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-   -   What if I use turbfoam instead of simplefoam what is steady turblence (https://www.cfd-online.com/Forums/openfoam-solving/59220-what-if-i-use-turbfoam-instead-simplefoam-what-steady-turblence.html)

lakeat January 18, 2008 03:13

I noticed in programmers guide
 
I noticed in programmers guide,
3.2 Steady turbulent flow over a backward-facing step.
So, my questions are
1. What if I use turbfoam instead of simplefoam?
2. How can turbulence be steady? what is steady turbulence?

Another two questions are
3. Can I use turbfoam to solve a laminar case, what will it happen?
4. Since Rans like turbFoam.c also contain the d/dt, so concerning the information we get from both RANS and LES, what's the differences between them?

Any reply is appreciated.

Daniel

hjasak January 18, 2008 03:58

Hi Daniel, Easy. You wish
 
Hi Daniel,

Easy. You wish to use turbFoam if the flow is genuinely unsteady in the mean variables. A good example would be turbulent vortex shedding behind a cylinder: Karman vortices, which are large structures are definitely unsteady, while smaller scale turbulence is "steady" in the mean. You can talk about this in terms of "separation of scales", but this is probably too advanced right now.

If you use turbFoam for a statistically steady flow, the solution will converge (= stop changing in time) and you will get the same answer as using simpleFoam. In the other direction, if you use simpleFoam on a flow which in fact is not statistically steady, the solver will "refuse to converge", which is an indication of existence of an unsteady component.

There are frther issues with numerical acuracy and discretisation errors, but more on tht some other time.

Enjoy,

Hrv

lakeat January 18, 2008 07:03

You mean 1. I can use turbFo
 
You mean
1. I can use turbFoam, and the answer is the same as using simpleFoam; But simpleFoam is faster and a better choice since it does not have to discretize or solve the temporal item d/dt.
2. backward-facing step flow in programmers guide is a turbulence flow which is statistically steady, and that is what is called steady turbulence.

Please correct me if i am wrong.

Thank you, Hrv.

Daniel

lakeat January 18, 2008 07:09

What about the last two questi
 
What about the last two questions?

1. What if I use a solver with RANS method to solve a laminar problem, what will happen? can I get the same result as with the laminar solver?

2. Since Rans solver also contain the discretization of d/dt, and then we get a series of time step result, just like LES method. So concerning the information of result we get from both RANS and LES, what are their differences? I mean concerning the information of the results.

Thanks a lot.

Daniel

hjasak January 18, 2008 12:13

Ad 1: Nothing. Choose a lamina
 
Ad 1: Nothing. Choose a laminar "turbulence model" and you're done. Have a look at the equations, ou will see the difference being only the Reynolds stress term.

Ad 2: The modelling assumptions are completely different: in RANS you are solving for mean values, whereas in LES you are looking at filtered values - the difference is fundamental.

Hrv


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