Single phase load 120 / 208V power question

[cppoly]

A 60/60/2 fused disconnect serves a UPS which has and 208V/120V, 1 phase, 3 wire output.

The UPS serves three separate 120V, 12A loads.

Is the total power consumed at the UPS: 120V x 12A = 1,440 VA x 3 = 4,320 VA?

Is the amps seen at the line side of the UPS 4,320 / 208V = 20.7 amps?

 

[petersonra]

I don't get your question.

Is this a single phase 208 volt input ups with a 120 vac output?

I am pretty sure the manual that came with it will tell you the maximum input current and voltage.

The total input VA will likely be more than the output VA because the battery charger has to be able to run while still providing full output current.

 

[wwhitney]

A 60/60/2 fused disconnect serves a UPS which has and 208V/120V, 1 phase, 3 wire output.

The UPS serves three separate 120V, 12A loads.

Is the total power consumed at the UPS: 120V x 12A = 1,440 VA x 3 = 4,320 VA?

I think you mean power exiting the UPS, it is "consumed" at the loads. But yes.

Is the amps seen at the line side of the UPS 4,320 / 208V = 20.7 amps?

That depends on what the UPS is doing. If it's doing 208V AC to DC to AC on the outputs, then it would be 20.7A divided by the efficiency of the above process. [So if that efficiency is 90%, it would be 20.7 / 0.9 = 23A. The UPS would be "consuming" the other 2.3A and generating 2.3A * 208V = 478W of heat.] If the DC bus is supported by batteries, then the battery charge rate (AC input) and the battery discharge rate (AC output) could be completely uncoupled, so the AC input current could be anything between 0 and the maximum battery charge rate.

But if it's operating in a pass through mode, no, it can't be drawing 20.7A on each line. It would be as if the UPS were a small panel and the 3 loads were on separate branch circuits. E.g. if the 3 loads are identical, and all 3 are on L1-N, then the currents would be L1 = N = 36A, L2 = 0. If they are split 2 on L1-N and 1 on L2-N, then the current would be L1 = 24A, L2 = 12A, N = 12*sqrt(3) = 20.8A.

Cheers, Wayne

 

[cppoly]

Thanks to clarify, the UPS is 208/120V, 1 phase 3 wire input.

The UPS output is supplying 120V single phase loads. In this case, three 120V, 12 amp loads.

In pass through mode, the UPS would be seeing 4,320 VA?

Thanks for the clarification on the current:

"But if it's operating in a pass through mode, no, it can't be drawing 20.7A on each line. It would be as if the UPS were a small panel and the 3 loads were on separate branch circuits. E.g. if the 3 loads are identical, and all 3 are on L1-N, then the currents would be L1 = N = 36A, L2 = 0. If they are split 2 on L1-N and 1 on L2-N, then the current would be L1 = 24A, L2 = 12A, N = 12*sqrt(3) = 20.8A."

 

[wwhitney]

In pass through mode, the UPS would be seeing 4,320 VA?

In pass through mode, ignoring the resistive losses within the UPS itself, it will be seeing 4,320VA in and 4,320 VA out. On the output side, you have only two wire circuits, so it's easy to say that each of the six conductor has a current of 12A. On the input side, those currents will combine according to the proper vector addition to give you just 3 currents on 3 wires.

Cheers, Wayne

 

[infinity]

The UPS output is supplying 120V single phase loads. In this case, three 120V, 12 amp loads.

Depending on how the branch circuits are wired you could end up with A-N (12+12) = 24 amps, B-N = 12 amps.

 

[wwhitney]

Depending on how the branch circuits are wired you could end up with A-N (12+12) = 24 amps, B-N = 12 amps.

That's true they way you state it (assuming identical loads), as you are giving currents for two overlapping two-wire subsets But if you actually want to know the current on N on the input side, you have to add those currents, and they are 120 degrees out of phase. So they add as 24 plus 12 gives 20.8A. E.g. via the law of cosines.

Cheers, Wayne

 

[infinity]

That's true they way you state it (assuming identical loads), as you are giving currents for two overlapping two-wire subsets But if you actually want to know the current on N on the input side, you have to add those currents, and they are 120 degrees out of phase. So they add as 24 plus 12 gives 20.8A. E.g. via the law of cosines.

Cheers, Wayne

Did he ask about the neutral current? I thought that the question was what is the current on the line side when the UPS is in bypass?

 

[wwhitney]

Did he ask about the neutral current? I thought that the question was what is the current on the line side when the UPS is in bypass?

Right, and the line side has 3 conductors, A, B and N (*); in your example, the line side currents would be A = 24, B = 12, N = 20.8. Same as if the UPS were a panelboard.

Cheers, Wayne

(*) Edit: that's not explicit in the OP. But if the line side were only A and B, the UPS can't serve 120V loads in bypass mode; it could only serve 208V loads in bypass mode.

 

[infinity]

Right, and the line side has 3 conductors, A, B and N (*); in your example, the line side currents would be A = 24, B = 12, N = 20.8. Same as if the UPS were a panelboard.

Cheers, Wayne

(*) Edit: that's not explicit in the OP. But if the line side were only A and B, the UPS can't serve 120V loads in bypass mode; it could only serve 208V loads in bypass mode.

I'm not trying to read too much into this other than he said 3-12 amp loads @120 volts.

 

[wwhitney]

I'm not trying to read too much into this other than he said 3-12 amp loads @120 volts.

OK. I was just expanding on your answer that one possibility is "A-N = 24A" and "B-N =12A" by computing the actual neutral line side current in that case.

Cheers, Wayne

 

[infinity]

OK. I was just expanding on your answer that one possibility is "A-N = 24A" and "B-N =12A" by computing the actual neutral line side current in that case.

Cheers, Wayne

You're probably correct that he would want to know that too.