electromagnetic torque

[Ryc92]

Hi,

Can someone good at maths explain to me why electromagnetic torque in a machine is proportional to:

MMFstator x MMFrotor x Sin delta

I don't understand why it is the product of the field and stator mmf? I have looked around on-line but all I see is this equation stated but not derived.

Many thanks,

Ryan

 

[electricpete]

It can be done from energy principles.
Specifically
T = d/d_delta (W) where W is total energy, primarily in the airgap

Let theta be angular coordinate and delta be angle between rotor and stator fields.
B is flux density
w is energy density in airgap
W is total energy in the airgap

Then
MMFstator ~ S*cos(theta)
MMFrotor ~ R*cos(theta + delta)
MMFtotal = MMFstator + MFFrotor ~ S*cos(theta) + R*cos(theta+delta)
B ~ MMFtotal ~ S*cos(theta) + R*cos(theta+delta)
w ~ B^2 = [S*cos(theta) + R*cos(theta+delta)]^2
w ~ S^2*cos^2(theta) + R*cos(theta+delta)^2 + 2*S*R*cos(theta)*cos(theta+delta)
w ~ w1 + w2 + w3 where w1, w2, w3 are the three terms above
W = integral(w(theta), theta = 0..2*pi)
W ~ W1 + W1 + W3 where W1, W2, W3 correspond to w1,w2, w3 above
T ~ dW / d_delta
T ~T1 + T2 + T3 where T1, T2, T3 correspond to W1,W2, W3 above.

T1 ~ d/d_delta ( Int {S^2*cos^2(theta), theta=0..2*pi})
T1 ~ 0 because the item being differentiated does not depend on delta

T2 ~ d/d_delta ( Int {R^2*cos^2(theta+delta), theta=0..2*pi}))
T2 ~ 0 for the same reason. Note that delta interoduce a shift in the sinusoidal function being integrated, but since it is still integrated over a full period 0..2*P, the result does not depend on that shift.


T3 ~ d/d_delta ( Int { 2*S*R*cos(theta)*cos(theta+delta), theta=0..2*pi})
Now you need to use some trig identities to simplify
Cos(A)Cos(B) = 0.5*[cos(A+B) + cos(A-B)]
cos(theta)*cos(theta+delta) = 0.5* [cos(2*theta + delta) + cos(delta)]
T3 ~ d/d_delta ( Int { 2*S*R*0.5* [cos(2*theta + delta) + cos(delta)], theta=0..2*pi})
Remove the constants (we’re talking proportionalities)
T3 ~ d/d_delta ( Int {[cos(2*theta + delta) + cos(delta)], theta=0..2*pi})
The first term again integrates to zero. The second term integrates to cos(delta)*2*pi
When we differentiate the 2nd term we end up with sin(delta)

In additional to throwing away the constants (working with ~ instead of =), I haven't been careful to keep track of the signs (+/-). That takes a little more thought to do correctly, but we already know the answer (rotor whose field is lagging stator field is pulled in direction of the stator field... stator sees equal/opposite torque)

=====================================
(2B)+(2B)' ?

 

[electricpete]

Note also of course my S*R would have made it into the proportionality which matches your formulation.

=====================================
(2B)+(2B)' ?

 

[7anoter4]

A modest demonstration may be the following one:
Let's say Capital Letter will be for rms value and small letter for instantaneous value.
f=b*i*lengthrotor [Laplace Law] f=force; b=instant. stator magn.flux density.; i=instant.rotor current
As b is a vector and i*length is another vector the vectorial product module will be:
|f|=|b|*|i*length|*sin(B,i*length)
If h in the gap it is perpendicular per i*length then sin(B,i*length)=1
b=Bm*sin(w*t) ; i=sqrt(2)*I*sin(w*t+fib) ;;fib=the rotor current lag angle referred to b ,or to h, or to MMFstator.
b=miuo*h ; Bm=miuo*H ;
t= Bm*sin(w*t)*sqrt(2)*I*sin(w*t+fib)*length
sin(wt)*sin(wt+fib)=[cos(-fib)-cos(2*w*t+fib)]/2
t=Bm*sqrt(2)*I*[cos(fib)-cos(2*w*t+fib)]/2
T=Bm*sqrt(2)*I*integral{[cos(fib)-cos(2*w*t+fib)]*dt}|from t=0 to t=1/2/frq (w*t=pi())
integral{[cos(fib)-cos(2*w*t+fib)]*dt}|t=0 - t=1/2/frq|=1/2*sin(fib)
T=Bm*sqrt(2)/2*I*SIN(fib)
H=MMFstator/(2*KC*d) neglecting the MMFstator drop through the laminates [Biot-Savart Law]
kc=Carter factor; d=gap distance
MMFrotor=rotor no.of turns*kwrot*Irot
kwrot=rotor winding factor
Irot=MMFrotor/rotor no.of turns/kwrot
T=miuo*MMFstator/(2*KC*d)*MMFrotor/rotor no.of turns/kwrot*length**r
T=Konst*MMFstator*MMFrotor*sin(fib)
But it is still complicated.

 

[Ryc92]

Yeah having looked at your derivation I realise I am not quite ready for that!

Thanks anyway, at my stage of learning I tend to accept more than I question.

I can understand it in terms of the usual cross product torque relating to mechanical force in terms of the quadurature component times the distance.

Soon as you guys are obviously well educated, anyone fancy taking a stab at my next problem... (substantially less complicated)

If in a synchronous generator (or transformer secondary, induction machine rotor) the armature MMF Fa (or flux) is opposing the flux which produced it, how comes the armature current Ia is shown in anti-phase with it? In my eyes the armature current is producing the mmf Fa and should therefore be shown IN phase with it, not anti phase as is shown. I see it referenced due to the fact that is a source or generator, the current must be in anti-phase with the flux. In the same way the secondary mmf in a transformer or induction machine rotor opposes the mutual flux and is created by the current I2, yet is shown in anti-phase with I2. (I am not reffering to the primary reaction to this mmf by the way, I understand that this must be in anti-phase with the secondary mmf Fs)

Thanks! Ryan

 

[7anoter4]

Some corrections :
In the first eq. t[torque]=f*r[rotor radius] so t= Bm*sin(w*t)*sqrt(2)*I*sin(w*t+fib)*length*r.
In the before last T=miuo*MMFstator/(2*KC*d)*MMFrotor/rotor no.of turns/kwrot*length*r*sin(fib)
I neglected also armature reaction –I think in an induction motor it is possible.

 

[Ryc92]

Thanks I'll have a think about it and see if I can make sense of it!

Any ideas on my next problem?

 

[electricpete]

Yeah having looked at your derivation I realise I am not quite ready for that!

Thanks anyway, at my stage of learning I tend to accept more than I question.

I can understand it in terms of the usual cross product torque relating to mechanical force in terms of the quadurature component times the distance.

I agree the math got messy. But one important takeaway imo is that it starts with an expression for energy. Torque is the derivative of energy with respect to angle. (just like force is derivative of energy with respect to distance... or for constant force: energy = force times distance).

in a synchronous generator (or transformer secondary, induction machine rotor) the armature MMF Fa (or flux) is opposing the flux which produced it, how comes the armature current Ia is shown in anti-phase with it? In my eyes the armature current is producing the mmf Fa and should therefore be shown IN phase with it, not anti phase as is shown. I see it referenced due to the fact that is a source or generator, the current must be in anti-phase with the flux. In the same way the secondary mmf in a transformer or induction machine rotor opposes the mutual flux and is created by the current I2, yet is shown in anti-phase with I2. (I am not reffering to the primary reaction to this mmf by the way, I understand that this must be in anti-phase with the secondary mmf Fs)

I will say for transformers and motors I have seen equivalent circuits where the rotor/secondary current arrow is drawing pointing either direction: towards the magnetizing branch or away from the magnetizing branch. You would adjust your equation for magnetizing branch current accordingly. It is the same current no matter which way you draw it so phase relationship to other quantities like stator current is not affected by which way you draw the current arrow. I may well have misunderstood your question and I apologize in advance if I did.


=====================================
(2B)+(2B)' ?

 

[Ryc92]

Ahh okay, I believe you are saying it is a question of convention rather than actual relationship?

I have attached a rudementary phasor diagram which explains what I mean, notice that the armature flux is pointing down towards the mutual flux, whereas the armature current which is producing the armature flux is pointing in the other direction, the direction of the induced EMF in the armature. I can understand that it is a generator and therefore the current is flowing out of the armature terminals as opposed to a motor. My brain is telling me though that if a current produces a flux, then it must be drawn IN phase with it.

This is a phasor diagram I have come across in many a respectable machine theory book.

 

[7anoter4]

I think you are right. Something is rotten... in this phasor diagram. I think it was still in work state when was exposed.
The EMF is leading the total flux and the rotor flux has no sense. If I well understood the problem the phasor diagram has to be as I attached it.

 

[7anoter4]

I have to apologize, it is still wrong phasor diagram. This could be better.