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.

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, $\frac{\partial v_x}{\partial x}$ and $v_y$ 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 $v_x\big|_{y = 0} = 0$ and $\frac{\partial v_x}{\partial y}\big|_{y=L} = 0$. However, I don't know why they just assume $\frac{\partial v_x}{\partial x}$ and $v_y$ are 0; where does that come from? The fluid is flowing under the influence of gravity so I would think $v_x$ 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 $v_z \big|_{z=0}=0$ or $\big|_{z=W}=0$ boundary conditions anyway. So the solution would be the same...

  • $\begingroup$ $\partial u/\partial x = 0$ probably comes from mass convervation (MC) and the long-wave approximation. For MC in incompressible flow in 2D cartesian system we have $\partial u/\partial x + \partial v/\partial y = 0$. The long-wave approximation is common in free-surface flows, but may also be applied here, and assumes that length-wise variations are small with respect to the depth-wise variations which comes from the geometry of the flow as $H/L \ll 1$, where $H$ and $L$ are depth-wise and length-wise length scales. Scaling MC then gives $u/L \ll v/H$, hence $\partial u/\partial x \sim 0$ $\endgroup$
    – Dai
    Jul 31, 2019 at 13:57

1 Answer 1


First of all I am going to assume from your question that the viscosity is large enough, so the flow would be laminar (if it is not the case then the best answer would be numerical simulation or estimate based on the law of the wall velocity profile). Furthermore we will also neglect surface tension effects.

Since we have a stationary problem the boundary conditions for the free surface boundary simply express the fact that this boundary must remain stationary.

For instance consider the following picture

enter image description here

Mass does not accumulate in the 'pillbox' object depicted, so the flow of mass through the bottom ($dA_2$) and top ($dA_1$) areas must be identical up to the sign. Since media 1 is vacuum we have $\mathbf{v}_\text{boundary} \cdot \mathbf{n} = 0$, where $\mathbf{n}$ if the unit vector normal to the boundary, i.e. the flow must be tangential to the boundary.

The same argument holds for momentum. Flow of momentum through void is zero so the flow of momentum through $dA_2$ must be zero also. That gives us $\tau_{ij} n_j = 0 $ at the free surface, where $\tau_{ij}$ is momentum tensor of viscous fluid. This reads as $$( - p \delta_{ij}+\sigma_{ij}) n_j = 0,\quad \text{for all }i,$$ where $\sigma_{ij}$ is viscous stress tensor.

If we have the flow with only one velocity component such as $v_x(y,z)$ and if $\mathbf{n}=\mathbf{\hat{y}}$ then this gives us the following conditions: $$ p \vert_\text{boundary} =0, \qquad \left. \frac{\partial v_x}{\partial y}\right\vert_\text{boundary} =0 .$$

The first condition comes from $\tau_{yy}=0$ component, the second from $\tau_{xy}=0$ components. So we have the same boundary condition at the free surface as in 2D problem. Add to this the no-slip conditions at the bottom and at the walls of the channel and we will have enough to solve the problem.

The y-component of the N-S equation gives us the pressure and confirms that fluid free surface is $y=\mathrm{const}$.

The x-component of the N-S reads $$ - g \sin \theta = \nu \left(\frac{\partial^2 v_x}{\partial y^2}+\frac{\partial^2 v_x}{\partial z^2}\right),$$ which could be solved (at least numerically with high precision).

As to why ansatz $\mathbf{v}=v_x(y,z) \mathbf{\hat{x}}$ work for this problem, we will have to analyze the symmetries of the system. We have the stationary flow problem, with the invariance wrt to translations along the x-axis. So this ansatz represents the simplest form that is compatible with the symmetries of the boundary conditions. Note that since the flow is incompressible this limits the possibilities.

In particular the dependence on x should be eliminated if we consider established flow in the channel. That way the energy gained from gravity is exactly equal to the energy dissipated through viscosity, so no 'speedup' occurs.


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