# High Re Flow over Cylinder

 Register Blogs Members List Search Today's Posts Mark Forums Read

 May 26, 2009, 01:56 High Re Flow over Cylinder #1 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Hi, I am trying to solve a Re ~ 10^7 flow around a circular cylinder. Based on reviews and talks in forums I chose k - omega SST model. From Coefficient of Drag Vs Reynolds number plots, I should be getting a Cd value of around 0.7. I am using this as a check if my turbulence model is predicting boundary layer separation properly. I refined the grid upto min(y+) =20. But The average value of Cd that I get is around 0.35. I tested the same grid with k epsilon and I was getting 0.72. Is K omega SSt highly dependant on freestream turbulence? If yes how do I arrive at the inlet turbulence quantities? Someone please shed some light. Thank you. Prapanj Last edited by prapanj; May 26, 2009 at 02:31.

 May 26, 2009, 02:01 Additional Info #2 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Let me also provide the problem boundary conditions etc... The dimension of the cylinder is 14 m diameter. I am doing the simulation for a height of 28 m ( hence the grid is also 28 m tall with top and bottom - symmetry boundary conditions) . The upstream velcity is 45 m/s, T.I 0.07. I calculated k according to T.I^2 x U^2 x 1.5 = 14.884 and the omega value is " Cu^(-0.25) x k^(0.5) / l " . I used 0.09 for Cu . Since k-omega is mostly recommended for these kinds of problems, I am sure I am doing something wrong. Please guide me. thank you Prapanj Last edited by prapanj; May 26, 2009 at 02:31.

 May 26, 2009, 04:38 #3 Member   Anton Lyaskin Join Date: Mar 2009 Location: Samara, Russia Posts: 66 Rep Power: 10 Usually all version of K-Omega model are Low Reynolds (High Reynolds modifications are also exist, but "original" K-Omega and K-Omega SST are not High Reynolds!), so you need to refine down to Y+=1

 May 26, 2009, 10:08 #4 Senior Member   N/A Join Date: Mar 2009 Posts: 189 Rep Power: 10 I guess you employ wall function at y^+=20 as BC. The wall functions is problematic for this problem since you have flow separation. Did you first match the Strouhal number ? K-omega problem may introduce slight grid dependence but it is not gonna have such a drastic effect on the computed results.

 June 5, 2009, 00:10 Strouhal Number #5 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Hi Harish, Could you please explain to me what you mean by matching the Strouhal Number? Because for my case I was getting the value (strouhal no. 0.2 for a reynold's number of ~10E7) when I looked at the oscillations. But the coefficient of drag was a low value 0.34 instead of a 0.7 expected when I used Kw SSt with a y+ of ~20. With k epsilon I got cd around 0.7, but no vortex shedding to be seen. Which confuses me because to predict the right Cd the turbulence model should be predicting the right transition, and if k-e predicted the transition, it should have simulated proper vortex shedding too. This looks like a seemingly simple problem compared to a lot of complex geometries that are solved out there. And I have followed a lot of suggestion on this from regular CFD users. I am beginning to be skeptic on the validity of the solutions now. Prapanj.

 June 5, 2009, 01:47 #6 Senior Member   N/A Join Date: Mar 2009 Posts: 189 Rep Power: 10 There is an empirical formula for prediction of the Strouhal number based on the Re. I do not remember it on the top of my head. First, the RANS models cannot accurately predict the vortex shedding as they do not capture the instantaneous fluctuations. Second when you use yplus of 20, you are employing a wall function BC at the first grid node. This can be problematic for the current study. You need to try to integrate the k-w model through the viscous region. I doubt that k-e can predict better than SST model (depends on k-e that you used). I think the k-omega model would have been applied for vortex shedding studies. Try to find previous studies and their relative degree of success for the problem.

 June 5, 2009, 05:05 #7 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Hi Harish, The strouhal number for this Reynolds number is 0.21. I don't get vortex shedding with k epsilon. I am using standard k epsilon by the way. When I used k omega SST, it had a hybrid wall function which clamined, I could use a grid with y+ upto 30. Unfortunately, with all my googling skills, I couldn't find one paper that does flow over cylinder for reynolds number > 1E7. Most papers call 10^6 as very high reynolds number. And I even found this paper that deals with flow in the drag bucket region with LES - http://wwwp.coc.ufrj.br/~alvaro/papers/vortex.pdf - which does not succeed in predicting the drag values. In fact the paper ends trivially, by stating that their errror in drag prediction is because they used a coarse grid. The difference in prediction is 200% in some cases. I think there is more work required in this area. Especially with K -w SST model. Any suggestions? Prapanj

 June 5, 2009, 15:51 #8 Senior Member   N/A Join Date: Mar 2009 Posts: 189 Rep Power: 10 Vortex shedding is not that accurate to predict with a RANS model. The hybrid wall function may be unsuitable for this problem and thats why you might want to integrate into the viscous layer. If you could not get results even after integrating through the viscous layer, you might want to look at what other people did. Also if its been done at 1e6 then 1e7 might not be a serious issue. I guess the drag crisis occurs at a much lower Re. What software are you using btw ?

June 6, 2009, 02:06
#9
Senior Member

andy
Join Date: May 2009
Posts: 130
Rep Power: 10
Quote:
 Originally Posted by prapanj Let me also provide the problem boundary conditions etc... The dimension of the cylinder is 14 m diameter. I am doing the simulation for a height of 28 m ( hence the grid is also 28 m tall with top and bottom - symmetry boundary conditions) . The upstream velcity is 45 m/s, T.I 0.07. I calculated k according to T.I^2 x U^2 x 1.5 = 14.884 and the omega value is " Cu^(-0.25) x k^(0.5) / l " . I used 0.09 for Cu . Since k-omega is mostly recommended for these kinds of problems, I am sure I am doing something wrong. Please guide me.
You have a cylinder of diameter 14 in a solution domain of size of 28 with symmetry BC on the top and bottom? If so, this will be significantly different to the flow over a cylinder in free space because the restricted flow area will force the mean flow to accelerate substantially more. To see the strength of the effect increase the height of the solution domain.

You may also have issues to do with simulating the boundary layer development and its separation plus capturing any non-turbulent flow instabilities (e.g. vortex shedding) when using a turbulence model which assumes the flow is steady. However, I would suggest sorting out the required size of the solution and external boundary conditions first.

 June 6, 2009, 02:09 #10 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Hi harish, I am using OpenFoam. By integrating through the viscous sub layer, do you mean, reducing my y+ down to 1 or less than 1? I am currently running a LES simulation with no upstream turbulence. The min y plus is 1. The simulation has been for 13 s so far and I see the amplitude of coeff of lift (cross) oscillations keep increasing so far.. I am expect it to stabilize. The Cd is around 0.9 now. What do you think?

 June 6, 2009, 16:04 #11 Senior Member   N/A Join Date: Mar 2009 Posts: 189 Rep Power: 10 I did not note your domain size until andy- quoted it in his last message. What is your upstream and downstream distance in terms of cylinder diameter ? In terms of D, the up and down sides are just 2D and this makes the problem totally different from the vortex shedding problem since the top and bottom may have significant influence like andy suggested. Yes reducing your yplus to less than 0.1. That usually guarantees a grid independent solution with RANS. Also 2D and LES ?

 June 7, 2009, 10:46 #12 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Hi Harish and Andy, Thanks for pointing out this. I guess I am doing something fundamentally wrong. I want a bit more clarification though, which will be of great help. So, the span for which I am simulating is 28 m long, whereas the diameter of the cylinder is 14 m. On the upstream, I have 1.5 D and the downstream is around 5 D. I am using openFoam. I had used symmetric boundary condition at the ends of the span. Could you explain how this would increase my meanflow? Is it because I had set symmetric boundary or is it because the span is short? In case of a 2D domain this span is not present. And still I see results that claim a coefficient drag that is expected. So by having a considerable span (2D) am I not giving room for the 3D eddies ? Please correct me if I am wrong. And if possible please explain to me how the mean flow is affected by the span or the symmetry boundary conditions. Now this is like a study on the flow around a chimney. I want the properties at about 2/3rd the height of the chimney. This is what I am trying to estimate roughly. And please suggest an alternative BC that I can use at the top and bottom ( the ends of the span). By span I therefore mean vertical height. Thank you . Prapanj

 June 7, 2009, 11:40 #13 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Hi, I think I know what you mean by ' symmetry BC resulting in increased mean flow'. At the boundary, when a flux is supposed to leave out, symmetry boundary would assume a ghost cell that is a mirror of the boundary cell, about the boundary. And this ghost cell will have a flux that is coming in, due to the symmetry condition imposed. This would mean cancellation of outgoing flux. Hence the mean flow inside would increase inside the boundary, because flow cannot escape. Right? Now, if this happens, wouldn't it be in violation of mass conservation? A question to Harish : When you mentioned that reducing y+ to less than 0.1 would make the solution grid independent, why did you put a "question mark" after mentioning 2D and LES ? Thank you. Your replies are helping me a lot. Prapanj.

June 8, 2009, 00:20
#14
Senior Member

Prapanch Nair
Join Date: Mar 2009
Location: Bangalore, India
Posts: 105
Rep Power: 10
I am attaching a picture showing my domain. Hope this would avoid any confusion.

Thank you
Prapanj
Attached Images
 text10491.png (31.6 KB, 69 views)

 June 8, 2009, 01:18 #15 Senior Member   N/A Join Date: Mar 2009 Posts: 189 Rep Power: 10 Is your cylinder mounted vertically upto the symmetry boundary or is it located horizontally ? Strictly speaking performing LES in 2D grid is not correct. I guess even openfoam gives a warning.

 June 8, 2009, 01:29 #16 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Harish, Please take a look at the picture posted above. My case is 3D. Of the 6 sides of the domain, 4 are symmetry boundaries. the other 2 are inlet and outlet respectively. Thanks, prapanj

 June 8, 2009, 01:34 #17 Senior Member   N/A Join Date: Mar 2009 Posts: 189 Rep Power: 10 There was an European benchmark workshop on a similar problem. I do not remember it on the top of my mind. That might be a good benchmark to first test your setup. I will update my msg once I find it in my bookmark.

June 8, 2009, 02:48
#18
Senior Member

andy
Join Date: May 2009
Posts: 130
Rep Power: 10
Quote:
 Originally Posted by prapanj I am attaching a picture showing my domain. Hope this would avoid any confusion. Thank you Prapanj
Assuming you are simulating an incompressible fluid, whatever mass flow you specify at the inlet will be constant on every transverse plane until you reach the exit. Your cylinder blocks the flow which has to deflected around it. If your top boundary condition is a symmetry condition, as you say, there can be no flow across the boundary and so the mean flow has to accelerate across the planes where the cylinder has reduced the flow area in order to maintain a constant mass flow. This blockage is a significant effect and, for example, people taking reasonably careful measurements in wind tunnels will correct for it.

If you look at numerical simulations performed by others of cylinders, aerofoils and the like in free streams you will normally see the boundaries placed several tens of body sizes away in order to make the effect negligible.

In order to quantify the effect in your case simply run a set of simulations where the only parameter changed is the distance to the symmetry boundaries. It is important to keep the grid resolution around the cylinder the same or else you will be changing more than one significant parameter and get confusing results. I would recommend doing this in 2D in order to get answers in a reasonable time scale.

A couple of questions related to the turbulence modelling:
- do you need to use something as computationally expensive as LES?
- is your Reynolds number always 10^7 and above or do you need to simulate the change in drag at lower Reynolds numbers?

 June 8, 2009, 03:00 #19 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 Hi Andy, Incompressible flow, yes. In a previous post, you had mentioned increase in mean flow. So when you say, the increase in velocity due to blockage, does it mean increase in mean-flow? By symmetry, I would not only prevent flow going out, I would also redirect some flow that has phi outward. Right? So you mean I could also use a wall condition here. This would then simulate a proper wind tunnel. In Openfoam, I now tried using inletOutlet velocity condition for the boundaries, which imposes a zero gradient when flux is outbound and a inlet flow when flux is inbound. But this has resulted in very high velocity regions in the domain. I hope you had taken a look at the picture attached above. It's been great help so far. I will try 2D to play with the boundary conditions. My reynold's number is always going to stay ~10^7. I am using LES only for a proper vortex shedding, because as a next stage of this project, I will be placing another cylinder in the wake to study the relative load on the second cylinder, and this will be performed for different angles of inflow. Hence using LES. With K epsilon and k omega SST I was not able to get a good vortex street. regards, Prapanj

 June 8, 2009, 03:11 #20 Senior Member   Prapanch Nair Join Date: Mar 2009 Location: Bangalore, India Posts: 105 Rep Power: 10 And thank you Harish. I would greatly appreciate it if you could send me a benchmark case like that. Thank you prapanj.

 Tags coefficient of drag, high re, k omega, turbulence

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post Phillips FLUENT 0 August 24, 2008 18:01 opnd FLUENT 5 October 26, 2007 04:25 Sawa FLUENT 3 January 14, 2003 02:10 hani Main CFD Forum 1 October 9, 2002 13:28 Axel Rohde Main CFD Forum 2 August 17, 2002 12:18

All times are GMT -4. The time now is 13:26.