# Flow of a Newtonian fluid down an inclined plane.

1. Aug 30, 2006

### siddharth

I'm basically reading on how the velocity profile is found for a laminar flow of a Newtonian fluid down an inclined plane surface. (x is along the incline, y is perpendicular to the incline)

- The fluid is Newtonian
- It's laminar
- It's fully developed
- It's incompressible

What the book did, was to take an infinitesimal control volume, find the forces acting on it, and equate it to the sum of the linear momentum flux and rate of accumulation of momentum in the c.v (along the x-direction)

I understand how the sum of the flux and the accumulation is zero. Next, the book evaluates the forces.
It says,

$$\sum F_x = P \Delta y|_x - P \Delta y|_{x+\Delta x} + \tau_{yx} \Delta x|_{y+\Delta y} - \tau_{yx} \Delta x|_y + \rho g \Delta x \Delta y \sin \theta$$

which I understand.

Then it says
Note that the pressure-force terms also cancel because of the presence of a free liquid surfaces.

This is what I don't understand. Why should the pressure be constant for a free liquid surface? For example, if we take a fluid between two cylinders, and rotate the inner cylinder (and make the same assumptions), then the centrifugal force (you know what I mean) would cause a pressure gradient along the radial direction. So, even at the free surface at the top, the pressure won't be constant.

Last edited: Aug 30, 2006
2. Aug 30, 2006

### Cyrus

Free liquid surfaces means it is at atm pressure, ie. zero pressure gauge.

3. Aug 30, 2006

### siddharth

Ouch. Yeah, it's kinda obvious now

Last edited: Aug 30, 2006
4. Aug 30, 2006

### Clausius2

That's not true. If you leave the top open, as Cyrus said, the pressure is the atmospheric one at the top of the upmost film. The centrifugal force will curve the shape of the surface as it were a paraboloid, such that the hydrostatic pressure balances the centrifugal overpressure generated.

In your problem, the book should say that pressure is consant along the streamlines, but it is holding hydrostatic equilibrium normal to the plane, but instead with the gravitational acceleration g, with $$gcos\theta$$. So it is not uniform in the normal direction to the plane.

5. Aug 30, 2006

### siddharth

Yeah, I get it. Thanks