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# Re: Torque calculation (2)

- To: femm@yahoogroups.com
- Subject: Re: Torque calculation (2)
- From: "vecerkaj" <vecerkaj@yahoo.com>
- Date: Tue, 21 May 2002 08:13:23 -0000
- In-reply-to: <090EBA399E3140489BBFC1235F43E28376D22C@garnet.finasmet.org>
- User-agent: eGroups-EW/0.82

    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

- Follow-Ups:
  - Re: [femm] Re: Torque calculation (2)
    - From: David Meeker <dmeeker@ieee.org>

- References:
  - Torque calculation (2)
    - From: <Inaki@inasmet.es>

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