The shock that passes without an echo

One diaphragm, every wave ends on a condition it names and does not compute. A shock tube’s incident shock reflects from the closed end of the tube, brings the gas it has just set moving back to rest, and leaves behind it a slug of hot, high-pressure, motionless gas — the reservoir that a reflected-shock tunnel expands through a nozzle, and that a chemical kineticist heats a fuel mixture in. The reflected shock then runs back up the tube and meets the contact surface, the boundary with the driver gas. What it sends back from that meeting decides how long the reservoir lasts.

A surface that remembers the diaphragm explains why the contact is there at all: it is made of fluid, it carries the entropy difference between shocked and expanded gas, and nothing in the pressure field reveals it. This essay puts a different gas on its far side and asks what the reflected shock finds there. The answer is a number, not a qualitative preference, and it is the same number whichever way it is derived.

The meeting is a second diaphragm #

When the reflected shock reaches the contact, the gas on one side of the contact is the air it has just processed — at rest, at the reservoir pressure $p_5$. The gas on the other side is helium that has not yet heard of the reflection: it is still moving towards the end wall at the speed the incident shock gave the air, $u_2$, and still at the pressure behind the incident shock, $p_2$. Two uniform states meet at a plane, with different pressures and different velocities. That is a Riemann problem — exactly the problem the original diaphragm posed, except that the two states are now moving relative to each other and are made of different gases.

Its solution has the same structure as before. One wave runs into each gas, and between them a new contact settles at a single pressure $p^*$ and a single velocity. The wave running into the helium must bring the helium from $u_2$ to the interface velocity. The wave running back into the reservoir must bring the air from rest to the same velocity. Whether that second wave is a shock or an expansion, and how strong, is fixed by $p^*$ against $p_5$.

The computation solves it the way any exact Riemann solver does. For each side there is a function giving the velocity change a wave produces in taking the gas from its own pressure to a trial $p^*$: the Rankine–Hugoniot branch if $p^*$ is higher, the isentropic expansion branch if it is lower. The two velocity changes must add up to the velocity difference $u_2$ that the two gases start with, and a bisection on $p^*$ finds the pressure at which they do. The ratio $p^*/p_5$ is the whole outcome. Below one, an expansion returns to the end wall; above one, a shock does.

Behind it is the incident shock itself: the fill Mach number $M_s$ gives $p_2$, $u_2$ and $T_2$ from the normal-shock relations, the driver’s own expansion gives the diaphragm pressure ratio $p_4/p_1$ needed to make that shock, and a second application of Rankine–Hugoniot, stopping gas moving at $u_2$ against a wall, gives $p_5$ and $T_5$. The whole chain is ideal-gas and one-dimensional, with each gas stated by its ratio of specific heats and its molar mass.

The one strength at which nothing returns #

Every light driver produces the same shape. At a weak incident shock the ratio sits below one, so an expansion returns, and it gets lower before it recovers — for helium it falls to 0.76 between Mach 1.8 and Mach 2. Then it rises steadily through one and keeps rising, so that a strong shock from a light driver sends back a strong shock. There is exactly one crossing for each driver, and it is what the literature calls the tailored condition: for helium driving air from 300 K, at an incident Mach number of 3.4085. At that strength the interaction has no reflected wave. The reflected shock simply continues into the helium, the interface velocity is zero to machine precision, and the reservoir behind it is untouched.

Either side of it the damage is quick. At nine-tenths of the tailored Mach number the ratio is 0.920, so an expansion arrives and removes 8% of the reservoir pressure; at eleven-tenths it is 1.090, so a shock arrives and adds 9%. The word used for the two sides records which way the facility is wrong. A tube running under-tailored has a driver too soft for the shock it is making and returns an expansion. One running over-tailored has a driver too stiff and returns a shock. Under-tailoring is the worse of the two in practice. A pressure increase merely changes the reservoir to a new, still steady state, which can be measured and allowed for. An expansion cools the test gas and sets it moving towards the nozzle.

The timing is set by the tube’s length and the fill gas’s speed of sound. With the diaphragm at x = 0 and the end wall at x = L, the incident shock at Mach 2.6 arrives at the wall at t = 0.385 $L/a_1$. The reflected shock, travelling at the speed Rankine–Hugoniot gives it into gas moving the other way, meets the contact 0.89 L from the diaphragm at 0.482 $L/a_1$. The head of the returning expansion runs at the reservoir’s own speed of sound, $a_5$, and reaches the wall at 0.539 $L/a_1$. That leaves a steady reservoir for 0.154 $L/a_1$ — for a ten-metre driven tube filled with air at 347 m/s, 4.4 milliseconds. The tailored tube sees no such arrival, and its reservoir lasts until something slower comes: the expansion from the far end of the driver, or the driver gas itself.

Two gases stopped at the same pressure #

There is a second way to state the condition, and it is the one that says why it is a single number. If nothing returns to the end wall, the reservoir is still at rest at $p_5$ after the meeting — so the wave that ran into the helium must also have left the helium at rest at $p_5$. It is a shock, transmitted into the helium, that stops helium moving at $u_2$ and brings it to pressure $p_5$.

But the reflected shock in the air also stopped gas moving at $u_2$ and brought it to $p_5$. So the tailored condition is that one velocity change, $u_2$, stops the shocked air and the expanded helium at the same pressure. A shock’s pressure rise per unit of velocity destroyed — $(p_5 - p_2)/u_2$ — is its shock impedance, and tailoring is the equality of the two gases’ shock impedances at exactly this amplitude. The two stopping pressures, computed independently, agree to better than one part in 10¹² at the tailored point. That is the check that the Riemann solver and the two-shock argument are the same statement.

The expected form of that statement is familiar from waves in any medium. A pressure wave crossing a boundary between two media is reflected with an amplitude set by the mismatch in their acoustic impedance, ρa, and passes through without reflection when the two are equal. That is the condition the pulse that grows as it leaves the heart uses for an arterial pulse meeting a branching vessel. It is also why an optical coating is designed the way it is, and why a transmission line is terminated in its own characteristic impedance. The shock-tube version is the same idea for a wave that is not small. The shock impedance depends on the wave’s strength, so the matching has to happen at the right amplitude, and that is why the condition picks out one Mach number rather than one pair of gases.

And the acoustic version is not far off. At the tailored point the acoustic impedances of the shocked air and the expanded helium, $\rho_2a_2$ and $\rho_3a_3$, differ by only 2.8% — although the reflected shock there is not weak, with $p_5/p_2$ = 5.47. Away from it they are far apart: 0.768 of each other at Mach 2.6, 1.79 at Mach 5. So a designer estimating the tailored point by matching ρa across the contact lands close, and the exact Riemann solution then moves the answer by a few per cent. The two gases are also far from alike in any other respect: the air behind the incident shock is at 957 K, and the helium, cooled by its own expansion from 300 K, is at 149 K and 11% less dense.

A driver mixed to the shock that is wanted #

A facility does not get to pick its test condition to suit its driver. The reservoir temperature and pressure are set by the experiment — a combustion chemist needs a particular ignition temperature, and a hypersonic tunnel needs a particular flight enthalpy. So the driver is adjusted instead, and the cheapest adjustment is its composition. Mixing a heavier gas into helium makes the driver denser and its sound speed lower. Its expanded gas is stiffer, and the crossing in the ratio figure moves to a weaker incident shock.

The mixtures are computed with each component’s molar heat capacity weighted by its mole fraction, so seventy per cent helium in nitrogen has γ = 1.556 and a molar mass of 11.2 g/mol. Against air, the tailored Mach number runs smoothly from 1.25 at thirty per cent helium to 3.41 at a hundred. With argon in place of nitrogen the same effect is sharper, because argon is heavier than air: half helium and half argon tailors at 1.07, eighty per cent helium at 1.75, and twenty per cent helium tailors at no Mach number at all. A driver of the driven gas itself is the limiting case. Air driving air returns a shock at every strength it can make — the interface pressure is 1.11 times the reservoir’s at Mach 1.5, 1.90 at Mach 3 and 3.33 at Mach 5. The expanded driver air is colder and therefore denser than the shocked test air, and a denser gas at the same pressure is the stiffer one. A single-gas shock tube is over-tailored by construction.

Heating the driver raises the point #

Mixing moves the point down; heating moves it up. A hotter driver has a faster sound speed, and a faster sound speed does two things at once. It lets the driver’s expansion produce a higher gas velocity for the same pressure ratio, so a stronger incident shock. It also makes the expanded driver gas less dense for the same pressure, so it is softer, and a stronger shock is needed before its shock impedance catches up with the air’s. For helium the tailored Mach number rises to 4.03 at 400 K, 4.56 at 500 K, 5.04 at 600 K and 5.89 at 800 K. For hydrogen it is 7.00, 7.87, 8.65 and 10.04 at the same temperatures.

These numbers explain the arrangement of the high-enthalpy facilities. A tunnel that needs the reservoir of a flight at Mach 10 or above cannot reach it tailored with a cold helium driver, which stops at 3.4. The two traditional routes are the ones the figure draws. One is to heat the driver, electrically or by combustion. The other is to compress it adiabatically, which is what a free-piston driver does: a heavy piston shoots down a compression tube and raises a helium–argon driver to several thousand kelvin in the moment before the diaphragm bursts. In both cases the composition is then adjusted to sit on the tailored point at the temperature achieved.

The reservoir a tailored tube delivers #

Each tailored point is a reservoir. With helium the air is brought to 1,777 K at 73.3 times the fill pressure, which is about the stagnation temperature of a flight at Mach 6 in the stratosphere. With hydrogen it is 5,032 K at 292 times the fill pressure. With seventy per cent helium in nitrogen it is 685 K at 12.2 times. These are the ideal-gas values, and the last of the three is the only one a calorically perfect model can honestly claim. At the helium point, vibration in nitrogen and oxygen is already well excited, and a gas that has not finished being shocked shows how long it takes to reach equilibrium behind a shock. At the hydrogen point, much of the oxygen is dissociated. The effect on the tailored Mach number goes in a definite direction. A real gas absorbs energy into internal modes, so its γ falls, as when gamma stops being a number computes. The shocked air then comes out denser and cooler than the ideal model says, its shock impedance rises, and the crossing moves.

The reservoir pressure is always slightly below the driver’s. At the three tailored points $p_5/p_4$ is 0.985 for the mixture, 0.954 for helium and 0.876 for hydrogen, and the reason is exact. Tailored, the driver gas has been expanded from rest at $p_4$ to $u_2$ and $p_2$ and then brought back to rest at $p_5$. Had it been brought back to rest by an isentropic compression travelling the same way, the Riemann invariant $u + 2a/(\gamma - 1)$ that its own expansion carried would be conserved. It would arrive at rest with its original sound speed, and so at $p_4$ exactly: an isentropic stop simply retraces the expansion. A shock stops it with an entropy rise and so short of $p_4$, by the stagnation-pressure loss that two totals, one of which a shock cannot touch tabulates. The stronger the reflected shock in the driver gas — and it is strongest for hydrogen — the larger the shortfall. So the driver’s starting pressure is an upper bound on any tailored reservoir, and the gap below it is the entropy the reflected shock left in the driver gas.

What the picture cannot show #

The distance–time diagram draws each wave as a line, and three things are lost in doing so.

The returning expansion is a fan, not a line. Its head travels at $a_5$ and its tail more slowly, so the reservoir pressure falls over a finite interval rather than at an instant. The line drawn is the head, which is what ends the steady period, and the 17% is the drop once the whole fan has arrived.

The contact is not a surface. A surface that remembers the diaphragm notes that the contact is unstable under acceleration, and the reflected shock is precisely such an acceleration. It is the Richtmyer–Meshkov instability in its classic form: an impulsive shock crossing a perturbed density interface. So the real interaction is a shock passing through a mixing layer a few centimetres thick, not a plane, and the mixed gas reaches the end wall earlier than the plane contact would. A tailored tube removes the reflected wave. It does not remove the driver gas, whose arrival — driver-gas contamination — is the next limit on the test time, and in practice often the binding one.

And the boundary layer is missing. Behind the incident shock the tube wall grows a boundary layer that the reflected shock then runs into, and at high enough Mach number the reflected shock bifurcates into a lambda foot that lets driver gas jet along the wall towards the end. The one-dimensional calculation is the ideal case, and the ideal case is still the one every facility designs from.

The convention the numbers depend on #

Regions are numbered as in every shock-tube text: 1 is the driven gas at rest before the diaphragm bursts, 2 the gas behind the incident shock, 3 the expanded driver gas behind the contact, 4 the driver at rest before the burst, and 5 the gas brought to rest behind the reflected shock. Pressures are ratios to $p_1$. Times are in units of $L/a_1$, where L is the driven tube’s length and $a_1$ the fill gas’s speed of sound, and distance is measured from the diaphragm. The Mach number is always the incident shock’s, relative to the gas at rest ahead of it. Both gases are calorically perfect. Air and nitrogen are taken at γ = 1.4, helium and argon at 5/3, hydrogen at 1.4, although hydrogen’s rotational heat capacity is not fully excited at room temperature and its γ there is a little above 1.4 — a difference this calculation does not pursue. Mixtures take their γ from mole-weighted molar heat capacities.

How each number was checked #

The chain is checked at each link rather than only at the end. The incident shock’s gas speed is recomputed from the other side, as the velocity the driver’s isentropic expansion produces in falling from $p_4$ to $p_2$, and the two agree to rounding error at Mach 1.5, 3 and 5. Mass flux into the reflected shock, in its own frame, equals mass flux out. The tailored Mach number is found by bisection on $p^*/p_5 - 1$ and then independently confirmed by solving for the pressure at which a shock stops the expanded helium, which must be $p_5$ and is. The ordering — hydrogen tailors above helium, the mixture below — is a prediction made before the numbers were computed, and so is the sign of the returning wave either side of the root. So is the claim that a single-gas tube never tailors. Each could have failed, and the calculation refuses an incident Mach number of one, a mole fraction outside zero to one, and a tolerance it cannot meet.

Who found it, and when #

The reflected-shock tunnel was developed at the Cornell Aeronautical Laboratory in the 1950s, and the tailored interface was stated there by Wittliff, Wilson and Hertzberg in 1959. Their paper gives the condition, shows that it lengthens the useful reservoir by a large factor, and sets the design practice — choose the driver to tailor at the test condition — that facilities have followed since. The Riemann problem at its heart is Riemann’s from 1860. The instability of the contact that the reflected shock drives is Richtmyer’s from 1960 and Meshkov’s from 1969. The free-piston driver that makes tailoring at high enthalpy practical is Stalker’s, from the late 1960s, and its tuning is exactly the mixture-and-temperature adjustment computed above.

The same test time is what makes the shock tube a chemical instrument. The zone of delayed reaction that a gas that has not decided to react yet describes behind a detonation front is measured in reflected-shock reservoirs. A long ignition delay can be measured only in a tube that holds its reservoir steady for longer than the delay, and that is a tailored tube.

Still open: the tailored point in real air, and the test time it buys #

Two calculations follow directly. The first is the tailored Mach number with equilibrium air in place of the ideal gas. Region 5 at the helium and hydrogen points is hot enough that vibration and dissociation change the reflected shock’s jump conditions. The direction of the shift is known from the argument about shock impedance, but its size is not, and it is what a facility actually has to set. The same Riemann solution with an equilibrium equation of state for the air would give it, and the γ-dependence of the tailored point already drawn for mixtures is the first estimate.

The second is the test time a tailored tube actually wins. Once the returning wave is removed, the reservoir ends when the reflected head of the driver’s own expansion arrives from the driver’s closed end, or when driver gas reaches the end wall — whichever is first. The first is a characteristic calculation in the driver, the method of the wall that cancels its own waves run in time instead of space. The second needs the contact’s mixing layer and the wall boundary layer, and so a model beyond one dimension. Comparing the two against the under-tailored 0.154 $L/a_1$ would say how much the condition is actually worth for a given tube.