# 2D cylinder Turbulent Flow

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February 20, 2017, 12:05
2D cylinder Turbulent Flow
#1
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Fredi Cenci
Join Date: Dec 2016
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Hello guys,

I am simulating a turbulent flow around a circular cylinder.
I ran 28500 steps already and the lift doesn’t converge to the experimental results. Some information:

Turbulence Model = k-w SST
Re = 60000
D = 1 m
nu = 1.5e-5 m2/s
I = 0.5%
Turbulent mixing length: D*0.07 = 0.07 m
rho = 1.225
U = 0.9

k = 1.5*(U*I)² = 0,00003
omega = Cu^(-1/4) *k^(0.5)/T length = 0,1437

I have meshed with blockMesh and a have a pretty refined mesh near to the cylinder y+ < 1.
The courant number is also generally bellow 1 (I had to reduce the time step to 0.002) . Look at the graphs.

Can someone look at my schemes and solution files? Give some tips about my set up and etc… I will try a 3 dimensional cylinder after this..

Link with the simulation (compacted with .tar): https://www.dropbox.com/s/2y5kyja3gu...on.tar.gz?dl=0

Thanks a lot.
Attached Images
 LIFT.jpg (52.6 KB, 94 views) DRAG.png (62.0 KB, 83 views) Screenshot from 2017-02-20 13-30-34.png (45.4 KB, 98 views) RESIDUALS.jpg (41.7 KB, 61 views)

 February 21, 2017, 00:09 #2 Senior Member     Uwe Pilz Join Date: Feb 2017 Location: Leipzig, Germany Posts: 742 Rep Power: 14 > pretty refined mesh near to the cylinder y+ < 1. As far as I understand the problem: The k-eps model makes some assumptions for the innermost layer, the laminar Prandtl layer. This layer reachces until around y+=30. Therefore the innermost nodal layer should be in this range (and not much finer). __________________ Uwe Pilz -- Die der Hauptbewegung überlagerte Schwankungsbewegung ist in ihren Einzelheiten so hoffnungslos kompliziert, daß ihre theoretische Berechnung aussichtslos erscheint. (Hermann Schlichting, 1950)

 February 21, 2017, 00:18 #3 Member   Fredi Cenci Join Date: Dec 2016 Posts: 38 Rep Power: 8 Thanks for your reply, So, I am not using wall functions. I am solving the viscous layer, as far as I know I should put a cell in Y+<1 so I can solve the viscous layer. Are you saying the problem is that I have too much cells near to the wall in some locations? It is hard to keep a constant number for Y+ once the velocity changes a lot around the cylinder. I am using k-omega SST..

 February 21, 2017, 01:08 #4 Senior Member   Mikko Join Date: Jul 2014 Location: The Hague, The Netherlands Posts: 243 Rep Power: 11 If you look at your experimental data, there is a sudden decrease in drag coefficient at around Re=300 000. This means that the boundary layer is mostly laminar below Re=300 000 and turbulent above it. Common turbulence models (k-omega SST, k-eps etc.) assume always turbulent boundary layer and therefore you cannot use them with low Reynolds numbers. So, try switching off the turbulence model. The mean lift coefficient for symmetric bodies such as cylinder is zero and therefore in your experimental data they give the root mean square of the lift. fredicenci and Leonardo.flores like this.

February 21, 2017, 08:32
#5
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Fredi Cenci
Join Date: Dec 2016
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Yehh, I am gonna try switch off the turbulence. I thought it was turbulent already because this picture (also from the Sumer book "Hydrodynamics around cylindrical structures").

Thank you! I will post the results when I finish it.
Attached Images
 Screenshot from 2017-02-21 10-31-22.png (69.9 KB, 69 views)

February 21, 2017, 10:51
#6
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Fredi Cenci
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Quote:
 Originally Posted by Flowkersma If you look at your experimental data, there is a sudden decrease in drag coefficient at around Re=300 000. This means that the boundary layer is mostly laminar below Re=300 000 and turbulent above it. Common turbulence models (k-omega SST, k-eps etc.) assume always turbulent boundary layer and therefore you cannot use them with low Reynolds numbers. So, try switching off the turbulence model. The mean lift coefficient for symmetric bodies such as cylinder is zero and therefore in your experimental data they give the root mean square of the lift.

The lift and drag curves are better.
Cd (mean) = 1.5... experimental is 1.25

Cl (peak) = 1.67... experimental is 0.4 r.m.s.. 0.4/0.7071 = 0.56

I don't know why the lift has such a big error... I checked the Diameter (D=1), reference area (A = 1) and reference length (L=1), everything looks fine. Any Idea?

I really appreciate the support.
Attached Images
 DRAG.png (50.7 KB, 55 views) LIFT.png (44.2 KB, 45 views)

 February 22, 2017, 00:24 #7 Senior Member     Uwe Pilz Join Date: Feb 2017 Location: Leipzig, Germany Posts: 742 Rep Power: 14 > I don't know why the lift has such a big error. As you can see in your force coefficient diagrams the solution is not stable, you have a transient case. The coefficents are changing forever. The best thing you may calculate is a time averaged value of it. __________________ Uwe Pilz -- Die der Hauptbewegung überlagerte Schwankungsbewegung ist in ihren Einzelheiten so hoffnungslos kompliziert, daß ihre theoretische Berechnung aussichtslos erscheint. (Hermann Schlichting, 1950)

 February 22, 2017, 04:13 #8 Senior Member   Mikko Join Date: Jul 2014 Location: The Hague, The Netherlands Posts: 243 Rep Power: 11 Try first copying fvScheme and fvSolution dictionaries from a tutorial case which uses pisoFoam. For instance, I think you should not use under relaxation factors with pisoFoam. Also, you cannot calculate rms from the mean value straightaway. Your mesh could be improved but probably it's good enough. I would start first with lower Reynolds number to minimize the effect of turbulence.

February 22, 2017, 06:46
#9
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Fredi Cenci
Join Date: Dec 2016
Posts: 38
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Quote:
 Originally Posted by Flowkersma Try first copying fvScheme and fvSolution dictionaries from a tutorial case which uses pisoFoam. For instance, I think you should not use under relaxation factors with pisoFoam. Also, you cannot calculate rms from the mean value straightaway. Your mesh could be improved but probably it's good enough. I would start first with lower Reynolds number to minimize the effect of turbulence.

I got the schemes and solution from a tutorial of a cylinder with lower Reynolds (Re = 100).
Yes I added Relaxation Factors because I was using k-omega SST but then I switched off the turbulence and I forgot to remove them. =)

And Yes, I probably should run more steps.. I will do it!

 September 24, 2017, 11:50 #10 New Member   Miad Al Mursaline Join Date: Nov 2016 Posts: 15 Rep Power: 8 Can you please tell me where you got experimental data for turbulent flow around cylinders?

 September 27, 2017, 07:38 #11 Member   Fredi Cenci Join Date: Dec 2016 Posts: 38 Rep Power: 8 http://proceedings.asmedigitalcollec...icleid=1732960 Here you have pretty much all you need.

 September 27, 2017, 09:08 #12 Senior Member     Uwe Pilz Join Date: Feb 2017 Location: Leipzig, Germany Posts: 742 Rep Power: 14 .. or here https://www.researchgate.net/profile...Validation.pdf without the need to pay anything. fredicenci likes this. __________________ Uwe Pilz -- Die der Hauptbewegung überlagerte Schwankungsbewegung ist in ihren Einzelheiten so hoffnungslos kompliziert, daß ihre theoretische Berechnung aussichtslos erscheint. (Hermann Schlichting, 1950)

 September 28, 2017, 13:10 #13 New Member   Miad Al Mursaline Join Date: Nov 2016 Posts: 15 Rep Power: 8 Thank you for the information

December 5, 2017, 18:10
#14
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Leonardo
Join Date: Nov 2017
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Quote:
 Originally Posted by fredicenci I = 0.5% Turbulent mixing length: D*0.07 = 0.07 m
Hello fredicenci, can you please explain me how do you choose this Turbulence Intesity and Lenght? I have similar trouble with Drag an Lift, but using Fluent 18. Thank you.