Hi all, I am presently working with an unstructured code which implements a Finite Volume, pressure-correction method using a co-located arrangement of the variables. The Rhie&Chow modification to the mass fluxes is implemented in order to suppress checkerboarding. My experience is that, while the R&C modification does its job on a quadrilateral mesh, it is not so effective on triangular meshes. The more regular the mesh is (particularly diamond shaped triangular ones), the worst it performs. I wondered if any of you you had a similar experience and a justification for the different behaviour between quadrilaterals and triangles.

(1). I kind of agree with you that a four-sided quad or rectangle is not the same as a three-sided triangle. (2). The difference can be seen on the boundary, on the computational domain, etc. (3). On a wall, one can pave a layer of rectangular cells evenly and nicely. (4). This is not possible, because you will end up with a layer of saw-tooth shaped triangles, with one side edge attached to the wall. And the funny thing is, in between these cells, there are triangular cells with only one vertex attached to the wall. This edge-point-edge-point boundary cells on the wall is quite different from that of the rectangular cells on the wall. (5). And if you have to put these triangular cells with different size and height on the wall, you will have very difficult and ugly boundary conditions on the wall. (6). Can you solve a simple flow through a channel with just one cell? For the rectangular cell, there is no problem, you can use just one cell in the flow field. And still have inlet, outlet, bottom wall, and the top wall. (7). This is not possible with triangular cell. You will have to use a upper left cell and a lower bottom cell to cover the computational domain. In this case, the upper wall will be one edge of the inlet cell, and the bottom wall will be one edge of the outlet cell. (8). So, at least with this two-to-one rule, you can say that both methods can handle the same problem. So, the minimum requirement is not the same. (9). Well, if you are borrowing a method from rectangular cell side, at lest this two-to-one rule need to be observed, in addition to the saw-tooth problem on the wall. (10). What I am saying is that: two triangular cells are equivalent to one rectangular cell. And my old mistake of treating a girl like a boy is not going to work, and I have changed my point of view since my college days. So, you can't solve a problem with a single triangular cell, because there is always something missing. And by the time you put two triangles together, you are back to the rectangular cell.

Dear Aldo,

Now I start to work in a Navier-Stokes and heat transmition code using Finite Volume Methods in unstructured triangular meshes. Earlier I made an extensive literature review about the FVM discretization schemes applied to the diffusive, convective and transient therms and I have many interesting conclusion about it. I am ready to discuss some questions with you and also change some experiences. I have some papers that mention the influencies of the unstructured behaviour of the mesh in the final solution accurace.

Anything else, you could ask to me

Marcio Arêdes Martins | Doctoral Student | Mechamical Eng. Dep. | UFMG - BRAZIL | aredes.bh@zaz.com.br |