# pressure boundary condition for a rotating wall

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March 13, 2015, 11:39
pressure boundary condition for a rotating wall
#1
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Hello everybody,
i have experienced some difficulties with the pressure boundary condition for a rotating wall, which can be explained by examining a couette flow with rotating inner wall and stationary outer wall, laminar flow, constant density.

The theoretical solution for the pressure gradient distribution is:
dp/dr= rho*u_theta^2/r
throughout the domain (see graph)

The flow setup in cfx is : symmetry boundary conditions in the direction of axis of rotation, rotational periodic boundary conditions over the circle segment, no-slip walls, stationary at the outside, rotating on the inside.
(see picture of the grid)

The problem occurs at the inner wall, where cfx takes the the normal of the pressure gradient constant across the first cell, (clearly shown on the blue line in the graph cfx_pressuregradient). This creates an error, since it is in disagreement with the theoretical solution. This problem is getting worse when the grid is refined towards the inner cylinder (red line in the graph), leading to erroneous pressure distributions (see graph)

This leaves me with two questions:

a) Is there a way to implement the right boundary condition: dp/dr = rho u_theta^2/r

b) Since the Navier Stokes equations for any rotating no-slip wall reduce to dp/dr = rho u_theta^2/r does the cfx boundary treatment always lead to incorrect results for rotating walls?

I hope i described my questions clearly, thank you in advance
Attached Images
 theoretical_pressure_gradient.png (6.9 KB, 30 views) theoretical_pressure_distribution.png (5.7 KB, 22 views) rheometer_steady_single_phase_129x129_inflationlayer_omega10_001.jpg (56.4 KB, 35 views) cfx_pressuregradient.jpg (38.2 KB, 19 views) cfx_pressure.jpg (49.5 KB, 19 views)

 March 13, 2015, 15:54 #2 Super Moderator   Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,372 Rep Power: 139 CFX assumes the pressure gradient normal to the wall is zero. This is a good assumption for moderate to high Re flow but for low Re flow it can be a problem (as you have seen). I know of no way to fix this. You should be able to minimise it by refining the mesh. Also I think you will find few, if any commercial CFD codes correctly implement this low Re pressure boundary condition as it only occurs at low Re number and few customers run at low Re number. meth likes this.

March 16, 2015, 08:59
#3
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Hi Glenn,

i have noticed that the grid refinement towards the stationary outer wall works for getting the solution more accurate.
However, at the rotating inner wall i haven't been able to get the solution more accurate.
It probably has to do with the fact that in theory, for laminar cylindrical couette flow, the second derivative of the pressure with respect to the radius does go to zero at the outer wall, but at the rotating inner wall it does not (see graph),

in this way the cfx approach (pressure gradient normal to the wall is zero) leads to a much larger error at the rotating cylinder than at the stationary one.

i think i will avoid using rotating walls for laminar flows in cfx. it is a pity there is no option for changing the pressure gradient normal to the wall to another constant than zero. I think it would not change the numerical solution procedure that much and i believe it would have solved my problem.

Attached Images
 analytical_second_derivative_of_pressure_couette_flow.png (6.3 KB, 8 views)

Last edited by barteljoris; March 16, 2015 at 12:05.

 March 16, 2015, 12:21 #4 Senior Member   Join Date: Jun 2009 Posts: 1,693 Rep Power: 30 Have you tried solving the problem in the rotating frame ? That is, the rotating wall is stationary in the rotating frame, and the stationary one is now counter rotating ? Such test may give some additional insight..

March 16, 2015, 14:50
#5
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Hi Opaque,

i tried this already, but it simply shifts the problem from the inner cylinder to the outer cylinder.(see graphs) The cfx solution in the rotating frame of reference is excellent close to the inner rotating wall, since the in the rotating frame the source term

gives me exactly the boundary condition at the inner cylinder that i want to have.
The momentum equation in radial direction in the rotating frame of reference becomes
-dp/dr + S_cfg = -rho*u_theta^2/r , so the cfx wall treatment which sets the wall pressure gradient to zero gives the exact solution at the inner wall in the rotating frame solution.

On the other side, at the stationary wall, where in the stationary frame the pressure normal gradient = zero is modelled correctly, the pressure gradient in the rotational frame becomes:
dp/dr = S_cfg=
and goes through the roof.

I cannot understand why the rotating wall in a stationary frame is treated differently than the same rotating wall in the rotational frame of reference. Why isnt the same boundary condition implemented, by modelling the pressure gradient normal to a rotating wall in a stationary frame as
dp/dr = rho*u_theta^2/r
just as in the case for the rotating frame, and make solutions in the stationary and rotational frame agree? It is just a constant value other than the value zero what it is now, so i think it wouldnt cause much numerical difficulties. What harm can it do? It seems that with the current implementation, for walls with different rotational velocities only one of them can be modelled correctly.

Isnt this problem occuring for any calculation with rotating walls with different rotational velocities?
Attached Images
 pressure_gradient_rotating_frame.jpg (39.4 KB, 10 views) pressure_rotating_frame.jpg (39.4 KB, 8 views)

Last edited by barteljoris; March 16, 2015 at 16:45.

March 16, 2015, 17:03
#6
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Glenn Horrocks
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Quote:
 It probably has to do with the fact that in theory, for laminar cylindrical couette flow, the second derivative of the pressure with respect to the radius does go to zero at the outer wall, but at the rotating inner wall it does not (see graph), in this way the cfx approach (pressure gradient normal to the wall is zero) leads to a much larger error at the rotating cylinder than at the stationary one.
Yes, that is correct.

Quote:
 it is a pity there is no option for changing the pressure gradient normal to the wall to another constant than zero. I think it would not change the numerical solution procedure that much
We cannot be sure as none of us has seen the CFX source code but I suspect it will be a major change.

That is very interesting that if you do this in a rotating frame of reference the problem transfers to the outside stationary wall.

So doesn't that mean you can work-around this problem by putting the inner rotating wall in a rotating frame of reference, the outer stationary wall in a stationary frame of reference and a GGI in the middle to connect things up? Not very elegant, but it might work and that is all which counts soemtimes.

 March 16, 2015, 17:21 #7 New Member   Join Date: Feb 2014 Location: Netherlands Posts: 4 Rep Power: 11 Hi Glenn, thanks for the quick reply, the GGI option is the solution which gives me the correct pressure boundary conditions at both sides, i didnt think about that yet, i will give it a try!