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 lost.identity November 25, 2009 09:42

Dual-time stepping CFL number

Hi there

I'm trying to use Jameson's Dual-time stepping technique in an implicit simulation.

I have a question regarding the time-step size. I know that I can use large CFL numbers for the pseudo-time step. But what about the physical time-step?

I've been told that I have to use a CFL number of 1 for the physical time-step for time accuracy and indeed this is what I observed from my simulations.

Can anyone give me any reasons why this is so?

Thanks.

 vinayender November 27, 2009 13:56

Hellow,

Answer is you cant take your physical time step very large, and it should be based oin the physical time step involved in your problem..

Reason is as follows....,

For an unsteady problem, as your solutions depends on the time step you cannot use an implicit scheme and get your solution at a given time faster by taking big time steps....and hence you have to use an exlicit scheme.

But if you are using an explicit sheme, your time step would be fixed by a numerical constriant (that is CFL for a given gird) and you will be forced to use lower time step.
ok.. if your time scales involved in your physical problem is also small and same in order as that you get from the CFL constriant, then there is no problem in using small time steps.

Suppose if you are solving a physical problem in which times scales involved are much larger then compared to that we get from the CFL number (a numerical constriant), then you would fell that its waste to use a very small time step to capture the right physics and wait for many iterations before you get the solution at a time step. In this case your time step has got a numerical constriant rather than a physical constirant. so to over come this problem a small change is done in numerics ( ref to Jamson's dual time stepping)

Here.., if you are sloving an unsteay problem governed by equation
dU/dt = f(U);

now solve this equaiton with explict scheme
(U_n+1 - U_n ) / ( del t) = F(U_n)
this eqation can be written as (U_n+1 - U_n ) / ( del t) - F(U_n) = 0

Since RHS is 0 .., we can replace with dU/dt_pesdo
and
new equatino is
(U_n+1 - U_n ) / ( del t) - F(U_n) = dU/dt_pesdo

now in the above equation you can fix del t based oin the physics of your problem and reulted equation can sloved by using an implicit scheme in t_pesdo time

More over to calculate RHS you can use any first order sechem to get your flux faster and your solution will not be effected as RHS becomes 0
Note:- you should use second order scheme for getting fluxes in LHS.

 lost.identity December 1, 2009 09:58

Thanks very much for the reply. That makes a lot of sense.

I was confused earlier as I thought people referred to Jameson's Dual time-stepping method as an implicit scheme.

I just have the following questions

1) I want my calculations to be done unsteadily and it has to be time-accurate as I want to capture the transients. So I understand that this sets a physical time constraint. I was told that this physical time constraint comes from the CFL where the flow should move one grid-cell within the time-step. This is how I have been fixing the physical time step. I don't really know what other physical time steps I could use as I don't know much about the turbulent time scales of the problem.

2) So in Jameson's scheme the inner iteration is solved explicitly like normal steady RANS?

Thanks again

 vinayender December 2, 2009 00:45

Certainly dual time stepping is not implicit time stepping, but it uses implicit time stepping in pusedo time step and explicit time stepping in real time marching,

comming to your questions,

1) for example, say you are solving for a problem in which physicals time scales are order of 0.1 sec, means what ever significant flow changes happens that will happen only in time scales of order 0.1 sec. In such senario your time scale in the solver is sufficient to be of same order to capture that significant physics.
If you are solving the same problem using the explicit scheme, then the time scale will be decided on numerical constriant that is using CFL(depends on the robostness of each scheme) and the grid size. and if this time step computed using this of the order of 0.01 sec. Using this time scale you will end up with increase in no of iterarations to atain the solution at a given time step.
To sumarise, your physical time steps of 0.1 sec is good enough to attain the solution at a given time level, but your numerics (scheme and grid) will restrict you to use that time step and ending up taking smaller time steps and needs to solve for more iterations (roughly 10 times more for this example)

2) outer or main loop uses expicit to march in real time
and inner main loop uses implicit time step to march in pesudo time step

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