Aerospace Engineering DISCUSSION

Why does a high-bypass turbofan burn less fuel than a turbojet producing the same thrust?

Started by sarikasourcecode propulsive efficiencybypass ratioturbofan vs turbojetjet velocitythrust equation
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Latest activity · 30 Sep 2026

Why does a high-bypass turbofan burn less fuel than a turbojet producing the same thrust?

sarikasourcecode Aerospace Engineering Forum
#1

Both a turbojet and a turbofan burn fuel in a gas turbine core, and thrust is mass flow times the change in velocity. So I would expect a small mass of air accelerated a lot to be equivalent to a large mass accelerated a little.

Why then is the high-bypass turbofan so much more efficient at airliner speeds? I would like to see it with numbers, for example 50 kN of thrust at a flight speed of 250 m/s.

Community replies 4

Re: Why does a high-bypass turbofan burn less fuel than a turbojet producing the same thrust?

#2

The two are equivalent in thrust but not in energy. Thrust is momentum: F = mass flow × (Vj - V0), where Vj is the jet velocity and V0 the flight speed. The power the engine has to put into the air is kinetic energy, which goes with velocity squared. A small mass flow with a large velocity increase takes much more energy for the same momentum change than a large mass flow with a small increase.

The useful output is thrust power, F × V0. Whatever kinetic energy is left in the jet relative to the still air, ½ × mass flow × (Vj - V0)², is wasted in the wake.

Re: Why does a high-bypass turbofan burn less fuel than a turbojet producing the same thrust?

#3

Numbers for 50 kN at 250 m/s. Thrust power is 50,000 × 250 = 12.5 MW in both cases. Turbojet with a jet velocity of 900 m/s: mass flow = 50,000 / 650 = 77 kg/s, wasted jet power = ½ × 77 × 650² = 16.3 MW. Turbofan with a mean jet velocity of 350 m/s: mass flow = 50,000 / 100 = 500 kg/s, wasted power = ½ × 500 × 100² = 2.5 MW.

Propulsive efficiency is useful over total: 12.5 / 28.8 = 43 percent for the turbojet and 12.5 / 15 = 83 percent for the turbofan. The shortcut formula, 2 / (1 + Vj / V0), gives the same two figures.

Re: Why does a high-bypass turbofan burn less fuel than a turbojet producing the same thrust?

#4

Propulsive efficiency is only half of the picture. Overall efficiency is thermal efficiency times propulsive efficiency. The thermal part, how much of the fuel energy the core turns into mechanical power, depends on the overall pressure ratio and the turbine entry temperature. The ideal Brayton cycle gives 1 - 1 / r^0.286 for air, about 65 percent at a pressure ratio of 40, and real engines are well below the ideal because of component losses and cooling air.

The turbofan lets the designer improve both: a hot, high-pressure core for thermal efficiency, and a large fan driven by that core to move a lot of air slowly.

Re: Why does a high-bypass turbofan burn less fuel than a turbojet producing the same thrust?

#5

There are limits to pushing the bypass ratio higher. The fan and nacelle get larger, which adds weight and drag and runs into ground clearance under the wing. A large fan turning at the speed the low-pressure turbine prefers would have very high tip speeds with high losses and noise, which is one reason geared fans exist.

Thrust also falls off faster with flight speed when the jet velocity is low, since thrust depends on Vj - V0. For supersonic aircraft a low-bypass engine with a high jet velocity is still the right choice, while a propeller, in effect a very high bypass ratio, wins at lower speeds.

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