# NCEES #524



## Rei (Mar 23, 2010)

Could someone tell me why they divide by 4? Thanks.


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## yellowjacket03 (Mar 23, 2010)

At 1/2 full-load, the load losses (I^2*R) is 1910W

Now, find the load losses at full-load. If R is constant and I goes up by two, then (I^2) * R will go up by (2^2) or 4.

At full-load, the load losses are 4*1910W = 7640W

Total Loss = Core Loss + Load Loss = 460W + 7640W = 8100W



Rei said:


> Could someone tell me why they divide by 4? Thanks.


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## Rei (Mar 23, 2010)

yellowjacket03 said:


> At 1/2 full-load, the load losses (I^2*R) is 1910WNow, find the load losses at full-load. If R is constant and I goes up by two, then (I^2) * R will go up by (2^2) or 4.
> 
> At full-load, the load losses are 4*1910W = 7640W
> 
> Total Loss = Core Loss + Load Loss = 460W + 7640W = 8100W


Thanks. When we talk about losses and efficiency, specifically the copper loss, we are talking about both the primary copper losses and secondary copper losses, right? NCEES 528 only considered the primary copper losses, what about the secondary?


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## Flyer_PE (Mar 23, 2010)

Rei said:


> Thanks. When we talk about losses and efficiency, specifically the copper loss, we are talking about both the primary copper losses and secondary copper losses, right? NCEES 528 only considered the primary copper losses, what about the secondary?


Both the primary and secondary resistances are represented in the 10.9 ohm value in the problem. When they say it's referred to the primary side, all that means is that the total of both as read from the primary is 10.9 ohms.

Impedance translates by the square of the turns ratio. Referred to the secondary of the 2300/230 V transformer, the impedance on the low side would be 10.9*(230/2300)^2 = 0.109 ohms. Referred to the secondary side, both the primary and secondary resistance would be contained in the 0.109 ohm value.


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## dianevp (Apr 7, 2011)

yellowjacket03 said:


> At 1/2 full-load, the load losses (I^2*R) is 1910WNow, find the load losses at full-load. If R is constant and I goes up by two, then (I^2) * R will go up by (2^2) or 4.
> 
> At full-load, the load losses are 4*1910W = 7640W
> 
> ...


This solution that NCEES has been bugging, so I approached it from an another angle to see if I would get the same answer. Let me know what you think.

Let's say No Load, 460 =X and 1/2 Load, 2370 =A. We know that A-X = Copper losses and the equation for Copper losses is I^2R.

So we have (A-X)=I^2R, and R isn't really going to change in full load status, R is constant. This leaves us with (A-X) = I^2

Take sq root and get Sqt root(A-X) = I -- and this is at half load.

Increase to full load, "double" the load, and 2I becomes I' and equation reads I' =2*sqt root(A-X)

Substitute I' into I^2R --- (I')^2 --- (2*sqt root(A-X))^2

With numbers supplied: (2*sqt root(2370-460))^2 = 7639 + No load (460) = 8099.99 --- Answer B


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