A big slow push

Thrust is a mass flow multiplied by a velocity change. That is the whole of it, and it means the same thrust can be produced in infinitely many ways: a tonne of air per second accelerated by ten metres per second, or ten kilograms per second accelerated by a kilometre per second, or anything between.

The choice is not free, and the reason is that the two sides of the product enter the power bill differently. Thrust goes as the velocity change to the first power. The energy left behind in the jet goes as the square. So the two arrangements cost wildly different amounts to produce the same push, and the number that says by how much has no engine in it at all.

The same disc, run the other way #

The previous rung built a disc that takes energy out of a stream and found a ceiling on how much. The propulsor is the identical object with the signs reversed: a surface across the flow that raises the pressure rather than lowering it, with no blades, no thickness and no viscosity in it.

The streamtube now contracts rather than widening, because the air is speeding up and the mass has nowhere to go. The induction factor is positive in the other sense: the speed at the disc is $U(1+a)$ and the far jet is at $U(1+2a)$, with the disc speed again the mean of the two ends. The thrust coefficient is $4a(1+a)$, which is the same quadratic on its other branch.

Write $\sigma$ for the ratio of jet speed to flight speed. Then

$$\sigma = \frac{u_{\text{jet}}}{U} = \sqrt{1 + C_T}$$

and the whole of the argument below is about that one number.

The audit, which is an identity #

Take the mass flow through the disc, $\dot{m}$, and follow the energy.

The power added to the stream is the rate at which its kinetic energy rises:

$$P_{\text{total}} = \tfrac{1}{2}\dot{m}\left(u_j^2 - U^2\right)$$

The useful power is the thrust multiplied by the speed the vehicle is travelling at, since that is the rate at which the thrust does work on the vehicle:

$$P_{\text{useful}} = \dot{m}(u_j - U)\,U$$

The wasted power is the kinetic energy of the jet as seen by somebody standing still — the air left behind moving at $u_j - U$ relative to the atmosphere it was taken from, and never getting it back:

$$P_{\text{wasted}} = \tfrac{1}{2}\dot{m}(u_j - U)^2$$

Add the last two and they give the first, exactly. It is an algebraic identity — a difference of two squares written as a product — and its exactness is the reason the efficiency has a closed form:

$$\eta = \frac{P_{\text{useful}}}{P_{\text{total}}} = \frac{2U}{U + u_j} = \frac{2}{1 + \sigma}$$

What was computed, and what the assertion catches #

The three powers are computed independently and the sum is asserted. That looks like checking that arithmetic works, and in a sense it is — but it is checking arithmetic that is very easy to get wrong, in a way no picture reveals.

The trap is the velocity that appears in the wasted term. It is tempting to write the jet’s kinetic energy as $\tfrac{1}{2}\dot{m}u_j^2$, which is the energy of the jet in the vehicle’s frame and is not what was wasted; or as $\tfrac{1}{2}\dot{m}(u_j - u_{\text{disc}})^2$, using the speed at the disc rather than the flight speed. Both give a number of the right order, both produce a smooth curve, and neither satisfies the identity. The assertion is the only thing standing between the figure and either of them.

The second thing checked is that the efficiency computed from the audit equals $2/(1+\sigma)$, which catches an error in the other direction, and the third is simply that it lies between nothing and everything — a propulsive efficiency above one is a machine getting more work out of the air than it put in, and it is the kind of result that a sign error produces cheerfully.

Why the trade is so lopsided #

Hold the thrust fixed and let the disc grow.

At a small disc the air has to be thrown hard: a large $\sigma$, a large velocity change, and a wasted term that goes as its square. At a large disc the same thrust is had from a velocity change so small that almost nothing is left in the wake, and $\eta$ approaches one.

The numbers on that curve are worth putting into the units an airline uses. A large turbofan at cruise moves roughly 500 kilograms of air per second and accelerates it by something like 60 metres per second, giving about 30 kilonewtons of thrust at a flight speed near 250 m/s. The jet-to-flight speed ratio is therefore about 1.24, and the propulsive efficiency about 89 per cent. A turbojet of the 1950s producing the same thrust from a fifth of the mass flow would have needed a velocity change five times larger, a $\sigma$ near 2.2, and a propulsive efficiency of 62 per cent — before any difference at all in the quality of the machinery.

That gap of nearly thirty points, taken on the fuel bill of every flight, is what the bypass ratio bought. It is also why the improvement stalled: the curve is flattening. Going from $\sigma = 2.2$ to 1.24 was worth 27 points; going from 1.24 to 1.10 is worth another five, and each of those points now costs a larger fan, a heavier nacelle and more room under a wing that is already close to the ground.

The asymptote is worth stating plainly. In the limit of an infinitely large disc, thrust is free of wake loss. Not free of power — the useful power $T U$ is still there and still has to come from somewhere — but free of the additional cost of having made the thrust in the first place. That limit is unreachable and the approach to it is what an engineer is buying.

Two consequences, both visible on any airfield:

• A propeller is a very large actuator disc. A 2-metre propeller on a light aircraft at 60 m/s is moving a column of air far wider than the engine, with a jet perhaps ten per cent faster than the flight speed, and a propulsive efficiency in the high eighties.
• A high-bypass turbofan is a propeller with a duct round it. The bypass ratio — the mass of air going round the core divided by the mass going through it — rose from zero in the 1950s to over ten today, and every increment of it is a move to the right on the trade curve above. The core got better too, but the core’s improvement is a thermal efficiency and it multiplies this one rather than replacing it.

The same identity, said as a momentum argument #

There is a second route to the result, and it is worth taking because it connects this essay to one of the site’s oldest arguments.

Air must be pushed down makes the case that the popular momentum account of lift names the right mechanism and does the accounting badly. The propulsive version is the same shape and the accounting comes out. A vehicle in steady flight is pushing a stream of air backwards; the rate of momentum given to that air is the thrust, by Newton’s third law and nothing else. The energy given to it, however, is not fixed by the thrust — it depends on how the momentum was delivered, and that is where all the choice lives.

Write it as a ratio and the identity falls out in one line. If the same momentum $\dot{m}\Delta u$ is delivered by doubling $\dot{m}$ and halving $\Delta u$, the useful power $\dot{m}\Delta u\,U$ is unchanged and the wasted power $\tfrac{1}{2}\dot{m}\Delta u^2$ is halved. Doubling the disc area halves the waste, every time, at every operating point. There is no diminishing return in the statement itself — the diminishing return is entirely in what a larger fan weighs, what it costs in drag, and whether it fits under the wing.

That is a satisfying place for a control-volume argument to end up: the thing the model cannot see — the mass, the nacelle, the ground clearance — is precisely what decides the answer in practice, and the thing it can see is exactly the part that no amount of engineering will move.

The case where the ratio has no meaning #

The whole essay is written in terms of $\sigma = u_j/U$, and there is an important machine for which that ratio is infinite and the efficiency is exactly zero: a rotor in a hover. A helicopter holding station does no work on itself, by the same argument that gives a rocket on the pad an efficiency of nothing, and yet the difference between a good hovering rotor and a bad one is the difference between an aircraft and a pile of parts.

So the identity has to be read differently when the flight speed vanishes, and what it gives is not an efficiency but a price. With no oncoming stream, the disc has to generate its own: the air arrives at the rotor at the induced velocity $v$, which by the momentum balance is

$$v = \sqrt{\frac{T}{2\rho A}}$$

and the power required is $T v$, so

$$P = \frac{T^{3/2}}{\sqrt{2\rho A}}.$$

The useful form of that is its reciprocal, the power loading — newtons of lift per watt burned:

$$\frac{T}{P} = \sqrt{\frac{2\rho}{T/A}}.$$

Everything about hovering is in that square root. The only property of the machine that appears is $T/A$, the disc loading, and the price of lift falls as its square root — which is the hover’s version of the trade curve above, with an exponent instead of a ratio.

The arithmetic is stark. A helicopter carries about 300 newtons per square metre of rotor, which gives an ideal power loading near 0.09 N/W — eleven watts to hold up each newton, so a five-tonne machine needs something over half a megawatt before any losses are counted. A jet-lift aircraft hovering on its exhaust is at perhaps thirty times that disc loading, so its power loading is worse by a factor of five or six, and it burns fuel at a rate that makes hovering something done for seconds rather than hours. Neither figure has anything to do with the quality of the engine; both are the same statement that a large slow push is cheap and a small fast one is not.

And this is why the hovering literature does not quote an efficiency at all. It quotes a figure of merit: the ideal power above divided by the power actually required, so that a rotor is measured against what the momentum theory says its own disc could achieve rather than against a ratio that is identically zero. Good rotors reach about 0.75 to 0.80 of it. That is a considered response to the identity rather than a way around it — the disc’s answer is kept as the benchmark, and the machine is scored against the one number in the problem that no machine can change.

Where the disc’s answer stops being the whole answer #

The efficiency above is not the efficiency of an engine. It is the efficiency with which power in the stream is converted into useful work on the vehicle, and an engine has a second efficiency in front of it — how much of the fuel’s chemical energy becomes power in the stream at all. The overall figure is the product, and the two pull in opposite directions with flight speed:

• Thermal efficiency rises with jet speed, because a hotter, faster core is a better heat engine.
• Propulsive efficiency falls with jet speed, by the identity above.

So there is an optimum, it moves with flight speed, and it is why the answer changes with the aircraft. At 250 m/s a large fan wins. At Mach 2 the jet has to be fast simply to exceed the flight speed at all, $\sigma$ cannot be small, and the propulsive efficiency of a supersonic engine is structurally poor — which is a fact about the identity rather than about Concorde.

That last case is worth pausing on, because it looks like a paradox and is not one. A rocket on the pad produces enormous thrust and does no work, because work is force times distance moved and nothing is moving. As it accelerates the efficiency climbs, reaches its maximum where the vehicle speed equals the exhaust speed — at which point $\sigma = 1$ and $\eta = 1$, and the exhaust is left at rest in the atmosphere with no kinetic energy at all — and falls again beyond. That maximum is a real and famous effect, and it is the same identity read at a different point.

Three further things the disc does not contain:

• It has no duct. A ducted fan and an open propeller have different streamtubes, and a duct that is doing work on the flow changes the momentum balance. The disc is the open case.
• Its wake does not spin. A real propeller applies a torque and leaves swirl behind, which is wasted in exactly the way the axial jet is wasted, and it is what the next rung computes.
• Nothing here is compressible. At tip speeds approaching the speed of sound a propeller’s blades develop their own wave drag, and that is the real reason propellers stop being used at high speed rather than any failure of this argument. The propulsive efficiency of a propeller at Mach 0.85 would be excellent if the blades survived.

Who found it, and when #

William Froude published the momentum theory of the screw propeller in 1878, and his son Robert Edmund Froude completed it in 1889 with the observation that half the acceleration happens ahead of the disc — the same statement that fixes the disc velocity in the extraction problem. Rankine had the outline in 1865.

All of it predates the aeroplane. The question was ships, the fluid was water, and the practical matter at hand was why a large slow screw beat a small fast one on the same engine. It is one of the tidiest cases in this subject of a result outliving its motivation completely: the argument was made about propellers in water in the 1870s, was applied unchanged to aircraft propellers in the 1920s, decided the architecture of the jet engine in the 1960s, and is the reason a modern airliner’s fan is so much larger than the machine turning it.

The elder Froude is better known on this site for a different number, which is also his and is also about ships. Both are ratios; neither is about the vessel.

There is a small historical irony in the sequence. The momentum theory was worked out for marine screws, where the fluid is a thousand times denser than air and a modest propeller therefore moves an enormous mass flow; ship propellers have always sat far to the right on the trade curve, at efficiencies in the seventies once the hull’s own interference is counted, and the theory was telling their designers something they already believed. It took the aeroplane — where the fluid is thin, the mass flow is hard-won, and the temptation to make a small fast jet is strong — to turn the same identity into a constraint that anybody had to argue with.

Where the ladder goes #

The disc has now been run both ways and the two answers have the same character: a ceiling in one direction and an efficiency in the other, both exact, both containing no machine.

The third rung breaks it. A disc cannot apply a torque — it has no blades to apply one with — and a real rotor must. The reaction leaves the wake spinning, that rotational energy is taken from the stream and never reaches the shaft, and the axial theory has nowhere to put it. Glauert’s optimum rotor, which does account for it, sits below Betz’s ceiling at every finite tip-speed ratio, and the gap is computed there rather than estimated.

That rung is also where the field’s method shows its limit. A control volume does not need to know what is inside it, which is what makes these results so strong — but it also reveals nothing the box’s faces do not carry. Angular momentum crosses those faces, and the moment it is put back in, the answer changes.