available arc fault current

[Lance Campbell]

On a GE Spectra series distribution with a 400 amp main I have a rating on the cabinet of 10 kA rms sym. The 150 kva transformer is 50 feet away and has 2.5% imp. We ran 600 mcm copper to the main. When we do the fault calcs, we get 15,685 amps fault available line to line, and 15k line to neut. Is there any way I can be in compliance with this scenario, like short time current limiting fuses mounted on the transformer?

 

[davidr43229]

we get 15,685 amps fault available line to line

If you have 15,685 available a RK-1 fuse will have a let-trough of 8,000 amps and a RK-5 will have a let through of 10,000amps.
Yo ualso may be able to use a CLass J or CLass T.
I have enclosed a link to find your existing breakers to "Series-Rate" and properly protect your breakers.
http://cooperbussmann.com/pdf/5eb98e2e-157d-4b59-943a-a01918f23a34..pdf

 

[steve66]

Most people do fault calculations assuming that infinite current is available on the primary of the transformer. This gives a conservative fault current.

If you can get the POCO to give you a little more info. on the primary, you might find your max. fault current is below 10KA.

For example, if you know you have x feet of number y wire on the primary of the transformer, you can move the inifinite current to the start of that wire.

Steve

 

[BLUEDEVIL]

provide a fuse or circuit breaker rated 22000 ahead and series rated with your panel

 

[FNCnca]

.

If you can get the POCO to give you a little more info. on the primary, you might find your max. fault current is below 10KA.



Steve

The company I work for, requires you, to make every effort, to get this info from the POCO. My POCO did so without hesitation.
They made the statement "We wonder why all customers do not ask for this information".
The information made a significant difference in our calculations.

 

[kingpb]

Without the voltage, it is not possible to determine fault current from the values provided. Working backyards, it looks like maybe 208Y/120V system?

If that is the case, then 150KVA @ 2.5% impedance, using an infinite bus is only 6MVA of short circuit capability. For a power company, they could easily have 100 times that, or 600MVA. It would be unlikely for them to be so low that the transformer @ 150KVA being so small, would ever look like anything but an infinite bus. Therefore, your best bet is to put protection on the LV side of the transformer, ahead of your panel with current limiting fuses.

 

[bob]

On a GE Spectra series distribution with a 400 amp main I have a rating on the cabinet of 10 kA rms sym. The 150 kva transformer is 50 feet away and has 2.5% imp. We ran 600 mcm copper to the main. When we do the fault calcs, we get 15,685 amps fault available line to line, and 15k line to neut. Is there any way I can be in compliance with this scenario, like short time current limiting fuses mounted on the transformer?

I ran a calculation using a primary available fault of 7000 amps and the fault at the main was 12941 amps. You would need to have the primary fault
to be less than 1000 amps to get the secondary fault down below 10 ka.
Thats not very likely

 

[templdl]

I'm responding to this based upon the title of this series of posts "available arc fault current"

It is of my understanding that there would be concern for available fault current with bolted faults where the current that flows into the fault is based upon the available fault current at the secondary of the service entrance transformer and the dynamics or impedance of the circuit to the location of the bolted fault. The OCPD must be rated to clear that fault.
The current of an arcing fault is often times much less than a bolted fault so much less that it may not even be of great enough magnetude to trip an OCPD. The OCPD may view it as a simple load.
Bolted and arcing faults are two different animals.

 

[kingpb]

Keep in mind that besides tripping capability ofthe OCPD, current would somehow need to be limited to meet withstand ratings. This cannot be achieved by simply putting a protection device ahead of the panel. It would need to be current limiting.

Another option is to install a current limiting reactor, although I have not seen one for such a simple application. You could also possibly have the utility change out the transformer and provide one with a higher impedance. Standard impedance is 5.75%, which would result in a worst case bolted fault of 7.8kA (given a 7.5% manufacturing tolerance)

Working backwards the lowest impedance you could have, and meet the 10kA would be approx. 4.5%.