Ex-2: - Two transformer in parallel with negligible source impedance.


Problem

Consider a S/S having two transformers of (a) 132/33kV 25 MVA having leakage reactance of 9.5% and (b) 132/33kV 50 MVA having leakage reactance of 12.5%. Let the S/S source impedance be negligible as compared to transformer leakage reactance. Find out the 3-phase and 1-phase fault current for 33kV Bus fault.



Solution
Consider 50 MVA as base MVA
Hence XT1-50MVA = XT1-25MVA*(50/25) = 9.5*2 = 19% ----- ( using formula – 2)
XT2 = 12.5%
X = XT1||XT2 = 19*12.5/(19+12.5) = 7.54%
3-Ph short circuit MVA = 50x100/7.54% = 663 MVA ---- (using formula – 3)
1-Ph short circuit MVA = 3x50x100/(7.54+7.54+7.54) = 663 MVA (using formula – 4)
(As for transformer X1 = X2 = X0)
For 33kV voltage 1 MVA = 17.5 Amp
Hence this current will be 17.5*663 = 11602 Amp.
80% of this current shall be 9281 Amp

Comments
Fault current for faults at transformer LV bus increases with transformer capacity. If transformer operates in parallel fault current will be still more.

1 comment:

  1. Generally 100 MVA selected as base MVA. Here higer rating transformer selected as base MVA

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