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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 7: CONTROL OF MOTORS
// EXAMPLE : 7.2 :
// Page number 798
clear ; clc ; close ; // Clear the work space and console
// Given data
W = 175.0 // Weight of multiple unit train(tonnes)
no = 6.0 // Number of motors
F_t = 69000.0 // Total tractive effort(N)
V = 600.0 // Line voltage(V)
I = 200.0 // Average current(A)
V_m = 38.6 // Speed(kmph)
R = 0.15 // Resistance of each motor(ohm)
// Calculations
alpha = F_t/(277.8*W) // Acceleration(km phps)
T = V_m/alpha // Time for acceleration(sec)
t_s = (V-2*I*R)*T/(2*(V-I*R)) // Duration of starting period(sec)
t_p = T-t_s // (sec)
energy_total_series = no/2*V*I*t_s // Total energy supplied in series position(watt-sec)
energy_total_parallel = no*V*I*t_p // Total energy supplied in parallel position(watt-sec)
total_energy = (energy_total_series+energy_total_parallel)/(1000*3600) // Energy supplied during starting period(kWh)
energy_waste_series = (no/2)/2*(V-2*I*R)*I*t_s // Energy wasted in starting resistance in series position(watt-sec)
energy_waste_parallel = no*(V/2)/2*I*t_p // Energy wasted in starting resistance in parallel position(watt-sec)
total_energy_waste = (energy_waste_series+energy_waste_parallel)/(1000*3600) // Total energy wasted in starting resistance(kWh)
energy_lost = (no*I**2*R*T)/(1000*3600) // Energy lost in motor resistance(kWh)
useful_energy = T*F_t*V_m/(2*3600**2) // Useful energy supplied to train(kWh)
// Results
disp("PART IV - EXAMPLE : 7.2 : SOLUTION :-")
printf("\nEnergy supplied during the starting period = %.2f kWh", total_energy)
printf("\nEnergy lost in the starting resistance = %.1f kWh", total_energy_waste)
printf("\nUseful energy supplied to the train = %.1f kWh", useful_energy)
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