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

Thrust — where it appears

The force a propulsor exerts on its vehicle, equal to the rate at which it adds momentum to the fluid. A large slow push is more efficient than a small fast one for the same thrust, since the wasted energy goes as the square of the velocity added.

Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.

The tube widens because the air slows. The streamtube through an actuator disc at an induction factor of 0.333. The three radii are not drawn to taste: each is fixed by requiring the same mass to pass every station, and the slowest station is therefore the widest. The tube widening in front of a wind turbine is why some of the wind goes round it rather than through it, and it is the whole reason a disc cannot take everything.

The most a disc can take

A wind turbine cannot extract more than sixteen twenty-sevenths of the energy passing through the circle its blades sweep. That is not a limit on turbines — it is a limit on anything at all, and it follows from three conservation laws and no engineering.

applied · Actuator disc Efficiency is decided before the engine is chosen. Froude's propulsive efficiency against the ratio of jet speed to flight speed. It is 2/(1+σ) and nothing else — no engine, no fuel, no combustion. A turbojet with a jet at three times flight speed cannot exceed 50% however good its core is, and a propeller moving a great deal of air slowly is above 90% before anybody has designed anything.

A big slow push

The same thrust can be had from a lot of air moved a little or a little air moved a lot, and the two are not equivalent. One number decides which, it contains no engine, and it is why every airliner built since 1970 has a fan far larger than the machine driving it.

applied · Actuator disc The jet divides 75% to 25%. A jet striking a plate at 60 degrees. Both sheets leave at the jet's own speed, because their surfaces are at ambient pressure and Bernoulli allows nothing else, and the plate can exert no force along itself because the fluid has no viscosity. Momentum along the plate then fixes the split at (1 + cos β)/2 = 0.7500, and the normal force at ṁV sin β = 0.8660. Nothing about the plate's material, size or roughness enters either.

What a jet cannot push sideways

A jet striking a plate divides in two, and how it divides is fixed by a single sentence — an inviscid fluid exerts no force along a surface. That one statement, plus mass, gives the split exactly — and the same sentence turns a flat plate into a bucket worth twice as much.

applied · Jet A cone sends 1.7 per cent of its jet's momentum sideways at 15°. A conical divergent section of 15° half-angle and exit area ratio 25, drawn to scale from its throat to its lip, with its virtual apex to the left. The gas leaves as a source flow: straight streamlines from the apex, each at its own angle, and a Mach number uniform on spheres centred there. On the spherical cap through the lip the flow is normal to the surface and uniform, at Mach 3.925 for a gas with γ = 1.2; its axial momentum is ρV² times the cap's projection onto the exit disc, while the mass crossing it is ρV times the cap itself. The ratio of the two areas is (1 + cos α)/2 = 0.9830, and it is the only thing the cone's shape does to the momentum. On the flat exit plane the flow is not uniform: its edge is further from the apex than its centre, and the Mach number there is higher.

The jet a cone sprays sideways

The one-dimensional nozzle sends all its gas straight out along the axis. A real divergent section is a cone, and the gas leaves it as a spray of straight lines from the cone's apex. Only the axial part of that momentum pushes, and the share that does is (1 + cos α)/2 — 98.3 per cent at fifteen degrees, 93.3 at thirty — whatever the gas, the Mach number or the area ratio, and it touches the momentum and never the pressure.

compressible · Area mach Momentum theory answers every descent rate except those between hover and twice the hover inflow. The induced velocity at a rotor disc against its climb speed, both in units of the hover induced velocity √(T/2ρA), at fixed thrust. The climb branch (thick) solves v(V + v) = 1 and is a streamtube for every climb and for hover, where v = 1. Continued into descent (dashed) it still has a root, but the air it describes leaves the tube at both ends. The windmill-brake branch (thin) solves v(V + v) = −1 and is real only for descent faster than two hover inflows, where it meets v = 1 again. Between V = −2 and V = 0 (shaded) neither is a streamtube. The faint diagonal is v = −V, where the rotor would need no power: it crosses the band and touches neither valid branch.

Between hover and twice the hover inflow

A rotor's momentum balance has an answer for every climb and for every fast descent, and none for descending at anything between zero and twice its own hover inflow. There one root sends air out of both ends of its streamtube and the other root is not a real number, and at each edge of the band one end of the tube stops moving — which is where the vortex ring state lives, and where a wind turbine's thrust coefficient of one sits.

applied · Actuator disc An open rotor's slipstream shrinks; a duct holds it open. The radius of the slipstream behind a hovering rotor of radius R, against the distance behind the disc: for an open rotor, from the vortex-cylinder model, contracting towards R/√2 so that the wake ends with half the disc's area; and for a rotor in a straight duct, whose slipstream leaves at the full area, and in a duct that widens to 1.3 times it. The same thrust from a wider jet needs a slower one, and a slower jet wastes less energy.

A duct is worth the square root of two

An open rotor squeezes its slipstream to half its own area and pays for the fast jet that results. Put the same rotor in a straight duct and the slipstream leaves at the rotor's full area, the jet is slower, and hovering costs 29 per cent less power. The duct is not a passive guard: it carries half the thrust itself, on the suction round its inlet lip. In cruise almost all of the advantage disappears.

applied · Actuator disc The wall lets go long before the textbook's shock arrives. Where the flow leaves the nozzle's wall, as a fraction of the divergent section's length, against the ratio of chamber to ambient pressure. The inviscid picture runs full to the exit until the pressure ratio falls to 24.5, where a normal shock stands at the exit, and then moves the shock inside. The wall flow separates instead, when its pressure falls to Summerfield's 0.4 of ambient — at a pressure ratio of 192 — or to Schmucker's Mach-dependent value, at 138: six to eight times higher. Across that whole band the textbook's nozzle is full and the real one is not.

An overexpanded nozzle lets go before its shock arrives

The textbook's overexpanded nozzle runs full until its back pressure is high enough to hold a normal shock at the exit, and then draws the shock walking inside. A real nozzle never shows that sequence. Its wall's boundary layer cannot climb the pressure rise a normal shock imposes; it separates at a wall pressure of about four-tenths of ambient, which for a rocket nozzle happens at six to eight times the pressure ratio the textbook's shock needs. The separation is not a failure: it is what keeps the nozzle's thrust, and it is why a sea-level engine can be built twice the size it expands to.

compressible · Area mach The exit area is a condition, not a second choice. Critical entrainment and critical compression ratio for a steam ejector with a mixing tube of 60 nozzle throats, against the nozzle's exit area as a multiple of the exit that delivers the jet at exactly the pressure it shares with the entrained gas. Both peak at the matched exit, at every suction drawn, so no exit area buys entrainment at the cost of compression or the other way. Halving the exit costs 2.4 per cent of the entrainment at a hundredth of the motive pressure, doubling it 3.2 per cent.

An ejector's nozzle exit is a condition, not a choice

A steam ejector has two areas a designer can pick, the mixing tube's and the nozzle exit's, and one of them looked like a second way round the trade between entrainment and compression. It is not. Both are largest with the exit matched to the pressure the jet meets, as a rocket's thrust is, and every other exit draws a curve inside the matched one. What moves the machine is loss, and the losses sort themselves: only the nozzle's reaches the entrainment, and the sharp edge of the characteristic belongs to the ideal machine alone.

applied · Ejector

Named alongside it

The objects these essays reach for when they reach for this one.

Control volumeModel limitMomentum theoremActuator discEfficiencyKinetic energyMomentum fluxStreamtubeBernoulli's equationThe Betz limitConservationMass flow

All concepts