Power factor / efficency of shaded pole motors

[mbrooke]

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.

 

[GoldDigger]

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.

Have you measured the power factor at 5-6A? If it is around .4 that would not be inconsistent with the 1.6A calculated draw.

Also, possibly not a practical consideration, but theoretically a shaded pole motor will develop more resistive losses because of losses in the shading coils than a similar power capacitor start motor.

 

[Phil Corso]

... 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.

The efficiency of shaded-pole motors are extremely low... typically less than 5% for extremely small units to about 30% for say, 1/20Hp! Conversely, its popularity is based on simplicity, ruggedness, and low initial cost! Larger air-gap dimensions and the fact that magnetic-flux is "squeezed" in the shading-coil accounts for higher magnetic-core loses than is found in other small fractional-Hp motor design!

Another distinction is that LR-Current is slightly less than FL-current! Another is that it can be used as a variable-speed motor (within limits, of course)!

Regards, Phil Corso

 

[Besoeker]

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.

If a lot of it was waste heat you'd have a very hot motor.

 

[Phil Corso]

If a lot of it was waste heat you'd have a very hot motor.

Typical... which is why they are often used in small-fan duty!

Phil

 

[Besoeker]

Typical... which is why they are often used in small-fan duty!

Phil

Hot is hot. Regardless of size.

 

[mbrooke]

Have you measured the power factor at 5-6A?

I have not, but its safe to say a SP motor can have a power factor other than unity?

If it is around .4 that would not be inconsistent with the 1.6A calculated draw.

So there is a possibility some of those 5 to 6 amps are indeed reactive?

Also, possibly not a practical consideration, but theoretically a shaded pole motor will develop more resistive losses because of losses in the shading coils than a similar power capacitor start motor.

Correct, but I am wondering how much of those 5 to 6 amps are also resistive. The motor does have a large fan fwiw.

 

[mbrooke]

If a lot of it was waste heat you'd have a very hot motor.

You do, not a very efficient motor. But what has me stumped is just how much of those 5 to 6 amps are going out as waste heat. Assuming unity power factor, that would be around 500-600 watts of heat!

 

[iwire]

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.

Any current not used for turning the motor must be going away as heat right?

 

[mbrooke]

Any current not used for turning the motor must be going away as heat right?

That would be correct if the motor had unity power factor, right? :?

 

[iwire]

That would be correct if the motor had unity power factor, right? :?

Even if the PF was .5 where else but heat would the power go?:huh::?

 

[mbrooke]

Even if the PF was .5 where else but heat would the power go?:huh::?

Power (watts) would go out as heat. But unless unity PF, not all the current (amps) is being put to work (watts)

 

[iwire]

Power (watts) would go out as heat. But unless unity PF, not all the current (amps) is being put to work (watts)

Current can't disappear can it?

It's either spinning the motor or heating things. Is there a third option?:huh:

 

[mbrooke]

Even if the PF was .5 where else but heat would the power go?:huh::?

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).


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.


Assuming a .5 power factor motor, Do I have this correct?

 

[mbrooke]

Current can't disappear can it?

You used the term power, now current :roll: But in any case current can't disappear, but its divided into two categories, active and reactive. reactive current does no work, and gives off no heat.

It's either spinning the motor or heating things. Is there a third option?:huh:

Being shuffled back and forth as reactive current.

 

[iwire]

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).


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.


Assuming a .5 power factor motor, Do I have this correct?

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

Is there any other way for the power to be consumed? :?

 

[mbrooke]

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

Power does not and never disappears. Neither does current. Now whether said current is being put toward actual work (watts) is a different story.

Is there any other way for the power to be consumed? :?

Watts will always result in energy being radiated into the environment be it heat, torque times speed, microwaves energy, light energy ect.

VA, or (reactive current) is current that does no work, only being shuffled back and forth going in and out of the motor.

 

[GoldDigger]

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

Is there any other way for the power to be consumed? :?

Consider an almost ideal inductor. It will have a very very low PF.
If you have 5A of current at 120V, roughly 4.97A will be reactive current and .05A will be resistive current.

Of the 600W of input VA only 6W will go into the combination of mechanical work and heat. The rest does not represent any net power transfer, just energy being stored in the magnetic field and being pumped back into the AC source every half cycle.

There is another very long thread on this subject which you may find entertaining.

 

[iwire]

There is another very long thread on this subject which you may find entertaining.

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.

 

[mbrooke]

Consider an almost ideal inductor. It will have a very very low PF.
If you have 5A of current at 120V, roughly 4.97A will be reactive current and .05A will be resistive current.

Of the 600W of input VA only 6W will go into the combination of mechanical work and heat. The rest does not represent any net power transfer, just energy being stored in the magnetic field and being pumped back into the AC source every half cycle.

There is another very long thread on this subject which you may find entertaining.

Thanks and I did find this:

http://www.tcf.com/docs/fan-enginee...irrel-cage-motors---fe-1100.pdf?Status=Master


So that SP motor in question most likely has a PF or 50 to 60%, which would mean 2 to 3 amps of current is appears across the amp meter without actually being converted to heat or torque?