Incompressible Flow over Airfoils Questions and Answers

Answer: A

Correct answer: A is false but R is true

The reason is true and it is exactly why the assertion fails. Thin airfoil theory assumes attached, inviscid flow, and stall is the breakdown of that assumption - the boundary layer separating under an adverse pressure gradient. A theory containing no boundary layer cannot predict when one will separate, which is why it gives an excellent lift curve slope and no stalling angle at all.

Answer: B

Correct answer: A is true but R is false

The assertion is correct: a slat raises the maximum lift coefficient substantially. The reason is not the mechanism. A slat works by passing high-energy air through the slot onto the upper surface so that the boundary layer resists separation to a higher incidence. Area increase is the Fowler flap’s distinguishing feature, not the slat’s.

Answer: C

Correct answer: 1 and 3 only

Statements 1 and 3 are correct: angle of attack is measured from the chord line to the relative wind, and exceeding its critical value causes the stall. Statement 2 confuses it with pitch attitude, which is referenced to the horizon. The two differ by the flight path angle, and an aircraft can be nose-down and stalled at the same time.

Answer: D

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: camber shifts the zero-lift angle negative so the section lifts at zero geometric incidence, and it raises the maximum lift coefficient. Statement 3 is nonsense - a symmetric section has zero camber by definition, and camber and thickness are independent parameters described by different digits of the NACA designation.

Answer: A

Correct answer: 0.10

The slope is the change in lift coefficient per degree: \[ a_0 = \frac{0.8 - 0.6}{6 - 4} = \frac{0.2}{2} = 0.10\ \text{per degree.} \] This is close to the thin airfoil value, since \( 2\pi \) per radian is about 0.11 per degree. Real sections fall a little below it because the boundary layer displacement slightly decambers the effective shape.

Answer: B

Working back from \(c_l = 0.6\) at \(4^\circ\), the section needs \(0.6/0.10 = 6\) degrees of effective incidence, so the zero-lift angle lies \(6\) degrees below \(4\), that is at \(-2^\circ\). The negative value confirms the section is cambered.

Answer: C

Correct answer: 1.73

Rearranging the pressure coefficient relation: \[ C_p = 1 - \left(\frac{V}{V_\infty}\right)^2 \Rightarrow \left(\frac{V}{V_\infty}\right)^2 = 1 - (-2.0) = 3.0, \] \[ \frac{V}{V_\infty} = \sqrt{3} = 1.73. \] So a \( C_p \) of minus two corresponds to the flow travelling about 73 per cent faster than free stream - a strong but entirely ordinary suction peak near the leading edge.

Answer: D

Correct answer: A plain flap

The plain flap is a trailing edge device, hinged so that deflecting it increases the camber of the rear of the section. The other three are all leading edge devices: a slat stands off ahead of the wing forming a slot, a Krueger flap hinges down from the lower surface, and a nose droop deflects the whole leading edge region downward.

Answer: A

By raising the maximum lift coefficient they lower the stall speed, so approach and take-off can be flown more slowly and the runway required is shorter. They add considerable drag, which is why they are retracted in cruise.

Answer: B

Air passing through the gap from the high-pressure side keeps the boundary layer on the deflected flap attached to larger angles, so more lift is gained before separation. A plain flap without the slot separates much earlier.