Phase VA contribution 1

[mikeengurs]

I'm configuring a 4 wire 120/208v panel and have this question. I have a 1581VA load, that I'd like to supply with a 2-pole 208v supply (Vac). I understand that a current of 7.6A would be seen by phases A and C. What does that mean VA wise to each phase? In other words, would each phase's load be 1581/2? Or that should follow the phase power equation: S=Vpahse*Iphase = 120*7.6?

 

[waross]

Your load current is in phase with the 208 volts from A to C.
The load current is not in phase with the voltages from A to the neutral or from B to the neutral.
In the case of a 120/208 Volt panel the phase current is not equal to the arithmetic sum of the line to line load currents and the line to neutral load currents.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

Why do you want to know.

You can't put a load "on a phase". The load must be connected phase to phase or phase to neutral so the load is across the source. You would be correct trying to determine this if considering the load on the source and trying to determine the capability of the source to supply the load. For example, if the source was 3 single phase transformers wye connected in a bank then you should make sure the load doesn't exceed the capacity of any single transformer. Using your example numbers, the load on a single transformer in this bank would be 120V*7.6A.

 

[mikeengurs]

LionelHutz, the Transformer example you've indicated is applicable here. The situation I have is simillar to 3 single phase transformers, each with 120v to neutral, and all connected in a Y configuration. The 1581VA load is across phases A and C, so the voltage across the load is 208v. This means, the Line current, which is equivelant to the phase current in this case = 1581/208= 7.6A. Now, if both Transformers (A & C) are seeing 7.6*120=912VA, this means that the total power supplied = 912*2=1824. This doesn't make sense, because the load is only drawing 1581VA. There is something I'm missing here.

 

[iop995]

Connecting load between A and C phase mean that current in load will be 1581/208 = 7.6A. A-C phase it's like a single source and can't be seen like 2 parallel source feeding power to load in this configuration. If load is two-phase then these two phase (A and C) may be seen like 2 source and can calculate power supplied by each phase.

 

[xnuke]

You can't add VA values together unless the phase angles are the same, which they aren't. Here is the proof (it helps to sketch out the circuit and the phasor diagrams):

Assume Vca = 208[∠]0[°] V and 1581[∠]0[°] VA (angles assumed for ease of calculation), then
Ic = S/Vca = 7.6[∠]0[°] A and
Ia = -Ic = 7.6[∠]180[°] A

With Vca defined above, then
Vc = 120[∠]-30[°] V and Va = 120[∠]-150[°] V

Finally,
S = Sa + Sc = VaIa* + VcIc* = 912[∠]30[°] VA + 912[∠]-30[°] [≈] 1581 VA.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[mikeengurs]

That makes sense! I failed to recognize the fact that Real power is also a vector! Thanks everyone for teh help!