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} \leq y \leq \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 $?

Question

cdquestions Admin · Accepted Answer

Step 1: Lift Calculation The total lift $ L $ on a lifting surface is related to the circulation distribution by the Kutta-Joukowski theorem: $$ L = \rho V_\infty \int_{-b/2}^{b/2} \Gamma(y) \, dy $$ However, in this problem, the circulation distribution $ \Gamma(\theta) = A \sin 3\theta $ is given with odd symmetry (because of the $ \sin 3\theta $ term), and when integrated over the span, the total circulation results in zero: $$ \int_{-b/2}^{b/2} \Gamma(y) \, dy = 0 $$ Therefore, the total lift $ L $ is zero. Step 2: Induced Drag Calculation The induced drag $ D_i $ is related to the downwash distribution, which is given by: $$ w(\theta) = V_\infty \left( \frac{3A \sin 3\theta}{\sin \theta} \right) $$ Since the downwash is nonzero, the interaction between the circulation and the downwash will produce a nonzero induced drag. The induced drag $ D_i $ is given by: $$ D_i = \int_{-b/2}^{b/2} \frac{\Gamma(y) w(\theta)}{V_\infty} \, dy $$ This results in a nonzero induced drag because the downwash $ w(\theta) $ is nonzero and varies along the span. Thus, $ L = 0 $ and $ D_i \neq 0 $.

Answer

$ L = 0 $ and $ D_i = 0 $

Answer

$ L = 0 $ and $ D_i \neq 0 $

Answer

$ L \neq 0 $ and $ D_i = 0 $

Answer

$ L \neq 0 $ and $ D_i \neq 0 $

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The Correct Option is B

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