path: root/Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter11.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter11-Heat recovery"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex1-pg308"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.1\n",
      "print('Example 11.1\\n\\n');\n",
      "print('Page No. 308\\n\\n');\n",
      "\n",
      "##given\n",
      "V = 205.;## Flow rate in m^3\n",
      "T1 = 74.;## in degree celcius\n",
      "T2 = 10.;## in degree celcius\n",
      "m = 1000.;## Steam in kg\n",
      "p = 950.;## Density of steam in kg/m^3\n",
      "C = 85.;## Cost in Pound per m^3\n",
      "C_V = 43.3*10**6;## Calorific value in J/kg\n",
      "Cp = 4.18*10**3;## heat capacity of water J/kg-K\n",
      "h = 2.33*10**6;## Heat of the steam in J/kg\n",
      "n = 0.65;## Average bolier efficiency\n",
      "\n",
      "S_cost = ((m*h*C)/(C_V*p*n));## Steam cost in Pound per  1000 kg\n",
      "E_save = V*m*(T1 - T2)*Cp;## Energy saving in J per day\n",
      "S_save = E_save/h;## in kg per day\n",
      "print'%s %.2f %s'%('the steam saving is ',S_save,' kg per day \\n')\n",
      "G_save = (S_cost*S_save)/m;## Pound per day\n",
      "print'%s %.2f %s'%('The gross saving is ',G_save,' Pound per day per year')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.1\n",
        "\n",
        "\n",
        "Page No. 308\n",
        "\n",
        "\n",
        "the steam saving is  23537.17  kg per day \n",
        "\n",
        "The gross saving is  174.34  Pound per day per year\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex2-pg313"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "import numpy\n",
      "#import tabulate\n",
      "\n",
      "## Example 11.2\n",
      "print('Example 11.2\\n\\n');\n",
      "print('Page No. 313\\n\\n');\n",
      "\n",
      "##given\n",
      "p1 = 10.;##heat-sensitive liquor percen\n",
      "p2 = 50.;##heat-sensitive liquor percent\n",
      "m = 0.28;## mass rate in kg/s\n",
      "t = 150.;## time in h per week\n",
      "\n",
      "## This question does not contain any calculation part in it.\n",
      "I = ['8250', '1150', '14850', '16500'];##Installation cost in Pound\n",
      "A = ['69300' ,'36800' ,'23600' ,'24600'];## Annual steam cost in Pound\n",
      "A_S =['0' ,'32500' ,'45700' ,'44700'];## Annual savings in Pound\n",
      "for column in zip(I, A, A_S):\n",
      "\tprint ' '.join(column)\n",
      "#from tabulate import tabulate\n",
      "#print tabulate([['single effect', I[0], A[0],[A_S[0]]], ['double effect', I[1], A[1],[A_S[1]]] ,['double effect+vapour compression',I[2], A[2],[A_S[2]]] ,['Triple effect',I[3], A[3],[A_S[3]]]], headers=['Installation cost', 'Annual steam cost','Annual saving'])\n",
      "\n",
      "\n",
      "\n",
      "#print'%s %.2f %s'%(' The results enable the return on investment to be assessed by one of the standard economic procedures and the final selsction made.')\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.2\n",
        "\n",
        "\n",
        "Page No. 313\n",
        "\n",
        "\n",
        "8250 69300 0\n",
        "1150 36800 32500\n",
        "14850 23600 45700\n",
        "16500 24600 44700\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex3-pg314"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.3\n",
      "print('Example 11.3\\n\\n');\n",
      "print('Page No. 314\\n\\n');\n",
      "\n",
      "##given\n",
      "f = 1.;## feed of sodium hydroxide in kg\n",
      "v = 0.5;## produed vapour in kg\n",
      "A = 30.;## in m^2\n",
      "T1 = 95.;## Temperature of boiling solution in deg C\n",
      "U = 3.*10**3;## heat transfer coefficent in W/m^2-K\n",
      "m = 1.;## feed rate in kg/s\n",
      "Tf = 70.;## Feed temperature in deg C\n",
      "h_f = 260.*10**3.;## Enthalpy of feed in J/kg\n",
      "h_b = 355.*10**3.;## Enthalpy of boiling solution in J/kg\n",
      "h_v = 2.67*10**6.;## Enthalpy of vapour in J/kg\n",
      "P1 = 0.6;## Pressure in vapour space in bar\n",
      "\n",
      "Q = (v*h_b) + (v*h_v) -(f*h_f);## in W\n",
      "print'%s %.2f %s'%('The total energy requirement is ',Q,' W \\n')\n",
      "\n",
      "## As Q = A*U*dT\n",
      "dT = Q/(U*A);## in degree celcius\n",
      "T2 = dT + T1;## in degree celcius\n",
      "##The temperature of the heating steam T2 corresponds to a pressure of 1.4 bar. Dry saturated steam at 1.4 bar has a total enthalpy of 2.69*10^6 J/kg\n",
      "##Assuming an isentropic compression of the vapour from 0.6 bar to 1.4 bar, the outlet enthalpy is 2.84*10^6 J/kg\n",
      "\n",
      "## from steam table\n",
      "P2 = 1.4## pressure in bar\n",
      "h_s = 2.69*10**6;## enthalpy of dry saturated steam in J/kg\n",
      "h_v2 = 2.84*10**6 ;## the outlet enthalpy of vapour in J/kg\n",
      "\n",
      "W = v*(h_v2 - h_s);## Work in W\n",
      "T_E = W + 60.*10**3;## in W\n",
      "print'%s %.2f %s'%('The total energy consumption is ',T_E,' W')\n",
      "\n",
      "\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.3\n",
        "\n",
        "\n",
        "Page No. 314\n",
        "\n",
        "\n",
        "The total energy requirement is  1252500.00  W \n",
        "\n",
        "The total energy consumption is  135000.00  W\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex4-pg316"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.4\n",
      "print('Example 11.4\\n\\n');\n",
      "print('Page No. 316\\n\\n');\n",
      "\n",
      "##given\n",
      "Cm_S = 10000.;## Company saving in Pound per annum\n",
      "S = Cm_S/12.;## Saving in Pound per months\n",
      "Ca_C = 10500.;## Capital cost in Pound\n",
      "Ins_C = 7500.;## Installation cost in Pound\n",
      "T_C = Ca_C + Ins_C;## Total cost in Pound\n",
      "T = T_C/S;## pay-back time in months\n",
      "print'%s %.2f %s'%('The pay-back period was ',T,' months\\n')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.4\n",
        "\n",
        "\n",
        "Page No. 316\n",
        "\n",
        "\n",
        "The pay-back period was  21.60  months\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex5-pg318"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.5\n",
      "print('Example 11.5\\n\\n');\n",
      "print('Page No. 318\\n\\n');\n",
      "\n",
      "##From the heat balance:- \n",
      "##Heat recovered in the boiler = heat gained by the air = heat lost by the flue gases\n",
      "##=> Q = m_a*Cp_a*dT_a = m_f*Cp_f*dT_f\n",
      "## As mass flow rate of air/flue gas is not given in the book\n",
      "##Assuming m_a = m_f = 2.273 kg/s & Cp_a = 1*10^3 J/kg-K\n",
      "\n",
      "m_a = 2.273;## in kg/s\n",
      "m_f = m_a;## in kg/s\n",
      "Cp_a = 1.*10**3;## Specific heat capacity of air in J/kg-K\n",
      "T1_a = 20.;## Entrance temperature of air in degree celcius\n",
      "T2_a = 130.;## Exit temperature of air in degree celcius\n",
      "dT_a = T2_a - T1_a;##in K\n",
      "T1_f = 260.;## Entrance temperature of flue gases in degree celcius\n",
      "T2_f = 155.;## Entrance temperature of flue gases in degree celcius\n",
      "dT_f = T1_f - T2_f;##in K\n",
      "\n",
      "##From heat balance:- Q = m_a*Cp_a*dT_a = m_f*Cp_f*dT_f\n",
      "Cp_f = ((m_a*Cp_a*dT_a)/(m_f*dT_f));## in J/kg-K\n",
      "Q = m_f*Cp_f*dT_f;## in W\n",
      "print'%s %.2e %s'%('The total heat recovered at full load if ',Q,' W')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.5\n",
        "\n",
        "\n",
        "Page No. 318\n",
        "\n",
        "\n",
        "The total heat recovered at full load if  2.50e+05  W\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex6-pg320"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.6\n",
      "print('Example 11.6\\n\\n');\n",
      "print('Page No. 320\\n\\n');\n",
      "\n",
      "C = 10000.;## Installation cost of the pump in Pound\n",
      "S = 3500.;## Saving in Pound per annum \n",
      "T = C/S;## in year\n",
      "print'%s %.2f %s'%('The pay back time is ',T,' year\\n\\n')\n",
      "\n",
      "## This question further does not contain any calculation part in it.\n",
      "print('In a heat-pump system the work input to drive the compressor,W, produces a heat absorption capacity,Q2,\\nand to balance the energy flow, a quantity of heat, Q1, must be dissipated.\\nThus the energy equation is\\n -> Q1 = W + Q2\\nand the coeffient of performance is \\nC.O.P. = Q1/W = Q1/(Q1 - Q2)\\n   Consequently the C.O.P. is always greater than unity.\\nThe maximum theoretical value of the C.O.P. is that predicted by the Carnot in chapter 2,namely :\\n -> (C.O.P.)max = T1/(T1 - T2)')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.6\n",
        "\n",
        "\n",
        "Page No. 320\n",
        "\n",
        "\n",
        "The pay back time is  2.86  year\n",
        "\n",
        "\n",
        "In a heat-pump system the work input to drive the compressor,W, produces a heat absorption capacity,Q2,\n",
        "and to balance the energy flow, a quantity of heat, Q1, must be dissipated.\n",
        "Thus the energy equation is\n",
        " -> Q1 = W + Q2\n",
        "and the coeffient of performance is \n",
        "C.O.P. = Q1/W = Q1/(Q1 - Q2)\n",
        "   Consequently the C.O.P. is always greater than unity.\n",
        "The maximum theoretical value of the C.O.P. is that predicted by the Carnot in chapter 2,namely :\n",
        " -> (C.O.P.)max = T1/(T1 - T2)\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex7-pg320"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.7\n",
      "print('Example 11.7\\n\\n');\n",
      "print('Page No. 320\\n\\n');\n",
      "\n",
      "##given\n",
      "T1 = 40.;## in degree \n",
      "T2 = 0.;## in degree celcius\n",
      "##As from carnot cycle, C.O.P = (T1/(T1 - T2)), where temperature are in degree celcius\n",
      "C_O_P1 = ((T1+273.)/((T1+273.) - (T2+273.)));\n",
      "print'%s %.2f %s'%('C.O.P. is ',C_O_P1,' \\n')\n",
      "\n",
      "## A secondary fluid as hot water at 60 deg C is used\n",
      "T3 = 60;##  Temperature of hot water in degree celcius\n",
      "C_O_P2 = ((T3+273.)/((T3+273.) - (T2+273.)));\n",
      "print'%s %.2f %s'%('C.O.P. when secondary fluid is used is ',C_O_P2,' \\n')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.7\n",
        "\n",
        "\n",
        "Page No. 320\n",
        "\n",
        "\n",
        "C.O.P. is  7.83  \n",
        "\n",
        "C.O.P. when secondary fluid is used is  5.55  \n",
        "\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex8-pg323"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.8\n",
      "print('Example 11.8\\n\\n');\n",
      "print('Page No. 323\\n\\n');\n",
      "\n",
      "## This question does not contain any calculation part in it.\n",
      "print('No calculation is required as not in shown in book')\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.8\n",
        "\n",
        "\n",
        "Page No. 323\n",
        "\n",
        "\n",
        "No calculation is required as not in shown in book\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex9-pg324"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.9\n",
      "print('Example 11.9\\n\\n');\n",
      "print('Page No. 324\\n\\n');\n",
      "\n",
      "##given\n",
      "T1 = 273.;## Measured temperature In degree celcius\n",
      "P = 1.;## Measured pressure in bar\n",
      "T2 = 290.;## initial temperature In degree celcius\n",
      "T3 = 1000.;## Final temperature In degree celcius\n",
      "T4 = 1150.;## Entering tempearture In degree celcius\n",
      "v1 = 7.;## in m^3/s\n",
      "v2 = 8.;## in m^s\n",
      "M = 22.7;## in kmol/m^3\n",
      "d = 0.1;## Diameter in m\n",
      "A = 0.01;## Surface area per regenerator channel in m^2\n",
      "u = 1.;## maximum velocity in m/s\n",
      "Cp_1 = 34.*10**3;## Heat capacity at T4 temperature in J/kmol-K\n",
      "Cp_2 = 32.*10**3;## Heat capacity at outlet temperature in J/kmol-K\n",
      "Cp_m = 30.*10**3;## Heat capacity at mean temperature in J/kmol-K\n",
      "\n",
      "m_c = v1/M;##  Molal air flow rate in kmol/s\n",
      "H_c1 = Cp_m*(T3 - T1);## Enthalpy of air at 1000K in J/mol\n",
      "H_c2 = Cp_m*(T2 - T1);## Enthalpy of air at 290 in J/mol\n",
      "Q = (m_c*(H_c1 - H_c2))/10**6;## in 10^6 W\n",
      "print'%s %.2f %s'%('The heat transfer, Q is ',Q,' *10^6 W \\n')\n",
      "\n",
      "m_F = v2/M;##  Molal flow rate of flue gas in kmol/s\n",
      "dH = (Q/m_F)*10**6;## enthaply chnage of the flue gas in J/kmol\n",
      "H_F1 = Cp_1*(T4 - T1);## Enthalpy of the flue gas at 1150 K in J/kmol\n",
      "H_F2 =H_F1 - dH;## Enthalpy at the exit temperature in J/kmol\n",
      "T_F2 = (H_F2/Cp_2) + T1;## in K\n",
      "print'%s %.2f %s'%('The exit tempearture of the flue gas is ',T_F2,' K \\n')\n",
      "S_R = v2/u;##cross sectional area of the regenerator in m^2\n",
      "N = S_R/A;\n",
      "print'%s %.2f %s'%('The number of channels required is ',N,'\\n')\n",
      "print('Consequently for this regenerator a square layout could be achieved with 40 channels arranged horizontally and 20 channels vertically.')\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.9\n",
        "\n",
        "\n",
        "Page No. 324\n",
        "\n",
        "\n",
        "The heat transfer, Q is  6.57  *10^6 W \n",
        "\n",
        "The exit tempearture of the flue gas is  622.39  K \n",
        "\n",
        "The number of channels required is  800.00 \n",
        "\n",
        "Consequently for this regenerator a square layout could be achieved with 40 channels arranged horizontally and 20 channels vertically.\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10-pg324"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "## Example 11.10\n",
      "print('Example 11.10\\n\\n');\n",
      "print('Page No. 324\\n\\n');\n",
      "\n",
      "##given\n",
      "Pr = 100.;## Production in tonnes per day\n",
      "p = 10.2;## percentage of sulphur dioxide\n",
      "T1 = 900.;##Burner temperature in degree celcius\n",
      "T2 = 425.;##Required temperature in degree celcius\n",
      "P = 10.;## Dry saturated steam pressure in bar\n",
      "T = 120.;## Dry saturated steam temperature in degree celcius\n",
      "##At the given Temperature =T and Pressure P, the required heat Qr to geberate steam from feed water is calculated from the steam table.\n",
      "Qr = 2.27*10**6;## in J/kg\n",
      "\n",
      "Sp_1 = 1.14*10**3;## Specific heat of the inlet gas in J/kmol-K\n",
      "Sp_2 = 1.03*10**3;## Specific heat of the outlet gas in J/kmol-K\n",
      "pr_rate = 1.2;## production rate in kmol/s\n",
      "\n",
      "##In the calculation part, the book has taken percentage of sulphur dioxide p = 10.6 in the place of p = 10.2, so there exists a deviation in answer\n",
      "Q_in = ((Pr*pr_rate)/p) * Sp_1 * T1;## Heat content of the inlet gas in J/s\n",
      "Q_out = ((Pr*pr_rate)/p) * Sp_2 * T2;## Heat content of the outlet gas in J/s\n",
      "Qa = Q_in - Q_out;## Heat available for steam\n",
      "S = Qa/Qr;## in kg/s\n",
      "print'%s %.2f %s'%('The steam production is ',S,' kg/s')##Deviation in answer is due to some wrong value substition as discussed above\n",
      "\n",
      "\n",
      "\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Example 11.10\n",
        "\n",
        "\n",
        "Page No. 324\n",
        "\n",
        "\n",
        "The steam production is  3.05  kg/s\n"
       ]
      }
     ],
     "prompt_number": 10
    }
   ],
   "metadata": {}
  }
 ]
}