Hydrostatics: Pascal's principle and ocean depth

[Chenkel]

TL;DR Summary

I'm a little confused about the application of Pascal's principle, and the consistency of it's application to varying depths

Hello everyone!

I've been learning about hydrostatics and one thing that I've heard is that the pressure is the same throughout confined incompressible fluids, this is one of the reasons hydraulics work, because when you have the same pressure per square meter, you can change the mechanical advantage of the system by changing the area for the input piston relative to the output piston.

I'm wondering what is the correct methodology for applying Pascal's principle, for example in a hydraulic system we can assume the same pressure throughout, but why can we do that with the hydraulic system and not with the ocean for example?

My intuition is telling me that for hydraulic systems that are very big, hundreds of meters tall, that Pascal's principle may not apply, but I really don't want to assume that before I get the assessment of someone who knows the answer. My intuition also says that Pascal's principle might apply in a hydraulic system of any size because it's a confined fluid, but it's hard for me to see why confinement of fluid causes uniform pressure, and why we might be able to ignore gravity.

If you can shed some light on the matter let me know what you think, thank you!

 

[Dale]

It is just an approximation based on the scale. For example, in my tractor the hydraulic fluid is about 90% as dense as water, and at full extension the maximum fluid height in the system is about 4 or 5 feet. That works out to about 2 psi due to hydrostatic pressure variation with height. My tractor operates at about 3000 psi, so 2 psi is negligible.

It is not that the 2 psi variation is not present. It is there, but it just doesn’t matter.

 

[Arjan82]

Please note that this is a statement about hydrostatics. So, no movement in the fluid. In this case the pressure is equal to $\rho g h$ But this pressure will never be able to produce any work in a confined system, therefore you can ignore this in actual computations. The ocean has a free surface which can deform under pressure. This means that whatever causes a local change in pressure, can cause a local change in surface height and thus a non-uniform pressure arises.

 

[nasu]

Pascal's principles does not state that the pressure is the same throughout the fluid. It states that an external "force" (or change in pressure) is transmitted equally throughout the fluid. For example, the atmospheric pressure acting on the surface of the fluid adds 1 ATM to the hydrostatic pressure for any point of the fluid, at any depth. This addition may be comparable with the hydrostatic pressure for water in a pool or may be negligible for the bottom of the ocean.

 

[vanhees71]

Formally the pressure for an incompressible fluid is a Lagrange multiplier in the formulation of hydrodynamics in terms of the stationary-action principle for the constraint that the density stays constant. That does not imply that the pressure is constant throughout the fluid. It's constant along a free surface of a fluid.

For an ideal fluid the equation of motion reads
$$\rho \mathrm{D}_t \vec{v}=\rho [\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}] = \vec{f}-\vec{\nabla} P,$$
where $\vec{f}$ is the external force per volume. For hydrostatics you have simply
$$\vec{f}=\vec{\nabla} P.$$
For the gravitational force of the Earth (close to Earth) you have
$$\vec{r}=\rho \vec{g}=\rho g \vec{e}_z.$$
Then you have
$$P=P_0+\rho g z,$$
i.e., as expected you have the usual hydrostatic pressure from gravity.

 

[Lnewqban]

The weight of the fluid induces the same exact static pressure for every horizontal layer, which value increases with depth.
Any additional external pressure on the fluid reaches the whole volume of it, in addition to any existing gravity induced static pressure.

This not only works for positive pressures respect to the atmospheric pressure (like in hydraulic actuators).
If you can create enough vacuum on the surface of a contained fluid, the whole mass of it could reach a vaporization point due to reduction of the pressure.
That is a problem with pumps and cavitation.

Please, see:
https://en.m.wikipedia.org/wiki/Cavitation

 

[Chenkel]

Because pressure varies at varying distances from the bottom of the ocean does that mean that there's a flow of liquid from high pressure (bottom) to lower pressure (top)?

Also, I notice Mr. Walter Lewin shows how barometric pressure is measured, and he submerges a tube, and holds one end tight with his finger, the fluid fills the tube as he lifts it vertically, I'm unclear why the fluid does this. He also says that if he lifts the tube higher than ten meters the cranberry juice will break free from the top of the tube, and there will be an empty area at the top of the tube, I'm wondering why there is an empty area, is that a vacuum, or is there air there? Also, what causes the separation of cranberry juice? One thing he says is that the pressure is 0 at the empty point, it is unclear to me why that is the case.

I know these are a lot of questions, but I'm willing to learn, I am just happy that there are so many on this forum that have insights and are willing to shed light.

Thanks again for the responses, they have been very helpful.

 

[Dale]

Because pressure varies at varying distances from the bottom of the ocean does that mean that there's a flow of liquid from high pressure (bottom) to lower pressure (top)?

No, the pressure difference counteracts gravity to keep it hydrostatic

 

[Chenkel]

No, the pressure difference counteracts gravity to keep it hydrostatic

By hydrostatic I'm guessing you mean that the elemental volumes in the tube have static equilibrium?

 

[Dale]

By hydrostatic I'm guessing you mean that the elemental volumes in the tube have static equilibrium?

Yes. Pascal's principle is based on the hydrostatic assumption. It doesn't apply to fluids that are not in static equilibrium, by definition.

 

[vanhees71]

Hydrostatics just meanx that the fluid is at rest. The pressure counteracts gravity everywhere such that the total force on each fluid element is zero.

 

[Chenkel]

Also, I notice Mr. Walter Lewin shows how barometric pressure is measured, and he submerges a tube, and holds one end tight with his finger, the fluid fills the tube as he lifts it vertically, I'm unclear why the fluid does this. He also says that if he lifts the tube higher than ten meters the cranberry juice will break free from the top of the tube, and there will be an empty area at the top of the tube, I'm wondering why there is an empty area, is that a vacuum, or is there air there? Also, what causes the separation of cranberry juice? One thing he says is that the pressure is 0 at the empty point, it is unclear to me why that is the case.

I watched the video again, and have come to the following theory: the liquid rises because there is the barometric force pushing against the weight of the column of water, and if you lift the tube high enough it will create a zero pressure point (because the barometric pressure at the bottom of the column can't push a bigger column of water, this will also create a vacuum), the larger the weight of the column of liquid, the smaller the distance is required to get the same barometric pressure reading.

I believe there is a flaw in my reasoning because if we increase the diameter of the tube, we increase the weight of the column of water, which means we'd get different barometric readings for different diameter pipes.

Mr Lewin uses the following logic: pressure at vacuum point (A) is 0, and the pressure at ground level (B) is one atmosphere, so the pressure is $P = \rho gh$, where h is the distance between the points.

 

[Chestermiller]

Pascal's principle states that the pressure of a fluid at a given depth acts equally in all directions ( horizontally and vertically and in arbitrary directions). It doesn't say anything about the absolute value of the pressure at that depth. So, pressure is "isotropic" at each location. For a deforming fluid, pressure represents the isotropic part of the stress in a fluid. In this case, the stress tensor is not necessarily isotropic.

 

[jbriggs444]

I believe there is a flaw in my reasoning because if we increase the diameter of the tube, we increase the weight of the column of water, which means we'd get different barometric readings for different diameter pipes.

This is a classic difficulty faced by many students. There are tons of textbook exercises designed to test for this confusion and (hopefully) exercise it away.

The easy case would be a U shaped tube with a fixed cross-section throughout. It is open on one end and closed on the other and initially filled with mercury to the top of the tube (maybe one meter high so you can get a vacuum up top).

If you increase the diameter of the tube, you are increasing the weight of the column of mercury. But you are also increasing the area which is exposed to atmospheric pressure. It is a wash -- changing the diameter does not affect the height of the column.

So then you think: "What if I changed the profile of the vertical tube? For instance, tapering it toward the top, tapering it toward the bottom or putting a bulge in the middle".

When you do that, you are just complicating the analysis. Now you have to account for the upward force supplied by the tube walls (if the tube widens on the way up) or the downward force supplied by the tube walls (if the tube narrows on the way up) or the net upward force from the tube walls (if there is a bulge).

You could even try to get super creative with a serpentine tube that goes up and down and up and down and up and down, all the time wide on the way up and narrow on the way down. Now your analysis gets super tricky because of the complicated shapes and the net force of mercury pressure on all of the curves of tubing without a fixed radius.

It will turn out that no matter how you manipulate the shape, the column height will remain unchanged.

Just as Pascal's principle demands.