From here I found out a lot about chevrons on the Boeing 787, but not enough to satisfy why it results in performance loss.

enter image description here

My thinking was originally, if they are "cutting triangular holes" in the "bell cone", then some part of the exhaust will expand sooner than others. (I'm more of a rocket guy myself). Or you could think of it as "adding triangular extensions" to the optimal bell-cone, in which case some parts of the exhaust will expand later than usual. Or you could meet in the middle, which is probably the best way, but will still result in suboptimal expansion.

But I find myself wondering about an even simpler analogy: If sound is wasted energy from an engine (that could have gone into thrust), then reducing sound somehow must mean that otherwise wasted energy has actually gone into thrust production and therefor would boost thrust performance?

So which is it really, and why?

  • 1
    $\begingroup$ I'd assume it'd become heat. $\endgroup$
    – bjb568
    Jul 1, 2015 at 21:02

1 Answer 1


The chevrons need a small pressure difference between the fan flow and the surrounding air. Then they will work like small delta wings and create two strong vortices per triangle, which will help to mix fan flow and outer flow. This mixing is responsible for the reduced noise, because it will distribute the kinetic energy of the fan flow over a bigger mass of air.

This mixing by itself uses up some of the energy of the fan flow, and thrust is produced by the conversion of the pressure increase due to the fan into kinetic energy. With the chevrons, a little of this kinetic energy is now spent for mixing and cannot contribute to thrust anymore. Hence the reduced efficiency.

You're right to assume that less sound should mean more thrust. In fact, the sound energy is not evenly distributed over all frequencies, and by adding some noise at frequencies with low noise contribution, the peaks at the lower frequencies which contain most of the noise can be lowered.

  • 1
    $\begingroup$ Also keep in mind that sound contains a very very small amount of energy, as shown by physicscentral.com/explore/poster-coffee.cfm $\endgroup$
    – ROIMaison
    Jul 2, 2015 at 7:16
  • 2
    $\begingroup$ That's correct; a 94dBSPL sound, roughly equivalent to the sound level of a circular saw as heard by the user, is a pressure wave of about one Pascal pressure difference, which is one Newton per square meter of surface area. The average jet engine produces about 200 Pa in sound level as measured through open air at about 30 meters away, which is instantly deafening. The 787's most powerful engine configuration, for the 787-10, produces 1,700 times that much thrust at the exhaust of each engine. $\endgroup$
    – KeithS
    Jul 7, 2015 at 22:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .