# NCEES (Power) Practice Problem #530



## nailbighter (Sep 20, 2009)

I cant for the life of me figure this problem out. The solution dosent describe how the get the .025 in the equation Isc = 1.0/(.04 + .025). Can anyone help me out here? I have been staring at this problem for hours.


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## juan (Sep 20, 2009)

0.025 is the pu fault duty impedance of 40MVA on 12.47KV system

Zsystem(pu) =1MVA(base)/(40MVA) = 0.025

If you take 40 MVA Base then:

Zsystem= 1pu on 40 MVA base

ZTransformer= ZT= 0.04*(40/1)= 1.6pu on 40 MVA base

= &gt; Isc=1/(1+1.6) = 0.385 pu on 40 MVA base

IBase480V=40*10^6/(sqr(3)*(480))= 48112.52A

Isc=(0.385)*(48112.52) = 18,504 A

Or you can try this way:

12.47KV system &amp; 12.47KV Transformer side are components connected in series

at the 12.47KV bus:

Z transformer =ZT=(4%/100)*(12.47x10^3)^2/(1000*10^3)= 6.22 ohms

Zsystem= 12470^2/ 40x10^6= 3.89 ohms

Z transformer + Zsystem= 10.11 ohms (at 12.47KV bus)

Converting to 480V bus: 10.11*480^2/12470^2= 0.015 ohms

Icc= 480/ (sqrt(3)*0.015)=18,475 Amps.

3 ways, same answer.

Or you can try the MVA method too.


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