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Navier-Stokes equation for parallel flow

  1. Nov 13, 2015 #1
    1. The problem statement, all variables and given/known data
    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:
    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 [itex]\vec{u}=u(z)\hat{e}^{(z)}[/itex].
    5. No fluid property varies in the direction parallel to the bottom[itex](x).[/itex]
    6. No fluid property varies in the cross-stream [itex](y)[/itex] direction.
    7. The gravity vector is [itex]-g\hat{e}^{'(z)},[/itex], where the prime indicates vertical in gravity aligned coordinates.
    8. At the river bottom, [itex]u=0[/itex].
    9. At the surface, [itex]du/dz = 0.[/itex]
    2. Relevant 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]

    3. 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!
     
  2. jcsd
  3. Nov 13, 2015 #2
    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
     
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