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# Re: [femm] Re: Torque calculation (2)

- To: femm@yahoogroups.com
- Subject: Re: [femm] Re: Torque calculation (2)
- From: David Meeker <dmeeker@ieee.org>
- Date: Wed, 22 May 2002 10:13:05 -0400
- References: <accvj3+lqhm@eGroups.com>
- User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.0rc2) Gecko/20020510

`vecerkaj wrote:`

> > `I have two questions related to the Torque calculation in a microengine that I presented in this forum some 10 days ago and that David Meeker helped with. I include his file for those interested.
> > First question: I have been informed that our microengine will have AC current. I guess it will be 50 Hz frequency. Are the calculation that I made valid? Should I change something?`

> `
> I guess, the calculations are correct, because you "stopped" the time and calculated the torque at this time. If the motor is vector controled, the torque could be constant though the time.`

`
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 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.`
`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. 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.`

> > `Second question: The current density that David chose was, by some kind of magic, very appropriate for our microengine: we are looking for a torque of 0.0020 N*m per unit meter length, and with the 10 MA/m^2 current density that David chose, we obtained a 0.0015 N*m per unit meter length torque. Quite close. However I am confused about the calculation of the intensity that will go through the coils. Is the intensity just the product of the current density by the area I have modeled in the FEMM program? This will be 10 MA/m^2 multiplied by 0.263873 mm^2 = 2.64 Amperes, which seems to be a lot. Am I missing something, perhaps the number of coils?
> > For instance, if the number of coils is 20, the intensity would be 0.132 Amperes? I have been told that this is a much more reasonable value of intensity for our microengine. I am maintaining the same geometric configuration`

> `
> If you use 8.66MA/m^2 instantaneous values, amplitude will be J_amp = 10MA/m^2. Let suppose, the cross-sectional area of the coil is S (0.263873mm^2), and number of turns N (20). AC current is the RMS value, but intensity 10MA/m^2 is the amplitude, you should use (amplitude /sqrt(2) )`

>     Current of the coil is:
>     I = J_amp * S / sqrt(2) / N
>     I = 10 * 0.263875 / 1.41 / 20 = 0.093 A

> `Note, that real RMS intensity in the copper is higher, because of the "filling" factor of the coil k_f (for example 0.5). (Jreal = J_amp / sqrt(2)/k_f = 10 / 1.41 /0.5 = 14.18 A/mm^2). If you use this intensity, you should use also different S_real=S*k_f ...... And the current will be the same as in the calculation above. :-)`
> `Jiri Vecerka`

`This is a good explanation. Thanks, Jiri.`
`Dave.`

Attachment: `zip00029.zip`
Description: Zip compressed data

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