NCEES #524

[robertplant22]

Why are the copper losses multiplied by 4 and not by 2 in the solution to this problem? The question is as follows:

A power transformer has the following test data at rated voltage:

% of Nameplate kVA Total Losses (W)

0 460

50 2,370

The total transformer loses (W) at 100% of nameplate kVA are most nearly?

In this question I understand that the Core losses are Pcore= 460W. I also understand that to find the Copper losses, Pcu,50% at 50% loading you need to subtract the Core losses (Pcore= 460W) from the Total losses (PT,50% = 2,370W); therefore:

Pcu,50%= PT,50% - Pcore= 2,370W - 460W = 1,910W

Here is the part I dont understand:

Pcu,100%= (4)(Pcu,50%) + Pcore = (4)(1,910W) + 460W = 8,100W

 

[robertplant22]

It seems as though the formatting got change from when I typed the message; the table read

At 0% kVA rating, Total losses are 460W

At 50% kVA ratin, Total losses are 2,370W

 

[Power12]

Look at this thread

524

 

[robertplant22]

Thanks. The thread clarifies this. How did you find the thread; I tried doing the search for "NCEES #524" and the serach returns nothing!

Thanks again.

 

[knight1fox3]

Thanks. The thread clarifies this. How did you find the thread; I tried doing the search for "NCEES #524" and the serach returns nothing!

Thanks again.

When I was studying, if EB.com didn't find anything I searched for, I tried Google where I would search for "NCEES 5## engineerboards.com" or a variation there of. And if something existed, it would usually come up there.

 

[robertplant22]

knight1fox3, thank you for the tip. It works like a charm!

 

[kris7o2]

The thread above no longer works. Does anyone have the new link?

 

[akyip]

Use this link:

https://engineerboards.com/threads/ncees-power-pe-exam-64.36947/
I posted my explanation of the solution to this problem on this link.