A compression that costs nothing in the end

Turning the other way is free states the asymmetry this subject runs on: a supersonic expansion costs no total pressure at all, and a compression does. This essay is about how much, which turns out to be so little for a weak shock that the asymmetry can be engineered away.

The entropy is a cube #

The asymmetry drawn out. Expansion costs exactly nothing at every angle — not approximately nothing, nothing, because a Prandtl–Meyer fan is a continuous isentropic turn with no discontinuity in it. Compression costs a loss that starts at zero and grows.

How it grows is the whole essay.

Writing $\epsilon = M^2 - 1$ for the strength, the entropy rise expands as

$$\frac{\Delta s}{R} = \frac{2\gamma}{3(\gamma+1)^2}\,\epsilon^3 + O(\epsilon^4),$$

and both the exponent and the coefficient are checked here rather than quoted. Over five decades of strength the measured exponent is 2.9989 and the coefficient 0.16216 against the closed form’s 0.162037.

Dividing by the cube isolates the constant, which approaches its value from below as the strength falls — the residual being the fourth-order term, which is what the approach is measuring.

One transcription note, because it cost a debugging session. The expression above is per unit gas constant; the form printed in most textbooks is per unit $c_v$, and $c_v = R/(\gamma-1)$, so the two differ by a factor of $\gamma - 1$ — which is 2.5 for air and is exactly the discrepancy this check first reported. A closed form quoted in the wrong units produces a curve of the right shape and the wrong height, which is the hardest kind of error to see.

It is positive everywhere, which is the second law doing its work, and vanishingly small when the shock is weak — $1.5\times10^{-4}$ at $M = 1.05$, which is where the whole design opportunity lies.

Total pressure is the currency the loss is usually priced in, and since $p_{02}/p_{01} = e^{-\Delta s/R}$ the loss is cubic too: 0.015 per cent at $M = 1.05$, seven per cent at $M = 1.5$.

What “free” means for an expansion #

Before spending the essay on compressions it is worth being precise about the other direction, because “exactly zero” is a strong claim.

A Prandtl–Meyer expansion is a continuous fan of Mach waves. Each wave is an infinitesimal disturbance, and an infinitesimal disturbance is a sound wave, and a sound wave is isentropic — so the entropy rise through the fan is a sum of zeroes rather than a small number. That is why the loss is identically zero at every angle including large ones: it is not that the turn is weak, it is that the turn is made of infinitely many weak pieces and each piece costs nothing to the order that would matter.

The corresponding compression cannot be built that way by itself, because a compression’s Mach waves converge and coalesce into a shock. That is the whole asymmetry. Expansions spread and compressions focus, and the direction of that is set by whether the flow being turned is speeding up or slowing down.

So the design problem is not to make a compression isentropic — that is already possible in principle, and this essay’s arithmetic is about how nearly it can be done in practice — but to get the flow where it is going before the waves it is made of arrive at one another. That reframing is what the wall that cancels its own waves is about on the expansion side, and it is the same geometry run backwards.

So split the turn #

The cube is not a curiosity. It is the reason a supersonic inlet has ramps.

Turn a stream through a total angle $\theta$ in $N$ equal steps. Each step is a weak oblique shock of strength proportional to $\theta/N$, costing entropy proportional to $(\theta/N)^3$, and there are $N$ of them — so

$$\Delta s \propto N\left(\frac{\theta}{N}\right)^3 = \frac{\theta^3}{N^2}.$$

Exactly one over $N$ squared. The measured exponent over ramp counts from one to sixty-four is $-2.00$.

Two ramps cost a quarter of one ramp; four a sixteenth; sixteen a two-hundred-and-forty-fifth; sixty-four a three-thousand-nine-hundred-and-twentieth. In the limit of infinitely many infinitesimal turns, the compression is isentropic — genuinely, not approximately — and that is the isentropic inlet.

The comparison is fair: every configuration turns the flow through twelve degrees and arrives at nearly the same downstream Mach number. What differs is only how much total pressure was spent getting there, which is the quantity an inlet’s whole design is about — an engine’s thrust is roughly proportional to the total pressure delivered to it, so a few per cent here is a few per cent of the aircraft’s range.

The oblique shock, and where the strength went #

One step in the ramp arithmetic deserves unpacking, because it is where “strength proportional to the turn” comes from.

An oblique shock at angle $\beta$ in a stream of Mach number $M$ behaves exactly like a normal shock at the normal component $M\sin\beta$, with the tangential component carried through unchanged. So the entropy rise depends on $M_n^2 - 1$ alone, and everything above applies to it directly.

For a small turn $\theta$, the shock angle is close to the Mach angle $\mu = \arcsin(1/M)$, and expanding the theta–beta–M relation gives $\beta - \mu$ proportional to $\theta$ — so $M_n^2 - 1$ is proportional to $\theta$ as well, and the entropy is proportional to $\theta^3$. That is the link between the geometry and the thermodynamics, and it is why the ramp argument works with the turn angle rather than with the Mach number.

It also explains something the ledger shows and the essay has not: the successive ramps are not equally weak. Each one leaves the flow slower, so the next one’s Mach angle is larger and the same turn produces a stronger shock. The equal-angle case computed here therefore under-performs the optimal one, in which the later ramps turn through smaller angles — and the correction is a second-order effect on top of an argument whose whole content is the third order.

And what the limit costs instead #

The arithmetic says the limit is free. It is not.

A concave wall emits a continuous fan of Mach waves rather than a discrete shock, and the waves converge. Each one leaves at the local Mach angle, the Mach angle opens as the flow is compressed, so a later wave is steeper than an earlier one and overtakes it. Where they cross, the compression has become a shock again — and the entropy the ramps saved is spent all at once.

The distance to the focus is a few wall lengths, and it moves the wrong way: a four-degree turn puts it at 3.54 wall lengths and a sixteen-degree turn at 0.89. Making the compression gentler does not push the focus away — the wall gets longer in proportion and the focus goes with it, so the ratio stays of order one and the focus is always a small multiple of the turning length away.

That is the real trade. The isentropic compression exists, it is exactly free, and it needs the flow to be captured before the waves collect — which for an inlet means the cowl lip has to be inside the focus, and therefore that a longer, gentler ramp needs a longer inlet. The limit is approached along a path that runs out of room.

One row of that table needs a comment. The saving from one ramp to sixty-four comes out at 3,920 rather than at $64^2 = 4096$, and the four per cent shortfall is not an error. The cube is the leading term of the entropy, and a twelve-degree turn done in one go is not a weak shock — it is at $M_n = 1.466$, where the fourth-order term is already contributing — so the single-ramp case costs a little less than the cube would predict and the ratio is correspondingly smaller.

That is the usual fate of an asymptotic argument evaluated at a finite value, and it is worth stating rather than rounding away. The exponent measured over the range where every shock is genuinely weak — sixteen ramps to sixty-four — is $-2.0000$; the four per cent is the price of the first point being outside that range.

How much of this survives a real inlet #

Three numbers put the arithmetic in proportion, because a design decision is not made on a cube alone.

A single normal shock at Mach 3 keeps 32.8 per cent of the total pressure. That is what a pitot-type inlet does, and it is why nobody builds one above about Mach 1.6.

One twelve-degree oblique ramp followed by a normal shock keeps considerably more, and each additional ramp adds less than the one before — because the shocks are getting weaker and the cube is already doing most of its work by the third or fourth. In practice inlets use two or three ramps, and the reason is not the cube running out but the two effects the cube does not include: the boundary layer, which thickens under each adverse gradient and eventually separates, and the length, which the focus argument above bounds.

And the last few per cent are worth chasing anyway. An engine’s thrust is close to proportional to the total pressure at its face, so one per cent of pressure recovery is roughly one per cent of thrust, and at cruise that is a measurable fraction of the range. The cube is why the first ramp is worth so much; the boundary layer is why the fifth is not.

Why the cube and not something else #

The exponent is worth understanding rather than accepting, because it is where all the leverage comes from.

A shock conserves mass, momentum and energy exactly, and those three are what fix the jumps. Entropy is not conserved and is not used to find them — it is computed afterwards from the state on either side. So the entropy rise is not a term in the equations; it is a consequence, and it turns out that the first two orders of it cancel.

The reason they cancel is that a shock of vanishing strength is a sound wave, and a sound wave is isentropic. So $\Delta s$ must vanish as $\epsilon \to 0$ — first order gone — and the reversibility of a linear wave means it must also vanish to second order, since a second-order term would give an entropy change of one sign for a compression and the opposite for an expansion, and there is no such thing as an expansion shock to have a negative one. Third order is the first term whose sign is the same for both, and it is where the irreversibility can live.

That argument is worth having because it says the cube is not special to shocks. Any weakly irreversible process whose reversible limit is a wave has the same structure, and the distance a finite-amplitude sound wave travels before it shocks is the same expansion read from the other end.

The same cube, three times over #

The exponent turns up in three places in this collection and it is the same exponent each time, which is worth collecting.

A weak shock’s entropy, which is this essay.

A weak shock’s failure to carry the Riemann invariants, which is two numbers that do not change: the invariants are exactly constant along characteristics in a smooth flow and are not carried across a shock, and the amount by which they fail is third order in the strength, with a measured exponent of 2.997.

And the distance a finite-amplitude sound wave travels before it steepens into a shock, which goes as the reciprocal of the amplitude — the same expansion inverted.

All three are the same statement: the deviation of a weak shock from a sound wave is second order in its strength in the quantities that are conserved and third order in the quantities that are not. That is a single fact about the Hugoniot curve’s contact with the isentrope, and every consequence in this part of the subject descends from it.

A fourth cube, in a different fluid and in closed form #

The three above are all gas dynamics, and there is a fourth that is worth adding because it is not an expansion at all.

The energy lost across a hydraulic jump in shallow water is exactly

$$\Delta E = \frac{(h_2-h_1)^3}{4h_1h_2},$$

a closed form with no small parameter anywhere in it, and cubic in the height rise. So the cube is not an artefact of expanding the gas relations near a sound wave: it is what happens whenever a discontinuity conserves mass and momentum while losing energy, and shallow water is the case where the whole thing can be written down rather than expanded.

The design consequence inverts. In an inlet the loss is the enemy, so the compression is taken in many weak steps and the cube rewards it. In a stilling basin below a dam the loss is the objective — the water has to arrive at the river slow — so the right move is a single strong jump, and the cube says why a sequence of small drops would fail to do the job that one big one does.

Same exponent, opposite intent, and hydraulic engineering uses both deliberately.

Where else the trick is used #

The same arithmetic appears in two other places worth naming, because seeing it three times is what turns it from a calculation into a tool.

Supersonic diffusers in wind tunnels use a contoured second throat for exactly this reason: the flow is decelerated through a series of weak compressions rather than through one normal shock, and the tunnel’s power requirement falls with the total-pressure recovery.

And a supercritical aerofoil’s upper surface does the same thing in reverse. The flow accelerates to supersonic over the crest and has to be returned to subsonic before the trailing edge, and a carefully shaped rear curvature does it through weak compressions rather than through a single strong shock — which is most of why a supercritical section’s drag rise is delayed. That is the transonic version of this essay’s argument, and it is fought over the same cube.

In both cases the constraint is the one the focus figure describes: the compression has to be finished before the waves it is made of catch each other, and the length available to do it in is set by the geometry rather than by the aerodynamics.

What is not claimed #

A normal shock’s entropy is used for the expansion coefficient. The cubic is derived and checked on a normal shock, and applied to oblique ones through their normal Mach number — which is exact, since an oblique shock is a normal shock with a tangential velocity added, and is the same decomposition a shock that leans is built on.

Perfect gas, constant specific heats. At the temperatures behind a strong shock, $\gamma$ is not a number, and when gamma stops being a number is what happens then. Nothing here applies above a few Mach numbers.

One-dimensional entropy accounting. The entropy quoted is the jump across each shock, summed. In a real inlet the shocks are not of uniform strength across the capture area, so the total-pressure recovery is an average over a distribution and the distortion of that distribution matters to the engine as much as its mean — which is a version of the mean is not the flow with expensive consequences.

Inviscid. The ramps carry boundary layers, and a shock impinging on one can separate it — which costs far more total pressure than the shock itself and is the actual limit on how many ramps an inlet can usefully have. That interaction is not modelled here.

The focus calculation is a wave construction, not a solve. Mach waves are drawn from the local Mach angle along a turning wall and intersected; nothing solves the flow past the focus, where the construction is no longer valid because the waves have crossed.

The focus is drawn for a straight-line wall turning at a constant rate. A real isentropic ramp is contoured — its curvature is chosen so that the waves arrive at the cowl lip together rather than before it — which is exactly the construction the wall that cancels its own waves does for a nozzle, run in the opposite direction. The essay’s claim is that a focus exists at a finite distance, not that it is where a designer would let it be.

And the ramps are all at the same angle. An optimally designed multi-ramp inlet does not use equal turns — the later shocks are in a slower stream and are stronger for the same turn — and its ramp angles are chosen to equalise the losses. The equal-angle case is the one whose arithmetic is clean, and it is a lower bound on how well the job can be done.