0
$\begingroup$

Good evening, kind forum users.

I'm trying to understand the microscopic mechanisms that justify the Bernoulli principle. I found an interesting discussion on the site of NASA that tries to describe, in a general way, the mechanism underlying the pressure reduction in favor of the increase in speed. I propose the text in question, and then I explain to you my doubt.

"We can make another interpretation of the Bernoulli equation by considering the motion of the gas molecules. The molecules within a fluid are in constant random motion and collide with each other and with the walls of an object in the fluid. The motion of the molecules gives the molecules a linear momentum and the fluid pressure is a measure of this momentum. If a gas is at rest, all of the motion of the molecules is random and the pressure that we detect is the total pressure of the gas. If the gas is set in motion or flows, some of the random components of velocity are changed in favor of the directed motion. We call the directed motion "ordered," as opposed to the disordered random motion. We can associate a "pressure" with the momentum of the ordered motion of the gas. We call this pressure the dynamic pressure. The remaining random motion of the molecules still produces a pressure called the static pressure. From a conservation of energy and momentum, the static pressure plus the dynamic pressure is equal to the original total pressure in a flow (assuming we do not add or subtract energy in the flow). The form of the dynamic pressure is the density times the square of the velocity divided by two."

[Text from: https://www.grc.nasa.gov/www/k-12/airplane/bern.html ]

From the text we learn that the pressure reduction with speed can be explained by thinking that part of the "chaotic" kinetic energy translates into "ordered" kinetic energy. The remaining chaotic kinetic energy gives rise to static pressure, while the ordered kinetic energy is the so-called "dynamic pressure". If a particle of fluid initially at rest is set in motion, then a force must have acted. This force comes from the pressure difference on the sides of the fluid particle. The text suggests that the motion of the particle is due to the conversion of a part of the chaotic kinetic energy into ordered kinetic energy, thanks to the pressure unbalance on the sides of the fluid particle. Therefore the speed increases and the static pressure decreases."

I'm trying to understand this better. The force acting on the fluid particle, performing work, should add kinetic energy to the system, not converting an energy already present in the particle into another. Why do we have a conversion of potential energy (static pressure) into ordered kinetic energy to the action of a force on our system (the fluid particle)? Should we not maintain the same chaotic kinetic energy and, if anything, have that extra due to the work of force?

Thank you all!

$\endgroup$
  • $\begingroup$ Please add the URL where you found the quoted text. $\endgroup$ – Ján Lalinský Apr 2 at 21:11
  • $\begingroup$ ok i will edit immediately! $\endgroup$ – Valerio Quattrini Apr 3 at 10:52
0
$\begingroup$

The quoted explanation is misleading, almost incorrect. The connection between pressure and chaotic motion is valid only in a limited way and only for gases; it is not valid for incompressible liquid flows such as water in pipes.

The force acting on the fluid particle, performing work, should add kinetic energy to the system, not converting an energy already present in the particle into another.

This is exactly right and this work of external forces on the fluid element is the origin of the pressure term in the Bernoulli equation. This equation is a convenient rewrite of the work-energy theorem applied to non-viscous fluid flows: work done by external forces equals change of kinetic energy of a body. If the work is done only by forces due to pressure gradients (no gravity or other kinds of external forces), then this work is entirely captured by the pressure term in the Bernoulli equation.

In this derivation of the Bernoulli equation from the work-energy theorem there is no chaotic motion involved; such motion is a microscopic detail that has no appearance or role in the theory behind the Bernoulli equation. For adiabatic gas flows, it is true that lowering pressure means comparable lowering of average kinetic energy of chaotic motion of molecules (indicated by lowered temperature in an adiabatic process of gas expansion), but for liquid flows such as water in pipes there is no universal basis for such effect (water expansion with pressure decrease is significantly smaller and decrease of the average kinetic energy of chaotic motion is negligible).

Bernoulli equation in its standard (simplest) form is most accurate model of incompressible flow of liquids, where there is no simple relation between adiabatic decrease of pressure and change of temperature. So it is not a good idea to explain the Bernoulli equation based on behaviour of ideal gases.

$\endgroup$
  • $\begingroup$ Thank you! Then how can we explain the reduction in pressure with speed at a microscopic level? I mean considering a smlal cube containing gas molecules. How it work the convertion of Chaotic Kinect Energy in Ordered Kinect Energy while a gradient of pressure is acting on the particle? $\endgroup$ – Valerio Quattrini Apr 3 at 8:01
  • $\begingroup$ It is explained in molecular statistical physics: the expanding gas works on its environment and therefore loses internal energy. This manifests as decreased temperature. $\endgroup$ – Ján Lalinský Apr 3 at 11:10
  • $\begingroup$ Ok. Would you like to describe this aspect in more detail? It is precisely the part that I want to understand. I imagine a fluid particle initially at rest, which is set in motion due to a pressure imbalance in its surroundings. How do you justify its conversion of energy from chaotic to ordinate, also taking into account the energy introduced by the work of the force acting as a pressure force? $\endgroup$ – Valerio Quattrini Apr 3 at 14:42
  • $\begingroup$ Perhaps it would be better if you asked that as a new question, without mentioning the misleading justification for the Bernoulli equation. $\endgroup$ – Ján Lalinský Apr 3 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.