# Wall boundary condition in cylindrical co-ordinate

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 December 2, 2009, 01:11 Wall boundary condition in cylindrical co-ordinate #1 New Member   Chandra Join Date: Jul 2009 Posts: 5 Rep Power: 10 Hi, I am a bit confused about application of the wall BC at solid walls in cylindrical co-ordinate on staggered grid. For the imaginary cells, the lateral velocity components v,w are set as negative of the adjacent inner cell to keep these zero on the wall. The normal component u of this imaginary cell is determined from "Div = 0" at the wall. i.e. DEL.V = 0 on the wall. However, v,w = 0 on the wall. So, this becomes: [d(ru)/dr] = 0 ---- (1) If r_i defines the cylindrical boundary and we discretize this, it gives: u_(i+1) = [r_(i-1) / r(i+1)].u_(i-1); which is quite right. However, if we take the equation (1) further, it gives: du/dr + u/r = 0. Since u = 0 at the wall, it becomes, du/dr = 0 i.e. u_(i+1) = u_(i-1), which is wrong in cylindrical co-ordinate. Where is the mistake? Thanks!

 December 2, 2009, 01:21 #2 Senior Member     p ding Join Date: Mar 2009 Posts: 337 Rep Power: 11 Since u = 0 du/dr .NE. 0 u has gradient near the wall.

 December 2, 2009, 03:28 #3 New Member   Chandra Join Date: Jul 2009 Posts: 5 Rep Power: 10 Thanks! What you are saying is surely the case. However, why does the divergence-free condition lead to that incorrect condition? What's wrong?

 December 2, 2009, 09:20 #4 Senior Member     p ding Join Date: Mar 2009 Posts: 337 Rep Power: 11 I think the problem is that the continuity equation is used for fluid. not for the wall

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