AIC rating formula???

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jahilliard

Senior Member
I have to show the fault current rating available for a breaker we are installing and the POCO claims they don't know what is available at their transformer. I just need the formula, with a little explanation of how it all comes together, to provide the inspector for the job. I have a little bit of an understanding of how to figure it out but it's been waaay too long to remember that! I appreciate it!
 

eric9822

Senior Member
Location
Camarillo, CA
Occupation
Electrical and Instrumentation Tech
I have to show the fault current rating available for a breaker we are installing and the POCO claims they don't know what is available at their transformer. I just need the formula, with a little explanation of how it all comes together, to provide the inspector for the job. I have a little bit of an understanding of how to figure it out but it's been waaay too long to remember that! I appreciate it!

Assume unlimited fault current ahead of the transformer. Divide the FLC of the secondary of the transformer by the impedance. Example 2000KVA, 480V, 5.75% impedance. [2,000,000/(480*1.732)]/(.0575) = 41,838 A.
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
I have to show the fault current rating available for a breaker we are installing and the POCO claims they don't know what is available at their transformer. I just need the formula, with a little explanation of how it all comes together, to provide the inspector for the job. I have a little bit of an understanding of how to figure it out but it's been waaay too long to remember that! I appreciate it!

If need to know how to calculate the fault at any point or for any breaker use below link it has given step by step procedure with excellent examples.

http://ecmweb.com/mag/electric_shortcircuit_calculation_methods/

or
http://www.bussmann.com/library/docs/EPR_Booklet.pdf
 

zog

Senior Member
Location
Charlotte, NC
I have to show the fault current rating available for a breaker we are installing and the POCO claims they don't know what is available at their transformer. I just need the formula, with a little explanation of how it all comes together, to provide the inspector for the job. I have a little bit of an understanding of how to figure it out but it's been waaay too long to remember that! I appreciate it!

An AIC rating is the maximum amount of fault current a breaker is capable of interupting and is determined though ectensive testing by the OEM. There is no formula.

Are you asking for a formula to calculate the available fault current for a specific point in your facility? The formula provided by eric will give you the fault current availabel at the secondary terminals of the transformer and will be higher than actual if you assume an infinite primary, for an accurate calculation you need the primary fault current (Your utility has that info, you are just not talking to the right person, most utilities have someone specifically assigned to doing this) and then you need to find the fault current at the point in question, there are several methods to do this, the easiest being the point to point method that you should find inthe busman link provided above by Hameedulla.

Do you need this calulation only to prove to the inspector that the AIC rating on a specific breaker exceeds your fault current available?
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
An AIC rating is the maximum amount of fault current a breaker is capable of interupting and is determined though ectensive testing by the OEM. There is no formula.

Are you asking for a formula to calculate the available fault current for a specific point in your facility? The formula provided by eric will give you the fault current availabel at the secondary terminals of the transformer and will be higher than actual if you assume an infinite primary, for an accurate calculation you need the primary fault current (Your utility has that info, you are just not talking to the right person, most utilities have someone specifically assigned to doing this) and then you need to find the fault current at the point in question, there are several methods to do this, the easiest being the point to point method that you should find inthe busman link provided above by Hameedulla.

Do you need this calulation only to prove to the inspector that the AIC rating on a specific breaker exceeds your fault current available?

Zog,

Please see this link there is a formula regarding to AIC in a discusion forum. I had not seen it before. please check it once

http://www.mathkb.com/Uwe/Forum.aspx/stat-consult/845/Correct-formula-for-AIC
 

skeshesh

Senior Member
Location
Los Angeles, Ca
I think zog is right they're talking about probability calculations in a non-linear regression. I guess they could be trying to model the probability of a fault occuring using a huge number of samples but that's a very long shot and I highly doubt that's what they're talking about. What it's certainly not is a calucation for finding AIC.
 

Hameedulla-Ekhlas

Senior Member
Location
AFG
Pretty sure that is nto electrically rated, they are discussing something else, one example has a negative AIC so that would not be very useful would it?

yeah, it is something else not Ampere interrupting Capacity.

but jahilliard has mentioned in his question title AIC rating Formula,
Indeed, AIC is the ampere interrupting capacity of a breaker. So, has to do fault calculation to find the maximium fault current and compare it to the breaker's AIC

here is a another example.

The following example illustrates a potential problem of installing underrated electrical equipment if X/R ratios and motor contribution aren't considered in the initial design.

Example design criteria:


?Utility transformer rating: 2,500kVA
?Utility transformer % impedance: 4.775%
?Service conductors: 10 sets of 600 MCM copper
?Available fault current at utility transformer secondary: 63,000A
?X/R ratio at the utility transformer secondary: 11
?Motor contribution: 400 hp
?Ampacity of service conductors: 4,000A
?Service gear tested X/R ratio: 4.9

A fault current of 62,321A is calculated at the switchgear. This value is based on the 2,500kVA utility transformer with 4.775% impedance and minimal impedance from the service conductors (11 feet of 10 sets of 600 MCM copper).

The simple form of this calculation, based on infinite bus theory, is indicated below:

2,500kVA?(√3?480V)?0.04775=62,975, or 63,000AIC at the utility transformer secondary

However, the AIC is further reduced at the service to 62,321A, based on the impedance of the service conductors. This assumes no contribution from the motors in the system or from the asymmetric component.
 

zog

Senior Member
Location
Charlotte, NC
Holy brain fart batman, I forgot all about motor contribution, been one of those days. Hey jahilliard, you need to figure in motor contribution too.
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
A more dangerous scenario is that given a utility transformer failure, the utility company replaces the transformer with one they have pulled out from another location and is sitting in their yard, the impedance happens to be Z=4.2%. PCO doesn't care, i.e. 2500KVA is 2500KVA.

Using the conventional worst case scenario calculation of infinite bus, the available fault current on the LV side (480V) of the transformer now goes from approx. 63kA to around 67kA. The breaker originally installed rated for 65kA is not rated properly now.

The moral of the story is: if the PCO replaces a transformer to your service, don't think there isn't anything to be checked on the Owner's side.

Food for thought: How many facilities do you think exist that have inappropriately rated breakers, simply because the original designer tried to save a couple bucks and go with the lower kA rated device instead of conservative choice, or they should have used fuses instead of a breaker. When it comes to AIC ratings always error on the conservative side, safety should never be compromised.
 

zog

Senior Member
Location
Charlotte, NC
Food for thought: How many facilities do you think exist that have inappropriately rated breakers, simply because the original designer tried to save a couple bucks and go with the lower kA rated device instead of conservative choice, or they should have used fuses instead of a breaker. When it comes to AIC ratings always error on the conservative side, safety should never be compromised.


It is pretty common, with all the plants doing the SC studies for the arc flash analysis and finding out thier breakers AIC ratings are not adequate. I get calls daily for upgrading breakers.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
It is pretty common, with all the plants doing the SC studies for the arc flash analysis and finding out their breakers AIC ratings are not adequate. I get calls daily for upgrading breakers.
I have seen similar. When I worked for Siemens, we got a contract to go into the NUMI automotive plant (New United Motors Inc., JV between GM and Toyota) to do NFPA 70E and SCCR studies on equipment, i.e. switchgear, switchboards, panelboards, MCCs, machine control panels, etc. The upshot was that they identified over 5,000 pieces of electrical equipment that were inadequately rated for the AFC. They just closed that plant, but if anyone buys it they are going to have to contend with that. It's a big deal all over and from what I can tell, inspectors are dealing with it by sort of looking the other way, until someone wants to upgrade something in an installation.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
I think zog is right they're talking about probability calculations in a non-linear regression. I guess they could be trying to model the probability of a fault occuring using a huge number of samples but that's a very long shot and I highly doubt that's what they're talking about. What it's certainly not is a calucation for finding AIC.

OMG, as soon as I saw that my head started to hurt remembering Statistics classes... and it's been 30 years since i took one. Still had a textbook though, don't ask me why. I'm putting them in the donation bin today.

AIC = Akaike?s Information Criterion, a method used in evaluating statistical data models for non-linear regressions.

Don't EVER make me think of that again...
 

MarkyMarkNC

Senior Member
Location
Raleigh NC
Anyone have links to cases where too low of an AIC rating directly led to fires / explosions etc.? I've tried to explain AIC rating problems on existing equipment to people before, and they never seem to give a hoot.
 

zog

Senior Member
Location
Charlotte, NC
Anyone have links to cases where too low of an AIC rating directly led to fires / explosions etc.? I've tried to explain AIC rating problems on existing equipment to people before, and they never seem to give a hoot.

Had a large power plant have an entire 15kV lineup burn down last week about an hour from you, can't name names but it happens often enough to be a major concern.
 

zog

Senior Member
Location
Charlotte, NC
OMG, as soon as I saw that my head started to hurt remembering Statistics classes... and it's been 30 years since i took one. Still had a textbook though, don't ask me why. I'm putting them in the donation bin today.

AIC = Akaike?s Information Criterion, a method used in evaluating statistical data models for non-linear regressions.

Don't EVER make me think of that again...

Still have my old stats book too, thought about looking this up but didn't want my head to explode. Stats was the most fun and hardest class I ever took. My professor could do cubed roots in his head, hard ones too, used to make me so mad.
 

tkb

Senior Member
Location
MA
Assume unlimited fault current ahead of the transformer. Divide the FLC of the secondary of the transformer by the impedance. Example 2000KVA, 480V, 5.75% impedance. [2,000,000/(480*1.732)]/(.0575) = 41,838 A.

I used the Bussman Point to Point method and get slightly different results.
The formulas is a little different.
If I was able to attach an Excel file I could show you.
 

Last Leg

Member
Location
Houston, TX
I come across this so often - needing to figure ISCA. I've got the Busman formula and various 'self-calculating forms'. What I can never figure is the impedence on the transformers. Typical line voltage transformer Z can be found, but even these vary greatly. However, I have seen Z for pole mount utiliy transformer figured from 1.5% to 6% Z. This is an extreme factor in the calculation. Does anyone have a source for these?
 

zog

Senior Member
Location
Charlotte, NC
I come across this so often - needing to figure ISCA. I've got the Busman formula and various 'self-calculating forms'. What I can never figure is the impedence on the transformers. Typical line voltage transformer Z can be found, but even these vary greatly. However, I have seen Z for pole mount utiliy transformer figured from 1.5% to 6% Z. This is an extreme factor in the calculation. Does anyone have a source for these?

It is stamped on the nameplate of the transformer
 

rcwilson

Senior Member
Location
Redmond, WA
"Anyone have links to cases where too low of an AIC rating directly led to fires /"

I believe that in most cases it would be difficult to prove that inadequate AIC was the cause of a burn down since the fault is initiated for some other reason. Improper AIC increases the damage but does not cause it.

I investigated MCC fires at a public college campus. Eveytime a small HVAC motor or other load shorted out the related MCC Buckets would completely burn up. Once the fire got ot some cabling above the MCC. I found the contractor had installed cheap, unmarked fuses in all MCC's (Assumed AIC = 5 kA). Actual SC level was about 35kA. The fuses blew open and started the control wiring on fire. The contractor was long gone and the fuse company was out of business.

I am sure there are similar stories out there.
 
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