Utility Impedance & Isc
Utility Impedance & Isc
(OP)
2000 kVA transformer
4160 / 480 5%IZ
What would be a reasonable assumption or WAG ..
for available Short Circuit Amps ... and impedance on the utility side (4160)
Thanks ..
jOmega
4160 / 480 5%IZ
What would be a reasonable assumption or WAG ..
for available Short Circuit Amps ... and impedance on the utility side (4160)
Thanks ..
jOmega





RE: Utility Impedance & Isc
As for the utility fault level any WAG would be absolutely that, except from the utility of course.
RE: Utility Impedance & Isc
Ignoring the data from the utility and figuring the theoretical worst case:
Isc utility = [VA / (V *sqrt(3))] * [1/%Z]
Isc utility = [2000000/ (480 * 1.73)] * [1/0.05]
Isc utility = 48,112 A (symetrical)
This is the maximum let-thru from the utility transformer. You also need to account for your load. Assuming all motor loads to a maximum loading of 80% of capacity:
Isc laod = [(VA * %load) / (V * sqrt(3))] * 6
where 6 is the average inrush of a motor
Isc load = [(2000000 * 0.80) / (480 * 1.73)] * 6
Isc load = 11,547 A (symetrical)
Isc = 48,112 + 11,547 = 59659 A (symetrical)
This is the worst case disign requirements. Your load will have some non-contributing loads and the utility will be less than this. In-addition, your feeders will limit the current due to resistance and length. See Bussmans website for some literature for additional hand calculations on feeders.
RE: Utility Impedance & Isc
But in absence of any data assuming anything less than 350 MVA could be a bold approach.
RE: Utility Impedance & Isc
No utility system impedance Zsys. The system impedance Zsys will make the transformer secondary short circuit current level somewhat lower.
RE: Utility Impedance & Isc