and check.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Example 2.4 pageno : 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p0 = 76.;\t\t\t#pressure at 0 deg.C in cm of hg\n",
+      "p1 = 228.;\t\t\t#pressure (76+152) at T deg.C in cm of hg\n",
+      "t0 = 273.;\t\t\t#temperature of bulb in K\n",
+      "\n",
+      "# Calculations\n",
+      "T = p1*t0/p0;\t\t\t#temperature at 228 cm of hg pressure in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of bulb is %3.2f K'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of bulb is 819.00 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.5 page no : 30"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 0;\t\t\t    #temperature in deg.C\n",
+      "t2 = 100;\t\t\t#temperature in deg.C\n",
+      "t3 = 208;\t\t\t#temperature in deg.C\n",
+      "r1 = 3.5;\t\t\t#resistance in ohms\n",
+      "r2 = 5.2;\t\t\t#resistance in ohms\n",
+      "r3 = 6.9;\t\t\t#resistance in ohms\n",
+      "r4 = 9.4;\t\t\t#resistance in ohms\n",
+      "\n",
+      "# Calculations\n",
+      "tpt = (r3-r1)*100/(r2-r1);\t\t\t#temperature in deg.C\n",
+      "d = round((t3-tpt)/(2.08*1.08),2);\t\t\t#deflection\n",
+      "tp = round((r4-r1)*100/(r2-r1),2);\t\t\t#temperature in deg.C\n",
+      "t6 = (3.5*(((tp/100)**2)-tp/100))+tp;\t\t\t#temperature in deg.C\n",
+      "t7 = (3.5*(((t6/100)**2)-t6/100))+tp;\t\t\t#temperature in deg.C\n",
+      "t8 = (3.5*(((t7/100)**2)-t7/100))+tp;\t\t\t#temperature in deg.C\n",
+      "t9 = (3.5*(((t8/100)**2)-t8/100))+tp;\t\t\t#temperature in deg.C\n",
+      "\n",
+      "#Result\n",
+      "print 'the temperature of the bath is %3.2f deg.C'%(t9)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of the bath is 385.50 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch3.ipynb b/Heat_And_Thermodynamics/ch3.ipynb
new file mode 100755
index 00000000..8becd279
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch3.ipynb
@@ -0,0 +1,941 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 3 : The mechanical equivalent of  heat"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.1 pageno : 44"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 20;\t\t\t#calorimeter of water equivalent in gm\n",
+      "n = 1030;\t\t\t#weight of water in gm\n",
+      "p = 2;\t\t\t#no.of paddles\n",
+      "a = 10;\t\t\t#weight of each paddle in kg\n",
+      "s = 80;\t\t\t#dismath.tance between paddles in m\n",
+      "g = 980;\t\t\t#accelaration due to gravity in cm/sec**2\n",
+      "\n",
+      "# Calculations\n",
+      "E = (p*a*1000*g*s*100);\t\t\t#potential energy in dyne cm\n",
+      "T = (E)/(1050*4.18*10**7);\t\t\t#rise in temperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the rise in temperature of water is %3.2f deg.C'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rise in temperature of water is 3.57 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.2 pageno : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "cp = 0.1;\t\t\t#specific heat of copper in kj/kg-K\n",
+      "w = 120;\t\t\t#weight of copper calorimeter in gm\n",
+      "a = 1400;\t\t\t#weight of paraffin oil in gm\n",
+      "cp1 = 0.6;\t\t\t#specific of parafin oil in kj/kg-K\n",
+      "b = 10**8;\t\t\t#force to rotate the paddle in dynes\n",
+      "T = 16;\t\t\t#rise in temperature in deg.C\n",
+      "n = 900;\t\t\t#no.of revolutions stirred \n",
+      "pi = 3.14;\t\t\t#value of pi\n",
+      "\n",
+      "# Calculations\n",
+      "c = 2*pi*b;\t\t\t#work done by a rotating paddle per rotation in dyne cm per rotation\n",
+      "d = c*n;\t\t\t#total work done in dyne cm \n",
+      "hc = w*cp*16;\t\t\t#heat gained by calorimeter in calories\n",
+      "hp = a*cp1*16;\t\t\t#heat gaained by paraffin oil in calories \n",
+      "J = d/(hc+hp);\t\t\t#mecanical equivalent of heat in erg/cal\n",
+      "\n",
+      "# Result\n",
+      "print 'mecanical equivalent of heat is %.2e erg/cal'%(J)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mecanical equivalent of heat is 4.15e+07 erg/cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.3 pageno : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "cp = 0.12;\t\t\t#specific heat of iron in kj/kg-K\n",
+      "m = 25;\t\t\t#mass of iron in lb\n",
+      "h = 0.4;\t\t\t#horse power developed in 3 min\n",
+      "t = 3;\t\t\t#time taken to develop the horse power in min\n",
+      "T = 17;\t\t\t#raise in temp in deg.C\n",
+      "\n",
+      "# Calculations\n",
+      "w = h*33000*t;\t\t\t#total work done in ft-lb\n",
+      "H = m*cp*T;\t\t\t#aount of heat developed in B.Th.U\n",
+      "J = (w)/H;\t\t\t#the value of mechanical equivalent of heat\n",
+      "\n",
+      "# Result\n",
+      "print 'the mechanical equivalent of water is %3.1f ft-lb/B.Th.U'%(J)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mechanical equivalent of water is 776.5 ft-lb/B.Th.U\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.4 pageno : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "n = 2.;\t\t\t#no.of lead blocks\n",
+      "m = 210.;\t\t\t#mass of each lead block in gm\n",
+      "v = 20000.;\t\t\t#velocity of block relative to earth in cm/sec\n",
+      "J = 4.2*10**7;\t\t\t#mechanical equivalent of heat in ergs/calorie\n",
+      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg-K\n",
+      "\n",
+      "# Calculations\n",
+      "E = (m*v**2)/2;\t\t\t#kinetic energy of each block in ergs\n",
+      "E2 = n*E;\t\t\t#total kinetic energy in ergs\n",
+      "T = E2/(J*m*n*cp);\t\t\t#mean rise in temperature in T\n",
+      "\n",
+      "# Result\n",
+      "print 'the mean rise in temperature is %3.1f deg.C'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mean rise in temperature is 158.7 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.5 pageno : 45"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "h = 150;\t\t\t#height froom which ball fallen in ft\n",
+      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg-K\n",
+      "J = 778;\t\t\t#mechanical equivalent of heat in ft lb/B.Th.U\n",
+      "\n",
+      "# Calculations\n",
+      "#work done in falling is equal to heat absorbed by the ball\n",
+      "T = 160./(J*cp)*(5./9);\t\t\t#the raise in temperature in T\n",
+      "\n",
+      "# Result\n",
+      "print 'the raise in temperature is %3.1f deg.C'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the raise in temperature is 3.8 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.6 pageno : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "# Variables \n",
+      "w = 26.6;\t\t\t#work done one horse in to raise the temperature in lb\n",
+      "T1 = 32.;\t\t\t#temperature at initial in deg.F\n",
+      "T2 = 212.;\t\t\t#temperature at final in deg.F\n",
+      "t = 2.5;\t\t\t#time to raise the tmperature in hrs\n",
+      "p = 25.;\t\t\t#percentage of heat lossed \n",
+      "\n",
+      "# Calculations\n",
+      "#only 75% of heat is utillised\n",
+      "x = w*180.*100.*778./((100-p)*150);\t\t\t#the rate at which horse worked\n",
+      "\n",
+      "# Result\n",
+      "print 'the rate at which horse worked is %3.0f ft-lb wt/min'%(x)\n",
+      "print \"Note : Answer in book is rounded off, Please calculate manually. This answer is accurate.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rate at which horse worked is 33112 ft-lb wt/min\n",
+        "Note : Answer in book is rounded off, Please calculate manually. This answer is accurate.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.7 pageno : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "l = 100.;\t\t\t#length of glass tube in cm\n",
+      "m = 500.;\t\t\t#mass of mercury in glass tube in gm\n",
+      "n = 20.;\t\t\t#number of times inverted i succession\n",
+      "cp = 0.03;\t\t\t#specific heat of mercury in cal/gm/deg.C\n",
+      "J = 4.2;\t\t\t#joule's equivalent in j/cal\n",
+      "g = 981.;\t\t\t#accelaration due to gravity in cm/s**2\n",
+      "\n",
+      "# Calculations\n",
+      "PE = m*g*l;\t\t\t#potential energy for each time in ergs\n",
+      "TE = PE*n;\t\t\t#total loss in ergs\n",
+      "T = TE/(m*cp*J*10**7);\t\t\t#rise in temperature in deg.C\n",
+      "#if T is the rise in temperature,then heat devoloped is m*cp*T\n",
+      "\n",
+      "# Result\n",
+      "print 'the rise in temperature is %3.2f deg.C'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rise in temperature is 1.56 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.8 page no : 46"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables \n",
+      "d = 0.02;\t\t\t#diameter of the copper wire in cm\n",
+      "i = 1;\t\t\t#current in amp\n",
+      "T = 100;\t\t\t#maximum steady temperature in deg.C\n",
+      "r = 2.1;\t\t\t#resistance of the wire in ohm cm\n",
+      "J = 4.2;\t\t\t#mechanical equivalent of heat in j/cal\n",
+      "a = 3.14*d**2/4;\t\t\t#area of the copper wire in sq.cm\n",
+      "a2 = 1;\t\t\t#area of the copper surface in sq.cm\n",
+      "\n",
+      "# Calculations \n",
+      "l = 1/(2*3.14*d/2);\t\t\t#length corresponding to the area in cm\n",
+      "R = r*l/a;\t\t\t#resistance of the copper wire in ohm\n",
+      "w = R*a2**2;\t\t\t#work done in joule\n",
+      "h = w/J;\t\t\t#heat devoleped in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the heat developed is %.f calories'%(round(h,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the heat developed is 25360 calories\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.9 pageno: 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "h = 10000;\t\t\t#vertical height of water fall in cm\n",
+      "v = 5;\t\t\t    #volume disharged per sec in litres\n",
+      "J = 4.18;\t\t\t#joule's constant in j/cal\n",
+      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
+      "\n",
+      "# Calculations\n",
+      "m = v*1000;\t\t\t#mass of water disharged per sec in gm\n",
+      "w = m*h*g;\t\t\t#work done in falling through 100m in erg\n",
+      "H = (v*10**7 *g)/(J*10**7);\t#quantity of heat produced in cal\n",
+      "T = H/m;\t\t\t#rise in temperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the quantity of heat produced is %3f cal  \\\n",
+      "\\nthe rise in temperature is %3.2f deg.C'%(H,T)\n",
+      "\n",
+      "print \"Note : Answer for part A in book is wrong. Please calculate manually.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the quantity of heat produced is 1173.444976 cal  \n",
+        "the rise in temperature is 0.23 deg.C\n",
+        "Note : Answer for part A in book is wrong. Please calculate manually.\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.10 page no : 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables \n",
+      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg.k\n",
+      "v = 10000;\t\t\t#initial velocity of bullet in cm/sec\n",
+      "J = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "ke = (v**2)/2;\t\t\t#kinetic energy of the bullet per unit mass in (cm/sec)**2\n",
+      "T = ke*95/(cp*J*100);\t\t\t#rise in temperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the rise in temperature is %3.1f deg.C'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rise in temperature is 37.7 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.11 page no : 47"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "h = 5000.;\t\t\t#height of the niagara falls in cm\n",
+      "J = 4.2*10**7;\t\t#joules constant in ergs per cal\n",
+      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "w = h*g;\t\t\t#work done per unit mass in ergs/gn\n",
+      "T = w/J;\t\t\t#rise in temperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the rise in temperature is %3.2f deg.C'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rise in temperature is 0.12 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.12 page no : 48\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables \n",
+      "E1 = 3.75;\t\t\t#potential difference in v\n",
+      "E2 = 3.;\t\t\t#potential differnce in v\n",
+      "i1 = 2.5;\t\t\t#current in amp\n",
+      "i2 = 2;\t\t\t    #current in amp\n",
+      "T = 2.7;\t\t\t#the rise in temperature of the water in deg.C\n",
+      "m1 = 48.;\t\t\t#water flow rate at 3 volts in gm/min\n",
+      "m2 = 30.;\t\t\t#water flow rate at 3.75volts in gm/min\n",
+      "s = 1;\t\t\t    #specific heat of the water kj/kg-K\n",
+      "\n",
+      "# Calculations\n",
+      "J = (E1*i1-E2*i2)/(s*T*(m1-m2)/60);\t\t\t#the mechanical equivalent in j/cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the mechanical equivalent is %3.3f j/cal'%(J)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mechanical equivalent is 4.167 j/cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.13 page no : 48"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables \n",
+      "R = 64*10**7;\t\t\t#mean radius of the earth in cm\n",
+      "cp = 0.15;\t\t\t#specific heat of earth in kj/kg-K\n",
+      "J = 4.2*10**7;\t\t\t#joules consmath.tant in erg/cal\n",
+      "\n",
+      "# Calculations\n",
+      "i = 2./5*R**2;\t\t\t#moment of inertia of the earth per unit mass in joules\n",
+      "w = (2*3.14)/(24*60*60);\t\t\t#angular velocity of the earth in rad/sec\n",
+      "T = (i*w**2)/(2*J*cp);\t\t\t#rise in temperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the rise in the temperature is %.1f deg C'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rise in the temperature is 68.7 deg C\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.14 page no : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "cp = 1.25;\t\t\t#specific heat of helium inkj/kg-K\n",
+      "v = 1000;\t\t\t#volume of the gas in ml\n",
+      "w = 0.1785;\t\t\t#mass of the gas at N.T.P in gm\n",
+      "p = 76*13.6*981;\t#pressure of the gas at N.T.P in dynes\n",
+      "T = 273;\t\t\t#temperature at N.T.P in K\n",
+      "\n",
+      "# Calculations\n",
+      "V = 1000/w;\t\t\t#volume occupied by the 1gm of helium gas in cc\n",
+      "cv = cp/1.66;\t\t#specific heat at constant volume it is monatomuc gas kj/kg-K\n",
+      "r = p*V/T;\t\t\t#gas constant in cm**3.atm./K.mol\n",
+      "J = r/(cp-cv);\t\t#mechanical equivalent of heat in erg/cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the mechanical equivalent of heat is %.2e ergs/calories'%(J)\n",
+      "print \"Note: answer slightly different because of rounding error.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mechanical equivalent of heat is 4.19e+07 ergs/calories\n",
+        "Note: answer slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "\n",
+      "Example 3.15 pageno : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables \n",
+      "n = 1./273;     \t\t\t#coefficent of expaaansion of air\n",
+      "a = 0.001293;\t    \t\t#density of air in gm/cc\n",
+      "cp = 0.2389;\t\t    \t#specific heat at consmath.tant pressure in kj/kg.K\n",
+      "p = 76*13.6*981;\t\t\t#pressure at 0 deg.C in dynes\n",
+      "\n",
+      "# Calculations\n",
+      "J = (p*n)/(a*(cp-(cp/1.405)));\t\t\t#mechanical equivalent of heat\n",
+      "\n",
+      "# Result\n",
+      "print 'mechanical equivalent of heat is %.2e ergs/cal'%(J)\n",
+      "print \"Note: answer slightly different because of rounding error.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mechanical equivalent of heat is 4.17e+07 ergs/cal\n",
+        "Note: answer slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.16 pageno : 49"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "# Variables \n",
+      "r = 120./60;\t\t\t#rate of flow of water in gm/sec\n",
+      "T1 = 27.30;\t\t\t#temperature at initial in deg.C\n",
+      "T2 = 33.75;\t\t\t#temperature at final in deg.C\n",
+      "v = 12.64;\t\t\t#potential drop in volts\n",
+      "s = 1.; \t\t\t#specific heat of water in kj/kg-K\n",
+      "i = 4.35;\t\t\t#current through the heating element in amp\n",
+      "\n",
+      "# Calculations\n",
+      "J = (v*i)/(r*s*(T2-T1));\t\t\t#the mechanical equivalent of heat in joule/calorie\n",
+      "\n",
+      "# Result\n",
+      "print 'the mechanical equivalent of heat is %3.2f j/cal'%(J)\n",
+      "print \"Note: answer slightly different because of rounding error.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mechanical equivalent of heat is 4.26 j/cal\n",
+        "Note: answer slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.17 page no : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables \n",
+      "cp = 6.865;\t\t\t#molar specific heat of hydrogen at consmath.tant pressure in kj/kg-K\n",
+      "cv = 4.880;\t\t\t#molar specific heat of hydrogen at consmath.tant volume in kj/kg-K\n",
+      "p = 1.013*10**6;\t\t\t#atmospheric pressure in dynes/cm**2\n",
+      "v = 22.4*10**3;\t\t\t#gram molar volume in ml\n",
+      "T = 273;\t\t\t#temperature at N.T.P in kelvins\n",
+      "\n",
+      "# Calculations\n",
+      "J = (p*v)/(T*(cp-cv));\t\t\t#mechanical equivalent of heat\n",
+      "\n",
+      "# Result\n",
+      "print 'the mechanical equivalent of heat is %.2e ergs/cal'%(J)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mechanical equivalent of heat is 4.19e+07 ergs/cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.18 page no : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math \n",
+      "# Variables\n",
+      "v = 1000.;\t\t\t#volume of hydrogen in ml\n",
+      "t = 273.;\t\t\t#tempature of hydrogen in kelvin\n",
+      "p = 76.;\t\t\t#pressure of hydrogen in mm of hg\n",
+      "w = 0.0896;\t\t\t#weigh of hydrogen in gm\n",
+      "cp = 3.409;\t\t\t#specific heat of hydogen in kj/kg-K\n",
+      "cv = 2.411;\t\t\t#specific heat of hydrogen in kj/kg-K\n",
+      "g = 981.;\t\t\t#accelaration due to gravity in cm/sec**2\n",
+      "a = 13.6;\t\t\t#density of mercury in gm/cm**2\n",
+      "\n",
+      "# Calculations\n",
+      "J = (p*v*g*a)/(w*t*(cp-cv));\t\t\t#mechanical equivalent of heat in ergs/cals\n",
+      "\n",
+      "# Result\n",
+      "print 'mechanical equivalent of heat is %.2e ergs/calorie'%(J)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mechanical equivalent of heat is 4.15e+07 ergs/calorie\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.19 page no : 50"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "cp = 0.23;\t\t\t#specific heat at constant pressure in kj/kg-K\n",
+      "a = 1.18;\t\t\t#density of air in gm/lit\n",
+      "J = 4.2*10**7;\t\t\t#mechanical equivalent of heat in ergs/cal\n",
+      "t = 300;\t\t\t#temperature of air in kelvin\n",
+      "p = 73*13.6*981;\t\t\t#pressure of air in dynes\n",
+      "\t\t\t#cp-cv = (r/J) = pv/(tj)\n",
+      "\n",
+      "#CALCULATON\n",
+      "cv = cp-(p*1000/(a*t*J));\t\t\t#specific heat at constant volume in calories\n",
+      "\n",
+      "# Result\n",
+      "print 'the specific heat at constant volume is %.4f calories'%(cv)\n",
+      "print \"Note: answer slightly different because of rounding error.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the specific heat at constant volume is 0.1645 calories\n",
+        "Note: answer slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.20 pageno : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 0;\t\t\t#temperature of water in deg.C\n",
+      "t2 = 0;\t\t\t#temperature of ice in deg.C\n",
+      "J = 4.18*10**7;\t\t\t#the joules thomson coefficent in erg/cal\n",
+      "l = 80;\t\t\t#latent heat og fusion kj/kg\n",
+      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
+      " \n",
+      "# Calculations\n",
+      "h = l*J/(15*g);\t\t\t#height from which ice has fallen\n",
+      "\n",
+      "# Result\n",
+      "print 'the height from which ice has fallen is %.2e cm'%(h)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the height from which ice has fallen is 2.27e+05 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.21 page no : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "T = 80;\t\t\t#temperature of bullet in deg.C\n",
+      "cp = 0.03;\t\t\t#specific heat of lead in kj/kg-K\n",
+      "J = 4.2;\t\t\t#mechanical equivalent of heat in j/cal\n",
+      "\n",
+      "# Calculations\n",
+      "h = T*cp;\t\t\t#heat developed per unit mass in calorie\n",
+      "v = (J*10**7*h*2/0.9)**0.5;\t\t\t#velocity of bullet in cm/sec\n",
+      "\n",
+      "# Result\n",
+      "print 'the velocity of bullet is %.1e cm/sec'%(v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the velocity of bullet is 1.5e+04 cm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 3.22 pageno : 51"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "w = 5.0;\t\t\t#weight of lead ball in lb\n",
+      "cp = 0.032;\t\t\t#specific heat of lead in Btu/lbdeg.F\n",
+      "h = 50;\t\t\t#height at which ball thrown in feets\n",
+      "v = 20;\t\t\t#vertical speed in ft/sec\n",
+      "g = 32;\t\t\t#accelararion due to gravity in ft/sec**2\n",
+      "\n",
+      "# Calculations\n",
+      "u = (v**2)+2*g*h\n",
+      "ke = (w/2*(u));\t\t\t#kinetic energy of the ball at ground\n",
+      "T = ke/(2*32*778*w*cp);\t\t\t#rise of temperature in deg.F\n",
+      "\n",
+      "# Result\n",
+      "print 'the rise in temperature is %.1f deg.F'%(T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rise in temperature is 1.1 deg.F\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch4.ipynb b/Heat_And_Thermodynamics/ch4.ipynb
new file mode 100755
index 00000000..6682bd50
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch4.ipynb
@@ -0,0 +1,1097 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 4 : Kinetic theory of gases"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.1 page no : 137\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 273;\t\t\t    #temperture of the oxygen molecule in K\n",
+      "m = 32;\t\t        \t#molecular mass of the gas in gm\n",
+      "r = 8.32*10**7;\t\t\t#molar gas consmath.tant in ergs per mole\n",
+      "v2 = 33200.;\t\t\t#velocity of the gas in cm/sec\n",
+      "\n",
+      "# Calculations\n",
+      "v1 = ((3*r*t)/m)**(1./2);\t\t\t#rms velocity of the molecule in cm/s\n",
+      "T = ((v2*v2*m)/(3*r));\t\t\t#temperature of the molecule with sound has velocity in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the rms velocity of the molecule is %.2e cm/s  \\\n",
+      "\\nthe temperature of the molecule is %3.0f K'%(v1,T)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rms velocity of the molecule is 4.61e+04 cm/s  \n",
+        "the temperature of the molecule is 141 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.2 page no : 137"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "t1 = 308.;\t\t\t#temperature of the nitrogen molecule in K\n",
+      "m1 = 28.;\t\t\t#molecular mass of the nitrogen in gm\n",
+      "m2 = 2.;\t\t\t#molecular mass of the hydrogen molecule in gm\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = (t1*m2/m1);\t\t\t#temperature of the hydrogen molecule in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of the hydrogen molecule is %3.0fK'%(t2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of the hydrogen molecule is  22K\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.3 pageno : 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "y = 0.00129;\t\t\t#density of the air in gm/cc\n",
+      "p = 76;\t\t\t#pressure of the nitrogen molecule in cm\n",
+      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
+      "m = 13.6;\t\t\t#density of the mercury in gm/cc\n",
+      "\n",
+      "# Calculations\n",
+      "v = ((3*p*g*m)/y)**(1./2);\t\t\t#rms velocity of air at ntp in cm/sec\n",
+      "\n",
+      "# Result\n",
+      "print 'the rms velocity of the air is %.2e cm/sec'%(v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rms velocity of the air is 4.86e+04 cm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.4 pageno : 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "d = 16*0.000089;\t\t\t#density of the oxygen molecule in gm/cc\n",
+      "p = 76;\t\t\t#pressure of the air in cm\n",
+      "g = 981;\t\t\t#gravitaitonal accelaration in cm/sec**2\n",
+      "m = 13.6;\t\t\t#density of the mercury in gm/cc\n",
+      "\n",
+      "# Calculations\n",
+      "v = ((3*p*g*m)/d)**(1./2);\t\t\t#velocuty of the oxygen molecule in cm/sec\n",
+      "\n",
+      "# Result\n",
+      "print 'velocity of oxygen molecule is %.2e cm/sec'%(v)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "velocity of oxygen molecule is 4.62e+04 cm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.5 pageno : 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 273;\t\t\t#temperature of the hydrogen molecule in K\n",
+      "n = 6.03*10**23;\t\t\t#1 mole of hydrogen molecules\n",
+      "r = 8.31*10**7;\t\t\t#universal gas consmath.tant in erg/K/mole\n",
+      "\n",
+      "# Calculations\n",
+      "e = (1.5*r*t)/n;\t\t\t#kinetic energy of the hydrogen molecule in erg\n",
+      "\n",
+      "# Result\n",
+      "print \"the kinetic energy of the hydrogen molecule is %.2e erg\"%e\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the kinetic energy of the hydrogen molecule is 5.64e-14 erg\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.6 page no : 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 1.;\t\t\t#mass of the oxygen in gm\n",
+      "r = 8.31*10**7;\t\t\t#universal gas consmath.tant in erg/K/mole\n",
+      "t = 320.;\t\t\t#temperature of the oxygen in K\n",
+      "\t\t\t#for 1gm mole k.e is 1.5rt then for 1 gm oxygen (1/32)(k.e)\n",
+      "\n",
+      "# Calculations\n",
+      "e = (m/32)*(3*r*t/2);\t\t\t#kinetic energy of the oxygen in erg\n",
+      "\n",
+      "# Result\n",
+      "print 'the kinetic energy of the oxygen is %.2e erg'%(e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the kinetic energy of the oxygen is 1.25e+09 erg\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.7 pageno : 138"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 273;\t\t\t#temperature at ntp in K\n",
+      "\t\t\t#rms velocity of oxygen is 3/2 times its rms velocity at ntp then e1 = (3/2)*e\n",
+      "\n",
+      "# Calculations\n",
+      "t1 = (9.*t/4.);\t\t\t#temperature of the oxygen molecule in K\n",
+      "\n",
+      "# Result\n",
+      "print 'temperature of the oxygen in %3.1f K'%(t1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature of the oxygen in 614.2 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.8 page no : 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p = 10;\t\t\t#pressure of the gas in atm\n",
+      "v = 5000;\t\t\t#volume of the gas in ml\n",
+      "l = 76;\t\t\t#length of the mercury in barometer in cm\n",
+      "g = 981;\t\t\t#accelaration due to gravity in cm/sec**2\n",
+      "d = 13.6;\t\t\t#density of the mercury in gm/cc\n",
+      "\n",
+      "# Calculations\n",
+      "e = 3*p*v*l*g*d;\t\t\t#kinetic energy of the molecule in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'the kinetic energy of the molecule is %.2e ergs'%(e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the kinetic energy of the molecule is 1.52e+11 ergs\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.9 page no : 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 323;\t\t\t#temperature of the hydrogen molecule in K\n",
+      "m1 = 1;\t\t\t#mass of the hydrogen molecule in gm\n",
+      "m2 = 2;\t\t\t#molecular weight of the hydrogen in gm\n",
+      "r = 8.3*10**7;\t\t\t#universal gas consmath.tant in erg/K/mole\n",
+      "\n",
+      "# Calculations\n",
+      "e = (m1*r*t*3/(m2*2));\t\t\t#kinetic enrgy of the hydrogen molecule in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'the kinetic energy of the molecule is %.0e ergs'%(e)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the kinetic energy of the molecule is 2e+10 ergs\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.10 page no : 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "t1 = 273;\t\t\t#temperature of the hydrogen molecule at n.t.p in K\n",
+      "\n",
+      "# Calculations\n",
+      "#rms value of hydrogen molecule is double to its rms value at n.t.p, so 3rt/m = 4(3rt/m)\n",
+      "t2 = 4*t1;\t\t\t#temperature of the hydrogen molecule in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of the hydrogen molecule is %.f K'%(t2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of the hydrogen molecule is 1092 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.11 page no : 139"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 273.;\t\t\t#temperature of the hydrogen molecule in K\n",
+      "t2 = 373;\t\t\t#temperature of the hydrogen molecule in K\n",
+      "d = 0.0000896;\t\t\t#density of the hydrogen molecule in gm/cc\n",
+      "p = 76*13.6*981;\t\t\t#pressure of the hydrogen molecule in gm/cm/sec**2\n",
+      "\n",
+      "# Calculations\n",
+      "v0 = (3*p/d)**(0.5);\t\t\t#rms velocity at 0deg.C\n",
+      "v100 = v0*(t2/t1)**(0.5);\t\t\t#rms velocity at 100deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the rms velocity at 0deg.C is %.2e cm/sec  \\\n",
+      "\\nthe rms velocity at 100deg.C is %.3e cm/sec'%(v0,v100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rms velocity at 0deg.C is 1.84e+05 cm/sec  \n",
+        "the rms velocity at 100deg.C is 2.154e+05 cm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.12 page no : 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "cp = 6.84;\t\t\t#specific heat at consmath.tant pressure in cal/gm mole/deg.C\n",
+      "r = 8.31*10**7;\t\t\t#universal gas constant in ergs/gm mole/deg.C\n",
+      "v = 130000;\t\t\t#velocity of sound in cm/sec\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "\n",
+      "#CALCULATION\n",
+      "cv = cp-(r/j);\t\t\t#specific heat at constant volume in gm-mole/deg.C\n",
+      "y = (cp/cv);\t\t\t#index of co-efficient\n",
+      "v1 = (3/y)**(0.5)*v;\t\t\t#rms velocity in cm/sec\n",
+      "\n",
+      "# Result\n",
+      "print 'the rms velocity of gas molecule is %.3e cm/sec'%(v1)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rms velocity of gas molecule is 1.898e+05 cm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.13 page no : 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 300;\t\t\t#temperature of the oxygen molecule in K\n",
+      "n = 6.02*10**23;\t\t\t#avagdrao's number\n",
+      "m = 32/n;\t\t\t#mass of each molecule in oxygen\n",
+      "k = 1.38*10**(-16);\t\t\t#boltzmann consmath.tant in erg/deg\n",
+      "\n",
+      "# Calculations\n",
+      "v = (8*k*t/(3.14*m))**(0.5);\t\t\t#average velocity of oxygen molecule in cm/sec\n",
+      "v2 = v*0.022384;\t\t\t#velocity in miles/hrs\n",
+      "\n",
+      "# Results\n",
+      "print 'the avg velocity of  oxygen molecule is %.f miles/hour'%(v2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the avg velocity of  oxygen molecule is 997 miles/hour\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.14 page no : 140"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "v1 = 2.4;\t\t\t#velocity of first particle in km/sec\n",
+      "v2 = 2.6;\t\t\t#velocity of second particle in km/sec\n",
+      "v3 = 3.7;\t\t\t#velocity of third particle in km/sec\n",
+      "\n",
+      "# Calculations\n",
+      "rv = ((v1**2+v2**2+v3**2)/(3))**(0.5);\t\t\t#rms velocity of the particles in km/sec\n",
+      "mv = (v1+v2+v3)/(3);\t\t\t                #mean velocity of the particles in km/sec\n",
+      "r = rv/mv;\t\t\t                            #ratio of the rms to mean velocity\n",
+      "\n",
+      "# Results\n",
+      "print 'the ratio of rms to mean velocity is %3.3f'%(r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the ratio of rms to mean velocity is 1.019\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.15 pageno : 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 2.76*10**19;\t\t\t#no.of molecules per cc\n",
+      "d = 3.36*10**(-8);\t\t\t#diameter of the helium molecule in cm\n",
+      "\n",
+      "# Calculations\n",
+      "mf = 1/((2**(0.5))*3.14*(d**2)*n)\n",
+      "\n",
+      "# Result\n",
+      "print 'the mean free path of the hydrogen molecue is %.2e cm'%(mf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mean free path of the hydrogen molecue is 7.23e-06 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.16 page no : 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "n = 85*10**(-6);\t\t\t#coefficent of vismath.cosity in dynes/cm**2/velocity gradient\n",
+      "c = 16*10**4;\t\t\t#velocity in cm/sec\n",
+      "p = 0.000089;\t\t\t#density in gm/cc\n",
+      "N = 6.06*10**23/22400;\t\t\t#avagadro number\n",
+      "a = (2)**(0.5)*(22./7);\t\t\t#constant\n",
+      "\n",
+      "# Calculations\n",
+      "mf = (3*n/(p*c));\t\t\t#mean free path in cm\n",
+      "cr = c/mf;\t\t\t#collision rate\n",
+      "d = (1/(a*N*mf))**(0.5);\t\t\t#molecular diameter of hydrogen gas in cm\n",
+      "\n",
+      "# Results\n",
+      "print 'the mean free path is %.2e cm  \\\n",
+      "\\nthe collision rate is %.1e  \\\n",
+      "\\nthe molecular diameter of hydrogen gas is %.1e cm'%(mf,cr,d)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mean free path is 1.79e-05 cm  \n",
+        "the collision rate is 8.9e+09  \n",
+        "the molecular diameter of hydrogen gas is 2.2e-08 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.17 page no : 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "d = 2*10**(-8);\t\t\t#diameter of the molecule in cm\n",
+      "k = 1.38*10**(-6);\t\t\t#boltzmann constant in ergs/deg\n",
+      "t = 273;\t\t\t#temperature at ntp in K\n",
+      "p = 76*13.6*981;\t\t\t#pressure at ntp in gm/cm/sec**2\n",
+      "\n",
+      "# Calculations\n",
+      "mf = ((k*t)/(2**(0.5)*3.14*(d**2)*p));\t\t\t#mean free path in cm\n",
+      "\n",
+      "# Result\n",
+      "print 'mean free path at ntp is %.1e cm'%(mf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "mean free path at ntp is 2.1e+05 cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.18 page no : 141"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 288;\t\t\t#temperature in K\n",
+      "k = 1.38*10**(-16);\t\t\t#boltzmann constant in erg/deg\n",
+      "N = 6.02*10**23;\t\t\t#avagadro number\n",
+      "m = 32/N;\t\t\t#mass of each oxygen molecule in gm\n",
+      "v = 196*10**-6;\t\t\t#viscosity in poise\n",
+      "\n",
+      "# Calculations\n",
+      "av = ((8*k*t/(3.14*m))**0.5);\t\t\t#average velocity in cm/sec\n",
+      "d = (m*av/(3*3.14*2**(0.5)*v))**0.5;\t\t\t#diameter of the molecule in cm\n",
+      "\n",
+      "# Results\n",
+      "print 'diameter of the molecule is %.1e cm'%(d)\n",
+      "print \"Note : answer is slightly different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "diameter of the molecule is 3.0e-08 cm\n",
+        "Note : answer is slightly different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.19 page no : 142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "mf = 15;\t\t\t#mean free path in cm\n",
+      "t = 300;\t\t\t#temperature of oxygen molecule in K\n",
+      "d = 3*10**(-8);\t\t\t#diameter of the molecule in cm\n",
+      "N = 6.02*10**23;\t\t\t#avagadro number\n",
+      "r = 8.32*10**7;\t\t\t#universal gas constant in ergs/mole/deg\n",
+      "a = (2**(0.5))*(22./7);\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "p = (r*t)/(N*a*(d**2)*mf);\t\t\t#pressure of the oxygen molecule in dynes/sq.cm\n",
+      "\n",
+      "# Result\n",
+      "print 'the pressure of the oxygen molecule is %3.3f dynes/sq.cm'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the pressure of the oxygen molecule is 0.691 dynes/sq.cm\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.20 pageno : 142"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "k = 5.64*10**-14;\t\t\t#kinetic energy of the hydrogen molecule ergs\n",
+      "t = 273;\t\t\t#temperature of the oxygen molecule in K\n",
+      "r = 8.32*10**7;\t\t\t#universal gas constant in ergs \n",
+      "\n",
+      "# Calculations\n",
+      "N = (3./2)*(r*t/k);\t\t\t#avagadro number\n",
+      "\n",
+      "# Result\n",
+      "print 'the avagadro number is %.2e'%(N)\n",
+      "print \"Note : answer is slightly different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the avagadro number is 6.04e+23\n",
+        "Note : answer is slightly different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.21 pageno : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "q = 5000;\t\t\t#total number of molecules\n",
+      "e = 2.7183;\t\t\t#constant value\n",
+      "t1 = 0.5;\t\t\t#distance travled to the mean free path\n",
+      "t2 = 1;\t\t\t#distance travelled to the mean free path\n",
+      "\n",
+      "#CALCULATONS\n",
+      "p1 = q*(e**-t1);\t\t\t#n0.of molecules having no collision in traversing a dismath.tance t1\n",
+      "p2 = q*(e**-t2);\t\t\t#n0.of molecules having no collision in traversing a dismath.tance t2\n",
+      "\n",
+      "#OUPUT\n",
+      "print 'the no. of molecules having no collision in traversing distance equal to 0.5 times the mean free path is %.f  \\\n",
+      "\\nthe no. of molecules having no collision in traversing a distance equal to 1 time the mean free path is %.f'%(p1,p2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the no. of molecules having no collision in traversing distance equal to 0.5 times the mean free path is 3033  \n",
+        "the no. of molecules having no collision in traversing a distance equal to 1 time the mean free path is 1839\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.22 page no  : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "t = 38380;\t\t\t#temperature of the molecule in K\n",
+      "k = 1.38*10**-16;\t\t\t#boltzman consmath.tant of one electron in ergs/K\n",
+      "e = 1.6*10**-12;\t\t\t#charge of one electron volts\n",
+      "\n",
+      "#CALCULATIOS\n",
+      "mk = 1.5*k*t/e;\t\t\t#mean kinetic energy per atom in ev\n",
+      "\n",
+      "# Result\n",
+      "print 'the mean kinetic energy of the molecule is %3.2f ev'%(mk) \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mean kinetic energy of the molecule is 4.97 ev\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.23 pageno : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "v = 1.7*10**-4;\t\t\t#vismath.cosity of the air molecule in cgs\n",
+      "d = 0.00129;\t\t\t#density of the molecule in gm/ml\n",
+      "p = 76*13.6*981;\t\t\t#pressure of the molecule in gm/cm/sec**2\n",
+      "\n",
+      "# Calculations\n",
+      "r = (3*p/d)**(0.5);\t\t\t#rms velocity of the molecule in cm/sec\n",
+      "mf = (3*v/(d*r));\t\t\t#mean free path in cm\n",
+      "cf = r/mf;\t\t\t#collision frequency\n",
+      "\n",
+      "# Result\n",
+      "print 'the mean free path is %.1e cm  \\\n",
+      "\\nthe collision frequency is %.e'%(mf,cf)\n",
+      "print \"Note : answer is slightly different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the mean free path is 8.1e-06 cm  \n",
+        "the collision frequency is 6e+09\n",
+        "Note : answer is slightly different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.24 page no : 143"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# The pressure of the gas\n",
+      "\n",
+      "# Variables\n",
+      "t2 = 296.4;\t\t\t#temperature of the first plate in K\n",
+      "t1 = 304.7;\t\t\t#temperature of the second plate in K\n",
+      "f = 1.6*10**-2;\t\t\t#force repelled cold is dynes/sq.cm\n",
+      "\n",
+      "# Calculations\n",
+      "p = (4*f*t2/(t1-t2));\t\t\t#pressure of the gas in dynes/sq.cm\n",
+      "\n",
+      "# Result\n",
+      "print 'the pressure of the gas is %3.3f dynes/sq.cm'%(p)\n",
+      "print \"Note : mistake in answer in book. Please calculate manually.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the pressure of the gas is 2.285 dynes/sq.cm\n",
+        "Note : mistake in answer in book. Please calculate manually.\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.25 page no : 144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "mf = 28.5*10**-6;\t\t\t#mean free path in cm\n",
+      "d = 0.000178;\t\t\t#density of helium in gm/ml\n",
+      "m = 6*10**-24;\t\t\t#mass of the helium atom in gm\n",
+      "a = (2**(0.5))*3.14;\t\t\t#constant\n",
+      "\n",
+      "# Calculations\n",
+      "d = (m/(a*d*mf))**(0.5);\t\t\t#diameter of the size in cm\n",
+      "\n",
+      "# Result\n",
+      "print 'the size of the helium atom is %.2e cm'%(d)\n",
+      "print \"Note : answer is slightly different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the size of the helium atom is 1.63e-08 cm\n",
+        "Note : answer is slightly different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.26 page no : 144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "a1 = 0*10**-4;\t\t\t#first horizontal print lacement in cm\n",
+      "a2 = 5.6*10**-4;\t\t\t#second horizontal print lacement in cm\n",
+      "a3 = -4.7*10**-4;\t\t\t#third horzontal print lacement in cm\n",
+      "a4 = -10.8*10**-4;\t\t\t#fourth horizontal print lacement in cm\n",
+      "a5 = 6.6*10**-4;\t\t\t#fifth horizontal print lacement print lacement in cm\n",
+      "a6 = -9.8*10**-4;\t\t\t#sixth horizontal print lacement in cm\n",
+      "a7 = -11.2*10**-4;\t\t\t#7th horizontal print lacement in cm\n",
+      "a8 = -4.0*10**-4;\t\t\t#8th horizontal print lacement in cm\n",
+      "a9 = 15.0*10**-4;\t\t\t#9thhorizontal print lacement in cm\n",
+      "a10 = 19.1*10**-4;\t\t\t#10th horizontal print lacement in cm\n",
+      "a11 = 16.0*10**-4;\t\t\t#11ht horizontal print lacement in cm\n",
+      "T = 293;\t\t\t#temperature of the particle in K\n",
+      "v = 0.01;\t\t\t#viscosity in cgs\n",
+      "r = 1.15*10**-5;\t\t\t#radius of the particle in cm\n",
+      "R = 8.32*10**7;\t\t\t#universal gas constant in kj/kg mole\n",
+      "t = 30;\t\t\t#time for observation of each in sec\n",
+      "\n",
+      "# Calculations\n",
+      "x = (a1**2+a2**2+a3**2+a4**2+a5**2+a6**2+a7**2+a8**2+a9**2+a10**2+a11**2)/11\n",
+      "n = R*T*t/(x*3*3.14*v*r);\t\t\t#no.of molecules in the observation \n",
+      "\n",
+      "# Result\n",
+      "print 'the value of n is %.1e'%(n)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of n is 5.7e+23\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 4.27 page no : 144"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 6.*10**-24;\t\t\t#mass of the helium atom in gm\n",
+      "k =  1.38*10**-16;\t\t\t#boltzmann consmath.tant in erg\n",
+      "t1 = 100.;\t\t\t#temperature in K\n",
+      "t2 = 900.;\t\t\t#temperature in K\n",
+      "\n",
+      "# Calculations\n",
+      "r = (t1/t2)**(3./2)*(2.7183**(m*(1./(2*k))*10**8*(1-(1./9))));\t\t\t#fractional change in the no.of helium atoms\n",
+      "\n",
+      "#OUPUT\n",
+      "print 'the fractional change in the no.of helium atoms %.3f'%(r)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the fractional change in the no.of helium atoms 0.256\n"
+       ]
+      }
+     ],
+     "prompt_number": 45
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch5.ipynb b/Heat_And_Thermodynamics/ch5.ipynb
new file mode 100755
index 00000000..13038a34
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch5.ipynb
@@ -0,0 +1,140 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : Equations of state"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.1 pageno : 167"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 304;\t\t\t#temperature of the gas in k\n",
+      "p = 73;\t\t\t#pressure of the gas in atm\n",
+      "r = 0.00366;\t\t\t#universal gas constant in j/K/mole\n",
+      "\t\t\t#ct = 8a/27br;cp = a/27b**2\n",
+      "\n",
+      "# Calculations\n",
+      "b = (t*r/(8*p));\n",
+      "a = p*27*b**2;\n",
+      "\n",
+      "# Result\n",
+      "print 'the value of the constant b is %.5f  \\\n",
+      "\\nthe value of the constant a is %3.5f'%(b,a)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of the constant b is 0.00191  \n",
+        "the value of the constant a is 0.00715\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4 pageno : 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "tc = 132;\t\t\t#critical temperature in K\n",
+      "pc = 37.2;\t\t\t#critical pressure in atm\n",
+      "r = 82.07;\t\t\t#universal gas constant in cm**3atm/mole/K\n",
+      "\n",
+      "# Calculations\n",
+      "a = 27*(r**2)*(tc**2)/(64*pc);\t\t\t#value of a in atm/cm**6/mol**2\n",
+      "b = r*tc/(8*pc);\t\t\t#value of b in cm**3/mol\n",
+      "\n",
+      "# Result\n",
+      "print 'the value of is %.2e atm/cm**6/mol**2  \\\n",
+      "\\nthe value of b is %3.2f cm**3/mol'%(a,b)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of is 1.33e+06 atm/cm**6/mol**2  \n",
+        "the value of b is 36.40 cm**3/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.5 pageno : 169"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p = 2.26*1.013*10**5;\t\t\t#critical pressure in N/m**2\n",
+      "v = 4./69;\t\t\t#critical volume in m**3/kmol\n",
+      "r = 8.31*10**3;\t\t\t#universal gas consmath.tant in J/kmol.K\n",
+      "\n",
+      "# Calculations\n",
+      "t = (8*p*v/(3*r));\t\t\t#critical temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'critical temperature of the given problem is %3.2f K'%(t)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "critical temperature of the given problem is 4.26 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch6.ipynb b/Heat_And_Thermodynamics/ch6.ipynb
new file mode 100755
index 00000000..2a03c0cd
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch6.ipynb
@@ -0,0 +1,228 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 6 : Change of state"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.1 page no : 194"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "vl = 1;     \t\t\t#volume of water in cc\n",
+      "vs = 1.0908;\t\t\t#volume of ice in cc\n",
+      "t = 273;\t    \t\t#temperature in k\n",
+      "p = 76*13.6*981;\t\t#pressure in dynes/sq.cm\n",
+      "l = 80;\t\t\t        #latent heat of fusion in cal\n",
+      "j = 4.2*10**7;\t\t\t#joules consmath.tant in erg/cal\n",
+      "\n",
+      "# Calculations\n",
+      "v = vl-vs;\t\t\t        #change in volume\n",
+      "T = (v*t*p)/(j*l);\t\t\t#change in melting point of water\n",
+      "\n",
+      "# Result\n",
+      "print 'the change in melting point of water is %.5f'%(T)\n",
+      "print \"there is wrong answer printed in book. Please calculate manually.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change in melting point of water is -0.00748\n",
+        "there is wrong answer printed in book. Please calculate manually.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.2 page no : 195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "vv = 1674.;\t\t\t#volume of vapour in cc\n",
+      "vl = 1.;\t\t\t    #volume of liquid in cc\n",
+      "p = 760.;\t\t\t#pressure of steam and water in mm\n",
+      "t = 373.;\t\t\t#temperature in K\n",
+      "p1 = 27.12;\t\t\t#superincumbent pressure in mm\n",
+      "\n",
+      "# Calculations\n",
+      "v = vv-vl;              \t\t\t#change in volume\n",
+      "l = (v*p1*t*0.024203/(p));\t\t\t#latent heat of vapourisation in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the latent heat of vapourisation is %3.1f cal'%(l)\n",
+      "print \"Note: Answer is slightly different because of rounding error.\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the latent heat of vapourisation is 539.0 cal\n",
+        "Note: Answer is slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.3 page no : 195"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "m = 1./(342*100);\t\t\t#molar concentration of water\n",
+      "t = 289.;\t\t\t        #temperature in K\n",
+      "p = 53.5*13.6*981;\t\t\t#pressure in dynes/sq.cm\n",
+      "\n",
+      "# Calculations\n",
+      "k = p/(t*m);\t\t\t    #the value of k in ergs/mol.deg\n",
+      "\n",
+      "# Result\n",
+      "print 'the value of k is %.1e ergs/mol.deg'%(k)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the value of k is 8.4e+07 ergs/mol.deg\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.4 pageno : 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 4.60;\t     \t\t    #presure at 0deg.C in mm per deg.C\n",
+      "p2 = 4.94;\t\t        \t#pressure at 1deg.C in mm per deg.C\n",
+      "t = 0.0072;\t\t\t        #lowering the melting point in deg.C\n",
+      "t1 = 7.1563979*10**(-3);\t\t\t#rise in melting point in deg.C\n",
+      "p = 760;\t\t\t        #atmospheric pressure in mm hg\n",
+      "\n",
+      "# Calculations\n",
+      "dp = p2-p1;\t\t\t        #rate of increase of pressure in mm per deg.C\n",
+      "p3 = (t1*p)/t;\t\t\t    #pressure in mm\n",
+      "dt = (755.4-p3)/dp;\t\t\t#tmperature for the triple point in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'temperature for the triple point is %3.6f deg.C'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "temperature for the triple point is 0.007188 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 6.5 pageno : 196"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "v = 21*10**4;\t\t\t#change in volume from vapour to liquid in cc\n",
+      "Ls = 687;\t\t\t#latent heat of sublimation in cal\n",
+      "lv = 607;\t\t\t#latent heat of vapourisation in cal\n",
+      "t = 273;\t\t\t#temperature of water in deg.C\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "sv = lv*j/(t*(v));\t\t\t#slope of vapourisation curve at 0 deg.C in dyne/sq.cm/deg.C\n",
+      "ss = Ls*j/(t*(v));\t\t\t#slope of sublimation curve at 0 deg.C in dyne/sq.cm/deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the slope of vapourisation curve is %.2e dyne/sq.cm/deg.C  \\\n",
+      "\\nthe slope of sublimation curve is %.2e dyne/sq.cm/deg.C'%(sv,ss)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the slope of vapourisation curve is 4.45e+02 dyne/sq.cm/deg.C  \n",
+        "the slope of sublimation curve is 5.03e+02 dyne/sq.cm/deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch7.ipynb b/Heat_And_Thermodynamics/ch7.ipynb
new file mode 100755
index 00000000..05fb5d54
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch7.ipynb
@@ -0,0 +1,334 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 7 : The joule thomson cooling efect"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.1 page no : 239"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 33.18;\t\t\t#critical temperature in K\n",
+      "pc = 12.80*76*981*13.6;\t\t\t#critical pressure in dynes/sq.cm\n",
+      "r = 83.15;\t\t\t#universal gas constant in kj/kg.K\n",
+      "d = 0.08987;\t\t\t#density of hydrogen in gm/lit\n",
+      "v = 2000/0.08987;\t\t\t#gram molecular volune of hydrogen in cc\n",
+      "\n",
+      "# Calculations\n",
+      "b = r*10**6*t/(8*pc);\t\t\t#vanderwaal constant in cm**3/mol\n",
+      "to = 2*27*t*(1-(b/v))/8;\t\t\t#inversion temperature of the hydrogen in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the inversion temperature of hydrogen is %3.2f K'%(to)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the inversion temperature of hydrogen is 223.70 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.2 pageno : 240"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "b = 0.00136;\t\t\t#vanderwaal constant in suv/gm\n",
+      "a = 0.011;\t\t\t#vanderwaal constant in atm(suv)**2/gm**2\n",
+      "r = 0.003696;\t\t\t#universal gas constant in atm(suv)/gm.deg\n",
+      "t = 423;\t\t\t#temperature of steam in K\n",
+      "cp = -0.674/0.024205;\t\t\t#specific heat at 423K in atm(cc)gm(deg)\n",
+      "\n",
+      "# Calculations\n",
+      "dt = (-b+(2*a/(r*t)))/cp;\t\t\t#change of temperature per atm drop of pressure in deg/atm\n",
+      "\n",
+      "# Result\n",
+      "print 'the change of temperature per atmosphere drop of pressure is %3.7f deg/atm'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change of temperature per atmosphere drop of pressure is -0.0004565 deg/atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.3 pageno : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "r = 8.3*10**7;\t\t\t#universal gas constant in ergs/deg.C\n",
+      "a = 1.36*10**6*76*13.6*981;\t\t\t#vanderwaal constant in atm.(suv**2)/(gm**2)\n",
+      "b = 32;\t\t\t#vanderwaal constant in cc\n",
+      "cp = 7.03;\t\t\t#specific heat at constant pressure in cal\n",
+      "j = 4.18*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "t = 273;\t\t\t#temperature of the gas in K\n",
+      "\n",
+      "# Calculations\n",
+      "dt = ((2*a/(r*t))-b)*10**6/(cp*j);\t\t\t#change of temperature in atmosphere drop of pressure in deg/atm/cm**3\n",
+      "\n",
+      "# Result\n",
+      "print 'the change of temperature in atmosphere drop of pressure is %3.2f deg C/atm/cm**2'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change of temperature in atmosphere drop of pressure is 0.31 deg C/atm/cm**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.4 pageno : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "u = 1.08;\n",
+      "cp = 8.6;\t    \t        \t#specific heat in kj/kg.K\n",
+      "j = 4.2;\t\t    \t        #joules constant in j/cal\n",
+      "p1 = 1*1.013*10**6;\t\t\t    #pressure at intial in N/sq.m\n",
+      "p2 = 20*1.013*10**6;\t\t\t#pressure at final in N/sq.m\n",
+      "\n",
+      "# Calculations\n",
+      "dh = -u*cp*j*(p1-p2);\t\t\t#change in enthalpy in joules\n",
+      "\n",
+      "# Result\n",
+      "print 'the change in enthalpy is %3.3e joules'%(dh)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change in enthalpy is 7.508e+08 joules\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.5 pageno : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "tc = 5.26;\t\t\t#critical temperature of the helium in K\n",
+      "\n",
+      "# Calculations\n",
+      "ti = 27*tc/4;\t\t\t#inversion temperature of the helium in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the inversion temperature of the helium is %3.2f K'%(ti)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the inversion temperature of the helium is 35.50 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.6 pageno : 241"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "a = 0.245*10**6*10**6;\t\t\t#vanderwaal constant in cm**4.dyne/mole**2\n",
+      "b = 2.67*10;\t\t\t#vanderwaal constant in cc/mole\n",
+      "r = 2*4.2*10**7;\t\t\t#universal gas constant in ergs/mole.K\n",
+      "\n",
+      "# Calculations\n",
+      "ti = 2*a/(b*r);\t\t\t#inversion temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'inversion temperature of hydrogen is %.f K'%(round(ti,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "inversion temperature of hydrogen is 220 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.7 pageno : 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "dp = 50*10**6;\t\t\t#change in pressure in dynes/sq.cm\n",
+      "cp = 7*4.2*10**7;\t\t\t#specific heat constant pressure in ergs/mole.K\n",
+      "a = 1.32*10**12;\t\t\t#vanderwaal constant in cm**4.dyne/mole**2\n",
+      "b = 31.2;\t\t\t#vanderwaal constant in cm**2/mole\n",
+      "t = 300;\t\t\t#inital temperature in K\n",
+      "r = 2*4.2*10**7;\t\t\t#ergs/mole.K\n",
+      "\n",
+      "# Calculations\n",
+      "dt = ((2*a/(r*t))-b)*dp/cp;\t\t\t#change in temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the change in temperature is %3.1f K'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change in temperature is 12.5 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 7.8 pageno : 242"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "p1 = 1.;\t\t\t    #inital pressure in atm\n",
+      "p2 = 51.;   \t\t\t#final pressure in atm\n",
+      "t1 = 300.;\t    \t\t#inital temperature in K\n",
+      "y = 1.4;\t\t    \t#coefficient of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = t1*(p2/p1)**((1-y)/y);\t\t\t#final temperature in K\n",
+      "dt = t1-t2;\t                \t\t#drop in temperature in K\n",
+      "\n",
+      "# Results\n",
+      "print 'the drop in temperature is %3.2f K'%(dt)\n",
+      "print \"Note : answer is slightly different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the drop in temperature is 202.45 K\n",
+        "Note : answer is slightly different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch8.ipynb b/Heat_And_Thermodynamics/ch8.ipynb
new file mode 100755
index 00000000..c82c1776
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch8.ipynb
@@ -0,0 +1,749 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 8 : First law of thermodynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.1 page no : 261\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "l = 80;\t\t\t#latent heat of fusion in cal\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "w = -0.092*10**6;\t\t\t#work done in changing phase change in ergs\n",
+      "\n",
+      "# Calculations\n",
+      "q = l*j;\t\t\t#heat added in ergs\n",
+      "du = q-w;\t\t\t#internal energy in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'the change in internal energy is %.1e ergs'%(du)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change in internal energy is 3.4e+09 ergs\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.2 page no : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 1.;\t        \t\t#mass in gm\n",
+      "l = 536.;\t\t    \t#latent heat in cal/gm\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "v = 1649;\t\t\t    #volume of water in cc\n",
+      "p = 76*13.6*981;\t\t#pressure of water in dynes/sq.cm\n",
+      "\n",
+      "# Calculations\n",
+      "dq = m*l*j;\t\t\t#heat supplied in ergs\n",
+      "dw = p*v;\t\t\t#work done in ergs\n",
+      "du = dq-dw;\t\t\t#internal energy developed in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'internal energy of water is %.2E ergs'%(du)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "internal energy of water is 2.08E+10 ergs\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.3 pageno : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "dv = 10;\t\t\t#ratio of original volume to final volume\n",
+      "t1 = 293;\t\t\t#inital temperature in K\n",
+      "y = 1.41;\t\t\t#coefficent of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = t1*(dv)**(y-1);\t\t\t#final temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the final temperature is %.f K'%(t2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final temperature is 753 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.4 pageno : 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 273;\t\t\t#temperature of earth at height h in K\n",
+      "p = 760;\t\t\t#pressure in mm of hg\n",
+      "dp = 1;\t\t\t#change in pressure in mm of hg\n",
+      "y = 1.418;\t\t\t#coefficient of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "dt = ((y-1)/y)*dp*t/p;\t\t\t#change in temperature in deg.C\n",
+      "\n",
+      "# Result\n",
+      "print 'the change in temperature is %3.3f deg.C'%(dt)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the change in temperature is 0.106 deg.C\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.5 pageno : 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 2;\t\t\t#pressure initial in atm\n",
+      "p2 = 1.;\t\t\t#pressure final in atm\n",
+      "t1 = 273 + 15;\t\t\t#inital temperature in K\n",
+      "y = 1.4;\t\t\t#coefficent of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = t1*(p2/p1)**((y-1)/y);\t\t\t#final temperature in K\n",
+      "dt = t1-t2;\t\t\t#drop in temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'drop in temperature is %3.2f K'%(dt)\n",
+      "print \"Note : answer is slightly different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "drop in temperature is 51.74 K\n",
+        "Note : answer is slightly different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.6 pageno : 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 288;\t\t\t#inital temperature in K\n",
+      "dv = 1./2;\t\t\t#ratio of inital to final volume\n",
+      "y = 1.4;\t\t\t#coefficient of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = t1*(dv)**(y-1);\t\t\t#final temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the final temperature is %3.1f K'%(t2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final temperature is 218.3 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.7 pageno : 262"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "y = 1.4;\t\t\t#coefficent of exapnsion\n",
+      "p1 = 1;\t\t\t#standard pressure in atm\n",
+      "dv = 50;\t\t\t#ratio of initial volume to final volume\n",
+      "t1 = 273;\t\t\t#standard temperature in K\n",
+      "\n",
+      "# Calculations\n",
+      "p2 = p1*dv;\t\t\t#final pressure when slowly compressed in atm\n",
+      "p3 = p1*(dv)**(y);\t\t\t#final pressure when suddenly compressed in atm\n",
+      "t2 = t1*(dv)**(y-1);\t\t\t#rise in temperature when it is suddenly compressed in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the final pressure when it is compressed slowly is %.f atm  \\\n",
+      "\\nthe final pressure when it is compressed suddenly is %.f atm  \\\n",
+      "\\nthe rise in temperature when it is suddenly compressed is %.0f K'%(p2,p3,t2)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final pressure when it is compressed slowly is 50 atm  \n",
+        "the final pressure when it is compressed suddenly is 239 atm  \n",
+        "the rise in temperature when it is suddenly compressed is 1305 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.8 pageno : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "y = 1.5;\t\t\t#coefficient of expansion\n",
+      "dp = 1./8;\t\t\t#ratio of inital pressure to final pressure\n",
+      "t1 = 300;\t\t\t#inital tempreature in K\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = t1*(dp)**((1-y)/y);\t\t\t#change in temperature in K\n",
+      "t3 = t2-t1;\t\t\t#rise in temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the rise in temperature is %3.2f K'%(t3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the rise in temperature is 300.00 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.9 pageno : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "t1 = 400;\t\t\t#inital temperature in K\n",
+      "dv = 2;\t\t\t#ratio of volumes final and inital\n",
+      "r = 8.31*10**7;\t\t\t#universal gas constant in ergs/kg.K\n",
+      "\n",
+      "# Calculations\n",
+      "w = r*t1*math.log(2);\t\t\t#work done in expanding isothermally in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'the work done in expanding isothermally is %.1e ergs'%(w)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the work done in expanding isothermally is 2.3e+10 ergs\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.10 page no : 263"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 76;\t\t\t#inital pressure in cm\n",
+      "t1 = 290;\t\t\t#inital temperature in K\n",
+      "y = 1.4;\t\t\t#coefficent of expansion\n",
+      "dv = 2;\t\t\t#ratio of inital to fianl volume when air expands isothermally \n",
+      "dv1 = 2;\t\t\t#ratio of inital to final volume when air expands adiabatically\n",
+      "\n",
+      "# Calculations\n",
+      "p2 = p1/dv;\t\t\t#final pressure when air expands isothermally in cm of hg\n",
+      "t2 = t1;\t\t\t#final temperature when air expands isothermally in K\n",
+      "t3 = t2*(1./dv1)**(y-1);\t\t\t#temprature when air expands adiabatically in K\n",
+      "p3 = p2*(1./dv1)**(y);\t\t\t#final pressure when air expands adiabatically in mm of hg\n",
+      "\n",
+      "# Result\n",
+      "print 'final pressure when air expands isothermally in cm of hg %3.2f mm of hg  \\\n",
+      "\\nfinal temperature when air expands isothermally is %3.2f K  \\\n",
+      "\\ntemprature when air expands adiabatically is %3.1f K  \\\n",
+      "\\nfinal pressure when air expands adiabatically is %3.2f cm of mercury'%(p2,t2,t3,p3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "final pressure when air expands isothermally in cm of hg 38.00 mm of hg  \n",
+        "final temperature when air expands isothermally is 290.00 K  \n",
+        "temprature when air expands adiabatically is 219.8 K  \n",
+        "final pressure when air expands adiabatically is 14.40 cm of mercury\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.11 pageno : 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p = 76*13.6*981;\t\t\t#pressure of air in dynes/sq.cm\n",
+      "v = 11100.;\t\t\t#volume expanded in ml\n",
+      "t1 = 273.;\t\t\t#inital temperature in K\n",
+      "t2 = 274.;\t\t\t#final temperature in K\n",
+      "cv = 2.411;\t\t\t#specific heat at constant volume in cal/K\n",
+      "j = 4.2*10**7;\t\t\t#joules constant in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "w = p*v*math.log(t2/t1);\t\t\t#work done in ergs\n",
+      "h = cv*(t2-t1)+w/j;\t\t\t        #heat supplied in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the work done is %3.3e erg  \\\n",
+      "\\nthe heat supplied is %3.3f cal'%(w,h)\n",
+      "print \"Note : answer is different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the work done is 4.115e+07 erg  \n",
+        "the heat supplied is 3.391 cal\n",
+        "Note : answer is different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.12 pageno : 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p = 10**6;\t\t\t#pressure of air in dynes\n",
+      "d = 0.0001293;\t\t\t#density of air in gm/cc\n",
+      "t1 = 273;\t\t\t#inital temperature in K\n",
+      "dv = 2;\t\t\t#ratio of inital volume to final volume\n",
+      "y = 1.4;\t\t\t#coefficient of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "r = p/(d*t1);\t\t\t#universal gas constant in dynes.cc/gm.K\n",
+      "t2 = round(t1*(dv)**(y-1));\t\t\t#final temperature in K\n",
+      "w = r*(t2-t1)/(y-1);\t\t\t#work done in adiabatic compression in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'work done in adiabatic compression is %.3e ergs'%(w)\n",
+      "print \"Note : answer is different because of rounding error\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "work done in adiabatic compression is 6.162e+09 ergs\n",
+        "Note : answer is different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.13 pageno : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "m = 5;\t\t\t#mass of air in gm\n",
+      "cv = 0.172;\t\t\t#specific heat at consmath.tant volume cal/gm\n",
+      "dt = 10;\t\t\t#changi in temperature in K\n",
+      "\n",
+      "# Calculations\n",
+      "ie = m*cv*dt;\t\t\t#change in internal energy in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'change in internal energy is %3.2f cal'%(ie)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "change in internal energy is 8.60 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.14 pageno : 265"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "v1 = 10**3;\t\t\t#inital volume in cc\n",
+      "v2 = 2*v1;\t\t\t#final volume in cc\n",
+      "p1 = 76*13.6*981;\t\t\t#pressure in dyne/sq.cm\n",
+      "t1 = 273;\t\t\t#intial temperature in K\n",
+      "d = 1.29;\t\t\t#density of the gas gm/lit\n",
+      "cv = 0.168;\t\t\t#specific heat at constant volume in cal/gm\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = (v2/v1)*t1;\t\t\t#final temperature in K\n",
+      "r = 0.068;\t\t\t#universal gas consmath.tant in cal\n",
+      "cp = cv+r;\t\t\t#specific heat at constant pressure in cal\n",
+      "q = d*cp*(t2-t1);\t\t\t#heat supplied in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the heat supplied to the gas is %3.2f cal'%(q)\n",
+      "print \"Note: answer is slightly different because of rounding error.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the heat supplied to the gas is 83.11 cal\n",
+        "Note: answer is slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.15 pageno : 165"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t = 303;\t\t\t#temperature of the one mole of the argon in K\n",
+      "v1 = 1;\t\t\t#intial volume in litres\n",
+      "v2 = 10;\t\t\t#final volume in litres\n",
+      "r = 8.31*10**7;\t\t\t#universal gas constant in ergs/K.mol\n",
+      "\n",
+      "# Calculations\n",
+      "w = r*t*math.log(v2/v1);\t\t\t#work done in isothermal expansion in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'the work done in isothermal expansion is %.1e ergs'%(w)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the work done in isothermal expansion is 5.8e+10 ergs\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.16 pageno : 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "dv = 4;\t\t\t#final volume of neon in lit\n",
+      "t = 273;\t\t\t#temperature of the gas in K\n",
+      "n = 2.6/22.4;\t\t\t#the no.of moles of neon\n",
+      "r = 1.98;\t\t\t#universal gas constant in cal/K.mol\n",
+      "\n",
+      "# Calculations\n",
+      "w = n*t*r*math.log(dv);\t\t\t#work done by gas in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'the work done by 2.6lit of neon is %3.2f ergs'%(w)\n",
+      "print \"Note: answer is slightly different because of rounding error.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the work done by 2.6lit of neon is 86.98 ergs\n",
+        "Note: answer is slightly different because of rounding error.\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.18 page no : 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "dv = 10**(-3);\t\t\t#ratio of initial and final volume\n",
+      "t1 = 10**5;\t\t\t#initial temperature in K\n",
+      "y = 1.66;\t\t\t#coefficient of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "t2 = t1*((4./3*math.pi*10**12)/(4./3*math.pi*10**15))**(y-1);\t\t\t#final temperature in K\n",
+      "\n",
+      "# Result\n",
+      "print 'final temperature of the gas is %3.2f K'%(t2)\n",
+      "print \"Note : Answer in book is wrong. Please calculate manually.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "final temperature of the gas is 1047.13 K\n",
+        "Note : Answer in book is wrong. Please calculate manually.\n"
+       ]
+      }
+     ],
+     "prompt_number": 46
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 8.19 pageno : 267"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 8.;\t\t\t#intial pressure in cm of hg\n",
+      "p2 = 6.;\t\t\t#final pressure in cm of hg\n",
+      "v1 = 1000.;\t\t\t#intial volume in cc\n",
+      "v2 = 1190.;\t\t\t#final volume in cc\n",
+      "\n",
+      "# Calculations\n",
+      "y = math.log(p1/p2)/math.log(v2/v1);\t\t\t#coefficient of expansion\n",
+      "\n",
+      "# Result\n",
+      "print 'the coefficent of expansion is %3.2f'%(y)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the coefficent of expansion is 1.65\n"
+       ]
+      }
+     ],
+     "prompt_number": 42
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
diff --git a/Heat_And_Thermodynamics/ch9.ipynb b/Heat_And_Thermodynamics/ch9.ipynb
new file mode 100755
index 00000000..00373b6e
--- /dev/null
+++ b/Heat_And_Thermodynamics/ch9.ipynb
@@ -0,0 +1,597 @@
+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : Second law of thermodynamics"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.1 pageno : 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t2 = 300;\t\t\t#temperature of the math.sink in K\n",
+      "n1 = 0.4;\t\t\t#efficiency of the engine\n",
+      "n2 = 0.6;\t\t\t#efficiency of the engine\n",
+      "\n",
+      "# Calculations\n",
+      "t1 = t2/(1-n1);\t\t\t#temperature of the source in K\n",
+      "t3 = t2/(1-n2);\t\t\t#temperature of the source in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the temperature of the source when 0.4 efficiency is %3.2f K  \\\n",
+      "\\nthe temperature of the source when 0.6 efficiency is %3.2f K'%(t1,t3)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the temperature of the source when 0.4 efficiency is 500.00 K  \n",
+        "the temperature of the source when 0.6 efficiency is 750.00 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.2 pageno : 308"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t2 = 273.;\t\t\t#temperature of the math.sink in K\n",
+      "t1 = 373.;\t\t\t#temperature of the source in K\n",
+      "q1 = 840.;\t\t\t#heat supplied in joules\n",
+      "j = 4.2;\t\t\t#joukes constant in erg/cal\n",
+      "\n",
+      "# Calculations\n",
+      "w = (q1/t1)*(t1-t2);\t\t\t#work done in joules\n",
+      "q2 = (q1/j)*(t2/t1);\t\t\t#heat rejected in calories\n",
+      "n = 1-(t2/t1);\t\t\t#efficiency of the engine\n",
+      "\n",
+      "# Result\n",
+      "print 'work done is %3.f j  \\\n",
+      "\\nheat rejected is %3.f cal  \\\n",
+      "\\nthe efficiency of the engine is %3.1f %%'%(w,q2,n*100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "work done is 225 j  \n",
+        "heat rejected is 146 cal  \n",
+        "the efficiency of the engine is 26.8 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.3 pageno : 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 90.;\t\t\t#temperature of the oxygen boils in K\n",
+      "t2 = 20.;\t\t\t#temperature of the liquid hydrogen in K\n",
+      "t3 = 300.;\t\t\t#temperature of the sink in K\n",
+      "\n",
+      "# Calculations\n",
+      "n = (t1-t2)/t1;\t\t\t#efficiency of the engine\n",
+      "t4 = t3/(1-n);\t\t\t#temperature of the source in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the efficiency of the engine is %3.2f  \\\n",
+      "\\nthe temperature of the source is %3.2f K'%(n,t4)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the efficiency of the engine is 0.78  \n",
+        "the temperature of the source is 1350.00 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.4 pageno : 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 373.;\t    \t    \t#temperature of the source in K\n",
+      "t2 = 273.;\t\t    \t    #temperature of the sink in K\n",
+      "w = 1200*10**5*980;\t\t\t#work done in ergs\n",
+      "j = 4.18*10**7;\t\t    \t#joules constant in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "q = (w/j)*(t1/(t1-t2));\t\t\t#heat added in cal\n",
+      "\n",
+      "# Result\n",
+      "print 'the heat added is %3.2f cal'%(round(q,-1))\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the heat added is 10490.00 cal\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.5 pageno : 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 273.;\t\t\t#temperature of the source in K\n",
+      "t2 = 290.;\t\t\t#temperature of the sink in K\n",
+      "l = 8*10.**11;\t\t\t#latent of fusion in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "n = (t2-t1)/t1;\t\t\t#efficiency of the engine\n",
+      "w = n*l;\t\t\t#energy to be supplied in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'efficiency of the engine is %.2f %%  \\\n",
+      "\\nenergy to be supplied is %.3e ergs'%(n*100,w)\n",
+      "print \"Note: answer in book are wrong please calculate manually.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "efficiency of the engine is 6.23 %  \n",
+        "energy to be supplied is 4.982e+10 ergs\n",
+        "Note: answer in book are wrong please calculate manually.\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.6 pageno : 309"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "\n",
+      "# Variables\n",
+      "t1 = 373;\t\t\t#temperature in K\n",
+      "t2 = 273;\t\t\t#temperature of math.sink in K\n",
+      "q = 10**4;\t\t\t#heat taken at higher temperature in cal\n",
+      "j = 4.2*10**7;\t\t\t#joules consmath.tant in ergs/cal\n",
+      "\n",
+      "# Calculations\n",
+      "w = q*j*(t1-t2)/t1;\t\t\t#work done in ergs\n",
+      "\n",
+      "# Result\n",
+      "print 'work done is %.1e ergs'%(w)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "work done is 1.1e+11 ergs\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.7 page no : 310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "p = 100*746/4.2;\t\t\t#power developed in cal/sec\n",
+      "t1 = 300.;\t\t\t#temperature of the sink in K\n",
+      "t2 = 500.\n",
+      "\n",
+      "# Calculations\n",
+      "te = 1 - (t1/t2)\n",
+      "Q1 = p * 100/40        # heat supplied\n",
+      "Q2 = Q1 * 0.6\n",
+      "\n",
+      "\n",
+      "# Result\n",
+      "print \"Thermal efficiency = %.f %%\"%(te*100)\n",
+      "print \"Power developed by the engine %.2f calories/sec\"%p\n",
+      "print \"If Q1 heat supplied , Q1 = %.2e cal/sec\"%Q1\n",
+      "print \"Q2 = %.2e cal/sec\"%Q2"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Thermal efficiency = 40 %\n",
+        "Power developed by the engine 17761.90 calories/sec\n",
+        "If Q1 heat supplied , Q1 = 4.44e+04 cal/sec\n",
+        "Q2 = 2.66e+04 cal/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.8 page no : 310"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "l = 964.8;\t\t\t#latent heat of steam in B.Th.U per lb\n",
+      "q = 4*15*l*778;\t\t\t#heat developed in ft lbs\n",
+      "w = 30000*60;\t\t\t#work done is ft lbs\n",
+      "pv = 12*1.013*10**6*10**3 \n",
+      "T = 600                # K\n",
+      "\n",
+      "# Calculations\n",
+      "n = (w/q)*100;\t\t\t#efficiency of the engine\n",
+      "p = 100-n;\t\t\t#percentage of heat wasted\n",
+      "T2 = 600./(6**.4)\n",
+      "R = pv/T\n",
+      "W = R * (T - T2) * 2.303 * math.log10(6)\n",
+      "e = 1 - (T2/T)\n",
+      "# Result\n",
+      "print \"Lowest temperature T2 = %.f K\"%T2\n",
+      "print \"Work done W = %.2e ergs\"%W\n",
+      "print \"Efficiency = %.1f %%\"%(e*100)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Lowest temperature T2 = 293 K\n",
+        "Work done W = 1.11e+10 ergs\n",
+        "Efficiency = 51.2 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.9 page no : 311"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "l=964.8;            #latent heat of steam in B.Th.U per lb\n",
+      "q=4*15*l*778;       #heat developed in ft lbs\n",
+      "w=33000*60;         #work done is ft lbs\n",
+      "\n",
+      "#CALCULATIONS\n",
+      "n=(w/q)*100;       #efficiency of the engine\n",
+      "p=100-n;          #percentage of heat wasted\n",
+      "\n",
+      "# Results\n",
+      "print ('the percentage of the heat wasted is %3.2f'%p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the percentage of the heat wasted is 95.60\n"
+       ]
+      }
+     ],
+     "prompt_number": 49
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.10 page no : 311"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "ip = 16.3*500*778/33000;\t\t\t# Variables power of the engine in HP\n",
+      "me = 0.72;\t\t\t#mechanical efficiency of the engine\n",
+      "bhp = 31;\t\t\t#brake horse power in b.h.p\n",
+      "ihp = bhp/me;\t\t\t#indicated horse power in HP\n",
+      "\n",
+      "# Calculations\n",
+      "i = ihp/ip;\t\t\t#indicated thermal efficiency\n",
+      "\n",
+      "# Result\n",
+      "print 'the indicted thermal efficiency is %3.2f %%'%(i*100)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the indicted thermal efficiency is 22.41 %\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.11 pageno : 312"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p = 200.;\t\t\t#horse power of steam engine in lbs coal per hour\n",
+      "j = 770.;\t\t\t#joules constant in ft lbs per B.Th.U\n",
+      "\n",
+      "# Calculations\n",
+      "w = 12500*p*j;\t\t\t#equivalent work in ft.lb.per.hr\n",
+      "hp = w/(60*33000);\t\t\t#horse power\n",
+      "\n",
+      "# Result\n",
+      "print 'horse power of the engine is %3.2f'%(hp)\n",
+      "print \"Note : answer in book is wrong. Please check manually.\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "horse power of the engine is 972.22\n",
+        "Note : answer in book is wrong. Please check manually.\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.12 pageno  : 312"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 340.;\t\t\t#temperature of the atmosphere in K\n",
+      "t2 = 612.;\t\t\t#temperature of the compression stroke in K\n",
+      "y = 1.39;\t\t\t#adiabatic expansion \n",
+      "t3 = 2040.;\t\t\t#temperature after consmtant volume ignition in K\n",
+      "\n",
+      "# Calculations\n",
+      "d = (t2/t1)**(1/(y-1))\t\t\t#density in gm/cc\n",
+      "n = 1-(1/d)**(y-1);\t\t    \t#efficiency of the engine\n",
+      "p = ((d)**(y))*(t3/t2);\t\t\t#maximum temperature of the temperature in atm\n",
+      "\n",
+      "# Result\n",
+      "print 'the maximum pressure of the engine is %3.f atm'%(p)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the maximum pressure of the engine is  27 atm\n"
+       ]
+      }
+     ],
+     "prompt_number": 43
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.13 pageno : 313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "t1 = 915;\t\t\t#temperature at the beggining in K\n",
+      "t2 = 2040;\t\t\t#temperature at the end in K\n",
+      "d = 12.6;\t\t\t#adiabatic expansion ratio\n",
+      "y = 1.39;\t\t\t#coefficent of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "x = t2/t1                               \t\t\t#ratio temparatures\n",
+      "n = 1-(1/d)**(y-1)*((x**y)-1)/(y*(x-1));\t\t\t#efficiency of the engine\n",
+      "\n",
+      "# Result\n",
+      "print 'the efficiency of the engine is %3.3f'%(n)\n",
+      "print \"Note : answer slighty different because of rounding error\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the efficiency of the engine is 0.566\n",
+        "Note : answer slighty different because of rounding error\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 9.14 pageno : 313"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "# Variables\n",
+      "p1 = 15.;\t\t\t#intial pressure in lb/sq.inch\n",
+      "dv = 15.;\t\t\t#ratio of intial to final volume\n",
+      "t1 = 520.;\t\t\t#temperature at intial in K\n",
+      "y = 1.4;\t\t\t#coefficient of expansion\n",
+      "\n",
+      "# Calculations\n",
+      "p2 = p1*(dv)**(y);\t\t\t#final pressure in lb/sq.inch\n",
+      "t2 = t1*(dv)**(y-1);\t\t\t#final temperatire in K\n",
+      "\n",
+      "# Result\n",
+      "print 'the final pressure is %3.2f lb/sq.inch  \\\n",
+      "\\nthe final temperature is %.f K'%(p2,t2)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the final pressure is 664.69 lb/sq.inch  \n",
+        "the final temperature is 1536 K\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file
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diff --git a/Heat_And_Thermodynamics/screenshots/chapter_6.png b/Heat_And_Thermodynamics/screenshots/chapter_6.png
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diff --git a/Heat_And_Thermodynamics/screenshots/temperature_of_the_source.png b/Heat_And_Thermodynamics/screenshots/temperature_of_the_source.png
new file mode 100755
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