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udf boundary condition (vorticity)

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Old   May 18, 2020, 08:37
Exclamation udf boundary condition (vorticity)
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Hi
I don't know how to write udf boundary condition for vorticity .I bring to your attention that my work has been done on the nanofluid and more precisely I am working on the configuration of Rayleigh Bénard (natural convection).

Fig. 1 shows a schematic diagram of the Rayleigh-Bénard (RB) problem. For the RB problem, the distance between the upper cold and lower hot plates is defined by H and the width of the top and bottom plates is defined by W. The plates’ width W is considered infinite and treated by a periodicity boundary condition. The bottom plate is maintained at a hot temperature TH=50 C whereas the top plate is maintained at a cold temperature TC=22 C. The fluid enclosed between the plates is water based nanofluid containing CuO nanoparticles. The nanofluid is assumed incompressible and the flow is assumed as laminar and twodimensional.
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Old   May 18, 2020, 09:08
Exclamation boundary using UDF -vorticity-
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Can anyone tell me how to apply dv/dx=0 and omega=du/dy at a boundary using UDF.
[v=delta(psi)/delta(x) u=-delta(psi)/delta(y)]
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Old   May 18, 2020, 09:15
Default Wall boundary
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Fluent uses momentum conservation and not vorticity or Poisson's equation. The equation you have is written in terms of streamfunction and implies no-slip condition. That is the default condition in Fluent and you don't need to do anything
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Old   May 18, 2020, 09:52
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Quote:
Originally Posted by vinerm View Post
Fluent uses momentum conservation and not vorticity or Poisson's equation. The equation you have is written in terms of streamfunction and implies no-slip condition. That is the default condition in Fluent and you don't need to do anything
But I fond in the results that dv/dx is not equal 0
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Old   May 18, 2020, 09:55
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That won't be 0 and should not be. Look at the document you shared. The condition is NOT

\frac{d^2\psi}{dy^2} = 0

but \psi = 0, implying velocity is 0.
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Old   May 18, 2020, 10:05
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Quote:
Originally Posted by vinerm View Post
That won't be 0 and should not be. Look at the document you shared. The condition is NOT

\frac{d^2\psi}{dy^2} = 0

but \psi = 0, implying velocity is 0.
I think psi=0 that mean (noslip) and omega(rotational flow)??
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Old   May 18, 2020, 10:15
Default Noslip
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That's correct and no-slip implies velocity is 0 at the wall, i.e., the default condition in Fluent. As far as rotation is concerned, in a 2D case, rotation is possible only around z-axis, so, \Omega = (0~0~\omega_z) where \omega_z is a function of velocity gradient.
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Old   May 18, 2020, 10:21
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Quote:
Originally Posted by vinerm View Post
That's correct and no-slip implies velocity is 0 at the wall, i.e., the default condition in Fluent. As far as rotation is concerned, in a 2D case, rotation is possible only around z-axis, so, \Omega = (0~0~\omega_z) where \omega_z is a function of velocity gradient.
so I don't need to do anything
okay thank you so much
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