A duct beats Betz only on the area it chooses

The most a disc can take proves that a wind turbine, idealised as a disc that drops the pressure of the air crossing it, can take at most 16/27 of the kinetic energy that would have flowed through its area had it not been there. The proof has two moves. The disc slows the air, so less air arrives at it than at an empty hole of the same size; and each kilogram that does arrive gives up only the energy between the wind’s speed and the wake’s. Take too little per kilogram and the power is small; take too much and the air goes round instead. The best compromise slows the wake to a third of the wind and passes air through the disc at two-thirds of it, and the product is 16/27.

A duct is worth the square root of two put a propeller in a duct and found the duct changing the answer by changing a boundary condition: the slipstream can no longer contract behind the rotor, so for a given thrust the jet is slower and the induced power smaller. Its last section named the same change for a turbine, and the claim that comes with it is among the most argued-over in wind energy: that a turbine in a flaring duct — a diffuser — can take more than 16/27. Measured per unit of rotor area, it can, and the measurements are real. This essay does the momentum theory and asks the question the claim leaves out: the limit is a statement about an area, and there are three areas in a ducted turbine, not one.

The one number momentum theory cannot supply #

The model is the actuator disc of a big slow push with a duct round it. Far upstream the wind blows at $U$. The rotor removes a head from every kilogram that crosses it, so that far downstream, once the air has returned to atmospheric pressure, it moves at a wake speed $u_w$; write $x = u_w/U$. The rotor sits in the duct’s throat, and the duct widens behind it to an exit $\sigma$ times the rotor’s area. Inside the duct nothing escapes, so continuity ties the speed at the rotor to the speed at the exit: $u_d = \sigma u_e$. Between the exit and the far wake the air is free again and, with no losses assumed, recovers to ambient pressure along its streamtube.

That leaves one quantity the balance does not fix: the pressure at the exit plane. For an open rotor the momentum theorem decides it, because nothing but the rotor pushes on the air. With a duct, the duct pushes too, and how hard depends on the flow round its outside — a flange, a cambered ring, a lip that the external stream curves over. So the exit pressure enters as an input, written here as a pressure coefficient $-c$: $c = 0$ is an exit at ambient pressure, $c > 0$ a suction held behind the exit.

With that stated, everything follows from Bernoulli’s equation along the streamtube, used on each side of the rotor, and the continuity. The exit speed is $u_e = U\sqrt{x^2 + c}$; the power the rotor takes is the head it removes times the volume flow through it; and measured per unit of the exit’s area the power coefficient is

$$C_{P,\text{exit}} = \sqrt{x^2 + c}\,\bigl(1 - x^2\bigr).$$

The area ratio $\sigma$ has dropped out, which is the first surprise and a clean one: per unit of exit area, the diffuser’s flare does not matter at all. It sets how fast the air goes through the rotor and therefore how small the rotor can be, but the power the device takes out of the wind is decided by the exit’s size, the wake speed chosen and the suction behind the exit.

Three ways through #

The first figure follows the pressure along the axis for three devices. The open disc is the faint line: the air approaching it slows and its pressure rises to 0.556 of the dynamic pressure above ambient, the disc drops it by 8/9 to −0.333, and the air behind recovers to ambient in the wake. The recovery behind an open disc is free — it happens in the expanding streamtube after the air has left — and the momentum theorem forbids the disc from arranging it any other way.

The diffuser with its exit at ambient pressure does something different in front of the rotor. The duct draws air in: the pressure ahead of the rotor is already below ambient, −0.333, because the air is accelerating into a throat half the exit’s size, and behind the rotor it reaches −1.0. The diffuser then climbs from −1.0 back to zero along its own length. That climb is the whole point of the device. It lets the rotor work across a pressure difference whose downstream side sits far below ambient, and it draws more air through the rotor than a bare rotor of the same size would pass. It is also a steep adverse pressure gradient inside a duct, which is where a boundary layer separates if it is asked for too much.

The third line holds a suction of a third behind the exit. The pressure behind the rotor then falls to −1.67, and the air leaves the exit still below ambient, recovering the last third outside the duct as the open disc’s wake does. The two diffusers draw the same shape; the second has simply been lowered at every station by what the outside of the shroud holds at its exit.

A plain diffuser is worse than its own exit #

The second figure plots the power against the wake speed, which is the turbine’s one free choice: how hard to load the rotor. Each curve has the shape of Betz’s, rising from zero with a stopped wake — here the exit speed is then $\sqrt{c}$, and with no suction nothing flows — through a maximum, and down to zero when the rotor takes nothing and the wake keeps the wind’s speed.

Setting the derivative to zero gives the best wake in closed form, $x^2 = (1 - 2c)/3$ while $c$ is less than a half, and the best power per exit area as

$$C_{P,\text{exit,max}} = \frac{2}{3\sqrt 3}\,(1 + c)^{3/2}.$$

With no suction at the exit this is 0.385. A diffuser that lets its air out at atmospheric pressure takes 65 per cent of what an open rotor as large as its exit would take. Its best wake is slower than the open disc’s in relative terms — 58 per cent of the wind against a third — and its rotor is loaded more lightly, a pressure drop of two-thirds of the dynamic pressure rather than eight-ninths. It passes a great deal of air through a small rotor, but the air that passes through the whole device is less than an open disc of the exit’s size would pass, because the diffuser, like the disc, has a pressure field that turns part of the wind aside before it arrives.

That is the answer to the claim in its simplest form. A turbine mounted in a plain diffuser is a smaller rotor in a larger hole, and the larger hole takes less power than a larger rotor would.

The threshold is the back face of Betz’s disc #

The third figure draws the best power against the suction. It rises steadily from 0.385, and it crosses Betz’s 16/27 at a suction of one third. The crossing is exact, and the arithmetic is short: setting $(2/3\sqrt3)(1+c)^{3/2} = 16/27$ gives $(1+c)^{3/2} = 8\sqrt3/9$, whose two-thirds power is $4/3$.

The number a third is not a coincidence, and the fourth figure shows what it is.

At the best loading, the speed through the exit plane rises with suction, from 58 per cent of the wind towards 71, and the rotor’s pressure drop rises from two-thirds to one. At a suction of a third the exit speed is exactly two-thirds of the wind and the rotor’s drop is exactly eight-ninths — the two numbers of Betz’s optimum. And the pressure on the back face of Betz’s open disc, the station just behind it in the first figure, is $1 - (2/3)^2 - 8/9 = -1/3$. At the threshold, the diffuser’s exit plane is an open Betz disc: the same speed across it, the same pressure behind it, the same wake downstream. The rotor and the flaring duct ahead of the exit are a device for reproducing, at the exit, the plane just behind an ideal disc of the exit’s size.

That reading makes the limit easy to state. An open disc is the member of this family with $\sigma = 1$ and no force on any shroud, and its own back face sits at −1/3 because the momentum theorem puts it there. A diffuser matches it on its exit area when the outside of the shroud holds the exit at the same pressure the open disc holds behind itself, and beats it only by holding more. Every per-exit-area gain over Betz is a suction bought from the flow outside the duct, beyond what an open disc makes for itself.

The model confirms the reading independently. Setting $\sigma = 1$ and choosing the exit pressure so that the shroud feels no force — a duct that is not there — recovers Betz’s disc at every loading, not only the best: its power $4a(1-a)^2$, its disc speed $1 - a$, and at the optimum an exit suction of exactly a third, produced by the model rather than put in.

The number that gets printed #

Measured on the rotor, the picture is different, and the fifth figure shows why the claim gets made. The power per rotor area is the power per exit area multiplied by $\sigma$, so it grows in proportion to how much the duct flares. A plain diffuser with twice the rotor’s area, exit at ambient, reports 0.770 on its rotor, thirty per cent above Betz — while taking less power than an open rotor the size of its exit. At a suction of a third it reports 1.185; at a half, 1.414. Nothing limits the number but how far the diffuser can flare.

What limits the flare is the adverse gradient seen in the first figure. An ideal diffuser of area ratio $\sigma$ recovers the fraction $1 - 1/\sigma^2$ of the dynamic pressure at its inlet: three-quarters at $\sigma = 2$. Real conical diffusers stay attached and recover most of that only at modest angles and area ratios; beyond an area ratio of about two the flow lets go of the wall unless the boundary layer is energised — by slots, by a second shroud, by suction. The band in the figure is where diffusers have actually been run attached. Inside it, the per-rotor-area coefficient comes out between one and two and a half times Betz for suctions up to a half, and every one of those numbers is a statement about how small the rotor is.

The per-rotor-area coefficient is not dishonest. A rotor is expensive, and a smaller rotor that delivers the same power is a real saving — the same trade the ducted propeller makes in reverse; that is the case for a diffuser, and it is a case about cost. It is not a case that the diffuser extracts more of the wind’s energy than the momentum theorem allows, and comparing it with 16/27 compares a coefficient on one area with a limit on another.

Where the suction comes from #

The comparison that is fair to the device measures its power against its frontal area — everything it puts in the wind’s way, shroud and flange included — since a larger open rotor could have filled the same area instead. The momentum balance cannot compute that directly, because the suction $c$ is set by the shroud’s outside flow, which the one-dimensional theory does not contain. It can say how much frontal area the suction is allowed to cost.

The sixth figure is that break-even area: the best power per exit area divided by 16/27. Below a suction of a third it is less than one — the exit alone is already too large, and no device of that kind can beat the limit on any area it occupies. Above it the margin is thin. At a suction of 0.4 the whole device may present 1.076 exit areas to the wind, which is a rim round the exit only 3.7 per cent of its radius. At a suction of a half, the most the formula covers, it may present 1.19 exit areas, a rim of 9 per cent.

A suction behind a bluff edge is a base pressure, the low pressure a separated wake holds behind a body, and base pressures of a third to a half of the dynamic pressure are what bluff rims and discs set up. The flanged diffusers that report the largest rotor coefficients use flanges for exactly this purpose, and a flange tall enough to hold a disc-like base pressure is not a rim of a few per cent. Whether any shroud can hold a suction above a third with a frontal area inside the break-even curve is not proved either way by anything here: the curve is a bound on the price, and the price is set by a flow this model leaves out. What the curve does say is that the question is about the outside of the shroud, and not about the rotor at all.

The load that comes with it #

The seventh figure is the cost the tower pays. The whole device’s thrust is the momentum the wind loses, the mass flow times the drop from wind speed to wake speed. An open disc at its optimum carries 8/9 of the dynamic pressure on its own area. The diffuser with its exit at ambient carries 0.488 per exit area — less, since it takes less power — and at a suction of a third it carries exactly 8/9, as the exit-is-a-Betz-disc reading requires, with half of it on the shroud. At a suction of a half it carries 1.41, 59 per cent more than the open disc, for 19 per cent more power.

The shroud is pushed downstream in every case drawn, and the share it carries grows with the suction: the suction behind the exit acts on the shroud’s own trailing surfaces as a drag. That is the other half of what the outside flow is doing. The low pressure behind the exit draws more air through the rotor and at the same time pulls the whole device downstream, and a structure sized for an open rotor of the same power is not sized for it.

What was checked #

The eighth figure is the ledger. The closed-form optimum agrees with a direct search over the wake speed to a part in $10^{16}$ at six suctions, including one past a half where the formula changes. The crossing of 16/27, found by bisection with no knowledge of the answer, lands on a third to twelve figures, with the exit speed and loading at Betz’s values. The open disc comes back as the duct with no shroud force at four loadings. And the power agrees with an energy balance made from quantities the rotor’s head never enters: the thrust times the wind speed, less the kinetic energy the wake carries off relative to the air.

What momentum theory cannot say here #

The suction. Everything in the essay is conditional on $c$, and $c$ is a property of the shroud’s outside flow — its camber, its flange, its angle to the wind. A ring-shaped aerofoil round the rotor carries circulation, and its bound vortex, as a wing’s does, induces the extra flow through the rotor and the suction at its trailing edge together; the one-dimensional balance can use the result but not produce it.

The mixing. The wake is assumed to recover to ambient pressure without loss. Behind a diffuser at a large suction the jet leaving the exit is fast and the air round it slow, and the shear layer between them mixes — a jet mixing as a pump does — which is a loss the ideal balance does not charge. At a suction of a half the best wake is at rest, and beyond it the formula asks for a wake moving upstream, which is where the ideal model has plainly stopped describing air.

The diffuser’s own losses. Friction and separation inside the duct lower the recovery the first figure assumes, and every loss inside the duct comes directly off the rotor’s pressure drop.

The swirl. A real rotor leaves a wake that has to spin, and the swirl’s energy is taken from the same head; a diffuser does not remove that cost, and a slow rotor in a fast duct pays more of it.

The yaw. A duct aligned with the wind is the best case. A diffuser turned a few degrees across the wind presents a different shape to the flow on either side, and its suction and its recovery both change.

The convention: three areas and a sign #

Power coefficients are power divided by $\frac12\rho U^3$ and an area, and the area is always named: the rotor’s, the exit’s $\sigma$ times that, or the frontal area of the whole device. Thrust coefficients are divided by $\frac12\rho U^2$ and the exit’s area unless stated. The exit suction $c$ is minus the pressure coefficient at the exit plane, referred to the free stream, so a positive $c$ is a pressure below ambient. The wake speed $x$ is the far-wake speed divided by the wind’s, after the wake has recovered to ambient pressure.

Who found it, and when #

Lilley and Rainbird, at the College of Aeronautics at Cranfield in 1956, made the first momentum analysis of a ducted windmill and found the duct raising the power through the rotor. Gilbert, Oman and Foreman at Grumman built and tested diffuser-augmented wind turbines in the 1970s and reported rotor-area coefficients well above the limit, which made the claim famous. Hansen, Sørensen and Flay in 2000 found the same with a numerical actuator disc in a diffuser and noted that the gain tracks the increase in mass flow through the rotor. van Bussel in 2007 and Jamieson in 2008 set out the momentum theory with the exit pressure as the free parameter; Ohya and his colleagues at Kyushu built flanged diffusers — the “wind lens” — from 2008 whose flange exists to deepen the base pressure behind the exit. The observation that the threshold is the open disc’s own back-face pressure follows from their equations and is set out here.

Still open: what the shroud’s outside can hold #

The whole argument hangs on a number the shroud’s external flow sets and this model takes as given. The next calculation replaces the given with a computed one: the shroud as a ring aerofoil — a vortex ring of distributed strength on a cambered section — with the rotor inside it as an actuator disc, solved together so that the ring’s circulation and the rotor’s loading are consistent. That gives the exit pressure as an output, for a stated section and flare, and the frontal area comes with the geometry. The question it answers is the one the break-even curve poses: whether any ring whose frontal area stays inside the curve holds a suction above a third, or whether every shroud that holds that much suction is larger than the open rotor that would have done the same work.