Moving on from my previous power factor thread

[Surefire01]

I have a small three phase test motor and on the nameplate it gives a power factor of 0.77. This equates to an angle of around 45 degrees. I hooked the motor up to an oscilloscope and measured around 4mS difference between the voltage and current.
Four milliseconds equates to around 72 degrees and the cosine of 72 gives a power factor of 0.34. That is some difference between the nameplate and the measured values.
One reason that I can think of for this discrepancy is that the value quoted on the nameplate is full load power factor and during my measurements the motor was running unloaded. Could this be the reason? If so it just goes to show how inefficient it is to run a motor partially loaded!

 

[sibeen]

Well as there going to be input power used and no output power used I'd have to say the efficiency would have to be 0

 

[Surefire01]

Well on the test motor I jut tried putting some load on the shaft and sure enough the displacement between voltage and current fell from 4mS to 3mS that equates to a change in power factor from 0.3 to 0.59. Still some way off the quoted nameplate value of 0.77 but getting better and proving that as the motor is loaded the power factor improves.

 

[jraef]

Yes it is because of the loading difference. The nameplate value is under full load, unloaded the PF can be anywhere from .2 to .72 depending on design.

"Will work for (the memory of) salami"

 

[squeeky]

All motor name plates are at full load. At full load, measure the voltage, current and power factor calculate the input power.using the input power against the rated power and you can calculate efficiency. Power factor and efficiency will drop off at low loads.