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September 13, 2001, 01:03 |
what's the initial condition of turbulence?
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#1 |
Guest
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Dear all:
I want to carry out a DNS of the Navier-Stokes equations. What's the commonly used initial condition which will lead to turbulence? Thanks in advence. With regards Paul |
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September 13, 2001, 11:44 |
Re: what's the initial condition of turbulence?
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#2 |
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1. What kind of physical problem do you consider to simulate?
2. Depends on your problem, initial condition could be different. 3. But, if you did not decide your problem, please refer 'Fluid Flow Phenomena, A numerical Toolkit' by PAOLO ORLANDI |
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September 14, 2001, 06:18 |
Re: what's the initial condition of turbulence?
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#3 |
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Thank you for your reply.
The problem is defined as following: 1) 2-D Navier-Stokes equation, incompressible. 2) Rectangular domain, periodic boundary condition. 3) What's the initial condition which will guarantee a fully developed turbulence? 4) Can the random noise be used as initial condition to generate turbulence? Are there some analytical initial conditions? |
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September 14, 2001, 09:37 |
Re: what's the initial condition of turbulence?
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#4 |
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1. Direct Numerical Simulation of "turbulent" flow means three dimensional analysis(usually) in order to consider vortex stretching which is 3-dimensional phenomenon.
2. Depends on your physical problem to simulate, your initial condition and boundary conditions could be decided. 3. In the case that your physical domain has the source of turbulence production(wall, etc), after some calculation from zero velocity flow field, your flow will reach to the turbulent region. Computational domain size should be modified carefully to include fully developed turbulence. 4. If your problem is to solve the monotonically decaying turbulent flow, some special initial condition is necessary. Please refer chapter 8 of Orlandi's book ( pp52, 8.8 Initial conditions). The book also has source codes which could be applied to your code easily. Jongdae Kim. |
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September 14, 2001, 10:06 |
Re: what's the initial condition of turbulence?
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#5 |
Guest
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Kim, thanks for your kind help.
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