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 Post subject: When Only Utility Short Circuit Current ProvidedPosted: Wed Mar 17, 2010 4:47 pm

Joined: Mon Mar 15, 2010 7:00 am
Posts: 3
I am struggling with how to calculate utility impedance (Zu) based on the short circuit current provided by the Utility or ,for my instance, from our sub station. Typically we're only provided the short circtuit current (Isc) at a bus location to start the calculation.

To solve for KVAsc I use the following equation: KVAsc=(sqrt(3)*Isc*V_ll)/1000 and to solve for impedance, I would use Z=((V_ll)^2)/(KVAsc*1000). Which I then combine and simplify but I am struggling with V_ll and which voltage I would substitute into the V_lls, or are they the same.

For instance lets say I had a 480V supply from the sub station but the fault I am analyzing is on the secondary side of the 480-240V transformer. In other words my fault is on a 240V bus. Wouldn't the KVAsc be based on 480V_ll and the Z be based on 240V_ll.

Also, I am trying to simplify the calculation by using only line-line voltage values, but this may be fouling me up. If there is a better way to approach this with logical assumptions then please let me know. This has been driving me nuts and I don't have a good resource to check my answer against since I have not been able to find information specifically about this. The information seems to talk around it but never directly upon the issue of only being provided Isc from a utilty with no supporting X/R or Z.

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 Post subject: Posted: Wed Mar 17, 2010 5:04 pm
 Plasma Level

Joined: Mon Sep 17, 2007 5:00 pm
Posts: 1635
Location: Scottsdale, Arizona
The transformer is the confusing part. Here are a few formulas.

Zsource = kVLL^2/MVAshort-circuit
kVLL is the line to line voltage at the SAME bus where the short circuit current/MVA is given. In your case if that is 480 volts it would be 0.48^2 / MVAshort-circuit

If the short circiut is on the 240 volts side but the utility provided the available short circuit current on the 480 volts side, a bit or work is needed. You are correct there are not a lot of good examples out there. I wrote a paper about short circuits involving transformers and source impedance a few years ago. You can find it in the "short circuit" section of

When the source X/R ratio is not given (which is most of the time) a common assumption is 12. The X/R is the tangent of the impedance angle and is used to break impedances down into their respective R and X quantities for more accurate vector addtion. The assumption of 12 gives an angle of 85.23 degrees which is pretty steep. When assuming a steep angle for the source impedance and adding it to another impedance with an angle that is probably not as steep, the total impedance factoring the different angles will be lower, more conservative and lead to a higher short circuit value.

I hope it helps. If you have any other questions, let me know!

_________________
Jim Phillips, P.E.
Brainfiller.com

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 Post subject: Posted: Wed Mar 17, 2010 5:30 pm
 Sparks Level

Joined: Sun Dec 23, 2007 1:44 pm
Posts: 348
Location: Charlotte, NC
To add to Jim's response, you need a really firm grasp on the sequence components and per unit calcs concepts. If you do not have a thorough understanding of those concepts, that is the first place I would recommend you study.

Good luck and come back as necessary.

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 Post subject: Posted: Wed Mar 17, 2010 8:35 pm
 Arc Level

Joined: Thu Jan 10, 2008 8:49 pm
Posts: 510
Location: New England
Since arc flash is only concerned with three phase bolted fault, wouldn't it be easier for most of us to learn the MVA method of fault currents. I use it a lot of arc flash as its very easy to understand and apply. I think there are some papers on the net explaining it.

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 Post subject: Posted: Wed Mar 17, 2010 9:09 pm

Joined: Mon Mar 15, 2010 7:00 am
Posts: 3
Thanks Jim. Can you clarify for me though?--in your paper on Short Circuit Calcs--Transf and Source Impedance, p2 of 3, under Derivation of Step 1, it says, Zsource ohms=(kV_ll2)/(MVAsc).

1) Isn't kV_ll2 supposed to be squared, or is this a misprint? It appears to conflict with your post.

2) I am assuming kV_ll2, in your paper, is the secondary voltage of the transformer, or the voltage of the circuit at the location of the short circuit fault of concern.

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