// 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")