figuring fault current

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arits74

Senior Member
Location
dixie arkansas
Occupation
working owner electrician
the utility has told me the size of the transformer,primary fault current,primary voltage,but they don't seem to know the impedance.the 1 guy that knows is never in the office.i know the length of the conductors and size.is this enough info to get close to figuring secondary fault current
 
iwire said:
I have no doubt we could all guess numbers. :D
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You can random guess if you want to I looked in my American Electricians Handbook.

arits74 said:
1289 primary amps fault current, single phase,25 Kva,7620 volts primary,240 secondary
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AEH, granted it is a 25-30 year old edition, states 1.91% impedance for a single phase 25 kVA. If you are concerned you could go a little lower with impedance to get a worst case scenario, but unless you are supplying something that is very close to the transformer you will in most cases have well less then 10kA of fault current on this one - that may be all that really matters.
 
electrofelon said:
The utilities pretty much always give you a guess, those figures are never accurate
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There is a difference.

I am assuming the OP has to provide the fault current to the electrical inspector and if the inspector asks where the figures come from I feel pretty confident the inspector is not going to be impressed if the answer is 'some random guys on the Internet'.

But to each their own, you, the OP and myself will do things as we each see fit. :)
 
You can always run the calculation assuming zero impedance on the transformer and only using the impedance of the service conductors.

Then it isn't a guess, but rather a _very_ conservative estimate. :) :)

arits74: it looks like your calculation is the available fault current at the transformer secondary, and does not include the secondary conductors.

-Jon
 
electrofelon said:
The utilities pretty much always give you a guess, those figures are never accurate
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They normally give you the infinite bus, based on the lowest impedance transformer in that size range that they use. Not really a guess, and on the conservative side, but good enough for equipment selection. Not suitable for use in incident energy calculations.
 
don_resqcapt19 said:
They normally give you the infinite bus, based on the lowest impedance transformer in that size range that they use. Not really a guess, and on the conservative side, but good enough for equipment selection. Not suitable for use in incident energy calculations.
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Perhaps, but often I wonder where they get these figures from, I don't really buy the "lowest impedence transformer they might use" argument. Recently I had the utility give me a value three times what it actually was ("actual" was assuming infinite primary). I can't see a different transformer having anywhere near 1/3 the impedence. Maybe the figure in a few sizes larger as well. But anyway, my overall point was that if we are being super nitpicky and technical, IMO 110.24 (A) requires the actual installed max value, not a "Max possible when the stars align future" value, and that would disallow the utility values that we often use.
 
electrofelon said:
Perhaps, but often I wonder where they get these figures from, I don't really buy the "lowest impedence transformer they might use" argument. Recently I had the utility give me a value three times what it actually was ("actual" was assuming infinite primary). I can't see a different transformer having anywhere near 1/3 the impedence. Maybe the figure in a few sizes larger as well. But anyway, my overall point was that if we are being super nitpicky and technical, IMO 110.24 (A) requires the actual installed max value, not a "Max possible when the stars align future" value, and that would disallow the utility values that we often use.
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Give the purpose of that section, I don't see an unreasonably high value as being a code violation....it may be costly, but not a code violation.
I don't really see where you have any choice other that to accept the value provided by the utility. The installer or design engineer, would almost never have the information required to even do an infinite bus short circuit calculation.
 
don_resqcapt19 said:
Give the purpose of that section, I don't see an unreasonably high value as being a code violation....it may be costly, but not a code violation.
I don't really see where you have any choice other that to accept the value provided by the utility. The installer or design engineer, would almost never have the information required to even do an infinite bus short circuit calculation.
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I also find it frustrating that padmount transformers do not come with an externally visible nameplate, for you to confirm this information. Rarely would I go on a site visit with a pentahead wrench, or with the authority to open up the transformer. It seems like if you can put an external nameplate on a disconnect, it should be possible to put an external nameplate on a transformer.

Sometimes you can look up a serial number or order sticker on the transformer, and ask the manufacturer. Although good luck finding someone in their general inquiry contact who will provide this information.
 
winnie said:
You can always run the calculation assuming zero impedance on the transformer and only using the impedance of the service conductors.

Then it isn't a guess, but rather a _very_ conservative estimate. :) :)

arits74: it looks like your calculation is the available fault current at the transformer secondary, and does not include the secondary conductors.

-Jon
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i did get it figured out including the conductors,utility pretty much said go with 2% impedance on transformer.
 
I've seen transformer impedances suddenly drop a lot in my area. I assume the utility specs have been changed to help eliminate wasted energy in the transformers.

I doubt if anyone will see a power transformer with a %Z below 1, but I was told one pad mount would have a 1.2% Z.

Long story short, I'd use 1% if I was going to make a worst case assumption.
 
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