Fault current question

[BJAP]

I had previously asked a question on finding the fault current for an inspector at a residence. I now have invormation faxed from the utility company and need some help.

The inspector is asking for the fault current from the transformer to the service, then from the service to the outlets.

This is what we have:

1 Phase 3W 120/240: General Service
Service type: 50 KVA
Fault current Inital 12,300 Ultimate 171,800
Ampres RMS Symmetrical at 240 Volts

Underground is 100 foot
600 amp service

 

[ron]

171,800A
I think there are drugs involved. Not possible with a 50kVA xfmr.
12kA at the secondary stabs of the xfmr is possible with an approx 2% internal impedance.
Fault current at end of 100' (2 sets of 350 AL) is approx 9,500A

 

[BJAP]

Ron,

thanks for the help. What would the approx. fault current be at the devices if it is 9500 at the service?

 

[LarryFine]

Ron,

thanks for the help. What would the approx. fault current be at the devices if it is 9500 at the service?

Underground is 100 foot

Depends. What's the conductor size?

 

[BJAP]

Conductor size is 12-2 romex

 

[bphgravity]

I have yet to see a utility company allow more than 10K amperes of fault current to a residential application. I even believe some are limited to that value per regulation...

 

[mo2004]

For 12/2

9500 amp available fault current at the service.

The fault current at the outlet will depend on how far is that outlet from the panel where the 9500 amp is available.

I calculated for 100 ft away from panel the fault current will be 691 amp this is due the high Z of the 12 awg wires.

If your concern is ARC FLASH your fault current will need to stay around 2000 amp for the circuit breaker to clear the fault in a short duration of time without causing high release on incident energy.

 

[kingpb]

using the MVA Method (see other post) and assuming (a lot) for one the romex is only 4' long (worst case) you might have 8300A at the panel board (L-L). At the receptacle, now your at 120V and a fault would be line to ground. With the voltage dropping to 120V I predict around 8600 A at the receptacle terminals with no contribution from anything plugged into the recep.

I don't think I have ever actually looked at fault current at this level in a system, comments please.

 

[kingpb]

For 12/2

9500 amp available fault current at the service.

The fault current at the outlet will depend on how far is that outlet from the panel where the 9500 amp is available.

I calculated for 100 ft away from panel the fault current will be 691 amp this is due the high Z of the 12 awg wires.

If your concern is ARC FLASH your fault current will need to stay around 2000 amp for the circuit breaker to clear the fault in a short duration of time without causing high release on incident energy.

Using your 100ft I cam up with 678A. I think that confirms we may be in the ballpark. Unfortunately, for short romex runs out of the panel you could be considerably higher.

 

[BJAP]

I have yet to see a utility company allow more than 10K amperes of fault current to a residential application. I even believe some are limited to that value per regulation...

I have a fax from them stating all this information. Maybe we are just screwed up down here

 

[coulter]

I have yet to see a utility company allow more than 10K amperes of fault current to a residential application. I even believe some are limited to that value per regulation...

Just curious - when was the last time you saw a 600A residential service? If you have seen one, what was the size on the POCO transformer?

carl

 

[jim dungar]

I have yet to see a utility company allow more than 10K amperes of fault current to a residential application. I even believe some are limited to that value per regulation...

Utilities in Wisconsin have had a design level of 22kSCA for residential services for at least 25 years.

 

[BJAP]

Ok, well the inspector wants us to use the 171,000 ISCA

So single phase short circuit calculation:

Line to Line:

f= ( 2 x L x ISCA ) / ( N x C x L-L )

m=1 / ( 1+f )

ISCA L-L = ISCA x m

Line to Neutral:

ISCA L-N = ( ISCA L-L x 1.5 )

f= [ 2 x L x ISCA L-N ] / ( N x C x L-N )

m = 1 / ( 1 + f )

ISCA L-N New = ISCA L-N x m

The under ground is parallel 250 copper mcm & 3/0 neutral at each run of 100 foot

Here are the values for pipe 1 & 2:

250 mcm in pvc x 2 : 18593 x 2=?
3/0 thhn in pvc x 1 : 13923 x 1

Can some one work this up for me? If not too much trouble the written work would be nice to try and help me understand the formula.

thanks

 

[kingpb]

You are wasting your time using 171 kA as the short circuit number. it is physically impossible to have that much current on the low side of a 50KVA transformer. Before you proceed , you need to confirm with the POCO, as to WHERE the current is present. From expereience, I will put money on it, that it's on the HV side of the 50KVA transformer.

 

[BJAP]

I have the paperwork from the utility company and the inspector insists that that area is set up at 171,800 ISCA. He will not move forward until I get the numbers from above. Any thoughts?

 

[jim dungar]

I have the paperwork from the utility company and the inspector insists that that area is set up at 171,800 ISCA. He will not move forward until I get the numbers from above. Any thoughts?

Make your service entrance equipment a device rated for 200kSCA. Have this equipment feed your very own 50kVA single phase 240-120/240 transformer with a minimum of 3%ZI. Use the secondary of this transformer to feed your building as a separately derived system which can only supply <10kSCA.

 

[BJAP]

I guess my real problem here is trying to figure out how to use the above formula with the numbers the inspector wants me to use. Is there a place online where I can input these numbers and get a fault current number to satisfy this inspector? He says use the 171,800 number. thanks

 

[BJAP]

Anyone? Please help, thying to get this done for months. Job is at a stand still.

 

[charlie b]

I had previously asked a question on finding the fault current for an inspector at a residence. I now have invormation faxed from the utility company and need some help.

I think that either you are mis-interpreting what the fax says, or the utility company mis-interpreted your request for information. I will send you my fax number via Private Message. Send me a copy, and I'll take a look at it.

 

[charlie b]

OK. I have looked at the utility's fax. It has blank spots that are filled in by hand. One of these is "ultimate" fault current.

I am not certain exactly what is represented by the phrase, ?ultimate 171,800 amps symmetrical at 240 volts.? But it certainly has nothing to do with the fault current available at a specific user?s transformer location. As evidence, I offer the fact that a city in Houston (served by the same utility company as the OP is discussing) has published a list of requirements related to permits and construction. Here it is (copy into browser to view the document):
[FONT='Arial','sans-serif']www.bunkerhill.net/Government/Permits/Contractor%20Permit%20Package%20122007.pdf[/FONT]

On page 18 of that document, it declares that the local utility?s value of ultimate fault current, for single or three phase, is 171,800. So clearly that number is not related to any given installation.

The number you should be using as the basis for the rating of any equipment within your building need be no higher than the 12,300 amps that the utility shows as ?initial fault current.?