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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)