NCEES #136

[Jabert]

Problem states impedance = 0.050+j0.050 ohms, 3 phase, and with a fixed 50A current. Question is to determine differences in voltage drop with unity pf, zero pf, and 0.707 lagging pf.

Solution uses NEC Chapter 9, table 9, V = I(XsinΦ + RcosΦ), and since X=R, voltage drop proportional to (sinΦ+cosΦ) only.

At 0.707 power factor, (sinΦ+cosΦ) = 1.414

Unity, (sinΦ+cosΦ) = 1.0

Zero pf, (sinΦ+cosΦ) = 1.0

Answer is at 0.707 pf... D

However, if you don't use NEC equation and work it out (solving for voltage drop), the answer is different!

At 0.707, V= |IZ| = (50<-45)*(0.05+j0.05) = 3.53V

At unity, V= |IZ| = (50<0)*(0.05+j0.05) = 3.53V

At zero pf, V= |IZ| = (50<90)*(0.05+j0.05) = 3.53V

I found that the power factor does not affect the voltage drop of the feeder (which according to NCEES, is wrong)

Can someone explain why calculating the voltage drop gives a different answer than using the NEC???

 

[iahim]

You results are correct - the magnitude of the voltage drop is equal for all 3 power factor. But the phase angle is different. If you include the phase angle in the result, you will see that for 0.707 pf, the angle of the voltage drop across the line is zero, so it is in phase with the line voltage. This will cause the largest voltage drop at the end of the feeder.

You can try to calculate the voltage at the end of the feeder for the 3 power factors using some random source voltage (208V, for example) if you want to convince yourself.

 

[EEpowerOK]

208/sr3 - 50<-45(.05+.05j) = 116.46 back to 3ph 201.72 (pf=.707)

208/sr3 - 50<0(.05+.05j) = 117.5 back to 3ph 203.6 (pf=1)

208/sr3 - 50<90(.05+.05j) = 117.5 back to 3ph 203.56 (pf=0)

 

[iahim]

208/sr3 - 50<-45(.05+.05j) = 116.46 back to 3ph 201.72 (pf=.707)

208/sr3 - 50<0(.05+.05j) = 117.5 back to 3ph 203.6 (pf=1)

208/sr3 - 50<90(.05+.05j) = 117.5 back to 3ph 203.56 (pf=0)

Exactly! So the load with the 0.707 will see the largest voltage drop.

 

[Jabert]

Makes sense... Guess I better keep that phase angle in there the whole way!

Thanks for the help