Current Unbalance Due To 1Phase Loading 1

[nightfox1925]

In a 3 phase 3 wire system, the voltages are measured at 242VAC. The currents are measured at 20A, 18A and 7A. If these are single phase line-to-line loads, in the absence of a panel schedule, can we say that the power consumption will be 1.732 x (20A + 18A + 7A)x 242V ?

It is also observed that the current unbalance due to actual loading is severe 100%-[(7/20)x 100%] = 65% unbalance. Is there a way to avoid this much of unbalance regardless of load switching arrangement or is it a fact that if the panelboard circuits were properly balanced, then there is no way of any unacceptable current unbalance regardless of how much loads are switched on?

 

[davidbeach]

You need to think about your formula a bit more. What would it look like if all three currents were the same, and you only wrote that number once?

 

[nightfox1925]

Thanks for the correction, for this case then,the three phase apparent power can be safely estimated using the highest current on the phase (1.732 x 242V x 20A).

My question is if there is another way of making the computations adding up the total power per phase in this unbalanced condition?

 

[davidbeach]

It really depends on what you are trying to accomplish. If you want to estimate a utility bill, the average current would probably be fine. If you are trying to figure out the capacity of the system in use then the maximum current would be better.

 

[nightfox1925]

Thanks david for the technical advise.

 

[peebee]

Um, sorry to disagree with y'all, but you have the 1.732 factor on top, and it should be on the bottom. If 242 is the L-L voltage, then the total power would be (20A + 18A + 7A)x 242V/(1.732).

Your sum of (20+18+7) = 45A = 3 TIMES the average phse current. The average phase current = 45/3=15A. If you use 45, then you need to divide by 1.732; if you use 1/3 x 45 = 15, then the 1.732 goes on top:

P=IV sqrt(3)
P=15x242x1.732 = 6.3kVA
P=45x242/1.732 = 6.3kVA
P=45x242x1.732/3 = 6.3kVA
P=(20A + 18A + 7A)x242/1.732 = 6.3kVA.

 

[davidbeach]

peebee, your formulas are correct. nightfox1925 could have modified his original formula to put the 1.732 on the bottom as in your fourth equation; or he could have just used a single current as he did in his second post and the 1.732 is correctly on the top there, as in your first equation.

 

[Senselessticker]

Depends on "where" you are measuring the power consumption.... If you want to look upstream of the xmfr then you may be in for a suprise especially if dealing with bad pf's and/or non-linear loads. Also, is the measured value a fundemental current measurement or does it include harmonic currents as well?