Panel listing

[4827]

We are looking to get a panel listed by Ul and I have a question? When you have vfd's in your motor control cabinet you go off the input current of the drive for FLA in all calculations not the motor correct? If so is this right? Power supply @2amps,6-drives @3.2amps protected by 4amp breakers

Branch circuit conductor and OCPD( 1.25*3.2 )=4amps
Feeder for motor control panel (1.25*3.2)+(3.2*5)+2=22amps
OCPD for motor control panel 4+(3.2*5)+2=22amps round down 20amp OCPD

So FLA of the panel would be labeled 22amps?

Seems off that the motor really means nothing when using a vfd

 

[bob]

We are looking to get a panel listed by Ul and I have a question? When you have vfd's in your motor control cabinet you go off the input current of the drive for FLA in all calculations not the motor correct? If so is this right? Power supply @2amps,6-drives @3.2amps protected by 4amp breakers

Branch circuit conductor and OCPD( 1.25*3.2 )=4amps
Feeder for motor control panel (1.25*3.2)+(3.2*5)+2=22amps
OCPD for motor control panel 4+(3.2*5)+2=22amps round down 20amp OCPD

So FLA of the panel would be labeled 22amps?

Seems off that the motor really means nothing when using a vfd

Your calculations are for the conductor sizing and not the OCP. The feeder OCP is(if you use the max %)

250% of the largest motor FLA + the FLA of the remaining motors. 2.5 x 3.2 + 5 x 3.2 + 2 power supply so we have 8 + 16 + 2 = 26 amps.

The motors as so small that there is not much difference.
I'm not sure what you are looking for.

 

[petersonra]

are you trying to calculate the nameplate full load current?

 

[4827]

I just want to know if this is correct when sizing a motor control panel for our customer and more important UL. They require a making for FLA of a panel on the outside of the enclosure and I would assume that would basically be the feeder conductor calculation. I sized the breaker for the drive according to the 80 percent rule. I do not need to size it 2.5 times the load. My questions are:
A.Do you always use the drive input current for FLC calculations and not the motor?
B.Is the feeder calculation basically the FLA for the motor control cabinet?

 

[Jraef]

I just want to know if this is correct when sizing a motor control panel for our customer and more important UL. They require a making for FLA of a panel on the outside of the enclosure and I would assume that would basically be the feeder conductor calculation. I sized the breaker for the drive according to the 80 percent rule. I do not need to size it 2.5 times the load. My questions are:
A.Do you always use the drive input current for FLC calculations and not the motor?
B.Is the feeder calculation basically the FLA for the motor control cabinet?

I don't quite understand.

1. If you are a UL 508A listed panel shop, you would know this because it is in your procedure: 49.2 for the panel FLA calculations, 50 for the individual circuit calculations, 52 for the labeling requirements.
2. If you are not a UL 508A listed shop and you are sending your panel to one for listing, THEY are the ones that are required to list the information, I would imagine that would be included in their price anyway (it always was with me).
3. If you are not a UL 508A listed panel shop and you think you can just build a panel to UL standards and that is OK, you are a little deluded. You will either have to chose 1 or 2 above, or pay UL to do a field inspection, which will cost you a LOT more than taking it to a panel shop that is already listed.