Per unit Impedance. CI: 2#51, CI: 3#54

[fetaker]

All Gurus,

CI: 2#51

3 phase transformer is provided with following information: 12MVA, 240/120kV, 4.6%Z

What is the new %Z at SB=50MVA and VB=230kV?

While calculating the new %Z, how do I know which voltage (120kV or 240kV) to use?

CI: 3#54

A 20MVA generator with 23% reactance has 12.5kV terminal voltage. This voltage is stepped up by a 35MVA transformer (13%Z) to 34.5kV. What is the fault current at the high voltage side of the transformer?

I used SB=35MVA and VB=34.5kV.

And generator per unit reactance would be Zg, new = 0.23 x (35/20) x (12.5 /34.5)2

Zpu, new = Zpu old x (SB new/SB old) x (VB old /VB new)2

But while analyzing the answer, the book does not use the voltage ratio - (12.5 /34.5)2 to get the answer.

Am I doing something wrong?

Thanks for your input.

 

[iahim]

Not a Guru, but here are my 2c:

1. It doesn't matter what voltage you use. If you calculate the ohmic value and use the primary voltage, you get the impedance reflected on the primary side. If you use secondary voltage, you get the impedance on the secondary.

2. I would solve this problem with the MVA method. It's much easier and faster.

 

[fetaker]

CI: 2#51.

I agree. To get the answer what the question was asking for, the question should have been more clear; specifying on which side it seeks to have the calculation done.

CI: 3#54

MVA method could be easier here but I need to know, if there is a flaw in my calculation or concepts. Or if it the book which is doing it wrong.

Thank you for your input on the first one...

 

[iahim]

I think your solution for 3#54 is correct. Perhaps they used 12.5kV as base in the book, so they only had to convert to a different Sbase.

I solved the problem with the MVA method and with pu, and I'm getting the same answer: 1,100A.

 

[fetaker]

My calc is as below.

SB=35MVA and VB=34.5kV

.IB = 585.7 Amp (calculated)

Zg, new = 0.23 x (35/20) x (12.5 /34.5)2

=

 

[fetaker]

oops....

continued...

Zg = 0.053pu (calculated)

And Ztransformer = 0.13pu

Then total Z = 0.183 pu

And Ipu = 5.47 (calculated)

Fault current = Ipu * IB = 3202 Amp.=Answer

The thing is if i get rid of the voltage ratio (12.5 /34.5)2 I get the correct answer. And i'm not convinced that this ratio should be omitted.

 

[iahim]

I agree with you up to this line:

"And Ipu = 5.47 (calculated)"

The source voltage is 12.5/34.5 pu = 0.36 pu, the total impedance is 0.183 pu, so Isc is 0.36/0.183 = 1.967 pu

Isc = 1.967 * 585.7 = 1152A

 

[fetaker]

Iahim, I appreciate your effort on explaining the concepts.I went through this problem with several methods to check my understanding.

I had forgotten the concept of per unit along my preparation. Based on my understanding there are n+1 base voltages in a per unit calculations (n being number of transformers). In my calc above, the old base voltage of the transformer is 12.5kV and using the transformation the new base voltage for the generator would be 12.5kV. So, the ratio still exist, which is 1. This gives the correct answer.

Just a week to go now....