Mystery on Power factor readings!

[pwcwatt]

To all interested parties:

I am running into some strange readings on power factor correction with a single phase motor. Here is the process in a nutshell:

I have an electric meter, a power strip, 200 feet of line, and a motor arranged in that order. With my Fluke I took the following readings:

At the motor -
Volts - 116.84
Amps - 4.11
kW - .15
kVA - .48
kVAR - .46
Power Factor - .313

At the meter -
Volts - 118.84
kW - .18
kVA- .48
kVAR - .45
Power Factor - .375

So far so good. It appears that there is .03kW (30 watts) of loss in the line. This computes to about 1.8 Ohms of resistance which fits fairly well with known data.

Then I hook up a capacitor at the motor and take all the readings again.

At the motor -
Volts - 117.34
Amps - 1.47
kW - .10
kVA - .18
kVAR - .14
Power Factor - .556

At the meter -
Volts - 119.24
Amps - 1.48
kW - .10
kVA - .18
kVAR - .15
Power Factor - .556

So I went from .18kW before correction at the meter to .10kW after correction at the meter. Yet this seems to be impossible. It appears that the wattage drop of 30 watts (line loss before correction) would be the maximum that could be saved, yet I am saving more like 80. It gets stranger. If I time the meter the savings comes out to be more like 20% instead of the 40% that the Fluke shows which is totally baffling.

By the way I hove done this test with 2 motors at the end of the line and the results are virtually identical.

Please help.

Thanks

 

[jghrist]

It appears that the motor is more efficient at the higher voltage.

Part of the difference could be meter error. Are you measuring power directly or calculating it from volts, amps, and power factor?

 

[bacon4life]

I will try a crack at this, but keep in mind I don’t have much real world experience with motors.
1. Is the motor loaded? I know motors often have bad pf, but 0.3 fully loaded?
2. Perhaps you have a type D motor with significant rotor losses, or are running a class B motor near the breakdown point torque region rather than the small slip region.

For a normal motor, losses would be reduced as the voltage decreased for unloaded motors. At no load the magnetizing losses would dominate the losses, reducing the voltage would lower the magnetizing losses. The rotor losses are a product of the slip and the power transmitted to the rotor, and thus are small. However for a D motor, I think the rotor losses would still be significant at no load.

At full load, the power transmitted to the rotor is much more, so the rotor losses will be much more significant. Also, as the voltage decreases the slip will increase, also increasing losses. It is easy to forget that decreasing the voltage can actually increase current, but that would be for voltages lower than 90 or 95% of nameplate. What voltage is the motor rated?

Hopefully someone else will chime in here, because this sure is intriguing.

 

[pwcwatt]

Some additional information:

The motor is not loaded, merely idling. Hope this helps.

 

[davidbeach]

Power factor on unloaded induction motors is usually bad. What kind of meter are you using for your tests, does it have decent accuracy for power measurements at low power factor?

 

[electricpete]

You had an inductive load (unloaded motor) and you added power factor correction which decreased the current by almost a factor of 3 (4.1A down to 1.5A).

So the losses in the 200' of line should go down by current squared or close to 1/9 of original line losses.

Original line losses were recorded at 0.03kw, so reduction in losses close to 0.03kw.

If we believe the 20% savings instead of the 40% savings, it's not too far off. 0.2* 0.18kw ~ 0.03kw.

Also there are always some errors in measurement, particlarly if the quantities measured are near the bottom of the range of the instrument. Less likely the motor efficiency may change slightly over time as it heats up (stator and rotor winding that is not too resistances increase).

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