THE PHYSICS OF... POWER

[Ingenieur]

It serves no useful purpose to describe electrical phenomena with Newtonian concepts

as was stated kvar do not morph into watts with load
once at rated speed kvar is generally constant with load

 

[Ingenieur]

take a 1:1 xfrmer that is pure inductive kVAR on both coils (no R), the device is 100 efficient. the secondary is open ckt (no load), jX_primary=1ohm, voltage = 100Vac_rms

its 100% kVAR, right? my generator is moving 100amps yet no power is dissipated anywhere, charge is just free-wheeling around the primary ckt at a rate of 100 coulombs/sec

now connect a 1ohm power resistor to the secondary, does the primary amps go up, down, or remain the same ?


if i have a fixed physical motor with R+jX and its free wheeling, i then put load on output shaft, how exactly does that make R go up ??

Primary amps go down 100/1.414 or 70.7 A
what is the point?

I haven't read the whole discourse but R in a motor is fixed (other than temp effect)
it is constant
Xl goes up since Xl = 2 Pi freq L
and L is primarily based on configuration/geometry

in a motor
R is constant more or less
Xl increases with speed and DECREASES with load

 

[GoldDigger]

its 100% kVAR, right? my generator is moving 100amps yet no power is dissipated anywhere, charge is just free-wheeling around the primary ckt at a rate of 100 coulombs/sec

now connect a 1ohm power resistor to the secondary, does the primary amps go up, down, or remain the same ?


if i have a fixed physical motor with Z_vector=R_constant+jX and its free wheeling, i then put load on output shaft, how exactly does that make R go up ??

As I suspected, you have the common confusion about just what an impedance of R + jX represents.
It is NOT a resistance, R, in parallel with a reactive impedance, X. It is a resistance, R, in series with a reactive impedance, X.

If you put a resistor in parallel with a reactive element which in isolation has a pure reactive impedance, X, then as the resistance decreases the value, B, in the impedance A + jB will decrease according to the parallel impedance formula.

The inverse (the conductance) that you work with just gets a little more complicated when complex numbers are involved. The transformation from parallel conductance back to the corresponding impedance involves taking the inverse of a complex number.

 

[FionaZuppa]

As I suspected, you have the common confusion about just what an impedance of R + jX represents.
It is NOT a resistance, R, in parallel with a reactive impedance, X. It is a resistance, R, in series with a reactive impedance, X.

If you put a resistor in parallel with a reactive element which in isolation has a pure reactive impedance, X, then as the resistance decreases the value, B, in the impedance A + jB will decrease according to the parallel impedance formula.

The inverse (the conductance) that you work with just gets a little more complicated when complex numbers are involved. The transformation from parallel conductance back to the corresponding impedance involves taking the inverse of a complex number.

Z=R+jX is the vector sum of pure resistance and inductive reactance, j is √-1 :thumbsup:


here's a riddle, the kVAR is used to make mag field, mag field energy is responsible for output, so free output energy if kVAR is not converted ??

Induction motors require both real and reactive power to operate. The real power (kW) produces work and heat. The reactive power (kVAR) establishes the magnetic field in the motor that enables it to operate.

Primary amps go down 100/1.414 or 70.7 A
what is the point?

I haven't read the whole discourse but R in a motor is fixed (other than temp effect)
it is constant
Xl goes up since Xl = 2 Pi freq L
and L is primarily based on configuration/geometry

in a motor
R is constant more or less
Xl increases with speed and DECREASES with load

loading a secondary of a xfrmer increases the load on the primary, input=output, increase load on sec then pri load has to increase ??

 

[Ingenieur]

Admittance Y and impedance Z are easy to work with in phasor notation

in fact in large power network analysis using linear algebra (matrices) it is easier and preferred

 

[Ingenieur]

Z=R+jX is the vector sum of pure resistance and inductive reactance, j is √-1 :thumbsup:


here's a riddle, the kVAR is used to make mag field, mag field energy is responsible for output, so free energy if kVAR is not converted ??

X can be L or C

the charging or establishing the base field is Q var with some P for windage, ir losses, inertia, friction
once increased emf is required due to shaft load real power is required
so Q var is constant and P increases

 

[FionaZuppa]

X can be L or C

the charging or establishing the base field is Q var with some P for windage, ir losses, inertia, friction
once increased emf is required due to shaft load real power is required
so Q var is constant and P increases

X is the vector sum of L & C with L&C 90° from R, and L is 180° from C, but when i said "inductive" that means their sum ended up more L :thumbsup:

 

[Ingenieur]

X is the vector sum of L & C with L&C 90° from R, and L is 180° from C, but when i said "inductive" that means their sum ended up more L :thumbsup:

you are stating the obvious
but seem to lack understanding of the basics

a riddle
what basic force generates torque in a motor
what imbalance causes this force?

 

[GoldDigger]

Z=R+jX is the vector sum of pure resistance and inductive reactance, j is √-1 :thumbsup:


here's a riddle, the kVAR is used to make mag field, mag field energy is responsible for output, so free energy if kVAR is not converted ??

Absolutely! And that is not what you get when you put two elements with impedances R and jZ in parallel.

if i have a fixed physical motor with Zvector=Rconstant+jX and its free wheeling, i then put load on output shaft, how exactly does that make R go up ??

In particular you asked why putting a load on the output shaft should make R go up.
The motor is NOT properly represented by a Zvector of Rconstant +jX.

The R_d I am talking about is best considered as a parallel resistance, not a series resistance.

The static circuit (no rotation) of a motor can be represented as R_winding +jZ_{winding inductive impedance} Note that the inductance of the motor winding in Henries is roughly constant while Z varies with frequency. And actually resistive losses in the rotor, in the case of an induction motor, form a separate term not covered by R_winding.

Once you have rotation you have a counter EMF to add into the picture, so the two lump model of the ideal motor is no longer valid.
You need a series voltage source whose value (coefficient of the trig term) will depend very strongly on rotation speed and slip.
Looking at the effect of the counter EMF, it should I think reduce both the resistive and the reactive current at the input terminals as the motor moves from LRA to the operating region. Starting from an unloaded motor shaft at equilibrium speed the counter EMF will go down as load is added and slip increases.

The dynamic effect of motor speed and loading on the impedance makes a model based on a transformer with a variable load inappropriate.

 

[GoldDigger]

you are stating the obvious
but seem to lack understanding of the basics

a riddle
what basic force generates torque in a motor
what imbalance causes this force?

:thumbsup::thumbsup::thumbsup::thumbsup:

FZ stating the obvious does not make it an effective argument when it does not mean what he thinks it means. :happysad:

 

[Ingenieur]

It helps to look at Z at 3 conditions
locked rotor
rated speed no load
rated speed full load

assume 460 vac
600 A
50 A
100 A

|Z| Ohm
0.76
9.2
4.6

 

[GoldDigger]

It helps to look at Z at 3 conditions
locked rotor
rated speed no load
rated speed full load

assume 460 vac
600 A
50 A
100 A

|Z| Ohm
0.76
9.2
4.6

But what is the phase angle for each of those? (If you have that information readily available)

 

[Ingenieur]

:thumbsup::thumbsup::thumbsup::thumbsup:

FZ stating the obvious does not make it an effective argument when it does not mean what he thinks it means. :happysad:

he needs to start with a permanent magnet stator/field, commutator and single loop rotor/armature motor type

I do not mean this as an insult
it is how we ALL started

once the concepts of the em field force and charge alignment/repulsion are grasped it gets much easier to build on
AC in lieu of a communtator, etc

 

[Ingenieur]

But what is the phase angle for each of those? (If you have that information readily available)

Did not want to get into that lol, pf etc may muddy the H2O
we know what is going on (or at least a good working knowledge)
others may be better served by concepts

but if we assume R is fairly constant, not unreasoble
then X = sqrt(Z^2 - R^2)
From that a phase relationship can be determined

the point is power P is added to satisfy the load, not kvar tranformed into watts

 

[mike_kilroy]

It serves no useful purpose to describe electrical phenomena with Newtonian concepts

as was stated kvar do not morph into watts with load
once at rated speed kvar is generally constant with load

Go further: kvar is almost a pure constant in a motor running at ANY speed at its designed v/Hz point! At 90 degrees off from the voltage, it certainly will not, and cannot Turn into kwatts.

It is nothing more than equiv to a permanent magnet, made from pure L, and is constant value.

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[Ingenieur]

Go further: kvar is almost a pure constant in a motor running at ANY speed at its designed v/Hz point! At 90 degrees off from the voltage, it certainly will not, and cannot Turn into kwatts.

It is nothing more than equiv to a permanent magnet, made from pure L, and is constant value.

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For the love of all that is Good and Holy PLEASE DO NOT throw variable speed into the mix!!!!
I implore you!!!

:lol:

 

[FionaZuppa]

part of kVAr is used to construct the mag field and it happens at frequency, the mag field is not free work. if we need to go there we can.

It helps to look at Z at 3 conditions
locked rotor
rated speed no load
rated speed full load

assume 460 vac
600 A
50 A
100 A

|Z| Ohm
0.76
9.2
4.6

last few posts suggests that kVAr remains constant?
ok, the R of the copper windings also remains constant.
so how is your Z changing? is there some other components not included in Z=R+jX ?

 

[Ingenieur]

part of kVAr is used to construct the mag field and it happens at frequency, the mag field is not free work. if we need to go there we can.

pretty much ALL of it goes to magnetization
less leakage and such

 

[mike_kilroy]

part of kVAr is used to construct the mag field and it happens at frequency, the mag field is not free work. if we need to go there we can.



last few posts suggests that kVAr remains constant?
ok, the R of the copper windings also remains constant.
so how is your Z changing? is there some other components not included in Z=R+jX ?

Perhaps that is where you should go next then? Not sure what you mean by "mag fld is not free work," cuz 1, there is never any work done by it, and 2, yes it is free,aka imaginary, and 3, CONSTANT.

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[mike_kilroy]

Suggest googling "equivalent circuit of a induction motor" to see the missing term that makes torque by increasing the IN PHASE current. Much more to it than simple STATOR R in your Z.

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