path: root/Fundamental_of_internal_combustion_engines/chap4.ipynb

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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 4:Fuel Air Cycles and their analysis"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.1 Page No 117"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=8.5                                #The compression ratio \n",
      "sv=1.4                               #The specific heat at constant volume in percent\n",
      "\n",
      "#Calculations\n",
      "import math\n",
      "n=1-(1/r)**(sv-1) \n",
      "ef=(((1-n)/n)*(sv-1)*(math.log(r))*(sv/100.0))*100\n",
      "\n",
      "#Output\n",
      "print\"The efficiency decreases by \",round(ef,3),\"percent\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The efficiency decreases by  0.885 percent\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.2 Page No 118"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=18.0                                      #The compression ratio \n",
      "l=6.0                                       #The cut off taking place corresponding of the stroke in percent\n",
      "sc=2.0                                      #The specific heat at constant volume increases in percent\n",
      "cv=0.717                                  #The specific heat at constant volume in kJ/kgK\n",
      "R=0.287                                   #Gas constant in kJ/kgK\n",
      "\n",
      "#Calculations\n",
      "import math\n",
      "Vs=(r-1)\n",
      "B=((l/100.0)*Vs)+1\n",
      "cp=cv+R\n",
      "R1=cp/cv\n",
      "n=1-(((((1/r)**(R1-1))*(B**R1-1))/(R1*(B-1)))) \n",
      "dn=(((1-n)/n)*((R1-1)*((math.log(r))-(((B**R1)*math.log(B))/(B**R1-1))+(1/B)))*(sc/100.0))*100\n",
      "\n",
      "#Output\n",
      "print\"The efficiency decreases by \",round(dn,1),\"percent\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The efficiency decreases by  1.1 percent\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.3 Page No 120"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=8                                    #The compression ratio\n",
      "af=15                                  #Air/fuel ratio\n",
      "p1=1                                   #The pressure at the beginning of a compression stroke in bar\n",
      "t=60                                   #The temperature at the beginning of a compression stroke in degree centigrade\n",
      "cv=44000                               #The calorific value of the fuel in kJ/kg\n",
      "n=1.32                                 #The index of the compression \n",
      "Cv=0.717                               #specific heat at constant volume in kJ/kgK\n",
      "\n",
      "#Calculations\n",
      "T1=t+273\n",
      "p2=p1*(r)**n\n",
      "T2=T1*r**(n-1)\n",
      "f=(1/(af+1))\n",
      "a=(af/(af+1))\n",
      "q23=cv/(af+1)\n",
      "T3=((-10430+((10430)**2+(4*494.8*10**5))**(1/2.0))/2.0)\n",
      "p3=(T3/T1)*(r)*p1\n",
      "T31=(q23/Cv)+T2\n",
      "p31=(T31/T1)*r*p1\n",
      "\n",
      "#Output\n",
      "print\"(a) The Maximum temperature in the cylinder  = \",round(T3,0),\"K\" \n",
      "print\"The Maximum pressure in the cylinder P3 = \",round(p3,0),\"bar\" \n",
      "print\"(b)With constant value of Cv \"\n",
      "print\"The Maximum temperature in the cylinder  = \",round(T31,0),\"K\" \n",
      "print\"The Maximum pressure in the cylinder P3 = \",round(p31,1),\"bar\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) The Maximum temperature in the cylinder  =  3541.0 K\n",
        "The Maximum pressure in the cylinder P3 =  85.0 bar\n",
        "(b)With constant value of Cv \n",
        "The Maximum temperature in the cylinder  =  4483.0 K\n",
        "The Maximum pressure in the cylinder P3 =  107.7 bar\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.4 Page No 121"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=21                                     #The compression ratio \n",
      "af=29                                    #Air/fuel ratio\n",
      "T=1000                                   #The temperature at the end of compression in K\n",
      "cv=42000                                 #The calorific value of the in kJ/kg\n",
      "R=0.287                                  #Gas constant in kJ/kgK\n",
      "\n",
      "#Calculations \n",
      "q23=cv/(af+1)\n",
      "T3=(-0.997+(((0.997)**2)+(4*2411*14*10**-6))**(1/2.0))/(28.0*10.0**-6)\n",
      "V3=(T3/T)\n",
      "Vs=(r-1)\n",
      "V=V3-1\n",
      "pc=(V/Vs)*100\n",
      "\n",
      "#Output\n",
      "print\"The percentage of stroke at which combustion is complete = \",round(pc,3),\"percent\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The percentage of stroke at which combustion is complete =  6.706 percent\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.5 Page No 122"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=16.0                                         #The compression ratio \n",
      "l=6.0                                          #The cut-off of the stroke in percent\n",
      "p3=70.0                                        #The maximum pressure obtained in bar\n",
      "p1=1.0                                         #The pressure at the beginning of compression in bar\n",
      "T1=(100.0+273.0)                                 #The temperature at the beginning of compression in K\n",
      "R=0.287                                      #Gas constant in kJ/kgK\n",
      "g=1.4                                        #Assume the isentropic index \n",
      "\n",
      "#Calculations\n",
      "T2=T1*(r)**(g-1)\n",
      "Cv=(1/(T2-T1))*(0.716+125*10**-6*(T2**2-T1**2))\n",
      "Cp=Cv+R\n",
      "g1=Cp/Cv\n",
      "T21=T1*(r)**(g1-1)\n",
      "Cv1=(1/(T21-T1))*((0.716+125*10**-6*(T21**2-T1**2)))\n",
      "Cp1=Cv1+R\n",
      "g2=Cp1/Cv1\n",
      "gm=1.358                       #mean value\n",
      "T22=T1*(r)**(gm-1)\n",
      "p2=(T22/T1)*r*p1\n",
      "T3=(p3/p2)*T22\n",
      "V=((l/100.0)*(r-1))+1\n",
      "T4=(V)*T3\n",
      "p4=p3\n",
      "g3=1.3\n",
      "V5=r/V\n",
      "T5=T4*(1/V5)**(g3-1)\n",
      "Cv2=((0.716*(T5-T4))+(62.5*10**-6*(T5**2-T4**2)))/(T5-T4)\n",
      "Cp2=Cv2+R\n",
      "g4=Cp2/Cv2\n",
      "T51=T4*(1/V5)**(g4-1)\n",
      "Cv3=((0.716*(T51-T4))+(62.5*10**-6*(T51**2-T4**2)))/(T51-T4)\n",
      "Cp3=Cv3+R\n",
      "g5=Cp3/Cv3\n",
      "T52=T4*(1/V5)**(g5-1)\n",
      "p5=(T52/T1)*p1\n",
      "\n",
      "#Output\n",
      "print\"The pressure and temperature at all points of the cycle \\nat point 2: Temperature T2 = \",round(T22,0),\"K and Pressure P2 = \",round(p2,2),\" bar\"  \n",
      "print\"at point 3 :Temperature T3 = \",round(T3,1),\" K  and Pressure P3 = \",p3,\" bar\" \n",
      "print\"at point 4 : Temperature T4 = \",T4,\" K    and Pressure P4 = \",p4,\"bar\" \n",
      "print\"at point 5 :Temperature T5 = \",round(T52,0),\" K  and Pressure P5 = \",round(p5,2),\"bar\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The pressure and temperature at all points of the cycle \n",
        "at point 2: Temperature T2 =  1006.0 K and Pressure P2 =  43.17  bar\n",
        "at point 3 :Temperature T3 =  1631.9  K  and Pressure P3 =  70.0  bar\n",
        "at point 4 : Temperature T4 =  3100.5625  K    and Pressure P4 =  70.0 bar\n",
        "at point 5 :Temperature T5 =  1698.0  K  and Pressure P5 =  4.55 bar\n"
       ]
      }
     ],
     "prompt_number": 32
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.6 Page No 125"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=8                                                #Compression ratio\n",
      "lcv=44000                                          #The lower heating value of the fuel in kJ/kg\n",
      "af=15                                              #The air/fuel ratio\n",
      "Cv=0.71                                            #The specific heat at constant volume in kJ/kgK\n",
      "p=1                                                #The pressure at the beginning of the compression in bar\n",
      "t=60                                               #The temperature at the beginning of the compression in degree centigrade\n",
      "Mo=32                                              #Molecular weight of oxygen\n",
      "Mn=28.161                                          #Molecular weight of nitrogen\n",
      "Mh=18                                              #Molecular weight of water \n",
      "n=1.3                                              #Polytrpic index\n",
      "\n",
      "#Calculations\n",
      "T1=(t+273)\n",
      "sa=(12.5*(Mo+(3.76*Mn)))/((12*8)+(1*Mh))\n",
      "Y=af*(((12*8)+(1*Mh))/(Mo+(3.76*Mn)))\n",
      "x=(12.5-Y)*2\n",
      "nb=1+Y+(Y*3.76)\n",
      "na=x+7.8+9+46.624 \n",
      "Me=((na-nb)/nb)*100\n",
      "T2=T1*(r)**(n-1)\n",
      "T3=(lcv/(af+1))*(1/Cv)+(T2)\n",
      "p3=r*(T3/T1)*p\n",
      "p31=p3*(na/nb)\n",
      "\n",
      "#Output\n",
      "print\"The percentage molecular expansion is \",round(Me,0),\"percent\"\n",
      "print\"(a) Without considering the molecular expansion \"\n",
      "print\" The maximum temperature is \",round(T3,0),\"K\" \n",
      "print\"The maximum pressure is \",round(p3,0),\"bar\"\n",
      "print\"(b) With molecular expansion \" \n",
      "print\"The maximum temperature is \",round(T3,0),\"K\" \n",
      "print\"The maximum pressure is \",round(p31,1),\"bar\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The percentage molecular expansion is  6.0 percent\n",
        "(a) Without considering the molecular expansion \n",
        " The maximum temperature is  4495.0 K\n",
        "The maximum pressure is  108.0 bar\n",
        "(b) With molecular expansion \n",
        "The maximum temperature is  4495.0 K\n",
        "The maximum pressure is  114.4 bar\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.7 Page No 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "f=0.03                          #The residual fraction of an engine\n",
      "e=1.2                           #The equivalence ratio\n",
      "F=0.0795                        #Fuel/air ratio for corresponding equivalence ratio \n",
      "\n",
      "#Calculations\n",
      "\n",
      "T=1+F\n",
      "fa=1-f\n",
      "ff=F*(fa)\n",
      "ra=f\n",
      "rf=ra*F\n",
      "\n",
      "#Output\n",
      "print\"Fresh air = \",fa,\"kg\" \n",
      "print\"Fresh fuel = \",ff,\"kg\" \n",
      "print\"Air in residual = \",ra,\"kg\" \n",
      "print\"Fuel in residual = \",rf,\"kg\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Fresh air =  0.97 kg\n",
        "Fresh fuel =  0.077115 kg\n",
        "Air in residual =  0.03 kg\n",
        "Fuel in residual =  0.002385 kg\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.8 Page No 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "T=800                          #The given temperature in K\n",
      "e=1                            #The equivalence ratio \n",
      "hi=154.723                     #Sensible Enthalpy for isooctane at 800 K in MJ/kmol \n",
      "ho=15.841                      #Sensible Enthalpy for oxygen at 800 K in MJ/kmol \n",
      "hn=15.046                      #Sensible Enthalpy for nitrogen at 800 K in MJ/kmol\n",
      "nc=0.00058                     #Number of kmoles of C8H18 for equivalence ratio for 1 kg of air \n",
      "no=0.00725                     #Number of kmoles of oxygen for equivalence ratio for 1 kg of air \n",
      "nn=0.0273                      #Number of kmoles of nitrogen for equivalence ratio for 1 kg of air \n",
      "R=8.314                        #Gas constant in kJ/kgK\n",
      "\n",
      "#Calculations\n",
      "\n",
      "Hs=(nc*hi)+(no*ho)+(nn*hn)\n",
      "Hs1=Hs*1000\n",
      "n=nc+no+nn\n",
      "Us=Hs-(n*R*10**-3*(T-298))\n",
      "Us1=Us*1000\n",
      "\n",
      "#Output\n",
      "print\"Total sensible enthalpy of reactants = \",round(Hs1,3),\"kJ/kg air\" \n",
      "print\"Sensible internal energy of reactants = \",round(Us1,3),\"kJ/kg air\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total sensible enthalpy of reactants =  615.342 kJ/kg air\n",
        "Sensible internal energy of reactants =  468.723 kJ/kg air\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.9 Page No 131"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "T=500.0                                 #The given temperature in K\n",
      "e=1.0                                   #Equivalence ratio \n",
      "Ai=0.0662                             #The amount of isooctane for 1 kg of air in kg\n",
      "Ta=298.0                                #Consider the ambient temperature in K \n",
      "R=8.314                               #Gas constant in kJ/kgK\n",
      "\n",
      "#Calculations\n",
      "import math\n",
      "E=((0.0662*((0.44*math.log(T/Ta))+(3.67*10**-3*(T-Ta))))+((0.921*math.log(T/Ta))+(2.31*10**-4*(T-Ta))))*1000\n",
      "Ri=Ri/114.0\n",
      "W=(0.5874-(0.662*Ri*math.log(T/Ta))-(0.287*math.log(T/Ta)))*1000\n",
      "\n",
      "#Output\n",
      "print\"The isentropic compression functions at 500 K for the unburned\" \n",
      "print\"isooctsne-air mixture are \",round(E,1),\"J/kg air and\",round(W,1),\"J/kg air\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The isentropic compression functions at 500 K for the unburned\n",
        "isooctsne-air mixture are  587.4 J/kg air and 438.9 J/kg air\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.10 Page No 133"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=7.8                               #Compression ratio \n",
      "p=1.0                               #The pressure at the start of compression in atm\n",
      "T1=335.0                              #The temperature at the start of compression in K\n",
      "W1=100                              #Isentropic compression function for T1 in J/kg air K \n",
      "T2=645                              #The temperature corresponding to isentropic compression function in J/kg air K \n",
      "U1=35                               #Internal energy corresponding to temp T1 in kJ/kg air \n",
      "U2=310                              #Internal energy corresponding to temp T2 in kJ/kg air \n",
      "E1=120                              #Isentropic compression function at T1 \n",
      "E2=910                              #Isentropic compression function at T2 \n",
      "\n",
      "#Calculations\n",
      "import math\n",
      "W2=W1-(292*math.log(1/r))\n",
      "V1=(292*T1)/(p*10.0**5)\n",
      "p2=p*(T2/T1)*r\n",
      "V2=V1/r\n",
      "W=U2-U1\n",
      "p21=(math.exp((E2-E1)/292.0)) \n",
      "\n",
      "#Output\n",
      "print\"(a)At the end of the compression stroke\"\n",
      "print\"The temperature is \",T2,\"K\" \n",
      "print\"The pressure is \",round(p2,0),\"atm\" \n",
      "print\"The volume per unit mass of air is \",round(V2,3),\"m**3/kg air\"\n",
      "print\"The pressure is \",round(p21,0),\"atm\" \n",
      "print\"(b)The work input during compression is \",W,\"kJ/kg air\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)At the end of the compression stroke\n",
        "The temperature is  645 K\n",
        "The pressure is  15.0 atm\n",
        "The volume per unit mass of air is  0.125 m**3/kg air\n",
        "The pressure is  15.0 atm\n",
        "(b)The work input during compression is  275 kJ/kg air\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.11 Page No 134"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "p=65                      #The pressure in the cylinder in bar\n",
      "r=10                      #The compression ratio \n",
      "V3=0.1                    #The volume per unit mass of air at the start of expansion in m**3/kg air \n",
      "p3=p*100                  #The pressure in the cylinder after the completion of combustion in kN/m**2\n",
      "T3=2240                   #The temperature from the chart corresponding to p3,V3 in K\n",
      "u3=-1040                  #The energy from the chart in kJ/kg air \n",
      "s3=8.87                   #The entropy from the chart in kJ/kg air K\n",
      "T4=1280                   #The temperature from the chart corresponding to p4,V4 in K \n",
      "u4=-2220                  #The energy from the chart in kJ/kg air \n",
      "p4=4.25                   #The pressure from the chart in bar \n",
      "\n",
      "#Calculations \n",
      "s4=s3\n",
      "V4=r*V3\n",
      "W=-(u4-u3)\n",
      "\n",
      "#Output\n",
      "print\"(a)At the end of expansion stroke\"\n",
      "print\"The pressure is \",p4,\"bar\" \n",
      "print\"The temperature is \",T4,\"K\" \n",
      "print\"The volume is \",V4,\"m**3/kg air\" \n",
      "print\"(b)The work during the expansion stroke is \",W,\"kJ/kg air\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)At the end of expansion stroke\n",
        "The pressure is  4.25 bar\n",
        "The temperature is  1280 K\n",
        "The volume is  1.0 m**3/kg air\n",
        "(b)The work during the expansion stroke is  1180 kJ/kg air\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.12 Page no 137"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given\n",
      "#From Table 4.4\n",
      "H1=-224.1*1000     #MJ/mol, Enthalpy of C8H18\n",
      "H2=-393.52*1000    #Enthalpy of CO2\n",
      "H3=-241.82*1000    #Enthalpy of H2O\n",
      "U1=-204.1*1000     #MJ/mol, Internal energyof C8H18\n",
      "U2=-393.52*1000    #Internal energy of CO2\n",
      "U3=-240.6*1000    #Internal energy of H2O\n",
      "\n",
      "\n",
      "M1=114.0       #g, molecular wt of C8H18\n",
      "M2=32.0       #g, molecular wt of O2\n",
      "M3=28.0       #g, molecular wt of N2\n",
      "M4=44.0       #g, molecular wt of CO2\n",
      "M5=18.0       #g, molecular wt of H2O\n",
      "#For 1 kg air, from the eq.,  the fraction of wt are\n",
      "x1=0.0661    \n",
      "x2=0.232\n",
      "x3=0.768\n",
      "x4=0.204\n",
      "x5=0.094\n",
      "\n",
      "#Calculation\n",
      "import sympy\n",
      "f=sympy.Symbol(\"f\")\n",
      "n1=(x1/M1)*(1-f)        #No. of kmoles of C8H18\n",
      "n2=x2/M2*(1-f)\n",
      "n3=x3/M3\n",
      "n4=x4/M4*f\n",
      "n5=x5/M5*f\n",
      "N1=(n1*H1+n4*H2+n5*H3)\n",
      "N2=n1*U1+n4*U2+n5*U3\n",
      "\n",
      "#Result\n",
      "print \"Standard Enthalpy of formation is\",N1\n",
      "print \"Internal energy of formation is\",N2\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Standard Enthalpy of formation is -2957.40091174907*f - 129.938684210526\n",
        "Internal energy of formation is -2962.62629186603*f - 118.342192982456\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.13 Page No 138"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "Tu=645.0                                #The temperature at the end of compression process in K\n",
      "usu=310.0                               #The internal energy at the end of compression process in kJ/kg air \n",
      "pu=(15.4*1.013)                       #The pressure at the end of the compression process in bar \n",
      "Vu=0.124                              #The volume at the end of the compression process in m**3/kg air \n",
      "e=1.0                                   #Equivalence ratio \n",
      "f=0.065                               #Burned gas fraction \n",
      "Tb=2820.0                               #The temperature for constant volume \n",
      "pb=6500.0                               #The pressure for constant volume \n",
      "hsu=440.0                               #The enthalpy from chart corresponding to temp Tu in kJ/kg air \n",
      "pb1=1560.0                              #The pressure for constant pressure adiabatic combustion in kN/m**2 \n",
      "ub1=-700.0                              #Trail and error along the pb internal energy in kJ/kg air\n",
      "Tb1=2420.0                              #The temperature for constant pressure \n",
      "\n",
      "#Calculations \n",
      "ufu=-118.5-(2963*f)\n",
      "ub=usu-ufu\n",
      "Vb=Vu\n",
      "hfu=-129.9-(2958*f)\n",
      "hb=hsu+hfu\n",
      "vb1=(118-ub1)/pb\n",
      "\n",
      "#Output\n",
      "print\"(a)For constant volume adiabatic combustion\"\n",
      "print\"The temperature is \",Tb,\"K\" \n",
      "print\"The pressure is \",pb,\"kN/m**2\"\n",
      "print\"(b)For constant pressure adiabatic combustion\"\n",
      "print\"The temperature is \",Tb1,\"K\" \n",
      "print\"The pressure is \",pb1,\"kN/m**2\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)For constant volume adiabatic combustion\n",
        "The temperature is  2820.0 K\n",
        "The pressure is  6500.0 kN/m**2\n",
        "(b)For constant pressure adiabatic combustion\n",
        "The temperature is  2420.0 K\n",
        "The pressure is  1560.0 kN/m**2\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4.14 Page No 139"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given\n",
      "r=8.0                               #The compression ratio \n",
      "T1=350.0                            #The given temperature at the start of compression in K\n",
      "p=1.0                               #The given pressure at the start of compression in bar\n",
      "f=0.08                              #The exhaust residual fraction \n",
      "cv=44000                            #The calorific value in kJ/kg\n",
      "W1=150                              #Isentropic compression functions for corresponding temp T1 in J/kg air K\n",
      "T2=682                              #The temperature corresponding to isentropic compression function in K \n",
      "us2=350                             #The internal energy corresponding to temp T2 in K\n",
      "us1=40                              #The internal energy corresponding to temp T1 in K \n",
      "T3=2825                             #The temperature at point 3 corresponding to u3,V3 on the burned gas chart in K\n",
      "p3=7100                             #The pressure at point 3 in kN/m**2 \n",
      "s3=9.33                             #Entropy at point 3 in kJ/kg air K \n",
      "u4=-1540                            #The internal energy at point 4 corresponding to V4,s4 in kJ/kg air \n",
      "p4=570                              #The pressure at point 4 in kN/m**2 \n",
      "T4=1840                             #The temperature at point 4 in K \n",
      "\n",
      "#Calculations\n",
      "import math\n",
      "W2=W1-(292*math.log(1/r))\n",
      "V1=(292*T1)/(p*10.0**5)\n",
      "p2=p*(T2/T1)*r\n",
      "V2=V1/r\n",
      "Wc=us2-us1\n",
      "ufu=-118.5-(2963*f)\n",
      "u3=us2+ufu\n",
      "V3=V2\n",
      "s4=s3\n",
      "V4=V1\n",
      "We=u3-u4 \n",
      "Wn=We-Wc\n",
      "nth=((Wn)/((1-f)*0.0662*cv))*100\n",
      "imep=((Wn*1000)/(V1-V2))/10.0**5\n",
      "nv=(((1-f)*287*298)/(1.013*10**5*(1-0.125)))*100\n",
      "\n",
      "#Output\n",
      "print\"(a)At point 2, \\nThe temperature is \",T2,\" K \\nThe pressure is \",round(p2,1),\"atm\"\n",
      "print\"At point 3, \\nThe temperature is \",T3,\" K \\nThe pressure is \",p3,\"kN/m**2\" \n",
      "print\"At point 4, \\nThe temperature is \",T4,\" K \\nThe pressure is \",p4,\"kN/m**2\" \n",
      "print\"(b)The indicated thermal efficiency is \",round(nth,1),\"percent\" \n",
      "print\"(c)The indicated mean effective pressure is \",round(imep,0),\" bar\" \n",
      "print\"(d)The volumetric efficiency is \",round(nv,2), \"percent\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)At point 2, \n",
        "The temperature is  682  K \n",
        "The pressure is  15.6 atm\n",
        "At point 3, \n",
        "The temperature is  2825  K \n",
        "The pressure is  7100 kN/m**2\n",
        "At point 4, \n",
        "The temperature is  1840  K \n",
        "The pressure is  570 kN/m**2\n",
        "(b)The indicated thermal efficiency is  45.7 percent\n",
        "(c)The indicated mean effective pressure is  14.0  bar\n",
        "(d)The volumetric efficiency is  88.77 percent\n"
       ]
      }
     ],
     "prompt_number": 4
    }
   ],
   "metadata": {}
  }
 ]
}