Fault Current Ratings

What's the transformer impedance? I = V/Z

Fault current ratings (as in NEMA, NEC, ASTM, NFPA or UL) are one thing, and actual fault currents (hopefully less than rating) are another.

Wow that was quick! I was going to assume 80% of max 'nominal' fault rating as a likely fault but how to calc the max fault to begin with?

Is there a 'typical' impedance value one can use as an assumption..... 5% Voltage?

You need to find the multiplier for that transformer. Take 100/%Z of the transformer. Then to find the maximum available fault current on the secondaries of the transformer, multiply the full load rated current of the transformer by the Multiplier above. In other words the multiplier for a 5% transformer impedance would be 100/5= 20. So the maximum fault let thru current (based upon an infinite bus on the primary) would be the FLA X 20. Best source to read up on this is the Bussman handbook.

So, is it....?

1.732 / V (11kV) / 10.9mVA = 572A as its working load..... then x 20 = 11.4kA max fault current

Assuming a 5% impedance (100/5=20). It looks about right.

Consult "Ugly's" page 51

Hi GHOG,

'Ugly's?? unfortunately, I don't have a copy....what is it?

Excuse my ignorance, I'm a mechanical bod who has been forced to dabble in this stuff that you can't see...called electricity!!...nasty stuff with loads of calcs every left, right and centre!!!

Ugly's Electrical Reference

Burleson Distributing Corp.
3501 Oak Forest Dr
Houston, TX 77018
800.531.1660
www.uglyselectrical.com

Thanks ol' goat, I'll pop an order into amazon.