6
$\begingroup$

According to Hoffman, the standard $k-\epsilon$ turbulence model incorporates a two-layer approach (inner region and outer region formulations to reperesent mixing length) when it comes to regions adjacent to wall surface $(y^+ \approx 30 - 50)$.

My understanding of the $k-\epsilon$ model is that a two additional equations are added to the system of flow governing equations, one equation for kinetic turbulence energy $k$ and the other is for the turbulence energy dissipation rate $\epsilon$, and the closure problem is solved by using dimensionless groups as boundary conditions such as turbulence intensity, turbulence length scale and hydraulic diameter.

My question is why the $k-\epsilon$ model uses zero equations models (algebraic models) for regions adjacent to the wall, what advantages do zero equations models have over $k-\epsilon$ if by definition zero equations models assume that the rate of production of turbulence and the rate of dissipation are approximately equal and they do not include the convection of turbulence?

$\endgroup$
  • $\begingroup$ Just to be sure. When you say adjacent to the wall this means close to the wall, right?! The one of the layers (with the algebraic model) will be from the wall to $y^+$ and then comes the turbulence model, right?! $\endgroup$ – rul30 Sep 23 '15 at 18:31
  • $\begingroup$ @rul30 correct. $\endgroup$ – Algo Sep 23 '15 at 18:33
2
$\begingroup$

There are two ways of "wall treatment" both feature two layers but at different $y^+$. The general idea is that the velocity gradients in the boundary layer are so high that one would need a very high number of grid cells in order to resolve those gradients. In order to overcome that the flow close to the wall is modeled by one of the following algebraic models (see cfd-online:

  1. Low Reynolds number treatment (LRN)
  2. High Reynolds number treatment (HRN)

Both treatments are based on the Law of the Wall. Numerous experiments have shown that the boundary layer of a flat plate consists of two distinct regions. From the wall until $y^+ \approx 4$ it is called viscous sublayer with a linear velocity gradient and a logarithmic region from $y^+ \approx 30$. Both layers are joint by a region called buffer-layer.

If one chooses a LRN approach the numerical grid needs to fully resolve the boundary layer in the log-region with (rule of thumb) the first grid-cell at $y^+ =1$. The viscous sub-layer is then modeled algebraically.

In case the HRN approach is chosen the first grid-cell should be in the log-region. Both layers viscous sublayer and log-layer are than modeled algebraically.

enter image description here
from Turbulence and Transport Phenomena by Hanjalic and Wikipedia

The answer to your question has three parts:

  1. Due to the high velocity gradients in the boundary layer it is beneficial with respect to the computational costs to model the flow close to the wall.

  2. The modeling of the boundary layer makes the simulations more robust.

  3. The error introduced into the solution by algebraically modeling the boundary layer is generally small (obviously depending on the simulation itself). Your mentioned neglect of the convective transport of turbulent properties is correct (Which in certain cases will introduce an error).

$\endgroup$
-1
$\begingroup$

I know that from a CFD point of view (or any time you need a numerical solution to a system of ODEs/PDEs), zero-equation models have two major benefits: 1. they don't need to be integrated by the numerical algorithm, which reduces the computational complexity of the simulation and may improve solution convergence and numerical accuracy, and 2. they don't require any kind of state or history to be maintained (which reduces memory consumption). The cost of this simplicity is, of course, accuracy, especially in regions of high turbulence, flow separation, etc.

Depending on the problem, I usually start with k-e because it's simple and numerically stable, and only after I get a good initial solution and work out all the other bugs in the simulation do I experiment with more complex turbulence models.

$\endgroup$
  • $\begingroup$ Actually the algebraic models are by definition incomplete models, even the essence of those models (which is the assumed analogy between gas free path and turbulent mixing length) is questionable, and they are by no means more accurate than two equations models. Not to mention the assumption of equality between production and dissipation of turbulence (and disregarding the past history of turbulence in the flow) by those models which is contrary to the nature of most flows. $\endgroup$ – Algo Sep 23 '15 at 18:28
  • $\begingroup$ Yes, the accuracy of some of the simpler turbulence models leaves a lot to be desired, but what I meant is that the accuracy of the solution in terms of numerical error can be improved if you can substitute algebraic equations for some differential equations. $\endgroup$ – Carlton Sep 23 '15 at 19:05
  • $\begingroup$ @Carlton, you are right that from a numerical point of view the solution of two additional DEs will add new numerical errors. However, this is a very academic point of view. Todays 2-eq-turbulence models are very powerful. In order to get physically meaningful simulation results one should happily accept those additional numerical errors in exchange for better prediction of the flow field. Your answer might give the impression that algebraic models (in general, not only in the boundary layer) will produce usable results, which they don't. $\endgroup$ – rul30 Sep 23 '15 at 21:03
  • $\begingroup$ Whether or not an algebraic model produces usable results is entirely case-dependant. I have modeled turbulence using the Baldwin-Lomax model in cases where near-wall turbulence was much less important than the overall energy loss to turbulence. The resulting simulations were sufficiently accurate. I will edit my answer to point out that there's a trade-off between accuracy and computational cost for any turbulence model. $\endgroup$ – Carlton Sep 24 '15 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.