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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "CHAPTER 2: THERMAL STATIONS"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.1, Page number 25-26"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "M = 15000.0+10.0        #Water evaporated(kg)\n",
      "C = 5000.0+5.0          #Coal consumption(kg)\n",
      "time = 8.0              #Generation shift time(hours)\n",
      "\n",
      "#Calculation\n",
      "#Case(a)\n",
      "M1 = M-15000.0\n",
      "C1 = C-5000.0\n",
      "M_C = M1/C1\n",
      "#Case(b)\n",
      "kWh = 0                                           #Station output at no load\n",
      "consumption_noload = 5000+5*kWh                   #Coal consumption at no load(kg)\n",
      "consumption_noload_hr = consumption_noload/time   #Coal consumption per hour(kg)\n",
      "\n",
      "#Result\n",
      "print('Case(a): Limiting value of water evaporation , M/C = %.1f kg' %M_C)\n",
      "print('Case(b): Coal per hour for running station at no load = %.f kg' %consumption_noload_hr)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Case(a): Limiting value of water evaporation , M/C = 2.0 kg\n",
        "Case(b): Coal per hour for running station at no load = 625 kg\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2, Page number 26"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "amount = 25.0*10**5          #Amount spent in 1 year(Rs)\n",
      "value_heat = 5000.0          #Heating value(kcal/kg)\n",
      "cost = 500.0                 #Cost of coal per ton(Rs)\n",
      "n_ther = 0.35                #Thermal efficiency\n",
      "n_elec = 0.9                 #Electrical efficiency\n",
      "\n",
      "#Calculation\n",
      "n = n_ther*n_elec                         #Overall efficiency\n",
      "consumption = amount+cost*value_heat      #Coal consumption i 1 year(kg)\n",
      "combustion = consumption*value_heat       #Heat of combustion(kcal)\n",
      "output = n*combustion                     #Heat output(kcal)\n",
      "kWh = output/860.0                        #Annual heat generated(kWh). 1 kWh = 860 kcal\n",
      "time = 365*24.0                           #Total time in a year(hour)\n",
      "load_average = kWh/time                   #Average load on the power plant(kW)\n",
      "\n",
      "#Result\n",
      "print('Average load on power plant = %.2f kW' %load_average)\n",
      "print('\\nNOTE: ERROR: Calculation mistake in the final answer in textbook')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Average load on power plant = 1045.32 kW\n",
        "\n",
        "NOTE: ERROR: Calculation mistake in the final answer in textbook\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.3, Page number 26"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "consumption = 0.5      #Coal consumption per kWh output(kg)\n",
      "cal_value = 5000.0     #Calorific value(kcal/kg)\n",
      "n_boiler = 0.8         #Boiler efficiency\n",
      "n_elec = 0.9           #Electrical efficiency\n",
      "\n",
      "#Calculation\n",
      "input_heat = consumption*cal_value                 #Heat input(kcal)\n",
      "input_elec = input_heat/860.0                      #Equivalent electrical energy(kWh). 1 kWh = 860 kcal\n",
      "loss_boiler = input_elec*(1-n_boiler)              #Boiler loss(kWh)\n",
      "input_steam = input_elec-loss_boiler               #Heat input to steam(kWh)\n",
      "input_alter = 1/n_elec                             #Alternator input(kWh)\n",
      "loss_alter = input_alter*(1-n_elec)                #Alternate loss(kWh)\n",
      "loss_turbine = input_steam-input_alter             #Loss in turbine(kWh)\n",
      "loss_total = loss_boiler+loss_alter+loss_turbine   #Total loss(kWh)\n",
      "output = 1.0                                       #Output(kWh)\n",
      "Input = output+loss_total                          #Input(kWh)\n",
      "\n",
      "#Result\n",
      "print('Heat Balance Sheet')\n",
      "print('LOSSES:  Boiler loss      = %.3f kWh' %loss_boiler)\n",
      "print('         Alternator loss  = %.2f kWh' %loss_alter)\n",
      "print('         Turbine loss     = %.3f kWh' %loss_turbine)\n",
      "print('         Total loss       = %.2f kWh' %loss_total)\n",
      "print('OUTPUT:  %.1f kWh' %output)\n",
      "print('INPUT:   %.2f kWh' %Input)\n",
      "print('\\nNOTE: Changes in the obtained answer from that of textbook is due to precision')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Heat Balance Sheet\n",
        "LOSSES:  Boiler loss      = 0.581 kWh\n",
        "         Alternator loss  = 0.11 kWh\n",
        "         Turbine loss     = 1.214 kWh\n",
        "         Total loss       = 1.91 kWh\n",
        "OUTPUT:  1.0 kWh\n",
        "INPUT:   2.91 kWh\n",
        "\n",
        "NOTE: Changes in the obtained answer from that of textbook is due to precision\n"
       ]
      }
     ],
     "prompt_number": 1
    }
   ],
   "metadata": {}
  }
 ]
}