COMPLEX IMAGINARY, VOLUME # 2, PROBLEM # 28 vs NCEES 517

[khatrib]

I thought I had the leading/lagging powerfactor figured out untill I came across this problem.

The syn motor apparent power is given with leading power factor. The solution shows 7kva<38.7

The Induction motor apparent power is given with lagging power factor. The solution shows 9kva<-49.5

Should it be opposite? If you look at a similar problem # 517 in NCEES exam, the signs are exactly opposite for leading/lagging power factor.

What am I missing here??

 

[stinkycheese]

I think that CI mixed the signs up. A leading angle for current is positive, which then (S = VI*) makes the angle for leading complex power negative.

So, lagging: positive S angle. Leading: negative S angle.

I get the combined power as 11.6<12.3 kVA -> .98 lagging pf, so the answer is still B.

 

[kwatson18]

So i've done my homework (I think) and i'm in the same boat as khatrib...one tweak away from having clarity and confidence on the test in regards to this concept...

My system before that seemed to work pretty flawlessly for most of my practice problems was:

ASSUMPTIONS:

LEADING PF: I< +@ , S< - @, Z < - @

LAGGING PF: I< -@ , S< + @, Z < + @

Every now and then i'll run into a practice problem (SpinUp, ComplexImag, NCEES) where it doesn't seem to apply and I always brush it off like i'll figure it out. Enough is enough, want that confidence during the test, if you would indulge me for a moment...

I've read through Flyer PE's explanation of NCEES #117:

"For an inductive load, the angle for " I " with respect to "V" is -@ and angle with respect to apparent power " S " is +@ and conversely

"For an capacitive load, the angle for " I " with respect to "V" is +@ and angle with respect to apparent power " S " is +-@

The ultimate test being the fact that S = V I * (conjugate, so in this relationship, I->V angle should have opposite phase angle as I->S)

So is it correct in the solution for 517, that per ASSUMPTIONS above:

Synch motor / S = 8kVA, 480V, 60Hz, PF = 0.70 leading

Induct motor / I= 14.43A, PF = 0.60 lagging

arc cos (0.70) = 45.6

arc cos (0.60) = 53.13

Smotor / S= 8000 VA < -45.6 (kosher with ASSUMPTIONS above)

Imotor / I = 14.43 A <-0.60 (kosher with ASSUMPTIONS above)

S= sqrt(3) x I x V <+0.60 (because S=VI* so you just flip the sign of the angle when you calc the apparent power?)

So after writing out this long winded question, I think I answered my own question but if anybody's bored, confirm.

Think i'm good to go with above ASSUMPTIONS and always double checking my answers/work with S=VI*

Hope that helped somebody else, love this site for others who are runnign into same concepts as myself, thanks in advance!!!

Good luck to everyone, can't wait to have a life again....

kW

 

[stinkycheese]

kW:

Imotor current would be I=14.43<-53.13

Imotor apparent power would be S=sqrt(3) * 14.43 * 480 = 12000<+53.13 (double check- correct as this is lagging)

An easier way (for me) than the NCEES solution here is to just add the S vector forms. 8000<-45.6 + 12000<+53.13 = 13,370<+16.9

cos(16.9) = 0.96 (lagging)

Good luck!