NCEES #512

[mhelms_2]

Okay, don't understand the solution to this one.

It wants to know the pahse to neutral value of the complex impedance of a 400 kcmil, 75 deg cable at 500 ft. I used table 9 of the NEC to get the resistance and reactance. Then I did the following:

Z=(0.035+0.049j)(500/1000)(2) , where the two takes into account going out and coming back.

There solution is as follows:

Z=(0.035+0.049j)(500 / (1000x2)

Why is the two on the bottom with the 1000 foot?

 

[knight1fox3]

When calculating the impedance for two parallel sets of conductors, you deal with it as two impedances (resistances) in parallel. In this case two identical impedances, therefore the resulting impedance is halved.

There have been a number of threads on this problem as well. When I was studying for the exam, I would always do a quick search for the problem I had questions on before starting a new thread. It usually lead to a discussion or two.

/>http://engineerboards.com/index.php?showtopic=17339

 

[mhelms_2]

Another question on this one. The power factor is 0.9, and the NEC says that any power factor other than 0.85 requires correct by

Rxpf + XL(sin(arccos(pf)

why didn't they use the correction factor?

 

[DK PE]

The "effective Z" must be corrected since it is dependent on power factor (see note 2 table 9). The conductor characteristics are what they are looking for, not effective Z so no correction factor is needed.