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diff --git a/Fundamental_of_internal_combustion_engines/chap3.ipynb b/Fundamental_of_internal_combustion_engines/chap3.ipynb new file mode 100755 index 00000000..a0a9064e --- /dev/null +++ b/Fundamental_of_internal_combustion_engines/chap3.ipynb @@ -0,0 +1,1164 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Reactive Sysyem"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.1 Page no 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "E=20.0 #Methanol burned with excess air in percentage \n",
+ "p=1.0 #Pressure of air in bar\n",
+ "t=27.0 #Temperature of air in degree centigrade\n",
+ "O=32.0 #The molecular weight of oxygen\n",
+ "N=28.0 #The molecular weight of nitrogen\n",
+ "R=8314.0 #Universal gas constant in Nm/kmolK\n",
+ "C=32.0 #Molecular weight of methanol\n",
+ "CO=44.0 #Molecular weight of the carbondioxide \n",
+ "H=18.0 #Molecular weight of the water\n",
+ "\n",
+ "#Calculations\n",
+ "S=((1.8*O)+(6.768*N))/C\n",
+ "A=((1.8*O)+(6.768*N))/C\n",
+ "M=1.8+6.768\n",
+ "V=(M*R*(t+273))/(p*10**5)\n",
+ "T=(1+1.8+6.768)\n",
+ "Cm=(1/T)\n",
+ "Om=(1.8/T)\n",
+ "Nm=(6.768/T)\n",
+ "Mr=(Cm*C)+(Om*O)+(Nm*N)\n",
+ "Tp=(1+2+6.768+0.3)\n",
+ "COm=(1/Tp)\n",
+ "Hp=(2/Tp)\n",
+ "Np=(6.768/Tp)\n",
+ "Op=(0.3/Tp)\n",
+ "Mp=(COm*CO)+(Hp*H)+(Np*N)+(Op*O)\n",
+ "Pp=(Hp*p)\n",
+ "D=60\n",
+ "\n",
+ "#Output\n",
+ "print\"(a) The volume of air supplied per kmole of fuel = \",round(V,3),\"m**3/kmole fuel\"\n",
+ "print\"(b) The molecular weight of the reactants = \",round(Mr,3)\n",
+ "print\"The molecular weight of the products =\",round(Mp,3)\n",
+ "print\"(c) The dew point of the products = \",D,\"degree centigrade\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) The volume of air supplied per kmole of fuel = 213.703 m**3/kmole fuel\n",
+ "(b) The molecular weight of the reactants = 29.171\n",
+ "The molecular weight of the products = 27.722\n",
+ "(c) The dew point of the products = 60 degree centigrade\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.2 Page no 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "C1=40.0 #The content of C7H16 in the fuel in percentage\n",
+ "C2=60.0 #The content of C8H18 in the fuel in percentage\n",
+ "d=0.12 #The diameter of the bore in m\n",
+ "l=0.145 #The length of the bore in m\n",
+ "r=8.5 #Compression ratio \n",
+ "p=1.1 #Pressure at exhaust stroke in bar\n",
+ "T=720.0 #The temperature at the exhaust stroke in K\n",
+ "O=32.0 #The molecular weight of oxygen\n",
+ "N=28.0 #The molecular weight of nitrogen\n",
+ "C3=100.0 #Molecular weight of C7H16\n",
+ "C4=114.0 #The molecular weight of C8H18\n",
+ "R=8314.0 #Universal gas constant in Nm/kmolK\n",
+ "CO2=44.0 #Molecular weight of the carbondioxide \n",
+ "C5=28.0 #Molecular weight of the carbonmonoxide\n",
+ "H=18.0 #Molecular weight of the water\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "N2=100-(12+1.5+2.5)\n",
+ "Y=84/3.76\n",
+ "X=13.5/7.6\n",
+ "Z=(22.34-15.25)*2\n",
+ "Hl=(6.4+10.8)/2.0\n",
+ "Hr=7.98\n",
+ "Hd=Hl-Hr\n",
+ "A=((12.58*(O+(3.76*N)))/(((C1/100.0)*C3)+((C2/100.0)*C4)))\n",
+ "Vs=(math.pi/4.0)*d**2*l\n",
+ "Vc=Vs/(r-1)\n",
+ "M=((6.757*CO2)+(0.8446*C5)+(1.408*O)+(47.3*N)+(8.6*H))/(6.757+0.8446+1.408+47.3+8.6)\n",
+ "R1=R/M\n",
+ "m=((p*10**5)*Vc)/(R1*T)\n",
+ "\n",
+ "#Output \n",
+ "print\"(a)The air/fuel ratio = \",round(A,2)\n",
+ "print\"(b)The mass of the exhaust gases in the clearance space = \",round(m*1000,3),\"*10**-3kg\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The air/fuel ratio = 15.93\n",
+ "(b)The mass of the exhaust gases in the clearance space = 0.114 *10**-3kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3 Page no 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "C=0.86 #The amount of carbon content in the 1kg of fuel by weight in kg\n",
+ "H=0.05 #The amount of hydrogen content in the 1kg of fuel by weight in kg\n",
+ "O=0.02 #The amount of oxygen content in the 1kg of fuel by weight in kg\n",
+ "S=0.005 #The amount of sulphur content in the 1kg of fuel by weight in kg\n",
+ "N=0.065 #The amount of nitrogen content in the 1kg of fuel by weight in kg\n",
+ "E=25.0 #The amount of excess air supplied in percentage\n",
+ "o=32.0 #Molecular weight of the oxygen\n",
+ "co=44.0 #Molecular weight of the carbondioxide \n",
+ "c=12.0 #Molecular weight of the carbon\n",
+ "s=32.0 #Molecular weight of the sulphur\n",
+ "so=64.0 #Molecular weight of sulphur dioxide\n",
+ "n=28.0 #Molecular weight of the nitrogen\n",
+ "\n",
+ "#Calculations\n",
+ "o1=(o/c)*C\n",
+ "coa=(co/c)*C\n",
+ "o2=(o/4.0)*H\n",
+ "h2=(36/4.0)*H\n",
+ "o3=(o/s)*S\n",
+ "s1=(so/s)*S\n",
+ "To=o1+o2+o3\n",
+ "Tt=To-O\n",
+ "As=(Tt*100)/23.0\n",
+ "as_=As*(1+(E/100.0))\n",
+ "o2a=0.23*(E/100)*As\n",
+ "n2a=0.77*(1+(E/100))*As\n",
+ "n2e=n2a+N\n",
+ "Tw=coa+n2e+o2a\n",
+ "pco=(coa/Tw)*100\n",
+ "pn=(n2e/Tw)*100\n",
+ "po=(o2a/Tw)*100\n",
+ "mco=(coa/co)\n",
+ "mn=(n2e/n)\n",
+ "mo=(o2a/o)\n",
+ "Tm=mco+mn+mo\n",
+ "vco=(mco/Tm)*100\n",
+ "vn=(mn/Tm)*100\n",
+ "vo=(mo/Tm)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)Stoichiometric air/fuel ratio = \",round(As,2) \n",
+ "print\"(b)The percentage of dry products of combustion by weight :\"\n",
+ "print\" CO2 = \",round(pco,2),\"percent\"\n",
+ "print\"N2 = \",round(pn,2),\"percent\"\n",
+ "print\"O2 = \",round(po,2),\"percent\"\n",
+ "print\"(c)The percentage of dry products of combustion by volume : \"\n",
+ "print\"CO2 = \",round(vco,2),\"percent\"\n",
+ "print\"N2 = \",round(vn,2),\"percent\"\n",
+ "print\"O2= \",round(vo,2),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Stoichiometric air/fuel ratio = 11.64\n",
+ "(b)The percentage of dry products of combustion by weight :\n",
+ " CO2 = 20.89 percent\n",
+ "N2 = 74.68 percent\n",
+ "O2 = 4.44 percent\n",
+ "(c)The percentage of dry products of combustion by volume : \n",
+ "CO2 = 14.47 percent\n",
+ "N2 = 81.3 percent\n",
+ "O2= 4.23 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.4 Page no 71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "CO=12.0 #The composition of carbondioxide of combustion by volume in percentage \n",
+ "C=0.5 #The composition of carbonmoxide of combustion by volume in percentage \n",
+ "O=4.0 #The composition of oxygen of combustion by volume in percentage \n",
+ "N=83.5 #The composition of nitrogen of combustion by volume in percentage \n",
+ "o=32.0 #Molecular weight of the oxygen\n",
+ "co=44.0 #Molecular weight of the carbondioxide \n",
+ "c=12.0 #Molecular weight of the carbon\n",
+ "s=32.0 #Molecular weight of the sulphur\n",
+ "so=64.0 #Molecular weight of sulphur dioxide\n",
+ "n1=28.0 #Molecular weight of the nitrogen\n",
+ "h=2.0 #Molecular weight of the hydrogen\n",
+ "\n",
+ "#Calculations\n",
+ "m=12+0.5\n",
+ "x=N/3.76\n",
+ "z=(x-(CO+(C/2)+O))*2\n",
+ "n=z*h\n",
+ "Af=((x*o)+(N*n1))/((m*c)+(n))\n",
+ "As=((18.46*o)+(69.41*n1))/173.84\n",
+ "Ta=(Af/As)*100\n",
+ "mc=((m*c)/173.84)*100\n",
+ "mh=(n/173.84)*100\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)The air/fuel ratio = \",round(Af,1)\n",
+ "print\"(b)The percent theoretical air = \",round(Ta,1),\"%\"\n",
+ "print\"(c)The percentage composition of fuel on a mass basis : \" \n",
+ "print\"C = \",round(mc,1),\"%\"\n",
+ "print\"H = \",round(mh,1),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The air/fuel ratio = 17.5\n",
+ "(b)The percent theoretical air = 120.3 %\n",
+ "(c)The percentage composition of fuel on a mass basis : \n",
+ "C = 86.3 %\n",
+ "H = 13.7 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.5 Page no 72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "C=86.0 #The composition of carbon in the fuel by weight in percentage\n",
+ "H=14.0 #The composition of hydrogen in the fuel by weight in percentage\n",
+ "e=1.25 #Equivalent ratio\n",
+ "o=32.0 #Molecular weight of the oxygen\n",
+ "co=44.0 #Molecular weight of the carbondioxide \n",
+ "c=12.0 #Molecular weight of the carbon\n",
+ "s=32.0 #Molecular weight of the sulphur\n",
+ "so=64.0 #Molecular weight of sulphur dioxide\n",
+ "n=28.0 #Molecular weight of the nitrogen\n",
+ "h2=2.0 #Molecular weight of the hydrogen\n",
+ "Fc=0.86 #Fraction of C\n",
+ "\n",
+ "#Calculations\n",
+ "Ra=1/Fc\n",
+ "x=2*(1+(0.9765/2.0)-(1.488*0.8))\n",
+ "Tm=0.5957+0.4043+4.476\n",
+ "vc=(0.5957/Tm)*100\n",
+ "vco=(0.4043/Tm)*100\n",
+ "vn=(4.476/Tm)*100\n",
+ "\n",
+ "#Calculations\n",
+ "print\"The percentage analysis of dry exhaust gas by volume : \"\n",
+ "print\"CO = \",round(vc,2),\"percent\" \n",
+ "print\"CO2 = \",round(vco,2),\"percent\" \n",
+ "print\"N2 = \",round(vn,2),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage analysis of dry exhaust gas by volume : \n",
+ "CO = 10.88 percent\n",
+ "CO2 = 7.38 percent\n",
+ "N2 = 81.74 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.6 Page no 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25.0 #The temperature of both reactants and products in degree centigrade\n",
+ "p=1.0 #The pressure of both reactants and products in bar\n",
+ "\n",
+ "#Calculations \n",
+ "h=0\n",
+ "hf1=-103.85\n",
+ "hf2=-393.52\n",
+ "hf3=-285.8\n",
+ "hf4=(3*hf2)+(4*hf3)\n",
+ "Q=hf4-hf1\n",
+ "\n",
+ "#Output\n",
+ "print\" The heat transfer per mole of fuel = \",Q,\"kJ/mol fuel\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The heat transfer per mole of fuel = -2219.91 kJ/mol fuel\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.7 Page no 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25.0 #The temperature of the air entering the diesel engine in degree centigrade \n",
+ "T=600.0 #The temperature at which the products are released in K\n",
+ "Ta=200.0 #Theoretical air used in percentage \n",
+ "Q=-93.0 #Heat loss from the engine in MJ/kmol fuel\n",
+ "f=1.0 #The fuel rate in kmol/h\n",
+ "\n",
+ "#Calculations \n",
+ "hfr=-290.97\n",
+ "h1=-393.52\n",
+ "h11=12.916\n",
+ "hfc=h1+h11\n",
+ "h2=-241.82\n",
+ "h22=10.498\n",
+ "hfh=h2+h22\n",
+ "h3=0 \n",
+ "h33=9.247\n",
+ "hfo=h3+h33\n",
+ "h4=0\n",
+ "h44=8.891\n",
+ "hfn=h4+h44\n",
+ "hfp=(12*hfc)+(13*hfh)+(18.5*hfo)+(139.12*hfn)\n",
+ "W=Q+hfr-hfp\n",
+ "W1=(f*W*10**3)/3600.0\n",
+ "\n",
+ "#Output\n",
+ "print\"The work for a fuel rate of 1 kmol/h is \",round(W1,0),\"kW\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The work for a fuel rate of 1 kmol/h is 1606.0 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.8 Page no 78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "P=600 #Power of an engine in kW\n",
+ "t=25 #Temperature at which fuel is used in degree centigrade\n",
+ "Ta=150 #Theoretical air used in percentage\n",
+ "T1=400 #The temperature at which air enters in K\n",
+ "T2=700 #The temperature at which the products of combustion leave in K\n",
+ "Q=-150 #The heat loss from the engine in kW\n",
+ "C=12 #Molecular weight of carbon\n",
+ "h=1 #Molecular weight of hydrogen\n",
+ "\n",
+ "#Calculations\n",
+ "hfc=-259.28\n",
+ "hfo1=3.029\n",
+ "hfn1=2.971\n",
+ "HR=(hfc)+(1.5*12.5*hfo1)+(1.5*12.5*3.76*hfn1)\n",
+ "hfco=-393.52\n",
+ "hfco1=17.761\n",
+ "hfh=-241.82\n",
+ "hfh1=14.184\n",
+ "hfo2=12.502\n",
+ "hfn2=11.937\n",
+ "HP=(8*(hfco+hfco1))+(9*(hfh+hfh1))+(6.25*hfo2)+(70.5*hfn2)\n",
+ "H=HP-HR\n",
+ "nf=((Q-P)*3600)/(H*10.0**3)\n",
+ "M=(8*C)+(18*h)\n",
+ "mf=nf*M\n",
+ "\n",
+ "#Output\n",
+ "print\" The fuel consumption for complete combustion is \",round(mf,1),\"kg/h\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The fuel consumption for complete combustion is 74.3 kg/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.9 Page no 79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25 #Temperature at which fuel is used for combustion in degree centigrade \n",
+ "p=1 #The pressure at which fuel is used in bar\n",
+ "T=400 #The temperature of the products of combustion in K\n",
+ "R=8.314*10**-3 #Universal gas constant\n",
+ "hfco=-393.52 #The enthalpy of the carbondioxide in MJ/kmol fuel\n",
+ "hfco1=4.008 #The change in enthalpy of the carbondioxide for the given conditions in MJ/kmol fuel\n",
+ "hfh=-241.82 #The enthalpy of the water in MJ/kmol fuel\n",
+ "hfh1=3.452 #The change in enthalpy of the water for the given conditions in MJ/kmol fuel\n",
+ "\n",
+ "#Calculations\n",
+ "hfc=-103.85 \n",
+ "HR=(1*(hfc-(R*(t+273))))+(5*(-R*(t+273)))\n",
+ "HP=(3*(hfco+hfco1-(R*T)))+(4*(hfh+hfh1-(R*T)))\n",
+ "Q=HP-HR\n",
+ "Q1=-Q\n",
+ "\n",
+ "#Output\n",
+ "print\"The heat transfer per mole of propane = \",round(Q1,1),\"kJ/mol propane\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat transfer per mole of propane = 2026.6 kJ/mol propane\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.10 Page no 82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=1500 #The given temperature in K\n",
+ "hfco=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+ "hf1=61.714 #The change in enthalpy for actual state and reference state in MJ/kmol\n",
+ "hfc=-110.52 #The enthalpy of formation for carbonmonoxide in MJ/kmol\n",
+ "hf2=38.848 #The change in enthalpy of CO for actual and reference state in MJ/kmol\n",
+ "hfo=0 #The enthalpy of formation for oxygen gas\n",
+ "hf3=40.61 #The change in enthalpy of oxygen for different states in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "HP=hfco+hf1\n",
+ "HR=(hfc+hf2)+(0.5*(hfo+hf3))\n",
+ "H=HP-HR\n",
+ "\n",
+ "#Output\n",
+ "print\" The enthalpy of combustion is \",round(H,1),\"MJ/kmol CO\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The enthalpy of combustion is -280.4 MJ/kmol CO\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 Page no 83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "E=30 #The amount of excess air in percentage\n",
+ "tp=400 #The temperature at which propane enters in K\n",
+ "ta=300 #The temperature at which air enters in K\n",
+ "T=900 #The temperature at which products leave in K\n",
+ "m=83.7 #The average molar specific heat of propane at consmath.tant pressure in kJ/kmolK\n",
+ "Mp=44 #The molecular weight of propane\n",
+ "hfc=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+ "hf1=28.041 #The change in enthalpy of CO2 for actual and reference state in MJ/kmol\n",
+ "hfh=-241.82 #The enthalpy of formation for water in MJ/kmol\n",
+ "hf2=21.924 #The change in enthalpy of water for actual and reference state in MJ/kmol\n",
+ "hfn=0 #The enthalpy of nitrogen gas \n",
+ "hf3=18.221 #The change in enthalpy of nitrgen for actual and reference state in MJ/kmol\n",
+ "hfo=0 #The enthalpy of oxygen gas \n",
+ "hf4=19.246 #The change in enthalpy of oxygen for actual and reference state in MJ/kmol\n",
+ "hfp=-103.85 #The enthalpy of formation for propane in MJ/kmol\n",
+ "R=0.0837 #Universal gas constant \n",
+ "hfo1=0 #The enthalpy of oxygen gas \n",
+ "hf11=0.054 #The change in enthalpy of oxygen gas for actual and reference state in MJ/kmol\n",
+ "hfn1=0 #The enthalpy of nitrogen gas\n",
+ "hfn22=0.054 #The change in enthalpy of nitrogen for actual and reference state in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "HP=(3*(hfc+hf1))+(4*(hfh+hf2))+(24.44*(hfn+hf3))+(1.5*(hfo+hf4))\n",
+ "HR=(1*(hfp+(R*(tp-ta))))+(6.5*(hfo1+hf11))+(24.44*(hfn1+hfn22))\n",
+ "Q=HP-HR\n",
+ "Q1=(-Q/Mp)\n",
+ "\n",
+ "#Output\n",
+ "print\" The amount of heat transfer per kg of fuel is \",round(Q1,3),\"MJ/kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The amount of heat transfer per kg of fuel is 32.0 MJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 Page no 84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "Ta=150 #The presence of Theoretical air\n",
+ "hfc=-393.52 #The enthalpy of formation for carbondioxide in MJ/kmol\n",
+ "hfh=-285.8 #The enthalpy of formation for water in MJ/kmol\n",
+ "hfon=0 #The enthalpy of formation for oxygen and nitrogen gas \n",
+ "hfch=-74.87 #The enthalpy of formation for methane in MJ/kmol\n",
+ "np=2 #Number of moles of product\n",
+ "nr=4 #Number of moles of reactant\n",
+ "R=8.314*10**-3 #Universal gas constant \n",
+ "t=298 #The temperature in K\n",
+ "hfh1=-241.82 #The enthalpy of formation for water in MJ/kmol\n",
+ "np1=4 #Number of moles of product\n",
+ "nr1=4 #Number of moles of reactant\n",
+ "\n",
+ "#Calculations\n",
+ "HP=(hfc)+(2*hfh)\n",
+ "HR=1*hfch\n",
+ "H=HP-HR\n",
+ "n=np-nr\n",
+ "U1=H-(n*R*t)\n",
+ "HP1=(1*hfc)+(2*hfh1)\n",
+ "H1=HP1-HR\n",
+ "n1=np1-nr1\n",
+ "U2=H1-(n1*R*t)\n",
+ "\n",
+ "#Output\n",
+ "print\"(a)The water as liquid\" \n",
+ "print\"The standard enthalpy of combustion is \",H, \"MJ/kmol\"\n",
+ "print\"The standard internal energy of combustion is \",round(U1,3),\"MJ/kmol\"\n",
+ "print\"(b)The water as a gas \"\n",
+ "print\"The standard enthalpy of combustion is \",H1,\"MJ/kmol\"\n",
+ "print\"The standard internal energy of combustion is \",U2,\"MJ/kmol\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The water as liquid\n",
+ "The standard enthalpy of combustion is -890.25 MJ/kmol\n",
+ "The standard internal energy of combustion is -885.295 MJ/kmol\n",
+ "(b)The water as a gas \n",
+ "The standard enthalpy of combustion is -802.29 MJ/kmol\n",
+ "The standard internal energy of combustion is -802.29 MJ/kmol\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.13 Page no 85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "cv=44000 #The lower calorific value of liquid fuel in kJ/kg\n",
+ "C=84 #The carbon content present in the fuel in percentage\n",
+ "H=16 #The hydrogen content present in the fuel in percentage\n",
+ "t=25 #The temperature in degree centigrade\n",
+ "hfg=2442 #The enthalpy of vaporization for water in kJ/kg\n",
+ "c=12.0 #Molecular weight of carbon \n",
+ "h=2 #Molecular weight of hydrogen\n",
+ "co2=44.0 #Molecular weight of carbondioxide\n",
+ "h2o=18 #Molecular weight of water \n",
+ "o2=32.0 #Molecular weight of oxygen\n",
+ "R=8.314 #Universal gas constant in J/molK\n",
+ "\n",
+ "#Calculations\n",
+ "CO2=(0.84*(co2/c))\n",
+ "H2O=(0.16*(h2o/h))\n",
+ "cvd=H2O*hfg\n",
+ "HHV=cv+cvd\n",
+ "np=3.08/co2\n",
+ "nr=3.52/o2\n",
+ "n=np-nr\n",
+ "HHVv=HHV+(n*R*(t+273))\n",
+ "LHVv=cv+(n*R*(t+273))\n",
+ "\n",
+ "#Output\n",
+ "print\" The higher calorific value at constant pressure = \",HHV,\"kJ/kg fuel\" \n",
+ "print\"The higher calorific value at constant volume = \",round(HHVv,0),\"kJ/kg fuel\"\n",
+ "print\"The lower calorific value at constant volume = \",round(LHVv,0),\"kJ/kg fuel\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The higher calorific value at constant pressure = 47516.48 kJ/kg fuel\n",
+ "The higher calorific value at constant volume = 47417.0 kJ/kg fuel\n",
+ "The lower calorific value at constant volume = 43901.0 kJ/kg fuel\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.14 Page no 87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "E=100 #The amount of excess air in percent\n",
+ "T=298 #The temperature of reactants in K\n",
+ "nc=1 #Number of moles of propane\n",
+ "hfch=-103.85 #Enthalpy of formation for propane in MJ/kmol fuel\n",
+ "hfc=-393.52 #Enthalpy of formation for carbondioxide in MJ/kmol fuel\n",
+ "hfh=-241.82 #Enthalpy of formation for water in MJ/kmol fuel\n",
+ "hfon=0 #Enthalpy of formation for both oxygen and nitrogen gas\n",
+ "T1=1500 #Assuming the products temperature for fist trail in K\n",
+ "hfc1=61.714 #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel\n",
+ "hfh1=48.095 #The change in enthalpy for water for trail temp in MJ/kmol fuel\n",
+ "hfo1=40.61 #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel\n",
+ "hfn1=38.405 #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel\n",
+ "T2=1600 #Assuming the products temperature for second trail in K\n",
+ "hfc2=67.58 #The change in enthalpy for corbondioxide for trail temp in MJ/kmol fuel\n",
+ "hfh2=52.844 #The change in enthalpy for water for trail temp in MJ/kmol fuel\n",
+ "hfo2=44.279 #The change in enthalpy for oxygen for trail temp in MJ/kmol fuel\n",
+ "hfn2=41.903 #The change in enthalpy for nitrogen for trail temp in MJ/kmol fuel\n",
+ "\n",
+ "#Calculations\n",
+ "HR=nc*hfch\n",
+ "x=HR-((3*hfc)+(4*hfh)+(5*hfon)+(37.6*hfon))\n",
+ "hfn=x/37.6\n",
+ "HP1=(HR-x)+(3*hfc1)+(4*hfh1)+(5*hfo1)+(37.6*hfn1)\n",
+ "HP2=(HR-x)+(3*hfc2)+(4*hfh2)+(5*hfo2)+(37.6*hfn2)\n",
+ "Te=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1\n",
+ "\n",
+ "#Output\n",
+ "print\" The adiabatic flame temperature for steady-flow process is \",round(Te,0),\"K\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The adiabatic flame temperature for steady-flow process is 1510.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 Page no 89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=600 #The initial temperature of air in K\n",
+ "p=1 #The initial pressure of air in atm\n",
+ "R=8.314 #Universal gas constant in J/molK\n",
+ "Tr=298 #The temperature of reactants in K\n",
+ "a=4.503 #Given Constants \n",
+ "b=-8.965*10**-3\n",
+ "c=37.38*10**-6\n",
+ "d=-36.49*10**-9\n",
+ "e=12.22*10**-12\n",
+ "hfc=-393.52 #Enthalpy of formation for carbondioxide in MJ/kmol fuel\n",
+ "hfh=-241.82 #Enthalpy of formation for water in MJ/kmol fuel\n",
+ "hfn=0 #Enthalpy of formation for nitrogen gas\n",
+ "hfc1=-74.87 #The enthalpy of formation for methane in MJ/kmol fuel \n",
+ "hfh1=9.247 #The change in enthalpy of the water in MJ/kmol\n",
+ "hfn1=8.891 #The change in enthalpy of nitrogen in MJ/kmol\n",
+ "Tc=3700 #The corresponding temperature for the enthalpy of guess nitrogen in K\n",
+ "T1=2800 #The temperature assumed for the first trail in K\n",
+ "hco1=140.444 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hh1=115.294 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hn1=85.345 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "T2=2500 #The temperature assumed for the second trail in K\n",
+ "hco2=121.926 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hh2=98.964 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hn2=74.312 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "T3=2600 #The temperature fo the third trail in K\n",
+ "hco3=128.085 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hh3=104.37 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hn3=77.973 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "Tc1=3000 #Assume temperature for first trail in K\n",
+ "hcoa1=146.645 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hha1=120.813 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hna1=89.036 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "Tc2=3200 #Assume temperature for the second trail in K\n",
+ "hcoa2=165.331 #The change in enthalpy for the assume temp for carbondioxide in MJ/kmol\n",
+ "hha2=137.553 #The change in enthalpy for the assume temp for water in MJ/kmol\n",
+ "hna2=100.161 #The change in enthalpy for the assume temp for nitrogen in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "HP=(1*hfc)+(2*hfh)+(7.52*hfn)\n",
+ "hch=(R*((a*(T-Tr))+((b/2.0)*(T**2-Tr**2))+((c/3.0)*(T**3-Tr**3))+((d/4.0)*(T**4-Tr**4))+((e/5.0)*(T**5-Tr**5))))/1000.0\n",
+ "HR=((hfc1+hch)+(2*hfh1)+(7.52*hfn1))\n",
+ "x=HR-HP\n",
+ "hfn2=x/7.52\n",
+ "HP1=hco1+(2*hh1)+(7.52*hn1)+(HR-x)\n",
+ "HP2=hco2+(2*hh2)+(7.52*hn2)+(HR-x)\n",
+ "HP3=hco3+(2*hh3)+(7.52*hn3)+(HR-x)\n",
+ "Ta1=(((HR-HP2)/(HP3-HP2))*(T3-T2))+T2\n",
+ "UR1=HR-(10.52*R*10**-3*T)\n",
+ "UP1=hcoa1+(2*hha1)+(7.52*hna1)+(HR-x)-(0.08746*Tc1)\n",
+ "UP2=hcoa2+(2*hha2)+(7.52*hna2)+(HR-x)-(0.08746*Tc2)\n",
+ "Tu=(((UR1-UP1)/(UP2-UP1))*(Tc2-Tc1))+Tc1\n",
+ "\n",
+ "#Output\n",
+ "print\"The adiabatic flame temperature at \"\n",
+ "print\"(a)Constant pressure process is \",round(Ta1,0),\"K\" \n",
+ "print\"(b)Constant volume process is \",round(Tu,3),\"K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The adiabatic flame temperature at \n",
+ "(a)Constant pressure process is 2550.0 K\n",
+ "(b)Constant volume process is 3089.34 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.16 Page no 91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T=600.0 #Temperature at constant pressure process in K\n",
+ "p=1.0 #The pressure in atm\n",
+ "E=50.0 #The amount of excess air in percent\n",
+ "L=20.0 #The amount of less air in percent\n",
+ "cp=52.234 #Specific constant for methane in kJ/kmolK\n",
+ "t=298.0 #Assume the normal temperature in K\n",
+ "hfch=-74.87 #The enthalpy of formation for carbondioxide in MJ\n",
+ "ho=9.247 #The change in enthalpy of oxygen in MJ\n",
+ "hn=8.891 #The change in enthalpy of nitrogen in MJ\n",
+ "hfc1=-393.52 #The enthalpy of formation of carbondioxide in MJ\n",
+ "hfh1=-241.82 #The enthalpy of formation of water in MJ\n",
+ "Tc=2800.0 #The corresponding temperature in K\n",
+ "T1=2000 #The temperature for first trail in K\n",
+ "hfc11=91.45 #The enthalpy for the assume temp for carbondioxide in MJ\n",
+ "hfh11=72.689 #The change in enthalpy for the assume temp for water in MJ\n",
+ "hfn11=56.141 #The change in enthalpy for the assume temp for nitrogen in MJ\n",
+ "hfo11=59.199 #The change in enthalpy for the assume temp for oxygen in MJ\n",
+ "T2=2100 #The temperature for second trail in K\n",
+ "hfc22=97.5 #The enthalpy for the assume temp for carbondioxide in MJ\n",
+ "hfh22=77.831 #The change in enthalpy for the assume temp for water in MJ\n",
+ "hfn22=59.748 #The change in enthalpy for the assume temp for nitrogen in MJ\n",
+ "hfo22=62.986 #The change in enthalpy for the assume temp for oxygen in MJ\n",
+ "hfchr=-74.87 #The enthalpy of formation for methane in MJ\n",
+ "hor=9.247 #The change in enthalpy for oxygen in MJ\n",
+ "hnr=8.891 #The change in enthalpy for nitrogen in MJ\n",
+ "hfcop=-110.52 #The formation of enthalpy for carbonmoxide in MJ\n",
+ "hfcp=-393.52 #The formation of enthalpy for carbondioxide in MJ\n",
+ "hfhp=-241.82 #The formation of enthalpy for water in MJ\n",
+ "Tp1=2000.0 #The temperature for first trail in K\n",
+ "hco11=56.739 #The change in enthalpy for CO in MJ\n",
+ "hco211=91.45 #The change in enthalpy for CO2 in MJ\n",
+ "hh11=72.689 #The change in enthalpy for water in MJ\n",
+ "hn11=56.141 #The change in enthalpy for nitrogen in MJ\n",
+ "Tp2=2400 #The temperature for second trail in K\n",
+ "hco22=71.34 #The change in enthalpy for CO in MJ\n",
+ "hco222=115.788 #The change in enthalpy for CO2 in MJ\n",
+ "hh22=93.604 #The change in enthalpy for water in MJ\n",
+ "hn22=70.651 #The change in enthalpy for nitrogen in MJ\n",
+ "Tp3=2300.0 #The temperature for first trail in K\n",
+ "hco33=67.676 #The change in enthalpy for CO in MJ\n",
+ "hco233=109.671 #The change in enthalpy for CO2 in MJ\n",
+ "hh33=88.295 #The change in enthalpy for water in MJ\n",
+ "hn33=67.007 #The change in enthalpy for nitrogen in MJ\n",
+ "hccc=-283.022 #The only combustible substance is CO in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "hch=cp*(T-t)*10**-3\n",
+ "HR=hfch+hch+(3*ho)+(11.28*hn)\n",
+ "HP=hfc1+(2*hfh1)\n",
+ "x=HR-HP\n",
+ "hn2=x/11.28\n",
+ "HP1=hfc11+(2*hfh11)+(11.28*hfn11)+(hfo11)+(HR-x)\n",
+ "HP2=hfc22+(2*hfh22)+(11.28*hfn22)+(hfo22)+(HR-x)\n",
+ "Ta1=(((HR-HP1)/(HP2-HP1))*(T2-T1))+T1\n",
+ "X=2*(2-1.6)\n",
+ "HRr=hfchr+hch+(1.6*hor)+(6.01*hnr)\n",
+ "HPp=(0.8*hfcop)+(0.2*hfcp)+(2*hfhp)\n",
+ "HPp1=(0.8*hco11)+(0.2*hco211)+(2*hh11)+(6.016*hn11)-HPp\n",
+ "HPp2=(0.8*hco22)+(0.2*hco222)+(2*hh22)+(6.016*hn22)+HPp\n",
+ "HPp3=(0.8*hco33)+(0.2*hco233)+(2*hh33)+(6.016*hn33)+HPp\n",
+ "Ta2=(((HRr-HPp3)/(HPp2-HPp3))*(Tp2-Tp3))+Tp3\n",
+ "Q=-0.8*hccc #The thermal energy loss in MJ/kmol fuel\n",
+ "\n",
+ "#Output\n",
+ "print\" The adiabatic flame temperature having \"\n",
+ "print\"(a)50 percent excess air is \",round(Ta1,1),\"K\"\n",
+ "print\"(b)20 percent less air is \",round(Ta2,2),\"K\"\n",
+ "print\"The loss of thermal energy due to incomplete combustion is \",round(Q,2),\"MJ/kmol fuel\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The adiabatic flame temperature having \n",
+ "(a)50 percent excess air is 2027.6 K\n",
+ "(b)20 percent less air is 2311.22 K\n",
+ "The loss of thermal energy due to incomplete combustion is 226.42 MJ/kmol fuel\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.18 Page no 98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "T1=3000 #Given temperature in K\n",
+ "T2=4000 #Given temperature in K\n",
+ "p=1 #The pressure in atm\n",
+ "KP1=1.117 #Natural logarithm of equilibrium constant at 3000 K \n",
+ "KP2=-1.593 #Natural logarithm of equilibrium constant at 4000 K\n",
+ "a1=0.4 #The dissociation of 1 mole of CO2 for the first trail\n",
+ "a2=0.5 #The dissociation of 1 mole of CO2 for the second trail \n",
+ "K1=3.674 #The value of equilibrium constant for the first trail \n",
+ "K2=2.236 #The value of equilibrium constant for the second trail\n",
+ "a3=0.9 #The dissociation of 1 mole of CO2 for the first trail\n",
+ "a4=0.89 #The dissociation of 1 mole of CO2 for the second trail\n",
+ "K3=0.1995 #The value of equilibrium constant for the first trail \n",
+ "K4=0.2227 #The value of equilibrium constant for the second trail \n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "Kp1=math.exp(KP1)\n",
+ "Kp2=math.exp(KP2)\n",
+ "a12=(((K1-Kp1)/(K1-K2))*(a2-a1))+a1\n",
+ "A12=a12*100\n",
+ "a23=(((Kp2-K4)/(K3-K4))*(a3-a4))+a4\n",
+ "A23=a23*100\n",
+ "\n",
+ "#output\n",
+ "print\"The percent dissociation of carbondioxide into carbonmonoxide and oxygen at \"\n",
+ "print\"(a) at 3000 K and 1 atm pressure = \",round(A12,3),\"percent\" \n",
+ "print\"(b) at 4000 K and 1 atm pressure = \",round(A23,1),\"percent\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percent dissociation of carbondioxide into carbonmonoxide and oxygen at \n",
+ "(a) at 3000 K and 1 atm pressure = 44.3 percent\n",
+ "(b) at 4000 K and 1 atm pressure = 89.8 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.19 Page no 99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "p=1 #Initial pressure in atm\n",
+ "T=300 #Initial temperature in K\n",
+ "Tc=2400 #To calculate the molefraction of the products at this temperature in K\n",
+ "KP1=3.866 #Natural logarithm of equilibrium constant at 2400 K for the equation\n",
+ "a=0.098 #The dissociation of 1 mole of CO2\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "K1=math.exp(KP1)\n",
+ "nr=1+0.5\n",
+ "Pp=(p*Tc)/(nr*T)\n",
+ "np=(a+2)/2.0\n",
+ "xco=(2*(1-a))/(2+a)\n",
+ "xc=(2*a)/(2+a)\n",
+ "xo=a/(2.0+a)\n",
+ "PP=5.333*np\n",
+ "\n",
+ "#output\n",
+ "print\"Mole fraction of the carbondioxide is \",round(xco,3)\n",
+ "print\"Mole fraction of the carbonmonoxide is \",round(xc,3)\n",
+ "print\"Mole fraction of oxygen is \",round(xo,3)\n",
+ "print\"Pressure of the product is \",round(PP,3),\"bar\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mole fraction of the carbondioxide is 0.86\n",
+ "Mole fraction of the carbonmonoxide is 0.093\n",
+ "Mole fraction of oxygen is 0.047\n",
+ "Pressure of the product is 5.594 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.20 Page no 100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "t=25.0 #The temperature of air in degree centigrade\n",
+ "p=1.0 #The pressure of air in atm\n",
+ "T1=2200.0 #Given first temperature in K\n",
+ "T2=2400.0 #Given second temperature in K\n",
+ "h1=59.86 #The change in enthalpy of hydrogen at 2200 K in MJ/kmol\n",
+ "h2=66.915 #The change in enthalpy of hydrogen at 2400 K in MJ/kmol\n",
+ "T=298.0 #The temperature of air in K\n",
+ "HR=0 #The total enthalpy on the reactants side since all the reactants are elements\n",
+ "Kp1=-6.774 #Natural logarithm of equilibrium constant at 2200 K for the equation \n",
+ "a1=0.02 #By trail and error method the degree of dissociation of H2O\n",
+ "hfh=-241.82 #The enthalpy of formation of water at both 2200 and 2400 K in MJ/kmol\n",
+ "hfh1=83.036 #The change in enthalpy of water at 2200 K in MJ/kmol\n",
+ "hfd1=59.86 #The change in enthalpy of hydrogen at 2200 K in MJ/kmol\n",
+ "hfo1=66.802 #The change in enthalpy of oxygen at 2200 K in MJ/kmol\n",
+ "hfn1=63.371 #The change in enthalpy of nitrogen at 2200 K in MJ/kmol\n",
+ "a2=0.04 #By trail and error method the degree of dissociation of H2O at 2400 K\n",
+ "hfh2=93.604 #The change in enthalpy of water at 2400 K in MJ/kmol\n",
+ "hfd2=66.915 #The change in enthalpy of hydrogen at 2400 K in MJ/kmol\n",
+ "hfo2=74.492 #The change in enthalpy of oxygen at 2400 K in MJ/kmol\n",
+ "hfn2=70.651 #The change in enthalpy of nitrogen at 2400 K in MJ/kmol\n",
+ "\n",
+ "#Calculations\n",
+ "import math\n",
+ "K1=math.exp(Kp1)\n",
+ "HP1=(0.98*(hfh+hfh1))+(0.02*hfd1)+(0.01*hfo1)+(1.88*hfn1) \n",
+ "HP2=(0.96*(hfh+hfh2))+(0.04*hfd2)+(0.02*hfo2)+(1.88*hfn2) \n",
+ "H1=HP1-HR\n",
+ "H2=HP2-HR\n",
+ "Tl=(((T2-T1)/(HP2-HP1))*(HR-HP1))+T1\n",
+ "\n",
+ "#Output\n",
+ "print\"The adiabatic flame temperature taking dissociation into account is \",T1+236,\"K\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The adiabatic flame temperature taking dissociation into account is 2436.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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