Flow of viscous fluid between two walls inclined at an angle

Question

I suppose this is a basic fluid mechanics problem but I have one thing I do not understand.

I am about to solve a problem with steady, incompressible, parallel, laminar flow of viscous fluid falling between two infinite walls, inclined at an angle $\theta$. There is no applied pressure driving the flow, it falls by gravity alone. And I want to calculate the velocity field.

In order to solve it I use Navier-Stokes equation for incompressible fluids:

$$\rho\left[\frac{\partial v}{\partial t}+\left(v\nabla \right) v\right]=-\nabla p + \eta\Delta v$$

and with the "keywords" describing the problem it can be simplified to

$$\nabla p=\eta \Delta v$$

The next step is to divide the equation in the $\hat{x}$ and $\hat{y}$ direction . When I do it I just got

$$\frac{\partial p}{\partial x}=\eta \frac{\partial^2v }{\partial y^2} $$

in $\hat{x}$ direction (parallel to the flow).

But there should be a term with the gravitational force component, which reads $\rho g \sin \theta$ here as well I think. Where does it come from? Someone who can help me?

Answer 1

As you say your flow is not driven by a stream-wise pressure gradient, i.e. $\partial_x p=0$. Instead it is driven by gravity, a body force i.e. $\vec{f}=\rho \vec{g}$ which you need to include in the Navier-Stokes equation inorder for something to drive the flow. The Navier-Stokes equations then become:

$$\rho\left[\frac{\partial \vec{v}}{\partial t}+\left(\vec{v}\vec{\nabla} \right)\vec{v}\right]=-\vec{\nabla} p + \eta\Delta \vec{v} + \vec{f}$$

and are simplified to:

$$0=\eta \frac{\partial^2v }{\partial y^2} + \rho g_x$$

where $g_x=\rho g \sin\theta$ with $g$ the gravitational constant. From here you should be able to solve this problem with the correct boundary conditions.

Note that while the stream-wise pressure gradient is zero, the gradient normal to the flow $\partial_y p$ is not. This is the result of the geometry of the system which is infinite in length and width such that this becomes a one-dimensional problem. In that case entrance and exit effects can be neglected and so can wall effects except for in the y-direction which under the influence of gravity have a hydrostatic pressure component found from: $$0=-\frac{\partial p}{\partial y}+ g_y$$