Outflow boundary condition

I know that in outflow boundary we assume a zero normal gradient condition and use upwind scheme for approximation. However, I saw this sentence in a book which I do not understand; "Convective fluxes are usually assumed to be independent of the coordinate normal to an outflow boundary." What does it exactly mean and why are we allowed to make that assumption?

Thanks.

• What book? Did the statement come with any more context? Apr 1, 2016 at 13:29
• @BillBarth Thanks for the reply. It is from Computational Methods for Fluid Dynamics by Joel H. Ferziger. You can see that page in the link below. [link] (books.google.de/…)
– Eman
Apr 1, 2016 at 22:11
• Both your assumption and the statement printed in bold are, as an aside, only true for scalar convection problems. Apr 7, 2016 at 20:44
• @WolfgangBangerth Thanks a lot for the comment. I am not sure if I quite understood what you mean. Because, for the transport of momentum, which is obviously a vector field, we can assume a zero normal gradient for streamwise velocity on the outflow surface provided that the flow is close to fully developed condition and the fluid is incompressible. My guess is, in this particular case, the streamwise component of velocity is a scalar field by itself, therefore, the assumptions above cannot be extended to vector fields. Hopefully, I got it right.
– Eman
Apr 8, 2016 at 11:56
• I was thinking of systems of equations, e.g., the Euler equations of gas dynamics. There, a boundary may be in- and out-flow at the same time. Apr 8, 2016 at 19:01

I understand it as a Neumann boundary condition $\frac{\partial F}{\partial n} = 0$ where $F$ is the convective flux and $n$ is the unit normal vector at the boundary. This is the most convenient boundary condition to solve numerically the problem.
"Independent of the coordinate normal to [the boundary]" means that, in the local coordinate system of the boundary defined by the vectors $(n, t_1, t_2)$, the field $F$ may locally vary only along the tangential coordinates $(t_1, t_2)$. Hence $\frac{\partial F}{\partial n} = 0$.