Short Circuit Current
- Thread starter MLbS
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I recognize this problem from the Cram for Exam volumes...
You would have to include the transformer if the secondary side also has an energized generator or motor... That is not clearly stated in this problem.
If the secondary does not have another energized generator or motor, then you wouldn't have to include the transformer, since the 3-phase fault closes the ground fault path between the generators and where the fault is located.
That is how I would normally understand it...
1. Yes, the transformer contributes to the three-phase fault on the delta (50kV) side.
2. I don't know. I thought I had the MVA method down pretty well, but I'm not getting any of the answers listed, although this appears to be a very basic MVA/3-phase fault type of problem.
The first thing I do is draw an MVA diagram. Calculate the MVA contributions for each component (see image below). We have 2 generators in parallel in series with a transformer. For parallel you just add them together and you get 100MVA for the 2 generators. Since they're in series with the transformer, you use the formula (MVA1xMVA2)/(MVA1+MVA2). I get 50MVA total.
Since the short circuit current fault contribution is on the delta side, Vp=Vl = 50kV, so I'm getting 50MVA/50kV (otherwise it would be 50MVA/(1.73x50kV). I get 1000A, but even if I used 1.73 I would not get any of the answer choices, so I'm a bit stumped on this one.
@Cram For The PE what am I missing here?
2. I don't know. I thought I had the MVA method down pretty well, but I'm not getting any of the answers listed, although this appears to be a very basic MVA/3-phase fault type of problem.
The first thing I do is draw an MVA diagram. Calculate the MVA contributions for each component (see image below). We have 2 generators in parallel in series with a transformer. For parallel you just add them together and you get 100MVA for the 2 generators. Since they're in series with the transformer, you use the formula (MVA1xMVA2)/(MVA1+MVA2). I get 50MVA total.
Since the short circuit current fault contribution is on the delta side, Vp=Vl = 50kV, so I'm getting 50MVA/50kV (otherwise it would be 50MVA/(1.73x50kV). I get 1000A, but even if I used 1.73 I would not get any of the answer choices, so I'm a bit stumped on this one.
@Cram For The PE what am I missing here?
Since the fault is in between the 2 generators in parallel and the transformer, the 2 gens and the transformer are in parallel with respect to the fault. So you have to add 100 MVA from the 2 parallel gens, and the 100 MVA from the transformer.
If 2 generators are in parallel, you add their MVA fault contributions together (MVA1 + MVA2), so 33.33MVA + 66.67MVA = 100MVA.
Oh, but you're saying since the 2 generators are in parallel with the fault (not in series), then you ADD them, so the total fault contribution is 100 + 100 = 200MVA
Then the short circuit current is 200MVA/(1.73 x 50kVA) = 2309.4 Amps.
Is that correct? My error was looking at the 2 generators as if they were in series instead of parallel.
Oh, but you're saying since the 2 generators are in parallel with the fault (not in series), then you ADD them, so the total fault contribution is 100 + 100 = 200MVA
Then the short circuit current is 200MVA/(1.73 x 50kVA) = 2309.4 Amps.
Is that correct? My error was looking at the 2 generators as if they were in series instead of parallel.
I agree with you. I was getting confused. I think we don’t need to see TR impedance as long as we don’t have generator or motor.
Thanks @Mbsamani! Yes see my post above. I was looking at the generators and transformers as being in series, not parallel. That was the main source of my error.
Also, it was tempting for me to look at the delta secondary and leave out the 1.73 (sq. rt. 3), but we're solving for current so we have to use it.
Thanks for posting!
Why did you contribute transformer in fault current?
actually I got your solution but I did get why you used the transformer because we don’t have generator on the secondary side.
You always include the transformer contribution (or any component along the fault).
See article below for reference.
https://arcadvisor.com/faq/mva-method-short-circuit-fault-current-calculations
It's kind of odd to me, I remember seeing a few practice questions in which a 3-phase fault is located in the middle of the line and the solution only considered the elements (generator, line, transformer) from the source to where the fault is. The other side of the fault is completely dismissed if there is no generator or energized element since the idea is that the other side won't contribute to the fault without an energized source...
When I have time, I'll try to pull out and show such an example question...
