From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 3472/CH45/EX45.2/Example45_2.sce | 40 ++++++++++++++++++++++++++++++++++++++++ 1 file changed, 40 insertions(+) create mode 100644 3472/CH45/EX45.2/Example45_2.sce (limited to '3472/CH45/EX45.2/Example45_2.sce') diff --git a/3472/CH45/EX45.2/Example45_2.sce b/3472/CH45/EX45.2/Example45_2.sce new file mode 100644 index 000000000..a6cc5dffe --- /dev/null +++ b/3472/CH45/EX45.2/Example45_2.sce @@ -0,0 +1,40 @@ +// 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) -- cgit