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February 18, 2009, 04:51 |
What Time Step?
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#1 |
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Hello
I'm trying to model a 2D Channel flow and I'm using SIMPLE algorithm. I have testet my code for the cavity problem and it looks good. In channel it also works but I want to get more accurate answers. As u know the velocity in the center line of the channel should be 1.5*U0 where U0 is the velocity at the inlet . I approached 1.47*U0 but I need 1.499*U0. reducing mesh size wont help too much, I noticed time step is more important. I know that CFL number should be <1 for stability, but what CFL number can be used for best answers? is it common to use small number of CFL number, like 0.05 to get better answers? or anyting less than 1 should give good answers? Reducing time step result in better answers but When I set it too small the answers will be bad again. can anyone help me? Cheers RAYAN |
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February 18, 2009, 07:04 |
Re: What Time Step?
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#2 |
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If you want a good time resolution, your CFL should be around 1, meaning the flow particle travels 1 cell at a time step. If you work with inhomogeneous mesh, you need to keep CFL low only at he place of interest. Having CFL 0.01 will not help. Well it depends on what are you computing, in LES 0.1 is about good, in RANS it is useless. In unsteady regime with SIMPLE it is important to have every time-step converged. The accuracy of the computations also depends on the discretisation in space and time. If you are not interested in unsteady solution, your CFL could by 5, 10 or more.
good luck matej |
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February 18, 2009, 09:09 |
Re: What Time Step?
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#3 |
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Matej thnx for the answer. Infact my regime is unsteady and laminar. im working with homogeneous, rectangular mesh.
Unfortunately CFL=1 does not give me good answers. U the x component of velocity at the center line will be something around U=1.2U0 , but I need U=1.499U0. CFL=0.1 give better answers, around U=1.47U0 but still not enough. and of course in each time step I have convergency. what else cna be the problem? |
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February 18, 2009, 09:10 |
Re: What Time Step?
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#4 |
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Matej thnx for the answer. Infact my regime is unsteady and laminar. im working with homogeneous, rectangular mesh.
Unfortunately CFL=1 does not give me good answers. U the x component of velocity at the center line will be something around U=1.2U0 , but I need U=1.499U0. CFL=0.1 give better answers, around U=1.47U0 but still not enough. and of course in each time step I have convergency. what else can be the problem? |
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February 18, 2009, 09:28 |
Re: What Time Step?
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#5 |
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Is your flow-field developed? Have you waited long enough time for the flow to develop in a channel?
If yes, then the problem would be your differencing schemes. In time and in space. What differencing are you using? Are you using Crank-Nicolson or other at least second-order accurate differencing? What about the space - also using some 2nd order scheme? The last and first chance is also - you have some bug in your code. matej |
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February 18, 2009, 14:17 |
Re: What Time Step?
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#6 |
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How well resolved is your mesh?
Remember you're trying to resolve a boundary layer that fills the entire (half) channel. You need to try meshes of a progressively smaller cell width to learn when you're resolved 'enough.' Also, if you're starting at the inlet with u(0,y) = U0, it'll take some distance downstream before the parabolic profile is fully developed. You can pick that distance out from solutions (not CFD) given in the classic Schlichting's Boundary Layer Theory. Good luck. |
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February 19, 2009, 06:38 |
Re: What Time Step?
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#7 |
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well actually the answers will change only very very small amounts after 3 sec. According to Patankar's book, I'm Using a power law model, which is actually a first order scheme but should give good results. about bug, well I don't know, bug is always waiting there for us but I don't think theres a bug, cuz its working good, the problem is somewhere in resolution or time step.
Otd, thnx for replying, in fact I'm using a 2D channel with an aspect ratio of 30-1 so I think the flow should be fully developed at the outlet. If I use bigger aspect ratios, run time will increase significiantly. thank you for answering me, I'll try again! |
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February 22, 2009, 23:00 |
Re: What Time Step?
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#8 |
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Hi RAYAN,
Some thoughts on a useful time-step rule can be found in these links: <http://adthermtech.com/documents/RAEFM2_paper_ID024.pdf> <http://adthermtech.com/documents/RAE...sent_ID024.pdf> Have fun. mw... Momentum waves <www.adthermtech.com/wordpress3> |
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