quiz - no-load losses, load-varying losses, and location of peak effic

[electricpete]

This question is just for fun.

Assuming that motor losses L can be expressed as Total losses L = A + BX^2
where:
- A is a component which is constant with respect to load (no-load losses)
- B*X^2 = a load-varying component which varies according to load squared
- X=load expressed as fraction of rated load
- B = the value of load-varying losses when load is at rated load.

Let's say I tell you that the peak efficiency for a motor occurs at rated load, can you tell me the ratio of A/B?

What if the peak efficiency for a motor occurs at 80% rated load, can you tell me the ratio of A/B?

I will share the answer if no-one responds. I just thought it would be more fun to challenge you guys to figure it out. It does give a small insight into the significance of location of peak power and its relationship to no-load and load varying losses (within the bounds of the assumption).

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[Skogsgurra]

Phew! Quite a long time since I differentiated a quotient of two functions...

Setting Efficiency = P/(P+L), differentiating and solving for zero gives me A = Bx^2. And that tells me A has to be equal to B if the relation shall hold for all x. So, I guess that A/B = 1.

It is close to midnight over here. So I leave the other part for toworrow. (Hoping for someone else to do it over night).

 

[aolalde]

I agree with Skosgurra, the maximum efficiency is when A=Bx^2

since x = motor current and assuming Ifl = 1.0 pu

For 100% current: A=B or A/B=1
for 80% current: A=B*.8^2 and A/B=0.64

 

[Skogsgurra]

Good morning!

Thanks aolalde. Yes, of course.

In the light of the new day, I see that I should not have said "hold for all x" but rather have plugged in a "1" for x. The consequence of "all x" would have been A/B = 0, which is wrong.

 

[electricpete]

Yup, I should have known it would be easy for you guys.

I thought it was an interesting result.

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[electricpete]

Out of curiosity I tried out the formula using an example motor:

http://www.reliance.com/pdf/pdf/aced/PD36-G4911.pdf

From the 0% load point: A = sqrt(3)*v*i*pf = sqrt(3)*460*32.2*0.0737 = 1891 watts

From the 100% load point: B = Pelec - Pout = sqrt(3)*v*i*pf - 100hp
B = (sqrt(3)*460*114*0.873) watts - (100hp * 745.7) w / hp = 4723.5 watts

As solved above:
A = B*Xpeak^2
Xpeak = sqrt(A/B) = sqrt(1891/4723) = 63%

Page 1 gives efficiency at 25,50,75,100 and 125%. Page 2 (curve 3) gives a curve of efficiency. Looking at these curves I would say peak efficiency is probably somewhat higher than 63%. It must either be an error in my calculations or deviation from the assumed form of the loss curve. Is there any error you guys can see?

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[electricpete]

I saw one error already.

B = Pelec-Pout-A

Let me try to recalculate it with Pa

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[electricpete]

B=Pelec - Pout-A
B=(sqrt(3)*460*114*0.873)watts-(100hp*745.7wattts/hp)-1891watts
B =2832. watts

Xpeak = sqrt(A/B) = sqrt(1891/2832)=82%

This looks more consistent with the efficiency curve and tabulated data.


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[Skogsgurra]

Interesting and refreshing excersise, pete.

A friend asked me about the general shape of the efficiency vs load curve last week. All I could say was that I thought that the peak efficiency probably was at about three quarter load. I will see that guy tomorrow - and I will inform him till he wishes he never asked... And I will bring those data sheets from Rockwell, too!

 

[electricpete]

For posterity, I will write out the way I solved it.

Start with skogsgurra's maximize efficiency = P/(P+L)
where P = Pout =X*P100 and where P100 is rated output power

This should be the same as minimizing the inverse.
ie. minimize 1/effic. = (L+P)/P = L/P + P/P = L/P+1
The 1 is a constant and doesnt affect the minimization.
=> minimize L/P

minimize: L/P = [A + B*X^2] / [X * P100%] = [A/X + B*X] / P100
Since P100 is a constant with respect to X, it doesn't affect the minimization.
=> minimize [A/X + B*X]

Set derivative =0
d/dx{[A/X + B*X]} = [-A*X^-2 + B] = 0
A*X^-2 = B
A = BX^2
Xpeak=sqrt(A/B)

I didn't bother to check 2nd derivative since we already know the shape of the curve.
OK, back to some real work now. Thanks for listening.

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