Single phase to ground fault calculation

[redox]

I am doing a three phase and a single phase to ground fault calculation on the secondary side of a transformer. But the answers I obtain from my software-program and the answers from my hand-calculations (invinite bus) are not the same - somewhere I am doing something wrong.

Trsf data:

Rating - 2MVA
Primary - 11kV
Secondary - 400V
%X - 6.56%
Dyn11 - secondary solidly grounded
Fault each time on the secondary bushings.

The three phase fault calculation is no problem, the hand-calcs and the software-calcs are almost the same:
Hand - 44kA
Software - 43kA


My problem is on the single phase to ground fault scenario.

With the software I obtain 56kA - This is way higher than the three-phase fault?
With hand calculations, asuming the positive, negative and zero sequence impedances are the same, I obtain 14.7kA?

Thanks for any help

Regards
R

 

[rcw retired EE]

redox- I think you calculated Io, the zero sequence current for a phase to ground fault. I got the same answer, 14.7 kA for Io.

But the actual fault current on a single phase to ground fault is 3Io = 3 x 14.7 = 44 kA.

The difference in our quickie hand calcs (I used the MVA method) and your program is probably due to how the program treats the transformer zero sequence impedance. A ground fault at a delta-wye transformer will usually be larger than the three phase since the high voltage system zero sequence impedance is not in the circuit.

Search the site, there are a lot of threads on this subject.

Good Luck, Bob W.

 

[victa1288]

Which Software did you use? I suggest you to check the data input on transformer page if you use ETAP.Usually the hand calculation result should be very close to Software calculation result. I will check with my software tomorrow to see if there is anythng wrong.

 

[redox]

Thanks for both of you.

The intention of my hand-calculations is to ensure the software calculations are correct. Kind of a double-check or a check-up.

Literature indicate:
Ia1 = Ia2 = Iao = Ea / (Z1 +Z2+ Zo)

Do I have to add the three together?
Ia1 + Ia2 + Iao = Ifault

And what will the actual fault-current be in the faulted phase?

Io or 3Io?

Thanks.

NS:
I am using Digsilent.

 

[jghrist]

Actual fault in the faulted phase is 3Io. Is there any motor contribution?

 

[redox]

jghrist:

No motor contributions. See it as a transformer feeding a bus, with the incomer to the bus open. Thus, just the transformer energized.

Now I understand that that the fault-current would be 3Io, but why are the software-calculated current so much higher than the hand-calculated current? Ain't the hand-calculations more conservative?

 

[wareagle]

Jcrist post "Actual fault in the faulted phase is 3Io."
J
I agree with with you answer but why is it 3 times Io?
Also if the is a grounding resistor in the system it is
3 times Zg.

 

[jghrist]

1. It is 3·Io because I1=I2=Io and Ia=I1+I2+Io.

2. OP shows secondary solidly grounded.

3. The program may be assuming a core form transformer with Zo=0.36·Z1. This seems awfully low. Westinghouse T&D Manual indicates a typical Zo/Z1 ratio of 0.85 for core form transformers. Three single phase or shell form three phase transformers would have Zo=Z1.

 

[wareagle]

Jgcrist
i know it is solidly grounded. however if you had a
Zg, the caculations would use 3Zg. Questions is about the
3Zg.

 

[victa1288]

Redox,

Sorry for the late response. I was swamped at work during the weekdays.

I did calc. by ETAP and by hands and got the same results, which are 44kA(3phase, L-G). I was assuming that you input a lower zero sequence impedance for the transformer. Check your inputs and do it again. the sooftware can't be wrong. Only your inputs can make a difference.