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Additional boundary conditions for inclined flow?

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data

    I am solving an inclined flow problem, and am stuck. The problem is to find the volumetric flow rate of inclined flow in a square channel. Once I have the velocity profile, I can just integrate over that to get the flow rate.

    2. The attempt at a solution

    Letting the x-axis be along the direction of flow, I start with Navier-Stokes:

    $$\rho v_x \frac{\partial v_x}{\partial x} + \rho v_y \frac{\partial v_x}{\partial y} = \rho g \sin \theta + \mu \left( \frac{\partial^2 v_x}{\partial x^2} + \frac{\partial^2 v_x}{\partial y^2} \right)$$

    $$\rho v_x \frac{\partial v_y}{\partial x} + \rho v_y \frac{\partial v_y}{\partial y} = \rho g \cos \theta + \mu \left( \frac{\partial^2 v_y}{\partial x^2} + \frac{\partial^2 v_y}{\partial y^2} \right)$$

    No pressure terms since it's a free fluid. I found in a textbook the solution for inclined plane flow (but my problem is inclined flow in a square channel). In that solution, [itex]\frac{\partial v_x}{\partial x}[/itex] and [itex]v_y[/itex] are assumed to be 0, which gives

    $$-\frac{\rho}{\mu} g \sin \theta = \mu\frac{\partial^2 v_x}{\partial y^2}$$

    which is easy enough to solve with the boundary conditions [itex]v_x\big|_{y = 0} = 0[/itex] and [itex]\frac{\partial v_x}{\partial y}\big|_{y=L} = 0[/itex]. However, I don't know why they just assume [itex]\frac{\partial v_x}{\partial x}[/itex] and [itex]v_y[/itex] are 0; where does that come from? The fluid is flowing under the influence of gravity so I would think [itex]v_x[/itex] is increasing along x (imagine if you roll a ball down an inclined plane), and I would also think a free flow would decrease in height (L) as it travels. I mean, if you continuously dump water at the top of the incline, you're going to have a thicker water "height" at the top I would think...

    And even if I accept that those assumptions are true, I don't know how to extend the problem from a plane to a square channel. There's no z-flow anyway, so I don't need any [itex]v_z \big|_{z=0}=0[/itex] or [itex]\big|_{z=W}=0[/itex] boundary conditions anyway. So the solution would be the same...
     
  2. jcsd
  3. May 2, 2013 #2
    This is not set up correctly. First of all, the coordinates should be taken parallel and perpendicular to the inclined plane. Secondly, the problem statement implies that the flow is fully developed (the length of the incline is very large compared to the side of the square channel). The coordinate along the incline should be z, and the two velocity components in the cross channel directions should be zero. Furthermore, the derivative on the down-channel velocity with respect to the down-channel coordinate z should be zero (again, fully developed). Write your partial differential equation for the velocity vz under these circumstances. You should end up with a Poisson equation in vz. Boundary conditions are that vz is zero at the walls of the square duct.

    Chet
     
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