Calculate current limiting effect of conductor length..

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1. An industrial control panel is labeled with an SCCR of 5kA
2. The feeder supplying it utilizes a J class 25 amp fuse with a let-through current of 5.5kA (UL 508A table SB4.2)
3. I want to prove that a 50 foot length of #10 cable supplying power to this 5 kA rated panel will effectively limit the available fault current to a value below 5kA.

Can it be as simple as taking the resistance of 50 feet of #10 at 0.05 ohms times 2 (out and back) for a total of 0.10 ohms, using I=E/R and coming up with 480 V / 0.1 Ohms = 4800 Amps?
 

bob

Senior Member
Location
Alabama
Rich,
That is basically it. I ran the calcullation on my computer and got a shade over the 5000. I would do as Augie and Sog said. Use the online calculators. You can include more information and get even a lower fault current. This installation is called a series rating where you have a fuse and a breaker in series. The combination is tested by the mfg to verify that it works correctly.
 

mull982

Senior Member
1. An industrial control panel is labeled with an SCCR of 5kA
2. The feeder supplying it utilizes a J class 25 amp fuse with a let-through current of 5.5kA (UL 508A table SB4.2)
3. I want to prove that a 50 foot length of #10 cable supplying power to this 5 kA rated panel will effectively limit the available fault current to a value below 5kA.

Can it be as simple as taking the resistance of 50 feet of #10 at 0.05 ohms times 2 (out and back) for a total of 0.10 ohms, using I=E/R and coming up with 480 V / 0.1 Ohms = 4800 Amps?

I would think that you need to take the source impedance into consideration. A 1500kVA would supply a higher fault current then a 500kVA transformer for example.

Maybe I am overlooking something but I dont believe you can just divide the bus voltage by the impedance without knowing the upstream system capacity and impedances?
 

kingpb

Senior Member
I don't think a transformer or system ahead of it comes into play since the assumption is that there is enough fault current ahead of the fuse, that the maximum 5.5KA is going to be available. That would be the conservative approach. In this case the fuse maximum let-through is acting the same as a transformer would do.

The cable will also act as a choke and have some limiting of the initial 5.5KA current. I modeled it in ETAP and get 2.7kA at the panel, and by hand I got 3.2kA. The hand calc is using the MVA method and it is on the conservative side, so the results make sense to me.
 

steve66

Senior Member
I don't think a transformer or system ahead of it comes into play since the assumption is that there is enough fault current ahead of the fuse, that the maximum 5.5KA is going to be available. That would be the conservative approach. In this case the fuse maximum let-through is acting the same as a transformer would do.

The cable will also act as a choke and have some limiting of the initial 5.5KA current. I modeled it in ETAP and get 2.7kA at the panel, and by hand I got 3.2kA. The hand calc is using the MVA method and it is on the conservative side, so the results make sense to me.
That's close to what I got with a spreadsheet that does a calculation based on the point to point method described in the Bussman SPD handbook. I got a little over 2.5KA.

The previous answer of 400 amps seems low, and the other answer of over 5KA seems high.
 
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