# Unbalanced Three Phase : How to calculate total power ?

Discussion in 'Homework Help' started by kbarb, Jul 17, 2015.

1. ### kbarb Thread Starter New Member

Dec 29, 2008
6
0
In our building, we have a four wire 208v/120v Wye service.

I'm trying to measure the watts used by a subpanel, but the loads are unbalanced.

For example, for the 208v lines (black, red, blue) I get :
-- 42A
-- 84A
-- 80A

I know about P = √3 Vl × Il × cos φ where cos φ = power factor, but that's for balanced scenarios, right ?

I'm going to have to guess at the power factor - most of the loads are lighting & computers/monitors.

But anyone know how to calculate the total power draw ?

If you look on this page, Three Phase Current - Simple Calculation and look at the "Balanced Voltages" section it implies that for unbalanced systems you can still calc as such :
• the line to neutral (phase) voltage VLN = 208/√3 = 120 V
• phase 1 apparent power = 42 x 120 = 5040 VA = 5.04 kVA
• phase 2 apparent power = 84 x 120 = 10080 VA = 10.08 kVA
• phase 3 apparent power = 80 x 120 = 9600 VA = 9.6 kVA
• Total three phase power = 5.04 + 10.08 + 9.6 = 24.72 kVA
Anyone know if this is correct ?

2. ### WBahn Moderator

Mar 31, 2012
18,096
4,920
Yes, that looks correct.

As you say, you would have to guess at the power factor and that could have a significant effect on your real power.

As a quick sanity check, you can take the average of the three currents (which would be 68.7 A) and throw that at the balanced three-phase formula and see if you get something that is in the ball park of when you do the per-leg unbalanced computation. In this case that comes out to 24.74 kVA, which is a pretty small ballpark.

Looking at the formulas, these work out exactly.

3. ### kbarb Thread Starter New Member

Dec 29, 2008
6
0
Thanks a lot for your very fast reply.
And I like your sanity check exercise - I can always use a bit of that.
Thanks again,

Kent

4. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,047
295
Yikes....this is not as simple as it appears. But yes....it is probably unsolvable without knowing the power factor.