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# Re: [femm] Torque calculation (3) Eddy currents

- To: femm@yahoogroups.com
- Subject: Re: [femm] Torque calculation (3) Eddy currents
- From: David Meeker <dmeeker@ieee.org>
- Date: Fri, 24 May 2002 15:31:29 -0400
- References: <090EBA399E3140489BBFC1235F43E2830B74AB@garnet.finasmet.org>
- User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.0rc2) Gecko/20020510

`Garmendia Azurmendi Iñaki wrote:`

>     > 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 have this number!`

`
Just look at the "steady-state" part of the torque result, from either Lorentz torque or stress tensor. The trick is to compare a run in which the conductivity = 0 versus a run in which the conductivity is 10 MS/m (conductivity for "cold-rolled steel" that I cribbed from the EIT review manual). The run with a zero conductivity yields 1.52 mN*m/m for the torque. The run with the 10 MS/m conductivity yields 0.96 mN*m/m for the torque at 50 Hz. The difference, 0.56 mN*m/m, is due to eddy current drag.`

>     > 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`

`
Whoops--I /meant/ 0.6 mN*m/m, as you suspect.`

> `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.`

`You can do this by bumping up the frequency in the "Problem" dialog box (select Problem off of the preprocessor main menu) to 1667 Hz. You get a cool-looking picture of the field in the stator getting dragged around by the eddy currents (see attached), but the drag torque is a lot higher than the motoring torque for this case. There are plenty of remedial actions that might be taken, e.g. build up the stator out of thin washers (rather than a solid cylinder).`
`Dave.`

- References:
  - Torque calculation (3) Eddy currents
    - From: Garmendia Azurmendi Iñaki <inaki@inasmet.es>

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