Hi Pulgar,

Perhaps the first thing to remember is that flows in the same geometry are equivalent if the non-dimensional numbers are the same. So, if you have incompressible flow around an object, if the Re number is the same it doesn’t matter if the fluid has the viscosity of air or water; the flow field will be the same.

Now forget about numerical parameters for the moment and just think about non-dimensionalasing the system. Take the inflow velocity U to be the characteristic velocity and the diameter of the object L to the characteristic length. The Reynolds number is Re=UL/nu, where nu is the (kinematic) viscosity.

We may as well non-dimensionalise the system so that the characteristic values are unity: divide the flow velocity u by U and the spacial dimensions x by L. The inflow velocity is now 1 in this scaled system, and the diameter of the object is 1 unit too. Since speed is distance/time you also have the scaled time. Since U=L=1, the non-dimesnional viscosity is nu=1/Re. Thus, flows in this geometry with the same Re (or same non-dim. viscosity) are the same. Of course the “physical” values are different (ie the scales are different), but the flow is not. If you want to go from non-dim values back to dimensional (physical) values you just do things in reverse: multiply the non-dim velocity by U, etc.

Since the flow is now governed totally by the Re number we may as well discretise the non-dim. system. Let’s say we want N points per unit length (ie in a characteristic length). The grid spacing is thus dx=1/(N-1). We also need to put the (unit) non-dim. velocity into the computational scale. Since the “grid speed” is c=dx/dt, we scale the (non-dim.) velocity by dividing it by c: uLB=1/c=dt/dx. The “grid Re number” is therefore Re=uLB*(N-1)/nu. nu is the lattice viscosity, if you like, and from this you can find tau.

You have to be careful of how you choose c (or equivalently, dt) because it defines the Mach number: Ma, and thus uLB, have to be small. But if you set Ra and Ma, and specify the grid size (ie, dx) you have everything you need.

In your specific example, Re=100. In the non-dim. system it doesn’t matter if the viscosity if for air or water or whatever - it’s the Re number that dictates everything (well, and the Ma number too). Think of your system above with Re=100, Umax=0.1 and Lobst=41 (say 41 for ease) as the discretisation of the non-dim. system. Your grid spacing is dx=1/40 and your timestep is dt=Umax*dx=0.1/40. The lattice viscosity is nuLB=(N-1)*uLB/Re, and the relaxation time follows simply. If you double the number of point in your unit length (ie make the obstacle twice as big) while keeping Re and Umax fixed, then what you are doing is making dx (and dt) smaller, but keeping exactly the same physics (assuming the aspect ratio is also fixed). In other words, this is a mesh refinement so you will (hopefully!) be getting a more accurate solution to the physical problem.

If you want to covert back to physical (dimensional) parameters, you just do things in reverse. You said that your fluid is water with nu=1e-6m^2/s and Re=100. Let’s say the diameter of the obstacle is L-meters. This means that the inflow velocity is u=(1e-4/L)m/s. To get from your LB predictions for velocity to physical values you thus divide them by Umax(=0,1 here) and multiply by the u just found.