Fault current calculations

Status
Not open for further replies.
J

jhuppert

Member
Inspector is requiring fault current calculation on all devices,we usually provide calcualtion for service. can not find the requirement
 
What kind of building is this, and what do you mean by "devices"? Are you talking, for example, about receptacles in a house? It would be absurd to ask for a fault calculation at the location of each receptacle outlet.
 
jhuppert

jhuppert

the building is a center with a gym and kithen area for a church's function
the inspector wants fault current calculations for items such as the ranges
in the kitchen regardless of what the service fault current availibilty is.
 
If this is a new building, or if you are adding a panel to an existing building, then it would be reasonable for an Inspector to ask for fault current calculations. The intent is to prove that the amount of fault current available to all panels in the new building, or to any new panel in an existing building, is lower than the fault current rating of the panel(s) and lower than the fault current ratings of the breakers installed in the panel(s).

However, kitchen equipment, such as ranges and dishwashers, do not have fault current ratings. Neither do the receptacles into which you would plug such equipment. Therefore, there is no need to calculate the amount of fault current that they might experience. I can do such a calculation, and I can give you a number, but there will be nothing to compare that number with, no way to determine if that number is too high.

A panel needs a fault current rating, because if the current passing through the panel exceeds that rated value the panel can literally explode (metal flying everywhere ? not good). A breaker needs a fault current rating, because if the current passing through the breaker exceeds that rated value the breaker?s contacts can melt and become welded together (the breaker could never trip ? also not good). Receptacle outlets and kitchen appliances do not have fault current ratings, because they don?t need them.
 
How do I determine the interupt rating of the breaker inside the panel in a house. I don't know the impedence or data of the upstream distribution transformer. What is the standard interupt rating panel that selling in hardware store can be used at home?
 
There are three things under discussion here. First, how much current can the utility supply to the building, under fault conditions? Secondly, how much fault current is the panel rated to be able to handle? Third, how much fault current is the type(s) of breaker(s) installed within the panel rated to be able to handle? The later two items have to be at least as high as the first item.

As to the common residential application, my ?guess? (don?t use this as though it were a calculation?s result) is that the panels and breakers are likely to be rated for at least 10,000 amps, and that the utility can supply nowhere near that value. That is why I am surprised than an Inspector would be looking for fault calculations for a residential design.

To answer your specific question, you need to ask the utility for the available fault current. If you knew the KVA and percent impedance of their service transformer, you could calculate the fault current by dividing the two. For example, if the transformer were 225 KVA, and if it had a 5.75% impedance, the fault current would be 225/.0575, or 3,913 amps. Also, to get the fault current ratings of the panel and the breakers, you need to find out their make and model numbers, and contact the manufacturer. You might be able to look this information up on the Internet.
 
charlie b said:
However, kitchen equipment, such as ranges and dishwashers, do not have fault current ratings. Neither do the receptacles into which you would plug such equipment. Therefore, there is no need to calculate the amount of fault current that they might experience. I can do such a calculation, and I can give you a number, but there will be nothing to compare that number with, no way to determine if that number is too high.

A panel needs a fault current rating, because if the current passing through the panel exceeds that rated value the panel can literally explode (metal flying everywhere ? not good). A breaker needs a fault current rating, because if the current passing through the breaker exceeds that rated value the breaker?s contacts can melt and become welded together (the breaker could never trip ? also not good). Receptacle outlets and kitchen appliances do not have fault current ratings, because they don?t need them.
Click to expand...

Charlie,

You are correct that 110.9 is basically addressing AIC, and as such only applies to things like breakers and fusible switches.

However, 110.10 says "... component short circuit ratings" so this article could be about "withstand" and SCCR values. For example: article 409.110 (yeah I know it is for industrial control panels) requires a SCCR marking, but it never specifically says why. - Is the answer 110.10?

I see no reasonable way the inspector could stretch 110.10 to required short circuit values at receptacles for plug connected devices.
 
charlie b said:
For example, if the transformer were 225 KVA, and if it had a 5.75% impedance, the fault current would be 225/.0575, or 3,913 amps.
Click to expand...
Oops! Sorry, Charlie, Starkist wants tuna that tastes good! :D

You dropped the K during your division. Your transformer would have a short-circuit capability of 3,913Ka (or, 3.913Ma), not 3,913a. Fortunately, we rarely get that much current at the load end of service conductors. :rolleyes:
 
225kva, 480V, 3ph, 5.75%Z: SCC = FLA/.0575 = 225x1000/480/1.732/.0575 = 4707A

225kva, 208V, 3ph, 5.75%Z:
SCC =225x1000/208/1.732/.0575 = 10862A

carl
 
Charlie, Larry,
You both have it wrong...the calculation is based on the secondary current, not the kVA. If that is a single phase transformer with a 240 volt secondary the available fault current would be ~16,304.
 
don_resqcapt19 said:
Charlie, Larry,
You both have it wrong...the calculation is based on the secondary current, not the kVA. If that is a single phase transformer with a 240 volt secondary the available fault current would be ~16,304.
Click to expand...
Don, please don't complicate things by contributing facts into the discussion. :wink:
 
I would like the formula for figuring fault current. Of the three pervious, which one is correct? If you would give an example that would be great!
 
paulnorris said:
I would like the formula for figuring fault current. Of the three pervious, which one is correct? If you would give an example that would be great!
Click to expand...
See post #9 for 3-phase.

For single-phase 225 kVA, 120/240 volt, 5.75 %Z:
ISC (L-L) = 225/(0.240 * 0.0575) = 16,304 amps (see post #10)

To get L-N you can estimate about 1.5 times the L-L
 
mivey said:
See post #9 for 3-phase.

For single-phase 225 kVA, 120/240 volt, 5.75 %Z:
ISC (L-L) = 225/(0.240 * 0.0575) = 16,304 amps (see post #10)

To get L-N you can estimate about 1.5 times the L-L
Click to expand...

Please remember that many members will be confused by your above calculations. I like the KISS program when explaining this to those without a good math background. You sorta changed horses in mid stream. I'm sure some are puzzled trying to figure out where you came up with those numbers. I'mm sure you can explain better than I your division of the kva and voltage by 1000.
 
RHJohnson said:
Please remember that many members will be confused by your above calculations. I like the KISS program when explaining this to those without a good math background. You sorta changed horses in mid stream. I'm sure some are puzzled trying to figure out where you came up with those numbers. I'mm sure you can explain better than I your division of the kva and voltage by 1000.
Click to expand...
OK

For single phase:
ISC (L-L) = kVA/(kV(L-L) * %Z/100)

[edit: and for 3-phase: kVA/(kV(L-L) * sqrt(3) * %Z/100) ]
 
Last edited:
Doesn't really apply to residential wiring but to really to a fault calculation you need to ADD the contribution of any motors and/or generators to the fault. A turning motor can act as a generator. These calculations are kinda tough by hand, easy if you have software. Look up the terms transient and subtransient reactance.

For the purposes of this thread, go to Mike Holt's "Free Stuff" page and download the fault calculator spreadsheet found here:

http://www.mikeholt.com/documents/calculations/formulas/FaultV6.2.xls

This spreadsheet does NOT account for motor contribution.

Here's another freebie that allows you to ESTIMATE motor contribution to the fault. This one is OK for ESTIMATING if you don't have generators or large synchronous motors on your system

https://www.sea.siemens.com/consultant/docs/DA_SC_Calculator_V1.1.xls

Here are some more "freebies" - I think?? You can download them for free but there is a copyright statement with them:

http://www.newcalc.com/engineering-software.htm
 
Last edited:
nawao said:
What will be ISC (L-N)?:confused:
Click to expand...
For single phase, you can approximate it at the transformer terminals by using 1.5 times the L-L (see post #13). According to the Bussman SPD manual it has a theoretical range of 1.33 to 1.67 times the L-L.
 
Detailed calculations are best done by software or, if it is short, do it by hand with the impedences if you want to have fun. I think it would probably be a good exercise to do at least one by hand using the impedences and symmetrical components.

I ran a two-bank substation with a split-bus with a parallel generator and isolation transformer. I ran scenarios for the bus tie open and closed, generator on & off & blackout mode, generator breaker open & closed, for L-G and 3-phase faults at the generator, bus, down-line, etc. Not something you want to do every day but a good one-time exercise. It took a huge chunk of amppad.
 
Status
Not open for further replies.
Top