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 Post subject: Transformer NameplatePosted: Fri Jun 19, 2009 1:54 pm

Joined: Fri Jun 19, 2009 1:42 pm
Posts: 2
Hi.

I'm looking at an installation with (3)500kVA single phase utility transformers. They are wired Delta-Wye 12470-120/208.

Our utilty quoted 80,000A symmetrical RMS available fault current. This is presumed to be based on infinite bus, 5.2%Z.

Yes, I know, pretty useless for arc flash.

I'm going to assume a 100MVA base for fault current and use actual nameplate impedance, +10%.

Where I am confused is how to apply the single phase %Z to a three phase value.

Thanks

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 Post subject: Posted: Sat Jun 20, 2009 5:42 am
 Plasma Level

Joined: Mon Sep 17, 2007 5:00 pm
Posts: 1637
Location: Scottsdale, Arizona
Source Impedance and Transformer Impedance

Since three phase calculations are typically on a per phase basis, you use the 5.2% To arrive at 80,000 Amps, use the following:

SCA = (FLA x 100) / %Z

SCA = Short Circuit Amps

FLA = 1500 kVA / (sqrt 3 x 0.208) = 4163.6 Amps (quite large)
Three Phase kVA = 3 x 500 kVA

SCA = (4163.6 x 100) / 5.2 = 80,069 Amps

This method is a variation of the per unit method that does not require converting or using 100 MVA base.

If you are using software for the analysis, perhaps use an infinite source like you already have and then as an option use 750, 500, 250, 150 and 100 MVA as a source contribution from the utility to see how it effects the results.

Here is a paper I wrote a while ago about the simplified method for this:

To determine the source impedance for 750 MVA you would take:
(1.5 MVA / 750 MVA) x 100 = 0.2%.
The 1.5 MVA is from 1500 kVA of the transformer

Add 0.2% to the 5.2 % and recalculate the short circuit current using 5.4%

Do the same for 500 MVA through 100 MVA adding the new source impedance to the transformer 5.2%. As the source MVA becomes lower (weaker source) the secondary short circuit current will also become lower and at some point the incident energy might jump due to low short circuit and high clearing time.

At that point, see what source contribution caused it and then decide if that is very realistic. At least you will have a range of utility contribution where your know your i.e. numbers are good and you will know how low the utility contribution has to be before it creates a problem.

_________________
Jim Phillips, P.E.
Brainfiller.com

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 Post subject: Posted: Mon Jun 22, 2009 9:46 am

Joined: Fri Jun 19, 2009 1:42 pm
Posts: 2
Jim-

Thanks for the thoughtful response and your illustrations.

Although I didn't state it very well, I was concerned about using a %Z based on the individual, single phase transformer nameplate, because it seems to be a different number than what would be produced from running the short circuit test with the transformers cabled together in a Delta-Wye bank. I was considering this number (6.1%) only because it was much larger than the utility furnished figure. (5.2%)

Also, I dismissed the utility's infinite bus, available fault current as unrealisatically high, as the 12kV switch feeding the transformer primary is rated 4k, which implies a less than 100MVA base.

The utility won't readily divulge additional information. Giving a very high fault current is convenient and also a "defensive engineering" posture for them. Unlike the utilty, I need to use engineering judgement on whether "overdutied" equipment needs to be replaced, and what typical and low end fault current is for incident energy calcs.

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 Post subject: Posted: Mon Jun 22, 2009 6:00 pm
 Plasma Level

Joined: Mon Sep 17, 2007 5:00 pm
Posts: 1637
Location: Scottsdale, Arizona
For a three phase calculation using three single phase transformers you use the impedance of one transfomer assuming the other 2 are the same (or very close to the same).

Yes, it is sometimes difficult to obtain information from the electric utility company. Sometimes they don't have an extensive data base or system model. Some utilities are still trying to determine how to handle arc flash with customers. Good Luck!

_________________
Jim Phillips, P.E.
Brainfiller.com

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