How To Determine FLA of a 508a Panel?

[Whorse]

I've searched around a bit and can't find a clear answer or much information on this anywhere.

UL508a Section 49.2 (Supply Ratings)

The full-load ampere rating of the panel shall, at a minumum, include the sum of the ampere ratings of all loads that are able to be operated simultaneously plus the primary ampere rating of all control transformers connected to the input voltage.

How does this apply to multi-phase systems? Currently, we just total up all the loads, regardless of phase, and list that as the FLA of the panel.

Main disconnect/breaker sizing is based on 80% of the highest FLA per phase, which leads to some angry installers, because the FLA is larger than the Main breaker size.

Thanks,

JB

 

[charlie b]

I am not sure I understand the situation, not being familiar the rules for establishing ratings.

I will say, however, in case it helps, that for 3-phase systems you can?t add amps to amps and get amps as a correct result. The reason is that current leaving the source on one phase will return to the source on both of the other two phases, and there will be a shift in phase angle from one to the others. An attempt to add two or more currents, each of which reaches its peak value (i.e., during the ups and downs of the sine wave pattern) is not an easy task, and one that easily leads a person astray.

It is always safer to convert each load to units of power (VA). You can add VA to VA and get a correct result for total VA. For single phase 120 volt loads, regardless of which phase (A, B, or C) powers the load, you take 120 volts times the load current. For single phase 208 volt loads (I don?t know if your system will have any of these), you take 208 volts times the load current. For three phase 208 volt loads, you take 208 times the load current times the square root of three. Then you add these three values to get a total VA. Finally you divide the total VA by 208 volts and divide again by the square root of three, and that is your load current.

Welcome to the forum.

 

[Whorse]

Good post. To add a bit more insight:

We do industrial controls for greenhouses, so many of our panels control 3 phase reversing motors, single phase reversing tube motors, 3 phase motors for pumps, transformers for control circuits, breakers for aux. receptacles, and an assortment of lighting systems.

When dealing with single phase 208V or 480V lighting systems, we do the root 3 calculation. We generally don't control much equipment that is single phase line to line, however. Mostly it is single phase line to neutral, so we try to balance the loads across all 3 phases.

The confusion is when we have a few 3 phase motors, plus a number of single phase motors, all from the same supply. A 60A breaker might be protecting a phase A of 36 amps, phase B of 32 amps, and a phase C of 40 amps (arbitrary numbers).

The breaker is sufficient for all the loads, but the FLA number would be more than 60 amps, if calculated the way 49.2 reads.

 

[LarryFine]

UL508a Section 49.2 (Supply Ratings)

Oh. I was wondering what a 508 amp panel was. :roll:

 

[charlie b]

. . . but the FLA number would be more than 60 amps, if calculated the way 49.2 reads.

Not true at all. That was the point of my earlier post. You do not take the three individual phase values (36, 32, and 40, to use your numbers), add them up as though you were adding apples to apples, and get a result of 108. You are not adding amps to amps to amps, and the answer is not in amps. You are adding a thing called "amps at a phase angle of zero degrees" to a thing called "amps at a phase angle of 120 degrees" to a thing called "amps at a phase angle of 240 degrees." You don't get the right answer by just looking at the numbers.

Thus, when 49.2 tells you to use the sum of the ampere ratings, you don't do the summation that way. You do it the way I described earlier. Using your numbers, you are going to wind up with a result of 36 (in this case, it works out to be the average of the three values).

 

[Whorse]

What you're saying makes sense, but don't all the inductive loads pull the phases out of sync?

I'm also a bit unclear on how the line to neutral calculations on different phases can be accurate when divided by root 3 at the end of the calculation.

 

[Smart $]

What you're saying makes sense, but don't all the inductive loads pull the phases out of sync?

To be accurate, the load current angles will not be the same as the supply voltage angles. To get a more precise result, you would have to include the power factor for each load into your calculations and use vector math to total. Being unfamiliar with establishing UL supply ratings, I have to ask how precise must your calculations be?

I'm also a bit unclear on how the line to neutral calculations on different phases can be accurate when divided by root 3 at the end of the calculation.

Pretty simple really...

xVA ? 208 ? √3 = xVA ? 120 ? 3​

However, there is some slight difference due to rounding: 120 ? √3 = 207.8460969082652752232935609807.

 

[Whorse]

To be accurate, the load current angles will not be the same as the supply voltage angles. To get a more precise result, you would have to include the power factor for each load into your calculations and use vector math to total. Being unfamiliar with establishing UL supply ratings, I have to ask how precise must your calculations be?

I don't think they need to be very precise at all. Usually calculations are based on a power factor of .9 unless otherwise stated in the scope of the job. The issue is when the numbers we show appear to cause an issue with the size of the main disconnect in the panel.

Pretty simple really...

xVA ? 208 ? √3 = xVA ? 120 ? 3​

However, there is some slight difference due to rounding: 120 ? √3 = 207.8460969082652752232935609807.

Will this still work if the single phase loads aren't balanced?
Quite often we'll end up with a panel that has a bunch of three-phase motors, and then one large single phase motor which puts one phase out of balance.

I think the calculations above might be more towards sizing the overcurrent protection than calculating the actual FLA rating of the panel. The UL code is a bit unclear so we choose our ratings by just totalling all the loads like it says, to be on the safe side.

 

[Smart $]

I don't think they need to be very precise at all. Usually calculations are based on a power factor of .9 unless otherwise stated in the scope of the job. The issue is when the numbers we show appear to cause an issue with the size of the main disconnect in the panel.

Will this still work if the single phase loads aren't balanced?
Quite often we'll end up with a panel that has a bunch of three-phase motors, and then one large single phase motor which puts one phase out of balance.

...and an averaging method, such as the one described by charlie b, could cause such an issue (sorry, charlie ).

I think the calculations above might be more towards sizing the overcurrent protection than calculating the actual FLA rating of the panel. The UL code is a bit unclear so we choose our ratings by just totalling all the loads like it says, to be on the safe side.

The way I would do it if I were you, would be to include pf into your calculations, and keep 3?, 1? L-L, and 1? L-N loads separate per line (phase), then use vector math to determine the highest of the three line currents.

The 3? and 1? L-N loads are the easiest to total per phase. The 1? L-L loads are a bit harder to figure in. I have often considered developing an Excel spreadsheet for the purpose. There is no sense of urgency for me to do so being a field electrician.