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simpleFoam - Convergence problem - Simple rectangular prism domain

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Old   April 26, 2013, 18:10
Default simpleFoam - Convergence problem - Simple rectangular prism domain
  #1
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Canakkale Dardanelspor
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Hi

There is a convergence issue that I suspect I had done something wrong with the boundary conditions.

=== 1 ===
Brief description of the domain:

Domain is a rectangular prism like a simple channel through which fluid flows. All lateral surfaces are set to slip boundary condition. Fluid flows from inlet to outlet.

Brief description of the numerical setting:

simpleFoam is used with k-omega SST turbulence model.

Simples!

=== 2 ===

Mesh is generated in GAMBIT. Exported with ".msh" extention. Converted with "fluentMeshToFoam".

checkMesh.log

=== 3 ===

RASProperties
transportProperties
controlDict
fvSchemes
fvSolution

In my opinion, the setting is OK.

=== 4 ===

Fluid flows at a constant speed of 1.73 m/s from inlet to outlet.
The "0" directory files are as follows:

U
p
omega
k

=== 5 ===

Residuals up to 400-time-step
2 representative time-iteration detail from log file

The velocity field is constant through entire volume of the domain. However, the pressure field varies towards outlet. It means to me, considering dynamic pressure is constant due to constant speed, the static pressure changes, IMHO.

Plus, in my guess, there is something wrong with the streamwise velocity predictions as the number of iterations shoots up to 1000 at each time step:

Code:
smoothSolver:  Solving for Uz, Initial residual = 0.000153982, Final residual = 7.208e-05, No Iterations 1000
=== 6 ===
I have found some other forum pages which consider the same error message in a slightly different context. Therefore, I somehow couldn't adapt the given answers to my case.

I appreciate any help.

Many thanks in advance.
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Old   April 29, 2013, 05:00
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Nima Samkhaniani
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which version of OpenFOAM do you use?
it seems there is a bug for simpleFoam at 2.0.1, so i suggest to update your version, if you had, then it would be solved!
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Old   May 7, 2013, 14:08
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Canakkale Dardanelspor
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Many thanks for your reply.

I have tried "fluent3DMeshToFoam" and it didn't work as well.

The version I am using is 2.1.0.

I suspect with the mesh itself. So that if anyone wants to have to look, I uploaded it to the post.

Rectangular prism GAMBIT mesh
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Old   May 8, 2013, 15:14
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Canakkale Dardanelspor
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Hi,

== 1 ==
I found the reason why I have been obtaining such a high level of residuals.

I have been using the script from "how to plot residuals". It seems the default script has been plotting the initial residuals rather than the final residuals. (In the script, -f13 needs to be used instead of -f9 to plot the final residuals.).

== 2 ==

This muddling, however, contributed to an interesting observation to me.

I have made 2 identical rectangular prism grids in GAMBIT and with blockMesh.

The grid has (height x width x length) = ( 10 x 10 x 100) and meshed with a node per 1m.

Case is simply steady-state and laminar flow where Re = 10 (as the characteristic length is 100).

This is the residual plot from GAMBIT grid case.



This is the residual plot from the blockMesh grid case.



I found it interesting and wonder why? Would anyone could comment on it? Otherwise, let's note it as an experience and leave it.

If anyone would glance at so that I uploaded both computed cases: GAMBIT&blockMeshGrid_Laminar_simpleFoam.

Many thanks.
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Old   May 9, 2013, 02:06
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Ali
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Dear Hakiki

Did you checked a positive initial value for the pressure?
I mean although you are running an incompressible flow, setting p=0 may cause some problem.
I am working on partly similar problem but compressible flow using rhoCenteralFoam. I have wrote some more details here.
Do you have any idea for me?
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Old   May 9, 2013, 05:59
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HECKMANN Frédéric
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Dear Ali, you cannot compare a compressible and an incompressible solver. The compressible solvers use the absolute pressure value while the incompressible solvers use the gauge value (or relative if you like).

Therefor, a compressible solver will crash if you set p=0 but not an incompressible solver as p=0 is actually equal to p=101325 (or whatever other pressure because the solver is independent of the ambient pressure).
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