Hello all,
I am a new user to the LBM and to this forum also. I like what the community is doing to help everybody come up to a level in the LBM.
I have a problem with the whole concept of boundary conditions. First, I want to know if “bounceback” is a boundary condition or just an operation. Also, is it good to say that no slip is applied where ever there is a solid obstacle bordering a fluid interface?
Hello,
at first: the assumption that there is no slip at fluid-solid boundaries is well-motivated, at least if you are above the nanometer scale. Significant slip velocities usually only appear when you have nanometer resolution. Maybe you can also see slip on larger scales (I am not sure), but those cases should be very rare.
Second: I strongly recommend that you read some literature first. You find some selected articles[/url] and [url=http://www.lbmethod.org/literature:books]books on this website.
The bounceback scheme is an operation which emulates no slip at walls to a good approximation. However, some basic rules have to be followed.
Good luck,
Timm
Well,I have read so many of the articles and this forum is an opportunity for clarifications where I have issues.
Now, I do understand that the bounceback mimicks a no slip condition in the physical sense. Thus, is it fair to say that where there is a slip velocity between a solid wall and the fluid, that the slip (reflection of distribution fxns) boundary condition should be applied?
I have never used slip boundary conditions, but I think that you are right. Maybe someone else can help here?
Well, I got his cuoette flow matlab code but the boundary that was implemented for the stationary wall looked odd. below is an extract from the code
C6 C2 C5 ^ y
% \ | / |
% C3-C0-C1 |
% / | \ |
% C7 C4 C8 -----> x
% p1=1:XMAX;
% p2=2:XMAX-1;
temp=f(1:XMAX,1,2);
f(1:XMAX,1,2)=f(1:XMAX,2,4);
f(1:XMAX,1,4)=temp;
% temp=f(1:XMAX,YMAX,4);
% f(1:XMAX,YMAX,4)=f(1:XMAX,YMAX,2);
% f(1:XMAX,YMAX,4)=temp;
f(1:XMAX,YMAX,4)=f(1:XMAX,YMAX,2);
temp=f(1:XMAX,1,5);
f(1:XMAX,1,5)=f(1:XMAX,1,7);
f(1:XMAX,1,7)=temp;
% temp=f(1:XMAX,YMAX,7);
% f(1:XMAX,YMAX,7)=f(1:XMAX,YMAX,5);
% f(1:XMAX,YMAX,5)=temp;
f(x,YMAX,7)=l*(f(x,YMAX,5))+m*(f(x,YMAX,6));
% f(1,YMAX,7)=lf(XMAX,YMAX-1,5)+mf(2,YMAX-1,6);
% f(XMAX,YMAX,7)=lf(XMAX-1,YMAX-1,5)+mf(1,YMAX-1,6);
temp=f(1:XMAX,1,6);
f(1:XMAX,1,6)=f(1:XMAX,1,8);
f(1:XMAX,1,8)=temp;
% temp=f(1:XMAX,YMAX,8);
% f(1:XMAX,YMAX,8)=f(1:XMAX,YMAX,6);
% f(1:XMAX,YMAX,6)=temp;
f(x,YMAX,8)=m*(f(x,YMAX,5))+l*(f(x,YMAX,6));
% f(1,YMAX,8)=lf(XMAX,YMAX-1,5)+mf(2,YMAX-1,6);
% f(XMAX,YMAX,8)=lf(XMAX-1,YMAX-1,5)+mf(1,YMAX-1,6);
The moving wall is located at Ymax while the stationary wall is at y=1 (thus the complete bounceback scheme). can somebody please explain the kind of boundary that was used for the moving wall.
Regards