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PDE mode in comsol 3.5(a)

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Hi

I want to know that whether we can use PDE mode in axisymmetric geometries in comsol 3.5(a). I think it is not available for axisymmetric geometries and only exist for 2D geometries. When I select axisymmetric dimension I cannot find the PDE mode in the model navigator at all in 3.5(a).

Thanks for any suggestions

2 Replies Last Post 21.08.2012, 05:30 GMT-4

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Posted: 1 decade ago 20.08.2012, 12:49 GMT-4
Hi,


If you can get hold of page 737 of the reference manual, ver 4.2a, it is clearly mentioned how to implement equations containing the gradient, divergence, or curl in an axisymmetric geometry. You will have to compensate for the missing terms related to the curvature of the coordinate system as stated in the manual.

I don't know what is best in 3.5a, I have never attempted.

But I guess you will have to select the "weak form" PDE for 2D, then change two things:

1. the weak form itself to be multiplied by 2*pi*r.
2. add the missing terms due to the curvature. For example, if you were to access page 737, it is stated that:

div(u) = del(ur)/delr + ur/r + deluz/delz

where r and z being the coordinates and u is the dependent variable.

Note that in 2D, div(u) would only have the first and the last term of the above equation. But when you write your full weak form, you should not forget the middle term.


Suresh
Hi, If you can get hold of page 737 of the reference manual, ver 4.2a, it is clearly mentioned how to implement equations containing the gradient, divergence, or curl in an axisymmetric geometry. You will have to compensate for the missing terms related to the curvature of the coordinate system as stated in the manual. I don't know what is best in 3.5a, I have never attempted. But I guess you will have to select the "weak form" PDE for 2D, then change two things: 1. the weak form itself to be multiplied by 2*pi*r. 2. add the missing terms due to the curvature. For example, if you were to access page 737, it is stated that: div(u) = del(ur)/delr + ur/r + deluz/delz where r and z being the coordinates and u is the dependent variable. Note that in 2D, div(u) would only have the first and the last term of the above equation. But when you write your full weak form, you should not forget the middle term. Suresh

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Posted: 1 decade ago 21.08.2012, 05:30 GMT-4
Many thanks suresh your reply has given me some vision. I will see what can be done in v 3.5a.

Many thanks suresh your reply has given me some vision. I will see what can be done in v 3.5a.

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