Advanced10 min read7 / 9

The efficiency chain & range

Every engine demo ends its readout with three bars — thermal, propulsive, overall — because jet performance is a relay race: the cycle turns fuel heat into jet kinetic energy, then the airframe turns jet kinetic energy into useful work. Each handoff loses something, the losses multiply, and one compact equation connects the result to how far an aircraft can fly.

The chain

Fig. 1. The chain drawn to scale for a modern turbofan: the width of each band is the fraction of the fuel's power still flowing. The cycle converts half of it to jet kinetic energy (the rest leaves as exhaust heat), and the airframe captures 80% of that as useful thrust power (the rest stays in the sky as wake motion). Stage one is thermodynamics (the Brayton article); stage two is mechanics (the momentum article) — and the survivor is the product of the two.

\eta_{th} = \frac{\tfrac12 \dot m\,(V_e^2 - V_0^2)}{\dot m_f\, h_{PR}}, \qquad \eta_p = \frac{F\,V_0}{\tfrac12 \dot m\,(V_e^2 - V_0^2)}, \qquad \eta_o = \eta_{th}\,\eta_p = \frac{F\,V_0}{\dot m_f\,h_{PR}}

(1)

Notice how cleanly the middle term cancels when you multiply: overall efficiency never mentions the jet at all — it is simply useful thrust power over fuel power. The two stages pull against each other through $V_e$ : a hotter, harder cycle makes $V_e$ (and $\eta_{th}$ ) bigger, but a bigger $V_e$ drags $\eta_p$ down. The turbofan is the industry’s resolution: keep the ferocious core for $\eta_{th}$ , launder its output through a fan for $\eta_p$ .

TSFC is the chain in disguise

Rearrange the last form of (1):

\text{TSFC} = \frac{\dot m_f}{F} = \frac{V_0}{\eta_o\,h_{PR}}

(2)

$h_{PR}$: fuel heating value (43 MJ/kg jet fuel) [J/kg]
$\eta_o$: overall efficiency: ~25% for 1960s turbojets, ~40% for modern fans [–]
$V_0$: flight speed — note TSFC grows with it even at fixed η [m/s]

Two lessons hide in (2). First, at fixed flight speed, TSFC is a pure inverse measure of $\eta_o$ — which is why engineers treat them as synonyms. Second, comparing TSFC across speeds is unfair: a Mach 2 engine hauls each newton through twice the distance per second, so it may burn more per newton while wasting less per metre. Between engines at the same cruise point, though, the demo’s TSFC bar is exactly the number airlines buy on.

From efficiency to range: Breguet

Why does a fraction of a percent of $\eta_o$ matter? Integrate fuel burn along a cruise at constant lift-to-drag ratio $L/D$ and constant $\eta_o$ , with the aircraft lightening as it burns (mass $m_i$ down to $m_f$ ), and you get the Breguet range equation:

R = \eta_o\,\frac{h_{PR}}{g}\,\frac{L}{D}\,\ln\frac{m_i}{m_f}

(3)

Range is a product of exactly three technologies: propulsion ( $\eta_o$ ), aerodynamics ( $L/D$ — the finite-wing article of the aerodynamics course), and structures (the mass ratio in the logarithm). Each community owns one factor; the logarithm explains why squeezing out fuel weight pays less and less while engine efficiency pays linearly, forever.