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How to deal with source term when using RKschemes?

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Old   May 10, 2006, 07:49
Default How to deal with source term when using RKschemes?
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There are a lot of numerical schemes to deal with the derivatives in the NS equation. My question is how to deal with the chemical raction source term? For example: in Rounge-Kutta schemes there are several intermediate 'times steps' between n to n+1 i.e. v(n+1)=a*v1(n+1/3)+b*v2(n+1/2)+c*v(n) my question is how to deal with the energy and species source terms between those small "fraction time steps" such as from "n to n+1/3" or form "n+1/3 to n+1/2", let them to be constant or let them advance with the dependent variabes such as temperatures and so on?
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Old   May 10, 2006, 09:46
Default Re: How to deal with source term when using RKsche
Jeff Moder
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If you are using the R-K scheme to march in "pseudo-time" to a steady-state solution, then you should try updating source terms (and possibly viscous fluxes) on only a "few" stages (maybe first and last, for example) to save on CPU time for your simulation. You can experiment yourself with updating source terms every R-K stage, versus only on some of the R-K stages. This is a very typical procedure and since you are just using the R-K procedure (in this case) as an iteration technique to find the steady-state solution, you can play around with what gets updated on each stage -- all you care about is that the simulation converges and converges in some "reasonable" number of iterations.

If you are using the R-K scheme to march explicity in physical time (ie, a time-accurate simulation), then you should update your source terms every R-K stage. Of course, you could evaluate the loss of accuracy in not doing so and make your own choice in accuracy loss versus CPU time reduction for a time-accurate simulation using R-K to march in physical time.

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Old   May 11, 2006, 10:34
Default Re: How to deal with source term when using RKsche
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Though the intermediate t* are not clearly known evaluating the source term at all the intermediate steps using the same t seems to work for me.

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