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Old   June 29, 2010, 15:47
Default zero gradient pressure B/C
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Srinath Madhavan (a.k.a pUl|)
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Can someone kindly help in explaining why setting the component of "pressure gradient" normal to the wall zero is an appropriate Boundary condition for walls?

Put simply, why should the pressure gradient (i.e. grad (p)) be orthogonal to the outward normal to the wall?
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Old   June 30, 2010, 13:49
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Anybody?? Does setting "grad(p) dot nhat = 0" mean that from the wall onward up till beyond the wall boundary (i.e. the region occupied by ghost CVs), there is no component of the "pressure gradient" in the direction of the outward wall normal?

In other words, physically speaking, we are enforcing the condition that there is no fluid flow through the wall?? Am I even close???
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Old   August 5, 2010, 23:52
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Found some free time. Thought I could use this to close my query. A little too embarrassing actually, but after a sharp reprimand from one of my profs. for not knowing something so trivial I have the answer:

http://www.ualberta.ca/~madhavan/zer...nt_at_wall.pdf
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Old   August 7, 2010, 00:16
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Thank you for sharing Srniath.
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Old   August 7, 2010, 00:26
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Glad you found it useful Have a good weekend!
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Old   August 7, 2010, 14:22
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Okay, I have mixed up x and n! I'm gonna to study this again, because it seems very interesting. Sorry for the rigor, but as a mathematician, one is compelled to investigate things more critically.

Cheers and have a nice one!
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Old   August 7, 2010, 15:16
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No problem. I for one welcome any criticisms. After all we are all here to share and learn
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Old   September 28, 2010, 15:31
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Is there a patch/method, in OpenFOAM, which extrapolates the pressure to the wall from the adjacent normal cells rather than forcing the approximation of dp/dy = 0 ?

That's my first question, my second being is it easily achievable to prescribe dp/dy = nu*d2v/dy2 as a custom patch?

I ask in the context of using the icoFoam solver on 2-D laminar boundary layers.

Any suggestions would be greatly appreciated. Thanks!
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Old   November 11, 2010, 10:58
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I'm also interested in this topic. If you have a look on the cavity tutorial, the pressure-gradient normal to the movingWall is certainly not zero.

Did you already come to a result, steph79?
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Old   November 15, 2010, 06:45
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Quote:
Originally Posted by AsafPa View Post
I'm also interested in this topic. If you have a look on the cavity tutorial, the pressure-gradient normal to the movingWall is certainly not zero.

Did you already come to a result, steph79?
Yes, PRESTO! pressure discretisation is what you're looking for but whether that can easily be implemented in OpenFOAM I'm not sure. Could someone with more experience please comment on that?

Thank you.
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Old   December 13, 2010, 20:43
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Hi, I have a question.
In page 2, line 6 where you say dv/dy=0. then d2v/dy2=0, I cannot see why is that? for other points, you have that because you can compare 2 points on the wall ( which is x direction) but for y direction, I don't know how to say that. Any way, thank you for this calculation, really helps understand what is going on in OpenFOAM.
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Old   December 14, 2010, 10:50
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I think there is another reason for zeroGradient pressure BC at walls. It comes from numerics. When you implement PISO or SIMPLE methods finally a Poisson equation is assembled for pressure. This equation requires at least a one point with prescribed pressure which is searched from BC. If there is not a prescribed pressure BC then the pRefValue is taking into account. Since pressure is not known at walls, the BC is set to take the value from the nearest point within the fluid, or in other words taking the zeroGradient perpendicularly to the wall.

Just my 2 cents.

Regards.
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Old   July 28, 2015, 11:02
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Quote:
Originally Posted by sahm View Post
Hi, I have a question.
In page 2, line 6 where you say dv/dy=0. then d2v/dy2=0, I cannot see why is that? for other points, you have that because you can compare 2 points on the wall ( which is x direction) but for y direction, I don't know how to say that. Any way, thank you for this calculation, really helps understand what is going on in OpenFOAM.
Yours.
Sorry to open this thread again. But this is exactly what I was asking myself: dv/dy = 0 => d2v/dy2 = 0 ...why?

I was really happy to find a such a simple answer to this question but I actually doubt it's as simple as your prof pretended.
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Old   January 20, 2016, 12:42
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The second derivative of velocity along the wall is zero because the value is constant. The second derivative normal to the wall is zero because of the linear velocity increase in the viscous sublayer... correct me if I'm wrong.
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Old   January 20, 2016, 13:05
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Quote:
Originally Posted by Luke.12 View Post
The second derivatives of velocity are zero because of the linear velocity increase in the viscous sublayer... correct me if I'm wrong.
You are now bringing turbulence into the problem, which will not make this more easy to solve However, the linear velocity profile in the viscous sublayer of turbulent flows refers to the parallel velocity component u --> du/dy = const. You can't simply apply this to v.

I still think that dv^2/dy^2 is not generelly zero. However, for high Reynolds numbers the influence can be neglected since it is part of the visous terms in the NSE. For low Reynolds numbers, the zero gradient condition is wrong and can therefore lead to unphysical simulation results.
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Old   January 20, 2016, 13:15
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Ok, thank you, I have edited my initial reply quite fast, but still it does not explain what you are pointing out.

But if the turbulent flow and the viscous sublayer is really present, the wall normal velocity should be zero in that layer. And if that layer has a finite thickness, the first and second derivatives are zero as well at the wall, right?

Again sorry for confusion and thank you for quick reply!
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Old   January 20, 2016, 14:23
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I am not an expert in turbulence, but I think you are right about the vertical velocity component. It should be zero in the sublayer. And because of this, the second derivative should also be zero. But this line of argumentation only makes sense to me if the first layer of cells is thinner than the viscous sublayer.

But all this things we know about the viscous sublayer were derived based on certain assumptions and maybe even empirical findings. So I wouldnt use the conclusion from this ( --> du^2/dy^2) and insert this into the derivation of this purely mathematical attempt to derive the pressure boundary condition. At least not without thoroughly thinking it through.
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Old   January 20, 2016, 14:49
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I agree, but it is better than nothing. I will keep an eye on this thread and keep the problem in mind to ask someone more experienced too if solution does not appear until I do so.

I have hands on low-Re SST k-w and wall resolved LES right now, so my mesh is far below (y+ = 1) the viscous sublayer thickness (y+ = 5), so it won't get me toss, turn and sweat in the night.

It is also not clear to me, why it would not be similar to laminar flow problem... The danger of getting thin boundary layer and wall normal velocity components is even smaller there. At least as far as I understand.

Thanks for discussion and the initial post of this thread especially!
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Old   January 20, 2016, 17:48
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Quote:
It is also not clear to me, why it would not be similar to laminar flow problem... The danger of getting thin boundary layer and wall normal velocity components is even smaller there. At least as far as I understand.
Actually, if you look at the boundary layer equations by Prandtl (which I guess should be valid for laminar flow), you can see that they even explicitly state that dp/dy = 0 in the whole boundary layer. But still, there are certain assumptions necessary to get from the NSE to the boundary layer equations (i.e. concerning the magnitude of Re). And I am pretty sure, that the dp/dy = condition is not generally valid for low Re-numbers flows. I already experienced convergence issues that are most likely caused by this.

Also as I just had a look at the boundary layer equations... If you apply them at the wall (y=0). Then you get du2/dy2 = 1/mu * dp/dx. So if we can really use du2/dy2 = 0, as described in the derivation linked in the third post in this thread, we would get dp/dx = 0 (basically the Euler equation without convection terms due to u=v=0 at the wall). And this really doesn't make sense.

But if you ever find the answer to this question, let me know!
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Old   June 1, 2016, 02:58
Default Physical significance and Maths behind boundary conditions
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Hello,

Does anyone have any material specifying the physical significance and maths behind different boundary conditions(fixedValue, zeroGradient) in OpenFOAM?

Thanks
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