# Vfd Effect On Motor Efficiency

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I'm stepping a bit out of my comfort zone here, but I have an application possibility for a regen VFD to be used on a small 25kW hydro generator that will be grid connected. The concept is to use the regen VFD so that the output of the generator can be maintained, albeit with reduced capacity, in low flow conditions. The question I'm not sure of is this:

If we reduce flow to where it is only getting 12kW of mechanical input, will the motor/generator be operating at peak efficiency because the VFD is reducing the synchronous speed to always be just below the mechanical input power, or would we consider the efficiency of the motor be operating at the lower end of the efficiency curve because the load is lower? I would imagine it wouldn't be any different that if it were a motor, so in reality I want to know is, when we reduce speed in a motor with a VFD how do we evaluate the motor efficiency at the reduced speed? is it based on it being a lower kW motor, or does the motor efficiency load curve stay the same regardless? To my mind, if we have reduced the speed to 50%, essentially we have changed the motor to a 12.5kW motor, so the efficiency would still be somewhere in the higher range.

Any thoughts?

"He's not dead, he's just pinin' for the fjords!"
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I'm stepping a bit out of my comfort zone here, but I have an application possibility for a regen VFD to be used on a small 25kW hydro generator that will be grid connected. The concept is to use the regen VFD so that the output of the generator can be maintained, albeit with reduced capacity, in low flow conditions. The question I'm not sure of is this:

If we reduce flow to where it is only getting 12kW of mechanical input, will the motor/generator be operating at peak efficiency because the VFD is reducing the synchronous speed to always be just below the mechanical input power, or would we consider the efficiency of the motor be operating at the lower end of the efficiency curve because the load is lower? I would imagine it wouldn't be any different that if it were a motor, so in reality I want to know is, when we reduce speed in a motor with a VFD how do we evaluate the motor efficiency at the reduced speed? is it based on it being a lower kW motor, or does the motor efficiency load curve stay the same regardless? To my mind, if we have reduced the speed to 50%, essentially we have changed the motor to a 12.5kW motor, so the efficiency would still be somewhere in the higher range.

Any thoughts?

Hi Jraef,

I think the below referenced report will answer some of your questions.

The report is entitled: Electric Motor Efficiency under Variable Frequencies and Loads

Also see: Measured System Efficiency vs Speed

Kind regards,

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• 2 weeks later...

Hi Jeff

In fairly basic terms;

the losses in the motor at full speed are primarily copper loss, Iron loss , windage and frictional loss.

The copper loss is highest, followed by the iron loss and a long way behind, the windage and frictional losses.

If we assume that the motor is operated at rated torque at the reduced speeds, and we have a constant V/Hz, then there are some fundemental statements that we can make that show that the efficiency will reduce with reducing speed.

FIrstly, the eficiency of an induction motor driven from a VFD will be lower than DOL.

Secondly, as we reduce the speed, but keep the torque constant, the current will remain the same (rotor and stator) and so the copper losses will be essentially speed independent.

The flux in the iron is constant, but the frequency of the flux reduces. The reducing frequency reduces that actual iron loss due to reducing flux reversals.

The reducing speed reduces the windage and frictional losses.

The motor rating reduces directly with speed.

If we assume that 50% of the losses are copper losses and that the copper losses are constant, the iron loss reduces with frequency and the rest reduce with the square of frequency, then the constant copper loss will result in an increasing loss per KW motor rating as the motor frequency is reduced.

THis results in a reducing efficiency.

Best regards,

Mark.

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