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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"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": {}
}
]
}
|
