Balancing Loads in Lighting panel

[Charz]

I have two type of Lighting fixtures fed from the same lighting panel. One has a voltage rating of 120 VAC (Line - Neutral) and the other has 208 VAC (Line-Line).
How to balance the loads in the panel?

 

[LarryFine]

That's easy: 120v from L1, 208v from L2 & L3.

 

[charlie b]

I think it's perhaps not quite as easy as that. Start with listing the 120V loads and the 208V loads. List them in terms of the VA drawn on each voltage level. If for example, the total 208V load is far greater than the total 120V load, then Larry's suggestion might not work as well as you could wish. I would try to put the 208V load on 3 circuits (or 6, or 9, as needed), and connect them as L1-L2, and L1-L3, and L2-L3. I would try to do the same with the 120V loads, with 1/3 of the total load on each phase.

 

[LarryFine]

I didn't realize we're talking about multiple lights of each type. :ashamed1: How many circuits of each load type?

It's still easy: presuming similarly-loaded circuits, stacking the 2p breakers down one side of the panel will equalize load power distribution.

To wit: A-B, C-A, B-C, A-B, C-A, B-C, etc.

The 1p breakers will similarly distribute loading, wherever they can also be installed, whether below or beside the others.

 

[Charz]

My concern is, If 120V fixture (Line - Neutral) is connected to A phase, then I would put the current drawn in the Ia column (current in A phase)
For 208V fixture (Line-Line) connected to B & C Phase, Should I mention the current in both B & C Phase?

 

[LarryFine]

My concern is, If 120V fixture (Line - Neutral) is connected to A phase, then I would put the current drawn in the Ia column (current in A phase)
For 208V fixture (Line-Line) connected to B & C Phase, Should I mention the current in both B & C Phase?

Yes.

 

[charlie b]

I agree with Larry's "Yes." But I disagree with the question. The information you insert into a panel schedule is the power, in units of VA, and not the current, in units of amps. As an example, if you have 10 amps worth of lighting load on a 208 volt, single phase circuit, then you convert the 10 amps to VA by multiplying 10 x 208, to get a total load of 2080 VA. Then you put half of that (1040 VA) on phase B and the other half on phase C.

 

[synchro]

I agree with Larry's "Yes." But I disagree with the question. The information you insert into a panel schedule is the power, in units of VA, and not the current, in units of amps. As an example, if you have 10 amps worth of lighting load on a 208 volt, single phase circuit, then you convert the 10 amps to VA by multiplying 10 x 208, to get a total load of 2080 VA. Then you put half of that (1040 VA) on phase B and the other half on phase C.

Charlie's way of using volt-amps is the best way to do it because power is conserved, and so it's less likely to make errors using VA's.
However, if you really want to do it with currents, then to do it correctly you need to add up the equivalent "line" currents on each phase. In Charlie's example the line current on both Phase B and on Phase C from the 208V load would be (sqrt(3)/2) x 10A = 0.866 x 10A = 8.66A. Note that this current develops the same VA as Charlie came up with because 120V x 8.66A = 1040 VA.
The Line-Neutral load currents need no correction to get their contribution to the line current on each phase, and therefore they can be added together directly on each phase.

 

[Hv&Lv]

Charlie's way of using volt-amps is the best way to do it because power is conserved, and so it's less likely to make errors using VA's.
However, if you really want to do it with currents, then to do it correctly you need to add up the equivalent "line" currents on each phase. In Charlie's example the line current on both Phase B and on Phase C from the 208V load would be (sqrt(3)/2) x 10A = 0.866 x 10A = 8.66A. Note that this current develops the same VA as Charlie came up with because 120V x 8.66A = 1040 VA.
The Line-Neutral load currents need no correction to get their contribution to the line current on each phase, and therefore they can be added together directly on each phase.

I have a question...
amps and va all good, either or..

my question to the OP. When getting the values of the amps, are you getting it from watts divided by voltage or by the manufactures amp draw rating on the device?
In 220 there is a piece on using actual amp values for the inductive lights or the LED drivers.

The reason I ask is because I have seen some that take the wattages of all their fixtures and simply divide by the voltage.
Then they are surprised when the amps is far greater than they calculated for because of PF.