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I'm reading the wikipedia page on turbofans, where it says turbojets are inefficient at low speeds because:

energy is wasted as the propelling jet is going much faster rearwards than the aircraft is going forwards, leaving a very fast wake. This wake contains kinetic energy that reflects the fuel used to produce it, rather than the fuel used to move the aircraft forwards.

Can someone please explain this a little bit? Why is there a better way of utilizing kinetic energy than simply expelling the mass backward?

It seems to me like attempting to "capture" the expelled air is akin to putting an air-powered generator behind a turboprop propeller and using the energy to spin an extra propeller. How is that possibly more efficient than just letting it be?

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    $\begingroup$ Umm just looking at the title alone I'd say that's more a question for Medical SE : ( $\endgroup$ Mar 26 at 0:45

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It depends on what you call efficient.

In principle, the force imparted on the airplane is equal to the momentum expelled backwards:

$$F=\frac{\Delta m}{\Delta t} \cdot v$$

The power requirement to deliver that force is equal to the kinetic energy in the exhaust,

$$P=0.5\frac{\Delta m}{\Delta t} \cdot v^2$$

So, because of the $v^2$, you obviously get more force for your energy input if you accelerate a lot of air by very little speed, instead of accelerating very little air by a lot.

This assumes that it doesn't cost energy to actually "get" all this air mass through your engine. In reality, friction losses start dominating at some point. For rockets, you need to actually bring all the mass up with you instead of just taking in some atmospheric air, so then it's actually more efficient to accelerate very little mass by a lot.

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    $\begingroup$ This also applies for example to helicopters, in which they build big rotors to accelerate large masses of air to slow speeds $\endgroup$
    – U_flow
    Mar 23 at 11:28
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    $\begingroup$ So in a world without fan blades, could a plain jet or rocket benefit from mixing the rapidly moving expelled gas, with slower moving normal air? Kind of like the way a "bladeless" dyson fan turns a small jet of fast air into a large jet of slower air. $\endgroup$ Mar 23 at 11:39
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    $\begingroup$ Yes, this is called an air augmented rocket. en.m.wikipedia.org/wiki/Air-augmented_rocket $\endgroup$
    – Sanchises
    Mar 23 at 12:33
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seems to me like attempting to "capture" the expelled air is akin to putting an air-powered generator behind a turboprop propeller to turn another propeller

You're close, but why not put the generator turbine directly behind the gasses being expelled from the engine?

This is how turbojets are designed. One "generator" powers the compressor, another can be added to run a fan or prop.

This wake contains kinetic energy that reflects the fuel used to produce it, rather than fuel used to move the aircraft forward.

This is utter nonsense. Burning fuel will never move an aircraft forward unless it is used to create some type of propulsive force.

The propulsive force per unit fuel burned is the efficiency factor. Dumping a 55 gallon drum of fuel next to an airplane and burning it (obviously) will have a thrust efficiency factor of 0.

Burning it in a tube and allowing the heated gasses to expand backwards will created an equal amount of momentum, as mv, forward.

Burning it in a tube (or cylinder) and using the expanding gasses to create mechanical force to turn a fan or propeller increases efficiency further, but only under certain conditions.

Propellers are only more efficient up to around Mach 0.4, while (ducted) fans (extracting energy from the expelled gasses of the jet engine core) remain more efficient at higher Mach numbers because they can create more momentum per unit fuel burned than the turbojet.

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  • $\begingroup$ "Propellers are only more efficient up to around Mach 0.4 ..." Is this why piston engines and turbo props fell out of favor for commercial passenger aviation, but why the "relatively slow" Lockheed C-130 still uses turbo props to this day? $\endgroup$
    – dgnuff
    Mar 23 at 20:50
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    $\begingroup$ @dgnuff ...which gives the C-130 the ability to take-off on much shorter runways for a given power and weight, which is important for a military transport aircraft. $\endgroup$ Mar 23 at 21:06
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    $\begingroup$ I think that the quoted comment about the wake being inefficient is a clumsy attempt at saying that turbulent air represents wasted energy? $\endgroup$
    – MikeB
    Mar 25 at 7:57
  • $\begingroup$ @MikeB The "wake" in this case is expelled engine gasses. These plumes become turbulent, but not at the exhaust (if the pipe is properly designed). Sanchises explains the physics mathematicly. I marveled at the logic, as fuel is burned for propulsion, and propulsion moves the plane forward (with varying degrees of efficiency). $\endgroup$ Mar 25 at 9:09
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    $\begingroup$ @MikeB, Put another way, the turbulence doesn't steal energy from the engine. It's more like, proof that the engine allowed the energy go out the tail pipe without using it to do any useful work. $\endgroup$ Mar 25 at 14:03

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