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-rw-r--r--3472/CH45/EX45.1/Example45_1.sce26
-rw-r--r--3472/CH45/EX45.2/Example45_2.sce40
-rw-r--r--3472/CH45/EX45.3/Example45_3.sce42
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diff --git a/3472/CH45/EX45.1/Example45_1.sce b/3472/CH45/EX45.1/Example45_1.sce
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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 : 7.1 :
+// Page number 798
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+no = 2.0 // Number of motors
+V_m = 48.0 // Uniform speed(kmph)
+t = 30.0 // Time(sec)
+F_t_m = 13350.0 // Average tractive effort per motor(N)
+
+// Calculations
+F_t = no*F_t_m // Average tractive effort(N)
+energy = t*F_t*V_m/(2*3600**2) // Useful energy for acceleration(kWh)
+energy_loss = energy/no // Approximate loss of energy in starting rheostats(kWh)
+
+// Results
+disp("PART IV - EXAMPLE : 7.1 : SOLUTION :-")
+printf("\nApproximate loss of energy in starting rheostats = %.3f kWh", energy_loss)
diff --git a/3472/CH45/EX45.2/Example45_2.sce b/3472/CH45/EX45.2/Example45_2.sce
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+++ b/3472/CH45/EX45.2/Example45_2.sce
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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)
diff --git a/3472/CH45/EX45.3/Example45_3.sce b/3472/CH45/EX45.3/Example45_3.sce
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index 000000000..774891b67
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+++ b/3472/CH45/EX45.3/Example45_3.sce
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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")