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Numerical diffusion, tet, hex, poly mesh

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Old   November 27, 2022, 13:18
Default Numerical diffusion, tet, hex, poly mesh
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Hi,

In many references from proponents of polyhedral meshing it is often said that polyhedral cells have 6 preferred directions (for a 12 face polyhedral cell). A hexahedral cell has 3 and a tetrahedral cell has zero. This is then used as an argument for numerical diffusion (or minimisation of). I do not understand how this works. It seems to me that while a polyhedral cell may have 6 preferred directions it will then also have 6 directions that may be oblique to the flow, which of course will cause numerical diffusion.

As I understand it, the only way of avoiding numerical diffusion is across cell faces that are normal to the velocity vector. Triangles can be used for this, just as well as quads, but most text-book examples use an unstructured triangular mesh as comparison compared to a structured quad mesh and therefore it seems that triangles cannot be used.

As a comparison, figure 1 is from ANSYS best practice guidelines, and figure 2 is just a structured triangular mesh with the same type of comparison (cell values plotted, not node values).

Information propagates in small wiggles for the triangles as the cell centers are not aligned perfectly as for the quads, but that does not matter since the information transport is contained in the same quad that was split to create the triangles.

Anyways, for the general case with a fully unstructured mesh it may be easier to have parts of the domain aligned with a certain flow direction if quad/hex mesh is being used, so perhaps the text-book examples are OK.

However, I still do not understand the argument about "preferred directions" when it comes to polyhedral cells. The best argument for polyhedral cells is that we have more neighbours when calculating gradients, but that has no connection to the numerical diffusion.

What do you think, do you have any better explanation for the argument of polyhedral cells' preferred direction?
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