2 Pole 208 V Load calculations.

[dennisohara2]

I have a 2 pole 208 volt L-L 120 volt L-N UPS system. The total power displayed is 1848 watts. Using Ohms Law I=P/E I get 8.9 amps. Does that mean I have 8.9 amps per phase average?

Should power be calculated by voltage L-L times average amps per phase 208*8.9 = 1848 watts or the L-N voltage (120) * amps on one phase (8.9) + L-N voltage (120) * amps on the other phase (8.9) = 2,136 watts?

Is there a difference if I had a 2 pole 240 volt L-L? Would I calculate by (120*I) + (120*I) = P or 240 * I avg = P

 

[ActionDave]

Welcome to the forum.

This just came up recently. Check out this thread and see if it helps.
http://forums.mikeholt.com/showthread.php?t=169084

 

[mivey]

I have a 2 pole 208 volt L-L 120 volt L-N UPS system. The total power displayed is 1848 watts. Using Ohms Law I=P/E I get 8.9 amps. Does that mean I have 8.9 amps per phase average?

Should power be calculated by voltage L-L times average amps per phase 208*8.9 = 1848 watts or the L-N voltage (120) * amps on one phase (8.9) + L-N voltage (120) * amps on the other phase (8.9) = 2,136 watts?

Is there a difference if I had a 2 pole 240 volt L-L? Would I calculate by (120*I) + (120*I) = P or 240 * I avg = P

You are missing the fact that if you are mixing L-L and L-N then there are phase angles to be dealt with, even with resistive loads. Reactive loads mean there are additional phase angles but let's discuss resistive loads to keep it simpler, at least to start with.

Complex power is given by the complex product S = VI* where we take the complex conjugate of the current. This gives use power in the form S = P + jQ. P is the real component and Q is the reactive component.

For balanced L-N loads, you would have 7.7 amps per phase. For example:
Ia = 7.7<0
Ia* = 7.7<0
Va = 120<0
Sa = Pa + jQa = 924 + j0

Ib = 7.7<240
Ib* = 7.7<120
Vb = 120<240
Sb = Pb + jQb = 924 + j0

Stotal = (Pa+Pb) +j(Sa+Sb) = 1848 +j0

It might be worth noting the neutral current: In = Ia + Ib = 7.7<-60


For a L-L load, you would have 8.8912 amps per phase. For example:
Iab = 8.8912<30
Iab* = 8.8912<-30
Vab = 207.846<30
Sa = Pa + jQa = 1848 + j0

but you can also look at the L-N values even with a L-L load:
Ia = Iab = 8.8912<30
Ia* = 8.8912<-30
Va = 120<0
Sa = Pa + jQa = 924 - j533.4716

Ib = Iba = 8.8912<210
Ib* = 8.8912<150
Vb = 120<240
Sa = Pa + jQa = 924 + j533.4716

Stotal = (Pa+Pb) +j(Sa+Sb) = 1848 +j0

Note that in this case we have a L-L load and In = Ia + Ib = 0

 

[Smart $]

I have a 2 pole 208 volt L-L 120 volt L-N UPS system. The total power displayed is 1848 watts. Using Ohms Law I=P/E I get 8.9 amps. Does that mean I have 8.9 amps per phase average?

Well by convention you only have one phase. But otherwise yes, your calculation is essentially an average. We cannot tell just by the UPS reporting 1848 watts whether it is all L-L, L-N, a combination thereof, or balanced or unbalanced if the latter. Additionally, your calculation assumes a power factor of 1 (i.e. purely resistive load).

Should power be calculated by voltage L-L times average amps per phase 208*8.9 = 1848 watts or the L-N voltage (120) * amps on one phase (8.9) + L-N voltage (120) * amps on the other phase (8.9) = 2,136 watts?

Is there a difference if I had a 2 pole 240 volt L-L? Would I calculate by (120*I) + (120*I) = P or 240 * I avg = P

You can do the calc' either way using the average amperes... but you have to include line power factor when calculating at 120V for the 120/208 system because though it is considered 1-phase, there are three different voltage angles.

208V x 8.9A x 1pf = 1851W
(2) 120V x 8.9A x [cos (?30)] pf = 925W

The power factor for the 120V calc' is one at +30 degrees, the other at -30 degrees, the cosines of which are equal.