Flow down an incline - Understanding boundary conditions

Question

After working with some problems regarding flow, I came up to a similiar problem as the one presented here:

In solving the problem, we assume a laminar flow in steady state.

When using Navier-Stokes equations, to fully determine the velocity profile, I get a differential equation of order two. Meaning I need two boundary conditions in order to fully describe the velocity profile. First of all, if we assume the incline is stationary, no-slip forces us to set $v_x(0) = 0$ where $v_x$ is the velocity in the $x$-direction and a function of $y$. Now, since I've seen examples of solving this before, they also assumed that $v_x'(h)=0$, and it was somehow related to the shear stress.

First, I'd like to understand how certain partial derivatives of the velocity with combinations of different directions relate to shear stresses (i.e. derivative of $v_x$ with respect to $y$ or $v_y$ with respect to $x$).

Secondly, I want to understand why this boundary conditions holds. I assume that it has to do with friction and viscosity of the fluid. Since no fluid is above that region, we will have no shear force from any layer.

Answer 1

Recall that the viscosity stress tensor for an incompressible Newtonian fluid is (by definition): $$ \sigma=\eta(\nabla\otimes v+\nabla\otimes v^T) $$ You can derive this from a microscopic theory (transport phenomena) or phenomenologically (assume linearity, isotropy, and only dependence in first order spatial derivatives).

Btw, the resulting viscous shear forces at the boundary of a region $\Omega$ is: $$ F=\int \sigma \cdot d^2x $$ You obtain the Navier-Stokes by considering momentum balance in an infinitesimal domain or equivalently using Stokes’ formula.

In your case, using the equations give the shear stress of the free surface to be: $$ \sigma_{xy}=\eta(\partial_xv_y+\partial_yv_x) $$ Using the laminar hypothesis, this simplifies to: $$ \sigma_{xy}=\eta v_x’ $$ So the boundary condition of the free surface is equivalent to: $$ \sigma_{xy}(y=h)=0 $$

As you’ve written, this is justified physically. It essentially assumes that the air is inviscid so cannot apply a shear stress at the interface.

Hope this helps.