# Boundary Conditions for Lid Driven Cavity case (SIMPLE Algorithm)

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 April 4, 2015, 02:44 Boundary Conditions for Lid Driven Cavity case (SIMPLE Algorithm) #1 New Member   Mandeep Deka Join Date: Jun 2013 Posts: 2 Rep Power: 0 Hello I am trying to write a code for solving a lid driven cavity case in finite volume method by SIMPLE algorithm using a staggered grid. I am slightly confused about the boundary conditions! Firstly, at the boundary, the pressure gradient normal to the wall is zero. How should I implement this condition? Since there is no discrete pressure equation I am solving, should I consider that the gradient of the pressure correction term near the wall be zero?? (i.e the same condition for pressure correction term as for the pressure term!) Secondly, are all velocities at the walls considered to be zero (except the u velocity at the top wall of course)? Is it necessary to implement any interpolation technique to obtain the velocity at the grid point near the wall? Thank you!

 April 7, 2015, 08:01 Look Buddy #2 New Member   Biswajit Ghosh Join Date: Oct 2014 Location: Durgapur, India Posts: 21 Rep Power: 11 the answer is as following: suppose the boundary is as given below | | p_out | p_in | then, always set p_out(n) = p_in(n) ------> for nth time step now for next time step as it need to be p_out(n+1) = p_in(n+1), it is obvious that the correction term at both side should be equal. so, I guess now u can solve the poisson equetion easily! for the second question: if the velocity is not available in the wall you need interpolation. And for stationary wall the velocities are always 0 still confused? mail me : mebiswajithit@gmail.com