A lifting surface has a spanwise circulation distribution of $\Gamma(\theta) = A \sin 3\theta$ (where $A \neq 0$) over its span $-\frac{b}{2} \le y \le \frac{b}{2}$, and $y = -\frac{b}{2} \cos \theta$ is the spanwise coordinate. Furthermore, the downwash varies along the span as $w(\theta) = V_\infty \left(\frac{3A \sin 3\theta}{\sin \theta}\right)$, where $V_\infty$ is the freestream velocity. Which one of the following options represents the total lift ($L$) and induced drag ($D_i$)?
Options:
(A) $L = 0$ and $D_i = 0$
(B) $L = 0$ and $D_i \neq 0$
(C) $L \neq 0$ and $D_i = 0$
(D) $L \neq 0$ and $D_i \neq 0$
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.36)**
**1. Intuitive Physical Setup & Given Parameters**:
A lifting surface has a spanwise circulation distribution:
\Gamma(\theta) = A \sin(3\theta) \quad (A \neq 0)
over its span $-b/2 \le y \le b/2$, where $y = -\frac{b}{2}\cos\theta$.
Downwash distribution is given by:
w(\theta) = V_\infty \left(\frac{3A\sin 3\theta}{\sin\theta}\right)
We need to determine the total lift $L$ and induced drag $D_i$.
**2. Step-by-Step Aerodynamic Derivation**:
1. **Fourier Series Expansion of Circulation (Prandtl Lifting-Line Theory)**:
In general Prandtl lifting-line theory, circulation is expanded in a Fourier sine series:
\Gamma(\theta) = 2 b V_\infty \sum_{n=1}^\infty A_n \sin(n\theta)
Comparing with $\Gamma(\theta) = A \sin(3\theta)$:
- $A_1 = 0$
- $A_2 = 0$
- $A_3 = \frac{A}{2 b V_\infty} \neq 0$
- $A_n = 0$ for all $n \neq 3$.
2. **Total Lift ($L$) Evaluation**:
The total aerodynamic lift is given by integrating circulation along the span:
L = \rho_\infty V_\infty \int_{-b/2}^{b/2} \Gamma(y) dy = \rho_\infty V_\infty \left(\frac{b}{2}\right) \int_0^\pi \Gamma(\theta) \sin\theta d\theta
Substitute $\Gamma(\theta) = A\sin(3\theta)$:
L \propto \int_0^\pi \sin(3\theta) \sin(\theta) d\theta
By the orthogonality of trigonometric sine functions:
\int_0^\pi \sin(m\theta)\sin(n\theta) d\theta = \begin{cases} 0 & m \neq n \\ \frac{\pi}{2} & m = n \end{cases}
Since $m = 3$ and $n = 1$ ($3 \neq 1$):
\int_0^\pi \sin(3\theta)\sin(\theta) d\theta = 0 \implies L = 0
3. **Induced Drag ($D_i$) Evaluation**:
The induced drag is given by:
D_i = \rho_\infty \int_{-b/2}^{b/2} w(y) \Gamma(y) dy
In terms of Fourier coefficients:
C_{Di} = \pi AR \sum_{n=1}^\infty n A_n^2 = \pi AR (3 A_3^2) > 0 \quad (\text{since } A_3 \neq 0)
Alternatively, integrating downwash times circulation:
D_i = \rho_\infty \frac{b}{2} \int_0^\pi w(\theta)\Gamma(\theta)\sin\theta d\theta = \rho_\infty \frac{b}{2} \int_0^\pi V_\infty \left(\frac{3A\sin 3\theta}{\sin\theta}\right) (A\sin 3\theta) \sin\theta d\theta
D_i = \frac{1}{2}\rho_\infty b V_\infty (3A^2) \int_0^\pi \sin^2(3\theta) d\theta = \frac{3}{2}\rho_\infty b V_\infty A^2 \left(\frac{\pi}{2}\right) \neq 0
Therefore:
L = 0 \quad \text{and} \quad D_i \neq 0
**3. Comprehensive Options & Distractor Analysis**:
- **(A) L = 0 and Di = 0**: ❌ **INCORRECT**. Induced drag is non-zero because downwash and circulation are non-zero.
- **(B) L = 0 and Di ≠ 0**: ✅ **CORRECT**. Lift depends strictly on the fundamental harmonic $A_1$, which is zero ($L = 0$), while induced drag sums all squared harmonics $\sum n A_n^2 > 0$ ($D_i \neq 0$).
- **(C) L ≠ 0 and Di = 0**: ❌ **INCORRECT**. Non-zero lift with zero induced drag is physically impossible for finite wings.
- **(D) L ≠ 0 and Di ≠ 0**: ❌ **INCORRECT**. Lift is zero due to sine orthogonality.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **B**
- **Exam Takeaway**: By Fourier orthogonality in Prandtl lifting line theory: $C_L = \pi AR A_1$ (only $n=1$ generates lift), while $C_{Di} = \pi AR \sum n A_n^2$ (every harmonic contributes positively to induced drag).