Jump to content
Engineer Boards

# NCEES #540

## Recommended Posts

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?

#### Share this post

##### Share on other sites
Flyer_PE    929

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.

#### Share this post

##### Share on other sites

^ 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.

#### Share this post

##### Share on other sites

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.

#### Share this post

##### Share on other sites
Flyer_PE    929

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.

#### Share this post

##### Share on other sites

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.

#### Share this post

##### Share on other sites

this one sort of lost me. what is the impedance dividing? is this just (V^2)/Z

#### Share this post

##### Share on other sites

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?

#### Share this post

##### Share on other sites

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.

#### Share this post

##### Share on other sites

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,

#### Share this post

##### Share on other sites

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!

#### Share this post

##### Share on other sites

thanks for your guys help

#### Share this post

##### Share on other sites

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?

#### Share this post

##### Share on other sites

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

#### Share this post

##### Share on other sites
Flyer_PE    929

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

#### Share this post

##### Share on other sites

Flyer_PE,

Thank you for the clarification.

#### Share this post

##### Share on other sites

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.

#### Share this post

##### Share on other sites

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!

#### Share this post

##### Share on other sites
Flyer_PE    929

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.

#### Share this post

##### Share on other sites

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?

#### Share this post

##### Share on other sites

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!

#### Share this post

##### Share on other sites
Flyer_PE    929

^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.

#### Share this post

##### Share on other sites

OK! that one made the light bulb turn on.

Thanks Flyer_PE!

#### Share this post

##### Share on other sites

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.

#### Share this post

##### Share on other sites
Flyer_PE    929

^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.

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account

## Sign in

Already have an account? Sign in here.

Sign In Now

×