real power in an unbalanced three phase load

[k2keylargo]

Well, only 9.4 days left.... gotta ask another one! (and I found the superscript/subscript buttons B)

In a three phase unbalanced delta load, the total real power is either:

the sum of (Vp-p2 / R) for each phase, or

the sum of the I2R of each phase or leg.

Both equations should be able to be done with just magnitudes, since we are only talking real power, right???

 

[k2keylargo]

to clarify, I should have added: The impedance in each leg could be complex impedances, but for the real power calc we can do our calcs with only the R

 

[Flyer_PE]

The following can be blamed on my nuclear background (I've been trained to be a detail freak):

I'm assuming "Vp-p" in your post is phase-to-phase voltage and is an RMS value. The only reason I ask is that the subscript 'P' causes me to think "peak' voltage.

You are correct. Real power is the only thing that can be dissipated in the 'R' component.

 

[k2keylargo]

Yes, Vp-p = phase to phase voltage, in RMS

 

[Art]

P = V I pf

pf = cos of the phase angle

phase angle is determined by all the impedences...both real R & reactive X...ie, the complex load

Z = sq rt (X^2 + R^2) the power triangle

pf = R/Z

if no X then Z = R and pf = 1

V = IZ

or substituting

P = (IZ) I (R/Z) = I^2 R

know the basics and their derivation

 

[k2keylargo]

Even if the impedance is complex, which will make the current out of phase with the voltage, the equations should both work - that is, the angles can be ignored to get magnitude - real power = P = V2/R = I2R. Shouldn't have to use complex math, is my point. Am I correct with these formulas? I think I am, but not certain if I'm forgetting something..... :f_115m_e45d7af:

 

[Flyer_PE]

^^ You are correct. Real power dissipated will be I2R or V2/R. The relative angle between voltage and current doesn't matter in this instance.

 

[k2keylargo]

Thanks Flyer.