When you move from fundamental principles (in physics) to calculating something ‘useful’ (in engineering), you seem to move from energy to enthalpy.

Enthalpy is measured in Joule, as well as energy. It is assigned to a ‘system’, a part of the physical world separated from other parts by interfaces. The canonical example is a vessel containing gas: The *energy* of this volume of gas is proportional to the average kinetic energy of the gas molecules.

The vessel shall have a movable lid which is kept in place by a weight. It shall have a volume V equal to base area A times height h. The pressure inside is denoted by p.

_0_ ||||| ||||| ||||| ----------- ^ | | | | | | | p,V | h | | | | | | |_________| v A

*Enthalpy* is the sum of energy plus the potential energy of this weight, relative to the base area of the vessel. Energy is the truly intrinsic property, enthalpy is in a vague sense characteristic of how the system is build up to finally fill a volume, given a constant pressure. In this article I am going to look at this from different angles – especially why this is useful when looking at machines that do work.

The canonical introduction to enthalpy:

Energy can be added to a system by heating it up or by doing mechanical work on it. If air is compressed in a vessel, work is done against the pressure p in the vessel. Force is applied, and pressure is force F over area A. If the lid is moved a tiny bit, the work done is the force times the distance the lid moved: W = Fdx. If the lid has an area A, then pressure p is F/A and work is Fdx = (F/A)^{.}Adx = pdV. If the volume becomes smaller, positive energy is added, so the correct sign is: -pdV. In total, the change in energy dE is heat dQ plus work:

dE = dQ – pdV

If the volume would not change, energy added would be equal to the heat added. But it is often more useful to talk about energy equal to heat when the pressure remains constant. For example, if a process is done in ambient air, then pressure is fixed by this condition. pdV is part of the differential of PV, because: d(PV) = pdV + Vdp. Now enthalpy – denoted H – is defined as:

H = E + pV

Then a change in enthalpy is dH = dE + d(pV). Inserting for dE makes pdV go away:

dH = dQ – pdV + pdV + Vdp = dQ + Vdp

If pressure is constant, the change in enthalpy is just the heat added (or removed).

Looking at the vessel with the lid again:

If the vessel has height h, then the potential energy of the weight with respect to the bottom of the vessel is the mass of the weight, times the acceleration in the earth’s gravitational field times the height: mgh. mg is the acting force, so this is Fh. The pressure p exerted on the lid is p = F/A, so potential energy Fh can be expressed as Fh = pAh = pV – pressure times volume! The term pV in the definition of enthalpy is the potential energy of the weight – the ‘thing’ that maintains the pressure.

Why does this become more natural in the context of real working machines?

Useful systems are things like turbines or heat exchangers. A liquid or gas enters the machines, its state is changed, it leaves the machine. It maybe does mechanical work. If you want to find out if a machine is efficient you need to follow the fluid through several connected components and calculate the changes of characteristic properties.

The engineer’s machine is not characterized by a single pressure, as the physicist’s vessel was. The fluid enters the machine via an inlet and leaves through an outlet. At these points, pressures of the working fluid can be different. Changing the pressure is the idea of apparatuses such as a compressors.

You need to follow a small piece of material through the machine: The working material may change from the liquid state to vapor and back, so it makes sense to consider a flow of *mass* than a flow of volume (with wildly varying density). I will call that a *packet of fluid*.

All additive (extensive) physical properties are obtained by multiplying mass with the appropriate density: Energy E is mass m times energy density e. The small packet of fluid in the flow of material has a mass dm, and its energy is dE = e^{.}dm. Volume is mass divided by density: V = m / (m/V) = m/ρ, so a small change of volume is related to a small change of mass via: dV = dm/ρ. Instead of the mass density ρ, its reciprocal is easier to use: the specific volume v. Then: dV = v^{.}dm.

When the fluid enters, it is pushed into or sucked into the machine. Work has to be done against the currently existing pressure at that point: an energy equal to pdV = pv^{.}dm. The entering packet of fluid dm carries with it an energy e_{1} dm, and in addition p_{1} v_{1} dm is needed to push it in. At the outlet it carries an energy e_{2} dm and you need p_{2} v_{2} dm to suck it out – or get back this energy. pv^{.}dm is positive if added to the system by pushing the packet in, negative when it leaves the machine.

In total, we are noticing a change of: dm ( e_{1} + p_{1}v_{1} – e_{2} – p_{2} v_{2} ). The property that emerges naturally in this energy balance is:_{ }

h = e + pv

Specific enthalpy shows up because the change in volume dV is expressed via the same packet of mass dm. If an actually useful machine would be investigated, there would be further terms in the energy balance as the mechanical work extracted, but these are not important here.

At first glance, the picture of the vessel with the lid and the machine with inlet and outlet may conjure up different ways to think about enthalpy – the weight on the lid of the pot versus this push in / suck out energy. It is as if something external (the weight) becomes ‘internal’ in the case of the flowing fluid.

A simple example of a ‘machine’ is an isenthalpic ‘expansion valve’. Model: Two thermally insulated pistons have a common interface, this interface has a small hole. Different pressures are maintained on both sides, therefore different densities:

//////////////////////////////// __________ ________________ || |_| || -->|| 1 _ 2 ||--> ||__________| |________________|| ////////////////////////////////

No heat can be exchanged with the environment, only mechanical work is done. Assuming that the piston is moved slowly, the ‘push in / suck out energy’ is the only contribution to the energy balance; nothing else happens. When piston 1 is pushed to the right, pressure 1 and specific volume 1 (density 1) remain constant. This is only possible because molecules can escape into space 2 through the hole. When piston 1 hits the wall with the hole, the work done is the sum of (integral over) all small pdV’s along the way. As pressure is constant, this is p_{1} times the initial volume V_{1}. As the second chamber is in parallel filled up with molecules, piston 2 moves at constant pressure p_{2} until the volume is finally expanded into the volume of final size V_{2}. If the fluid is an ideal gas made up of non-interacting point-like particles, their kinetic energy would not change on their way from volume 1 into volume 2. The internal energy of a gas depends on temperature only, so temperature would not change. pV is proportional to the temperature and the number of particles (ideal gas law), so the product pV would not change. As both internal energy and pV do not change, enthalpy is also unchanged.

Removing the wall with the hole: Even this would count as a machine in this very general sense:

//////////////////////////////// _____________________________ || || -->|| ||--> ||_____________________________|| ////////////////////////////////

Looking again at the weight on the lid and comparing it with these simple ‘machines’:

Energy pV = mgh was obtained by taking the baseline of zero energy at the *bottom of the vessel*. The energy pV is built up in this way: Start with an *empty* vessel, lid on the ground. Let a gas flow into the vessel *at constant pressure*. The lid is lifted and the volume expands. When the lid has reached its final height the work done is pV. At each instant of time the constant pressure in the pot balances the weight exactly. In contrast to the expansion valve, the weight still sits on the lid then the process has been completed *as there is no outlet*.

Pressures seems to be kept constant automagically – when there is no weight. How can you say that the pressure at a certain point in the system is ‘maintained’? There are several interconnected components and after some non-stationary cycles you expect the whole system to reach a stationary state – to exhibit constant pressures at some interesting junction points in the system. Each component is impacting its neighbor. There is no magic piston or weight that tries to keep pressure constant, but in a sense each component works as such for the next one.

At college we were taught the formula H = U + pV where U is the internal energy of the system. This raised the question that if pV was not part of the internal energy of the system, then it must be external in some way. Our teacher tried to explain it by saying that pV could be thought of as the work done pushing the surroundings aside to make space for the system, or something like that.

“Making space” is a great way to put it, thanks!