Turboprops and Propeller Theory

Shaft-power engines, actuator-disk momentum theory, ideal propeller efficiency, power loading and static thrust

Hovering on a shaft

To land vertically, the F-35B takes 29,000 horsepower off its engine's fan shaft. An extension shaft, a clutch and a bevel gearbox carry it to a two-stage, contra-rotating fan 50 inches (1.3 m) across, buried in the fuselage behind the cockpit. The fan gives about 20,000 lbf (89 kN) of cold thrust, almost half of the 41,900 lbf the whole lift system produces in the hover (the LiftSystem). Why a fan, and not just the engine's hot jet pointed downward? For the reason this lesson quantifies: a large disk pushing a lot of air gently gives far more thrust per unit power than a small, fast jet. It is the reason helicopters have big rotors, and the reason turboprops fly most short routes.

Some gas turbines deliver almost all of their power as shaft power instead of jet velocity. In a turboprop a power turbine — often a separate, free turbine on its own shaft — drives a propeller through a reduction gearbox; the exhaust still gives a little jet thrust. A turboshaft drives a helicopter rotor; the F-35B's lift fan is the same idea built into a fighter.

The core of such an engine is analysed exactly like a turbojet's. The new question is the propeller: how much thrust a disk makes from a given power.

💡The actuator disk

Replace the propeller by a thin disk of area

A

that adds a uniform pressure jump to the air passing through it. The air speeds up from

V_0

far ahead to

V_p

at the disk and

V_w

in the far wake, and the stream tube contracts as it does. Three results follow from momentum, energy and mass alone:

The velocity at the disk is the mean of the far-upstream and far-wake velocities: $V_p = (V_0 + V_w)/2$ . Half the speed-up happens before the air reaches the disk.
The thrust is $T = \rho A V_p(V_w - V_0) = 2\rho A V_p\,v$ , where $v = V_p - V_0$ is the induced velocity.
The power absorbed is $P = T V_p$ , so the ideal efficiency is $\eta_{ideal} = V_0/V_p$ .

\eta_{ideal} = \frac{T V_0}{P} = \frac{2}{1 + V_w/V_0} = \frac{2}{1 + \sqrt{1 + C_T}}, \qquad C_T = \frac{T}{\tfrac12\rho V_0^2 A}

The ideal propeller efficiency: the Froude efficiency of Lesson 1-2 written for a disk, and in terms of the thrust coefficient CT.

Derivation

Momentum theory of the actuator disk

Momentum. The mass flow through the disk is

\rho A V_p

; it leaves at

V_w

, so

T = \rho A V_p(V_w - V_0)

Pressure. Bernoulli holds separately ahead of and behind the disk (energy is added only at the disk). Ahead:

p_0 + \frac12\rho V_0^2 = p_1 + \frac12\rho V_p^2

. Behind:

p_2 + \frac12\rho V_p^2 = p_0 + \frac12\rho V_w^2

. Subtracting, the pressure jump is

p_2 - p_1 = \frac12\rho(V_w^2 - V_0^2)

, and

T = A(p_2 - p_1)

Equate the two thrusts:

\rho A V_p (V_w - V_0) = \tfrac12\rho A (V_w - V_0)(V_w + V_0) \quad\Longrightarrow\quad V_p = \frac{V_0 + V_w}{2}.

The power is the kinetic energy added per second,

\frac12\rho A V_p(V_w^2 - V_0^2) = T V_p

, which gives

\eta_{ideal} = V_0/V_p = 2/(1 + V_w/V_0)

. Writing

T

in terms of

C_T

gives

V_w = V_0\sqrt{1 + C_T}

. Lab 3 does the whole derivation in SymPy.

📐The deck's propeller

A propeller 1.8 m in diameter makes 6.2 kN at 75 m/s near sea level (

\rho = 1.22

kg/m³). With

A = 2.545

m²,

T = 2\rho A V_p(V_p - V_0)

is a quadratic in

V_p

(Lab 1):

$V_p = 86.5$ m/s at the disk; the far wake leaves at 98.1 m/s; the induced velocity is 11.5 m/s.
Shaft power $P = T V_p = 536.5$ kW; ideal efficiency $75/86.5 =$ 86.7 %.

The same disk making the same thrust at rest needs an induced velocity of 31.6 m/s and 196 kW — and in flight terms its efficiency is zero, because it is not going anywhere.

💡Power loading and static thrust

At rest,

T = 2\rho A v^2

and

P = T v

, so

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

Thrust per unit power grows with disk area: the same thrust from twice the area needs

1/\sqrt2

of the power. That is why rotors and lift fans are as big as the aircraft allows. The F-35B's fan makes 4.1 N per kilowatt; an ideal open rotor of its size would need 15.1 MW instead of 21.6, and an ideal ducted fan, whose duct holds the wake at the fan's own area, only 10.7 MW (Lab 1).

📐A turboprop's thrust lapse

Hold the shaft power at 536.5 kW and vary the flight speed (Lab 2, a cubic solved with numpy.roots):

At rest: 12.1 kN.
50 m/s: 7.8 kN, ideal efficiency 73 %.
75 m/s: 6.2 kN, 86.7 %.
125 m/s: 4.1 kN, 96 %.

At constant power, thrust falls roughly as

1/V_0

once the aircraft is moving: superb take-off thrust, fading with speed. Real propellers add blade drag, swirl and tip losses, reaching 80–88 % in cruise, and their tips approach the speed of sound as the aircraft speeds up — which is why most turboprops cruise below about Mach 0.65.

For two reasons we decided to use two propellers. In the first place we could, by the use of two propellers, secure a reaction against a greater quantity of air, and at the same time use a larger pitch angle than was possible with one propeller; and in the second place by having the propellers turn in opposite directions, the gyroscopic action of one would neutralize that of the other.

– Orville Wright, 'How We Made the First Flight', in The Early History of the Airplane (Wikiquote)

💡The same disk, run backwards

Let the disk take energy out of the wind instead of adding it, and the same equations describe a wind turbine. Maximising the power extracted gives Betz's limit: no turbine can capture more than 16/27 ≈ 59 % of the kinetic energy flowing through its disk (Betz's law).

💡Where you will use this

Lesson 10-2 applies momentum theory to geared fans and open rotors, the next steps beyond the turbofan. The same theory sizes helicopter rotors and drone propellers, and — reversed — wind turbines.

📚Further reading

J. Gordon Leishman, Principles of Helicopter Aerodynamics (2nd ed., Cambridge University Press, 2006), chapter 2, on momentum theory.
Barnes W. McCormick, Aerodynamics, Aeronautics, and Flight Mechanics (2nd ed., Wiley, 1995), on propellers.
Momentum theory on Wikipedia; NASA Glenn Research Center, Propeller analysis.

↓But there is a problem

Every engine in this unit was perfect: compressors and turbines without losses, burners without pressure drop, nozzles expanding exactly to ambient pressure, propellers with no blade drag. None of that exists. For decades the obstacle to the gas turbine was not thrust at all: early machines could not even make enough power to drive their own compressors. The next unit puts the losses in.