In an incompressible Ideal fluid, can the pressure increase with depth?

Question

Ideal Fluid is defined as an "In-Compressible Fluid". Without taking "Compressibility" into account, is it really possible that pressure increases with the depth?

When we consider compressibility in the fluid then I understand to some extent that pressure will increase with depth, but in the case of In-Compressible fluid how do we justify that density is uniform everywhere but still the pressure is increased?

I request the responder to kindly give the answer at the molecular level reasoning, whichever case the responder supports.

Answer 1

The pressure increase with depth has nothing to do with compressibility nor molecular level physics.
It is simply the total weight of the fluid column above you that increases with depth. Take a water column from Earth, put it on Mars, and on the bottom you will have less pressure, due to the lower surface gravity, hence weight of the column.

The hydrostatic pressure $P$ at any point $z$ in a fluid is given by integrating the z-component of the Navier-Stokes equations, which is $$\frac{\partial P}{\partial z} = -g\rho \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$$ with $g$ being the surface gravity and $\rho$ the fluid density. Integrating this equation requires knowledge about $\rho(z)$, which is just a constant for incompressible fluids. You see, the physics is the same no matter if you're compressible or incompressible.

You mention ideal gases in your self-answer: Whether a gas is ideal or not, changes only the relation between $P$, $\rho$ and possibly $T$, called equation of state. This is where microphysics is actually encapsulated, but is an independent physics ingredient that you use in order to solve (1).

Answer 2

I think both the answers given so far are incomplete. You can give a microscopic answer to why pressure increases with depth with simplified models. I have written a lengthy post on this just now but I will list the main concepts here as well. Keep in mind that in the end all of the models, kinetic theory of gases as well as any continuum science are simplifications and approximations of an inherently complex nature. (In particular I share the view that physics are deterministic but we just do not have sufficient initial data. Thus in particular any probabilistic method is only a way to describe the missing information that we are simply lacking.)

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Pressure on a macroscopic level

Interactions in actual fluids can be very complicated - a balance of contracting and repulsive forces that vary with distance of the particles from each other. This makes such phenomena very unattractive (and complicated) to model from a microscopic perspective. Somewhat surprisingly the macroscopic laws take the same form for dense liquids and moderately dilute gases. For instance the Navier-Stokes equations in their traditional form are valid in the continuum limit for Newtonian fluids. Liquids and gases only differ in terms of order of magnitude of dimensional numbers and equation of state. Both are nothing but viscous dampers.

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Kinetic theory of gases

In the kinetic theory of gases you model the a gas as a multi-body system of particles interacting in collisions (or even complicated far-field interactions). For comparably simple elastic collisions one can find

$$\lambda = \frac{m_P}{\sqrt{2} \pi d^2 \rho}$$

for the mean free path where the density $\rho$ is coupled to the static pressure by the equation of state of an ideal gas

$$p \, v = \frac{p}{\rho} = R_m T.$$

This means that static pressure, which takes into account the hydrostatic pressure, is inversely proportional to the mean free path and thus with increasing pressure particles are more closely packed. They exert a bigger force on their surroundings as simply more particles collide with the wall (collisions happen more often).

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Simple analogy for liquids

This simplified view can also be transferred to a solid where the spheres are so densely packed that they can't really be compressed anymore (incompressible fluid). The force and thus the pressure in between the particles increases as they are pressed together more strongly.

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Misuse of incompressibility

An ideal gas law, incompressibility and hydrostatic pressure are incompatible concepts. If you fix the density (which is by definition of incompressibility) for an increasing static pressure (which is the natural consequence of hydrostatic pressure) the temperature has to increase as well, meaning particles would have to move faster in areas of higher pressure which would equilibrate and thus such a configuration would not be stable. This means a flow of an ideal gas with a large pressure gradient can never be assumed incompressible. Such a simplification is incompatible with the equation of state! Nonetheless the flow around a car might be assumed incompressible as hydrostatic pressure is approximately equal for all points and the change of static pressure due to dynamic pressure (Bernoulli's principle) is comparably small.

This is differently for liquids such as water where the governing equation of state is the Tait equation

$$p - p_0 = C \left[ \left( \frac{\rho}{\rho_0} \right)^m - 1 \right], $$

which clearly allows for large pressure ranges with small changes in density for the typical value of $m \approx 7$. As a consequence water can be assumed as incompressible over a wide pressure range.

Answer 3

Without taking "Compressibility" into account, is it really possible that pressure increases with the depth?

Yes. Compressibility has nothing to do with the cause of pressure at a particular depth, which is given solely by the weight of the fluid above.

If the fluid above is compressible, that might increase the weight above a particular depth, but the pressure at depth $D$ does not in any way depend on the compressibility at depth $D$. To see this, consider a bag of oil, a bag of water and a bag of air (a balloon) all under 3m of water: The pressure in the bags will be the same, even though their compressibility is different.

The questioner seems to want a "microscopic" explanation, but there is no microscopic explanation for the pressure at depth $D$ in terms of microscopic phenomena at $D$; it's determined by the total material above.

What can be considered microscopically, hence locally, is how pressure equilibrates in a microscopic volume: If the pressure at $D$ is $p(D)$, how does the fluid arrange to have pressure $p(D+\epsilon)$ at depth $D+\epsilon$? But the mechanism for that doesn't depend on compressibility per se at all; compressibility actually cancels out in the result. Rather, the fluid rearranges itself (one molecule at a time) until the pressure below plus the weight of the small volume of molecules matches the pressure above, and therefore the flux of molecules up vs down matches and equilibrium is reached.

Although that equilibrium will demonstrate the compressibility of the fluid, it doesn't require it: the same mechanism will work for real and ideal (incompressible) fluids.

Answer 4

See it like this. The water molecules are all already packed. Their movement toward each other can be seen like charging a spring with a non constant rising k. So after reaching a certain level even a small compression in water density will lead to a very high pressure force. This non constant increasing constant of the simulated spring is because of electrons defying each other.