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September 22, 2015, 18:49 |
stress free boundary condition
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#1 |
New Member
Arianna
Join Date: Sep 2015
Posts: 9
Rep Power: 11 |
Hi everyone, does anyone know what is a "stress free" boundary condition? and how do you set that in FLUENT? I'm going to simulate the blood flow in a model of thoracic aorta. I've read some papers in which they talk about this stress free outflow BC to impose at the outlet. I've searched around but could't figure out what stress free boundary condition exactly is and how to impose it on fluent.
Any help would be really appreciated! thanks! |
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September 23, 2015, 04:23 |
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#2 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,896
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Quote:
it is often equivalent to set to zero the normal derivative of the velocity components |
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September 23, 2015, 17:32 |
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#3 | |
New Member
Arianna
Join Date: Sep 2015
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Quote:
I also found this kind of explanation, but it still not clear to me how I'm supposed to impose this boundary condition at the outlet of my domain. I found some people talking about the use of symmetry condition in fluent but this is not clear to me since it seems to be more related to the no-slip condition at the wall and not at the outlet... |
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December 19, 2023, 13:26 |
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#4 | |
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Matthew
Join Date: Mar 2022
Location: United Kingdom
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Quote:
[LaTeX Error: Syntax error] I have a sheet of compressible elastic material that is resting on one edge, and the others are free to move. I know that there is going to be no stress on the boundary and no shear. Would the above BC cover the conditions? |
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December 19, 2023, 13:46 |
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#5 |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
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Latex formula showing but,
Normal derivative equal to zero is the impenetrable BC for a slip-wall. If you have a free surface of an elastic solid (which I suspect you do from your bajillion other posts) then you need the entire stress tensor going to zero. For the butted end, normal derivative of velocity going to zero is okay because that is also the impenetrable BC. |
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December 19, 2023, 14:12 |
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#6 | |
Senior Member
Matthew
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Quote:
Cheers. |
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December 19, 2023, 21:12 |
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#7 |
Senior Member
Lucky
Join Date: Apr 2011
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Your case is 1D, your "entire" stress tensor is only one derivative
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December 20, 2023, 07:00 |
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#8 | |
Senior Member
Matthew
Join Date: Mar 2022
Location: United Kingdom
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Quote:
I'm thinking about a 2D compressible sheet with a free surface. I know that on that free surface there is no stress(or traction?) and I have the usual free surface equation derived by saying that particles on the free surface remain on the free surface. |
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