Induction Motor - constant Torque 4

[CDG16]

What is the effect of a constant low voltage on a Induction Motor (constant Torque, 100Hp)? Tell me if I am wrong, but it will draw more current to compensate for the lower voltage, therefore more I(sqr)R Rotor losses and therefore more heat. Eventually the winding insulation will break down and the motor will fail. Is this all correct? If not, could someone please explain what will happen in terms of the current drawn and how it will effect losses in the motor.

Thanks for all your comments.

 

[DickDV]

You haven't given us enough information to answer. What is the frequency of this low voltage? And, what is the load on the shaft of the motor?

 

[CDG16]

The frequency does not change for the drop in voltage and remains 60Hz. It's a 100Hp motor and the load on the shaft does not change either. The only paramter that change is the Voltage, dropping from 480V to 420V.

 

[electricpete]

As a rough approximation:

If you were near full mechanical load, current would increase by a factor of 48/42. (if at lower load, the fractional increase in current is smaller)

Stator i^2*r and rotor i^2*r losses and stray losses (maybe 60% of total losses) would go up by a factor of approx (48/42)^2

iron losses (maybe 20% of total) would go down by a factor of approx 42/48

friction losses and (maybe 10% of total) would remain approximately constant

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[electricpete]

All of the above assumes that you don't exceed breakdown torque.

I think that's a safe assumption. The torque speed curve decreases by approx (42/48)^2. If breakdown torque is 200% at full voltage, it is maybe 160% at the reduced voltage.

Note in the US, the nameplate voltage of motors is 460, not 480 (even though they are used on 480 volt system). So the correct ratio to use may involve 460/420 if you want to compare to nameplate conditions.

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[Marke]

Hello CDG16

If you reduce the voltage applied to an induction motor running at a constant torque, you will reduce the flux in the iron. This will reduce the maximum torque available, and if you reduce the voltage too much, the torque available will be less than the torque drawn by the load and the motor will stall.
Assuming the voltage reduction is not sufficient to induce a stall, the following will happen:
The magnetizing current of the motor will drop.
The work current of the motor will increase.
The slip of the rotor will increase.

If the work current of the motor is significantly higher than the magnetizing current, you will see an increase in line current to the motor.
If the work current is significantly less than the magnetizing current of the motor, you will see a reduction in the line current to the motor.
The copper loss in the stator will vary with the square of the line current.
The iron loss in the stator will reduce.
The slip losses in the rotor will increase.

If the motor is very lightly loaded, the energy loss in the motor will reduce. If the motor is moderately to heavy loaded, the energy loss in the motor will increase.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[jraef]

I think a key point here however is what electricpete pointed out earlier. If it is a true NEMA MG-1 designed motor, the design voltage should be 460V. Not only that, but the voltage tolerance should be +- 10% of design and maintain torque and speed within normal, so 460 x .9 = 414V. Considering that then, 420V should not be detrimental to the application.