I want to figure out how much energy it would take to lift a turbofan-helicopter.

For reference I started with a rotor helicopter and assumed $T=100\ \mathrm{kN}$ gross weight and $d = 16\ \mathrm m$ rotor diameter (compare S92 or AH64).

Using the propeller-formula $$ T = ( 0.5*P^2*\eta_\text{tot}^2*\pi*d^2*\rho_\text{air})^{1/3}. $$

with $\eta_\text{tot}^2=0.58$ (source) I get an engine-output of $P=1\,870\ \mathrm{kW}$

If I now take a typical turbofan and a take-off $tsfc$ of $10\ \mathrm{g/(kNs)}$ I get a fuel consumption of $1\ \mathrm{kg/s}$ for my $100\ \mathrm{kN}$ helicopter. If I furthermore assume 30% overall efficiency and $E_\text{Jetfuel}=43.5\ \mathrm{MJ/kg}$ I get an energy demand of $13\,050\ \mathrm{kW}$ to run the fan (neglected all the other stuff taking energy from the shaft).

Does this mean it would take 7 times the amount of energy to lift a turbofan-helicopter (with engines running at full throttle)? Or a magnitude if I assume cruise tsfc...

Does this suit with your experience? Or is there a big error/misconception in my short calculation?

  • 2
    $\begingroup$ Yes, that's precisely the reason why vertical take-off in jets is limited to fighters, which have insane engine power for their weight, and very rare even there. $\endgroup$
    – Jan Hudec
    Oct 30 '17 at 16:16

Helicopter rotors are very large precisely for that reason. Lift from a rotor (or thrust from a propeller or fan) is obtained by accelerating a given mass of air m to a velocity v. The thrust obtained is proportional to the increase of momentum ∆mv. Obviously, you can get the same ∆mv by accelerating a given mass of gas to a high speed (propeller/fan) or by accelerating a larger mass to a lower speed (rotor). But the energy needed is proportional to v squared... Hence, and for the same thrust/lift, it's much more expensive, in energy or power terms, to use a propeller/fan than a big rotor ...

  • $\begingroup$ Using this logic, would it make sense to build planes with massive rotors (maybe retract/move them when near the ground)? $\endgroup$ Oct 29 '17 at 11:33
  • $\begingroup$ The Osprey V-22, when in airplane mode, has very big propellers, and flies quite well... $\endgroup$
    – xxavier
    Oct 29 '17 at 11:46
  • 3
    $\begingroup$ The complication is that an airplane's propulsion system faces an already moving (relative to the airplane) air stream. Big props work well at low speed but not so well as the aircraft gets faster. AIUI the ospreys proprotors are a compromise, they are larger than ideal for airplane mode and smaller than ideal for helicopter operation. $\endgroup$ Oct 29 '17 at 13:51
  • $\begingroup$ @peter green The size is also infuenced by the US Navy's storage requirements. $\endgroup$
    – T.J.L.
    Oct 29 '17 at 17:25
  • 1
    $\begingroup$ @xxavier your answer explains exactly why the AeroVelo Atlas human powered helicopter uses four huge rotors (the aircraft is larger than a 737) with low speed. $\endgroup$ Oct 29 '17 at 21:43

A turbofan helicopter must also take off by producing at least the thrust equal to the TO weight. With turbofans, the static thrust at zero speed is very much determined by the bypass ratio. You're mentioning a typical turbofan, so let's take a CFM56-3 with a bypass ratio of 6, this is a 100 kN engine.

  • The UH-60 has a max. TO weight of 22,000 lb and a rotor diameter of 16.4 m, so this would be your target helicopter. Its installed max. TO power is 3,086 SHP = about 2,300 kW. It needs excess power to climb, so I would say your required engine power is in the ballpark.
  • The CFM56 has a mass flow rate through the gas generator 53.5 kg/s (source an old textbook), so total mass flow = (1+6)*53.5 = 374.5 kg/s. Average exhaust velocity $V_e$= T / $\dot{m}$ = about 265 m/s. Power = $\frac {1}{2}\cdot \dot{m} \cdot {V_e}^2 $ = 13.2 kW

So I get roundabout the same numbers that you get: your statement is valid, a helicopter powered by a CFM56 engine takes about 7 times the installed power to lift off.

  • $\begingroup$ From dimensional considerations, the power $P = k\cdot T^{3/2}\cdot D^{-1}\cdot \rho^{-1/2}$. Hence, for the same thrust and density, $P = k\cdot D^{-1}}$ where k is some constant... $\endgroup$
    – xxavier
    Oct 29 '17 at 8:32
  • 1
    $\begingroup$ Try a Rolls Royce Pegasus rather than a CFM56! $\endgroup$ Oct 29 '17 at 13:56
  • 2
    $\begingroup$ @BrianDrummond Royce Pegasus is of course a successful vertical lifter of airframes. It runs 81.1 kg/s through the gas generator and has a bypass ratio of 1:1.2. If we make the same calculation as above, we get a supersonic exhaust velocity, which can only be accommodated by a convergent-divergent exhaust pipe. This the Harrier does not have, so the excess energy causes a pressure rise, which delivers thrust as well. But much more detail of the geometry of the Pegasus is required to properly make the computation. The CFM56 case allows this method because average exhaust speed is subsonic. $\endgroup$
    – Koyovis
    Oct 31 '17 at 3:13

You face another problem. Helicopter rotors are also used to maintain control of the aircraft, by varying the pitch of the rotor blades on a portion of their arc.

With a ducted turbofan, you'd have to find another way to control it. Probably bleed air exhausted on the roll and pitch axes, which is how the Harrier and F35 do it when in hover.

So you would need to add to your power budget, enough reserve bleed air for control.


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.