A tailored tube buys its test time with its driver

The shock that passes without an echo found the one incident Mach number at which a shock tube’s reflected shock crosses the contact surface and sends nothing back. Below it the contact returns an expansion that lowers the reservoir; above it, a shock that raises it; at it, the reflected shock passes into the driver gas as though the interface were not there, and the reservoir at the end wall holds. For helium driving air at room temperature that Mach number is 3.41, and a tube running a little under it — at Mach 2.6 — heard its echo 0.154 driven-tube crossing times after the reservoir formed.

That essay ended by naming what a tailored tube waits for instead. The first candidate is the driver’s own expansion. When the diaphragm bursts, the driver gas starts to expand towards the driven tube, and the head of that expansion runs the other way, back into the driver at the speed of sound there. It reflects from the driver’s closed end and follows the rest of the gas down the tube. Nothing stops it, and when it reaches the end wall the reservoir’s pressure begins to fall. The second candidate is the driver gas itself, mixed across the contact and carried along the wall, and one dimension cannot see it. This essay computes the first, exactly, and uses it to say what tailoring buys and how long a driver it needs.

The tube and its waves #

The tube is the one diaphragm that makes every wave: a driver of length $\ell$ filled with helium, hydrogen or a helium–nitrogen mixture at high pressure, a driven tube of length $L$ filled with air at 300 K, a diaphragm between them and a closed end wall at the far end. Everything is ideal: calorically perfect gases, one dimension, no viscosity. Lengths are in units of the driven tube’s length and times in its crossing time at the air’s sound speed, $L/a_1$.

When the diaphragm bursts, five things leave it. The incident shock runs into the air; the contact surface follows it, separating shocked air from expanded driver gas; and the driver’s expansion is a centred fan, its head running back into the driver at $-a_4$ and its tail following more slowly. At the end wall the incident shock reflects and brings the air to rest in the reservoir. The reflected shock meets the contact, and — tailored — passes into the driver gas as a transmitted shock that brings that gas to rest too. The contact stops where it stands.

The head’s path is exact #

The piece that needs care is the reflected head. After it leaves the driver’s closed end, at $t_0 = \ell/a_4$, it runs back into the fan that is still streaming towards the wall, and a wave crossing a fan is usually a numerical problem: the region behind it is no longer a simple wave, and its characteristics have to be traced step by step. The wall that cancels its own waves did exactly that kind of tracing in space.

The head itself is the exception. It is the first characteristic of the reflected wave, so the flow ahead of it is still the undisturbed incident fan, a simple wave in which the Riemann invariant $u + 2a/(\gamma_4 - 1)$ keeps its driver value everywhere and every state lies on a straight line $x/t = u - a$ from the origin. Along the head, $dx/dt = u + a$. Writing $\xi = x/t$ and using the invariant to express $u$ and $\xi$ through the sound speed $a$ gives

$$t\,\frac{d\xi}{dt} = 2a, \qquad \frac{d\xi}{da} = -\frac{\gamma_4 + 1}{\gamma_4 - 1},$$

so the sound speed along the head falls as a power of time,

$$a = a_4\left(\frac{t}{t_0}\right)^{-2(\gamma_4-1)/(\gamma_4+1)},$$

and the head leaves the fan’s tail, where the sound speed has fallen to $a_3$, at $t_0\,(a_3/a_4)^{-(\gamma_4+1)/2(\gamma_4-1)}$. For helium at its tailored point the exponent is $-2$ and $a_3/a_4 = 0.705$, so the head spends about as long crossing the fan as it took to reach the driver’s end.

The factor is worth a paragraph, because it is where the drivers differ most. The head is slowed inside the fan for two reasons at once: the gas there is still moving towards the diaphragm at a speed that grows along the head’s path, and it is colder than the driver gas at rest, so its sound speed is lower. Both are fixed by how far the driver gas has been expanded, which is the ratio $a_3/a_4$, and the closed form raises that ratio to a power set by the driver’s ratio of specific heats. For helium, with $\gamma_4 = 5/3$, the power is $-2$ and the fan crossing takes 2.01 times the time to the driver’s end; for hydrogen, a diatomic gas with $\gamma_4 = 7/5$, it is $-3$, and although hydrogen is expanded less at its own tailored point, $a_3/a_4 = 0.743$, the crossing takes 2.43 times as long. The mixture, expanded least, crosses in 1.60. A gas whose molecules store energy in rotation cools more slowly as it expands and holds the head in its fan for longer. A fourth-order march of $dx/dt = u + a$ through the simple wave agrees with the closed form to two parts in $10^7$ for all three drivers, which is the first check.

From the fan’s tail onwards the head runs through uniform gas and its path is straight lines: through the expanded driver gas at $u_3 + a_3$; across the transmitted shock into the driver gas at rest, at $a_6$; across the stopped contact into the reservoir, at $a_5$; and so to the wall. If the driver is short and the head reaches the contact before the reflected shock does, it goes through the shocked air at $u_2 + a_2$ instead and meets the reflected shock on the way. The head carries no strength of its own, so its path needs only the speeds of the regions it crosses and the positions of the waves between them.

Tailored, the reservoir waits for the driver #

What the end wall records is the plainest way to see the difference tailoring makes. In an untailored tube the pressure there steps up when the incident shock reflects, holds, and then steps again — down by 8 per cent at nine-tenths of the tailored Mach number, up by 9 at eleven-tenths — when the contact’s echo arrives, and it goes on stepping as the echo rattles between the wall and the contact. In a tailored tube it steps up once and holds until the reflected head arrives, and then falls through the whole of the reflected fan. What a shock costs is the entropy each of those steps leaves behind; the tailored reservoir has paid only for the first.

The first figure is the tailored tube, helium driving air at Mach 3.41, with a driver half as long as the driven tube. The incident shock reaches the end wall at 0.293 $L/a_1$ and the reservoir forms. The reflected shock meets the contact 0.92 of the way down the tube at 0.354, and the contact stops there; the transmitted shock carries on into the helium at 1.83 $a_1$ back towards the diaphragm. Nothing returns to the wall.

Meanwhile the fan’s head has reached the driver’s closed end at 0.170 and started back. It leaves the fan’s tail at 0.343, crosses the transmitted shock at 0.460, passes the stopped contact at 0.522 and reaches the end wall at 0.555. The reservoir has lasted 0.261 $L/a_1$ — for a ten-metre driven tube, 7.5 milliseconds. That is about seventy per cent more than the under-tailored tube’s 0.154, which is a smaller gain than the idea of tailoring suggests, and the reason is that the driver in the figure is short.

A tailored tube buys its test time with driver length #

The second figure varies the driver’s length. With the contact silent, the reflected head is the only thing in one dimension that ends the reservoir, and every stage of its journey scales with the length it had to go: the time to the driver’s end, the fan crossing, which the closed form makes a fixed multiple of it, and the straight runs after, most of which were set by where the fan left the head. So the tailored test time grows linearly with the driver’s length: 0.69 $L/a_1$ per driven length of driver for helium, 0.56 for hydrogen, 1.19 for the 70 per cent helium mixture.

The ordering is the drivers’ sound speeds. Hydrogen’s is highest, its expansion head runs fastest, and a hydrogen driver of a given length holds the reservoir for the least time; the mixture is heaviest and slowest and holds longest. That is the price paid for what each driver is chosen for. Hydrogen tailors at Mach 6.0 and gives a reservoir at 5,000 K; the mixture tailors at Mach 1.88 and gives 685 K. The same property — a light, fast driver gas — makes the strong shock and ends it early.

Tailoring pays only past a break-even driver #

The comparison that says what tailoring is worth sets the tailored tube’s test time against the echo time of the same tube run off its tailored Mach number. The third figure finds the driver length at which the two are equal: shorter than that, a tailored tube’s own reflected expansion reaches the end wall before an untailored tube’s echo would have, and tailoring gains nothing at all. It is a break-even driver, and it is not short. Near tailoring it is about a quarter of the driven tube for helium, 0.26; about a sixth for hydrogen, 0.15; two-fifths for the mixture, 0.38. At nine-tenths of the tailored Mach number, where the echo would remove 8 per cent of the reservoir pressure, the helium break-even is 0.29.

In a facility’s units that is a helium driver of 2.6 metres on a ten-metre driven tube before tailoring is worth anything, and every metre beyond it buys two milliseconds that an untailored tube would not have. The hydrogen break-even is shorter, a metre and a half on the same tube, only because hydrogen’s untailored echo comes early too: its strong shock crosses the tube fast and its reservoir’s sound speed is high, so everything in a hydrogen-driven tube happens sooner, the waves that end the reservoir as well as the ones that make it. Measured against its own echo, each driver needs a driver section of the same order — between a sixth and two-fifths of the driven tube — before tailoring changes the answer.

The curves change slowly with how far off tailoring the comparison is made, because the echo’s timing is set by how fast the reflected shock and the reservoir’s sound speed cross the tube — by the shock’s own strength — and not by how strong the echo is. The echo’s strength is what tailoring removes. Its timing is what the driver has to beat. That is a design constraint rather than a detail: a driver built short to save pressure vessel and cost makes tailoring pointless, and the saving is paid for in reservoir time the tube was built to deliver.

A hot driver tailors higher and holds for less #

The fourth figure asks the question a facility asks next. Heating the driver gas raises its sound speed, and the earlier essay found that a faster driver tailors at a stronger shock: helium at 600 K tailors at Mach 5.04 and delivers a reservoir at 3,600 K against 1,777 K from room temperature. The same speed shortens the test time. The expansion head runs faster at every stage of its journey, and the test time per driver length falls from 0.69 to 0.49 $L/a_1$.

Enthalpy is bought with test time. Every route to a hotter reservoir in a tailored tube — a lighter driver, a hotter one — makes the driver’s expansion faster and the reservoir shorter per metre of driver. The only way to keep both is a longer driver, and that is why the facilities built for high enthalpy are long ones, or replace the driver’s expansion altogether with a free piston that compresses the driver gas just before the diaphragm breaks, so that the gas behind the diaphragm is hot while its expansion wave meets a moving piston rather than a closed end.

Milliseconds per metre of driver #

The fifth figure turns the result into a facility’s units. A ten-metre driven tube of air has a crossing time of 29 milliseconds at 347 metres a second. With a ten-metre helium driver, tailored at Mach 3.41, the reservoir holds for 17.5 milliseconds; with a ten-metre hydrogen driver at Mach 6.0, 14.9; with the mixture, 30. Every metre of helium driver buys about two milliseconds.

These are the numbers the one-dimensional ideal gives, and they are upper bounds, since the one-dimensional ideal has no mechanism for the driver gas to reach the end wall at all. In practice it does. The reflected shock meets the boundary layer the incident shock left on the tube’s wall and bifurcates, and cold driver gas jets along the wall under the bifurcated foot towards the end wall; the contact is not a plane but a mixing layer, which a surface that remembers the diaphragm shows growing from the start. Measured contamination times in reflected-shock tunnels are commonly a fraction of the ideal wave-limited times at high enthalpy. The wave calculation says what the tube can do; the contamination says what it will.

What the calculation was checked against #

The ledger holds three checks and two results. The head’s exit from the fan in closed form matches a march through the simple wave to $2\times10^{-7}$ for three drivers. Off tailoring, the same wave diagram reproduces the earlier essay’s echo — 0.1543 $L/a_1$ at Mach 2.6 against 0.154 — which says the reflected shock, the contact and the reservoir’s sound speed are placed exactly as before. And at the tailored point the transmitted shock leaves the driver gas at rest to $2\times10^{-15}$, which is what tailoring means in this construction: the interface stops dead.

What one dimension leaves out #

Driver-gas contamination. The one-dimensional contact stops and stays; a real one is a growing mixing layer, and driver gas reaches the end wall along the boundary layer under the reflected shock’s bifurcated foot. It is usually the binding limit at high enthalpy.

The head is only a head. The reflected expansion’s head starts the reservoir’s decline; the pressure falls over the time the whole reflected fan takes to arrive. Where the reservoir is called steady here, it is steady to the arrival of the first weak wave.

Real gas. At the helium and hydrogen points the air in the reservoir is hot enough to vibrate and dissociate, which moves the tailored Mach number and the reservoir’s sound speed; neither is in a calorically perfect gas. When gamma stops being a number computes how far the ratio of specific heats falls, and a gas that has not finished being shocked how long the air behind a shock takes to reach it.

Diaphragm opening. A real diaphragm takes a finite time to open, and the fan is not quite centred. The effect on the head’s arrival is a small shift in its origin.

The convention: the driven tube’s crossing time #

Times are in $L/a_1$, the driven tube’s length over the air’s sound speed at the fill temperature; lengths in $L$, with the diaphragm at zero and the end wall at one; the driver’s length $\ell$ is a fraction of $L$. The test time is the interval between the incident shock’s reflection at the end wall and the first arrival of any wave there. The drivers are pure helium, pure hydrogen and 70 per cent helium in nitrogen by mole, each at 300 K unless heated.

Who worked it out #

The reflected-shock tunnel was developed at the Cornell Aeronautical Laboratory in the 1950s, where Wittliff, Wilson and Hertzberg used tailored interfaces to extend its running time; the wave-diagram analysis of shock-tube test time, including the reflected expansion from the driver’s end, is laid out in Glass and Hall’s handbook of 1959 and in Gaydon and Hurle’s book on the shock tube of 1963. The free-piston driver that heats the driver gas just before the diaphragm breaks is Stalker’s, from the 1960s.

Still open: the driver gas that arrives first #

The one-dimensional reservoir is ended by the reflected head, and a real one is usually ended sooner by driver gas. The next calculation gives the incident shock a turbulent boundary layer on the tube wall, lets the reflected shock bifurcate over it with the foot height the interaction predicts, and follows the cold driver gas that the bifurcation drives along the wall towards the end wall, asking at what incident Mach number and tube diameter the contamination arrives before the reflected head — the diameter at which a tailored tube stops being wave-limited and becomes wall-limited, which is the number that decides whether a longer driver buys anything at all.