Operating electric motor at reduced voltage

[maypot]

Hi,

What are the consequences of operating an electric motor , say 4.0 MW 11 k V delta connected, at 6.6 k V keeping the same delta configuration. It is understood that the 4.0 MW motor is not provided with six leads at the terminal box.
I understand that so long that the Imax of the motor is not exceeded, there is no risk of damaging the motor.
I also understand that the shaft power will only be one third of 4.0 MW .
It will be simpler if the motor was 11 k V star connected.
My concern is about the fluxing of the motor, will the reuced voltage sufficient to provide the required flux ?

Bob

 

[Marke]

Hi maypot

Operating the motor at a lower voltage will reduce the flux in the gap and thereby reduce the torque capacity of the motor. You will have no problems provided that the shaft power is reduced. The operating efficiency will not be optimum as the iron loss will be less but the copper loss will be higher than for a motor wound for that voltage.
Just be careful that the motor does not operate at increased slip as you will damage the rotor.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[jraef]

I agree, as long as you are clear that the output torque of the motor will be roughly 1/3 of normal, thoughout the torque speed curve. One hidden consequence of that may be recovery after a load change. If your application has any kind of step change in load, you will not have the full benefit of all the original breakdown torque available to reaccelerate it. You may want to consider using a lower Overload setting as well just to protect the motor better.

"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more."
Nikola Tesla

 

[aolalde]

As far as the connected load does not exceeds the nameplate full load current for 11 kV the windings are safe. I assume you will keep the same overcurrent protection as that for the original 11 kV condition.

Your flux will drop proportional to the voltage reduction and the torque will be reduced with the squared ratio of voltage reduction (6.6/11)^2*100 = 36%

 

[mc5w]

There are also variable frequency drive and autotransformer controllers that do this for industrial motors that are oversized for some reason. Reasons can be extra starting torque using a standard design B motor rather than a harder to obtain and less efficient design D motor. However, most of these only do a reduction to 75% to 90% of full voltage as a lower voltage increases copper losses and impairs recover torque.

Other reasons can be for worst case load such a vibratory finishing using steel shot ( which uses more power than stones or corn cob meal ) or the largest possible die that can fit into a punch press. When operating at a lower load reduced voltage running can make a 20 horsepower motor act as a 10 horsepower motor if a machine is that overmotored for that much - if the load can span 7 to 20 horsepower for example than reduced voltage running is feasible if you already are paying for a VFD or autotransformer starter.

 

[jraef]

Just to be clear, you can't do that with a typical autotransformer starter. The transformers are rated for maybe 15 seconds of operation, not continuous. As a custom designed system where all factors are considered, maybe.

I just wanted to point this out in case someone unfamiliar with how specialized that is might try it with an off-the-shelf autotransformer starter.

And of course if he had a VFD, he would most likely not have asked the question. A VFD for a 4MW 11kV motor would be in the neighborhood of US$750,000.00+, and I would think anyone who was thinking of spending that much would not be bothering with free advice (at least I would hope not).

"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more."
Nikola Tesla

 

[maypot]

Obviously I do not have a vfd at hand. I understand the fact that reducing the voltage from 11 to 6.6 kV will reduce the torque by three. What about the output shaft power ?

Rgds.

Bob

 

[Marke]

Hi Bob

As the torque is one third, and the speed is the same, the power must also be one third. Power equals torque times speed.

Best regards,


Mark Empson

http://www.lmphotonics.com