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 Soham June 7, 2012 10:32

Turbulence vs Laminar flow

Hi everyone,

I am a novice in CFD so my question might sound very basic.

We have been trying to run a CFD model of vertical flow between two parallel plates. The aim was to model the pressure drop and compare it with our plant observed value.

When a k-epsilon turbulence model was used, the pressure drop predicted was much higher than we observe but when a laminar model was used, the pressure drop was very similar to what we observe. As you will agree, these are expected results.

My question is that why should CFD generate two very different results for the same system? If the actual system in the plant is being subjected to laminar flow, my expectation is that even if a turbulence model is used to describe it, the effect of turbulence should be negligible and the value thrown up by the turbulence model should be similar to what is observed in the plant ?

Thank you

 FMDenaro June 7, 2012 11:58

Quote:
 Originally Posted by Soham (Post 365279) Hi everyone, I am a novice in CFD so my question might sound very basic. We have been trying to run a CFD model of vertical flow between two parallel plates. The aim was to model the pressure drop and compare it with our plant observed value. When a k-epsilon turbulence model was used, the pressure drop predicted was much higher than we observe but when a laminar model was used, the pressure drop was very similar to what we observe. As you will agree, these are expected results. My question is that why should CFD generate two very different results for the same system? If the actual system in the plant is being subjected to laminar flow, my expectation is that even if a turbulence model is used to describe it, the effect of turbulence should be negligible and the value thrown up by the turbulence model should be similar to what is observed in the plant ? Thank you
I think you need some more reading on turbulence/modelling issues .... first, there is no "laminar model", but only a NS system of equation that can give you laminar solution or turbulent solution depending on the problem. The turbulent model is supplied when you have an "unresolved" situation in which you are not able (or you don't want) to solve all the scales of the motion. Actually, the term "model" in turbulence is similar in RANS/URAN and LES but the meaning is very different. I suppose you are using RANS, that means that the variable you solve is a statistically steady-state function. That means that is not automatically true that the turbulence model "disappears" when the solution is laminar ....
The problem is that the in RANS approach all the scale of the motion are "affected" by the action of the model...

 Soham June 7, 2012 12:41

Re:

Filippo

Thank you for the answer. I agree; definitely need to read up.

But help me understand; you mean to say that the difference in the pressure drop values between the laminar & turbulent form of Navier-Stokes is due to the fact that the turbulent form includes scales of motion not considered by the laminar form (from your last line).

Also, when you say scales of motion , are you referring to local/bulk velocities like eddies, vortices etc.

 FMDenaro June 7, 2012 12:57

Quote:
 Originally Posted by Soham (Post 365303) Filippo Thank you for the answer. I agree; definitely need to read up. But help me understand; you mean to say that the difference in the pressure drop values between the laminar & turbulent form of Navier-Stokes is due to the fact that the turbulent form includes scales of motion not considered by the laminar form (from your last line). Also, when you say scales of motion , are you referring to local/bulk velocities like eddies, vortices etc.

As "characterisc scales of the motion" I mean the broad range of vortical energetic flo structures that you can see in the physical space and you can also locate in the energy spectrum depending on the wavelenghts.

My focus point is that using the "laminar equations" (that I call simpy NS equations) you have a certain class of deterministic solution. Using the RANS equations your solution is different, it satisfies a modified set of equation and is a statistical solution. Depending on the type of modelling the statistical solution in RANS can driven to modify a laminar solution. Of course that means that the model is not able to "understand" the fact that there are no fluctuations at all. This was one of the reasons to switch to LES formulations many years ago.
I don't know if your case is model-dependent or other problems can be due to the coding. I suggest testing the classical channel flow problem (laminar and turbulent Reynolds number) and check for the pressure drop with and without modelling. In turbulent channel flows, the statistically averaged pressure gradient depends on the average of the fluctuations, too.

PS: for compressible flows, also the pressure needs modelling ...

Hope to have someway helped...

 Martin Hegedus June 7, 2012 13:17

The difference between the NS and RANS equation is, at least to first order, the introduction of eddy viscosity to model turbulence. In the RANS equations the viscous term is modified to include eddy viscosity, in other words, mu(RANS) = mu(lam) + mu(turb). For the NS equations, mu(NS) = mu(lam). If mu(turb) goes to zero, then NS and RANS will give the same result, at least for the linear model. (The nonlinear models have terms which are a function of the kinetic energy.)

However, if you specify a turbulence model then mu(turb) will not be zero next to the wall. One of the things to try is to plot the eddy viscosity for your problem.

 Soham June 7, 2012 16:57

The eddy viscosity values are around 0.005 - 0.007 Pa.s I do not think they are insignificant though I do not know how to determine how much of an effect they have. Any thought?

Another related question: In the k-epsilon model, there is a collection of five constants to determine turbulent kinetic energy and dissipation rate. Do they also serve the purpose of damping ? In other words, can I choose the constants in such a way so as to come close to the value predicted by laminar form of NS i.e. can the pressure drop reduce by varying the constants? For you to know (maybe I should have told this before), the CFD simulation is only over a 4" length. I am extrapolating this 4" to 144" by multiplying it. That might work for laminar but listening to you guys, is it possible that the eddy viscosity is being over predicted in the small 4" range but might actually be less over the entire 144" length of the plate I am trying to model. a

 FMDenaro June 7, 2012 17:01

normalize the eddy viscosity by the value of your molecular viscosity ... this way, you have idea of the effect of the model

 Martin Hegedus June 7, 2012 17:32

I concur with FMDenaro.

In regards to the second part, I can't say anything about the coefficients. However, you should make the length longer. This will probably add more eddy viscosity into the problem but may distribute more of it to the center line. The velocity profile will be determined by the distribution of the eddy viscosity. Hypothetically, if the eddy viscosity is distributed uniform from wall to wall, then you would get a laminar profile. However, the boundary conditions force the eddy viscosity to go to zero next to the wall, so you'll never get a completely uniform result. But, maybe, the gradient from wall to wall won't be as severe further downstream.

 FMDenaro June 8, 2012 03:24

What I think is really complicating the task is the fact that we must not see "turbulent vs laminar flows" because laminar, transitional and turbulent conditions often cohexist in the flow... this makes RANS coefficients very difficult (if not impossible) to be properly tuned ...

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