Fault Current

[Hileg]

Could anyone tell me what the rated fault current would be with six 250Kva single phase networked transformers each with a 2.7% impedence. The secondary voltage is 120/208.
Thanks

 

[musicguy1800]

Where is the location of the fault?

 

[Deansharafi]

It depends on the connection of the transformers, but as a rule of thumb for a (three phase) transformer:
Short circuit Kva=((rated Kva/(% impedance/100))
and Short circuit current=(short circuit kva/((sqrt3)*rated voltage KV))
Thus you know short circuit current for on of the transformers. The combination can be calculated by knowing how they are connected.
Hope it helps.

 

[jbartos]

Suggestion: Please, describe how they are networked.
A connection can be two transformers per phase; however, then the proper question would be what is the short circuit of the two transformers, Z=2.7% each, in parallel?
Six transformers of one phase in parallel would have Z/6=2.7/6=.45% output impedance. Then
250kVA/.0045=555.55kVAsc
and
Isc=555.55kVA/(sqrt3 x .208kV)=1542A

 

[rbulsara]

How are the transformers connected?

In order to get 208/120 secondary system, it has to be connected for 3 phase configuration.

Six single phase units connected in 3 phase formation will provide Two banks of 750kVA, 3 phase system, each with 2.7% Z.

In that case the SCC will be 2*(750/1.732*0.208)/.027=154,000A

jbartos:
There is a decimal error in your post. It will be 55555 kVAsc or 154,000A on the secondary of the transformers.

 

[jghrist]

musicguy's question on the location of the fault is relevant for networked transformers (as opposed to paralleled at the same location) because there will be some secondary conductor in the path regardless of the location. The impedance of the secondary must also be known.

 

[Hileg]

Three transfomres are connected in wye and the the other 3 parrallel them. So 2-250Kva per phase.
Thanks for your help.

 

[jbartos]

Thanks to rbulsara for the correction (I beg your pardon)
Six transformers of one phase in parallel would have Z/6=2.7/6=.45% output impedance. Then
250kVA/.0045=55555kVAsc
and
Isc=55555kVA/(sqrt3 x .208kV)=154207A