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Transient Simulations - Effect of number of grid points |
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April 16, 2018, 07:03 |
Transient Simulations - Effect of number of grid points
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#1 |
Member
Thamilmani M
Join Date: Sep 2017
Location: IIT Bombay, Mumbai
Posts: 52
Rep Power: 9 |
Hello everyone,
I am working on a transient 2D laminar flow over two tandem cylinders in ANSYS. I used finer mesh by using different grid points on each side of cylinder. Let that be a. I used values of 40, 60 and 80. so that the total number of cells also keep increasing as we increase a. Now, I got the a 80 converges to a steady solution after around 5 lakh iterations. But a 40 or a 60 has run even more than that many iterations, but didn't converge to a steady solution, but the Cl plots are just growing. We know that, increasing number of grid points increase the fineness of the mesh, which improves the accuracy of the solution, however also increase the computational time greatly. But, Here we see that decreasing the number of grid points didn't actualy reduce the computational time as expected. Though a 80 would still be at higher accuracy. Why? Following case conditions are same of all those meshes: Upstream distance is 10D, and Downstream distance is 20D and the walls are at 10D each. Re is 100. With PISO solver. All discretizations are Second order. Can anyone reason out this Why? Or is there something fundamentally wrong in me? Thanks for all. Always Thedal...
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Always Thedal |
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April 16, 2018, 08:00 |
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#2 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,896
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What is wrong is that at Re=100 the solution is physically unsteady. Are you using second order upwind?
I can suggest to try the NITA setting the second order central discretization. Then, be careful about the grid resolution around the cylinders |
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April 16, 2018, 08:23 |
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#3 | |
Member
Thamilmani M
Join Date: Sep 2017
Location: IIT Bombay, Mumbai
Posts: 52
Rep Power: 9 |
Quote:
Thanks for the reply. I am using Second Order Implicit scheme for Transient formulations sir. Yes, sir Solution is physically unsteady, But why does a lesser number of cells take more time to converge than a higher one is my primary doubt?
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Always Thedal |
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April 16, 2018, 09:06 |
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#4 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,896
Rep Power: 73 |
Quote:
As the case is unsteady, I assume that for "convergence" you do not mean the steady state but the convergence of the iterative methods at each time step. The slower convergence can be due to high local truncation error on the coarse grid as well as to the guess solution that is far from the finaln one. |
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April 16, 2018, 09:54 |
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#5 | |
Member
Thamilmani M
Join Date: Sep 2017
Location: IIT Bombay, Mumbai
Posts: 52
Rep Power: 9 |
Quote:
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Always Thedal |
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April 17, 2018, 08:33 |
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#6 | |
Member
Thamilmani M
Join Date: Sep 2017
Location: IIT Bombay, Mumbai
Posts: 52
Rep Power: 9 |
Quote:
__________________
Always Thedal |
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April 17, 2018, 08:36 |
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#7 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,896
Rep Power: 73 |
Quote:
The number of time steps, or better the final time to reach a developed flow, generally depends on the initial condition and on the Reynolds number. The time step is a parameter depending on the mesh size when the numerical stability constraint must be fulfilled. |
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Tags |
ansys, computational time, cylinders, meshing, unsteady |
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