# time step, Incompressible

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 September 18, 2005, 08:05 time step, Incompressible #1 Agyeya Guest   Posts: n/a Could anybody inform me what is supposed to be the best way to find the time step for time-accurate numerical solution of the incompressible NS equation say on a Cartesian grid?

 September 19, 2005, 00:29 Re: time step, Incompressible #2 Mani Guest   Posts: n/a That depends on your problem and your accuracy requirements. What is the smallest time scale that you would like to resolve?

 September 19, 2005, 07:12 Re: time step, Incompressible #3 Agyeya Guest   Posts: n/a Suppose the flow is laminar which involves relatively larger time scales.

 September 19, 2005, 09:24 Re: time step, Incompressible #4 Renato N. Elias Guest   Posts: n/a You could select the smallest time step computed from the element Courant Number (dt = (he*cfl)/||u||), where he is the element lenght, cfl is the Courant Number desired and ||u|| is the Euclidean norm of the local velocity vector. I suggest some papers that discuss this topic better: Valli AMP, Carey GF, Coutinho ALGA, Control strategies for timestep selection in finite element simulation of incompressible flows and coupled reaction-convection-diffusion processes, INTERNATIONAL JOURNAL FOR NUMERICAL METHODS IN FLUIDS 47 (3): 201-231 JAN 30 2005 Valli AMP, Carey GF, Coutinho ALGA, Control strategies for timestep selection in simulation of coupled viscous flow and heat transfer, COMMUNICATIONS IN NUMERICAL METHODS IN ENGINEERING 18 (2): 131-139 FEB 2002 Regards Renato N. Elias High Performance Computing Center NACAD/COPPE/UFRJ Rio de Janeiro, Brazil http://www.nacad.ufrj.br

 September 19, 2005, 12:53 Re: time step, Incompressible #5 Mani Guest   Posts: n/a Your time step should be chosen small enough to resolve time scales of interest/importance. Example: Let's say you're looking at laminar 2D vortex shedding over a cylinder. The solution is periodic of a certain frequency. In that case you would choose your time step small enough to resolve the period accurately, e.g. choose some 30 time steps per oscillation period, depending on your scheme's order of accuracy. Accuracy is the only consideration, when you're using an implicit method. However, with explicit time marching, the time scale of interest may so large that your algorithm cannot handle it. In that case, you'd need to choose a time-step according to the stability conditions (CFL condition) of your explicit scheme. So there are two conditions: Accuracy and stability. Both conditions dictate a certain maximum time step. Between those two, you have to choose the smaller one to insure that both conditions are met. This should make clear that the actual time step you are going to choose will depend on both the flow of interest and the numerical scheme.

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