# How energy efficient is an aircraft take off?

How much work is generally done by an aircraft (e.g. B737NG) in order to reach V_rotate, and how does this compare to the energy content of the fuel that is used?

• Welcome to aviation.SE. This looks like an assignment. What have you tried? What exactly do you need help with? what do you mean "how does this compare to the energy content of the fuel" ? – Federico Mar 12 '19 at 14:24
• You can determine this by measuring fuel consumption to reach Vrotate (factoring in head, tail, or no wind). Power would be full or close to full, so check fuel burn rate under those conditions and average time to reach rotation speed. That should get you close. – Robert DiGiovanni Mar 12 '19 at 16:07

You can make a decent approximation with very basic numbers. Take for example the CFM56-7B (which is incidentally also used on the 737NG).

• Maximum thrust specific fuel consumption $$TSFC$$ at takeoff: ~10.9 g/s/kN
• Maximum thrust $$F$$ at takeoff: 121 kN
• Jet fuel specific energy $$e$$: 43 MJ

Combining these figures gives a fuel burn of 1.4 kg/s at takeoff, or 57 MJ of chemical energy released every second ($$P=57MW$$).

To get an airplane of mass $$m$$ to a certain speed $$v$$, it needs $$E=\frac{1}{2}mv^2$$ of kinetic energy. It does so in $$t=\frac{v}{a} = \frac{vm}{F}$$ seconds (remember, $$F=ma$$). The efficiency $$\eta$$ is useful work done $$E$$ divided by input energy ($$Pt$$), $$\eta =\frac{mv^2}{2Pt} = \frac{vF}{2P} = \frac{v}{2\cdot TSFC \cdot e}$$

Immediately, we see why TSFC is an important figure. Plugging in some numbers, let's say $$V_R =145\mathrm{kts} \approx 75\mathrm{m/s}$$, we get a total efficiency of $$\eta \approx 8\%$$ until rotation.

Note that this calculation assumes constant TSFC while in reality it is very much a function of $$v$$ (indeed, with this calculation you could have $$\eta>1$$ if $$v\to\infty$$ which is clearly impossible).

• Now you've made me wonder what would happen if someone were to combine these calculations with the special relativity colinear velocity addition formula... – user Mar 12 '19 at 16:33
• @aCVn Nothing interesting I'm afraid. The situation I sketch is similar to a small rocket being fed fuel from an impractically huge rocket travelling alongside it - it would approach light speed but never reach it, and be infinitely efficient because I'm not counting the energy expended by the impractically huge rocket. – Sanchises Mar 12 '19 at 16:39
• What would happen with the efficiency if an electric engine would be used instead of the CFM motor? (considering instant available torque and higher efficiency of the motor) – JNld Mar 22 '19 at 10:43
• @JNld Your batteries run out quickly. But apart from that, it's mostly down to the difference between a propeller engine (constant output power at constant input power) and a jet engine (constant output thrust at constant input power). Again, those are first-order approximations. You will find that a propeller, electric or not, is much more efficient at low speed. You can find the equation for propeller efficiency here – Sanchises Mar 22 '19 at 16:00