Fluid Flow & Aerodynamics
why things fly, and why the usual explanation is wrong
Concept

Aerodynamic heating — where it appears

The heating of a surface by the fluid's own kinetic energy being dissipated in the layer beside it. It grows as the square of the speed and is what sets the material of a high-speed vehicle, arriving as a flux driven from the recovery temperature rather than from the air's.

Named by 7 essays across 2 fields — each of them below, with the objects they name alongside it.

What the skin settles at, before anything is done to it. The adiabatic wall temperature against Mach number, in air at 216.7 K, with the stagnation temperature above it. The gap between the two is the recovery factor, which is 0.8417 here and stays there at every Mach number — it is a property of the Prandtl number and not of the speed. At Mach 2 the skin sits at 363 K, at Mach 3 at 545 K, and at Mach 5 at 1128 K, which is past what aluminium will do. Nothing has been burnt and nothing has been rubbed: the air was brought to rest, and this is where its kinetic energy went.

The wall that heats itself

A surface told nothing about its temperature does not settle at the air's. It settles most of the way to the stagnation temperature, and the heat flux is driven from that invented temperature rather than from the free stream's — so a wall hotter than the air can be being heated by it.

compressible · Recovery A skin lags the air by a time its own thickness sets. A skin held at one flight condition, relaxing towards the adiabatic wall temperature. The approach is exponential with a time constant ρc τ / h — 8.37 s at 2 mm, 25.1 s at 6 mm, 50.2 s at 12 mm — so the temperature a steady recovery calculation gives is reached after several minutes rather than at once. The fluid supplies one number to this calculation, the adiabatic wall temperature, and the structure supplies everything else.

The skin that lags the flight

A wall can be told its temperature or told nothing, and both are solved problems. A real skin is told neither. It has heat capacity, so its temperature is a transient whose time constant is its own thickness divided by what the layer delivers — and the number the steady calculation returns is an upper bound a short exposure never collects.

compressible · Recovery A cooled wall at β = 1: the linear relation is 158 K out inside the layer. The static temperature across a laminar layer at Mach 5 with an edge temperature of 220 K, over a wall whose total enthalpy is 0.5 of the edge's, at a pressure gradient β = 1: exact (thick) and from the Crocco–Busemann linear relation (thin), at a Prandtl number of one with constant properties. The wall is at 660 K in both. The exact profile peaks at 684 K and the linear one at 759 K; the largest difference, −158.3 K, is at η = 0.81.

The gradient the heat never hears

On a flat plate at a Prandtl number of one, a boundary layer's total enthalpy is a straight-line function of its velocity, whatever the wall's temperature. Put the same layer in a pressure gradient and the straight line fails everywhere except on an insulated wall, because the gradient enters the velocity's equation and not the enthalpy's — and a favourable gradient can leave a band of gas colder than the free stream above a wall three times hotter than it.

compressible · Recovery The worst jet amplifies the stagnation pressure by about the Mach number. The largest amplification of the stagnation pressure, over every incident turn, against the free-stream Mach number, on a logarithmic axis: the type IV jet, the best single turning shock followed by a normal shock, and the lossless ceiling. The jet's peak runs close to the line equal to the Mach number itself, from 3.5 at Mach 4 to 12.4 at Mach 12. The ceiling grows as the Mach number to the power of three and a half and is never approached. The estimate with one turning shock falls further behind the jet as the Mach number rises.

The spot a local theory cannot see

Newtonian theory gives every panel of a hypersonic vehicle a pressure set by its own angle to the stream, and no panel more than the stagnation pressure behind a normal shock. Let a shock from one part cross the bow shock of another and a supersonic jet forms that reaches the surface through weaker shocks. At Mach 8 it stagnates at 8.6 times the ceiling — and the worst amplification at every Mach number is close to the Mach number itself.

misconceptions · Newtonian A cold wall holds its layer on longer; a hot one lets it go sooner. The pressure-gradient parameter β at which the wall shear vanishes, against the wall's total enthalpy over the edge's. With the velocity deaf to the temperature it is −0.1988 for every wall. Coupled, it is -0.3264 for a wall at absolute zero, -0.2623 for a wall at half the edge's total enthalpy, -0.1988 at one, -0.1573 at one and a half and -0.1295 at two. Cooling delays separation by more than the same heating hastens it.

A cold wall holds its layer on longer

A compressible boundary layer's pressure gradient pushes on its density, and its density is its temperature. Solve the velocity and the enthalpy together and the wall's temperature reaches the velocity: a cold wall holds its layer on to an adverse gradient two-thirds steeper, a hot one lets it go a third sooner, the stagnation point's heat-to-friction ratio more than doubles between them, and the cold band an uncoupled layer put above a cooled nose disappears.

compressible · Recovery A cone swallows its sheath as the cube root of the Reynolds number; a cylinder in proportion to it. The swallowing distance in nose radii against the Reynolds number on the nose radius at Mach 15: a 7° sphere-cone, laminar and turbulent, and the sphere-cylinder of the same nose, laminar. At Reₙ = 10⁵ the cone's laminar layer swallows the sheath at 160 nose radii and its turbulent one at 28.1; the cylinder's laminar layer at 1.36·10⁵.

A blunted cone eats its nose's sheath in a few hundred radii

A sphere-cylinder's boundary layer takes a third of a Reynolds number of diameters to get through the hot sheath its nose leaves, which on any real body is never. Put the same nose on a slender cone and the answer collapses: the cone's widening perimeter feeds the boundary layer so fast that it swallows the sheath in tens to hundreds of nose radii, growing only as the cube root of the Reynolds number. On a two-metre cone at Mach 15 that makes a nose of about two centimetres the line between a base still bathed in the nose's gas and one that has forgotten it.

compressible · Blast wave Behind a strong shock the layer is an exact rotational flow, squeezed into a tenth of the shock's radius. Streamlines of the constant-density shock layer between a spherical shock (radius 1) and a sphere at density ratio ε = 0.167, from the solution ψ = sin²χ(ar⁴ + br² + c/r): the body sits at 0.8944 of the shock's radius, a standoff of 0.118 body radii. The gas enters along the shock, turns, and runs round the body in a layer whose vorticity over r sin χ is the same everywhere.

The shock layer is Hill's vortex, and it heats the nose a tenth harder than Newton says

Behind a strong bow shock the gas is squeezed six-fold or more and turned round the nose in a thin layer, and if its density is taken as constant there the layer has an exact solution: the same family of rotational flows as Hill's spherical vortex, fitted to the shock's three conditions. It stands off further than a mass balance says, and its gas accelerates away from the stagnation point a fifth faster than Newton's pressure law implies. Since a nose's heating goes as the square root of that acceleration, the layer heats the nose about ten per cent harder than the estimate every first calculation of re-entry uses.

compressible · Shock layer

Named alongside it

The objects these essays reach for when they reach for this one.

Boundary layerHeat transferMach numberTotal enthalpyAdiabatic wallAdverse pressure gradientAnalogyBoundary conditionEnergy equationFalkner–SkanHypersonicModel limit

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