# 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.
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!