Navier-Stokes equation for parallel flow

[krabbie]

Homework Statement

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Find an equation for the flow velocity of a river that is parallel to the bottom as a function of the perpendicular distance from the surface. Apply the boundary conditions given and solve, and find the velocity at the surface. Note that the coordinates are: x is the direction the water is flowing, z is perpendicular to the river bed, and y is perpendicular to x (so, y is the cross-current). Consider the surface of the river to be z=0, and the bottom to be z = -h.

The river is has a grade (slope) of 43cm per km. The water depth is 2m. The kinematic viscosity \mu/\rho = 10^{-6}m^2/s. The vertical gravitational acceleration is g = 9.8 m/s^2.

The given assumptions are:

1. The flow obeys the Navier-Stokes equation for an incompressible flow.
2. The current is steady.
3. The density is uniform.
4. The current is a parallel shear flow \vec{u}=u(z)\hat{e}^{(z)}.
5. No fluid property varies in the direction parallel to the bottom(x).
6. No fluid property varies in the cross-stream (y) direction.
7. The gravity vector is -g\hat{e}^{'(z)},, where the prime indicates vertical in gravity aligned coordinates.
8. At the river bottom, u=0.
9. At the surface, du/dz = 0.

Homework Equations

Navier stokes for incompressible fluid:
\rho \frac{Du(z)}{Dt} =\rho g \hat{e}^{(x)}- \frac{\partial p}{\partial x} + \mu \nabla^2 u(z)

The Attempt at a Solution

So, I've managed to simplify N-S by noting that
\frac{\partial u(z)}{\partial t} = 0, \ \ \ \frac{\partial u(z)}{\partial x} = 0, \ \ \ \text{ so } \ \ \ \rho \frac{Du(z)}{Dt} =0,
\rho g_x = -\rho g \sin{\theta},
(I used \theta as the grade/slope for now, so I wouldn't have to deal with messy numbers)
\frac{\partial p}{\partial x} = 0,
since pressure is a fluid property and does not vary in the parallel stream, and finally
\mu \nabla^2 u(z) = \mu \frac{\partial^2}{\partial z^2}u(z).
So, my pde is:
\frac{\partial ^2}{\partial z^2} u(z) = \frac{\rho}{\mu}g\sin{\theta},
which I have solved as
u(z) = \frac{1}{2}\frac{\rho}{\mu}z^2 + k_1z + k_2.
Then, I found that the constants were: k_2 = u(0), and k_1 = \frac{1}{2}\frac{\rho}{\mu}g \sin{\theta}h + \frac{1}{h}u(0), which makes my final equation
u(z) = \frac{1}{2}\frac{\rho}{\mu}g\sin{\theta}z^2 + \left[\frac{1}{2}\frac{\rho}{\mu}g\sin{\theta}h + \frac{1}{h}u(0)\right]z + u(0).
Here is my problem: I next need to find u(0). But when I sub in 0 into my equation, I get:
u(0) = u(0),
which is not very illuminating. Did I mess up my solution, or is there another way to find u(0) that I am stupidly missing?

Thanks!

 

[Chestermiller]

I didn't check every last detail, but you need to make use of the fact that the shear stress is equal to zero at z = 0.

Chet