NCEES #524

[pbo064]

For NCEES sample exam problem #524 the problem gives the % of Nameplate kVA and Total Losses as follows:

% of Nameplate kVA, Total Losses (W)

0 , 460

50 , 2,370.

It asks "The total transformer losses (W) at 100% of nameplate kVA is most nearly:

(B) 8,100 MW

The problem solves this by subtrcting no load losses from 2,370 to get 1,910 MW. It then multiples this number by 4 to get 7,640. It re-adds in the no load losses to get 8,100MW - answer B.

My question is why is multiply by 4 and not by 2. What is the convention that leads to multiply by 4. And does anybody know of a good refecence to read up on this.

Thanks!

 

[Flyer_PE]

The losses at power are primarily I2R. Current at 100% loading is twice that at 50%. 22=4

If they had given the losses at 33% power instead of 50%, the multiplication factor would be 9.

 

[pbo064]

I think i found my answer later in the same test for 527.

1) no load or core losses are due to hystersis and eddy currents and are sustanitally constant

2) load losses (or I(^2)R losses) are proportioanl to the load current squared.