# How can Implicit Time Marching algorithms NOT be Time Accurate?!

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 July 30, 2010, 07:19 How can Implicit Time Marching algorithms NOT be Time Accurate?! #1 New Member   Freddie Sizer Join Date: Apr 2009 Posts: 19 Rep Power: 10 This is probably a very simple question for someone in the know, but I'm trying to simulate flow in a Scramjet for my final year project and can not reach a converged solution. The nature of the scramjet combustion chamber gives rise to an oscillatory motion. Using the sceme that I have I am unable to reach a well converged solution. The algorithm is Line Implicit with a choice of local or global time steps. My question is that I am unsure how a code that time marches can be time inaccurate? Can one of the numerical methods experts please explain this to me? Thank you.

 July 30, 2010, 08:55 #2 Senior Member   Join Date: Jul 2009 Posts: 247 Rep Power: 12 Why should it necessarily be time accurate? Time-marching implies that you are essentially extrapolating forward in time. The accuracy of that extrapolation is going to be determined by the starting point of the approximate solution, the slope of the function (again approximate), and how far forward you extrapolate (the time step). This of course does not even bring to consideration the approximation error due to such things as round-off error or the implementation of boundary conditions or other possible sources of error.

 July 30, 2010, 09:56 Unsteady case? #3 New Member   Freddie Sizer Join Date: Apr 2009 Posts: 19 Rep Power: 10 Thank you for your reply agd. So in essence, you are using values in the current cell and surrounding cells, such as rho, u,v,w etc. at (t) to then calculate the new values at (t+1). If this is the case my logic says that when you then recalculate at (t+1) the values should then be reasonably accurate, especially in an inviscid example. I don't think I'm explaining myself very well, so I'll try to explain what I am doing in a bit more detail: I have a paper that reports an experiment carried out in a gun tunnel. The paper gives schlieren images at different time steps during the duration of the wind tunnel run time (of order ~15ms). I am trying to numerically replicate these results using a line implicit, second order, time marching solution. The code seeks a steady solution as far as I understand. How can I use this type of code to obtain an accurate representation of the Schlieren images. Particularly if the solution is unsteady. I have thought that I might be able to stop the code at a certain time to produce a snapshot, but if it is not time accurate, then this snapshot will not be accurate. Otherwise if I wait for the code to converge I can't understand what it will be converging to (again assuming it is unsteady). An explanation to this would be most helpful because at the moment my head is unsteady!!!

 August 3, 2010, 18:33 Thank you! #6 New Member   Freddie Sizer Join Date: Apr 2009 Posts: 19 Rep Power: 10 Thank you both so much. Your help is much appreciated!

 August 5, 2010, 11:06 #7 Senior Member   Join Date: Feb 2010 Posts: 145 Rep Power: 10 agd's recommendation to drop the time step down is a good one. f_sizer, my guess is that you might be confusing the concepts of stability versus accuracy. In general, implicit time integration algorithms are stable. This is in contract to explicit time integration which has a stability criteria. However, "grid independence" for the time step must also be demonstrated just as grid independence needs to be demonstrated for the mesh. Good luck!

 August 19, 2010, 09:32 #8 New Member   Freddie Sizer Join Date: Apr 2009 Posts: 19 Rep Power: 10 Thank you Jade M. I don't suppose you know how to estimate cfl numbers for use with implicit algorithms do you? I understand fully the explicit way of life, but am a bit fuzzy on why implicit algorithms can handle much higher cfl numbers?

 August 19, 2010, 09:40 #9 Senior Member   Join Date: Feb 2010 Posts: 145 Rep Power: 10 Which software are you using, CFX? I'm sorry as I cannot tell which forum this is posted in. What type of analysis are you conducting, LES? As far as implicit algorithms being able to handle higher CFL numbers, I suppose it is probably really due to numerical stability issues. Implicit methods simply have less restrictions on them. I know this is not a precise explanation. I'll see if I can find some information. It looks like there is some good discussion at http://www.cfd-online.com/Forums/mai...condition.html In particular, for practical use, see the very last post by John Chien. There is also some good information on wikipedia at http://en.wikipedia.org/wiki/Courant...Lewy_condition Last edited by Jade M; August 19, 2010 at 11:24.

 August 19, 2010, 10:14 #10 Senior Member   Join Date: Feb 2010 Posts: 145 Rep Power: 10 For LES with implicit time integration, the CFX documentation states • For accuracy, the average Courant (or CFL) number should be in the range of 0.5-1. Larger values can give stable results, but the turbulence may be damped. For compressible flows where the acoustic behavior is being modeled (e.g., for noise calculations), this conclusion still holds, but for the CFL number based on the acoustic velocity as well as the convective velocity. • 1,000 - 10,000 time steps are typically required for getting converged statistics. More steps are required for second order quantities (for example, variances) than for means. Check the convergence of the statistics. For a vortex-shedding problem, several cycles of the vortex shedding are required. • The implicit coupled solver used in CFX requires the equations to be converged within each time step to guarantee conservation. The number of coefficient loops required to achieve this is a function of the time step size. With CFL numbers of order 0.5-1, convergence within each time step should be achieved quickly. It is advisable to test the sensitivity of the solution to the number of coefficient loops, to avoid using more coefficient loops (and hence longer run times) than necessary. LES tests involving incompressible flow past circular cylinders indicate that one coefficient loop per time step is sufficient if the average CFL number is about 0.75. If the physics or geometry is more complicated, additional coefficient loops (3-5) may be required. Time step sizes which require more coefficient loops than this will tend to degrade solution accuracy. Since the Courant number is equal to U*Dt/Dx, the time step Dt can be calculated based on the grid spacing Dx and the characteristic velocity U. Demonstrating grid independence prior would probably minimize computational time. Last edited by Jade M; August 19, 2010 at 12:15.

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