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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 : 4.1 :
// Page number 510-512
clear ; clc ; close ; // Clear the work space and console
// Given data
MVA = 15.0 // Generator rating(MVA)
kV = 6.9 // Generator voltage(kV)
X_1 = 25.0 // Positive sequence reactance(%)
X_2 = 25.0 // Negative sequence reactance(%)
X_0 = 8.0 // Zero sequence reactance(%)
X = 6.0 // Reactor placed in line(%)
// Calculations
a = exp(%i*120.0*%pi/180) // Operator
Z_1 = %i*X_1/100 // Positive sequence impedance(p.u)
Z_2 = %i*X_2/100 // Negative sequence impedance(p.u)
Z_g0 = %i*X_0/100 // Impedance(p.u)
Z = %i*X/100 // Impedance(p.u)
Z_0 = Z_g0+3*Z // Zero sequence impedance(p.u)
E_a = 1.0 // Voltage(p.u)
E_b = a**2*E_a // Voltage(p.u)
// Case(a)
I_a0_a = 0 // Current(A)
I_a1_a_pu = 1.0/(Z_1+Z_2) // Current(p.u)
I_a1_a = I_a1_a_pu*MVA*1000/(3**0.5*kV) // Current(A)
I_a2_a = -I_a1_a // Current(A)
I_b0_a = 0 // Current(A)
I_b1_a = a**2*I_a1_a // Current(A)
I_b2_a = a*I_a2_a // Current(A)
I_a_a = I_a1_a+I_a2_a // Line current(A)
I_b_a = I_b1_a+I_b2_a // Line current(A)
I_c_a = -I_b_a // Line current(A)
I_g_a = 0 // Ground wire current(A)
V_a_a = (E_a-I_a1_a*Z_1-I_a2_a*Z_2-I_a0_a*Z_0)*kV*1000/3**0.5 // Voltage(V)
V_b_a = (a**2*E_a+%i*3**0.5*I_a1_a_pu*Z_1)*kV*1000/3**0.5 // Voltage(V)
V_c_a = V_b_a // Voltage(V)
// Case(b)
I_a1_b_pu = E_a/(Z_1+(Z_2*Z_0/(Z_2+Z_0))) // Current(p.u)
I_a1_b = I_a1_b_pu*MVA*1000/(3**0.5*kV) // Current(A)
I_a2_b_pu = -Z_0*Z_2/(Z_2*(Z_0+Z_2))*I_a1_b_pu // Current(p.u)
I_a2_b = -Z_0*Z_2/(Z_2*(Z_0+Z_2))*I_a1_b // Current(A)
I_a0_b_pu = -Z_0*Z_2/(Z_0*(Z_0+Z_2))*I_a1_b_pu // Current(p.u)
I_a0_b = -Z_0*Z_2/(Z_0*(Z_0+Z_2))*I_a1_b // Current(A)
I_a_b = I_a0_b+I_a1_b+I_a2_b // Line current(A)
I_b_b = I_a0_b+a**2*I_a1_b+a*I_a2_b // Line current(A)
I_c_b = I_a0_b+a*I_a1_b+a**2*I_a2_b // Line current(A)
I_0_b = 3*I_a0_b // Current in the ground resistor(A)
V_a_b_pu = E_a-I_a1_b_pu*Z_1-I_a2_b_pu*Z_2-I_a0_b_pu*Z_0 // Voltage(p.u)
V_a_b = abs(V_a_b_pu)*kV*1000/(3**0.5) // Voltage(V)
V_b_b = 0 // Voltage(V)
V_c_b = 0 // Voltage(V)
// Results
disp("PART III - EXAMPLE : 4.1 : SOLUTION :-")
printf("\nCase(a): Initial symmetrical rms line current when ground is not involved in fault, I_a = %.f A", abs(I_a_a))
printf("\n Initial symmetrical rms line current when ground is not involved in fault, I_b = %.f A", real(I_b_a))
printf("\n Initial symmetrical rms line current when ground is not involved in fault, I_c = %.f A", real(I_c_a))
printf("\n Ground wire current = %.f A", I_g_a)
printf("\n Line to neutral voltage, V_a = %.f V", real(V_a_a))
printf("\n Line to neutral voltage, V_b = %.f V", real(V_b_a))
printf("\n Line to neutral voltage, V_c = %.f V", real(V_c_a))
printf("\nCase(b): Initial symmetrical rms line current when fault is solidly grounded, I_a = %.f A", abs(I_a_b))
printf("\n Initial symmetrical rms line current when fault is solidly grounded, I_b = (%.f+%.fj) A", real(I_b_b),imag(I_b_b))
printf("\n Initial symmetrical rms line current when fault is solidly grounded, I_c = (%.f+%.fj) A", real(I_c_b),imag(I_c_b))
printf("\n Ground wire current = %.fj A", imag(I_0_b))
printf("\n Line to neutral voltage, V_a = %.f V", V_a_b)
printf("\n Line to neutral voltage, V_b = %.f V", V_b_b)
printf("\n Line to neutral voltage, V_c = %.f V\n", V_c_b)
printf("\nNOTE: Changes in the obtained answer from that of textbook is due to more precision here and approximation in textbook")
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