Thermal, Propulsive and Overall Efficiency

Where the fuel's energy goes, the Froude efficiency, the thrust–efficiency trade and the case for bypass

12 March 1961

American Airlines puts a new version of the Boeing 707 into service: the 707-123B. Its engines are Pratt & Whitney JT3Ds. The company had taken the gas generator of its JT3 turbojet — the civil J57 — and added a two-stage fan in front of it, blowing a second stream of air around the core. Pratt & Whitney even sold kits so that airlines could convert their existing JT3C turbojets in the overhaul shop (the Smithsonian's JT3D). The converted engines gave more take-off thrust, burned less fuel and made less noise — from the same core. The fan's jet was much slower than the turbojet's. The last lesson ended with thrust rising with jet velocity. This one explains why the industry spent the next sixty years making jets slower.

Follow the energy. The fuel brings chemical energy

\dot m_f h_{PR}

, where

h_{PR}

is the heating value (about 43 MJ/kg for Jet A). The engine turns part of it into kinetic energy of the flow — the jet leaves faster than the air arrived. Part of that kinetic energy becomes thrust power

F V_0

, the useful work done on the aircraft each second; the rest stays behind in the wake as moving air.

Two conversions, two efficiencies, and their product.

\eta_{th} = \frac{\tfrac12 \dot m_9 V_9^2 - \tfrac12 \dot m_0 V_0^2}{\dot m_f\, h_{PR}}, \qquad \eta_p = \frac{F\,V_0}{\tfrac12 \dot m_9 V_9^2 - \tfrac12 \dot m_0 V_0^2}, \qquad \eta_0 = \eta_{th}\,\eta_p = \frac{F\,V_0}{\dot m_f\, h_{PR}}

Thermal efficiency (fuel → kinetic energy), propulsive efficiency (kinetic energy → thrust power) and overall efficiency (fuel → thrust power), for a perfectly expanded nozzle.

The thermal efficiency is a property of the thermodynamic cycle: how much of the fuel's energy appears as kinetic energy instead of leaving as heat in a hot exhaust. Unit 3 shows that it is set mainly by the compressor pressure ratio.

The propulsive efficiency is a property of the jet: how much of the kinetic energy given to the flow actually pushes the aircraft. It has a famous simple form.

\eta_p \approx \frac{2}{1 + V_9/V_0}

The Froude propulsive efficiency, for f ≪ 1 and a perfectly expanded nozzle. It tends to 1 as the jet velocity approaches the flight speed.

Derivation

The Froude efficiency

With the fuel neglected (

f \ll 1

, so

\dot m_9 \approx \dot m_0 = \dot m

) and

p_9 = p_0

, the thrust is

F = \dot m (V_9 - V_0)

and the kinetic power added to the flow is

\tfrac12 \dot m V_9^2 - \tfrac12 \dot m V_0^2 = \tfrac12 \dot m (V_9 - V_0)(V_9 + V_0).

Divide the thrust power by it:

\eta_p = \frac{\dot m (V_9 - V_0)\,V_0}{\tfrac12 \dot m (V_9 - V_0)(V_9 + V_0)} = \frac{2 V_0}{V_9 + V_0} = \frac{2}{1 + V_9/V_0}.

The mass flow and the velocity jump cancel: propulsive efficiency depends only on the ratio of jet to flight velocity. At

V_9 = 3V_0

it is 50 %; at

V_9 = 1.5 V_0

it is 80 %.

📐The Mach 2 engine's efficiency chain

Continue the engine of Lesson 1-1:

\dot m_0 = 25

kg/s,

f = 0.03

V_0 = 598.9

m/s,

V_9 = 1050

m/s,

F = 12\,066

h_{PR} = 42.8

MJ/kg.

Fuel power: $0.75 \times 42.8 \times 10^6 = 32.1$ MW
Kinetic power added: $\tfrac12 (25.75)(1050)^2 - \tfrac12 (25)(598.9)^2 = 9.71$ MW → $\eta_{th} = 30.3\,\%$
Thrust power: $12\,066 \times 598.9 = 7.23$ MW → $\eta_p = 74.4\,\%$ (Froude's formula, which ignores the fuel's mass, says 72.6 %)
Overall: $\eta_0 = 7.23/32.1 = 22.5\,\%$

Seven-tenths of the fuel's energy leaves as heat in the exhaust; of the rest, a quarter is left behind as a fast wake.

💡The trade you cannot escape

For a given mass flow, a faster jet gives more thrust — the specific thrust

F/\dot m = V_9 - V_0

rises — but a lower propulsive efficiency. You cannot have both from the same air. Turn it around. An aircraft in cruise needs a fixed thrust (equal to its drag). With

F = \dot m (V_9 - V_0)

fixed, the kinetic power left in the wake is

P_{wake} = \tfrac12 \dot m (V_9 - V_0)^2 = \frac{F^2}{2\,\dot m}.

Double the mass flow and you halve the energy thrown away. The only route to high propulsive efficiency at a given thrust is to move more air.

Derivation

The wake power at fixed thrust

The wake carries the kinetic energy of the air relative to the undisturbed atmosphere. Far behind the engine the jet moves backward at

V_9 - V_0

relative to still air (in the ground frame), so each second it leaves behind

P_{wake} = \tfrac12\,\dot m\,(V_9 - V_0)^2 .

Substitute the thrust,

V_9 - V_0 = F/\dot m

P_{wake} = \tfrac12\,\dot m\left(\frac{F}{\dot m}\right)^2 = \frac{F^2}{2\dot m}.

The kinetic power the engine gives the flow is the useful part plus this loss,

F V_0 + F^2/(2\dot m)

, and dividing the useful part by the total gives the Froude efficiency again:

\eta_p = 1/\left(1 + F/(2\dot m V_0)\right)

It is possible to fly without motors, but not without knowledge and skill.

– Wilbur Wright, letter to Octave Chanute, 13 May 1900 (Wikiquote)

The bypass idea. A turbojet makes its power and spends it on the same small stream of air, which leaves very fast. A turbofan separates the two jobs. The core — compressor, combustor, turbine — is a gas generator. An extra turbine stage takes power out of its hot gas and drives a fan, which gives a modest velocity increase to a much larger stream of air that bypasses the core. The bypass ratio is the bypass mass flow divided by the core mass flow.

The JT3D's bypass ratio was about 1.4. Modern airliner engines run at 10 to 12.5: the GE9X at 10, Pratt & Whitney's geared PW1100G at 12.5. Each step slows the average jet, raises the propulsive efficiency, and cuts noise, which grows with a high power of jet velocity (Lesson 5-4).

🕰Two levers, sixty years

Overall efficiency of airliner engines has roughly doubled since the first jets, from below 20 % to around 40 % in cruise. Two levers did it. Thermal efficiency rose with core pressure ratio and turbine entry temperature (Units 3, 4 and 6). Propulsive efficiency rose with bypass ratio. Today the propulsive lever is running out of room — a bigger fan means a heavier engine and a draggier nacelle — which is why the newest designs gear the fan (Pratt & Whitney's GTF) or take the nacelle off altogether (open rotors, Lesson 10-2).

💡Where you will use this

Lesson 3-2 derives the ideal turbofan and its optimum fan pressure ratio from exactly this trade. Unit 4 puts component losses into the thermal efficiency. Lesson 10-1 follows adaptive-cycle military engines that switch between a high-specific-thrust mode and a high-efficiency mode in flight.

📚Further reading

Nicholas Cumpsty and Andrew Heyes, Jet Propulsion (3rd ed., Cambridge University Press, 2015), chapter 2, on propulsive efficiency and why bypass wins.
Alan H. Epstein, 'Aeropropulsion for Commercial Aviation in the Twenty-First Century and Research Directions', AIAA Journal 52 (5), 2014 — the efficiency levers left, by a former Pratt & Whitney chief technologist.
Philip G. Hill and Carl R. Peterson, Mechanics and Thermodynamics of Propulsion (2nd ed., Addison-Wesley, 1992), chapter 5.

↓But there is a problem

Efficiencies are ratios. An airline's accountants see something else: kilograms of fuel per hour, per tonne of thrust, and the distance between two cities. In October 2019 an almost empty Boeing 787 took off from New York with its tanks full, to see whether it could reach Sydney without stopping. Whether it could depended on three numbers — and only one of them belonged to the engines.