AU-MECH Spring 2 0 1 1 M E C H 3 0 3 0

F L U I D M E C H A N I C S S O L U T I O N T O Q U I Z 00

Maximum: 18.5 Average: 11.82 Minimum:6 Problem 1 (4 Points): (a) The PV work is determined by multiplying the pressure difference between the inside and outside of the module ( P ) and the final volume, i.e.: W = P V = (101,000-1000) Pa (50) m3 = 5x10 6 J . (2 p ts ) (b) Dividing the amount of work by the time duration needed for filling up the module gives the power: Power = W /t = 5 x 106 J / 250 s = 20,000 W = 26.81 hp. (2 pts)

Problem 2 (4 Points):

Integrate once: 22

xC

dxdyC

dxdyx , with C being a constant. (2 pts)

Integrate again, .DxCy (2 pts)

with D being another constant. Problem 3 (3 Points): Adding the two equations gives Q2 = 4, which further leads to Q= 2. (1.5 pts) Substituting Q back to either equation gives P3 =-8, which gives P= -2. (1.5 pts) Problem 4 (3 Points): The answer is (d) because the function V is independent of variable t. (3 pts) Problem 5 (6 Points): (a) The free body diagram is shown. Notice that W is acting downward, whereas and D are acting to the right and left, respectively. (4 pts)

(b) There is also a buoyancy force due to the air surrounding the aircraft (acting upward). (1 pt) (c) The aircraft will slow down as long as the thrust force is smaller than the drag, that is < D. (1 pt)

W

D