THE PHYSICS OF... POWER

[Ingenieur]

Good question. I have not defined it. It could also be defined in terms of (synchronous speed -rotation speed)/sync speed I suppose.
I see it used in one of models, but without a definition attached.

yepper
same thing

 

[FionaZuppa]

Wrong again. When you add more resistance in parallel the j term in the overall Z actually decreases. Look at my example again (j compared to j/2).

i add R' in parallel to an existing Z. what is I and X_L in right gram?
so are you saying that the R' that shows up related to 1/s reduces jX in some way?

 

[GoldDigger]

yepper
same thing

It seems to me that the definition you posted would be affected by motor efficiency unless you are very careful how you define Welec.

 

[Ingenieur]

It seems to me that the definition you posted would be affected by motor efficiency unless you are very careful how you define Welec.

Wmech = 2 Pi freq
Wmech = 2 Pi n, n is mech rev/sec of rotor
elec is field rotational speed 120 x freq/poles in rev/min
mech is actual rotor rotational speed in rev/min

 

[FionaZuppa]

W = 2 Pi f
elec is field rotational speed 120 x freq/poles
mech is actual rotor rotational speed

In physics, angular frequency. ω. (also called the angular speed, radial frequency, and radian frequency) is a measure of rotation rate. A high rate of angular frequency means something is turning very fast. The angular frequency is the magnitude of the vector quantity angular velocity which is also known as the angular frequency vector

.

 

[GoldDigger]

Wmech = 2 Pi freq
Wmech = 2 Pi n, n is mech rev/sec of rotor
elec is field rotational speed 120 x freq/poles in rev/min
mech is actual rotor rotational speed in rev/min

Sorry. I forgot that there is a problem with Greek letters on the forum and W has nothing to do with Work.

 

[Ingenieur]

Sorry. I forgot that there is a problem with Greek letters on the forum and W has nothing to do with Work.

No problem, understandable
you should see my hand writing lol

 

[wwhitney]

i add R' in parallel to an existing Z. what is I and X_L in right gram?

Do you need help with the math?

I'm going to use Y for the inverse of impedance Z, and I'm going to change the X_L in the left diagram to 60 ohms for computational convenience. Then the question is, what is the impedance of Za = 60 + 60j ohms in parallel with Zb = 85 ohms?

1/(1 + j) = (1 - j) /2, so Ya = 1/120 - 1/120 j
Yb = 1/85
Ya + Yb = (1/85 + 1/120) - (1/120 j)

Using a calculator, 1/(Ya + Yb) = 42.5 + 17.6 j.

So yes, putting a resistance in parallel with the given impedance results in an impedance with a smaller imaginary component. Plus it reduces the angle of the complex impedance.

Cheers, Wayne

 

[FionaZuppa]

i add R' in parallel to an existing Z. what is I and X_L in right gram?
so are you saying that the R' that shows up related to 1/s reduces jX in some way?

i just not that good with this stuff. can you help me solve this (above)

this R' that you speak of, looks like it shows up in induction motor models, but it seems to be a series R', not a shunt, thus this R' seems like it decreases as motor load goes up? a reduction in this R' increases real power via I²R' for fixed voltage

or

 

[GoldDigger]

It is in series with X2 but in parallel with the whole left side of the model.

The network on the left side of the transformer(including the inductance of the transformer primary) is the only impedance seen when slip is zero, and therefore can be roughly considered to represent the no-load impedance of the motor, which is very crudely modeled as a fixed inductance in series with a fixed resistance (mostly the winding resistance).

What the network is approximated by in a given motor depends on the relative magnitude of the various R and X terms.

 

[FionaZuppa]

It is in series with X2 but in parallel with the whole left side of the model.

The network on the left side of the transformer(including the inductance of the transformer primary) is the only impedance seen when slip is zero, and therefore can be roughly considered to represent the no-load impedance of the motor, which is very crudely modeled as a fixed inductance in series with a fixed resistance (mostly the winding resistance).

What the network is approximated by in a given motor depends on the relative magnitude of the various R and X terms.

but that term shows s in bottom. when motor s goes down the torque goes up, so why is it R₂'/s ??

 

[GoldDigger]

but that term shows s in bottom. when motor s goes down the torque goes up, so why is it R₂'/s ??

As slip goes down the work done by the motor decreases. This is consistent with the 1/s resistive term approaching infinity.
How that corresponds to the available torque is slightly more complex, but basically torque also decreases to zero as the speed approaches synchronous speed (for an induction motor).
If you do not put a load on the motor, it cannot be supplying any torque, right?

How the torque varies in the vicinity of locked rotor (with s approximately 1) is an entirely different question.

 

[FionaZuppa]

interesting how reactance changes.

In the condition of induction motor slip, the voltage remains the maximum and slip is 100% as the motor starts rotating but both slip as well as voltage reduce as soon as the rotor starts to turn. Frequency is directly proportional to the slip, i.e. frequency decreases with decrease in slip. Inductive reactance of an induction motor depends on both frequency as well as slip. When the rotor is stationary, the frequency, slip, and inductive reactance are at the maximum level. When the rotor turns, the inductive reactance remains low and power factor reaches to 1. The inductive reactance changes with slip because the rotor is the summation of constant resistance and changeable inductive reactance.

 

[GoldDigger]

interesting how reactance changes.

What is that quote from?

It seems to me that the "frequency" being referred to is the rotational speed of the magnetic field relative to the rotor material, not a frequency in the same sense as the applied line frequency.

In that sense, at zero slip the rotor sees a static magnetic field in its own rotating reference frame and therefore there are no magnetic or circulating current losses in the rotor.

 

[mike_kilroy]

As slip goes down the work done by the motor decreases. This is consistent with the 1/s resistive term approaching infinity.
How that corresponds to the available torque is slightly more complex, but basically torque also decreases to zero as the speed approaches synchronous speed (for an induction motor).
If you do not put a load on the motor, it cannot be supplying any torque, right?

How the torque varies in the vicinity of locked rotor (with s approximately 1) is an entirely different question.

Variations in coupled reactances should be left out of this discussion. Not important. That leakage reactance changes are small percentages of totals and do not effect these discussions.

How it all changes while across the line starting also should be ignored - that has no bearing to these discussions.

All that should effect these discussions is 1) the magnetizing imaginary current from Xm/Rc is constant and 90 electrical degrees from voltage, and 2) r'/s current (in phase doing work making torque) is effectively LINEAR with load from 0 to motor rated torque. Since R' is simply the effective rotor resistance reflected to the motor input wires, then s slip just varies linearly with load from 0 to motor torque rating.

Simple as that for a good understanding of how induction motor works.

 

[FionaZuppa]

What is that quote from?

It seems to me that the "frequency" being referred to is the rotational speed of the magnetic field relative to the rotor material, not a frequency in the same sense as the applied line frequency.

In that sense, at zero slip the rotor sees a static magnetic field in its own rotating reference frame and therefore there are no magnetic or circulating current losses in the rotor.

all textbooks about induction motors, thats where. the input reactance will decrease when motor is loaded, thus why a motor with no load can have very poor PF, but when loaded the PF can approach 1. see Steinmetz model.

 

[mike_kilroy]

all textbooks about induction motors, thats where. the input reactance will decrease when motor is loaded, thus why a motor with no load can have very poor PF, but when loaded the PF can approach 1. see Steinmetz model.

NO!! Change your word reactance to resistance and it is correct!

that is the whole point! You keep wanting to morph the imaginary, 90 degree, Xm reactance into resistance to change it into in phase current increase with load. NO.

we keep telling u Xm, REACTANCE does NOT change with load.

R'/s is pure simple RESISTANCE! IT CHANGES WITH LOAD: NOTHING ELSE CHANGES! (for these discussions we ignore the very small percentage changes in the other terms.)

Sent from my SM-G900V using Tapatalk

 

[GoldDigger]

NO!! Change your word reactance to resistance and it is correct!

that is the whole point! You keep wanting to morph the imaginary, 90 degree, Xm reactance into resistance to change it into in phase current increase with load. NO.

we keep telling u Xm, REACTANCE does NOT change with load.

OK, Mike, let's also make a small tweak to what you are saying.
The inductances (and therefore component reactances) in the motor model do not change, and the amount of reactive current does not change significantly. That does not, however, mean that the reactive term of the overall input impedance, Reff + jZeff, does not change.
See the discussion of the effect of parallel resistance on the coefficient of j in a simplified situation.

Part of the apparent conflict between some of FZ's statements and those of Mike and others, including me, is that there is no agreement of what the simple word reactance refers to. It can either refer to the lumped component reactances in the model, which do not change, or the reactive term of the overall input impedance, Z, which does change.

Also, FWIW, getting back to FZ, I do not see that changing the "frequency" and therefore the relation between the inductances of the components and their inductive reactances, is a valid way to analyze the model or the real motor. Something is missing in the context of FZ's quote.

 

[Besoeker]

interesting how reactance changes.

Or maybe not. Look at the spreadsheet I offered.
Real machine, real application.

 

[GoldDigger]

Or maybe not. Look at the spreadsheet I offered.
Real machine, real application.

By spreadsheet, do you mean the almost unreadable mass of numbers you posted in #380? Or did you link to an actual spreadsheet at some point that I missed?

I have some problems with some of the values I think I see in #380 in any case. Specifically, how do you get such a high torque and such a high mechanical load figure with almost zero slip? That seems to contradict my understanding of the effects of slip.
I also do not see a term in that mass that directly represents the coefficient of j in Zinput.