Power factor / efficency of shaded pole motors

[mbrooke]

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

But isn't PF; watts/VA=PF?

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

So if I had a 2.75amp light bulb in parallel with a 2.75amp reactor, I would have...? (not saying you are wrong here however)

 

[junkhound]

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?

What Sahib said, you cannot add arithmetically, have to use vectors.

I'm too buzy (or lazy) to do the math and draw it out at the moment, but if you take the test data and draw the vectors for the 5.32 A at 0.43 PF and then add 25W/116.8 V to the real vector, you will likely see the 5.33 A appear and the phase shift to a 0.47 PF.

The math is left as an exercise for the student <G>

 

[mbrooke]

What Sahib said, you cannot add arithmetically, have to use vectors.

Not saying you are wrong, but why does the NEC allow arithmetic summing?

If we know the shaft power, and the motor efficiency, why can't we derive the reactive current draw of said motor?

I'm too buzy (or lazy) to do the math and draw it out at the moment, but if you take the test data and draw the vectors for the 5.32 A at 0.43 PF and then add 25W/116.8 V to the real vector, you will likely see the 5.33 A appear and the phase shift to a 0.47 PF.

The math is left as an exercise for the student <G>

When the math is ready baking, Id like to give it a whirl!

 

[wwhitney]

But isn't PF; watts/VA=PF?

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

So if I had a 2.75amp light bulb in parallel with a 2.75amp reactor, I would have...? (not saying you are wrong here however)

The magnitude of the current would be sqrt(2.75^2 + 2.75^2) = 2.75*sqrt(2) = 3.89 amps. That's what Sahib's phrase "sum up in quadrature" means.

Cheers, Wayne

 

[mbrooke]

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

Oh, ok, that makes sense now! Need some help applying it but its clearing up.

The magnitude of the current would be sqrt(2.75^2 + 2.75^2) = 2.75*sqrt(2) = 3.89 amps. That's what Sahib's phrase "sum up in quadrature" means.

Cheers, Wayne

So, lets say I had 5.5 amps at 120 volts. How would I work backwards to find out the motors restive and reactive quantities?

 

[wwhitney]

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)

Thanks for clarifying that, your estimate is based on the approximation that (electrical input power - mechanical output power) will be constant for small values of mechanical output power.

Since I'm not very familiar with motors, a basic question: is the electrical power drawn when the motor is unloaded going to be a lower bound for (electrical input power - mechanical output power) over the range of the motor's operation?

Cheers, Wayne

 

[Sahib]

Not saying you are wrong, but why does the NEC allow arithmetic summing?

Perhaps the code assumes same power factor loads so that their VA's can be added arimetically?

 

[wwhitney]

Not saying you are wrong, but why does the NEC allow arithmetic summing?

Arithmetic summing is conservative by the triangle inequality, the magnitude you get will be at least as big as the vectorial sum. So if you size your wires using an arithmetic sum, they won't be undersized.

Cheers, Wayne

 

[GoldDigger]

Oh, ok, that makes sense now! Need some help applying it but its clearing up.




So, lets say I had 5.5 amps at 120 volts. How would I work backwards to find out the motors restive and reactive quantities?

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.
The "reactive power" is a fictitious power intended to make another slightly less fictitious power, "apparent power" work out right when added in quadrature to the real power.

If you have two resistors in series, the real power dissipated by each is in fact I²R₁ + I²R₂
No quadrature there. You can't, however, draw the real and reactive power numbers as being at right angle in the complex plane.
That only works for currents.

 

[wwhitney]

So, lets say I had 5.5 amps at 120 volts. How would I work backwards to find out the motors restive and reactive quantities?

First, let me say that as my background is not in physics, I can't say whether the terms "reactive current" and "real current" are standard or not. But certainly you can decompose the current sinewave into the sum of two sinewaves, one in phase with voltage and one 90 degrees out of phase with the voltage. That is tantamount to modeling the load impedance as a pure resistance and a pure inductance in parallel. [The voltage over each component is the same, and the total current is the sum of the currents in the two components.]

It may be more typical to decompose the voltage waveform into two components, one in phase with the current and one 90 degrees out of phase with the current. That would be tantamount to modeling the load impedance as pure resistance and a pure inductance in series. [The current in each component is the same, and the total voltage drop is the sum of the voltage drops in the two components.]

Let me continue this answer shortly.

Cheers, Wayne

 

[Sahib]

Arithmetic summing is conservative by the triangle inequality, the magnitude you get will be at least as big as the vectorial sum. So if you size your wires using an arithmetic sum, they won't be undersized.

Cheers, Wayne

Splendid explanation.:thumbsup:

 

[mbrooke]

First, let me say that as my background is not in physics, I can't say whether the terms "reactive current" and "real current" are standard or not.

Well, technically they are incorrect (in my view) because there is only one current wave. However, I sometimes breaker the two apart to help me understand better. Learning power calc via "equivalent" circuits has polluted my real world perception in that respect.

But certainly you can decompose the current sinewave into the sum of two sinewaves, one in phase with voltage and one 90 degrees out of phase with the voltage. That is tantamount to modeling the load impedance as a pure resistance and a pure inductance in parallel. [The voltage over each component is the same, and the total current is the sum of the currents in the two components.]

I like to break it into an ideal equivalent reactor and ideal equivalent resistor.

It may be more typical to decompose the voltage waveform into two components, one in phase with the current and one 90 degrees out of phase with the current. That would be tantamount to modeling the load impedance as pure resistance and a pure inductance in series. [The current in each component is the same, and the total voltage drop is the sum of the voltage drops in the two components.]

Let me continue this answer shortly.

Cheers, Wayne

Sure no problem

 

[mbrooke]

Arithmetic summing is conservative by the triangle inequality, the magnitude you get will be at least as big as the vectorial sum. So if you size your wires using an arithmetic sum, they won't be undersized.

Cheers, Wayne

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.

 

[mbrooke]

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.
The "reactive power" is a fictitious power intended to make another slightly less fictitious power, "apparent power" work out right when added in quadrature to the real power.

Whoa! boil this down for me :dunce:

If you have two resistors in series, the real power dissipated by each is in fact I²R₁ + I²R₂
No quadrature there. You can't, however, draw the real and reactive power numbers as being at right angle in the complex plane.
That only works for currents.

What would a complex plane look like for a single phase motor?

 

[junkhound]

Here is a quick visual with close to your 2.75 Arms values at 120 Vac rms 60 Hz, you can add currents, but only the instantaneous values, not rms values.
Add the instantaneous currents, inductor current (red) with the instantaneous resistor current (yellow) to get the instantaneous total current ( blue) --edit, colors shift when posted, I am color blind but think what I call yellow shifted to what others call green?
However, Note that the rms values (straight lines) add by the vector sum (sq rt of sum of squares), not straight addition. Also note the phase shifts.

Hope that helps you visualize the relationships.

 

[GoldDigger]

What would a complex plane look like for a single phase motor?

Just like the complex plane for each phase individually for a three phase motor.

 

[mbrooke]

.....

 

[mbrooke]

Just like the complex plane for each phase individually for a three phase motor.

Ahhh, :lol: Silly me :slaphead: But, I am still lack luster even with 3 phase :angel:

 

[junkhound]

This is just a reactor and resistor in series and not a motor being drawn as equivalent?

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.

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.

 

[mbrooke]

Here is a quick visual with close to your 2.75 Arms values at 120 Vac rms 60 Hz, you can add currents, but only the instantaneous values, not rms values.
Add the instantaneous currents, inductor current (red) with the instantaneous resistor current (yellow) to get the instantaneous total current ( blue) --edit, colors shift when posted, I am color blind but think what I call yellow shifted to what others call green?
However, Note that the rms values (straight lines) add by the vector sum (sq rt of sum of squares), not straight addition. Also note the phase shifts.

Hope that helps you visualize the relationships.

View attachment 15841

View attachment 15842

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)?