Take the transformer VA rating, divide it by the % impedance. then divide that by the voltage and 1.732. Result is the maximum three-phase fault current through the transformer assuming an infinite source.
(I'm making an slightly skeptical assumption that this is not a homework problem.)
Thanks for the info. And no, its not a homework problem. I assume I would use the LV value? hence that would provide the total available fault current? I'm a T-line engineer trying to assist some overworked substation co-workers!
It works in either direction, just use the voltage for the side you are interested in. Normally, for a step-down transformer, you'd be interested in the low side fault current that the transformer could produce.
This result is the three-phase symmetrical fault current through the transformer for a fault on the low side. It does not included any contributions from motors on the low side, or other local sources.
Dpc, your calculation is correct, and that is the maximum amount of fault current the trandformer can produce. This fault current would be limited by the source feeding the primary side of the transformer.
Downstream of the Secondary side, the fault current would be reduced due to the impedance in the cable, thus the motors and other components in the circuit would not add to the fault current.
So, if trying to protect downstream devices, ensure the device you are using to clear the fault current (fuse, breaker) has an adequate interrupt rating.
Just wanted to add some info from my experience working at a fuse manufacturer.
