The gradient the heat never hears

The wall that heats itself rested on one coincidence. At a Prandtl number of one, the equation for a flat-plate boundary layer’s total enthalpy is the equation for its velocity, term for term. An insulated wall therefore recovers the whole stagnation temperature, and a wall held at any other temperature carries a total enthalpy that is a straight-line function of the velocity across the layer, from the wall’s value where the fluid is at rest to the edge’s value where it moves at the edge speed. That straight line is the Crocco–Busemann relation. It is the reason a compressible laminar layer’s temperature can be written down once its velocity is known, and it is where most hand estimates of aerodynamic heating still begin: solve for the velocity, then read the temperature off a line.

The flat plate is the one boundary layer with no pressure gradient, and every surface such an estimate is actually wanted for has one — a nose, a leading edge, a nozzle throat, the back of a wing. This essay puts the same layer into the Falkner–Skan family, where the edge speed grows or falls as a power of the distance along the surface, and asks whether the line survives. It does not. It survives at zero gradient and on an insulated wall and nowhere else, and the reason is a single term that is present in one equation and absent from the other. What follows measures the consequence in the three places an estimate would use it: the profile across the layer, the heat flux at the wall, and the static temperature a hot layer actually reaches.

The enthalpy’s equation has no pressure in it #

Three conventions come before any number. The pressure gradient is the Falkner–Skan parameter β = 2m/(m + 1) for an edge speed growing as $x^m$: β = 0 is the flat plate, β = 1 the plane stagnation point, and a negative β an adverse gradient, with the attached family ending at β = −0.1988, the steepness How much uphill a layer can take derives. The similarity variable η is scaled so that the momentum equation reads $f''' + f f'' + \beta(1 - f'^2) = 0$, which puts the flat plate’s wall shear at $f''(0) = 0.4696$ rather than the 0.332 of Blasius’s own scaling; it is the same layer stretched by $\sqrt{2}$, and every wall slope below is in this scaling. The wall’s state is $g_w$, its total enthalpy as a fraction of the edge’s, and at a Prandtl number of one $g_w = 1$ is an insulated wall.

In that variable, with constant density and viscosity and a Prandtl number of one, the two equations are

$$u'' + f\,u' = -\beta\,(1 - u^2), \qquad G'' + f\,G' = 0,$$

where $u = f'$ is the velocity as a fraction of the edge’s and $G$ is how far the total enthalpy has come from the wall’s value towards the edge’s. Both are zero on the wall and one outside the layer. Both are carried outward by the same convection $f$ and spread by the same diffusion, because a Prandtl number of one makes momentum and heat diffuse at the same rate — the condition under which The other layer found the velocity and thermal profiles on a flat plate to be identical rather than merely similar. The difference is on the right-hand side. At β = 0 the two equations are one equation with one set of boundary values, so $G = u$ exactly and the total enthalpy is linear in velocity for any wall. At β ≠ 0 the velocity is pushed by the pressure and the enthalpy is not.

The absence has a plain physical reason. In steady flow a pressure gradient does no net work on a particle’s total enthalpy: pushing a particle along a falling pressure turns its static enthalpy into kinetic energy, and the total is the sum of the two. Only conduction and viscous work move the total, and at a Prandtl number of one those two combine into the same diffusion operator the velocity carries. So the pressure gradient reaches the enthalpy only second-hand, through the stream function $f$ that convects it.

The enthalpy equation also has a closed solution once $f$ is known: $G'$ is proportional to $\exp(-\int_0^\eta f)$, so $G$ is a quadrature of the velocity field’s own stream function and needs no iteration at all. The velocity comes from a shooting solve of the momentum equation. The two are found by different methods from different equations, which is what makes their agreement at zero gradient a test rather than a tautology.

The first figure carries the argument before any departure is computed. The coloured velocity profiles spread across the plot; the faint enthalpy profiles barely move. Their wall slopes run from 0.409 at β = −0.15 to 0.570 at β = 1, a factor of 1.39, while the velocity’s wall slopes over the same four gradients run from 0.216 to 1.233, a factor of 5.7. The enthalpy does feel the gradient, through the thicker or thinner stream function that carries it, but the velocity feels it directly, and a straight-line relation between two profiles that respond to the same cause at such different strengths cannot hold.

Uphill the enthalpy runs ahead of the velocity, downhill it lags #

The difference $G - u$ is the whole of the departure. Multiplied by the difference between the edge’s total enthalpy and the wall’s, it is exactly the error the linear relation makes in total enthalpy at each height, so it is the quantity to draw.

In the adverse gradient $G - u$ is positive, peaking at +0.125 at η = 1.32 for β = −0.15; at the gentler β = −0.1 it peaks at +0.070 and at β = −0.05 at +0.031. In a favourable gradient it is negative, −0.114 at β = 0.3 and −0.240 at β = 1, and the peak moves in towards the wall as the layer thins, to η = 0.96 and 0.81. Past the stagnation point it keeps growing: at β = 2 the peak is −0.330 at η = 0.70. None of these is a small correction. A departure of 0.24 in a quantity that runs from zero to one is a quarter of the whole enthalpy difference misplaced.

The wall is where the mechanism is plainest, and it follows in one line from the two equations. At the wall the stream function is zero, so the momentum equation says $u''(0) = -\beta$: the pressure gradient sets the curvature of the velocity profile at the wall, which is the observation The gradient that does both builds both separation and instability out of. The enthalpy equation at the same place says $G''(0) = 0$ at every β. A favourable gradient bends the velocity profile towards the edge value from the first step off the wall while the enthalpy profile leaves the wall straight, so the velocity gets ahead; an adverse gradient bends the velocity back, and the enthalpy, still straight, gets ahead of it. Everything further out follows from that start, and both profiles are pulled back together at the edge, which is why every departure curve has returned to zero by η ≈ 4.

The direction also has a reading in terms of what a particle carries. Near the wall in a favourable gradient the slow fluid is being accelerated by the pressure, so its momentum is supplied locally as well as diffused in from above. Its total enthalpy has no local supply and can only be conducted and convected in from the hotter fluid farther out, so it lags. In an adverse gradient the pressure is taking momentum from the slow fluid, and the enthalpy, which nothing is taking, is left ahead. Four profiles, one drag found two guessed velocity profiles that satisfy the right conditions at the wall giving drags within three per cent of each other. The enthalpy and velocity profiles here share their end conditions exactly and still differ by a quarter, because the condition that separates them is the curvature at the wall, and only one of them is told it.

Friction moves fourteen-fold while heat transfer moves by half #

The wall is where the numbers engineers use are read. The wall shear is proportional to $f''(0)$ and the heat flux to $G'(0)$ times the enthalpy difference across the layer, in the same similarity units, so the ratio $G'(0)/f''(0)$ is exactly the Reynolds-analogy factor 2St/Cf — the heat transfer measured against what the skin friction predicts. On the flat plate it is one. That is the statement The number that is an answer makes about St = Cf/2 at a Prandtl number of one, and it is why a measured friction has been used to estimate heating for as long as either has been measured.

Across the attached family, from β = −0.19 to β = 1, the wall shear rises from 0.0857 to 1.233, a factor of 14.4, and the heat transfer rises from 0.365 to 0.570, a factor of 1.56. The analogy factor falls from 4.26 at β = −0.19 through 2.96 at −0.18, 1.37 at −0.1 and 1.14 at −0.05 to one at the flat plate, then on through 0.836 at 0.1, 0.671 at 0.3 and 0.581 at 0.5 to 0.463 at the stagnation point and 0.359 at β = 2. It falls monotonically, and it passes through one only at zero.

Both ends carry a practical warning. At a stagnation point the heat transfer is 0.463 of what the friction implies, so a heating estimate taken from a skin friction through the flat-plate analogy is 2.16 times too high — the error is on the safe side, which is part of why it survives in practice. Towards separation the error reverses and has no bound. At β = −0.19, just short of the family’s end, the friction has fallen to 18 per cent of its flat-plate value while the heat transfer is still 78 per cent of its own. A wall about to lose all its shear keeps most of its heating, and an analogy-based estimate would report it nearly cold. At separation itself the friction is zero and the heat flux is not, which is the limit in which the analogy factor is infinite.

This part of the result needs no Mach number. It holds at any speed at which the Prandtl number is one and the properties are constant, and in a low-speed layer it is the whole of the story: a heated plate in an accelerating stream loses less heat than its friction says and a heated plate in a decelerating one loses more, in the ratios the third figure draws.

A layer colder than the stream above a wall three times as hot #

Static temperature is where the relation was meant to be used, and turning enthalpy into temperature needs a speed. With $k = (\gamma - 1)M^2/2$ at the edge Mach number, the static temperature as a fraction of the edge’s is the total-enthalpy ratio times (1 + k), less k times the square of the velocity fraction. Take a Mach 5 layer with its edge at 220 K, so a total temperature of 1,320 K, over a wall held at half of that — 660 K, three times the free stream’s static temperature — in a stagnation-point gradient.

The linear relation puts the temperature peak at 759 K, a hundred kelvin above the wall, at η = 0.27. The solved profile peaks at 684 K, only 24 K above the wall, and at η = 0.13. The largest disagreement is 158 K at η = 0.81, where the linear relation reads the gas as hotter than it is by nearly a quarter of the wall’s own temperature. In a gentler favourable gradient the same wall and stream give a solved peak of 710 K against the same 759 K at β = 0.3, and in an adverse gradient the error changes sign: at β = −0.1 the solved peak is 802 K, 43 K above the line.

What the linear relation cannot produce at all lies farther out. At η = 2.11 the solved temperature is 205.9 K — 14 K colder than the free stream, in a layer whose wall is 440 K hotter than the free stream. The linear relation can dip below the edge temperature only when the wall is colder than the edge’s static temperature. Its excess over the edge temperature factors as $(1 - u)$ times an expression linear in $u$, and that expression is positive at both ends whenever the wall’s total enthalpy exceeds the edge’s static enthalpy, which a 660 K wall under a 220 K stream does threefold.

The undershoot is the lag of the second section made visible. At η = 2 the velocity is within a few per cent of the edge speed, so almost all of the edge’s kinetic energy is present; the total enthalpy has not caught up, and the static enthalpy is total minus kinetic. The favourable gradient has accelerated the gas faster than conduction from the hot wall and the hot core of the layer can refill its energy, and the shortfall appears as cold. Cooling the wall harder deepens it. At $g_w = 0.2$, a 264 K wall in the same stream, the solved profile peaks at 330 K against the linear relation’s 517 K, the largest error is 253 K, and the cold band reaches 139 K at η = 1.46, 81 K below the free stream. That is the surprising reading of the calculation: the coldest gas over a cooled stagnation point is not at the wall and not in the stream, but in a band between them that a straight line through the two end temperatures never contains.

Only an insulated wall escapes, and a heated wall is wrong the other way #

The departure in total enthalpy is the edge-to-wall enthalpy difference times $G - u$, and $G - u$ depends on the gradient alone. So at a given gradient the temperature error has the same shape at every wall temperature, scaled by $1 - g_w$, and it vanishes only when the wall’s total enthalpy equals the edge’s.

At Mach 5 a wall at half the edge total enthalpy gives largest errors of −75.5 K at β = 0.3, −158.3 K at β = 1 and +46.1 K at β = −0.1. At $g_w = 0.2$ the stagnation-point error is −253.2 K, exactly 1.6 times the half-enthalpy value, as the ratio (1 − 0.2)/(1 − 0.5) requires. A wall hotter than the recovery temperature — $g_w = 1.5$, a surface heated from inside or one lying downstream of something hotter — reverses every sign, to +158.3 K at the stagnation point. The wall’s temperature sets the size of the error; the gradient sets its shape and its sign.

This is also why the relation’s failure is so easy to miss. The insulated wall, the case The wall that heats itself and The thermometer that heats itself are built on, is the one case in which it cannot fail. At a Prandtl number of one an insulated wall carries a uniform total enthalpy across the layer at every gradient, and a straight line between two equal end values is a constant that any profile satisfies. So the recovery temperature a stagnation point reaches is right whatever the gradient. The heat flux into a cooled nose is not, and neither is the temperature above it — and a cooled wall is what The skin that lags the flight showed a real skin to be for most of any short exposure, since its temperature spends the exposure climbing towards the recovery value from below.

The linearity in the wall’s enthalpy has a use of its own. Because the error is exactly proportional to $1 - g_w$, a single calculation at one wall temperature fixes it at all of them, and a design sweep over wall temperatures needs the departure profile once per gradient rather than once per case. The constant-property layer earns that proportionality honestly; the next-to-last section says why a real layer does not.

Speed multiplies the error without changing its shape #

The error is an error in enthalpy, and a fixed fraction of the edge’s total enthalpy is a fixed fraction of (1 + 0.2M²) times the edge’s static temperature. At a fixed wall ratio the temperature error therefore grows exactly with that factor, and the Mach number decides how many kelvin the departure is worth without touching where it sits.

At the stagnation point over a wall at half the edge total enthalpy the largest error is −26.4 K at Mach 0, −47.5 K at Mach 2, −158.3 K at Mach 5 and −364.0 K at Mach 8. The factor is 1.8 at Mach 2, 5.05 at Mach 4.5 and 13.8 at Mach 8, and the ratios are exactly those. The favourable β = 0.3 curve reaches −173.8 K at Mach 8 and the adverse β = −0.1 curve +106.0 K.

Two readings keep this honest. At low speed the relation is simply the statement that temperature is linear in velocity across a heated or cooled layer, and at a stagnation point that statement misplaces 24 per cent of the wall-to-stream difference — the 26.4 K out of 110 K at Mach 0. That error is a heat-transfer error in a flow with no compressibility in it at all, which is where the analogy factor of the third section lives. At the other end, the edge total temperature at Mach 8 is 3,036 K, where the ratio of specific heats is no longer 1.4, as When gamma stops being a number showed behind a strong shock, and where a layer with constant density and viscosity is not a model of anything. The right-hand end of the figure is the scaling carried past the gas that obeys it, drawn to show the shape of the growth rather than to quote its value.

The identity checked by two methods that share nothing #

Every departure above is measured against a zero. The zero is only as good as the agreement at β = 0 between two calculations that have nothing in common but the stream function: a shooting solve of a third-order nonlinear equation for the velocity, and a double quadrature of an exponential for the enthalpy.

At a step of 0.008 the two profiles differ by at most 1.8 × 10⁻⁶; at 0.004 by 1.1 × 10⁻⁶; at 0.002 and 0.001 by 1.0 × 10⁻⁶, and the analogy factor at 0.002 is 0.9999992. The floor does not move with the step because it is not the quadrature’s. The velocity profile the shooting solve returns ends 1.0 × 10⁻⁶ above the edge value at every step tried, while the enthalpy fraction reaches exactly one there by construction. The wall shear it returns is 0.4696007 against 0.4695999 for Blasius’s 0.332057 stretched by $\sqrt{2}$, a difference of the same size. A departure of 10⁻⁶ at zero gradient against departures of 0.03 to 0.33 in a gradient is a margin of four to five orders of magnitude.

The rest of the calculation is checked the same way, each claim against a route that does not share its algebra. The enthalpy profile satisfies its own equation, by finite differences on the grid, to better than 10⁻⁴ at β = 0.5 and β = −0.1. The analogy factor has converged: halving the step moves it from 2.9627445 to 2.9627442 at β = −0.18 and from 0.46281931 to 0.46281923 at the stagnation point. It falls monotonically across eight gradients from −0.18 to 1, above one for every adverse gradient and below for every favourable one. An insulated wall carries a temperature error below 10⁻⁹ K at β = −0.1 and β = 1, and the error at two wall enthalpies scales as $1 - g_w$ to one part in a billion. And the calculation refuses what it has no answer for: a gradient past separation, a wall with no enthalpy, a negative Mach number and a step of zero.

What the constant-property layer cannot show #

The largest omission is the one that couples the two equations. A real compressible layer at Mach 5 spans more than a factor of three in temperature, so its density and viscosity change across it by the same order. With viscosity proportional to temperature — a Chapman–Rubesin parameter of one — the Illingworth–Stewartson transformation restores the incompressible form with one change, and the change is in the momentum equation: the pressure-gradient term becomes β times the total-enthalpy ratio less the square of the velocity, where the constant-property layer has β times one less the square. The enthalpy equation is unchanged at a Prandtl number of one. So at β = 0 nothing moves, and the Crocco–Busemann relation is exact for a real compressible laminar flat-plate layer under those two conditions; that classical result stands. In a gradient the velocity now depends on the enthalpy. Near a cooled wall the enthalpy ratio is below one, so the pressure gradient pushes the slow gas less hard; near a heated wall it pushes harder. Both the friction and the departure drawn here move with the wall’s temperature, and the proportionality in $1 - g_w$ is lost.

The second is the Prandtl number itself. Air’s is about 0.71, and then the enthalpy equation carries a dissipation term the velocity’s does not, the insulated wall no longer holds a uniform total enthalpy, and the recovery factor falls to roughly $\sqrt{\mathrm{Pr}}$ as The wall that heats itself computed. The straight line is then approximate even at zero gradient, and the form used in practice replaces it with a quadratic built on the recovery temperature. Every number here is for a Prandtl number of one, which is the condition that makes the separation of the gradient’s effect clean; for air the gradient’s departure adds to a Prandtl-number departure that this calculation does not include.

The third is self-similarity. The Falkner–Skan family holds only where the edge speed is a power of distance, which near a real body is the stagnation region and not much else. A layer whose gradient changes along the surface carries the history of the gradients upstream of it, so the departure at a station depends on more than the local β. The figures also cannot show a turbulent layer, where the analogy is a correlation rather than an identity, nor a dimensional heat flux: every wall slope here is in similarity units, and the flux in watts depends on the distance from the stagnation point and the Reynolds number built on it.

Still open: the layer whose wall temperature moves its velocity #

The next calculation couples the two equations: the family of similar compressible solutions in which the momentum equation carries β times the enthalpy ratio and is solved simultaneously with the enthalpy equation for a wall held at a given $g_w$. It answers three questions the uncoupled layer leaves open, each with a figure it could fail. Whether the cold band above a cooled stagnation point survives when the cold gas is also dense and slow. How far a cooled wall moves the gradient at which the layer separates, which the uncoupled layer pins at −0.1988 for every wall, and whether a heated wall separates earlier by the same amount a cooled one delays it. And whether the analogy factor at a stagnation point stays near 0.46 when the wall is cold, or is itself a function of the wall’s temperature.

Beside it is the zero-gradient layer at a Prandtl number of 0.71, where the straight line becomes the recovery-temperature quadratic and the question is how large that departure is against the gradient’s. The air-data reduction in Three readings, and the one each answer leans on leaned on a recovery factor near 0.98 known to half a per cent; the corresponding question for a heated or cooled wall is how well the temperature profile above it is known, and this essay has shown that the answer depends first on the pressure gradient and only second on the gas.