Fundamentals8 min read1 / 9

Thrust is momentum

Every jet engine — turbojet, giant fan, ramjet, all of them — does one thing: it takes air moving at flight speed and throws it out the back faster. Newton does the rest. Before any thermodynamics, this article derives the thrust equation and the single trade-off that explains why six different engine families exist at all.

The control volume

Draw a box around the engine and do the momentum bookkeeping (Newton’s second law for a steady flow: force equals the rate of change of momentum passing through). Air enters the front at the flight velocity $V_0$ and leaves at the exhaust velocity $V_e$ , with a mass flow $\dot m$ kilograms per second:

F = \dot m\,(1+f)\,V_e - \dot m\,V_0 \;\approx\; \dot m\,(V_e - V_0)

(1)

$\dot m$: air mass flow through the engine [kg/s]
$f$: fuel/air ratio — ~2% for a jet engine, hence the ≈ [–]
$V_0$: flight velocity (freestream) [m/s]
$V_e$: exhaust velocity, fully expanded to ambient pressure [m/s]

Fig. 1. Momentum in, momentum out. The imbalance is a forward force on whatever caused it — the engine. (A pressure term appears if the nozzle exit pressure differs from ambient; the ideal analysis expands the jet perfectly, so it vanishes.)

The price of the shove

Thrust is momentum, but fuel buys energy. The engine’s useful output is thrust power $F\,V_0$ ; what it had to generate is the change of kinetic energy of the stream. Their ratio is the propulsive efficiency:

\eta_p = \frac{F\,V_0}{\tfrac12\dot m (V_e^2 - V_0^2)} = \frac{\dot m (V_e - V_0)\,V_0}{\tfrac12\dot m (V_e - V_0)(V_e + V_0)} = \frac{2}{1 + V_e/V_0}

(2)

(The difference of squares factorises, $\dot m$ and $(V_e - V_0)$ cancel, and a famously clean result drops out.) Read it carefully:

$V_e \gg V_0$ : lots of thrust per kg/s, terrible efficiency — most of the fuel’s work is left behind as a hot, fast wake still churning after the aircraft has gone.
$V_e \to V_0$ : efficiency → 100%, but thrust per kg/s → 0. To get useful thrust you must move an enormous mass flow — a bigger fan, a bigger propeller.

Fig. 2. Equation (2). The three markers are real design points: a pure turbojet throws a small flow out ~3× faster than flight; a modern high-bypass fan barely 15% faster — and collects the efficiency reward.

Feel the trade yourself:

Flight speed

V_0

250 m/sExhaust speed

V_e

600 m/s

F/\dot m = 600 - 250

350\ \tfrac{\text{N}}{\text{kg/s}},\quad \eta_p = 59\,\%

Small velocity jump: less thrust per kg/s of air, but most of the jet energy becomes useful work. To keep total thrust up, move more air — the whole case for the big fan.

Where the rest of the course goes

Equation (1) turned propulsion into a question: how fast can you afford to throw, and how much can you afford to swallow? The next articles supply the machinery — the atmosphere the engine breathes, the Brayton cycle that generates $V_e$ from fuel, and then each engine family as a different answer to the same question.