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egge24 September 8, 2011 20:27

Modelling a cylinder in a tank

I'm trying to model (2DH) a cylinder (3m diameter) inside a tank (width 50m and length 250m) with a free stream of 2.5m/s.
I dont have any problem modelling and running the simulation, but once I see the results I can't see the Von Kármán eddies. I think it is something related with boundary conditions.
The pilar has a wall boundary condition type, with no slip. Something that surprise me is that the walls of the tank and the pilar aren't separate, all appear as wall. When I did the mesh I separate walls and pilar...

The mesh discretization is fine to ensure solve the scale of the eddies, each element in the wake is 1/12 of the pilar diameter.

Does anyone knows what I can do?

Thanks in advance,


Philipov September 9, 2011 09:50

1. Be sure your Re is not laminar or strong turbulent;
2. extend the domain behind the cylinder and may be you will see results;
3. Such kind of analysis require transient case to be calculated, not steady-state.

m2montazari September 9, 2011 10:08

hi eduardo,
if I understand your problem correctly, you have a 2-dimensional cylinder in a box and front is velocity inlet and back is pressure outlet and up,down are walls as well as cylinder wall. so check if in the Reynolds no. of the flow you should have vortex shedding or not? if you must have and the results dosn't show such flow, check your viscous model. if flow is laminar, choose laminar and if flow is turbulent, choose k-w or k-e. another thing in turbulent flow is checking the y+ of cylinder-wall. check that in contours from turbulence menu. it should not be higher than 300 in standard wall function and higher than 10 in enhanced wall treatment.
tell me about the results,

egge24 September 9, 2011 21:17

Dear friends,

Thanks for your replies.

To help clarify my modelling scenario:

- A cylinder (3m diameter) inside a tank (width 50m and length 250m)
- Free stream = 2.5m/s (I also tried increasing velocity but nothing happens only that the wake extends longer downstream)
- Domain mesh is 2 dimensional horizontal (there is no depth!)
QUESTION = even do I'm modelling a 2DH case, how can I introduce the depth of the tank in order to obtain vertical integrated results?

Define > Material:

- liquid water

Define > Solver

- time = unsteady
- other paramenters set as default

Define > Viscous

- K-epsilon > Standard > standard wall functions
(@ m2montazari, I don't find in viscous menu the "y+ of cylinder-wall" where is it?)
- other paramenters set as default

Define > Boundary Conditions:

- Inlet = velocity-inlet
- Outflow = outflow
- Fluid = water
- Walls = wall (shear condition = no slip)
- Default interior = interior

Solve > iterate

- time step size= 1s
- Number of time steps = 20s
- Max. iterations per time step = 1000s

I still not able to see the V-K eddies with fluent 6.3.26....

I include a solution using an environmental hydrodynamic model (
The domain has the same dimensions as stated above but with depth 10m. Pilar boundary has shear velocity = 0

m2montazari September 10, 2011 02:22

first of all I think the main problem is the timestep size you specified. 1sec for vonkarman vortex? you should use 0.1 or 0.01 sec to have an accurate solution.
y+ is a variable that depends on mesh size and velocity and viscosity. you can find it in display-contour-velocity-wall yplus and check the y+ on cylinder walls. y+ is the main thing that make change in wall treatment in turbulent models.
additionally you can try other turbulent models such as k-w sst and changing grid sizes specifically grids near cylinder. also check the grid quality(skewness and aspect ratio) and be sure that they are in acceptable range.

egge24 September 10, 2011 18:39

Thanks Mohammad,

I'm going to try your advices and I will let you know the results.



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