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September 8, 2010, 14:26 |
Boundary Conditions. Simulation cylinder.
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#1 |
Member
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Hello! I just started with OpenFOAM.
I use version 1.7.0. I'm trying to simulate a cylinder in which the laminar (or turbulent) incompressible fluid flow. And the bottom of the cylinder rotates. Like this: What is the boundary condition I set on the bottom and how to do it? I would be grateful for any assistance conditional. Maybe someone has already done this? Sorry for my English A! And what solver shall i use? |
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September 11, 2010, 13:30 |
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#2 |
Member
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Surely no one can help? It's very, very important to me, please...
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September 17, 2010, 09:42 |
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#3 | |
Senior Member
maddalena
Join Date: Mar 2009
Posts: 436
Rep Power: 23 |
Hi David and welcome! Hope to be not too late...
Everything depends on what simulation you want to perform. I guess you want a steady state laminar or turbulent simulation. In that case I would use simpleFoam. Quote:
Enjoy mad |
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September 19, 2010, 11:04 |
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#4 |
Member
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Thank you very much, maddalena, I built something like the geometry and boundary conditions, but I used iсoFoam not simpleFoam. What is the difference between them?
I apologize for maybe a stupid question. I just sooo hard to read literature in English |
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September 20, 2010, 04:01 |
simpleFoam v.s. icoFoam
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#5 |
Senior Member
ata kamyabi
Join Date: Aug 2009
Location: Kerman
Posts: 323
Rep Power: 18 |
Hi David
How are you? I hope you are doing very well. As you can see in the source of the solvers icoFoam is a transient solver for incompressible, laminar flow of Newtonian fluids and simpleFoam is a steady-state solver for incompressible, turbulent flow. Good luck Best regards Ata |
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September 23, 2010, 09:42 |
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#6 |
Member
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Now I'm faced with this problem. I can not define the boundary condition at the upper edge of the cylinder (i mean the top sureface of my fluid). Maybe someone knows how to do this? Uz = 0, dUx / dx = 0, dUy / dy = 0
Last edited by davidmd; September 23, 2010 at 10:16. |
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September 23, 2010, 09:47 |
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#7 |
Member
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Hi, ata!
Thank you for help, I hope you are all well too! davidmd |
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September 23, 2010, 11:14 |
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#8 |
Member
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I'm trying to simulate simple mixer. It looks like cylinder with rotating bottom and open top filled with fluid. The other cylinder is inserted into the first.
So something like this: So, on the walls of this cylinders i set the fixedValue BC On the bottom - rotatingWallVelocity. And to top I want to set the boundary condition such as on the surface of the liquid ... At first I tried to do zeroGradient boundary condition. But then it turns out like this: It turns out that the fluid enters the center and comes out at the edges, but this is not correct. I thought that it was possible to establish a boundary condition like Uz = 0 Ux = dUx / dx Uy = dUy / dy Maybe i need another BC... I don't know. The main thing that it was just on the surface of the liquid. Once again: sorry for my English :-) |
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September 23, 2010, 11:48 |
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#9 |
Senior Member
maddalena
Join Date: Mar 2009
Posts: 436
Rep Power: 23 |
Mmm...
I am still confused by your geometry... there are two cylinders, one inside the other, filled with fluid. The external one has a rotatingWall BC, while the internal one is fixed. The bottom of the internal cylinder is open. You want to study the effect of the bottom surface rotation of the external cylinder into the fluid. Am I right? What about your pressure BC? As a first idea, I would use an inletOutlet BC for U and a fixedValue for p. But not sure about that. mad |
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September 23, 2010, 15:34 |
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#10 |
Member
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maddalena, you are right!
My preasure BC is set to fixedValue and it is 0... |
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September 23, 2010, 16:16 |
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#11 |
Senior Member
maddalena
Join Date: Mar 2009
Posts: 436
Rep Power: 23 |
So, this is fine:
if you fix pressure, you cannot fix velocity, and your gradient should be specified instead. I would use an inletOutlet with an inletValue (0 0 0): it will behave as a fixedValue on the inner cylinder and as a zeroGradient on the fluid between the two cylinder. what you think about that? mad |
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September 24, 2010, 04:33 |
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#12 |
Member
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You mean the boundary condition inletOutlet with inletValue (0 0 0) on the upper surface of the cylinder?
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September 24, 2010, 04:57 |
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#13 |
Senior Member
maddalena
Join Date: Mar 2009
Posts: 436
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September 24, 2010, 12:52 |
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#14 |
Member
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September 26, 2010, 08:44 |
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#15 |
Senior Member
maddalena
Join Date: Mar 2009
Posts: 436
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October 6, 2010, 10:42 |
cylindrical shapes
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#16 |
Senior Member
Join Date: Sep 2010
Posts: 226
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Hi guys
since you worked with cylindrical shapes i thought you might help me please i need to do a flow between two parallel disks the bottom is stationary and the on eon top rotates with a certain omega. How to draw? Help please, and what about the BCs? thanks a lot |
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October 6, 2010, 15:49 |
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#17 |
Member
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Hi, T.D.!
I don't understand your geometry maybe you can draw it? It's very difficult to me to wrote on english
__________________
davidmd |
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October 6, 2010, 18:15 |
Drawing attached
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#18 |
Senior Member
Join Date: Sep 2010
Posts: 226
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hi
this is the drawing attached thanks a lot |
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October 6, 2010, 19:31 |
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#19 |
Member
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i'm not sure, but you probably should use BoundaryFOAM
And create such a geometry will be no problems. If you want me to send you an example of my case with rotating bottom. But you have to fundamentally alter it.
__________________
davidmd |
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October 7, 2010, 03:33 |
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#20 |
Senior Member
ata kamyabi
Join Date: Aug 2009
Location: Kerman
Posts: 323
Rep Power: 18 |
Hi T.D.
How are you brother? I hope you are doing very well. May be you can use gambit for mesh generation. Good luck Best regards Ata |
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Tags |
boundary conditions, simulation cylinder |
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