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# Re: J-Square integral

- To: femm@yahoogroups.com
- Subject: Re: J-Square integral
- From: "David Meeker" <dmeeker@ieee.org>
- Date: Thu, 02 Oct 2003 02:00:42 -0000
- In-reply-to: <blfekb+q1lo@eGroups.com>
- User-agent: eGroups-EW/0.82

    --- In femm@yahoogroups.com, "geezer_gatech" <gte584m@p...> wrote:
    > I need to compute the integral of Integ(J^2 rdrdz) over an rz block
    > (in axisymmetric geometry). Is there a convenient way to do this?

    The "resistive losses" volume integral computes:
    Integ(J^2 rdrdz)/Conductivity
    for DC problems and
    (1/2)Integ(J^2 rdrdz)/Conductivity
    for AC problems. If you have a region of interest with a non-zero
    conductivity, you can perform the resistive losses integral and
    multiply by conductivity to get your desired result. If the region of
    interest has been assigned a zero conductivity, you could always
    re-analyze the problem with a conductivity of 1 S/m (which is nearly a
    zero conductivity) for the purposes of performing the computation.

    Dave.
    --
    David Meeker
    http://femm.berlios.de/dmeeker

- References:
  - J-Square integral
    - From: "geezer_gatech" <gte584m@prism.gatech.edu>

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