Wall boundary condition in cylindrical co-ordinate

shekharc · December 2, 2009, 01:11

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!

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

December 2, 2009, 01:11	Wall boundary condition in cylindrical co-ordinate	#1
shekharc New Member Chandra Join Date: Jul 2009 Posts: 5 Rep Power: 17	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!

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
How to set rotating wall as boundary condition	babu	FLUENT	7	December 14, 2013 05:48
Change wall boundary condition in multiphase?	Jacques	FLUENT	0	November 27, 2008 07:26
Wall boundary condition – UDF	Ssn	FLUENT	0	January 25, 2008 04:43
Variable temperature wall boundary condition	Virginie	FLUENT	2	October 2, 2007 10:23
the boundary condition of permeable wall	lisa	FLUENT	12	May 5, 2005 21:58

December 2, 2009, 01:21		#2
ztdep Senior Member p ding Join Date: Mar 2009 Posts: 427 Rep Power: 19	Since u = 0 du/dr .NE. 0 u has gradient near the wall.

December 2, 2009, 03:28		#3
shekharc New Member Chandra Join Date: Jul 2009 Posts: 5 Rep Power: 17	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
ztdep Senior Member p ding Join Date: Mar 2009 Posts: 427 Rep Power: 19	I think the problem is that the continuity equation is used for fluid. not for the wall