Unsteady 2nd order Dual Time-Step Diverges When Restarted From Steady Solution
Hello,
I am currently running trying to run a unsteady case in SU2 v7.1.1 and I want to restart from a converged steady solution that I already have. I used the 2nd order dual time-step method. Since I need two solutions to restart, I just made two copies of the steady solution, but SU2 diverges before any iterations start. Below is the unsteady paramaters part of the config file. This google drive link has my case files. https://drive.google.com/drive/folde...J8SK9aW2xwMNeq Code:
% ------------------------- UNSTEADY SIMULATION -------------------------------% |
This UNST_CFL_NUMBER= 1 gives you a variable time step which is not correct for the second order dual time approach, because the time derivative is based on finite differences.
Set UNST_CFL_NUMBER to 0 and use TIME_STEP instead. |
Hello,
Oh, so specifying the CFL number makes it like local time stepping? But then what is its role for an unsteady problem? EDIT : and then why does this config work when I am not restarting? |
Hi,
You have a point... I looked at the code again and it looks like a bug. For TIME_STEPPING the unsteady CFL is used at each time iteration to determine the time step. For DUAL_TIME_x it looks like the intention is to determine an automatic time step on the first iteration, and then fix it (because it cannot be variable for these methods) however the logic for this seems to be broken when the simulation is restarted. I'll see about fixing this, but for now set it to zero and use TIME_STEP. In dual time methods, the pseudo time step determined for inner iterations is based on the normal CFL, this UNST_CFL_NUMBER option is only for the "real" time step. That is, dual time is just steady state with a source term. |
Hello,
Thank you for the clarification. I had a question. When you say that the time step is determined by the cfl number, do you mean that su2 chooses the minimum timestep available as the cell sizes are different and the timesteps calculated from them would all be different? |
Yes correct, minimum timestep over the entire problem.
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