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How to calculate the SLG kA contributes from Utility

How to calculate the SLG kA contributes from Utility

How to calculate the SLG kA contributes from Utility

(OP)
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.

RE: How to calculate the SLG kA contributes from Utility

Hint: It is a single phase circuit.
Use single phase voltage, L-N and formula.

Rafiq Bulsara
http://www.srengineersct.com

RE: How to calculate the SLG kA contributes from Utility

ciy,
Use the same equation but with 3000 MVA/ 230,000 V.Then the 1-phase to ground fault current will be 7530.8 A.

RE: How to calculate the SLG kA contributes from Utility

Kiribanda:
You may want to try that again...with L-G voltage.

Rafiq Bulsara
http://www.srengineersct.com

RE: How to calculate the SLG kA contributes from Utility

ciy,
Could you please give us the single phase fault current which is given out from your software program when you key-in single phase fault level of 3000 MVA?

RE: How to calculate the SLG kA contributes from Utility

MVA3sc= 8000 MVA
MVA1sc= 3000 MVA
Vb=230kV
calculation symmetrical components reactance in per unit base 100MVA-230kV:
X1= X2=100/8000= 0.0125 pu
Xg=3*100/3000=   0.1000
X0= 0.1000 - 2*0.0125 = 0.0750 pu
conclusion:
X1= X2= 0.0125 pu base 100MVA-230kV
X0= 0.0750 pu base 100MVA-230kV
 

RE: How to calculate the SLG kA contributes from Utility

Is a single phase amp calculation from a given MVA and voltage that complex and difficult?

 

Rafiq Bulsara
http://www.srengineersct.com

RE: How to calculate the SLG kA contributes from Utility

ciy,
Sorry for coming back little late.
I donot know how your program has calculated SLG=22654 A when 1-phase fault MVA fault=3000 MVA only.For me,MVA rating is just a value and has no practical sense like short circuit current rating.Therefore,whether it is 1-phase fault MVA or 3-phase fault MVA,the equation is used with line quantities and NOT with phase qauntities.Based on above,if the 1-ph fault MVA=3000 MVA @ 230 kV,then corresponding 1-ph fault current should be 7530 A only.
Also if I use the X/R=10.618 which is given in the PDF attachment,my calculation gives R0=0.0070324 pu and X0=0.084 pu,whereas your program had given R0 and X0 values with an additional zero.Please double check the default settings in your program or contact tech support to get a final reply is my advice.

RE: How to calculate the SLG kA contributes from Utility

Quote:

Use the same equation but with 3000 MVA/ 230,000 V.Then the 1-phase to ground fault current will be 7530.8 A.  
3000/[230/sqrt(3)] = 22.59

RE: How to calculate the SLG kA contributes from Utility

First of all, I dislike using MVA ratings for fault duty because I find it not very intuitive.

But in my experience, the SLG MVA rating is based on a three phase equivalent, sort of like having three SLG faults occurring in rapid succession. In either the 3LG or SLG case, multiply the current by the line to ground voltage times 3.

I believe this method was necessary when breakers were rated by MVA, simply use the higher value.  

RE: How to calculate the SLG kA contributes from Utility

rbulsara,
I am sorry for that calculation.
I only use SCC contribute from Utility in MVA to calculate short circuit in MY system. So, I have to convert to impedance.
I did not see SKM attached. With X/R we can change:

MVA3sc= 8000; % MVA
MVA1sc= 3000; % MVA
Vb=230; %kV
XR=10.618;% (X/R)

% calculation symmetrical components impedance in per unit base 100MVA-230kV:
Z1= 100/8000;
Z2=Z1;
Zg=3*100/3000;
Z0= Zg-Z2-Z1;

% Z1^2 = R1^2 + X1^2;
% Z1^2/R1^2= R1^2/R1^2 + X1^2/R1^2;
% R1^2 = Z1^2 / (1 + XR^2) ;
R1 = sqrt(Z1^2 / (1 + XR^2));
X1 =XR*R1;

% Z02 = R0^2 + X0^2;
% Z0^2/R0^2= R0^2/R0^2 + X0^2/R0^2;
% R0^2 = Z0^2 / (1 + XR^2) ;
R0 = sqrt(Z0^2 / (1 + XR^2));
X0 =XR*R0;

%base 100MVA-230kV
Z1= R1+ j * X1;=== 0.0012 + j0.0124 pu
Z0= R0+ j * X0;===>  0.0070 + j0.0747 pu
Z2=Z1;


 

RE: How to calculate the SLG kA contributes from Utility

odlanor,
Thanks odlanor for your calculation.It shows that my calculated values are in consistence with yours.Now the question is the pdf attachment posted by ciy gives different impedance values with an additional zero for the same X/R=10.618.Please check the pdf at post#6.

RE: How to calculate the SLG kA contributes from Utility

All,
See what jghrist posted. I have been trying to hint at that all along. Not sure why that was so difficult.

Single phase to ground voltage is =VLL/1.732.

3-phase short circuit analysis is always done on per phase basis.

Rafiq Bulsara
http://www.srengineersct.com

RE: How to calculate the SLG kA contributes from Utility

For 8000 MVA 3-phase and 3000 MVA SLG,

EasyPower calculation gives:  

3-Phase: 20,082 A

SLG:  7531 A

I used X/R = 50

The convention normally used for SLG MVA fault current is:

MVASLG=SQRT(3)*VLL*ILG

Note the line-to-line voltage is used.

I generally try to get information in amps or ohms to avoid this confusion.


 

David Castor
www.cvoes.com

RE: How to calculate the SLG kA contributes from Utility

I see that ETAP does the same as what dpc indicated. Not sure why, but I can understand the source of confusion.

But to me SLG or Line to Ground Fault MVA means a product of Single Line to Ground fault current and L-G voltage. Never gave it a second thought. I am with SKM, I guess.

I also agree that dealing with amps values is the way to go to avoid confusion.

Rafiq Bulsara
http://www.srengineersct.com

RE: How to calculate the SLG kA contributes from Utility

Note that sqrt(3)*Vll*I=3*Vlg*I, so dpc, Kiribanda, and I agree.

3000*MVA/(sqrt(3)*230*kV)= 7.53 kA

Counterintuitive, but see my post above for what I believe to be the history behind it. Aspen One-liner uses this convention.  

RE: How to calculate the SLG kA contributes from Utility

I suspect stevenal is correct regarding the link to the old MVA-based breaker ratings.  I had never considered that.  

 

David Castor
www.cvoes.com

RE: How to calculate the SLG kA contributes from Utility

I agree and understand that breaker MVA rating has to be SQRT(3)*VLL*Isc(where Isc is the fault line current, SLG or 3 phase), whichever is higher.

Still breaker MVA ratings and "fault MVA" could be kept separate, but as long as the manual explains what they mean by their displays I am OK.

Rafiq Bulsara
http://www.srengineersct.com

RE: How to calculate the SLG kA contributes from Utility

Kiribanda,
for me there is an inconsistency in the X0 / R0 ratio.
DPC post with X/R=50 is more radical!

RE: How to calculate the SLG kA contributes from Utility

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
www.cvoes.com

RE: How to calculate the SLG kA contributes from Utility

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

RE: How to calculate the SLG kA contributes from Utility

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

 

David Castor
www.cvoes.com

RE: How to calculate the SLG kA contributes from Utility

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
 

RE: How to calculate the SLG kA contributes from Utility

Quote:

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
www.cvoes.com

RE: How to calculate the SLG kA contributes from Utility

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

RE: How to calculate the SLG kA contributes from Utility

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
www.cvoes.com

RE: How to calculate the SLG kA contributes from Utility

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

RE: How to calculate the SLG kA contributes from Utility

Quote:

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.
 

RE: How to calculate the SLG kA contributes from Utility

(OP)
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.
 

RE: How to calculate the SLG kA contributes from Utility

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.
 

RE: How to calculate the SLG kA contributes from Utility

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
 

RE: How to calculate the SLG kA contributes from Utility

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.

RE: How to calculate the SLG kA contributes from Utility

Quote:

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.

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