The wall that heats itself

A wall in a stream can be told its temperature, and then heat flows in or out to maintain it. Or it can be told nothing — insulated, or moving fast enough that whatever heat it exchanges is negligible — and then it settles wherever the flow puts it.

Where the flow puts it is not the free-stream temperature. The fluid next to the surface has been brought to rest, its kinetic energy has gone somewhere, and the somewhere is heat — which is the compressible energy equation’s whole content. A wall told nothing about its temperature ends up most of the way to the stagnation temperature, and every consequence of high-speed flight that involves a material property follows from that number.

The equation with the term put back #

The other layer solved the energy equation with its right-hand side set to zero, which is legitimate at low speed and throws away the whole of this essay. Restored, it reads

$$T'' + \tfrac{Pr}{2}\,f\,T' = -Pr\,\frac{U^2}{c_p}\,(f'')^2 .$$

The source term is the dissipation: the rate at which the shear in the layer is converting mechanical energy into heat. It is proportional to the square of the velocity gradient, so it is largest where the shear is largest, which is at the wall — and its size relative to the temperatures already present is set by $U^2/c_p$, which is the stagnation temperature rise.

At low speed that term is nothing. At Mach 0.85 it is five kelvin of difference between two ways of computing a skin temperature, which nobody would notice. At Mach 3 it is sixty, and at Mach 5 it is a hundred and seventy, and by then it is the design.

The wall condition is $T'(0) = 0$: no heat crosses. The equation is still linear in $T$ once $f$ is known, so there is still no shooting and still exactly one answer.

Exactly one, at exactly one Prandtl number #

At $Pr = 1$ the recovery factor is 1.0000000 — the solve returns 0.99999984 — and the reason is worth more than the number.

When $Pr = 1$ the total enthalpy $h + \tfrac{1}{2}u^2$ satisfies the same equation as the velocity, with the same boundary conditions, so it is uniform across the whole layer. Every parcel carries the same total enthalpy it had in the free stream, whatever it has done with the split between temperature and speed. A parcel at rest at the wall has all of it as temperature; therefore the wall is at the stagnation temperature.

That is an exact statement about a conserved quantity, and this site’s solve — which was told nothing about enthalpy, and integrates a temperature equation with a dissipation source — reaches it to seven figures.

√Pr is a fit. It is quoted in every handbook as though it were a result, and it is a curve that happens to pass through the one point where an exact argument exists and to stay close over the range where gases live. This site’s habit is to compute the thing and draw the fit beside it, and the gap between them is the whole of the value in doing so.

The temperature a wall settles at #

$$T_{aw} = T_\infty\left(1 + r\,\frac{\gamma-1}{2}M^2\right),$$

with $r = 0.8417$ for air in a laminar layer. At the tropopause, where $T_\infty = 216.65$ K:

Two aircraft make the numbers concrete. Concorde cruised at Mach 2 and its skin ran at about 90 °C at the nose, which is 363 K — the value in the table. The airframe grew about 25 centimetres in length in cruise and shrank again on descent, and the fuel was pumped fore and aft partly to manage the resulting trim change. The SR-71 cruised above Mach 3, was built of titanium for the reason in the fourth row, and leaked fuel on the ground because its tank seals were designed to close up when the structure expanded.

The sign of the heat flux is not what it looks like #

Here is the part that reads as a paradox and is arithmetic.

The heat flux into a wall is proportional to $(T_{aw} - T_w)$ — not to $(T_\infty - T_w)$. So the flux changes sign at the temperature the flow invents, rather than at the free stream’s.

A skin at 400 K in air at 217 K sounds as though it must be losing heat to the air. It is not: it is 145 kelvin below the temperature the flow wants it at, and heat is flowing into it. Every cooling system on a high-speed aircraft is fighting the recovery temperature, not the ambient one, and the difference between those two numbers at Mach 3 is larger than the whole ambient temperature.

This is also why thermometers on aircraft do not read the air temperature. A total-air-temperature probe is a small body in the flow, and an instrument reads what the flow does to it with a recovery factor of its own, and what it measures is $T_\infty(1 + r(\gamma-1)M^2/2)$ with its $r$ rather than the layer’s. Probes are calibrated to get $r$ close to one on purpose, so that the reading is nearly the stagnation temperature and the static temperature can be recovered from the Mach number. A probe with an unknown recovery factor is an unknown thermometer.

There is a symmetry in that band worth noticing. Its width is $T_{aw} - T_\infty = r(\gamma-1)M^2/2$ times the static temperature, so it grows as the square of the Mach number while the ambient temperature stays where it is. At Mach 1 the band is 36 kelvin; at Mach 3 it is 328; at Mach 5 it is 912. The range of wall temperatures over which intuition about hot and cold is exactly backwards grows as the square of the speed, which is a fair summary of why aerodynamic heating surprises people who are used to subsonic aircraft.

Where the heat goes, and the Reynolds analogy again #

The same solve gives the heat transfer to a wall held at some other temperature, and the result is the analogy of the essay before this one with one change: the driving temperature difference is measured from $T_{aw}$.

$$q_w = h\,(T_{aw} - T_w), \qquad St\,Pr^{2/3} \approx \frac{c_f}{2}.$$

So the whole machinery of skin-friction estimation carries across to heat transfer at high speed, with the recovery temperature doing the work the free-stream temperature does at low speed. Aerodynamic heating is a friction problem in disguise, and a surface that is aerodynamically clean is thermally clean for the same reason.

That connection has a blunt design consequence. Tripping a boundary layer turbulent triples the skin friction, and it triples the heating with it; the turbulent recovery factor is also higher, near $Pr^{1/3} = 0.89$ rather than $\sqrt{Pr} = 0.84$. A re-entry vehicle’s transition location is therefore a structural question rather than a drag question, and predicting it is the hardest unsolved problem in the subject.

One more number is worth putting in the same units as everything else on this site. At Mach 3 and laminar conditions, the heat flux into a wall held at 300 K is proportional to $(545 - 300) = 245$ kelvin; into a wall held at 500 K it is proportional to 45. Letting the skin get hot is by far the cheapest cooling strategy available, because the driving difference falls as the wall approaches the recovery temperature and the flux with it. Every high-speed aircraft is designed around that: the structure is allowed to run hot, and the effort goes into materials rather than into refrigeration.

Where the wall beats the stagnation temperature #

The recovery factor is introduced as a number a little less than one, and for gases it always is. It is worth asking what decides which side of one it falls on, because the answer is not “some of the energy is lost” — an adiabatic wall loses nothing, and the layer over it is in a steady balance where every joule the shear dissipates is eventually conducted away outward.

The balance is between two transports and the Prandtl number is their ratio. Dissipation deposits heat where the shear is, which is at the wall. Conduction carries it out, at a rate set by the thermal diffusivity $\alpha$, while the shear that produced it is arranged by the momentum diffusivity $\nu$. If heat diffuses more readily than momentum — $Pr < 1$ — some of what is dissipated near the wall escapes outward before it can accumulate, and the wall settles short of the stagnation temperature. The shortfall is the whole of $1 - r$. At $Pr = 1$ the two transports are identical, nothing preferentially escapes, and the wall gets everything, which is the enthalpy argument above seen from the other side.

Run the reasoning past one and it does not stop. At $Pr > 1$ heat leaves more reluctantly than the shear arranges it, the near-wall fluid keeps more than its own share, and the recovery factor exceeds one — the wall becomes hotter than any parcel of that fluid could be made by bringing it to rest. The solved curve in the figure above crosses one at exactly $Pr = 1$ and keeps climbing; at $Pr = 10$ it runs 6.3 per cent under the $\sqrt{Pr}$ fit, which puts it near three.

That is not a curiosity of the equation. It is the reason a lubricating film gets hot.

A journal bearing runs an oil with a Prandtl number in the hundreds through a gap a few tens of microns wide, sheared at a speed of metres a second. The film’s own recovery factor is enormous, the dissipation has almost nowhere to conduct to except into the metal on either side, and the temperature rise across a film is far larger than any argument from $U^2/2c_p$ would suggest — which is why bearing design is a thermal problem rather than a hydrodynamic one, and why the viscosity in a bearing calculation has to be solved for rather than looked up. The same mechanism sets the melt temperature in a polymer extruder, where the material’s Prandtl number is larger still and the whole process is dissipation heating deliberately arranged.

The group that governs it in general is not $Pr$ but the Brinkman number $Br = \mu U^2/(k\,\Delta T)$ — the dissipated power measured against the conducted power, for whatever temperature difference the problem happens to impose. A boundary layer over an adiabatic wall is the special case in which there is no imposed temperature difference at all, so the flow has to supply its own: the only scale available is $U^2/c_p$, the Brinkman number becomes $Pr$ exactly, and the recovery factor is what is left. The reason $Pr$ alone decides the answer here is that the wall has been told nothing, so the problem has one temperature scale rather than two.

Which puts the aerodynamic case in its proper place. Air’s Prandtl number is 0.71 and has been since the subject began, so the recovery factor is 0.84 and every high-speed vehicle is designed against a temperature slightly under the stagnation value. That closeness is an accident of what air is made of, not a general property of insulated surfaces, and reading it as one is how the sign error in the previous section gets made a second time: the flow does not merely invent a temperature for the wall, it invents one that is not bounded by anything the free stream contains.

What the picture cannot show #

It is laminar and the properties are constant. A real high-speed layer has a temperature ratio of two or three across it, so the viscosity and conductivity are not the same at the wall as at the edge, and the profile is distorted. The standard repair is a reference-temperature method — evaluate the properties at a weighted mean of wall and edge conditions — which is a correlation, is quoted here as one, and is not computed.

It is a flat plate with no pressure gradient. A leading edge has an enormous one, and the stagnation region’s heating is a different calculation entirely — Fay and Riddell’s, which this collection names and does not derive.

There is no radiation. Above about 1500 K a surface loses heat to space fast enough to matter, and the equilibrium skin temperature of a hypersonic vehicle is set by a balance between convective input and radiative output rather than by the recovery temperature alone.

And there is no chemistry. Past Mach 6 or so the air behind a shock dissociates, $\gamma$ stops being 1.4, and the stagnation temperature computed from a perfect gas is a substantial over-estimate — which is a mercy, since the perfect-gas value at Mach 20 is above ten thousand kelvin and the real one is not.

The turbulent case, which is the one that flies #

Everything above is laminar, and a real high-speed vehicle is not, so the two changes that matters are worth naming even though this collection does not compute them.

The recovery factor rises, from $\sqrt{Pr} = 0.84$ to about $Pr^{1/3} = 0.89$ for air. That is a small change with a visible consequence: the adiabatic wall temperature at Mach 3 rises from 545 K to about 562 K, and at Mach 5 from 1,129 K to about 1,170 K. The reason for the different exponent is the same reason the heat-transfer exponent changes at transition — turbulent eddies carry momentum and energy in the same parcels, so the two transports become more alike and the recovery gets closer to complete.

The heat flux rises by a factor of three to five, because the friction does. That is the number that decides a structure. A laminar re-entry and a turbulent one are different vehicles, and predicting which one occurs is the least reliable calculation in the whole of high-speed aerodynamics — transition on a real surface depends on roughness, on the wall’s own temperature, on the noise in the free stream, and on shapes that no criterion captures.

So the honest summary of the laminar arithmetic in this essay is that it gives the temperature well and the flux only in the regime it was solved in. The recovery temperature is a property of the flow’s energy balance and moves by three per cent between laminar and turbulent; the heat transfer is a property of the layer’s structure and moves by four hundred.

Who found it, and the connection worth keeping #

The recovery factor was measured before it was computed: Eckert and Weise reported adiabatic wall temperatures in 1940 and found them short of the stagnation value by a consistent fraction. Pohlhausen had already solved the equation that explains it — the same 1921 solution that gives the $Pr^{1/3}$ law at low speed — and the $\sqrt{Pr}$ result comes out of it as an approximation nobody has ever needed to improve for gases.

The connection worth keeping is with the essay that runs the same equation without its source term. Both are the second boundary condition on the second field; the difference is one term on the right-hand side, and that term is negligible below Mach 0.5 and decisive above Mach 2. A term that is dropped for good reasons in one regime is the whole subject in another, which is this collection’s most repeated lesson and is why every figure here states the regime it was drawn at.

Where the ladder goes next #

The rung above asks what the $Pr = 1$ argument becomes for a wall that is not insulated. On a flat plate it becomes the Crocco–Busemann relation — total enthalpy linear in velocity across the layer, for a wall at any temperature — and that is the reason compressible boundary-layer solutions can be written down at all. It is a zero-gradient result rather than a general one: in a pressure gradient the straight line fails everywhere except on an insulated wall, and The gradient the heat never hears measures by how much.

The one beside it is what happens when the wall is neither insulated nor held: an aircraft skin has heat capacity, so its temperature is a transient that lags the flight profile. That is a thermal-structural problem rather than a fluid one — but the boundary condition on the fluid side is the one computed here, and it is what a structural analysis is given.