path: root/Thermodynamics/Chapter5.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 5:FIRST LAW,INTERNAL ENERGY AND ENTHALPY"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Exapmle 5.1, Page No:169"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "Q12=-250; # Heat transfer during Discharging of battery in kcal\n",
      "W21=-0.53; # Consumption of electricity dring Charging process in kWh\n",
      "\n",
      "#Calculation\n",
      "Q21=(W21*3600)-(Q12*4.1868); # First law of thermodynamics\n",
      "\n",
      "#Result\n",
      "print \"Heat loss from battery during charging process = \",Q21,\"kJ\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat loss from battery during charging process =  -861.3 kJ\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.2, Page No:173"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "m=5; # Mass of water in a tank in kg\n",
      "T1=30; # Temperature of water at initial state (1) in degree celcius\n",
      "T2=95; # Temperature of water at final state (2) in degree celcius\n",
      "Qout=70; # Heat transfer from the water tank to the surrounding air in kJ\n",
      "W=75; #Electric energy input to a stirrer inside water in kJ\n",
      "mf=32.3; # Mass of fel in bomb in grams\n",
      "u1=125.78; # Internal energy of water from steam table (uf at T1) in kJ/kg\n",
      "u2=397.88; # Internal energy of water from steam table (uf at T2) in kJ/kg\n",
      "\n",
      "#Calculation\n",
      "Qf=m*(u2-u1)-W+Qout; # From First law of thermodynamics\n",
      "qf=Qf/(mf*10**-3); # Heat consumption per unit mass of fuel \n",
      "\n",
      "#Result\n",
      "print \"Heat consumption per unit mass of fuel =\",round(qf,0),\"kJ/kg\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat consumption per unit mass of fuel = 41966.0 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.3, Page No:173"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V=50; # Volume of water in a tank in litres\n",
      "T1=120; # Temperature of water at initial state (1) in degree celcius\n",
      "x1=0.6; # Dryness fraction at initial state (1)\n",
      "T2=-10; # Temperature of water at final state (2) in degree celcius\n",
      "vf1=0.00106; # specific volume of water from steam tables at T1 in m^3/kg\n",
      "vg1=0.8919; # specific volume of water from steam tables at T1 in m^3/kg\n",
      "v1=(1-x1)*vf1+x1*vg1; # Specific volume of misture of liquid and water at state (1)\n",
      "m=(V*10**-3)/v1; # Mass of water in the tank\n",
      "vs2=0.0010891; # Specific volume of saturated ice at T2 in m^3/kg\n",
      "vg2=466.7; # Specific volume of water vapour at T2 in m^3/kg\n",
      "v2=v1; # constant specific volume during cooling process\n",
      "x2=(v2-vs2)/(vg2-vs2); # Dryness fraction at state (2)\n",
      "uf1=503.5; # Specific internal energy at state (1) in kJ/kg\n",
      "ug1=2529.3; # Specific internal energy at state (1) in kJ/kg\n",
      "us2=-354.09; # Specific internal energy at state (2) in kJ/kg\n",
      "ug2=2361.4; # Specific internal energy at state (2) in kJ/kg\n",
      "\n",
      "#Calculation\n",
      "u1=(1-x1)*uf1+x1*ug1; # Total Specific internal energy at state (1) in kJ/kg\n",
      "u2=(1-x2)*us2+x2*ug2; # Total Specific internal energy at state (2) in kJ/kg\n",
      "Q12=m*(u2-u1); # Heat transfer during cooling pocess\n",
      "\n",
      "#Result\n",
      "print \"Heat transfer during cooling pocess = \",round(Q12,1),\"kJ   (round off error)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat transfer during cooling pocess =  -193.3 kJ   (round off error)\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.4, Page No:174"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V1=0.3; # Initial volume of water upto stop 1 in m^3\n",
      "p1=1; # Initial pressure of water in bar\n",
      "x1=0.2; # Dryness fraction at initial state (1)\n",
      "p2=3; # Pressur required to lift the piston in bar\n",
      "V4=0.45; # Volume of water upto stop 2 in m^3\n",
      "vf1=0.001043; # Specific volume at state (1) from steam table in m^3/kg\n",
      "vg1=1.694; # Specific volume at state (1) from steam table in m^3/kg\n",
      "\n",
      "#Calculation for (a)\n",
      "v1=vf1+x1*(vg1-vf1); # Total Specific volume at state (1) from steam table in m^3/kg\n",
      "m=V1/v1; # Mass of water\n",
      "v3=V4/m; # Specific volume at stop 2\n",
      "v2=v1; p3=p2; v4=v3; V3=V4; V2=V1; # From process diagram\n",
      "#  (a)\n",
      "p4=0.361; # Final Pressure at v4 from steam table in Mpa\n",
      "\n",
      "#Result for (a)\n",
      "print \"(a)\",\"\\nFianl pressure = \",p4,\"MPa\"\n",
      "\n",
      "#Calculation for (b)\n",
      "#  (b)\n",
      "W14=p2*10**2*(V3-V2); # Work done in process \n",
      "uf1=417.36; # Specific internal energ at initial state in kJ/kg\n",
      "ufg1=2088.7; # Specific internal energ at initial state in kJ/kg\n",
      "u1=uf1+x1*ufg1;  # Total Specific internal energr at initial state in kJ.kg\n",
      "u4=2550.2; # Specific internal energ at final state in kJ/kg\n",
      "Q14=m*(u4-u1)+W14; # From first law of thermodynamics\n",
      "\n",
      "#Result for (b)\n",
      "print \"\\n\\n(b)\",\"\\nWork done during the process = \",W14,\"kJ\",\"\\nHeat transfer during the process = \",round(Q14,1),\"kJ\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) \n",
        "Fianl pressure =  0.361 MPa\n",
        "\n",
        "\n",
        "(b) \n",
        "Work done during the process =  45.0 kJ \n",
        "Heat transfer during the process =  1560.0 kJ\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.5, Page No:177"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V1=0.01; # Initial Volume of Freon 12 vapour in cylinder in m^3\n",
      "T1=15; # Initial Temperature of Freon 12 vapour in degree celcius\n",
      "p1=4.914; # Initial pressure (Psat at T1) in bar\n",
      "p2=9; # Final pressure of Freon 12 vapour after compression in bar\n",
      "T2=65; # Final temperature of Freon 12 vapour after compression in degree celcius\n",
      "Q=-0.5; # Heat lost to surroundings during compresson process in kJ\n",
      "v1=0.035413; # Initial specific volume of Freon 12 vapour from table in m^3/kg\n",
      "\n",
      "#Calculation\n",
      "m=V1/v1; # Mass of vapour\n",
      "hg1=193.644; # specific enthalpy of Freon 12 vopour at state 1 in kJ/kg\n",
      "u1=hg1-(p1*10**2*v1); # Total Specific internal energy at state 1\n",
      "h2=222.9; # specific enthalpy of Freon 12 vapour at state 2 in kJ/kg\n",
      "v2=0.022537; # specific volume of Freon 12 vapour at state 2 in m^3/kg\n",
      "u2=h2-(p2*10**2*v2); # Total Specific internal energy at state 2\n",
      "W=-m*(u2-u1)+Q; # From first law of thermodynamics\n",
      "\n",
      "#Result\n",
      "print \"Work of compression = \",round(W,2),\"kJ\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Work of compression =  -7.95 kJ\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.6, Page No:181"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "tS=-10; # initial temperature of ice in degree celcius\n",
      "ts=0; # Melting temperature of ice in degree celcius\n",
      "Cs=1.94; # Specific heat of ice in kJ/kg k\n",
      "m=1000; # Mass of ice in kg\n",
      "tF=10; # Temperature of water after 24 hours in degree celcius\n",
      "tf=0; # Freezing temperature of water in degree celcius\n",
      "Cf=4.1868; # Specific heat of water in kJ/kg K\n",
      "hsg=335; # Latent heat of fusion of ice in kJ/kg\n",
      "\n",
      "#Calculation\n",
      "Q=m*(Cs*(ts-tS)+hsg+Cf*(tF-tf)); # Heat gain of water\n",
      "Q_dot=Q/(24*3600); # Rate of cooling\n",
      "\n",
      "#Result\n",
      "print \"Rate of cooling = \",round(Q_dot,3),\"kW\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rate of cooling =  4.586 kW\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.7, Page No:188"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "T1=300; # Temperature of air at state 1 in Kelvin\n",
      "T2=500; # Temperature of air at state 2 in Kelvin\n",
      "m_=28.966; # Molecular weight oh air in kg\n",
      "Cpoav=1.017; # Average value of specific heat of air in kJ/kg K\n",
      "\n",
      "#Calculation for (a)\n",
      "# (a). change in enthalpy\n",
      "h_=27.43*(T2-T1)+3.09*10**-3*(T2**2-T1**2)-0.2296*10**-6*(T2**3-T1**3); #change in enthalpy during process in kJ/kmol\n",
      "h=h_/m_; # change in enthalpy during process in kJ/kg\n",
      "\n",
      "#Result for (a)\n",
      "print \"(a).change in enthalpy during process = \",round(h,2),\"kJ/kg\"\n",
      "\n",
      "#Calculation for (b)\n",
      "# (b).change in enthalpy\n",
      "h=Cpoav*(T2-T1); # change in enthalpy in kJ/kg\n",
      "\n",
      "#Result for (b)\n",
      "print \"(b).change in enthalpy during process with average specfic heat = \",h,\"kJ/kg\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a).change in enthalpy during process =  205.69 kJ/kg\n",
        "(b).change in enthalpy during process with average specfic heat =  203.4 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.8, Page No:189"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "m=0.1; # mass of nitrogen gas in kg\n",
      "V1=0.1; # Initial volme of nitrogen gas in m^3\n",
      "p1=1.2; # Initial pressure of nitrogen gas in bar\n",
      "V2=0.075; # Final volume of nitrogen gas in m^3\n",
      "Cpo=1.041; # Specific heat at constant pressure of nitrogen in kJ/kg K\n",
      "R=0.2969393; # Characteristic gas constant of nitrogen in Kj/kg K\n",
      "\n",
      "#Calculation\n",
      "T1=(p1*10**2*V1)/(m*R); # Initial temperature of nitrogen gas \n",
      "T2=T1*(V2/V1); # Final temperature of nitrogen gas (constant pressure process)\n",
      "Q=m*Cpo*(T2-T1); # Heat transfer to surroundings\n",
      "W=p1*10**2*(V2-V1); # Work done\n",
      "\n",
      "#Result\n",
      "print \"Final Temperature of nitrogen gas = \",round(T2,0),\"K\",\"\\nHeat transfer to surroundings = \",round(Q,1),\"kJ\"\n",
      "print \"Work done = \",W,\"kJ\",\"  The Work is done on the gas\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Final Temperature of nitrogen gas =  303.0 K \n",
        "Heat transfer to surroundings =  -10.5 kJ\n",
        "Work done =  -3.0 kJ   The Work is done on the gas\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.9, Page No:190"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "p=1; # pressure inside piston cylinder arrangement in MPa\n",
      "# stae 1 = saturated liquid\n",
      "# state 2 = saturated vapour\n",
      "# state 3 = superheated vapour\n",
      "v1=0.001127; # specific volume at state 1 in m^3/kg\n",
      "v2=0.19444; # specific volume at state 2 in m^3/kg\n",
      "v3=0.4011; # specific volume at state 3 in m^3/kg\n",
      "u1=761.68; # specific internal energy at state 1 in kK/kg\n",
      "u2=2583.6; # specific internal energy at state 2 in kK/kg\n",
      "u3=3296.8; # specific internal energy at state 3 in kK/kg\n",
      "h1=762.81; # specific enthalpy at state 1 in kJ/kg\n",
      "h2=2778.1; # specific enthalpy at state 2 in kJ/kg\n",
      "h3=3697.9; # specific enthalpy at state 3 in kJ/kg\n",
      "\n",
      "#Calculation for (Method I)\n",
      "w12=p*10**3*(v2-v1); # Work done during process 1-2\n",
      "w23=p*10**3*(v3-v2); # Work done during process 2-3\n",
      "wtotal=w12+w23; # Total work done\n",
      "# Calculation of heat transfer\n",
      "# Method I\n",
      "q12=(u2-u1)+w12; # Heat transfer during process 1-2\n",
      "q23=(u3-u2)+w23; # Heat transfer during process 2-3\n",
      "qtotal=q12+q23; # Total Heat transfer\n",
      "\n",
      "#Results for (Method I)\n",
      "print \"Work done = \",round(wtotal,0),\"kJ/kg\"\n",
      "print \"\\n\\nCalculation of Heat Transfer\",\"\\nMethod I\",\"\\nHeat Transfer = \",round(qtotal,1),\"kJ/kg\"\n",
      "\n",
      "#Calculation for (Method II)\n",
      "# Method II\n",
      "q12=h2-h1; # Heat transfer during process 1-2\n",
      "q23=h3-h2; # Heat transfer during process 2-3\n",
      "qtotal=q12+q23; # Total Heat transfer\n",
      "\n",
      "#Result for (Method II)\n",
      "print \"\\nMethod II\",\"\\nHeat Transfer = \",round(qtotal,1),\"kJ/kg\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Work done =  400.0 kJ/kg\n",
        "\n",
        "\n",
        "Calculation of Heat Transfer \n",
        "Method I \n",
        "Heat Transfer =  2935.1 kJ/kg\n",
        "\n",
        "Method II \n",
        "Heat Transfer =  2935.1 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.10, Page No:191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "#Variable declaration\n",
      "p1=1; # initial pressure of air in piston cylinder arrangement in bar\n",
      "T=300; # Temperature of air in piston cylinder arrangement in kelvin\n",
      "p2=10; # Final pressure of air in piston cylinder arrangement in bar\n",
      "R=0.287; # Characteristic gas constant of air in kJ/kg K\n",
      "\n",
      "#Calculation\n",
      "w=R*T*math.log(p1/p2); # Work done\n",
      "q=w; # From first law of thermodynamics\n",
      "\n",
      "#Results\n",
      "print \"The change in internal energy during the isothermal process is zero\"\n",
      "print \"Work done = \",round(w,1),\"kJ/kg\",\"\\nHeat transfer = \",round(q,1),\"kJ/kg\"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The change in internal energy during the isothermal process is zero\n",
        "Work done =  -198.3 kJ/kg \n",
        "Heat transfer =  -198.3 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.12, Page No:195"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#Variable declaration\n",
      "p1=65;# (Error in textbook) # Pressure of air at state 1 in bar\n",
      "v1=0.0135; # Volume of air at state 1 in m^3\n",
      "v2=0.1; # Volume of air at state 2 in m^3\n",
      "R=0.287; # Characteristic gas constant of air in kJ/kg K\n",
      "Cvo=0.7165; # Specific heat at constant volume in kJ/kg K\n",
      "\n",
      "#Calculation for (a)\n",
      "# (a).Adiabatic process\n",
      "k=1.4; # Index of adiabatic process\n",
      "p2=p1*(v1/v2)**k; # Pressure of air at state 2 \n",
      "T1=p1*10**2*v1/R;# Tempewrature of air at state 1\n",
      "T2=p2*10**2*v2/R;# Tempewrature of air at state 2\n",
      "w=R*(T2-T1)/(1-k); # work done\n",
      "q=0; # Adiabatic expansion process\n",
      "delta_u=Cvo*(T2-T1); # Change in internal energy of air\n",
      "\n",
      "#Result for (a)\n",
      "print \"(a).Adiabatic Process\",\"\\nFinal Temperature = \",round(T2,1),\"K\",\"\\nWork done = \",round(w,1),\"kJ\"\n",
      "print \"Change in internal energy of air = \",round(delta_u,1),\"kJ\",\"\\nHeat Ineraction = \",round(q,1),\"kJ\"\n",
      "\n",
      "#Calculation for (b)\n",
      "# (b).Polytropic  process\n",
      "n=1.3; # Index of adiabatic process\n",
      "p2=p1*(v1/v2)**n; # Pressure of air at state 2 \n",
      "T1=p1*10**2*v1/R;# Tempewrature of air at state 1\n",
      "T2=p2*10**2*v2/R;# Tempewrature of air at state 2\n",
      "w=R*(T2-T1)/(1-n); # work done\n",
      "delta_u=Cvo*(T2-T1); # Change in internal energy of air\n",
      "q=delta_u+w; # Adiabatic expansion process\n",
      "\n",
      "#Result for (b)\n",
      "print \"\\n\\n(b).Polytropic Process\",\"\\nFinal Temperature = \",round(T2,1),\"K\",\"\\nWork done = \",round(w,1),\"kJ\"\n",
      "print \"Change in internal energy of air = \",round(delta_u,1),\"kJ\",\"\\nHeat Ineraction = \",round(q,1),\"kJ\"\n",
      "\n",
      "#Calculation for (c)\n",
      "# (c).Isothermal process\n",
      "T1=p1*10**2*v1/R;# Tempewrature of air at state 1\n",
      "T2=T1; # Tempewrature of air at state 2\n",
      "p2=p1*(v1/v2); # Pressure of air at state 2\n",
      "w=R*T1*math.log (v2/v1); # work done\n",
      "delta_u=Cvo*(T2-T1); # Change in internal energy of air\n",
      "q=delta_u+w; # Adiabatic expansion process\n",
      "\n",
      "#Result for (c)\n",
      "print \"\\n\\n(c).Isothermal Process\",\"\\nFinal Temperature = \",round(T2,1),\"K\",\"\\nWork done = \",round(w,1),\"kJ\"\n",
      "print \"Change in internal energy of air = \",round(delta_u,1),\"kJ\",\"\\nHeat Ineraction = \",round(q,1),\"kJ\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a).Adiabatic Process \n",
        "Final Temperature =  137.2 K \n",
        "Work done =  120.9 kJ\n",
        "Change in internal energy of air =  -120.7 kJ \n",
        "Heat Ineraction =  0.0 kJ\n",
        "\n",
        "\n",
        "(b).Polytropic Process \n",
        "Final Temperature =  167.7 K \n",
        "Work done =  132.1 kJ\n",
        "Change in internal energy of air =  -98.9 kJ \n",
        "Heat Ineraction =  33.2 kJ\n",
        "\n",
        "\n",
        "(c).Isothermal Process \n",
        "Final Temperature =  305.7 K \n",
        "Work done =  175.7 kJ\n",
        "Change in internal energy of air =  0.0 kJ \n",
        "Heat Ineraction =  175.7 kJ\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.13, Page No:201"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "m=10; # mass flow rate of chilled water in kg/s\n",
      "V1=50; #velocity of chilled water at section 1 in m/s\n",
      "z1=30; # Elevation of section 1 in m\n",
      "V2=10; #velocity of chilled water at section 2 in m/s\n",
      "z2=60; # Elevation of section 2 in m\n",
      "h1=21; # Enthalpy of chilled water at section 1 in kJ/kg\n",
      "h2=43; # Enthalpy of chilled water at section 2 in kJ/kg\n",
      "W=35; # Rate of work done by pump in kW\n",
      "g=9.80665; # Acceleration due to gravity in m/s^2\n",
      "\n",
      "#Calculation \n",
      "Q=m*(h2-h1)+(m*(V2**2-V1**2)/2000)+(m*g*(z2-z1)/1000)-W; # Steady flow energy equation\n",
      "\n",
      "#Result\n",
      "print \"The rate of Heat Transfer From Building (Error in textbook)= \",round(Q,2),\"kW\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The rate of Heat Transfer From Building (Error in textbook)=  175.94 kW\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.14, Page No:206"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#Variable declaration\n",
      "p1=1; # Suction pressure of air in bar\n",
      "p2=5; # Delivery pressure of air in bar\n",
      "T1=310; # Suction Temperature of air in kelvin\n",
      "Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K\n",
      "# (a) Polytropic compression\n",
      "T2=475; # Delivery Temperature of air in kelvin\n",
      "Q=-0.15; # Heat loss to the cooling water in kW\n",
      "Wpoly=-5.3; # Power consumption of compressor in kW\n",
      "\n",
      "#Calculation for (a)\n",
      "m=(-Wpoly+Q)/(Cpo*(T2-T1)); # mass flow rate of air from SSSF energy equation\n",
      "n=1/((1-((math.log (T2/T1))/(math.log (p2/p1))))); # Index of polytropic process\n",
      "\n",
      "#Result for (a)\n",
      "print \"(a).Polytropic compression\",\"\\nmass flow rate of air = \",round(m,3),\"kg/s\",\"\\nIndex of polytropic process = \",round(n,4)\n",
      "\n",
      "#Calculation for (b)\n",
      "# (b) Adiabatic compression\n",
      "k=1.4; # Index of adiabatic process\n",
      "Wad=-m*Cpo*T1*((p2/p1)**((k-1)/k)-1); # power consumption of compressor\n",
      "\n",
      "#Result for (b)\n",
      "print \"\\n(b).Adiabatic compression\",\"\\nPower consumption of compressor = \",round(Wad,2),\"kW\"\n",
      "\n",
      "#Calculation for (c)\n",
      "# (c).Difference between specific work\n",
      "wdiff=(abs(Wad)-abs(Wpoly))/m; # Difference between specific work\n",
      "\n",
      "#Result for (c)\n",
      "print \"\\n(c).Difference between specific work\",\"\\nDifference between specific work = \",round(wdiff,2),\"kJ/kg   (roundoff error)\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a).Polytropic compression \n",
        "mass flow rate of air =  0.031 kg/s \n",
        "Index of polytropic process =  1.3608\n",
        "\n",
        "(b).Adiabatic compression \n",
        "Power consumption of compressor =  -5.65 kW\n",
        "\n",
        "(c).Difference between specific work \n",
        "Difference between specific work =  11.22 kJ/kg   (roundoff error)\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.15, Page No:208"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "W=500; # Power output from steam turbine in MW\n",
      "Q=10; # Heat loss to surroundings in MW\n",
      "p1=12.5; # Pressure of staem at steam turbine inlet in MPa\n",
      "p2=10; # Pressure of staem at steam turbine outlet in kPa\n",
      "V1=50; # Velocity of steam at steam turbine inlet in m/s\n",
      "V2=100; # Velocity of steam at steam turbine outlet in m/s\n",
      "x2=0.85; # Quality of steam at steam turbine outlet\n",
      "h1=3341.8; # Specific enthalpy of staem at inlet from steam table in kJ/kg\n",
      "hf2=191.83; hg2=2584.7;# Specific enthalpies of fluid and steam at outlet from steam table in kJ/kg\n",
      "\n",
      "#Calculation\n",
      "h2=(1-x2)*hf2+x2*hg2;# Specific enthalpy of staem at outlet in kJ/kg\n",
      "m=(W-Q)*10**3/((h1-h2)+(V1**2-V2**2)/2000); # Mass flow rate of steam\n",
      "\n",
      "#Result\n",
      "print \"Mass flow rate of steam = \",round(m,0),\"kg/s\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Mass flow rate of steam =  441.0 kg/s\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.16, Page No:212"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#Variable declaration\n",
      "p1=3; # Pressre of air at state 1 in bar\n",
      "p2=p1; # constant pressure process\n",
      "T1=450; # Temperature of air at state 1 in kelvin\n",
      "T2=1250; # Temperature of air at state 2 in kelvin\n",
      "T3=1000; # Temperature of air at state 3 in kelvin\n",
      "V3=50; # Velocity of air at state 3 in m/s\n",
      "T4=800; # Temperature of air at state 4 in kelvin\n",
      "Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K\n",
      "\n",
      "#Calculation for (a)\n",
      "# (a).Combustion chamber\n",
      "q=Cpo*(T2-T1); # Heat added to air\n",
      "\n",
      "#Result for (a)\n",
      "print \"(a).Combustion chamber\",\"\\nHeat added to air = \",q,\"kJ/kg   (round off error)\"\n",
      "\n",
      "#Calculation for (b)\n",
      "# (b).Turbine \n",
      "k=1.4; # Index of adiabatic process\n",
      "w=Cpo*(T2-T3)-V3**2/2000; # Work done \n",
      "\n",
      "#Result for (b)\n",
      "print \"\\n(b).Turbine\",\"\\nWork done = \",w,\"kJ/kg   (round off error)\"\n",
      "\n",
      "#Calculation for (c)\n",
      "# (c).Nozzle\n",
      "V4=math.sqrt (2*Cpo*10**3*(T3-T4)+V3**2); # Velocity of air leaving the nozzle\n",
      "\n",
      "#Result for (c)\n",
      "print \"\\n(c).Nozzle\",\"\\nVelocity of air leaving the nozzle = \",round(V4,0),\"m/s   (round off error)\"\n",
      "\n",
      "#Calculation for (d)\n",
      "# (d).Pressure drop\n",
      "p3=p2*(T3/T2)**(k/(k-1)); # Pressure of air leaving turbine\n",
      "p4=p3*(T4/T3)**(k/(k-1)); # Pressure of air leaving nozzle\n",
      "\n",
      "#Result for (d)\n",
      "print \"\\n(d).Pressure drop\",\"\\nPressure of air leaving turbine = \",round(p3,3),\"bar\"\n",
      "print \"Pressure of air leaving nozzle = \",round(p4,2),\"bar\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a).Combustion chamber \n",
        "Heat added to air =  802.8 kJ/kg   (round off error)\n",
        "\n",
        "(b).Turbine \n",
        "Work done =  249.625 kJ/kg   (round off error)\n",
        "\n",
        "(c).Nozzle \n",
        "Velocity of air leaving the nozzle =  636.0 m/s   (round off error)\n",
        "\n",
        "(d).Pressure drop \n",
        "Pressure of air leaving turbine =  1.374 bar\n",
        "Pressure of air leaving nozzle =  0.63 bar\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.17, Page No:213"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "V=1000; # Speed of aircraft in kmph\n",
      "p1=0.35; # Ambient pressure in bar\n",
      "T1=258; # Ambient temperature in kelvin\n",
      "V1=V*1000/3600; # unit conversion kmph into m/s\n",
      "p=1.01325; # Atmospheric pressure in bar\n",
      "Cpo=1.0035; # Specific heat at constant pressure in kJ/kg K\n",
      "k=1.4; # Index of compression process\n",
      "\n",
      "#Calculation\n",
      "T2=T1+(V1**2)/(2*Cpo*10**3); # The temperature after leaving inlet diffuser\n",
      "p2=p1*(T2/T1)**(k/(k-1)); # Pressure after leaving inlet diffuser\n",
      "r=p/p2; # Pressre ratio of compressor required for pressurization\n",
      "\n",
      "#Result\n",
      "print \"The temperature after leaving inlet diffuser = \",round(T2,1),\"K\"\n",
      "print \"Pressre ratio of compressor required for pressurization = \",round(r,2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The temperature after leaving inlet diffuser =  296.4 K\n",
        "Pressre ratio of compressor required for pressurization =  1.78\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.18, Page No:214"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "p1=50; # Pressre of steam at diffuser inlet in kPa\n",
      "T1=150; # Temperature of steam at diffuser inlet in degree celcius\n",
      "V1=180; # Velocity of steam at diffuser inlet in m/s\n",
      "A1=1000; # area of diffuser inlet in cm^2\n",
      "V2=90; # Velocity of steam at diffuser outlet in m/s\n",
      "p2=1; # Pressre of steam at diffuser outlet in bar\n",
      "Q=120; # Heat loss to the surroundings in kW\n",
      "v1=3.24; # Specific volume of steam from superheated steam table in m^3/kg at inlet\n",
      "h1=2645.9; #  # Specific enthalpy of steam from superheated steam table in m^3/kg at inlet\n",
      "\n",
      "#Calculation\n",
      "m=V1*A1*10**-4/v1; # Mass flow rate of steam\n",
      "q=Q/m; # Heat transfer per unit mass of steam\n",
      "h2=q+h1+(V1**2-V2**2)/2000; #  Specific enthalpy of steam from SSSF energy equationat outlet\n",
      "v2=1.704; # Specific volume of steam from superheated steam table in m^3/kg at outlet\n",
      "A2=m*v2/V2; # Area of diffuser exit and Answer given in the textbook is wrong\n",
      "\n",
      "#Result\n",
      "print \"Area of diffuser exit = \",round(A2*10**4,0),\"cm^2   (Answer was wrong in textbook because that\"   \n",
      "print \"the value considered in the textbook for V2 is incorrect. So there is a difference in solution.)\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Area of diffuser exit =  1052.0 cm^2   (Answer was wrong in textbook because that\n",
        "the value considered in the textbook for V2 is incorrect. So there is a difference in solution.)\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.19, Page No:217"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from numpy import matrix\n",
      "from numpy import linalg\n",
      "\n",
      "#Variable declaration\n",
      "p=0.15; # Pressure of Freon-12 gas in MPa\n",
      "T1=10; # Temperature of Freon-12 gas in degree celcius\n",
      "T2=35; # Temperature of Freon-12 gas in degree celcius\n",
      "h1=196.762; # Specific enthalpy of Freon-12 from table at 1 in kJ/kg \n",
      "h2=69.49; # Specific enthalpy of Freon-12 from table at 2 in kJ/kg \n",
      "h3=178.54; # Specific enthalpy of Freon-12 from table at 3 in kJ/kg \n",
      "\n",
      "#Calculation\n",
      "# For solving Conservation of mass and SSSF energy equations\n",
      "A=[[1,-1],[h3,-h2]];\n",
      "B=[[1] ,[h1]];\n",
      "M=linalg.solve(A, B) \n",
      "\n",
      "#Result\n",
      "print \"Mass of saturated liquid at 35 oC = \",round(M.item(1),3),\"kg/s\"\n",
      "print \"Mass of saturated vapour at 0.15 MPa = \",round(M.item(0),3),\"kg/s\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Mass of saturated liquid at 35 oC =  0.167 kg/s\n",
        "Mass of saturated vapour at 0.15 MPa =  1.167 kg/s\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.20, Page No:219"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "p1=15;# Inlet pressure of condenser in kPa\n",
      "x=0.92; # Quality of steam\n",
      "ms=25; # mass flow rate of steam in kg/s\n",
      "d_t=12; # temperature rise of water \n",
      "Cpw=4.1868; # Specific heat of water in kJ/kg K\n",
      "hfg=225.94; h4_3=2599.1; # specific enthalpy in kJ/kg\n",
      "\n",
      "#Calculation\n",
      "h1=(1-x)*hfg+x*h4_3; # specific enthalpy \n",
      "h2=225.94; # specific enthalpy in kJ/kg\n",
      "Q=ms*(h1-h2); # Heat transfer rate in condenser\n",
      "mw=Q/(Cpw*d_t); # Mass flow rate of water \n",
      "\n",
      "#Result\n",
      "print \"Heat transfer rate in condenser = \",round(Q,0),\"kW\",\"\\nMass flow rate of water = \",round(mw,1),\"kg/s\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat transfer rate in condenser =  54583.0 kW \n",
        "Mass flow rate of water =  1086.4 kg/s\n"
       ]
      }
     ],
     "prompt_number": 19
    }
   ],
   "metadata": {}
  }
 ]
}