# Flow of viscous fluid between two walls inclined at an angle

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?

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: $$0=-\frac{\partial p}{\partial y}+ g_y$$

• But i f the pressure gradient is zero, the equation in y-direction, perpendicular to the plane , will reduce to gy=0, which can not bu true? I don't understand why it must be zero just because there is no applied pressure driving the fluid. – C.Andersson Sep 24 '16 at 17:09
• @C.Andersson good question! I should have been clearer that the pressure gradient in the stream wise direction is zero but the gradient normal to the flow is not. This has to do with the geometry of the system, which is infinite in length and width such that entrance, exit and wall effects are negligible and this becomes a one-dimensional problem. However, the top and bottom of the channel are walls and under the influence of gravity a hydrostatic pressure is produced such that dp/dy=gy. I hope that clears it up. I am on my phone atm so I will make an edit later to clarify this in the answer. – nluigi Sep 24 '16 at 18:25