# Why is a higher bypass ratio more efficient at lower speeds?

Why is it that a larger volume of air moving slower is a more efficient means of propulsion than a smaller volume of air moving faster? (This is true up to a given speed, obviously, as supersonic aircraft use lower BPR engines).

Because bigger reaction mass with lower speed carries away less energy for the same momentum.

To produce thrust the engine applies force to the exhaust gasses and air. This force accelerates the air. To produce thrust $F$, the engine has to impart momentum $p = \frac{F}{t}$ per unit of time on the air. Momentum is proportional to mass and first order of speed. $p = mv$ where $m$ is mass of the gasses (reaction mass) affected by the engine per unit of time and $v$ is speed of those gasses.

The gasses however also take away some kinetic energy. Kinetic energy is also proportional to mass, but also to second order of speed. $E_k = \frac12mv^2$.

This energy has to be provided by the engine. Since the fuel can only provide so much energy (it's heating value) per unit of mass, an engine that needs less energy needs less fuel. And that is achieved by moving more air. Which means higher bypass ratio for jet and bigger propeller for engines with propellers.

Note, that this does not say high-bypass engine would be more efficient at slow speed than at high speed. Because it isn't. All jet engines are more efficient at high speed. This is because the power the engine produces is $P = Fv$. So at high speed, larger part of the power output is the useful work moving the aircraft.

In high-bypass engine the maximum power output of the fan is limited though (similar to propeller), so compared to a turbojet the difference in efficiency is higher at low speed.

• Interesting parallel: The same goes for lift: It is more efficient to produce the same lift by moving a lot of air by a little. Like a high bypass ratio engine, a high aspect ratio wing is more efficient. The physics is the same in both cases. – Peter Kämpf Jul 28 '14 at 7:42