Unit 1 · Lesson 1-1
Thrust
2h 35m1300 XP
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Thrust

The momentum balance on an engine: gross thrust, ram drag, pressure thrust, installed thrust and rocket thrust

Kona, Hawaii — August 2024
A Boeing 777-9 lands in Seattle after a test flight from Kona, Hawaii. In the inspection that follows, mechanics find a broken thrust link — the part between the engine and its pylon through which the engine's thrust reaches the airframe. Two more test aircraft turn out to have cracked thrust links, and Boeing grounds the whole 777-9 test fleet. Flight testing resumes only in January 2025 (AvWeb's report). Each of those engines is a GE9X, rated at 105,000 lbf of take-off thrust — about 470 kN, enough to hold up 48 tonnes, the weight of some thirty family cars. All of it reaches the wing through parts like that thrust link. This lesson is about that force: where it comes from, how to calculate it from what goes in and out of the engine, and why the number on the data plate is not the number the wing feels.
An aircraft engine sells exactly one thing to the aircraft: a force along the flight path. It produces it the only way a machine can push on air — by changing the air's momentum. Air enters at the flight speed V0V_0 and leaves at a higher jet speed V9V_9. To accelerate it, the engine pushes the air backward; by Newton's third law the air pushes the engine forward, and the engine passes that push through its mounts to the airframe.
A rocket does the same with something it carries: it throws its own propellant backward. An air-breathing engine borrows almost all of its working fluid from the sky — about 98 % of the mass leaving a turbojet's nozzle is air that was there already, and only 2 % is fuel. That is why air-breathers are so much more economical in the atmosphere, and why they cannot work outside it.
💡Station numbers
Engine engineers label the flow path by station numbers, the same in every textbook and every engine data sheet (SAE ARP755):
  • 0 — free stream, far ahead of the engine
  • 2 — fan or compressor face
  • 3 — compressor exit (the highest pressure in the engine)
  • 4 — combustor exit, turbine entry (the highest temperature)
  • 5 — turbine exit
  • 9 — core nozzle exit; 19 — fan (bypass) nozzle exit of a turbofan
So V9V_9 is the core jet velocity, Tt4T_{t4} the turbine entry temperature, pt3/pt2p_{t3}/p_{t2} the compressor pressure ratio. The whole course speaks this language.
To find the thrust we draw a control volume around the engine: a large box whose surface is far enough away that the air on it is at ambient pressure p0p_0, except across the nozzle exit plane, where the jet may leave at a different pressure p9p_9. The pylon crosses the box. The force the pylon transmits is the thing we want.
Air enters the box through the front at m˙0\dot m_0 and V0V_0, fuel enters through the pylon at m˙f=f m˙0\dot m_f = f\,\dot m_0 with no momentum along the flight direction, and the mixture leaves through the nozzle at m˙9=(1+f)m˙0\dot m_9 = (1+f)\dot m_0 and V9V_9. The momentum balance then gives the thrust equation.
F=m˙0[(1+f) V9−V0]+(p9−p0) A9F = \dot m_0\left[(1+f)\,V_9 - V_0\right] + (p_9 - p_0)\,A_9

The uninstalled thrust of an air-breathing engine. The first term is the momentum thrust, the second the pressure thrust; ṁ0 is the air mass flow, f the fuel–air ratio, A9 the nozzle exit area.

Derivation
The thrust equation from a momentum balance
For a steady flow, Newton's second law applied to a control volume says: the net force on the fluid inside equals the momentum flux leaving minus the momentum flux entering,
∑Fx=m˙outVout−m˙inVin.\sum F_x = \dot m_{out} V_{out} - \dot m_{in} V_{in}.
Two forces act on the fluid along the flight direction xx. The engine pushes on it with a force FF toward the rear — the reaction to the thrust. And the pressure on the control surface: p0p_0 everywhere, except over the exit area A9A_9 where it is p9p_9. A uniform pressure p0p_0 over a closed surface gives no net force, so only the difference over A9A_9 remains, acting on the fluid as (p0−p9)A9(p_0 - p_9)A_9 in the flight direction. With xx pointing backward along the jet:
F+(p0−p9) A9=m˙9V9−m˙0V0.F + (p_0 - p_9)\,A_9 = \dot m_9 V_9 - \dot m_0 V_0 .
The fuel enters through the pylon with no xx-momentum, so it adds to m˙9=(1+f)m˙0\dot m_9 = (1+f)\dot m_0 but not to the inflow term. Rearranging,
F=m˙0[(1+f)V9−V0]+(p9−p0) A9.F = \dot m_0\left[(1+f)V_9 - V_0\right] + (p_9 - p_0)\,A_9 .
By Newton's third law the fluid pushes the engine forward with the same FF, and the engine passes it to the pylon. The assumptions are the ones in the sketch: steady flow, uniform properties across the inlet and exit planes, and a control surface far enough away to sit at p0p_0.
💡Gross thrust, ram drag, net thrust
The thrust equation splits into three named pieces:
  • Gross thrust Fg=m˙9V9+(p9−p0)A9F_g = \dot m_9 V_9 + (p_9 - p_0)A_9 — what the nozzle produces. It depends only on the jet.
  • Ram drag Dr=m˙0V0D_r = \dot m_0 V_0 — the momentum the engine must first give to the air it swallows, which arrives at the flight speed. It grows with both mass flow and speed.
  • Net (uninstalled) thrust Fn=Fg−DrF_n = F_g - D_r.
Installed thrust is less again: the nacelle and pylon have their own drag, and the inlet and nozzle disturb the flow around them (spillage drag at the inlet lip, boat-tail drag at the nozzle). Engine makers quote uninstalled thrust; aircraft designers book the installation losses — typically a few per cent.
📐The deck's Mach 2 engine
An air-breathing engine flies at M0=2.0M_0 = 2.0 where T0=−50 ∘T_0 = -50\,^\circC and p0=10p_0 = 10 kPa. It takes in 25 kg/s of air, burns fuel at 3 % of the air flow, and its nozzle is perfectly expanded (p9=p0p_9 = p_0) with V9=1050V_9 = 1050 m/s. Speed of sound: a0=1.4×287×223.15=299.4a_0 = \sqrt{1.4 \times 287 \times 223.15} = 299.4 m/s, so V0=598.9V_0 = 598.9 m/s.
  • Gross thrust: Fg=25×1.03×1050=27 038F_g = 25 \times 1.03 \times 1050 = 27\,038 N
  • Ram drag: Dr=25×598.9=14 972D_r = 25 \times 598.9 = 14\,972 N
  • Net thrust: Fn=12 066F_n = 12\,066 N ≈ 12.1 kN
Ram drag eats 55 % of the gross thrust. At Mach 2 the engine spends more than half of its effort just bringing the incoming air up to speed — which is why supersonic engines need very fast jets, and why the same jet gives less net thrust the faster you fly.
To every action there is always opposed an equal reaction; or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.
– Isaac Newton, Principia (1687), Law III, in Andrew Motte's 1729 translation
Take-off thrust. On the runway V0≈0V_0 \approx 0, so ram drag vanishes, and with a perfectly expanded nozzle the pressure term goes too:
Ftakeoff≈m˙9V9.F_{takeoff} \approx \dot m_9 V_9 .
Take-off thrust is proportional to the air the engine can swallow. For a turbofan the two streams add, F=m˙19V19+m˙9V9F = \dot m_{19} V_{19} + \dot m_9 V_9, and the fan stream dominates: in a big modern engine nearly nine-tenths of the take-off thrust comes from air that never goes near the combustor. That is why fans keep growing — the GE9X's is 134 inches (3.4 m) across, with 16 carbon-fibre composite blades.
Rocket thrust. A rocket's control volume has no inlet: nothing enters, so there is no ram drag, and the thrust barely depends on flight speed.
F=m˙pVe+(pe−pa) Ae,Isp=Fm˙p g0.F = \dot m_p V_e + (p_e - p_a)\,A_e, \qquad I_{sp} = \frac{F}{\dot m_p\, g_0}.
The specific impulse IspI_{sp}, in seconds, is the thrust per unit weight flow of propellant. The pressure term changes with altitude: as the ambient pressure pap_a falls, the thrust rises. The Space Shuttle main engine, in rounded numbers, burns 514 kg/s of hydrogen and oxygen with Ve=4300V_e = 4300 m/s and pe=16.6p_e = 16.6 kPa across a 4.17 m² exit. At sea level the pressure term is negative, (16.6−101.3)×103×4.17=−353(16.6 - 101.3)\times 10^3 \times 4.17 = -353 kN, so F=2.210−0.353=1.857F = 2.210 - 0.353 = 1.857 MN and Isp=368I_{sp} = 368 s; in vacuum the same term adds 69 kN, for 2.279 MN and 452 s.
Derivation
Rocket thrust: the same balance with nothing coming in
Take the same control volume around a rocket. Nothing crosses the front of the box, so m˙0=0\dot m_0 = 0 and the inflow momentum term disappears; all the propellant m˙p\dot m_p was already inside. The balance becomes
F+(pa−pe) Ae=m˙pVe⟹F=m˙pVe+(pe−pa) Ae.F + (p_a - p_e)\,A_e = \dot m_p V_e \quad\Longrightarrow\quad F = \dot m_p V_e + (p_e - p_a)\,A_e .
Dividing by the weight flow of propellant m˙pg0\dot m_p g_0 gives the specific impulse,
Isp=Fm˙pg0=Veg0+(pe−pa)Aem˙pg0,I_{sp} = \frac{F}{\dot m_p g_0} = \frac{V_e}{g_0} + \frac{(p_e - p_a)A_e}{\dot m_p g_0},
which is the effective exhaust velocity measured in units of g0g_0. With pa=0p_a = 0 (vacuum) the pressure term is largest: a rocket is at its best above the atmosphere.
💡Specific impulse of an air-breather
Air-breathing engines can be rated by specific impulse too, counting only the fuel: Isp=F/(m˙fg0)=1/(g0⋅TSFC)I_{sp} = F/(\dot m_f g_0) = 1/(g_0 \cdot \mathrm{TSFC}). The Mach 2 engine above has Isp=12 066/(0.75×9.81)≈1640I_{sp} = 12\,066/(0.75 \times 9.81) \approx 1640 s. A modern high-bypass turbofan in cruise reaches about 6500 s. The best chemical rocket, burning hydrogen and oxygen in vacuum, manages about 450 s. Taking the oxidiser from the air, and most of the working mass with it, is worth more than a factor of ten.
💡Where you will use this
The thrust equation is the bottom line of every analysis in this course. Units 3 and 4 compute V9V_9 and ff from the engine cycle; Unit 5 decides p9p_9 and A9A_9 in the nozzle; Unit 8 shows how the mass flow changes with altitude and throttle. In a rocket propulsion course the same equation, without ram drag, sizes every launch vehicle stage.
📚Further reading
  • Jack D. Mattingly and Keith M. Boyer, Elements of Propulsion: Gas Turbines and Rockets (2nd ed., AIAA, 2016), chapter 1 — the thrust equation and its installed corrections.
  • Nicholas Cumpsty and Andrew Heyes, Jet Propulsion (3rd ed., Cambridge University Press, 2015), chapters 1–2.
  • NASA Glenn Research Center, Beginner's Guide to Propulsion: the thrust equation — short and free.
  • George P. Sutton and Oscar Biblarz, Rocket Propulsion Elements (9th ed., Wiley, 2017), chapter 2 — rocket thrust and specific impulse.
↓But there is a problem
Thrust grows with the jet velocity. Double V9−V0V_9 - V_0 and you double the thrust of every kilogram of air. So the best engine has the fastest jet? In 1961, Pratt & Whitney took a perfectly good turbojet, bolted a large fan to its front that slowed the average jet right down — and airlines lined up to buy it. The next lesson is about why.