NCEES #540

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

ElecPwrPEOct11

Well-known member
Joined
Sep 22, 2011
Messages
116
Reaction score
0
Location
Virginia
Hopefully another easy one. I understand the math that produces the answer. But isn't the 2291 MVA answer the total short circuit MVA on the 230 kV bus? The problem asks for the MVA contribution from just G1. Am I thinking about this wrong?

 
Last edited by a moderator:
I think you're talking about NCEES #540.

If the only short circuit contributor to the bus is the generator, then the contribution from G1 is the total available short circuit current at the bus. The implication of the problem is that there are other generators or loads that will also contribute to that value. They just don't show them.

 
Last edited by a moderator:
^ yes you're right, I changed the title thanks. Your explanation makes sense as well.

How would you approach the problem if there were a large motor or another source connected directly to the 230 kV bus? I think I would find Stotal and then do a weighted fraction to find each contribution.

 
Also another 'what-if' question. What if they gave you the generator values of both Xd and X'd. Which would you use for the generator's impedance? Seeing X'd threw me for a loop in this problem.

 
Also another 'what-if' question. What if they gave you the generator values of both Xd and X'd. Which would you use for the generator's impedance? Seeing X'd threw me for a loop in this problem.
The type of fault analysis you are performing determines which value you use.

Xd is the direct reactance. It's pretty much a steady state value.

X'd is the transient reactance. This value is used for the transient fault current that will last for the first few cycles of the fault. The current value using the transient reactance is the one you will likely have to interrupt.

X''d is the sub-transient reactance. This value is used to determine the maximum instantaneous fault current. The momentary/withstand ratings of the connected equipment need to be greater than the fault current.

 
That makes sense, thanks flyer_PE. I was surprised for this type of short circuit problem not to be given Xd. I guess use what you're given.

 
is the solution wrong by saying "Ssc = 1/Zpu = 2.75pu" - isn't this Isc unless it is V2/Z

i didn't use Isc i used MVAsc = MVA base / Zpu to get the answer.

can someone please explain this simple thing to my crazy brain?

 
Not sure where you're coming from Rick. Ssc = 1/ Zpu is also = Spu/Zpu for this problem. This is since Spu on its own base = 1.

I don't love their method either, I just pick a Sbase and then convert both of the Zpu to the new base. Then Ssc = Sbase/Zpu-total.

 
This types of problem has simple solution as under:

MVA (G1-SC)= 834/0.23=3626, MVA (T1-SC)=933/0.15=6220 and finally, MVA (SC)= (3626*6220)/(3626+6220)=2291 (Answer)

Thanks,

 
Yea I have the MVA method down though want to make sure I understand calculating using per-unit also.

There are a lot of formulas using per-units that I'm not familiar with - like Ssc = Spu/Zpu. Is there a good reference showing how these are derived? Though I think I'll just keep with what I know as we are getting close!

 
This types of problem has simple solution as under:

MVA (G1-SC)= 834/0.23=3626, MVA (T1-SC)=933/0.15=6220 and finally, MVA (SC)= (3626*6220)/(3626+6220)=2291 (Answer)

Thanks,
I'm still confused here. Isn't that the total MVA(SC)? The problem is asking for just the contribution from the generator. That's why when i initially did this problem i just calculated MVASc(G1) = 834/0.23 = 3626. This is one of the answers, albeit the incorrect one. I just don't understand why the correct answer is the total MVASc and not just the MVASc of the generator. The other fundamental question with that too is does MVA in parallel add, and MVA in series acts like resistors in parallel?

 
BH_Cubed,

I have the same question. Why is the short circuit contribution from the transformer being considered, when the question clearly asks for the Generator contribution only. Flyer_PE tried to answer this question at the beginning of the thread but I still don't understand. Can anyone elaborate on this?

I can speak for the second part of your question. Yes, you are correct! when using the MVA method "components" that are in parallel are added as series resistors. "Components" that are in series are added as parallel combination of resistors.

You may want to check the paper "Short Circuit ACB-Learn it in an Hour, Use it Anywhere, Memorize no Formula" is very usefull. Here is a link to it:

www.arcadvisor.com/pdf/ShortCircuitABC.pdf

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

 
BH_Cubed,

I have the same question. Why is the short circuit contribution from the transformer being considered, when the question clearly asks for the Generator contribution only. Flyer_PE tried to answer this question at the beginning of the thread but I still don't understand. Can anyone elaborate on this?

I can speak for the second part of your question. Yes, you are correct! when using the MVA method "components" that are in parallel are added as series resistors. "Components" that are in series are added as parallel combination of resistors.

You may want to check the paper "Short Circuit ACB-Learn it in an Hour, Use it Anywhere, Memorize no Formula" is very usefull. Here is a link to it:

www.arcadvisor.com/pdf/ShortCircuitABC.pdf
Thanks, i've been trying to find a copy of that paper. Haven't had much luck.

 
Flyer_PE,

Thanks for the explanation, and thank you for helping not just me but every one else in this forum in passing the test. I hate to sound like an idiot but I'm still not getting it. I understand the transformer is not a contributor; a contributor is a motor or a generator. However, the MVA rating of the transformer is considered in the computation at the fault location regardless of the method being used (Per Unit, or MVA). I personally like the MVA because is quick and easy.

I think I could accept the fact that this is how the computation works and do it this way every single time, and I will. I would, however; like to understand "why" the MVA rating of the generator is considered. Can you refer me to a paper or an article that explains this; you know something that would make the light bulb turn on.

Thanks again!

 
The only reason to really care about the MVA and voltage ratings of the generator is that the transient reactance is given as a per unit value on those base values. Using either the MVA method or pu analysis, one way or another, the generator, transformer, and system impedances have to be evaluated on a common base.

Unfortunately, I don't have anything to simply describe this stuff beyond the paper mentioned above for the MVA method.

 
I think I could accept the fact that this is how the computation works and do it this way every single time, and I will. I would, however; like to understand "why" the MVA rating of the generator is considered. Can you refer me to a paper or an article that explains this; you know something that would make the light bulb turn on.

Thanks again!
Flyer gave you a good answer but I interpreted your question a bit differently... Let's say for simplicity we have two generators on the system and they each have 10% reactance on their respective bases. One is rated 100kVA and the other 5MVA. It should intuitively make sense that the capacity to deliver energy through the short circuit is much larger for the larger machine. Help any?

 
Back
Top