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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 : 4.11 :
// Page number 520-521
clear ; clc ; close ; // Clear the work space and console
// Given data
n = 6.0 // Number of alternator
kV_A = 6.6 // Alternator rating(kV)
X_1 = 0.9 // Positive sequence reactance(ohm)
X_2 = 0.72 // Negative sequence reactance(ohm)
X_0 = 0.3 // Zero sequence reactance(ohm)
Z_n = 0.2 // Resistance of grounding resistor(ohm)
// Calculations
E_a = kV_A*1000/3**0.5 // Phase voltage(V)
// Case(a)
Z_1_a = %i*X_1/n // Positive sequence impedance when alternators are in parallel(ohm)
Z_2_a = %i*X_2/n // Negative sequence impedance when alternators are in parallel(ohm)
Z_0_a = %i*X_0/n // Zero sequence impedance when alternators are in parallel(ohm)
I_a_a = 3*E_a/(Z_1_a+Z_2_a+Z_0_a) // Fault current assuming 'a' phase to be fault(A)
// Case(b)
Z_0_b = 3*Z_n+%i*X_0 // Zero sequence impedance(ohm)
I_a_b = 3*E_a/(Z_1_a+Z_2_a+Z_0_b) // Fault current(A)
// Case(c)
Z_0_c = %i*X_0 // Zero sequence impedance(ohm)
I_a_c = 3*E_a/(Z_1_a+Z_2_a+Z_0_c) // Fault current(A)
// Results
disp("PART III - EXAMPLE : 4.11 : SOLUTION :-")
printf("\nCase(a): Fault current if all alternator neutrals are solidly grounded, I_a = %.f A", imag(I_a_a))
printf("\nCase(b): Fault current if one alternator neutral is grounded & others isolated, I_a = %.1f∠%.1f° A", abs(I_a_b),phasemag(I_a_b))
printf("\nCase(c): Fault current if one alternator neutral is solidly grounded & others isolated, I_a = %.2fj A\n", imag(I_a_c))
printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
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