A lifting surface has a spanwise circulation distribution of Ī(š) = A sin 3š (where
A ā 0) over its span -
b
2 ā¤ y ā¤
b
2, and y = -
b
2 cos š is the spanwise coordinate.
Furthermore, the downwash varies along the span as w(š) = Vā (
3A sin 3š
sin š ), where
Vā is the freestream velocity. Which one of the following options represents the
total lift (L) and induced drag (Di)?
Options:
(A) L = 0 and Di = 0
(B) L = 0 and Di ā 0
(C) L ā 0 and Di = 0
(D) L ā 0 and Di ā 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**: ā **INCORRECT**. Induced drag is non-zero because downwash and circulation are non-zero.
- **(B) L = 0 and Di ā 0**: ā **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 ā 0 and Di = 0**: ā **INCORRECT**. Non-zero lift with zero induced drag is physically impossible for finite wings.
- **(D) L ā 0 and Di ā 0**: ā **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).
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