From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- Thermodynamics_by_C_P_Arora/Chapter15.ipynb | 361 ++++++++++++++++++++++++++++ 1 file changed, 361 insertions(+) create mode 100755 Thermodynamics_by_C_P_Arora/Chapter15.ipynb (limited to 'Thermodynamics_by_C_P_Arora/Chapter15.ipynb') diff --git a/Thermodynamics_by_C_P_Arora/Chapter15.ipynb b/Thermodynamics_by_C_P_Arora/Chapter15.ipynb new file mode 100755 index 00000000..537afb59 --- /dev/null +++ b/Thermodynamics_by_C_P_Arora/Chapter15.ipynb @@ -0,0 +1,361 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:aba24805b98ee0919f2ef39d5927b5007d2ce98264adcd2e65263a9fbffd1f67" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15:CHEMICAL EQUILIBRIUM" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.1, Page No:676" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "# (b).Number of moles of each constituents\n", + "nCH4=2; # Number of moles of CH4\n", + "\n", + "#Calculation\n", + "E=3-nCH4; # Amount of reaction from (a) and refer example 15.1 (a)\n", + "nH2O=1-E;# Number of moles of H2O\n", + "nCO=1+E;# Number of moles of CO\n", + "nH2=4+3*E;# Number of moles of H2\n", + "\n", + "#Results\n", + "print \"(b).Number of moles of each constituents\",\"\\nNumber of moles of H2O = \",nH2O\n", + "print \"Number of moles of CO = \",nCO,\"\\nNumber of moles of H2 = \",nH2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(b).Number of moles of each constituents \n", + "Number of moles of H2O = 0\n", + "Number of moles of CO = 2 \n", + "Number of moles of H2 = 7\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.2, Page No:680" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "T0=298; # Given temperature in kelvin\n", + "R_1=8.314; # Universal gas constant in kJ/kg mol K\n", + "# (a).CO+1/2 O2 = CO2\n", + "# From table of properties of combustion\n", + "del_hfco2=-393509;# Enthalpy of heat \n", + "del_hfco=-110525;# Enthalpy of heat \n", + "s_co2=213.795;# Entropy of heat \n", + "s_co=197.652;# Entropy of heat \n", + "s_o2=205.142;# Entropy of heat \n", + "\n", + "#Calculation for (a)\n", + "del_Ga=(del_hfco2-del_hfco-T0*(s_co2-s_co-(1/2*s_o2)));\n", + "Ka=math.exp (abs (del_Ga)/(R_1*1000*T0));\n", + "\n", + "#Result for (a)\n", + "print \"(a).CO+1/2 O2 = CO2\"\n", + "print (\" The equilibrium constant at 298 K = %0.3f (Error in textbook) \")%Ka\n", + "\n", + "#Calculation for (b)\n", + "# (b).2CO + O2 = 2CO2\n", + "Kb=math.exp (2*abs (del_Ga)/(R_1*1000*T0));\n", + "\n", + "#Result for (b)\n", + "print \"\\n(b).2CO + O2 = 2CO2\"\n", + "print (\"The equilibrium constant at 298 K = %0.3f (Error in textbook)\")%Kb\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a).CO+1/2 O2 = CO2\n", + " The equilibrium constant at 298 K = 1.123 (Error in textbook) \n", + "\n", + "(b).2CO + O2 = 2CO2\n", + "The equilibrium constant at 298 K = 1.262 (Error in textbook)\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.3, Page No:686" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "T0=298; # Temperature of surroundings in kelvin\n", + "R_1=8.314; # Universal gas constant in kJ/kg mol K\n", + "T=2800; # Given Temperature in kelvin\n", + "\n", + "#Calculation\n", + "# From table of properties of combustion\n", + "del_hfco2=-393509; # Enthalpy of heat \n", + "del_hfco=-110525; # Enthalpy of heat \n", + "del_H=del_hfco2-del_hfco; # Standard enthalpy of reaction\n", + "Ka=1.202*10**45; # The equilibrium constant From the example 15.2\n", + "K1=math.log (Ka);\n", + "K=math.exp(-(del_H/R_1)*((1/T)-(1/T0))+K1);\n", + "\n", + "#Result\n", + "print\"K =\",round(K,3),\" (roundoff error)\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "K = 5.687 (roundoff error)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.5, Page No:689" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "T=2800; # Temperature of combustion in kelvin\n", + "p=1; # Pressure of combustion in atm\n", + "# For this reverse reaction at 2800K and 1atm, from Table 15.1\n", + "K=44.168; # K=e^3.788;\n", + "\n", + "#Calculation\n", + "K=math.sqrt (K); # For stoichiometric equation CO+1/2 O2 = CO2 which is halved\n", + "# From equation 15.24a and by the iteration process we get the following\n", + "a=0.198;\n", + "b=(1+a)/2;\n", + "c=1-a;\n", + "\n", + "#Results\n", + "print \"The balance for the actual reaction equation CO + O2 \u2192 aCO + bO2 + cCO2 is given by \"\n", + "print \"a =\",a,\"\\nb =\",b,\"\\nc =\",c\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The balance for the actual reaction equation CO + O2 \u2192 aCO + bO2 + cCO2 is given by \n", + "a = 0.198 \n", + "b = 0.599 \n", + "c = 0.802\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.6, Page No:691" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "# By driving the equation for equilibrium constant as shown in example 15.6 we get 6.646(6)^(1/6)=((1-a)/a)((3+a)/(1+a))^1/2\n", + "# by simple iteration process we get\n", + "a=0.095;\n", + "\n", + "#Calculations\n", + "b=(1+a)/2;\n", + "c=1-a;\n", + "\n", + "#Results\n", + "print \"The equilibrium composition of CO = \",a,\"mol (roundoff error)\"\n", + "print \"The equilibrium composition of O2 = \",b,\"mol (roundoff error)\"\n", + "print \"The equilibrium composition of CO2 = \",c,\"mol (roundoff error)\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equilibrium composition of CO = 0.095 mol (roundoff error)\n", + "The equilibrium composition of O2 = 0.5475 mol (roundoff error)\n", + "The equilibrium composition of CO2 = 0.905 mol (roundoff error)\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.7, Page NO:691" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "T=2800; # Temperature of combustion in kelvin\n", + "p=1; # Pressure of combustion in atm\n", + "# For this reverse reaction at 2800K and 1atm, from Table 15.1\n", + "K=44.168; # K=e^3.788;\n", + "\n", + "#Calculations\n", + "K=math.sqrt (K); # For stoichiometric equation CO+1/2 O2 = CO2 which is halved\n", + "# From equation 15.24a and by the iteration process we get the following\n", + "a=0.302;\n", + "b=(1+a)/2;\n", + "c=1-a;\n", + "\n", + "#Results\n", + "print \"The balance for the actual reaction equation CO + 1/2O2 + 1.88N2 \u2194 aCO + bO2 + cCO2 +3.76N2 is given by \"\n", + "print \"a =\",a,\"\\nb =\",b,\"\\nc =\",c" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The balance for the actual reaction equation CO + 1/2O2 + 1.88N2 \u2194 aCO + bO2 + cCO2 +3.76N2 is given by \n", + "a = 0.302 \n", + "b = 0.651 \n", + "c = 0.698\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.8, Page No:693" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "T=3000; # Temperature of combustion in kelvin\n", + "p=1; # Pressure of combustion in atm\n", + "T0=298; # Temperature of surroundings in kelvin\n", + "R_1=8.314; # Universal gas constant in kJ/kg mol K\n", + "# Gibbs functions at 298K from Table 14.1\n", + "del_gNO=86550; # In kJ/kmol\n", + "del_gNO2=51310; # In kJ/kmol\n", + "# From table of properties of combustion\n", + "del_hfNO=90250; # Enthalpy of heat \n", + "del_hfNO2=33180; # Enthalpy of heat \n", + "\n", + "#Calculations\n", + "K1=math.exp (-(del_hfNO/R_1)*((1/T)-(1/T0))-((del_gNO)/(R_1*T0)));\n", + "K2=math.exp (-(del_hfNO2/R_1)*((1/T)-(1/T0))-((del_gNO2)/(R_1*T0)));\n", + "# By solving equilibrium equations by iteration method\n", + "E1=0.228; E2=0.0007;\n", + "yNO=E1/4.76; # Mole fraction of NO in exhaust gas\n", + "yNO2=E2/4.76; # Mole fraction of NO2 in exhaust gas\n", + "\n", + "#Results\n", + "print \"Percentage of NOx present in the exhaust gas \",\"\\nMole fraction of NO in exhaust gas = \",round(yNO*100,2),\"%\"\n", + "print \"Mole fraction of NO2 in exhaust gas = \",round(yNO2*100,4),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of NOx present in the exhaust gas \n", + "Mole fraction of NO in exhaust gas = 4.79 %\n", + "Mole fraction of NO2 in exhaust gas = 0.0147 %\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit