GATE AE 2026 • Q56 NAT • 2 Marks Aerodynamics & Fluid Mechanics

Aerodynamics & Fluid Mechanics

Airfoil & Wing Theory

Subtopic: Finite Wing Theory & Induced Drag

Question Statement:

Consider a finite wing of aspect ratio 10 with span effectiveness factor 0.95. Its airfoil section has a lift slope of 0.106 per degree and a zero-lift angle of attack of -1.5°. The lift coefficient of the wing at an angle of attack of 3.5° is ________ (rounded off to 2 decimal places).

Official Answer Key:

0.44

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2026 Q.56)** **1. Intuitive Physical Setup & Given Parameters**: - Finite wing aspect ratio: $AR = 10$ - Span effectiveness factor: $e = 0.95$ - Airfoil lift slope: $a_0 = 0.106\text{ per degree}$ - Zero-lift angle of attack: $\alpha_{L=0} = -1.5^\circ$ - Operating angle of attack: $\alpha = 3.5^\circ$ We need to calculate the wing lift coefficient $C_L$ (rounded off to 2 decimal places). **2. Step-by-Step Aerodynamic Derivation**: 1. **Finite Wing Lift Curve Slope Formula**: Using Helmbold/Prandtl finite wing theory, the 3D lift curve slope $a$ (per degree) is given by: a = \frac{a_0}{1 + \frac{57.3 a_0}{\pi AR e}} where the factor $57.3 \approx \frac{180}{\pi}$ converts radians to degrees in the induced downwash term. 2. **Compute Denominator Term**: \frac{57.3 \times 0.106}{\pi \times 10 \times 0.95} = \frac{6.0738}{29.8451} \approx 0.20351 a = \frac{0.106}{1 + 0.20351} = \frac{0.106}{1.20351} \approx 0.088076\text{ per degree} 3. **Calculate Lift Coefficient $C_L$**: The lift coefficient is: C_L = a (\alpha - \alpha_{L=0}) \alpha - \alpha_{L=0} = 3.5^\circ - (-1.5^\circ) = 5.0^\circ C_L = 0.088076 \times 5.0 \approx 0.44038 \approx 0.44 **3. Comprehensive Options & Distractor Analysis**: - **Evaluation Criteria**: Numerical Answer Type (NAT). - **Exact Analytical Result**: $C_L \approx 0.440$. - **Rounding Requirement**: Two decimal places $\implies \mathbf{0.44}$. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **0.44** (Acceptable Range: **0.42 to 0.46**) - **Exam Takeaway**: When airfoil slope is given per degree, convert the induced drag correction term via $\frac{180 a_0}{\pi^2 AR e} \approx \frac{57.3 a_0}{\pi AR e}$ before computing $C_L = a(\alpha - \alpha_0)$.
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