[kmgraju]

Dear all
I have come across stress free boundary condition at the outlet in a pressure drop analysis in many journals ( Transient analysis).
I have confused with 0 Pa setup at the outlet with stress free boundary condition.

for stress free boundary condition, I assume that the initial condition pressure setup and outlet boundary condition must be the same.
Please correct me if I am wrong.

Thank you

Regards

Govind

[oj.bulmer]

From what I understand, the stress-free boundary condition must be "shear-stress free" boundary condition which can be symmetry or free slip wall

Analogically, since three is no change in velocity perpendicular to the boundary (unlike wall where we have significant velocity gradient), the shear stress value should be zero/negligible.

Though, you may want to confirm this with a bit of literature search.

OJ

[kmgraju]

Originally Posted by oj.bulmer

From what I understand, the stress-free boundary condition must be "shear-stress free" boundary condition which can be symmetry or free slip wall

Analogically, since three is no change in velocity perpendicular to the boundary (unlike wall where we have significant velocity gradient), the shear stress value should be zero/negligible.

Though, you may want to confirm this with a bit of literature search.

OJ

Thank you for your reply

so, stress free boundary condition doesn't mean that outlet pressure equals to zero . Am I correct?

I know the symmetry boundary condition where all the normal component should be zero. consider a pipe , how do you apply symmetry boundary condition at the outlet face?
if so what does it mean?

when we apply free slip boundary condition at the outlet face?

Thank you

Regards

Govind

[ghorrocks]

This question has been asked a few times on the forum. Do a search of the forum for more details. I recall the stress free BC is zero normal gradient at the outlet. CFX does not implement the outlet boundary exactly this way, read the documentation for what it does do. And implementing a zero stress BC is going to be difficult, and frankly why bother when the built in outlet boundary is better in many ways.

[oj.bulmer]

so, stress free boundary condition doesn't mean that outlet pressure equals to zero . Am I correct?

The stress free boundary (or symmetry) condition should have tangential (shear) stresses and normal velocity as zero. This is what precisely the characteristic of symmetry.

Ref: Houseman, G. A. "Boundary Conditions and Efficient Solution Algorithms For the Potential Function Formulation of the 3‐D Viscous Flow Equations." Geophysical Journal International 100.1 (1990): 33-38.

I know the symmetry boundary condition where all the normal component should be zero. consider a pipe , how do you apply symmetry boundary condition at the outlet face?
if so what does it mean?

Whether stress-free boundary condition is appropriate for you or not, depends on your flow physics. You can have a model with symmetry boundary condition and outlet pressure as zero. Except, you can't specify symmetry at the outlet of your pipe. If you do that, the fluid can't flow out, nor can it flow parallel to outlet towards pipe walls as there won't be any motivation for it. The whole situation would be nonphysical.

when we apply free slip boundary condition at the outlet face?

Free slip wall condition is different than symmetry. Here, there can be normal components of gradients. Moreover, you can have curved surface as free slip, unlike symmetry where you need planar surfaces. Consequently, you can't have outlet, with free-slip wall, because fluid can't go out!!

OJ

[randolph]

Originally Posted by oj.bulmer

The stress free boundary (or symmetry) condition should have tangential (shear) stresses and normal velocity as zero. This is what precisely the characteristic of symmetry.

Ref: Houseman, G. A. "Boundary Conditions and Efficient Solution Algorithms For the Potential Function Formulation of the 3‐D Viscous Flow Equations." Geophysical Journal International 100.1 (1990): 33-38.



Whether stress-free boundary condition is appropriate for you or not, depends on your flow physics. You can have a model with symmetry boundary condition and outlet pressure as zero. Except, you can't specify symmetry at the outlet of your pipe. If you do that, the fluid can't flow out, nor can it flow parallel to outlet towards pipe walls as there won't be any motivation for it. The whole situation would be nonphysical.


Free slip wall condition is different than symmetry. Here, there can be normal components of gradients. Moreover, you can have curved surface as free slip, unlike symmetry where you need planar surfaces. Consequently, you can't have outlet, with free-slip wall, because fluid can't go out!!

OJ

Would you mind explain why Free slip wall condition is different than symmetry?

[ghorrocks]

Didn't he answer your question?

Free slip wall condition is different than symmetry. Here [free slip], there can be normal components of gradients. Moreover, you can have curved surface as free slip, unlike symmetry where you need planar surfaces.

[randolph]

Originally Posted by ghorrocks

Didn't he answer your question?

can you have dirichlet and neumann boundary condition at same boundary at the same time?

[ghorrocks]

Have a think about what they mean mathematically - the answer is no. You can define the boundary to have, in general, either a prescribed value at the boundary or a prescribed gradient, but you can't define it to have both a prescribed value and gradient.

[randolph]

Originally Posted by ghorrocks

Have a think about what they mean mathematically - the answer is no. You can define the boundary to have, in general, either a prescribed value at the boundary or a prescribed gradient, but you can't define it to have both a prescribed value and gradient.

Isn't setting up the reference value (for example pressure) is somewhat like a dirichlet boundary condition?

[ghorrocks]

No. It just offsets the entire pressure field. It has nothing to do with defining boundary conditions.

[Asura]

Originally Posted by ghorrocks

Have a think about what they mean mathematically - the answer is no. You can define the boundary to have, in general, either a prescribed value at the boundary or a prescribed gradient, but you can't define it to have both a prescribed value and gradient.

The boundary condition for u_2 in the Orr-Sommerfeld equation is u_2=partial_y u_2 =0 at walls. Therefore, yes you can have both Dirichlet and Neumann boundary condition simultaneously.

[ghorrocks]

Good point. You are correct, the direct answer to the question is that yes, you can have both a Dirichlet and Neumann boundary condition in place simultaneously.

It gets a bit more complex in CFD because you have boundary conditions which need to be applied to both momentum and pressure equations simultaneously. So yes, you can apply Dirichlet and Neumann boundary conditions on, say, the momentum equation simultaneously, but what do you apply to the pressure equation? Also you need to ensure that all boundaries are mathematically well posed across the entire domain, with issues like how the pressure needs to be set somewhere and they cannot be all velocity boundaries in incompressible flow and so on.

So while the answer you the question is yes, if you consider implementing it things get more complex as CFD is never simple