path: root/Thermodynamics:_From_concepts_to_applications/Chapter19_3.ipynb

{
 "metadata": {
  "name": "",
  "signature": "sha256:9d79eec6ce58c29eedc2099ff08aa61dc95a4086cf18a82c8aaaa70d00549f3b"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter19-Chemical reactions"
     ]
    },
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Example1-pg 528"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#caluclate fuel ratio and excess air and emass air-fuel ratio\n",
      "##initialisation of variables\n",
      "pN2= 79. ##percent\n",
      "VN2= 82.3 ##m^3\n",
      "VCO2= 8. ##m^3\n",
      "VCO= 0.9 ##m^3\n",
      "M= 32. ##gms\n",
      "M1= 28. ##gms\n",
      "##CALCULATIONS\n",
      "P= (pN2/(100-pN2))\n",
      "z= VN2/P\n",
      "x= VCO2+VCO\n",
      "w= VCO2+(VCO/2)+(VCO2/10)\n",
      "y= 2*w\n",
      "r= y/x\n",
      "TO= x+(y/4)\n",
      "X= (z/TO)-1\n",
      "AF= z*(M+P*M1)/(12*x+y)\n",
      "##RESULTS\n",
      "print'%s %.3f %s'%('fuel ratio=',r,'')\n",
      "print'%s %.3f %s'%('excess air=',X,'')\n",
      "print'%s %.2f %s'%('emass air-fuel ratio=',AF,'')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "fuel ratio= 2.079 \n",
        "excess air= 0.618 \n",
        "emass air-fuel ratio= 23.98 \n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example2-pg 534"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate heat interaction\n",
      "##initialisation of variables\n",
      "m1= 24. ##kg\n",
      "M1= 32. ##kg\n",
      "m2= 28. ##kg\n",
      "M2= 28. ##kg\n",
      "e= 0.5\n",
      "T3= 1800. ##C\n",
      "T0= 25. ##C\n",
      "T1= 25. ##C\n",
      "T2= 100. ##C\n",
      "R= 8.314 ##Jmol K\n",
      "cp= 4.57 ##J/mol K\n",
      "cp1= 3.5 ##J/mol K\n",
      "cp2= 3.5 ##J/mol K\n",
      "hCO2= -393522. ##J\n",
      "hCO= -110529. ##J\n",
      "##CALCULATIONS\n",
      "n1= m1/M1\n",
      "n2= m2/M2\n",
      "N= n1-0.5*e\n",
      "N1= n2-e\n",
      "N2= e\n",
      "N3= N+N1+N2\n",
      "y1= N/N3\n",
      "Q= ((N*cp+N1*cp1+N2*cp2)*R*(T3-T0)-(n1*cp*(T1-T0)+n2*cp2*(T2-T1))+N*(hCO2-hCO))/60.\n",
      "##RESULTS\n",
      "print'%s %.f %s'%(' Heat interaction=',Q,'kW ')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Heat interaction= -940 kW \n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg536"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "##initialisation of variables\n",
      "hCO2= -393520. ##kJ/kg mol\n",
      "hH2O= -285840. ##kJ/kg mol\n",
      "hC7H16= -187820. ##kJ/kg mol\n",
      "M= 100\n",
      "hH2O1= -241830. ##kJkg mol\n",
      "##CALCULATIONS\n",
      "HHV= -(7*hCO2+8.*hH2O-hC7H16)/M\n",
      "LLV= -(7*hCO2+8.*hH2O1-hC7H16)/M\n",
      "#RESULTS\n",
      "print'%s %.2f %s'% (' Higher heating vlue= ',HHV,' kJ/kg mol ')\n",
      "print'%s %.2f %s'%  ('  Lower heating vlue= ',LLV,' kJ/kg mol ')\n",
      "#round off error\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Higher heating vlue=  48535.40  kJ/kg mol \n",
        "  Lower heating vlue=  45014.60  kJ/kg mol \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example4-pg 537"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate adiabatic flame temperature\n",
      "##initialisation of variables\n",
      "T0= 25. ##C\n",
      "T1= 220. ##C\n",
      "hCO2= -393520 ##kJ/kg\n",
      "hH2O= -241830 ##kJ/kg\n",
      "hC3H8= -103850 ##kJ/kg= 1.4\n",
      "R= 8.314 ##Jmol K\n",
      "k= 1.4\n",
      "k1= 1.29\n",
      "##CALCULATIONS\n",
      "T= T0+((15*(R*(k/(k-1)))*4.762*(T1-T0)-(3*hCO2+4*hH2O-hC3H8))/(R*((3+4)*(k1/(k1-1))+(10+56.43)*(k/(k-1)))))\n",
      "##RESULTS\n",
      "print'%s %.1f %s'%('adiabatic flame temperature=',T,'C ')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "adiabatic flame temperature= 1142.4 C \n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example6-pg 415"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate enthalpy formation\n",
      "##initialisation of variables\n",
      "T= 25. ##C\n",
      "hfT= -241820 ##kJ/kmol\n",
      "R= 8.314 ##J/mol K\n",
      "k= 1.4\n",
      "cpH2O= 4.45\n",
      "cpO2= 3.5\n",
      "T1= 1000. ##C\n",
      "##CALCULATIONS\n",
      "S= (cpH2O-k*cpO2)\n",
      "hfT1= hfT+S*(T1-T)\n",
      "##RESULTS\n",
      "print'%s %.f %s'%('enthalpy formation=',hfT1,'kJ/kmol ')\n",
      "#there is error because of round off error \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "enthalpy formation= -242259 kJ/kmol \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example7-pg 545"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate equlibrium constant at K and KT1\n",
      "##initialisation of variables\n",
      "R= 8.314 ##J/mol K\n",
      "T= 25. ##C\n",
      "gf= 16590. ##kJ/kmol\n",
      "T1= 500. ##C\n",
      "Cp= 4.157 ##J/mol K\n",
      "hf= -46190 ##kJ/kmol\n",
      "e=0.5\n",
      "##CALCULATIONS\n",
      "K=math.pow(math.e,gf/(R*(273.15+T)))\n",
      "r= (1-((273.15+T)/(273.15+T1)))*((hf/(R*(273.15+T)))+(R/Cp))-2*math.log((273.15+T1)/(273.15+T))+0.6\n",
      "KT1= K*math.pow(math.e,r)\n",
      "##RESULTS\n",
      "print'%s %.1f %s'%('equilibrium constant=',K,'bar^-1 ')\n",
      "print'%s %.5f %s'%('equilibrium constant=',KT1,'bar^-1 ')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "equilibrium constant= 806.5 bar^-1 \n",
        "equilibrium constant= 0.00797 bar^-1 \n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example8-pg 546"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#what equilibrium constant at T1 and T2\n",
      "##initialisation of variables\n",
      "uCO2= -394374 ##J/mol\n",
      "uCO= -137150 ##J/mol\n",
      "uO2= 0.\n",
      "R= 8.314 ##J/mol K\n",
      "T= 25. ##C\n",
      "cpCO2= 4.57 ##J/mol K\n",
      "cpCO= 3.5 ##J/mol K\n",
      "cpO2= 3.5 ##J/mol K\n",
      "T1= 1500. ##C\n",
      "hf= -393522 ##kJ/kmol\n",
      "gf= -110529 ##kJ/kmol\n",
      "T2= 2500. ##C\n",
      "##CALCULATIONS\n",
      "r= -(uCO2-uCO-0.5*uO2)/(R*(273.15+T))\n",
      "s= (cpCO2-cpCO-0.5*cpO2)\n",
      "r1= (1-((273.15+T)/(273.15+T1)))*((hf-gf)/(R*(273.15+T))-s)+s*math.log((273.15+T1)/(273.15+T))\n",
      "KT1= math.pow(math.e,r+r1)\n",
      "r2= (1-((273.15+T)/(273.15+T2)))*((hf-gf)/(R*(273.15+T))-s)+s*math.log((273.15+T2)/(273.15+T))\n",
      "KT2= math.pow(math.e,r+r2)\n",
      "##RESULTS\n",
      "print'%s %.f %s'%('equilibrium constant at T1=',KT1,'C ')\n",
      "print'%s %.3f %s'%('equilibrium constant at T2=',KT2,'C ')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "equilibrium constant at T1= 3477 C \n",
        "equilibrium constant at T2= 2.635 C \n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example9-pg548"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#what is maximum work of given variable \n",
      "##initialisation of variables\n",
      "Wc= 12. ##kg\n",
      "hf= -393520 ##kJ/kmol\n",
      "gf= -394360 ##kJ/kmol\n",
      "##CALCULATIONS\n",
      "Wmax= -gf/Wc\n",
      "##RESULTS\n",
      "print'%s %.f %s'%('maximum work=',Wmax,'kJ/kg of carbon ')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "maximum work= 32863 kJ/kg of carbon \n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example10-pg549"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate the outlet temperature and energy of formation and energy out let and energy of the products\n",
      "##initialisation of variables\n",
      "T= 25 ##C\n",
      "R= 8.314 ##Jmol K\n",
      "k= 1.27\n",
      "k1= 1.34\n",
      "hf= -393520 ##kJ/kmol\n",
      "M= 28 ##gms\n",
      "gf= -394360 ##kJ/kmol\n",
      "M= 12 ##gms\n",
      "##CALCULATIONS\n",
      "T1= T+(-hf/((R)*((k/(k-1))+(0.2+4.5144)*(k1/(k1-1)))))\n",
      "Bin= 0\n",
      "dh= (k1*R/(k1-1))*(T1-T)\n",
      "dh1= (k1*R/(k1-1))*math.log((273.15+T1)/(273.15+T))\n",
      "H= dh-(273.15+T)*dh1\n",
      "h= (k*R/(k-1))*(T1-T)+hf\n",
      "h1= (k*R/(k-1))*math.log((273.15+T1)/(273.15+T))+((hf-gf)/(273.15+T))\n",
      "h2= h-(273.15+T)*h1\n",
      "Bout= (h2+(0.2+4.5144)*H)/M\n",
      "##RESULTS\n",
      "print'%s %.2f %s'%('outlet temperature=',T1,'C')\n",
      "print'%s %.f %s'%('energy of formation=',Bin,'J')\n",
      "print'%s %.f %s'%('energy at outlet=',H,'kJ/kmol')\n",
      "print'%s %.f %s'%('energy of the products=',Bout,'k')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "outlet temperature= 2057.82 C\n",
        "energy of formation= 0 J\n",
        "energy at outlet= 46519 kJ/kmol\n",
        "energy of the products= -9961 k\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example11-pg553"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#calculate the change in energy and amount of air and gas and netchange in energy and percent change in energy\n",
      "##initialisation of variables\n",
      "b= 1475.30 ##kJ/kg\n",
      "b0= 144.44 ##kJ/kg\n",
      "h2= 3448.6 ##kJkg\n",
      "h1= 860.5 ##kJ/kg\n",
      "k= 1.27 \n",
      "k1= 1.34\n",
      "R= 8.314 ##J/mol K\n",
      "hf= -393520 ##kJ/kmol\n",
      "hg= 72596 ##kJ/kmol\n",
      "Mc= 12 ##kg\n",
      "n= 1.2 ##moles\n",
      "n1= 3.76 ##moles\n",
      "M= 32. ##gms\n",
      "M1= 28. ##gms\n",
      "M2= 44. ##gms\n",
      "n2= 0.2 ##moles\n",
      "n3= 4.512 ##moles\n",
      "B1= 25592. ##kJ/kmol C\n",
      "B2= 394360. ##kJ/kmol C\n",
      "e= 0.008065\n",
      "##CALCULATIONS\n",
      "B= b-b0\n",
      "Q= h2-h1\n",
      "CpCO2= k*R/(k-1)\n",
      "CpO2= k1*R/(k1-1)\n",
      "Qcoal= (hg+hf)/Mc\n",
      "mcoal= Q/(-Qcoal)\n",
      "ncoal= mcoal/Mc\n",
      "r= (n*M+n1*M1)/Mc\n",
      "r1= (M2+n2*M+n3*M1)/Mc\n",
      "mair= r*mcoal\n",
      "mgas= r1*mcoal\n",
      "Bfuel= (B1-B2)*e\n",
      "Bnet= Bfuel+B\n",
      "p= B*100/(-Bfuel)\n",
      "##RESULTS\n",
      "print'%s %.2f %s'% ('change in energy=',B,'kJ/kg ')\n",
      "print'%s %.3f %s'%('amount of air=',mair,'kg/kg ')\n",
      "print'%s %.3f %s'%('amount of gas=',mgas,'kg/kg ')\n",
      "print'%s %.3f %s'%('net change in energy=',Bnet,'kg/kg steam ')\n",
      "print'%s %.2f %s'%('percent energy in original fuel=',p,'percent ')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "change in energy= 1330.86 kJ/kg \n",
        "amount of air= 1.159 kg/kg \n",
        "amount of gas= 1.425 kg/kg \n",
        "net change in energy= -1643.254 kg/kg steam \n",
        "percent energy in original fuel= 44.75 percent \n"
       ]
      }
     ],
     "prompt_number": 13
    }
   ],
   "metadata": {}
  }
 ]
}