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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 5 : The thermodynamic web"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.2  Page No : 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "T_c = 370. ; \t\t\t#[K]\n",
+      "P_c = 41.58 * 10**5 ;\t\t\t#[N/m**2]\n",
+      "R = 8.314 ;\n",
+      "V1 = 0.001 ;\t\t\t#[m**3]\n",
+      "V2 = 0.04 ; \t\t\t#[m**3]\n",
+      "q = 10400. ; \t\t\t#[J]\n",
+      "\n",
+      "# Calculations\n",
+      "a = 27./64 * (R **2)*(T_c)**2 / P_c ;\n",
+      "del_U = -0.96 * (1 / V2 - 1 / V1) ; \n",
+      "W = round(del_U - q,-1) ;\n",
+      "\n",
+      "# Results\n",
+      "print \"     Work done for the expansion = %.f J/mol\"%(W) ;\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "     Work done for the expansion = -9460 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.4  Page No : 282"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "from scipy.integrate import quad\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "P_1 = 9.43 * 10**5 ; \t\t\t#[N/m**2]\n",
+      "P_2 = 18.9 * 10**5 ; \t\t\t#[N/m**2]\n",
+      "T_1 = 80 + 273 ; \t\t\t#[K]\n",
+      "T_2 = 120 + 273 ; \t\t\t#[K]\n",
+      "A = 1.935 ;\n",
+      "B = 36.915 * 10**-3 ;\n",
+      "C = -11.402 * 10**-6 ;\n",
+      "T_c = 425.2 ; \t\t\t# [K]\n",
+      "P_c = 37.9 * 10**5 ;\t\t\t#[N/m**2]\n",
+      "R = 8.314 ;\n",
+      "\n",
+      "\n",
+      "del_h_1 = 1368 ; \t\t\t#[J/mol]\n",
+      "del_h_3 = -2542 ; \t\t\t#[J/mol]\n",
+      "Ws = 2100. ; \t\t\t#[J/mol]\n",
+      "\n",
+      "a = 0.42748 * R**2 * T_c**2.5 / P_c ;\n",
+      "b = 0.08664 * R * T_c / P_c ;\n",
+      " \n",
+      "\n",
+      "def f1(v):\n",
+      "    return R * T_1 / (v - b) - a / (math.sqrt(T_1) * v *(v + b)) - P_1;\n",
+      "\n",
+      "za=  fsolve( f1,0.001)\n",
+      "\n",
+      "def f2(v):\n",
+      "    return R * T_2 / (v - b) - a / (math.sqrt(T_2) * v *(v + b)) - P_2;\n",
+      "\n",
+      "zb=  fsolve( f2,0.001)\n",
+      "v1 = zb\n",
+      "v2 = zb\n",
+      "def f3(v):\n",
+      "    T = 353.15\n",
+      "    return b*R*T/(v-b) + a/T**0.5 * (3./(2*b) * math.log(v/(v+b)) - 1./(v+b))\n",
+      "                                      \n",
+      "delta_h1 =  quad(f3,za,zb)\n",
+      "\n",
+      "    \n",
+      "def f(T):\n",
+      "    return R * ( A * T + B/2 * T**2 + C/3 * T**3) ;\n",
+      "\n",
+      "del_h_2 = f(T_2) - f(T_1) ;\n",
+      "del_h_total = del_h_1 + del_h_2 + del_h_3 ;\n",
+      "q = del_h_total - Ws ;\n",
+      "\n",
+      "# Results\n",
+      "print \"v1 = %.2e m**3/mol,          v2 = %.2e m**3/mol\"%(za,zb ) ;\n",
+      "print \"The heat input = %.f J/mol\"%(q);\n",
+      "\n",
+      "# Note : answer is slightly different because of rounding error."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "v1 = 2.60e-03 m**3/mol,          v2 = 1.27e-03 m**3/mol\n",
+        "The heat input = 1421 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.6  Page No : 289"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from sympy import log\n",
+      "from sympy.solvers import solve\n",
+      "from sympy import Symbol\n",
+      "\n",
+      "#Variables\n",
+      "A = 2.104 ; \n",
+      "B = 2.98 * 10**-3 ;\n",
+      "R = 8.314 ;\n",
+      "Beta = 3.5 * 10**-5 ;\n",
+      "v = 7.1 * 10**-6        # m**3/mol\n",
+      "P2 = 10000*10**5 ; \t\t#[N/m**2]\n",
+      "P1 =1 ;\t\t\t        #[N/m**2] \n",
+      "\n",
+      "#Calculations\n",
+      "T2 = Symbol('T2')\n",
+      "q = solve(A*R*log(T2/1000)+B*R*(T2-1000)-(Beta*v*(P2-P1)),T2)\n",
+      "\n",
+      "#Result \n",
+      "print \"Heat input = \",q,\"J/mol\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Heat input =  [1005.88624122037] J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.7 page no : 296"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from scipy.integrate import quad\n",
+      "\n",
+      "# Variables  ( same as example 5.4)\n",
+      "R = 8.314\n",
+      "Tc = 425.2      # K \n",
+      "Pc = 37.9      # bar \n",
+      "w =  0.199\n",
+      "T1 = 353.15     #  K\n",
+      "Tc = 425.2      # K\n",
+      "P1 = 9.47       # bar\n",
+      "Pc = 38.        # bar\n",
+      "T2 = 393.15     #  K\n",
+      "P2 = 18.9        # bar\n",
+      "R = 8.314\n",
+      "\n",
+      "# calculations\n",
+      "T1r = T1/Tc\n",
+      "P1r = P1/Pc\n",
+      "T2r = T2/Tc\n",
+      "P2r = P2/Pc\n",
+      "\n",
+      "def f(T):\n",
+      "    return 1.935 + 36.915 * 10**-3 * T - 11.402 * 10**-6 * T**2\n",
+      "\n",
+      "cpDt = R * quad(f,353,393)[0]\n",
+      "h2h1 = .536 * R*Tc +cpDt - .969*R*Tc\n",
+      "\n",
+      "# Results\n",
+      "print \"the heat capacity = %.f J/mol\"%cpDt\n",
+      "print \"h2 - h1 = %.f J/mol\"%h2h1\n",
+      "\n",
+      "# Note : answer is slightly differnt becuase of rounding off error. and this problem is same as 5.4"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the heat capacity = 4695 J/mol\n",
+        "h2 - h1 = 3164 J/mol\n"
+       ]
+      }
+     ],
+     "prompt_number": 19
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 5.10  Page No : 302"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "from scipy.integrate import quad\n",
+      "\n",
+      "# Variables\n",
+      "T_c = 126.2; \t        \t\t#[K] , From appendix A.1\n",
+      "P_c = 33.8 * 10**5 ; \t\t\t#[N/m**2] , From appendix A.1\n",
+      "w = 0.039 ;     \t\t    \t# From appendix A.1\n",
+      "enth_dep_1 = -2.81 ; \t\t\t# From table C.1 Appendix C\n",
+      "A = 3.28 ;\t\t\t            # From Appendix A.2\n",
+      "B = 0.593 * 10**-3 ;\t\t\t# From Appendix A.2\n",
+      "del_h_dep_l = -5.1 ; \n",
+      "del_h_dep_v = -0.1 ;\n",
+      "T1 = 151.            \t\t\t#[K]\n",
+      "P1 = 100. * 10**5 ; \t    \t\t#[N/m**2]\n",
+      "P2 = 1. * 10**5 ; \t\t    \t#[N/m**2]\n",
+      "T2_r = 0.61 ; \t\t\t        # From figure 5.4\n",
+      "\n",
+      "# Calculations\n",
+      "T1_r = T1 / T_c ;\n",
+      "P1_r = round(P1 / P_c) ;\n",
+      "P2_r = P2 / P_c ;\n",
+      "T2 = round(T2_r * T_c) ; \t\t\t#[K]\n",
+      "\n",
+      "def f(T):\n",
+      "    #return A * T + B/2 * T**2 \n",
+      "    return A + B*T\n",
+      "\n",
+      "ans = quad(f,151,T2)[0]\n",
+      "\n",
+      "x = 1. / T_c *ans\n",
+      "y = enth_dep_1 - x ;\n",
+      "\n",
+      "# Results\n",
+      "X = ( y - del_h_dep_l) / (del_h_dep_v - del_h_dep_l);\n",
+      "print \"Quality = %.2f\"%(X) ;\n",
+      "\n",
+      "# Note: answer is different because of quad is giving different value -1.9 instead of -1.28. "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Quality = 0.85\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file