Short Circuit MVA Method

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Byk

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I am wondering if problem below can be solved using MVA method? Can we assume that Sbase is the actual value of S?

A customer is to be served from a 50 MVA, 115/13.8 kV transformer which has an 
impedance of 7.5% on a 50 MVA base. The 115 kV bus has a system equivalent 
impedance of 1.5 + j3.2 % on a 100 MVA base. The distribution line serving the 
customer has an impedance of 5.0 + j15.0 % on a 100 MVA base. Assuming the 
customer short-circuit contribution to be nil, using a 100 MVA base determine the 
minimum short-circuit rating in MVA for the circuit breaker to be installed by the 
customer.


 
I am wondering if problem below can be solved using MVA method? Can we assume that Sbase is the actual value of S?

A customer is to be served from a 50 MVA, 115/13.8 kV transformer which has an 
impedance of 7.5% on a 50 MVA base. The 115 kV bus has a system equivalent 
impedance of 1.5 + j3.2 % on a 100 MVA base. The distribution line serving the 
customer has an impedance of 5.0 + j15.0 % on a 100 MVA base. Assuming the 
customer short-circuit contribution to be nil, using a 100 MVA base determine the 
minimum short-circuit rating in MVA for the circuit breaker to be installed by the 
customer.
I'm not sure if you can use the S base as the rated S when using the S SC = S rated / Z pu formula for the MVA method...

But you are given the rated S and pu impedance of the transformer. The percent impedance can easily be converted to pu impedance. So that takes care of S xfmr SC.

And for the lines... Typically you can calculate the short circuit VA for the lines using S line SC = V line LL^2 / Z line actual. It's not 100% clear to me, but I think the distribution line that feeds the customer would be 13.8 KV based on the transformer secondary...

Those are my two cents based on a quick glance of this problem statement.

 
Though the more I think about it, since they give you the percent impedances, I think it might be a bit easier to solve this using the per-unit method. They state a 100 MVA base, so you would have to convert the 50 MVA rated transformer impedance to a new p.u. impedance based on a 100 MVA base.

Hope my 2 posts help any.

 
@akyip That's exactly what I did, I converted the xfmr impedance to 100MVA base and just added them.

I just wanted to pick someone else's brain to see if they would think of going for the MVA method on such a problem.

 
@akyip That's exactly what I did, I converted the xfmr impedance to 100MVA base and just added them.

I just wanted to pick someone else's brain to see if they would think of going for the MVA method on such a problem.
The more I think about it... I don't think it's appropriate to use the S base as the rated S in the S SC = S rated / Z pu formula. The reason I say that is because the S base is not necessarily equal to the rated S of a piece of equipment.

 
Byk, I agree with what akyip said above.  Another thing to keep in mind is the MVA works for three-phase symmetrical faults.  Other faults (single line to ground, line to line, double line to ground) will require the per-unit method.

See below link to a very good article/video from @Zach Stone, P.E.:






 
@DuranDuran that is an excellent point! I do not know why I simply assumed it was 3phase.
That's an easy trap to fall into.  In regards to fault current analysis, we tend to favor one method over another (MVA vs PU).  It's tempting to see a problem which (at first) might scream "MVA Method!" when you first see it, but it will lead you to the wrong answer.  Zach Stone posted an example of this once.  So yes, please make sure the problem states "3 phase fault" or "3 phase bolted fault".  

NOTE:  I think the MVA method can be used to solve for phase to ground faults.  Could someone weigh in on this?  Thanks!

https://www.jmpangseah.com/wp-content/uploads/2003/01/chapter-5.pdf

 
Byk, I agree with what akyip said above.  Another thing to keep in mind is the MVA works for three-phase symmetrical faults.  Other faults (single line to ground, line to line, double line to ground) will require the per-unit method.

See below link to a very good article/video from @Zach Stone, P.E.:
DuranDuran,

I definitely agree with you that the MVA method is used only for 3-phase faults.

But something I picked up and thought about in the question/problem statement: the question asks for the MINIMUM SHORT-CIRCUIT rating of the short circuit breaker. I would think you actually want to size the minimum short-circuit rating of the breaker based on the 3-phase fault conditions, because the 3-phase fault is the most severe and worst-case fault.

I would think you want to size the minimum service breaker based on the maximum available fault current from a 3-phase fault.

Though I should clarify: you can still use the per-unit method for a 3-phase fault. And I think in this problem that is the better way to go since it gives the per-unit/percent impedances...

 
Last edited by a moderator:
DuranDuran,

I definitely agree with you that the MVA method is used only for 3-phase faults.

But something I picked up and thought about in the question/problem statement: the question asks for the MINIMUM SHORT-CIRCUIT rating of the short circuit breaker. I would think you actually want to size the minimum short-circuit rating of the breaker based on the 3-phase fault conditions, because the 3-phase fault is the most severe and worst-case fault.

I would think you want to size the minimum service breaker based on the maximum available fault current from a 3-phase fault.
Yes, it does seem counter-intuitive.  You want to solve for worst case.  So yes 3-phase fault is worst case (and also very uncommon, maybe the most uncommon if I remember correctly?  But also most potentially damaging?)  So let's say you're doing a fault current analysis and you come up with 9,500 A for the short circuit current.  You would want to choose a 10kAIC (10,000A Available Interrupting Current) rated breaker, right? 10kAIC is a standard rating for panelboards/CBs. You could choose a 22kAIC or 65kAIC (also common), but if the problem asks for the minimum, go with the 10kAIC.

I hope my thinking is correct.  Anyone please weigh-in if otherwise.

 
Yes, it does seem counter-intuitive.  You want to solve for worst case.  So yes 3-phase fault is worst case (and also very uncommon, maybe the most uncommon if I remember correctly?  But also most potentially damaging?)  So let's say you're doing a fault current analysis and you come up with 9,500 A for the short circuit current.  You would want to choose a 10kAIC (10,000A Available Interrupting Current) rated breaker, right? 10kAIC is a standard rating for panelboards/CBs. You could choose a 22kAIC or 65kAIC (also common), but if the problem asks for the minimum, go with the 10kAIC.

I hope my thinking is correct.  Anyone please weigh-in if otherwise.
I agree with what you said.

Regarding the least severe vs. the most severe fault:

Single line to ground fault is least severe, yet most common.

Three-phase fault is most severe, yet least common. (Also, a 3-phase fault and a 3-phase-to-ground fault are practically the same fault.)

Double-line-to-ground fault and line-to-line fault are somewhere in between. From what I remember, I think their severity depends on several parameters such as the impedances and the voltage source. I don't remember if they are fixed in stone, I just know that they are generally more severe than a single-line-to-ground fault but less severe than a three-phase fault.

 
First let me correct a common misconception. The three phase fault in MOST cases is the most severe. This is not always true.

The SLG fault will be worse in cases where:
1-the generators have a solidly grounded neutral or low impedance neutral.
2-on the Y grounded side of a delta-wye grounded transformer.

For practical purposes when solving these problems or when sizing your equipment, you can assume the three phase fault to be the worst. However, it is technically incorrect to say this is always true.

Whenever your trying to size something for short circuit ratings, you assume worse case that the transformer is supplying all of its possible fault current. This is very conservative. But when you solve these problems this is what you do.

 
Why is the question asking for the minimum short circuit rating in MVA? Is that the right unit?
I have seen the 3-phase short-circuit rating expressed in both VA (volt-amperes) and A (amperes).

For VA, I guess it's because you may want to see the total available short-circuit fault power flowing from a source to a fault at any given point. And remember, the short-circuit apparent power in VA is the same amount on both sides of a given transformer (S1 = S2 for a transformer applies to fault or short-circuit power).

 
First let me correct a common misconception. The three phase fault in MOST cases is the most severe. This is not always true.

The SLG fault will be worse in cases where:
1-the generators have a solidly grounded neutral or low impedance neutral.
2-on the Y grounded side of a delta-wye grounded transformer.

For practical purposes when solving these problems or when sizing your equipment, you can assume the three phase fault to be the worst. However, it is technically incorrect to say this is always true.

Whenever your trying to size something for short circuit ratings, you assume worse case that the transformer is supplying all of its possible fault current. This is very conservative. But when you solve these problems this is what you do.
Thanks for this clarification, Cram for Exam.

Now that you mention it, I do remember the Wildi book also mentioned this in regards to generators - why they are not solidly grounded, due to the possible very large amount of fault current.

 
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