# Propulsive Efficiency

KleptoBear
Hello everyone,

Long time follower, first time user here. I have got a Q about propulsive efficiency.

My understanding of the jet engine in a nutshell is that air is compressed in the compressor (pressure and temp go up), energy added in the combustor (large increase in temp), work extracted by the turbine (pressure and temp decrease) and then the air goes out the exhaust with a velocity higher than inlet velocity.

Now I thought that the higher the difference between inlet and exit velocities of air, the more engine thrust there is. However, I read this line in Mattingly's book "Elements of Propulsion": "The turbofan also accelerates a larger mass of air to a lower velocity than a turbojet for a higher propulsive efficiency". I am little puzzled by this. According to this statement, the ideal jet engine would be one that actually has zero exit velocity, wouldn't it?! Surely that can't be right. Any help is appreciated.

Staff Emeritus
Hello everyone,

Long time follower, first time user here. I have got a Q about propulsive efficiency.

My understanding of the jet engine in a nutshell is that air is compressed in the compressor (pressure and temp go up), energy added in the combustor (large increase in temp), work extracted by the turbine (pressure and temp decrease) and then the air goes out the exhaust with a velocity higher than inlet velocity.

Now I thought that the higher the difference between inlet and exit velocities of air, the more engine thrust there is. However, I read this line in Mattingly's book "Elements of Propulsion": "The turbofan also accelerates a larger mass of air to a lower velocity than a turbojet for a higher propulsive efficiency". I am little puzzled by this. According to this statement, the ideal jet engine would be one that actually has zero exit velocity, wouldn't it?! Surely that can't be right. Any help is appreciated.

The higher propulsive efficiency comes from moving more air at a lower velocity and not having to compress (do work on it) as much. Thrust is proportional to $\dot{m}$ ve, ve = exhaust velocity. I'm ignoring the change in pressure. It's a balance of how much to maximize the mass flow rate and the exhaust velocity of that mass.

http://www.grc.nasa.gov/WWW/k-12/airplane/rktthsum.html

KleptoBear
Oh I see. So the statement isn't about air exiting through the exhaust but about air being compressed in the compressor. That makes sense. Thank you

Homework Helper
THe purpose of a jet engine is to produce thrust. From Newton's laws of motion, the way to produce thrust is to change the momentum of the air flowing through the engine.

To you change the momentum of the air, you have to change its velocity. That means you also have to change its kinetic energy, but the amount of energy change isn't useful in itself.

So to make an efficient engine, you need to figure out how to get the biggest change of momentum for one unit of energy, and the best way is to give a big mass of air a small increase in velocity, rather than giving a small mass of air a big increase in velocity.

The maths that leads to this conclusion is straightforward. The basic reason is that momentum depends on velocity, but KE depends on velocity squared.

Pkruse
At cruise, the ideal would be to have an exit velocity one mph higher than the air speed. We can’t quite do that, but that is the theory. The most subsonic engine would be to have a large fan with as high of a bypass ratio as possible. Then nearly all the thrust will be from the huge mass of air blowing out the back, most of which will come from the fan. Only a very small part of the forward thrust comes from the mass flow of exhaust gas from the turbine.

KleptoBear
Then nearly all the thrust will be from the huge mass of air blowing out the back, most of which will come from the fan. Only a very small part of the forward thrust comes from the mass flow of exhaust gas from the turbine.
That is true for the turbofan but what about a turbojet engine?

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Aj83
That is true for the turbofan but what about a turbojet engine?

Turbojet engine has no fan so all the thrust is produced by the engine's core (compressor, combustor and turbine) hence they are a lot less efficient than turbofan engine but only at sub sonic speeds. At supersonic speeds the large frontal area of turbofan engine becomes major contributor to the drag which is a function of velocity squared so any gains in efficiency are nullified due to the increase in drag.

The Jericho
Thrust is produced by the reaction of the change in momentum of the air exiting the nozzle.
A turbofan that you may see on a 747 for instance will accelerate a large mass of air at a relatively low velocity, whereas a turbojet engine (no bypass), will accelerate a relatively low mass of air at a high velocity, maybe Mach 6 in some cases, to achieve an adequate amount of momentum for the air exiting the nozzle to produce thrust. Remember that momentum is proportional to the product of mass and velocity. An explicit example of this mass-velocity balance would be that rocket engines are designed to achieve (and maintain) thrust by accelerating/jettisoning combusted propellant and oxidiser out of the nozzle at the highest possible velocity (about Mach 30) in order to produce adequate thrust whilst keeping propellant flow rate at a minimum so as to maximise fuel efficiency (specific impulse). In summary, it's the the momentum of the air that's important, not just the exit velocity.

Sincerely, The Jericho.

KleptoBear
Thank you everyone for your time. Above post along with all the other posts did direct me in the right direction. I have learned so much about the topic since I opened this thread. What really put me off was that I lacked a solid understanding of thrust at the time I read the statement in the introduction part of the book. Perhaps, it could have been omitted there or maybe I should have trusted that the author would adequately elaborate later in the book.

Mithrandir13
basically... thrust is a function of Mass and Acceleration...

the energy used is a function of mass and Acceleration^2

so maximizing the mass of air accelerated and minimizing the Delta-V is more efficient...