Power factor / efficency of shaded pole motors

[mbrooke]

Inductor and resistor in PARALLEL, see schematic below the graph, the little 1 millohm series resisitor is to just keep the Pspice program from having to deal with a zero value.

Parallel... again, its wrong word day for me

It would be a motor model if you could build a motor with NO leakage reactance. I just drew out a parallel circuit since it was easy to calculate values in my head for about 2.7 A each leg.
Intent is to illustrate the phase shift (90 deg for the inductor) and why you need to do vector sums when dealing with rms values of current.

Understand. Let me digest all this... :thumbsup:

 

[wwhitney]

a2+b2=c2? Or am I way off when you say triangle inequality? But that make sense now that I think about it regarding code sizing.

Triangle inequality:

For a triangle with sides A, B, and C and side lengths a, b, c, respectively, then c < a + b. Geometrically, if you stick the vertex AC (where sides A and C intersect) at the origin, then C = A + B as vectors. So the triangle inequality can be written as |A + B| < |A| + |B|, which compares arithmetic summing with vectorial summing.

Cheers, Wayne

 

[mbrooke]

Triangle inequality:

For a triangle with sides A, B, and C and side lengths a, b, c, respectively, then c < a + b. Geometrically, if you stick the vertex AC (where sides A and C intersect) at the origin, then C = A + B as vectors. So the triangle inequality can be written as |A + B| < |A| + |B|, which compares arithmetic summing with vectorial summing.

Cheers, Wayne

Super dumb question... more my own curiosity at play... why absolutes?

 

[Phil Corso]

Gentlepeople...

Like others, this thread too, has taken off into the Wild Blue Yonder!

Phil Corso

 

[Sahib]

What makes it all the more frustrating to understand is that in an LR circuit the real power is indeed I_real²R, but the "reactive power" is not necessarily I_reactive²X_L. .

In a series RL circuit current is same through resistance and reactance

 

[mbrooke]

Gentlepeople...

Like others, this thread too, has taken off into the Wild Blue Yonder!

Phil Corso

:lol: :thumbsup: Its ok, all for the sake of knowledge :angel:


But in all seriousness its not far off from the original question. The shaded pole motor in question can be overly simplified as a resistor and a reactor in parallel to help understand what we are seeing.

 

[wwhitney]

Yes, watts/VA = PF, but you can't cancel V from the the numerator and denominator to conclude that A_real / A_apparent = PF.

Upon reflection, this is not true for an LR parallel circuit:

Let's use phasors for voltage V and power I, but scalars for average power P and of course for apparent power P_apparent. The magnitude of a phasor V is written as |V|.

So if an inductance L and a resistance R are in parallel, let V be the common voltage drop and call I_real the current through the resistance, I_reactive the current through the inductor, and I the total current. I = I_real + I_reactive (as phasors) and V = I_real R (as phasors)

As no power is dissipated through the inductance, the real power P = |I_real| |V|. The apparent power P_apparent = |I| |V|.

So power factor is given by P / P_apparent = |I_real| |V| / |I| |V| = |I_real| / |I|.

The issue is just that I_reactive = I - I_real as phasors, but the analogous equation doesn't hold for magntitudes.

Cheers, Wayne

 

[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).

So you are right that the real current is 2.75 amps, but the reactive current is sqrt(5.5^2 - 2.75^2) = 4.76 amps.

Cheers, Wayne

 

[wwhitney]

Super dumb question... more my own curiosity at play... why absolutes?

|V| is the magnitude or length of the vector V.

If you can consider real numbers as one dimensional vectors, this definition coincides with the usual absolute value. Likewise for complex numbers.

Cheers, Wayne

 

[wwhitney]

In a series RL circuit current is same through resistance and reactance

Correct, so since Golddigger is splitting the current into two pieces, he must be talking about a parallel LR circuit.

Cheers, Wayne

 

[GoldDigger]

Super dumb question... more my own curiosity at play... why absolutes?

For vector quantities the | | sign indicates the magnitude of the vector, not just the making of a negative quantity (-1 = +1 angle 180 degrees, so that is a special case.)

Note that I am used to using bold text for a vector quantity and the corresponding normal text for its magnitude.

So for any vector A the quantity |A| =- A.
And the triangle inequality states that for any trio of points, a, b and c, ab +bc +ca = 0 and
ab + bc > ac.

 

[Phil Corso]

:lol: :thumbsup: Its ok, all for the sake of knowledge :angel: But in all seriousness its not far off from the original question. The shaded pole motor in question can be overly simplified as a resistor and a reactor in parallel to help understand what we are seeing.

But Shaded-Pole theory is nothing like general induction-motor theory!

o There is no rotating field in the stator.

o The current in the shading-coil delays the change of flux inducing it!

o The change in flux occurs twice per cycle!

o The shifting of flux causes torque, analogous to a rotating stator-field!

o The methodology is transitory in nature... hence, doesn't follow simple R-L-C circuit theory!

Phil

 

[junkhound]

I think we need to get mbrooke past basic phase relationships prior to individual motor operation, aka following post in response to an earlier question.

 

[junkhound]

Ok, so increasing the size (current magnitude) of the resistor can shift the blue to the left, and increasing the size of the reactor will shift the blue to the right? And this shift is condensed to power factor in relation to the voltage wave (cosine)?

Not correct.

Regardless of magnitude, the inductor current and the resistor current do not shift with relationship to the voltage in the parallel circuit used for an example.
The magnitudes do change; hence, you can see by adding points along the curves, as the resistor current increases in relationship to the voltage, the total current (green) curve does shift to be more in phase with the voltage. Which means the the power factor increases (cosine of that phase shift)

 

[mbrooke]

But Shaded-Pole theory is nothing like general induction-motor theory!

o There is no rotating field in the stator.

o The current in the shading-coil delays the change of flux inducing it!

o The change in flux occurs twice per cycle!

o The shifting of flux causes torque, analogous to a rotating stator-field!

o The methodology is transitory in nature... hence, doesn't follow simple R-L-C circuit theory!

Phil

Why twice per cycle? And is that why they are so fast?

 

[kwired]

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.

You came across as not having a clue of what was being talked about, yet with your experience I figured you knew at least a little on the topic. Then you claim that this topic is not agreed on by many, I don't get that at all, you either don't understand it fully or are not serious with the response. Reactive current is not that misunderstood by those that know their electrical theory well. It is misused, misunderstood, or a little of both by those selling the magic boxes that are supposed to save you on your electric bill though.

 

[Johnnybob]

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

So just a little basic trig, if I subtract the true power (watts) sqrd from apparent power (va) sqrd = sqr rt looks like 694.48 VARs inductive. Yes?
Correct?

 

[kwired]

So just a little basic trig, if I subtract the true power (watts) sqrd from apparent power (va) sqrd = sqr rt looks like 694.48 VARs inductive. Yes?
Correct?

Correct formula as you described it, not sure exactly where your values come from or if you calculated correctly.

 

[Phil Corso]

Why twice per cycle? And is that why they are so fast?

What do you mean by "Fast"? An 8-pole version, Ns=900rpm, as well as a reversible-version were also available.

Phil

 

[Johnnybob]

Correct formula as you described it, not sure exactly where your values come from or if you calculated correctly.

I took my numbers from wwhitney's post, which were taken, it looks like, from the original OP. The basic calcs look correct at a glance