path: root/Thermodynamics_An_Engineering_Approach/Chapter8.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 8: Exergy: A Measure of Work Potential"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-1 ,Page No.426"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi\n",
      "\n",
      "# given data\n",
      "D=12.0;#diameter of rotor in m\n",
      "V=10.0;#average velocity in m/s\n",
      "\n",
      "# density of air at 25C & 1atm\n",
      "p=1.18;\n",
      "\n",
      "#calculations\n",
      "ke=(V**2)/2/1000;#factor of 1000 for converting J into kJ\n",
      "m=p*pi*(D**2)*V/4.0; \n",
      "MP=m*(ke);\n",
      "print'Maximum power %f kW'%round(MP,1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum power 66.700000 kW\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-2 ,Page No.427"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given values\n",
      "TH=2000.0;#temperature of large furnace in R\n",
      "T0=77+460.0;#temperature of enviroment in R\n",
      "Qin=3000.0;#heat rate in Btu/sec\n",
      "\n",
      "#calculation\n",
      "nth=1-(T0/TH);\n",
      "Wmax=nth*Qin;\n",
      "Wmax=round(Wmax)\n",
      "print'the rate of energy flow %i Btu/s'%round(Wmax)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rate of energy flow 2195 Btu/s\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-3 ,Page No.429"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "Tsink=300.0;#temperature of sink in K\n",
      "Tsource=1200.0;#temperature of source in K\n",
      "Qin=500.0;#heat rate in kJ/s\n",
      "Wuout=180;#power output in W\n",
      "\n",
      "#calculations\n",
      "Wrev=(1-Tsink/Tsource)*Qin;\n",
      "print'The reversible power %i kW'%round(Wrev);\n",
      "I=Wrev-Wuout;\n",
      "print'the irreversiblity rate %i kW'%round(I)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The reversible power 375 kW\n",
        "the irreversiblity rate 195 kW\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-4 ,Page No.429"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from scipy.integrate import quad \n",
      "from pylab import *\n",
      "\n",
      "#given data\n",
      "m=500.0;#mass of iron block in kg\n",
      "T1=473.0;#temperature of iron block in K\n",
      "T0=300.0;#temperature of surrondings in K\n",
      "\n",
      "\n",
      "#from Table A-3\n",
      "cavg=0.45;\n",
      "\n",
      "#calculations\n",
      "def intgrnd1(T): \n",
      " return ((1-T0/T)*(-m*cavg))#intergrant\n",
      "Wrev, err = quad(intgrnd1,T1,T0) ;\n",
      "Qtotal=m*cavg*(T1-T0);\n",
      "print'The reversible power %i kJ'%round(Wrev);\n",
      "Wu=0;\n",
      "I=Wrev-Wu;\n",
      "print'the irreversiblity rate %i kJ'%round(I)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The reversible power 8191 kJ\n",
        "the irreversiblity rate 8191 kJ\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-5 ,Page No.431"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "Wrev=8191.0;#reversible work from Ex - 8.4\n",
      "Wtotal=38925.0;#total work is equivalent to total heat lost to heat engine from Ex - 8.4\n",
      "TL=278.0;#outdoor temperature in K\n",
      "TH=300.0;#temperature of house in K\n",
      "\n",
      "#calculations\n",
      "Wrm=Wtotal-Wrev;#work remaining\n",
      "COPHP=1/(1-TL/TH);\n",
      "Wd=COPHP*Wrev;#work delivered\n",
      "PS=Wd+Wrm;\n",
      "PS=round(PS/1000);#factor of 1000 for converting kJ into MJ\n",
      "print'Maximum amount of heat that can be supplied %i MJ'%PS"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum amount of heat that can be supplied 142 MJ\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-6 ,Page No.434"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "COP=1.0;# the efficiency that the dealer is referring to is the first law efficiency, meaning that for each unit of electric energy (work) consumed, the heater will supply the house with 1 unit of energy (heat). That is,the advertised heater has a COP of 1\n",
      "TL=283.0;#outdoor temperature in K\n",
      "TH=294.0;#indoor temperature in K\n",
      "\n",
      "#calculations\n",
      "COPHP=1/(1-TL/TH);\n",
      "nII=COP/COPHP;\n",
      "print'the second law efficiency %f'%round(nII,3)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the second law efficiency 0.037000\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-7 ,Page No.438"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "\n",
      "#given data\n",
      "P1=1000.0;#pressure of compressed air in KPa\n",
      "V=200.0;#volumne of rigid tank in m^3\n",
      "T1=300.0;#temperature of enviroment in K\n",
      "T0=T1;#state of air in tank\n",
      "P0=100.0;#pressure of enviroment in KPa\n",
      "\n",
      "#constants used\n",
      "R=0.287;#in kPa m^3/kg K\n",
      "\n",
      "#calculations\n",
      "m1=P1*V/(R*T1);\n",
      "O1=R*T0*(P0/P1-1)+R*T0*log(P1/P0);# O refers to exergy\n",
      "X1=m1*O1/1000;#factor of 1000 for converting kJ into MJ\n",
      "print'work obtained %i MJ'%round(X1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "work obtained 281 MJ\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-8 ,Page No.439"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "T0=20+273;#temperature of enviroment in K\n",
      "P1=0.14;#intail pressure in MPa\n",
      "T1=-10;#intail temperature in K\n",
      "P2=0.8;#final pressure in MPa\n",
      "T2=50;#final temperature in K\n",
      "\n",
      "#the properties of refrigerant\n",
      "#at inlet\n",
      "h1=246.36;\n",
      "s1=0.9724;\n",
      "#at outlet\n",
      "h2=286.69;\n",
      "s2=0.9802;\n",
      "dO=h2-h1-T0*(s2-s1);# O refers to exergy\n",
      "print'the exergy change of the refrigerant %i kJ/kg'%round(dO)\n",
      "wmin=dO;\n",
      "print'the minimum work input that needs to be supplied is %i kJ/kg'%wmin"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the exergy change of the refrigerant 38 kJ/kg\n",
        "the minimum work input that needs to be supplied is 38 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-10 ,Page No.447"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given values\n",
      "Q=1035.0;#rate of heat transfer in W\n",
      "T0=273.0;#outdoor temperature in C\n",
      "Tin=293.0;#inner wall surface temperature in K\n",
      "Tout=278.0;#outer wall surface temperature in K\n",
      "T1=300.0;#indoor temperature in C\n",
      "\n",
      "#calculations\n",
      "#Xin - Xout - Xdestroyed = dX/dt\n",
      "Xdestroyed=Q*(1-T0/Tin)-Q*(1-T0/Tout);\n",
      "Xdestroyed=round(Xdestroyed);\n",
      "print'the rate of exergy destroyed %i W'%round(Xdestroyed);\n",
      "#the total rate of exergy destroyed\n",
      "Xdestroyed=Q*(1-T0/T1)-Q*(1-T0/T0);\n",
      "print'the total of exergy destroyed %f W'%round(Xdestroyed,1);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the rate of exergy destroyed 52 W\n",
        "the total of exergy destroyed 93.100000 W\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-11 ,Page No.448"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "m=0.05;#mass of steam in kg\n",
      "P1=1000;#intial pressure in MPa\n",
      "T1=300+273;#intial temperature in K\n",
      "P2=200;#final pressure in MPa\n",
      "T2=150+273;#final temperature in K\n",
      "P0=100;#pressure of surroundings in KPa\n",
      "T0=25+273;#temperature of surroundings in K\n",
      "Qout=2;#heat losses in kJ\n",
      "\n",
      "#from Table A-6 & A-4\n",
      "u1=2793.7;\n",
      "v1=0.25799;\n",
      "s1=7.1246;\n",
      "u2=2577.1;\n",
      "v2=0.95986;\n",
      "s2=7.2810;\n",
      "u0=104.83;\n",
      "v0=0.00103;\n",
      "s0=0.3672;\n",
      "\n",
      "#calculations\n",
      "X1=m*(u1-u0-T0*(s1-s0)+P0*(v1-v0));\n",
      "X2=m*(u2-u0-T0*(s2-s0)+P0*(v2-v0));\n",
      "print'exergy of intial state %i kJ'%round(X1);\n",
      "print'exergy of final state %f kJ'%round(X2,1);\n",
      "dX=X2-X1;\n",
      "print'exergy change in system %f kJ'%round(dX,1);\n",
      "Wout=-Qout-m*(u2-u1);\n",
      "Wu=Wout-P0*m*(v2-v1);\n",
      "Xdestroyed=X1-X2-Wu;\n",
      "print'the exergy destroyed %f kJ'%round(Xdestroyed,1);\n",
      "nII=Wu/(X1-X2);\n",
      "print'second law efficiency of this process is %f'%round(nII,3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "exergy of intial state 35 kJ\n",
        "exergy of final state 25.400000 kJ\n",
        "exergy change in system -9.700000 kJ\n",
        "the exergy destroyed 4.300000 kJ\n",
        "second law efficiency of this process is 0.551000\n"
       ]
      }
     ],
     "prompt_number": 42
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-12 ,Page No.451"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from scipy.integrate import quad \n",
      "from pylab import *\n",
      "\n",
      "#given data\n",
      "m=2.0;#mass of air in lbm\n",
      "T0=70+460.0;#intial temperature in R\n",
      "P1=20.0;#intial pressure in psia\n",
      "T1=70+460.0;#temperature of surroundings in R\n",
      "T2=130+460.0;#final temperature in R\n",
      "\n",
      "#constants used\n",
      "Cv=0.172;#in Btu/lbm - F\n",
      "\n",
      "#calculations\n",
      "Xdestroyed=T0*m*Cv*log(T2/T1);\n",
      "print'exergy destroyed %f Btu'%round(Xdestroyed,1);\n",
      "def intgrnd1(T): \n",
      " return ((1-T0/T)*m*Cv)#intergrant\n",
      "Wrev, err = quad(intgrnd1,T1,T2) ;\n",
      "print'the reversible work %f Btu'%round(Wrev,1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "exergy destroyed 19.600000 Btu\n",
        "the reversible work 1.100000 Btu\n"
       ]
      }
     ],
     "prompt_number": 45
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-13 ,Page No.453"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "T0=20+273.0;#temperature of surrounding in K\n",
      "P0=100.0;#pressure of surrounding in kPa\n",
      "Tiw=30+273.0;#temperature of water in K\n",
      "mw=100.0;#mass of water in kg\n",
      "Tii=350+273.0;#temperature of block in K\n",
      "mi=5.0;#mass of block in kg\n",
      "\n",
      "#constants used(Table A-3)\n",
      "cw=4.18;#specific heat of water in kJ/kg C\n",
      "ci=0.45;#specific heat of iron in kJ/kg C\n",
      "\n",
      "#calculations\n",
      "Tfk=(mi*ci*Tii+mw*cw*Tiw)/(mw*cw+mi*ci);\n",
      "Tfc=Tfk-273;#in C\n",
      "print'the final equilibrium temperature %f C'%round(Tfc,1);\n",
      "X1i=mi*ci*(Tii-T0-T0*log(Tii/T0));\n",
      "X1w=mw*cw*(Tiw-T0-T0*log(Tiw/T0));\n",
      "X1t=X1i+X1w;#total exergy \n",
      "print'intial exergy of combined systems %i kJ'%round(X1t);\n",
      "X2i=mi*ci*(Tfk-T0-T0*log(Tfk/T0));\n",
      "X2w=mw*cw*(Tfk-T0-T0*log(Tfk/T0));\n",
      "X2t=X2i+X2w;#total exergy \n",
      "print'intial exergy of combined systems %f kJ'%round(X2t,1);\n",
      "Xdestroyed=X1t-X2t;\n",
      "print'the wasted work %f kJ'%round(Xdestroyed,1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the final equilibrium temperature 31.700000 C\n",
        "intial exergy of combined systems 315 kJ\n",
        "intial exergy of combined systems 95.800000 kJ\n",
        "the wasted work 219.100000 kJ\n"
       ]
      }
     ],
     "prompt_number": 46
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-14 ,Page No.455"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "TR=1200.0;#temperature of furnace \n",
      "T0=300;#temperature of surrounding in K\n",
      "P0=100;#pressure of surrounding in kPa\n",
      "Tsys=400.0;#temperature of argon in K\n",
      "P1=350;#temperature of argon in K\n",
      "V1=0.01;#intail volumne in m^3\n",
      "V2=2*V1;#final volumne\n",
      "\n",
      "#calculations\n",
      "W=P1*V1*log(V2/V1);\n",
      "Wsurr=P0*(V2-V1);\n",
      "Wu=W-Wsurr;\n",
      "print'the useful work output %f kJ'%round(Wu,2);\n",
      "# Qin - W = m*Cv*dT, Since dt=0\n",
      "Q=W;\n",
      "Sgen=Q/Tsys-Q/TR;\n",
      "Xdestroyed=T0*Sgen;\n",
      "print'the exergy destroyed %f kJ/K'%round(Xdestroyed,2);\n",
      "Wrev=T0*Q/Tsys-Wsurr+(1-T0/TR)*Q;\n",
      "print'the reversible work is done in the process %f kJ'%round(Wrev,2);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the useful work output 1.430000 kJ\n",
        "the exergy destroyed 1.210000 kJ/K\n",
        "the reversible work is done in the process 2.640000 kJ\n"
       ]
      }
     ],
     "prompt_number": 48
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-15 ,Page No.460"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculation error in textbook in part - b which changes all the following answers\n",
      "\n",
      "#given data\n",
      "m=8.0;#mass flow rate in kg/s\n",
      "T0=298.0;#temperature of surrounding in K\n",
      "P0=100.0;#pressure of surrounding in kPa\n",
      "P1=3000.0;#inlet pressure in kPa\n",
      "T1=450.0;#inlet temperature in K\n",
      "P2=200.0;#outlet pressure in kPa\n",
      "T2=150.0;#outlet temperature in K\n",
      "Qout=300.0;#heat rate in kW\n",
      "\n",
      "#from Table A-6 and A-4\n",
      "h1=3344.9;\n",
      "s1=7.0856;\n",
      "h2=2769.1;\n",
      "s2=7.2810;\n",
      "h0=104.83;\n",
      "s0=0.3672;\n",
      "\n",
      "#calculations\n",
      "# Ein = Eout\n",
      "Wout=m*(h1-h2)-Qout;\n",
      "Wout=round(Wout)\n",
      "print'the actual power output %i kW'%Wout\n",
      "# Xin = Xout\n",
      "Wrev=m*(h1-h2)-(T0*(s1-s2));\n",
      "Wrev=round(Wrev);\n",
      "print'the maximum possible work output %i kW'%Wrev;\n",
      "nII=Wout/Wrev;\n",
      "print'second law efficiency is %f'%round(nII,3);\n",
      "Xdestroyed=Wrev-Wout;\n",
      "print'the exergy destroyed %i kW'%round(Xdestroyed);\n",
      "X1=h1-h0-T0*(s1-s0);\n",
      "print'the exergy of the steam at inlet conditions %i kJ/kg'%round(X1);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the actual power output 4306 kW\n",
        "the maximum possible work output 4665 kW\n",
        "second law efficiency is 0.923000\n",
        "the exergy destroyed 359 kW\n",
        "the exergy of the steam at inlet conditions 1238 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-16 ,Page No.462"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "T0=70+460;#temperature of surroundings in R\n",
      "T1=50;#temperature of water in F\n",
      "T2=240;#temperature of steam in F\n",
      "T3=130;#outlet temperature in F\n",
      "#as dicussed in example 7-20\n",
      "m1=300;#mass flow rate of water in lbm/min\n",
      "m2=22.7;#mass flow rate of steam in lbm/min\n",
      "m3=322.7;#outlet mass flow rate  in lbm/min\n",
      "\n",
      "#from steam tables\n",
      "h1=18.07;\n",
      "s1=0.03609;\n",
      "h2=1162.3;\n",
      "s2=1.7406;\n",
      "h3=97.99;\n",
      "s3=0.18174;\n",
      "\n",
      "#calculations\n",
      "Wrev=m1*(h1-T0*s1)+m2*(h2-T0*s2)-m3*(h3-T0*s3);\n",
      "Wrev=round(Wrev);\n",
      "print'the reversible power %i Btu/min'%round(Wrev)\n",
      "Xdestroyed=Wrev;\n",
      "print'the rate of exergy destruction %i Btu/min'%Xdestroyed\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the reversible power 4588 Btu/min\n",
        "the rate of exergy destruction 4588 Btu/min\n"
       ]
      }
     ],
     "prompt_number": 56
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8-17 ,Page No.463"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "V=200.0;#volumne of rigid tank in m^3\n",
      "P1=100.0;#intial surroundings in kPa\n",
      "P2=1000.0;#final pressure in kPa\n",
      "P0=100.0;#pressure of surroundings in kPa\n",
      "T=300.0;#temperature of surroundings in K\n",
      "\n",
      "#constants used\n",
      "R=0.287;#in kPa m^3/kg K\n",
      "\n",
      "#calculations\n",
      "#Xin - Xout = Xdestroyed = X2 - X1\n",
      "m2=P2*V/(R*T);\n",
      "X2=R*T*(log(P2/P0)+P0/P2-1);\n",
      "Wrev=m2*X2/1000;\n",
      "Wrev=round(Wrev);\n",
      "print'Work requirement %i MJ'%Wrev\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Work requirement 281 MJ\n"
       ]
      }
     ],
     "prompt_number": 58
    }
   ],
   "metadata": {}
  }
 ]
}