Fault current

[dave121]

The po. Co. tells me I have 10,011 amps at the top of the costumer pole. From there 2 to 3 ft. Open air 3/0 cu thwn stranded, then in rigid alu. conduit run with the neutral for 10 ft to the top of the service disc. 277/480 v 3 phase, 4 wire. Then more 3/0 in PVC for 2 ft, then the 3/0 continues inside the starter panel for 5-6ft to the top of a pump starter panel disc. After the starter disc 1 ft. Of stranded #2 cu. The serv. disc. has dual el. current limiting time del. fuses. 200 amp. The starter panel has 175a fuses of the same make. I want to know what the fault current would be at the end of the #2 cu is. From previous threads I see wire length can lower the FC. But how much? I have yet to find a calculator on line to help.
Help would be appreciated. Thanks

 

[don_resqcapt19]

Bussmann has some software you can download and they also have smart phone apps to do that type of calculation.

 

[charlie b]

Is it not true that all electrical panels have a fault current rating of at least 10,000 amps? If so, then you don't really need to do the calculation, as it is clear that the fault current available at the main panel will be lower than 10K.

 

[jim dungar]

Is it not true that all electrical panels have a fault current rating of at least 10,000 amps?

That is pretty much true for power equipment like circuit breaker and fusible panelboards and disconnects.

However for motor control panels (e.g. those built to UL508A), 5kA is not uncommon as it can be achieved with minimal effort.

 

[dave121]

Fault current

Its not the panels, they are all service rated. We are looking at installing 3 pole, 277/480 breakers that have a 10K rating, tapping off the end of the #2 cu wire.
I assume that even though the breaker is 10K, it should be able to handle 10,011 amps. How ever, knowing there is more than 10K I'm not willing to install the breaker. The reason for breakers instead of fused disconnects are cost.

 

[charlie b]

You have at least 20 feet of wire between the point at which the utility gave you the 10,011 amp figure and the location of the breakers. That wire is going to knock more than 11 amps off the fault current available at the breaker. I would not be concerned about using a 10K rated breaker. But here is how it can be calculated:

• At the utility connection point, you have 480 volts and a fault current of 10,011 amps.
• The resistance associated with that fault current is therefore 480/10,011, or .04795 ohms.
• From Table 8, 3/0 copper has a DC resistance of .0766 ohms per 1000 feet.
• 20 feet of this conductor would then have a resistance of .00153 ohms.
• The total resistance is therefore .04795 plus .001532, or about .04948 ohms.
• The available fault current at the breaker is 480/.04948, or 9700 amps.
• This is well below the 10K rating of the breaker.
• Q.E.D.

Note that I used the DC resistance instead of the AC resistance. I did that because I knew it would be lower, and would therefore give me a more conservative (i.e., higher) fault current at the breaker location. Note also that I used the one-way distance of the run. I did that for the same reason.

 

[charlie b]

Its not the panels, they are all service rated.

Being "service rated" does not mean that they are rated to handle the amount of fault current that would be imposed on them at their location within the circuit. You need to show that the available fault current is lower than their rated fault current. But as I mentioned earlier, the panels are going to be rated for at least 10K amps, and I have just shown that the available fault current is lower than that value. So the panels should be fine.

 

[dave121]

The panels are rated at a minimum of 200,000 amps with the fuse rejection clips and class r fuses installed.

Thank you for the help.

 

[Jraef]

I want to know where you found a 480V rated breaker that only has a 10kAIC rating on it! That must be some cheap chit, are you using those little European style breakers that clip onto DIN rail? be careful with those, many of them are not UL listed as "Circuit Breakers", they are called "Supplemental Circuit Protectors" under UL1077 and you are not allowed to use them like this, they must have a UL489 listed Branch Circuit Breaker ahead of them anyway, which makes them pointless.

Maybe that's the real problem and you should consider a regular US style molded case circuit breaker and not worry about it. The cheapest ones I know of will have at least 14kAIC.

 

[dave121]

Yes they are, the breaker says UL 489 rated circuit breaker. 10kA. I am meeting with their rep. and will pose these concerns to them.

 

[Jraef]

Yes they are, the breaker says UL 489 rated circuit breaker. 10kA. I am meeting with their rep. and will pose these concerns to them.

Well if they say UL489 on them, that's OK, it's just that not all of them do, you have to be careful.

I wasn't aware of any of the UL489 versions being only 10kAIC, but in snooping around a little I see some are. Changing brands may be another solution, I saw some that were 14KAIC - 15kAIC.

 

[dave121]

I am also going to be looking at some 14k's. It sounds like we are on the right track. Thanks for the input.

 

[Julius Right]

Bussmann has some software you can download and they also have smart phone apps to do that type of calculation.

Do you mean this:
http://www.cooperindustries.com/con...ch_Lib_Short_Circuit_Current_Calculations.pdf

 

[kwired]

I think with simple ohms law calculations we have determined the current will easily be below 10kA at the equipment in question.

But for future reference Mike Holt has an Excel spreadsheet fault current calculator in the free stuff section of his website.

http://www.mikeholt.com/freestuff-shortcircuitcalculations.php

 

[shortcircuit2]

The last 2 links are really helpful. They don't consider motor contribution though.

Wouldn't motor contribution be important when there is a multi tenant building with a lot of motor loads?

Would one tenants business with many motor loads affect an adjacent tenant's available fault current?

 

[hmspe]

Yes, motors affect fault current, but generally not much. A motor will contribute its nameplate full load amps at most, but the contribution will die off quickly. With this 200 amp service motor contribution should be less than 100 amps based on the 175 amp fuses. As Charlie noted your fault current at the switchgear will be 9700 amps worst case based on 20' of service conductors. 9700 amps + 100 amps is still under 10,000 amps. In this case motor contribution to fault current will not be an issue.

 

[jim dungar]

Along with motor current, you also need to consider the X/R ratio of the circuit.

At this point you are stepping into some mildly technical engineering, therefore many power study people recommend simply adding a 'safety margin' of 10-20% to their device selection or fault current estimation. for example if the calculated fault current was 9kA, they would not use 10kAIC devices.

I know this goes against what most sales people say and contractors do, actually it seems most ignore it entirely, leading to many 'closely sized' devices being judged inadequate during a formal power system study.

 

[kwired]

I know this goes against what most sales people say and contractors do, actually it seems most ignore it entirely, leading to many 'closely sized' devices being judged inadequate during a formal power system study.

Aren't sales people going to push for higher rated (and more profitable) gear when the numbers are close? It is the bean counters that will want to be conservative as much as possible.

 

[GoldDigger]

To an engineer conservative is overbuilding. To a bean counter it is under building and praying. To a sparky conservative is doing it the way they were first taught, with "tried and true" materials.

Sent from my XT1080 using Tapatalk

 

[jim dungar]

Aren't sales people going to push for higher rated (and more profitable) gear when the numbers are close?

When was the last time a contractor got a price from only one sales person?

After selling for 20+ years, I believe AIC ratings are among the first causalities during 'value engineering' and competitive bidding.