NCEES Power Practice Problem #524

[nimnoo]

(Practice exam book printed in 2014)





Can someone tell me where the multiplication factor of 4 comes from when calculating the 100% rated load loss? And where that line of calculation even comes from?

Also, with the given information, are the I^2*R losses calculated the following way?:

I^2*R = (P_50% - P_NoLoad) / 50% = (2370 - 460)/0.5 = 3820 W

 

[Dude99]

Losses are I^2 R

so 2 times the I therefore 2^2 = 4 times the losses

460 are no load losses, independent of load

50% I2R losses = 2370 - 460 = 1910

I2R at 100% = 2^2 x 1910 = 7640

 

[nimnoo]

Ok, I figured out my mistake. The rated percentages should also be squared with the current. I was using:

P_50% = 0.5 * I^2 * R + P_NoLoad

When it's supposed to be:

P_50% = (0.5 * I)^2 * R + P_NoLoad

So then I^2*R should be:

I^2*R = (P_50% - P_NoLoad) / 0.5^2 = (2370 - 460)/0.25 = 7640 W

And so total power loss at 100% rating is:

P_100% = (1.0 * I)^2 * R + P_NoLoad = 7640 + 460 = 8100 W