# Slip Boundary Condition for Moving Boundary

 Register Blogs Members List Search Today's Posts Mark Forums Read

 November 11, 2005, 07:12 Slip Boundary Condition for Moving Boundary #1 Shukla Guest   Posts: n/a Hi, I am testing my Euler solver for moving cascade case, and have some doubts about the moving wall slip boundary condition. It u and v are x and y velocities and grid moving with velocity 'grid' in y direction, I am assuming that normal velocity in relative frame is zero, is this correct or I have take it in absolute frame. Any guidance would be helpfull. Thanks Shukla

 November 11, 2005, 07:51 Re: Slip Boundary Condition for Moving Boundary #2 ag Guest   Posts: n/a The proper condition is zero mass flow through the boundary. Thus relative to the boundary the fluid has the same normal velocity as the boundary. If the boundary is fixed the normal component of velocity of the fluid is zero.

 November 11, 2005, 08:49 Re: Slip Boundary Condition for Moving Boundary #3 Shukla Guest   Posts: n/a Can you tell me the mathematical formulation, at present I am using Vn = u.nx + (v-vgrid).ny, grid is moving only in y direction. To get the dummy cell variables I use Vdummy = V - 2Vn. Is this correct. Thanks Shukla

 November 11, 2005, 15:02 Re: Slip Boundary Condition for Moving Boundary #4 Mani Guest   Posts: n/a If you are using dummy cells, the formulation is very simple. For a steady boundary you used to specify the dummy velocity such that the average normal velocity on the boundary is zero, and the average tangential velocity is the same as in the interior cell. The flow velocity Vb at the boundary is then Vb = ut*nt +0*nn, where ut is the tangential velocity component of the enterior cell, nt, nn are unit vectors in tangential, normal directions. The dummy velocity Vd to provide this average velocity is always defined by solving Vb = (Vd +V)/2 for Vd, where V is the flow velocity of the interior cell. In the general inviscid case, the normal flow velocity must be equal to the normal velocity of the boundary (which is indeed 0 in the steady case): Vb = ut*nt +ubn*nn, where ubn is the wall velocity (!) in normal direction. As before, Vb = (Vd +V)/2 is solved for dummy velocity Vd. You can figure out for yourself if this is consistent with your formulation.