[HectorRedal]

Hi,

I would like to know if the following boundary conditions are well possed:

Domain: rectangle

Left wall: inlet velocity (Vx, 0)
Right wall: dVx/dx = dVy/dx = 0 and P = 0.
Top and bottom walls: dVy/dy = dVx/dy = dP/dy = 0

The point is that if I run the simulation with this boundary conditions, it starts obtaining strange results, flow entering and exiting through the upper and bottom walls.

But if I use the following boundary conditions:
Left wall: inlet velocity, Vx = constant and Vy = 0
Right wall: dVx/dx = dVy/dx = 0 and P = 0.
Top and bottom walls: Symetry wall, Vy = 0 (But letting Vx free)

Every thing goes smoothly (well).

So, I am wondering if the first set of boundary conditions is well posed.

Best regards,
Hector.

[FMDenaro]

If I understand your nomenclature, for a 2D problem you are implicitly setting the constraint dVx/dx=0 on top and bottom. Check it.

[HectorRedal]

Originally Posted by FMDenaro

If I understand your nomenclature, for a 2D problem you are implicitly setting the constraint dVx/dx=0 on top and bottom. Check it.

Yes, that it is. For an incompressible flow (dVx/dx + dVy/dy = 0), if I set the boundary condition at the bottom and at the top as dVx/dy = dVy/dy = 0, then it means that dVx/dx = 0 (dVx/dx = - dVy/dy = 0).

So, in some sense, if dVx/dx = 0 along the top and bottom domain walls, Vx does not vary along the botton and top of the domain. So in some sense it is equivalent to the second type of boundary conditions (Vx = constant, specified value).

My question is "Is this a well-posed problem"?

[FMDenaro]

Originally Posted by HectorRedal

Yes, that it is. For an incompressible flow (dVx/dx + dVy/dy = 0), if I set the boundary condition at the bottom and at the top as dVx/dy = dVy/dy = 0, then it means that dVx/dx = 0 (dVx/dx = - dVy/dy = 0).

So, in some sense, if dVx/dx = 0 along the top and bottom domain walls, Vx does not vary along the botton and top of the domain. So in some sense it is equivalent to the second type of boundary conditions (Vx = constant, specified value).

My question is "Is this a well-posed problem"?

It depends on how do you set these conditions... For example, if you are using a projection method you have to consider the relation between pressure gradient and velocity field. I think you have some redundance and you have to properly set the BC.s for the pressure equation.
Could you give details for the setting in the momentum and in the pressure equations?

[HectorRedal]

Yes, I am using a projection method.
I recalled you had explained me n another thread the relation between the pressure gradient and the velocity field:
d2p/dx^2 = (1/dt) du*/dx (equation coming from Grad p = (v* -v)/dt)

In the momentun equation, I am assuming du/dx = 0 in the boundary.
In the pressure equation, dp/dx = constant (dp/dx = integral{(1/dt) du*/dx})

[FMDenaro]

Your assumption provides directly d2p/dx2= 0 on the boundary

[HectorRedal]

But, d2p/dx^2 = (1/dt) d(v*-v)/dx, being v* the estimation of v obtained in the first step of the algorithm (where the pressure is not considered).
d2p/dx2 would be zero if the estimation of the first step is equal to the actual velocity field.

May I be wrong?

[FMDenaro]

Well, if you assume du/dx=0 there is no other congruent assumption that also prescribing du*/dx=0.
Furthermore, from the continuity dv/dy=0, therefore if v=0 on the upper o lower outflow corner, it must be zero along the outflow.

[sbaffini]

May I ask what set of equations you are solving and what method are you using (finite volume, difference, etc.)? And what do you mean by "symmetry wall" in terms of conditions?

If you consider a classical, cell centered, finite volume scheme, your two approaches are identical from the programming point of view.

[HectorRedal]

Originally Posted by sbaffini

May I ask what set of equations you are solving and what method are you using (finite volume, difference, etc.)? And what do you mean by "symmetry wall" in terms of conditions?

If you consider a classical, cell centered, finite volume scheme, your two approaches are identical from the programming point of view.

I am solving the Navier Stokes equations for incompressible fluids.
Symmetry wall: I am assuming that the velocity is parallel to the wall at that point (uy=0).

In my humble opinion, if uy=0 then duy/dy = 0, but if duy/dy=0, that does not mean that uy=0.

[HectorRedal]

Originally Posted by FMDenaro

Well, if you assume du/dx=0 there is no other congruent assumption that also prescribing du*/dx=0.
Furthermore, from the continuity dv/dy=0, therefore if v=0 on the upper o lower outflow corner, it must be zero along the outflow.

See it this way, it is as you mention. This is something that I have overlooked.

[sbaffini]

Consider a cell centered fv framework and a cell next to the top boundary.

Incompressible FV requires you to fix Vy and dVx/dy.

Symmetry sets both to 0.

But note that, in fv, not fixing a certain flux term really means setting it to 0.

So, the dVy/dx which is missing from above, is actually set to 0.

Analogously, the Vy that you don't set in your first bc, it is actually set to 0 in a cell centered fv framework.

So, according to the framework and how you set the symmetry, the two bc may actually end up being the same, that's why I asked.