NCEES #540

[robertplant22]

I apologize. On DK PE's quote of my previous message above I meant to say the MVA rating of the transformer.

DK PE, your example does make sense. Where I get confused on this question is that the MVA rating of the transformer is accounted for in the computation even though the question asked for the contribution from the generator.

Transformers are not contributors to short circuit current. The transformer impedance acts to reduce the available short circuit current at the low-side bus.

I think if I understand this correctly; the transformer needs to be included in the computation because even though there is more available short circuit current upstream from the transformer, the amount of available short circuit current at the fault location will be limited by the amount of MVA and impedance the transformer can handle, since the short circuit current will have to go through the transformer in order to get to the fault location.

I'm not sure if that is correct. Do I make any sense on what I stated above, or am I making up a bunch of gibberish?

Thanks!

 

[Flyer_PE]

^The only correction I would offer is that the available fault current at the bus is not limited by the MVA rating of the transformer. The only limiting factor is the transformer impedance. The only reason you care about the MVA rating is that the impedance is not expressed in ohms. It is expressed as a per-unit value that is tied to the voltage and MVA ratings of the transformer.

 

[robertplant22]

OK! that one made the light bulb turn on.

Thanks Flyer_PE!

 

[bdoug]

I understand the explanations provided, but have one additional question:

Let's say there is an additional generator, we'll call it "G2", but we still want to find the contribution only from G1. Can you simply ignore the G2 source to find the contribution of G1, or would the solution be more complicated?

Thanks.

 

[Flyer_PE]

^The solution would be the same. The contribution from G1 will be the same regardless of contribution from any other source. Say you had three generators in parallel on the system. The fault contributions from each of the three could be calculated independently such that the you have X amps from Gen 1, Y amps from Gen 2, and Z amps from Gen3. The total fault current is simply X+Y+Z.

 

[katag]

This thread is very helpful, I just wanted to bring it back for some clarification if possible. In the above post Flyer_PE you said that if there was another Generator "G2" in parallel with G1 and we wanted to know the contribution of just G1 we would completely ignore G2 and calculate the same as done in the actual problem, but if we wanted the total MVAsc then we would include G2? Just want to make sure I have this down. What if there was a motor connected at the load and we wanted just G1 contribution? Same thing? And finally, does everyone feel that just knowing the MVA method for Short Circuit Calcs is sufficient and can be used regardless or should I brush up on Per Unit Method as well? Thanks for all the help!

 

[Flyer_PE]

The contribution to a fault from any individual machine is not affected by the contribution by any other machine. The contribution from G1 will be the same regardless of how many other generators or motors are connected to the system. Adding or removing fault contributing devices will change the total fault current.

My opinion on the MVA method: The MVA method is just fine for calculating fault currents. Like any other analysis tool, it doesn't cover every situation. Personally, I wouldn't feel comfortable going into the exam without being fluent in PU analysis.

 

[Spark-E]

Can anyone explain how they are able to use the equation Ssc = Spu/Zpu? I can't find it in any of my references or online.

 

[Blink]

This formula is used to solve short circuit problems with the MVA method.

 

[StinkyTofu]

Do we need to convert all per unit impedances to a common base? It doesn't look like we do.