NCEES 530

[dayrongarcia]

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

 

[knight1fox3]

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

 

[kris7o2]

Does anyone have the link above?

 

[akyip]

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.