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SimpleFoam / Laminar flow / Why k & w are still required?

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Old   May 8, 2013, 06:27
Default SimpleFoam / Laminar flow / Why k & w are still required?
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Hi,

== 1 ==

A steady-state & laminar (Re << 100) flow case is run with simpleFoam. (Considering icoFoam is suggested to be used for only transient laminar flows in the User Guide). The turbulence modelling is kOmegaSST.

== 2 ==

I wonder why I need to set up k and w parameters in the numerical setting even though the flow is laminar?

== 3 ==

RASProperties/transportProperties/turbulenceProperties
k&Omega Settings

Many thanks in advance for any help.
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Old   May 8, 2013, 09:50
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why do you want to use the kOmegaSST RASmodel for your case?

just replace

"RASModel kOmegaSST;"

with


"RASModel laminar;"

then it should work.

regards
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Old   May 8, 2013, 10:19
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== 1 ==
Thank you very much for your answer. Simply I am confused how to properly set the laminar case. This case, however, has been computed without any error. Nevertheless, the numerical setting was not the right one to model a laminar flow as far as "k" and "omega" were input.

For a future reference for the novices like me, the right setting is -the remaining above is identical-:

RASProperties/transportProperties under "constant" directory.

In addition, "turbulenceProperties" under "constant" directory and "k" & "omega" under "0" directory are omitted.

== 2 ==

This arised another question, though. Why do I need to input "turbulence" keyword in the "RASProperties" file although I input "RASModel laminar;" ?

When I do not input this keyword, the terminal gives me an error message:

Code:
--> FOAM FATAL IO ERROR: 
keyword turbulence is undefined in dictionary
If I need to input the keyword just for completeness; why "turbulence on;" could also work well (at least in terms of numericals - the simulation can be run) ? Could anyone tell me what this means?

Many thanks in advance for any reply.
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