Incompressible Flow over Airfoils Questions and Answers

Answer: C

Correct answer: A symmetric airfoil has zero pitching moment about its quarter chord

This follows from the symmetry of the pressure distribution: with no camber there is nothing to produce a couple about the quarter chord, so \( c_{m,c/4} = 0 \). The other three are all false - a cambered section has a negative zero-lift angle, thin airfoil theory is inviscid and says nothing about profile drag, and a symmetric section produces no lift at zero incidence.

Answer: D

The theory is entirely inviscid, which is why it gives no drag and cannot predict the stall. The other three statements describe the theory correctly, and its accuracy at low incidence is remarkable given how much it leaves out.

Answer: A

The fixed point at the quarter chord is the aerodynamic CENTRE, not the centre of pressure, and confusing the two is the classic error. The centre of pressure moves continually with incidence, which is why the other point is preferred.

Answer: B

Correct answer: The camber line of the airfoil

Thin airfoil theory replaces the section by its mean camber line and distributes a sheet of vorticity along it, choosing the strength so that the camber line is a streamline and the Kutta condition is satisfied at the trailing edge. Because the model is inviscid it contains no boundary layer and no wake, which is precisely why it predicts lift accurately and drag not at all.

Answer: C

Panelling the true surface removes the thin-airfoil restriction and handles thickness properly, giving accurate pressure distributions for real sections. It remains inviscid, so it still predicts no drag and cannot find the stall.

Answer: D

Moving the point of maximum thickness aft keeps the pressure falling for a longer distance, which delays transition and preserves laminar flow with its much lower friction. The benefit is lost if the surface is rough or contaminated.

Answer: A

The favourable gradient only holds if the surface is smooth and clean, so insects, rain or manufacturing waviness can trip the layer early and return the drag to normal levels. This is why the gains are often smaller in service than in the tunnel.

Answer: B

A higher Reynolds number gives a fuller, more energetic boundary layer that runs further against the adverse gradient before separating, so the section reaches a higher angle and a higher maximum coefficient. This is one reason model results understate full-scale performance.

Answer: C

Flow accelerating around the nose produces the largest negative pressure coefficient just aft of the leading edge, and the pressure then recovers towards the trailing edge. It is that recovery, an adverse gradient, which eventually causes separation.

Answer: D

Raising the incidence sharpens the suction peak, so the pressure must recover over the same distance from a lower value. The gradient therefore steepens and separation moves forward, which is what limits the maximum lift coefficient.