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Utility of providing the CFL and the time-step simultaneously |
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January 29, 2017, 17:09 |
Utility of providing the CFL and the time-step simultaneously
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#1 |
New Member
Lise Amaniou
Join Date: Jan 2017
Location: France
Posts: 1
Rep Power: 0 |
NOTE: sorry in advance if my english isn't perfect, I swear I do my best
Hi everyone! I'm working on the Von Karman Vortex Street appearing behind a cylinder put in an airflow. Used parameters : - Unsteady flow (implicit scheme + second order scheme) - Ideal gas - Turbulent k-omega SST model - Coupled flow I'm wondering why I have to enter the time-step AND the CFL, since the velocity is calculated and the dx is provided with the mesh. Why doesn't the time-step set the CFL? Moreover, I thought an implicit scheme implied no CFL condition (always stable), why is there one in this case? Last question: my best results come with a CFL = 40, and the solution fully diverges when CFL=1, I would have think the opposite would occur (by definition of the CFL). Can someone give me a clue about the reason for that? Thanks in advance! |
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January 31, 2017, 09:22 |
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#2 |
Senior Member
Ping
Join Date: Mar 2009
Posts: 556
Rep Power: 20 |
for unsteady the cfl sets the pseudo time step for the inner iterations within the outer time steps - see the user guide under pseudo time step
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Tags |
cfl, star ccm+, time-step |
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