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diff --git a/Fundamentals_Of_Thermodynamics/Chapter16_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter16_6.ipynb new file mode 100755 index 00000000..c155fdc6 --- /dev/null +++ b/Fundamentals_Of_Thermodynamics/Chapter16_6.ipynb @@ -0,0 +1,377 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2573953a96bd6739252db580058edb8938919cec53d4f1d76f9484cfa4f2a1d5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 16:INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.1:676"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ques1\n",
+ "# pressure required to make diamonds from graphite\n",
+ "import math\n",
+ "T=25 # temp in degree C\n",
+ "ggrp= 0 # g for fraphite\n",
+ "gdiamnd=2867.8 # g for diamond in kJ/mol\n",
+ "vgrp=0.000444 # specific volume of graphite in m^3/kg\n",
+ "vdiamnd=0.000284 # specific volume of graphite in m^3/kg\n",
+ "BetaTgrp=0.304e-6 # beta for graphite in 1/MPa\n",
+ "BetaTdiamnd=0.016e-6 # beta for diamond in 1/MPa\n",
+ "\n",
+ "P=-(-2*(vgrp-vdiamnd)+math.sqrt((2*vgrp-2*vdiamnd)**2-4*(vgrp*BetaTgrp-vdiamnd*BetaTdiamnd)*(2*gdiamnd/(12.011*1000))))/(2*(vgrp*BetaTgrp-vdiamnd*BetaTdiamnd))\n",
+ "print int(P),\" MPa is the pressure at which possibility exists for conversion from graphite to diamonds\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1493 MPa is the pressure at which possibility exists for conversion from graphite to diamonds\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.2:PG-681"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ques2\n",
+ "#to determine change in gibbs free energy\n",
+ "\n",
+ "#1-H2\n",
+ "#2-O2\n",
+ "#3-H2O\n",
+ "\n",
+ "#at T=298 K\n",
+ "T1=298.0;#K\n",
+ "Hf1=0;#Enthalpy of formation of H2 at 298 K\n",
+ "Hf2=0;#Enthalpy of formation of O2 at 298 K\n",
+ "Hf3=-241826;#enthalpy of formation of H2O at 298 K in kJ\n",
+ "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",
+ "Sf1=130.678;#Entropy of H2 at 298 K n kJ/K\n",
+ "Sf2=205.148;#Entropy of O2 at 298 K in kJ/K\n",
+ "Sf3=188.834;#entropy of H2O at 298 K in kJ/K\n",
+ "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",
+ "dG1=dH-T1*dS;#change n gibbs free energy in kJ\n",
+ "print\" Change in gibbs free energy at\",T1,\"kelvin is\",round(dG1),\"kJ \"\n",
+ "#at T=2000 K\n",
+ "T2=2000.0;#K\n",
+ "Hf1=52942-0;#Enthalpy of formation of H2 at 2000 K\n",
+ "Hf2=59176-0;#Enthalpy of formation of O2 at 2000 K\n",
+ "Hf3=-241826+72788;#enthalpy of formation of H2O at 2000 K in kJ\n",
+ "dH=2*Hf1+Hf2-2*Hf3;#Change in enthalpy in kJ\n",
+ "Sf1=188.419;#Entropy of H2 at 2000 K n kJ/K\n",
+ "Sf2=268.748;#Entropy of O2 at 2000 K in kJ/K\n",
+ "Sf3=264.769;#entropy of H2O at 2000 K in kJ/K\n",
+ "dS=2*Sf1+Sf2-2*Sf3;#Change in entropy in kJ/K\n",
+ "dG2=dH-(T2*dS);#change n gibbs free energy in kJ\n",
+ "print\" Change in gibbs free energy at\",T2,\"kelvin is\",round(dG2),\" kJ \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Change in gibbs free energy at 298.0 kelvin is 457179.0 kJ \n",
+ " Change in gibbs free energy at 2000.0 kelvin is 271040.0 kJ \n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.3:PG-683"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ques3\n",
+ "#calculating equilibrium constant \n",
+ "\n",
+ "dG1=-457.166;#change in gibbs free energy at temp 298 K from example2 in kJ\n",
+ "dG2=-271.040;#change in gibbs free energy at temp 2000 K from example2 n kJ\n",
+ "T1=298;#K\n",
+ "T2=2000;#K\n",
+ "R=8.3145;#gas constant\n",
+ "K1=dG1*1000/(R*T1);\n",
+ "K2=dG2*1000/(R*T2);\n",
+ "print\" Equilibrium constant at \",T1,\"K = \",round(K1,2)\n",
+ "print\" Equilibrium constant at \",T2,\"K = \",round(K2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Equilibrium constant at 298 K = -184.51\n",
+ " Equilibrium constant at 2000 K = -16.299\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.4:Pg-684"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# The example is about proving that Equlibrium constant can be found using table hence doesn't require solution in python "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.5:Pg-685"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# ques 5\n",
+ "# To determine Heat Transfer\n",
+ "# The process is two step as:\n",
+ "# Combustion: C + O2--->CO2\n",
+ "# Dissociation reaction: 2CO2---->2CO + O2\n",
+ "# overall process : C + O2 \u2192 0.5622 CO2 + 0.4378 CO + 0.2189 O2\n",
+ "nCO2=0.5622 # moles of CO2\n",
+ "nCO=0.4378 # moles of CO\n",
+ "nO2=0.2189 # moles of NO2\n",
+ "\n",
+ "# from Table A.9:\n",
+ "hfCO2=0 # enthalpy of formation\n",
+ "hfC=0 # enthalpy of formation\n",
+ "hfCO=0 # enthalpy of formation\n",
+ "hfO2=0 # enthalpy of formation\n",
+ "hfCO23000=-393522 # enthalpy @ 3000K\n",
+ "hfCO3000=-110527 # enthalpy @ 3000K\n",
+ "hfO23000=0 # enthalpy @ 3000K\n",
+ "hfCO2298=-152853 # enthalpy @ 298K\n",
+ "hfCO298=-93504 # enthalpy @ 298K\n",
+ "hfO2298=-98013 # enthalpy @ 298K\n",
+ "\n",
+ "Hr=hfC+hfO2 # enthalpy of reactants\n",
+ "Hp=nCO2*(hfCO2+hfCO23000-hfCO2298)+nCO*(hfCO+hfCO3000-hfCO298)+nO2*(hfO2+hfO23000-hfO2298)\n",
+ "\n",
+ "Qcv=Hp-Hr # using first law\n",
+ "print round(Qcv),\"kJ/kmol C is the heat transfer \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "-121302.0 kJ/kmol C is the heat transfer \n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.6:Pg-687"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# ques 6\n",
+ "# to determine the composition\n",
+ "# the standard equation is C + 2O2 \u2192 CO2 + O2\n",
+ "# from equilibrium C + 2O2 \u2192 (1 \u2212 2z)CO2 + 2zCO + (1 + z)O2\n",
+ "T=3000 # temp in K\n",
+ "P=0.1 # prssure in MPa\n",
+ "z = 0.1553 # from equilibrium equation using table\n",
+ "yCO2=(1-2*z)/(2+z) # mole fraction of CO2\n",
+ "yCO = 2*z/(2+z) # mole fraction of CO\n",
+ "yO2=(1+z)/(2+z) # mole fraction of O2\n",
+ "print \"The mole fraction is \",round(yCO2,3),\" for CO2 \\n\"\n",
+ "print \"The mole fraction is \",round(yCO,3),\" for CO \\n\"\n",
+ "print \"The mole fraction is \",round(yO2,3),\" for O2 \\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mole fraction is 0.32 for CO2 \n",
+ "\n",
+ "The mole fraction is 0.144 for CO \n",
+ "\n",
+ "The mole fraction is 0.536 for O2 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.7:Pg-691"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# ques 7\n",
+ "# to determine the equilibrium composition\n",
+ "# The reaction equation is:\n",
+ "# (1): 2.H2O -->2H2 + O2\n",
+ "# (2): 2 H2O -->H2 + 2OH\n",
+ "# the equilibrium equation is \n",
+ "# H2O \u2192 (1 \u2212 2a \u2212 2b)H2O + (2a + b)H2 + aO2 + 2bOH\n",
+ "P= 0.1 # pressure in MPa\n",
+ "T=3000 # temp in Kelvin\n",
+ "a=0.0534 # using value of K from Table A.11 @ 3000k\n",
+ "b=0.0551 # using value of K from Table A.11 @ 3000k\n",
+ "nH2O= (1-2*a-b) # moles of H2O\n",
+ "nH2=2*a+b# moles of H2\n",
+ "nO2=a# moles of O2\n",
+ "nOH= 2*b# moles of OH\n",
+ "X=nH2O+nH2+nO2+nOH\n",
+ "yH2O=nH2O/X # mole fraction\n",
+ "yH2=nH2/X # mole fraction\n",
+ "yO2=nO2/X # mole fraction\n",
+ "yOH=nOH/X # mole fraction\n",
+ "print \"The mole fraction of H2O is\",round(yH2O,2),\"\\n\"\n",
+ "print \"The mole fraction of H2 is\",round(yH2,2),\"\\n\"\n",
+ "print \"The mole fraction of O2 is\",round(yO2,2),\"\\n\"\n",
+ "print \"The mole fraction of OH is\",round(yOH,2),\"\\n\"\n",
+ "# the answers are slightly different due to approximation in textbook while here the answers are precise"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mole fraction of H2O is 0.72 \n",
+ "\n",
+ "The mole fraction of H2 is 0.139 \n",
+ "\n",
+ "The mole fraction of O2 is 0.046 \n",
+ "\n",
+ "The mole fraction of OH is 0.095 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.8:pg-696"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# ques 8\n",
+ "# determine the the equilibrium composition\n",
+ "P= 1 # pressure in Kpa\n",
+ "T=10000 # temp in Kelvin\n",
+ "z=0.2008 # using k from table \n",
+ "# the chemical equation is Ar -->Ar^(+) + e\u2212\n",
+ "\n",
+ "yAr=(1-z)/(1 + z) # mole fraction of Ar\n",
+ "yArpositive= z/(1+z) # mole fraction of Ar(+)\n",
+ "yenegative=z/(1+z) # mole fraction of ye-\n",
+ "print \"The mole fraction of Ar is\",round(yAr,3),\"\\n\"\n",
+ "print \"The mole fraction of Ar+ is\",round(yArpositive,3),\"\\n\"\n",
+ "print \"The mole fraction of e- is\",round(yenegative,3),\"\\n\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mole fraction of Ar is 0.666 \n",
+ "\n",
+ "The mole fraction of Ar+ is 0.167 \n",
+ "\n",
+ "The mole fraction of e- is 0.167 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |