Hello, I am writing a finite volume code incorporating the k-epsilon turbulence model (so far), and in the source term of the k and epsilon equation the dissipation function (GEN) sometimes gets negative values in the beginning of the simulation (compressible fluid), when the residuals are quite high still. My question is, should I 'let' these negative values enter the Su and Sp, or when GEN becomes negative should I consider it as zero and do not add it in the source terms of the equations?

Thank you in advance!

George,

For a k-eps model it is very important to have a positive scheme. Instead of introducing a negative source term in the k and eps equation (on the right hand side) you can have a positive matrix coef using the following trick:

In the k - equation you have the term -eps. Suppose you multiply and divide this term with k:

- eps*k/k

This will not help much, but if you use k on two different time steps:

-eps*k(n+1)/k(n)

(where n is the previous and n+1 the next time step) you have changed the explicit term into an implicit positive matrix coeficient, namely +eps/k(n)

The same procedure can be used with the source term in the eps equation.

No more negative values!

Succes,

Cees

You might want to look at a few of the positivity preserving schemes out there, too. The work by Ilinca and Pelletier comes to mind. "Positivity Preservation and Adaptive Solution for the k-epsion model of Turbulence" in AIAA Journal Vol 36, 1998. Ilinca and Pelletier have several other papers out there too, Web of Science or some other paper database should turn up several papers. They all list most of the details of their scheme.

Basically the idea is to write equation for the log of k & e, not k & e themselves.

Another try is to just take the absolute values of the results the first few steps. If you don't need the transient data, this method sometimes works too, though I would consider this a method of absolute last resort.

Thanks for your answer.

You mean to use the k in two different time-steps (k(n+1, k(n)) or in two different iterations? Can you explain a little further this trick?

Thank you in advance!

George,

In two different time steps.

regards,

Cees