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.
This is a good explanation. Thanks, Jiri.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
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