Fault Current
Fault Current
(OP)
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
Thanks
When was the last time you drove down the highway without seeing a commercial truck hauling goods?
Download nowINTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS Come Join Us!Are you an
Engineering professional? Join Eng-Tips Forums!
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail. Posting GuidelinesJobs |
|
RE: Fault Current
RE: Fault Current
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.
RE: Fault Current
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
RE: Fault Current
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.
RE: Fault Current
RE: Fault Current
Thanks for your help.
RE: Fault Current
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