What is pressure accourding to Bernoulli's theorem?

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Hello everyone,
In Bernoulli's theorem, I understand Potential energy (because of height) and Kinetic energy (because of velocity), but I don't understand pressure [energy]; Is it something like the vibration of molecules and bumping them into each other (in simple words).
Any help or simulation link would be greatly appreciated.
Thanks,
Behrouz
 

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  • #2
russ_watters
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Yes, pressure in gases arises due to the momentum transfer of molecules bouncing off of each other and the walls of the container. Increase the number of collisions (by compressing the gas and pushing the molecules close together) and the pressure goes up. Increase the number and speed of the collisions (by increasing the temperature) and it also goes up.
 
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  • #3
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n Bernoulli's theorem, I understand Potential energy (because of height) and Kinetic energy (because of velocity), but I don't understand pressure [energy]; Is it something like the vibration of molecules and bumping them into each other (in simple words).

Let's say we have an evacuated vacuum chamber with valve that we can use to let air into the chamber.

Now we open the valve just a little bit. The kinetic energy of the air seems to come from the potential energy of the atmosphere. Because the atmosphere is descending. Right?

Okay now we turn the valve completely open. Now the kinetic energy of the air seems to come from the surrounding air. Because the pressure of the surrounding air decreases. Right?


Addition: Same story as above but the chamber is now located at the bottom of the sea. There is one difference: air cools when its pressure decreases, but water doesn't.
 
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Let's say we have an evacuated vacuum chamber with valve that we can use to let air into the chamber.

Now we open the valve just a little bit. The kinetic energy of the air seems to come from the potential energy of the atmosphere. Because the atmosphere is descending. Right?

Okay now we turn the valve completely open. Now the kinetic energy of the air seems to come from the surrounding air. Because the pressure of the surrounding air decreases. Right?


Addition: Same story as above but the chamber is now located at the bottom of the sea. There is one difference: air cools when its pressure decreases, but water doesn't.
None of this makes much sense to me, particularly since we do not need gravitational potential energy to have pressure or to change pressure. Maybe you can explain in more detail.

@russ_watters provides a coherent explanation of what pressure actually is.
 
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  • #5
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Let's say we have an evacuated vacuum chamber with valve that we can use to let air into the chamber.

Now we open the valve just a little bit. The kinetic energy of the air seems to come from the potential energy of the atmosphere. Because the atmosphere is descending. Right?

Okay now we turn the valve completely open. Now the kinetic energy of the air seems to come from the surrounding air. Because the pressure of the surrounding air decreases. Right?


Addition: Same story as above but the chamber is now located at the bottom of the sea. There is one difference: air cools when its pressure decreases, but water doesn't.
Yes, pressure in gases arises due to the momentum transfer of molecules bouncing off of each other and the walls of the container. Increase the number of collisions (by compressing the gas and pushing the molecules close together) and the pressure goes up. Increase the number and speed of the collisions (by increasing the temperature) and it also goes up.
Thank you, that makes sense. But please kindly note that Bernoulli's theorem refers to in-compressible fluids.
 
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But please kindly note that Bernoulli's theorem refers to in-compressible fluids.
Oh, I was not aware of that. So now I know that what we are interested about is:

pressure [energy] of some liquid

Liquids under pressure do not have much energy. They are like an extremely stiff spring that has been compressed by a large force.
 
  • #7
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None of this makes much sense to me, particularly since we do not need gravitational potential energy to have pressure or to change pressure. Maybe you can explain in more detail.
Well we are interested about "pressure [energy]", as OP said.

Pressure [energy] of liquid is not quite the same as pressure [energy] of gas. Maybe that is why Bernoulli's principle only applies to incompressible flows.

Pressure [energy] of gas is thermal energy. Pressure [energy] of liquid is mechanical energy, like energy in a compressed spring.
 
  • #8
russ_watters
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Thank you, that makes sense. But please kindly note that Bernoulli's theorem refers to in-compressible fluids.
Yes, (though there is a compressible version), but your question wasn't really about Bernoulli's principle.
 
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  • #9
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Yes, (though there is a compressible version), but your question wasn't really about Bernoulli's principle.
Thanks again. The reason I asked about concept of pressure in Bernoulli is that in Bernoulli, each section has energy nature (i.e. Potential, Kinetic, and, Pressure energy). Potential and Kinetic energy is understandable to me, but not Pressure energy. With your explanation, I'd assume the pressure energy is coming from the movement of molecules and their bumping into each other.
Regarding the Compressible version Bernoulli, I'd appreciate if you could give me a link, when you get a chance, as I couldn't find any online.
Thanks for your help again. Truly appreciated.
 
  • #10
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Thanks again. The reason I asked about concept of pressure in Bernoulli is that in Bernoulli, each section has energy nature (i.e. Potential, Kinetic, and, Pressure energy). Potential and Kinetic energy is understandable to me, but not Pressure energy. With your explanation, I'd assume the pressure energy is coming from the movement of molecules and their bumping into each other.
Regarding the Compressible version Bernoulli, I'd appreciate if you could give me a link, when you get a chance, as I couldn't find any online.
Thanks for your help again. Truly appreciated.
The Bernoulli equation is basically the fluid equivalent of the work-energy theorem. The pressure term represents the applied force part of the work-energy theorem which changes the kinetic energy and potential energy of the fluid.

For a reference to the compressible version of the Bernoulli equation, see Transport Phenomena by Bird, Stewart, and Lightfoot.
 
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  • #11
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Well we are interested about "pressure [energy]", as OP said.

Pressure [energy] of liquid is not quite the same as pressure [energy] of gas. Maybe that is why Bernoulli's principle only applies to incompressible flows.

Pressure [energy] of gas is thermal energy. Pressure [energy] of liquid is mechanical energy, like energy in a compressed spring.
During my long professional career, my main area of expertise was fluid mechanics. In all the 50 years since I received my PhD (doing a fluid mechanics thesis), I have never heard the term "pressure energy." Please precisely define this term, using something other than "the energy of pressure."
 
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  • #12
Lord Jestocost
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In all the 50 years since I received my PhD (doing a fluid mechanics thesis), I have never heard the term "pressure energy."
The term "pressure energy" is sometimes used as a synonym for "flow work". But, as M.J. McPherson remarks in the book "Subsurface ventilation and enviromental engineering":

"As fluid continues to be inserted into the pipe to produce a continuous flow, then each individual plug must have this amount of work done on it. That energy is retained within the fluid stream and is known as the flow work. The appearance of pressure, P, within the expression for flow work has resulted in the term sometimes being labelled "pressure energy". This is very misleading as flow work is entirely different to the "elastic energy" stored when a closed vessel of fluid is compressed."
 
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  • #13
russ_watters
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During my long professional career, my main area of expertise was fluid mechanics. In all the 50 years since I received my PhD (doing a fluid mechanics thesis), I have never heard the term "pressure energy." Please precisely define this term, using something other than "the energy of pressure."
I don't necessarily like it, but here it is:

http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html
 
  • #14
vanhees71
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Thank you, that makes sense. But please kindly note that Bernoulli's theorem refers to in-compressible fluids.
In an in-compressible fluid pressure can be understood as a Lagrange parameter to meet the boundary conditions of a fluid at a wall in Hamilton's principle of least action. See: A. Sommerfeld, Lectures on Theoretical Physics, vol. 2.
 
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  • #15
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During my long professional career, my main area of expertise was fluid mechanics. In all the 50 years since I received my PhD (doing a fluid mechanics thesis), I have never heard the term "pressure energy." Please precisely define this term, using something other than "the energy of pressure."
By pressure energy I mean the energy of elastic deformation of fluid parcels. And by that I mean that the 'deformed' parcels are squeezed into a smaller volume.

Like, if a hole appears suddenly on a submarine's wall, the energy of the water stream must come from somewhere, it does not come from potential energy because there is no net downwards flow of water, so the energy comes from the "pressure energy". (If we want a steady state situation, we can say that a pump is pumping the water out of the submarine)

If we decide to consider water to be incompressible, then I might say: "water has no pressure energy at all".

(If we want to use Bernoulli's theorem in the case of punctured submarine, then we assume that water is incompressible and the energy of the stream comes from the potential energy of water as the sea level starts to fall immediately when the hole on the submarine wall appears. Right?)
 
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  • #16
vanhees71
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Well, pressure is (part of) the stress tensor, i.e., the space-space component of the energy-momentum tensor. Of course, as such pressure occurs in the energy-momentum balance equation, but it's not an energy (density).
 
  • #17
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By pressure energy I mean the energy of elastic deformation of fluid parcels. And by that I mean that the 'deformed' parcels are squeezed into a smaller volume.

Like, if a hole appears suddenly on a submarine's wall, the energy of the water stream must come from somewhere, it does not come from potential energy because there is no net downwards flow of water, so the energy comes from the "pressure energy". (If we want a steady state situation, we can say that a pump is pumping the water out of the submarine)

If we decide to consider water to be incompressible, then I might say: "water has no pressure energy at all".

(If we want to use Bernoulli's theorem in the case of punctured submarine, then we assume that water is incompressible and the energy of the stream comes from the potential energy of water as the sea level starts to fall immediately when the hole on the submarine wall appears. Right?)

But, in the context of gases I do not always mean by pressure energy the energy of elastic deformation of fluid parcels. In that context I mean by pressure energy the thermal energy of the gas, which may change without any deformation gas parcels, as happens in incompressible flow of gas.

@Behrouz: Bernoulli's theorem does actually apply to compressible fluids like gases, if the densities of gas parcels stay constant, which means that volumes of gas parcels stay constant, which is possible even if gas parcel's pressures change, if temperatures of gas parcels change too by a correct amount.
 
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  • #18
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But, in the context of gases I do not always mean by pressure energy the energy of elastic deformation of fluid parcels. In that context I mean by pressure energy the thermal energy of the gas, which may change without any deformation gas parcels, as happens in incompressible flow of gas.

@Behrouz: Bernoulli's theorem does actually apply to compressible fluids like gases, if the densities of gas parcels stay constant, which means that volumes of gas parcels stay constant, which is possible even if gas parcel's pressures change, if temperatures of gas parcels change too by a correct amount.
Thank you.
 
  • #19
FactChecker
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Thanks again. The reason I asked about concept of pressure in Bernoulli is that in Bernoulli, each section has energy nature (i.e. Potential, Kinetic, and, Pressure energy). Potential )and Kinetic energy is understandable to me, but not Pressure energy. With your explanation, I'd assume the pressure energy is coming from the movement of molecules and their bumping into each other.
Regarding the Compressible version Bernoulli, I'd appreciate if you could give me a link, when you get a chance, as I couldn't find any online.
In preparation for understanding Bernoulli, you should start to distinguish between pressure from molecules going in random directions, with equal probability in each direction, versus molecules tending to go in a particular direction causing a flow velocity. And the velocity of any molecule can be separated into a component equal to the flow velocity (if any) and the remaining, random-direction velocity component.

PS. It always surprises me how high the average velocity of a molecule in the atmosphere is. "For typical air at room conditions, the average molecule is moving at about 500 m/s (close to 1000 miles per hour)." (from https://pages.mtu.edu/~suits/SpeedofSound.html )
 
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  • #20
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\

(If we want to use Bernoulli's theorem in the case of punctured submarine, then we assume that water is incompressible and the energy of the stream comes from the potential energy of water as the sea level starts to fall immediately when the hole on the submarine wall appears. Right?)
This is totally and utterly incorrect, and borders on misinformation. Bernoulli's equation works fine for a punctured submarine even if gravitational terms are omitted from the equation. The cause of the pressure being high outside the submarine is incidental.
 
  • #21
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Thank you all. I can this the beauty of nature in balancing the 'assets' of molecules at each moment; The summation of velocity, distance from the centre of the earth, and 'vibrations and bumping to other molecules around' should be constant. Fair enough :)
 
  • #22
boneh3ad
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I have become completely and utterly lost in this thread, which is really a testament to some of the strange explanations in here given that I do this for a living.

Let's start with some basic facts here:
  • Bernoulli's equation applies traditionally to incompressible flows, not just incompressible fluids. This is an important distinction, as it applies equally to liquids and gases provided that the velocities involved are small enough that the flow remains incompressible in a gas.
  • The terms in Bernoulli's equation can be written such that they are all in the units of pressure, which in turn, is equivalent (in terms of units) to an energy per volume.
There are three terms, typically. It seems everyone here has a pretty good grasp on two of them: the gravitational term (hydrostatic pressure) and the velocity term (dynamic pressure). The first of these is related to gravitational potential energy per volume, and the second is the bulk kinetic energy of the flowing fluid per volume.

The static pressure term seems to be the problematic one, yes? It is perhaps best illustrated if you go look into kinetic theory. If you do that, it should be reasonably clear that, if you want to think in terms of energy, the static (thermodynamic) pressure is essentially a measure of the kinetic energy per volume due to the random motions of the molecules that comprise the fluid. Note that this motion is random, so the fluid is not actually moving since the average of all the molecules' velocities is zero, but all those velocities squared, then averaged, is nonzero.
 
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  • #23
vanhees71
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Bernoulli's equation is for the stationary flow of non-relativistic ideal fluids (no matter if compressible or incompressible). It's derivable from the continuity equation (local equation for mass conservation) and the Euler equation of motion. For stationary flows one has ##\partial_t =0## for all fields. Thus the continuity equation is
\begin{equation}
\label{1}
\vec{\nabla} \cdot (\rho \vec{v})=0
\end{equation}
and the equation of motion
\begin{equation}
\label{2}
\rho (\vec{v} \cdot \vec{\nabla}) \vec{v}=-\vec{\nabla} p +\rho \vec{g},
\end{equation}
where ##p## is the pressure and ##\vec{g}## the gravitational acceleration close to Earth.

Since for ideal fluids there's no heat exchange by assumption the entropy is conserved. From now on it's convenient to write everything concerning extensive thermodynamical quantities in terms of their specific values, i.e., the quantity per unit mass of the fluid. So we define ##s## and ##h## as the entropy and enthalpy per unit mass. Then the 1st law of thermodynamics reads
\begin{equation}
\label{3}
\mathrm{d} h=T \mathrm{d} s + \frac{1}{\rho} \mathrm{d} p=\frac{1}{\rho} \mathrm{d} p \; \Rightarrow \; \vec{\nabla} h = \frac{1}{\rho} \vec{\nabla} p.
\end{equation}
The final step uses the fact that the ideal-fluid flow is adiabatic, i.e., ##\mathrm{d} s=0##.

Now the trick is to rewrite the left-hand side of (\ref{2}) with help of the identity
\begin{equation}
\label{4}
\frac{1}{2} \nabla \vec{v}^2=\frac{1}{2} \vec{\nabla} v^2 = \vec{v} \times (\nabla \times \vec{v}) + (\vec{v} \cdot \vec{\nabla}) \vec{v},
\end{equation}
After some algebra using this equation in (\ref{2}) one gets using (\ref{4})
$$\vec{v} \times (\vec{\nabla} \times \vec{v})=\vec{\nabla} \left (h + \frac{v^2}{2} +g z\right).$$
Here we have chosen the coordinate system such that ##\vec{g}=-g \vec{e}_z##. This implies that
\begin{equation}
\label{5}
\vec{v} \cdot \vec{\nabla} \left (\frac{v^2}{2} +h + g z \right )=0.
\end{equation}
This tells you that the expression
$$h+\frac{v^2}{2}+g z=\text{const along stream lines},$$
and this is Bernoulli's equation for the compressible fluid.

For the incompressible fluid, one can use (\ref{3}) again in (5) and write ##\nabla h=(\nabla p)/\rho=\nabla (p/\rho)## since then by definition ##\rho=\text{const}##, and the Bernoulli equation takes the usual form for an incompressible fluid, i.e.,
$$\frac{p}{\rho} + \frac{v^2}{2} + g z=\text{const along stream lines}.$$
 
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  • #24
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Bernoulli's equation is for the stationary flow of non-relativistic ideal fluids (no matter if compressible or incompressible). It's derivable from the continuity equation (local equation for mass conservation) and the Euler equation of motion. For stationary flows one has ##\partial_t =0## for all fields. Thus the continuity equation is
\begin{equation}
\label{1}
\vec{\nabla} \cdot (\rho \vec{v})=0
\end{equation}
and the equation of motion
\begin{equation}
\label{2}
\rho (\vec{v} \cdot \vec{\nabla}) \vec{v}=-\vec{\nabla} p +\rho \vec{g},
\end{equation}
where ##p## is the pressure and ##\vec{g}## the gravitational acceleration close to Earth.

Since for ideal fluids there's no heat exchange by assumption the entropy is conserved. From now on it's convenient to write everything concerning extensive thermodynamical quantities in terms of their specific values, i.e., the quantity per unit mass of the fluid. So we define ##s## and ##h## as the entropy and enthalpy per unit mass. Then the 1st law of thermodynamics reads
\begin{equation}
\label{3}
\mathrm{d} w=T \mathrm{d} s + \frac{1}{\rho} \mathrm{d} p=\frac{1}{\rho} \mathrm{d} p \; \Rightarrow \; \vec{\nabla} h = \frac{1}{\rho} \vec{\nabla} p.
\end{equation}
The final step uses the fact that the ideal-fluid flow is adiabatic, i.e., ##\mathrm{d} s=0##.

Now the trick is to rewrite the left-hand side of (\ref{2}) with help of the identity
\begin{equation}
\label{4}
\frac{1}{2} \nabla \vec{v}^2=\frac{1}{2} \vec{\nabla} v^2 = \vec{v} \times (\nabla \times \vec{v}) + (\vec{v} \cdot \vec{\nabla}) \vec{v},
\end{equation}
After some algebra using this equation in (\ref{2}) one gets using (\ref{4})
$$\vec{v} \times (\vec{\nabla} \times \vec{v})=\vec{\nabla} \left (h + \frac{v^2}{2} +g z\right).$$
Here we have chosen the coordinate system such that ##\vec{g}=-g \vec{e}_z##. This implies that
\begin{equation}
\label{5}
\vec{v} \cdot \vec{\nabla} \left (\frac{v^2}{2} +h + g z \right )=0.
\end{equation}
This tells you that the expression
$$h+\frac{v^2}{2}+g z=\text{const along stream lines},$$
and this is Bernoulli's equation for the compressible fluid.

For the incompressible fluid, one can use (\ref{3}) again in (5) and write ##\nabla h=(\nabla p)/\rho=\nabla (p/\rho)## since then by definition ##\rho=\text{const}##, and the Bernoulli equation takes the usual form for an incompressible fluid, i.e.,
$$\frac{p}{\rho} + \frac{v^2}{2} + g z=\text{const along stream lines}.$$
My understanding is that the flow being adiabatic is not a requirement for the Bernoulli equation to apply. That certainly isn't a feature of the Euler equation or the continuity equation.
 
  • #25
vanhees71
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Of course it is. Euler's equation describes ideal fluids, i.e., no shear or bulk viscosity and no heat transfer. This implies the adiabacity. The ideal-fluid equations of motion follow from kinetic theory under the assumption of 0 mean-free path, i.e., the fluid is in local thermal equilibrium at any time. This implies that the collision term in the Boltzmann equation vanishes and this is the case if and only if the system is in local thermal equilibrium and then and only then the entropy stays constant.

A very clear treatment of fluid dynamics as well as of kinetic theory can be found in Landau and Lifshitz vol. VI and X respectively.
 

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