# wall velocity in Halfway bonce back

Hi Dear friends
I have solved a simple 2D channel flow and used halfway bounce back for the walls.
I used uniform grid with dx=dy=1(lu) for all nodes.
you can see shcematic for showing bounceback:(vertical nodes on the upper wall are shown by numbers)

``````1
``````
• -|- -physical boundary(upper wall)
2
|
3
:
:

by halfway bounce back:(upperwall: y=1): f(x,1,i)=f(x,1,opp(i))
and perform other steps of LB algorithm.
At the end of program we have (for example at x=somewhere in fully developed region): u(x,1)=-0.16 and u(x,2)=0.05 ,u(x,3)=0.15
But we know that the node number 1 =(x,1) is located at distance of dx/2 from the wall and u(x,1) is not real and its velocity
useless. But for presentation the velocity profile the first node we need is the wall node that is located
between node 1 & 2 in the figure and this node doesn’t contribute in our calculation.
So how we calculate the wall velocity(that should be very close to 0). Can we extrapolate it
by using u(x,2) & u(x,3)?
or we should consider the wall node in a grid with dx/2 distance between first and second node?

Is the my procedure right?

Hi freinds
It seems that my question was not clear, I repet it:
when we use halfway bounce back for the stationary wall, How can we ensure that the wall velocity will be zero(or very close to zero)?
By averaging u(1) and u(2) or extrapolate from fluid velocity(u(2) and u(3))?
In other words how can we see the exact value of wall velocity by halfway bounce back?
Thanks