Thrust & Propulsive Efficiency — Newton's Laws at 35,000 Feet
Thrust & Propulsive Efficiency — Newton's Laws at 35,000 Feet
How does burning fuel in a tube push an aircraft weighing 400 tonnes through the air at 900 km/h? The answer is Newton's third law — but quantified with momentum equations that turn into elegant engineering formulas.
Thrust from Momentum Change
A jet engine is a momentum machine. Air enters at flight velocity $V_0$, is heated and accelerated, and exits at jet velocity $V_e$. The thrust force is simply the rate of momentum change:
$$F = \dot{m}_a (V_e - V_0) + \dot{m}_f V_e \approx \dot{m}_a (V_e - V_0)$$
where $\dot{m}_a$ is the air mass flow rate (kg/s) and $\dot{m}_f$ is the fuel mass flow rate. For most purposes, since the fuel-to-air ratio is small ($f \approx 0.02$–$0.04$), we simplify to:
$$F \approx \dot{m} (V_e - V_0)$$
Specific Impulse and Specific Thrust
Specific thrust normalises thrust by air mass flow: $F_{sp} = F/\dot{m}$ (N·s/kg or m/s). Higher specific thrust means a smaller, lighter engine for the same thrust — useful for fighters. Lower specific thrust (high bypass turbofans) means better fuel economy — the right choice for airliners.
Specific impulse normalises thrust by fuel weight flow:
$$I_{sp} = \frac{F}{\dot{m}_f g_0}$$
Units are seconds — a universal metric for all propulsion systems from rockets to turbofans. A modern turbofan has $I_{sp} \approx 5{,}000$–$7{,}000$ s; a rocket is only $\approx 300$–$450$ s. Air-breathing engines are dramatically more fuel-efficient because they don't carry their own oxidiser.
Propulsive Efficiency — Why Slow Jets Beat Fast Jets
The fraction of jet kinetic energy that actually becomes useful thrust power is the propulsive efficiency:
$$\eta_P = \frac{2V_0/V_e}{1 + V_0/V_e}$$
This has a beautiful implication: $\eta_P \to 1$ as $V_e \to V_0$. The most efficient engine ejects air at just barely above flight speed — moving a huge mass of air by a tiny amount, rather than a small mass at high speed. This is why modern turbofans have bypass ratios of 10:1 or higher, moving enormous quantities of air around the core at low excess velocity.
Aircraft at $V_0 = 250$ m/s. Turbojet: $V_e = 600$ m/s. High-bypass turbofan: $V_e = 350$ m/s. Compare propulsive efficiencies.
- Turbojet: $\eta_P = \frac{2 \times 250/600}{1 + 250/600} = \frac{0.833}{1.417} = 58.8\%$
- Turbofan: $\eta_P = \frac{2 \times 250/350}{1 + 250/350} = \frac{1.429}{1.714} = 83.3\%$
- The turbofan is 42% more propulsively efficient — which translates directly into fuel savings.
- That's why every modern airliner uses high-bypass turbofans.
Overall engine efficiency = thermal efficiency × propulsive efficiency × combustion efficiency. Modern turbofans achieve overall efficiencies approaching 50% — extraordinary for any heat engine. Every improvement of 1% in efficiency saves a long-haul airline roughly $1 million per aircraft per year in fuel costs.