Short ckt Calculation

I have a Bussman pamphlet, SPD84, from 1984. Its section 5.7 in system analysis mentions the following things you need: power source's transformer full load amperes and transformer impedance. Your available short circuit amps from the transformer will not be larger than (FLA)*(100)/(Impedance). Then you need to add to this any additional contribution from larger motors in your system, because they can act as generators when you have a short on the system feeding them. An approximation of their contribution is to use four times the sum of the motor full load currents.

Assume your transformer impedance is 5%, its short circuit current could be 60,000 Amps. If you have 1000 amps of motor load, the short circuit current could be as high as 64,000 amps.

This gives a fair estimate of the real values. It assumes an infinite supply available at the line side of your transformer, and has to be adjusted downward for the impedance of the feeder conductors from the transformer to the point where your equipment is. I won't give the added formulas here, but for carrying the 3000A at 415V down 40-feet (11-meters) of busway, the available short circuit current is only about 5% lower.

Engineers have "fun" with this stuff, so ordinary guys like you and me need all the tricks and help we can get!

--John M.

Thanks a lot for your clarification.

also clarify me that if i had some generators connected at down stream ,how they will be considered and also for the load of ups & lighting whether contribution will be taken same as of motor.

thanks& reg

You can greatly simplify your S/C calcs by assuming:

1) Infinite Buss,

2) The fault is at the service transformer secondaries

3) The transformer impedance, Z, is 2%

The Isc = Transformer KVA Rating /( SQRT(3)* V Line-to-Line* Z)

So assuming your service transformer is a 1500 KVA the maximum available fault current would be 104KA.