




#175905  03/14/08 07:34 AM
Re: Total power usage?
[Re: renosteinke]

Member
Registered: 09/15/03
Posts: 652
Loc: boston, ma

Reno,
I may disagree with your analysis.
The original question is not clear; just what does one mean by 'A phase = 2568w'?
If this means that there is a 2568w load from supply leg A to supply leg B, and similarly the B wattage is B to C, and C wattage is C to A, then I'm pretty sure that your analysis is correct.
But if the _total_ load connected to leg A is 2568w, including A to B loads, A to C loads, and A to N loads, then your analysis is not correct, and in fact I don't believe we know enough to answer the question.
Finally, if we are talking all line to neutral loads, with nothing connected leg to leg, then the total power is simply the sum of the power on the three legs.
I completely agree with your basic points: for any 'line to line' loads, you get a certain amount of 'double counting' going on.
Jon

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#175977  03/16/08 04:21 PM
Re: Total power usage?
[Re: smellelectric]

Broom Pusher and
Member
Registered: 10/19/00
Posts: 2724
Loc: Anaheim, CA. USA

When I added things up, it appeared that the compiled values were Apparent Power across a 3 Phase 3 Wire Circuit  as the total sum is 7190 VA.
Seeing that Ã˜A and Ã˜B draw 2568 Watts, and Ã˜C draws 2054 Watts makes the Load look Reactive. Could be the True Power (1.0 PF) is 2500 Watts across Lines A & B, but this would mean the Line C load would be LN.
If the Balanced Load is 1666.667 Watts per Line, then that portion would be 1.0 PF (all True Power), and the remainder would be something else.
Using the values given by the O.P., here are some figures:
*A Total Apparent Power for the Load: 7,190 VA (VoltAmps)
*B Power Factor = 0.695 (69.5%)
*C Total VARs (VoltAmps Reactive) for this Load: 5,160 VARs
This is figuring the Total True Power (Watts) at 5,000 Watts.
Scott
_________________________
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!

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