|
[Sponsors] |
Differences between steady and transient simulations |
|
LinkBack | Thread Tools | Search this Thread | Display Modes |
April 27, 2012, 20:47 |
Differences between steady and transient simulations
|
#1 |
Member
Join Date: Jan 2011
Posts: 45
Rep Power: 15 |
Hi,
I'm simulating a cylinder in a rectangular tank and I would like to know what represents the steady solution. At high Reynolds numbers in the transient simulation a von Kármán vortex street is formed, so at sucessive time steps vortex tavel and dissipate. But in the steady simulation, does the solution represents a time average calculation? or what exactly represents? Thanks in advance |
|
April 28, 2012, 01:09 |
|
#2 | |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,761
Rep Power: 66 |
Quote:
In a steady simulation, there is no time dependence of any variables. They are annihilated from the start. That is, it is assumed that there is no time-dependence. The governing equations in this case does not include time (transient effects). In a transient simulation, the time dependence is kept in the variables. The governing equations contain the transient terms and the variables are time-dependent. A steady simulation will tend towards the steady-state solution (if one exists). A transient simulation will solve for the time-dependent solution. The time-averaged representation of the transient solution will yield a time-averaged solution. Only in the case of when the original problem is steady will the steady and transient simulations yield the same results. i.e. if there is no unsteadiness in the problem then the two will yield the same results. The time-averaged solution of the transient simulation will be about the same as the steady solution. do not confuse unsteadiness with variables changing with time. All flows are inherently variant in time but steady flows will have simple fluctuations about the average whereas unsteady flows will have fluctuations superimposed on top of a time-varying variable. The von Kármán vortex street is not a steady flow. If you ran a steady simulation (and somehow got it to converge, which would be difficult) it would probably be garbage results and not accurately reflect the true time-averaged solution. To get the proper time-averaged flow field, you need to run an unsteady simulation and then take the average of the variables in time. Last edited by LuckyTran; April 28, 2012 at 22:57. |
||
April 28, 2012, 21:52 |
|
#3 |
Member
Join Date: Jan 2011
Posts: 45
Rep Power: 15 |
Thanks LuckyTran! Your explanation help me to clarify my doubts.
Kind Regards |
|
|
|
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Laminar gas flow under slight suction - transient works but not steady | audrey | CFX | 1 | September 8, 2011 20:43 |
Steady solution from Transient simulations | wawa | FLUENT | 2 | November 9, 2010 18:44 |
Best practice for transient simulations? | siw | CFX | 5 | October 30, 2010 06:45 |
transient phenomenon using steady state simulations | beguxa | FLUENT | 2 | April 15, 2010 03:22 |
transient simulations vs steady state for bouyancy | Drauss | Main CFD Forum | 8 | October 29, 2005 14:31 |