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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 9 : The Chemical Reaction Equation and Stoichiometry"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 9.1  Page no. 228\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Balancing a reaction for a Biological Reaction\n",
+      "# Solution\n",
+      "\n",
+      "# variables\n",
+      "# Given \n",
+      "#Main eqn. C6H12O6 + aO2 ---> bCO2 + cH2O\n",
+      "# By carbon balance\n",
+      "b = 6 ;\n",
+      "\n",
+      "#By hydrogen balance\n",
+      "c = 6;\n",
+      "\n",
+      "# calculation\n",
+      "#Balancing oxygen in reaction\n",
+      "a = (c*1+b*2-6)/2.0;\n",
+      "\n",
+      "#result\n",
+      "print 'Value of a is  %i'%a\n",
+      "print 'Value of b is  %i'%b\n",
+      "print 'Value of c is  %i'%c"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of a is  6\n",
+        "Value of b is  6\n",
+        "Value of c is  6\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 9.2  Page no. 229\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Use of Chemical Reaction to Calculate the Mass of Reactants given the Mass of Products\n",
+      "# Solution\n",
+      "\n",
+      "# Variables\n",
+      "m_CO2 = 44.0 ;          #molecular wt-[g]\n",
+      "m_C7H16 = 100.1 ;       #molecular wt-[g]\n",
+      "p_con = 50. ;           # percentage conversion of CO2 to dry ice\n",
+      "amt_di = 500. ;         # amount of dry ice to be produce per hour-[kg]\n",
+      "\n",
+      "# Calculation\n",
+      "# By using the given equation \n",
+      "amt_C7H16 = (amt_di*m_C7H16)/((p_con/100.)*m_CO2*7) ;# [kg]\n",
+      "\n",
+      "# Result\n",
+      "print 'Amount of heptane required per hour to produce 500kg dry ice per hour is  %.1f kg.'%amt_C7H16"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Amount of heptane required per hour to produce 500kg dry ice per hour is  325.0 kg.\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 9.3  Page no. 230\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Application of Stoichiometry when more than one Reaction occurs\n",
+      "# Solution\n",
+      "\n",
+      "# Variables\n",
+      "m_CaCO3 = 100.1 ;              #molecular wt-[g]\n",
+      "m_MgCO3 = 84.32 ;              #molecular wt-[g]\n",
+      "m_CaO = 56.08 ;                #molecular wt-[g]\n",
+      "m_MgO = 40.32 ;                #molecular wt-[g]\n",
+      "m_CO2 = 44.0 ;                 #molecular wt-[g]\n",
+      "\n",
+      "# Limestone analysis\n",
+      "p_CaCO3 = 92.89 ;              # percentage of CaCO3\n",
+      "p_MgCO3 = 5.41 ;               #  percentage of MgCO3 \n",
+      "inrt = 1.7 ;                   #percentage of inert\n",
+      "\n",
+      "# Calculation and Results\n",
+      "#(a)\n",
+      "amt_CaO  = (((p_CaCO3/100)*m_CaO)/m_CaCO3)*2000 ;      #Pounds of CaO produced from 1 ton(2000lb) of limestone\n",
+      "print ' Amount of CaO produced from 1 ton(2000lb) of limestone is  %.0f lb.'%amt_CaO\n",
+      "\n",
+      "#(b)\n",
+      "mol_CaCO3 = (p_CaCO3/100)/m_CaCO3 ;      # lb mol of CaCO3\n",
+      "mol_MgCO3 = (p_MgCO3/100)/m_MgCO3 ;      # lb mol of MgCO3\n",
+      "total_mol = mol_CaCO3+mol_MgCO3;\n",
+      "amt_CO2 = total_mol*m_CO2 ;              # Amount of CO2 recovered per pound of limestone-[lb]\n",
+      "print '  Amount of CO2 recovered per pound of limestone is  %.3f lb.'%amt_CO2\n",
+      "\n",
+      "#(c)\n",
+      "amt_CaO = m_CaO*mol_CaCO3 ;              # since lb mol of CaO  =  CaCO3\n",
+      "amt_MgO = m_MgO*mol_MgCO3 ;              # since lb mol of MgO  =  MgCO3\n",
+      "total_lime =  amt_CaO+amt_MgO+(inrt)/100 ;     # total amount of lime per pound limestone\n",
+      "amt_lmst = 2000/total_lime ;              # Amount of limestone required to make 1 ton(2000lb) of lime \n",
+      "print '  Amount of limestone required to make 1 ton(2000lb) of lime   %.1f lb.'%amt_lmst"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Amount of CaO produced from 1 ton(2000lb) of limestone is  1041 lb.\n",
+        "  Amount of CO2 recovered per pound of limestone is  0.437 lb.\n",
+        "  Amount of limestone required to make 1 ton(2000lb) of lime   3550.7 lb.\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 9.4  Page no. 235\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation of extent of Reaction\n",
+      "# Solution\n",
+      "\n",
+      "# Variables\n",
+      "f_NH3 = 5. ;        # NH3 in feed-[g]\n",
+      "f_N2 =  100. ;      # N2 in feed-[g]\n",
+      "f_H2 =  50. ;       # H2 in feed-[g]\n",
+      "p_NH3 = 90.         # NH3 in product-[g]\n",
+      "m_NH3  = 17. ;      # Molecular wt. of NH3-[g]\n",
+      "m_N2  = 28. ;       # Molecular wt. of N2-[g]\n",
+      "m_H2  = 2. ;        # Molecular wt. of H2-[g]\n",
+      "\n",
+      "# Calculations \n",
+      "# Extent of reaction can be calculated by using eqn. 9.3 \n",
+      "# For NH3\n",
+      "ni = p_NH3/m_NH3 ;       #[g mol NH3]\n",
+      "nio = f_NH3/m_NH3 ;      #[g mol NH3]\n",
+      "vi = 2 ;                 # coefficint of NH3\n",
+      "ex_r =  (ni-nio)/vi      # Extent of reaction - moles reacting\n",
+      "\n",
+      "#Determine H2 and N2 in product of reaction by Eqn. 9.4\n",
+      "# For N2\n",
+      "nio_N2 = f_N2/m_N2 ;     #[g mol N2]\n",
+      "vi_N2 = -1 ;             # coefficint of N2\n",
+      "ni_N2 = nio_N2 + vi_N2*ex_r ;      #N2 in product of reaction-[g moles ]\n",
+      "m_N2 = ni_N2*m_N2 ;                # mass of N2 in product of reaction-[g]\n",
+      "\n",
+      "# Results\n",
+      "print '  N2 in product of reaction is  %.2f g moles  '%ni_N2\n",
+      "print '   Mass of N2 in product of reaction is  %.2f g   '%m_N2\n",
+      "# For H2\n",
+      "nio_H2 = f_H2/m_H2 ;               #[g mol H2]\n",
+      "vi_H2 = -3 ;                       # coefficint of H2\n",
+      "ni_H2 = nio_H2 + vi_H2*ex_r ;      #H2 in product of reaction-[g moles ]\n",
+      "m_H2 = ni_H2*m_H2 ;                # mass of H2 in product of reaction-[g]\n",
+      "print '   H2 in product of reaction is  %.2f g moles  '%ni_H2\n",
+      "print '  Mass of H2 in product of reaction is  %.2f g   '%m_H2\n",
+      "\n",
+      "# ARP\n",
+      "m_SO2 = 64. ;                      # Molecular wt.of SO2-[g]\n",
+      "mol_SO2 =  2. ;                    # moles of SO2\n",
+      "ARP = (1./m_NH3)/(mol_SO2/m_SO2);\n",
+      "print '  ARP is  %.2f    '%ARP"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "  N2 in product of reaction is  1.07 g moles  \n",
+        "   Mass of N2 in product of reaction is  30.00 g   \n",
+        "   H2 in product of reaction is  17.50 g moles  \n",
+        "  Mass of H2 in product of reaction is  35.00 g   \n",
+        "  ARP is  1.88    \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 9.5  Page no. 238\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation of Limiting and Excess Reactants\n",
+      "# Solution\n",
+      "\n",
+      "# Variables\n",
+      "f_N2 = 10. ;          # N2 in feed-[g]\n",
+      "f_H2 = 10. ;          # H2 in feed-[g]\n",
+      "m_NH3 = 17.02         # Molecular wt. of NH3-[g]\n",
+      "m_N2 = 28. ;          # Molecular wt. of N2-[g]\n",
+      "m_H2 = 2. ;           # Molecular wt. of H2-[g]\n",
+      "\n",
+      "# Calculations\n",
+      "# Extent of reaction can be calculated by using eqn. 9.3 \n",
+      "# Based on N2\n",
+      "nio_N2 = f_N2/m_N2         #[g mol N2]\n",
+      "vi_N2 = -1 ;               # coefficint of N2\n",
+      "ex_N2 = -(nio_N2)/vi_N2 ;  # Max. extent of reaction based on N2\n",
+      "\n",
+      "# Based on H2\n",
+      "nio_H2 = f_H2/m_H2 ;       #[g mol H2]\n",
+      "vi_H2 = -3 ;               # coefficint of H2\n",
+      "ex_H2 = -(nio_H2)/vi_H2    # Max. extent of reaction based on H2\n",
+      "\n",
+      "#(a)\n",
+      "vi_NH3 = 2 ;               # coefficint of NH3\n",
+      "mx_NH3 = ex_N2*vi_NH3*m_NH3 ;    # Max. amount of NH3 that can be produced\n",
+      "\n",
+      "# Results\n",
+      "print ' (a) Max. amount of NH3 that can be produced is %.1f g'%mx_NH3\n",
+      "\n",
+      "#(b) and (c)\n",
+      "if (ex_H2 > ex_N2 ):\n",
+      "    print '  (b) N2 is limiting reactant  '\n",
+      "    print '  (c) H2 is excess reactant  '\n",
+      "    ex_r = ex_N2\n",
+      "else:\n",
+      "    print '  (b) H2 is limiting reactant  '\n",
+      "    print '  (c) N2 is excess reactant  '\n",
+      "    ex_r = ex_H2 ;"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " (a) Max. amount of NH3 that can be produced is 12.2 g\n",
+        "  (b) N2 is limiting reactant  \n",
+        "  (c) H2 is excess reactant  \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 9.6  Page no. 242\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Yeilds in the Reaction of Glucose to produce Ethanol\n",
+      "# Solution\n",
+      "\n",
+      "# variables\n",
+      "#(a)\n",
+      "mol_bms =  0.59 ;      # Biomass produced per g mol of glucose-[g mol biomass/ g mol glucose]\n",
+      "mw_bms =  23.74 ;      # molecular wt. of biomass -[g]\n",
+      "mw_gls =  180.0 ;      # molecular wt. of glucose -[g]\n",
+      "\n",
+      "# calculations and Results\n",
+      "ms_bms =  (mol_bms*mw_bms)/mw_gls ;     # Biomass produced per gram of glucose-[g biomass/ g glucose]\n",
+      "print '(a) Biomass produced per gram of glucose is %.4f g biomass/ g glucose.'%ms_bms\n",
+      "\n",
+      "#(b)\n",
+      "mol_etol = 1.3 ;       #Ethanol produced per g mol of glucose-[g mol ethanol/ g mol glucose]\n",
+      "mw_etol =  46.0 ;      # molecular wt. of ethanol -[g]\n",
+      "ms_etol =  (mol_etol*mw_etol)/mw_gls ;     # Ethanol produced per gram of glucose-[g ethanol/ g glucose]\n",
+      "print ' (b) Ethanol produced per gram of glucose is %.3f g ethanol/ g glucose.'%ms_etol"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Biomass produced per gram of glucose is 0.0778 g biomass/ g glucose.\n",
+        " (b) Ethanol produced per gram of glucose is 0.332 g ethanol/ g glucose.\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 9.7  Page no. 243\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Selectivity in the Production of Nanotubes\n",
+      "# Solution\n",
+      "\n",
+      "# Variables\n",
+      "# By using reaction (a)\n",
+      "H2_a = 3-0.50        # H2 produced in reaction (a)\n",
+      "\n",
+      "# Calculations\n",
+      "C_a = (2./3)*H2_a ;  # Nanotubes(the C) produced by reaction (a)\n",
+      "sel = C_a/0.50 ;     # Selectivity of C reletive to C2H4-[g mol C/ g mol C2H4]\n",
+      "\n",
+      "# Results\n",
+      "print 'Selectivity of C reletive to C2H4 is %.2f g mol C/ g mol C2H4.'%sel"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Selectivity of C reletive to C2H4 is 3.33 g mol C/ g mol C2H4.\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      " Example 9.8  Page no. 244\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculation of various terms Pertaning to Reaction\n",
+      "# Solution\n",
+      "\n",
+      "# Variables\n",
+      "m_C3H6 = 42.08          # molecular wt. of propene-[g]\n",
+      "m_C3H5Cl = 76.53 ;      # molecular wt. of C3H5Cl-[g]\n",
+      "m_C3H6Cl2 = 112.99 ;    # molecular wt. of C3H6Cl2-[g]\n",
+      "\n",
+      "# Product analysis\n",
+      "pml_Cl2 = 141.0 ;       # [g mol]\n",
+      "pml_C3H6 = 651.0 ;      #[g mol]\n",
+      "pml_C3H5Cl = 4.6 ;      # [g mol]\n",
+      "pml_C3H6Cl2 = 24.5 ;    # [g mol]\n",
+      "pml_HCL = 4.6 ;         #[g mol]\n",
+      "\n",
+      "# Calculation & Results\n",
+      "#(a)\n",
+      "a_Cl = pml_C3H5Cl;      # Chlorine reacted by eqn.(a)\n",
+      "b_Cl = pml_C3H6Cl2 ;    # Chlorine reacted by eqn.(b)\n",
+      "fed_Cl = pml_Cl2+a_Cl+b_Cl ;    # Total chlorine fed to reactor-[g mol]\n",
+      "\n",
+      "#by analysing reaction (a) and (b)\n",
+      "a_C3H6 = a_Cl+b_Cl ;     # C3H6 reacted by reaction (a)\n",
+      "fed_C3H6 = pml_C3H6+a_C3H6 ;      #Total C3H6 fed to reactor-[g mol]\n",
+      "\n",
+      "\n",
+      "print '(a) Total chlorine fed to reactor is %.2f  g mol '%fed_Cl\n",
+      "print '     Total C3H6 fed to reactor is %.2f  g mol '%fed_C3H6\n",
+      "\n",
+      "#(b) and (c)\n",
+      "# Extent of reaction can be calculated by using eqn. 9.3 \n",
+      "# Based on C3H6\n",
+      "nio_C3H6 = fed_C3H6 ;       #[g mol C3H6]\n",
+      "vi_C3H6 = -1 ;              # coefficint of C3H6\n",
+      "ex_C3H6 = -(nio_C3H6)/vi_C3H6 ;     # Max. extent of reaction based on C3H6\n",
+      "\n",
+      "# Based on Cl2\n",
+      "nio_Cl2 =  fed_Cl;          #[g mol Cl2]\n",
+      "vi_Cl2 = -1 ;               # coefficint of Cl2\n",
+      "ex_Cl2 = -(nio_Cl2)/vi_Cl2 ;# Max. extent of reaction based on Cl2\n",
+      "\n",
+      "if (ex_Cl2 > ex_C3H6 ):\n",
+      "    print '  (b) C3H6 is limiting reactant  '\n",
+      "    print '    (c)Cl2 is excess reactant  '\n",
+      "    ex_r = ex_C3H6;\n",
+      "else:\n",
+      "    print '  (b) Cl2 is limiting reactant  '\n",
+      "    print ' (c) C3H6 is excess reactant  '\n",
+      "    ex_r = ex_Cl2;\n",
+      "\n",
+      "#(d)\n",
+      "fr_cn = pml_C3H5Cl/fed_C3H6 ;      #Fractional conversion of C3H6 to C3H5Cl\n",
+      "print '  (d) Fractional conversion of C3H6 to C3H5Cl is %.2e  '%fr_cn\n",
+      "\n",
+      "#(e)\n",
+      "sel = pml_C3H5Cl/pml_C3H6Cl2 ;     # Selectivity of C3H5Cl relative to C3H6Cl2\n",
+      "print '  (e) Selectivity of C3H5Cl relative to C3H6Cl2 is %.2f g mol C3H5Cl/g mol C3H6Cl2  '%sel\n",
+      "\n",
+      "#(f)\n",
+      "yld = (m_C3H5Cl*pml_C3H5Cl)/(m_C3H6*fed_C3H6)   # Yield of C3H5Cl per g C3H6 fed to reactor\n",
+      "print '  (f) Yield of C3H5Cl per g C3H6 fed to reactor is %.3f g C3H5Cl/g C3H6  '%yld\n",
+      "\n",
+      "#(g)\n",
+      "vi_C3H5Cl = 1  ;                 # coefficint of C3H5Cl\n",
+      "vi_C3H6Cl2 = 1  ;                # coefficint of C3H6Cl2\n",
+      "ex_a = (pml_C3H5Cl-0)/vi_C3H5Cl ;      # Extent of reaction a as C3H5Cl is produced only in reaction a\n",
+      "ex_b = (pml_C3H6Cl2-0)/vi_C3H6Cl2 ;    # Extent of reaction b as C3H6Cl2 is produced only in reaction b\n",
+      "print '  (g) Extent of reaction a as C3H5Cl is produced only in reaction a is %.1f   '%ex_a\n",
+      "print '     Extent of reaction b as C3H6Cl2 is produced only in reaction b %.1f   '%ex_b\n",
+      "\n",
+      "#(h)\n",
+      "in_Cl = fed_Cl*2 ;        #Entering Cl -[g mol]\n",
+      "out_Cl  = pml_HCL ;       # Exiting Cl in HCl-[g mol]\n",
+      "ef_w = out_Cl/in_Cl ;     # Mole efficiency of waste\n",
+      "ef_pr =  1-ef_w ;         # Mole efficiency of product\n",
+      "print ' (h) Mole efficiency of product is %.3f '%ef_pr"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "(a) Total chlorine fed to reactor is 170.10  g mol \n",
+        "     Total C3H6 fed to reactor is 680.10  g mol \n",
+        "  (b) Cl2 is limiting reactant  \n",
+        " (c) C3H6 is excess reactant  \n",
+        "  (d) Fractional conversion of C3H6 to C3H5Cl is 6.76e-03  \n",
+        "  (e) Selectivity of C3H5Cl relative to C3H6Cl2 is 0.19 g mol C3H5Cl/g mol C3H6Cl2  \n",
+        "  (f) Yield of C3H5Cl per g C3H6 fed to reactor is 0.012 g C3H5Cl/g C3H6  \n",
+        "  (g) Extent of reaction a as C3H5Cl is produced only in reaction a is 4.6   \n",
+        "     Extent of reaction b as C3H6Cl2 is produced only in reaction b 24.5   \n",
+        " (h) Mole efficiency of product is 0.986 \n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "code",
+     "collapsed": true,
+     "input": [],
+     "language": "python",
+     "metadata": {},
+     "outputs": []
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file