GATE AE 2026 • Q24 MSQ • 1 Mark Propulsion

Propulsion

Thermodynamic Cycles of Gas Turbines

Subtopic: Thrust Equations & Specific Fuel Consumption

Question Statement:

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?

Options:

  • (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

Official Answer Key:

A, B

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2026 Q.24)** **1. Intuitive Physical Setup & Given Parameters**: For a given air-standard thrust power (or kinetic power added to the jet), the propulsive efficiency $\eta_p$ of a turbofan engine is higher than that of a turbojet engine. We are asked to identify the physical reasons behind this higher propulsive efficiency (MSQ). **2. Step-by-Step Propulsion Derivation**: 1. **Definition of Thrust and Kinetic Energy Flux**: - Thrust generated by a jet engine is approximately: T = \dot{m}(v_j - v_0) - The rate of kinetic energy imparted to the flow (wasted kinetic energy in the exhaust wake) is: \Delta \dot{KE} = \frac{1}{2}\dot{m}(v_j^2 - v_0^2) 2. **Propulsive Efficiency Formulation**: - The Froude propulsive efficiency is given by: \eta_p = \frac{T v_0}{T v_0 + \frac{1}{2}\dot{m}(v_j - v_0)^2} = \frac{2 v_0}{v_j + v_0} = \frac{2}{1 + \frac{v_j}{v_0}} - To achieve high propulsive efficiency (bringing $\eta_p \to 1$), the ratio $v_j/v_0$ must be as close to 1 as possible, meaning the jet exit velocity $v_j$ must be as low as possible. 3. **Turbofan Operating Mechanism**: - A turbofan bypasses a large mass of air around the core: - It accelerates a **larger mass flow rate** $\dot{m}$ of air. - To a **lower exit velocity** $v_j$. - This produces the required thrust $T = \dot{m}(v_j - v_0)$ while minimizing the wasted kinetic energy $\frac{1}{2}\dot{m}(v_j - v_0)^2$, substantially boosting $\eta_p$. - Therefore, the reasons are: higher mass flow rate (Option A) and lower exit velocity (Option B). **3. Comprehensive Options & Distractor Analysis**: - **(A) The mass flow rate is more for a turbofan engine**: ✅ **CORRECT**. The bypass stream dramatically increases total air mass flow rate $\dot{m}_{\text{total}} = \dot{m}_{\text{core}} + \dot{m}_{\text{fan}}$. - **(B) The exit velocity is lower for a turbofan engine**: ✅ **CORRECT**. Lower jet velocity $v_j$ reduces wake dissipation and maximizes $\eta_p = 2/(1 + v_j/v_0)$. - **(C) A turbofan engine operates at a lower altitude**: ❌ **INCORRECT**. Turbofans power high-altitude commercial transports at 30,000–40,000 ft; altitude is not the governing physical reason for higher $\eta_p$. - **(D) The fan of a turbofan engine consumes lesser power**: ❌ **INCORRECT**. The fan absorbs massive turbine shaft power (often $70\text{--}80\%$ of the core turbine power) to accelerate the bypass stream. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **A, B** - **Exam Takeaway**: "Accelerate a large mass of air gently rather than a small mass violently" is the golden rule of high propulsive efficiency, explaining why high-bypass turbofans dominate commercial aviation.
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