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Old   March 6, 2007, 02:34
Default BC for none vertical outflow
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CH
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for a FVM, if the outflow BC is du/dx=0, u(i)=u(i+1).

however, if the edge (for 2d) is not vertical, how should the BC du/dx be imposed? i believe it should be different since in this case, du/dx /= du/dn.

can someone enlighten me? thank you
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Old   March 6, 2007, 09:41
Default Re: BC for none vertical outflow
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Harish
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Do you perform coordinate transformation or solve in the original domain?

If you perform coordinate transformation then du_new/dx_new=0 and you use the definition of contravariant velocity components to satisfy the boundary bondition.If you are solving in the original cartesian coordinates, you can write the equation in original form V.n=0 and expand them and implement the boundary condition.
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Old   March 6, 2007, 11:04
Default Re: BC for none vertical outflow
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CH
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hi harish, i'm solving in the original structured grids. the BC at outlet is du/dx & dv/dx=0. how did u get V.n=0? are V & n vectors? in that case, are u saying V is perpendicular to the edge? but that is not the case...

am i missing something?

hope you can clarify...

thanks
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Old   March 6, 2007, 11:21
Default Re: BC for none vertical outflow
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Harish
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I think I got the original idea wrong.I assume that you set both du/dx=0 and dv/dx=0 on the assumption that flow perturbations decay in far field.By the way is your solver compressible or incompressible.? For compressible flows this condition might not always work.

Regarding curved surface you can assume similarly that du/dn=0 and dv/dn=0. Then use the chain rule to transform the boundary condition to the original cartesian direction.

du/dn=du/dx*dx/dn+du/dy*dy/dn

dv/dn=dv/dx*dx/dn+dv/dy*dx/dn

calculate the value of dx/dn and dy/dn.If dx/dn is smaller then

du/dy=-(du/dx*dx/dn)/dy/dn and so on.

This has worked for me in few cases.

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Old   March 6, 2007, 11:48
Default Re: BC for none vertical outflow
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CH
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oh no, i'm getting confused. let me clarify my qn:

flow is incompressible

initially the outflow far field boundary is vertical (in c-grid), du/dx & dv/dx =0 are the BC used.

now due to rotation of the grid, outflow boundary is no longer vertical, but slanted. what should the Bc for out flow be now?

is it still du/dx & dv/dx =0? or du/dn, dv/dn=0. i believe the 2 are different.

the next qn is of course how to discretize them...
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Old   March 6, 2007, 11:57
Default Re: BC for none vertical outflow
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Harish
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You can still try du/dx=0 and dv/dx=0 and check if it works.It might for few cases.Depending on what order you use for discretization you can use

f_n=f_n-1 (first order ) f_n=2*f_n-1 - f_n-2 (second order)

nope du/dn and dv/dn are different from du/dx and dv/dx.Here n is the normal direction to the curved surface.

check the book of hirsch.I think he has a very good discussion on the implementation of the boundary condition.
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Old   March 20, 2007, 15:54
Default Re: BC for none vertical outflow
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ramp
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The outflow BC "A ZERO DIFFUSISION FLUX in the direction normal to the boundary for ALL flow variables" is generally used.

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