132kV Substation-B having two transformers is being feed by 132kV single circuit radial line A-B
having line length km. Line +Ve sequence impedance is 0.16+J0.40 ohms/km.
Source Substation-A has 132kV Bus 3-Ph fault level as MVA.
Its 1-Ph fault level is MVA.
For Substation-B; 25 MVA transformer has percentage leakage reactance %; and 50 MVA transformer has percentage leakage reactance % Find out the Substation-B 33kV Bus fault current for 3-Ph and 1-Ph fault.
(Note: Consider line zero sequence impedance as 3 times +Ve sequence impedance; and Base MVA as 100).
Section-A: 33kV Bus fault current for LLL fault.
1) 3-Ph Fault MVA = BaseMVAx100/X1S; Thus: X1S=
%
2a) For km line ohmic reactance =
Ohms
2b) and its +Ve Seq. % reactance = %
2b) and its +Ve Seq. % reactance = %
3a) 50 MVA % impedance (X1T1) with base MVA 100
%
3b) 25 MVA % impedance (X1T2) with base MVA 100 %
3b) 25 MVA % impedance (X1T2) with base MVA 100 %
4) +Ve sequence impedance between infinite source and fault point would be
X1S + X1L + (X1T1 || X1T2) = %
5) Thus for S/S-B 33kV Bus LLL Fault MVA = MVA.
6) S/S-B 33kV Bus LLL Fault Current = AMP.
Section-B: 33kV Bus fault current for SLG fault.
7) For 1-Ph fault by reverse engineering; for 132kV Bus at S/S-B we can write
X1S + X2S + X0S = %
8) In above formula X1S = X2S and from (1) we have X1S; Thus X0S = %
9) For line X0L = %
10a) 50 MVA % impedance X1T1 = X2T1 = X0T1 =
%
10b) 25 MVA % impedance X1T2 = X2T2 = X0T2 = %
10b) 25 MVA % impedance X1T2 = X2T2 = X0T2 = %
11a) Equivalent sequence impedance between fault point and infinite source will be
X1 = X1S + X1L + (X1T1 || X1T2) = %
11b) X2 = X1 = %
11c) X0 = X0S + X0L + (X0T1 || X0T2) = %
12) As per above diagram we know values of X1, X2 and X0 between fault point and infinite source. Thus SLG fault level at 33kV Bus would be = %.
13) S/S-B 33kV Bus SLG Fault Current = AMP.
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