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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 II : TRANSMISSION AND DISTRIBUTION
// CHAPTER 18: POWER DISTRIBUTION SYSTEMS
// EXAMPLE : 18.8 :
// Page number 442-443
clear ; clc ; close ; // Clear the work space and console
// Given data
V = 440.0 // Voltage between outer(V)
I_pos = 210.0 // Ligting load current on positive side(A)
I_neg = 337.0 // Ligting load current on negative side(A)
I_power = 400.0 // Power load current(A)
P_loss = 1.5 // Loss in each balancer machine(kW)
// Calculations
P = I_power*V/1000.0 // Power(kW)
load_pos = I_pos*V*0.5/1000.0 // Load on positive side(kW)
load_neg = I_neg*V*0.5/1000.0 // Load on negative side(kW)
loss_total = 2*P_loss // Total loss on rotary balancer set(kW)
load_main = P+load_pos+load_neg+loss_total // Load on main machine(kW)
I = load_main*1000/V // Current(A)
I_M = I-610.0 // Current through balancer machine(A)
I_G = 127.0-I_M // Current through generator(A)
output_G = I_G*V*0.5/1000.0 // Output of generator(kW)
input_M = I_M*V*0.5/1000.0 // Input to balancer machine(kW)
// Results
disp("PART II - EXAMPLE : 18.8 : SOLUTION :-")
printf("\nLoad on the main machine = %.2f kW", load_main)
printf("\nOutput of generator = %.2f kW", output_G)
printf("\nInput to balancer machine = %.2f kW", input_M)
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