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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:

Flow down a Slope in two dimensions

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.

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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.

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  • $\begingroup$ 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 $\endgroup$
    – Reety
    Dec 24, 2018 at 5:21
  • $\begingroup$ @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. $\endgroup$
    – RRL
    Dec 24, 2018 at 5:26
  • $\begingroup$ so the final answer is $u=\frac{g}{2v}sina(2dy-y^2)$? and is this realistic? $\endgroup$
    – Reety
    Dec 24, 2018 at 5:34
  • $\begingroup$ 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 $10 m$. $\endgroup$
    – RRL
    Dec 24, 2018 at 5:45
  • $\begingroup$ aah right okay, ill try and figure it out now. Thank you very much! $\endgroup$
    – Reety
    Dec 24, 2018 at 5:47

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