How to deal with the turbulent kinetic engery
 

Hi, I am using the S-A turbulence model for the simulation of compressible flow. As is well known, the S-A turbulence model doesn't give the value of turbulence kinetic energy (k), while in the energy equation there are the terms of k. Then my question is how to deal with these terms? Ignore them,or get k from the S-A model by some special treatment? Is there a relation between k and the modified eddy viscosity which is the variable in S-A model. I will be grateful for any response and suggestion! Thanks! Maximus

Re: How to deal with the turbulent kinetic engery
 

A first approximation could be:

k = (uprime**2 + vprime**2 + wprime**2) / 2

uprime = (u_instantaneous - u_avg) ...and so on for the other components

uavg could be built in time if you're doing URANS.

Re: How to deal with the turbulent kinetic engery
 

Can you clearly define 'instantaneous' and 'average'?

In steady-state computations, if by 'instantaneous' you mean the velocity of the current iteration step, then that's not necessarily a velocity with physical meaning, since in most cases your iteration is not time-accurate. Likewise, the 'average' velocity is hard to define, it's actually a result that you expect once your steady-state computation is fully converged. You might rely on your inaccurate (nonphysical) TKE to converge along with the flow solution, but you have to be careful not to introduce new sources of instability and stiffness to your system. Does this work well in practice, and how large is the contribution of TKE to the total energy?

In unsteady computations you have the same problem as above, and in addition you will need to come up with an appropriate definition of the time-dependent 'average' velocity (small time-scale, ensemble?). How exactly do you do that?

Re: How to deal with the turbulent kinetic engery
 

"In steady-state computations, if by 'instantaneous' you mean the velocity of the current iteration step, then that's not necessarily a velocity with physical meaning, since in most cases your iteration is not time-accurate"

If velocity does not have any "physical meaning" during iterations, does anything else has to have a physical meaning during iterations?

"Does this work well in practice, and how large is the contribution of TKE to the total energy? "

Yes

"...you will need to come up with an appropriate definition of the time-dependent 'average' velocity (small time-scale, ensemble?). How exactly do you do that?"

The answer to this question comes from the answer to the question: How do you define Unsteady RANS? If you know this answer to what is Unsteady and Reynolds Averaged NS, you'll know the answer to unsteady (i.e. time dependent) 'average' velocity.

Re: How to deal with the turbulent kinetic engery
 

One question you haven't answered: How large is the contribution of TKE to the total energy in this kind of flow?

In other words, how large would be the error by neglecting TKE in the energy equation, with regard to the accuracy you expect from a low order turbulence model?

In yet other words: Is it even worth bothering about TKE in this case?

Re: How to deal with the turbulent kinetic engery
 

Also, what was left unanswered are the questions regarding the implementation. Let's say you consider the velocity of the current iteration step as the currently best approximation to your "average" velocity (which it is). Then how do you evaluate the "instantaneous" velocity? My problem is that I don't see how in RANS you could get a physical representation of the turbulent velocity fluctuation, if your turbulence model doesn't provide it. Please provide more detail on how you do that, or maybe a reference. Thanks.

Re: How to deal with the turbulent kinetic engery
 

"How large is the contribution of TKE to the total energy in this kind of flow?"

Answer to this depends on the "kind of flow". Highly turbulent flows will have a large contributions, laminar flows will have none.

"how large would be the error by neglecting TKE in the energy equation"

This is again problem dependent.. for weakly turbulent flows, you'll have no visible effect but you can have a large error for highly turbulent flows. This question is more like, although not to same scale as what happens if you ignore the diffusion term in the momentum equation.