Torque calculation (3) Eddy currents

[Garmendia Azurmendi Iñaki <inaki@xxxxxxxxxx>]

Hello,
> As Jiri has noted, it is important that the motor be 
> vector-controlled 
> to actually be able to run the machine like this in steady-state.
> 
> In the last message about this, I'd mentioned that you'd 
> probably want 
> to get a handle on the eddy current drag as well. I've attached a 
> geometry where I did as I'd described in the last message, 
> replacing the 
> magnet with some current sheets and posing the thing as a 
> time-harmonic 
> problem. (In this one, you'd want to evaluate torque either 
> with stress 
> tensor, or by evaluating Lorentz torque over both the coils /and/ the 
> iron). Originally, you'd used "Pure Iron" as the stator 
> material, which 
> probably wouldn't be realistic--not many machines are 
> actually made out 
> of pure iron, and it's actually somewhat nontrivial to even obtain 
> really pure iron. I'd changed it to 1018 steel, which is a readily 
> available low carbon steel.
> 
> If the material really is 1018, you may have some eddy 
> current problems. 
> The analysis predicts an eddy current drag on the order of 
> 0.5 mN*m when 

How do you get this number? I have looked at the results on Torque calculation (both line Integral and Lorentz torque) and can not imagine how to combine the "net" part of the torque and the 2xomega part of the torque to havethis number!

> the shaft is spinning at 3000 rpm (i.e. 50 Hz), which is about 1/3 of 
> the force that we calculated for the same sort of currents in the 
> absence of the eddy currents. Typically, these small machines 
> run a lot 
> faster than 50 Hz. For example, the micromotors described at 
> www.smoovy.com run between 20,000 and 100,000 rpm, depending 
> on the motor.


You are right, once more. The real target is to have our micromotor runningat 100,000 rpm (in vacuum) and with a lower velocity in charge.


> 
> If the rotor is spinning slow enough that you can ignore skin 
> effects, 
> without going into the details of how you'd derive it, you could 
> estimate the eddy current drag torque to be:
> 
> Tdrag = (B^2 * d^3 * sigma * Pi * R * omega) / 3
> 
> where B is the peak flux density in the worst cross-section 
> of iron, d 
> is the thickness of the iron, sigma is the conductivity of 
> the iron, R 
> is the mean radius of the iron, and omega is the frequency in 
> rad/s. For 
> the case we'd been looking at, B is about 0.9 Tesla, d is 0.5 
> mm, sigma 
> is about 10^7 S/m, R is about 1.85 mm, and omega is 100*Pi 
> rad/sec. This 
> yields 6.16233 mN*m, which is pretty good for a simple approximation. 

I guess there is an operating error. Tdrag should be 0.616233 mN*m, which compares quite well with the previous 0.5 mN*m


> What this shows is that to reduce the eddy current losses, 
> you'd first 
> want to pick a material with low conductivity (e.g. some kind 
> of silicon 
> iron, which would reduce eddy current drag torque by a factor of 3 or 
> 4). Since the flux density in the iron isn't really that 
> high, you could 
> stand to make the iron thinner. If you halved the back-iron (which is 
> probably about as much thinning as you'd want to do), you'd 
> get about an 
> additional factor of 2 reduction in drag torque.
> 

I imagine I have to work with the new frecuency (1666.66 Hz) to calculate the Eddy currents influence. A little bit surprised by the importance of them.

Thanks again