[HectorRedal]

Hello,


I would like to know which boundary condition to use at the exit of an outer flow.
Suppose you have a pipe with an inlet boundary condition and an outlet boundary condition. Up / Down are no slip conditions.


When simulating, I fix the pressure at the exit at all points of the mesh. Then, the simulation is fine, and the flow at the exit shows clearly that the velocity goes outward (showing a parabolic profile). At the inlet the velocity is contant at all the points.



If I set the pressure only at one point of the exit, for simplicity, let put it in the middle of the vertical exit, part of the flow goes outward, but other part of the flow goes inward.



Why?


Am I using a wrong pressure boundary condition?


Thanks for your support.


Best regards,
Hector.

[LuckyTran]

If you fix the pressure only at one point, what is the boundary condition at all other points? The problem isn't even fully defined yet to even have any solution. Clearly, there is already some confusion about what is even the definition. You need boundary conditions on the entire boundary.

[FMDenaro]

I am supposing you are working using the incompressible flow model.



When you see the correction equation along the streamwise direction you have something like


u^n+1=u* - dt* dp/dx


If for some reason (but is not mandatory to fix a Dirichlet BC for the pressure) you set the pressure value at the outlet, you have to be careful to be congruent with the above equation (Hodge decomposition). Of course, you don't know u^n+1 at the outlet, therefore you need to think a way to evaluate u* at the whole outlet.


I suggest to check:
1) the divergence-free condition in each cell. Is it satisfied? At the level of the round-off or local truncation error?
2) Try the Neumann BC at the outlet and check if the solver converges.

[HectorRedal]

If I fix the pressure at one point, I am fixing the zero traction boundary conditon at the exit for the rest of points.
If we take t as the traction, p as the pressure, n as the outward normal to the exit surface and tau as the deviatoric stress, then we have:
t = -p * n + tau * n.
If t = 0, then p * n = tau * n.
This gives me the value of p at the exit boundary for the rest of points (BC at the exit).


The scheme I am solving is:
u* = f(u^n)
tau = f(u^n)

p = f(u*, u^n)
u^(n+1) = f(u^n, u*, p)


In my humble opinion, this scheme is congruent with the Hodge descomposition
The u* is estimated without taking into account the pressure term, but the rest of variables. Then it is corrected when knowing the pressure.



In regards with your suggestion:
2) Try the Neumann BC at the outlet and check if the solver converges.



Which variable are you suggesting for the Neumann BC? Are you referring to the pressure? dp/dx=0?

[FMDenaro]

Originally Posted by HectorRedal

If I fix the pressure at one point, I am fixing the zero traction boundary conditon at the exit for the rest of points.
If we take t as the traction, p as the pressure, n as the outward normal to the exit surface and tau as the deviatoric stress, then we have:
t = -p * n + tau * n.
If t = 0, then p * n = tau * n.
This gives me the value of p at the exit boundary for the rest of points (BC at the exit).


The scheme I am solving is:
u* = f(u^n)
tau = f(u^n)

p = f(u*, u^n)
u^(n+1) = f(u^n, u*, p)


In my humble opinion, this scheme is congruent with the Hodge descomposition
The u* is estimated without taking into account the pressure term, but the rest of variables. Then it is corrected when knowing the pressure.



In regards with your suggestion:
2) Try the Neumann BC at the outlet and check if the solver converges.



Which variable are you suggesting for the Neumann BC? Are you referring to the pressure? dp/dx=0?

First of all, for incompressible flows you have no physical meaning for the pressure. Call it Phi, the result is that this scalat function does not enter in to the equations by means of a value but only by means of its gradient. The Hodge decomposition can be seen as


v^n+1 = v* + v'


and that says you have to sum to the intermediate velocity a velocity coming from a potential. The only goal of that is to ensure the divergence-free condition.


You can write the Neumann condition as


dp/dx=1/dt*(u*-u^n+1)



and if you substitute this relation into the pressure equation you see that u^n+1 at the exit is not available. Thus, you need to recast differently the convective conditio, for example by assuming


d/dx[dp/dx=1/dt*(u*-u^n+1)]=0


This way the pressure equation depends only on the vertical gradient.

[HectorRedal]

I think that I have not included all the information, and this has caused confusion.
The algorithm I am solving is the following:


Step 1: u* = u^n+1 - d/dx (u rho u) - d/dx(tau) - rho g + B.C.
Bear in mind that P has been ruled out from step 1.


Step 2: 1 / c^2 delta p = - delta T * (d/dx (u^n+1 + u* - d/dx(p)) + B.C.

Bear in mind that I am resolving a Poisson equation.
Then, the P is entering into the equations as gradient and not as scalar, as you comment (So, we agree on this)



So, based on your explanation, it is in Step 2 where Neumann BC should be set.


But the point is that B.C are set by means of an integral component:
integral over boundary (u* - delta T dp/dx)
And, I am not sure how to enforce the type of boundary condition you are referring to: d/dx[dp/dx=1/dt*(u*-u^n+1)]=0


I understand you would like to enforce divergence free of the velocity field: d u^n+1 / dx = 0 = du*/ dx - dt dp/dx.
But, du*/ dx - dt dp/dx = 0 does not mean u* - delta T dp/dx = 0, as the B.C integral is expecting.
So, not sure how to enforce this B.C at the pressure equation (Poisson equation).


Thanks.

[FMDenaro]

First, do not write the pressure equation in terms of a Laplacian operator but in terms of the DivGrad operator and discretize the two operators in sequence. I am also assuming you work on staggered grid.

Assuming a convective-type BC at the outlet means that you can assume


du^n+1/dx=0, du*/dx=0 -> d/dx(dp/dx)=0


Note that the last relation is congruent to a fully developed flow, like Poiseuille, showing a linear decreasing pressure.
Therefore, the pressure equation at the outlet has no derivatives along the x-direction (streamwise):


d/dy(dp/dy) =(1/dt) dv*/dy


BC: dp/dy=(1/dt) (v*-v^n+1)


That ensures you satisfy also the integral compatibility condition.


More theoretical details for the pressure equation can be found in my paper:


https://www.researchgate.net/publica...f3ddKwvRpF382A

[HectorRedal]

I write the pressure equation as a divgrad operator, but as commented, the boundary conditions appear as functon of u - dt * dp/dx.
I am not using staggered grid, since I am using FEM instead of FDM.


On the other hand, I am interested in transient problems, where the exit oscilated, such as a Karman vortex sheet. At the exit, we cannot assume Poiseuille flow.


I will take a look at the paper you are referring to. Thanks.

[FMDenaro]

The BC I wrote above does not imply only the poiseuille solution! It can be prescribed for unsteady outlet conditions. I used that for LES.