Re: Torque calculation (2)

["vecerkaj" <vecerkaj@xxxxxxxxx>]

Hello

> 
> 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 frecuency. Are the calculation 
that I made valid? Should I change sometthing?

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. 

> 
> Second question: The current density that David chosed was, by some 
kind of magic, very appropiate for our microengine: we are looking 
for a torque of 0.0020 N*m per unit meter lenght, and with the 10 
MA/m^2 current density that David chosed, we obtained a 0.0015 N*m 
per unit meter lenght 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 micro engine. I am mantaining the same geometric configuration
---
If you use 8.66MA/m^2 instantaneous values, amplitude will be J_amp = 
10MA/m^2.
Let suppose, the crosssectional 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 
intesity, you should use also different S_real=S*k_f ...... And the 
current will be the same as in the calculation above. :-)

Jiri Vecerka