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 August 5, 2007, 10:26 Time-marching for steady flows #1 Gui Guest   Posts: n/a Sponsored Links I'd really like to know what's the time marching procedure used by CFX for steady flows cause it's not written in the manual. Thank you

 August 5, 2007, 16:09 Re: Time-marching for steady flows #2 Harun Guest   Posts: n/a I believe CFX employs artificial time stepping to a steady state. My question then becomes: Will the transient term be retained once steady state is reached?

 August 5, 2007, 16:58 Re: Time-marching for steady flows #3 Gui Guest   Posts: n/a yes it is artificial time stepping, it's written in the manual but they don't say much about it... But I still don't understand that principle : what methods does it involve? if Robin could be there I'm sure I'd know...

 August 5, 2007, 18:24 Re: Time-marching for steady flows #4 Peter Attar Guest   Posts: n/a Once steady state is reached the transient term is zero....

 August 6, 2007, 04:26 Re: Time-marching for steady flows #5 Harun Guest   Posts: n/a Okay, then what happens in transient simulations? How are the inner iterations marched in time? If the transient term equals zero when steady state is reached, how is the the transient term updated in transient flows once the solution is converged at each timestep?

 August 6, 2007, 06:53 Re: Time-marching for steady flows #6 Gui Guest   Posts: n/a with the first or second order Euler Backward method... wha's artificial timestepping?

 August 6, 2007, 18:38 Re: Time-marching for steady flows #7 Glenn Horrocks Guest   Posts: n/a Hi guys, This is discussed in the manual. It is an implicit time differencing scheme of first or second order. Some people think of "time marching schemes" as explicit schemes and CFX is not explicit, so just be careful to not get confused there. In steady state the transient term goes to zero. In steady state simulations a number of the non-linear parts of the true transient scheme are removed, but most of the terms remain - hence it is a "psuedo-transient" scheme. Glenn Horrocks

 August 6, 2007, 20:07 Re: Time-marching for steady flows #8 Gui Guest   Posts: n/a Thank you I see more what's the idea. But how does it come that you don't have to specify the time discretisation method (first or second order backward euler) when you perform a steady state simulation?

 August 6, 2007, 23:51 Re: Time-marching for steady flows #9 Peter Attar Guest   Posts: n/a The solution does not need to be "time accurate" hence the accuracy of the time discretization does not matter. The only thing you are trying to do is drive the time derivative to zero as fast as possible which in theory should recover the steady solution. Most every book written on CFD will have covered this....perhaps it is time to visit the library....

 August 7, 2007, 04:51 Re: Time-marching for steady flows #10 Gui Guest   Posts: n/a I have the book Introduction to CFD of Versteeg. It's written nowhere about this artificial time stepping. I'm not new to numerical methods, I already had classes on that, and I never heard of artificial timestepping. In the manual it's written nowhere either. Sorry but for me steady state is steady state, so the transient term is zero and I don't see why it wouldn't, and so I didn't understand why it was still discretized in time.

 August 7, 2007, 10:37 Re: Time-marching for steady flows #11 Robin Guest   Posts: n/a Hi Gui, Basically the transient term is left in the steady state equations as a form of relaxation, which is why there are no relaxation coefficients for users to modify in CFX. Although the steady state behavior may be time-like, it is not an accurate transient (unless you were to run with a very, very small timestep), because it is only a first order backward Euler scheme and the non-linear coefficients are not updated within the timestep. Furthermore, the calculation of residuals in a steady state calculation does not include the influence of the transient term. Since the objective is to reach steady state, the residual is calculated as: What goes in - What goes out = Steady State Residual In a transient calculation it is important to accurately resolve the time-evolution of the flow. This requires a second order treatment of the transient term and also requires that the non-linear coefficients be updated within the timestep, wheras the transient term is not updated until the next timestep (thus advancing the solution). For a transient timestep to be converged the time accurate residual must be small (although not usually as small as required by a steady solution), but keep in mind that this includes the transient term, thus: What goes in + What accumulates - What goes out = Transient Residual Where the accumulation is the change in momentum, mass, etc. within the control volume. This often results in confusion, because users wonder why a transient calculation appears to converge, whereas a steady state one does not. The reality is that the residuals are representing different (steady vs. transient) convergence criteria. Regards, Robin Quasar_89 likes this.

 August 7, 2007, 14:36 Re: Time-marching for steady flows #12 Gui Guest   Posts: n/a Thank you for all these precisions. Maybe I'm asking too much,but what do you mean by the non linear coefficients?

 August 7, 2007, 16:36 Re: Time-marching for steady flows #14 Gui Guest   Posts: n/a Robin, you're the man ! Thanks a lot Quasar_89 likes this.

 August 7, 2007, 18:03 Re: Time-marching for steady flows #15 Harun Guest   Posts: n/a Robin, Looking at transient simulations now. How are the inner iterations (coefficient iterations) marched in time? Does CFX employ some sort of dual-time marching? Thanks in advance.

 August 7, 2007, 18:07 Re: Time-marching for steady flows #16 Robin Guest   Posts: n/a Inner iterations are purely coefficient loops and aren't marched in time.

 August 7, 2007, 18:23 Re: Time-marching for steady flows #17 Harun Guest   Posts: n/a Robin, Are you saying then, that the resulting equation set is no longer hyperbolic/parabolic in time? What is the nature of the equation set that is being solved during the inner iterations? Thanks in advance.

 August 8, 2007, 09:32 Re: Time-marching for steady flows #18 HekLer Guest   Posts: n/a No, the equations are still hyperbolic/parabolic. The transient term is still assembled during each coefficient loop. It's not like it's magically deleted or something. The multiple coefficient loops are needed to update "lagged" non-linearities. The advecting mass flows/velocity Robin previously mentioned are one example. Another example is if you run the high-resolution advection model (the default BTW), the control volume gradients, used for the second order correction terms, are under relaxed so you do not get their full effect until the time step converges. In some cases, with a small enough time step (CFL~1) and particular physical models (say single phase compressible laminar or turbulent flow), you can get away with one coefficient loop per timestep, which really speeds things up.

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