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 Post subject: Transformer Fault CurrentPosted: Mon Dec 21, 2009 3:29 pm

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
I have a questions about transformer fault current. I am trying to model a system using Arcalc by SKM. The transformer is a 75kva. I know that a study is not needed for this small system but I am trying to understand concepts with this system. The utility gave me the secondary fault current of 14264 3 phase amps. The transformer is 75KVA, 1.48 IMP, 208v sec 12470 Pri. The source imp on the secondary is %R 1267 + %jX 1477 on a 100 MVA base. I know that equates to a 1.165 X/R ratio. My question, Is there a way, using this info, to get the available fault current on the primary side if the transformer? The primary fusing is bayonet 8 amp. I also have a 500 MCM 160 feet of parallel cabel to the panel board. Thanks for any info.

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 Post subject: Posted: Tue Dec 22, 2009 6:44 am
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Joined: Mon Sep 17, 2007 5:00 pm
Posts: 1637
Location: Scottsdale, Arizona
Actually you do need to include this transformer. Even though it is less than 125 kVA, it has a low impedance. The problem with the way the standard is written is it does not elaborate on what IS a low impedance.

The second part is a bit more complex. Yes you can calculate the primary short circuit current based on the data provided. I have based the calculations on the following information that you provided:

Transformer: 75 kVA 1.4% impedance (transformer X/R unknown)
Primary Voltage = 12.47 kV, Secondary Voltage = 208 Volts
Thevenin Equivalent Impedance on the Secondary = 1267 + j 1477%
(Includes source and transformer)
100 MVA Base
Short Circuit Amps on the Secondary 14262.3 Amps

Here is what I have:

Total impedance magnitude Z = sqrt(1267^2 + 1477^2) = 1945.97%

Convert from 100 MVA base to .075 MVA (75 kVA) base so we can subtract the transformer impedance.

1945.97% x (0.075/100) = 1.459 % on .075 MVA base

Sanity check for short circuit current:

Short Circuit Current on Secondary = (FLA secondary x 100) / Total Impedance

FLA secondary = Full Load Amps = kVA / (sqrt(3) x kVsecondary)
FLA secondary = 75 kVA / sqrt(3) x 0.208kV
FLA secondary = 208.18 Amps

Short Circuit Current on Secondary = (208.18 x 100) / 1.459
Short Circuit Current on Secondary = 14268 Amps (your problem had 14262 – rounding)

The 1.459% now represents the transformer and source impedance using the transformer base (which is what the transformer 1.4% is based on)

How much of this impedance is from the source? Take 1.459 and subtract the transformer impedance of 1.4.

Source Impedance = 1.459 – 1.4 = 0.059%

I ignored the X/R since the transformer’s was unknown and this would have been a more complex problem. If you know the X/R transformer, then just break the impedances into respective X and R components and add them using vector addition.

For the next part, reference this article I wrote several years ago.

[url="http://www.brainfiller.com/documents/TransformerandSourceImpedance_000.pdf"]Short Circuit Calculations - Transformer and Source Impedances[/url]

Here is the bottom line:

Z source = (MVA transformer / MVA source) x 100

Rearrange it to solve for MVA source:

MVA source = (MVA transformer / Z source) x 100
MVA source = 0.075MVA / 0.059) x 100
MVA source = 127.1 MVA

Short Circuit Current on Primary

SC Primary = (127.1 MVA X 1000 kVA/MVA) / (12.47 x sqrt(3)
SC Primary = 5885 Amps at 12.47 kV

Of course, this is just how to work the problem, you will need to verify all of this yourself for any study or analysis.

Hope it helps!

_________________
Jim Phillips, P.E.
Brainfiller.com

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 Post subject: Posted: Tue Dec 22, 2009 2:24 pm

Joined: Thu Apr 02, 2009 4:33 pm
Posts: 29
Location: OH
Awesome! Thanks, Brainfiller.

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