B C for 2D axisymmetric FV case
Hello,
I am modelling flow over a sphere in a duct by finite volume method solving 2D axisymmetric Euler equations. Does anybody know what are the boundary conditions to be applied to the axis of symmetry and the wall ? Thanks in advance, Amith. |
Re: B C for 2D axisymmetric FV case
On the wall, you can applied no-slip boundary conditions, radial velocity is zero, z-direction velocitiy is set to reflection boundary conditions.
Kim |
Re: B C for 2D axisymmetric FV case
at the axis of symmetry, the normal gradient must be zero ( dP/dn=0, dV/dn=0 )..Regarding the wall you have to use slip condition because of the inviscid Euler equations.
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Re: B C for 2D axisymmetric FV case
yes, you are right, I thought it was Navier-Stokes equation.
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Re: B C for 2D axisymmetric FV case
Thanks Jim and Junseok for your help. I have another problem at the axis of symmetry. The surface areas of the bottom faces of the cells at the axis of symmetry are zero. Hence I am not able to implement dP/dn = 0 & dV/dn = 0. Even though I set the pressure and velocity in the ghost cells, it is actually ineffective. Please suggest.
Thanks again, Amith |
Re: B C for 2D axisymmetric FV case
It depends on how you define values on cell centers or cell edges.
For example, if you define velocity values at cell centers, then the ghost cell center velocity values are u_ghost = -u_nearest_to_axis and v_ghost = v_nearest_to_axis, where (u, v) are radial and axial velocities. Junseok |
Re: B C for 2D axisymmetric FV case
I am setting the velocities in ghost cells as told by you, but the bottom face areas at axis of symmetry are zero. And hence the setting of velocity and pressure in ghost cells is ineffective as they are getting multiplied by the bottom areas.
Amith. |
Re: B C for 2D axisymmetric FV case
Area should not be zero. I suggest you the simplest for of the zero gradient ..at axis of symmetry extrapolate from the interior of the domain i.e. P(i,1)=P(i,2), v(i,1)=0 & u(i,1)=u(i,2)...This called zero order extrapolation.Good Luck..
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