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NCEES #540

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#1 ElecPwrPEOct11

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Posted 19 October 2011 - 02:54 PM

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?

#2 Flyer_PE

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Posted 19 October 2011 - 03:03 PM

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.

#3 ElecPwrPEOct11

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Posted 19 October 2011 - 03:48 PM

^ 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.

#4 ElecPwrPEOct11

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Posted 20 October 2011 - 05:05 PM

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.

#5 Flyer_PE

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Posted 20 October 2011 - 05:17 PM

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.

#6 ElecPwrPEOct11

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Posted 20 October 2011 - 05:31 PM

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.

#7 rick.conner

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Posted 24 October 2011 - 09:30 PM

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

#8 rick.conner

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Posted 26 October 2011 - 08:12 PM

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?

#9 ElecPwrPEOct11

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Posted 26 October 2011 - 08:29 PM

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.

#10 Insaf

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Posted 26 October 2011 - 11:50 PM

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,

#11 rick.conner

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Posted 27 October 2011 - 12:10 AM

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!

#12 rick.conner

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Posted 27 October 2011 - 12:10 AM

thanks for your guys help

#13 BH_Cubed

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Posted 03 April 2012 - 01:03 AM

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?

#14 robertplant22

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Posted 03 April 2012 - 02:32 AM

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

#15 Flyer_PE

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Posted 03 April 2012 - 11:30 AM

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

#16 BH_Cubed

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Posted 03 April 2012 - 11:31 PM

Flyer_PE,

Thank you for the clarification.

#17 BH_Cubed

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Posted 03 April 2012 - 11:37 PM

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.

#18 robertplant22

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Posted 04 April 2012 - 12:15 AM

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!

#19 Flyer_PE

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Posted 04 April 2012 - 03:01 AM

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.

#20 DK PE

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Posted 04 April 2012 - 03:11 AM

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?

#21 robertplant22

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Posted 04 April 2012 - 02:44 PM

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!

#22 Flyer_PE

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Posted 04 April 2012 - 02:53 PM

^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.

#23 robertplant22

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Posted 04 April 2012 - 04:57 PM

OK! that one made the light bulb turn on.

Thanks Flyer_PE!

#24 bdoug

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Posted 04 April 2012 - 06:38 PM

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.

#25 Flyer_PE

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Posted 04 April 2012 - 06:48 PM

^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.

#26 katag

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Posted 26 September 2012 - 02:42 PM

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!

#27 Flyer_PE

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Posted 26 September 2012 - 04:23 PM

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.
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#28 Spark-E

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Posted 24 October 2013 - 12:55 AM

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.



#29 Blink

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Posted 24 October 2013 - 12:58 AM

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







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