# Why is Pressure as energy density not considered in an energy balance for a closed system?

When considering a closed system in thermodynamics the energy balance is often described as:

Just as an enthalpy considers the term PV as a measure of the energy it requires for a system to push the atmosphere aside and occupy physical space, why is the term PV not included in an energy balance for a closed system? And this leads me to my final question: Would it be possible to have flow energy in an open system and no kinetic energy in a fluid?

EDIT: Basically I mean if pressure can be expressed as energy density like:

then it follows that a fluid in a closed system has a certain amount of energy inherent in its pressure alone. Why is this only considered in the case of open systems? and not closed systems as well?

The PV quantity is dimensionally (has the units of) energy, but it is not energy. Dimensional analysis can lead to related concepts, and PV is related to energy, however, you cannot call them the same. Setting KE and PE aside as understood, the remaining energy in a substance is the so called "internal energy" U.

The quantity enthalpy is defined as the internal energy U plus the PV quantity. Enthalpy is dimensionally equivalent to energy, but again, it isn't energy. Perhaps the most useful aspect of enthalpy and interesting aspects of PV that is seen in introductory thermodynamics is the "flow work" that it sounds like you are talking about. In this case, as a working fluid flows through a control volume, in addition to all the energy that it carries across the boundary, it works as it crosses the boundary. The flow work done when crossing the boundary is the pressure times the volume that crosses the boundary at each inlet or outlet. With enthalpy, we can basically remove the actual work from consideration and reduces the problem to a flow problem for the property enthalpy. Assuming no other work sources

$$\Delta E_{T} = W_{f,in} - W_{f,out} + E_{in} - E_{out}$$

where $$E_{T}$$ means total energy $$E_{i} = U_{i} + KE_{i} + PE_{i}$$ at each crossing, and $$W_{f,i}$$ is the flow work at the crossing. Basically, if you are doing mass flow, there is no "work" concept, it is just all the mass flowing in and all the mass flowing out. If you are doing energy, this initially does not work out the same way. When the boundary is crossed by something carrying energy (as $$U$$ or $$PE$$ or whatever), there is an additional effect by the nature of its own motion across the boundary. The substance does work on the system due to its pressure and motion (if pressure is 0, then the substance can enter the volume without this additional work). However, this complication can be removed by moving that PV term over and calling that a new property, enthalpy. The energy transfer to such a system (in essence, a control volume with steady flow), is the enthalpy flux as a property only.

In the above equation we need only add a term to remove shaft work or whatever, and we can calculate the power that an engine develops as a function of enthalpy flow across the system alone. This is useful, but don't fall into the trap of having a "conservation of enthalpy" law or anything like that. There is no such law in thermodynamics.

It is also worth noting that the supposed "flow work" across a boundary is just a matter of perspective. If, instead of the fixed control volume, we have the boundaries of our system move precisely with the flow (a so called "control mass"), then what we were calling the "flow work" crossing the boundary is now the so called "moving boundary work" at the boundary. In this case, there is no flow of anything, and so all changes can be understood as

$$\Delta E_{T} = \int P d\vec{x}_{b} \cdot d\vec{A}$$

which is all the moving boundary work. In our case, it would just be the inlet and outlet areas moving at the flow velocity. This particular change is often called the Reynolds transport theorem.

then it follows that a fluid in a closed system has a certain amount of energy inherent in its pressure alone.

Sort of but be careful, I would go so far as to say "no" in fact. This is because, in the example of an incompressible fluid, changes in pressure lead to no changes in energy. You can change the pressure of an incompressible fluid without doing any work, so if you did this in an adiabatic system it would violate energy conservation (unless energy and enthalpy are different of course). Notice that as you squeeze something incompressible, you can change its pressure to anything without changing the volume. In this case, U remains the same (because you are not doing any work), but H is made to vary (by changing the PV through changes in P alone).

When you increase the pressure of say, a gas, by compressing it (think of squeezing a balloon smaller) you are doing work on it. In the balloon squeezing example, the natural boundary is the rubber balloon, which is, in effect, a control mass. The work in that perspective is moving boundary work. However, just because balloons are natural control masses, and turbines are natural control volumes, neither is necessary. The Reynolds transport theorem always connects the two (as do calculus of moving surfaces techniques for more generality), and you could understand the balloon relative to a fixed volume, and then there would be flow work at certain areas.

When we look at the $$PV$$ quantity, its total derivative is $$PdV + VdP$$. Only $$PdV$$ contributes to energy changes, $$VdP$$ does not, but the total derivative is needed for enthalpy changes.

Would it be possible to have flow energy in an open system and no kinetic energy in a fluid?

By definition, the energy flow across the boundary must be kinetic, but it is often small and neglected. A lot of 1st order power plant analysis ignores the kinetic energy. However, all flow across the boundary must carry kinetic energy for nonzero density. Wind turbine design does not use internal energy or pressure, but rather kinetic energy (the air in the wind is not much of a different temperature or pressure than other air, it is moving). This lead to the Betz law, which is not exact, but the point still stands that when one is extracting KE from a fluid in steady state operation, one must consider how the slower fluid is evacuated from the control volume.

Note that for an ideal gas, both U and H are functions of temperature alone. That U is a function of temperature alone is a pretty deep physical result, but that H is then is just math:

$$H = U + PV = U + NRT$$

so if U is a function of temperature alone so is H. Again, the two are not the same. U is the energy to take all the stationary gas particles and bring them to that state. H adds an additional factor of $$NRT$$. The energy you would get from the gas by reducing all the particles to 0 speed is U.

• Thank you for the detailed answer, however I can find several sources that claim pressure can be thought of as energy density and as such, when multiplied by a volume it becomes a measure of energy. If im still mistaken can you explain how? hyperphysics.phy-astr.gsu.edu/hbase/press.html#ed phys.libretexts.org/Courses/University_of_California_Davis/…General_Physics/5%3A_Flow_Transport_and_Exponential-_working_copy/5.02%3A_Static_Fluids Commented Feb 24, 2023 at 22:04
• That says "Pressure in a fluid may be considered to be a measure of energy per unit volume or energy density." We agree it is "a measure" in certain circumstances. They give the example of the Bernoulli equation. The Bernoulli equation is what in thermodynamics is the PE + KE + H (enthalpy) flow relation. However, for an incompressible fluid, it is U that is constant and so drops out. While H is "a measure" of energy density, it is not energy density for the reason I described. If you increase the pressure of an incompressible fluid, you do no work, so the energy must have not changed. Commented Feb 24, 2023 at 22:09
• The second link did not work for me. Commented Feb 24, 2023 at 22:12
• I don't know about incompressible fluids ill admit. But if you took a sphere of gas which is compressible and teleported it into outer space it would expand due to its internal pressure. If you put a grain of sand on the border of the sphere of gas it would experience an increase in kinetic energy due to the gas doing work on it. THat energy must come from the pressure. or am I wrong? Commented Feb 24, 2023 at 22:24
• And im pretty sure my steam tables in my book says that the internal energies for saturated water rises with increasing pressure. even though its incompressible? Commented Feb 24, 2023 at 22:26

Basically I mean if pressure can be expressed as energy density like ... then it follows that a fluid in a closed system has a certain amount of energy inherent in its pressure alone. Why is this only considered in the case of open systems? and not closed systems as well?

No, this is not true in general. Incompressible fluid can be put to immense pressure without doing any work. Just put a big heavy weight on the piston. Since the fluid is incompressible, the piston with weight does not move at all, and thus no work is done on the gas. This is also why the pressure term in the Bernoulli equation cannot be understood as "pressure energy". The pressure term there quantifies work of external pressure forces that would be done on a liquid element flowing from one to the other point in the flow, but the element itself has no pressure energy.

In practice all fluids are compressible somewhat, but typically only gases and thermal radiation are compressible so well so that increasing pressure means comparable increase in energy density in them (with some factor of proportionality close to 1).

For solids and liquids compressibility is low, and increasing pressure means only many times smaller increase in energy density.

• But if you eject water into a vacuum it would evaporate? Then it would be capable of doing work. Also if you had a cylinder full of liquid water at say, and kept increasing the pressure untill it exploded, would that not mean the water had some energy tied to its pressure? Commented Feb 27, 2023 at 21:42
• Yes, water both in liquid state and gas state can do work. Yes, liquid water is compressible and can store some energy due to being compressed. But the amount is many times lower than $pV$. Change of pressure from $p_1$ to $p_2$ stores less energy than $p_2|\delta V|$ and $\delta V$ in the usual case won't be more than 1/100 of the initial volume. The cylinder would explode because of the work of the agent that was maintaining the pressure, water energy is only marginal for described reason. Commented Feb 27, 2023 at 22:44

Very good question.

Basically, it boils down to where the notion of $$PV$$ comes from.

Closed system has its full energy as you gave in your question. If we start pumping fluid into the closed system - thus, making it an open system - we must add the energy of the fluid that we've pumped into the system in order to get the total energy of the system.

The energy of the pumped fluid - that we can call a flow - is given by the sum of the flow's internal energy, kinetic energy, potential energy and the work it is doing to push the substance in the system away to clear the space for itself to sit into.

That work is given by $$PV$$. Why is it given by $$PV$$?

Here's a very non-rigorous and not-in-depth explanation.

The general formula for work is $$W=FS$$, here $$F$$ is force, $$S$$ is displacement.

Force, in turn is pressure multiplied by area: $$F=PA$$. Plug it into the work equation: $$W=FS=PAS=PV$$, here $$AS=V$$ is volume because volume is area multiplied by displacement.

Thus the work of the flow is $$W=PV$$.

Since the substance in a closed system doesn't push some other substance - like in the case of the open system, when the substance of the flow is pushing the substance of the system - it doesn't do work which is $$W=PV$$ and, therefore, we don't include $$PV$$ into energy balance for closed systems.

When a substance is flowing it always has kinetic energy. Thus, it is impossible to have a flow without kinetic energy. Also, a flow always has $$PV$$. I don't know how to explain that. To explain that we need a rigorous derivation of $$PV$$ which I don't know.

When a closed system in a cylinder pushes a piston, we say that it exerts work on the piston $$W=P \Delta V$$. We plug that expression in the first law: $$Q=E+W => E = Q - W = Q - P \Delta V$$. But the total energy consists of kinetic energy, potential energy and internal energy. Thus, we have $$U + mv^2/2 + mgh = Q - P \Delta V$$.

Therefore, we include $$P \Delta V$$ in the energy balance for a closed system sometimes, but $$P \Delta V$$ is not the same as $$PV$$ of a flow.

• Thanks for the answer, I can follow what you are saying, but pressure can also be expressed as energy density. Which means in a fluid that is NOT flowing anywhere will still have energy tied to its pressure. I have edited my original question to include what I mean. So why do we not include that? Commented Feb 24, 2023 at 19:20
• @JanF.S, let me disagree with you about your statement that pressure can always be represented as energy density. Your units are correct. But your interpretation of $J$ as energy doesn't seem that obvious to me. Instead, I would interpret $J$ as work. And then use my derivation to show that $P = W/V = J/m^3$. Thus, if there is no work, there is no $P = J/m^3$. And a substance in a closed system that isn't contracting or expanding isn't doing work. Whereas flow always does work. Commented Feb 24, 2023 at 21:06
• Well work is just a transfer of energy - either to or from something - so it's ok to call it energy density. Just like if I took a sphere of gas with an internal pressure P and teleported it into a vacuum - like in outer space the gas would expand due to its pressure. If I laid a grain of sand on the sphere of gas it would accelerate it, by doing work. This work must come from the inherent pressure of the gas. Commented Feb 24, 2023 at 21:45
• I would disagree with you. Energy doesn't include work. It is evident from the first law: energy is on the left hand side and work is on the right hand side. It is a very peculiar case, though, that we include that work into energy when speaking of the total energy of the flow. I don't know how to explain that and it's good you brought it up. Honestly, if a student asks me this, I wouldn't know how to explain it. Commented Feb 25, 2023 at 2:00
• Which part don't you agree with? Work has the [J] as unit - if I do work on something to increase its speed I have increased its kinetic energy - ergo the work must be a representation of energy transfer. Otherwise I don't know what to make of it. Commented Feb 25, 2023 at 10:06