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What's Axial symmetry 1?

Posted 10 sept. 2012, 04:24 UTC−4 Version 4.2a 4 Replies

Pu
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Hi, everyone

I think people using 2D axisymmetry mode know the default Axial symmetry 1 condition. I feel this condition strange, as most of time there's no variable, no weak expression or no constraint. Then what's imposed on this boundary? Does anyone know this?

Thanks a lot!
4 Replies Last Post 10 sept. 2012, 17:23 UTC−4
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 10 sept. 2012, 07:53 UTC−4
Hi

its the boundary by default on the axis of symmetry, indeed nothing special to sayabout it, what do you expect.
You could want to impose some other conditio n to it, cte T or I do no what

I see it as to remind you with where it is, and the flux thereon has a internal normal condition, even if not explicit, coming from the symmetry

--
Good luck
Ivar
Hi its the boundary by default on the axis of symmetry, indeed nothing special to sayabout it, what do you expect. You could want to impose some other conditio n to it, cte T or I do no what I see it as to remind you with where it is, and the flux thereon has a internal normal condition, even if not explicit, coming from the symmetry -- Good luck Ivar
Pu
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Posted: 1 decade ago 10 sept. 2012, 08:42 UTC−4
Thanks for replying!

I want to make some generalization to include non-zero order dependence on the angular variable \phi, such as exp(i*m*\phi). I just want to see the weak expression for the axis and figure out if the generalization will affect the condition on the axis.
Thanks for replying! I want to make some generalization to include non-zero order dependence on the angular variable \phi, such as exp(i*m*\phi). I just want to see the weak expression for the axis and figure out if the generalization will affect the condition on the axis.
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 10 sept. 2012, 17:14 UTC−4
Hi

well then I would say try support, they are the only knowing 100% what is behind. But does your changes really respect cylindrical symmetry ?

--
Good luck
Ivar
Hi well then I would say try support, they are the only knowing 100% what is behind. But does your changes really respect cylindrical symmetry ? -- Good luck Ivar
Pu
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Posted: 1 decade ago 10 sept. 2012, 17:23 UTC−4
Thanks anyway for your reply!

I will try to ask the support people.

Regarding your concern, the dependence on \phi can be separated explicitly leaving the other part independent of \phi. This part follows some PDE only having r and z, thus equivalent to an axisymmetric problem.
Thanks anyway for your reply! I will try to ask the support people. Regarding your concern, the dependence on \phi can be separated explicitly leaving the other part independent of \phi. This part follows some PDE only having r and z, thus equivalent to an axisymmetric problem.

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