Calculating what percent load maximum efficiency occurs at

[rockman7892]

I have seen on many motor datasheets that efficiency in a motor increases slightly from no load/small slip then peaks at a maximum around some slip value before it starts to decerase again for higher slip values.

I wanted to analytically calculate for what % loading the maximum efficiency occurs at. To do this I recognize the following:

% eff = Pout / Pin

= [I2^2(R2/s) - I2^2R2] / [(3Is^2R2)/Ws]

Can I then take this above equation find the deriviive of deff/ds and set this to 0 so solve for s will this give me the slip that the maximum eff will occur at? Can I then take this slip value and covert it to a %load based upon the full slip range?

My calculus is a bit rusty, but I just wanted to make sure I was on the right track analytically.

Does the max efficiency usually occur around the same % loading for all motors? If so why?

Is it possible that a motor that is more heavily loaded and operating at a better efficiency will draw less current/power then a motor that is more lightly loaded operating a a lesser efficiency?

 

[electricpete]

First of all, as you know the most precise answer is look at the data sheet. But you have the data sheets and you want a math model to explain it. I propose the following will be pretty close:

let losses be L = A + BX^2 where X is power.
maximize efficiency = Pout/(Pout+L)
where Pout =X*P100 and where P100 is rated output power

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

minimize: L/Pout = [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)

What are A and B physically
A =NLL = no-load losses loss
B = LL = the load-dependent portion of full-load losses (i.e. full-load losses minus no-load loasses)

For max E at 100%, X=1, NLL=LL motor
For NLL>LL, max E at higher than 1
For LL>NLL, max E at lower than 1

Given peak occurs at specific fraction of full load for example Xp, what is the ratio A/B?
A/B = X^2
Example – max efficiency occurs at 90% of full load. Xp = 0.9
A/B = 81%. NLL = 81% of LL
NLL = 0.81/(1.81) = 45% of full load losses
LL (load dependent losses at full load) = 1/(1.81) = 55% of full load losses

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[waross]

I'll leave the math for E-Pete.
"Does the max efficiency usually occur around the same % loading for all motors? If so why?"
I understand that the maximum efficiency point is a result of the inevitable design compromises. I expect maximum efficiency at 60% to 80% full load.
"Is it possible that a motor that is more heavily loaded and operating at a better efficiency will draw less current/power then a motor that is more lightly loaded operating a a lesser efficiency?"
Efficiency is a combination of fixed losses and load dependent losses. A big factor in both is the winding resistance and the resulting I²R losses.
The answer is yes, BUT.
Motor efficiency of even poor efficiency motors is pretty good. I don't expect the difference to be great. Consider the difference in efficiencies of the two motors at the expected operating point. Look at the difference in losses of the two motors. You can take that as a rough guide to the amount of extra work the more efficient motor may do on the same or less energy.
Remember the losses of both motors are on curves and the different curves will put the motors at different efficiencies.
Put it another way:
With the same partial load a more efficient motor will be expected to draw less energy than a less efficient motor. As the load on the more efficient motor is increased the energy draw will increase. At some point the energy draws of the two motors will be equal but the more efficient motor will be doing more work. Within those limits the answer to your question is yes.
I should have said that first!

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

To circle back over the questions:

Can I then take this above equation find the deriviive of deff/ds and set this to 0 so solve for s will this give me the slip that the maximum eff will occur at?

No, I don't think your approach will work because it neglects no-load losses. The estimate of power level where peak efficiency occurs will be falsely low.

Does the max efficiency usually occur around the same % loading for all motors? If so why?

No. Traditional wisdom was that peak efficiency occurred around 75% load. Now with higher efficiency I think it was pointed out on the forum peak efficiency is tpyically closer to 50%. But certainly there are many variations... and using my model above the ratio of no-load losses to load-related losses dictates what power the highest efficiency will occur at.

Is it possible that a motor that is more heavily loaded and operating at a better efficiency will draw less current/power then a motor that is more lightly loaded operating a a lesser efficiency?

If two identical motors (with all conditions such as voltage identical except for load), then no (because input power is a monotonically increasing function of output power). If two different motors, yes as explained by Bill.

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[rockman7892]

Electricpete

My equation for Vout accounts for rotor losses but I guess does not account for stator losses and magnetizing losses is that what you are saying?

So what you and bill are saying for the exact same motor if we run it more lighly loaded at a lower efficiency we will still use less power then if we run it more heavily loaded operating at a higher efficiency?

However if we are using two different motors with different efficiencies then operating the higer efficiency motor at a heavier load will better efficiency will consume less power?

What did you mean by "input power increasing monotonically with output power?

 

[electricpete]

My equation for Vout accounts for rotor losses but I guess does not account for stator losses and magnetizing losses is that what you are saying?

Yes - more specifically no-load losses include core losses, friction and windage. Lastly there is also a small contribution from no-load stator I^2*R, but my model does not properly capture that last component either (would need to get a little more complicated to capture that).

So what you and bill are saying for the exact same motor if we run it more lighly loaded at a lower efficiency we will still use less power then if we run it more heavily loaded operating at a higher efficiency?

Yes that's what I'm saying. And my reasoning was that "input power is a monotonically increasing function of output power". i.e. if you draw a curve with output power on the horizontal axis and input power on the vertical axis, the slope of the curve is always positive. (because the input power includes output plus losses and losses increase with power). So I will not be able to find any point where I get higher input for lower output (regardless of the fact that efficiency might be lower at the lower power level).. that would be a negative slope.

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