# Thread: Reducing AIC at a machine.

1. The AIC rating of the breaker has nothing to do with the available fault current, other than the breaker rating has to exceed the available fault current. You need a short circuit current study to find out how much current is actually available at the machine.

2. you may be alright
you may not
you need a fault study before you do anything

do you have any info on the service xfmr? kva, pu Z, etc?

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Originally Posted by Jraef
• AFC = Available Fault Current, also known as Short Circuit Amps (SCA); the term for how much fault current (typically in kA or thousands of amps) that are available at the line side terminals of something. It's derived from the transformer size, the capacity of the primary circuit of that transformer, the impedance of it and any other impedance values between that and the line terminals, typically the resistance of the length of cables (and anything like reactors added).

I would add that the available fault current has to be obtained from the utility and be careful that you are not provided with the standard utility response of an infinite bus fault current based on transformer size.

If there was an arc flash study done, the information you are seeking should be available.
Last edited by charlie b; 01-09-18 at 06:49 PM. Reason: Repaired the quotation

4. I know I'll get guff over this
op, please do not do this, no offense but you lack the fundemental understanding

I do this as a sanity check
start the machine, control panel, etc
measure v at cb and at machine connection point
messure all 3 ph and avg
do the same for current, doesn't matter which point

example:
cb 480 vac
i = 40 A
v drop = 10
line Z = 10/40 = 0.25 Ohm

max avail fault i inf bus = 480/0.25 = 1920 A
it will be less because under a fault due to the high i lowering the 480 v but Z will be the same
depending on loads you may need to add a motor fault i contribution, I use 4 x sum(connected fla)

5. Originally Posted by Ingenieur
I know I'll get guff over this
op, please do not do this, no offense but you lack the fundemental understanding

I do this as a sanity check
start the machine, control panel, etc
measure v at cb and at machine connection point
messure all 3 ph and avg
do the same for current, doesn't matter which point

example:
cb 480 vac
i = 40 A
v drop = 10
line Z = 10/40 = 0.25 Ohm

max avail fault i inf bus = 480/0.25 = 1920 A
it will be less because under a fault due to the high i lowering the 480 v but Z will be the same
depending on loads you may need to add a motor fault i contribution, I use 4 x sum(connected fla)
Hmmm...
Having the correct SCCR at the panel and having it meet or exceed the AFC is now a CODE REQUIREMENT.
Your methodology, while probably good for a guesstimate (I'm not doing the math), is NOT going to suffice in lieu of a proper fault study and the required labeling.

6. Originally Posted by wbdvt
[/LIST]
I would add that the available fault current has to be obtained from the utility and be careful that you are not provided with the standard utility response of an infinite bus fault current based on transformer size.

If there was an arc flash study done, the information you are seeking should be available.
Good points.

7. SCC is limited by impedance of the conductors between the source and the point of interest. Impedance of source factors in also.
As mentioned it wont take all that much conductor length for 30-60 amp conductors to significantly reduce available SCC at the load end.

Also note available current at panel in question could be more then 14 kA, but the 14kA breakers might be series rated with whatever is ahead of them.

8. Originally Posted by Jraef
Hmmm...
Having the correct SCCR at the panel and having it meet or exceed the AFC is now a CODE REQUIREMENT.
Your methodology, while probably good for a guesstimate (I'm not doing the math), is NOT going to suffice in lieu of a proper fault study and the required labeling.
and I've used it to satisfy equipment supplier concerns
the numbers are easy to validate

assume a 500 kva, xfmr 480, Z 5% pu
Z = 0.05 x sqrt3 480^2 / 500000 = 0.0399 ohm
i fault = 500000/(sqrt 3 480 0.05) = 12028 A
same as 480/0.0399 = 12028

assume we add the 0.25 as determined above
40 x 0.25 = 10 or 10/480 x 100 = 2.08% drop, within a common range

ignoring the feeder to the pnl the cb is in and only using the xfmr and load ckt
0.0399 + 0.25 = 0.2899
i fault at load = 480/0.2899 = 1656 A < 1920 A
Last edited by Ingenieur; 01-09-18 at 08:51 PM.

9. Originally Posted by Ingenieur
and I've used it to satisfy equipment supplier concerns
the numbers are easy to validate
Point taken. The OP is in Indiana, so they are still on the 2008 code and the latest changes wouldn't apply, unless the machine is considered "industrial" (Article 409 covers "Industrial" control panels and was introduced in the 2005 code).

PS: Actually, the relevant wording hasn't really changed since the 2005 code. 409 requires it for "industrial controls", then defines that. But 430.8 requires it as well, for "motor controllers" in general.
430.8 Marking on Controllers. A controller shall be marked
with the manufacturerâ€™s name or identification, the voltage,
the current or horsepower rating, the short-circuit current rating,
and such other necessary data to properly indicate the
applications for which it is suitable.
Last edited by Jraef; 01-09-18 at 09:02 PM.

10. now to get really crazy lol
if you know
and vdrop 10/480
i fault (ignoring any upstream Z) = 40/(10/480) = 1920 A lol
a pu calc using the 40 A as base

pu Z % = prim v required to get sec rated current / rated prim v
if pu 5% on a 12470:480...623.5 prim volt produces rated sec i..623.5/12470 = 0.05

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