Power factor / efficency of shaded pole motors

[junkhound]

Ok, so basically I've come across conundrum where I have a shaded pole pump motor that draws about 5 to 6 amps when running, yet for the work it does it should only be drawing about 1.6 amps. My question is from those 6 amps how much of that current ends up as waste heat and how much of it is just reactive power? I'm at a loss to explain this exceptionally high current draw.

Good question - realized and am embarrassed that in 7 decades have not thought to ever actually measure one of these little motors.
Just always took them for granted and that as per textbook the efficiency was under 50% and PF under 0.7
So, learn some facts every day.

grabbed an old GE fan motor, 1/6 HP shaded pole, 6.0Amp, 6 pole, 120 Vac 60 Hz motor from the stockroom.

Bench test data:
Unloaded
116.8 Vac
5.32 Amps rms
269 Watts
PF 0.43
Math check: 116.8 * 5.32* .43 = 267W, close, within 1%

Small fan load
116.8 Vac
5.33A rms
294 Watts
PF =0.47
Math check: 293W, within 1%
efficiency = 25Wload/294W = a whopping 8.5% efficiency, not very 'green', eh?

My dynamometer is hooked to a small micro-hydro turbine at the moment, so cannot load the motor for a better load measurement, but 1/6Hp shaft = 124W, so working thru the math the best efficiency would be about 40% full load. PF estimate max about 0.6 full load.

 

[kwired]

No way, :happyno: that is an engineering black hole and if you guys with the letters after your names can't seem to agree about the topic I will consider it as undetermined.

Assuming your response was serious

power factor correction capacitors draw current, don't do any work, and the only power drawn is because of any resisitive component that exists - though that is usually low enough to be considered negligible. All conductors have some amount of resistance.

There are two kinds of power:

true power - the power that is doing work, which also includes power lost as inefficiencies from friction or resistance in the conductors.

apparent power - the power that is reactive in nature - you only see this in inductors and capacitors, this power flows between the source and the inductive or reactive load, but does no work, because of increase in current however it will increase resistance losses, though those are usually relatively low, but do add up after you get enough reactive loads or even have a long circuit run.

By placing a power factor capacitor near the motor (on the same circuit) we allow that reactive current normally between the source and the motor to take place between the capacitor and the motor. This is why the main supply current reduces when you add a power factor capacitor, yet if you can clamp the motor leads you see the motor is still drawing approximately what it would draw without the capacitor.

 

[kwired]

How often does one see a shaded pole motor driving something other then a fan? If you do it is usually something very limited in duty cycle and/or a very low torque demanding application.

If it is not driving a fan as primary application it will still have a cooling fan attached to the shaft if it has much of a duty cycle at all. I recall a drain pump on a washing machine that was shaded pole motor - did have fairly limited run times, but did also have a cooling fan on the shaft.

 

[mbrooke]

Good question - realized and am embarrassed that in 7 decades have not thought to ever actually measure one of these little motors.
Just always took them for granted and that as per textbook the efficiency was under 50% and PF under 0.7
So, learn some facts every day.

grabbed an old GE fan motor, 1/6 HP shaded pole, 6.0Amp, 6 pole, 120 Vac 60 Hz motor from the stockroom.

Bench test data:
Unloaded
116.8 Vac
5.32 Amps rms
269 Watts
PF 0.43
Math check: 116.8 * 5.32* .43 = 267W, close, within 1%

Small fan load
116.8 Vac
5.33A rms
294 Watts
PF =0.47
Math check: 293W, within 1%
efficiency = 25Wload/294W = a whopping 8.5% efficiency, not very 'green', eh?

My dynamometer is hooked to a small micro-hydro turbine at the moment, so cannot load the motor for a better load measurement, but 1/6Hp shaft = 124W, so working thru the math the best efficiency would be about 40% full load. PF estimate max about 0.6 full load.

Awesome, I mean just awesome! I needed exact numbers And no, certainly not green.


But in any case that is exactly what had me wondering. Even if my motor in question is 40% efficient, it still did does not explain those 2-3 extra amps, but when PF is taken into account it makes perfect sense now. Either that thermodynamics had taken a major polar shift

 

[mbrooke]

Good question - realized and am embarrassed that in 7 decades have not thought to ever actually measure one of these little motors.
Just always took them for granted and that as per textbook the efficiency was under 50% and PF under 0.7
So, learn some facts every day.

grabbed an old GE fan motor, 1/6 HP shaded pole, 6.0Amp, 6 pole, 120 Vac 60 Hz motor from the stockroom.

Bench test data:
Unloaded
116.8 Vac
5.32 Amps rms
269 Watts
PF 0.43
Math check: 116.8 * 5.32* .43 = 267W, close, within 1%

Small fan load
116.8 Vac
5.33A rms
294 Watts
PF =0.47
Math check: 293W, within 1%
efficiency = 25Wload/294W = a whopping 8.5% efficiency, not very 'green', eh?

My dynamometer is hooked to a small micro-hydro turbine at the moment, so cannot load the motor for a better load measurement, but 1/6Hp shaft = 124W, so working thru the math the best efficiency would be about 40% full load. PF estimate max about 0.6 full load.

Ok, so my 120 volt 5.5 amp motor. Assuming a PF of 50%:


330 VA, or 2.75 amps of reactive current,

330 W, or 2.75 amps of active current.

Assuming 30% efficiency that would give me 100 watts of shaft power and 230 watts of dissipated heat.



Does my math look right?

 

[Sahib]

Now, mbooke, you find the mistake in your calculation in post14/25 or not?

 

[mbrooke]

Now, mbooke, you find the mistake in your calculation in post14/25 or not?

No, not found, need some help getting there.

 

[wwhitney]

I am asking you, I really do not know. But I am fairly certain power does not just disappear.

Right, power does not disappear.

But the difference between VA and Watts is that VA is not power. VA is just "possible power if the power factor were 1.0," i.e. if the current and voltage were in phase.

So when the OP says the current is 6A, then if the voltage is 120V, the VA is 720. From the OP's info, the mechanical power provided by the motor is then supposedly 1.6A * 120V = 190W.

The discrepancy between 190W mechanical power and 720VA "apparent power" is then somewhere between these two extreme cases:

1) The power factor is 1.0, so the electrical power delivered is the full 720W, and as you say, the missing 720W - 190W = 530W must be dissipated as heat. That will be one hot motor!
2) The power factor is very low, the motor is 100% efficient, and the electrical power delivered is only 190W. The discrepancy is all due to reactive power, the "extra" current is flowing but not providing any power to the load, and no heat is developed (other than the I^2 R losses in the conductors attributable to that extra current.)

Hope that helps.

Cheers, Wayne

 

[Sahib]

Mbrooke You summed up active and reactive currents arithmetically whereas in reality they sum up in quadrature.

 

[mbrooke]

You summed up active and reactive currents arithmetically whereas in reality they sum up in quadrature.

I mean they do, but for the sake of the discussion, and circuit sizing to the NEC, basic addition is sufficient.

 

[wwhitney]

efficiency = 25Wload/294W = a whopping 8.5% efficiency, not very 'green', eh?

Just curious, how did you determine for this test that the mechanical power developed was 25W?

Cheers, Wayne

 

[iwire]

Assuming your response was serious

Man you and I can never seem to communicate well. We are both intelligent guys with diverse backgrounds and I bet we are both good at our jobs but here on the forum we are like oil and water.

You seemed to understand I was serious about this topic being a black hole I am not interested in pursuing due to the fact that so many folks with much more education than I cannot agree to much about any of it. :huh:

Yet you still felt the need to try to drag me into it.

 

[mbrooke]

Right, power does not disappear.

But the difference between VA and Watts is that VA is not power. VA is just "possible power if the power factor were 1.0," i.e. if the current and voltage were in phase.

So when the OP says the current is 6A, then if the voltage is 120V, the VA is 720. From the OP's info, the mechanical power provided by the motor is then supposedly 1.6A * 120V = 190W.

The discrepancy between 190W mechanical power and 720VA "apparent power" is then somewhere between these two extreme cases:

1) The power factor is 1.0, so the electrical power delivered is the full 720W, and as you say, the missing 720W - 190W = 530W must be dissipated as heat. That will be one hot motor!
2) The power factor is very low, the motor is 100% efficient, and the electrical power delivered is only 190W. The discrepancy is all due to reactive power, the "extra" current is flowing but not providing any power to the load, and no heat is developed (other than the I^2 R losses in the conductors attributable to that extra power.)

Hope that helps.

Cheers, Wayne

Around there lol. The new replacement motor for this application is listed at 1.5amps (1.8amps on the machine label), rated 3150rpm and is a permanent split capacitor. It is intended to provide similar if not identical results. Considering a PSC is more efficient with good power factor required output power on the shaft should be around 150 watts of power.

 

[Sahib]

I mean they do, but for the sake of the discussion, and circuit sizing to the NEC, basic addition is sufficient.

The way you split current using power factor is not correct and I can't believe NEC supports it.

 

[wwhitney]

If said shaded pole motor is drawing 5.5 amps at 120 volt with a PF of .5, that would mean about 2.75 amps of reactive current is being drawn. The remaining 2.75amps of the 5.5 amps being drawn is being put to work (watts or active power).

Since you asked, that's not correct, as Sahib pointed out.

Those 330 watts of active power fork out into two places. 100 watts are being spent on driving the output shaft (and in turn fan/impeller), while the remaining 130 watts are being given off as waste heat.

130 + 100 = 230, so you probably meant 230 watts of waste heat. Otherwise correct, and you can do the power accounting like this without have to calculate reactive currrents.

Cheers, Wayne

 

[mbrooke]

The way you split current using power factor is not correct and I can't believe NEC supports it.

Its more to help me understand whats going on with the current magnitude entering the motor at any given point in time. Seeing it in scalar quantaties helps me grasp the concept better and from my understanding is still correct at the fundamental level.


AS is, in the NEC when we size wire we do it in VA instead of watts.

 

[Sahib]

Its more to help me understand whats going on with the current magnitude the motor at any given point in time. Seeing it in scalar qunataties helps me grasp the concept better and from my understanding is still correct at the fundamental level.

No it won't help you to match your calculated results with junk hound experimental results.

 

[mbrooke]

Since you asked, that's not correct, as Sahib pointed out.


130 + 100 = 230, so you probably meant 230 watts of waste heat. Otherwise correct, and you can do the power accounting like this without have to calculate reactive currrents.

Cheers, Wayne

typo, I meant 230 watts of waste heat like in this post:

http://forums.mikeholt.com/showthread.php?t=179439&page=3&p=1772948#post1772948

Still checks out ok?

 

[wwhitney]

Still checks out ok?

Not sure what you mean, your power accounting is fine, your calculation of 2.75 amps as the real and reactive current is not.

You can say that 2.75 amps is the current that would be drawn if the real power was still 330W and the power factor was 1.0. But if you add a current waveform that has a magnitude of 2.75 amps to a current waveform of the same magnitude that is 90 degrees out of phase, the amplitude of the resulting waveform is not 5.5 amps.

Cheers, Wayne

 

[junkhound]

Just curious, how did you determine for this test that the mechanical power developed was 25W?

Cheers, Wayne

Since the current is almost the same, I^2*R losses are nearly the same, and since the voltage is the same the eddy and hysteresis losses in the core are similar, hence one can stick their thumb in the air and take the wattage difference as the power going to the load (admit that is a rough approximation, the core losses do change somewhat with the load as more current in the rotor, etc, and that I should do a current ratio squared corrections also, but sufficient for this discussion)
To be accurate, would need to hook up the dynamometer, but close enough for this discussion.

edit: for completeness of the test data, values taken from readout of a Geist BRNC-100 panel load center.