NCEES Power PE exam # 64

[yaoyaodes]

Can anyone explain this to me? No sure what's going on.

 

[DLD PE]

If you do a search, you can find where these problems have been discussed at one time or another. Just click "Search", then "NCEES #____" Also, previous practice exam versions used a different numbering method. For example, problem #11 used to be #111, and problem 65 used to be 525

Someone also put together an index of some of these problems. I will copy the link here:

Updated NCEES Sample Exam Index (and Cram)

 

[akyip]

Attached is my solution for this problem (this was previously question number 524 on the pre-updated NCEES practice exam).

For transformers, the power loss equation is:

P total loss = P no load loss + P copper loss

P copper loss = (n * I)^2 * R = n^2 * I^2 * R

P copper loss at 100% = (1 * I)^2 * R = I^2 * R

In this problem, you are given the total power losses at 0% and 50%. You need to break down the total power loss into the no load loss and copper loss.

P total loss at n% load = P no load loss + P copper loss at n% load

P total loss at n% load = P no load loss + n^2 * I^2 * R

The total loss at 0% load is the same as the no-load loss, because:

P total loss at 0% load = P no load loss + (0 * I^2 * R) = P no load loss = 460 W in this problem

The total loss at 50% load is:

P total loss at 50% load = P no load loss + (0.5^2 * I^2 * R) = P no load loss + 0.25 * I^2 * R

You need to algebrically solve for I^2 * R, which corresponds to 100% copper losses, using these 2 equations (total loss at 0% load and total loss at 50% load).

Then, the total loss at 100% load is:

P total loss at 100% load = P no load loss + I^2 * R

P total loss at 100% load = P no load loss + P copper loss at 100% load