# phase power LL vs LN

[LowSpark]

I will begin with some facts and then ask the question.

Phase to Phase Voltages Lead Phase to Neutral voltages by 30 degrees.

Three Phase Apparent Power Equations:
If using Phase to Phase voltages:
S=1.732*V*I'

If using Phase to Neutral Voltages:
S=3*V*I'

Now when solving for the power and you know lets say the Phase to neutral voltage and its angle, why don't you multiply by 1.732 and add 30 degrees to it and stick that value into the Phase to Phase 3 phase apparent power formula?

For example:
V_L-n=120 @ an angle of 31 degrees
I*= 22 @ an angle of 20 degrees

V_L-L= (120*1.732) @ an angle of 61 degrees

Apparent Power Calc:

Using Phase to Neutral:
S=3*(120@31)*(22@20)= 7920@51degrees

Using Phase to Phase:
S=1.732*(208@61)*(22@20)= 7920@81degrees

As you can see the two apparent power calculations do not agree. The 7920@51degrees is in fact the correct answer. My question is, if Phase to Phase voltages lead Phase to neutral voltages, why don't you use this fact when solving three phase power with the Phase to Phase three phase formula? I mean you are supposed to put in the Phase to Phase voltage magnitude, why not the angle as well?

 

[electricpete]

Clarification in bold:

If you assume balanced scenario then you can use per-phase analysis to develop suitable equation which involves VLL and then substitute VLL=sqrt(3)VLN. But the Phi is still the same Phi as we used in per-phase analysis. See Shooter's first post.

=====================================
(2B)+(2B)' ?

 

[electricpete]

Sorry spraymax, I was late to the party and should have read your later comments. I agree with those. We could derive the balanced three-phase relationship from the "single-phase" S=V I* relationship two ways:
Start with the per-phase V-I pair (VLN, Iphase)
or
Start with the phase-to-phase V-I pair (VLL, Idelta).

The angle between VLN and Iphase would turn out to be the same as the angle between VLL and Idelta (call it Phi in either case).

I don't think I've added anything new here. just summarizing for myself.

=====================================
(2B)+(2B)' ?

 

[LowSpark]

Yes,the summary is very accurate. That was my question from the beginning. I know that the angle should be the same and therefore the powers be the same but by simply trying to insert the relationship off LL voltage in the equation was throwing me off because of the 30 degrees. I'm just going to stick with my old textbook and just remember, only use phase voltage angles in wye system, and only use phase current angles for delta systems. Another way said, if you are specially told to use the "LL" 3 phase apparent power formula, and they ask for the angle of apparent power not just the magnitude, you can use the "LL" magnitudes but just use the "phase value" angles. I wish my textbook went into more depth in that paragraph that I posted but I'll just take it at face value.

 

[SparkyAt10]

Aren't you just missing another square root of 3? You have the sum of three (current vector * voltage vector) for power; it looks like you are effectively taking current vector * voltage magnitude.