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Old   August 10, 2005, 07:03
Default mesh in polar coordinates
  #1
t.v.b.m
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guys, can someone give me an advice? I am to solve an axisymmetric flow (symmetry about z axis in r,z,theta coordiante system). So, I took r-z plane as my computational domain which I'd like to mesh. due to certain reasons, the mesh points are to be defined in polar coordinates. I chose the origin at one side of my geometry lying on the centerline. now mesh: one set of lines is clear, r=const., but how about second one? theta=const is not acceptable for me becuse: - it may intersect my geometry where I don't want it, - it may result in triangular cells (and I need all of them to be quadrilateral) I went through maybe 30 books and web pages on cfd and mesh generation but found nothing that would suit me.....
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Old   August 10, 2005, 09:36
Default Re: mesh in polar coordinates
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Jonas Holdeman
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If I understand your problem, it seems like you are starting from the wrong 3D coordinate system. Suppose you started from the spherical coordinate system and imposed the same cylindrical symmetry about the axis phi=0 or pi. Then you would be doing your computations in a plane with theta=constant and the variables would be r and phi, with the coordinate lines radial and circular.
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Old   August 10, 2005, 10:27
Default Re: mesh in polar coordinates
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andy
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Your description does not make sense. In the r-z plane (planes with theta=constant) there are no theta=constant grid lines. There are r=constant grid lines, z=constant grid lines or curvilinear grid lines.

What shape is your geometry? Is it the same shape on every theta=constant plane? If not, your flow is not axisymmetric. If it is, then you have to grid the 2D shape.
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Old   August 10, 2005, 11:09
Default Re: mesh in polar coordinates
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t.v.b.m
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oh, now I see I'm talking on two thetas each of which has another meaning. So: - my flow is in a pipe, thus axisymmetric and can easily be desribed by cylindrical coordinates r-z-theta - r-z (theta=const) plane has been chosen as a domain - in this plane I'm to solve the flow in polar coordinates, i.e. I have geometry defined by r-z coordinates (cartesian) and into this geometry I have to fit a mesh defined by polar coordinates R-phi; the origin of this polar coord system is on the centerline on one end of the pipe.

One set of lines is R=const, no problem; but the second set? phi=const is not acceptable because of the reasons mentioned above. the mesh is to be as simple as possible, containing quadrilateral cells only...

I'm thinking of I'll divide all R=const arcs into the same number of segments - thus every arc would have two boundary points lying on the pipe walls and, say, 10pts between these two. the interior pts on the arcs would be connected by straight lines

has anyone better idea (doubt about mine)?
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Old   August 11, 2005, 04:38
Default Re: mesh in polar coordinates
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andy
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For a single structured grid block there are two common ways to grid a cylindrical pipe:

(1) Cylindrical polar grid with a singularity at the centre-line. For unsteady 3D flows using grid-orientated velocity components (typical of many LES codes) the singularity can be very difficult to handle in a reasonable manner. For codes which solve for Cartesian velocity/stress components the singularity is not a significant problem. This grid resolves the wall boundary layer well but uses far too many points in the centre of the pipe.

(2) Deform a rectangular grid to fit the circular shape. This places the 4 corners of the square on the wall. The 4 corners become singularities because the two (different) grid lines that meet at the point are aligned. However, this does cause any significant problems in practice. This gives a more uniform grid disribution throughout the pipe but the grid distribution near the walls is far from ideal particularly if resolving high gradients near the wall is required.

If your code can handle structured grid blocks and the structured blocks can butt in a general manner then the almost universal approach is to use 5 blocks. A centre block surrounded by a ring of 4 blocks which each connect to the centre block, the outside wall and the two neighbours in the ring. This gives an even grid distribution and has grid lines aligned with the outer wall to resolve the boundary layer. If your code assumes a curvilinear coordiante system there are 4 singularities on the 4 corners of the centre block but this is not the case for an unstructured grid approach to differencing.
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Old   September 1, 2005, 08:49
Default Re: mesh in polar coordinates
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Angelina
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