Oblique shock — the series · Fluid Flow & Aerodynamics

A 10° wedge at Mach 2.00 has two shocks that solve it. The wedge turns the flow through a fixed angle, and the θ–β–M relation offers two shock angles that achieve it. The weak solution, drawn steeply forward, leaves the flow supersonic and is what a wedge in a free stream produces. The strong solution leaves it subsonic and appears where downstream pressure forces it. Nothing local to the wedge chooses between them.

A shock that leans

Tilt a shock and only the velocity component across it is changed — the component along it passes through untouched. That single observation turns every oblique shock into a normal shock in disguise, and it is why a wedge at Mach 2 leaves the flow supersonic while a blunt nose does not.

part 1 · compressible

Past 23.0° at Mach 2.00 there is no attached shock. The same wedge at two half-angles. On the left the θ–β–M relation has a root and the shock sits on the nose. On the right it has none, and the solver throws rather than returning the nearest thing — which matters, because a solver that quietly clamped to the maximum would draw a neat attached shock on a body that cannot carry one. The bow shock on the right is indicative: its shape is not solved here.

When the wedge is too blunt

Every curve of shock angle against deflection has a maximum. Past it there is no attached shock at any angle — the solver has no root to return and must say so, rather than quietly handing back the nearest thing and drawing a picture that cannot exist.

part 2 · compressible

A 20° cone at Mach 2: the shock at 37.80°, and a flow still compressing behind it. The conical flow round a cone of 20° half-angle at Mach 2, from the Taylor–Maccoll equation. The shock sits at 37.80° from the axis and turns the flow crossing it through only 8.57°, leaving it at Mach 1.693. Between the shock and the surface every ray from the apex carries its own state: the Mach number falls from 1.693 just behind the shock to 1.568 at the surface and the pressure rises from 1.586 to 1.912 times the free stream's — the rest of the turn, made without a shock. Nothing depends on the distance from the apex, so the rays are lines of constant state.

A cone finishes its turn after the shock

A wedge turns a supersonic stream all at once, at its shock. A cone of the same angle does not: its shock turns the flow only part of the way and leaves the rest to a smooth compression between the shock and the surface. Solved from Taylor and Maccoll's equation, the cone's shock is weaker, keeps more of the total pressure, carries less than half the wedge's surface pressure, and stays attached to 40.7° at Mach 2 where the wedge gives up at 23°.

part 3 · compressible

A cone hands the terminal shock a spread of Mach numbers. At Mach 2, the Mach number across the annulus a cowl on the conical shock captures, from the cone's surface to the shock, for cones of 15°, 22° and 28°, the last the best for recovery. Behind a wedge the flow is uniform; behind a cone it is fastest just behind the shock and slowest at the surface, 1.43 to 1.32 for the best cone. The terminal shock acts on all of it.

A cone intake's shock sees the whole capture

A cone's shock keeps far more total pressure than a wedge's at the same angle, because the cone finishes its turn in an isentropic compression after the shock. An intake then needs a terminal normal shock, and that shock meets not one Mach number but the whole spread the cone leaves across the captured annulus — fastest by the shock, where most of the air is, and slowest at the surface, where little is. Averaged over the mass, the cone's advantage over a wedge at the best angle for each is under two points of recovery, and the usual estimate from the surface Mach number nearly doubles it.

part 4 · compressible

Each oblique shock added before the normal one buys less than the last. The best total-pressure recovery of a train of oblique shocks ending in a normal shock, against the number of oblique shocks, at Mach 2, 2.5, 3 and 4. At Mach 3 one oblique shock recovers 58.1%, two 74.4%, four 88.9% and eight 96.4%; a normal shock alone keeps 32.8%.

The last shock should be the weakest

An intake slows a supersonic stream through a train of oblique shocks and one normal shock, and the rule for laying out the train is that its shocks should be equally strong. Solved exactly rather than quoted, the rule is right for the oblique shocks to four figures and wrong for the last one: the terminal normal shock should be about five per cent weaker than the others, and the turns that give equal strengths are not equal turns but grow along the train.

part 5 · compressible

At Mach 3 the cowl caps external compression at 88 per cent. Best recovery at Mach 3 against the number of oblique shocks, free and with the total turn capped at the free stream's detachment angle, 34.07°, which is as far as the cowl's outer surface can turn. Free, six shocks reach 94.2%; capped, 86.7%, and no number of shocks, nor a continuous isentropic ramp turning the same angle, can pass 88.5%.

The cowl caps what the ramps can buy

Every ramp in front of a supersonic intake turns the flow towards the cowl, and the cowl's outer surface has to turn the free stream at least as far. Past the free stream's own detachment angle it cannot, and that one angle caps every external intake: at Mach 3 no number of ramps, nor a smooth ramp, recovers more than 88 per cent, and the shock that is left at the throat is the strongest in the train. Above Mach 3 the cap falls fast, which is why fast intakes compress inside.

part 6 · compressible

Behind the lip the ceiling is set by starting, not by the cowl. The best total-pressure recovery with every compression isentropic but the terminal normal shock, against free-stream Mach number: an external ramp turning to the free stream's detachment angle; the same ramp followed by the largest internal contraction a fixed duct can start through, Kantrowitz's limit; and the same ramp followed by a contraction to a throat Mach number of 1.2, which only a duct that opened its throat to start can hold. At Mach 3 the three are 88.5%, 95.1% and 99.3%; at Mach 4, 71%, 82.5% and 99.3%.

Compression behind the lip is bought by opening the throat

The cowl caps how far an intake's ramps can turn the flow, and moving some of the compression inside the duct escapes the cap. It does not escape the duct's other limit: a duct that contracts a supersonic stream has to swallow the shock that stands at its lip while it starts. With fixed walls that allows a contraction of about a tenth, and at Mach 3 six shocks with three inside recover exactly what six outside do. A duct that opens its throat to start and closes it after recovers 94 per cent — the internal compression is paid for by the moving wall, not by the walls being inside.

part 7 · compressible