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Mesh and initial conditions for k-omega model

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Old   September 12, 2008, 16:14
Default Mesh and initial conditions for k-omega model
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Martin
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I wish to simulate the Re 100 flow around a sphere using Menter's SST k-omega model. This is my first turbulence calculation ever, and I have some beginners questions I hope you can help me with:

1) The model uses wall functions, and I am not sure if those wall functions require me to put the first off-the-wall grid point in the viscous sublayer or above. But let's assume I figure out a y+ for the first grid point off the wall. Then how do I figure out what that y+ corresponds to in real wall distance?

With y+ = u_tau*y/nu, I need the friction velocity to get y. If all I have is the Reynolds number and the kinematic viscosity, then how can I calculate u_tau - I would need the wall shear stress for this, which I don't have - neither directly nor indirectly via a skin friction coefficient.

2) Assuming I solve (1), nect problem is: how do I mesh the boundary layer? Should cell size increase in some non-heuristic manner away from the wall (i.e. in proportion to how u+ grows according to the law of the wall) ? And how can I know a priori what the thickness of the boundary layer will be; I only want a fine mesh in the BL, not in the core flow.

3) About initial conditions for k and omega. I don't know how to specify them intelligently. My own thoughts went like this:

Initially I just assume isotropic turbulence (u_x)' = (u_y)' = (u_z)' at 5% of the free stream velocity. Then I get k from k = 0.5*((u_x)'² + (u_y)'² + (u_z)'²).

For omega I went through a lot of anguish. I used the fact that w = e/k to change the problem into specifying an initial value for e. Then, assuming we are not near the wall, ie. for example in the log-law region, I know the dissipation rate e scales like u²/l where u and l are velocity and length scales of the energy containing motions. Hence, it is reasonable to model e as e = C*k^1.5/l_m where C is a constant and l_m is the mixing length. I also know that in the log-law, l_m = kappa*y and that turbulence production and dissipation are about equal. From that I found that

e = c³k^1.5/l_m

and worked out c at about 0.55. For the mixing length I would then just replace it by, say, the diameter of the sphere.

But all this assumes I am in the log-law region and it also appears very heuristic to me.

Please, can someone explain how to solve these three issues.

Thanks, Martin
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