A cold wall holds its layer on longer

The gradient the heat never hears put a hot boundary layer into a pressure gradient and found that the total enthalpy stops following the velocity. The pressure gradient appears in the momentum equation and not in the energy equation, so it bends the velocity profile and leaves the enthalpy’s alone, and the straight-line Crocco–Busemann relation between the two survives only at zero gradient or over an insulated wall. That calculation held the density constant. Its velocity layer was the textbook Falkner–Skan layer, the same over a hot wall as over a cold one, and so it pinned the separating gradient at −0.1988 for every wall.

It closed by naming the coupling it had left out. In a real gas the pressure gradient acts on the local density, and the local density is set by the temperature: near a cold wall the gas is dense and slow to accelerate, near a hot one it is light and quick. It asked three things, each with a figure that could fail it — whether the cold band it had found above a cooled stagnation point survives when the cold gas is also dense, how far a cooled wall moves the separating gradient and whether a heated wall moves it the other way by as much, and whether the analogy between heat transfer and friction at a stagnation point stays near its uncoupled 0.46. All three answers turn out to be large.

The two equations, coupled #

The standard route to a compressible laminar layer goes through two simplifications. The Prandtl number is taken as one, which makes the energy equation one for the total enthalpy alone, and the product of density and viscosity is taken as constant across the layer — the Chapman–Rubesin constant set to one — which with the Stewartson–Illingworth transformation removes the variable properties from the momentum equation’s viscous term. What survives is the pair

$$f''' + ff'' + \beta\,(g - f'^2) = 0, \qquad g'' + fg' = 0,$$

where $f'$ is the velocity over the edge’s, $g$ is the total enthalpy over the edge’s, and $\beta$ is the Falkner–Skan pressure-gradient parameter. The wall holds $g$ at $g_w$; the edge holds both at one.

The coupling is the single $g$ in the first equation. In the incompressible layer it is a one: the pressure gradient’s push, $\beta$, is balanced against the fluid’s inertia, $\beta f'^2$, with the same density everywhere. Here the push acts per unit of the local density, which is lower where the gas is hot, so it is weighted by the local enthalpy. Where $g < 1$, near a cooled wall, the gradient pushes the dense gas less, accelerating it less in a favourable gradient and decelerating it less in an adverse one. At $g_w = 1$ — a wall at the edge’s total enthalpy, very nearly an insulated one — $g$ is one everywhere and the layer is Falkner–Skan’s. At zero gradient the $\beta g$ term vanishes and the layer is Blasius’s for any wall, which is why the wall that heats itself could do without the coupling on a flat plate.

The pair is integrated outwards from the wall by fourth-order Runge–Kutta and shot on the two unknown wall values, the shear $f''(0)$ and the enthalpy gradient $g'(0)$, by Newton’s method. Separation is found directly, by fixing the wall shear at zero and shooting on $\beta$ and $g'(0)$ instead.

Where the layer lets go #

How much uphill a layer can take found the incompressible answer: −0.1988, the most adverse self-similar gradient a laminar layer survives. Coupled, that number belongs to one wall only. A wall at absolute zero holds its layer on to −0.3264, two-thirds more adverse. A wall at half the edge’s total enthalpy holds it to −0.2623. A wall at one and a half times the edge’s lets go at −0.1573, and one at twice, at −0.1295.

The mechanism is the coupling term read in an adverse gradient. A rising pressure decelerates the slow gas next to the wall until it stops; that is separation. Cold gas is dense, and the same pressure rise decelerates it less, so a cooled layer can climb further before its wall shear reaches zero. Hot gas is light and gives way sooner.

The two directions are not symmetric. Cooling from one to a half moves the separating gradient by 0.064; heating from one to one and a half moves it by 0.041. Cooling buys more than heating costs, which is the direction hypersonic practice has long found — wall cooling shrinks a laminar separation — with the asymmetry that the coupling explains: a cooled model gains more than a heated one loses.

The push on light and heavy gas #

The whole family shows the same thing across every gradient. All three walls share the Blasius shear, 0.4696, at zero gradient, where the momentum equation cannot see them. In a favourable gradient the hot wall’s layer is pushed harder and its shear climbs faster: at a stagnation point, 1.74 for the hot wall against Hiemenz’s 1.233 for the adiabatic-like wall and 0.806 for the cold one. In an adverse gradient the order reverses and the hot wall’s shear collapses first.

The stagnation-point numbers matter for anything blunt in a hot stream. A cooled nose — every re-entry body, every hypersonic leading edge whose structure is kept below the recovery temperature — has a skin friction at its stagnation point barely two-thirds of the textbook Hiemenz value in these transformed units, and a hot one has half as much again.

Friction moves, heat transfer hardly does #

The previous essay’s analogy factor, $2St/C_f$, was 0.463 at a stagnation point for every wall, because the velocity it was computed on did not depend on the wall. Coupled, it runs from 0.781 over a wall at absolute zero through 0.568 at half the edge’s enthalpy to 0.354 at twice it — more than a factor of two. At β = 0.5 the spread is smaller and the same way round.

Almost all of that is the friction. The enthalpy equation is the same equation in both calculations; what changes is the velocity profile it is solved on, through $f$, and that changes the heat transfer modestly. The friction changes a great deal, as the previous figure showed. So a heat-transfer estimate made from a measured or computed friction through a constant analogy factor — a common way of getting heating rates from a skin-friction code — is wrong at a stagnation point by as much as the wall’s temperature is far from the edge’s, and cold walls, the ones heating estimates are usually wanted for, are where the factor is largest.

The cold band was the uncoupling’s #

The previous essay found something surprising above a cooled stagnation point at Mach 5: a band in the layer colder than the free stream, over a wall three times hotter than the stream’s static temperature. It arose because in a favourable gradient the velocity outruns the enthalpy, and the static temperature, the total enthalpy minus the kinetic energy, dips where the gas has already gained its speed and not yet its enthalpy. Over a wall at a quarter of the edge’s total enthalpy that dip reached 155 K under a 220 K stream, 65 degrees colder.

Coupled, it is gone. The cold, dense gas near the wall is accelerated less by the same pressure drop, the velocity no longer runs ahead of the enthalpy, and the dip is 0.024 K. The band was a property of a velocity that could not see the temperature. That answers the first of the previous essay’s questions with a firm no, and it is a warning about the reverse inference too: a measured static temperature that does not dip is not evidence against a constant-property velocity profile, it is evidence for the coupling.

Over a hot wall the layer outruns the stream #

The same coupling read the other way produces something no constant-density layer can. Over a hot wall in a favourable gradient the light gas near the wall is accelerated more than the dense gas above it, and it can go faster than the stream. The velocity profiles show it: the cold layers are thick and slow near the wall, the hot ones thin, and over a wall at two and a quarter times the edge’s enthalpy the velocity peaks 2.3 per cent above the edge speed before falling back to it.

The overshoot is small and grows smoothly. At a stagnation point it passes a hundredth of a per cent near a wall enthalpy ratio of 1.38, reaches 0.11 per cent at 1.5 and 2.3 per cent at 2.25; at β = 0.5 it starts later and reaches 0.13 per cent. Small as it is, it changes what the edge of the layer means. A Pitot traverse through such a layer reads a maximum inside it, and an edge chosen as the point of maximum velocity — a common practical choice — puts the edge in the wrong place and the edge velocity too high.

A cooled nose, worked #

Put the stagnation-point numbers on one body: the nose of a vehicle at Mach 5 in a 220 K stream, its wall held at 330 K, a quarter of the edge’s total enthalpy. The uncoupled layer would give it Hiemenz’s transformed wall shear, 1.233, and a heat-transfer number of 0.571. The coupled layer gives 0.806 and 0.526. The heating is eight per cent lower than the constant-density layer predicts; the friction is thirty-five per cent lower. An engineer who had measured the friction on such a nose and converted it to heating with the textbook stagnation-point factor would under-predict the heating by nearly thirty per cent — 0.46 used where 0.65 holds — and the error would be on the unsafe side.

Heat a wall instead, to twice the edge’s enthalpy, as a nose that has soaked in a long cruise might be relative to a cooler descent, and the signs reverse: the friction rises forty per cent above Hiemenz’s, the heat-transfer number eight per cent, and the analogy factor falls to 0.354. Whichever way the wall departs from the stream’s stagnation temperature, a constant analogy factor fails in proportion. And because the wall’s own temperature changes through a flight — the wall that heats itself is what an insulated one does, the thermometer that heats itself what a probe does — a factor fitted at one point of a trajectory is wrong at the next.

The separation shift has the same practical shape. A flap or a compression corner on a cooled body tolerates a steeper adverse gradient before its laminar layer lets go, so a hot-wall wind-tunnel test of a model that will fly cold over-predicts the separation it will have. Behind a blunt nose the gas the layer swallows has come through the entropy layer the sheath leaves, hot and slow, which is a coupling of exactly this kind applied to the edge rather than the wall.

Finding the right root #

The coupled equations hid a trap that the uncoupled ones did not, and it is worth recording because it would have produced a wrong figure with every residual small. Over a hot wall in a favourable gradient, shooting from an ordinary guess found a second root on the finite domain: a profile with a smaller wall shear, no overshoot, and a stream function that turned negative near the edge, creeping up to the edge velocity only at the very end of the domain. It satisfied both boundary conditions to ten figures. It was not a boundary layer. Lengthening the domain moved it by 0.15 in wall shear; the physical root did not move at all. Every solution here is therefore accepted only if its stream function is positive at the edge, its shear has vanished there, and nowhere does it reverse, and the hot-wall family is reached by continuation in the wall temperature from the decoupled layer.

Five checks on the family #

At $g_w = 1$ the equations decouple and must give Falkner–Skan’s layer: the wall shear is 0.469600 at zero gradient against Blasius’s 0.469600, 1.232588 at a stagnation point against Hiemenz’s 1.232588, and the separating gradient is −0.198838 against Falkner–Skan’s. At zero gradient over a cold wall and a hot one, the computed enthalpy equals Crocco–Busemann’s straight line to $10^{-14}$, as it must when $\beta g$ vanishes. Halving the step and lengthening the domain moves the wall values by $10^{-11}$. A hot-wall layer computed on domains of eight and ten moves by $10^{-13}$, the test that the physical branch was found. The tests also refuse a negative wall enthalpy and a gradient parameter outside the family.

What the two unities hide #

A Prandtl number of one. Air’s is 0.71. At 0.71 the total enthalpy is not conserved across an insulated layer — the recovery factor drops below one — and the enthalpy equation gains a dissipation term. The coupling here is unchanged in form; its numbers move by a few per cent.

A Chapman–Rubesin constant of one. It assumes viscosity proportional to temperature. Sutherland’s law makes $\rho\mu$ vary across a layer with a large temperature ratio, by tens of per cent between a cold wall and a hot edge, and the momentum equation’s viscous term then carries it. The separation and overshoot trends survive; their numbers depend on it.

Similarity. The layers are the self-similar ones, which need the edge speed to vary as a power of distance and the wall to be at a uniform enthalpy ratio. A real nose has neither exactly; the similar solutions are the local estimate a marching calculation starts from.

Laminar flow. All of it. A cold wall also stabilises a laminar layer against the first-mode instability, which is a separate reason cold-wall models stay attached, and destabilises the second mode at hypersonic speeds, which is a reason they sometimes do not stay laminar; where transition happens is a number that is not a number, and a turbulent layer’s response to wall temperature is a different calculation with its own closure.

The convention: enthalpy ratios, and transformed distances #

The wall’s temperature is given as $g_w$, its total enthalpy over the edge’s, so $g_w = 0$ is absolute zero, $g_w = 1$ is a wall at the edge’s stagnation temperature, and $g_w = 0.25$ at Mach 5 is a 330 K wall under a 220 K stream. Distances across the layer are in the transformed coordinate, in which the heights of a cold and a hot layer are not comparable directly; the physical height is $\int (T/T_e)\,d\eta$. The analogy factor is the transformed heat flux over the transformed shear, which is the physical ratio for a Chapman–Rubesin constant of one.

Cohen and Reshotko, 1956 #

The coupled similar layers were tabulated by Cohen and Reshotko for NACA in 1956, for wall enthalpy ratios from zero to two and the whole Falkner–Skan range, and the separation shift with wall temperature, the analogy factor’s dependence on it and the hot-wall overshoot are all in their tables; their −0.326 for the coldest wall is the −0.3264 here. Stewartson and Illingworth had given the transformation in 1949. What this essay adds is the reading against the uncoupled layer of the essay before it — that its cold band, its single separation gradient and its constant analogy factor were each an artefact of one missing $g$ — and the spurious root a shooting code finds over a hot wall.

Still open: the wall that is cooled only near the nose #

Every layer here sits over a wall at one temperature. A real cooled nose is cooled where the heating is worst and left to run hot downstream, and the layer then carries the cold gas of the nose into a region with a hot wall and an adverse gradient. The next calculation marches the coupled equations along a body from a cooled stagnation point, with the wall’s enthalpy ratio rising downstream, and asks whether the cold gas the layer brings with it delays separation on the hot afterbody — whether a wall’s cooling is remembered downstream over something like the entry length that how far before the heat arrives computed for a duct, or forgotten within a few layer thicknesses.