AIC at panel?

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ecetcellgirl

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so i've re-educated myself on fault currents and short circuit calculations, but everytime i make a calc, i get slightly different answers. this is what i know:

existing tx - 25kva, 120/240v 1ph.
utility states available fault = 42000A
my run (tx to meter) is approx. 200'

if i use 1#4/0, i get a result of 8400A. is this accurate?

i have an example i've tried to reverse engineer and it makes my eyes cross.

help? thanks.
(long time reader, first time poster)
 
AL of Cu

Using Bussman's program with AL in non-mag raceway I get 3182
 
AIC At panel

AIC At panel

bob said:
You can not make the calculation with out knowing the transformer %Z.
Click to expand...

Assuming the 42K value the utility provided is on the secondary, which I assume it is, I got the same answer as Augie47 on the Bussman site. I always assume an infinite primary since you never know when the primary system may be updated by the utility.
 
I'll jump in with my usual statement. You are trying to determine the available SCA (short circuit amps) at your panel so that you can determine the AIC (amps interrupting capacity) your protective device needs to be rated for.

Be careful about that the formula you are using is for single phase 3-wire circuits.

Because you are not addressing the X/R ratio of your system or the protective device I would recommend that you leave at least a 25% margin between your available SCA and the AIC you chose for your devices.
 
grant said:
Assuming the 42K value the utility provided is on the secondary, which I assume it is, I got the same answer as Augie47 on the Bussman site. I always assume an infinite primary since you never know when the primary system may be updated by the utility.
Click to expand...

OK. I'll buy that. I do not pick up on that part of your post. Also If you use 240 volts when calculating the fault, you need to know that the fault is higher when using a phase to neutral (120 volts).
 
augie47 said:
AL of Cu

Using Bussman's program with AL in non-mag raceway I get 3182
Click to expand...
using Cu, in non-mag.

bob said:
You can not make the calculation with out knowing the transformer %Z.
Click to expand...
i dont know or even think i can obtain the %Z.

bob said:
OK. I'll buy that. I do not pick up on that part of your post. Also If you use 240 volts when calculating the fault, you need to know that the fault is higher when using a phase to neutral (120 volts).
Click to expand...
yes. i am aware of phase to neutral difference. i do believe i'm looking for the phase to phase

jim dungar said:
I'll jump in with my usual statement. You are trying to determine the available SCA (short circuit amps) at your panel so that you can determine the AIC (amps interrupting capacity) your protective device needs to be rated for.

Be careful about that the formula you are using is for single phase 3-wire circuits.

Because you are not addressing the X/R ratio of your system or the protective device I would recommend that you leave at least a 25% margin between your available SCA and the AIC you chose for your devices.
Click to expand...

i'm working w/ standard cell site equipment that is supposed to have an AIC rating of no less than 65KAIC (per the carrier). the plan checker wants to see the fault calcs on the plans. hence my post.

using the Bussman Short Circuit calcs book, i'm using
f=(2*length*AIC)/(C*n*E(L))
so f=(2*200*42000)/(16673*1*240) = 4.2
M=1/(1+4.2) = .2
Isca=42000*.2 = 8400.

is there something inherently wrong w/ this calc?
thanks for all your help. it's been a while since i've had to crunch numbers. :cool:
 
I don't think that a 42000A fault level from a 25KVA 240V transformer is very likely. That implies a transformer impedance of much less than 1%. Am I misunderstanding something, or are the numbers wrong? (eg a 4200A fault level, or a 250KVA transformer?)

-Jon
 
winnie said:
I don't think that a 42000A fault level from a 25KVA 240V transformer is very likely. That implies a transformer impedance of much less than 1%. Am I misunderstanding something, or are the numbers wrong? (eg a 4200A fault level, or a 250KVA transformer?)

-Jon
Click to expand...

information was supplied by the utility as 25 kva tx and 42000A fault.
 
winnie said:
I don't think that a 42000A fault level from a 25KVA 240V transformer is very likely. That implies a transformer impedance of much less than 1%. Am I misunderstanding something, or are the numbers wrong? (eg a 4200A fault level, or a 250KVA transformer?)

-Jon
Click to expand...

Typically you must always use the fault current provided by the utility, even if it seems outrageous. If the utility provides a value on the secondary, it makes no difference what the transformer size and %Z actually are. The normal reason is "you never know when the utility may change the transformer to a different one".

I may not agree with this rationale but our state AHJ does.:mad:
 
Available fault current from the utility

Available fault current from the utility

I think the attached link lists "available fault" current at secondary of transformer! What the original poster meant in "42kA" is the "Utility Fault Current", the available fault current that will be coming from the system if faults occur at the service.

Using the point to point method for short-circuit calculations (done at three-decimals), the fault available at 25 kVA service tx is 8,079 amps (copper in non-metallic raceway, 1-phase, 240V, 200 feet long 4/0 cable.

8400 amps figure came out when the computed value of F factor was taken as 4.2 when it was 4.198. M then was taken as 0.2 when the more precise value should have been 0.192.
 
winnie said:
I don't think that a 42000A fault level from a 25KVA 240V transformer is very likely. That implies a transformer impedance of much less than 1%. Am I misunderstanding something, or are the numbers wrong? (eg a 4200A fault level, or a 250KVA transformer?)

-Jon
Click to expand...

I'd put my moneyy on 250 kva also, but since the OP had a 42k value given by POCO, I used that to make my calculations.
I don't know that I've ever bothered with AIC on a 25 kva transformer.:smile:
 
topgone said:
I think the attached link lists "available fault" current at secondary of transformer! What the original poster meant in "42kA" is the "Utility Fault Current", the available fault current that will be coming from the system if faults occur at the service.

Using the point to point method for short-circuit calculations (done at three-decimals), the fault available at 25 kVA service tx is 8,079 amps (copper in non-metallic raceway, 1-phase, 240V, 200 feet long 4/0 cable.

8400 amps figure came out when the computed value of F factor was taken as 4.2 when it was 4.198. M then was taken as 0.2 when the more precise value should have been 0.192.
Click to expand...

The link lists the fault current at the terminals of a typical 25 kva transformer.
9200 amps is a long way form the 42000 amps the OP was given by the utility. They made a mistake either by using the wrong transformer size or the wrong transformer impedance. The %z would have to be 0.25% to get a 42 ka fault. 9200 amps is close to your 8079 if you include the impedance of the service.
 
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