# NCEES 530



## dayrongarcia (Mar 17, 2015)

Can anyone explain to me how they got 0.025 impedance, and where it comes from?


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## knight1fox3 (Mar 17, 2015)

http://engineerboards.com/?showtopic=23295


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## kris7o2 (May 24, 2021)

Does anyone have the link above?


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## akyip (May 25, 2021)

Attached is my work on how I solved problem 530 using 2 different methods.

The 0.025 is the per-unit impedance of the utility. It comes from the following:


The utility has a line-to-line voltage of 12.47 KV and a 3-phase fault duty (available 3-phase fault short circuit power) of 40 MVA. So it can be represented by an impedance of:
ZU = VU LL^2 / SU, SC = (12.47 KV)^2 / (40 MVA) = 3.888 ohms
The base values for the utility correspond to the primary side of the given transformer. So:
V Base1 LL = 12.47 KV
S Base 3-ph = 1000 KVA
Z Base 1 = V Base 1 LL^2 / S Base 3-ph = (12.47 KV)^2 / (1000 KVA) = 155.501 ohms
The p.u. impedance of the utility is then:
ZU pu = ZU / Z Base 1 = 3.888 / 155.501 = 0.025 p.u.


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