# Low Reynolds number question

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 August 10, 2013, 10:49 Low Reynolds number question #1 Senior Member   Join Date: Jul 2013 Posts: 124 Rep Power: 12 Does anyone know why using small Reynolds numbers ( < 5 ) requires continuously smaller time steps? Even with the cavity tutorial, using Reynolds numbers less than one requires very small time steps to prevent the Courant number from exploding. There is probably something fundamental I'm missing. Is icoFoam not the best choice for low Reynolds numbers? Thanks for your help.

 August 10, 2013, 13:13 #2 Member   Glenn Carlson, PE, PhD (ret) Join Date: Oct 2012 Location: US Posts: 49 Rep Power: 13 Sorry. I'm not sure what you mean by "small Reynolds numbers." Do you mean the velocity of the wall is low? Or are you referring to the behavior of turbulence models close to a wall? Of course, very fine meshes can require small time steps to avoid large Courant numbers, but it doesn't sound like this is what you are referring to.

 August 10, 2013, 13:20 #3 Senior Member   Join Date: Jul 2013 Posts: 124 Rep Power: 12 Hi, thanks for you reply. I mean Reynolds numbers less than 5. I run into this problem with every mesh while using icoFoam, even the cavity tutorial. For example, in the cavity tutorial, the only thing I am changing is the Reynolds number, not the grid size or the velocity. I am controlling the Reynolds number by changing the kinematic viscosity. As far as I am aware, you only need to decrease the time step if you decrease the grid size in the direction of the velocity, or increase the velocity in the direction of the grid. However, when I start decreasing the Reynolds number below 5 or below 1, etc... I have to significantly decrease the time step in order to preserve stability.

 August 10, 2013, 13:57 #4 Member   Glenn Carlson, PE, PhD (ret) Join Date: Oct 2012 Location: US Posts: 49 Rep Power: 13 Sorry again. How are you calculating Reynolds number? Re = UD/nu, where U is velocity of moving wall and nu is kinematic viscosity? What are you using for D? The length of a side of the cavity? Something else? Are you modeling turbulence? If so, what model are you using? Also, what version of OpenFOAM are you using?

 August 10, 2013, 14:25 #5 Senior Member   Join Date: Jul 2013 Posts: 124 Rep Power: 12 I'm calculating the Reynolds number as you said, Re = U * d / nu, where U = 1, d = 0.1. Thus my Re = 0.1 / nu. Just as they do in the cavity tutorial. I'm not modelling turbulence, as icoFoam is a laminar solver, and I'm using OpenFOAM 2.2.0. Everything is fine for Reynolds numbers from say 5 to 100, but with Reynolds numbers less than 5 or 1, I have to decrease the time step to preserve stability. I should also note that I am not using implicit schemes.

 December 12, 2013, 09:47 #6 New Member   Otto S. Join Date: Oct 2013 Location: Germany Posts: 2 Rep Power: 0 Dear wildfire230, I have a similar problem: With increasing viscosity I have to lower the time steps. Do you found a solution for this problem?

 December 13, 2013, 01:55 #7 Senior Member   Karl-Johan Nogenmyr Join Date: Mar 2009 Location: Linköping Posts: 279 Rep Power: 21 Ferziger and Peric discusses this. In 2nd edition you'll find it in section 6.3.1. When diffusion starts to dominate over convection, you have another instability to cater for than the one set by the Courant number. You'll trigger the instability by refining mesh, or increasing viscosity. And if your solver only sets time step according to Courant number, you'll trigger it when the Courant number allows you to take large time steps. deltaT < 1 / ( 2*Gamma/(rho*dx*dx) + u/dx ) Courant number is only the last term above. (Gamma is diffusion coefficient (viscosity)) tfuwa, lev, SHUBHAM9595 and 1 others like this.

 December 13, 2013, 04:34 #8 New Member   Otto S. Join Date: Oct 2013 Location: Germany Posts: 2 Rep Power: 0 Hey kalle, thank you for your answer! This is the explanation I was looking for.