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 },
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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 13: Gas Mixtures"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13-1 ,Page No.683"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "mO2=3.0;#moles of oxygen\n",
      "mN2=5.0;#moles of nitrogen\n",
      "mCH4=12.0;#moles of methane\n",
      "\n",
      "#molecular masses in kg\n",
      "MO2=32.0;\n",
      "MN2=28.0;\n",
      "MCH4=16.0;\n",
      "\n",
      "#constants used\n",
      "Ru=8.314;#in kJ/kg - K\n",
      "\n",
      "#calculations\n",
      "\n",
      "#part - a\n",
      "mm=mO2+mN2+mCH4;\n",
      "mfO2=mO2/mm;\n",
      "mfN2=mN2/mm;\n",
      "mfCH4=mCH4/mm;\n",
      "print'mass fraction of oxygen is %f'%round(mfO2,2);\n",
      "print'mass fraction of nitrogen is %f'%round(mfN2,2);\n",
      "print'mass fraction of methane is %f'%round(mfCH4,2);\n",
      "\n",
      "#part - b\n",
      "NO2=mO2/MO2;\n",
      "NN2=mN2/MN2;\n",
      "NCH4=mCH4/MCH4;\n",
      "Nm=NO2+NN2+NCH4;\n",
      "yO2=NO2/Nm;\n",
      "yN2=NN2/Nm;\n",
      "yCH4=NCH4/Nm;\n",
      "print'mole fraction of oxygen is %f'%round(yO2,3);\n",
      "print'mole fraction of nitrogen is %f'%round(yN2,3);\n",
      "print'mole fraction of methane is %f'%round(yCH4,3);\n",
      "\n",
      "#part - c\n",
      "Mm=mm/Nm;\n",
      "print'average molecular mass %f kg/kmol'%round(Mm,1);\n",
      "Rm=Ru/Mm;\n",
      "print'gas constant of mixture %f kJ/kg - K'%round(Rm,3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "mass fraction of oxygen is 0.150000\n",
        "mass fraction of nitrogen is 0.250000\n",
        "mass fraction of methane is 0.600000\n",
        "mole fraction of oxygen is 0.092000\n",
        "mole fraction of nitrogen is 0.175000\n",
        "mole fraction of methane is 0.734000\n",
        "average molecular mass 19.600000 kg/kmol\n",
        "gas constant of mixture 0.425000 kJ/kg - K\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13-2 ,Page No.687"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "NN2=2.0;#moles of nitrogen\n",
      "NCO2=6.0;#moles of carbon dioxide\n",
      "Tm=300.0;#temperature of gases in K\n",
      "Pm=15000.0;#pressure of gases in kPa\n",
      "\n",
      "#constants used\n",
      "Ru=8.314;#in kJ/kmol - K\n",
      "\n",
      "#calculations\n",
      "\n",
      "#part - a\n",
      "Nm=NN2+NCO2;\n",
      "Vm=Nm*Ru*Tm/Pm;\n",
      "print'the volume of the tank on the basis of the ideal-gas equation of state %f m^3'%round(Vm,3);\n",
      "\n",
      "#part - b\n",
      "#from Table A-1\n",
      "#for nitrogen\n",
      "TcrN=126.2;\n",
      "PcrN=3390;\n",
      "#for Carbondioxide\n",
      "TcrC=304.2;\n",
      "PcrC=7390;\n",
      "yN2=NN2/Nm;\n",
      "yCO2=NCO2/Nm;\n",
      "Tcr=yN2*TcrN+yCO2*TcrC;\n",
      "Pcr=yN2*PcrN+yCO2*PcrC;\n",
      "Tr=Tm/Tcr;\n",
      "Pr=Pm/Pcr;\n",
      "#from Fig A-15b\n",
      "Zm=0.49;\n",
      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
      "print'the volume of the tank on the basis Kay\u2019s rule %f m^3'%round(Vm,3);\n",
      "\n",
      "#part - c\n",
      "#for nitrogen\n",
      "TrN=Tm/TcrN;\n",
      "PrN=Pm/PcrN;\n",
      "#from Fig A-15b\n",
      "Zn=1.02;\n",
      "#for Carbondioxide\n",
      "TrC=Tm/TcrC;\n",
      "PcrC=Pm/PcrC;\n",
      "#from Fig A-15b\n",
      "Zc=0.3;\n",
      "Zm=yN2*Zn+yCO2*Zc;\n",
      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
      "print'the volume of the tank on the basis compressibility factors and Amagat\u2019s law %f m^3'%round(Vm,3);\n",
      "\n",
      "#part - d\n",
      "VRN=(Vm/NN2)/(Ru*TcrN/PcrN);\n",
      "VRC=(Vm/NCO2)/(Ru*TcrC/PcrC);\n",
      "#from Fig A-15b\n",
      "Zn=0.99;\n",
      "Zc=0.56;\n",
      "Zm=yN2*Zn+yCO2*Zc;\n",
      "Vm=Zm*Nm*Ru*Tm/Pm;\n",
      "#When the calculations are repeated we obtain 0.738 m3 after the second iteration, 0.678 m3 after the third iteration, and 0.648 m3 after the fourth iteration.\n",
      "Vm=0.648;\n",
      "print'compressibility factors and Dalton\u2019s law the volume of the tank on the basis %f m^3'%round(Vm,3);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the volume of the tank on the basis of the ideal-gas equation of state 1.330000 m^3\n",
        "the volume of the tank on the basis Kay\u2019s rule 0.652000 m^3\n",
        "the volume of the tank on the basis compressibility factors and Amagat\u2019s law 0.639000 m^3\n",
        "compressibility factors and Dalton\u2019s law the volume of the tank on the basis 0.648000 m^3\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13-3 ,Page No.691"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "mN=4.0;#mass of nitrogen in kg\n",
      "T1N=20.0;#temperature of nitrogen in K\n",
      "P1N=150.0;#pressure of nitrogen in kPa\n",
      "mO=7.0;#mass of oxygen in kg\n",
      "T1O=40.0;#temperature of oxygen in K\n",
      "P1O=100.0;#pressure of oxygen in kPa\n",
      "\n",
      "#molecular masses in kg\n",
      "MO=32.0;\n",
      "MN=28.0;\n",
      "\n",
      "#constants used\n",
      "Ru=8.314;#in kJ/kg - K\n",
      "\n",
      "#from Table A-2a\n",
      "CvN=0.743;\n",
      "CvO=0.658;\n",
      "\n",
      "#calculations\n",
      "\n",
      "#part - a\n",
      "#Ein - Eout = dEsystem\n",
      "# (m*cv*dT)N2 + (m*cv*dT)= 0;\n",
      "Tm= (mN*CvN*T1N+ mO*CvO*T1O)/(mN*CvN+mO*CvO);\n",
      "print'the mixture temperature %f C'%round(Tm,1);\n",
      "\n",
      "#part - b\n",
      "NO=mO/MO;\n",
      "NN=mN/MN;\n",
      "Nm=NO+NN;\n",
      "VO=NO*Ru*(T1O+273)/P1O;\n",
      "VN=NN*Ru*(T1N+273)/P1N;#Exergy Destruction during Mixing of Ideal Gases\n",
      "Vm=VO+VN;\n",
      "Pm=Nm*Ru*(Tm+273)/Vm;  \n",
      "print'the mixture pressure after equilibrium has been established %f kPa'%round(Pm,1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mixture temperature 32.200000 C\n",
        "the mixture pressure after equilibrium has been established 114.500000 kPa\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13-4 ,Page No.692"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "\n",
      "#given data\n",
      "NO=3.0;#moles of oxygen \n",
      "NC=5.0;#moles of carbondioxide\n",
      "T0=25+273.0;#temperature of gasses in K\n",
      "\n",
      "#constants used\n",
      "Ru=8.314;#in kJ/kg - K\n",
      "\n",
      "#calculations\n",
      "Nm=NO+NC;\n",
      "yO=NO/Nm;\n",
      "yC=NC/Nm;\n",
      "#dSm= -Ru*(NO*log(yO)+NC*log(yC))\n",
      "Sm=-Ru*(NO*log(yO)+NC*log(yC));\n",
      "print'the entropy change %f kJ/K'%round(Sm);\n",
      "Xdestroyed=T0*Sm/1000;\n",
      "print'exergy destruction associated %f MJ'%round(Xdestroyed,1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the entropy change 44.000000 kJ/K\n",
        "exergy destruction associated 13.100000 MJ\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13-5 ,Page No.694"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "T1=220;#intial temperature in K\n",
      "T2=160;#final temperature in K\n",
      "Pm=10;#pressure of air in MPa\n",
      "yN=0.79;#mole fraction of nitrogen\n",
      "yO=0.21;#mole fractions of oxygen\n",
      "\n",
      "\n",
      "#critical properties\n",
      "#for Nitrogen\n",
      "TcrN=126.2;\n",
      "PcrN=3.39;\n",
      "#for Oxygen\n",
      "TcrO=154.8;\n",
      "PcrO=5.08;\n",
      "\n",
      "#constants used\n",
      "Ru=8.314;#in kJ/kg - K\n",
      "\n",
      "#from Tables A-18 & 19\n",
      "#at T1\n",
      "h1N=6391;\n",
      "h1O=6404;\n",
      "#for T2\n",
      "h2N=4648;\n",
      "h2O=4657;\n",
      "\n",
      "#calculations\n",
      "#part - a\n",
      "qouti=yN*(h1N-h2N)+yO*(h1O-h2O);\n",
      "print'the heat transfer during this process using the ideal-gas approximation %i kJ/kmol'%round(qouti);\n",
      "\n",
      "#part - b\n",
      "Tcrm=yN*TcrN+yO*TcrO;\n",
      "Pcrm=yN*PcrN+yO*PcrO;\n",
      "Tr1=T1/Tcrm;\n",
      "Tr2=T2/Tcrm;\n",
      "Pr=Pm/Pcrm;\n",
      "#at these values we get\n",
      "Zh1=1;\n",
      "Zh2=2.6\n",
      "qout=qouti-Ru*Tcrm*(Zh1-Zh2);\n",
      "print'the heat transfer during this process using Kay\u2019s rule %i kJ/kmol'%round(qout);\n",
      "\n",
      "#part - c\n",
      "#for nitrogen\n",
      "TrN1=T1/TcrN;\n",
      "TrN2=T2/TcrN;\n",
      "PrN=Pm/PcrN;\n",
      "#from Fig A-15b\n",
      "Zh1n=0.9;\n",
      "Zh2n=2.4;\n",
      "#for Oxygen\n",
      "TrO1=T1/TcrO;\n",
      "TrO2=T2/TcrO;\n",
      "PcrO=Pm/PcrO;\n",
      "#from Fig A-15b\n",
      "Zh1O=1.3;\n",
      "Zh2O=4.0;\n",
      "#from Eq 12-58\n",
      "h12N=h1N-h2N-Ru*TcrN*(Zh1n-Zh2n);# h1 - h2 for nitrogen\n",
      "h12O=h1O-h2O-Ru*TcrO*(Zh1O-Zh2O);# h1 - h2 for oxygen\n",
      "qout=yN*h12N+yO*h12O;\n",
      "print'the heat transfer during this process using Amagat\u2019s law %i kJ/kmol'%round(qout);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the heat transfer during this process using the ideal-gas approximation 1744 kJ/kmol\n",
        "the heat transfer during this process using Kay\u2019s rule 3502 kJ/kmol\n",
        "the heat transfer during this process using Amagat\u2019s law 3717 kJ/kmol\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13-6 ,Page No.705"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "\n",
      "#13.6 (d) answer not matching as float datatype is giving more accurate answer in comparison to textbook that has given approximate due to rounding off to two decimal places\n",
      "\n",
      "#given data\n",
      "mfs=0.0348;#salinity mass fraction\n",
      "mfw=1-mfs;\n",
      "T0=288.15;#temperature of sea water in K\n",
      "\n",
      "#constants used\n",
      "Mw=18;\n",
      "Ms=58.44;\n",
      "Rw=0.4615;\n",
      "pm=1028;\n",
      "Ru=8.314;\n",
      "\n",
      "#calculations\n",
      "#part - a\n",
      "Mm=1/((mfs/Ms)+(mfw/Mw));\n",
      "yw=mfw*Mm/Mw;\n",
      "ys=1-yw;\n",
      "print'the mole fraction of the water is %f'%round(yw,4);\n",
      "print'the mole fraction of the saltwater is %f'%round(ys,5);\n",
      "\n",
      "#part - b\n",
      "wmin=-Ru*T0*(ys*log(ys)+yw*log(yw));\n",
      "wm=wmin/Mm;\n",
      "print'the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts %f kJ'%round(wm,2);\n",
      "\n",
      "#part - c\n",
      "wmin=Rw*T0*log(1/yw);\n",
      "print'the minimum work input required to obtain 1 kg of fresh water from the sea %f kJ'%round(wmin,2);\n",
      "\n",
      "#part - d\n",
      "Pmin=pm*Rw*T0*log(1/yw);\n",
      "print'the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes %i kPa'%round(Pmin)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the mole fraction of the water is 0.989000\n",
        "the mole fraction of the saltwater is 0.010980\n",
        "the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts 7.850000 kJ\n",
        "the minimum work input required to obtain 1 kg of fresh water from the sea 1.470000 kJ\n",
        "the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes 1510 kPa\n"
       ]
      }
     ],
     "prompt_number": 1
    }
   ],
   "metadata": {}
  }
 ]
}