THE PHYSICS OF... POWER

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mbrooke said:
Can someone explain that to me, why PF changes as motors are loaded up/down?
Click to expand...

pf = P/S
S = sqrt(P^2 + Q^2)
Q is constant (more or less)

assume
50% load P = 50
100% load P = 100
Q is 40 either point

at 50% load
pf = 50/64 = 0.78

at 100% load
pf = 100/108 = 0.93

as the motor loads up more P is required as it increases pf approaches 1 but never gets there due to Q
 
FionaZuppa said:
ok, show me.
i have a no-R motor, purely inductive motor. if i had electrical R in my motor that would just waste energy, this one is a real motor with copper that is super conducting. input amps is governed by 2πfL (only the jX portion of Z).

the input is only reactive, explain how power is transformed from electrical input amps to mechanical work output @ 100% efficiency. also show what VAr looks like for poco. me wait while you drum up the answer.
Click to expand...

Correct me if I'm wrong
The variable R he references represents the variable load in the equivilent ckt model
R is used since it is active power = shaft doing work based on rpm
the power disappated across it is the motor HP output
 
Ingenieur said:
pf = P/S
S = sqrt(P^2 + Q^2)
Q is constant (more or less)

assume
50% load P = 50
100% load P = 100
Q is 40 either point

at 50% load
pf = 50/64 = 0.78

at 100% load
pf = 100/108 = 0.93

as the motor loads up more P is required as it increases pf approaches 1 but never gets there due to Q
Click to expand...


Essentially, the inductance that causes the reactive effects is a value that depends on the geometry of the windings and the iron core's magnetic properties. It is the same value whether the motor is at a low power output or a high power output.


Therefore, the reactive power associated with the inductance, is constant. The real power, associated with winding resistive losses, mechanical losses, and mechanical output, does depend on the power output. So the power factor varies due to the real power changing without the reactive power changing along with it.
 
GoldDigger said:
I am not going to go through the full modelling of what you describe.
I will just reiterate that the current through the motor will be affected by the back EMF produced by the combination of the rotation and the induced magnetism in the rotor.
And that when you load the motor down, even though if it could be statically represented as a pure inductor there will be a resistive component to the input current when it is loaded. That resistive input current is in addition to the magnetizing current, not the result of the magnetizing current somehow producing work.
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??
changing R, not from mag field. interesting. how about this one.
i have a 0.1Hz AC voltage on small coil, it produces a mag field that follows current and separated from voltage by Φ. the coil is fixed on a horizonal table. just in front of the coil is a perm magnet which sits upon a frictionless oil surface. when the AC current is on the magnet moves back and forth and the 1way trip is 2.5cm. there's work being done, but from where? where is the energy coming from that creates the required force to move an object of mass? energy is being transformed from mag field into work. if i add a little mass to the magnet then what? well, the magnet still moves back and forth yet the distance has decreased slightly, hmmm, same amount of work though. the reactive energy within the inductor is in the mag field, some of that energy is coupled out to real work. but wait a sec, the input side of the coil has Z=R+jX, but the VAr (energy, or imaginary power) associated with jX is partially (or fully) transferred to the magnet movement.
 
Carultch said:
Essentially, the inductance that causes the reactive effects is a value that depends on the geometry of the windings and the iron core's magnetic properties. It is the same value whether the motor is at a low power output or a high power output.


Therefore, the reactive power associated with the inductance, is constant. The real power, associated with winding resistive losses, mechanical losses, and mechanical output, does depend on the power output. So the power factor varies due to the real power changing without the reactive power changing along with it.
Click to expand...
Yes.
And better expressed than my #377 post.
 
wwhitney said:
I'm pretty sure your computation is wrong, and that the line experiences 10,000 W of heating. Your vector math is wrong.

Look at it this way: while the circuit current is out of phase with the circuit voltage, the voltage drop that occurs along the pure resistance of the wire is caused by the current and hence is in phase with the current.

That is, my understanding is that for resistance losses along a conductor, P = I^2 * R is a scalar equation. I is just the RMS current, and R is the resistance, not the complex impedance.

Cheers, Wayne
Click to expand...

I believe you are correct. He did not say what the load was, other than PF=.90 or angle was 25 degrees... So assuming it is a motor for this example,

YES, clamp on ammeter shows 100 amps thru wires to load, so I^2R loss in 1 ohm cable is 10kw.

Now if we want to know how much of this 10kw heat loss is contributed by each of the two components...

Only 90.63amp is in phase current making torque in the motor. So THAT portion produces 8.2kw of the cable heat, and the remaining 1.8kw is from the reactive 42.26amp current 90 degree out of phase with the motor. Imagine that! Imaginary current making real heat-em-up watts of heat.

Now the magnetic field produced in the motor from this 42.26amps is slightly LESS than expected, since the voltage delivered to the motor is now 100a*1ohm=100v LESS than supplied... But its imaginary stuff now averages to 0, just at a slightly smaller total imaginary level than if we had 0 ohm cables.

It all still works out fine.

Why we have to take the total vector sum of 100 amps into account sizing the wire.
 
FionaZuppa said:
??
changing R, not from mag field. interesting. how about this one.
i have a 0.1Hz AC voltage on small coil, it produces a mag field that follows current and separated from voltage by Φ. the coil is fixed on a horizonal table. just in front of the coil is a perm magnet which sits upon a frictionless oil surface. when the AC current is on the magnet moves back and forth and the 1way trip is 2.5cm. there's work being done, but from where? where is the energy coming from that creates the required force to move an object of mass? energy is being transformed from mag field into work. if i add a little mass to the magnet then what? well, the magnet still moves back and forth yet the distance has decreased slightly, hmmm, same amount of work though. the reactive energy within the inductor is in the mag field, some of that energy is coupled out to real work. but wait a sec, the input side of the coil has Z=R+jX, but the VAr (energy, or imaginary power) associated with jX is partially (or fully) transferred to the magnet movement.
Click to expand...
What you are missing, plain and simple, is that when the coil is causing a magnet with non-zero mass to oscillate back and forth the magnetic field coupling between the coil and the permanent magnet results in the coil no longer acting as a pure inductor.
It can be a pure inductor ONLY if the magnetic field within the coil is solely that produced by the coil itself. When you add a moving magnet the magnetic field within the coil is changed by the motion of the magnet.
(Well OK, actually what I should have said is that the non-constant magnetic field in the coil is only that produced by the coil itself. Nailing down a magnet inside the coil will not change its behavior as a perfect inductor unless your magnet is not a perfect magnet and exhibits hysteresis losses.)
 
FionaZuppa said:
??
just in front of the coil is a perm magnet which sits upon a frictionless oil surface. when the AC current is on the magnet moves back and forth and the 1way trip is 2.5cm. there's work being done
Click to expand...
What work is being done? The surface is frictionless, so moving an object on it does not require work.

Cheers, Wayne
 
wwhitney said:
What work is being done? The surface is frictionless, so moving an object on it does not require work.

Cheers, Wayne
Click to expand...

OOPS. There is work being done to accelerate the mass, even on a frictionless surface. But the magnet will also do work back on the coil during its deceleration part of the cycle and so the time average of the work is zero.

In many ways like the situation with an ideal inductor.

For FZ: The current in the coil is simultaneously converting input energy into energy stored in the magnetic field and energy in the form of kinetic energy in the magnet. Both of those will be returned later in the cycle. The acceleration of the mass will be greatest when the current is at a maximum but the work done on the mass will depend also on its velocity and hence not be in phase with the acceleration.
So both energies will come back, but the phasing of the stored inductive energy and the stored kinetic energy will be different. The stored kinetic energy will be at its maximum when the current is zero making it look a lot like an alternating load and generator on alternate quarter cycles.
 
Ingenieur said:
pf = P/S
S = sqrt(P^2 + Q^2)
Q is constant (more or less)

assume
50% load P = 50
100% load P = 100
Q is 40 either point

at 50% load
pf = 50/64 = 0.78

at 100% load
pf = 100/108 = 0.93

as the motor loads up more P is required as it increases pf approaches 1 but never gets there due to Q
Click to expand...


ah, so we have fixed Q, and for the most part fixed electrical R
so when you load down the motor how does real power increase and lower Φ ? if Q is fixed and your real power increases this implies lower R (more amps but less real watts), which makes Φ go up?
 
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FionaZuppa said:
ah, so we have fixed Q, and for the most part fixed electrical R
so when you load down the motor how does real power increase and lower Φ ? if Q is fixed and your real power increases this implies lower R (more real amps), which makes Φ go up?
Click to expand...
Didn't you understand my spreadsheet?
 
FionaZuppa said:
ah, so we have fixed Q, and for the most part fixed electrical R
so when you load down the motor how does real power increase and lower Φ ? if Q is fixed and your real power increases this implies lower R (more real amps), which makes Φ go up?
Click to expand...

If you assume a resistive load represents Φ = 0, then more resistive load will make R decrease and that will in turn make Φ decrease. It will also make Zvector more resistive, but by decreasing the i term more than it decreases the real term.

You are once again confusing the effects of parallel impedances and series impedances. I know it is hard to overcome that problem, but please try harder!
 
GoldDigger said:
If you assume a resistive load represents Φ = 0, then more resistive load will make R decrease and that will in turn make Φ decrease. It will also make Zvector more resistive, but by decreasing the i term more than it decreases the real term.

You are once again confusing the effects of parallel impedances and series impedances. I know it is hard to overcome that problem, but please try harder!
Click to expand...

"then more resistive load will make R decrease"
how's that? only two ways to make a load look more resistive. add R, or decrease jX
if you add R in parallel the R component goes down, more amps, more real power, but the ratio of R to jX makes it look more reactive, not less.

R is fixed in a motor, so that doesnt really change as to load down the motor. but you mention decreasing jX (is that what you mean by "i")? decreasing jX means less total Z means more amps, this more real power, but, decreasing jX means Q decreases. is there another term in Z not accounted for here?

i'm still listening.
 
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mike_kilroy said:
I believe you are correct. He did not say what the load was, other than PF=.90 or angle was 25 degrees... So assuming it is a motor for this example,

YES, clamp on ammeter shows 100 amps thru wires to load, so I^2R loss in 1 ohm cable is 10kw.

Now if we want to know how much of this 10kw heat loss is contributed by each of the two components...

Only 90.63amp is in phase current making torque in the motor. So THAT portion produces 8.2kw of the cable heat, and the remaining 1.8kw is from the reactive 42.26amp current 90 degree out of phase with the motor. Imagine that! Imaginary current making real heat-em-up watts of heat.

Now the magnetic field produced in the motor from this 42.26amps is slightly LESS than expected, since the voltage delivered to the motor is now 100a*1ohm=100v LESS than supplied... But its imaginary stuff now averages to 0, just at a slightly smaller total imaginary level than if we had 0 ohm cables.

It all still works out fine.

Why we have to take the total vector sum of 100 amps into account sizing the wire.
Click to expand...

ref post 329

http://forums.mikeholt.com/showthread.php?t=178681&page=33&p=1765904#post1765904
 
FionaZuppa said:
"then more resistive load will make R decrease"
how's that? only two ways to make a load look more resistive. add R, or decrease jX

R is fixed in a motor, so that doesnt really change as to load down the motor. but you mention decreasing jX (is that what you mean by "i")? decreasing jX means less total Z means more amps, this more real power, but, decreasing jX means Q decreases. is there another term in Z not accounted for here?

i'm still listening.
Click to expand...

The R that is changing is in fact a parallel R component of the input Z.
X can decrease without Q of the inductor changing at all.

Example:

1. One ohm in parallel with an inductor that has an X value of 1 and a series R value of zero. The resultant Z is NOT R + jX.
It is R || jX = 1 || j, which is 1/(1/1 + 1/j)
That will be 1/(1 -j), which is (1+j)/(1+j)(1-jX) = (1+j)/2 = 1/2 + j/2
2. Infinite ohms in parallel with that same inductor.
Z = 1/(1/infinity + 1/j) = 1/(0 + 1/j) = j.

By decreasing R (case 1) and keeping X constant we have increased the ratio of the resistive term of Z to the inductive term of Z by comparison to case 2. And the j term got smaller just by decreasing R.

The R in R+ jX is not fixed for a motor. The series resistance that is part of the inductor is fixed, but the dynamic resistance is NOT fixed.
You can have an R component in Z even for an ideal motor as long as it is loaded.
 
GoldDigger said:
The R that is changing is in fact a parallel R component of the input Z.
X can decrease without Q of the inductor changing at all.

Example:

1. One ohm in parallel with an inductor that has an X value of 1 and a series R value of zero. The resultant Z is NOT R + jX.
It is R || jX = 1 || j, which is 1/(1/1 + 1/j)
That will be 1/(1 -j), which is (1+j)/(1+j)(1-jX) = (1+j)/2 = 1/2 + j/2
2. Infinite ohms in parallel with that same inductor.
Z = 1/(1/infinity + 1/j) = 1/(0 + 1/j) = j.

By decreasing R (case 1) and keeping X constant we have increased the ratio of the resistive term of Z to the inductive term of Z by comparison to case 2. And the j term got smaller just by decreasing R.

The R in R+ jX is not fixed for a motor. The series resistance that is part of the inductor is fixed, but the dynamic resistance is NOT fixed.
You can have an R component in Z even for an ideal motor as long as it is loaded.
Click to expand...

ok, i follow you. adding parallel R' makes more resistive amps. so now show me where this R' term suddenly comes from.

and for clarity, you dont really decrease R, you added R in parallel which makes Z decrease while the jX component remains constant.
 
FionaZuppa said:
ok, i follow you. adding parallel R' makes more resistive amps. so now show me where this R' term suddenly comes from.

and for clarity, you dont really decrease R, you added R in parallel which makes Z decrease while the jX component remains constant.
Click to expand...

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).

Furthermore, I did decrease R because the R value in the expression for Zmotor is not the same R that appears in the characterization of the stator winding in isolation.

I did not decrease Rinductor because that is not what happens when the motor is loaded.

We need to keep track of more subscripts if we are going to use the letter R for everything resistive.
There are lots of subscripts in each of the submitted models, without any explanation of what physical quantity they are associated with.
 
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