Induction machine pull up torque

[jaydesai07ee738]

Hello,


I was going over the equivalent circuit model and realized that there is no explanation for why there is pull up torque for certain induction machines.

Sure, we have the definition of the pull up torque.

I would really appreciate if someone could pin point the exact reason for the dip in the torque with the increase in speed to a certain level.

Thank you,
Sincerely,
Jay

 

[7anoter4]

The induction motor torque is equal with the "transferred "power from stator to rotor divided by synchronous speed.
The power it is Pe =k1*Rrot*Irot^2/s
Then the torque is: Pe/w w=2*pi*f
Irot=s*Erot/sqrt(Rrot^2+s^2*Xrot) where Irot = rotor current at slip s; Xrot it is rotor reactance at start [s=1]; Erot =EMF in rotor at s=1;Rrot=rotor [winding] resistance at slip s
Rrot=k(s)*Ro [approx.] k(s) is the skin and proximity effect -more if rotor frequency is elevated[s elevated]. Ro=rotor resistance at s=sn~=0
If k(s)=~1+a*s and Tq=k1*Ro/(2*pi*f)*Erot^2*s^2*k(s)/(k(s)^2*Ro^2+s^2*Xrot)/s
then d(Tq)/ds=K*{[(1+2a*s)*[(1+2*a*s+a^2*s^2)*Ro^2+s^2*Xrot]-(s+a*s^2)*[(2*a+2*a^2*s)*Ro^2+2*s*Xrot^2]}/[(1+2*a*s+a^2*s^2)*Ro^2+s^2*Xrot]^2
where K=k1*Ro/(2*pi*f)*Erot^2
We get a equation of second degree Ro^2*(1+2*a*s+a^2*s^2)-Xrot^2*s^2=0
If d(Tq)/ds=0 –in order to find the minimum and the maximum Tq we shall get only 2 point-the maximum in motor regime and in generator regime.
If we take k(s)=1+a*s+b*s^2+c*s^3 we shall get 4 points –two maximum and two minimum[the slip of the pull-up torque].

 

[cswilson]

It's been a while since I looked deeply into the physics of induction motors, but I believe the reason is this:

If you use a basic model of the induction motor that assumes the rotor resistance is constant over slip frequencies, your torque versus speed curve does not show this effect. For large values of slip, the greater the slip frequency (so the lower the rotor speed at a fixed electrical frequency input), the lower the torque.

But many rotors are designed to exploit the "skin effect". This effect means that at higher AC frequencies, the inside of the conductor becomes less and less able to conduct any current - only the outside of the conducting member effectively conducts current. (This is a key reason why with high-power electrical transmission lines, you cannot just increase cable diameter to increase transmission capacity.) With a smaller cross-section conducting current at higher slip frequencies, the resistance goes up.

With higher resistance at higher slip frequencies, the L/R time constant of the rotor is decreased, and the torque generated is higher than it would be with higher resistance. These designs are intended to increase the startup torque.

Curt Wilson
Delta Tau Data Systems

 

[waross]

The torque depends on the current through the rotor bars.
The current depends on the induced voltage and on the impedance of the rotor circuit.
As the speed of the rotor changes the induced voltage changes, and the impedance changes.
Dual squirrel cage motors are designed to optimize the effect of changing impedance as a rotor accelerates.
Most of the current flows in the fairly high resistance surface squirrel cage at slow speeds when the rotor frequency is high.
At higher speeds the rotor frequency is lower and progressively more of the rotor current flows in the deeper, lower resistance squirrel cage.
Bottom line:- Motor starting torque is non linear.
Note;- Many double squirrel cage designs use figure 8 shaped rotor bars rather than two separate sets of rotor bars.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Sparweb]

7anoter4: ...2 point-the maximum in motor regime and in generator regime...

These equations are valid when slip is negative and it turns faster than synchronous speed?


STF

 

[7anoter4]

These equations are valid when slip is negative and it turns faster than synchronous speed?

Yes.

 

[LiteYear]

For what it's worth, I re-did 7anoter4's derivation and got fundamentally the same thing. However:

Irot=s*Erot/sqrt(Rrot^2+s^2*Xrot)

I think that should be Xrot^2.

If k(s)=~1+a*s and Tq=k1*Ro/(2*pi*f)*Erot^2*s^2*k(s)/(k(s)^2*Ro^2+s^2*Xrot)/s

Again, Xrot^2.

then d(Tq)/ds=K*{[(1+2a*s)*[(1+2*a*s+a^2*s^2)*Ro^2+s^2*Xrot]-(s+a*s^2)*[(2*a+2*a^2*s)*Ro^2+2*s*Xrot^2]}/[(1+2*a*s+a^2*s^2)*Ro^2+s^2*Xrot]^2

Again, Xrot^2 in the denominator.

We get a equation of second degree Ro^2*(1+2*a*s+a^2*s^2)-Xrot^2*s^2=0

My Xrot^2 terms didn't cancel quite as well:
Ro^2*(1+2*a*s+a^2*s^2)-Xrot^2*s^2*(2a^3+a^2-2a-2)=0

But yes, still quadratic in s, when k(s) = 1+a*s.

If we take k(s)=1+a*s+b*s^2+c*s^3 we shall get 4 points

Seems plausible although I didn't do the derivation.

So I guess in amongst all the equations (cue request for LaTeX rendering on Eng-Tips!), the point you're making is that the torque has two turning points for s ∈ (0, 1), because the rotor resistance changes in a quartic fashion due to skin and proximity effect. Quite a complex phenomena, isn't it?