What is pressure accourding to Bernoulli's theorem?

In summary, 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. In Bernoulli's theorem, potential and kinetic energy are easy to understand, but pressure energy can be a bit more complex. It is essentially the applied force that changes the kinetic and potential energy of the fluid. This also applies to incompressible fluids, though there is a separate compressible version of the Bernoulli equation.
  • #1
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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
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
Behrouz said:
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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  • #4
jartsa said:
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
jartsa said:
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.
russ_watters said:
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.
 
  • #6
Behrouz said:
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
Chestermiller said:
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
Behrouz said:
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
russ_watters said:
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
Behrouz said:
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
jartsa said:
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
Chestermiller said:
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
Chestermiller said:
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
Behrouz said:
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
Chestermiller said:
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
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
jartsa said:
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
jartsa said:
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
Behrouz said:
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
jartsa said:
\

(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
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
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
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
vanhees71 said:
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
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.
 
  • #26
Since when is an ideal fluid assumed to be adiabatic (I.e., have zero thermal conductivity)? For purposes of mechanical energy balance (I.e., Bernoulli equation), such an assumption is totally unnecessary.
 
  • #27
Ideal fluids were always considered to flow adiabatically. It's the definition of an ideal fluid. As I said, it's all very clearly explained in Landau&Lifshitz' textbooks.
 
  • #28
vanhees71 said:
Ideal fluids were always considered to flow adiabatically. It's the definition of an ideal fluid. As I said, it's all very clearly explained in Landau&Lifshitz' textbooks.
The fluid mechanics textbooks we engineers us have no such defonition.
 
  • #29
boneh3ad said:
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.
Well I can agree with that, because of the following scenario:

Rigid container A contains air at room temperature and 100 atm's pressure. We open a valve on the container wall. Almost all air comes out at high speed. The valve gets cold. Randomly moving air molecules used their kinetic energy to do work. But, then there is this scenario:

Rigid container B contains water at room temperature and 100 atm's pressure. We open a valve on the container wall. Only a small fraction of the water comes out at high speed. The valve does not get cold, even if the container is so huge that the the amount of water coming out is large.
 
  • #30
jartsa said:
Well I can agree with that, because of the following scenario:

Rigid container A contains air at room temperature and 100 atm's pressure. We open a valve on the container wall. Almost all air comes out at high speed. The valve gets cold. Randomly moving air molecules used their kinetic energy to do work.But, then there is this scenario:

Rigid container B contains water at room temperature and 100 atm's pressure. We open a valve on the container wall. Only a small fraction of the water comes out at high speed. The valve does not get cold, even if the container is so huge that the the amount of water coming out is large.
So, what is the point of this? Can you identify the mechanistic reason why the valve gets cold when gas is coming out?
 
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  • #31
Chestermiller said:
The fluid mechanics textbooks we engineers us have no such defonition.
Interesting. Maybe that's for books on hydrodynamics of liquids (water), where you can assume incompressibility, but I'm sure that you need an equation of state and the fact that ideal fluid-flow is adiabatic to close the equations.

In relativistic fluid dynamics the usual derivation is via the energy-momentum-stress tensor for ideal fluid dynamics and energy-momentum conservation. If I think about it, that's an approach which is even easier to understand than the usual way used in non-relativistic fluid-dynamics textbooks, how and where the enthalpy and internal energy enter the equations. The ideal-fluid ems tensor is easily derived. Consider a fluid cell. In the local rest frame of this cell you have
$$\bar{T}^{\mu \nu}=\mathrm{diag}(\epsilon,P,P,P),$$
where ##\epsilon## is the internal-energy density and ##P## the pressure of the gas. For an arbitrary frame you get the tensor components by introducing the fluid-four-velocity vector (I use units with ##c=1## to save some work), which in the local rest frame is
$$\bar{u}^{\mu}=(1,0,0,0).$$
In this way we get
$$T^{\mu \nu} = (\epsilon+P) u^{\mu} u^{\nu}-P \eta^{\mu \nu}$$
with the Minkowski pseduo-metric tensor ##\eta^{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)##. Note that ##\epsilon## and ##P## are still defined as their values in the local rest frame. This has the advantage that they are all Lorentz scalar fields.

If you now apply energy-momentum conservation (not considering external forces) you get the energy-balance and the relativistic version of the ideal-fluid Euler equation
$$\partial_{\mu} T^{\mu \nu}=0.$$
Now we have to introduce some conserved particle-number like quantity. In relativistic plasma physics one usually uses the electric charge in relativistic heavy-ion collisions the net-baryon number (i.e., the quantity (number of baryons) minus (number of antibaryons), which is conserved under the strong interaction). Then we have the continuity equation
$$\partial_{\mu} (n u^{\mathrm{\mu}})=0,$$
where ##n## is the electric-charge or net-baryon-number density in the local restframe of the fluid cell. Note that we here tacitly assume that the net-charge flow and the energy-momentum flow are both determined by the same four-velocity ##u^{\mu}##. That's also only true for ideal fluids. For non-ideal (viscous) fluids you have to distinguish these four-velocities, and it's a matter of the physical situation which you define as the fluid four-velocity. If you use the energy-momentum flow velocity it's called the Landau framework if you choose a conserved-charge flow it's the Eckart framework. But here let's stick to the ideal fluid, where all the flow velocities are the same.

Then you can write the thermodynamic identities which are for a fixed quantity of substance for the enthalpy (because that's the quantity ##h=\epsilon+p## which occurs in the ideal-fluid ems-tensor above)
$$\mathrm{d}(h/n)=T \mathrm{d}(s/n) + \frac{1}{n} \mathrm{d} P.$$
Here ##s## is the entropy density (as measured in the local restframe of the fluid, i.e., a Lorentz scalar field again) and ##T## the temperature (also measured in the local fluid restframe and thus a scalar).

Writing out the equation for energy-momentum conservation,
$$\partial_{\mu} T^{\mu \nu}= u_{\nu} \partial_{\mu}(h u^{\mu}) + h u^{\mu} \partial_{\mu} u^{\nu} + \partial^{\nu} P=0.$$
Multiplying with ##u^{\mu}## (with the Einstein summation convention effective) we find
$$\partial_{\mu}(h u^{\mu})-u^{\mu} \partial_{\mu} P=0,$$
where we have used ##u_{\nu} u^{\nu}=1##, i.e., ##u_{\nu} \partial_{\mu} u^{\nu}=0##.

Using in the above equation ##h=n (h/n)## and also the continuity equation ##\partial_{\mu} (n u^{\mu})=0## we get
$$u^{\mu} \left (\partial_{\mu}(h/n)-\frac{1}{n} \partial_{\mu}P \right)=0.$$
Using the above thermodynamic relation, we can write this equation as
$$u^{\mu} \partial_{\mu} (s/n)=0,$$
i.e., indeed the entropy per charge is conserved along the stream-worldlines, and this means the fluid flow is adiabatic.
 
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  • #32
vanhees71 said:
Interesting. Maybe that's for books on hydrodynamics of liquids (water), where you can assume incompressibility, but I'm sure that you need an equation of state and the fact that ideal fluid-flow is adiabatic to close the equations.
Are you saying that I can't solve the flow and heat transfer equations for an ideal inviscid gas being heated or cooled (not adiabatic) in a tube?
 
  • #33
As soon as you have transport phenomena, it's no more an ideal fluid. Of course, you can solve the above mentioned problems, but it's no longer ideal fluid dynamics. You have to introduce new quantities, the transfport coefficients. The usual treatment is to start from kinetic theory and to a systematic moment expansion. In leading order (first order in gradients) you get the shear and buld viscosities related to momentum transport and heat conductivity etc.
 
  • #34
vanhees71 said:
As soon as you have transport phenomena, it's no more an ideal fluid. Of course, you can solve the above mentioned problems, but it's no longer ideal fluid dynamics. You have to introduce new quantities, the transfport coefficients. The usual treatment is to start from kinetic theory and to a systematic moment expansion. In leading order (first order in gradients) you get the shear and buld viscosities related to momentum transport and heat conductivity etc.
So, if I have steady state flow and heat transfer to a gas flowing through a tube (where the wall temperature is different from the inlet temperature), it would be a no-no to treat it as an ideal gas with zero viscosity, but with non-zero thermal conductivity because it would (a) be poorly defined mathematically, be a bad approximation, or (c) not be conforming to the precise definition of what physicists call an "ideal fluid."
 
  • #35
This is an interesting discussion for sure. Most sources I have available to me would say that an ideal fluid is one that is inviscid and incompressible and they make no mention of a requirement of adiabaticity. However, I also know that, from Crocco's theorem in a steady flow,
[tex]T\nabla s = \nabla h_0 - \vec{v}\times\vec{\zeta},[/tex]
a diabatic flow should have a total enthalpy gradient, which introduces vorticity, which in turn means the flow is no longer irrotational. So, if you consider irrotationality to be an important consideration (i.e., do you feel that an ideal flow should admit a potential function?), then it would seemingly need to be adiabatic.

Bernoulli's equation also technically requires the flow to be adiabatic. If you look at the energy equation for a steady, incompressible flow with zero heat conduction or generation and zero shaft work, then the energy entering/leaving a control volume through the control surfaces is
[tex]0 = \oint\limits_{CS}\rho\left(e+\dfrac{U^2}{2} + \dfrac{p}{\rho} + gz\right)\vec{U}\cdot\hat{n}\;dA.[/tex]
If you treat the CV as a streamtube so that you have a single inlet and single exit, you can use the energy equation in this way to derive Bernoulli's equation. You simply need a way to eliminate changes in internal energy, and in this case, this implies the flow must be incompressible and adiabatic. I obviously skipped a lot of steps there, but hopefully it is at least emblematic of the problems applying Bernoulli's equation to a flow with heat transfer.

Oddly enough, I have a PF Insight I was working on that would cover this topic and I left it around 75% finished and got too busy with work to finish it yet.
 

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