|
[Sponsors] | |||||
Utility of providing the CFL and the time-step simultaneously |
![]() |
|
|
LinkBack | Thread Tools | Search this Thread | Display Modes |
|
|
|
#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! |
|
|
|
|
|
|
|
|
#2 |
|
Senior Member
Ping
Join Date: Mar 2009
Posts: 556
Rep Power: 21 ![]() |
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
|
|
|
|
|
|
![]() |
| Tags |
| cfl, star ccm+, time-step |
| Thread Tools | Search this Thread |
| Display Modes | |
|
|