# NCEES Power 517



## FatDirk (Sep 14, 2011)

Question:

A 3-phase induction motor and a 3-phase synchronous motor are connected to a 480-V system. The synchronous motor, operating at 8 kVA, 480 A, 60 HZ, is overexcited producing a leading power factor of 0.70. The induction motor draws a current of 14.43 A and has a lagging power factor of 0.60. The system load power factor is most nearly:

my question is, why do you use the sqrt 3 on the current in this equation for the induction motor? I know why, but I'm not sure how I'm supposed to know that from the information provided.


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## Flyer_PE (Sep 15, 2011)

FatDirk said:


> Question:*A 3-phase induction motor* and a 3-phase synchronous motor are connected to a 480-V system. The synchronous motor, operating at 8 kVA, 480 A, 60 HZ, is overexcited producing a leading power factor of 0.70. The induction motor draws a current of 14.43 A and has a lagging power factor of 0.60. The system load power factor is most nearly:
> 
> my question is, why do you use the sqrt 3 on the current in this equation for the induction motor? I know why, but I'm not sure how I'm supposed to know that from the information provided.


The information given is that it is a three-phase motor with a given line current. Total power S for this motor is then: S=sqrt(3)*VL-L*I*Line


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## peterdubro (Oct 26, 2011)

I always hate an answer that just recites this formula because it is a confusing placement of the sqrt in this formula. As a power engineer the important formulas are only really S=VI* and V=IR with everything else derived.

I think the better way of thinking about it is that the current given is a phase current. Since this is a 3 phase problem, power is made up of 3X the phase power. Phase power is Vp*Ip = VLL/sqrt(3) * Ip . Total power becomes 3 times that and when you multiply by sqrt(3)/sqrt(3) you cancel the 3 in the numerator but are left with the sqrt(3) in the demoninator.

I notice that this problem solution doesn't use the complex conjugate I* as it is not necessary for the math - formulas in the book don't have it either even though it is the basic power calculation. Any comments on this?


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## ElecPwrPEOct11 (Oct 26, 2011)

peterdubro-

Good explanation of where the sqrt(3) comes from. It's standard for this type of problem to be given line voltage so S= sqrt(3)* V * I is very good to remember.

If the voltage angle= 0 the conjugate of I isn't needed. So S= V I is a special case of S= V I*. I always use V I* to be safe.


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## DK PE (Oct 26, 2011)

I would bet next week's paycheck if a problem similar to this is on the exam and the correct solution for the system power factor is 0.9 (for example) that both 0.9 lagging and 0.9 leading are possible solutions so you'd better know how to choose correctly.


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## ElecPwrPEOct11 (Oct 26, 2011)

I agree DK. What's always confused me is that a lagging load has a current with a negative angle (if voltage is at 0 degrees). However the same lagging load has POSITIVE Q. The same with a leading load that has positive current angle but negative Q. It seems backwards, but isn't. Confuses me every time.


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