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VFD readings

VFD readings

VFD readings

(OP)
I have a motor nameplate of:

Voltage = 460-480v
HP = 15HP
FLA = 17.1

HZ = 60hz

Meter readings were taken on the output of teh VFD operating at 60HZ with the following results:

Phase to Phase voltage = 465v
Load current in each phase = 14.7 amps

Similar results where seen at in bypass mode on the output of the bypass contactor.

In addition, readinsg where taken on teh supply side with teh following results:

Phase to Phase voltage = 474v
Current in each phase = 12.07 amps

I am trying to determine possible causes why the phase current on the output is lower than the nameplate rating and why teh supply side and load side results vary.

Thanks.
b


RE: VFD readings

The motor current is the vector sum of the magnetizing current, the load dependent current and the losses. At less than full load the load component of the current will be less than maximum. The load current is is the major component of the motor current at or near full load.
Rated current is the maximum current that the motor may be allowed to draw at rated ambient temperature without overheating.
For more see the Cowern Papers:
http://www.baldor.com/pdf/manuals/PR2525.pdf
From the readings on the output it would appear that your motor is running about 85% loaded.
With the bypass contactor closed either you have serious measurment errors or for some reason the bypass contactor did not close. I suspect (and your readings indicate) the latter.
By the way, given that the output of the VFD is not a sine wave but a pulse width modulated waveform, normal instruments are not accurate. To determine the output of the VFD you should use the VFD display.

Bill
--------------------
"Why not the best?"
Jimmy Carter

RE: VFD readings

Remember, too, that the input to a modern VFD is essentially a high-efficiency switching power supply. As such, the input has a very high power factor. Typical is 0.98, often adjustable. The output is supplying the VARs that the motor needs (uncorrected power factor). The output should always be higher, in spite of the VFD's inefficiency.

As Bill said, the motor nameplate amps will not be reached until the motor is fully loaded.

What sort of load do you have on your motor?

Best to you,

Goober Dave

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RE: VFD readings

(OP)
Thanks for the helpful information. The load is a supply fan integral to an air handling unit. Are there any technical documents that discuss this sort of topic on vfd testing and interpreting values?

RE: VFD readings

How are you measuring the output current and voltage? Standard digital multi-meters are pretty much useless on AFD outputs.

RE: VFD readings

Waross' and DRWeig's post boiled down:

The motor is likely at a .8 power factor, but the drive is at a .95 or so.

There aren't any papers I'm aware of on testing procedures, although if any exist, they are likely to be found at Fluke. But most drive data sheets will tell you that the drive corrects the power factor seen by the line to .95 or better, the rest is just how you apply that information.

By the way the voltage is different because some PROGRAMS the maximum output voltage of the drive. A VFD only uses the line power as a resource for energy. Everything going to the motor is essentially manufactured as a new power source created from that resource. So the output of a drive is whatever you tell it to be, up to the limits of what it was given of course. The VFD cannot create voltage potential that was not there to begin with. So in your case the drive COU
D HAVE been programmed to put out 474V at 60Hz, but that would not have been a good idea.

"Will work for (the memory of) salami"

RE: VFD readings

Just make sure you don't use the lower input current readings as an indicator that the VFD is saving power compared to running on the bypass contactor. A good power meter would make the readings much clearer. You'll find the input watts is greater then the output watts even though the input VA is less then the output VA.

RE: VFD readings

When the drive is controlling the motor, it operates in a manner to maintain constant flux in the iron. This in turn requires that as the output frequency reduces, the voltage also reduces. The voltage on the line side if fixed, but the voltage on the motor side changes with frequency. If you look at a transformer, it acts as a constant VA device, reduce the voltage and increase the current.
In the case of a VFD, in addition to the voltage transformation, the VFD strips the VARs that flow into the motor, from the supply current effectively bringing the displacement power factor back to better than 0.95. So rather than VA in equals VA out as in the case of a transformer, it is KW in equals KW out (minus losses) in the case of a VFD.
At reduced speed the input current is less than the output current due to a) the "power factor correction" and b) the voltage transformation.

Mark Empson
Advanced Motor Control Ltd

RE: VFD readings

I have been testing power measurement on the input of the VFD and made a first order rough correction for
the VFD's own power consumption. That worked quite well. If needed, one can add second order effects that are load dependent. Notes here:

http://www.gke.org/rapporter/files/Power%20measure...

Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

RE: VFD readings

Nice, Gunnar.

I may make a training episode out of that whole concept for my power measurement class, with your permission.

Best to you,

Goober Dave

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