Ncees 517

[ros]

According to solution they have

SM s=8k<-45.6

& IM s=1.73*480*14.43<53.13

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

 

[Flyer_PE]

According to solution they have
SM s=8k<-45.6

& IM s=1.73*480*14.43<53.13

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

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.

 

[Dolphin P.E.]

According to solution they have
SM s=8k<-45.6

& IM s=1.73*480*14.43<53.13

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

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.

 

[CntrSnr2001]

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?

 

[stinkycheese]

CntrSnr2001, I get:

293<-34.9 + 1131<+25.8 kVA = 1300<+14.5 kVA

PF = 0.968

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