|
[Sponsors] |
March 6, 2007, 02:34 |
BC for none vertical outflow
|
#1 |
Guest
Posts: n/a
|
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 |
|
March 6, 2007, 09:41 |
Re: BC for none vertical outflow
|
#2 |
Guest
Posts: n/a
|
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. |
|
March 6, 2007, 11:04 |
Re: BC for none vertical outflow
|
#3 |
Guest
Posts: n/a
|
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 |
|
March 6, 2007, 11:21 |
Re: BC for none vertical outflow
|
#4 |
Guest
Posts: n/a
|
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. |
|
March 6, 2007, 11:48 |
Re: BC for none vertical outflow
|
#5 |
Guest
Posts: n/a
|
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... |
|
March 6, 2007, 11:57 |
Re: BC for none vertical outflow
|
#6 |
Guest
Posts: n/a
|
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. |
|
March 20, 2007, 15:54 |
Re: BC for none vertical outflow
|
#7 |
Guest
Posts: n/a
|
The outflow BC "A ZERO DIFFUSISION FLUX in the direction normal to the boundary for ALL flow variables" is generally used.
|
|
|
|
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Outflow boundary Control Volume SIMPLE Help !!! | Mizerable | Main CFD Forum | 4 | August 22, 2011 14:00 |
No outflow - Error | sErik | OpenFOAM Running, Solving & CFD | 1 | November 18, 2009 12:10 |
mass flux correction at outflow boundaries | Subhra Datta | Main CFD Forum | 2 | November 24, 2003 14:11 |
outflow boudary condition in SIMPLE algorithm | dallybird | Main CFD Forum | 0 | March 6, 2003 12:55 |
Outflow Bocos | Romuald Skoda | Main CFD Forum | 3 | August 6, 1999 04:08 |