# CI Q3-54 MVA vs PU method



## supra33202 (Oct 6, 2017)

I understand the MVA method, but I want to solve it with PU method.

Q3-54) Using the MVA method, find the 3-phase short-circuit fault current for ta transformer's secondary terminal. The low-side of the transformer is served by a generator. The generator's nameplate data indicates ratings of: 20 MVA, 23% reactance. The transformer's ratings are: 35 MVA, 13% impedance, 12.5/34.5 kV. Assume no short-circuit contributions downstream of the transformer.

The answer is D. 1,100 A, which I agree with MVA method.

My PU method:

Set Sb = 35 MVA, Vb = 34.5 kV

Iactual = Ibase * Ipu

Ibase = 35 MVA / [ sqrt(3) * 34.5 kV ] = 585.7 A

Ipu = Vpu / Zpu

Vpu = 34.5 kV / 34.5 kV = 1 pu

Zg, pu = 0.23 pu * (35 MVA/ 20 MVA) * (12.5 kV / 34.5 kV)^2 = 0.0528 pu (need to change the generator's bases from 20MVA to 35MVA and from 12.5 kV to 34.5 kV)

Zxfmr, pu = 0.13 pu

Since this is in series, Zpu = Zg, pu + Zxfmr, pu = 0.1828 pu

Ipu = 1 pu / 0.1828 pu = 5.47 pu

Iactual = 585.7 A * 5.47 pu = 3245 A ( which is not 1,100A)

What did I do wrong with my PU method?

Thanks!


----------



## rg1 (Oct 6, 2017)

supra33202 said:


> I understand the MVA method, but I want to solve it with PU method.
> 
> Q3-54) Using the MVA method, find the 3-phase short-circuit fault current for ta transformer's secondary terminal. The low-side of the transformer is served by a generator. The generator's nameplate data indicates ratings of: 20 MVA, 23% reactance. The transformer's ratings are: 35 MVA, 13% impedance, 12.5/34.5 kV. Assume no short-circuit contributions downstream of the transformer.
> 
> ...


----------



## supra33202 (Oct 6, 2017)

To get the correct answer 1100A.

Zg, pu = 0.23 pu * (35 MVA/ 20 MVA) = 0.4025 pu

Since this is in series, Zpu = Zg, pu + Zxfmr, pu = 04025 pu + 013 pu = 0.535 pu

Ipu = 1 pu / 0.5325 pu = 1.878 pu

Iactual = 585.7 A * 1.878 pu = 1099.94 A

Why do we NOT to worry about the voltage for the generator?

The generator has 12.5KV, and Vbase that I chose was 34.5 kV.

Please advise.


----------



## rg1 (Oct 6, 2017)

supra33202 said:


> To get the correct answer 1100A.
> 
> Zg, pu = 0.23 pu * (35 MVA/ 20 MVA) = 0.4025 pu
> 
> ...


It is,  because across the transformer the MVAs do not change but the Voltages and currents change. The other way is to transfer the actual Z of Genr on 34.5KV side of the transformer, convert it into pu again and do it, you will get the same answer. If it does not make much sense then, it will in fact require an in depth discussion in understanding the pu concept and I would like to avoid to spend time on this, at this moment.


----------



## supra33202 (Oct 6, 2017)

http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

"





……………………………….(3)​
The voltage ratio in equation (3) is not equivalent to transformers voltage ratio. It is the ratio of the transformer’s voltage rating on the primary or secondary side to the system nominal voltage on the same side.​
"

I don't fully understand it, but it seems like in most cases, no need to worry about the voltage ratio.

Just worry about the Sbase old and Sbase new.


----------



## rg1 (Oct 6, 2017)

supra33202 said:


> http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/
> 
> "
> 
> ...


You are absolutely right. But just keep an eye on the transformer ratio, sometimes they give; for example, in your question Transformer ratio 34.5/12 and Genr at 12.5kV. In that case as you have assumed 34.5kV as base the base voltage at Genr will be 12kV and not 12.5 kV. Here you have to use the full formula to get new Zpu of Gen. Remember the Gen Zpu is on 12.5Kv base and new Voltage base will be 12kV. So the idea is change the base according to transformer ratio, after assumption of Base Voltage at one place and keep the MVA base same.


----------



## supra33202 (Mar 4, 2018)

@Zach Stone, P.E.

Could you help me with this one?

+





In your website electricalpereview.com Video Example 8 - Generator and Transformer Per Unit Method, it seems like you also ignore the whole "voltage base old and voltage base new" term.

Zg,new = 5.6% (20MVA/55MVA) = 0.0206 pu

Are the following correct for the generator?

Vbase, old = 13.8KV and Vbase, new = 69KV

Question: when we try to find the Z,pu new, how do we know if should just ignore the whole "voltage base old and voltage base new" term?

Thanks!


----------



## Zach Stone P.E. (Mar 5, 2018)

supra33202 said:


> @Zach Stone, P.E.
> 
> Could you help me with this one?
> 
> ...


Hi Tommy, 

For the most part, the voltage term is going to cancel.  Just use the base change formula with the voltage squared term as 1 and you will be good to go.


----------

