A heavy vapour has shocks that expand

The only law that forbids it took an expansion shock, a jump in which a gas’s pressure and density fall, and showed that it satisfies mass, momentum and energy conservation exactly, and is refused only by the second law, with an entropy that would fall. The jump the equations allow gave the mechanical reason. A compression’s rear is hotter and it signals faster, so it catches up with its front and steepens. An expansion’s rear is cooler and signals slower, so it spreads. These two essays fit together, and they are correct for air, steam, helium and essentially every gas that anyone meets.

They are not correct for every fluid. Each argument depends on a quantity nobody bothers to name in a perfect gas, because it’s always positive there, and it has a sign that’s a property of the fluid. Hans Bethe spotted that in 1942, in a wartime report on the theory of shocks in a general fluid. He showed that the sign of $\partial^2p/\partial v^2$ at constant entropy is what controls whether a compression shock is the stable one, and that nothing in thermodynamics forces that sign. Zel’dovich said the same in 1946. Thompson named the quantity and calculated its value for real fluids in 1971, and his name for it is the fundamental derivative of gas dynamics. Together they give their initials to the fluids that have the unusual sign: BZT fluids.

The number in the proof #

The speed of a simple wave is $u+c$, and its steepening depends on how that combination changes with density along an isentrope. A short calculation gives

$$\frac{\partial (u+c)}{\partial \rho}\bigg|_s=\frac{c}{\rho}\,\Gamma, \qquad \Gamma=\frac{v^3}{2c^2}\left(\frac{\partial^2p}{\partial v^2}\right)_s .$$

In a perfect gas, $\Gamma=(\gamma+1)/2$: 1.2 for air, and always greater than one. So denser fluid in a wave always carries a faster signal, compressions always steepen, and a textbook can just say so. Everything in the two earlier essays, the expansion wave spreading and the compression shock’s stability, is really the statement that $\Gamma>0$. And its second-order statement is the same. The entropy rise across a weak shock of strength $\Delta p$ is the classic

$$\Delta s=\frac{\Gamma\,v^3}{6\,T\,c^4}\,(\Delta p)^3,$$

so it’s the third power of the jump, and its sign is that of $\Gamma$ times that of $\Delta p$. Second law or no second law, the sign of $\Gamma$ decides which direction of jump the second law allows. Reverse $\Gamma$ and expansion shocks are permitted, and compression shocks aren’t.

So the question is whether $\Gamma$ can in fact go negative. Thermodynamics allows it. The second derivative at constant entropy isn’t constrained by stability, which only needs the first derivative to have the right sign. All that’s needed is a fluid whose isentropes have an inflection in the $p$–$v$ plane. Near a critical point isotherms have that inflection, by definition. Isentropes would have it too if they ran close to the isotherms. Isentropes run near isotherms when the fluid can absorb compression’s work without warming much, and that means a large heat capacity.

Almost everything met in practice has a value above one, which is why the question seldom arises, and why a compression that costs nothing in the end could leave it unstated. Water as a liquid has a $\Gamma$ of about four. A perfect monatomic gas has 4/3. Only a few heavy vapours near saturation go below one at all.

A van der Waals fluid can do it #

The simplest fluid with a critical point is enough. In the units of its critical point, van der Waals’s equation is

$$p=\frac{8T}{3v-1}-\frac{3}{v^2},$$

and with a constant heat capacity, the isentropes are $T(3v-1)^{R/c_v}=\text{constant}$. Everything else is differentiation. For a fluid with $c_v/R$ of 50, $\Gamma$ dips to about $-1.5$, near the saturated vapour line, and it recovers as the fluid expands. The dip gets larger with heat capacity, and it vanishes altogether at a clean threshold.

The crossing at $c_v/R=16.68$ reproduces the value in Thompson and Lambrakis’s paper of 1973. Air has 2.5. Steam is about 3 or 4. The threshold needs a molecule with a great many internal modes active at the fluid’s temperature, which means a long or heavy one. Large hydrocarbons, fluorocarbons such as perfluoro-perhydrophenanthrene, known as PP10, and long siloxanes like D6 have heat capacities in that range when near their critical points, and the van der Waals estimate of the threshold has been refined by better equations of state, which reduce the number somewhat but leave the qualitative picture alone.

Where’s the fluid when it does this? The figure below shows that it’s a small island.

It’s a thin region just above the vapour’s saturation line, near but below the critical pressure. That is what makes this hard to reach experimentally: a fluid at very nearly its critical pressure and temperature, at a density where it’s a dense vapour, is hot, close to decomposing for many of the fluids concerned, and easily thrown across the saturation line by the very wave being studied.

What the jump does there #

A direct calculation of the Rankine–Hugoniot jump gives what theory predicts. At each volume on the far side, the two conservation equations and the energy give a state, and its entropy can be compared with the start.

So in the island the signs are the reverse of a textbook’s. A weak compression shock would destroy entropy, and it can’t exist. A weak expansion shock gains it, and it can. A compression of a fluid in that state spreads out into a smooth fan with no shock at all, as an expansion does in air. Both essays’ reasoning stays valid, with the sign of $\Gamma$ changed: the entropy condition still selects the physical jump and the speeds still order themselves. Only the direction is different.

It’s worth noting the rest of that figure. A strong compression, down past a volume ratio of about 0.75, regains positive entropy and so is permitted, since it carries the state out of the island on to its far side, where $\Gamma$ is positive again. And a further condition, Lax’s speed ordering, requires a jump’s upstream and downstream states to have characteristics running into it, which restricts the allowed Hugoniot further and gives “mixed” waves, with a fan attached to a shock. The composite waves of a fluid with a $\Gamma$ changing sign, which are analysed by Cramer and Kluwick from 1984, include double sonic shocks. Those are a kind of wave with no analogue in a perfect gas.

The wave speeds, seen directly #

This is the mechanism in one curve. Over the range where the sound speed is rising as the pressure falls, the expanded fluid behind an expansion wave’s front has a faster sound speed than the front itself, so that the rear catches up, and the wave steepens. Run the expansion wave in air and the front is the fastest. So the mechanical story applies unchanged with the words interchanged.

There’s also the striking size of that dip. The sound speed in the island is about 0.3, in units where a perfect gas at the same reduced state has $\sqrt{\gamma p v}$, nearly 1. A fluid near its critical point has a very low sound speed anyway, and its speed-of-sound behaviour is at the opposite extreme from a stiff liquid’s. It’s a place where the Mach number is high at modest speeds, so non-ideal wave effects can be obtained in a small rig without the kind of power a hypersonic tunnel needs.

Why it takes a large molecule #

The threshold has a plain reading. A molecule of $N$ atoms has $3N-6$ vibrational modes on top of its translation and rotation, and each contributes up to one gas constant to the heat capacity when fully excited. So a heat capacity of 16.7 gas constants needs about seven atoms’ worth of fully excited modes. Vibrations are only partly active at ordinary temperatures, so in practice a molecule of twenty to forty atoms is what’s wanted, at a temperature near its critical point, where many of its modes are in play. Such molecules have critical temperatures of several hundred degrees Celsius, and they tend to be decomposing by then. That’s the whole experimental difficulty in one sentence.

There’s also a reason why the isentrope matters and not the isotherm. A van der Waals isotherm has an inflection at the critical point for any heat capacity, since it’s what makes the critical point critical. For a small molecule, compressing it heats it strongly, and the isentrope runs steeply across the isotherms and misses the inflection entirely. Only if the fluid soaks up compression’s work in its internal modes and hardly warms does the isentrope lie along an isotherm for long enough to inherit its shape. A large heat capacity makes the isentrope behave like an isotherm, and that is why the threshold is a heat capacity and not a pressure or a molecular weight.

And the same number has a milder consequence, far outside the island. Air’s $\Gamma$ is 1.2, and it’s a good approximation to think that a Mach number always rises in an expansion. For $\Gamma<1$, which many heavy fluids have over a much wider range of states than the island, the sound speed rises as the pressure falls, and in an expansion the Mach number’s rate of rise is reduced. In a steady flow it goes as

$$\frac{dM}{d\ln\rho}=-M\left(1-\Gamma+\frac{1}{M^2}\right),$$

so that at a high enough Mach number, once $\Gamma<1-1/M^2$, the Mach number falls along an expansion. Nothing about that needs a shock or the island. Cramer and Best predicted it in 1991, and it has been measured in a nozzle. It’s the mildest experimental evidence that this quantity is more than a curiosity.

Where it might pay #

It’s more than an oddity. Organic Rankine cycle turbines, which generate power from low-grade heat by expanding a heavy organic fluid in place of steam, often operate in a dense-vapour regime near the saturation line and have a small $\Gamma$ even if it isn’t negative. A turbine blade in a fluid with low $\Gamma$ has a weak, smeared shock in place of a sharp one, because the wave’s steepening and its entropy go as $\Gamma$. Its shock loss goes with $\Gamma$ too, at a fixed pressure ratio. Theoretical work by Brown and Argrow, Monaco and Cramer, and Colonna’s group at Delft has predicted smaller wave-drag and lower losses in supersonic blade rows. So $\Gamma$ is something a designer can choose, by choosing the fluid and its state.

Wave drag follows it directly, since a supersonic wing’s drag rises from the same wave-steepening physics. A wing flying through a fluid with $\Gamma$ close to zero in that sense would have a far lower wave drag than its Mach number suggests, in a gas nobody would want to breathe.

Experimentally, though, an unambiguous expansion shock in a controlled test has proved very hard to demonstrate. Borisov and colleagues reported one in 1983 in Freon-13 near its critical point, but the interpretation has been disputed, partly because near-critical effects, heat transfer and slow relaxation could produce similar signals. Fergason and his colleagues in Colorado tried it with a shock tube in the late 1990s and early 2000s, and they had problems with their fluids decomposing at the required conditions. Colonna’s FAST, and later Delft and Milan experiments, have measured non-ideal behaviour directly: a Mach number that behaves differently from a perfect gas’s in a nozzle’s expansion, for instance. So what’s been measured is non-ideal, and the sharpest form of the prediction remains a design for experiments still being built. That’s a good place to say what isn’t known.

What the picture cannot show #

A real fluid’s equation of state. Van der Waals’s equation is a qualitative caricature near a critical point, and it gives a critical compressibility of 0.375 against about 0.27 for most fluids. Real calculations use multi-parameter equations of state, and the island’s size and depth differ by a good deal. Colonna and others have found that the island shrinks in better models, and that for some candidate fluids it may vanish. The number 16.68 applies to this equation, and it doesn’t apply to any real molecule.

A variable heat capacity. A constant $c_v$ is an approximation that’s especially poor for large molecules, with their vibrational modes activating gradually with temperature. A real isentrope’s shape depends on this. It’s the same physics as the gas that hasn’t caught up with its shock, and a large molecule’s vibrational modes may not equilibrate within a shock, which would mean that the effective heat capacity across a fast wave is lower than the equilibrium value. It could remove a shock’s supposed negative-$\Gamma$ character. Nobody has settled how much.

Shock structure. The jump conditions state what’s on either side. They don’t say what happens inside, and the structure of a double sonic shock isn’t a thin, simple Taylor layer. Viscosity and heat conduction in a near-critical fluid are themselves unusual, and transport coefficients vary strongly near the critical point.

Two phases. The saturation line lies just next to the island. Most of the waves that are large enough to be interesting will enter the two-phase region, where the fluid condenses. Condensation has its own wave structure, and it isn’t in the model at all.

The convention #

Everything’s in van der Waals’s reduced units: pressure, volume and temperature over their critical values. The gas constant is $8/3$ in these units. The heat capacity is $c_v/R$, taken as constant; $\delta=R/c_v$ is the isentrope’s exponent. The sound speed is in units of $\sqrt{p_cv_c}$. “Above saturation” means at volumes greater than the vapour’s value by Maxwell’s construction. Entropy changes are per unit of the reduced gas constant, at $c_v/R=50$. A jump’s volume ratio is downstream volume over upstream.

How it was checked #

The dilute limit reproduces a perfect gas’s value, and the second derivative agrees with a direct finite difference, to four figures. The check with most content is the cubic law for a weak jump: a half-per-cent expansion from the ordinary state has $\Delta s=-2.46\times10^{-8}$ along the Hugoniot, and the formula with $\Gamma$ gives $-2.44\times10^{-8}$; from the island, $+6.33\times10^{-9}$ against $+6.49\times10^{-9}$. Those are two separate computations, one through the jump conditions and one through a derivative of an isentrope, and they agree in both size and sign. The saturation pressure from Maxwell’s construction agrees with the tabulated van der Waals value at 0.9 of the critical temperature. The threshold comes from a bisection on the lowest value of $\Gamma$ over the vapour region, and it matches the published 16.7. The model refuses a volume below the excluded volume, a temperature at or below zero, and a saturation line above the critical temperature.

Who found it, and when #

Bethe’s report on shocks in a general fluid is from 1942, Zel’dovich’s from 1946. Thompson’s fundamental derivative is from 1971, and Thompson and Lambrakis’s finding of candidate fluids from 1973. Cramer and Kluwick’s analysis of mixed waves appeared in 1984, and Borisov’s disputed observation in 1983. Colonna’s work on real fluids in Organic Rankine cycle expanders began around 2000 and continues.

Still open: a shock tube’s diaphragm in the island #

The waves here were analysed as isolated jumps and fans. The next calculation would march the one-dimensional Euler equations with this equation of state through a shock-tube’s diaphragm burst, starting in the island, with a shock-capturing scheme that respects the non-convex flux. It would ask which of the possible composite waves forms, how the expansion shock’s strength depends on the starting state, what a pressure trace at a fixed station would look like if a real experiment could get there, and so how much of a real measurement’s ambiguity could be resolved by comparing those traces with a fluid at $\Gamma$ just above zero, where the same tube would show no shock at all.