NCEES #511 (Power Depth)

[Kuku]

Can someone walk me through this problem. Thanks.

 

[Flyer_PE]

There is a pretty good discussion of this one here. Let us know if you need more.

 

[Kuku]

There is a pretty good discussion of this one here. Let us know if you need more.

Thanks for the link. I actually did search for this topic prior to posting, but the search returned absolutely nothing for me. Go figure.

 

[Flyer_PE]

No problem. There's a lot of good stuff on this board. I also had the advantage of knowing almost exactly where to look.

 

[jproctor6]

Can someone walk me through this problem. Thanks.

This may not make much sense without a picture of the line and load but here goes:

VAB = 12.5 <0 degrees (kV)

so

VAN = 12.5/sqrt3 <-30 degrees (kV) = 7.22 <-30 degrees (kV)

now

Van = VAN + IaA(5+j10) = (7.22 <-30degrees) + (70/1000 <-20degrees)(5+j10) = 7.48 <-24degrees

|Vab| = sqrt3*|Van| = (sqrt3)(7.48) = 12.95

Basically, you take the phase-to-phase load voltage "VAB" given to find the phase-to-neutral voltage "VAN" seen at load point "A".

Then you determine the phase-to-neatral voltage "Van" seen at the source by adding the load voltage "VAN" to the line voltage drop given by ohm's law = line current * line impedance.

Now the magnitude of the phase-to-phase voltage "Vab" seen at the source is just the phase-to-neutral voltage you just calculated "Van" multiplied by the square root of 3.

I hope that makes sense.

Sorry if my explaination or method is not helpful.