# Applying Navier Stokes to Fluid on Inclined Plane

I am trying to solve the well known problem of identifying the velocity profile of a fluid on an inclined plane. The assumptions we know are that the flow is Newtonian, incompressible, with a low reynolds number. I have been following my professor's notes on the problem, but ran into an assumption that I do not understand.

In evaluating Navier Stokes along an axis in line with the incline, we assume that the change in pressure along the incline $$\cfrac{\partial P}{\partial z}$$ is 0, since the flow is gravity driven. I do not understand why it is reasonable for us to make this assumption.

By evaluating the continuity equation, and Navier stokes along the other two axes, we discover that the velocity along the incline does not change (no acceleration). In addition, there is no pressure change out of the page and perpendicular to the incline axis (duh). However, I don't see any mathematical expression that leads to this conclusion.

Firstly, i want to set the coordinate system so that there is no confusion:

• I assume the x-coordinate is along the incline wall and the y-coordinate is perpendicular to the incline wall.
• In such a coordinate system, the acceleration due to gravity $\boldsymbol{g}$ is expressed as:

$$\boldsymbol{g}=\left(g_x,g_y\right)=\left(g\sin\theta,g\cos\theta\right)$$

where $g$ is the gravitational acceleration constant and $\theta$ is the angle of the incline.

Secondly, these typical text-book questions on velocity profiles for inclined planes always have a number of implicit assumptions:

• The flow is fully developed - This means that any transient effects have vanished and we are dealing with a steady-state system.
• The flow is laminar - This means that the fluid flow parallel to the incline wall in organized layers such that the y-component of the velocity is zero and the x-component of the velocity only depends on the y-coordinate.
• Entrance/exit effects are negligible - This means that any open boundaries to the flow are very far away from the section of the incline we are dealing with such that they do not interfere with the flow.

With these assumptions the Navier-Stokes equations reduce to:

$$0=-\frac{1}{\rho}\partial_xp + \nu\partial_y^2u_x + g_x \qquad 0=-\frac{1}{\rho}\partial_yp + g_y$$

Note that the continuity equation is satisfied identically with the above assumptions.

I do not understand why it is reasonable to assume that there is no pressure gradient along the incline wall, i.e. $\partial_xp=0$?

Consider the ways in which a pressure gradient may form:

• By an applied pressure at the in/out-flow boundaries
• By a hydrostatic contribution due to fluid resting on a wall.

Neither of these cases apply here as no applied pressure is specified and the in/out-flow boundaries are open (otherwise no flow) such that a hydrostatic pressure cannot be generated. Conclusion the pressure along the incline wall must vanish.

In addition, there is no pressure change out of the page and perpendicular to the incline axis (duh)

Actually, since $g_y\neq0$, there exists a pressure gradient in the y-direction and it is purely hydrostatic:

$$\partial_yp = \rho g_y = \rho g \cos\theta$$

If this may seem unintuitive, consider the case where the incline is perfectly horizontal ($\theta=0$). In that case there is no flow ($g_x=0$) and the fluid is simply resting on the wall exerting a hydrostatic pressure due to gravity ($g_y=g$).

• Thanks for the awesome explanation! It really makes sense now. The idea of pressure wasn't very clear to me before. – Austin Nov 20 '16 at 23:20
• @Austin Happy to help! As an addendum; you may be interested in reading up on Deen's Analysis of Transport Phenomena, specifically the topic on dynamical pressure. – nluigi Nov 23 '16 at 17:56