Boundary Layer and Drag Questions and Answers

Answer: B

Both are true and the second explains the first. The extra momentum near the surface lets the layer push further into a rising pressure before the wall shear vanishes. The same mechanism raises the velocity gradient at the wall and so costs more skin friction, which is the standing trade of the subject.

Answer: C

Both statements are true and the second is the reason for the first. Deliberately bringing on the drag crisis at a lower Reynolds number collapses the pressure drag, which dominates for a bluff body. The friction penalty is trivial beside the saving, and the ball flies about twice as far.

Answer: D

Both are true and the second explains the first. The gentler rear taper keeps the flow attached and leaves a narrow wake, and the pressure drag saved far exceeds the extra friction the longer surface costs. On a bluff body the pressure term is the one worth attacking.

Answer: A

Both statements are true and the second explains the first. Splitting the problem in two allowed potential-flow theory to be used where it works while viscosity was handled only where it matters. Before 1904 the two descriptions could not be reconciled, and d'Alembert's paradox stood unresolved.

Answer: B

Both are true and the second is the reason for the first. Outside the drag bucket the stagnation point moves and transition jumps forward, and in practice insects, rain and manufacturing waviness destroy the laminar run in any case. The measured benefit is often far below the wind-tunnel promise.

Answer: C

The assertion is false and the reason is true, and the reason is what refutes it. So long as the roughness is smaller than the sublayer the surface is hydraulically smooth and no drag penalty appears. Only once the elements protrude does the friction climb.

Answer: D

The assertion is too strong while the reason is correct. An adverse gradient is necessary for separation but not sufficient: a turbulent layer routinely negotiates a mild rise without separating. Separation requires the gradient to be steep enough to drive the wall shear to zero.

Answer: A

Both statements are true and the second explains the first. Because they are small, cheap and external, they can be added after flight testing reveals a separation problem, which is exactly how many transport aircraft acquired them. The cost is a small permanent drag increment.

Answer: B

Both are true and the second is the reason for the first. Shear stress at the wall is the viscosity times the velocity gradient there, and the fuller turbulent profile makes that gradient much steeper. On a long streamlined body the difference can be a factor of five or more.

Answer: C

Both statements are true and the second explains the first. Because the layer thickens the effective body, the outer flow must be recomputed about that enlarged shape, and iterating between the two is how viscous-inviscid interaction methods work.