AIC Ratings, and Available Fault Current at Service

[IllinoisElectrician]

This is kind of a followup to my last thread. I am supposed to install a generator 10kw generator, the ATS will be fed by the utility from a 50A breaker off the main panel.

My predicament is, there are 18,332A in available fault current from the 75kw transformer that supplies this house (200A Service, 120/240 3W 1P). The transfer switch has a "Amps Withstanding" rating of only 10,000kA.. the next one that that will work that is 22.5kA is ungodly expensive.

My question: Is the available fault current at the feeders different than at the service? And, by feeding the ATS with a breaker that has a 22.5kAIC rating, does that mean it isn't a problem that I use a pan only rated to withstand 10kA? I'm assuming no, but I really don't understand why not.

Oh, and... can someone explain available fault current to me so I have a more fundamental understanding of it?



THANKS!

 

[ozark01]

I am defintely not an expert but I have used current-limiting fuses to solve this problem. The disconnect and fuses are not that expensive compared to the higher rated ATS.

 

[dkarst]

My predicament is, there are 18,332A in available fault current from the 75kw transformer that supplies this house (200A Service, 120/240 3W 1P). more fundamental understanding of it?

THANKS!

The first question is where is the 18,332 A available... at the output terminals of the transformer or at the location of the service panel/proposed ATS? You need the available fault current at the ATS location. A 75KVA transformer is quite large.

 

[raider1]

This is kind of a followup to my last thread. I am supposed to install a generator 10kw generator, the ATS will be fed by the utility from a 50A breaker off the main panel.

My predicament is, there are 18,332A in available fault current from the 75kw transformer that supplies this house (200A Service, 120/240 3W 1P). The transfer switch has a "Amps Withstanding" rating of only 10,000kA.. the next one that that will work that is 22.5kA is ungodly expensive.

My question: Is the available fault current at the feeders different than at the service? And, by feeding the ATS with a breaker that has a 22.5kAIC rating, does that mean it isn't a problem that I use a pan only rated to withstand 10kA? I'm assuming no, but I really don't understand why not.

The available fault current will decrease with the impedance of the feeder conductors. You will need to calculate the available fault current at the ATS.

110.10 requires that the equipment used in a premise wiring system have a short circuit rating that is equal or greater than the available fault current at the point of use. So a 10,000 amp SCCR rated ATS would be a violation of 110.10 if the available fault current exceeded 10,000 amps.

You may be able to limit the available fault current by the use of current limiting fuses upstream of the ATS.[/quote]

Oh, and... can someone explain available fault current to me so I have a more fundamental understanding of it?



THANKS!

Available fault current is the amount of current that the utility can provide during a bolted fault condition. This will represent the amount of current that may flow through the equipment for a very short duration until the overcurrent protective device opens the circuit.

Chris

 

[don_resqcapt19]

I am defintely not an expert but I have used current-limiting fuses to solve this problem. The disconnect and fuses are not that expensive compared to the higher rated ATS.

The code does not permit the use of current limiting fuses for this purpose.

 

[Girl Engineer]

I think the rating you are talking about is the withstand rating. This rating takes into acount the forces on the bus during a fault.
From what I have read, if you install a breaker that is rated for the fault current before the transfer switch, you are protecting that transfer switch, and can take into account the inturrupting rating of the breaker. (See EC&M Electrical Calculations Handbook chapter 5 and IEEE Red Book 5.1.2)
Also, be sure to include the contribution from the generator you are installing.

 

[jim dungar]

The transfer switch manufacturer will be able to tell you what protective devices their product has been tested with, if any. You may not create your own combination ratings.

 

[ozark01]

The code does not permit the use of current limiting fuses for this purpose.

I believe you but can you point me to the code section? The only time I have done this was on a 1200 amp 3ph/4wire panel with the main fused with current limiting fuses that fed a switch gear.

 

[don_resqcapt19]

I believe you but can you point me to the code section? The only time I have done this was on a 1200 amp 3ph/4wire panel with the main fused with current limiting fuses that fed a switch gear.

For circuit breakers you find that in 240.86. I am not sure where you find something that tells you you can't do it for transfer switch.

 

[zog]

The code does not permit the use of current limiting fuses for this purpose.

Current limiting fuses are perhaps one of the most misunderstood and misapplied devices related to the OP's question. I saw a auto plant replace 30,000 fudes with current limiting ones thinking that was all they needed to do.

 

[jim dungar]

For circuit breakers you find that in 240.86. I am not sure where you find something that tells you you can't do it for transfer switch.

UL White book Guide WPTZ requires a listed switch, without integral protection, to be marked with its intended protective device. This means the listing procedure includes the only valid combination ratings.

 

[LEO2854]

Current limiting

Current limiting

I believe you but can you point me to the code section? The only time I have done this was on a 1200 amp 3ph/4wire panel with the main fused with current limiting fuses that fed a switch gear.

You need to read 110.10 and the comments after this will help you.
look at this link there are some good stuf about the subject.
http://205.243.100.155/frames/longarc.htm#480_volt_arc_flash

 

[ozark01]

Thanks for the link! I read the code section and the info at the link but I could not find where the use of current limiting fuses was prohibited for the application listed in the first post.

 

[jghrist]

My interpretation is that current-limiting protective devices can by used to limit short-circuit current to below the withstand rating of equipment, but that overcurrent protective devices such as breakers will not necessarily interrupt the reduced current because of the complex interactions during interruption. See Rule 240.2 and the comments following in the NEC Handbook (at least in the 2005 Handbook).

Properly selected current-limiting overcurrent protective devices, such as the one shown in Exhibit 240.1, limit the let-through energy to an amount that does not exceed the rating of the components, in spite of high available short-circuit currents.

Rule 240.86 makes it clear that circuit breakers have to be tested in combination with the current-limiting device. I don't think this applies to devices that don't interrupt short-circuit current, such as switches and bus.

 

[wirenut1980]

Hmm, my interpretation is you cannot use a current limiting fuse to limit fault current to the transfer switch unless it has been tested and is an approved combination. I would have no idea whether or not the current limiting fuse would blow in time before letting through enough energy to damage the transfer switch.

 

[jghrist]

UL White book Guide WPTZ requires a listed switch, without integral protection, to be marked with its intended protective device. This means the listing procedure includes the only valid combination ratings.

The guide says

Transfer switches without integral overcurrent protection, are marked to indicate the maximum rating of overcurrent protection to be provided ahead of the transfer switch.

To me, this means the load rating, not the short-circuit withstand rating or a particular device in combination with the switch.

Switches without integral overcurrent devices are tested for 3 cycles at their short-circuit rating. If the current is limited by a current-limiting fuse, it is interrupted in 1/2 cycle.

 

[jim dungar]

The guide says

To me, this means the load rating, not the short-circuit withstand rating or a particular device in combination with the switch.

Switches without integral overcurrent devices are tested for 3 cycles at their short-circuit rating. If the current is limited by a current-limiting fuse, it is interrupted in 1/2 cycle.

Have you read the entire UL White Book section? My reference is directly related to the short circuit rating of the switch.

In my experience. very few transfer switches, less than 800A, are rated above 10kA except when they are paired with a listed protective device (and yes type J fuses are common).

 

[rcwilson]

Oh, and... can someone explain available fault current to me so I have a more fundamental understanding of it?

It's ohms law. I = V/Z, where Z is the impedance of the entire power system from the generators, through transmission lines, transformers and cables to the bus of your 240/120V panel.

Available fault current is calculated by assuming a dead short circuit at the panel location (0 ohms) so the only limit to the short circuit current is the impedance of the upstream power system. More powerful electrical systems will have more available short circuit current.

In your case, you have a 240V panel and 18,332 amps: Z= 240V/18332 A = 0.01309 ohms. The utility has calculated that the equivalent 240V impedance of the entire North American power system is 13 milliohms as seen from your panel.

Sometimes the fault level is given in kVA or MVA. The short circuit kVA or power at your panel is 240V x 18,332A /1000 = 4,400 kVA = 4.4 MVA.

The available fault current is just the maximum current that could flow to a 0 ohm short circuit at the location of interest.

It takes some engineering math tricks to add and combine all the utility impedances, convert between voltage ratings, etc. But the utility probably didn?t do that. Since 90% of the impedance is in the low voltage transformer and wires, they probably just took typical values for a 75 kVA transformer that has about 1.7% impedance and calculated the short circuit power.

kVA short Circuit = KVA/%Z = 75/1.7% = 4,411 kVA. Or maybe it was a 1.5% transformer and they counted the impedance of the service drop.

[For example: 75kVA/1.5% = 5,000 kVA, Ohms = VxV/(kVA) = (240 x240/(5,000x1000)= 0.01152 ohms is the transformer impedance. The cable impedance would be = 0.01309-.01152= 0.00157 ohms. #2/0 cable is about 0.110 ohms/1000 feet so .00157 ohms is about 14 feet of #2/0.]

Another way to look at it is how much impedance needs to be added to drop the current to the 10 KA rating?

240V /10,000A = .024 ohms. We calculated impedance at the panel to be 0.01309 ohms so you need (0.024-0.01309) = 0.0109 ohms more. A 50 Amp circuit might be #6 awg which is .513 ohms/1000 ft. Length needed = 0.0109 x 1000/0.513 = 21.3 feet.

A 12 foot run of #6 awg will add 24 feet of wire to the circuit and drop the 240V fault current to under 10 kA at the ATS.

Bottom line, the ATS is probably OK, but .......we can?t say that.

Now for the buts:

120V fault level will be higher. (Same short circuit kVA, lower voltage means more amps. The utility transformer impedance for 120V will be higher also, typically around 20-50% per IEEE Std 142).

Maybe there are some motors that will increase the fault level. Motors act like generators during a fault and increase the current. (The standby generator will have no impact on this fault level because it is not on line with the utility).

I can?t tell you it will be OK. You need to provide the inspector with some calculations to convince him. Find a friendly engineer or switchgear salesman or fuse salesman who has the software or time to make the calculation look good for you. Or look at some of the short circuit calculations in Ugly?s page 53-55 or similar references.

Good Luck.

 

[Girl Engineer]

The IEEE violet book has some pretty clear calculations that show the withstand rating does not take into account the circuit breakers or fuses. It only takes into account the impedance of the line for reduction of fault current ratings. (My post earlier was incorrect)

Also, if the fault was at the switch, the generator would have a contribution. The fault pulls from all sources, including the utility, motors, and generators.

 

[jghrist]

Also, if the fault was at the switch, the generator would have a contribution. The fault pulls from all sources, including the utility, motors, and generators.

If the fault is in the switch, it is destroyed anyway and its withstand rating is not material. If the fault is next to the switch, then either the utility or the generator contributes to the fault, not both.