I'm interested in how the thrust of a turbofan engine is affected at higher airspeeds (TAS). I know (I believed) that engine thrust(at constant N1) was relatively constant like in the following graph (only slight deviations):

enter image description here

This graph is usually in the books/manuals describing engine performance with reference to speed.

Then I came across the data from CFM56-5C turbofan engine which states that max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N. That's almost 5 times more power on the ground than in cruise. Here is the link: How much air, by mass, enters an average CFM56 turbofan engine cruising per minute?

These are apparently contradictory statements or I am missing something. Which one is corrrect and why? Why is engine thrust being changed with speed? Also, on the graph above what are those two curves that when added form a net engine thrust?

After I did a few calculations using the thrust equation (F= mass flow * difference in exhaust and inlet velocities denoted as delta V -> we will disregard the fuel mass flow and assume exit pressure is equal to the free stream pressure thanks to a nozzle) and following data mentioned above in the link, I found out that delta V term in cruise and on takeoff is constant (at full power) and its value is 295 m/s, which states that exhaust velocity of the engine will always be 295 m/s faster from the inlet velocity(for a maximum power setting at any speed). I think that's logical because work done by the engine is used to increase kinetic energy (delta Ek) of the airflow which increases speed always by a constant amount at specific power/N1 setting (of course less power equals less delta V).

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    $\begingroup$ Please note the following: you state that max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N. That's almost 5 times more power on the ground than in cruise... That's wrong, because thrust is not power. There exist a relation between thrust, speed, and power, but the stationary case is an special one... $\endgroup$
    – xxavier
    Commented Mar 29, 2018 at 20:49
  • $\begingroup$ In terms of useful work, $energy = force \cdot distance$, $power = force \cdot velocity$ (vector dot products). At zero speed, using power to blow all that air behind you costs a lot of fuel energy but isn't doing anything useful. Even when you release the brakes and start to roll, it's not much power. (Jet engines are very inefficient at low speed.) $\endgroup$ Commented Sep 14, 2021 at 7:45

2 Answers 2


The first diagram you link to shows three lines but does not indicate what they represent. I guess the bold line is thrust over speed. Then this diagram is correct for a turbojet.

Thrust over airspeed diagram

Thrust $T$ is the difference between the engine's exit impulse minus the entry impulse: $$T = (\dot{m}_{air} + \dot{m}_{fuel})\cdot v_{exit} - \dot{m}_{air}\cdot v_{entry}$$ The exit speed $v_{exit}$ of a turbojet engine is almost constant over flight speed (relative to the engine of course), so as the engine accelerates, a larger entry impulse must be subtracted from a nearly constant exit impulse. Thrust drops slightly over speed.

At higher Mach numbers, precompression from the ram effect at the intake raises the pressure level (and hence the mass flow $\dot{m}_{air}$) inside the engine, so it will develop more thrust than in static conditions. This effect causes the thrust line to bend upwards at higher speed, and since the precompression grows nonlinearly with speed, the initial drop in thrust is soon reversed. Of course, now the fuel mass flow $\dot{m}_{fuel}$ will grow in the same way, so the fuel efficiency (thrust per fuel used) will continue to drop as speed increases.

Only when flight speed approaches the exit speed of the jet will thrust go down again. The typical exit speed of a turbojet is easily supersonic, therefore this type of engine is well suited for supersonic flight.

max engine thrust at cruise is approximately 29,360 Newtons while it's max thrust when stationary is 140,000 N

Here you have two effects combining to lower thrust. One is the reduction in the difference between entry and exit speed. This is more pronounced in a turbofan engine because the bypass flow will be accelerated much less than the core flow, and a higher flight speed will cause a proportionally bigger drop in thrust.

The second effect comes from the difference in air density between ground and cruise: Air density at a typical cruise altitude of 35,000 ft is only 0.38 kg/m³ or 31% of the air density at sea level. The original source for the cruise thrust number does not say for which altitude the figure is valid, but you can be sure that it is for about one third of ground density. Mass flow $\dot{m}_{air}$ is directly proportional with ambient density, and both effects combine. However, most sources give only a drop to a quarter of static thrust - the last table in the linked answer looks like someone mixed the values for the CFM56-5A and the CFM56-5C.

  • $\begingroup$ Just to clarify, by a turbojet's exit speed being "supersonic" is that meaning within the exhaust (with a higher speed of sound because it's hot) the actual flow is supersonic, or is it subsonic but faster than the speed of sound in the surrounding air? $\endgroup$
    – Talisker
    Commented Feb 24, 2019 at 22:04
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    $\begingroup$ @Talisker: Yes, supersonic means more than the speed of sound of this hot air. There are supersonic aircraft with subsonic engine exit speed, and that only works because of the higher speed of sound in hot air, but those are the exception. $\endgroup$ Commented Feb 25, 2019 at 6:18
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    $\begingroup$ @AbanobEbrahim: Yes, very good question! The exit speed is approx. constant relative to the aircraft, so this puts an upper limit on the maximum speed which can be achieved. $\endgroup$ Commented Mar 27, 2019 at 22:00
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    $\begingroup$ @PeterKämpf: Thank you. I'm new here on Aviation and I already admire your awesome answers. But just to make sure I understand you correctly, let's assume the exhaust velocity when the engine is stationary is 500 $m/s$. Now by approx. constant you mean that when the aircraft is moving at 200 $m/s$, the exhaust speed relative to the aircraft is still 500 $m/s$ but to someone standing on the ground the exhaust speed is just 300 $m/s$ in the opposite direction of the aircraft. So is that correct? $\endgroup$ Commented Mar 27, 2019 at 23:24
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    $\begingroup$ @AbanobEbrahim: Yes, that is correct. For the nitpickers: The actual numbers might vary a bit between the different scenarios. $\endgroup$ Commented Mar 28, 2019 at 6:17

Air density or pressure altitude must be factored in when comparing engine thrust figures. At cruise, FL400, air density will be about one fifth. Max thrust (or max HP for piston engines) will also be one fifth, as will air resistance/parasite drag. Air mass flow will be one fifth, so will fuel flow/burn. Even lift will be one fifth for a given air speed (TAS).

I assume you meant TAS in your question, however, if we consider IAS/CAS, that is a different story. At FL400, you would have numbers like Mach 0.82, 470kts TAS and 250kts IAS. Airliners rarely do level flight except at cruise altitude, however it would be possible to maintain 250kts IAS/CAS with 20% of static thrust (give or take) at altitudes lower than cruise, such as in a hold.

Playing with these might be helpful:

IAS/TAS vs Altitude: https://aerotoolbox.net/airspeed-conversions/

Engine power vs Altitude: http://www.csgnetwork.com/relhumhpcalc.html

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    $\begingroup$ I'm not sure that this answers the question that was asked. It's not bad information, but it's about how thrust varies with altitude, rather than what the question asks, which is why thrust varies with airspeed, per se. Also, I'd question your assertion that an airliner could ever maintain speed with only 20% of available power. Maybe on an incredibly light 747 or A-380, but on the 737's that I'm familiar with, that statement seems unsupportable. $\endgroup$
    – Ralph J
    Commented Mar 3, 2020 at 20:44
  • $\begingroup$ Ralph J, you are misunderstanding me if you say "20% of available power". I am addressing the question which references one fifth of static thrust or 29,360N. By definition of what IAS is and what it implies, how can anyone take issue with my statement? If thrust is 29,360N in cruise at a particular indicated airspeed, both will be approximately the same for any other altitude (except of course for speeds and altitudes out of limits). $\endgroup$ Commented Mar 4, 2020 at 23:57

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