What is the meaning of the free surface in fluid mechanics?

[Adesh]

I’m currently studying Fluid Mechanics, during an analysis I came across this

We now consider an example that combines centrifugal force and gravity: a liquid in a drum (centrifuge) rotates with constant angular velocity $\omega$ about a vertical axis. The centrifugal force per unit of volume is $$ \mathbf F = F_{r} \hat r ; F_r = \rho r \omega ^2 ; \hat r \text{is a unit vector} \perp \text{to the axis}$$
Its potential is given by $$ U = - \frac{1}{2} \rho r^2 \omega ^2 $$
The total potential of gravity and centrifugal force is (except for a constant) given by
$$ U = - \rho g z - \frac{1}{2} \rho r^2 \omega ^2 = - \gamma \left (
z + \frac{ r^2 \omega ^2}{2 g} \right ) $$ From the equation $p = -U + constant$ we can write $$ p = \gamma \left (
z + \frac{r^2 \omega^2 }{2g} \right) + constant $$
[we can find the constant by doing few simple things and after getting it we can write] $$ p = \gamma \left(
z-z_0 + \frac{r^2 \omega ^2}{2g} \right)$$ The free surface, characterised by $p=0$, has the equation $$ z_0 -z = \frac{r^2 \omega ^2}{2g} $$

My problem what does that line “the free surface, characterised by $p=0$” means? I tried searching Wikipedia and found that free surface means where there is no shear stress but I couldn’t connect it to here. How zero shear stress would make $p=0$ in the above equations?

P.S. : the book from which I have quoted is Sommerfeld’s Lectures on Theoretical Physics Vol II.

 

[etotheipi]

This is somewhat speculative, since I haven't read the book, however might $p$ be defined as the hydrostatic pressure wrt the surface of the liquid? The surface of zero pressure would then give you the surface of the liquid, which as far as I can recall is indeed the expression you have in your last line.

 

[Adesh]

This is somewhat speculative, since I haven't read the book, however might $p$ be defined as the hydrostatic pressure wrt the surface of the liquid? The surface of zero pressure would then give you the surface of the liquid, which as far as I can recall is indeed the expression you have in your last line.

$p$ denotes the gauge pressure, but I thought it is only at $r=0$ and on the surface that the gauge pressure would be zero because at some $r \neq 0$ we would have centrifugal pressure.

 

[etotheipi]

$p$ denotes the gauge pressure, but I thought it is only at $r=0$ and on the surface that the gauge pressure would be zero because at some $r \neq 0$ we would have centrifugal pressure.

I don't know enough to provide a good answer, I'm not sure what centrifugal pressure is (it might well be a thing!). To my mind there is in this case only hydrostatic pressure which is a function of position i.e. $p(r,z)$. Even at $r=0$ I would think that this pressure varies linearly with $z$.

 

[vanhees71]

I don't know how to translate "free surface". Here it's just the surface of a fluid in a rotating bucket (to take Newton's famous (gedanken) experiment). Obviously here you consider hydrostatics of a rotating fluid under homogeneous gravity of the Earth. The free surface then must be an isobaric surface, and that's what's calculated here.

 

[Adesh]

I don't know how to translate "free surface". Here it's just the surface of a fluid in a rotating bucket (to take Newton's famous (gedanken) experiment). Obviously here you consider hydrostatics of a rotating fluid under homogeneous gravity of the Earth. The free surface then must be an isobaric surface, and that's what's calculated here.

I’m sorry, but I request for a little more explanation.

 

[jbriggs444]

I’m sorry, but I request for a little more explanation.

You have water in a bucket. The bucket is spinning. The surface of the water assumes a curved shape. The "free surface" is the upper surface of the water. This is a "free" surface because the water is unconstrained there. It is free to flow. By contrast, the water at the bucket walls and bottom is constrained by the walls and is not [completely] free to flow.

Clearly, the atmospheric pressure is constant just above the surface. At equilibrium, the fluid pressure is also constant just below the surface. It is a surface of constant pressure -- isobaric.

 

[Dale]

The free surface is the surface of the fluid which is not touching the rotating drum. I.e. the surface touching the atmosphere.

 

[Adesh]

You have water in a bucket. The bucket is spinning. The surface of the water assumes a curved shape. The "free surface" is the upper surface of the water. This is a "free" surface because the water is unconstrained there. It is free to flow. By contrast, the water at the bucket walls and bottom is constrained by the walls and is not [completely] free to flow.

Clearly, the atmospheric pressure is constant just above the surface. At equilibrium, the fluid pressure is also constant just below the surface. It is a surface of constant pressure -- isobaric.

Is gauge pressure zero everywhere on the free surface?

 

[jbriggs444]

Is gauge pressure zero everywhere on the free surface?

Yes.

At equilibrium, it has to be. If the fluid pressure were less than ambient at some place on the surface, air would accelerate in, forming an indentation. If the fluid pressure were more than ambient at some place on the surface, the fluid would accelerate out, forming a bulge. If there is an equilibrium, there can be no acceleration. Everything has stabilized. So one can conclude that the pressure must have equalized.

 

[Adesh]

Yes.

At equilibrium, it has to be. If the fluid pressure were less than ambient at some place on the surface, air would accelerate in, forming an indentation. If the fluid pressure were more than ambient at some place on the surface, the fluid would accelerate out, forming a bulge. If there is an equilibrium, there can be no acceration. Everything has stabilized. So one can conclude that the pressure must have equalized.

Thank you so much, my doubt has been cleared.

 

[Chestermiller]

A free surface is the interface between a liquid and a gas.