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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 III : SWITCHGEAR AND PROTECTION
// CHAPTER 1: SYMMETRICAL SHORT CIRCUIT CAPACITY CALCULATIONS
// EXAMPLE : 1.2 :
// Page number 467-468
clear ; clc ; close ; // Clear the work space and console
// Given data
kV = 33.0 // Transmission line operating voltage(kV)
R = 5.0 // Transmission line resistance(ohm)
X = 20.0 // Transmission line reactance(ohm)
kVA_tr = 5000.0 // Rating of step-up transformer(kVA)
X_tr = 6.0 // Reactance of transformer(%)
kVA_A = 10000.0 // Rating of alternator A(kVA)
X_A = 10.0 // Reactance of alternator A(%)
kVA_B = 5000.0 // Rating of alternator B(kVA)
X_B = 7.5 // Reactance of alternator B(%)
// Calculations
kVA_base = kVA_A // Base rating(kVA)
X_gen_A = X_A*kVA_base/kVA_A // Reactance of generator A(%)
X_gen_B = X_B*kVA_base/kVA_B // Reactance of generator B(%)
X_trans = X_tr*kVA_base/kVA_tr // Reactance of transformer(%)
X_per = kVA_base*X/(10*kV**2) // X(%)
R_per = kVA_base*R/(10*kV**2) // R(%)
Z_F1 = (X_gen_A*X_gen_B/(X_gen_A+X_gen_B))+X_trans // Impedance upto fault(%)
kVA_F1 = kVA_base*(100/Z_F1) // Short-circuit kVA fed into the fault(kVA)
R_per_F2 = R_per // R(%)
X_per_F2 = X_per+Z_F1 // X(%)
Z_F2 = (R_per_F2**2+X_per_F2**2)**0.5 // Total impedance upto F2(%)
kVA_F2 = kVA_base*(100/Z_F2) // Short-circuit kVA fed into the fault at F2(kVA)
// Results
disp("PART III - EXAMPLE : 1.2 : SOLUTION :-")
printf("\nCase(a): kVA at a short-circuit fault between phases at the HV terminal of transformers = %.f kVA", kVA_F1)
printf("\nCase(b): kVA at a short-circuit fault between phases at load end of transmission line = %.f kVA \n", kVA_F2)
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here & approximation in textbook")
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