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August 5, 2007, 10:26 
Timemarching for steady flows

#1 
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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: Timemarching for steady flows

#2 
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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: Timemarching for steady flows

#3 
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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: Timemarching for steady flows

#4 
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Once steady state is reached the transient term is zero....


August 6, 2007, 04:26 
Re: Timemarching for steady flows

#5 
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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: Timemarching for steady flows

#6 
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with the first or second order Euler Backward method...
wha's artificial timestepping? 

August 6, 2007, 18:38 
Re: Timemarching for steady flows

#7 
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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 nonlinear parts of the true transient scheme are removed, but most of the terms remain  hence it is a "psuedotransient" scheme. Glenn Horrocks 

August 6, 2007, 20:07 
Re: Timemarching for steady flows

#8 
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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: Timemarching for steady flows

#9 
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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: Timemarching for steady flows

#10 
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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: Timemarching for steady flows

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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 timelike, 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 nonlinear 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 timeevolution of the flow. This requires a second order treatment of the transient term and also requires that the nonlinear 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 

August 7, 2007, 14:36 
Re: Timemarching for steady flows

#12 
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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, 15:24 
Re: Timemarching for steady flows

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A very simplified way to think of it is this: If you consider the transport of momentum, for instance, you have a term which is the mass flow through the face times the velocity (momentum per unit mass) which it is tranporting (i.e. m_dot*v). But, the mass flow itself is a function of the flowfield, hence the term is nonlinear, because it depends on itself.
In order to solve this, some coefficients need to be "lagged", which is to say it uses the old value. For advection of momentum you break it up into "advected" and "advecting" components. The advected component is the velocity you are solving for whereas the advecting component is the velocity used to calcuale mass flow, which is lagged. So in a steady state calculation, the advecting velocity is always lagged, whereas the advected velocity is active (i.e. solve for). This means they are out of sync, but the disparity between the two will eventually go away when a steady state is reached. For a timeaccurate transient having such a disparity will not do. To resolve this, the solver will take additional "coefficient loops", where it will update the advecting velocity to use the new value of advected velocity acheived by the previous coefficient loop. While these updates are ocurring, the transient term is kept the same, which prevents the flow from advancing in time and allows the nonlinear terms to converge within the timestep. In reality it is a lot more complicated, but this is where we sit back and be grateful for masochistic solver developers, who worry about the details. Regards, Robin 

August 7, 2007, 16:36 
Re: Timemarching for steady flows

#14 
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Robin, you're the man ! Thanks a lot


August 7, 2007, 18:03 
Re: Timemarching for steady flows

#15 
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Robin,
Looking at transient simulations now. How are the inner iterations (coefficient iterations) marched in time? Does CFX employ some sort of dualtime marching? Thanks in advance. 

August 7, 2007, 18:07 
Re: Timemarching for steady flows

#16 
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Inner iterations are purely coefficient loops and aren't marched in time.


August 7, 2007, 18:23 
Re: Timemarching for steady flows

#17 
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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: Timemarching for steady flows

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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" nonlinearities. The advecting mass flows/velocity Robin previously mentioned are one example. Another example is if you run the highresolution 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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