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## Homework Statement

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...