# new ncees #530



## cabby (Feb 3, 2009)

I can come up with the 480v Base Isc of 1203A. However, I do not understand how they are obtaining the p.u. Isc? 1MVA is given, transformer has a 4% impedence, I assume that is the .04. But where are they getting the .025?

thanks,


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## Flyer_PE (Feb 3, 2009)

Since you are given the fault capability of the system in MVA the impedance is 1p.u. on a 40 MVA base. The value needed is the utility impedance on the transformer base (1MVA).

Zpu-tx = Zpu-utility*MVAtx/MVAutility

Zpu-tx = 1*1MVA/40MVA = 0.025

Iscpu = Vpu/(Ztx-pu+Zutility) = 1/(.04+.025) = 15.4 pu


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## cabby (Feb 3, 2009)

Thank you. I got it now.



Flyer_PE said:


> Since you are given the fault capability of the system in MVA the impedance is 1p.u. on a 40 MVA base. The value needed is the utility impedance on the transformer base (1MVA).
> Zpu-tx = Zpu-utility*MVAtx/MVAutility
> 
> Zpu-tx = 1*1MVA/40MVA = 0.025
> ...


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