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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.

Devansh Mittal
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    Downvoted because the questioner gave a wrong answer to his own question, then selected it as the correct answer. – Bob Jacobsen Dec 12 '19 at 16:22
  • You argue that pressure does not increase with depth in an incompressible fluid. This is physically wrong. It doesn’t matter what viewpoint is considered. – Bob Jacobsen Dec 12 '19 at 16:30
  • I have considered your answer to be the right answer. I thought more on my line of thought and I realised that if we go to the zero kelvin then the thermal motion of molecules would vanish but the idea of pressure would still remain, so thermal motion is not considered while discussing the idea of pressure. – Devansh Mittal Dec 13 '19 at 05:48

4 Answers4

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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).

Community
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AtmosphericPrisonEscape
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  • This is not true. The fluid column presses molecules together more closely. This can solidly argued with the relation between mean free path, density and static pressure in the kinetic theory of gases. The difference between incompressible and compressible is lumped into the equations of state and the flow. Refer to my answer below for a more detailed argument. – 2b-t Dec 16 '19 at 23:36
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    @2b-t: "The fluid column presses molecules together more closely" is exactly what (1) expresses. How the molecules then react to this change in hydrostatic pressure, is given by an independent expression, the equation of state. Your confused answer that throws static and hydrostatic pressure somehow into different concepts doesn't change that at all. – AtmosphericPrisonEscape Dec 17 '19 at 11:27
  • Sorry, I misunderstood your argument in that way that there is no explanation of pressure increase with depth on a molecular level. In only tried to point out in my answer that there is. Thanks for your feedback, I'd be glad if you could give me some advice on how to improve my answer. Nonetheless static and hydrostatic pressure are indeed two different concepts: The hydrostatic pressure is only the part of the static pressure stemming from gravity. The static pressure of a confined fluid on the other hand can also increase due to temperature or amount of material. – 2b-t Dec 17 '19 at 21:36
  • Assume a closed tank, filled with a compressible fluid, that is sufficiently high such that the change in pressure due to the fluid column above can't be neglected. At the top there will be the static pressure $p_1$ and at the bottom the static pressure $p_2$. The static pressure $p_1$ has nothing to do with an increase of pressure with height, it is a result of squashing a liquid in a confined space. The difference between the two $p_2 - p_1$ will be the hydrostatic pressure $\frac{\partial p}{\partial z} = - \rho g$, which clearly is only a possible contribution to the static pressure. – 2b-t Dec 17 '19 at 21:43
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    @2b-t The pressure $p_1$ contributes equally to the pressure above 'squaching' the fluid or gas at the bottom of the container. Take a needle, make a hole into the bottom, and measure the ejected fluid velocity, as proxy for the total static pressure at the bottom - you will get incorrect values if you neglect $p_1$. – AtmosphericPrisonEscape Dec 17 '19 at 22:04
  • This is exactly what I said, the static pressure $p_2$ in point 2 is the static pressure $p_1$ in point 1 plus the hydrostatic pressure between the two points (whatever form it may take, that of course depends on the equation of state). Thus I differentiated between static and hydrostatic pressure: The hydrostatic pressure is only the difference in pressure between the points 2 and 1 and not the entire static pressure you measure in 2. – 2b-t Dec 17 '19 at 22:15
  • The hydrostatic pressure is only the integral of your equation (1) without a constant so without (!) $p_1$. It only takes into account gravity effects (the part that is contributed to the weight of the fluid) and not the effects of being locked up in a container. For an incompressible fluid the hydrostatic pressure is only $\rho g z$ and not $p_1 + \rho g z$. The latter is the entire static pressure. – 2b-t Dec 17 '19 at 22:24
  • As a consequence the hydrostatic pressure in point 1 is zero and in point 2 it is for an incompressible fluid $\rho g h_{1-2}$. The static pressure in point 1 is $p_1$ and the static pressure in point 2 $p_2 = p_1 + \rho g h_{1-2}$. The hydrostatic pressure by definition does not equal the static pressure. – 2b-t Dec 17 '19 at 22:32
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    @2b-t which has nothing to do at all whether you're compressible or not. Because you still have to integrate $dp/dz=-g\rho$ from $0$ to $\infty$. Your integral always splits into two parts, from 0 to $z_{top,container}$ over the incompressible fluid and then from $z_{top,container}$ to $\infty$ over air, which gives you $p1$. No matter if the fluid in your container is incompressible or not, $p1$ is always there. – AtmosphericPrisonEscape Dec 17 '19 at 22:39
  • Where did I say so this would make a difference? I said you can't combine an ideal gas with incompressibility and hydrostatic pressure. If there is hydrostatic pressure and an ideal gas law there will be compressibility (as the fluid has to fulfill the equation of state), if you combine hydrostatic pressure and incompressibiity this violates the equation of state of an ideal gas. – 2b-t Dec 17 '19 at 22:46
  • I have seen three or four posts on this platform (also one by the author of this one) where people claim that incompressibility and hydrostatic pressure are incompatible by observing that this leads to a temperature gradient for an ideal gas law. I have only tried to point out that the combination of all of these concepts is impossible. An ideal gas might be assumed incompressible for low Mach number flow but not when calculating the static pressure profile in the atmosphere. – 2b-t Dec 17 '19 at 22:47
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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.)

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.

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).

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.

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.

2b-t
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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.

Bob Jacobsen
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  • All of which has nothing to do with either compressibility or change of pressure with depth. – Bob Jacobsen Dec 13 '19 at 05:25
  • I have considered your answer to be the right answer. I thought more on my line of thought and I realised that if we go to the zero kelvin then the thermal motion of molecules would vanish but the idea of pressure would still remain, so thermal motion is not considered while discussing the idea of pressure. – Devansh Mittal Dec 13 '19 at 05:45
  • Now since I have removed my response and also changed my view, so now would you consider to change your vote so that others too can benefit from this question? – Devansh Mittal Dec 13 '19 at 05:46
  • @BobJacobsen Compressible or not is mainly a question of the mean free path on the microscopic level. In macroscopic modelling microscopic interactions are simply lumped into the equation of state. An ideal gas can never be combined with large changes in hydrostatic pressure and incompressibility. This is incompatible with the equation of state. On a microscopic level one can argue with the kinetic theory of gases that due to the column of fluid above molecules are pushed more closely together and thus compressed. This is reflected by an inverse proportionality of mean free path and pressure. – 2b-t Dec 16 '19 at 23:43
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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.

babak yousefi
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