CI # 3 Quest 79

[iwire]

I think I m pretty sure on the MVA method but I still not sure how to handle the PU unit impedance.

Q: 3 phase generator serves a 5/25 kV transformer rated 6 MVA. The transformer subtransient impedance is j0.39 pu. The transformer (steady state) rated impedance is 6%. If the generator is rated 11 MV, its contribution (MVA) under fault conditions at transformer secodary terminal is?

Answer 15.2k A

Z= Zpu X Z base

 

[iahim]

For generator:

they give you X"=j0.39 pu, so Z=0.39 pu

Contribution = 11/0.39

For transformer:

Z=6%=0.06pu

Contribution = 6/0.06

Total contribution: 22MVA

I get 508A. I assume, CI is wrong, as usual...

 

[iwire]

For generator:

they give you X"=j0.39 pu, so Z=0.39 pu

Contribution = 11/0.39

For transformer:

Z=6%=0.06pu

Contribution = 6/0.06

Total contribution: 22MVA

I get 508A. I assume, CI is wrong, as usual...

Ya throw me off

 

[EEpowerOK]

The transformer usually carries the j

For generator:

Contribution = 11/0.06

For transformer:

Contribution = 6/0.39

Total contribution: 14MVA

get 16.2 KA. using the 5K, close

 

[iahim]

The transformer subtransient impedance is j0.39 pu. The transformer (steady state) rated impedance is 6%.

The transformer was listed twice, so I assumed the first value is for generator, since it's subtransient (transformers don't have a different subtransient value).