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Supply Network/vsd/motor Power Ratio


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#1 miki83

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Posted 13 October 2009 - 03:58 PM

Hi guys.

I have a certain problem, and I was wondering if anyone could help me?

I need to select a proper VSD and a motor for a constant torque application. Since it would involve lowering the speed to approx. 25% (~750rpm, 12.5 Hz) I contacted our motors supplier to request a special motor with improved insulation&cooling, and he just gave me a headache.

The problem is he is trying to convince me to take a 132 kW motor instead of a 110 kW one we are using to run the same load in a constant speed application (star-delta or softstarter starting) because he is assuring me that I would have a smaller power output on the shaft when using a VSD instead of a classical starter. Now I know that in order to get my motor to reach its nominal power when using a VSD I would have to draw more power from the supply network meaning I would have to slightly oversize the VSD (nominal current of the VSD has to be bigger than the nominal current of the motor in order to avoid the VSD tripping due to an overload error, right?), but I can't get to understand why would I need a bigger motor in this case? If a VSD can substand a high enough current drawn from the network in order to supply the nominal current to the motor, I should have the same power output on the shaft as I would have if I was using a Star-Delta or a softstarter, right?

If anyone could clear this up a bit for me I would greatly appreciate it. Thanx in advance to all of You.

Best regards, Miki.


#2 marke

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Posted 13 October 2009 - 06:06 PM

Hello Miki

If you are going to operate your motor at a reduced speed, then the effective power that you have available from the motor reduces.

At rated speed, you have rated power and rated torque available. If you reduce the speed by a mechanical means, you will have rated power and higher than rated torque at the lower speed. Power = Torque x Speed.
In the case of a VFD, the maximum torque that you have available is limited by the flux in the iron and this is limited by saturation effects of the iron. The result is that the torque at speeds less than rated speed is limited by the flux to be equal to rated torque.

At very low speeds, the flux in the iron drops off due to the resistance of the winding becoming significant relative to the reactance of the winding and this causes a drop off in torque. Properly tuned open loop vector systems lower the frequency where this becomes an issue, but do not eliminate it.
Additionally, at low speeds, you have cooling issues with the motor. I personally would try to avoid operating the motor for any period of time below 20Hz, especially if rated torque is required.

In situations like this, I would tend to go for a lower speed motor and overspeed it at full speed.
In your case, you appear to be using a 2 pole motor and operating it at 12.5 Hz, if you use a 4 pole motor, you would operate it at 25Hz.

Do you need to operate at full speed also?

If you are concerned about full power at rated speed, the VFD would typically be more than 95% efficient, so the current into the VFD would not be significantly higher, however there are a lot of harmonics on the input current and due to this you will need to oversize the supply components by as much as 35%.
The output from modern VFDs is pretty good but there are additional losses in the motor due to harmonics. Perhaps the motor supplier is concerned about additional motor heating when operating on a VFD. With early VFDs, this was a significant issue, but with modern VFDs, the additional losses in the motor are small.

Best regards,
Mark.

#3 miki83

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Posted 13 October 2009 - 08:13 PM

Hi Marke and thank You very much for a swift reply.

Let me clarify the situation a bit more if You don't mind. The application is a screw compressor, and the idea is to make a feedback loop (feedback provided by a pressure transmitter) with a tuned PI controller to enable the VSD to adjust the speed of the motor (it is a 2-pole motor) hence adjusting the intake of the compressor to fit the current consumption. This taken in consideration the motor would run on a certain range - it will most likely be 40-100% of the rated speed (20-50 Hz, we are planning to try and lower the frequency to 12.5 and see how things would develop). Control method would be sensorless vector. Since I'm a process control engineer my knowledge of motors is somewhat limited, but the supplier tried to convince me that I have to use a bigger motor due to some losses in the VSD, and that's what didn't make sense to me. I'm aware that we have to watch out for the increased temperature of the motor due to lower speed which degrades the ventilation, but as far as I know I should have the rated torque available on the shaft for this speed range? Please correct me if I'm wrong on this? By the way we are planning to use a Danfoss FC 302 VSD which has a rated current of 212 Amps, while the rated motor current is 201 Amps, so I think we're good on this matter.

Best regards,

Miki.





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