Question

For a given air-standard power, the propulsive efficiency of a turbofan engine is more than that of a turbojet engine. Which of the following is/are the reason(s) for this?

• A. The mass flow rate is more for a turbofan engine
• B. The exit velocity is lower for a turbofan engine
• C. A turbofan engine operates at a lower altitude
• D. The fan of a turbofan engine consumes lesser power

Hint:

Propulsive efficiency rises as the jet exit velocity gets closer to the flight speed; a bigger mass flow rate needs a smaller velocity increment for the same thrust.

Answer 1

The Correct Option is A, B

Step 1: Recall the Propulsive Efficiency Formula.
For an air breathing jet engine, ignoring the small fuel mass added, propulsive efficiency compares the useful thrust power to the total kinetic energy rate given to the jet:
\[ \eta_p = \frac{2}{1 + \dfrac{V_e}{V_\infty}} \]
where \(V_e\) is the jet (exit) velocity and \(V_\infty\) is the flight speed. This shows that \(\eta_p\) is highest when \(V_e\) is close to \(V_\infty\), that is, when the jet velocity is only a little more than the flight speed.

Step 2: Relate Thrust, Mass Flow and Exit Velocity.
Thrust for a given air-standard power comes from the momentum change of the working fluid:
\[ T \approx \dot{m}(V_e - V_\infty) \]
For a fixed thrust or a fixed power requirement, a larger mass flow rate \(\dot{m}\) needs only a small velocity increment \((V_e - V_\infty)\) to produce that same thrust, while a smaller mass flow rate needs a much larger velocity increment.

Step 3: Apply This to a Turbofan.
A turbofan takes in a large bypass mass flow through the fan in addition to the core flow, so its total mass flow rate \(\dot m\) is much larger than that of a turbojet producing the same thrust. Because of Step 2, this larger \(\dot m\) means the turbofan only needs to accelerate the air by a small amount, so its exit velocity \(V_e\) stays much closer to \(V_\infty\) than a turbojet's does. From Step 1, a \(V_e\) closer to \(V_\infty\) directly gives a higher \(\eta_p\). So both a higher mass flow rate (option A) and a lower exit velocity (option B) are the reasons for the turbofan's higher propulsive efficiency, and they are really two sides of the same cause.

Step 4: Why C and D Are Wrong.
Option (C), operating at a lower altitude, is not a general reason for higher propulsive efficiency at all. Propulsive efficiency depends on the ratio \(V_e/V_\infty\), not on altitude; turbofans are in fact widely used at high cruise altitudes. Option (D), the fan consuming lesser power, is also not the cause. The fan actually absorbs a large share of the engine's shaft power, that is exactly why the bypass flow gets its extra mass and momentum, so the fan does not consume lesser power; and even if it did, that would be a statement about internal power distribution, not about the propulsive efficiency definition, which depends only on \(V_e\) versus \(V_\infty\).

Final Answer:
\[ \boxed{\text{Higher mass flow rate and lower exit velocity, i.e. options (A) and (B)}} \]