smaller timestep leads not to converge
Dear all,
I got a transient run with constant timestep 0.001s, smoothly done by CFX (ver 13.0). Other I input BCs: Laminar, isothermal. inlet flowrate around 0.003 kg/s; outlet is about 10000Pa For first two iteration , the courant number is about 200s, then reduced to the level of 50s.(I was noticed the CFX is fully implicit so greater courant number is not issue once converged. ) I monitor the outflow and inlet pressure, everything seems alright for this run. Then I tried smaller timestep 0.0001s but the it couldn't get convergence for the first iteration. The solver manager shows the linear solution failed in equations of U-Mom, V-Mom, and W-Mom; then fatal error, Floating point exception: Overflow. I haven't checked the mesh yet. I ticked the CFX-Pre->Solver control->Basic Setting->Timestep Initialisation-> Upper courant Number, and gave value to 10. The first iteration just flow through and running and running. I am wondering if this is right. Or there's another way around, I could use 0.0001s without divergence. THANKS |
hey,
really not sure about that but i think it could be a problem if the time step isnīt sufficient for letting the flow pass the very first volume with given velocity. neewbie |
You should set the outlet to 0Pa pressure and use the reference pressure to give the correct absolute pressure. Using a large outlet pressure causes large round-off errors which will get worse when you decrease the time step.
Also running double precision numerics might help. |
Quote:
I almost have the run finished I will check any diffrerence between two cases(0.001s and 0.0001s) Thanks! |
I am no expert on FSI but I am sure it is smart enough to be able to handle a reference pressure. This is really basic stuff. And very important for exactly the reason you have found.
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I I were you, I would use a steady state first.
I would define the reference pressure and I would defined the outlet pressure as a relative pressure, as a Glenn said. Besides, I would start as a Steady state with and Auto time scale = 1 (conservative), and let the program to make some iteration, as a result you will get an "accurate" time step. After you del with steady jump to transient.... Good luck!! |
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