the criterium of convergence for the SIMPLE algo
Hi all,
The SIMPLE algo is like that: 1. Set the boundary conditions. 2. Solve the discretized momentum equation to compute the intermediate velocity field. 3. Compute the mass fluxes at the cells faces. 4. Solve the pressure equation and apply under-relaxation. 5. Correct the mass fluxes at the cell faces. 6. Correct the velocities on the basis of the new pressure field. 7. Update the boundary conditions. 8. Repeat till convergence. But what does "till convergence" mean? norm[U_(n+1) - U_(n)]/norm[U_(n)] < tol is a good criterium? I need some advice, thanks in advance, Lam |
Re: the criterium of convergence for the SIMPLE al
Hi Lam,
I first thought that someone would have a good answer to give, but as nobody answered, I will try to help. "But what does "till convergence" mean? " "norm[U_(n+1) - U_(n)]/norm[U_(n)] < tol " Well... It's difficult to say, but I would think that that criterion is not really good. Personally, I would think that the best way to evaluate the convergence of a CFD software is evaluating Navier Stokes and mass conservation FULLY EXPLICITLY. I mean: you take every terms of NS at one side of the equality, which gives rho . v . grad(v) + grad(p) - mu . laplacian(v) - rho . f For an "exact" solution This shoud give zero. But if at some iteration you evaluate ALL these terms explicitly (with known values) it will give you a non null number, which is a good approximation of the error. Unfortunately, it strongly depends on your discretization scheme (finite volume finite elements or whatever) In my personal CFD code, this gives some strong errors, even for a very good looking (physical) solution. But now it depends on how you evaluate the total error on the entire domain. Personally, I sum all these errors on each finite volume. Another possibility is to take total error = sqrt ( sum(e_i) ) where e_i is the error of the i th control volume or finite element. But actually, there is no perfect way to know if a numerical solution is a good approx. of the Navier Stokes continuous solution. Hope it helps. |
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