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shekharc December 2, 2009 01:11

Wall boundary condition in cylindrical co-ordinate

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?


ztdep December 2, 2009 01:21

Since u = 0
du/dr .NE. 0

u has gradient near the wall.

shekharc December 2, 2009 03:28

Thanks! What you are saying is surely the case. However, why does the divergence-free condition lead to that incorrect condition? What's wrong?

ztdep December 2, 2009 09:20

I think the problem is that the continuity equation is used for fluid. not for the wall

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