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Joined: Feb 2008
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I am hoping someone can help me out with this. I am having a bit of a brain fart. Assuming 208v 3p 5w: A phase = 2568w B phase = 2568w C phase = 2054w PF @ Unity What is the total power usage and how? Please help.




Joined: Feb 2008
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I know it sounds stupid: power is cumulative i.e. p1+p2+p3=pT but this circuit isn't really drawing 7190w is it?




Joined: Jan 2005
Posts: 5,370 Likes: 1
Cat Servant Member

The figures are a bit misleading, because eahc wire shares the load with the other two. In effect, you're counting everything twice.
Add up the watts. Divide by 3 to get an average. Then multiply the average by the square root of three. This ought to be your actual usage.




Joined: Mar 2004
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If watts equal volts x amps, what is the voltage of just one phase?




Joined: Jan 2005
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Cat Servant Member

twh, it doesn't really work that way. Indeed, the usual math we use is close enough for daily work  but you've hit on one of the areas where things just don't seem to add up. There's a reason for this.
Remember, we're talking AC. The voltage is constantly changing. For the math to add up, you need to use calculus  especially when you have two 'legs' that are not exactly 180 degrees out of phase.
That's also why there sometimes appears to be a difference between VA and watts.




Joined: Sep 2003
Posts: 650
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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




Joined: Mar 2004
Posts: 947
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Does this help? http://www.yokogawa.com/tm/tr/tmtr0605_02.htmIt looks to me like there are two methods: a) Three watt meter method, where the voltage reference is from phase to neutral  the three meter readings are added. b) Two watt meter method, where the voltage reference is phase to phase  the two meter readings are added




Joined: Feb 2008
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Unfortunately I am getting my information third hand at best, so I am unsure of how the loads were determined. I ended up just adding them and divding by 1.73 and it seems to have been close enough for the person who had originally asked me. Thanks for all the help.




Joined: Oct 2000
Posts: 2,722
Broom Pusher and Member

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 ØA and ØB draw 2568 Watts, and ØC 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!



Posts: 61
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