It is said that quad/hex grids provide better accuracy than tri/tet grids. Can someone substantiate this?

(1). I am just guessing that you are probably talking about the finite-volume methods. (2). " In finite-volume methods, is the solution obtained on quad/hex grids more accurate than the solution on tri/tet grids? " This is a very important question. (3). The traditional mesh is the quad/hex mesh. It is consistent with how we derive the governing equations in differential form. As a result, there are East, West, South, North, Front and Rear surfaces of the cell. The center is always in the middle of the cell. Near the boundary, the surface can be aligned with the wall and a locally 1-D flat cell can be formed to handle the boundary conditions. The heat flux will be transimitted from one surface to the opposite surface smoothly with the quad/hex meshes. (4) In other word, the cell has the properties of representing both the streamline of the flow and the boundary conditions, without special treatment. And it can be viewed from either the differential or integral formulations. ( either finite-difference or finite-volume) (5). The tri/tet mesh has different kind of properties. One can grow the complex meshes easily with the tri/tet mesh approach. Only one additional point ( or vertex) is required to create a new tri/tet cell. One can fill the complex flow field with this type of mesh easily by adding vertices and cells. And this is purely a property of geometry. It is not derived from the traditional finite difference approach, or the conventional control volume approach. The concept of the East, West, South, and North cell surface no longer exists. (6). There is one more difference between a rectangle and a triangle. The area change from one side of a rectangle to the opposite side is zero. On the other hand, there is a linear area distribution from one edge of a triangle to the opposite side of a vertex. This area distribution plays an important role in getting approximate representation of solution in the cell. (7). For an equal-sided tri/tet cells, the slight shift in the center of the cell may not present a problem. But for a very thin or highly skewed tri/tet cells, the off-set could be very large. If one divides a thin rectangle into two thin triangles, then the centers of the new triangles will be off-set to the short edge sides. So, a seemingly very simple thin rectangle cell is now represented identically by two non-symmetric, off-set centered triangles. (8). Suppose that we have analytical solutions over the rectangular cell, can we easily obtain the identical solution using the solutions over the two triangular cells? (9). You don't have the final proof here. Because, it depends on how you derive the finite-volume formulation over the tri/tet mesh cells. The assumption here is that both mesh cells cover the same identical area (volume). ( two triangles for every rectangle) (10). You may want to try different approximation schemes over the triangle cells ( a pair) to see whether you can get back the analytical solution over the rectangle cell. (11). It may be easier to do this home work using existing codes which allow both types of meshes to be used.

Hy friends,

one hex-cell can be build up with six tet-cells. Can we say because of this, that for the same geometry you need 6x less hex than tets (for the same accuracy) ? Or is it different because the 6 tets have more faces ?

andi

(1). If the total cell volume is the same and the total number of data points is also the same, then it takes more work to use tet-mesh. If both results are the same, they both have the same accuracy. (2). If tet-cell also create additional data points, then it is hard to make comparison with different number of data points. (3). The accuracy will depends on several issues, the degree of approximation over the tet-cell certainly will affect the accuracy, especially when the cells are highly skewed, or not aligned with flow direction, etc..