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September 16, 2014, 02:13 |
LES wall model
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#1 |
Member
M. Nabi
Join Date: Jun 2009
Posts: 44
Rep Power: 17 |
I am writing a code for LES. For wall modelling, I am using zonal-approach double-layer model. As I insert a fine grid in the viscous sub-layer, I used the velocities at the first grid point of the coarse grid, as boundary condition for the fine grid. From other side, I used non-slip boundary condition (Dirichlet) beside the wall. Using this method, enables us to find the local shear velocity.
Now, the shear velocity has to be used as a boundary condition for the coarse grid. I wonder, how can I apply the shear velocity as a boundary condition for the coarse grid? Thanks Last edited by mnabi; September 16, 2014 at 06:25. |
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September 19, 2014, 06:30 |
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#3 | |
Member
M. Nabi
Join Date: Jun 2009
Posts: 44
Rep Power: 17 |
Quote:
The code is Finite Volume. The point is, I have the wall shear stress on the boundary (calculated by the fine grid). How can I use this wall shear stress as a boundary condition ? |
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September 19, 2014, 13:53 |
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#4 |
Senior Member
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Well, the boundaries for FV codes are just faces as the interior ones, thus they just need the viscous and the convective fluxes.
For the convective flux, it is simply zero. For the viscous one you need (incompressible case): n_j 2 mu S_ij Now, S_ij is: S_ij = 1/2 (du_i/dx_j + du_j/dx_i) and n_j du_j/dx_i = d (n_j u_j)/dx_i is zero at walls (for incompressible flows). Thus, what remains for the viscous flux is: n_j mu du_i/dx_j which, besides mu, is the wall-normal derivative of the three velocity components. Again, when working in wall parallel/normal coordinates, the wall normal velocity component has a null wall normal gradient (for incompressible flows). Thus, what actually remains is the wall normal gradient of the wall parallel velocity component. As a consequence, your wall function is supposed to take as input the wall-parallel velocity in the cell center ut and to return the wall parallel velocity gradient in the wall normal direction dut/dn_w. Hence, we finally get: n_j 2 mu S_ij = n_j mu du_i/dx_j = (mu dut/dn_w) (- ut_i) where ut_i is the versor of the wall parallel velocity (notice that the stress acts in the opposite direction with respect to ut). In conclusion, what actually changes for Wall Functions is just the black-box which for given ut returns dut/dn. |
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