The lift per unit span for a spinning circular cylinder in a potential flow is 6 N/m.
The free-stream velocity is 30 m/s, and the density of air is 1.225 kg/m3. The
circulation around the cylinder is ____ m2/s (rounded off to two decimal places).
Official Answer Key:
0.165
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.34)**
**1. Intuitive Physical Setup & Given Parameters**:
A spinning circular cylinder in 2D potential flow generates lift per unit span $L' = 6\text{ N/m}$.
Given parameters:
- Free-stream velocity: $V_\infty = 30\text{ m/s}$
- Free-stream air density: $\rho_\infty = 1.2\text{ kg/m}^3$
We need to compute the circulation $\Gamma$ (in $\text{m}^2/\text{s}$) around the cylinder.
**2. Step-by-Step Mathematical Derivation**:
1. **Governing Aerodynamic Theorem (Kutta-Joukowski)**:
By the Kutta-Joukowski lift theorem, the aerodynamic lift per unit span on any 2D body generating bound circulation $\Gamma$ is:
$L' = \rho_\infty V_\infty \Gamma$
2. **Rearrange for Circulation $\Gamma$**:
$\Gamma = \frac{L'}{\rho_\infty V_\infty}$
3. **Substitute Given Numerical Values**:
$\Gamma = \frac{6\text{ N/m}}{(1.2\text{ kg/m}^3)(30\text{ m/s})} = \frac{6}{36} = \frac{1}{6} \approx 0.1667\text{ m}^2/\text{s}$
Rounding to three decimal places gives **0.167** (acceptable interval: $[0.16, 0.17]$).
**3. Comprehensive Options & Distractor Analysis**:
- Numerical Answer Type (NAT) Question: No options provided.
- Computed Value: $0.167\text{ m}^2/\text{s}$.
- Official Range: `[0.16, 0.17]`.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **0.165** (Acceptable Range: `[0.16, 0.17]`)
- **Exam Takeaway**: The Kutta-Joukowski theorem $L' = \rho_\infty V_\infty \Gamma$ applies universally to rotating cylinders (Magnus effect) as well as arbitrary lifting airfoils. Check units: $\text{N/m} / (\text{kg/m}^3 \cdot \text{m/s}) = \text{m}^2/\text{s}$.
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