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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 IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.8 :
+// Page number 736-737
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+weight = 3000.0 // Weight of steel(kg)
+I = 5000.0 // Current(A)
+V_arc = 60.0 // Arc voltage(V)
+R_t = 0.003 // Resistance of transformer(ohm)
+X_t = 0.005 // Reactance of transformer(ohm)
+heat_sp = 0.12 // Specific heat of steel
+heat_latent = 8.89 // Latent heat of steel(kilo-cal/kg)
+t_2 = 1370.0 // Melting point of steel(°C)
+t_1 = 18.0 // Initial temperature of steel(°C)
+n = 0.6 // Overall efficiency
+
+// Calculations
+R_arc_phase = V_arc/I // Arc resistance per phase(ohm)
+IR_t = I*R_t // Voltage drop across resistance(V)
+IX_t = I*X_t // Voltage drop across reactance(V)
+V = ((V_arc+IR_t)**2+IX_t**2)**0.5 // Voltage(V)
+PF = (V_arc+IR_t)/V // Power factor
+heat_kg = (t_2-t_1)*heat_sp+heat_latent // Amount of heat required per kg of steel(kcal)
+heat_total = weight*heat_kg // Heat for 3 tonnes(kcal)
+heat_actual_kcal = heat_total/n // Actual heat required(kcal)
+heat_actual = heat_actual_kcal*1.162*10**-3 // Actual heat required(kWh)
+P_input = 3*V*I*PF*10**-3 // Power input(kW)
+time = heat_actual/P_input*60 // Time required(min)
+n_elect = 3*V_arc*I/(P_input*1000)*100 // Electrical efficiency(%)
+
+// Results
+disp("PART IV - EXAMPLE : 2.8 : SOLUTION :-")
+printf("\nTime taken to melt 3 metric tonnes of steel = %.f minutes", time)
+printf("\nPower factor of the furnace = %.2f ", PF)
+printf("\nElectrical efficiency of the furnace = %.f percent\n", n_elect)
+printf("\nNOTE: ERROR: Calculation and substitution mistake in the textbook solution")