PSE

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A good rule of thumb is that the available fault current at the secondary terminals of a transformer can be estimated by taking the full load current of the secondary side and dividing by the "per unit" impedance of the secondary windings. That later value is often on the order of 5.75%, so the value you divide by is 0.0575. Different transformers will have different values. In the example I gave earlier, the full load current for a 120/240V single phase 100 KVA transformer is 100,000/240, or 417 amps. Dividing that by 0.0575 give an AIC of 7,246 amps.

Most panels have a minimum AIC of 10,000 amps. To get that from the transformer, it would need to have an impedance of 4.17% or lower. It is possible that the utility transformer's impedance could be lower than that. Even still, the impedance of the secondary conductors will significantly reduce the available fault current at the building's main breaker.

So don't take the EC's statement at face value. Ask for the exact number that PSE provided to him (or her). Then you can do a calculation of the impact of the secondary conductor's impedance on the value PSE gives you. I am willing to predict that the final result is below 10,000 amps, and that you therefore do not have a problem in need of a solution.

Charlie, I find your %Z value to be to high. Mid 5's would be typical for a three phase pad of around 500KVA, but smaller single phase units are quite a bit less. I have some 15KVA single phase pads here that are a little under 2%. The utility 25KVA pole unit serving me is 2.3%. PSE's published fault current values can be found here on page 27:

https://www.pse.com/-/media/PDFs/Co...hash=261DD91382D90EF7192C85C21413E4DC06D07E63

Note also that L-N faults are higher than L-L (as they show in their tables).
 

tortuga

Code Historian
Location
Oregon
Occupation
Electrical Design
If I recall most 'Public Utility Commissions' or NESC regulations set a maximum available fault current for any residential service point. I might be mistaken though.
 
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