Boundary Layer and Drag Questions and Answers

Answer: C

With \( \delta^* = 1.721x/\sqrt{Re_x} \) and \( \theta = 0.664x/\sqrt{Re_x} \), the ratio is \( 1.721/0.664 = 2.59 \). A turbulent layer comes in nearer 1.4, and a laminar layer climbing past about 3.5 is on the point of separating.

Answer: D

Rearranging, \( x = Re\,\mu/(\rho V) = 5 \times 10^5 \times 1.8 \times 10^{-5} / (1.225 \times 30) = 9/36.75 = 0.245 \) m. Roughly the first quarter-metre is laminar and everything downstream turbulent, which is typical for air at ordinary speeds.

Answer: A

The dynamic pressure is \( 0.5 \times 1.225 \times 60^2 = 2205 \) Pa, so the drag is \( 0.0030 \times 2205 \times 12 = 79.4 \) N. Note that skin friction is referred to wetted area, not to planform area, which is a frequent source of confusion.

Answer: B

The ratio is \( 0.10/0.50 = 0.2 \), so four fifths of the drag has gone. That a mere change in the state of a layer a millimetre thick can remove eighty per cent of the drag is one of the more striking results in the subject.

Answer: C

The drag goes as \( C_F \times V^2 \), and \( C_F \) varies as \( Re^{-1/2} \), that is as \( V^{-1/2} \). The product therefore scales as \( V^{1.5} \), so doubling the speed multiplies the drag by \( 2^{1.5} = 2.83 \) rather than by four.

Answer: D

With \( \delta^* = 1.721x/\sqrt{Re_x} \) and \( \delta = 5x/\sqrt{Re_x} \), the ratio is \( 1.721/5 = 0.344 \). Roughly a third, in other words: the outer streamlines are pushed out by about a third of the layer's own thickness.

Answer: A

Using \( \delta = 0.37x/Re_x^{0.2} \), the denominator is \( (4 \times 10^6)^{0.2} = 21.0 \), so \( \delta = 0.37 \times 2/21.0 = 0.035 \) m. That is roughly seven times the laminar value at a comparable station, which is why turbulent layers matter so much for ducts and intakes.

Answer: B

The dynamic pressure is \( 0.5 \times 1.225 \times 80^2 = 3920 \) Pa, so the drag is \( 0.025 \times 3920 \times 30 = 2940 \) N. In steady level flight the thrust required is exactly this, so the power needed is the drag times the speed, about 235 kW.

Answer: C

The Blasius value of 2.59 applies with no pressure gradient; as an adverse gradient drains the near-wall momentum the factor climbs, and separation follows around 3.5. A turbulent layer holds on until about 2.4, a lower figure because its profile is inherently fuller.

Answer: D

Friction drag is proportional to wetted area at a fixed coefficient, so halving the area halves the drag. This linearity is why reducing wetted area is such a direct lever on a streamlined body, where friction rather than pressure drag dominates.