Hi out there,

I'd like to visualise the dissipation energy of a 3D turbulent flow field, calculated with CFD using the standard k,eps-turbulence model.

As far as I can the, the dissipation energy can be computed as the summ of the laminar dissipation function and the turbulent dissipation function. While the laminar disfct can be computed from the velocity gradients of the CFD results I do not know, how to calculate the turbulent disfct. The rate of turbulent dissipation epsilon seems to represent only a part of the turbulent dissipation energy.

Can anybody give me some advice ? Is there a way to assess, whether the missing energy part may be negligible ?

The same again in equations: e_diss = e_diss_lam + e_diss_turb e_diss_lam = viscosity*0.5*dissipation_function_laminar disfct_lam = (d(v_bar)i/dxj+d(v_bar)j/dxi)^2 e_diss_tur = viscosity*0.5*dissipation_function_turbulent disfct_tur = [(d(v_prime)i/dxj+d(v_prime)j/dxi)^2]_bar disfct_tur = part1 + part2 part1 = 2*[d(v_prime)i/dxj*d(v_prime)i/dxj]_bar part1 = 2*epsilon part2 = 2*[d(v_prime)i/dxj*d(v_prime)j/dxi]_bar

-> How to calculate part2 ?

I'd be grateful for any help. Max

(1) Use the effective viscosity in place of laminar viscosity (k production term is source of energy for turbulence).

(2) Alternatively, sum the work done on each cell.

Comparing (1) and (2) is a sensitive and general measure of grid dependence (though it is rarely used).

PS The k-e model assumes isotropic dissipation which can be a source of confusion.

You're right by saying that the k produktion term is the source of energy for turbulence, i.e. PRO = eddyviscosity * 0.5 * (d(v_bar)i/dxj+d(v_bar)j/dxi)^2 But should'nt you consider that locally the production of turbulent kinetic energy does not automatically imply the dissipation of kinetic energy? I mean first energy is transfered from the mean flow to the inner energy of the flowing media and in addition to that to the turbulent kinetic energy of the flow. Then after some cascades the turbulent kinetic energy also dissipates and transfers energy to the inner energy of the flowing media. But this does not happen at the same spot, meaning if you want to calculate local dissipation you have to calculate the direct dissipation from the mean flow and the dissipation of kinetic energy from the fluctuating velocities, as Max was stating.

What you describe is, of course, physically correct. The transfer of the turbulent motion to internal energy usually occurs downstream of where it was generated. In addition, energy can also be extracted from the turbulent motion and put back into the mean flow.

However, because of the eddy viscosity model used by the k-e model for the Reynolds stress, once energy has been transfered from the mean flow to the turbulence it has gone for ever. There is no mechanism for turbulent energy to be transferred back into the mean flow. The turbulent dissipation term for the mechanical energy equation is always a sink.

Perhaps I was answering the wrong question, but I had assumed the task was to plot where the useful work was being lost in the k-e prediction rather than to plot where the flow was being heated up. As you point out, they are different.