# Source Impedance of a Utility



## cos90 (Jul 11, 2017)

I'll post the exact question from CI when I get home, but I had a problem with calculating the source impedance of a utility.

I found a link with the formula. http://iaeimagazine.org/magazine/2015/07/07/calculating-short-circuit-current/

The link says the source impedance is 


%Z Utility


=  KVA Transformer  × 100


 


(Isc Utility) × (√3) × (kV Primary)


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## cos90 (Jul 11, 2017)

My question is where does this formula come from?

Additionally this formula:


Isc


= (Transformer KVA) × 100)


 


(√3) × (Secondary KV) ×  [(%Ztransformer)+(%Z Utility)]

I get the feeling because I dont understand this there are still some concepts of Per-Unit calculations I don't understand.


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## cos90 (Jul 11, 2017)

I think a similar statement of the formula is in this SEL paper, formula 1: https://cdn.selinc.com/assets/Literature/Publications/Technical Papers/6250_ApplicationExisting_JB-KZ_20070214_Web.pdf?v=20160511-160915

This is the change of base formula 







But there is a twist. Why is Z,PU,given 100????

Does anyone have a reference that covers this subject well? Nothing I can find on the internet or my books.


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## rg1 (Jul 11, 2017)

cos90 said:


> I'll post the exact question from CI when I get home, but I had a problem with calculating the source impedance of a utility.
> 
> I found a link with the formula. http://iaeimagazine.org/magazine/2015/07/07/calculating-short-circuit-current/
> 
> ...


Let %Zu= %Z of Utilty; Za=Actual Z of Utilty; Vbase= Vb=Primary V of Xmer; Ibase=Ib= Primary I of Xmer; Zbase=Vb/Ib; Isc= Shot Ckt. current of Utilty at Primary of Xmer

Now

Za=Vb/Isc

%Z= Za/Zb= (Vb/Isc)/(Vb/Ib)=Vb*Ib/Vb*Isc=Sqrt3*Vb*Ib/Sqrt3*Vb*Isc

=KVA of Xmer/Isc*sqrt3*kV Primary

Does it make some sense. I am not able to upload my hand made notes.


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## rg1 (Jul 11, 2017)

cos90 said:


> My question is where does this formula come from?
> 
> Additionally this formula:
> 
> ...


Here

Isc= Fault Current at Secondary of Xmer; Zu= Z of Utulity on secondary side of Xmer; Zt= Z of Xmer;  V1, I1=Primary of Xmer and V2, I2= Secondary of Xmer; Zactual= %Z*Zbase;

Zbase=V2/I2

Total Z(Actual)= Zu+Zt (Actuals)

Isc (Secondary of Xmer)= V2/Total Z (Actual)=V2/%(Zu+Zt)*Zbase=V2/%(Zu+Zt)*(V2/I2)=

=V2*I2/%(Zu+Zt)*V2=  Sqrt3*V2*i2/sqrt3*V2*%(Zu+Zt)

=KVA of Xmer/sqrt3*Secondary kV* %(Zu+Zt)

Could you get it, I can add further explanation to it.


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## TNPE (Jul 11, 2017)

If you're looking from the point of view of the secondary of the station XFMR down line, you have to develop the zero sequence impedance for the ground grid.

The ISC for the transformer you gave above concerns a bolted 3-phase fault at the secondary terminals of the XFMR, and that is the correct expression.


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## rg1 (Jul 11, 2017)

cos90 said:


> I think a similar statement of the formula is in this SEL paper, formula 1: https://cdn.selinc.com/assets/Literature/Publications/Technical Papers/6250_ApplicationExisting_JB-KZ_20070214_Web.pdf?v=20160511-160915
> 
> This is the change of base formula
> 
> ...


The explanation of pu is similar everywhere, its application is a bit tricky. I understand it like this. Bases are assumed, conventionally they are rated  values of V and I. PU or percent of any quantity is ratio of its actual value to its base value.

I have solved equations given in your previous posts from these basics only. I think they are enough for deriving all formula pertaining to pu system. Some mention of pu  is there in Wildi book as is there in any other book. May be that is useful to you.


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## rg1 (Jul 11, 2017)

TNPE said:


> If you're looking from the point of view of the secondary of the station XFMR down line, you have to develop the zero sequence impedance for the ground grid.
> 
> The ISC for the transformer you gave above concerns a bolted 3-phase fault at the secondary terminals of the XFMR, and that is the correct expression.


I picked up the formula from the link shared by @cos90 and yes in that link, author means Isc as fault current at Secondary of a Xmer. Author has assumed one utility, connected to a Xmer. He has mentioned the formula without explaining it. I suppose as of now Cos90 is interested only in understanding the pu system. Yes, I agree,  unbalanced faults will involve sequence components.


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## TNPE (Jul 11, 2017)

And every fault outside of a bolted 3-phase fault is unsymmetric, which means almost every single fault on a real power system will be of the unsymmetric type.


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## cos90 (Jul 12, 2017)

Thank you very much rg1! Your derivation is very helpful. I will look in the Wildi book. I have not used it much and can never find what I'm looking for in there.

This is from CI Test 4 Question 78 in case anyone is wondering.


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## cos90 (Jul 12, 2017)

TNPE said:


> If you're looking from the point of view of the secondary of the station XFMR down line, you have to develop the zero sequence impedance for the ground grid.
> 
> The ISC for the transformer you gave above concerns a bolted 3-phase fault at the secondary terminals of the XFMR, and that is the correct expression.


What's the best resource for this intuition of modeling faults?

The books either don't go enough into detail or see the problem as a good excuse to practice matrix computation.

Are there professional seminars on this subject which don't involve going back to graduate school?

I'm don't need this knowledge for my job because I don't work in power system protection, but I don't like having this gap in knowledge when I work in the utility field.


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## cos90 (Jul 12, 2017)

I think I should order a copy of Blackburn for after the Exam. I hope it doesn't just fill the pages with matrix manipulation like Power Systems Analysis by Grainger and Stevenson.


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## rg1 (Jul 12, 2017)

Cos90 -The explanation of pu is similar everywhere, its application is a bit tricky. I understand it like this. Bases are assumed, conventionally they are rated  values of V and I. PU or percent of any quantity is ratio of its actual value to its base value.

I do it only from above understanding. The concept is this much only, rest is maths, so I am not sure whether any book will help you.

If you can start with this and start doing the stuff your self, you may yourself understand it. You can start doing the problem yourself and when you get stuck, can ask me. I will clear your concept. I am sure. 

Now for unsymmetrical faults PM me your contact details I may share something which may be useful to you. I would have loved to post the stuff here but for my inability to do it.


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## TNPE (Jul 12, 2017)

Let's plug in some numbers for a bolted 3-phase fault condition at the secondary terminals:

161 kV - 13.2 kV, 20 MVA, 7%Z

20MVA/(sqrt3*13.2) = 874.77 A @ full load on secondary 

IFLA/.07 = 12496.71 A of available fault current at the secondary terminals

As for your concerns about resources, Chapman does a good job of building the backbone for fault analysis, and subsequently applying these fundamentals to symmetric and asymmetric faults.  I used his text and Graffeo almost exclusively when I took the PE, other than NEC and NESC where necessary.  Chapman provides the fundamentals and applicable proofs, but it is not heavy on cumbersome, academic theory that is customarily expected.  That said, it was one of my college texts and is a well-written book.  You can find a solution for most applications involving power systems and electric machines in this book.


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