While deriving Bernoulli's Theorem, our teacher said that the sum of KE, PE and Pressure Energy per unit volume remains constant at any two points.

$$P + \rho g h + \frac{\rho v^2}{2} = \text{Constant}$$

In this, he stated that the first term is Pressure Energy per unit Volume. What exactly is meant by Pressure Energy?

I know we can write:

$$P = \dfrac{F}{A} = \dfrac{F\cdot d}{A\cdot d} = \dfrac{W}{V} = \dfrac{\text{Energy}}{V} $$

What is physical significance and expression of pressure energy?

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    $\begingroup$ We can say that the sea level standard pressure is 1 atm = 101325 Pa or 1 atm = 101325 J/m^3. Some teachers prefer to use unusual units like J/m^3 (energy per volume) instead of Pa or N/m^2. It is just a matter of taste. $\endgroup$ – Energizer777 Nov 3 '15 at 18:57
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    $\begingroup$ I need help.. I have a doubt. I think potential energy per unit volume should be dgh/2 (d is density). My logic : take a cuboid container of base area A and fill it up to height h with liquid of density d. The mass of liquid is dAh and its center of mass is at height h/2. So PE = (dAh)*g*h/2. So, according to me, PE per unit volume is PE/Ah = dgh/2. Please help asap.. I have exams coming up. ;-; $\endgroup$ – Anurag B. Apr 17 '18 at 14:13

The technical answer

First I need to explain entropy. Suppose you have any system which you can only see at a certain granularity: we say that you see its "macrostate" but this could be any set of "microstates" which all "look alike". Since particle-interactions tend to multiply and distribute our uncertainty about a system, we could imagine under certain circumstances (constant total energy, constant volume, constant number of particles) that we are selecting a microstate essentially at random: hence the "largest" macrostate is the one that the system naturally "wants" to be in due to uncertainty-multiplication. ("Largest" here means: it contains the most distinct microstates. We would say that it has the most "phase space volume" where phase space adjoins momentum-space to real-space.) This additive measure of how large a macrostate is, is known as "entropy."

If a system has constant numbers of particles and is transitioning between states of equivalent entropy (so that randomness doesn't drive the transition), then the pressure is the rate of change of the total internal energy of the system, with respect to the volume. This could happen for whatever reason; it is all lumped together into the term "pressure."

Since energy is usually globally conserved by Noether's theorem, this is equivalent to saying "pressure is the capacity for a closed system to do work when it changes volume." This is probably the hardest theoretical thing to understand about pressure in the long run. It has a dual status: we can speak of the pressure at a point, but also distributed throughout a system. The full reconciliation of this dual nature involves treating each small cube of the fluid as a "grand canonical ensemble" which is sharing its particles and temperature with the much larger "system", and we can nevertheless ask how its "free energy" would vary with its volume, if it were to expand and then its internal energy and particle numbers equilibrated with its surroundings. That's a little more intense than this section allows.

But yeah, each point has a capacity to do work (in the form of a change in local free energy per change in local volume) and added up appropriately this means that a closed system has a capacity to do work (in the form of a change in total internal energy per change in total volume). Anything which contributes to this capacity is called "pressure".

What effects are lumped together under pressure?

The ideal gas does not have any self-interaction terms, and the pressure is solely a measure of kinetic energy per unit volume. This kinetic energy density causes the particles to push against the walls all the time, so that is how the pressure can do work. As we've said, the capacity to do work, from any source, shows a pressure.

Now let's take an ideal gas and turn on some particle-particle interactions. Let's consider a repulsive one as the easiest: imagine that we just suddenly gave each particle one electron of negative charge, so that they were all repelling each other.

The first effect, which has no effect on pressure, is that we had to secretly pump in a lot of internal energy to do this: we put a total charge $Q = N~e^-$ into the system, which changed the voltage of the box to $V$, so there may have been something like $VQ/2$ energy dumped into the box simply in this off-hand "throw some charge on it!".

But the secondary effects are more interesting. In conductors, charge tends to pile up on the edge of the box: so the center of the box now has a much lower density, the outside has a much higher density, and so we roughly would expect that the added "push" of the system outwards manifests as a higher total pressure. Repulsive particle interactions increase pressure, attractive particle interactions reduce it. You can similarly imagine that the attractive interaction means that when you increase volume, you get a "bump" from kinetic energy but you have to "tear apart" the potential energy holding these guys together, if that helps you visualize why the force on the external world is weaker.

Finally, it's worth considering diatomic compounds like $O_2$. These things can be treated a lot like ideal gases, but they have an internal energy (rotational kinetic energy) which doesn't tend to contribute to the pressure. This is to encourage you to forget the fallacy "average internal energy per unit volume" or some such; it's a rate-of-change, not an average.

Example: van der Waals equation of state

Probably the most famous example of the effects of particle-interactions on pressure is the so-called van der Waals equation. This is a simple, early heuristic to capture the non-ideal effects of a changing volume and pressure on a real fluid. It turns out that it contains a liquid-gas phase transition at a certain temperature, so it is our first stop also when we want to introduce phase transitions to our students. Actual fluids have been fitted to the following equation for parameters $(a, b)$:$$ \left( p + a~\left(\frac {n}{V}\right)^2\right)~\big(V - b~n\big) = n~R~T. $$See for example, Wikipedia's data page of these constants $a$ and $b.$ Now I can roughly explain that this is trying to equate a "total energy" (left) with a "thermal energy" (right). The term $b$ refers to a repulsive short-range potential which keeps the particles from occupying the same location: in real atoms this is because the electron clouds do not want to overlap; it is being modeled by pretending that the particles are secretly hard spheres and therefore the "available" volume is not the "total" volume $V$ but instead decreases proportional to $n$. The term $a$ refers to an attractive longer-range potential which makes the particles want to stick together; as I said above, attractive forces should reduce pressure in favor of some sort of internal energy density. In detail, we can see that we modify $p \mapsto p + \alpha \cdot (\text{# of handshakes}),$ if we envision the particles as "shaking hands" with each other: so for calculating the equation of state, we take the lowered-pressure and bolster it by the amount it was reduced. Since this attractive force drops with distance, the number-of-handshakes is not calculated like $n^2/2$ (total number of handshakes throughout the volume) but instead $n^2 / V^2,$ (handshakes with nearby neighbors, for some definition of "nearby").

Then the $p-V$ diagram for lower temperatures has a clear "dip" where you have to increase the pressure to compress it (fighting $b$) but also to expand it (fighting $a$). This manifests as a sudden amount of energy you'd get once you put enough pressure on it, as it all transitions from a gas to a liquid.

So what is pressure energy?

Summing this all together, pressure energy is the energy contained in each unit of the fluid due to the effects of thermal kinetic motions of the atoms lessened by the attractive forces of the fluid molecules on each other. Even if the fluid is viewed as incompressible from the point of view of the flow (i.e. the fluid flow is much less than the speed of sound in the fluid) it still answers the question of "if we changed the local volume, how would the local free energy change?". More importantly, it drives the motion of particles from one place in the fluid to another: if you pressurize this side by having it in contact with a fixed volume of air which you're pumping more air into with a bike pump, the added pressure gradient in the fluid causes the fluid to flow out of the reservoir faster, and into whatever else the system is connected to.

  • $\begingroup$ Why didn't you include "increased by the repulsive forces"? Aren't water molecules closer together the deeper you get? $\endgroup$ – CoolHandLouis Jul 27 '16 at 3:18
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    $\begingroup$ @CoolHandLouis I don't know exactly what you mean by that, but it turns out that pressure has almost nothing to do with pressing water molecules closer together. Molecules get slightly closer together (0.1% to 0.5% more dense) in a surface layer (0-250m in freshwater lakes, can go as deep as 700m in oceans?) or so called the "pycnocline," usually due to either a thermal (temperature) or salinity (salt concentration) gradient with depth, then afterwards this density remains constant with increasing depth even as pressure spikes through-the-roof! $\endgroup$ – CR Drost Jul 27 '16 at 16:11
  • $\begingroup$ "Pressure has almost nothing to do with pressing water molecules closer together". I hear this often, but after much research on this, I don't think this is correct. I think it actually has everything to do with it! Bouancy effects can be observed in a column of pure water at constant temperature. The pressure in an ideal gas is said to be mediated by the kinetic energy of molecules. What is the force carrier in a liquid, if not the "very large resistance to compression"? $\endgroup$ – CoolHandLouis Jul 29 '16 at 12:43
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    $\begingroup$ @CoolHandLouis I mean, I don't think you have formed a clear idea of what you mean yet. It might help if we start with a really simple example, bricks on a table. The brick exerts a pressure on the table. We stack another brick on top, more pressure on the table. Is that pressure increased caused by the first brick being closer to the table? I mean, maybe we can meaningfully talk that way--but we're being pedantic if we don't say that the more important thing is that a second brick was piled on top of the first as the real reason for the pressure increase. $\endgroup$ – CR Drost Jul 29 '16 at 17:13
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    $\begingroup$ This might be true practically, however if you teach that to a kid and then teach them what an ideal gas is (all of the atoms are allowed to pass through each other) then they will take your explanation to heart and predict that ideal gases must be locked to $P = 0$ because as Mr. CoolHand just told us, pressure only comes about by interatomic forces. And then $PV = nRT$ will seem quite the surprise. So it is important that when we graduate to the full definition of pressure we emphasize the word "anything" in the sentence "Anything which contributes to this capacity is called 'pressure'." $\endgroup$ – CR Drost Aug 1 '16 at 15:11

When a fluid is squeezed, as in a cylinder by a piston, work is done on the fluid. This work 1) elevates the pressure (pressure energy), and 2) the temperature (heat energy). (If the cylinder is insulated, this is called "adiabatic".)

In an ideal gas, these are all related by the ideal gas law, which says roughly that volume times pressure equals heat.

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    $\begingroup$ Thanks Mike, but squeeze would mean compression. We consider incompressible and inviscid fluid initially in derivation of the principle. $\endgroup$ – Max Payne Nov 3 '15 at 14:33
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    $\begingroup$ @TimKrul: No fluid is truly incompressible. If it were, the speed of sound in it would be c. Incompressibility only means it is quite stiff, in which case it can have high pressure, but not much pressure energy. i.e. d=0 in your equation. $\endgroup$ – Mike Dunlavey Nov 3 '15 at 21:45
  • $\begingroup$ Does that mean in an ideal fluid, the change in Mechanical Energy of the system ( K.E. + Gravitational P.E.) equals Work done by Pressure which in turns raises/decreases the Temperature of the liquid at different points? Meaning Change in Mechanical Energy = Change in Thermodynamic Internal Energy? $\endgroup$ – Tony Stark Dec 30 '20 at 18:01

The pressure energy is the energy in/of a fluid due to the applied pressure (force per area). So if you have a static fluid in an enclosed container, the energy of the system is only due to the pressure; if the fluid is moving along a flow, then the energy of the system is the kinetic energy as well as the pressure.

Because of the unit breakdown you have shown, I think it's better to view pressure as an energy density. For example, the energy density that prevents a star's collapse is the radiation pressure.

  • $\begingroup$ would it mean that even if a liquid is lying flat on a surface, or in outer space isolated from other entities, it will always have some pressure energy? What would be a possible expression for pressure energy as $\frac{mv^2}{2}$ is for KE? $\endgroup$ – Max Payne Nov 3 '15 at 14:35
  • $\begingroup$ Zero pressure indicates a vacuum, so there will always be pressure & hence, in your fluids, an energy density. The formula you are wanting is the equation of state and depends on the fluid you are modeling (e.g., for an ideal gas, $p(\gamma-1)=\rho e$). $\endgroup$ – Kyle Kanos Nov 3 '15 at 14:41

Pressure "energy" P is simply the measured pressure anywhere in the system.

For example, if we have a venturi whose entry and exit cross section are 1m^2 and velocity 1m/s, but it has a throat of cross section 0.1m/s, the gas will speed up through the throat, reaching a velocity of 10m/s.

The measured pressure on a pressure gauge with an inlet perpendicular to the flow (static pressure) will be lower in the throat due to the conversion of pressure energy to velocity energy (dynamic pressure.)

Note that the measured pressure on a pressure gauge facing into the flow will always show the "total pressure" (sum of the static pressure and dynamic pressure) and will be constant throughout the venturi, if we ignore friction. One way of understanding why this is, is that the flow gets stopped in the mouth of the pressure gauge.

Mike Dunlavey gives good answer explaining what pressure energy is: the energy stored in a fluid under pressure, due to its compressibility, like fluid compressed by a piston.

In practice in a flowing system, significant compression is pretty unusual, unless the speeds are close to the speed of sound, and therefore it can often be neglected.

For example the gases in the throat of a rocket nozzle are under high pressure and expand significantly as they reach the outlet of the nozzle, converting their pressure energy into kinetic energy. But there is no need to consider compression when analysing the flow of a desk fan.

From https://en.wikipedia.org/wiki/Bernoulli%27s_principle at the start of the section on compressible flow Bernoulli developed his principle from his observations on liquids, and his equation is applicable only to incompressible fluids, and compressible fluids up to approximately Mach number 0.3 Therefore further details of the compressiblity of the fluid are a later extension of the original Bernoulli Theorem.

  • $\begingroup$ But Steve dont we consider flow of incompressible and inviscid fluid while deriving the bernoulli's principle? $\endgroup$ – Max Payne Nov 3 '15 at 14:31
  • $\begingroup$ @TimKrul Sorry I got a bit carried away with compressible flow. I've added a bit more about P to my answer. $\endgroup$ – Level River St Nov 3 '15 at 15:14
  • $\begingroup$ Tim Krul, "Pressure energy per volume" measured in N x m/m^3 is the same as "pressure" measured, as you know, in N/m^2. The people, teachers, who use the term "pressure energy per volume" (rarely used in the context of deriving Bernoulli's Theorem) just want to impress the audience, the students. $\endgroup$ – Energizer777 Nov 3 '15 at 18:37

Your teacher refers to a common and simple but incorrect understanding of the Bernoulli theorem.

Unfortunately, it is quite common to understand the equation

$$P + \rho g h + \frac{\rho v^2}{2} = \text{Constant}$$

in the following way:

Total energy of a fluid element is $\text{Constant}$ times its volume and besides kinetic energy $\frac{\rho v^2}{2}\Delta V$ and potential energy $\rho g h\Delta V$ there is third contribution, $P\Delta V$, which depends on the pressure, so let's interpret it as pressure energy, and this energy has volume density $P$. (false)

The problem with this is there is no valid reason why the left-hand side of the equation should be interpreted as total energy of a fluid element. There is also no reason to think total energy of an element is constant as it moves along its streamline.

On the contrary, total energy of a fluid element generally changes, as there is work being done on it by the surrounding liquid via pressure forces. The pressure term in the Bernoulli equation has units of energy/volume, but pressure enters the equation through work of the liquid surrounding the element, not through total energy of the element.

You can find derivation of the Bernoulli theorem here: http://feynmanlectures.caltech.edu/II_40.html#Ch40-S3

Notice how pressure enters through work done on the liquid element and also how Feynman says The energy per unit mass of the fluid can be written as

$$ E=1/2v^2+ϕ+U... $$

with no pressure term in there.


You seem to understand the $\rho g h$ term, so I will explain the pressure energy in terms of that. Really the $P$ term and the $\rho g h$ term are very similar.

For now lets just ignore the $P$ term and focus on the $\rho gh + \frac{1}{2} \rho v^2 = \textrm{constant}$. This is basically just conservation of energy. It says that if you throw an object up into the air, it will experience a downward force, which causes it to lose speed. In fact, the equation tells you that you can calculate exactly how much speed it has lost just from its position. This is because you can find a potential ($\rho gh$) for the gravitational force. The change in potential then gives you a the change in the square of the speed.

Now the situation with pressure is exaclty analogous to this. So now lets imagine that we are in a zero-gravity environment, but we do consider pressure. So our equation now looks like $P + \frac{1}{2} \rho v^2 = \textrm{constant}$. When trying to understand this equation, the first thing to realize is that an object in a non-uniform pressure field experiences a force proportional to the gradient in the pressure. This force is analogous to the gravitational force from the prevoius paragraph. Now we can ask the question, "Is there a potential for this force?" The answer is yes: since the force is the already the gradient of the pressure, and the potential is also something you take the gradient of to get the force, the potential must be the pressure.

Now hopefully it makes more sense. The analogy to throwing a ball up in the air would be to move a bit of fluid up through a pressure gradient. The particle will experience a force which will tend to slow it down, so it will lose kinetic energy, but we can account for this by adding the pressure energy in. When this particle moves back into an area of low pressure, it will experience a force to speed it back up and recover the kinetic energy it originally had while losing the pressure potential energy. This is exactly analogous to throwing a ball in the air and having it come back down and it is exactly how potential energy is supposed to work. I hope that made sense.


Consider a gas that apparently moves in the $x$-direction with speed U. The molecules are also in random motion with velocities (u',v',w'), and they also have internal energy $i$ per unit mass, coming from rotation or vibration.

The energy of an individual molecule is $$(m/2 )((U+u')^2+v'^2+w'^2+i)=(m/2)(U^2+u'^2+2Uu'+v'^2+w '^2+i)$$ On average the term $2Uu'$ will cancel (because on average $u'=0$), so the average of the molecular energy per unit volume is $$(\rho/2)(U^2+(u'^2+v'^2+w'^2)+i)$$ displaying three kinds of energy. The term $U^2$ represents gross kinetic energy. The term $ (u'^2+v'^2+w'^2)$ is proportional to pressure. The term $(u'^2+v'^2+w'^2)+i$ is proportional to temperature. When a fluid is in motion in a way that does not lose energy these forms of energy get traded. Bernouillis equation specifies how the "pressure energy" gets traded with kinetic energy. The exact form of the trade can be derived from $F=ma$.

Also, it is the random collisions of molecules with the walls of the container that exert the force per unit area that we call pressure. This is given by the "pressure energy" per unit volume. So it makes perfect sense to regard pressure as a force per unit area, or as form of energy per unit volume, with the same dimensions in each case. A similar explanation can be given for a liquid, although the molecules have less freedom of movement.

This is, I admit, an oversimplified account, meant to do no more than remove some confusion in terminology, and give some intuitive understanding to the OP.


This site: Bernoulli Equation also uses the term "Pressure Energy". The pressure energy per unite volume is measured in N x m / m^3 = N / m^2. So this pressure energy per unite volume is in fact a pressure. Instead of the word "pressure" you can use the expression "pressure energy per volume". They are equivalent.

  • $\begingroup$ This basically rephrases what OP already knows and does not seem to provide the physical intuition OP requested. $\endgroup$ – Kyle Kanos Nov 3 '15 at 13:03
  • $\begingroup$ Kyle Kanos, My answer is the right one. You do not know what the opener knows or do not know. Do not speak in his name. I want to hear his opinion not yours. There is no distinction between "pressure energy per volume" (a term that is not too much used) and "pressure". $\endgroup$ – Energizer777 Nov 3 '15 at 18:27

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