A symmetrical circular-arc biconvex airfoil with thickness-to-chord ratio $\tau = t/c = 0.08$ flies at zero angle of attack $(\alpha = 0)$ in a supersonic airstream at Mach $M_\infty = 2.0$. According to Ackeret's linearized supersonic theory, the wave drag coefficient $c_{d,w}$ of the airfoil is closest to:

(A) 0.0197

(B) 0.0098

(C) 0.0394

(D) 0.0275

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles From linearized supersonic airfoil theory (Ackeret theory), the wave drag coefficient at zero angle of attack for a symmetrical profile of thickness-to-chord ratio $\tau = t/c$ is given by: $$c_{d,w} = \frac{4}{\sqrt{M_\infty^2 - 1}} \int_0^1 \left(\frac{dy}{dx}\right)^2 \, d\left(\frac{x}{c}\right)$$ For a parabolic circular-arc biconvex airfoil, the upper profile is $y(x) = 2t \left(\frac{x}{c} - \frac{x^2}{c^2}\right)$. Integrating $(dy/dx)^2$ yields the standard closed-form formula: $$c_{d,w} = \frac{16}{3} \frac{\tau^2}{\sqrt{M_\infty^2 - 1}}$$ ### 2. Analytical Formulation & Calculation Given: - $\tau = 0.08 \implies \tau^2 = (0.08)^2 = 0.0064$ - $M_\infty = 2.0 \implies \sqrt{M_\infty^2 - 1} = \sqrt{2.0^2 - 1} = \sqrt{3} \approx 1.73205$ ### 3. Step-by-Step Calculation $$c_{d,w} = \frac{16}{3} \times \frac{0.0064}{\sqrt{3}} = \frac{0.1024}{3 \times 1.73205} = \frac{0.1024}{5.19615} \approx 0.019707 \approx 0.0197$$ ### 4. Distractor Forensics - **Option A (0.0197):** Correct wave drag coefficient. - **Option B (0.0098):** Calculated for only one surface (forgot factor of 2 for upper and lower surfaces). - **Option C (0.0394):** Used wedge formula factor $4 \tau^2 / \sqrt{M^2-1}$ without the $4/3$ parabolic integration coefficient. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Remember the shape factor for a parabolic biconvex airfoil is $16/3$, whereas for a double-wedge diamond airfoil it is $4$. - **Standard Reference:** Anderson, J.D., *Modern Compressible Flow*, Chapter 9 (Linearized Supersonic Flow).

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