# Ncees 517



## ros (Mar 29, 2011)

According to solution they have

SM s=8k&lt;-45.6

&amp; IM s=1.73*480*14.43&lt;53.13

My question is, since syncronous motor has Leading PF, shoudn't SM angle be positive and induction motor angle negative? Thanks


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## Flyer_PE (Mar 29, 2011)

ros said:


> According to solution they have
> SM s=8k&lt;-45.6
> 
> &amp; IM s=1.73*480*14.43&lt;53.13
> ...


Which quantity are you asking about? Remember that S=VI*

For an inductive load, the angle for I with respect to V is negative and the angle applied to apparent power (S) will be positive.

For a capacitive load (such as an over-excited synchronous machine), the opposite is true (Positive current angle, negative power angle).

The NCEES solution is dealing with S so their sign convention is correct.


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## Dolphin P.E. (Mar 29, 2011)

ros said:


> According to solution they have
> SM s=8k&lt;-45.6
> 
> &amp; IM s=1.73*480*14.43&lt;53.13
> ...


you may need to differenciate between current angle and aparent power angle. if Synch Motor is over-excited, then the current angle is possitive (leading PF) and since S=VxI* then the aparent power angle is negative.


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## CntrSnr2001 (Apr 7, 2012)

could someone please confirm this topic to NCEES 523? if i wanted to find the combined PF of the two generator system, then what would I have?


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## stinkycheese (Apr 8, 2012)

CntrSnr2001, I get:

293&lt;-34.9 + 1131&lt;+25.8 kVA = 1300&lt;+14.5 kVA

PF = 0.968

S has a positive angle, so I has a negative angle; current lags voltage, so PF is lagging.


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