10
$\begingroup$

What's the scientific background behind a jet engine's top speed? Jet engines increase efficiency at higher airflow rates, so why does the efficiency stop increasing at some point?

$\endgroup$
8
  • 1
    $\begingroup$ Do you have a specific example? $\endgroup$ Dec 28, 2022 at 23:46
  • 3
    $\begingroup$ Remember that every technical design is a compromise between different parameters. Any specific design will have a possible speed range depending on a number of design parameters. For extremely high speed, say Mach number above 4, the engine will look very different from a typical jet engine and will most probably not generate any thrust a speeds below 1 Mach. $\endgroup$
    – ghellquist
    Dec 29, 2022 at 12:37
  • 2
    $\begingroup$ "Jet engines increase efficiency at higher airflow rates, so why does the efficiency stop increasing at some point?" It doesn't - your premise is completely flawed as the answers demonstrate. $\endgroup$
    – Ian Kemp
    Dec 29, 2022 at 15:18
  • $\begingroup$ Do you mean a maximum airspeed (forward velocity), or a maximum RPM of the blades? $\endgroup$
    – Daniel K
    Dec 30, 2022 at 20:39
  • 1
    $\begingroup$ Recommended reading here. It seems converting turbine energy into better "mass movers" creates much more specific impulse than simply ejecting it out the rear of the jet. (Rockets must do this as there is no air to move in space). $\endgroup$ Dec 30, 2022 at 23:36

5 Answers 5

16
$\begingroup$

Temperature and stability limits.

Thrust is created by accelerating a working mass in opposite direction. Net thrust is the difference between the impulse of the air flowing towards the engine and the combined impulse of burnt fuel and the air exiting the engine, derived according to time.

Impulse is mass times speed, so the exit speed of an engine needs to be larger than the entry speed to create positive thrust (assuming that the mass of the fuel in the exit stream is insignificant). Entry speed is flight speed, so the exit speed must be higher than flight speed.

The air in a jet is accelerated by heating it, so the gas expands and picks up speed in an isobaric process. The heat is provided by adding fuel, and since the fuel-air mixture in a jet is very lean, burning more fuel will produce higher exit speeds. The jet will still produce positive thrust at faster speeds.

However, adding more heat will also increase the thermal load on the turbine. By using better materials, the maximum heat in jet engines could be increased by several hundred degrees C in the past decades, but is still limited so the mixture ratio in the combustion process also must be limited. Add to that the better efficiency of higher compression, and you will have a high air temperature at the start of the combustion process already from compression, so less combustion heat can be added. That (and higher intake precompression) is the reason why modern civilian engines for subsonic flight use compression ratios of up to 50:1 while supersonic engines rarely exceed 30:1.

To get around that limitation, afterburners are used, but even here you cannot add unlimited amounts of fuel. With too much fuel the combustion process will become unstable (afterburner rumble or screech). The J-58 engine of the Mach 3.2 SR-71 even piped compressed air around the combustion chamber directly into the afterburner, so cooler air could be used to burn more fuel there. Follow the last link for more information on how the initial J-58 had to be modified to make Mach 3+ flight possible.

Ramjets run into the same temperature limit from intake compression alone: Above Mach 4 to 5 the temperature of the compressed air is so high that adding fuel will not heat the air much but will lead to dissociation as temperature increases further, using up the chemical energy to split molecular bonds instead of adding heat. Therefore, flight at Mach 5 and above requires scramjets where the air in the combustion area is still supersonic.

$\endgroup$
7
  • $\begingroup$ I believe that the questioner wants to know why we can't keep on increasing the TAS to obtain ever increasing efficiency at a constant engine fuel flow (because the efficiency does increase initially as speed increases). However, your first two sentences do answer that so +1 from me. $\endgroup$ Dec 29, 2022 at 12:25
  • $\begingroup$ Maybe you could add also the contribution due to the different pressure between inlet and outlet $\endgroup$
    – sophit
    Dec 29, 2022 at 17:08
  • $\begingroup$ Or alternatively you use smart air dam design to reduce the airspeed as it enters the turbine. That's how Concorde could supercruise without afterburners at the same speed as the then-fastest jet interceptor with afterburners. $\endgroup$
    – Graham
    Dec 29, 2022 at 17:29
  • $\begingroup$ @Graham: what's a smart air dam? $\endgroup$
    – sophit
    Dec 29, 2022 at 19:29
  • $\begingroup$ @sophit How they designed the air intakes. heritageconcorde.com/air-in-take-system $\endgroup$
    – Graham
    Dec 29, 2022 at 20:26
10
$\begingroup$

If a jet engine is operating at 1000 KTAS and its exhaust gases are also exiting the engine at 1000 KTAS, it is producing no thrust for the fuel it is burning - its propulsive efficiency is 0%. (If it operates at a speed higher than 1000 KTAS, it will produce drag instead of thrust - this engine is not capable of accelerating the exhaust gases to speeds any higher than 1000 KTAS, and so this speed is its "speed limit"). Turbofans achieve this limit at lower speeds, turboprops at even lower speeds.

The following equation explains why the propulsive efficiency initially increases with TAS:

$Power = work/time$

$Power = force × distance/time$

$Power = force × speed$

The force is thrust, and the speed is TAS.

Clearly, increasing the speed will increase power (while the fuel consumption remains the same). However, as mentioned earlier, thrust reduces as speed increases. This means that there will be one particular combination of speed and thrust at which their product (power) will be maximum. The speed at which this occurs is the speed for maximum propulsive efficiency, and going any faster (or slower) will reduce efficiency.

When it is said that "jet engines obtain their peak propulsive efficiency at higher speeds", it is being compared to other engines; increasing the speed doesn't necessarily mean increasing the efficiency - the efficiency initially increases, reaches it's peak, and then decreases. This is true for all air breathing engines, just that different engines attain their peak efficiency at different speeds.

$\endgroup$
6
$\begingroup$

To optimise fuel economy.

The jet engine needs to provide thrust that is higher than drag, the exit speed of the gas stream $V_e$ must be higher than the air speed $V_0$. For the thrust $T$ in N generated at full expansion to ambient air pressure, we can write:

$$T = \dot m\cdot \Delta V$$

with:

  • $\dot m$ = mass flow through the engine [kg/s]
  • $\Delta V = V_e - V_0$ = exhaust speed minus air speed [m/s].

The internal kinetic power of the gas stream $P_g$, provided by burning fuel, is:

$$P_g = \frac{1}{2} \cdot \dot m \cdot V_e^2$$

To generate a given amount of thrust, if we choose $\dot m$ as high as possible and $V_e$ as low as possible, we have generated the required thrust with the minimal amount of fuel required, since the $V_e$ term occurs squared in the required power equation. So we want to choose the jet exhaust speed as low as possible above the required cruise speed.

The jet engine itself does not care too much about how fast the aeroplane is flying, as long as the intake air is slowed down sufficiently and the exhaust air accelerated to above airspeed. For this we need to consider the jet engine as the core of the total engine installation.

enter image description here

The main design parameter for the type of jet engine to be used is the cruise speed. The image above (ref) is from prof. Wittenberg's college handout VTH-D 14, and depicts seven types of jet engine:

  • Ramjet
  • Pulse jet
  • Two types of turbojet
  • Three types of turbofans

If the aeroplane is to fly hypersonic the ramjet is the most suitable type, no turbine required since the incoming air is compressed adequately by the intake. However, the ramjet does not work from a standstill.

Turbojets and bypass turbofans are used in supersonic aeroplanes: most thrust is generated by the hot air exhaust, which exits at supersonic speed.

Turbofans are best for high subsonic aeroplanes: most of the thrust is generated by the fan, some by the hot exhaust.

enter image description here

The role of the intake and exhaust in thrust generation as a function of airspeed can be seen in picture above, earlier used in this answer. What is not shown here is the fuel economy: bypass & fan thrust is way more economical than exhaust gas thrust, but can only be used at lower speeds. The lower the cruise speed, the higher the fan thrust proportion should be. And fan thrust is subject to a speed limit, and above certain cruise speeds the intake-exhaust provides all the thrust.

$\endgroup$
0
3
$\begingroup$

Several points limit the top speed of the jet engine

  • Efficiency equation you're probably referring to is: $\eta_p = \frac{2}{1 + \frac{v_9}{v_0}}$
    • This never does stop increasing
    • Yet the growth is logarithmic with speed ~0.4*ln(v0)
    • Trying to fly 2x your speed only gains 1.33x efficiency
    • This type of efficiency only measures how much of the engine's fuel burn effort is applied to the airframe, rather than just moving air.
  • This efficiency matters, but it also needs Specific Impulse
  • Finally, all these considerations also ignore drag
    • The airplane is flying through air
    • The air has density and inertia.
    • To put the airplane where the air used to be you have to do something with the air
    • Generally means diverting or deflecting the air, or piping it through the plane
      • Busemann Biplanes are interesting variation on the last idea, although they don't produce lift.
      • Drag is generally thought of as two terms (Lift Drag and Parasitic Drag)
      • Lift Drag, drag because you're redirecting airflow to create lift, conveniently goes down with speed.
      • However, Parasitic Drag, drag because the airframe shape is diverting / deflecting air and the air has a friction effect on the skin, effectively rises as $V^2$
      • You put them together, and you get a drag curve that looks like this: DragCurves
      • This is one of the main reasons that most airplanes have a cruising speed.
  • Ideally, what you would want (if it was not highly complex) is an engine that acts like a turboprop, then a turbofan, then a ramjet, and then a scramjet.
    • Move a lot of air slowly at low speeds, and then move less air with a much higher exit velocity at high speeds.
$\endgroup$
1
$\begingroup$

Here is a thrust equation from the NASA website, which seems to have perked up a bit since the success of the Artemis mission.

We see thrust is produced by net momentum change and net pressure differential from a given exhaust area.

With turbojets, higher pressure is created by compressing ambient air. Heating it (by the relationship of Pressure and Volume being proportional to Temperature) allows both the exhaust area to be larger than the inlet and the exhaust velocity to be greater, producing net thrust. There is only a slight increase in mass flow in the engine core (but adding fans or props greatly improves this in subsonic flight regimes).

The issue is that drag is created by the inlet and temperature increase is caused by compression, robbing the engine of its net thrust until a thrust/drag breakeven point is reached. This is why some people claiming "hypersonic" aircraft these days are simply flying horizontal rockets, which are unburdened of the task of compressing adequate amounts oxidizer/mass from the atmosphere. But, then again, that is exactly how the X-15 managed hypersonic speeds back in the 1960s.

The limiting factor for jet propulsion is inlet temperature. Use of cryogenic fuel and/or oxidizer to cool inlet air is one way of improving overall efficiency.

$\endgroup$

You must log in to answer this question.

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