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Laminar pipe flow using 2D axis symmetric geometry |
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March 27, 2013, 15:27 |
Laminar pipe flow using 2D axis symmetric geometry
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#1 |
Senior Member
Vishal Nandigana
Join Date: Mar 2009
Location: Champaign, Illinois, U.S.A
Posts: 208
Rep Power: 18 |
Hi All,
I would like to know, if it is possible to solve a laminar flow in a cylindrical pipe using 2d axis symmetric simulations in openfoam. If so, please can anyone let me know how to construct the geometry. Further, should we change anything in the icoFoam solver. Thanks Regards Vishal |
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March 28, 2013, 02:06 |
Hello
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#2 | |
Senior Member
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Quote:
IcoFoam is a "Transient solver for incompressible, laminar flow of Newtonian fluids". Now, if you are using a Newtonian fluid for your case then you are progressing in a correct manner and you don't have to change anything in the solver. About the geometry you can construct any geometry from 2D -to- 3D using the polyline code (especially for cylindrical geometry) in openFOAM. |
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March 28, 2013, 02:17 |
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#3 | |
Senior Member
Vishal Nandigana
Join Date: Mar 2009
Location: Champaign, Illinois, U.S.A
Posts: 208
Rep Power: 18 |
Quote:
Thanks for your reply. Can you please illustrate more about the polyline code. I would like to create a cylinder of length (say) 30 nm and diameter 5 nm. The inlet of the cylinder has a uniform velocity and the walls have no slip. How can we create such a geometry. Can you provide a sample blockmesh file for this geometry along with the patch definitions. Thanks Vishal |
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March 28, 2013, 03:20 |
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#4 | |
Senior Member
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Quote:
Try making something similar to the blockMesh attached. Best of Luck. |
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April 1, 2013, 01:48 |
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#5 | |
Senior Member
Vishal Nandigana
Join Date: Mar 2009
Location: Champaign, Illinois, U.S.A
Posts: 208
Rep Power: 18 |
Quote:
I was trying to understand the blockMeshDict file you had sent and also started looking at pipecyclic tutorial. I understand that we convert the geometry in cylindrical coordinates into Cartesian coordinates. I would like to know, what does minY correspond to. Since minY = -y = -r*sin(pi). Wouldn't y and minY become zero. Please comment on this. Thanks Regards Vishal |
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