UL508A Full Load Current Calculation (UL508A 49.2 Interpretation)

[JaguarRD]

Hello,

I have a client who is questioning the procedure for determining Full Load Amps / Current for a UL508A panel and I was wondering if there was a UL-posted example (or advice!) on how this is done.

For a simple example, the panel (please see wiring schematic attached) is fed by 120VAC, and has loads of Utility Light (datasheet rating of 0.25A Max @ 120VAC), Duplex Receptacle (datasheet rating of 15A Max @ 120VAC) and Simplex Receptacle (datasheet rating of 15A Max @ 120VAC). The UPS plugged into the Simplex Receptacle is 8.3A Maximum Load at 120VAC.

I have always performed the calculation by adding the Maximum current values the panel could draw- so I would add the 0.25A (Utility Light), Duplex Receptacle rated amps (15A), and because the UPS is connected to the Simplex Receptacle I used the rated current of the UPS (8.3A) for a total of 23.55A to be noted on the UL508A label affixed to the control panel (and recommended an increase on CB-1 from 15A to 30A)

My client performs this on a measurement basis (putting a meter in series and measuring amp draw) and determined the FLA to be 1.6A (since nothing was plugged into the Duplex Receptacle and the PLC and other downstream devices of the UPS were not using the maximum power available from the UPS).

1.) Which is the correct way for interpreting UL508A 49.2 and determining the UL Label FLA Value?
2.) How should convenience receptacles be treated for FLA calculation? Is it correct to use the maximum rating of the receptacle (or current limiting device, for example if a 5A CB was ahead of the Duplex Receptacle you could use 5A for calculation), or since nothing is plugged into it normally is this not factored in?

Thanks for providing some info before I challenge this labeling issue with my client.

 

[iwire]

I don't know about doing it for UL but doing it for the NEC would involve following Article 220 to calculate the load or using 220.87 to record the existing load.

 

[Jraef]

You might be over thinking it a bit.

Look at 49.2 again. The second section of the first sentence is "...at a minimum,...". So there is nothing wrong with rating it at the MAXIMUM supply current it can handle.

In your drawing (assuming that is everything), I would just put the rating of the first protective device; 15A as shown, or 30A if you change it. You cannot load it any more than that, which is the purpose of the rating nameplate on the panel. It's there so that someone connecting it in the field can determine his MINIMUM circuit size. If you over think it and try to say it is 1.6A or something more like 2.4A or 5.9A, would it matter? He can't run a circuit in the field less than 15A anyway.

Why is really matters to calculate the actual FLA is when you have a motor controller where there is a separate OL relay, because for various reasons the circuit breaker may end up being significantly larger than the actual motor FLA (i.e. to prevent nuisance tripping but provide for short circuit protection). None of that applies here.

 

[JaguarRD]

You might be over thinking it a bit.

Look at 49.2 again. The second section of the first sentence is "...at a minimum,...". So there is nothing wrong with rating it at the MAXIMUM supply current it can handle.

In your drawing (assuming that is everything), I would just put the rating of the first protective device; 15A as shown, or 30A if you change it. You cannot load it any more than that, which is the purpose of the rating nameplate on the panel. It's there so that someone connecting it in the field can determine his MINIMUM circuit size. If you over think it and try to say it is 1.6A or something more like 2.4A or 5.9A, would it matter? He can't run a circuit in the field less than 15A anyway.

Why is really matters to calculate the actual FLA is when you have a motor controller where there is a separate OL relay, because for various reasons the circuit breaker may end up being significantly larger than the actual motor FLA (i.e. to prevent nuisance tripping but provide for short circuit protection). None of that applies here.

Thanks Jraef- I agree we could use the breaker size as the "default" value- I like to provide a more accurate value based on a calculation. Do you see any issues with the way I presume to calculate FLA?

The main issue here is that the client would like to use a "measured" value (reasons unknown) and I think that is dangerous as in my opinion others use the FLA value from the UL Label to plan their branch circuits (sometimes not looking at the drawings!) and based on that may supply a 3A or 5A breaker- which of course will pop when someone plugs a cordless drill in the convenience receptacle and turns it on!

 

[ron]

Thanks Jraef- I agree we could use the breaker size as the "default" value- I like to provide a more accurate value based on a calculation. Do you see any issues with the way I presume to calculate FLA?

The main issue here is that the client would like to use a "measured" value (reasons unknown) and I think that is dangerous as in my opinion others use the FLA value from the UL Label to plan their branch circuits (sometimes not looking at the drawings!) and based on that may supply a 3A or 5A breaker- which of course will pop when someone plugs a cordless drill in the convenience receptacle and turns it on!

Is the receptacle rated for more than 15A.

It would seem that there is no reason to indicate the FLA for be more than 15A, since that is the breaker size.

 

[jim dungar]

... may supply a 3A or 5A breaker- which of course will pop when someone plugs a cordless drill in the convenience receptacle and turns it on!

I have seen panels were the convenience receptacle has a label indicating there is a load restriction (i.e. 2Amps). If the breaker trips, it is the problem of the person using the receptacle not that of the panel manufacturer.

Usually I find receptacles wired like this to be inconvenient and a waste of panel space. Can you imagine having to tell someone that you caused their control panel to 'trip off-line' during production. Also, during LOTO the de-energized receptacle won't be of much use at all.

 

[JaguarRD]

I agree- when doing my own panel designs I always limit the receptacles through use of a CB (5A, for example) to prevent the entire panel from tripping if too much current is drawn.

Back to the OP, are we in agreement that simply measuring the current drawn by the panel at production is an incorrect way of determining FLA for UL508A calculations? Should a provided receptacle which is not marked be calculated as its rated current (unless current limited through a separate breaker)?

Thank you!

 

[petersonra]

On another issue, is the UPS appropriate?

I am not in my office so have no access to the standard but I seem to recall that generic UPS units (like you might buy at Walmart) do not have the proper listing to be used in a UL508a panel.

 

[JaguarRD]

On another issue, is the UPS appropriate?

I am not in my office so have no access to the standard but I seem to recall that generic UPS units (like you might buy at Walmart) do not have the proper listing to be used in a UL508a panel.

Hi Bob,

Normal "off the shelf" UPS units are not listed- however the 120VAC ones we typically use in panels carry UL 1778 listing, true on-line double conversion. Wherever possible I like to use 24VDC battery backup as most of the components are typically 24VDC so this is more efficient.

 

[Jraef]

Although my copy is too old to quote the exact section, I was told the newest version of 508A now allows inclusion of UL 1778 listed UPS units in panels.