Re: J-Square integral

["David Meeker" <dmeeker@xxxxxxxx>]

--- In femm@xxxxxxxxxxxxxxx, "geezer_gatech" <gte584m@xxxx> 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