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May 6, 2006, 02:47 |
time step??????????
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#1 |
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hi friends. What is this time step? and why it is required? what is its role in a steady state problem? these are my questions in this message. Please find the answers and help me out. thank you.
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May 6, 2006, 06:17 |
Re: time step??????????
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#2 |
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Answer: [b]D o c u m e n t a t i o n[b]
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May 6, 2006, 19:26 |
Re: time step??????????
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#3 |
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Hello Vijesh,
What is the difference between a mouse ? D. |
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May 8, 2006, 05:18 |
Re: time step??????????
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#4 |
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It has to do with accelerating the solution for steady state problems. U can just ignore it by setting it to autoscale time step......
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May 8, 2006, 15:28 |
Re: time step??????????
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#5 |
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hi vijesh, i also had the same problems like you are having with the physical time step. i think this has to do with adavancing the solution within the compuational domain. cfx advises you to take one quarter or half of the typical residence time of the flow in your domain. simply put: for a quasi- 1D problem with vel. u traversing a domain with length scale l, the residence time is then approximated t as l/u, and the physical timestep is betwen 0.25*t and 0.5*t. best way to get round it is to try out different values. the convergence of your solution may depend on that. hope this helps. ben.
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May 9, 2006, 10:17 |
Re: time step??????????
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#6 |
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Hi Ben and Ihab. Thanks for replying. Please be in touch. We can learn together. From your pleasant response, I feel our thinking styles match with each other. Regards VVJ.
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