15–38
Solution A plastic sphere is dropped into water. The terminal velocity of the sphere in water is to be determined.
Assumptions 1 The fluid flow over the sphere is laminar (to be verified). 2 The drag coefficient remains constant.
Properties The density of plastic sphere is 1150 kg/m3. The density and dynamic viscosity of water at 20C are
=
998 kg/m3 and
= 1.00210-3 kg/ms, respectively. The drag coefficient for a sphere in laminar flow is CD = 0.5.
m/s 0.155
0.53
1)–8m)(1150/99 006.0)(m/s 81.9(4
3
)1/(4 2
D
fs
C
gD
V
6 mm
15–22
15–39
Solution A spherical hot air balloon that stands still in the air is subjected to winds. The initial acceleration of the
be 1.20 kg/m3.
Analysis The frontal area of a sphere is A =
D2/4. Noting that 1 m/s = 3.6 km/h, the drag force acting on the balloon
is
N 1.570
m/skg 1
N 1
2
m/s) 6.3/40)(kg/m 20.1(
]4/m) 7()[2.0(
22
23
2
2
V
ACF DD
Then from Newton’s 2nd law of motion, the initial acceleration in the direction of the winds becomes
2
m/s 1.63
N 1
m/skg 1
kg 350
N 1.570 2
m
F
aD
15-40
Solution The flat mirror of a car is replaced by one with hemispherical back. The amount of fuel and money saved
per year as a result are to be determined.
Assumptions 1 The car is driven 24,000 km a year at an average speed of 95 km/h. 2 The effect of the car body on the
N 10.6
m/s kg1
N 1
km/h6.3
m/s 1
2
km/h)95)( kg/m20.1(
4
m) 13.0(
1.1 2
2
232
D
F
PROPRIETARY MATERIAL. © 2017 McGraw-Hill Education. Limited distribution permitted only to teachers and educators for course preparation. If
you are a student using this Manual, you are using it without permission.
15–41C
Solution We are to define and discuss the average skin friction coefficient over a flat plate.
15–42C
Solution We are to discuss the fluid property responsible for the development of a boundary layer.
15-43C
Solution We are to define and discuss the friction coefficient for flow over a flat plate.
15–25
15-44
Solution Laminar flow of a fluid over a flat plate is considered. The change in the drag force is to be determined
5.0
5.0
2/3
2
5.0
1
2
5.0
1
5.0
2
1
664.0
2
328.1
get welation, number reReynolds ngSubstituti
2
Re
328.1
Therefore
Re
328.1
where
2
L
AV
V
A
VL
F
V
AF
C
V
ACF
D
D
ff
D
When the free-stream velocity of the fluid is tripled, the new value of the drag
force on the plate becomes
5.0
5.0
2/3
2
5.0
2)3(664.0
2
)3(
)3(
328.1
L
AV
V
A
LV
FD
5.2033/2 2/3
2/3
1
2)3(
V
V
F
F
D
D
V
L
15–45
Solution Air flows over a plane surface at high elevation. The drag force acting on the top surface of the plate is to be
3 kg/m9751.0
K) K)(298/kgm kPa(0.287
Analysis (a) If the air flows parallel to the 8 m side, the
Reynolds number becomes
6
3
m) m/s)(5 9()kg/m 9751.0(
VL
9 m/s
2.5 m
15–27
15–46
Solution A train is cruising at a specified velocity. The drag force acting on the top surface of a passenger car of the
train is to be determined.
15–47E
Solution Air flows over a flat plate. The local friction coefficients at intervals of 1 ft is to be determined and plotted
1.643×10–4 ft2/s .
Analysis For the first 1 ft interval, the Reynolds number is
5
25 ft/s 1 ft
VL .
Air
15–48E
Solution Light oil flows over a flat plate. The total drag force per unit width of the plate is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Recr = 5 × 105. 3 The surface of
15–30
15–49E
Solution A refrigeration truck is traveling at a specified velocity. The drag force acting on the top and side surfaces
1.697×10–4 ft2/s .
15-50E
Solution The previous problem is reconsidered. The effect of truck speed on the total drag force acting on the top and
rho=0.07350 “lbm/ft3″
nu=0.6110/3600 “ft2/s”
V=Vel*1.4667 “ft/s”
g=32.2 “ft/s2”
F=Cf*A*(rho*V^2)/2/32.2 “lbf“
Pdrag=F*V/737.56 “kW”
V, mph
Re
Fdrag, lbf
Pdrag, kW
0
10
20
30
40
50
60
70
80
90
100
0
1.728E+06
3.457E+06
5.185E+06
6.913E+06
8.642E+06
1.037E+07
1.209E+07
1.382E+07
1.555E+07
1.728E+07
0.00
0.51
1.79
3.71
6.23
9.31
12.93
17.06
21.69
26.82
32.42
0.000
0.010
0.071
0.221
0.496
0.926
1.542
2.375
3.451
4.799
6.446
Discussion The required power increases rapidly with velocity – in fact, as velocity cubed.
15–32
15-51
Solution Air is flowing over a long flat plate with a specified velocity. The distance from the leading edge of the
= 1.184 kg/m3 and
= 1.562×10–5 m2/s .
Analysis The critical Reynolds number is given to be Recr = 5105. The
distance from the leading edge of the plate where the flow becomes turbulent is
the distance xcr where the Reynolds number becomes equal to the critical
cm 0.678 m 00678.0
)10(5
m) 976.0(91.4
Re
91.4
Re
91.4
2/152/1
,
2/1
,
cr
cr
crv
x
xv
x
x
15-52
Solution Water is flowing over a long flat plate with a specified velocity. The distance from the leading edge of the
plate where the flow becomes turbulent, and the thickness of the boundary layer at that location are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The critical
mm 0.39 m 00039.0
)10(5
m) 056.0(91.4
Re
91.4
Re
5
2/152/1
,
2/1
,
cr
cr
crv
x
xv
x
x
xcr
15-53
Solution Wind is blowing parallel to the side wall of a house. The drag force acting on the wall is to be determined
for two different wind velocities.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Recr = 5105. 3 Air is an ideal
002730.0
10105.1
1742
–
)10105.1(
074.0
Re
1742
–
Re
074.0
75/175/1
L
L
f
C
10 m
55 km/h
15–34
15–54
Solution The weight of a thin flat plate exposed to air flow on both sides is balanced by a counterweight. The mass of
the counterweight that needs to be added in order to balance the plate is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Recr = 5105. 3 Air is an ideal
15–57C
Solution We are to discuss why the drag coefficient suddenly drops when the flow becomes turbulent.
15–58
Solution A spherical dust particle is suspended in the air at a fixed point as a result of an updraft air motion. The
1 atm and 25C are
f = 1.184 kg/m3 and
= 1.84910-5 kg/ms.
Analysis The terminal velocity of a free falling object is reached (or the suspension of an object in a flow stream is
6
)(3 3
3
D
gVDggVD f
s
f
s
VV
22 2 3
-5
()
(9.81 m/s )(0.0001 m) (2100-1.184) kg/m
0.6186 m/s
18 18(1.849 10 kg/m s)
sf
gD
V
0.62 m/s
The Reynolds number in this case is
0.4
m/skg 101.849
m) 1m/s)(0.000 )(0.619kg/m (1.184
Re 5–
3
VD
which is in the order of 1. Therefore, the creeping flow idealization and thus Stokes law is applicable, and the value
calculated is valid.
Discussion Flow separation starts at about Re = 10. Therefore, Stokes law can be used as an approximation for
Reynolds numbers up to this value, but this should be done with care.
0.1 mm
V
Air
Dust
15–59
Solution A pipe is exposed to high winds. The drag force exerted on the pipe by the winds is to be determined.
Assumptions 1 The outer surface of the pipe is smooth so that Fig. 15-34 can be used to determine the drag coefficient. 2
Air flow in the wind is steady and incompressible. 3 The turbulence in the wind is not considered. 4The direction of wind is
15–37
15-60
Solution Spherical hail is falling freely in the atmosphere. The terminal velocity of the hail in air is to be determined.
Assumptions 1 The surface of the hail is smooth so that Fig. 15-34 can be used to determine the drag coefficient. 2 The
variation of the air properties with altitude is negligible. 3 The buoyancy force applied by air to hail is negligible since air
<< hail (besides, the uncertainty in the density of hail is greater than the density of air). 4 Air flow over the hail is steady
and incompressible when terminal velocity is established. 5 The atmosphere is calm (no winds or drafts).
3
4
624
2
2
3
2
2
2D
gVC
D
g
V
D
CW
V
AC s
f
Ds
f
D
f
D
D
D
f
D
s
C
V
C
C
Dg
V662.8
) kg/m269.1(3
m) )(0.008 kg/m)(910m/s 81.9(4
3
4
3
32
(1)
V
V
VD 578.9Re
/sm 10382.1
m) (0.008
Re 25
(2)
V = 13.7 m/s
The corresponding Re and CD values are Re = 7930 and CD = 0.40. Therefore, the velocity of hail will remain constant
Hail
D = 0.8 cm
D
15-61E
Solution A pipe is crossing a river while remaining completely immersed in water. The drag force exerted on the
pipe by the river is to be determined.
62.30 lbm/ft3 and
= 2.36 lbm/fth = 6.55610-4 lbm/fts.
15–39
15-62
Solution Dust particles that are unsettled during high winds rise to a specified height, and start falling back when
things calm down. The time it takes for the dust particles to fall back to the ground and their velocity are to be determined
Analysis The terminal velocity of a free falling object is reached when the drag force equals the weight of the solid
object less the buoyancy force applied by the surrounding fluid,
BD FWF
where
VDFD
3
(Stokes law),
VV
gFgW f
Bs
and ,
m/s 0.168
m/s 1676.0
s)kg/m 1018(1.872
kg/m 1.164)–(1600m) 106)(m/s 81.9(
18
)(
5–
3252
2
fs
gD
V
min 19.9 s 1194
m/s 0.1676
m 200
t V
L
0.06 mm
V
Dust
15–63
Solution A cylindrical log suspended by a crane is subjected to normal winds. The angular displacement of the log
and the tension on the cable are to be determined.
Assumptions 1 The surfaces of the log are smooth so that Fig. 15-34 can be used to determine the drag coefficient (not a