AIC Rating

[hl53us]

How do I determine what AIC Rating I need for a proposed breaker.

There is an existing main panel (120/240V, 450A 1Ph, 3W) feeding an existing subpanel. I have to remove the subpanel and replace it w/ a 250A 2 pole breaker to feed a new panel. How do I determine the interrupting rating for my breaker? My contract plans states that the (N) breaker shall match the existing interrupting rating.

 

[charlie b]

The only thing I can think of is to look for the make and model number on the existing breaker, and contact its manufacturer.

But I must say that this is not a very helpfull requirement to put into a contract. A better approach would have been to determine the available fault current (an engineer's task), and require that the replacement meet that value.

 

[LarryFine]

A better approach would have been to determine the available fault current (an engineer's task), and require that the replacement meet that value.

Sidebar, please. Charlie, I agree with your statement, but wouldn't it be acceptable to say "an engineering task," since I am permitted to do this on many types of jobs in my area?

 

[kingpb]

I would have to agree with Larry, it does not take an engineer to determine fault currents, just someone trained in how to calculate them.

In this case for the OP find the transformer ahead of the panel, take the nameplate KVA and divide by the impedance. This gives you the short circuit MVA (or KVA) then simply divide by the voltage and sqrt 3 (if three phase) and you have just determined the maximum possible available fault current. Make your short circuit AIC rating equal to or higher.

That's all there is to it.

CYA statement: This method assumes that motor contribution from loads on the panel side of the transformer are negligible and cross connection to other busses with fault contribution is not possible

 

[mayanees]

worst-worst case SCA

worst-worst case SCA

In this case for the OP find the transformer ahead of the panel, take the nameplate KVA and divide by the impedance. This gives you the short circuit MVA (or KVA) then simply divide by the voltage and sqrt 3 (if three phase) and you have just determined the maximum possible available fault current. Make your short circuit AIC rating equal to or higher.

That's all there is to it.

CYA statement: This method assumes that motor contribution from loads on the panel side of the transformer are negligible and cross connection to other busses with fault contribution is not possible

In place of the CYA statement, you could add a worst-case motor contribution of 25% of the fault current calculated using the above method. That covers a motor contribution equal to the KVA rating of the transformer, i.e. 2500 kVA/480,5.75%Z, no-motor fault calc of 52,300 amps. Add 25% or 13075 amps for a total of 65,375 amps. This number is certainly much higher than a real-life field value because of the assumed inifnite primary contribution, and the addition of 2500 kVA of motors, but safely represents a worst-worst case value.
JM