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April 2, 2013, 13:07 |
Dump question about turbulence viscosity
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#1 |
New Member
Arthur Piquet
Join Date: Mar 2013
Posts: 18
Rep Power: 13 |
Hi,
I've got a question about RANS equation. It's maybe a dumb question but I've to make sure that I understand the concept. http://en.wikipedia.org/wiki/Reynold...okes_equations So, because of the non-linearity of the convection term in NS equation and of the mesh size, this term won't be solve accurately enough for a good solution. In RANS modeling we try simulate this lost in the convection term by a term (ui'uj') with a turbulent viscosity coefficient (calculate by k-eps for example). But, what I don't understand is that if you have an inviscid flow, (or EULER eq.) you will still need to resolve this term (with k-eps)!! A long time ago I was thinking that the ui'uj' term that we want to simulate, was coming from the tensor term and not the convection term.... so now I don't understand anymore! Sorry for my bad spelling and english, I'm french!! Thanks |
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April 2, 2013, 13:31 |
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#2 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,833
Rep Power: 73 |
Quote:
some suggestions to help in your understanding: - in RANS v' is the residual defined by v(x,t) - <v>(t) and does not depends on the mesh size but depends on the statistical average. - for inviscid flows, one can not speak of "real turbulence" since the energy cascade is never terminated by the physical dissipation. However, the physical mechanism that onset turbulence is retained in the quadratic non-linear term. You can not resolve all the scales inifinitely generated by the non linearity therefore you have to model the behavior behind some threshold. This means to se some turbulence model also for inviscid flows ... |
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April 2, 2013, 13:55 |
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#3 |
New Member
Arthur Piquet
Join Date: Mar 2013
Posts: 18
Rep Power: 13 |
Thank you, It's more clear now.
But it's because of the mesh size that we have to consider this additional term ? In DNS, the mesh is enough thin to calculate directly the convection term and so not use a turbulence model. Just what happens if you use a DNS mesh with a RANS case? The turbulence term will tend to zero? so what's the difference with LES model? Thanks anyway! |
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April 2, 2013, 14:05 |
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#4 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,833
Rep Power: 73 |
Quote:
In other words, DNS is not realizable for inviscid flows... |
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April 11, 2013, 19:15 |
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#5 |
New Member
Arthur Piquet
Join Date: Mar 2013
Posts: 18
Rep Power: 13 |
I've got another dumb question about k-eps model for example.
Let's say a flow perfectly non-turbulent reach a flat plate wall. According to the theory, the fluid will be laminar until Re~300,000 when it will be transitory and then fully turbulent. Now, I want to simulate this with k-eps and without wall treatment. My initial condition for k and eps are 0 right? (non turbulent flow). So, if I take the k equation and epsilon equation and I look up all the term inside, I don't understand where did the production of turbulence come from? because all the term are equal to 0 (because k=0 and eps=0). So I was thinking, maybe the source term produce something near the wall? do you know the actual equation for this (without wall treatment)? Thank you and sorry again for my spelling. |
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April 12, 2013, 03:21 |
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#6 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,833
Rep Power: 73 |
I don't think is correct to simulate a transitional flow with RANS formulation...
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April 13, 2013, 08:46 |
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#7 |
Senior Member
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April 13, 2013, 08:54 |
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#8 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,833
Rep Power: 73 |
Quote:
In my idea there is no sense in a statistical RANS formulation wherein you have simultaneously laminar, transitional and turbulent conditions. But someone else could disagree ... |
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