CFD Online Discussion Forums (https://www.cfd-online.com/Forums/)
-   Main CFD Forum (https://www.cfd-online.com/Forums/main/)
-   -   Boundary Conditions for Euler (https://www.cfd-online.com/Forums/main/3195-boundary-conditions-euler.html)

 Kevin March 3, 2001 01:46

Boundary Conditions for Euler

Hello!

I'm writing a 2nd-order accurate FVM Euler code to compute a flow around an airfoil, on triangular grids. I have two questions concerning the boundary conditions.

(1)Solid body: I store solutions at a cell center. So, on the airfoil, I need to compute the flux. How can I compute this flux? Perhaps I need to set up a ghost cell inside the body, and use the interior scheme to find the flux. But what values should I put in the ghost cells? Or is there any other ways to compute the boundary flux?

(2)Far-Field: To compute a lifting flow, I need to place a point vortex at a quarter chord of the airfoil. But I don't know how to determine the strength of the vortex. This seems impossible to me because I wouldn't know the value of the lift, which is directly related to the circulation(or vortex strength), until I finish computing the solution. Does any body know how to determine the strength?

Any comments are welcome. Thank you!

Kevin

 O March 3, 2001 16:15

Re: Boundary Conditions for Euler

Concerning (1):

If I have this correctly you want to compute a flux using collocated FV at a solid boundary?

First the boundary cells usually have zero volume so the velocity in this cell is zero.

For the u-,v-, and w-momentum the velocity normal to the wall will (in most cases) be zero, thus a zero flux. If you are ignoring diffusion then just set your coefficient (A_i) normal to the wall equal to zero.

 Kevin March 4, 2001 23:53

Re: Boundary Conditions for Euler

Thank you for your comments! But I'm not sure waht you mean by "the boundary cells usually have zero volume so the velocity in this cell is zero". Are these cells are those having a boundary face? Then, they all have a nonzero volume. I think they usually are, for it wouldn't make sense to introduce cells with zero volume around an airfoil.

Right, the fluxes across the boundary must be zero. But pressure remains in the flux even on the boundary. Maybe I could extrapolate the pressure to the boundary from the interior. I think this is worth trying. Thank you!

Kevin

 Frank March 6, 2001 12:27

Re: Boundary Conditions for Euler

Hi Kevin

I actually don't understand, why the volume at boundary cells should be zero ?!

If you use a cell centered finite volume scheme with unstructured meshes I would actually not use any ghost cells. For a second order scheme, you have to reconstruct the variables anyhow. (in my experience it is better to reconstruct the primitive variables, than the conservative ones. Matter of choice ?!)

That means you will determine the gradients in the cell centers of each variable and then extrapolate the solution on the faces, where you solve the riemann problems for the convective terms. I would also not use a central scheme with artificial dissipation but use riemann solvers. At boundaries, all that changes is the stencil for the reconstruction, which becomes more one sides. One can simply extrapolate the solution on the boundary and then the flux is imply given exactly by the pressure in the momentum equations. If using riemann solvers, you can also extrapolate all variables on the boundary, compute the pressure (put it as left and right state of the riemann problem) and then before solving the riemann problem set the velocity normal to the solid wall it zero. When solving the riemann problem, this wil nothing but result in a contribution of the pressure again. This formulation of boundary conditions is only a weak one, since it is just via fluxes.

For the vortex: Look at a paper of Thomas and Salas: Thomas, J.L., Salas, M.D., Far--Field Boundary Conditions for Transonic Lifting Solutions to the Euler Equations AIAA J., 24, 1986, pp. 1074--1080

There is a formulation, how to obtain the circulation as a function of the lift. So, first you compute the lift by integrating the pressure around the airfoil (considering the normal vector on the surface and the angle of attack) , and from that you get the circulation. Of course, it is an iterative process. And the lift as well as the circulation will get its true value, when the solution is converged. At the beginning of the computation, both can get any value...

Hope it helps somewhat.

Frank ;-)

 Kevin March 6, 2001 21:54

Re: Boundary Conditions for Euler

Thank you very much! I will try solving Riemann problem at boundary. For vortex, I didn't know it was an iterative process. I'll look at the paper.

Kevin

 Frank March 7, 2001 08:15

Re: Boundary Conditions for Euler

Hi Kevin,

solving Riemann problems at boundaries is of course only useful, if you use riemann solvers anyway.

I personally favor the HLLC scheme by Batten and Leschziner (Journal of computational physics.. end of 97 or 98). It is nicer and easier than Roe. The AUSMDV also needs entropy and shock fix, which the HLLC does not require. But this a personal choice.

I just find it easier to program, if I do not need to replace the flux at the wall boundary by the flux given by the pressure. I mean: In this case one has to use a special treatment for these walls. What I do is just to copy the extrapolated values on the side, which is interior of the field, to the solution exterior of the field (one always needs 2 states for the riemann problem). And then set both normal velocities (of the left and right state of the riemann problem) to zero. But AFTER I have computed the pressure ! (In case you extrapolate the conservative variables. I first get the pressure and then set the velocity to zero). And then one just do one loop over all boundary faces, including far field boundaries and wall boundaries.

Since the whole procedure is an iteration, I would suppose that the vorticity should get out correctly at the end of the computation.

In my opinion, Euler wall conditions are harder to get working really nicely than Navier Stokes. And everybody does something different.

Cheers,

Frank

 All times are GMT -4. The time now is 07:22.