The sheath outlasts the body

The gas beside a hypersonic body remembers its nose followed each streamline round a blunt body from where it crossed the bow shock. Near the axis the shock is normal and the crossing costs the gas a great deal of entropy; far out the shock has weakened to a Mach wave and costs almost none. Expanded to the pressure beside the afterbody, the gas that came through the nose’s normal shock is ten to twenty times the free stream’s temperature at Mach 15, it never cools below a floor set by the nose alone, and the sheath that carries it thickens to more than the body’s own diameter.

Its last section named the question that decides whether any of that matters. The body’s own boundary layer — the thin layer next to the wall where the gas is slowed to rest and where the heat reaches the body — grows along the afterbody by drawing in the gas at its edge. That gas is the sheath’s. The boundary layer’s heating and its transition to turbulence are both set by its edge conditions, so what matters is how long the sheath lasts: how far down the body the boundary layer has to grow before it has taken in the sheath’s air and starts to feed on the cooler gas outside. The distance is called the swallowing distance, and it is usually said to be short. The calculation here grows the boundary layer through the sheath and measures it.

Growing a boundary layer through the sheath #

The body is the previous essay’s: a sphere-cylinder of diameter d at Mach 15 in a perfect gas, its bow shock Billig’s measured one, each streamline taking the normal-shock entropy for the local shock angle and expanding to the afterbody pressure the blast analogy gives. The streamline that starts a distance y from the axis upstream carries, between itself and the axis, the mass flux $\pi y^2\rho_\infty U$. The sheath’s edge is the streamline carrying half the axis’s entropy, and at Mach 15 it started 1.22 diameters out: the sheath is the air of a circle two and a half diameters across, spread round the body.

The boundary layer is laminar and locally similar, in the form Lees and Dorodnitsyn gave it. Its whole history of edge conditions is carried in one integral, $\xi = \int \rho_e\mu_e u_e r_b^2\,dx$ — density, viscosity and speed at the edge, times the body’s radius squared, summed along the body — and the mass it holds is $2\pi f\sqrt{2\xi}$, with f the Blasius stream function at the layer’s outer edge, 2.26. The boundary layer’s edge therefore sits at the streamline that carries the same mass, and that streamline’s density, viscosity and speed set how fast ξ grows next. The calculation marches along the body, updating the edge as the layer eats outward, until the edge streamline is the sheath’s. Viscosity goes as the temperature to the 0.7.

A third of a Reynolds number #

The first figure is the march. Each curve is the upstream radius of the streamline at the boundary layer’s edge, rising along the body as the layer thickens; the horizontal line is the sheath’s edge. At a Reynolds number of a million on the nose diameter the laminar layer has reached the streamline that started 0.12 diameters out by ten diameters back, 0.17 by a hundred and 0.27 by a thousand. It reaches the sheath’s edge at 340,000 diameters.

The distance is almost exactly proportional to the Reynolds number. At 10⁴ it is 3,300 diameters; at 10⁷, 3.4 million. The reason is in the laminar layer’s own scaling. Its thickness grows as the square root of distance over the Reynolds number, so the mass it holds does too, and to hold a fixed mass — the sheath’s — the distance has to grow as the Reynolds number itself. The constant, a third at Mach 15, is set by how much mass the sheath holds and by how little of it the hot, thin gas at the edge brings in per unit length.

The dashed curve is a turbulent boundary layer at a Reynolds number of a million, estimated from the one-seventh power law with the local edge conditions and assumed turbulent from the shoulder. It entrains far faster, and swallows the sheath at 536 diameters. That is a rough number — the incompressible law, uncorrected for the density variation across a hypersonic layer or a cooled wall, each of which moves it by factors of order two — but its order is not in doubt.

The edge stays hot #

The second figure is what the boundary layer sees. The gas next to the body — the axis streamline, which came through the normal part of the shock — cools from twenty times the free stream’s temperature just behind the shoulder towards its floor of nine as the afterbody pressure falls. The gas outside the sheath, which crossed the shock where it had weakened, is barely warmer than the free stream: 1.4 times ten diameters back, 1.06 a thousand back. The laminar boundary layer’s edge follows the first curve, not the second. Ten diameters back it is at 11.4 times the free stream’s temperature; a thousand back, still 8.0.

Even the turbulent layer, which gets through the sheath in hundreds of diameters, is feeding on sheath gas for the whole of any body of ordinary length. The body the boundary layer is growing on is, at every station, a body immersed in its own nose’s wake.

Linear in Reynolds number, a quarter power turbulent #

The third figure puts the scalings side by side. Laminar, the swallowing distance is a straight line of slope one on the log-log axes: 0.34 times the Reynolds number at Mach 15. Turbulent, it rises as the Reynolds number’s quarter power, because a turbulent layer’s mass grows as distance to the four-fifths over the Reynolds number to the fifth: from 162 diameters at 10⁴ to 961 at 10⁷. The two faint lines bracket the lengths of real bodies, 5 to 50 of their own nose diameters.

No curve comes near them. A laminar boundary layer on a body of practical length never gets through the sheath at any Reynolds number drawn. A turbulent one needs a body several hundred nose diameters long — which is to say, a nose far smaller than the body behind it. That is why the effect is measured and discussed on slender cones with slightly blunted tips, whose noses are a few millimetres across on bodies of a metre, and not on a blunt capsule, where the whole question is academic because the whole body is sheath.

Put into metres #

The scalings become concrete once a body is put in the atmosphere. Take a blunt nose half a metre across at Mach 15. At 50 kilometres, where the air is a thousandth of its sea-level density and the flight speed about 4.9 kilometres a second, the Reynolds number on the nose diameter is about 150,000. The laminar swallowing distance is then 51,000 diameters — 25 kilometres of body — and even a boundary layer turbulent from the shoulder would need 330 diameters, 165 metres. At 30 kilometres the air is eighteen times denser, the Reynolds number 2.8 million, and the laminar distance 960,000 diameters; the turbulent, 700 diameters, 350 metres.

A re-entry capsule is one to three of its own diameters long. A slender body with a half-metre nose might be ten. On every one of them the boundary layer’s edge is sheath gas from the nose to the tail, and the numbers are not close. The slender cone with a nose a centimetre across does better in proportion — a metre is a hundred of its nose diameters — but at 30 kilometres, on the sphere-cylinder’s terms, it would still need 256 nose diameters turbulent and 19,000 laminar. The sheath outlasts the body by a long way in every case except the slender, turbulent one, and there it comes within a factor of a few.

The contrast with a body that makes no sheath is the one the previous essay drew. A sharp cone’s shock lies close along its surface and every streamline crosses it at the same angle, so the gas beside it is uniformly warm and there is nothing for the boundary layer to swallow. The sheath exists only because the nose is blunt, and it is exactly the bluntness that makes the sheath’s gas hot and the swallowing slow.

Faster bodies carry the sheath further #

The fourth figure varies the Mach number. The laminar coefficient rises from 0.09 at Mach 8 to 0.34 at Mach 15 and 1.05 at Mach 25, and the turbulent distance at a Reynolds number of a million from 180 diameters to 1,350. Both rise for two reasons that work together. The sheath holds more air at higher Mach numbers, because the curved part of the shock that makes entropy reaches further out — the half-entropy streamline started 0.90 diameters out at Mach 8 and 1.58 at Mach 25. And its gas is hotter and thinner, so the boundary layer draws in less of it per unit length. The fastest bodies, whose heating matters most, carry their sheath furthest.

What the sheath does to the heating #

The sheath is not only a curiosity of the flow field. Lees’s local-similarity law makes the laminar heat flux proportional to $\rho_e\mu_e u_e r_b/\sqrt{2\xi}$, times the difference between the gas’s total enthalpy and the wall’s — and the total enthalpy is the same for every streamline, since the shock conserves it. So the sheath changes the heating only through the edge gas’s density, viscosity and speed. The fifth figure takes the ratio: the flux with the sheath at the edge over the flux the same boundary layer would receive if its edge gas were the gas outside. At a Reynolds number of a million it settles near 0.71, from 0.69 two diameters back to 0.72 far downstream. At 10⁴ the boundary layer is thicker, reaches further into the sheath and so meets cooler gas sooner, and the ratio climbs from 0.73 two diameters back to 0.84 a thousand back: the shielding fades as the layer approaches the sheath’s edge, and it would reach one where the sheath is swallowed.

The hot, thin sheath gas brings less mass flux to the boundary layer than its higher viscosity makes up, and the layer is thicker and conducts less. A blunt nose’s sheath takes about three-tenths off the afterbody’s laminar heating, for the whole length of any practical body. That is a second reason, after the stagnation-point argument the previous essay gave, that a nose too blunt for its shock to stay attached is the right choice for a body that has to survive: the curved shock that makes the sheath shields the body behind it, and the shielding lasts.

It also bears on transition. Transition to turbulence is set by the boundary layer’s edge conditions, and a sheath of hot, low-density gas lowers the local Reynolds number the layer sees. That is the mechanism usually offered for the measured fact that a small bluntness on a slender cone moves transition aft — and the calculation says the sheath is still there, at any practical length, to do it.

Why the sheath lasts #

The sheath lasts because it is large and the boundary layer is slow to eat. A hypersonic body leaves a line explosion behind it found that the body deposits its drag in the air as a cylindrical blast, and the sheath is the core of that blast, where the energy stays: the air that crossed the strong part of the shock. The spherical blast of a radius that gives the energy away set the pattern — a thin shell of dense gas at the front and a hot, nearly empty interior behind it. The fireball was hollow for the same reason — the air that met the strongest shock ends up hottest and thinnest, and the core of a blast is where most of its volume is and least of its mass. A boundary layer entrains mass, not volume, and the sheath’s small mass is spread over a large, hot, low-density volume. The boundary layer has to grow through all of it.

What the nose paid at the start, the cost of the shock in irreversible entropy, is therefore paid out along the body in the boundary layer’s edge conditions — slowly, and never completely on a body of ordinary length. The analogy’s error, that it put the energy at an impossible temperature on the axis, and the previous essay’s correction, that the energy is in a sheath round the body, lead here to one more fact: the sheath is still there at the tail.

What was checked #

The Blasius solution in the Lees–Dorodnitsyn scaling, found by shooting, gives the wall shear 0.46960 to a hundred-millionth and puts the 99 per cent edge at η = 3.47, where the stream function is 2.26. With uniform edge conditions the march has a closed form — ξ grows linearly and the swallowing distance is the Reynolds number times $y^4/(8f^2\rho\mu u\, r_b^2)$ — and the march finds it to two parts in a million. Doubling the number of steps moves the swallowing distance by less than a part in a million. The checks refuse a Mach number below one and a Reynolds number of zero.

What the calculation leaves out #

The pressure beside the body. The afterbody pressure is the blast analogy’s plus the ambient, which the previous essay named as a patch; the pressure sets every streamline’s temperature and so every edge condition.

Local similarity. The boundary layer is taken as similar at each station with its current edge conditions carried in ξ. That is exact for edge conditions that change slowly compared with the layer’s own growth and is the standard engineering form, but a layer eating into a gradient of entropy has a profile the similar one does not quite have.

Vorticity and a real gas. The sheath carries vorticity, since its entropy varies across it, and the boundary layer’s edge is not at uniform conditions. At Mach 15 and above, air dissociates behind the normal shock and the perfect-gas temperatures here are upper bounds, as the previous essay noted.

The nose’s own run. The march starts at the shoulder, half a diameter behind the nose, with the boundary layer’s growth over the hemisphere taken at the axis streamline’s conditions. The boundary layer there is thin and holds almost none of the sheath’s mass, so the swallowing distance does not feel the choice, but the edge conditions in the first diameter or two do.

The turbulent estimate. The one-seventh law with local edge properties and no correction for compressibility or wall temperature is an order-of-magnitude estimate. Turbulent heat transport at hypersonic speeds is itself a subject.

Who worked it out #

The entropy layer and its swallowing were identified in the hypersonic research of the late 1950s and 1960s, when blunt re-entry bodies made them unavoidable; Lees’s local-similarity heating law is from 1956. Rotta and others estimated swallowing distances in the 1960s, and Stetson’s experiments on slightly blunted cones in the 1980s tied the movement of transition to the swallowing length. The calculation here grows the boundary layer through the sheath on its own edge conditions and gives the distance’s scaling with Reynolds number and Mach number.

Still open: a sharp cone with a slightly blunted tip #

Every number here is for a sphere-cylinder, whose nose is as wide as the body. The case the swallowing distance matters for is a slender cone whose tip is blunted by a small radius, a hundredth of its base or less: its sheath is small, its body long in nose radii, and a turbulent boundary layer could swallow the sheath on the body. The next calculation puts the same entropy layer on a cone, with the cone’s own surface pressure in place of the blast analogy’s, and asks at what nose radius the swallowing distance equals the cone’s length — the bluntness that marks where a vehicle’s afterbody stops being shielded by its nose.