# Problem 540 NCEES



## roadrunner (Mar 30, 2010)

Can anyone give me the solution using the MVA method for this problem?!?!?

Quick description if you don't have the book:

A generating unit is shown in the figure. The 3 phase, short-circuit MVA contribution of G1 for a 230 kV bus is most nearly: Answer is 2290

figure layed out as G1-------------T1 (xfmr)---------(230 kv Bus)------Load

G1: 3phase, 834 MVA, 22 kV, X'd = 23%, 0.85 lagging

T1: 3phase, 933 MVA, 22/230 kV, Z = 15%, Delta/GRD-WYE


----------



## nuclear bus (Mar 30, 2010)

Try using the same method as on problem 530, but instead of %Z you're having to use subtransient reactance.


----------



## roadrunner (Mar 30, 2010)

Thanks but that is what I am doing but not getting the correct answer. I do most of my short circuit calcs using the MVA method but this one will not come out correctly. Could it be because I am given a impedance value for the transformer and a reactance value for the generator?? I thought i had the MVA method down locked but now i am simply confused


----------



## nmh0408 (Mar 31, 2010)

roadrunner said:


> Can anyone give me the solution using the MVA method for this problem?!?!?
> Quick description if you don't have the book:
> 
> A generating unit is shown in the figure. The 3 phase, short-circuit MVA contribution of G1 for a 230 kV bus is most nearly: Answer is 2290
> ...


MVA Method:

During fault condition:

G1: MVA1=834/.23=3626

T1: MVA2=933/.15=6220

Now two MVA in series, the equivalent MVA=(MVA1xMVA2)/(MVA1+MVA2)=3626x6220/(3626+6220)=2290.65 MVA, Answer is B

If you had to find the three phase fault, you would take the fault MVA and divide by square root of three multiplied by the line voltage.

Hope this helps!


----------



## threatta (Apr 11, 2010)

Can someone explain how this is just the short-circuit contribution of the generator and not the entire available short-circuit current available at the 230-kV bus? Are they one and the same?


----------



## nuclear bus (Apr 11, 2010)

threatta said:


> Can someone explain how this is just the short-circuit contribution of the generator and not the entire available short-circuit current available at the 230-kV bus? Are they one and the same?


I think it's one and the same. From the information given, G1 is only source feeding the 230 kV bus, therefore the only current available is originating from the G1 feed.


----------



## pelaw (Apr 11, 2010)

I think this problem is solved using MVA method.

1. Both generators are treated as reactors: only reactive components are considered in impedance.

2. For generators/transformers in series, reactive impedances are in series.

3. Arbitrary MVA is chosen equal to G1's 834 MVA.

4. For G1, the impedance is: Z new = Z old 0.23 (MVA arb 834 / MVA old 834) = 0.23

5. For T1: Z new = Z old 0.15 x (MVA arb 834 / MVA old 933) = 0.134

6. Z total = in series 0.23 of G1 + 0.134 of T1 = 0.364

7. Power contribution S pu= Vpu^2/Zpu = 1^2 / 0.364 = 2.747

8. S actual = S pu * S base = 2.747 * 834 = 2291


----------



## pelaw (Apr 11, 2010)

threatta said:


> Can someone explain how this is just the short-circuit contribution of the generator and not the entire available short-circuit current available at the 230-kV bus? Are they one and the same?



1. the problem asks for G1's MVA contribution, which is MVA flowing from the G1 only.

2. The load is not considered, which would be the case at the bus; and that load would be paralleled with G1 so it would be a separate source if motors were on the load side.

3. If the problem asked for MVA at the bus, then the bus voltage 230kV would be base since the current is transformed at the transformer. The result should be the same since no load information is provided.


----------



## gt2690b (Apr 14, 2010)

nmh0408 said:


> roadrunner said:
> 
> 
> > Can anyone give me the solution using the MVA method for this problem?!?!?
> ...


The problem does ask for the three phase fault


----------



## cbinla (Oct 10, 2010)

I take it that the information they give on the generator regarding the .85 lagging power factor is not important. And this is because this is the power factor of the system without a short circuit, correct?

They give the transient reactance of the generator and the impedance of the transformer. Therefore it is necessary to assume the impedance of the transformer is all reactance, correct? Therefore the angle of the impedances of both the generator and the transformer is the same and is &lt;90 degrees, and therefore you can solve it with just using the p.u. magnitudes without considering the phase angles of the impedances.

You have to make this assumption because they don't give you the impedances phase angles. Is this correct?

Thanks


----------



## ryanbabs (Oct 25, 2010)

To me, this problem is worded very poorly. The way it reads, the MVA contribution is taken only for G1, which would be 834/23% = 3626MVA. The way I understand the question is that they are asking for the G1 MVA contribution ONLY for the fault at the 230kV bus. Of course the total fault includes both G1 and T1.

However, the solution includes the transformer contribution also.


----------



## megavar (Oct 25, 2010)

ryanbabs said:


> To me, this problem is worded very poorly. The way it reads, the MVA contribution is taken only for G1, which would be 834/23% = 3626MVA. The way I understand the question is that they are asking for the G1 MVA contribution ONLY for the fault at the 230kV bus. Of course the total fault includes both G1 and T1.
> However, the solution includes the transformer contribution also.



the following article help me.

http://www.brainfiller.com/documents/Trans...pedance_000.pdf


----------



## dianevp (Apr 1, 2011)

pelaw said:


> threatta said:
> 
> 
> > Can someone explain how this is just the short-circuit contribution of the generator and not the entire available short-circuit current available at the 230-kV bus? Are they one and the same?
> ...


Just out of curiosity, what would be the case if there was a load(motor) provided in the info? How would the MVA method differ?

Thanks for your time!


----------



## Dolphin P.E. (Apr 1, 2011)

dianevp said:


> pelaw said:
> 
> 
> > threatta said:
> ...


In this case, the total MVAsc = (MVAsc,G // MVAsc,T) + MVAsc,M


----------



## dianevp (Apr 1, 2011)

dolphin said:


> dianevp said:
> 
> 
> > pelaw said:
> ...


Thank you for the reply!

So I understand, in other words, the MVAsc, M is added because it "feeds" into the other side of the bus, as if it was parallel to (MVAsc,G //MVAsc,T)?

On more question, is there ever a case where the MVAsc's are subtracted?

Thanks again!


----------



## Dolphin P.E. (Apr 1, 2011)

dianevp said:


> dolphin said:
> 
> 
> > dianevp said:
> ...


MVAsc,m is actually in parallel to the equivanent MVAsc of the generator and the transformer. So, it has to be added to the equivanent MVAsc of the generator and the transformer (which is true as the motor acts as a generator and contributes to the fault current). The fault is at the bus which has the generator and the transformer on one side and the Motor on the other side, so they are in parallel with respect to the fault location.

No subtraction in the MVA method. But keep in mind when you have the MVAsc's in series, the equivalent MVAsc is less than the smallest.

It opens up a point here, providing additional 1:1 ration transformer just to bring down the KAIC rating of the downstream electrical equipment. It could be a very cost effective way, specially when you have enormous amount of downstream panels and breakers along with a limited space available. As the KAIC rating goes higher, the cost and physical dimentions increase significantly.


----------

