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

Blast wave — the series

7 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The shock radius against time, from an equation that was given no exponent. The thin-shell energy balance integrated forward from a small initial radius, on logarithmic axes. The two-fifths power is not put in: the ordinary differential equation is Ṙ = √(E/AρR³), and the straight line is what it does. The fitted slope is 0.39983 and the fitted prefactor is 0.90721 against a closed form of 0.90702 — the small residuals being the integration's memory of where it started, which the similarity solution has no equivalent of.

    A radius that gives the energy away

    Four quantities, three dimensions, one group. The radius of a strong blast must be a constant times (Et²/ρ)¹⁄⁵, and nothing about the device, the chemistry or the initial size can appear. The exponent is free and the constant is not.

    part 1 · compressible
  2. Inside a strong blast: a shell of gas around a hot, nearly empty core. Left, the density, velocity and pressure inside a spherical strong blast wave at γ = 1.4, each as a fraction of its value just behind the shock, against the distance from the centre as a fraction of the shock radius. The density falls to nothing well inside the shock while the pressure levels off at 0.37 of its post-shock value. Right, the temperature, which is the pressure over the density and so rises without limit towards the centre.

    The fireball is hollow

    Inside a strong blast wave there is almost nothing. Half the air the shock has swept up lies in the outer four per cent of its radius; at half the radius the density is a hundredth of the air's; the centre is empty and, formally, infinitely hot. Integrating Sedov's equations to see this also shows that a published table of his constants was wrong at three of its four entries.

    part 2 · compressible
  3. A bow shock that grows as a line explosion does. The radius of the bow shock around a hemisphere-nosed cylinder, in body diameters, against distance behind the nose: the blast-wave analogy, R/d = 0.795 C^¼ (x/d)^½ with C, the drag coefficient, 0.919, which contains no Mach number, and Billig's correlation of measured bow shocks on spheres at Mach 5, 10 and 20, anchored at the nose and continued as a hyperbola to the Mach cone. Both grow as the square root of the distance; at Mach 20 the analogy's shock is a steady 0.71 of Billig's.

    A hypersonic body leaves a line explosion behind it

    A blunt body at hypersonic speed does work on the air at a rate equal to its drag, and each slice of air it passes through is struck once and left to expand. Seen from the ground, that is a line explosion, and Sedov's cylindrical blast wave gives the bow shock's width and the pressure on the afterbody without a Mach number in either. The analogy's classical constants come straight out of the blast solution. So do its limits: it holds only while its own shock stays strong, over a length that grows as the square of the flight Mach number, and it puts the body inside a core hotter than anything the flow can reach.

    part 3 · compressible
  4. Next to the body the gas remembers the nose. The gas temperature, over the free stream's, against distance from the axis of a hemisphere-nosed cylinder at Mach 15, at 2, 20 and 200 diameters behind the nose; the body's surface is at a half. The gas next to the body crossed the nearly normal part of the bow shock and carries its entropy: 16.7 times the free stream's temperature at 2 diameters, 9.29 at 200, falling outwards to the gas that crossed the weaker, oblique shock. The blast-wave analogy's core at 2 diameters, dashed, runs off the top of the frame on its way to infinity.

    The gas beside a hypersonic body remembers its nose

    The blast-wave analogy gets a blunt body's bow shock and afterbody pressure right and its temperature absurdly wrong: it puts the body in the empty core of an explosion, where the temperature has no bound. The real gas beside the body crossed the nearly normal shock at the nose and carries that crossing's entropy all the way down. Expanded to the afterbody's pressure, it is ten to twenty times the free stream's temperature at Mach 15, it never falls below a floor set by the nose alone, and the sheath that carries it is wider than the body.

    part 4 · compressible
  5. A laminar boundary layer eats through the sheath a Reynolds number of diameters back. How far out the boundary layer has eaten into the entropy layer, as the upstream radius of the streamline at its edge in nose diameters, against distance behind the nose, at Mach 15 for Reynolds numbers on the diameter of 10⁴ to 10⁷, laminar, and 10⁶ turbulent. The sheath's edge is the streamline whose entropy is half the axis's, 1.22 diameters out. The laminar layer reaches it at 3.3·10³ diameters for the smallest Reynolds number and 3.4·10⁶ for the largest; turbulent at 10⁶, at 536.

    The sheath outlasts the body

    A blunt hypersonic body wraps itself in a sheath of hot, thin gas from its nose's shock, and its boundary layer grows by eating that gas from the inside. The usual picture has the boundary layer through the sheath within a few nose diameters. It is not: a laminar boundary layer takes about a third of a Reynolds number of diameters to swallow the sheath at Mach 15, which is thousands to millions of diameters, and a turbulent one hundreds. On any body of ordinary length the boundary layer never sees the cooler gas outside, and it is heated by the sheath's gas all the way down.

    part 5 · compressible
  6. 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.

    part 6 · compressible
  7. Bluntness delays a cone's transition by a factor that peaks, and the correlation's Mach exponent sets how high. The distance at which a blunted 8° cone's layer transitions, over the sharp cone's 0.3 m, against the nose Reynolds number, at Mach 6 and 2.6·10⁷ per metre, for three correlations of the form Reθ/Mₑⁿ = C. Reθ alone: a largest delay of 4.96 at a nose of 43.9 mm, nose Reynolds number 1.14·10⁶; Reθ / √Mₑ: a largest delay of 2.57 at a nose of 20.1 mm, nose Reynolds number 5.23·10⁵; Reθ / Mₑ: a largest delay of 1.41 at a nose of 7.29 mm, nose Reynolds number 1.9·10⁵. Past a nose Reynolds number of 8.81·10⁵ the Mach-ratio criterion transitions within ten nose radii of the tip, where the model's edge is least trustworthy.

    A blunted nose delays its cone's transition only so far, and the correlation decides how far

    A blunted tip leaves a sheath of hot, slow gas along a hypersonic cone, and the boundary layer growing inside it sees a lower Reynolds number and a lower Mach number than a sharp cone's. That is why bluntness is said to push transition back. Read the layer's edge through the sheath and the push has a limit: the delay grows with the nose, peaks, and falls again once the transition point lies inside the sheath. How large the peak is depends on whether the transition correlation divides by the edge Mach number — five times the sharp distance at Mach 6 for the momentum-thickness Reynolds number alone, 1.4 for its ratio to the Mach number — and every peak sits forty to sixty-five nose radii from the tip.

    part 7 · compressible

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