I have found out that a turbofan engine of say an F-16 is producing zero power when stationary since $Power = Thrust × Velocity$. But assuming that the engine is producing the same thrust, it appears that regardless of whether the F-16 is stationary or moving, the engine is still consuming the same amount of fuel and therefore there is power output that is constant in both cases. I also found that the power I tried to calculate from the equation above is the propulsive power, so there must be something else that is consuming the rest of the power.

To make this simple, I will use some numbers. Let's assume a turbofan engine is capable of producing 100 kN of maximum thrust with no afterburner while consuming 4 kg of fuel per second. For the sake of simplicity I'm gonna use a figure of 30% thermal to mechanical efficiency for the engine that is constant through all of the following cases and which means that there is 30% $×$ 4 kg $×$ 42 MJ/kg = $50 MW$ of mechanical power available at all times. With 43 MJ/kg being the heat value of the jet fuel.

Case 1: The engine is stationary and producing max thrust and therefore the propulsive power is zero so I assume that the 50 MW available is completely used in heating and accelerating the exhaust.

Case 2: The aircraft is accelerating and is now at half its max speed at sea level which is say 200 $m/s$. The propulsive power is now = 100,000 kN × 200 = $20$ $MW$ and so I'm also assuming that the other 30 MW remaining of the 50 MW available are still being consumed in accelerating and heating the exhaust.

Case 3: The aircraft reached its max speed at 400 $m/s$ at sea level. The propulsive power is now = 100,000 kN × 400 = $40$ $MW$ and as before, 10 MW are consumed by the exhaust.

So, if I understand this correctly, the power distribution goes from 0% propulsive power and 100% exhaust power when the engine is stionary to 40% propulsive power and 60% exhaust power when the aircraft is at half full speed and continues to shift the power available to propulsive power at higher speed until it is 80% propulsive power and 20% exhaust power. My question now is: In simple terms and with acceptable approximations, does this shift that redistributes the power from exhaust to the propulsive power as the speed increases really happen?

  • $\begingroup$ Increasing speed increases the aircraft "power" by definition. Constant thrust (from the jet engine) will make the aircraft accelerate until drag brings it to a steady state speed. This is why we can differentiate engine output as thrust/time compared with fuel burn/time and "power" = Thrust x Velocity as its energy state. $\endgroup$ – Robert DiGiovanni Mar 27 '19 at 18:36
  • $\begingroup$ @RobertDiGiovanni, jet engines don't produce anything close to constant thrust by long shot. For pure turbojets and low bypass turbofans the increase in efficiency due to pressure recovery is however significant enough that the peak thrust is not at standstill. High-bypass turbofans behave more like props. $\endgroup$ – Jan Hudec Mar 27 '19 at 20:33
  • $\begingroup$ @Jan Hudec never said they always did, see sentence 3 paragraph 2 of my answer. The point is: that even with perfectly constant thrust, the Power, by definition, increases as velocity. Without going back to James Watt, or treating jet propulsion differently (why should we?), thrust will always be energy released from fuel burn (with an efficiency factor)- friction and drag from running the engine. This is different from the energy state A x V of the object being propelled and should not be confused. The Force from drag/friction resulting from movement brings it to steady state velocity. $\endgroup$ – Robert DiGiovanni Mar 27 '19 at 22:42

Your definitions are correct and your derivations correctly use the law of conservation of energy, so yes, the power distribution change does happen.

The more important question is what does this actually mean. Energy is a rather curious quantity in that while it is conserved in each reference frame (including non-inertial as long as appropriate potential is included), the values of kinetic and potential energy will be different in each!

You can look at all the cases in the reference frame of the aircraft and then the propulsive power will always be zero, because the velocity is—by definition of the reference frame—zero. But the very same situation viewed from the reference frame of the Earth it will be as you describe, and from the reference frame of the air mass it will be similar, but the values will differ slightly as the reference frame move relative to each other at the speed of the wind.

Note that enthalpy does not vary in coordinate transformation, so the sum of the power towards moving the aircraft and towards accelerating the air stream is will be the same in all the reference frames. Most easily calculated in the reference frame of the aircraft from difference in kinetic energy between the intake and the exhaust. This is the output power of the engine itself, and comparing this with the heating value of the fuel will give you the thermodynamic efficiency. Thermodynamic efficiency of turbine engines increases with flight speed due to pressure recovery, i.e. the ram air pressure increasing the effective compression ratio.

Also note that this variance in propulsive power is there for all kinds of propulsion systems, including propellers and wheels of ground vehicles. The propulsive power is always zero at standstill, and how quickly it increases depends on the reactive mass they have available. The more reactive mass, the larger the propulsive efficiency (= propulsive power to output power) will be, and the faster it will increase with velocity. Propellers have some more reaction mass available due to larger diameter, and wheels have whole Earth available, but their thrust is still limited by the surface friction and the maximum torque of the engine (and the friction of the clutch for engines that don't produce torque at zero RPM).


In your calculations, you multiply the jet engine thrust with the airspeed in order to get to propulsive power. Jet engines produce relatively constant thrust at all airspeeds - it goes against our intuition to define propulsive power as zero with the engine at full thrust on the runway before take-off, and I assume that that is the basis of your question.

In order to get from thrust in [N] to power in [W] we indeed can multiply a force with a velocity, but it makes much more sense to take the velocity of the outflowing exhaust gas stream (relative to the aircraft exhaust) for this calculation. You'll find a much more consistent efficiency: the internal fuel energy is now converted into heat + kinetic energy of the exhaust gas.

  • $\begingroup$ Actually, jet engines don't produce anything close to constant thrust by long shot. For turbojets, the pressure recovery is very significant, which means their thrust first declines a bit from standstill to about 0.3M, then increases again, peaks somewhere between maybe M1.5 and M2.5 and only then falls off as the flight speed approaches the exhaust speed. Turbofans behave more like props and their thrust declines faster due to the much lower exhaust speed. $\endgroup$ – Jan Hudec Mar 27 '19 at 20:40

This is where physics can be really fun. And there certainly are many possible interpretations here. The definition of Power = Thrust x Velocity = Mass x Acceleration x Velocity may more correctly apply to the energy of the F-16 as an impactor crashing into something rather then it's power output from burning fuel.

Engine power output may be more aptly described as Thrust + Compressor Drag. This is a much simpler addition of forces. Notice once the plane is moving and achieves steady state velocity, efficiency may increase due to ram air unloading the compressor, but the thrust required to maintain that speed will still be equal to the drag of the aircraft.

Notice, as we move to fans and turboprops, fan or prop "drag" becomes the dominant propulsive source. (Prop airfoil "lift" = "drag").

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    $\begingroup$ Power and force are different dimensions, so you can't make the definition you are trying to in the second paragraph. The definition in the question is the correct one anyway (mind the “propulsive” qualification). $\endgroup$ – Jan Hudec Mar 27 '19 at 19:54
  • $\begingroup$ In reality an engine produces thrust. Rocket, jet, prop, you name it. An examination of fuel burn vs thrust is efficiency. Movement and energy state of the propelled object are covered by A x V and are expressed as power. Because power can vary with speed at constant thrust, and drag factors are involved with movement, these are two different subjects. That is the point. $\endgroup$ – Robert DiGiovanni Mar 27 '19 at 23:09
  • $\begingroup$ Keep in mind that efficiency is a well defined term in physics that always means ratio of powers. Thrust specific fuel consumption may be a better comparison of jet engines, especially turbojets, than efficiency, and is usually used for the purpose, but it shouldn't be called efficiency. And a power output still has to be power, not force (power output of jet engine is the change of energy of the working fluid, power output of the core is the change in energy of the working fluid in the core plus torque times rpm on the fan). $\endgroup$ – Jan Hudec Mar 28 '19 at 6:31
  • $\begingroup$ I will go with thrust specific fuel consumption for all engines, not just jets. Reading Koyovis as well, thanks. See Kampf. $\endgroup$ – Robert DiGiovanni Mar 28 '19 at 6:45

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