Flow of a Newtonian fluid down an inclined plane.

AI Thread Summary
The discussion focuses on the velocity profile of laminar flow for a Newtonian fluid on an inclined plane, emphasizing the assumptions of incompressibility and fully developed flow. The participants analyze the forces acting on an infinitesimal control volume, particularly the balance of momentum flux and accumulation. A key point of contention is the constancy of pressure at a free liquid surface, with clarification that while pressure is atmospheric at the surface, it varies along the streamlines due to hydrostatic equilibrium. The discussion concludes that pressure remains constant along streamlines but is influenced by gravitational acceleration in the normal direction to the incline. Understanding these dynamics is crucial for accurately modeling fluid behavior in inclined planes.
siddharth
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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 assumptions being made are
- 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.
 
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Free liquid surfaces means it is at atm pressure, ie. zero pressure gauge.
 
Ouch. Yeah, it's kinda obvious now :blushing:
 
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siddharth said:
w 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.

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.
 
Clausius2 said:
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.

Yeah, I get it. Thanks
 
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