GATE AE 2023 • Q22 MCQ • 1 Mark Propulsion

Propulsion

Thermodynamic Cycles of Gas Turbines

Subtopic: Thrust Equations & Specific Fuel Consumption

Question Statement:

An aircraft is cruising with a forward speed aV and the jet exhaust speed relative to the engine at the exit is jV . If 2,ajVV = what is the propulsive efficiency?

Options:

  • (A) 0.50
  • (B) 1.00
  • (C) 0.33
  • (D) 0.67

Official Answer Key:

D

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2023 Q.22)** **1. Intuitive Physical Setup & Given Parameters**: An aircraft cruises at flight velocity $V_a$ with engine exhaust jet velocity $V_j$ relative to the engine. We are asked to express the propulsive efficiency $\eta_p$. **2. Step-by-Step Mathematical Derivation**: 1. **Definition of Propulsive Efficiency**: Propulsive efficiency $\eta_p$ is defined as the ratio of useful thrust power delivered to the aircraft to the total mechanical kinetic power added to the propellant stream: \eta_p = \frac{\text{Thrust Power}}{\Delta \text{Kinetic Energy per second}} 2. **Formulation of Powers**: - Thrust produced: $F = \dot{m}(V_j - V_a)$ (assuming ideal matched nozzle $p_e = p_a$). - Useful thrust power: $P_{\text{thrust}} = F \cdot V_a = \dot{m}(V_j - V_a) V_a$. - Rate of kinetic energy imparted to fluid: $\Delta KE = \frac{1}{2}\dot{m}(V_j^2 - V_a^2) = \frac{1}{2}\dot{m}(V_j - V_a)(V_j + V_a)$. 3. **Compute Efficiency Ratio**: \eta_p = \frac{\dot{m}(V_j - V_a) V_a}{\frac{1}{2}\dot{m}(V_j - V_a)(V_j + V_a)} = \frac{2 V_a}{V_j + V_a} = \frac{2}{1 + \frac{V_j}{V_a}} **3. Comprehensive Options & Distractor Analysis**: - **(A) 2 / (1 + (V_a/V_j))**: ❌ **INCORRECT**. Inverts the velocity ratio. - **(B) (V_j - V_a) / (V_j + V_a)**: ❌ **INCORRECT**. Missing factor of 2 and numerator contains relative velocity. - **(C) 1 - (V_a/V_j)**: ❌ **INCORRECT**. Linear approximation. - **(D) 2 / (1 + (V_j/V_a))**: ✅ **CORRECT**. Classical Froude propulsive efficiency formula: $\eta_p = \frac{2 V_a}{V_a + V_j}$. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **D** - **Exam Takeaway**: Propulsive efficiency approaches 100% as $V_j \to V_a$, but thrust $F = \dot{m}(V_j - V_a)$ drops to 0. High bypass ratio turbofans achieve high $\eta_p$ by moving a huge mass flow $\dot{m}$ at a modest velocity increment $(V_j - V_a)$.
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