VFD and induction motor power factor 1

[eeprom]

Hello,
I have a VFD driving a 50 HP induction motor (3 phase, 460V). At 60 Hz, the motor draws around 70 A. If I bypass the VFD and connect the motor directly to line voltage, the current draw is about 62 A. I have already taken into account the difficulty of measuring current from a VFD (I use the internal readout). I think that the reason for the difference in current draw has to do with the high frequency noise coming from the VFD. The VFD noise is effectively reducing the motors power factor, and so the motor requires more current to do the same amount of work. Does anyone know if this is true? Does a VFD have an effect on the power factor of a motor?

thanks,
Andy

 

[Skogsgurra]

It is a little simpler than that. A VFD has a power factor close to 1. So, the VFD takes less current from the mains than it outputs to the motor. The reactive power circulates between motor and DC link - never needs to go "on line".

When you connect motor to mains, you see full current. Including reactive current.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[dcasto]

Glad you agree with the PF being close to 1. My EE says you can get the PF to 1.1 if you specify it that way. In another post I was told the PF was always less than 1....

 

[Skogsgurra]

PF > 1? That means perpetual motion - or something...

And, please, I do know the difference between PF and eta...

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[dpc]

Displacement power factor is the cosine of the angle between the voltage and current, so it can never be more than 1.0. It can be lagging or leading, but never more than unity.

 

[Skogsgurra]

Yes. And PF, be it displacement or distortion, is P/S. And since S = sqrt(P^2 + Q^2) it also follows that no PF, how ever defined, can be more than 1.


Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[dcasto]

With the PF more than 1. in your plant, you can pull up the overall PF. This is an alternative to adding capacitors.

 

[LionelHutz]

The difference can depend on the line voltage vs the VFD output voltage. If you set the VFD to 460V motor it will output 460V even with a different line voltage (within limits of course). The change in voltage could affect the output current. I've seen this mostly be a factor when the motor voltage setting is set too high.

Of course, I've just seen higher currents on VFD's before and it is likely due to the extra losses from the carrier frequency. More current capacitive couples to ground too.

 

[Skogsgurra]

dcasto,

We just said that you cannot get higher than 1.

Having leading PF is another thing. But it cannot be higher than 1.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[rbulsara]

I would say Lionelhutz's response is more accurate and in-line with what the OP asked. Some of otherwise more knowledgeble respondents appear to have misread the question, even if their statements may be true on their own.

If pf were to deteriorate when "bypassing" the VFD, the current would increase. OP said that the current when motor is directly on line is less than while on VFD. It is not to be confused with input and output current of the VFD.

 

[jraef]

Does a VFD have an effect on the power factor of a motor?

(underline emphasis added).

I agree. Ihe question, as asked, relates to the motor's power factor, not what the line sees. To that question, the answer is; maybe, but does it matter?. If anything, the VFD can improve the effective pf of the motor when running at reduced speeds because it will be altering the motor power all together to match the load. At full speed however you may be right, but most likely the display of the load current is going to take that consideration into it's calculation, or at least it should. The only one who could really answer that however would be the engineer at the VFD manufacturer who was charged with designing that algorithm.

Bottom line though, I seriously doubt that this is the reason why you have two different readings. You are off by about 10-1/2%. Figure 3% error factor in the VFD's display reading (if that low) and a potential of 2% error in whatever you used for the across-the-line reading, and now you are only off by 5-1/2%. A VFD couple easily have that much in switching losses at full speed, depending on the technology, age, design, if there are reactors etc.

 

[Skogsgurra]

Yes, I did read the OP as a question about the installation. Not the motor as such.

Yes. Of course, the output of the VFD has a deep impact on the PF of the motor. As said by others. And jraef's error calculation is also something worth considering.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[jraef]

Don't know where that word "couple" came from, should have been "could". Must have been a screwy spell checker correction, sometimes I whip through those too fast I guess.

 

[aolalde]

When a motor is supplied by the utility the voltage and current are close to a sinus wave. When a VFD feeds the motor an integrated wave, some times far apart from a sinus wave is fed to the motor. Distortion could be the reason for increased current figures. The type of measuring instrument could play a role too. Could you sample the wave form with a scope?

 

[itsmoked]

Sinus wave??

Take two aspirin and call us in the morning.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Marke]

At ful line frequency output of the VFD, there is commonly some distortion of the output waveform resulting in some flat topping, plus the output voltage is typically a little lower than the line voltage, plus there is a lot of capacitive current flowing at the output of the drive due to the capacitance of the cable and motor. Essentially, all these factors will tend to increase the actual output current of the VFD at line frequency. Drop the frequency a little and you get a better sine wave output current and no voltage limiting.
Additionally, you have inaccuracies of measurement. Some VFDs measure the DC current and use that to determine the output current. This can lead to high inacuracy.

The Cos(thi) of the input to the VFD is better than 0.95, but the true power factor is typically less than 0.8 due to the distortion power factor.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[waross]

It's good to see you drop in, Marke. Your knowledge of drives is always valuable to the forum.
Respectfully

 

[Skogsgurra]

For Smoked: Yes. Sinus is the original Latin form for sine. That is what this function is called in most of the world outside US/UK/OZ.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[itsmoked]

Whoa! Really. Most interesting. Thanks Skogs.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Valvecrazy]

I agree with jraef. 5% loss in the drive itself is common. I have experienced this many times. The motor pulls less amps when running across the line, than when controlled by a drive. The output voltage of the drive cannot be higher than the input voltage. When using reactors or line filters, even less voltage makes it to the input of the drive. If the motor is running on 480 volts when supplied across the line, and 460 volt through the drive, the lower voltage must draw higher amperage to supply the same power. If actual energy used is volts X amps X power factor, then a motor running on a drive always uses more power than the same motor running across the line.