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authorprashantsinalkar2017-10-10 12:27:19 +0530
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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 : 7.3 :
+// Page number 799
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+W = 132.0 // Weight of electric train(tonnes)
+no = 4.0 // Number of motors
+V = 600.0 // Voltage of motor(V)
+I = 400.0 // Current per motor(A)
+F_t_m = 19270.0 // Tractive effort per motor at 400A & 600V(N)
+V_m = 39.0 // Train speed(kmph)
+G = 1.0 // Gradient
+r = 44.5 // Resistance to traction(N/tonne)
+inertia = 10.0 // Rotational inertia(%)
+R = 0.1 // Resistance of each motor(ohm)
+
+// Calculations
+W_e = W*(100+inertia)/100 // Accelerating weight of train(tonne)
+F_t = F_t_m*no // Total tractive effort at 400A & 600V(N)
+alpha = (F_t-W*r-98.1*W*G)/(277.8*W_e) // 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)
+V_transition = alpha*t_s // Speed at transition(km phps)
+t_p = T-t_s // (sec)
+loss_series = (no/2*((V-2*I*R)/2)*I*t_s)/(1000*3600) // Energy lost during series period(kWh)
+loss_parallel = (no*(V/2)/2*I*t_p)/(1000*3600) // Energy lost during parallel period(kWh)
+
+// Results
+disp("PART IV - EXAMPLE : 7.3 : SOLUTION :-")
+printf("\nCase(i) : Duration of starting period, t_s = %.1f sec", t_s)
+printf("\nCase(ii) : Speed of train at transition, αt = %.1f sec", V_transition)
+printf("\nCase(iii): Case(a): Rheostatic losses during series starting = %.2f kWh", loss_series)
+printf("\n Case(b): Rheostatic losses during parallel starting = %.2f kWh\n", loss_parallel)
+printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")