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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 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS
// EXAMPLE : 4.5 :
// Page number 514-515
clear ; clc ; close ; // Clear the work space and console
// Given data
V = 6600.0 // Alternator voltage(V)
kVA = 10000.0 // Alternator rating(kVA)
x_1 = 15.0 // Reactance to positive sequence current(%)
x_2 = 75.0 // Reactance to negative sequence current(%)
x_0 = 30.0 // Reactance to zero sequence current(%)
R_earth = 0.3 // Earth resistance(ohm)
// Calculations
a = exp(%i*120.0*%pi/180) // Operator
E_g = V/3**0.5 // Phase voltage(V)
// Case(a)
I = kVA*1000/(3**0.5*V) // Full load current of each alternator(A)
X = x_1*V/(100*3**0.5*I) // Positive sequence reactance(ohm)
Z_g1 = %i*X // Equivalent positive sequence impedance(ohm)
Z_g2 = Z_g1*x_2/100 // Equivalent negative sequence impedance(ohm)
Z_g0 = Z_g1*x_0/100 // Equivalent zero sequence impedance(ohm)
Z_1 = Z_g1/3 // Positive sequence impedance(ohm)
Z_2 = Z_g2/3 // Negative sequence impedance(ohm)
Z_0 = Z_g0/3 // Zero sequence impedance(ohm)
I_a_a = 3*E_g/(Z_1+Z_2+Z_0) // Fault current(A)
// Case(b)
Z_0_b = Z_g0 // Impedance(ohm)
I_a_b = 3*E_g/(Z_1+Z_2+Z_0_b) // Fault current(A)
// Case(c)
Z_0_c = R_earth*3+Z_g0 // Impedance(ohm)
I_a_c = 3*E_g/(Z_1+Z_2+Z_0_c) // Fault current(A)
// Results
disp("PART III - EXAMPLE : 4.5 : SOLUTION :-")
printf("\nCase(a): Fault current if all the alternator neutrals are solidly earthed, I_a = %.fj A", imag(I_a_a))
printf("\nCase(b): Fault current if only one of the alternator neutrals is solidly earthed & others isolated = %.fj A", imag(I_a_b))
printf("\nCase(c): Fault current if one of alternator neutrals is earthed through resistance & others isolated = %.f A\n", abs(I_a_c))
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here")
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