Constraint of Reynold number in LBM

Hi everyone.

I am new with LBM and I need help from you guys to answer this question.

Why does the standard LBM work only at low Re? Are there any constraint around this?

Many thanks for any help.

As far as I can tell, LBGK will work for any Reynolds number, but due to certain boundary conditions, the stability range may be constrained, thereby limiting stable simulations to low Reynolds numbers. You can cure this by choosing a finer mesh, but this will of course also increase the computation time when the mesh is uniform.

Thanks buddy. Got it.

Dear all,

It’s not only the stability but as well the derivation of the Navier-Stokes equation from the LBM which is legible for small Mach numbers. Basically by having a small Mach number you insure that the equation which the LBM simulates is indeed the Navier-Stokes equation.

Cheers,
Alex

True, the Mach number should be kept small to comply with the assumptions made during the derivations, but even at low Mach number, the Reynolds number (u_lb * N / nu_lb) can be high by having a low viscosity nu_lb. The viscosity is directly related to the relaxation frequency omega in the LBGK equation and for stability it is required that this parameter be not too close to 2.

So, for a critical value of the relaxation frequency, there will be a lowest possible lattice viscosity one can use in the simulations and thus, to increase the Reynolds number even further (having fixed the viscosity due to stability and the velocity due to the small Mach number assumption) the only way is to refine the mesh, as the typical length scale will then be represented by a greater number of nodes N.