A duct is worth the square root of two

The most a disc can take and a big slow push treat a rotor as an actuator disc: a surface that adds a pressure jump to the air passing through it and nothing else, analysed with a control volume that needs to know nothing about blades. The second essay’s conclusion is that thrust is cheapest when it is made by giving a large mass of air a small push, and that a propulsor’s efficiency is set by how fast its jet is compared with the speed it flies at. A disc that knows no blades adds, in one sentence, that a ducted fan is a disc with a boundary condition changed.

This essay works out what that change is worth. It turns out to be a factor of $\sqrt2$ in the power to hover, half the thrust transferred from the rotor to the duct, and almost nothing in cruise — three results from the same four lines of momentum theory.

What an open rotor throws away #

A hovering open rotor draws air in from all around and pushes it down. Continuity and the momentum balance say that the air passing through the disc at the induced velocity $v_i$ leaves far below at $2v_i$, so the slipstream, carrying the same mass at twice the speed, has half the disc’s area. Its thrust is the mass flow times the jet speed, $T = 2\rho A v_i^2$, and its power is the kinetic energy it leaves in the jet each second, $P = Tv_i = T^{3/2}/\sqrt{2\rho A}$.

That contraction is the waste. The rotor accelerates a stream of air whose far-wake area is half its own, so for a given thrust the jet is faster than it would need to be if the same mass left at the disc’s full area, and the kinetic energy in a jet goes as the square of its speed. Anything that keeps the slipstream from contracting lets the same thrust be made with a slower jet.

Holding the slipstream open #

A duct does exactly that. The rotor sits inside a ring whose exit has an area $\sigma A$, and the jet leaves the exit at the ambient pressure, so the slipstream’s area is set by the duct rather than by the free flow. The momentum theory is the same four balances with one condition changed. Mass: $\dot m = \rho A v_d = \rho\sigma A v_e$, with $v_d$ the speed through the disc and $v_e$ at the exit. Momentum on everything inside the control volume: $T = \dot m v_e$ in hover. Energy: $P = \tfrac12\dot m v_e^2$. And Bernoulli along the lossless paths before and after the disc gives the disc’s own pressure jump, $\Delta p = \tfrac12\rho v_e^2$.

Eliminating the velocities at a fixed thrust gives

$$P_{\text{ducted}} = \frac{T^{3/2}}{2\sqrt{\rho A\sigma}}, \qquad \frac{P_{\text{ducted}}}{P_{\text{open}}} = \frac{1}{\sqrt{2\sigma}}.$$

For a straight duct, $\sigma = 1$, the hover power falls by a factor of $\sqrt2$ — 29 per cent. A duct whose exit is half as large again as its disc, $\sigma = 1.5$, cuts it to 0.577 of the open rotor’s, a saving of 42 per cent. The open rotor itself is the case $\sigma = \tfrac12$, where the exit area is the one its free slipstream would have chosen, and there the formula returns the open rotor’s power exactly.

In numbers: a rotor a metre across lifting a hundred newtons — a ten-kilogram load — in sea-level air needs 721 watts of induced power in the open, 510 in a straight duct, and 447 with an exit 1.3 times the disc. Those are ideal figures; a real rotor needs more in every case, for profile drag and swirl, but the ratio is what the duct changes.

Where the saving comes from #

The saving has the same source as every efficiency gain in a big slow push: more air, pushed less hard. For the same thrust through the same disc, the open rotor passes air at $v_i = \sqrt{T/2\rho A}$ and the straight-ducted one at $v_d = \sqrt{T/\rho A}$, so the duct draws $\sqrt2$ times as much air through the disc. Its jet leaves at $v_d$ where the open rotor’s leaves at $2v_i$, which is $\sqrt2$ times slower. Thrust is mass flow times jet speed and is unchanged; the energy left in the jet is mass flow times jet speed squared, and falls by $\sqrt2$.

The duct does not add energy to the air or remove it; it changes the shape of the streamtube so that the same rotor works on a larger stream. That is also why the result is independent of the rotor’s details: whatever the blades do, the control volume sees a disc passing a mass flow into a jet of a given area, and the energy follows.

A convenient way to state the result for a designer is the power loading, the thrust obtained per watt. For the metre rotor lifting a hundred newtons it is 0.139 newtons per watt in the open and 0.196 in a straight duct. Since the ideal power loading of an open rotor rises only as the square root of its area, getting the same improvement without a duct would take a rotor with twice the disc area — forty-one per cent larger in diameter.

The duct carries half the thrust #

The pressure jump across the disc is set by the exit speed, $\Delta p = \tfrac12\rho v_e^2$, so the thrust the disc itself carries is $A\Delta p = \tfrac12\rho A v_e^2$. The whole system’s thrust is $\dot m v_e = \rho\sigma A v_e^2$. Their ratio is

$$\frac{T_{\text{disc}}}{T} = \frac{1}{2\sigma},$$

and for a straight duct it is exactly one half. The other half is carried by the duct.

The duct’s half is not mysterious once it is asked where on the duct a force can act. Air drawn into the duct from all round has to turn sharply over the rounded inlet lip, and air turning over a curved surface at speed has a low pressure there — the lip is a ring of aerofoil leading edges, and its thrust is their leading-edge suction. The suction acts on the lip’s outer, forward-facing curve, so it pulls the duct forward — upward, in hover — and it is as large as the rotor’s own thrust. A ducted fan whose lip is sharp instead of rounded separates at the lip, loses the suction, and loses most of the duct’s benefit, which is why the inlets of ducted fans are fat.

The split also explains the duct’s structural demands. Half the lifting force of a ducted-fan vehicle arrives through the duct, not through the rotor shaft, so the duct is a primary structure and not a fairing. And the split depends on the exit: a diffusing duct puts more of the thrust on itself, $1 - 1/2\sigma$ = 62 per cent at $\sigma = 1.3$, while a duct that contracts its exit hands the thrust back to the rotor and, at the open rotor’s own $\sigma = \tfrac12$, carries none.

A lightly loaded rotor #

Carrying only half the thrust changes what the rotor itself has to do. Its pressure jump, $\tfrac12\rho v_e^2$, is half the open rotor’s jump for the same total thrust, so its blades work at half the loading, and a lightly loaded rotor can turn slower or use smaller blades for the same job. Tip speed is what sets a rotor’s noise, and a ducted tail rotor — the fenestron fitted to many helicopters — is noticeably quieter than an open one partly for this reason and partly because the duct screens the blade tips from the side.

The same light loading makes the duct forgiving in one respect and demanding in another. Forgiving, because blade stall, which limits an open rotor’s thrust, arrives later. Demanding, because the half of the thrust that the duct carries depends on the lip’s flow staying attached, and a lip that separates takes that half with it while the rotor, sized for half the load, cannot make up the difference.

The open rotor, recovered #

The formulas were written for a duct, and they contain the open rotor as a special case — a check that they are the same theory and not a parallel one. In hover, setting $\sigma = \tfrac12$ reproduces the open rotor’s power and puts the whole thrust on the disc, to $2 \times 10^{-16}$. In axial flight the check is sharper. An open rotor’s far wake has an area that depends on its loading, $(V + v_i)/(V + 2v_i)$ of the disc’s, and giving the duct formulas that area at each thrust coefficient must return Froude’s efficiency for an open propeller.

It does, at every loading, to the last digit. An open rotor is a ducted one whose duct is its own slipstream. What the physical duct changes is only which area the jet leaves at, and the whole of its advantage is the difference between that area and the one the free wake would have chosen.

In cruise the advantage almost vanishes #

In axial flight at speed $V$ the same balances give an exit-to-flight speed ratio of $\tfrac12(1 + \sqrt{1 + 2C_T/\sigma})$ and an ideal propulsive efficiency $\eta = 4/(3 + \sqrt{1 + 2C_T/\sigma})$, against Froude’s $2/(1 + \sqrt{1 + C_T})$ for the open disc, where $C_T$ is the thrust over the dynamic pressure and the disc area.

The difference depends on the loading. A cruising aircraft propeller works at thrust coefficients of order a tenth to one, where the open disc’s ideal efficiency is already high — 0.976 at 0.1 and 0.828 at 1 — and the straight duct improves it to 0.977 and 0.845. Two per cent at best, and a real duct has a wetted area and a drag of its own that easily costs more. At heavy loading the gap opens: at a thrust coefficient of ten the open disc manages 0.46 and the straight duct 0.53, and as the loading grows without limit the ratio of the two tends to $\sqrt2$, which is hover.

That is why ducts appear where rotors work hard at low speed and not on cruising aircraft. A tug’s propeller in a Kort nozzle, a hovercraft’s lift fan, a helicopter’s ducted tail rotor, a drone’s guarded fans, a vertical-take-off aircraft’s lift fans: all run at high thrust coefficient for much of their lives. A tug is the clearest case, because its hardest job is pulling at a standstill — bollard pull — which is hover with water for air. There the useful comparison is thrust at fixed power rather than power at fixed thrust. Since the ideal power goes as thrust to the three-halves over the square root of $\sigma$, a fixed power buys a thrust proportional to $\sigma^{1/3}$, and a straight nozzle gives $2^{1/3}$ = 1.26 times the open propeller’s pull: 26 per cent more, from a ring of steel with no moving parts, which is the order of the gain Kort nozzles are fitted for.

The turbofan is the apparent exception and is not one, since its duct exists to contain a compressor and the bypass stream’s efficiency comes from its mass flow, not from the duct’s effect on the wake’s area.

In cruise the thrust split also moves back onto the rotor. At a thrust coefficient of one, the straight duct’s rotor carries 87 per cent of the thrust and the duct only 13; the lip suction that carried half in hover weakens as less air has to be drawn in round the lip from the sides.

What a real duct costs #

The ideal saving is an upper bound, and each of the things the momentum theory leaves out takes some of it.

The gap at the blade tips. The theory assumes the rotor fills the duct. A real rotor has a clearance between its tips and the duct wall, air leaks back through it, and the leak grows quickly with the gap: a clearance of a per cent or two of the radius can cost a noticeable fraction of the duct’s gain, which is why ducted fans are built to tight tolerances.

The duct’s own drag. In hover the duct’s skin friction is small; in forward flight it is not, and its frontal area adds form drag. On a cruising vehicle that is usually enough to make the duct a net loss, which is the practical form of the last figure.

Diffusion has a limit. An exit larger than the disc means a diffuser, and the flow in a diffuser separates if it is asked to slow down too quickly — how much uphill a boundary layer can take is finite. Real ducted fans rarely diffuse beyond about 1.2 or 1.3 times the disc’s area without losing more to separation than they gain.

The lip in a crosswind. The duct’s half of the thrust depends on attached flow round the whole lip. In forward flight or a crosswind the windward side of the lip sees air arriving from outside and the leeward side sees it arriving at a steep angle, and the lip can separate on one side, which costs thrust and produces a pitching moment that a ducted-fan vehicle has to be designed to live with.

What the picture cannot show #

One-dimensional momentum theory. Velocities are averaged over each section, and the flow is steady, inviscid and without swirl. The duct’s shape enters only through its exit area; the lip’s shape, which decides whether the duct’s half of the thrust is actually delivered, does not enter at all.

No blades. The rotor is a disc. Blade loading, tip losses and the swirl a real rotor leaves in its wake are all absent, and the spinning wake costs a real rotor power whether it is ducted or not.

An exit at ambient pressure. The jet leaves the duct at the surrounding pressure. That is a good description of a duct long enough for its exit flow to be parallel, and a poorer one of a short ring round a propeller.

The convention the numbers depend on #

$A$ is the disc area and $\sigma$ the ratio of the duct’s exit area to it; $\sigma = \tfrac12$ reproduces an open rotor in hover. Power is the ideal induced power, the kinetic energy left in the jet per second. The thrust coefficient is the thrust divided by the flight dynamic pressure and the disc area. The worked rotor is a metre in diameter, in air of density 1.225 kg/m³. The duct’s share of the thrust is everything the control volume carries that the disc’s pressure jump does not.

Who found it, and when #

The ducted propeller was tried on aircraft by Luigi Stipa in Italy in 1932, whose tubular fuselage was a duct round the propeller, and developed for ships by Ludwig Kort, whose nozzle of 1934 is still the standard on tugs and trawlers. The momentum theory of the duct is part of the general theory of propulsion set out by Küchemann and Weber in 1953, and the split of thrust between rotor and duct follows directly from Rankine’s and Froude’s actuator-disc arguments of the 1860s and 1880s, applied with the wake’s area fixed.

Still open: whether a duct beats Betz #

The same change of boundary condition applied to a wind turbine produces the most argued-over claim in wind energy: that a turbine in a diffusing duct can extract more than Betz’s 16/27 of the wind’s power through its rotor area. The duct lowers the pressure behind the rotor and draws more air through it, and measured per unit of rotor area the power can indeed exceed the limit. The calculation that follows does the turbine’s momentum theory with a diffuser of exit area $\sigma A$ and an exit pressure the duct sets, and asks the question the claim avoids: measured per unit of the duct’s own exit area, which is the area the device actually occupies in the wind, does the limit still hold — and if it does, what the duct is buying, apart from a smaller rotor for the same power.