{
 "metadata": {
  "name": "Chapter 13"
 }, 
 "nbformat": 2, 
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown", 
     "source": [
      "# Chapter 13 :- Reacting Mixtures and Combustion"
     ]
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.1 Page no-624"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Part(a)", 
      "# The combustion equation can be written in the form of ", 
      "# C8H18 + a(O2 + 3.76N2) --> b CO2 + c H2O + d N2", 
      "# Using conservation of mass principle", 
      "b = 8.00", 
      "c = 18.00/2.00", 
      "a = (2.00*b+c)/2.00", 
      "d = 3.76*a", 
      "", 
      "# The air\u2013fuel ratio on a molar basis is", 
      "AFbar = a*(1+3.76)/1.00", 
      "Ma = 28.97                                                                      # molar mass of air", 
      "MC8H18 = 114.22                                                                 # molar mass of C8H18", 
      "# The air\u2013fuel ratio expressed on a mass basis is", 
      "AF = AFbar*(Ma/MC8H18)", 
      "", 
      "# Result", 
      "print '-> The air\u2013fuel ratio on a molar basis is:  '", 
      "print AFbar", 
      "print '-> The air\u2013fuel ratio expressed on a mass basis is:  '", 
      "print round(AF,2)", 
      "", 
      "# Part(b)", 
      "# For 150% theoretical air, the chemical equation for complete combustion takes the form", 
      "# c8H18 + 1.5*12.5*(O2 + 3.76N2) --->  b CO2 + c H2O + d N2 + e O2", 
      "# Using conservation of mass", 
      "# Calculations", 
      "b = 8.00", 
      "c =18.00/2.00", 
      "e = (1.5*12.5*2 - c -2*b)/2.00", 
      "d = 1.5*12.5*3.76", 
      "# The air\u2013fuel ratio on a molar basis is", 
      "AFbar = 1.5*12.5*(1+3.76)/1", 
      "# The air\u2013fuel ratio expressed on a mass basis is", 
      "AF = AFbar*(Ma/MC8H18)", 
      "", 
      "# Results", 
      "print '-> The air\u2013fuel ratio on a molar basis is:  '", 
      "print AFbar", 
      "print '-> The air\u2013fuel ratio expressed on a mass basis is:  '", 
      "print round(AF,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The air\u2013fuel ratio on a molar basis is:  ", 
        "59.5", 
        "-> The air\u2013fuel ratio expressed on a mass basis is:  ", 
        "15.09", 
        "-> The air\u2013fuel ratio on a molar basis is:  ", 
        "89.25", 
        "-> The air\u2013fuel ratio expressed on a mass basis is:  ", 
        "22.64"
       ]
      }
     ], 
     "prompt_number": 18
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.2 Page no-626"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Part(a)", 
      "# The chemical equation", 
      "# a CH4 + b*(O2 + 3.76N2)  --->  9.7CO2 + .5CO + 2.95O2 + 86.85N2 + cH2O", 
      "# Calculations", 
      "# Applying conservation of mass ", 
      "a = 9.7 + 0.5", 
      "c = 2.0*a", 
      "b = ((9.7)*(2.0)+(0.5)+((2.0)*(2.95))+c)/2.00", 
      "Ma = 28.97                                                                      # molar mass of air", 
      "MCH4 = 16.04                                                                    # molar mass of methane", 
      "# On a molar basis, the air\u2013fuel ratio is", 
      "AFbar = (b*(1+3.76))/a", 
      "# On a mass basis", 
      "AF = AFbar*(Ma/MCH4)", 
      "", 
      "# Results", 
      "print '-> The air-fuel ratio on a molar basis is:  '", 
      "print AFbar", 
      "print '-> The air-fuel ratio on a mass basis is:  '", 
      "print round(AF,2)", 
      "", 
      "# Part(b)", 
      "# The balanced chemical equation for the complete combustion of methane with the theoretical amount of air is", 
      "# CH4 + 2(O2 + 3.76N2)  --->  CO2 + 2H2O + 7.52N2", 
      "# The theoretical air\u2013fuel ratio on a molar basis is", 
      "# Calculations", 
      "AFbartheo = 2.00*(1+3.76)/1.0", 
      "# The percent theoretical air is", 
      "Ta = AFbar/AFbartheo", 
      "# Result", 
      "print '-> The percent theoretical air is:  '", 
      "print round(Ta*100,2)", 
      "", 
      "# Ppart(c)", 
      "# The mole fraction of the water vapor is", 
      "yv = 20.4/(100+20.4)", 
      "pv = yv*1", 
      "# Interpolating in Table A-2,", 
      "T = 57                                                                          # in degree celcius", 
      "# Result", 
      "print '-> The dew point temperature of the products, in \u0004C, if the mixture were cooled at 1 atm is:  '", 
      "print T"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The air-fuel ratio on a molar basis is:  ", 
        "10.78", 
        "-> The air-fuel ratio on a mass basis is:  ", 
        "19.47", 
        "-> The percent theoretical air is:  ", 
        "113.24", 
        "-> The dew point temperature of the products, in \u0004C, if the mixture were cooled at 1 atm is:  ", 
        "57"
       ]
      }
     ], 
     "prompt_number": 19
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 13.3 Page no-628"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Part(a)", 
      "# The chemical equation", 
      "# (.8062CH4 + .0541C2H6 + .0187C3H8 + .0160C4H10 + .1050N2) + a(O2 + 3.76N2)  ---->  b(.078CO2 + .002CO + .07O2 + .85N2) + c H2O", 
      "# Calculations", 
      "# Using mass conservation", 
      "b = (0.8062 + 2*.0541 + 3*.0187 + 4*.0160)/(.078 + .002)", 
      "c = (4*.8062 + 6*.0541 + 8*.0187 + 10*.0160)/2", 
      "a = (b*(2*.078+.002+2*.07) + c)/2", 
      "# The air\u2013fuel ratio on a molar basis is", 
      "AFbar = a*(1+3.76)/1", 
      "# Result", 
      "print '-> The air-fuel ratio on a molar mass basis is:  '", 
      "print round(AFbar,2)", 
      "", 
      "# Part(b)", 
      "p = 1.0                                                                           # in bar", 
      "V = 100.0                                                                         # in m^3", 
      "Rbar = 8314.0                                                                     # in N.m/kmol.K", 
      "T = 300.0                                                                         # in kelvin", 
      "# Calculations", 
      "# The amount of fuel in kmol", 
      "nF = (p*10**5*V)/(Rbar*T)", 
      "# The amount of product mixture that would be formed from 100 m3 of fuel mixture is", 
      "n = nF*(b+c)", 
      "# Result", 
      "print '-> The amount of products in kmol that would be formed from 100 m3 of fuel mixture at 300 K and 1 bar is:  '", 
      "print round(n,2)", 
      "", 
      "# Part(c)", 
      "# The balanced chemical equation for the complete combustion of the fuel mixture with the theoretical amount of air is", 
      "# (10.8062CH4 + 0.0541C2H6 + 0.0187C3H8 + 0.0160C4H10 + 0.1050N2) + 2(O2 + 3.76N2)   ----> 1.0345CO2 + 1.93H2O + 7.625N2", 
      "# Calculations", 
      "# The theoretical air\u2013fuel ratio on a molar basis is", 
      "AFbartheo = 2*(1+3.76)/1", 
      "# The percent theoretical air is", 
      "Ta = AFbar/AFbartheo", 
      "# Result", 
      "print '-> The percent of theoretical air is:  '", 
      "print round(Ta*100,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The air-fuel ratio on a molar mass basis is:  ", 
        "13.76", 
        "-> The amount of products in kmol that would be formed from 100 m3 of fuel mixture at 300 K and 1 bar is:  ", 
        "59.58", 
        "-> The percent of theoretical air is:  ", 
        "144.58"
       ]
      }
     ], 
     "prompt_number": 20
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.4 Page no-633"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# The balanced chemical equation for complete combustion with the theoretical amount of air is obtained from the solution to Example 13.1 as", 
      "# C8H18 +12.5O2 + 47N2 -----> 8CO2 + 9H2O + 47N2", 
      "# From tabel A-25", 
      "hRbar = -249910                                                                 # in kj/kmol", 
      "mfdot = 1.8e-3                                                                  # mass flow rate of liquid octane in kg/s", 
      "M = 114.22                                                                      # molar mass of octane", 
      "Wcvdot = 37                                                                     # power output of the engine in kw", 
      "", 
      "# Calculations", 
      "# With enthalpy of formation values for CO2 and H2O(g) from Table A-25, and enthalpy values for N2, H2O, and CO2 from Table A-23", 
      "hpbar = 8*(-393520 + (36876 - 9364)) + 9*(-241820 + (31429 - 9904)) + 47*((26568 - 8669))", 
      "nFdot = mfdot/M                                                                 # molar flow rate of the fuel in kmol/s", 
      "Qcvdot = Wcvdot + nFdot*(hpbar-hRbar)                                           # in kw", 
      "", 
      "# Result", 
      "print '-> The rate of heat transfer from the engine, in kW is:  '", 
      "print round(Qcvdot,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The rate of heat transfer from the engine, in kW is:  ", 
        "-23.19"
       ]
      }
     ], 
     "prompt_number": 21
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 13.5 Page no-635"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# When expressed on a per mole of fuel basis, the balanced chemical equation obtained in the solution to Example 13.2  takes the form", 
      "# CH4 + 2.265O2 + 8.515N2  ------>  .951CO2 + .049CO + .289O2 + 8.515N2 + 2H2O", 
      "cpbar = 38.00                                                                   # specific heat in KJ/kmol.K", 
      "# From table A-25", 
      "hfnotbar = -74850.00                                                            # enthalpy of formation for methane", 
      "# From table A-23", 
      "deltahbarO2 = 14770-8682", 
      "deltahbarN2 = 14581-8669", 
      "", 
      "# Calculations", 
      "hRbar = hfnotbar + cpbar*(400-298) + 2.265*deltahbarO2 + 8.515*deltahbarN2      # in kj/kmol", 
      "# With enthalpy of formation values for CO2, CO, and H2O(g) from Table A-25 and enthalpy values from Table A-23", 
      "hpbar = .951*(-393520 + (88806 - 9364)) + .049*(-110530 + (58191 - 8669)) + .289*(60371 - 8682) + 8.515*(57651 - 8669) + 2*(-241820 + (72513 - 9904))", 
      "Qcvdot = hpbar - hRbar                                                          # in kj/kmol", 
      "", 
      "# Result", 
      "print '-> The rate of heat transfer from the combustion chamber in kJ per kmol of fuel is:  '", 
      "print round(Qcvdot,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The rate of heat transfer from the combustion chamber in kJ per kmol of fuel is:  ", 
        "-221235.72"
       ]
      }
     ], 
     "prompt_number": 22
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.6 Page no-637"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "nCH4 = 1.00                                                                        # moles of methane in kmol", 
      "nO2 = 2.00                                                                         # moles of oxygen in kmol", 
      "T1 = 25.00                                                                         # in degree celcius", 
      "p1 = 1.00                                                                          # in atm", 
      "T2 = 900.00                                                                        # in kelvin", 
      "Rbar = 8.314                                                                       # universal gas constant", 
      "# The chemical reaction equation for the complete combustion of methane with oxygen is", 
      "# CH4 + 2O2   ---->   CO2 + 2H2O", 
      "", 
      "# Part(a)", 
      "# with enthalpy of formation values from table A-25", 
      "hfbarCO2 = -393520", 
      "hfbarH2O = -241820", 
      "hfbarCH4 = -74850", 
      "# Calculations", 
      "# with enthalpy values from table A-23", 
      "deltahbarCO2 = 37405-9364", 
      "deltahbarH2O = 31828-9904", 
      "Q = ((hfbarCO2 + deltahbarCO2)+2*(hfbarH2O + deltahbarH2O) - hfbarCH4) + 3*Rbar*(T1+273-T2)", 
      "# Result", 
      "print '-> The amount of heat transfer in kJ is:  '", 
      "print round(Q,2)", 
      "", 
      "# Part(b)", 
      "p2 = p1*(T2/(T1+273))                                                             # in atm", 
      "# Result", 
      "print '-> The final pressure in atm is:  '", 
      "print round(p2,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The amount of heat transfer in kJ is:  ", 
        "-745436.08", 
        "-> The final pressure in atm is:  ", 
        "3.02"
       ]
      }
     ], 
     "prompt_number": 23
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.7 Page no-639"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# The combustion equation is", 
      "# CH4 + 2O2 + 7.52N2  ---->  CO2 + 2H2O + 7.52N2", 
      "", 
      "# Part(a)", 
      "# With enthalpy of formation values from Table A-25", 
      "hfbarCO2 = -393520                                                              # in kj/kmol", 
      "hfbarH2O = -285830                                                              # in kj/kmol", 
      "hfbarCH4 = -74850                                                               # in kj/kmol", 
      "M = 16.04                                                                       # molar mass of CH4 in kg/kmol", 
      "# Calculations", 
      "hRPbar = hfbarCO2 + 2*hfbarH2O - hfbarCH4                                       # in kj/kmol", 
      "hRP = hRPbar/M                                                                  # in kj/kg", 
      "# Result", 
      "print '-> Part(a)the enthalpy of combustion of gaseous methane,  fuel is:  ',hRP,'kJ/kg.'", 
      "", 
      "# Part(b)", 
      "hfbarCO2 = -393520                                                              # in kj/kmol", 
      "hfbarH2O = -241820                                                              # in kj/kmol", 
      "hfbarCH4 = -74850                                                               # in kj/kmol", 
      "# Calculations", 
      "hRPbar = hfbarCO2 + 2*hfbarH2O - hfbarCH4                                       # in kj/kmol", 
      "hRP = hRPbar/M                                                                  # in kj/kg", 
      "# Result", 
      "print '-> Part(b)the enthalpy of combustion of gaseous methane, fuel is:  ',hRP,'kJ/kg'", 
      "", 
      "# Part(c)", 
      "# From table A-23", 
      "deltahbarO2 = 31389-8682                                                        # in kj/kmol", 
      "deltahbarH2O = 35882-9904                                                       # in kj/kmol", 
      "deltahbarCO2 = 42769-9364                                                       # in kj/kmol", 
      "", 
      "# Using table A-21", 
      "# Calculations", 
      "# function cpbar = f(T)", 
      "T=298                                                                           # in kelvin", 
      "from scipy import integrate", 
      "cpbar = lambda T: (3.826 - (3.979e-3)*T + 24.558e-6*T**2 - 22.733e-9*T**3 + 6.963e-12*T**4)*8.314", 
      "deltahbarCH4 = integrate.quad(cpbar,298,1000)", 
      "var = deltahbarCH4[0]", 
      "", 
      "hRPbar = hRPbar + (deltahbarCO2 + 2*deltahbarH2O - var -2*deltahbarO2)", 
      "hRP = hRPbar/M                                                                  # in kj/kg", 
      "# Result", 
      "print '-> Part(c)the enthalpy of combustion of gaseous methane, per kg of fuel is ',hRP,'kJ/kg'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> Part(a)the enthalpy of combustion of gaseous methane,  fuel is:   -55506.8578554 kJ/kg.", 
        "-> Part(b)the enthalpy of combustion of gaseous methane, fuel is:   -50019.3266833 kJ/kg", 
        "-> Part(c)the enthalpy of combustion of gaseous methane, per kg of fuel is  -49909.7030379 kJ/kg"
       ]
      }
     ], 
     "prompt_number": 24
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 13.8 Page no-643"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Part(a)", 
      "# For combustion of liquid octane with the theoretical amount of air, the chemical equation is", 
      "# C8H18(l) + 12.5 O2 + 47N2   -------> 8 CO2 + 9 H2O(g) + 47N2", 
      "# with enthalpy of formation data from Table A-25", 
      "hfbarC8H18 = -249910.0                                                            # in kj/kmol", 
      "hfbarCO2 = -393520.0", 
      "hfbarH2O = -241820.0", 
      "", 
      "# Calculations", 
      "RHS = hfbarC8H18 -(8*hfbarCO2 + 9*hfbarH2O)                                     # in kj/kmol", 
      "# at temperature 2400k", 
      "LHS1 = 5089337.0                                                                  # in kj/kmol", 
      "# at temperature 2350 k", 
      "LHS2 = 4955163.0                                                                  # in kj/kmol", 
      "# Interpolation between these temperatures gives", 
      "Tp = 2400.00 + ((2400.0-2350.0)/(LHS1-LHS2))*(RHS-LHS1)", 
      "# Result", 
      "print '-> The temperature in kelvin with theoretical amount of air is:  '", 
      "print round(Tp,2)", 
      "", 
      "# Part(b)", 
      "# For complete combustion of liquid octane with 400% theoretical air, the chemical equation is", 
      "# C8H18(l) + 50O2 + 188N2  -------->  8CO2 + 9H2O + 37.5O2 + 188N2", 
      "", 
      "# Proceeding iteratively as part(a)", 
      "Tp = 962                                                                        # in kelvin", 
      "", 
      "# Result", 
      "print '-> The temperature in kelvin using 400 percent theoretical air is:  '", 
      "print round(Tp,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The temperature in kelvin with theoretical amount of air is:  ", 
        "2394.52", 
        "-> The temperature in kelvin using 400 percent theoretical air is:  ", 
        "962.0"
       ]
      }
     ], 
     "prompt_number": 25
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.9 Page no-649"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "import math", 
      "# Part(a)", 
      "Tp = 2395                                                                       # in kelvin, from example 13.8", 
      "# For combustion of liquid octane with the theoretical amount of air, the chemical equation is", 
      "# C8H18(l) + 12.5O2 + 47N2  ---->  8CO2 + 9H2O(g) + 47N2", 
      "", 
      "# From table A-25", 
      "sFbar = 360.79                                                                  # absolute entropy of liquid octane in kj/kmol.K", 
      "", 
      "# From table A-23", 
      "# For reactant side", 
      "sbarO2atTref = 205.03                                                           # in kj/kmol.K", 
      "sbarN2atTref = 191.5                                                            # in kj/kmol.K", 
      "Rbar = 8.314                                                                    # universal gas constant in SI units", 
      "yO2 = 0.21", 
      "yN2 = 0.79", 
      "# For product side", 
      "yCO2 = 8.0/64.0", 
      "yH2O = 9.0/64.0", 
      "yN2p = 47.0/64.0", 
      "", 
      "# Calculations", 
      "sbarO2 = sbarO2atTref - Rbar*math.log(yO2)                                      # in kj/kmol.K", 
      "sbarN2 = sbarN2atTref - Rbar*math.log(yN2)                                      # in kj/kmol.K", 
      "# With the help from table A-23", 
      "sbarCO2 = 320.173 - Rbar*math.log(yCO2)", 
      "sbarH2O = 273.986 - Rbar*math.log(yH2O)", 
      "sbarN2p = 258.503 - Rbar*math.log(yN2p)", 
      "sigmadot = (8*sbarCO2 + 9*sbarH2O + 47*sbarN2p) - sFbar - (12.5*sbarO2 + 47*sbarN2)", 
      "", 
      "# Result", 
      "print '-> The rate of entropy production, in kJ/K per kmol of fuel with theoretical amount of air is:  '", 
      "print round(sigmadot,2)", 
      "", 
      "# Part(b)", 
      "# The complete combustion of liquid octane with 400% theoretical air is described by the following chemical equation:", 
      "# C8H18(l) + 50 O2 + 188N2  ----->  8 CO2 + 9H2O(g) + 37.5O2 + 188N2", 
      "", 
      "# For product side ", 
      "yCO2 = 8.0/242.5", 
      "yH2O = 9.0/242.5", 
      "yO2 = 37.5/242.5", 
      "yN2p = 188.0/242.5", 
      "# Calculations", 
      "# With help from table A-23", 
      "sbarCO2 = 267.12 - Rbar*math.log(yCO2)", 
      "sbarH2O = 231.01 - Rbar*math.log(yH2O)", 
      "sbarO2p = 242.12 - Rbar*math.log(yO2)", 
      "sbarN2p = 226.795 - Rbar*math.log(yN2p)", 
      "sigmadot = (8.0*sbarCO2 + 9.0*sbarH2O + 37.5*sbarO2p +188.0*sbarN2p) -sFbar - (50.0*sbarO2 + 188.0*sbarN2)", 
      "", 
      "# Result", 
      "print '-> The rate of entropy production, in kJ/K per kmol of fuel with 400 percent theoretical air is:  '", 
      "print round(sigmadot,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The rate of entropy production, in kJ/K per kmol of fuel with theoretical amount of air is:  ", 
        "5404.17", 
        "-> The rate of entropy production, in kJ/K per kmol of fuel with 400 percent theoretical air is:  ", 
        "9754.75"
       ]
      }
     ], 
     "prompt_number": 26
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.10 Page no-653"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "import math", 
      "Rbar = 8.314                                                                    # universal gas constant in SI units", 
      "# The chemical equation for the complete combustion of methane with oxygen is", 
      "# CH4 + 2O2 ---->  CO2 + 2H2O", 
      "yCH4 = 1.0/3.0", 
      "yO2 = 2.0/3.0", 
      "yCO2 = 1.0/3.0", 
      "yH2O = 2.0/3.0", 
      "# From table A-25", 
      "sbarCH4atTref = 186.16                                                          # in kj/kmol.K", 
      "sbarO2atTref = 205.03                                                           # in kj/kmol.K", 
      "p2 = 3.02                                                                       # in atm", 
      "pref = 1.0                                                                      # in atm", 
      "", 
      "# Calculations", 
      "sbarCH4 = sbarCH4atTref - Rbar*math.log(yCH4)", 
      "sbarO2 = sbarO2atTref - Rbar*math.log(yO2)", 
      "# With help from table A-23", 
      "sbarCO2 = 263.559 - Rbar*math.log(yCO2*p2/pref)                                 # in kj/kmol.K", 
      "sbarH2O = 228.321 - Rbar*math.log(yH2O*p2/pref)                                 # in kj/kmol.K", 
      "deltaS = sbarCO2 + 2*sbarH2O - sbarCH4 -2*sbarO2                                # in kj/K", 
      "", 
      "# Result", 
      "print '-> The change in entropy of the system  is:  ',round(deltaS,2),'kJ/K'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The change in entropy of the system  is:   96.41 kJ/K"
       ]
      }
     ], 
     "prompt_number": 27
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.11 Page no-654"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Methane is formed from carbon and hydrogen according to", 
      "# C + 2H2 ------>  CH4", 
      "", 
      "# In the present case, all substances are at the same temperature and pressure, 25\u0004C and 1 atm, which correspond to the standard reference state values", 
      "hCbar = 0", 
      "hH2bar = 0", 
      "gRbar = 0", 
      "# With enthalpy of formation and absolute entropy data from Table A-25", 
      "hfbarCH4 = -74850", 
      "sbarCH4 = 186.16", 
      "sbarC = 5.74", 
      "sbarH2 = 130.57", 
      "Tref = 298.15                                                                   # in kelvin", 
      "", 
      "# Calculation", 
      "gfbarCH4 = hfbarCH4 -Tref*(sbarCH4-sbarC-2*sbarH2)                              # in kj/kmol", 
      "", 
      "# Result", 
      "print '-> The gibbs function of formation of methane at the standard state is:  ',gfbarCH4,'kJ/mol'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The gibbs function of formation of methane at the standard state is:   -50783.332 kJ/mol"
       ]
      }
     ], 
     "prompt_number": 28
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 13.12 Page no-662"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Complete combustion of liquid octane with O2 is described by", 
      "# C8H18(l) + 12.5O2   ------>   8CO2 + 9H2O", 
      "import math", 
      "# Part(a)", 
      "Rbar = 8.314                                                                    # universal gas constant in SI units", 
      "Tnot = 298.15                                                                   # in kelvin", 
      "# From table A-25", 
      "gbarC8H18 = 6610.0", 
      "gbarO2 = 0", 
      "gbarCO2 = -394380", 
      "gbarH2O = -228590", 
      "yO2 = 0.2035", 
      "yCO2 = 0.0003", 
      "yH2O = 0.0312", 
      "M = 114.22                                                                      # molecular weight of liquid octane", 
      "", 
      "# Calculations", 
      "ech = ((gbarC8H18 + 12.5*gbarO2 -8*gbarCO2 -9*gbarH2O) + Rbar*Tnot*math.log(yO2**12.5/(yCO2**8*yH2O**9    )))/M", 
      "# Result", 
      "print '-> Part(a) the chemical exergy obtained on a unit mass basis is:  ',round(ech,2),'kJ/K'", 
      "", 
      "# Part(b)", 
      "# With data from Table A-25 and Model II of Table A-26", 
      "gbarH2O = -237180.0", 
      "ebarCO2 = 19870.0", 
      "ebarH2O = 900.0", 
      "ebarO2 = 3970.0", 
      "", 
      "# Calculation", 
      "ech = ((gbarC8H18 + 12.5*gbarO2 -8*gbarCO2 - 9*gbarH2O) + 8*ebarCO2 + 9*ebarH2O - 12.5*ebarO2)/M", 
      "# Result", 
      "print '-> Part(b) chemical exergy on a unit mass basis is:',round(ech,3),'kJ/K'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> Part(a) the chemical exergy obtained on a unit mass basis is:   47345.85 kJ/K", 
        "-> Part(b) chemical exergy on a unit mass basis is: 47397.172 kJ/K"
       ]
      }
     ], 
     "prompt_number": 29
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.13 Page no-665"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "import math", 
      "Rbar = 8.314                                                                # universal gas constant in SI units", 
      "Tnot = 298.0                                                                # in kelvin", 
      "# With data from the steam tables", 
      "h = 2939.9                                                                  # in kj/kg", 
      "hnot = 104.9                                                                # in kj/kg", 
      "s = 7.2307                                                                  # in kj/kg", 
      "snot = 0.3674                                                               # in kj/kg", 
      "# With data from Table A-25", 
      "gbarH2Oliq = -237180.0", 
      "gbarH2Ogas = -228590.0", 
      "yeH2O = 0.0303", 
      "M =18.0                                                                     # molar mass of steam", 
      "", 
      "# Calculations", 
      "ech = (1.0/M)*(gbarH2Oliq-gbarH2Ogas + Rbar*Tnot*math.log(1/yeH2O))         # in kj/kg", 
      "ef = h-hnot-Tnot*(s-snot) + ech                                             # in kj/kg", 
      "", 
      "# Result", 
      "print '-> The flow exergy of the steam, in kJ/k is:  '", 
      "print round(ef,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The flow exergy of the steam, in kJ/k is:  ", 
        "793.8"
       ]
      }
     ], 
     "prompt_number": 30
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.14 Page no-665"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# For 140% theoretical air, the reaction equation for complete combustion of methane is", 
      "# CH4 + 2.8(O2 + 3.76N2)  ------->  CO2 + 2H2O + 10.53N2 + .8O2", 
      "", 
      "# For product side", 
      "yCO2p = 1.0/(1.0+2.0+10.53+.8)", 
      "yH2Op = 2.0/(1.0+2.0+10.53+.8)", 
      "yN2p = 10.53/(1.0+2.0+10.53+.8)", 
      "yO2p = 0.8/(1.0+2.0+10.53+.8)", 
      "", 
      "Rbar = 8.314                                                                    # universal gas constant in SI units", 
      "Tnot = 298.15                                                                   # in kelvin", 
      "", 
      "yeN2 = 0.7567", 
      "yeO2 = 0.2035", 
      "yeH2O = 0.0303", 
      "yeCO2 = 0.0003", 
      "", 
      "# Calculations", 
      "import math", 
      "ebarch = Rbar*Tnot*(math.log(yCO2p/yeCO2) + 2*math.log(yH2Op/yeH2O) + 10.53*math.log(yN2p/yeN2) + .8*math.log(yO2p/yeO2))", 
      "", 
      "# with data from tables A-23 at 480 and 1560 kelvin,the thermomechanical contribution to the flow exergy, per mole of fuel, is", 
      "contri480 = 17712.0                                                               # kJ per kmol of fuel", 
      "contri1560 = 390853.0                                                             # kJ per kmol of fuel", 
      "efbar480 = contri480 + ebarch                                                     # kJ per kmol of fuel", 
      "efbar1560 = contri1560 + ebarch                                                   # kJ per kmol of fuel", 
      "", 
      "# Results", 
      "print '-> At T= 480k, the flow exergy of the combustion products, in kJ per kmol of fuel is:  '", 
      "print round(efbar480,2)", 
      "print '-> At T = 1560K, the flow exergy of the combustion products, in kJ per kmol of fuel is:  '", 
      "print round(efbar1560,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> At T= 480k, the flow exergy of the combustion products, in kJ per kmol of fuel is:  ", 
        "35461.61", 
        "-> At T = 1560K, the flow exergy of the combustion products, in kJ per kmol of fuel is:  ", 
        "408602.61"
       ]
      }
     ], 
     "prompt_number": 31
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 13.15 Page no-667"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "mFdot = 1.8e-3                                                              # fuel mass flow rate in kg/s", 
      "ech = 47346.0                                                               # in kj/kg, from example 13.12(a)", 
      "Wcvdot = 37.0                                                               # power developed by the engine in kw", 
      "", 
      "# Calculations", 
      "Efdot = mFdot*ech                                                           # rate at which exergy enters with the fuel in kw", 
      "epsilon = Wcvdot/Efdot                                                      # exergetic efficiency", 
      "", 
      "# Result", 
      "print '-> The exergetic efficiency is:  '", 
      "print round(epsilon,3)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The exergetic efficiency is:  ", 
        "0.434"
       ]
      }
     ], 
     "prompt_number": 32
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 13.16 Page no-668"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "Tnot = 298                                                                       # in kelvin", 
      "", 
      "# For the case of complete combustion with the theoretical amount of air", 
      "sigmadot = 5404.0                                                                # rate of entropy production from example 13.9, in kj/kmol.K", 
      "Efdot = 5407843.0                                                                # rate at which exergy enters with the fuel from example 13.12, in kj/kmol", 
      "# Calculations:-", 
      "Eddot = Tnot*sigmadot                                                            # in kj/kmol", 
      "epsilon = 1-Eddot/Efdot", 
      "# Result", 
      "print '-> The exergetic efficiency with theoretical amount of air is:  '", 
      "print round(epsilon,3)", 
      "", 
      "# For the case of combustion with 400% theoretical air", 
      "sigmadot = 9754.0                                                                 # rate of entropy production from example 13.9, in kj/kmol.K", 
      "# Calculations", 
      "Eddot = Tnot*sigmadot                                                             # in kj/kmol", 
      "epsilon = 1-Eddot/Efdot", 
      "# Result", 
      "print 'The exergetic efficiency with 400 percent theoretical amount of air is:  '", 
      "print round(epsilon,3)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The exergetic efficiency with theoretical amount of air is:  ", 
        "0.702", 
        "The exergetic efficiency with 400 percent theoretical amount of air is:  ", 
        "0.463"
       ]
      }
     ], 
     "prompt_number": 33
    }
   ]
  }
 ]
}