path: root/Fundamental_of_internal_combustion_engines/chap3.ipynb

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  {
   "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": {}
  }
 ]
}