PE Power 522 pf

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Mr-T_Swift

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The pf is cos() = .95 and lagging. The solution to this problem uses a -18.19 angle instead of a positive.

1650832065607.png

1650832090442.png
 
This link might help:

https://engineerboards.com/threads/ncees-prob-522.36286/
The NCEES solution is confusing in regards to the first line where is says Is = 1.05<-18.19 degrees. I used the positive angle. Take the synchronous reactance into account and subtract the positive angle from 90 degrees.

V=Vi + IR
1<0 = Vi + Is x jXs
1<0 = Vi + 0.9x1.05<(90-18.19)
1.14<-51.8 = Vi
 
With pf = 0.95 lag, the power angle is 18.19 degrees.

power angle = voltage angle - current angle

If the voltage angle is referenced as 0 degrees, then current angle must be -18.19 degrees.

0 - -(18.19) = 18.19
 
With pf = 0.95 lag, the power angle is 18.19 degrees.

power angle = voltage angle - current angle

If the voltage angle is referenced as 0 degrees, then current angle must be -18.19 degrees.

0 - -(18.19) = 18.19
Thanks! I realized that last night. Mistakenly used power angle instead of the current angle.
 

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