Chapter 7

Fluid Mechanics

Flow of a viscous fluid on an inclined plane

Take 20 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

If you have more questions, feel free to ask them on the forum.

One consider a steady flow of layer of incompressible and viscous fluid, of height , on an inclined plane.

It is assumed one-dimensional flow : the velocity field is parallel to the axis Ox and will depend only on the variable .

At the free surface, the pressure is uniform and is .

The density of the fluid, assumed Newtonian, is and its viscosity .

It will be recognized that because of the low viscosity of the air above the fluid, there is no shear stress at .

Flow of a viscous fluid on an inclined plane

Given the Navier-Stokes equation for a Newtonian fluid :


Simplify and project the Navier-Stokes.



In a permanent regime :

Convective acceleration is equal to , with the hypotheses of the statement : ( )

Therefore :


Determine the velocity field by taking into account boundary conditions.



The previous equation gives, in projection ( and does not depend on ) :

The second equation given by integration :

The first becomes, noting that  :

The velocity is zero at and the air does not exert tangential force on the liquid at , thus :


So :

A parabolic velocity profile with a maximum at is obtained.


We are interested in a wide flow along the axis Oy, with , in order to neglect edge effects.

Calculate the volume flow rate and deduce the average fluid velocity.



The volumetric flow rate is :

And the average velocity of the flow is :


We are interested in a layer of glycerin to which :


and of thickness . The angle is .

Calculate the average speed and comment on the result using the Reynolds number.



We get :

And the Reynolds number is :

Viscosity forces are sufficient to impose a field of laminar speeds.

Navier-Stokes equation
Reynolds number