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, (A: Eom\\r\~o. C m\f\ «\(ﬁxg _ HZ. 614 Solutions Manual 0 Fluid Mechanics, Fifth Edition Solution: For sea—level air, take ,0 = 0.00237 slug/ft3. Convert 0.7 lbf per mi/h rolling
friction to 0.477 lbf per ft/s of speed. Then the power relationship for the cycle is ' Power = (Fd, + quw = [CD/igvz + Cm,le V, ftlbf
s or: 115*550 3
= [(5.5 ft2)—0'00237251“g/ft V2 + [0.477 %] V]V
S Solve this cubic equation, by iteration or EES, to ﬁnd Vmax z 192 ft/s z 131 mi/h. Ans. The heliumﬁlled balloon in
Fig. P7.75 is tethered at 20°C and 1 atm
with a string of negligible weight and drag.
The diameter is 50 cm, and the balloon
material weighs 0.2 N, not including the
helium. The helium pressure is 120 kPa. Estimate the tilt angle 6 if the airstream
velocity U is (a) 5 m/s or (b) 20 m/s. Solution: For air at 20°C and 1 atm, take p = 1.2 kg/m3 and ,u = 1.8E—5 kg/ms. For
helium, R = 2077 J/kg~°K. The helium density = (120000)/[2077(293)] z 0.197 kg/m3. The balloon net buoyancy is independent of the flow velocity:
7f 3 7T 3
BM = (pmr — pHe)ggD = (1.2— 0.197)(9.81)g(0.5) z 0.644 N The net upward force is thus Fz = (Bnet — W) = 0.644 — 0.2 = 0.444 N. The balloon drag
does depend upon velocity. At 5 m/s, we expect laminar ﬂow: F, 7/6
W25 2. Reb=%=mm Blimpm
S . — Drag = CD §U2 $1? = 0.4%?) (5)2 $0.5? z 1384 N Then Q, = tan“ { Drag = tan—1 = 72° Ans. (a)
1 F2 1 0.444 Chapter 7 o Flow Past Immersed Bodies 615 (b) At 20 m/s, Re = 667000 (turbulent), Table 7.3: CD z 0.2: 1.2 7: 9.43
D =0.2[—] 202— 0.52:9.43N, a =t "[—]=87° A . b
mg 2 ( ) 4( ) b an 0.444 "S 0 These angles are too steep——the balloon needs more buoyancy and/or less drag. P7.76 The recent movie The World ’3 Fastest Indi tells the story of Burt Munro,
a New Zealander who, in 1937, set a motorcycle recor of 201 mi/h on the Bonneville
Salt Flats. Using the data of Prob. P7.74, (a) estimate he horsepower needed to drive this
fast. (1)) What horsepower would have gotten Burt u to 250 mi/h? Solution: Prob. P7.74 suggests CDA = 5.5 ft2 and Fromng = 0.7 lbf per mi/h of speed.
Convert 201 mi/hr to 295 ft/s. Bonneville is at 43 0 ft altitude, so take p = 1.0784 kg/m3
= 0.00209 slug/ft3 from Table A.6. Now compu the total resistance force: F=Fdrag+Fr 011mg = (CDA)£2)—V2 + '7Vmi/h 3
= (5.5ﬁ2)(W)(295 /s)2 + 0.7(201) = 500 +141 = 64llbf Power = FV = (64llbf)(295ft/s) = 189,000 ﬁ—lbf/s +550 = 343hp Ans.(a) A lot of power! Presumably Burt di some streamlining to reduce drag.
([2) Repeat this for V: 250 mi/h = 67 ft/s to get F = 914 lbf, Power = 610 hp. Ans. (b) P7.77 To measure the rag of an upright person, without violating humansubject
protocols, a lifesized m nequin is attached to the end of a 6m rod and rotated at Q =
80 rev/min, as in Fig. P .77. The power required to maintain the rotation is 60 kW. By
including roddrag p er, which is signiﬁcant, estimate the drag—area CDA of the mannequin, in m2. Chapter 7 0 Flow Past Immersed Bodies 607 _ . V A sphere of density ps and diameter
D is dropped from rest in a ﬂuid of density p and Viscosity y. Assuming a constant DmgrBUOYancyl drag coefficient Cdo, derive a differential D
equation for the fall velocity V(t) and show
that the solution is
1/2
4 D — I
V = —g—(S—1) tanh Ct PM
3ng (5—1) “2
C = ———°2 La
45 D where S = ps/p is the speciﬁc gravity of the sphere material.
Solution: Newton’s law for downward motion gives ‘ V
deOWanadown’ or: W_B_CD—p—V2A=Ed—, Where A=£D2
‘ 2 g dt 4
and W—B=p(S—l)g%D3. Rearrangeto i—Y=IB—avz,
1] ngDA
= 1—— and (2:
ﬂ g[ S W Separate the variables and integrate from rest, V = 0 at t = 0: i dt = i dV/(ﬂ — aVz), or: V = F tanh (tﬁ) = Vﬁnaltanh(Ct) Ans.
a 1/2 1/2
where Vﬁnal = w and C = , s = 10—5 > 1
3CD 48 D ,0 P7.68 baseball weighs 145 g and is 7.35 cm in diameter. It is dropped from rest from
a 35—m—h1gh tower at approximately sea level. Assuming a laminar—flow drag coefﬁcient, estimate (a) its terminal velocity and (b) whether it will reach 99 percent of its terminal
veloc1ty before it hits the ground. Chapter 7 o Flow Past Immersed Bodies 609 Solution: For sea—level air, take ,0 = 1225 kg/m3 and y = 1.78E—5 kg/ms. Assume a
laminar drag coefﬁcient CD z 0.47 from Table 7.3. The terminal velocity is VFW]: ———2W 2= mﬂomxggm 2~34.1E Ans. (a)
CDp(7r/4)D O.47(l.225)(ﬂ/4)(0.0735) s Now establish the “specific gravity” of the ball, relative to air: pball =E=——w——=697.4 E “S”: pba” =69_7'4____569
U (7r/6)(0.0735)3 m3 ’ p,ir 1.225 Then the constant C from Prob. 7.66 gives the time history of velocity and displacement: C I [3ch6 —1)]”2 = [3(9.81)(0.47)(569 — 1) 1/2
2 2 a 0.287 s", v = vf tanh(Ct),
45 D 4(569) (0.0735) 34.1
0.287 Check ReD (max) = 1.225(34.1)(0.0735)/(1.78E—5) z 172000 (OK, CD z 0.47) or: V = 34.1 tanh (0.287t), Z =J th = ln[cosh (02870] We can now find the time and velocity when the balls hits Z = 35 m: Z = 35 = 34'1 ln[cosh(0.287t)], solve for t z 2.81 s, whence
0.287
V(at Z = 35 m) = 34.1 tanh[0.287(2.81)] z 22.8 E Ans. (b) S This is only 67% of terminal velocity. If we try the formulas again for V = 99% of
terminal velocity (about 33.8 m/s), we find that tz 9.22 s and Z w 230 m. P7.69 Two baseballs from Prob. 7.68
are connected to a rod 7 mm in diameter
and 56 cm long, as in Fig. P7.69. What
power, in W, is required to keep the
system spinning at 400 r/min? Include the
drag of the rod, and assume sea—level
standard air. ...
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