The jet engine
How does a jet engine push a plane?
The question
A jet engine has no propeller you can see, and it pushes only on air. How does it shove a plane down the runway?
Have a guess, then show the answer
It throws a huge amount of air backwards, and the air pushes the engine forwards. In a modern turbofan most of that air never goes near the fire. A big fan blows it round the outside, the bypass air, and it gives most of the push. The burning core mainly exists to spin the fan.
Play
Open the throttle
This is a turbofan cut open from the side. Blue dots are cold air, red dots are hot gas from the burner. Move the throttle and watch how the bypass air and the core air add up to the total push.
Tap any part of the engine to see what it does.
- Bypass air
- 33.4 kN
- Core air
- + 10.3 kN
- Thrust
- = 43.7 kN
176 kg of air a second goes round the core and 34 kg through it (5.1 to 1). The bypass air gives 76% of the push.
Full throttle matches the CFM56-7B27 take-off rating, 121.4 kN. The air flows and speeds are rounded examples, not published figures.
What's happening
- 1
The fan. A ring of wide blades at the front, 61 inches (1.55 m) across on a CFM56-7B, pulls air in. Most of it goes round the outside of the engine, in the bypass duct.
- 2
The compressor. The rest of the air goes into the core, where rows of small blades squeeze it into a much smaller space, so it is dense and hot.
- 3
The combustor. Fuel is sprayed into the squeezed air and burns all the time, like a blowtorch, not in bangs like a car engine.
- 4
The turbine. The hot gas rushes out through more rows of blades and spins them. They sit on shafts that turn the compressor and the fan. So the fire drives the fan.
- 5
The nozzle. Both streams leave out of the back, fast. Throwing air backwards pushes the engine forwards. On a CFM56-7B about 5 kg of air go round the core for every 1 kg through it.
The maths
Thrust is air thrown back each second times how fast it leaves. Add the two streams. With example flows at full power,
That is the take-off rating of a CFM56-7B27, 27,300 pounds of force. The bypass air gives about three quarters of it.
Simplified, and sources
The bypass ratio (5.1) and the take-off rating (27,300 lbf, 121.4 kN) are the published figures for the CFM56-7B27, and the 61-inch fan is the published size. The air flow (350 kg a second) and the two exit speeds (317 and 500 m/s) are round example values picked so the sum lands on that rating; they are the right size for this engine but are not published numbers, so the three-quarters split is approximate too. The formula leaves out the plane's own speed and the pressure at the nozzle, which is fine for an engine standing still. In the widget, thrust grows with the throttle squared, which is only roughly true, and the drawing is not to scale. Real engines have two shafts, one for the fan and one for the compressor.