How to calculate the SLG kA contributes from Utility

[ciy]

Hi,

I run the short circuit in SKM.

And, I fill in the SCC contribute from Utility in MVA as below.

3 phase = 8000MVA

Line to Ground= 3000MVA

I got the kA value from SKM are 20.082kA and 22.654kA base on 230kV respectively.

If(3p)=MVA(3p)/(1.732*VL-L)=8000/(1.732*230)=20.082kA---OK!

But, I am not sure how to calculate the fault current of Line to Ground.

Please advise! Thanks in advance.

 

[dpc]

The X/R ratio does not effect the magnitude of the symmetrical fault current calculations. It only impacts the asymmetrical current (and breaker SC ratings).

David Castor

http://www.cvoes.com

 

[odlanor]

dpc,
you're right. I thought X / R was taken from the utility, without discussion.

 

[dpc]

It does have to provided by the utility, but its impact is limited to the asymmetrical current for short-circuit calculations.



David Castor

http://www.cvoes.com

 

[VTer]

As others posted above, voltage line to line is 230kV so the LG voltage is 230kV/sqrt(3)=132.8kV
3000MVA/132.8kV=22.6kA

I don't understand the confusion?

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla

 

[dpc]

I don't understand the confusion?

The confusion is that the convention (in the US) for defining utility SLG fault MVA does not agree with your equation. General consensus is that the correct current is 7.53 kA.

This is consistent with results obtained from several reputable software programs using the same data input.



David Castor

http://www.cvoes.com

 

[rbulsara]

Well not quite, with all due respect.

The consensus is that many (not all) Power Analysis software "display" 3 phase equivalent fault MVA ratings even for a SLG fault (to check breaker ratings), which for a given SLG ground fault current will be 3 times the SLG fault MVA or SQRt(3)*VLL*I.

This does not make it a correct formula to calculate MVA or current of a single phase circuit, faulted or not.

This also does not make the 7.5 kA to be the "correct" value, automatically in this case. It depends upon what the supplier of the MVA value meant. This is no doubt a unnecessary confusion.

As for the software, their manual should and do explain what they mean.








.

Rafiq Bulsara

http://www.srengineersct.com

 

[dpc]

I agree it is a source of confusion.

FWIW, in Conrad St. Pierre's book A Practical Guide to Short Circuit Calculations, he uses the same equation I used above to calculate SLG Fault MVA.

Dave



David Castor

http://www.cvoes.com

 

[stevenal]

No controversy in '07
thread238-194085

 

[rbulsara]

It is more a case of improper naming of the quantity. Those software need to rename their fields to "3-phase Equivalent of SLG MVA" rather than calling it Single Phase MVA. The two are very distinct.






Rafiq Bulsara

http://www.srengineersct.com

 

[jghrist]

And, I fill in the SCC contribute from Utility in MVA as below. 3 phase = 8000MVA Line to Ground= 3000MVA

IMHO, the fault is with the utility giving the SCC contribution in this manner. They should give complex fault currents or current magnitude with X/R ratio and/or complex system impedances with the base voltage and power indicated.

 

[ciy]

Dear All,

Thanks for your advices.

Please refer to the attached that calculated method come from SKM.

And I transfer to the mentioned 230kV case, the result as below is very close to SLG value shown on the SKM map.

Utility per unit impedance on a 100 MVA base.

Positive sequence = Z1 = Z2 =0.001172+j0.012445

Zero sequence = Z0 = 0.000781 +j0.008297

Zt = Z1 + Z2 + Z0 = 0.003125 +j0.033187 =0.033334?

Base current at the Utility is { 100MVA/(1.732)(230 kV)} = 0.25102922kA

SLG = {3*1/(Z1+Z2+Z0)} = 89.99872497

I_SLG = Ib*pu = 22.59230971kA

For reference.

 

[jghrist]

ciy,
This is the correct calculation. If you have the system impedance from the utility, why did you use fault MVA in your first post and where did that come from?

Note that the example in your attachement has pos-, neg-, and zero-sequence impedances equal so 3Ø and 1Ø faults are identical - not a very useful example.

 

[VTer]

jghrist, I think the system impedance has been calculated by SKM based on the input MVA and X/R with respect to base MVA and V.

"Throughout space there is energy. Is this energy static or kinetic! If static our hopes are in vain; if kinetic — and this we know it is, for certain — then it is a mere question of time when men will succeed in attaching their machinery to the very wheelwork of nature". – Nikola Tesla

 

[stevenal]

It matters not how SKM interprets the data. What matters is how the utility arrived at it. Go back and ask. Also ask for fault current in amps and impedance data just to be sure.

If SLG amps are higher than 3LG, or zero sequence impedance is lower than positive; you had better be close to a delta-wye transformation.

 

[jghrist]

jghrist, I think the system impedance has been calculated by SKM based on the input MVA and X/R with respect to base MVA and V.

I see that now. I fall back on my earlier statement that you shouldn't get the fault information from the utility in terms of fault MVA - it leads to confusion. It is still dumb of SKM to illustrate the calculation of SLG fault current with an example where Z1 = Z2 = Z0 and 3P = SLG.