Phase to Phase Fault calculation

[ChipNJ]

What would the correct formula to use to determine the multiples of a transformers rated OA current and then calculate into actual ampheres?
Transformer:
10 MVA
impedance 8.5%
rated phase to phase voltage (secondary side) 12.47 kV
Supply side capacity 650 MVA
assume impedence angles are the same
assume phase-to-phase voltage at the secondary terminals (prior to the fault) is 12.47kV.

 

[mykh]

I use MVA method for this. Check

http://www.arcadvisor.com

for the procedure and more the method is capable of doing:

Transformer short circuit MVA :

10MVA / 0.085Zpu = 117.65 SC MVA

650MVA entrance in series with the 117.65 SCMVA is capable of supplying 99.62 SC MVA (X/R assumed to be equal)

Assuming equal 1 and 0 sequence network impedances and X/R ratios, both SCMVA1 and SCMVA2 equal to the 99.62 SCMVA.

The series combination of SCMVA1 and SCMVA2 results in 99.62 / 2 = 49.81 SCMVA.

Ia1 = 49.81 / (1.73 x 12.47) = 2.3kA.

Ia2 = - Ia1 for phase to phase fault.

Assuming B to C fault:

Ib = -Ic = a x a x Ia1 + a x Ia2 + Ia0 =
(-0.5 - j0.866) x Ia1 - (-0.5 + j0.866) x Ia1 + 0 =
-j x 1.73 x Ia1 = 3.98 kA

 

[ChipNJ]

Thanks. I was unsure if I needed to account for the supply side capacity for a fault on the secondary side. Thanks.

 

[tinfoil]

If you assume that the supply side is an infinite bus, then your answer will always be conservative (i.e. the resulting fault current value will be somewhat larger than if the actual supply impedance is allowed for).