Hi,
let’s assume that characteristic size of area is L=1 metre.
Take the space step as dx= 10[sup]-3[/sup].
Since dt should be an order of dx[sup]2[/sup], so it is necessary to take dt=10[sup]-6[/sup] second.
Where here there is an error or it is true?
Thanks.
Hello,
this is not correct. For a given spatial resolution you can compute the time step from the viscosity. Assume a kinematic viscosity of nu[sub]p[/sub] = 10[sup]-6[/sup]m[sup]2[/sup]/s and a numerical viscosity of, say, nu[sub]l[/sub] = 0.1. The grid size is 10[sup]-3[/sup]m. According to nu[sub]p[/sub] = nu[sub]l[/sub] dx[sup]^2[/sup]/dt you find dt = 0.1s.
Timm
Thanks, I’ve forgotten about viscosity.
How to delete this topic?
Dear Timm:
May I ask why the numerical viscosity is set nu[sub]l[/sub]=0.1?
How th choose the value?
Thank you.
Regards
Peterson
Hello Peterson,
the value nu[sub]l[/sub] = 0.1 corresponds to tau = 0.8 (relaxation time in collision step). Although you can in principle use any value larger 0.5 for tau, it is wise to choose it carefully. In our simulations we have found that a value of tau = 0.8…0.9 leads to minimum deviations compared to analytic results. Have a look here. However, this result is not new. Whenever I start with a simulation, I always set tau to a value close to 0.9, except there are good reasons for not doing so. Keep in mind that you can take smaller values of nu[sub]l[/sub], but you will find decreased accuracy.
Timm
I might mention that Sukop and Thorne’s LB book noted that tau = 1.0 was the “safest”, I think that refers to the stability of the model. tau<1 gives an overrelaxation scheme where the collision step overshoots the equilibrium distribution. Timm, thanks for the note on the best tau for smallest deviations!
Sigvat
Zachary, no need to delete the topic I think, this will probably be helpful for someone.
Yes!
So good to discuss here.
I am not lonely.
Thanks Timm and all of you!
Regards
peterson
Hi
Thanks Timm and all of you!
it was very usefull