Dear All,

I am dealing with Backward facing step problem. So far what I know from the papers, the boundary conditions which should be applied is only for the velocity, i.e. using no-slip boundary condition. What about for the pressure? can anyone tell me?

Thank you in advance.

The pressure boundary conditions are derived from consis tency with the momentum equations and no-slip conditions. To derive them take the dot product of the Navier-Stokes equations with the normal to your surface and evaluate on the boundary - this gives you an equation for dp/dn which is you boundary condition.

eg. if boundary condition is on y=0 the normal is in the y(=n) direction and you have

dp/dy = nu.d^2u/dy^2 evaluated on y=0

where nu is the viscosity.

normal derivative of pressure is equal to zero at all boundaries except outflow where p=0

In general it is not zero - if the flow is nonparallel (i.e. it's not a simple pipe flow) then there will be a nontrivial vertical velocity, so that v is not identically zero, evaluating the v-momentum equation on the surface then gives dp/dy = nu.d^2v/dy^2 which is in general nonzero (note I wrote u instead of v in my previous post).

what will happen to the profile of velocity, just after the inlet boundary, and just before the outlet boundary? Is there any possibility that the v (NOT u) velocity will be negative or positive instead of zeros?

I tried to simulate on simpler domain. Instead of using domain of backward facing step, I use pipe 2D domain. And what I get is, the profile of the velocity near the inlet boundary, for u is positive, but for v, in the upper part is negative, but in the lower part is positve. And near the outlet boundary, the profile for v is on the other hand, i.e. in the lower part is negative and in the upper part is positve.

Is it really like that???

I suspect that this depends upon what your inlet velocity and oulet conditions?

If the inlet velocity is v=0 and u is the fully developed velocity then at the steady state the solution will be every equal to the parabolic profile and v will be identically zero. Any deviation from the fully developed profile will give you a nonzero value of v. Now in this case, when v is nonzero, the noslip conditions on the walls means that you will have, or expect to have, both positive and negative values of v at a single x-location. Far downstream, near the outlet, you would expect the behaviour near the inlet to persist but with smaller vertical velocities - if the pipe is long enough v will be almost zero.