We already have a few questions and answers about the qualitative effects of altitude on fuel economy:

I am interested in understanding how the changes in pressure and temperature at different altitudes affect the fuel economy of a turbofan powered aircraft quantitatively. In the end, I would like to make a plot of relative fuel economy vs. altitude that takes all of these effects into account, but I don't know how to quantitatively combine these effects.

Some notes:

  • By fuel economy I mean fuel required per distance traveled, not time.
  • I am not interested in absolute numbers for the fuel per distance, which would require specifying a particular aircraft. I am only interested in how the fuel per distance figure would relatively change with altitude, e.g. normalized to 1 at sea level.
  • I assume ISA (International Standard Atmosphere) profiles for pressure and altitude.
  • I assume the no wind case. Different winds at different altitudes will of course have an effect on the result, but it is easy to take this into account after the no wind case is understood.
  • Let's assume a typical climb profile for a short- to medium-haul jet airliner: 250/280/0.78

    TAS and Mach for a typical climb profile

    You can see that the TAS increases until reaching Mach 0.78, then decreases due to the lower temperatures causing a lower speed of sound and then remains constant above the tropopause. I am particularly interested in how the fuel economy will behave around these altitudes.

  • $\begingroup$ Fuel economy per hour, per mile? Set speed or set AoA? A lot depends on gross weight. My flight planning cruise tables for a 737-400 indicate that best economy is found around 2000-5000 feet below service ceiling when operating m.74-m.78. $\endgroup$ – Max Power Sep 23 at 2:12
  • $\begingroup$ @MaxPower It's in the question: "By fuel economy I mean fuel required per distance traveled, not time." and "Let's assume a typical climb profile for a short- to medium-haul jet airliner: 250/280/0.78" (that is 250KIAS below FL100, 280KIAS above, until reaching Mach 0.78). $\endgroup$ – Bianfable Sep 23 at 7:05

I am adding this community wiki answer to show my current state of research and to provide the plot vs. altitude, which I will update when I learn more. Comments are welcome.

Thermal Efficiency

From this answer by Peter Kämpf, we know that the thermal efficiency for jet engines is given by

$$ \eta = \frac{T_\text{max} - T_\text{amb}}{T_\text{max}} $$

where $ T_\text{amb} $ is just the ambient temperature (from ISA) and $ T_\text{max} $ is the temperature resulting from the combustion. If I understand the answer correctly, this should be about 1100 K above ambient temperature, so I am currently using this term to describe the impact of thermal efficiency on fuel economy:

$$ \epsilon_\text{T} \propto \frac{1}{\eta} = \frac{T_\text{max}}{T_\text{max} - T_\text{amb}} = \frac{T_\text{amb} + 1100 \, \mathrm{K}}{1100 \, \mathrm{K}} $$

I am not sure if the increase in temperature of 1100 K is constant with altitude, so please correct me if this is wrong.


From another answer by Peter Kämpf, we know that induced drag is proportional to dynamic pressure

$$ q = \frac{v^2}{2} \cdot \rho $$

with $ v $ being the TAS and $ \rho $ the density (known from ISA). Since work required to overcome the drag per distance is proportional to the force, the fuel economy should just scale with

$$ \epsilon_\text{drag} \propto \text{TAS}^2 \cdot \rho $$


For the combined (relative) fuel economy term, I just multiply all previous terms:

$$ \epsilon = \epsilon_T \cdot \epsilon_\text{Drag} $$

The following plot now shows the relative fuel required per distance. Each curve has been normalized to 1 at sea level.

Fuel Economy vs. altitude

You can see that the drag term dominates over the thermal efficiency term. Since drag is constant at the same indicated airspeed, the resulting fuel economy looks roughly proportional to IAS. This result would imply that jets could just as well fly at sea level, which does not make any sense, so I am pretty sure that I am still missing at least one contribution!

| improve this answer | |
  • $\begingroup$ Watch out. The “drag is proportional to dynamic pressure” is sloppy wording from Peter. It is a function of dynamic pressure, but there is a dynamic pressure where drag is minimal. And there is another indicated airspeed a little above that where the power of drag (drag times speed is the rate of energy loss due to drag) and these only depend on the dynamic pressure, which means indicated airspeed, and not altitude (much; see that Peter's answer). But the conclusion that drag is basically constant is correct. $\endgroup$ – Jan Hudec Jun 6 at 12:04
  • $\begingroup$ Your conclusion is actually correct for propeller-driven aircraft. With decent constant speed propeller the propulsive efficiency, that is fuel consumed for unit of energy actually given to the aircraft is fairly flat. And because constant drag means constant energy for travelling given distance, the fuel consumption varies little with altitude. That is, however, not the case with turbojet and turbofan engines. Their propulsive efficiency increases quite significantly with forward speed. Unfortunately I didn't find good quantification anywhere. $\endgroup$ – Jan Hudec Jun 6 at 12:20
  • $\begingroup$ @JanHudec Thanks for your comments. I knew I was missing at least one term. I will have a closer look at the jet propulsive efficiency... $\endgroup$ – Bianfable Jun 6 at 14:11
  • $\begingroup$ the problem is that good data is difficult to find. Long ago I found some NASA lectures with a Java applet that is supposed to calculate this. The four links I wrote down are: <grc.nasa.gov/WWW/k-12/airplane/specth.html>, <grc.nasa.gov/WWW/k-12/airplane/turbfan.html>, <grc.nasa.gov/WWW/k-12/airplane/ngnsim.html> and <grc.nasa.gov/WWW/K-12/airplane/EngineTheory.pdf>. Back then I didn't see sources of the applet and I didn't get around to read through the paper, but sources seem to be there now. $\endgroup$ – Jan Hudec Jun 6 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.