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Old   April 19, 2010, 09:30
Default turbulence model required?
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abcdef123
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Hi, I am new to CFD and I have a 2D code working using the SIMPLE method. This code works for Re < ~200. I think the reason my code becomes unstable past this Re number is because of turbulence? So I guess I was wondering if anyone could tell me if I need a turbulence model for this to simulate high(er) Re numbers, or if I could do this without a turbulence model (without make my mesh very very fine). Any comments or references would be very much appreciated, thank you in advance.
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Old   April 19, 2010, 09:47
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You do not need a turbulence model to simulate a flowfield at a higher Re in a stable fashion. You will, however, need a turbulence model to simulate flowfields accurately at Reynolds numbers which are high enough to indicate that the flow is turbulent. I would guess that your stability issues are not connected to your lack of a turbulence model.
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Old   April 19, 2010, 10:22
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Hallo abcdef123,

as agd said, you need turbulence model to get correct solution, it does not solve stability problems in general. Althought turbulence model can increase the range of your code (because of additional dissipation), it is not the right way.

Lunak
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Old   April 19, 2010, 19:25
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Thank you both for your replies. I will investigate my code more to find my stability issues then. If I understand correct, that the code should be stable at high Re even without a turbulence model, then the problems are elsewhere in the code.

CfdLunak, can you explain what you mean by "increase the range of your code". Sorry, my english is not great.

Thank you again.
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Old   April 20, 2010, 00:46
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Dear abcdef123
"increase the range of your code" means "increase the stability range of your code".
stability can be achieved by incorporating a turbulence model to your code via damping a fraction of fluid flow energy (refer to energy cascade concept). but stability of your code can be also affected by other factors such as: mesh size, mesh distribution, discretization method, relaxation factors and chosen time step (for unsteady problems).
in order to get a converged solution, you should check them all.

regards
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