draw a curve of efficiency 1

[faycalleroi]

hello
in the aim to view the efficiency of induction motor variation with load variation i would know the formula of n(efficiency) and the P(power in shaft or load) to draw this curve example: for 10% 15% 20% 25% 30% 35%........OF P power what is the efficiency
best regards

 

[electricpete]

We can make certain generalization and guesses but there are plenty of exceptions and generally the manufacturer is the best source of efficiency vs load info for a given motor.
Example:

http://www.reliance.com/pdf/pdf/aced/L3517A-V.pdf

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(2B)+(2B)' ?

 

[waross]

A motor has fixed losses and load dependent losses. Load dependent losses vary with the square of the current.
The current has a magnetizing component that may be taken as fixed for a simple graph and a real component that varies with the load.
There is a "sweet spot" of greatest efficiency normally at about 50% to 75% load but often closer to 75%.


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

 

[electricpete]

There is an interesting mathematical excercize to find the location of max efficiency if we ASSUME that the losses are divided into two categories:
A - losses that are constant with respect to load
B - losses that vary with load^2 (I^2*R losses vary approximately with load^2 at high load... at lower load the approximation is not as good but will not matter if we're evaluating behavior at high load).

This suggests the model:
L = A + BX^2
where
L = total losses
X is power as fraction of full load power
A is no-load losses, excluding small losses due to stator I^2*R at no-load
B is load-related losses at 100% power (total 100% power losses minus no-load losses)

How do we 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)

IF no-load losses were half the total losses, then A=0.5, B=0.5, peak efficiency would occur at 100% load.

Typically no-load losses are a smaller fraction. Let's say A = 0.33, B=0.67, then X = sqrt(A/B) = sqrt(1/2) = 1/sqrt(2) = 70.7%. Max efficiency would occur at 71%.

A useful model to think about, but not exact.


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(2B)+(2B)' ?