GATE AE 2025 • Q52 NAT • 2 Marks Aerodynamics & Fluid Mechanics

Aerodynamics & Fluid Mechanics

Airfoil & Wing Theory

Subtopic: Finite Wing Theory & Induced Drag

Question Statement:

A jet-powered airplane is steadily climbing at a rate of 10 m/s. The air density is 0.8 kg/m^3, and the thrust force is aligned with the flight path. Using the information provided in the table below, the airplane's thrust to weight ratio is ______ (rounded off to one decimal place). Airplane speed 80 m/s Lift coefficient, CL 0.8 Aspect ratio 6 Oswald efficiency factor 0.8 Zero-lift drag coefficient 0.02
Question figure for GATE AE 2025 Q52

Scientific schematic diagram reproduced from official IIT paper

Official Answer Key:

0.2

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2025 Q.52)** **1. Intuitive Physical Setup & Given Parameters**: Steady climbing flight: - Rate of climb: $R/C = V_c = 10\text{ m/s}$ - Airspeed: $V = 80\text{ m/s}$ - Air density: $\rho = 0.8\text{ kg/m}^3$ - Thrust aligned with flight path ($\alpha_T = 0$) - Lift coefficient: $C_L = 0.8$ - Wing aspect ratio: $AR = 6$ - Oswald efficiency: $e = 0.8$ - Zero-lift drag coefficient: $C_{D0} = 0.02$ We need to find the airplane's thrust-to-weight ratio $T/W$ (rounded off to one decimal place). **2. Step-by-Step Flight Mechanics Derivation**: 1. **Climb Angle ($\gamma$)**: \sin\gamma = \frac{R/C}{V} = \frac{10}{80} = 0.125 \cos\gamma = \sqrt{1 - \sin^2\gamma} = \sqrt{1 - 0.015625} \approx 0.99215 \approx 1 2. **Total Drag Coefficient ($C_D$)**: Induced drag coefficient: C_{Di} = \frac{C_L^2}{\pi AR e} = \frac{0.8^2}{\pi \times 6 \times 0.8} = \frac{0.64}{4.8\pi} = \frac{0.64}{15.0796} \approx 0.04244 Total drag coefficient: C_D = C_{D0} + C_{Di} = 0.02 + 0.04244 = 0.06244 3. **Lift-to-Drag Ratio**: \frac{L}{D} = \frac{C_L}{C_D} = \frac{0.8}{0.06244} \approx 12.812 \frac{D}{L} = \frac{1}{12.812} \approx 0.07805 4. **Equation of Motion in Climb**: Along the flight path: T = D + W\sin\gamma Perpendicular to the flight path: L = W\cos\gamma \implies W = \frac{L}{\cos\gamma} Dividing by $W$: \frac{T}{W} = \frac{D}{W} + \sin\gamma = \left(\frac{D}{L}\right)\cos\gamma + \sin\gamma \frac{T}{W} \approx 0.07805(0.992) + 0.125 \approx 0.0774 + 0.125 = 0.2024 \approx 0.2 **3. Comprehensive Options & Distractor Analysis**: - **Evaluation Criteria**: Numerical Answer Type (NAT). - **Exact Analytical Result**: $T/W \approx 0.202$. - **Rounding Requirement**: One decimal place $\implies \mathbf{0.2}$. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **0.2** (Acceptable Range: **0.18 to 0.22**) - **Exam Takeaway**: Thrust-to-weight ratio in steady climb is $T/W = \frac{D}{L} + \sin\gamma$. Here $\frac{C_D}{C_L} \approx 0.078$ and $\sin\gamma = 0.125$, summing to $0.20$.
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