PE Power 522 pf

[Mr-T_Swift]

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

 

[DLD PE]

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

 

[akyip]

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

 

[Mr-T_Swift]

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.

 

[Zach Stone P.E.]

Thanks! I realized that last night. Mistakenly used power angle instead of the current angle.

Hi @Mr-T_Swift,

Here is a quick article you'll like to help you remember the difference between the power angle and the current angle (and how not to mix them up):

Leading and Lagging Cheat Sheet! Printable Reference