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Over excited motor 2

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sasice

Electrical
Feb 2, 2005
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I am being asked to run some small motors (0.75 to 4 kW) at up to 100hZ.
My question is what is the long term effect of this on the motor, and on the frequency invertor driving it?

Many thanks
 
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if your motors are rated at 60Hz and you happen to use the V/Hz type drives, you might need to modify some of the parameters on the drives so that the motor would not be operated under much-higher-than-rated voltage for long time. check a so called 'field weakening' operation of the induction motor at high speed. The key is that above the rated speed, the induction motors can be operated under 'constant power' mode, such that the voltage applied on the motor won't be too excessive but the magnetic field in the motor must be reduced.

if you are using more advanced drives ( vector type, direct torque control type), you shouldn't worry too much then. Those drives can handle the motor at that speed properly.
 
Thanks. I feel easier now, and can go back and insist that we get new drives (I've wanted to replace them for a while!!)
 
Be careful. It is not the drive type that determines the "constant power" application above base speed, it is the inherent design of an AC motor.

To vary the speed of an AC motor and provide constant torque throughout, you must maintain the Volts-per-Hertz ratio at a constant. So for instance, a 460V 60Hz motor must get 7.6 V/hz, derived from deviding 460 by 60. However, if you drive the motor to 100Hz, and you want the same torque at that speed, you would need to apply 760V. No VFD is capable of INCREASING the maximum output voltage beyond the input voltage, vector or not.

So once you get beyond the 60Hz range, your V/Hz ratio is going to go down, essentially creating a "field weakend" state where the HP remains constant, but the speed increases. Since Tq = 5250 x HP / RPM, as the RPM goes up with no increase in HP, the Tq will go down. If your load remains constant at that speed, you motor will begin to overload. There are tricks you can use to help compensate, but you need to describe your application a little better to get the right recommendation.

"Venditori de oleum-vipera non vigere excordis populi"


 
Oops, just noticed that you indicated power in kW, not HP so the formula looks a little different (although the results would remain the same). Tq(N-m) = 9550 x kW / RPM

"Venditori de oleum-vipera non vigere excordis populi"


 
jraef is absolute right in the explanation. One thing I forgot to mention is the load. if the load in your application is always less than your rated W/(100*2pi) N-m, it won't trigger an overload on those advanced drives, which will start the 'field weakening' operation above the base speed(so that the voltage is not exceeding your max rated voltage but the field is reduced, most could be found in traction applications). But for those low-end V/Hz drives, the V/Hz parameters needs to be manually set up(or take the default ones), then you need to take care of the above-the-base-frequency parameters properly(i.e. reduce the field,or V/Hz).

If you have a load that needs to be run at 100hz but above rated W/(100*2pi) N-m, jraef is right that no matter what kind of drive you have, the motor is over-loaded.
 
Great guys, thanks.
A little more info. I am driving a rotary valve through a gearbox, so the load will remain pretty much constant.
The torque should decrease slightly as the speed increases, but I dont want to rely on that. Most of the motor/gearbox drives are 0.75KW so I'm looking at using a 1.5kw motor (ie doubling the motor over the rated kw of the drive).
If I understood what you all said, then by doing this I should be OK, yes?
 
The fall off in torque is not linear, so you should get away with it by doubling the power. Are you doing this to attemp to make the valve close faster? If so make sure there are no unintended mechanical issues as a result. Too rapid of a pressure drop accross a valve can cause things like water hammer etc.

"Venditori de oleum-vipera non vigere excordis populi"


 
The reason I have to try this is to extend the delivery speed range of the valve.
The valve is a vane rotor that delivers a given amount of powder per revolution, at 50Hz this equates to approx 50kg/h. Although the relationship is not quite linear, I need to modulate the speed of the valve to deliver anywhere between 15 and 100 kg/h. As all the valves I am dealing with are already controlled by an invertor (for reasons I wont go into), speed control and overexcitation would seem to be the first step if it can be done.
 
Sounds to be relatively valid. A positive displacement scenario as you have described is a good application of this capability of a VFD. As long as you stick to the oversized motor it should work.

Watch out for increased friction issues however. I did something like this on a gravel feed out system, and the rotary valve box overheated. The original design assumed never exceeding the maximum flow with the motor running at 60Hz, so the case could not dissipate additional heat from the gravel friction fast enough. That may not be an issue in yours depending on the makeup of the powder.

Good luck, and let us know how it turned out.


"Venditori de oleum-vipera non vigere excordis populi"


 
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