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// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART I : GENERATION
// CHAPTER 2: THERMAL STATIONS
// EXAMPLE : 2.3 :
// Page number 26
clear ; clc ; close ; // Clear the work space and console
//Given data
consumption = 0.5 // Coal consumption per kWh output(kg)
cal_value = 5000.0 // Calorific value(kcal/kg)
n_boiler = 0.8 // Boiler efficiency
n_elec = 0.9 // Electrical efficiency
//Calculations
input_heat = consumption*cal_value // Heat input(kcal)
input_elec = input_heat/860.0 // Equivalent electrical energy(kWh). 1 kWh = 860 kcal
loss_boiler = input_elec*(1-n_boiler) // Boiler loss(kWh)
input_steam = input_elec-loss_boiler // Heat input to steam(kWh)
input_alter = 1/n_elec // Alternator input(kWh)
loss_alter = input_alter*(1-n_elec) // Alternate loss(kWh)
loss_turbine = input_steam-input_alter // Loss in turbine(kWh)
loss_total = loss_boiler+loss_alter+loss_turbine // Total loss(kWh)
output = 1.0 // Output(kWh)
Input = output+loss_total // Input(kWh)
//Results
disp("PART I - EXAMPLE : 2.3 : SOLUTION :-")
printf("\nHeat Balance Sheet")
printf("\nLOSSES: Boiler loss = %.3f kWh", loss_boiler)
printf("\n Alternator loss = %.2f kWh", loss_alter)
printf("\n Turbine loss = %.3f kWh", loss_turbine)
printf("\n Total loss = %.2f kWh", loss_total)
printf("\nOUTPUT: %.1f kWh", output)
printf("\nINPUT: %.2f kWh\n", Input)
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