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--- a/Elements_of_Thermodynamics_and_heat_transfer/Chapter_8.ipynb
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@@ -1,142 +0,0 @@
-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:efb3cb597d2918b83a867fe2c379b3952acba398aec14a9edf762b5c40043583"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 8 - Second and third law topics"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 1 - Pg 125"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the value of dp/ds at constant volume\n",
-      "#Initialization of variables\n",
-      "P=500 #psia\n",
-      "T=700. #F\n",
-      "J=778.\n",
-      "#calculations\n",
-      "dpds=1490 *144/J\n",
-      "#results\n",
-      "print '%s %d %s' %(\"dp by ds at constant volume =\",dpds,\" F/ft^3/lbm\")\n",
-      "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "dp by ds at constant volume = 275  F/ft^3/lbm\n",
-        "The answer is a bit different due to rounding off error in textbook\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 2 - Pg 131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the thermal efficiency\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "cp=0.25 #Btu/lbm R\n",
-      "T0=520. #R\n",
-      "T1=3460. #R\n",
-      "#calculations\n",
-      "dq=cp*(T0-T1)\n",
-      "ds=cp*math.log(T0/T1)\n",
-      "dE=dq-T0*ds\n",
-      "eta=dE/dq*100.\n",
-      "#results\n",
-      "print '%s %.1f %s' %(\"Thermal efficiency =\",eta,\"percent\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Thermal efficiency = 66.5 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3 - Pg 134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#calculate the Loss of available energy\n",
-      "#Initialization of variables\n",
-      "import math\n",
-      "cp=0.25 #Btu/lbm R\n",
-      "T0=520. #R\n",
-      "T1=3460. #R\n",
-      "dG=21069. #Btu/lbm\n",
-      "dH=21502. #Btu/lbm\n",
-      "#calculations\n",
-      "dq=cp*(T0-T1)\n",
-      "ds=cp*math.log(T0/T1)\n",
-      "dE=dq-T0*ds\n",
-      "eta=dE/dq\n",
-      "dw=eta*dH\n",
-      "de=-dG+dw\n",
-      "#results\n",
-      "print '%s %d %s' %(\"Loss of available energy =\",de,\"Btu/lbm\")\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Loss of available energy = -6774 Btu/lbm\n"
-       ]
-      }
-     ],
-     "prompt_number": 3
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file