path: root/Chemical_Engineering_Thermodynamics_by_Y_V_C_Rao/ch5.ipynb

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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 5 : Second law of thermodynamics and its applications"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.2  Page No : 161"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math \n",
      "\n",
      "# Variables\n",
      "Q = 1000.;\t\t    \t #amount of energy absorbed by the heat engine in kJ/s\n",
      "W = 650.;\t\t\t     #work delivered by the heat engine in kW\n",
      "T_source = 500. \t\t #temperature of the source in degree celsius\n",
      "T_sink = 25.\t\t\t #temperature of the sink in degree celsius\n",
      "\n",
      "# Calculations\n",
      "n_claimed = W/Q\n",
      "T1 = T_source+273.15\n",
      "T2 = T_sink+273.15\n",
      "n_carnot = 1-(T2/T1)\n",
      "\n",
      "# Results\n",
      "print \" The efficiency of the Carnot engine = %0.3f \"%(n_carnot);\n",
      "print \" The efficiency of the engine claimed by the inventor = %0.2f \"%(n_claimed);\n",
      "if n_claimed<n_carnot:\n",
      "    print \" The claimed heat engine is theoretically feasible as the efficiency of the engine is lesser than that of a Carnot engine\";\n",
      "else:\n",
      "    print \" The claimed heat engine is not theoretically feasible as the efficiency of the engine is greater than that of a Carnot engine\";\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The efficiency of the Carnot engine = 0.614 \n",
        " The efficiency of the engine claimed by the inventor = 0.65 \n",
        " The claimed heat engine is not theoretically feasible as the efficiency of the engine is greater than that of a Carnot engine\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.3  Page No : 165"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T_source_summer = 42; #temperature in the summer months in degree celsius\n",
      "T_sink_winter = 0;\t  #temperature in the winter months in degree celius\n",
      "T_amb = 25;\t\t\t  #temperature at which the house is to be maintained during both the months in degree celsius\n",
      "energy_loss = 0.5;\n",
      "\n",
      "# Calculations\n",
      "T_H_summer = T_source_summer+273.15\n",
      "T_L_summer = T_amb+273.15\n",
      "T_H_winter = T_amb+273.15\n",
      "T_L_winter = T_sink_winter+273.15\n",
      "W_summer = (energy_loss*((T_H_summer-T_L_summer)**2))/(T_L_summer)\n",
      "W_winter = (energy_loss*((T_H_winter-T_L_winter)**2))/(T_H_winter)\n",
      "\n",
      "# Results\n",
      "print \" The minimum power required to operate the device in summer = %.4f kW \"%(W_summer);\n",
      "print \" The minimum power required to operate the device in winter = %f kW \"%(W_winter);\n",
      "\n",
      "# Note: Answer in book is wrong. Please calculated manually."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The minimum power required to operate the device in summer = 0.4847 kW \n",
        " The minimum power required to operate the device in winter = 1.048130 kW \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.4  Page No : 166"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T_L = 4.25 \t\t\t #normal boiling point of helium in K\n",
      "Q_L = 0.083\t\t\t #latent heat of vaporization of helium in kJ/mol\n",
      "n = 1.  \t\t\t #amount of liquid helium to be produced in kmol\n",
      "T_amb = 42.\t\t\t #ambient temperature in summer in degree celsius\n",
      "\n",
      "# Calculations\n",
      "T_H = T_amb+273.15\n",
      "COP = (T_L)/(T_H-T_L)\n",
      "W = (Q_L)/COP;\t\t\n",
      "\n",
      "# Results\n",
      "print \" The maximum possible COP of the unit = %0.4f \"%(COP);\n",
      "print \" The minimum amount of work to be done on the refrigerating unit = %f kJ \"%(W);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The maximum possible COP of the unit = 0.0137 \n",
        " The minimum amount of work to be done on the refrigerating unit = 6.071694 kJ \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.5  Page No : 166"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T_ice = 0.;\t\t        \t #temperature of the ice to be produced in degree celsius\n",
      "m = 5000.;\t\t\t         #rate at which ice is to be produced in kg/hour\n",
      "T_water = 0.;\t    \t\t #temperature of water used to produce ice in degree celsius\n",
      "T_amb = 40.;\t\t    \t #ambient temperature in degree celsius\n",
      "T_source = 100.;\t\t\t #temperature of the source for operating heat engine in degree celsius\n",
      "lambda_fusion = 6.002;\t\t\t #latent heat of fusion of water in kJ/mol at 0 degree celsius\n",
      "molar_mass = 18*10**-3;\t\t\t #molar mass of water in kg/mol\n",
      "\n",
      "# Calculations\n",
      "T_L = T_water+273.15\n",
      "T_H = T_amb+273.15\n",
      "COP = (T_L)/(T_H-T_L)\n",
      "Q_L = ((m/3600)/molar_mass)*(lambda_fusion)\n",
      "W = (Q_L)/(COP);\t\t\t \n",
      "T1 = T_source+273.15;\t\t\n",
      "T2 = T_amb+273.15\n",
      "n_heatengine = (T1-T2)/T1\n",
      "Q1 = W/n_heatengine;\t\n",
      "energy_ratio = (Q1+Q_L)/Q_L;\n",
      "\n",
      "# Results\n",
      "print \" The minimum power required to operate the refrigerator = %0.2f kW\"%(W);\n",
      "print \" The maximum possible efficiency of the heat engine = %0.4f \"%(n_heatengine);\n",
      "print \" Ratio of the energy rejected to the ambient atmosphere to the\\\n",
      " energy absorbed from the water = %0.4f \"%(energy_ratio);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The minimum power required to operate the refrigerator = 67.82 kW\n",
        " The maximum possible efficiency of the heat engine = 0.1608 \n",
        " Ratio of the energy rejected to the ambient atmosphere to the energy absorbed from the water = 1.9107 \n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.6  Page No : 169"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T1 = 800.\t\t\t #temperature of reservoir 1 in K\n",
      "T2 = 400.\t\t\t #temperature of reservoir 2 in K\n",
      "Q1 = 1000.\t\t\t #energy absorbed from reservoir maintained at T1 in kJ\n",
      "Q2 = 400.\t\t\t #energy absorbed from reservoir maintained at T2 in kJ\n",
      "W = 1000.\t\t\t #work delivered by the heat engine in kJ\n",
      "T3 = 300.\t\t\t #temperature of the sink in K\n",
      "\n",
      "# Calculations\n",
      "Q3 = (Q1+Q2)-W\n",
      "clausius_inequality = (Q1/T1)+(Q2/T2)-(Q3/T3)\n",
      "\n",
      "# Results\n",
      "print \" The LHS of the Clausius inequality = %0.4f \"%(clausius_inequality);\n",
      "if clausius_inequality<0 or clausius_inequality == 0:\n",
      "    print \" The given process does not violate the second law of thermodynamics, therefore the claim is correct\"\n",
      "else:\n",
      "    print \" This is a violation of the second law of thermodynamics, and hence the claim cannot be justified\"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The LHS of the Clausius inequality = 0.9167 \n",
        " This is a violation of the second law of thermodynamics, and hence the claim cannot be justified\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.7  Page No : 172"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T_system = 200.;\t\t\t #temperature of the contents of reactor in degree celsius\n",
      "t = 15.         \t\t\t #operation time of agitator in minutes\n",
      "P = 750.    \t    \t\t #power of the operating motor in W\n",
      "\n",
      "# Calculations\n",
      "dQ = P*t*60*10**-3\n",
      "T = T_system+273.15\n",
      "del_S = dQ/T;\t\t\n",
      "\n",
      "# Results\n",
      "print \" The change in the entropy of the reactor contents = %0.4f kJ/K \"%(del_S);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The change in the entropy of the reactor contents = 1.4266 kJ/K \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.8  Page No : 172"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "\n",
      "# Variables\n",
      "P = 0.101325;\t\t\t #pressure in the piston cylinder assembly in MPa\n",
      "T1 = 300.;\t\t\t #temperature of the piston cylinder assembly in K\n",
      "T2 = 400;\t\t\t #final temperature of the piston cylinder assembly in K\n",
      "a = 45.369;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
      "b = 8.688*10**-3;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
      "e = 9.619*10**5;\t\t\t #coefficients to compute isobaric molar heat capacity of CO2(g) in J/molK\n",
      "\n",
      "# Calculations\n",
      "del_S = (a*math.log(T2/T1))+(b*(T2-T1))+((e)*((1./T2**2)-(1./T1**2)))\n",
      "\n",
      "# Results\n",
      "print \" The change in entropy of CO2 = %.4f J/molK\"%(del_S);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The change in entropy of CO2 = 9.2447 J/molK\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.9  Page No : 173"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "m = 1.      \t\t\t #amount of saturated liquid water in kg\n",
      "T_initial = 100.\t\t #initial temperature of water in degree celsius\n",
      "T_body = 500.\t\t\t #temperature of body which is brought into contact with the cylinder in degree celsius\n",
      "hfg = 2256.94\t\t\t #enthalpy of vaporization taken from steam tables corresponding to T1 in kJ/kg\n",
      "\n",
      "# Calculations\n",
      "T = T_initial+273.15\n",
      "del_S = hfg/T;\t\t\n",
      "\n",
      "# Results\n",
      "print \" The change in entropy of water = %0.4f kJ/kgK\"%(del_S);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The change in entropy of water = 6.0483 kJ/kgK\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.10  Page No : 173"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "m_steel = 10.   \t\t\t #mass of steel casting in kg\n",
      "T_steel = 800.\t    \t\t #temperature of steel casting in degree celsius\n",
      "m_water = 100.\t\t    \t #mass of water used for quenching in kg\n",
      "T_water = 30.\t\t\t     #temperature of water used for quenching in degree celsius\n",
      "Cp_steel = 0.461;\t\t\t #heat capacity of steel in kJ/kgK\n",
      "Cp_water = 4.23;\t\t\t #heat capacity of water in kJ/kgK\n",
      "\n",
      "# Calculations\n",
      "Ti_steel = T_steel+273.15\n",
      "Ti_water = T_water+273.15\n",
      "T_final = ((m_steel*Cp_steel*Ti_steel)+(m_water*Cp_water*Ti_water))/((m_steel*Cp_steel)+(m_water*Cp_water));\n",
      "del_S_steel = m_steel*Cp_steel*math.log(T_final/Ti_steel)\n",
      "del_S_water = m_water*Cp_water*math.log(T_final/Ti_water)\n",
      "\n",
      "# Results\n",
      "print \" The change in entropy of steel  =  %0.4f kJ/K\"%(del_S_steel);\n",
      "print \" The change in entropy of water  =  %f kJ/K\"%(del_S_water);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The change in entropy of steel  =  -5.7031 kJ/K\n",
        " The change in entropy of water  =  11.427392 kJ/K\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.11  Page No : 175"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "V = 2.\t    \t\t #volume of insulated tank in m**3\n",
      "Ta = 400.\t\t\t #temperature of gas in compartment (a) in K\n",
      "Pa = 3.\t\t    \t #pressure of gas in compartment (a) in MPa\n",
      "Tb = 600.\t\t\t #temperature of gas in compartment (b) in K\n",
      "Pb = 1.\t\t\t     #pressure of gas in compartment (b) in MPa\n",
      "R = 8.314;\t\t\t #universal gas constant in J/molK\n",
      "\n",
      "# Calculations\n",
      "Va = V/2\n",
      "Vb = V/2\n",
      "Na = (Pa*10**6*Va)/(R*Ta)\n",
      "Nb = (Pb*10**6*Vb)/(R*Tb)\n",
      "T = ((Na*Ta)+(Nb*Tb))/(Na+Nb)\n",
      "N = Na+Nb;\t\t\t \n",
      "P = ((N*R*T)/V)*10**-6\n",
      "Cp = (5./2)*R;\t\t\t\n",
      "del_S = ((Na*((Cp*math.log(T/Ta))-(R*math.log(P/Pa))))+(Nb*((Cp*math.log(T/Tb))-(R*math.log(P/Pb)))))*10**-3; # Calculations of the change in entropy using Eq.(5.43) in kJ/K\n",
      "\n",
      "# Results\n",
      "print \" Entropy change of the gas = %0.2f kJ/K\"%(del_S);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Entropy change of the gas = 2.19 kJ/K\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.12  Page No : 177"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "N = 1.\t    \t\t #amount of air to be separated into its components in kmol\n",
      "P = 0.1\t\t    \t #pressure of air in MPa\n",
      "T = 300.\t\t\t #temperature of air in K\n",
      "per_oxygen = 21.\t\t\t #percentage of oxygen in air\n",
      "per_nitrogen = 79.\t\t\t #percentage of nitrogen in air\n",
      "R = 8.314;\t\t\t         #universal gas constant in J/molK\n",
      "\n",
      "# Calculations\n",
      "\n",
      "x1 = per_nitrogen/100\n",
      "x2 = per_oxygen/100;\n",
      "W = (T*N*10**3*R*((x1*math.log (x1))+(x2*math.log (x2))))*10**-3\n",
      "\n",
      "# Results\n",
      "print \" Minimum work to be done to separate 1kmol of air at 0.1MPa and 300K into pure oxygen\\\n",
      " and nitrogen at the same temperature and pressure = %0.2f kJ\"%(abs(W));"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " Minimum work to be done to separate 1kmol of air at 0.1MPa and 300K into pure oxygen and nitrogen at the same temperature and pressure = 1281.91 kJ\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.13  Page No : 179"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "m_ice = 10.\t    \t\t #mass of the block of ice in kg\n",
      "T_ice = 0.\t\t    \t #temperature of the ice in degree celsius\n",
      "m_water = 100.\t\t\t #mass of watre in the tank in kg\n",
      "T_water = 30.\t\t\t #temperature of the water in the tank in degree celsius\n",
      "Cp = 4.23;\t\t\t     #heat capacity of water in kJ/kgK\n",
      "lambda_melting = 333.44\t\t\t #latent heat of melting of ice in kJ/kg\n",
      "\n",
      "# Calculations\n",
      "Ti_ice = T_ice+273.15\n",
      "Ti_water = T_water+273.15\n",
      "T_final = ((m_water*Cp*Ti_water)+(m_ice*Cp*Ti_ice)-(m_ice*lambda_melting))/((m_ice*Cp)+(m_water*Cp))\n",
      "del_S_ice = ((m_ice*lambda_melting)/(Ti_ice))+(m_ice*Cp*math.log (T_final/Ti_ice));\t\t\t \n",
      "del_S_water = m_water*Cp*math.log (T_final/Ti_water)\n",
      "del_S_G = del_S_ice+del_S_water;\t\t\t \n",
      "\n",
      "# Results\n",
      "print \" The change in entropy of ice  =  %f kJ/K\"%(del_S_ice);\n",
      "print \" The change in entropy of water  =  %f kJ/K\"%(del_S_water);\n",
      "print \" The entropy generated =  %f kJ/K\"%(del_S_G);\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The change in entropy of ice  =  15.211650 kJ/K\n",
        " The change in entropy of water  =  -14.035033 kJ/K\n",
        " The entropy generated =  1.176617 kJ/K\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.14  Page No : 182"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "P = 3.\t\t\t         #pressure of superheated steam in MPa\n",
      "T_enter = 300.\t\t\t #entrance temperature of superheated steam in degree celsius\n",
      "T_exit = 45.\t\t\t #final temperature at which the steam leaves in degree celsisus\n",
      "m = 1.      \t\t\t #mass flow rate of steam in kg/s\n",
      "\n",
      "# Calculations\n",
      "\n",
      "si = 6.5422\n",
      "hi = 2995.1\n",
      "sf = 0.6383\n",
      "hf = 188.35\n",
      "sg = 8.1661;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
      "hg = 2583.3;\t\t\t #entahlpy of saturayed vapour in kJ/kg\n",
      "\n",
      "Xe = (si-sf)/(sg-sf)\n",
      "he = ((1-Xe)*hf)+(Xe*hg)\n",
      "Ws = -m*(he-hi);\t\t\n",
      "\n",
      "# Results\n",
      "print \" The power  Results from the turbine = %0.1f kW\"%(Ws);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The power  Results from the turbine = 928.4 kW\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.15  Page No : 183"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "Pi = 3.\t\t\t #pressure of dry saturated steam when it enters the nozzle in bar\n",
      "Pe = 2.\t\t\t #pressure of dry saturated steam at the exit in bar\n",
      "\n",
      "# Calculations\n",
      "#From steam tables corresponding to Pi\n",
      "si = 6.9909;\t\t\t #entropy of steam at the entrance in kJ/kgK\n",
      "hi = 2724.7;\t\t\t #entahlpy of steam at the entrance in kJ/kg\n",
      "\n",
      "#From steam tables corresponding to Pe\n",
      "sf = 1.5301;\t\t\t #entropy of saturated liquid in kJ/kgK\n",
      "hf = 504.70;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
      "sg = 7.1268;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
      "hg = 2706.3;\t\t\t #entahlpy of saturayed vapour in kJ/kg\n",
      "\n",
      "se = 6.9909\n",
      "Xe = (se-sf)/(sg-sf)\n",
      "he = ((1-Xe)*hf)+(Xe*hg)\n",
      "Ve = math.sqrt (2*(hi-he)*10**3)\n",
      "\n",
      "# Results\n",
      "print \" The exit velocity of steam = %f m/s\"%(Ve);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The exit velocity of steam = 379.103198 m/s\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.16  Page No : 183"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Variables\n",
      "N_glycerol = 100.\t\t\t #molar flow rate of glycerol in mol/s\n",
      "Ti_gly = 227.\t\t\t #inlet temperature of glycerol in degree celsius\n",
      "Te_gly = 40.\t\t\t #outlet temperature of glycerol in degree celsius\n",
      "Ti_water = 25.\t\t\t #inlet temperature of cooling water in degree celsius\n",
      "Te_water = 50.\t\t\t #outlet temperature of cooling water in degree celsius\n",
      "Cp_gly = 280.\t\t\t #heat capacity of glycerol in J/molK\n",
      "Cp_water = 77.\t\t\t #heat capacity of water in J/molK\n",
      "\n",
      "# Calculations\n",
      "Ti_gly = Ti_gly+273.15\n",
      "Te_gly = Te_gly+273.15\n",
      "Ti_water = Ti_water+273.15\n",
      "Te_water = Te_water+273.15\n",
      "\n",
      "N_water = -(N_glycerol*Cp_gly*(Te_gly-Ti_gly))/(Cp_water*(Te_water-Ti_water));\n",
      "del_S_gly = N_glycerol*Cp_gly*math.log (Te_gly/Ti_gly)*10**-3\n",
      "del_S_water = N_water*Cp_water*math.log (Te_water/Ti_water)*10**-3\n",
      "S_G = del_S_gly+del_S_water\n",
      "\n",
      "# Results\n",
      "print \" The rate at which entropy is generated in the heat exchanger = %0.3f kJ/K s\"%(S_G);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The rate at which entropy is generated in the heat exchanger = 3.754 kJ/K s\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.17  Page No : 183"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "T_i = 150.\t\t\t #temperature of saturated steam taken up by the device in degree celsius\n",
      "T_e = 200.\t\t\t #temperature of superheated steam delivered by the device in degree celsius\n",
      "P_e = 0.2\t\t\t #pressure of superheated steam delivered by the device in MPa\n",
      "me2 = 0.949\t\t\t #mass of superheated steam leaving the device in kg\n",
      "me1 = 0.051\t\t\t #mass of saturated liquid leaving the device in kg\n",
      "T_liq = 100.\t\t #temperature of saturated liquid leaving the device in degree celsius\n",
      "mi = 1.     \t\t #mass of saturated steam fed to the device in kg\n",
      "\n",
      "# Calculations\n",
      "#From steam tables corresponding to T_i\n",
      "hi = 2745.4 \t\t\t #enthalpy of saturated vapour in kJ/kg\n",
      "si = 6.8358;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
      "\n",
      "#For saturated liquid at T_liq\n",
      "he1 = 419.06;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
      "se1 = 1.3069;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
      "\n",
      "#For superheated steam at P_e and T_e\n",
      "he2 = 2870.5;\t\t\t #enthalpy of superheated steam in kJ/kg\n",
      "se2 = 7.5072;\t\t\t #entropy of superheated steam in kJ/kgK\n",
      "\n",
      "LHS = mi*hi;\n",
      "RHS = (me1*he1)+(me2*he2);\n",
      "\n",
      "S_G = (me1*se1)+(me2*se2)-(mi*si);\n",
      "\n",
      "# Results\n",
      "print \" The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied =  %0.1f kJ\"%(LHS);\n",
      "print \" The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = %0.1f kJ\"%(RHS);\n",
      "print \" The entropy generated by applying the second law of thermodynamics to the flow device = %0.4f kJ/kgK\"%(S_G);\n",
      "if int(LHS) ==  int(RHS) and S_G>0 or S_G == 0 :\n",
      "    print \" As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \"\n",
      "else:\n",
      "    print \" As both the first and second law or either the first or second law of thermodynamics \\\n",
      "     are not satisfied, the device is not feasible \"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The LHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied =  2745.4 kJ\n",
        " The RHS of the equation applied to the flow device to check if the first law of thermodynamics is satisfied = 2745.5 kJ\n",
        " The entropy generated by applying the second law of thermodynamics to the flow device = 0.3552 kJ/kgK\n",
        " As the first and second law of thermodynamics are satisfied, the device is theoretically feasible \n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.18  Page No : 185"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Pi = 30.\t\t\t #pressure of superheated steam entering the turbine in bar\n",
      "Ti = 300.\t\t\t #temperature of superheated steam entering the turbine in degree celsius\n",
      "Pe = 0.1\t\t\t #pressure at which steam exits the turbine in bar\n",
      "Xe = 0.9\t\t\t #quality of steam at the exit (no unit)(for the actual turbine)\n",
      "\n",
      "# Calculations\n",
      "#For superheated steam at Pi and Ti\n",
      "hi = 2995.1;\t\t\t #enthalpy of superheated steam at the entrance in kJ/kg\n",
      "si = 6.5422;\t\t\t #entropy of superheated steam at the entrance in kJ/kgK\n",
      "\n",
      "#For steam at Pe\n",
      "hf = 191.83;\t\t\t #enthalpy of saturated liquid in kJ/kg\n",
      "hg = 2584.8;\t\t\t #enthalpy of saturated vapour in kJ/kg\n",
      "sf = 0.6493;\t\t\t #entropy of saturated liquid in kJ/kgK\n",
      "sg = 8.1511;\t\t\t #entropy of saturated vapour in kJ/kgK\n",
      "\n",
      "X2 = (si-sf)/(sg-sf)\n",
      "h2 = (hf*(1-X2))+(X2*hg)\n",
      "he = (hf*(1-Xe))+(Xe*hg)\n",
      "n_T = (hi-he)/(hi-h2)\n",
      "\n",
      "# Results\n",
      "print \" The isentropic efficiency of the turbine = %f \"%(n_T);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The isentropic efficiency of the turbine = 0.703395 \n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.19  Page No : 186"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Ti = 25.\t\t\t #temperature of air taken in by the adiabatic air compressor in degree celsius\n",
      "Pi = 0.1\t\t\t #pressure of air taken in by the adiabatic air compressor in MPa\n",
      "Pe = 1.\t    \t\t #discharge pressure of air in MPa\n",
      "n_c = 0.8\t\t\t #isentropic efficiency of the compressor (no unit)\n",
      "gaamma = 1.4\t\t #ratio of molar specific heat capacities (no unit)\n",
      "R = 8.314\t\t\t #universal gas constant in J/molK\n",
      "\n",
      "# Calculations\n",
      "Ti = Ti+273.15\n",
      "Te = Ti*(((Pe*10**6)/(Pi*10**6))**((gaamma-1)/gaamma))\n",
      "W_s = (((R*gaamma)/(gaamma-1))*(Te-Ti))*10**-3;\t\t\t\n",
      "Ws = W_s/n_c\n",
      "Te_actual = ((Ws*10**3*(gaamma-1))/(R*gaamma))+Ti\n",
      "\n",
      "# Results\n",
      "print \" The exit temperature of air = %0.2f K\"%(Te_actual);\n",
      "print \" The power consumed by the compressor  = %f kW/mol\"%(Ws);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The exit temperature of air = 645.01 K\n",
        " The power consumed by the compressor  = 10.093262 kW/mol\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.20  Page No : 187"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Ti = 30.\t\t\t #temperature of saturated liquid water in degree celsius\n",
      "m = 500.\t\t\t #mass flow rate of water being pumped in kg/s\n",
      "P2 = 3. \t\t\t #preesure maintained in the boiler in MPa\n",
      "n_p = 0.75;\t\t\t #isentropic efficiency of the pump (no unit)\n",
      "\n",
      "# Calculations\n",
      "#For saturated liquid water at Ti\n",
      "vf = 0.0010043\n",
      "P1 = 4.241;\t\t\n",
      "\n",
      "Ws_m = (vf*((P2*10**6)-(P1*10**3)))*10**-3\n",
      "Ws_act_m = Ws_m/n_p;\t\t\t \n",
      "P = ((Ws_act_m*10**3)*m)*10**-6;\n",
      "\n",
      "# Results\n",
      "print \" The power consumed by the pump  =  %d MW\"%(P);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The power consumed by the pump  =  2 MW\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.21  Page No : 188"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Pi = 3.\t\t\t #pressure of dry saturated steam entering the nozzle in bar\n",
      "Xe = 0.98\t\t #quality of steam exiting the nozzle (no unit)\n",
      "Pe = 2.\t\t\t #pressure of steam exiting the nozzle in bar\n",
      "\n",
      "# Calculations\n",
      "#For steam at Pi\n",
      "hi = 2724.7\n",
      "he = 2652.8\n",
      "V2_2_s = hi-he\n",
      "\n",
      "#For steam at Pe\n",
      "hf = 504.70\n",
      "hg = 2706.3\n",
      "he_act = ((1-Xe)*hf)+(Xe*hg)\n",
      "V2_2 = hi-he_act;\t\t\t\n",
      "n_N = (V2_2)/(V2_2_s)\t\t\n",
      "\n",
      "# Results\n",
      "print \" The isentropic efficiency of the nozzle = %0.3f \"%(n_N);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " The isentropic efficiency of the nozzle = 0.868 \n"
       ]
      }
     ],
     "prompt_number": 25
    }
   ],
   "metadata": {}
  }
 ]
}