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Is there any specific reason for pressure to be constant along the x axis, which is represented on this picture below.

If I think about an elemental volume of fluid with its base parallel to the x axis I see mainly 5 forces, which are the viscous shear forces on the upper and lower faces, two pressure forces on the right and left faces, and the x component of the force of gravity.

Now, apparently the two pressure forces cancel and the gravity alone will balance the shear forces, but how can we reach that conclusion?

It is easy for me to understand that a pressure gradient must exist on a horizontal flow with viscous stresses because it is what must balance those friction forces. But now the balance is made only due to gravity? Wouldn't it be plausible to assume that it would be a contribution from both gravity and pressure gradient?

Thank you!

Flow on an inclined plane in contact with air at a height h relative to the x axis directed along the inclined plane

Normal forces acting on an elemental fluid volume

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2 Answers 2

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This is a problem with a free surface. At the free surface, the pressure is equal to that of the air outside the fluid. The pressure is continuous across the free surface, so immediately below the free surface, the pressure is also atmospheric. This is true at all values of x. The fluid velocity normal to the incline (and normal to the free surface is zero). So,in the y direction, the pressure has to vary hydrostatically, with a pressure gradient in the y direction determined by $\rho g \sin{\theta}$. This happens at each value of x. So there is no pressure gradient in the x direction. All that is left to cause the fluid to flow (at constant thickness) down the incline is the gravitational component of force in the x direction.

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  • $\begingroup$ But does continuous pressure mean constant pressure? I agree with all the reasoning you presented except when you say that because it is a continuous function, the value of pressure immediately below the surface is the same as on the surface. And that would mean that the pressure at a random x section would be all equal to the atmospheric pressure? $\endgroup$
    – George Sailor
    Oct 27, 2017 at 17:50
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    $\begingroup$ Pressure is continuous at the boundary between two materials (actually, normal stress and shear stress). If not, if you do a force balance on a segment of the boundary, it will have infinite acceleration (since the interface has zero mass). Regarding a random x section, the pressure will be a function of y (only), given by $$p=p_{atm}+(h-y)\rho g \cos{\theta}$$ (in my previous comment, I wrote sin when I should have written cos). So, the pressure will be a function of y, but not x. $\endgroup$
    – Chet Miller
    Oct 27, 2017 at 18:43
  • $\begingroup$ Having constant pressure at the free surface immediately means that all the surfaces parallel to that one will have the same pressure? I'm posting an image above to add some content to my question. Basically all the forces that act on the elemental volumes are the same so it's the reason why the pressure will be constant on parallel surfaces? $\endgroup$
    – George Sailor
    Oct 27, 2017 at 20:42
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    $\begingroup$ In this problem, it means that all the surfaces parallel to the free surface will have the same pressure. In other problems, where, say the free surface is moving or non-flat, that would not be the case. To your second question, yes, the same differential force balance applies in the y direction at all values of x. So, at any depth (in the direction normal to the free surface), the pressure doesn't vary with x. $\endgroup$
    – Chet Miller
    Oct 27, 2017 at 21:04
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    $\begingroup$ It depends on the specific problem and on the taste of the person solving the problem. In this problem, it grabbed me that it was better to recognize from the get-go that the pressure is dictated by the air pressure at the free surface, and that the derivative of pressure in the x direction is zero. Experience, I guess. $\endgroup$
    – Chet Miller
    Oct 27, 2017 at 21:15
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Because this is a question similar to the one that I answered in Flow on an inclined plane and Bernoulli's principle, so I just copy Part II of my answer here:
Acheson = Acheson, D.J.: Elementary Fluid Dynamics, Oxford: Clarendon Press, 2005.
First we notice the notes [p.24 in https://projects.exeter.ac.uk/fluidflow/Courses/FluidDynamics3211-2/slides/lecture3-slides.pdf] for the figure in p.23 of the same webpage. Note 3 [No pressure gradient] is our final result for $p$ . It is important because it facilitates calculating the fluid velocity $\pmb{u}$ . The proof of $\nabla p=\pmb{0}$ is given in [Acheson, p.39, l.6--l.− 7]. §1.12.(A) in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/quantum-mechanics may help you read [Acheson].

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