According to Ackeret's linearized supersonic theory, a flat plate airfoil of chord $c$ at an angle of attack $\alpha$ in Mach $M_\infty > 1$ flow has its center of pressure located at:

(A) The mid-chord point ($x_{cp} = 0.50 c$)

(B) The quarter-chord point ($x_{cp} = 0.25 c$)

(C) The leading edge ($x_{cp} = 0$)

(D) The three-quarter chord point ($x_{cp} = 0.75 c$)

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles In linearized supersonic flow over a flat plate airfoil at angle of attack $\alpha$: - The pressure coefficient on the lower surface is uniform: $C_{p,l} = +\frac{2\alpha}{\sqrt{M_\infty^2 - 1}}$. - The pressure coefficient on the upper surface is uniform: $C_{p,u} = -\frac{2\alpha}{\sqrt{M_\infty^2 - 1}}$. - Consequently, the net lifting pressure difference $\Delta C_p = C_{p,l} - C_{p,u} = \frac{4\alpha}{\sqrt{M_\infty^2 - 1}}$ is **completely uniform and constant** along the entire chord from leading edge to trailing edge. ### 2. Center of Pressure Calculation Since the loading is rectangular (uniform), the centroid of the loading lies at the exact geometric midpoint: $$x_{cp} = \frac{\int_0^c \Delta C_p \cdot x \, dx}{\int_0^c \Delta C_p \, dx} = \frac{\Delta C_p \left[\frac{x^2}{2}\right]_0^c}{\Delta C_p [x]_0^c} = \frac{c^2 / 2}{c} = 0.50 c$$ ### 3. Distractor Forensics - Subsonic flow concentrates vorticity near the leading edge, placing $x_{cp}$ at $0.25 c$. In supersonic flow, uniform loading shifts $x_{cp}$ aft to $0.50 c$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Subsonic center of pressure is at $0.25c$; supersonic center of pressure is at $0.50c$ (causing substantial nose-down pitching trim changes during transonic transition). - **Standard Reference:** Anderson, J.D., *Modern Compressible Flow*, Chapter 9.

← Prev Question Next Question →