Power consumption=Kinetic energy=$1/2mv^2$.
Thrust=momentum change=$mv$.

I keep being told that aircraft engines are most efficient when their exhaust velocity closely matches the freestream.


enter image description here

The second graph shows the propulsive efficiency vs exhaust/freestream velocity ratio

I can understand that higher bypass engines have greater intake losses than narrower ones. But the ratio of airspeed to exhaust speed being the defining factor, as opposed to the absolute difference?? I have only one way to justify this - as speed increases, mass flow rate through an engine of given cross sectional area also increases.

Research on propfans, and on supersonic through-flow turbofans calls into question the idea that lower bypass ALWAYS = higher high-speed efficiency. And the above equations don't help either.

What mistake am I making?

  • 3
    $\begingroup$ related: In what way are the Concorde's engines considered efficient? $\endgroup$
    – Manu H
    Commented Jun 11, 2020 at 18:22
  • $\begingroup$ the concorde engines had exceptional pressure ratio - and thus thermal efficiency - at speed. but this here is about propulsive efficiency. $\endgroup$ Commented Jun 12, 2020 at 9:53
  • $\begingroup$ Thrust is proportional to speed difference and area. But the propulsion efficiency of the engine drops as exhaust velocity increases due to losses in the exhaust stream. That means larger area and slower speed is inherently more efficient than small area and large speed. $\endgroup$
    – Jan
    Commented Jun 12, 2020 at 10:19
  • 1
    $\begingroup$ @Jan I can understand that kinetic energy is not equal to thrust, but momentum is. And that is why higher BPR is beneficial. But why would we ever say that a lower BPR is more propulsive-efficient under certain circumstances? $\endgroup$ Commented Jun 12, 2020 at 12:29
  • $\begingroup$ The first graph you posted shows that propulsive efficiency of turbofan engines increases up to the point they no longer have enough exhaust velocity to produce thrust, contrary to your question. Also, what is the second graph supposed to show? $\endgroup$
    – Jan Hudec
    Commented Jun 14, 2020 at 9:46

2 Answers 2


Power consumption=Kinetic energy=$1/2mv^2$.
Thrust=momentum change=$mv$.

This is only true if the engine is perfectly efficient internally and the air is still relative to the aircraft.

If the air is moving relative to the aircraft (and the engine is still perfectly efficient internally) then.

Power consumption=Kinetic energy=$\frac{1}{2}mv_e^2 - \frac{1}{2}mv_a^2$ = $\frac{1} {2}m((v_a + v_\Delta)^2 - v_a^2) = \frac{1} {2}m(2v_av_\Delta + v_\Delta^2) $
Thrust=momentum change=$m(v_e-v_a) = mv_\Delta$

Where $v_a$ is the ambient velocity (relative to the aircraft), $v_e$ is the exhaust velocity (relative to the aircraft) and $v_\Delta$ is the difference between the ambient velocity and the exhaust velocity ($v_e - v_a$).

If the exhaust velocity (relative to the aircraft) is smaller than the ambient velocity (relative to the aircraft) then you have negative thrust (aka drag).

If the exhaust velocity (relative to the aircraft) is much larger than the ambient velocity (relative to the aircraft) then most of the power is wasted (the $v_\Delta^2$ term is larger than the $2v_av_\Delta$ term)

If the exhaust velocity is only slightly higher than the ambient velocity then most of the power goes into useful thrust (the $2v_av_\Delta$ term is larger than the $v_\Delta^2$ term)

What this means if that if the engine were perfectly efficient internally then the advantages of a higher bypass ratio would drop off with speed, but there would still be a slight advantage to the engine with higher bypass ratio.

But that is a big if, in particular we assume that all the energy in the intake stream is captured and returned to the exhaust stream.

Lets consider what happens if we instead assume that the energy in the intake stream is lost.

Our power consumption equation becomes.

Power consumption=Kinetic energy=$\frac{1}{2}mv_e^2$ = $\frac{1} {2}m(v_a + v_\Delta)^2) = \frac{1} {2}m(v_a^2 + 2v_av_\Delta + v_\Delta^2) $

For a given thrust $m$ is proportional to $\frac{1}{v_\Delta}$ so our goal is to minimise. $\frac{1}{v_\Delta}(v_a^2 + 2v_av_\Delta + v_\Delta^2)$ = $\frac{v_a^2}{v_\Delta} + 2v_a + v_\Delta$ differentiating with respect to v_\Delta and setting to 0 gives $-\frac{v_a^2}{v_\Delta^2} + 1$ which (given that v_\Delta must be positive) gives $v_a = v_\Delta$

Having considered an optimistic and a pessimitic case we can conclude that the optimal exhaust velocity for efficency is greater than the ambient velocity (otherwise you would not produce thurst) but probably less than twice the ambient velocity.

  • $\begingroup$ Actually the energy of the intake stream is not only not lost, it increases the pressure ratio and therefore thermodynamic efficiency of the engine. $\endgroup$
    – Jan Hudec
    Commented Jun 15, 2020 at 18:48

First, thrust = ma = mv/s

Next, what type of engine are you referring to? Throw out the first graph, it's all about propulsive efficiency vs speed (props, fans, etc.), not about engine efficiency.

Considering exhaust only, the engine has to work to push out the exhaust. This is why rockets have more thrust in space. Also known as "backpressure". Efficiency will improve as exhaust velocity approaches freestream velocity (and will continue to improve if freestream velocity is greater). A large widening exhaust pipe that exits to the free stream helps reduce backpressure.

But this is not good if you are trying to use your exhaust to make thrust. However, with jets, the free stream also helps air intake as well.

As far as the title to your question, now take the first graph and look at props. A "high bypass fan" has the same issue as a prop at higher speeds, it's own disk drag. This has nothing to do with engine efficiency.


You must log in to answer this question.

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