What a shock costs

The first thing to get straight about a shock is what it does not do. It does not destroy energy. Every joule that went in comes out, and the stagnation temperature — which is the energy, expressed as a temperature — is identical on both sides.

What it destroys is the flow’s ability to give that energy back as ordered motion.

Two totals, and only one of them is a conserved quantity #

The energy equation gives the stagnation enthalpy, and hence the stagnation temperature, as constant along a streamline through anything adiabatic — smooth or discontinuous, reversible or not.

The stagnation pressure is a different animal. It is the pressure the flow would reach if brought to rest isentropically, and that word is doing all the work. A flow that has produced entropy cannot be brought back to its original pressure, however carefully it is decelerated, because the process that would take it there is not available.

assertTotalsAcrossShock checks both halves. It recomputes the stagnation temperature independently on each side of the jump from that side’s own static state and Mach number, and requires them equal to within $10^{-12}$; and it requires the total pressure ratio to lie strictly between zero and one. The rejection test moves the temperature ratio by one part in a thousand, an amount no figure would render visibly, and requires the refusal.

The currency #

Total pressure is the working currency of compressible aerodynamics, and it is worth understanding why rather than accepting it as convention.

Everything a gas can be asked to do — accelerate through a nozzle, push a turbine, produce thrust — depends on the pressure ratio available between where it is and where it is going. A stream with a high total pressure and a stream with a low one at the same total temperature carry the same energy and are not equally useful, and the difference is precisely what an engine sells.

The relation to entropy is exact and short:

$$\frac{p_{02}}{p_{01}} = e^{-\Delta s / R}$$

so total-pressure ratio is the entropy rise, exponentiated. Two ways of saying one thing, and the pressure form is preferred by engineers because a pitot tube measures it.

The numbers, and where they become intolerable #

A single normal shock at the flight Mach number gives these total-pressure recoveries:

Below about Mach 1.5 a normal shock is nearly free and nobody bothers avoiding it. At Mach 2 losing 28 per cent of the total pressure is a serious hit but survivable. At Mach 3 a normal-shock intake throws away two thirds of what the flight condition offered, and no engine downstream can recover it.

At Mach 5 the recovery is six per cent, which means an engine fed by a single normal shock has essentially nothing to work with. That is a hard bound on the architecture and it is why hypersonic propulsion looks the way it does — and, past a point, why it stops decelerating the flow at all.

The cube law, which is the way out #

The escape is the fact that entropy production goes as the cube of shock strength.

The exponent is fitted rather than quoted. weakShockEntropyExponent computes the entropy rise at four small turning angles at a given Mach number, fits a straight line through log $\Delta s$ against log $\theta$, and returns the slope; assertWeakShockEntropyIsCubic requires it to be 3 within 0.06. At Mach 2 the fitted value is 2.985, the departure from exactly 3 being the higher-order terms that a finite fitting range still contains.

The consequence, in numbers: at Mach 2, a fifteen-degree compression taken as a single oblique shock keeps 95.9 per cent of the total pressure. The same fifteen degrees taken as four shocks of 3.75° each keeps 99.4 per cent. Three quarters of the loss has been avoided by changing nothing but the number of steps.

Which is why intakes look like that #

Every supersonic intake is a device for taking the compression in as many small bites as possible.

A pitot intake — a plain hole facing forward — has one normal shock and takes the whole loss. It is used up to about Mach 1.5, where the loss is under seven per cent and the simplicity is worth it.

An external-compression intake puts a wedge or a cone ahead of the opening, so the flow crosses one or more weak oblique shocks before meeting the final normal shock at a much reduced Mach number. Two oblique shocks plus a normal one at Mach 2 recover around 92 per cent against the single normal shock’s 72.

An isentropic intake takes the compression through a continuously curved surface, which is a compression fan of infinitely many infinitesimal waves and produces, in principle, no entropy at all. In practice those waves converge and coalesce into a shock a short distance out, so the ideal is approached and not reached.

The progression is a direct reading of the cube law, and every step of it costs geometry, weight and off-design behaviour. That is engineering: the second law hands over a graded penalty, and the design is a purchase of how much of it to pay.

The efficiency that is not a percentage #

A recurring confusion is worth heading off, because it makes the numbers in this essay sound worse than they are and occasionally better.

Total-pressure recovery is not an efficiency in the sense of “what fraction of the energy is usefully employed”. No energy has been lost at all — the stagnation temperature is unchanged — so an intake with 72 per cent recovery has not thrown away 28 per cent of the incoming energy.

What it has thrown away is 28 per cent of the pressure ratio it could have delivered, and the consequence for the engine’s output is neither 28 per cent nor nothing. It depends on the cycle: an engine with a high pressure ratio of its own is relatively insensitive to losing some of the intake’s contribution, while a ramjet, which has no compressor and lives entirely on ram compression, loses almost proportionally.

So the same shock loss is a minor nuisance in one machine and fatal in another, which is why the quantity tabulated is the recovery rather than any efficiency: recovery is a property of the flow alone, and what it is worth is a property of the machine attached to it.

Where the lost availability actually ends up #

Saying that the ability to give the energy back has been destroyed is correct and slightly abstract, and the concrete version is worth having because it is measurable.

The entropy rise means the gas can no longer reach its original pressure by any adiabatic deceleration. Read the same statement the other way and it says the flow, brought to any given static pressure downstream, arrives there hotter than it would have without the shock. The energy is all present and some of it is now in a form that is thermally degraded — at a higher temperature and a lower pressure than the same energy occupied before.

That has a direct consequence in a combustion system, and it is the reason intake losses are counted so carefully. A combustor is limited by the temperature its turbine can survive. Air arriving hotter because it has crossed a shock leaves less room between its own temperature and that limit, so less fuel can be burned in it — and the engine’s output falls for a reason that is not a pressure ratio at all. The same shock loss appears twice in the cycle, once as a pressure the nozzle does not have and once as a temperature margin the combustor does not have.

The overall sensitivity is the number a propulsion engineer carries, and it is unforgiving. For a supersonic turbojet, one per cent of intake total-pressure recovery is worth something like one to one and a half per cent of net thrust; for a ramjet, which has no compressor and lives entirely on ram compression, it is worse and approaches proportional. Against the table above — twenty-eight per cent lost to a single normal shock at Mach 2 — that is the whole argument for the geometry of a supersonic intake in one line.

Why total pressure is the currency and not merely a quantity #

There is a second reason the loss is tracked in this form rather than as an entropy or an efficiency, and it is the practical one. Losses in total pressure add.

A duct’s shock system, the friction on its walls, the mixing of a boundary layer with the core, the wake of a strut, the turning loss in a bend: each of them produces entropy by its own mechanism, and each appears as a decrement of $p_0$. They can therefore be measured separately, tabulated as loss coefficients, and summed — which is exactly what an internal-aerodynamics calculation does, and why every compressor cascade, diffuser and duct in the literature is characterised by a number of the form

$$\omega = \frac{p_{01} - p_{02}}{p_{01} - p_1},$$

a total-pressure loss normalised by an inlet dynamic pressure.

Nothing else in the problem has that property. Static pressures do not add — they are related by the geometry. Efficiencies multiply and are defined against different ideal processes in different components. Entropies do add, and are inconvenient to measure. Total pressure is the one quantity that is both directly measurable with a probe and additive across mechanisms, and that combination is why an entire branch of engineering is written in it.

The other cost, which is drag #

A shock takes total pressure, and it also pushes.

The pressure rise across a shock acts on whatever is behind it, and if that is a body, the result is a force along the flow direction: wave drag. It is a drag that exists in a fluid with no viscosity whatever, which makes it a genuine exception to d’Alembert’s paradox and one of the few places on this site where the ideal theory’s famous result fails on its own terms.

The energy accounting behind the two costs is the same accounting. Wave drag times flight speed is the rate at which the vehicle does work on the air, and that work reappears as the energy carried away in the wave system — which is exactly the availability the total-pressure loss measures. The drag and the entropy are two faces of one transaction, and the whole ladder that follows from it is a field of its own.

What it costs a nozzle, which is the same sum backwards #

The accounting works in the other direction too, and in a nozzle the loss is easier to see because it shows up directly as thrust that was not produced.

A nozzle whose shock stands at an area ratio of 2.43 meets flow at Mach 2.41 and loses 46.6 per cent of its total pressure, and the exit velocity it then achieves is a fraction of the design value. The propellant burned is unchanged; the thrust is not. Everything about how a nozzle’s shock position is set by its back pressure is therefore also a statement about how much of the chamber’s energy reaches the exhaust.

That is why an overexpanded rocket engine at sea level is worse than a smaller nozzle would have been, and why staging exists: it is cheaper to change the nozzle than to fly the wrong one for the whole ascent.

What the shock does to the boundary layer #

The pressure rise a shock imposes is, from the point of view of the thin layer on a surface underneath it, an adverse pressure gradient — a violently adverse one, applied over a distance of almost nothing.

A boundary layer can only climb so much of a pressure rise before separating, and a shock delivers more of it faster than any smooth deceleration would. So a shock standing on a surface usually separates the layer beneath it, and the separation feeds back and changes the shock, which changes the separation.

That interaction is where supersonic and transonic aircraft actually get into difficulty: buffet, control reversal, and intake instability are all shock–boundary-layer interaction rather than shock loss. This site’s solver does not compute it. What it can say honestly is the size of the pressure rise being applied — 4.5 times at Mach 2 — against the pressure rise a laminar layer can survive at all, which is a fraction of that.

What the solver computes, and how it is checked #

Three assertions carry this essay.

assertTotalsAcrossShock proves that the stagnation temperature does not move and the stagnation pressure does, by recomputing each on both sides independently. assertShockRaisesEntropy proves that the loss is in the direction the second law requires. assertWeakShockEntropyIsCubic fits the exponent rather than assuming it, which is the difference between a claim and a measurement.

The third is the one worth dwelling on, because the whole intake argument rests on it and it would be easy to state without checking. A fitted exponent of 2.985 is evidence; “entropy goes as the cube of shock strength, as is well known” is not. The rejection test hands the check an exponent of 2.0, which is what a plausible algebra slip would produce and which would make the multi-shock intake a far weaker idea than it is, and requires the refusal.

Where the model stops #

Adiabatic, and no work done. Total temperature is conserved because nothing adds heat and nothing does shaft work. In a real intake there is heat transfer to the structure, and downstream of it there is a compressor doing exactly the work this analysis excludes.

Uniform flow on both sides. The recovery figures above treat the intake shock system as a set of uniform states. Real intakes have non-uniform exit flow, and the distortion it presents to the compressor is a design constraint at least as serious as the mean recovery.

No viscous loss. Everything counted here is shock loss. Skin friction in the duct and separation losses add to it and are frequently comparable, so an intake’s measured recovery is always below the shock-only prediction.

Steady operation only. An intake that unstarts — losing its swallowed shock, which then pops out in front of the inlet — does so violently and unsteadily, and none of this analysis describes the event.

What the picture cannot show #

The recovery curve is a clean falling line, and it hides the thing that makes total pressure worth tracking: that it is irrecoverable.

Nothing in a plot of $p_{02}/p_{01}$ against $M_1$ says that the loss is permanent. A reader could reasonably take it for a resistance that a sufficiently good diffuser might undo, in the way a pressure drop through a bend can be partly recovered downstream. It cannot. The lost total pressure is not stored anywhere and there is no process that returns it.

That permanence is a thermodynamic statement, and — like everything else on the previous rung — it has no picture at all. The curve shows the size of the loss. Its irreversibility is carried entirely by the caption.

Who found it, and when #

The total-pressure accounting is Prandtl’s school, in the 1900s and 1910s, arriving from thermodynamics rather than from hydrodynamics — which is why the vocabulary is a thermodynamicist’s.

The multi-shock intake is Oswatitsch, in 1944, in a report that is one of the cleanest pieces of applied second-law reasoning in aerodynamics: it works out the optimum distribution of turning among a fixed number of oblique shocks, and finds that the shocks should be of equal strength. That result follows directly from the cube law — equal cubes minimise the sum for a fixed total — and it is still how intakes are laid out.

The delay between the theory and the hardware was short here, for the usual reason: the aircraft that needed it were being built at the time.

Where the ladder goes next #

Every shock in this ladder has been normal to the flow, and a normal shock is the most expensive kind there is. A shock that leans takes the same flow through the same conservation laws with only its normal component changed, which makes it weaker, cheaper, and able to leave the flow supersonic — and that is the next ladder.