Help needed for NCEES #530

AD_power · Sep 19, 2012

If somebody knows from what magic hat they pulled additional impedance of 0.025??? Transformer connected to the bus, it should be pretty simple and straightforward solution:

Isc= I on 480v side (1203A) / Z of transformer (0.04). I'm confused.

DK PE · Sep 19, 2012

Your solution is only valid if for the "infinite bus" on the utility side. In this case they give you the capability of that side.

Flyer_PE · Sep 19, 2012

There are a couple of threads that cover this: Thread 1, Thread 2

AD_power · Sep 20, 2012

thanks a lot. Now I get it.

danderson · Sep 20, 2012

As stated by other members here, the MVA is SO much easier on these type problems.

See...

http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf

JB66money · Apr 9, 2014

When using the per-unit method there is an impedance associated with the fault duty. This obtained first by getting the per-unit apparent power which is found by using the 1000kVA rating of the transformer the system's base apparent power and converting it into MVA, which is 1MVA. Next the fault duty's per-unit value is S=40MVA / 1MVA = 40p.u. . Next the per-unit impedancnce associated with the fault duty is z = 12 / 40 = .025pu. Next we find the per-unit fault current is

if = 1 / (.04+.025) = 15.38pu.Next the base current on the secondary side of the transformer where the fault occurs is Ib = (1000*103) / (sqrt(3) * 480 ) = 1202.81Amps

Finally we determine the acctual fault current by multiplying the per-unit fault current by the base current on the secondary side of the transformer.

IF = if * Ib = (15.38)*(1202.81) = 18499 Amps.