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September 4, 2015, 08:45 |
RANS model
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#1 |
New Member
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Hi everyoone,
I'm using a RANS model, with k-epsilon turbulence model (IH2VOF). One of the options in the model setup is to 'activate turbulence' and then specify 3 parameters (turbulence seed parameter, eddy viscosity and a parameter for boundary layer turbulence). My question is if I choose not to include turbulence what is happening to the equations? they became NS? Many thanks |
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September 10, 2015, 23:40 |
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#2 |
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Jingchang.Shi
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Location: Hang Zhou, China
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September 11, 2015, 03:40 |
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#3 | |
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Filippo Maria Denaro
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Quote:
that's not always true... without any turbulence model you have the original NS equations and turbulent flows obey to such equations...and DNS allows us to simulate turbulence from NS without any "artificial" inclusion of turbulence in the equations |
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September 11, 2015, 05:39 |
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#4 |
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Hi,
Thanks for the reply. The model is RANS but I'm not activating the turbulence option. The model runs well and of course no turbulence output files are created. However, when I plot the velocity files (sqrt(U^2+V^2)), figure in attach, vortices clear appear near the structure that I'm modeling. So my question is, since I didn´t activate the turbulence, these vortices are only due to molecular diffusion? I'm supposing that the equations remain RANS but the velocity components will not include the fluctuations (Reynolds tensor will be zero)? Another doubt:the molecular diffusion is not part of the fluctuations (I´m not sure on this)? Finally, even if vortices appear in the figure probably I'll have to conclude that these velocities are overestimated since dissipation due to turbulence was not included in the simulation? Thanks in advance for your help! |
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September 11, 2015, 05:44 |
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#5 | |
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Quote:
Please check my reply above/below |
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September 11, 2015, 05:45 |
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#6 | |
Senior Member
Filippo Maria Denaro
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Quote:
if you are using the steady 2D RANS formulation you have necessarily activated a turbulence model.... otherwise your solution has no physical meaning |
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September 11, 2015, 05:49 |
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#7 | ||
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Quote:
Quote:
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September 11, 2015, 05:50 |
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#8 |
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Jingchang.Shi
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According to definition, vortices are generated due to velocity gradients, instead of diffusion.
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September 11, 2015, 06:41 |
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#9 |
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Filippo Maria Denaro
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September 11, 2015, 06:50 |
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#10 |
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Jingchang.Shi
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I guess FMDenaro means that you need to supply some terms like mixing length scale to close the equations since RANS introduces new terms.
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September 11, 2015, 06:50 |
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#11 |
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Dear Filippo, I have posted my second reply (with a figure) but I guess you have already seen it.
If turbulence has to be supplied I still don't understand why in the model interface is available the option 'activate turbulence' |
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September 11, 2015, 06:58 |
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#12 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
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Quote:
I don't know the code you are using, it depends on what "activate turbulence" means in the user guide ... that's for sure that RANS without any turbulence modelling has no physical meaning |
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rans modelling, turbulence model |
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