Hello,

to calculate the convective fluxes of the momentum equations within the finite volume method one can often choose between upwind (1st order) or central (2nd order) or have a mix. So if I understood it right higher order means faster convergence to grid independent solution. (Right ?) Now my question. Has the choice of the differencing scheme the same influence concerning the k and eps equations ? Are there some numerical reasons why the differencing scheme in the turbulence equations is less important than in the momentum equations ?

Cheers,

Ron

There is a physical reason - the k and epsilon equations are very source-term dominated and hence the accuracy of the convective & diffusive parts of k and epsilon is often not as critical as for the momentum equations. You can often run k and epsilon with plain 1st order upwind without loosing much accuracy. The scheme for k and epsilon is important in some applications though, one typical example is transition-predictions.

Please excuse my perhaps somewhat ignorant question, but is it really possible to predict laminar-turbulent transition with k-e models?

Lars Ola

It is possible to predict by-pass transition with two-equation models. By-pass transition is caused/triggered by diffusion and convection of free-stream turbulence into the boundary layers. It is not possible to predict natural transition caused by growing instabilities in the boundary layers with a two-equation model.

However, in practise it is very difficult to predict by-pass transition with two-equation models. You can get good results on simple cases like flat-plate flows, but for more complex engineering flows you are usually not able to predict by-pass transition. If you want to predict by-pass transition you need a very good grid and you need to use good numerics. A typical example where it is very difficult, if not impossible, to predict by-pass transition with two-equation models is turbine blades (where the transition location is very important). Yet you can find many papers where people claim to predict transition on turbine blades.

I think that this source term dominance argument has been used before but may not be justified. The practice of using a first order scheme on the k-eps equations is more out of a need to avoid wiggles that result due to the high gradients in these functions when solved on the same grid density as for velocity. If wiggles do occur the potential exists for negative values which are both physically impossible and will destabalize the momentum equations by giving a negative viscosity. Some suggestions have been made that the k-eps should infact be solved on a finer grid than the momentum/continuity equations. See Ferziger and Peric

Further, most people wind-up blaming the turbulence model itself to explain discrepancies with experiments and may not have solved the modeled equations accuratly. It is noteworthy that many comparisons of "suposedly" the same turbulence model as applied by different authors show discrepancies of the order of those between different turbulence models. Maybe the discretization is playing a part here.

Regards,

Duane

Interesting comments. Can you give a good argument as to why you should use a finer grid for k and epsilon? Is it due to accuracy or due to stability?

I've done a lot of turbulence model studies for turbine blade heat-transfer computations and from my experience the accuracy of the scheme you use for k and epsilon is not that critical for the overall solution. What is critical though is that the scheme is stable enough. If you have numerical instabilities in your k and epsilon solutions that will of course affect the overall solution badly. I found that a quite simple 2:nd order TVD scheme is sufficient for k-eps, while using a more accurate 3:rd order scheme for the other equations.

Hi Jonas,

It is clear that since k is proportional to u^2 and epsilon (du/dx)^2 both are more rapidly varying that the basic variables. Assuming you design your grids for the most rapidly varying (in space i mean) primary variable, it will fall short for k and eps. If you for example plot the Fourier spectrum of k then there will be wavenumbers greater than for say u.

Personally, I would agree with you over the impact of this improvement over the primary varible. It might not be significant. But, people have already misunderstood the purpose of k-eps models and are forever trying to match real turbulent statistics using them and not just focusing on the question "Is the mean velocity acceptable?"

chidu... PS: I take this chance to thank you for this site, as it has been of immense use to me for around 5 years now.

Hi Chidu, thanks for your response... I think I have something new to learn here. I have a one question - is the "u" you are referring to the instantaneous velocity or the Reynolds-averaged velocity?

PS.. Thanks for the nice words about CFD Online. I'm glad that you find it useful.

hi Jonas,

Well I was a little casual about that. The point is k is the sum of the square of all Fourier modes of u (the instantaneous velocity) except for the mean. So in that sense it is a more rapidly varying function than umean. For example take the variation of k near the wall. It changes orders of magnitude that too in a very short interval of yplus. Suppose you want to capture two variables on the same grid. One is a super smooth function and the other a discontinuous one. The grid should then be chosen according to the discontinous variable requirements right.

My observation has been that mostly in routine calculation people make the grid thinking about the mean velocity and hence might not be totally appropriate for k or eps. I got the same impression when I was working in a company as expressed by the previous poster about solving it in a finer grid. But then never got down to trying anything.

regards, chidu...

(1). I think, you are on the right track. (2). My only suggestion to those who are solving turbulent flows is: MAKE SURE THAT YOU HAVE SOLVED THE 1-D FULLY DEVELOPED TURBULENT CHANNEL FLOWS AT SEVERAL REYNOLDS NUMBERS, USING A LOW REYNOLDS NUMBER MODEL AND OBTAINED MESH INDEPENDENT SOLUTIONS. (3). I would say 99% of the users have not done this necessary step. All of the questions will be answered when you are through this exercise.(4). The mesh independent solution of the fully-developed turbulent channel flow with a low Reynolds number model is only the first step.

I understand your argument in boundary-layers close to walls - the rapid variation of both k and epsilon close to walls is what is "dimensioning" for how fine grid you need there. Is the same argument also valid in non-wall-bounded flows in general?

Hi Jonas,

Now this is a tough one! Lets take a mixing layer such that the mean profile is say some kind of tanhyp(y). In this case the k peaks at the center and is an even function of y. I dont know, but this can be checked as to whether k has larger gradients than U. In fact like you mention, the production term for example is quite a crazy non-linearity which can quickly go from zero to very large values ((dU/dy)^2) and likewise epsilon which tries to match the production.

Looks like it could be true in non-wall bounded flows too.

chidu...

But of course, if one obtains a grid-converged solution for all the unknowns then there is nothing more that can be done. My only point then is that in order to speed up this procedure, one needs to pay due attention to the expected variation of k and eps.

I am quite sure that in most industrial applications a grid-convergence study is not performed. Generally due to time and resource constraints. In such situations the better the initial grid the better the solution.

regards, chidu...

Hi Jonas,

A good argument for using a finer grid (or a better match of node number and distribution to the variations in the k/eps field) has already been posted: the k and eps fields can vary more rapidly in space than the mean velocity and scalars. The variations are most extreme near walls and in free shear layers.

Is this for better accuracy or stability? Well I guess both. If the scheme is prone to dispersive errors (wiggles) in regions of high gradients then it is the lack of stability that has forced people to drop to UDS. But just like the long period that it took to make everyone aware of the dangers with UDS on the primary variables...because the solution looks believable with UDS and it will converge with iterative solvers....the errors in the k-eps variables may not have been addressed.

If you think a bit about the way people often do a grid convergence study it is not as obvious what the effect of a first order error in the k-eps will cause in the solution.

If you have a solution on one grid, plot u,v,w,p,T etc. and then double the nodes in each dimension, plot and compare. How do the discretization errors in these coupled nonlinear equations interact. Well if you have a low order scheme for the k-eps, which results in an error in the eddy viscosity...how does this affect the mean velocity field? Would we see the difference on one grid refinement or would we likely say it looks about the same....grid independant?

Further, and probably the strongest argument against UDS, is that when we are on a coarse grid and a long way off of the assymptotic convergence region, UDS is not usually monotonic. So the error can increase on one grid refinement, then decrease, then increase again...maybe back to the level of the first grid! This has been documented by a number of people, I saw a good one at a conference by Pat Roache and it is covered well in Ferziger and Peric's text.

Regards..................Duane

This is all a moot point. It's a model in which the turbulent viscosity dominates over the numerical viscosity that one gets using upwind differencing for the convection terms. You don't need > 1st order differencing for this sucker.