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Clarification for the RANS-based turbulence model

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Old   April 27, 2020, 08:12
Default Clarification for the RANS-based turbulence model
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I have some confusion w.r.t. the involvement of Reynolds stress from RANS model and hope to seek some clarification.
Is the Reynolds stress used to damp the high frequency numerical results from the RANS solver? In that case, using the mixing length model as a basis, is it right to say that a higher turbulent viscosity translates to higher Reynolds stress which then increase dissipation on the flow?

OR

Is the Reynolds stress use to produce high frequency numerical result to the flow, thus making the flow more turbulent? Thus, the higher the turbulent viscosity (larger mixing length), the flow become more turbulent due to stronger effects from the Reynolds stress? --> This makes less sense as the turbulent viscosity is supposed to dissipate the flow. However, this interpretation stems from my misunderstanding as to how time-averaged quantity affect the flow results from RANS solver which I imagine the result to be of low frequency (less turbulent).

May I know which one is correct? Sorry for this confusion as I have never seen the code before. Will be great if someone has any sample code that may help me ease up my confusion too
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Old   April 27, 2020, 10:32
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You didn't state which model, but when you apply say a one/two-equation model, you replace the Reynolds stresses with your eddy viscosity model. After that, there are no more Reynolds stresses appearing in your governing equations but just the regular diffusive stresses + a turbulent viscosity. You solve now an equation with the mean flow as a parameter and everything has frequency of 0 Hz.

What is this high frequency numerical result you speak of? They would be purely numerical and not related to flow, whether laminar or turbulent? Or you are referring to the presence or lack of physical turbulent fluctuations?

Last edited by LuckyTran; April 27, 2020 at 11:48.
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Old   April 27, 2020, 11:08
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Let's focus on the most adopted way of doing RANS, URANS and LES (yes, I need to put them in as well to give you more context), so that we avoid going into the details of what actually RANS or LES is and if they are done correctly or not. I'll just consider what most codes actually do.

Also, for the moment, let's just focus on eddy viscosity (Boussinesq) models for turbulence, which are the norm as well in both LES and RANS/URANS. I'll talk about other models later.

1) First of all let's stay on URANS and LES. They actually solve the exact same equations, except for the turbulence model. And as we only consider (for the moment), eddy viscosity models, they just differ by the way the eddy viscosity is defined. When you move from URANS to RANS the only thing that changes, in practical terms, is that all the time derivatives are dropped from all the equations.

2) From 1, and if you know a bare minimum of how LES, RANS and URANS fields look alike, comes out that the single eddy viscosity can regulate the amount of small scales that your flow simulation will develop. I expect you to receive this information a la "no shit sherlock", because this is how turbulence also works in general with the fluid viscosity... but you never know, so better be clear. It is secondary with respect to what we want to discuss here, but just note that LES achieves a smaller eddy viscosity by bounding it to the local cell size, while in RANS/URANS it depends from global stuff and generally has much higher values. When the eddy viscosity goes to 0, you ideally recover a DNS like behavior (ideally because your underlying grid and method should be properly equipped to make a DNS, but you get the point).

3) As you aready noticed, the turbulence aspects are embedded in a scalar that is a viscosity (and for the present discussion it is indeed bounded to be positive). So, will it actually produce higher frequencies from the smaller resolved ones? Of course not. It is just that and acts like that, the bigger it is the eddy viscosity the smoother will be the field.

4) As the equations are the same for DNS, LES and URANS and they only differ by the amount of eddy viscosity, which is only dissipative, so can't actually produce "turbulence" in the way you imagine it, what happens is that it is always the non-linear convective term the responsible for creating turbulence, but the three approaches respond differently to it. In DNS we have no additional eddy viscosity, so you get as much "turbulence" the convection is going to give you before (in the sense of the scale at which) the molecular viscosity can kill it. In LES the eddy viscosity is O(1-100) the molecular one, and it is meant to give you "turbulence" down to somewhere in the inertial range. In URANS (I'll cover RANS in a second), you don't actually have any "turbulence" as you imagine it. The eddy viscosity is so high and so non-linear that it affects the average flow field without actually producing any fluctuation. Imagine it like a very slow flow of a non-Newtonian fluid. The only time variations you expect to see in URANS are those directly linked to your boundary conditions (e.g., if you slowly change them in time). Everything else you see is, one way or the other, garbage.

5) Now, if your URANS has no garbage and your boundary conditions are steady, there is no reason (in theory) you should not be able to achieve a steady state, eventually. RANS is just pretending you're lucky, so you don't use the time derivatives at all and just look for the steady state. Now, if RANS were any different from URANS (and it isn't, except for the time derivative), it should have an even higher eddy viscosity.

So, take home message: the RANS model indeed damps the oscillations (by comparison with the other approaches, you should now understand that it is a sort of minimum requirement, because the non linear term is going to produce them) but it is not actually devised for that purpose. The eddy viscosity of RANS is instead coming out from one or more quantities related to the turbulence in the domain, and for which you solve an additional equation (k, eps, etc.) or you have an algebraic relation (e.g., mixing length). In general terms, however, the higher the eddy viscosity, the more energy you can expect to be in the missing scales covered by the model. In general, however, I would avoid identifying the model with the actual Reynolds stresses, because you are tempted to see a cause-effect relation which is not there (because you are using just a model).

Secondary thing to notice: not all the turbulence models are of eddy viscosity type. In LES some of these models actually inject "turbulence" in the way you imagined it, but not all of them, and certainly not in RANS/URANS, where this never happens.
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Old   April 27, 2020, 11:52
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Previous answers already raised the issues.
I just want to add an imaginary example. Consider the historical O. Reynolds experiment of a turbulence in a pipe. Imagine to repeat the experiment for N times (N very large) and imagine you have an experimental device able to make the ensemble averaging of the N experiments.

What do you imagine to see from the averaged measure? Each single measure produces a field that is reach of a wide range of frequencies but the averaged data is not, it shows a smooth and steady field. This is the meaning of RANS that you want to obtain solving directly the steady equation supplied by a trubulence model.
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Old   April 27, 2020, 14:03
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Quote:
Originally Posted by sbaffini View Post
4) As the equations are the same for DNS, LES and URANS and they only differ by the amount of eddy viscosity, which is only dissipative, so can't actually produce "turbulence" in the way you imagine it, what happens is that it is always the non-linear convective term the responsible for creating turbulence, but the three approaches respond differently to it. In DNS we have no additional eddy viscosity, so you get as much "turbulence" the convection is going to give you before (in the sense of the scale at which) the molecular viscosity can kill it. In LES the eddy viscosity is O(1-100) the molecular one, and it is meant to give you "turbulence" down to somewhere in the inertial range. In URANS (I'll cover RANS in a second), you don't actually have any "turbulence" as you imagine it. The eddy viscosity is so high and so non-linear that it affects the average flow field without actually producing any fluctuation. Imagine it like a very slow flow of a non-Newtonian fluid. The only time variations you expect to see in URANS are those directly linked to your boundary conditions (e.g., if you slowly change them in time). Everything else you see is, one way or the other, garbage.

5) Now, if your URANS has no garbage and your boundary conditions are steady, there is no reason (in theory) you should not be able to achieve a steady state, eventually. RANS is just pretending you're lucky, so you don't use the time derivatives at all and just look for the steady state. Now, if RANS were any different from URANS (and it isn't, except for the time derivative), it should have an even higher eddy viscosity.

So, take home message: the RANS model indeed damps the oscillations (by comparison with the other approaches, you should now understand that it is a sort of minimum requirement, because the non linear term is going to produce them) but it is not actually devised for that purpose. The eddy viscosity of RANS is instead coming out from one or more quantities related to the turbulence in the domain, and for which you solve an additional equation (k, eps, etc.) or you have an algebraic relation (e.g., mixing length). In general terms, however, the higher the eddy viscosity, the more energy you can expect to be in the missing scales covered by the model. In general, however, I would avoid identifying the model with the actual Reynolds stresses, because you are tempted to see a cause-effect relation which is not there (because you are using just a model).
Thanks so much as these really help me a lot!
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Old   April 27, 2020, 14:04
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Quote:
Originally Posted by FMDenaro View Post
Previous answers already raised the issues.
I just want to add an imaginary example. Consider the historical O. Reynolds experiment of a turbulence in a pipe. Imagine to repeat the experiment for N times (N very large) and imagine you have an experimental device able to make the ensemble averaging of the N experiments.

What do you imagine to see from the averaged measure? Each single measure produces a field that is reach of a wide range of frequencies but the averaged data is not, it shows a smooth and steady field. This is the meaning of RANS that you want to obtain solving directly the steady equation supplied by a trubulence model.
Thanks for the kind response but I am quite confused by this. Performing the simulation n times before averaging all of them seems to incur a lot of computational cost. Although it allows one to use lower density mesh, is this reduction significant to produce a lower net computational cost compared to solving the problem on a finer mesh but just once?
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Old   April 27, 2020, 14:10
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Quote:
Originally Posted by cfdnewb123 View Post
Thanks for the kind response but I am quite confused by this. Performing the simulation n times before averaging all of them seems to incur a lot of computational cost. Although it allows one to use lower density mesh, is this reduction significant to produce a lower net computational cost compared to solving the problem on a finer mesh but just once?



I am not talking about numerical simulations but about the physics you see in each real experiment (with a wide range of high frequencies) and the result you get from the ensemble averaging of real experiments (with no range of frequencies). RANS means just that.
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