{
 "metadata": {
  "name": "Chapter 11"
 }, 
 "nbformat": 2, 
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown", 
     "source": [
      "#Chapter 11 :- Thermodynamic Relations"
     ]
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 11.1 Page no-491"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "m = 4.00                                                                        # mass of carbon monoxide in kg", 
      "T = 223.00                                                                      # temperature of carbon monoxide in kelvin", 
      "D = 0.2                                                                         # inner diameter of cylinder in meter", 
      "L = 1.00                                                                        # length of the cylinder in meter", 
      "pi=3.14", 
      "# Analysis", 
      "M = 28.00                                                                       # molar mass in kg/kmol", 
      "# Calculations", 
      "V = (pi*D**2.00/4.00)*L                                                         # volume occupied by the gas in m^3", 
      "vbar = M*(V/m)                                                                  # The molar specific volume in m^3/kmol", 
      "", 
      "# Part(a)", 
      "# From Table A-1 for CO", 
      "Tc = 133                                                                        # in kelvin", 
      "Pc = 35                                                                         # in bar", 
      "Tr = T/Tc                                                                       # reduced temperature", 
      "Rbar = 8314                                                                     # universal gas constant in N.m/kmol.K", 
      "Z = 0.9", 
      "# Calculations", 
      "vrdash = (vbar*Pc*10**5)/(Rbar*Tc)                                              # pseudoreduced specific volume", 
      "p = (Z*Rbar*T/vbar)*10**-5                                                      # in bar", 
      "# Result", 
      "print '-> part(a)the pressure in bar is:  '", 
      "print round(p,2)", 
      "", 
      "# Part(b)", 
      "# The ideal gas equation of state gives", 
      "# Calculations", 
      "p = (Rbar*T/vbar)/10**5                                                          # in bar", 
      "# Result", 
      "print '-> Part(b)the pressure in bar is: '", 
      "print round(p,2)", 
      "", 
      "# Part(c)", 
      "# For carbon monoxide, the van der Waals constants a and b can be read directly from Table A-24", 
      "a = 1.474                                                                        # in (m^3/kmol)^2", 
      "b = 0.0395                                                                       # in m^3/kmol", 
      "# Calculations", 
      "p = (Rbar*T/(vbar-b))/10**5 - a/vbar**2", 
      "# Result", 
      "print '-> Part(c)the pressure in bars is:  '", 
      "print round(p,2)", 
      "", 
      "# Part(d)", 
      "# For carbon monoxide, the Redlich\u2013Kwong constants can be read directly from Table A-24", 
      "a = 17.22                                                                        # in m^6*K^.5/kmol^2", 
      "b = 0.02737                                                                      # in m^3/kmol", 
      "# Calculations", 
      "p = (Rbar*T/(vbar-b))/10**5 - a/(vbar*(vbar+b)*T**.5)", 
      "# Result", 
      "print '-> Part(d)the pressure in bar is:  '", 
      "print round(p,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> part(a)the pressure in bar is:  ", 
        "75.92", 
        "-> Part(b)the pressure in bar is: ", 
        "84.35", 
        "-> Part(c)the pressure in bars is:  ", 
        "72.32", 
        "-> Part(d)the pressure in bar is:  ", 
        "75.12"
       ]
      }
     ], 
     "prompt_number": 1
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "%matplotlib inline"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "", 
        "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].", 
        "For more information, type 'help(pylab)'."
       ]
      }
     ], 
     "prompt_number": 2
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 11.3 Page no-501"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      " ", 
      "# Given:-", 
      "# Part(a)", 
      "v = 0.4646                                                                       # specific volume in in m^3/kg", 
      "M = 18.02                                                                        # molar mass of water in kg/kmol", 
      "# At the specified state, the temperature is 513 K and the specific volume on a molar basis is", 
      "vbar = v*M                                                                       # in m^3/kmol", 
      "# From Table A-24", 
      "a = 142.59                                                                       # (m^3/kmol)^2 * K^.5", 
      "b = 0.0211                                                                       # in m^3/kmol", 
      "", 
      "Rbar = 8314.0                                                                    # universal gas constant in N.m/kmol.K", 
      "T = 513.0                                                                        # in kelvin", 
      "delpbydelT = (Rbar/(vbar-b) + a/(2*vbar*(vbar+b)*T**1.5)*10**5)/10**3             # in kj/(m^3*K)", 
      "", 
      "# By The Maxwell relation", 
      "delsbydelv = delpbydelT", 
      "# Result", 
      "print '-> The value of delpbydelT in kj/(m^3*K) is:   ',delpbydelT", 
      "", 
      "# Part(b)", 
      "from pylab import *", 
      "# A value for (dels/delv)T can be estimated using a graphical approach with steam table data, as follows: At 240\u0004C, Table A-4 provides the values for specific entropy s and specific volume v tabulated below", 
      "T = 240.0                                                                         # in degree celcius", 
      "# At p =1, 1.5, 3, 5, 7, 10 bar respectively", 
      "y = [7.994, 7.805, 7.477, 7.230, 7.064, 6.882]            # in kj/kg.k                                                   # in kj/kg.k", 
      "x = [2.359, 1.570, 0.781, 0.4646, 0.3292, 0.2275]               # in m^3/kg                                                # in m^3/kg", 
      "plot(x,y)", 
      "xlabel(\"Specific volume\")", 
      "ylabel(\"Specific entropy\")", 
      "show()", 
      "# The pressure at the desired state is 5 bar.The corresponding slope is", 
      "delsbydelv = 1                                                                    # in kj/m^3.K", 
      "print '-> From the data of the table,delsbydelv = ',delsbydelv"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The value of delpbydelT in kj/(m^3*K) is:    1.00430251045"
       ]
      }, 
      {
       "output_type": "display_data", 
       "png": 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TqIYNHRGRMlVJMCxYsICYmJgS7x86dMgpLPz8/MjMzCwRDFOnTnX8OSIigoiI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\n"
      }, 
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> From the data of the table,delsbydelv =  1"
       ]
      }
     ], 
     "prompt_number": 3
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 11.4 Page no-506"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Analysis", 
      "# For comparison, Table A-2 gives at 100\u0004C,", 
      "hgf =2257.00                                                                    # in kj/kg", 
      "ugf = 2087.6                                                                    # in kj/kg", 
      "sgf = 6.048                                                                     # in kj/kg.K", 
      "# Values", 
      "print '-> From table, hg-hf = ',hgf", 
      "print '-> From table, ug-uf = ',ugf", 
      "print '-> From table, sg-sf = ',sgf", 
      "", 
      "# Part(a)", 
      "T = 373.15                                                                      # in kelvin", 
      "# If we plot a graph between temperature and saturation pressure using saturation pressure\u2013temperature data from the steam tables, the desired slope is:", 
      "delpbydelT = 3570.00                                                            # in N/(m^2.K)", 
      "vg = 1.673                                                                      # in m^3/kg", 
      "vf = 1.0435e-3                                                                  # in m^3/kg", 
      "# Calculations", 
      "# From the Clapeyron equation", 
      "hgf = T*(vg-vf)*delpbydelT*10**-3                                               # in kj/kg", 
      "# Result", 
      "print '-> Part(a)using Clapeyron equation, hg-hf = ', round(hgf,2)", 
      "", 
      "# Part(b)", 
      "psat = 1.014e5                                                                  # in N/m^2", 
      "hgf = 2256.00                                                                   # can be obtained using IT software in kj/kg", 
      "# Calculations", 
      "ugf = hgf - psat*(vg-vf)/10**3                                                  # in kj/kg", 
      "# Result", 
      "print '-> Part(b)ug-uf = ', round(ugf,2)", 
      "# Part(c)", 
      "# Calculation", 
      "sgf =hgf/T                                                                      # in kj/kg.K         ", 
      "# Result", 
      "print '-> Part(c)sg-sf = ', round(sgf,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> From table, hg-hf =  2257.0", 
        "-> From table, ug-uf =  2087.6", 
        "-> From table, sg-sf =  6.048", 
        "-> Part(a)using Clapeyron equation, hg-hf =  2227.29", 
        "-> Part(b)ug-uf =  2086.46", 
        "-> Part(c)sg-sf =  6.05"
       ]
      }
     ], 
     "prompt_number": 4
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 11.6 Page no-517"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Part(a)", 
      "v = 1.00/998.21                                                             # specific volume of water in m^3/kg", 
      "T = 293.00                                                                  # given temperature in kelvin", 
      "beta = 206.6e-6                                                             # volume expansivity in /K", 
      "k = 45.90e-6                                                                # isothermal compressibility in /bar", 
      "# Interpolating in Table A-19", 
      "cp = 4.188                                                                  # in kj/kg.k", 
      "# Calculations", 
      "cpv = (v*T*beta**2.00/k)*10**2                                              # in kj/kg.k", 
      "cv = cp-cpv                                                                 # in kj/kg.k", 
      "errorPercentage = 100*(cp-cv)/cv", 
      "# Result", 
      "print '-> The percentage error is:  ',round(errorPercentage,2)", 
      "", 
      "# Part(b)", 
      "# Calculations", 
      "K = cp/cv                                                                   # specific heat ratio", 
      "c = ((K*v/k)*10**5)**0.5                                                      # velocity of sound in m/s", 
      "# Result", 
      "print '-> The velocity of sound is:  ',round(c,2),'m/s'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The percentage error is:   0.66", 
        "-> The velocity of sound is:   1482.19 m/s"
       ]
      }
     ], 
     "prompt_number": 5
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 11.8 Page no-526"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "p1 = 100.00                                                                     # in bar", 
      "T1 = 300.00                                                                     # in kelvin", 
      "p2 = 40.00                                                                      # in bar", 
      "T2 = 245.00                                                                     # in kelvin", 
      "", 
      "", 
      "# From table A-23", 
      "h1starbar = 8723.00                                                             # in kj/kmol", 
      "h2starbar = 7121.00                                                             # in kj/kmol", 
      "# From Tables A-1", 
      "Tc = 126.00                                                                     # critical temperature in kelvin", 
      "pc = 33.9                                                                       # critical pressure in bar", 
      "M = 28.00                                                                       # molar mass in kg/kmol", 
      "Rbar = 8.314                                                                    # universal gas constant in kj/(kmol.K)", 
      "Term1 = 0.5                                                                      ", 
      "Term2 = 0.31", 
      "", 
      "# Calculations", 
      "TR1 = T1/Tc                                                                     # reduced temperature at the inlet", 
      "PR1 = p1/pc                                                                     # reduced pressure at the inlet", 
      "TR2 = T2/Tc                                                                     # reduced temperature at the exit", 
      "PR2 = p2/pc                                                                     # reduced pressure at the exit", 
      "wcvdot = (1.00/M)*(h1starbar-h2starbar-Rbar*Tc*(Term1-Term2))                      # in kj/kg", 
      "", 
      "# Result", 
      "print '-> The work developed, in kJ per kg of nitrogen flowing is :  '", 
      "print round(wcvdot,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The work developed, in kJ per kg of nitrogen flowing is :  ", 
        "50.11"
       ]
      }
     ], 
     "prompt_number": 6
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 11.9 Page no-529"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Part(a)", 
      "# With values from Table A-23", 
      "sT2bar = 185.775                                                                # in kj/(kmol.K)", 
      "sT1bar = 191.682                                                                # in kj/(kmol.K)", 
      "Rbar = 8.314                                                                    # universal gas constant", 
      "M = 28.00                                                                       # molar mass in kg/kmol ", 
      "p2 = 40.00                                                                      # in bar", 
      "p1 = 100.00                                                                     # in bar", 
      "Term1 = 0.21", 
      "Term2 = 0.14", 
      "", 
      "# Calculations", 
      "import math", 
      "S2StarBarMinusS1StarBar = sT2bar-sT1bar-Rbar*math.log(p2/p1)                    # The change in specific entropy in kj/(kmol.K)", 
      "sigmacvdot = (1.00/M)*(S2StarBarMinusS1StarBar-Rbar*(Term2-Term1))", 
      "# Result", 
      "print '-> the rate of entropy production in kj/kg.K is:  '", 
      "print round(sigmacvdot,2)", 
      "", 
      "# Part(b)", 
      "# From Table A-23,", 
      "h2starbar = 6654.00                                                             # in kj/kmol", 
      "h1starbar = 8723.00                                                             # in kj/kmol", 
      "Tc = 126.00                                                                     # critical temperature in kelvin", 
      "Term2 = 0.36", 
      "Term1 = 0.5", 
      "wcvdot = 50.1                                                                   # from example 11.8", 
      "", 
      "# Calculations", 
      "wcvdots = (1.00/M)*(h1starbar-h2starbar-Rbar*Tc*(Term1-Term2))                  # isentropic work in kj/kg", 
      "etat = wcvdot/wcvdots                                                           # turbine efficiency", 
      "", 
      "# Result", 
      "print '-> The isentropic turbine efficiency is: '", 
      "print round(etat,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> the rate of entropy production in kj/kg.K is:  ", 
        "0.08", 
        "-> The isentropic turbine efficiency is: ", 
        "0.73"
       ]
      }
     ], 
     "prompt_number": 7
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 11.10 Page no-533"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Analysis", 
      "V = 0.241                                                                        # volume of the mixture in m^3", 
      "T = 511.00                                                                       # temperature of the mixture in kelvin", 
      "n1 = 0.18                                                                        # number of moles of methane in kmol", 
      "n2 = 0.274                                                                       # number of moles of butane in kmol", 
      "Rbar = 8314                                                                      # universal gas constant in (N.m)/(kmol.K)", 
      "", 
      "# Calculations", 
      "n = n1 + n2                                                                      # The total number of moles of mixture", 
      "y1 = n1/n                                                                        # mole fraction of methane", 
      "y2 = n2/n                                                                        # mole fraction of butane", 
      "vbar = V/(n)                                                                     # The specific volume of the mixture on a molar basis in m^3/kmol", 
      "", 
      "# Part(a)", 
      "p = (Rbar*T/vbar)*10**-5                                                         # in bar", 
      "# Result", 
      "print '-> The pressure in bar obtained using ideal gas equation is:  '", 
      "print round(p,2)", 
      "", 
      "# Part(b)", 
      "# From table A-1", 
      "Tc1 = 191.00                                                                     # critical temperature for methane in kelvin ", 
      "Pc1 = 46.4                                                                       # critical pressure for methane in bar", 
      "Tc2 = 425.00                                                                     # critical temperature for butane in kelvin", 
      "Pc2 = 38.00                                                                      # critical pressure for butane in bar", 
      "Z = 0.88", 
      "", 
      "", 
      "# Calculations", 
      "Tc = y1*Tc1 + y2*Tc2                                                             # critical temperature in kelvin", 
      "Pc = y1*Pc1 + y2*Pc2                                                             # critical pressure in bar", 
      "TR = T/Tc                                                                        # reduced temperature of the mixture", 
      "vRdash= vbar*Pc/(Rbar*Tc)", 
      "p = ((Z*Rbar*T)/vbar)*10**-5                                                     # mixture pressure in bar", 
      "# Result", 
      "print '-> Pressure obtained using Kay\u2019s rule together with the generalized compressibility chart, is:  '", 
      "print round(p,2)", 
      "", 
      "# Part(c)", 
      "# Table A-24 gives the following van der Waals constants values for methane", 
      "a1 = 2.293                                                                       # in (m^3/kmol)^2", 
      "b1 = 0.0428                                                                      # in m^3/kmol", 
      "# Table A-24 gives the following van der Waals constants values for butane", 
      "a2 = 13.86                                                                       # in (m^3/kmol)^2", 
      "b2 = 0.1162                                                                      # in m^3/kmol", 
      "", 
      "a = (y1*a1**.5 + y2*a2**.5)**2                                                   # in bar*(m^3/kmol)^2", 
      "b = y1*b1+y2*b2                                                                  # in m^3/kmol", 
      "# From van der Waals equation", 
      "p = ((Rbar*T)/(vbar-b))*10**-5 - a/(vbar**2)", 
      "print '-> The pressure in bar from van der Waals equation is:   '", 
      "print round(p,2)", 
      "", 
      "# Part(d)", 
      "# For methane", 
      "TR1 = T/Tc1", 
      "vR1dash = (.241/.18)*10**5*Pc1/(Rbar*Tc1)", 
      "Z1 = 1.00", 
      "# For butane", 
      "TR2 = T/Tc2", 
      "vR2dash = (.88*10**5*Pc2)/(Rbar*Tc2)", 
      "Z2 = 0.8", 
      "Z = y1*Z1 + y2*Z2", 
      "# Accordingly, the same value for pressure as determined in part (b) using Kay\u2019s rule results:", 
      "p = 70.4", 
      "", 
      "# Result", 
      "print '-> The pressure in bar obtained using the rule of additive pressures employing the generalized compressibility chart is:  '", 
      "print round(p,2)"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The pressure in bar obtained using ideal gas equation is:  ", 
        "80.03", 
        "-> Pressure obtained using Kay\u2019s rule together with the generalized compressibility chart, is:  ", 
        "70.43", 
        "-> The pressure in bar from van der Waals equation is:   ", 
        "66.97", 
        "-> The pressure in bar obtained using the rule of additive pressures employing the generalized compressibility chart is:  ", 
        "70.4"
       ]
      }
     ], 
     "prompt_number": 8
    }
   ]
  }
 ]
}