A shock that lies on the body

Every ratio across a normal shock has a limit as the Mach number rises, and the limits are functions of $\gamma$ alone:

$$\frac{\rho_2}{\rho_1}\to\frac{\gamma+1}{\gamma-1} = 6\ \text{for air},\qquad C_{p_0}\to 1.8394,\qquad \frac{p_2}{p_1}\to\infty.$$

The third of those is why the second is the right quantity to quote. A pressure ratio grows without bound, so it says nothing about whether anything has settled down; a pressure coefficient, which divides by the dynamic pressure, reaches a limit.

By Mach 5 the pressure coefficient is within three per cent of its limit and by Mach 20 within a tenth of a per cent. Above that, the flow round a blunt body has stopped depending on how fast it is going.

What Mach independence actually says #

The name is older than the understanding, and it is worth being exact about the claim, because the claim is not that nothing changes.

The pressure distribution, in coefficient form, stops changing. The density ratio stops changing. The shape of the shock stops changing. So the lift coefficient, the drag coefficient and the shape of the flow field are all fixed once the Mach number is high enough.

What does not stop changing is the temperature, and therefore everything thermal. $T_2/T_1$ goes as $M^2$ and does not saturate, which is why a hypersonic vehicle’s aerodynamic shape can be designed once and flown over a wide speed range while its thermal protection cannot.

That split is not a coincidence: the pressure comes from mass and momentum and the temperature comes from energy, and the first two contain no thermodynamics.

Deriving the standoff rather than fitting it #

The shock ahead of a blunt nose stands off by a fraction of the nose radius, and the reason is mass conservation in a layer that is six times denser than the free stream.

Take the shock as concentric with the body and the layer as uniform. The mass crossing the shock within an angle $\theta$ of the axis must leave through the annulus at that angle, and the tangential velocity survives the shock unchanged, so

$$\rho_\infty U\,\pi (R+\Delta)^2\sin^2\theta = \rho_2 (U\sin\theta)(2\pi R\sin\theta)\Delta.$$

The $\theta$ cancels — which is why the model works at all — leaving

$$\varepsilon(1 + \Delta/R)^2 = 2\Delta/R,\qquad \varepsilon = \rho_\infty/\rho_2,$$

a quadratic whose smaller root is the standoff. At $\varepsilon = 1/6$ it gives $\Delta/R = 0.101$.

What the model is worth #

Against Billig’s correlation of sphere measurements, $\Delta/R = 0.143\exp(3.24/M^2)$, the derived model is between seventy-one and eighty-two per cent of the measurement across Mach 4 to 20 — low by about a third, consistently.

It is low for reasons that are easy to name. A real shock layer is not concentric with the body — the shock is flatter than the nose near the axis — and it is not uniform, since the density falls as the gas accelerates round the shoulder. Both errors go the same way.

What the model gets right is the dependence, and that is the part this essay is about.

The dependence, in two sweeps #

Sweep the Mach number at fixed $\gamma$. Between Mach 2 and Mach 8 the standoff falls by a factor of three. Between Mach 8 and Mach 40 — a fivefold change of speed — it falls by 8.5 per cent.

Sweep $\gamma$ at fixed Mach number. From $5/3$ down to $1.1$ the standoff falls from $0.173$ to $0.026$, and between $1.4$ and $1.2$ it halves — a ratio of $1.991$.

So above Mach eight the answer to how far does the shock stand off is a question about the gas and not about the speed. That is the plainest statement of what hypersonic means.

Why γ is the whole story #

The chain is short enough to write out.

$$\frac{\rho_2}{\rho_1} = \frac{(\gamma+1)M_n^2}{(\gamma-1)M_n^2 + 2}\ \xrightarrow{M\to\infty}\ \frac{\gamma+1}{\gamma-1}.$$

The Mach number cancels. What is left counts the ways a molecule can hold energy: a monatomic gas with $\gamma = 5/3$ compresses by 4, a diatomic one with $\gamma = 1.4$ by 6, and a gas with vibration and dissociation active, whose effective $\gamma$ is nearer $1.15$, by 15.

The reason is thermodynamic rather than mechanical. A shock converts directed kinetic energy into random thermal energy, and the more internal modes a molecule has to put that energy into, the less of it goes into raising the temperature, so the less the gas expands against its own pressure rise, and the more it compresses. The shock layer’s thickness is a measurement of how many degrees of freedom the molecule has.

What that costs a re-entry vehicle #

The consequences of a factor of two in standoff are not academic.

The shock is closer, so the gas next to the body is hotter and the radiative heating from the shock layer — which goes as the layer’s own thickness and temperature — is a different number.

The stagnation-point velocity gradient changes, and the convective heat transfer at a stagnation point goes as the square root of it, so a thinner layer means higher heating for the same free-stream conditions.

And the pressure distribution barely moves, because the pressure comes from momentum and is nearly insensitive to the caloric behaviour. So a perfect-gas calculation gets the forces essentially right and the heating badly wrong, which is exactly the split this collection’s real-gas essay measures.

Why a blunt nose at all #

There is a design question sitting under this whole essay and it is worth answering, because the answer is counter-intuitive and it is the reason blunt bodies are worth a field’s attention.

A sharp nose has less drag, and every subsonic and supersonic instinct says to use one. A re-entry vehicle has a blunt one, and the reason is heating rather than drag.

The convective heat flux at a stagnation point goes as the inverse square root of the nose radius. So sharpening the nose increases the heating without bound, and a sharp leading edge at orbital speed melts. Blunting it spreads the stagnation region over a larger radius, lowers the velocity gradient there, and cuts the heat flux — at the cost of a great deal of drag, which for a vehicle that wants to decelerate is not a cost at all.

That argument is Allen and Eggers’ and it is why every returning capsule is a blunt shape. The standoff computed in this essay is the geometry that follows from it, and the entropy layer of the previous essay is the aerodynamic price. All three — the heating, the standoff and the entropy layer — are consequences of one decision about the nose.

The pressure coefficient’s own limit #

The nose pressure coefficient deserves its own paragraph, because 1.8394 is a number worth recognising and because the route to it says something.

$C_{p_0}$ is the stagnation pressure behind the bow shock, minus the free-stream static pressure, divided by the free-stream dynamic pressure. Incompressibly it is exactly 1 — that is what a pitot tube measures. Compressibly it grows, because the stagnation pressure a compressible flow reaches is higher than $\tfrac12\rho U^2$ above the static, and then it turns over as the shock loss starts eating the gain.

The limit is $\big((\gamma+1)^2/4\gamma\big)^{\gamma/(\gamma-1)}\cdot 4/(\gamma+1)$, evaluated at 1.8394 for air, and it is approached from below. So the pressure on the nose of a hypersonic body is $1.84 q_\infty$ whatever the Mach number, and every hypersonic force estimate starts from that.

It is also the number Newtonian theory is normalised against. The modified Newtonian rule replaces the 2 in $C_p = 2\sin^2\theta$ with the actual stagnation value, giving $C_p = 1.84\sin^2\theta$, and that one change accounts for most of the improvement modified Newtonian theory has over the original.

Reading a hypersonic result #

Three habits follow, and they are the practical content of this essay.

Quote coefficients, not ratios. A pressure ratio at Mach 20 is a large number that says nothing; $C_p$ at Mach 20 is 1.84 and says everything.

State γ, always. In subsonic aerodynamics γ can be left implicit because nothing depends on it much. Here it is the whole answer, and a result quoted without it is a result whose sensitivity cannot be judged.

And check whether the quantity is a momentum quantity or an energy quantity. The first kind saturates and the second does not, and knowing which is which decides whether a Mach-independent estimate is available at all. Forces, yes. Heating, no. That single distinction is worth more than most of the correlations.

Where Newton’s theory comes back #

There is a piece of this collection that has been waiting for these limits. Newton’s impact theory — $C_p = 2\sin^2\theta$ — was refuted there as an account of lift at ordinary speeds, and it is nearly right at hypersonic ones.

The reason is now visible. The theory assumes the oncoming gas loses its normal momentum entirely at the surface and slides along it, which is what happens when the shock layer is thin enough to lie on the body. At $\varepsilon = 1/6$ the layer is a tenth of a nose radius thick, and as $\gamma\to1$ it becomes arbitrarily thin.

So Newtonian theory is the $\varepsilon\to0$ limit, and its accuracy at a given condition is set by the same $\varepsilon$ that sets the standoff. A theory refuted in one field and recovered as a limit in another is a reasonable definition of what a limit is for.

The thin-layer approximation this licenses #

Once the layer is a tenth of a nose radius thick, a whole family of approximations becomes available, and they are the reason hypersonic aerodynamics has closed-form answers where supersonic aerodynamics does not.

The pressure is constant across the layer, to the order of the layer’s thickness times the streamline curvature. So the pressure on the body is the pressure just behind the shock, and the shock’s shape determines the body’s pressure distribution directly — which is the content of the Newtonian rule and of every tangent-wedge and tangent-cone method built on it.

The shock and the body are nearly parallel, so the local shock angle is the local body angle, and a surface pressure can be computed point by point without solving a field problem at all. That is why hypersonic force estimates can be made on a spreadsheet and transonic ones cannot.

And the shock layer is thin compared with everything else, so the boundary layer growing inside it can occupy a substantial fraction of it. At high altitude the two merge — the viscous layer fills the shock layer — and the whole inviscid–viscous split this collection is organised around stops being available. That regime has its own name, viscous interaction, and its own parameter, and it is where hypersonic flow stops being a branch of gas dynamics.

Why this is a hand calculation and not a solve #

A concentric uniform layer is a crude model, and the obvious question is why the standoff is not simply computed. The answer is that for fifteen years nobody could, and the reason is the most instructive thing about the blunt-body problem.

Look at where the flow is fast. Immediately behind the near-normal part of the bow shock it is subsonic — a strong normal shock always leaves subsonic flow behind it — and it then accelerates round the shoulder and passes back through Mach one somewhere on the body. So the shock layer contains a sonic line, with subsonic flow on one side of it and supersonic on the other.

That is fatal for the steady equations, because they change type across it: elliptic where the flow is subsonic, hyperbolic where it is supersonic. An elliptic problem is a boundary-value problem needing conditions all round its domain; a hyperbolic one is marched forward from initial data. The blunt-body problem is both at once, in one region, with the dividing line’s position unknown in advance — and with the shock itself a free boundary whose shape is part of the answer rather than an input to it.

There is no way to pose that as a well-behaved calculation. The methods of the 1950s went round it rather than through it: the inverse approach assumed a shock shape, marched inwards, and found out afterwards which body it belonged to — which works, and gives no way to compute the flow over the body somebody actually wants.

The resolution, in 1966, was to change the question. The unsteady equations are hyperbolic in time whatever the local Mach number, so the type problem simply does not arise: start from any guess, march forward in time with the shock captured rather than fitted, and let the solution settle to the steady state. The mixed-type boundary-value problem became a well-posed initial-value problem, and the blunt body stopped being the outstanding difficulty in the subject.

That method then outgrew its origin entirely. Time-marching to a steady state is now how nearly every steady compressible flow is computed, including transonic aerofoils — which have exactly the same embedded-subsonic-region difficulty, in a different geometry, for exactly the same reason.

So the model above is a hand balance because the honest alternative is a time-dependent solver, and this collection has none. What the balance buys is the dependence — on $\varepsilon$, and therefore on the molecule — which is the part that survives whatever the geometry is refined to.

The one thing the model cannot supply #

The mass balance above needs the density ratio, and it gets it from the normal-shock relations at a stated $\gamma$. That is the whole of the input, and it is where every remaining difficulty sits.

What $\gamma$ should be is not a property of air; it is a property of air at the temperature behind the shock, which is itself what is being computed. At Mach 20 the temperature behind a normal shock in a perfect gas is 20,000 K, which is far beyond dissociation, so the perfect-gas value that produced that temperature was never applicable.

The honest way out is an equilibrium calculation with a composition solved for, which this collection does not carry. The honest way to use these figures is as a sensitivity study: the standoff is a function of $\varepsilon$, $\varepsilon$ is a function of the gas, and here is how much the answer moves over the plausible range. That range is a factor of two, and it is the result.

What is not in the flow #

γ — that is, the molecule.

Below Mach three the answer depends on the Mach number and the gas together, and the Mach number is doing most of the work. Above Mach eight the Mach number has dropped out of every quantity in this essay except the temperature, and what is left is a single number counting the internal degrees of freedom of the molecules.

That is an unusual position for a fluid-mechanical result. Most of this collection’s answers depend on a ratio of two effects in the flow; this one depends on a property of the material and on nothing about the flow at all. The transition between the two regimes is what the flat part of the Mach sweep is.

The model limit #

Four, and they are worth listing because the number 0.101 gets quoted.

The layer is not concentric with the body, and the shock is flatter than the nose near the axis. This is the largest of the errors and it is why the model is low.

The layer is not uniform. The density falls as the gas accelerates round the shoulder, so a single $\varepsilon$ overstates the compression away from the axis.

The tangential velocity is taken as $U\sin\theta$, which is the free-stream tangential component and is right at the shock and not on the body.

And γ is constant. At every Mach number in the sweeps above except the lowest, the gas behind the shock is hot enough for that to be false — which is not a defect of this calculation so much as the subject of the next one.

The last of those is worth one more sentence, because it changes how the γ sweep should be read. That sweep is not a comparison of four different gases; it is a comparison of four models of air, at temperatures where none of the four is right and the true effective ratio lies somewhere inside the range. Its value is as a bound: whatever air really does behind a Mach 20 shock, the standoff is between 0.05 and 0.10 nose radii, and no amount of refinement to the geometry will move it outside that.