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February 1, 2002, 06:06 |
Number of cells
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#1 |
Guest
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I have been wondering. When do you know that your model has enough cells? I mean, I know that the solution should be "grid independent", but in praxis there are not always time to re-calculate the case with an increase in cells and if the model is large/complicated a restriction in cell number is sometimes required due to calculation time.
I know that if you are looking for separation, the cell-density near the walls should be dense, i.e. the number of cells should be increased in areas where you expect large gradients. BUT….. When do you know when there are enough cells. Christian (Maybe the answer is: "When the solution looks as expected" or "Judge independently from case to case based on the physical design and expected flow pattern"). |
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February 1, 2002, 10:30 |
Re: Number of cells
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#2 |
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just try it!
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February 1, 2002, 11:55 |
Re: Number of cells
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#3 |
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If you can capture what you are looking for, then the grid size is sufficient. Otherwise increase grid size. Grid should be clustered at walls because of steep gradients. After you do a lot of simulations it also comes by judgement.
Anup. |
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February 1, 2002, 12:10 |
Re: Number of cells
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#4 |
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Hi! Christian,
My topic is Large Eddy Simulation of turbulent flow around square prism. Re=22000 based on the inlet velocity and prism dimension. Depends on the objectives, I run my code with different input data files. For the simple mean quantities such as mean drag and lift coefficients, mean velocities and turbulent intensities, I use somewhat coarse grid (even though in this case, the # of grid points are almost 500,000). For the calculation of fully three dimensional flow characteristics, I need more grid points. In this case I'm using almost one million cells. For the bodies with complex geometry, it is almost impossible to get grid independent solution. Let's consider the trubulent flow around bridge deck, we should consider several things as follows : the cell sizes on the surface of deck, streamwise/lateral/spanwise computational domain sizes. So we should compromise with the reality. 1. We have to consider what we want to get from the simulation. Physical viewpoint. 2. We have to consider how the results should accurate. 3. Considering the computer you can use, We should decide the number of cells. This is an economical viewpoint. Conclusively, trial and error is the only choice. Why? Nobody knows the results before running the case. |
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February 4, 2002, 06:09 |
Re: Number of cells
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#5 |
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In the DNS or LES method, the number of the grid should be in the order of the Reynolds number power 9/4 to insure the grid independence.
In the Reynold time-averaged method, it is difficult to judge if the grid is independent. As the small vortex is omitted in the time-averaged method. With the refinement of the grid, more and more detailed flowfield information can be captured. Therefore, the criteria may be the accuracy of the solution that you wish. |
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February 4, 2002, 08:53 |
Re: Number of cells
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#6 |
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i´m using the reynolds averaged navier-stokes equations with the k-eps-modell and a logarithmical wall function for turbulent internal flows. I never reach the results desired in the first run. So therefore i try first a coarse mesh with a maximum of 50000 nodes. Then i plot the residuals of pressure and the other solution variables for locating areas of mesh refinement. After the second run the results sometimes match, if not i put even more nodes in the modell.
But if the simulations are similar (physically) i use in the first run the mesh-topology used in modells before. |
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February 4, 2002, 23:12 |
Re: Number of cells
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#7 |
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A number of respondents have concentrated on the grid-independent aspect of this. But there is another.
Implicit in their assumptions is that the governing equations/physical models are known and/or exact or at least good enough. This is a good assumption for aerodyanmic flows or flows with heat transfer etc. However, in many multiphase/chemical processes the physical models themselves are often poorly understood. In this case, uncertainty in these aspects may dominate the problem, whether you have a grid-independent solution or not. I suppose, what I'm getting at it, achieving a grid-independent solution is still advisable, however, it may not be the only aspect that's important in the model. Greg |
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February 6, 2002, 11:48 |
Re: Number of cells
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#8 |
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As I have heard, there is a algorithm based on Richardson Extrapolation. With this method, you can compare the solution of 3 gridsize. From them, you can get the order of accuracy of the solution. Then you can get Grid Convergence Index (GCI) for your method. This can be done on any simple benchmark problem to test your program. For this case, there is no specific requirement on grids unless you provided the detail of the objects. That is all I can provide. If you find more information on grid dependence evaluation, please let me know. I am also in need of it. Tingguang
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February 20, 2002, 07:37 |
Re: Number of cells
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#9 |
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For some applications of Richardson extrapolation an easy introduction with some examples would be: http://www.iihr.uiowa.edu/gothenburg...F/iihr_407.pdf, but on the net you can find a lot of resources about cfd validation.
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