Flow Over Immersed Bodies Questions and Answers

Answer: B

The propulsive power is 20 kW.

Relative to the boat, water enters at the boat speed, 5 m/s, and leaves at 9 m/s. The thrust is the rate of change of its momentum:

\[ \begin{aligned} F &= \rho Q(V_{j} - V) = 1000(1)(9 - 5) = 4000\ \text{N} \\ P &= FV = 4000 \times 5 = 20000\ \text{W} = 20\ \text{kW} \end{aligned} \]

A common slip is to ignore the momentum the water already has relative to the boat when it enters, which inflates the thrust to 9000 N. Only the momentum difference times the boat speed gives the useful propulsive power, and 10, 50 and 90 kW do not follow from that balance.

Answer: C

The settling velocity is \(\frac{g(\rho_{p} - \rho_{l})d_{p}^{2}}{18\mu_{l}}\), Stokes' law.

A very small particle settles in the laminar (Stokes) regime. At terminal velocity the net weight balances the Stokes drag \(3\pi\mu_{l} d_{p} v\):

\[ \begin{aligned} \frac{\pi d_{p}^{3}}{6}(\rho_{p} - \rho_{l})g &= 3\pi\mu_{l} d_{p} v \\ v &= \frac{g(\rho_{p} - \rho_{l})d_{p}^{2}}{18\mu_{l}} \end{aligned} \]

The tank depth \(L\) plays no part in a terminal velocity, which rules out every option containing \(L\). Those options also fail dimensionally: only \(d_{p}^{2}\) gives a velocity with the other factors.

Answer: C

The settling velocity is \(\frac{1}{18}\) m/s.

With \(c_{D} = 24/Re = 24\mu/(\rho V d)\), the drag reduces to Stokes' law, \(D = 3\pi\mu V d\); the liquid density cancels. At constant velocity the drag balances the weight (buoyancy is neglected):

\[ \begin{aligned} 3\pi\mu V d &= \rho_{s}\,g\,\frac{\pi d^{3}}{6} \\ V &= \frac{\rho_{s} g d^{2}}{18\mu} = \frac{11000 \times 10 \times 10^{-8}}{18 \times 1.1 \times 10^{-3}} \\ &= \frac{1}{18}\ \text{m/s} \end{aligned} \]

The other fractions do not satisfy the force balance; 1/24 is the trap of using the 24 of the drag law in place of the 18 of Stokes' settling velocity.

Answer: D

Drag acts along OB and lift along OD.

By definition, drag is the component of the aerodynamic force parallel to the free-stream velocity, and lift is the component perpendicular to it. The free stream approaches at 30 degrees to the airfoil's x axis, and OB is drawn at 30 degrees to x, so OB is the free-stream direction and drag acts along it. OD is drawn at right angles to OB, so it is the lift direction.

OA and OC are the body (chord) axes. Forces along them are the axial and normal components, which coincide with drag and lift only at zero angle of attack.

Answer: A

The match is P-3, Q-2, R-1, S-4.

At high Reynolds number drag on bluff bodies is mostly pressure drag, set by how abruptly the flow separates. The more streamlined the face toward the flow, the lower the drag coefficient.

• R, the half tube with its convex side to the flow, is the most streamlined: \(C_{D} \approx 1.2\).
• Q, the square cylinder with a corner to the flow, splits the flow gradually: \(C_{D} \approx 1.6\).
• P, the square cylinder with a flat face to the flow, separates at sharp corners: \(C_{D} \approx 2.1\).
• S, the half tube with its concave side to the flow, traps the flow like a cup: \(C_{D} \approx 2.3\), the highest.

Options putting the concave half tube at 1.2 (S-1) or the convex half tube at 1.6 or 2.3 reverse the effect of streamlining.

Answer: D

The lowest pressure drag is on Body S.

At Reynolds numbers of \(10^{4}\) and above, form drag is set mainly by flow separation at the rear. A rounded nose lets the flow follow the surface at the front, and a long, gradually tapering tail lets the flow decelerate slowly along it, so it stays attached and the low-pressure wake is narrow. Body S has both, a streamlined shape.

Body P, a blunt block, separates at its sharp front corners and leaves a wide wake, the highest drag. Body Q rounds the nose but still ends in a blunt base. Body R rounds the rear corners too, but its rear is still bluff compared with S's long taper, so its wake is wider than S's.