5-unsymmetrical Fault Analysis

UNIT : 6 Unsymmetrical Fault Analysis - I 1. Different types of unsymmetrical faults • Single line – to – ground (L – G ) fault • Line – to – line (L – L ) fault • Double line – to – ground (L – L – G ) fault 2. Analysis of single line – to – ground (L – G ) fault (unloaded generator)

L – G Fault Important Equations : Va1 = Ea – Ia1Z1 Va = 0 Va2 = - Ia2 Z2 Ib = 0 Va0 = - Ia0Z0 Ic = 0 From above equations Ia1 = Ia2 = Ia0 = Ea /(Z1 + Z2 + Z0) If = Ia = 3Ia1 = 3Ea / (Z1 + Z2 + Z0) All three sequence networks are connected in series

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NU/IT/EE/B. Tech. Sem.V/EPS-II/UAP

Example: A 25 MVA 13.2 kV, alternator with solidly grounded neutral has sub transient reactance of 0.25 pu. The negative and zero sequence reactances are 0.35 and 0.1 pu respectively. A single line to ground fault occurs at the terminals of an unloaded alternator. Determine the fault current and line to line voltage. [4685A, 7.717 kV, 15.087 kV, 7.717 kV] Example : A generator with grounded neutral has sequence impedances of Z 1, Z2 and Z0 and generator emf Ea. If a single line to ground fault occurs on the terminal ‘a’, find the expressions for Vb and Vc [Vb =Ea(α2-α)Z2 + (α2-1)Z0 / (Z1 + Z2 + Z0) Vc = Ea [(α-∝2) Z2 + (α-1)Z0 / (Z1 + Z2 + Z0)]

UNIT: 7 Unsymmetrical Fault Analysis - II 1. Line – to – line fault (L - L)

Conditions: (i) Vb = Vc (ii) Ia = 0 (iii) Ib + Ic = 0 Conclusion: 2

NU/IT/EE/B. Tech. Sem.V/EPS-II/UAP

(i) Ia1 = -Ia2 (ii) Ia0 = 0 (iii) Va1 = Va2 & Va0 = 0 i.e positive and negative sequence networks are in parallel 2. Sequence networks:

From Fig. Ia1 = Ea / (Z1+ Z2) If = Ib = Ib1 + Ib2 + Ib0 = α2Ia1 + α Ia2 + Ia0 = Ia1 [α2 - α] = Ia1 [-0.5 – j 0.866 + 0.5 – 0.866 ] = -j√3 Ia1 = -j√3Ea / (Z1+ Z2) 3. Double line – to – ground (L – L – G ) fault

Conditions: (i) Ia = 0 (ii) Vb = 0 (iii) Vc = 0 Conclusion: (i) Va1 = Va2 = Va0 4. sequence networks: i.e. all sequence networks are in parallel 3

NU/IT/EE/B. Tech. Sem.V/EPS-II/UAP

Ia = Ea / [Z1 + Z2Z0 / (Z2+Z0)] − 3Z 2 E a If = Z1 Z 2 + Z 2 Z 0 + Z1 Z1 5. Analysis of symmetrical fault using symmetrical components

Condition: (i) Va = Vb = Vc (ii) Ia + Ib + Ic = 0 Since Ia = Ib = Ic Ib = α2Ia Ic = αIa Ia1 = Ia Ia2 = 0 Ia0 = 0 Va1 = Va2 = Va0 = 0 Therefore sequence network

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NU/IT/EE/B. Tech. Sem.V/EPS-II/UAP

∴Ia1 =

Ea Z1

UNIT : 8 Unsymmetrical Fault Analysis - III (1) Comparison of single line – to – ground (L-G) fault and 3-phase (L-L-L) fault Comments: (i) Single line – to ground fault at the terminals of a generator with solidly grounded neutral is more severe than a 3-phase fault at the same location (ii) Fault at the generator terminals with neutral grounded through Xn If Xn is very small, single L-G fault current may be more than that for a 3phase fault and vice versa, as a term 3Xn appears in the denominator of a expression for a fault current 1 If Xn = (X1 – X0) : Line currents for the two cases are equal 3 1 If Xn < (X1 – X0) : single L-G fault current is more severe 3 1 If Xn > (X1 – X0) : 3-phase fault is more severe 3 Example: 1: A 50 MVA, 11kV 3-phase alternator was subjected to different types of faults. The fault currents are (1) 3- phase fault = 1870A (2) Line – to- line fault = 2590A (3) Single line – to – ground fault = 4130A Alternator neutral is solidly grounded. Find the values of three sequence reactances of the alternators [X1 = 4.247 Ohm, X2 = 0.851 Ohm, X0 = 0.366 Ohm] Example : 2: A 15 MVA, 13.2 kV, 3-phase generator has a solidly grounded neutral. Its sequence reactances are X1 = 40%, X2 = 30%, X0 = 5% (i) Determine the value of reactance to be connected in the neutral circuit so that the fault current for a single line – to – ground fault does not exceed the rated current. (ii) Find the value of resistance to be connected in the neutral circuit for the same purpose. [Xn = 8.712 Ohm, Rn = 11.247 Ohm] *****

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NU/IT/EE/B. Tech. Sem.V/EPS-II/UAP