// 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 5: ELECTRIC TRACTION-SPEED TIME CURVES AND MECHANICS OF TRAIN MOVEMENT // EXAMPLE : 5.9 : // Page number 783 clear ; clc ; close ; // Clear the work space and console // Given data V_A = 48.0 // Speed(kmph) t_1 = 24.0 // Time taken to accelerate from rest to speed(sec) t_2 = 69.0 // Coasting time(sec) r = 58.0 // Constant resistance(N/tonne) beta = 3.3 // Retardation(km phps) t_3 = 11.0 // Retardation time(sec) t_iii_a = 20.0 // Station stop time(sec) t_iii_b = 15.0 // Station stop time(sec) I = 10.0 // Rotational inertia(%) // Calculations alpha = V_A/t_1 // Acceleration(km phps) V_B = beta*t_3 // Speed at B(km phps) beta_c = (V_A-V_B)/t_2 // Retardation during coasting(km phps) distance_acc = 1.0/2*t_1*V_A/3600 // Distance covered during acceleration(km) distance_coasting = (V_A**2-V_B**2)/(2*beta_c*3600) // Distance covered during coasting(km) distance_braking = t_3*V_B/(3600*2) // Distance covered during braking(km) distance_total = distance_acc+distance_coasting+distance_braking // Total distance(km) speed_iii_a = distance_total*3600/(t_1+t_2+t_3+t_iii_a) // Scheduled speed with a stop of 20 sec(kmph) speed_iii_b = distance_total*3600/(t_1+t_2+t_3+t_iii_b) // Scheduled speed with a stop of 15 sec(kmph) // Results disp("PART IV - EXAMPLE : 5.9 : SOLUTION :-") printf("\nCase(i) : Acceleration, α = %.f km phps", alpha) printf("\nCase(ii) : Coasting retardation, β_c = %.2f km phps", beta_c) printf("\nCase(iii): Scheduled speed with a stop of 20 seconds = %.2f kmph", speed_iii_a) printf("\n Scheduled speed with a stop of 15 seconds = %.2f kmph\n", speed_iii_b) printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")