Turbojet Thrust⚠ unverified
Aerospace / Propulsion · Compute the net thrust of a turbojet engine (simplified model)
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| mdot | mdot | kg/s | 1.0 | Air mass flow rate through the engine |
| Ve | Ve | m/s | 1.0 | Exhaust exit velocity |
| V0 | V0 | m/s | 1.0 | Free-stream (flight) velocity |
| mfuel | mfuel | kg/s | 0.0 | Fuel mass flow rate, in kilograms per second (kg/s). Default is 0 |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | F | N | Net thrust, in newtons (N) |
The science & history
Understanding the Parameters
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Air mass flow $\dot m$ — the air the engine processes per second, set by inlet area, flight speed, and density. It is the raw quantity of "working fluid," and thrust scales directly with it — which is why high-thrust engines have large frontal areas (and why turbofans move even more air at lower velocity for efficiency).
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Exhaust velocity $V_e$ vs flight velocity $V_0$ — the engine earns thrust only from the difference $V_e - V_0$. This is the crucial air-breathing distinction: a rocket's exhaust momentum is all thrust, but a jet's incoming air already carries momentum $\dot m V_0$ that must be subtracted. As the aircraft speeds up, $V_0$ rises toward $V_e$ and the momentum term shrinks — thrust falls with airspeed, and would vanish if $V_0$ ever reached $V_e$.
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Fuel mass flow $\dot m_f$ — the small extra mass added in the combustor (typically $\dot m_f/\dot m \sim 0.02$–$0.04$). It leaves with the exhaust, contributing its own momentum $\dot m_f V_e$. Often neglected to first order, but the registry keeps it explicit; its default is $0$, giving the "cold" air-only estimate.
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Uninstalled, perfectly expanded. This form omits the pressure-thrust term $(P_e - P_0)A_e$ (assumes the nozzle expands to ambient) and installation/inlet drag. Real installed net thrust is lower.
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Default caveat. All flow defaults are $1.0$ (fuel $0$), giving $F = 1\cdot(1-1) + 0 = 0\,\text{N}$ — a placeholder until real values are entered.
Derivation (Approaching a Proof)
Apply a steady momentum balance to a control volume around the engine, cutting the free stream far upstream (air enters at $V_0$) and the nozzle exit (gas leaves at $V_e$). The net axial force — the thrust reacted onto the airframe — equals momentum out minus momentum in.
Air enters at rate $\dot m$ carrying momentum flux $\dot m\,V_0$. Fuel is injected (with negligible axial momentum of its own). The total gas leaving is $(\dot m + \dot m_f)$ at velocity $V_e$, carrying momentum flux $(\dot m + \dot m_f)V_e$. Therefore
$$F = (\dot m + \dot m_f)\,V_e - \dot m\,V_0 = \dot m\,(V_e - V_0) + \dot m_f\,V_e. \qquad\blacksquare$$
The grouping in the registry form separates the air momentum increment $\dot m(V_e - V_0)$ from the fuel-addition term $\dot m_f V_e$. Neglecting the small fuel term recovers the familiar approximation $F \approx \dot m(V_e - V_0)$, which already captures the essential physics: thrust is air flow times velocity increment.
This also exposes the propulsive-efficiency trade. Thrust power is $F V_0$; the kinetic energy dumped into the jet is wasted. Producing a given thrust with a large $\dot m$ and small $(V_e - V_0)$ wastes less energy than a small $\dot m$ and large excess velocity — the entire rationale for the high-bypass turbofan.
Dimensional check. $$[\dot m (V_e - V_0)] = \frac{\text{kg}}{\text{s}}\cdot\frac{\text{m}}{\text{s}} = \frac{\text{kg}\cdot\text{m}}{\text{s}^2} = \text{N}.\ \checkmark$$ Both terms are momentum-flux forces in newtons.
History and Development
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Whittle and von Ohain. The turbojet was invented independently by Frank Whittle in Britain (patent 1930, first run 1937) and Hans von Ohain in Germany (first flight, the He 178, 1939). Both recognised that a gas turbine could compress air, add heat, and expand it through a nozzle to make a high-velocity jet — the momentum equation above made flesh.
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The airspeed penalty and the turbofan. The $(V_e - V_0)$ term means a pure turbojet loses thrust and efficiency at high subsonic speed. The turbofan (1960s) fixed this by using a large fan to accelerate a big bypass mass of air to modest velocity, maximising $\dot m$ while minimising wasted jet kinetic energy — transforming airline economics.
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Ram limits and the ramjet. As $V_0$ climbs supersonically, ram compression alone becomes enough to run an engine with no turbomachinery — the ramjet — while the turbojet's rotating machinery eventually limits it. The two engines are points on the same air-breathing momentum principle.
Related Concepts: Ramjet Efficiency, Rocket Nozzle Thrust, Specific Impulse, Mass Flow Rate, Thrust, Conservation Of Momentum, Brayton Efficiency, Thrust-to-Weight Ratio
Notes: Registry calculator turbojet-thrust (unverified). Uninstalled, perfectly-expanded momentum form —
omits pressure thrust $(P_e-P_0)A_e$ and installation/inlet drag. Thrust earned from the increment $V_e-V_0$, so
it falls with airspeed. Fuel term $\dot m_f V_e$ often neglected (default $0$). High $\dot m$ / low excess velocity
= high propulsive efficiency (the turbofan principle).