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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 2: FAULT LIMITING REACTORS
// EXAMPLE : 2.4 :
// Page number 481
clear ; clc ; close ; // Clear the work space and console
// Given data
kVA = 20000.0 // Rating of generator(kVA)
f = 50.0 // Frequency(Hz)
V = 11.0*10**3 // Voltage of generator(V)
X_G = 20.0 // Generator short-circuit reactance(%)
x = 60.0 // Reactance falls to 60% normal value
// Calculations
kVA_base = 20000.0 // Base kVA
X = poly(0,"X") // Reactance of each reactors E,F,G & H(%)
X_AE = X+X_G // Reactances of A & E in series(%)
X_BF = X+X_G // Reactances of B & F in series(%)
X_CD = X+X_G // Reactances of C & D in series(%)
X_eq = X_AE/3 // X_eq = X_AE*X_BF*X_CD/(X_BF*X_CD+X_AE*X_CD+X_AE*X_BF). Combined reactances of 3 groups in parallel(%)
X_f = X_eq+X // Reactances of these groups to fault via tie-bar(%)
X_sol = roots(6.66666666666667-(100-x)/100*(X_f)) // Value of reactance of each reactors E,F,G & H(%)
I_fl = kVA_base*1000/(3**0.5*V) // Full load current corresponding to 20000 kVA & 11 kV(A)
X_ohm = X_sol*V/(3**0.5*100*I_fl) // Ohmic value of reactance X(ohm)
// Results
disp("PART III - EXAMPLE : 2.4 : SOLUTION :-")
printf("\nReactance of each reactor = %.4f ohm \n", X_ohm)
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
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