# doubt in B.C

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 November 14, 2008, 12:26 doubt in B.C #1 student Guest   Posts: n/a Sir, I have one doubt the boundary conditin given at the walls eg bottom wall u[i][j]=-u[i][j+1] and top wall u[i][j]=-u[i][j-1] please what is the name of this boundary condition and when it should use ?

 November 14, 2008, 14:51 Re: doubt in B.C #2 mettler Guest   Posts: n/a if it is a wall and the no-slip BC is used, there is no velocity. I have used the u[i][J] = u[i][J-1], where J = max(j) as an exit boundary condition for the flow assuming fully developed flow

 November 14, 2008, 20:03 Re: doubt in B.C #3 Paolo Lampitella Guest   Posts: n/a This seems to be a case with a staggered structured grid; in this case the wall parallel component of velocity has the first and the last component (j=1 and j=ny+2) which are outside of the domain and are used exclusively to enforce the no slip condition; in your case the bottom wall is exactly between the j(=1) and j+1 nodes, the top wall is between the j(=ny+2) and j-1 nodes (with ny i mean the number of grid points which are actually inside your domain). This is because in a staggered grid the velocities are not defined in the centers of the volumes but on different faces of the volumes so, actually, you can't say that on a wall the velocity is zero and you need some arbitrary points outside your domain to enforce this. This is of big concern also for the outflow condition, even if in a more subtle way. In this case, obviously, is the v velocity component which is affected.

 November 15, 2008, 00:53 Re: doubt in B.C #4 student Guest   Posts: n/a Dear Paolo Thank you for your nice explanation. Yes it is staggered grid. I didnt understand how u[i][j]= - u[i][+1] ensures noslip boundary condition.

 November 15, 2008, 06:44 Re: doubt in B.C #5 Paolo Lampitella Guest   Posts: n/a Actually this is the simplest way you can do it (but i don't know if there are more sophisticated ones...i guess yes). In your case the bottom wall is located at j+1/2 so you want to set the value of the velocity at j such that: 1/2 * [ U(j) + U(j+1) ] = Uwall that is a simple linear interpolation between the nodes j and j+1. With Uwall = 0 your case is recovered.

 November 23, 2008, 03:40 Re: doubt in B.C #6 student Guest   Posts: n/a Thank you so much for a clear reply. I understood.

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