# NCEES 2009 Power Question 503



## nuclear bus (Jan 23, 2010)

I just can't get this one to work out and I hope somebody here can help.

3-phase 460 V, 25 hp induction motor, 34 A, 0.75 lagging p.f.

I understand that the Smotor = 28.3 angle 41.4 degrees KVA. Based on that the following should be true.

Pmotor = 21.23

Qmotor = 18.72

because P = S cos (angle), and Q = S sin (angle) where the angle = 41.1 degrees.

So...if we want to improve the p.f. to 0.90 lagging, then the new S would be still 28.3 but at an angle 25.84 degrees KVA right?

thus the new Pmotor = 25.47 and new Qmotor = 12.33 right?

The solution shows the new Qmotor = 10.3, and that obviously throws everything off. Not sure where they get the "21.7" value for "P" in the solution either.

This seems like it should be pretty straightforward but I just don't get it. Does anybody have any clues they can offer? Thanks in advance for your help.


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## Flyer_PE (Jan 24, 2010)

nuclear bus said:


> So...if we want to improve the p.f. to 0.90 lagging, then the new S would be still 28.3 but at an angle 25.84 degrees KVA right?


Nope.

The number that stays constant in this problem isn't apparent power, it's real power. The real power drawn by the motor is determined by the mechanical load which doesn't change.

Pmotor = 21.2 kW

Qold = 18.7 kVAR

You need to determine a new value for apparent power from the following:

Pmotor = 21.2 kW @ 0.9 pf

S = 21.2/0.9 = 23.6 kVA

Qnew = sqrt (S^2 - P^2) = 10.3 kVAR

Qcorrection = Qnew - Qold = 10.3 - 18.7 = -8.4 KVAR


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## nuclear bus (Jan 24, 2010)

Flyer_PE said:


> nuclear bus said:
> 
> 
> > So...if we want to improve the p.f. to 0.90 lagging, then the new S would be still 28.3 but at an angle 25.84 degrees KVA right?
> ...


That is so simple it hurts!  I knew I was almost there. Thanks so much for your help.


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