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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 8: BRAKING
// EXAMPLE : 8.3 :
// Page number 810
clear ; clc ; close ; // Clear the work space and console
// Given data
W = 400.0 // Weight of train(tonne)
G = 100.0/70 // Gradient(%)
t = 120.0 // Time(sec)
V_1 = 80.0 // Speed(km/hr)
V_2 = 50.0 // Speed(km/hr)
r_kg = 5.0 // Tractive resistance(kg/tonne)
I = 7.5 // Rotational inertia(%)
n = 0.75 // Overall efficiency
// Calculations
W_e = W*(100+I)/100 // Accelerating weight of train(tonne)
r = r_kg*9.81 // Tractive resistance(N-m/tonne)
energy_recuperation = 0.01072*W_e*(V_1**2-V_2**2)/1000 // Energy available for recuperation(kWh)
F_t = W*(r-98.1*G) // Tractive effort during retardation(N)
distance = (V_1+V_2)*1000*t/(2*3600) // Distance travelled by train during retardation period(m)
energy_train = abs(F_t)*distance/(3600*1000) // Energy available during train movement(kWh)
net_energy = n*(energy_recuperation+energy_train) // Net energy returned to supply system(kWh)
// Results
disp("PART IV - EXAMPLE : 8.3 : SOLUTION :-")
printf("\nEnergy returned to lines = %.2f kWh\n", net_energy)
printf("\nNOTE: ERROR: Calculation mistakes & more approximation in textbook solution")
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