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 X_T1-50MVA = X_T1-25MVA*(50/25) = 9.5*2 = 19% ----- ( using formula – 2)

X_T2 = 12.5%

X = X_T1||X_T2 = 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.