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Losses in Induction Motors


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#1 marke

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Posted 09 July 2002 - 08:45 PM

Hi all.
I have recently been involved in discussions in regards to the losses in induction motors and it has been suggested that "Theory and Practice don't always match".
My understanding is this:
The losses in an induction motor operating from a constant supply voltage and frequency are:

  • Iron loss
  • Copper loss
  • frictional loss
  • windage loss
  • other minor losses

Of these, copper loss is proportional to the current squared, frictional and windage losses are speed dependant, and iron loss is dependant on the flux density in the iron.
This would suggest that copper loss only is dependent on load, decreasing with decreasing load.
Iron loss reduces with reducing voltage and increases dramatically if the voltage is elevated to the point of saturation.
The major losses in the motor are the copper and iron losses which are both in the same order of magnitude, plus or minus.
From the above, one would expect that the total losses in the motor would reduce as the load is reduced and be lowest at minimum load. This certainly is in accordance with my experience.
I recently had a case put to me of a motor where the off load losses were three times the full load losses. This was reported to have been verified by a certified laboratory and really destroys the classical equivilent circuit theory above.
Can anybody shed any light on why or how this can be so and under what circumstances it can be repeated??
;p;

#2 BigMax

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Posted 18 July 2002 - 03:40 AM

To someone who believes he/she has a good working knowledge of AC induction motors, a report like that is similar to being told the earth is square!

To my way of thinking™, the only losses that would actually increase with a DROP in motor shaft load would be windage and bearing friction due to the slight increase in speed as the slip decreases. I couldn't imagine that these losses would be noticable above the drop in I2R losses though.

Perhaps the report was badly conveyed, maybe the losses versus %efficiency was actually being discussed? Certainly the losses as a percentage of total power INCREASE as the load decreases, reaching 0% efficiency at no-load with all true power being due to losses.

Is this helpfull marke?

BigMax:cool:

#3 karthi

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Posted 30 June 2003 - 07:49 AM

energy saving in induction motor

sir,
I am student.I am doing my project in energy saving in induction motor.
we loaded the 5H.P machine at half full load.
the supply to the motor is given through the generator.
The generator output voltage is varied by using 3 phase variac.
The frequency of the generator is varied by varying the speed of the prime mover.
i thing in this method the current waveform may not distorted
Is method of finding the energy saving in the induction motor is correct.
I also agree that for this test we have to go for a alternator and a prime mover.
My question is this way of proceeding is correct .

#4 marke

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Posted 12 July 2003 - 02:18 AM

Hello karthi

This looks fine to me.
Best regards,

#5 karthi

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Posted 12 July 2003 - 05:49 AM

Hello marke,
karthi here.Thank you for your reply.
I have some doubts in calculating the equivalent circuit parameters.
Normally the equivalent circuit parameters are calculated by noload and blocked rotor test.
My doubt is the rotor parameters obtained from these tests(R2,X2) are referred in stator or rotor?
If the results obtained from the above are reffered to rotor how to transform them into stator?
We can trnsform this rotor to stator by multiplying the turns ratio.
How to calculate the turns ratio for cage motor?

#6 Guest__*

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Posted 19 October 2004 - 03:37 PM

can i ask people here how can one go about to measure the windage/frictional losses? someone suggested using a run-down test, but since the loss is speed dependent then its not a constant during the test, and so how cna one tell?

regards

#7 marke

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Posted 19 October 2004 - 06:16 PM

I would suggest drivng the motor in question with another motor and measure the input power with the driving motor open shaft, and with the test motor coupled. The losses are the difference between the two. - I would expect that you would use a driving motor much small er (around 10%) of the size of the test motor.
If you drive the motor with a variable speed drive, you will be able to make the test at different speeds.

Best regards,

#8 jraef

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Posted 19 October 2004 - 08:36 PM

Mark,
Back to your original post...
"I recently had a case put to me of a motor where the off load losses were three times the full load losses. This was reported to have been verified by a certified laboratory and really destroys the classical equivilent circuit theory above.
Can anybody shed any light on why or how this can be so and under what circumstances it can be repeated??"

Was this in the form of a marketing claim? I am very curious about this because, as you mention, it flies in the face of what I currently believe. Were they referring to actuall losses or rate of losses, which when compared to normal lower consumption at lower loads may appear to increase?

I'm thinking it is kind of like the inverse of way politicians tell us that something is costing us less, while they continue to take more money out of our pockets. It costs less in terms of it's RATE of increase, but it still costs more than it did last year. In this case, maybe they are saying that as a percentage of power consumed there is more loss at no load, but the losses are still much less they were at full load.
"He's not dead, he's just pinin' for the fjords!"

#9 marke

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Posted 20 October 2004 - 06:56 PM

Hi Jeff

This was a "certified" test result from a "reputable" laboratory and was produced to prove that my theories were totally incorrect. I dare say it has been produced to many people as a means of overcoming reservations.
I never did get an answer from them as to how the iron loss could increase dramatically at zero load.

I get quite a few arguments and threats in regards to my energy saving paper, but none that have held any water.

Best regards,




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