Reducing AIC at a machine.

[derelectric]

Hello,
Our company was ask to install a 3 phase 480/277v GE panel about 4 months ago. We installed several 30 & 60 amp type TEY 14K AIC 3 phase breakers. Ran the conduit and wire to where each new machine. Each machine has 3 fuses that our feeds connect to. Just received a call from the company that the machine manufacture wants the AIC on some of the machines to be 10K AIC. I couldn't find any 10K AIC 480 breakers at any supply houses. I called GE. They said they don't make them. My question is does anyone know how to reduce the AIC from 14K to 10K. Could I possible change the fuses on the machines that need to be 10K AIC. If so what style of fuse? Thanks, in advance!

 

[petersonra]

it would not take much wire to change a 14 kA SCC to 10 kA. Has anyone calculated what the actual SCC is at the panels?

In any case, it won't do you any good to change out the 14 kAIC breakers. the SCC will stay the same.

 

[charlie b]

. . . the machine manufacture wants the AIC on some of the machines to be 10K AIC.

Let me take Bob’s answer a step further. Let’s start by getting our terms right. AIC is the amount of current that an overcurrent protection device (i.e., fuse/holder or breaker) can interrupt, without being destroyed in the process. The term can only apply to an overcurrent device. AIC has nothing to do with the amount of fault current that can be imposed at the location of a component (i.e., your machines or the panel that feeds them).

If the manufacturer wants assurance that the amount of short circuit current available (SCCA – that is the correct term to use here) is no higher than 10,000 amps, then someone needs to perform a fault calculation to determine the SCCA.

 

[Ingenieur]

first, doesn't matter what the CB's are rated as long as > than the available fault current
so the 14k are fine

you need to reduce the fault current at the machine
as noted, you must first calculate what it currently is
then explore options
current limiting fuses
possible a current limiting reactor (CLR) http://www.lcmagnetics.com/inductors/current-limiting-reactor/, introduces a little vdrop but will add Z to drop current

 

[derelectric]

Ok, thanks for the quick replies. I will contact someone at the plant to see if they can calculate the Fault current. Just a note, I talk to Cooper Bussman Technical group. They informed me that Fuses are not rated AIC.

 

[petersonra]

Ok, thanks for the quick replies. I will contact someone at the plant to see if they can calculate the Fault current. Just a note, I talk to Cooper Bussman Technical group. They informed me that Fuses are not rated AIC.

I suspect you did not hear what they actually said. You cannot reduce the available short circuit current by the use of current limiting fuses. Many times though you can put them in series with another device and get a series rating of the devices as a pair that exceeds what the other component's rating is. But this has to be an NRTL tested pairing.

Just out of curiosity, how is that you selected the 14kAIC breakers in the first place?

 

[charlie b]

OJust a note, I talk to Cooper Bussman Technical group. They informed me that Fuses are not rated AIC.

I didn't think they were. I was not certain whether the combination of fuse and fuse holder has an AIC rating, as my designs almost never include fuses. Anyone know?

 

[drktmplr12]

Ok, thanks for the quick replies. I will contact someone at the plant to see if they can calculate the Fault current. Just a note, I talk to Cooper Bussman Technical group. They informed me that Fuses are not rated AIC.

you need a calculation that has been signed and sealed by an engineer regularly engaged in the practice of calculating fault currents.

you want to ask if they have an existing short circuit study or set of plans listing the amount of fault current available at the equipment.

 

[infinity]

If you assume that the 14kAIC breakers are the correct size for the available fault current at the panel would it be very hard to calculate the fault current at the fuses? Isn't there an app for that.

 

[Jraef]

Actually, the entire concept, as you have posted it, is preposterous. So either you heard and recorded it incorrectly, or the machine manufacturer is way off base.

We can ASSUME that what they MIGHT have MEANT was that the AVAILABLE FAULT CURRENT at their machine must be 10kA or less. That seems far more likely, but is not what you posted. So first off you must get clarity on exactly what they want using appropriate terminology.

• AIC = Amps Interrupting Capacity; a term for PROTECTIVE DEVICES that will interrupt the flow of current in the event of a fault. Breakers have an AIC rating based on the maximum amount of current they can handle going through them during a fault, because UNTIL the breaker interrupts the flow of current, ALL of the AFC will be attempting to flow through it and the mechanical stresses that represents (magnetic repulsion / attraction) can literally make it explode..
• IR = Interrupting Rating. Fuses have an Interrupting Rating which is technically the same, so it's a little odd that Cooper was being so literal. But yes, technically it's not an interrupting CAPACITY I guess.
• 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).
• SCCR = Short Circuit Current Rating; the TESTED value of a piece of equipment to withstand the mechanical forces that might take place during a fault of the rated current. This is typically a combination OF the AIC / IR of the protective device, the Let-Through current of that device and mechanical strength of the power devices down stream.
• Let-Through = the amount of current that will get through a protective device (CB or fuses) in the time it takes to stop current from flowing during a fault.

If it IS the case that what they meant was 10kA Available Fault Current (AFC), then their asking YOU to do something about it via component selection in the panels you installed is bogus. That's not how that would work. Attempting to limit the AFC at a machine is a tricky business and often not realistic, but is something that would more likely involve the services of a registered PE. Machine OEMs tend to not like to hear that though, they want to build the cheapest thing they can get away with and try to shift the burden off to someone else. The CORRECT way to do this would have been for them to ASK their customer what the AFC already IS, then build their control panel accordingly, along with a UL listing that showed an official SCCR (Short Circuit Current Rating) on the label as is now required by code. After the fact is not the way to approach that issue.

 

[don_resqcapt19]

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.

 

[Ingenieur]

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?

 

[wbdvt]

• 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.

 

[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
load 470 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)

 

[Jraef]

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
load 470 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.

 

[Jraef]

[/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. :thumbsup:

 

[kwired]

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.

 

[Ingenieur]

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.

depends what edition your juristiction adopted
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

 

[Jraef]

depends what edition your juristiction adopted
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 manufacturer’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.

 

[Ingenieur]

now to get really crazy lol
if you know
load 40 A
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