engineering recuitment GATE EE 2023-24 Test Series Power Systems PU System and Symmetrical Components Symmetrical Networks
The relevant data for the given system is as follows.
Generator: X1 = X2 = j0.1 pu, X0 = j0.020 pu, Rn = 1 \(\Omega\), (11kV and 10 MVA are the base value)
11/66kV Transformer: X1 = X2 = X3 = j0.1 pu on 10MVA base.
3-winding transformer (1) and (2):
66 kV winding: X1 = X2 = X3 = j0.05 pu (10 MVA base).
11 kV winding: X1 = X2 = X3 = j0.02 pu, Rn = 1 \(\Omega\) (10 MVA base).
3.3 kV winding: X1 = X2 = X3 = j0.05 pu (10 MVA base).
Find zero sequence impedance at the fault point.
1
(0.30 + j0.060) pu
2
(0.0575 + j0.25) pu
3
(0.25 + j0.0575) pu
4
(0.20 + j0.050) pu