- #1

krabbie

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## Homework Statement

[/B]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: [itex]x[/itex] is the direction the water is flowing, [itex]z[/itex] is perpendicular to the river bed, and [itex]y[/itex] is perpendicular to [itex]x[/itex] (so, [itex]y[/itex] is the cross-current). Consider the surface of the river to be [itex]z=0[/itex], and the bottom to be [itex]z = -h.[/itex]

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

The given assumptions are:

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

## Homework Equations

Navier stokes for incompressible fluid:

[tex]\rho \frac{Du(z)}{Dt} =\rho g \hat{e}^{(x)}- \frac{\partial p}{\partial x} + \mu \nabla^2 u(z) [/tex]

## The Attempt at a Solution

So, I've managed to simplify N-S by noting that

[tex]\frac{\partial u(z)}{\partial t} = 0, \ \ \ \frac{\partial u(z)}{\partial x} = 0, \ \ \ \text{ so } \ \ \ \rho \frac{Du(z)}{Dt} =0,[/tex]

[tex]\rho g_x = -\rho g \sin{\theta},[/tex]

(I used [itex]\theta[/itex] as the grade/slope for now, so I wouldn't have to deal with messy numbers)

[tex]\frac{\partial p}{\partial x} = 0,[/tex]

since pressure is a fluid property and does not vary in the parallel stream, and finally

[tex]\mu \nabla^2 u(z) = \mu \frac{\partial^2}{\partial z^2}u(z).[/tex]

So, my pde is:

[tex]\frac{\partial ^2}{\partial z^2} u(z) = \frac{\rho}{\mu}g\sin{\theta},[/tex]

which I have solved as

[tex]u(z) = \frac{1}{2}\frac{\rho}{\mu}z^2 + k_1z + k_2.[/tex]

Then, I found that the constants were: [itex]k_2 = u(0),[/itex] and [itex]k_1 = \frac{1}{2}\frac{\rho}{\mu}g \sin{\theta}h + \frac{1}{h}u(0),[/itex] which makes my final equation

[tex]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).[/tex]

Here is my problem: I next need to find [itex]u(0).[/itex] But when I sub in [itex]0[/itex] into my equation, I get:

[tex]u(0) = u(0),[/tex]

which is not very illuminating. Did I mess up my solution, or is there another way to find [itex]u(0)[/itex] that I am stupidly missing?

Thanks!