# Viscous Fluids at a Slope (Navier-Stokes)

An in-compressible viscous fluid flows down a flat slope of angle θ to the horizontal under the force of gravity, with g the acceleration due to gravity. What are the boundary conditions for the fluid at the point of contact with the slope and at the free surface?

What I know: $$u=0$$ at $$y=0$$ and $$\frac{du}{dy}=0$$ at y=d. Then $$\vec {g}=[g sin(a),-gcos(a),0]$$

Using orthogonal coordinates with the x-axis pointing down the slope and the y-axis perpendicular to the slope, find a solution to the Navier-Stokes equation for a flow of depth d down the slope under the assumptions that the flow is steady and uniform in the x-direction, including an expression for the pressure.

This is how i interpreted the question in a figure: A river descends by 100m over a distance of 100km. Given that the dynamic viscosity of water is approximately $$\mu=10^{−3}kgm^{−1}s^{−1}$$ estimate the predicted speed of the river using your own estimates for any other parameters involved.

Is it unrealistic? If so give possible reasons for the lack of realism.

What I know: is the Navier-Stokes Equation is... $$\dfrac{d\vec{v}}{d t}+ \vec{v} .\nabla \vec{v} = \vec{F} - \dfrac{1}{\rho} \nabla p + \nu \nabla^2 \vec{v}$$

Now I'm a bit unsure but I think due to the fluid being in-compressible we now have the equation: $$\dfrac{d\vec{v}}{d t}+ \vec{v} .\nabla \vec{v} = \vec{F} - \dfrac{1}{\rho} \nabla p$$

Really stuck from here, on this question don't really know how to approach it as I'm fairly new to fluid dynamics any help would be greatly appreciated.

You are conflating incompressible with inviscid. Here we have flow of an incompressible viscous fluid and the term $$\nu \nabla^2 \mathbf{v}$$ may not be neglected.

You are asked to find "a solution" under the assumptions that the flow is steady and uniform in the x-direction. This implies fully-developed unidirectional flow where the only non-vanishing component of velocity is $$u$$ (i.e., the component in the x-direction).

For an incompressible fluid, the equation of continuity gives

$$\nabla \cdot \mathbf{v} = \frac{\partial u}{\partial x} = 0$$

Thus, $$u$$ is a function only of the y-coordinate and the Navier-Stokes equations reduce to

$$0 = -\frac{1}{\rho}\frac{\partial p}{\partial x} +g \sin \alpha + \nu \frac{d^2u}{dy^2}, \\0 = -\frac{1}{\rho}\frac{\partial p}{\partial y} -g \cos \alpha$$

Since the flow is gravity-driven, we can neglect the x-component of the pressure pressure gradient $$\frac{\partial p}{\partial x}$$ in the first equation and find $$u$$ by applying the two boundary conditions to the solution of the second-order differential equation

$$\nu \frac{d^2 u}{dy^2} = -g \sin \alpha$$

The second equation allows for solution of the pressure as a function of the y-coordinate.

See if you can finish now.

• I got the answer after integrating twice: $u=-\frac{gy^2}{2v}sin(a)+Ay+B$, How would I go about calculating the speed with $\mu$ given? @RRL – Reety Dec 24 '18 at 5:21
• @Reety: Correct. The boundary conditions you identified above are correct as well , so you can solve for A and B. We get $B = 0$ etc. – RRL Dec 24 '18 at 5:26
• so the final answer is $u=\frac{g}{2v}sina(2dy-y^2)$? and is this realistic? – Reety Dec 24 '18 at 5:34
• So you can set $y = d$ to get the maximum velocity at the surface. They want you to plug in values for the river and see if you get a realistic speed. They specified the viscosity $\mu$ so you need to get the kinematic viscosity $\nu = \mu/\rho$ by dividing by the density which is about $1 g/cm^3$ for water. They give you enough information to find the angle. They don't tell you the depth $d$ but you can use a number like $10 m$. – RRL Dec 24 '18 at 5:45
• aah right okay, ill try and figure it out now. Thank you very much! – Reety Dec 24 '18 at 5:47