Incompressible Flow over Airfoils Questions and Answers

Answer: C

Both statements hold and the neglect of viscosity is precisely why no drag appears: without shear there is no friction drag, and without a boundary layer there is no separation to spoil pressure recovery. This is d'Alembert's paradox.

Answer: D

Both are true and connected: because the cambered section already lifts at zero incidence, it must be pitched nose-down before the lift vanishes, giving a negative zero-lift angle. A symmetric section reaches zero lift at zero degrees.

Answer: A

Both statements are true and the downwash is the cause: part of the geometric incidence is used up countering the induced flow angle, so more of it is needed for the same lift. The effect vanishes as aspect ratio tends to infinity.

Answer: B

The first statement is true, but the second is false: it is viscosity acting at the sharp trailing edge during the starting process that establishes the circulation in the first place. The Kutta condition is an inviscid statement of a viscous outcome.

Answer: C

Thin airfoil theory gives \(c_l = 2\pi\alpha\) with the angle in radians. Here \(\alpha = 5^\circ = 0.0873\) rad, so \(c_l = 2\pi \times 0.0873 = 0.55\). The common slip is to use degrees directly, which gives a value about fifty-seven times too large.

Answer: D

The effective angle is measured from the zero-lift line, so it is \(4 - (-3) = 7^\circ\), which is 0.1222 rad. Then \(c_l = 2\pi \times 0.1222 = 0.77\). Using the geometric angle alone would give 0.44 and understate the lift.

Answer: A

Correct answer: 6125 Pa

\[ q = \tfrac{1}{2}\rho V^2 = 0.5 \times 1.225 \times 100^2 = 6125\ \text{Pa.} \] Dropping the factor of one half gives 12 250 Pa, which is the distractor most often chosen. Dynamic pressure is the quantity that scales every aerodynamic force, so an error here propagates straight into the lift and drag.

Answer: B

Correct answer: 49 kN

\[ L = q\,S\,C_L = 6125 \times 20 \times 0.4 = 49\,000\ \text{N} = 49\ \text{kN.} \] The three factors are independent, so the lift equation is a straightforward product once the dynamic pressure is known. Note that the reference area is the plan area of the wing, not its wetted area - using the wrong one is a common source of a factor of two.

Answer: C

Correct answer: -0.69

For incompressible flow the pressure coefficient follows from the velocity ratio alone: \[ C_p = 1 - \left(\frac{V}{V_\infty}\right)^2 = 1 - 1.3^2 = 1 - 1.69 = -0.69. \] The negative sign says the local pressure is below free stream, which is what an accelerated flow must produce. Anywhere the flow speeds up, \( C_p \) is negative; only at a stagnation point, where \( V = 0 \), does it reach its maximum of \( +1 \).

Answer: D

Correct answer: 4

Lift is proportional to the square of the speed through the dynamic pressure, so with the lift coefficient, density and area unchanged, doubling the velocity multiplies the lift by \( 2^2 = 4 \). This square law is why stalling speed changes only with the square root of weight, and why a small speed margin on approach buys a large margin of lift.