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Unformatted text preview: Problem 9.1 Given: Model of riverboat Find: [2] Distance at which transition occurs Solution:
ρ⋅ U ⋅ x
U⋅ x
=
μ
ν Basic equation Rex = For water at 10oC ν = 1.30 × 10 Hence xp = For the model xm = 2
−6 m ν⋅ Rex
U
xp
18 ⋅ s and transition occurs at about Rex = 5 × 10 5 (Table A.8) and we are given xp = 0.186 m xp = 18.6⋅ cm xm = 0.0103 m xm = 10.3⋅ mm U = 3.5⋅ m
s Problem 9.2 Given: Minivan traveling at various speeds Find: [2] Plot of boundary layer length as function of speed Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
ρVL crit/μ =
500000 Re crit =
The critical length is then L crit = 500000 μ/V ρ
Tabulated or graphical data:
μ=
3.79E07
ρ=
0.00234
(Table A.9, 68oF) lbf.s/ft2
slug/ft3 Computed results:
V (mph) L crit (ft) 10
13
15
18
20
30
40
50
60
70
80
90 5.52
4.42
3.68
3.16
2.76
1.84
1.38
1.10
0.920
0.789
0.690
0.614 Length of Laminar Boundary Layer
on the Roof of a Minivan
6
5
4
L crit (ft) 3
2
1
0
0 10 20 30 40
50
V (mph) 60 70 80 90 100 Problem 9.3 [3] Given: Boeing 757 Find: Point at which transition occurs; Same point at 10,000 m Solution:
ρ⋅ U ⋅ x
U⋅ x
=
μ
ν Basic equation Rex = For air at 20oC ν = 1.50 × 10 Hence xp = At 10,000 m T = 223.3⋅ K 2
−5 m ν⋅ Rex
U ⋅ s 5 and transition occurs at about Rex = 5 × 10 (Table A.10) and we are given xp = 0.104 m xp = 10.4 cm (Table A.3) T = −49.8 °C U = 260⋅ km
hr ρ = 0.414 kg We need to estimate ν or μ at this temperature. From Appendix A3
μ= b⋅ T
S
T 1+ Hence μ= b⋅ T
1+ S
T b = 1.458 × 10 −6 kg ⋅ m⋅ s⋅ K
μ = 1.458 × 10 1
2 S = 110.4⋅ K − 5 N⋅ s
2 m For air at 10,000 m (Table A.3)
ρ
ρSL = 0.3376 xp = ρ = 0.3376⋅ ρSL 3 m μ
ν=
ρ
Hence kg ρSL = 1.225⋅
ν = 3.53 × 10 ν⋅ Rex
U 2
−5m xp = 0.0747 m 3 m
s and we are given xp = 7.47 cm U = 850⋅ km
hr Problem 9.4 [2] Given: Flow around American and British golf balls, and soccer ball Find: Speed at which boundary layer becomes turbulent Solution:
ρ⋅ U ⋅ D
U⋅ D
=
μ
ν Basic equation ReD = For air ν = 1.62 × 10 − 4 ft For the American golf ball D = 1.68⋅ in ⋅ 5 ReD = 2.5 × 10 and transition occurs at about 2 s (Table A.9) Hence U= For the British golf ball D = 41.1⋅ mm Hence U= For soccer ball D = 8.75⋅ in Hence U= ν⋅ ReD
D
ν⋅ ReD
D
ν⋅ ReD
D U = 289⋅ ft
s U = 197 mph U = 88.2 m
s U = 300⋅ ft
s U = 205 mph U = 91.5 m
s ft
s U = 37.9 mph U = 16.9 m
s U = 55.5⋅ Problem 9.5 [2] Given: Experiment with 1 cm diameter sphere in SAE 10 oil Find: Reasonableness of two flow extremes Solution:
ρ⋅ U ⋅ D
U⋅ D
=
μ
ν Basic equation ReD = For SAE 10 ν = 1.1 × 10 For ReD = 1 For ReD = 2.5 × 10 Note that for ReD = 2.5 × 10 For water ν = 1.01 × 10 For ReD = 2.5 × 10 2
−4 m ⋅ s and transition occurs at about (Fig. A.3 at 20oC)
we find 5 U= 5 5 ν ⋅ ReD
D
ν ⋅ ReD
D U = 0.011⋅
U = 2750 m
s m
s U = 1.10⋅ cm
s which is reasonable which is much too high! we need to increase the sphere diameter D by a factor of about 1000, or reduce the viscosity ν b
the same factor, or some combination of these. One possible solution is 2
−6 m ⋅ U= D = 1⋅ cm and s (Table A.8 at 20oC)
we find U= D = 10⋅ cm and
ν⋅ ReD
D Hence one solution is to use a 10 cm diameter sphere in a water tank. U = 2.52⋅ m
s which is reasonable Problem 9.6 [2] Given: Sheet of plywood attached to the roof of a car Find: Speed at which boundary layer becomes turbulent; Speed at which 90% is turbulent Solution:
ρ⋅ U⋅ x
U⋅ x
=
μ
ν Basic equation Rex = For air ν = 1.62 × 10 For the plywood x = 8⋅ ft − 4 ft ⋅ Rex = 5 × 10 and transition occurs at about 5 2 s (Table A.9) Hence When 90% of the boundary layer is turbulent x = 0.1 × 8⋅ ft U= ν ⋅ Rex Hence x ft
s
ν⋅ Rex U = 10.1⋅ U = 6.90⋅ mph U= U = 101⋅ x ft
s U = 69.0⋅ mph Problem 9.7 Given: Aircraft or missile at various altitudes Find: Plot of boundary layer length as function of altitude Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
Re crit = ρUL crit/μ = 500000 The critical length is then
L crit = 500000 μ/U ρ
Let L 0 be the length at sea level (density ρ0 and viscosity μ0). Then
L crit/L 0 = (μ/μ0)/(ρ/ρ0)
The viscosity of air increases with temperature so generally decreases with elevation;
the density also decreases with elevation, but much more rapidly.
Hence we expect that the length ratio increases with elevation
For the density ρ, we use data from Table A.3.
For the viscosity μ, we use the Sutherland correlation (Eq. A.1)
μ = bT 1/2/(1+S /T )
b=
S= 1.46E06
110.4 kg/m.s.K1/2
K [2] Computed results:
z (km) T (K) ρ/ρ0 μ/μ0 L crit/L 0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
6.0 288.2
284.9
281.7
278.4
275.2
271.9
268.7
265.4
262.2
258.9
255.7
249.2 1.0000
0.9529
0.9075
0.8638
0.8217
0.7812
0.7423
0.7048
0.6689
0.6343
0.6012
0.5389 1.000
0.991
0.982
0.973
0.965
0.955
0.947
0.937
0.928
0.919
0.910
0.891 1.000
1.04
1.08
1.13
1.17
1.22
1.28
1.33
1.39
1.45
1.51
1.65 7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0
16.0
17.0
18.0
19.0
20.0
22.0
24.0
26.0
28.0
30.0 242.7
236.2
229.7
223.3
216.8
216.7
216.7
216.7
216.7
216.7
216.7
216.7
216.7
216.7
218.6
220.6
222.5
224.5
226.5 0.4817
0.4292
0.3813
0.3376
0.2978
0.2546
0.2176
0.1860
0.1590
0.1359
0.1162
0.0993
0.0849
0.0726
0.0527
0.0383
0.0280
0.0205
0.0150 0.872
0.853
0.834
0.815
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.795
0.800
0.806
0.812
0.818
0.824 1.81
1.99
2.19
2.41
2.67
3.12
3.65
4.27
5.00
5.85
6.84
8.00
9.36
10.9
15.2
21.0
29.0
40.0
54.8 Length of Laminar Boundary Layer
versus Elevation
60
50
40
L/L 0
30
20
10
0
0 10 20
z (m) 30 Problem 9.8 [2] Given: Laminar boundary layer (air & water)
Find: Plot of boundary layer length as function of speed (at various altitudes for air)
Solution:
Governing equations:
The critical Reynolds number for transition to turbulence is
Re crit = UL crit/μ = 500000
The critical length is then
L crit = 500000 μ/U ρ
For air at sea level and 10 km, we can use tabulated data for density ρ from Table A.3.
For the viscosity μ, use the Sutherland correlation (Eq. A.1)
μ = bT 1/2/(1+S /T )
b = 1.46E06 kg/m.s.K1/2
S = 110.4 K
Air (sea level, T = 288.2 K):
ρ= 1.225 kg/m3 (Table A.3)
μ = 1.79E05 N.s/m2
(Sutherland) Water (20 oC): Air (10 K, T = 223.3 K):
ρ= 0.414 kg/m3 ρ = 998 slug/ft3 μ = 1.01E03 N.s/m2
μ = 1.46E05 N.s/m
(Table A.8)
(Sutherland)
(Table A.3) 2 Computed results: 0.05
0.10
0.5
1.0
5.0
15
20
25
30
50
100
200 Water Air (Sea level) Air (10 km)
L crit (m)
L crit (m)
L crit (m)
10.12
146.09
352.53
5.06
73.05
176.26
1.01
14.61
35.25
0.506
7.30
17.63
0.101
1.46
3.53
0.0337
0.487
1.18
0.0253
0.365
0.881
0.0202
0.292
0.705
0.0169
0.243
0.588
0.0101
0.146
0.353
0.00506
0.0730
0.176
0.00253
0.0365
0.0881 1000 0.00051 U (m/s) 0.0073 0.0176 Length of Laminar Boundary Layer
for Water and Air
100.0 1.0
L crit (m)
0.0 0.0
1.E02 Water
Air (Sea level)
Air (10 km)
1.E+00 1.E+02
U (m/s) 1.E+04 Problem 9.9 [2] Problem 9.10 [2] Problem 9.11 Given: Laminar boundary layer profile Find: [2] If it satisfies BC’s; Evaluate δ*/δ and θ/δ Solution:
3 u 3 y 1⎛ y⎞
=
− ⎜ ⎟ for which u = U at y = δ
U 2 δ 2 ⎝δ ⎠
du
u (0) = 0
=0
dy y =δ The boundary layer equation is The BC’s are u3
13
= (0) − (0) = 0
U2
2
⎛ 3 1 3 y2 ⎞
⎛3 1 3δ2 ⎞
du
⎟
=U⎜
−
=U⎜
⎜2δ 2δ3 ⎟
⎜2δ − 2δ3 ⎟ =0
⎟
dy
⎝
⎠ y =δ
⎝
⎠ At y = 0
At y = δ ∞ ⎛
0⎝ δ u⎞
u⎞
⎛
⎟dy = ∫ ⎜1 − ⎟dy
U⎠
U⎠
0⎝ For δ*: δ * = ∫ ⎜1 − Then 1
1
δ* 1δ⎛ u⎞
u ⎞ ⎛ y⎞
u⎞
⎛
⎛
= ∫ ⎜1 − ⎟dy = ∫ ⎜1 − ⎟d ⎜ ⎟ = ∫ ⎜1 − ⎟dη
δ δ 0⎝ U⎠
U ⎠ ⎝δ ⎠ 0 ⎝ U ⎠
0⎝ 1
u3
= η − η3
2
U2
1
1
1
δ*
u⎞
1 3⎞
3 2 1 4⎤ 3
⎛
⎛3
⎡
= ⎜1 − dη = ∫ ⎜1 − η + η ⎟dη = ⎢η − η + η ⎥ = = 0.375
δ ∫⎝ U ⎠
2
2⎠
4
8 ⎦0 8
⎣
0
0⎝ with Hence ∞ u
U
0 δ u⎞
u
⎛
⎜1 − ⎟dy = ∫
U
⎝ U⎠
0 u⎞
⎛
⎜1 − ⎟dy
⎝ U⎠ For θ: θ =∫ Then 1
1
θ 1δ u⎛ u⎞
u⎛
u⎞
u⎛
u ⎞ ⎛ y⎞
= ∫ ⎜1 − ⎟dy = ∫ ⎜1 − ⎟d ⎜ ⎟ = ∫ ⎜1 − ⎟dη
δ δ 0U ⎝ U⎠
U ⎝ U ⎠ ⎝δ ⎠ 0 U ⎝ U ⎠
0 Hence θ 1u
=
δ ∫U
0 1 1 1 3 ⎞⎛ 3
1 3⎞
9 2 1 3 3 4 1 6⎞
u⎞
⎛
⎛3
⎛3
⎜ 1 − ⎟ d η = ∫ ⎜ η − η ⎟ ⎜ 1 − η + η ⎟ d η = ∫ ⎜ η − η − η + η − η ⎟ dη
2
2 ⎠⎝ 2
2⎠
2
4
2
2
4⎠
⎝ U⎠
0⎝
0⎝ θ ⎡3 2 3 3 1 4 3 5 1 7⎤
39
= 0.139
=⎢ η − η − η + η − η ⎥ =
δ ⎣4
4
8
10
28 ⎦ 0 280
1 Problem 9.12 Given: Laminar boundary layer profile Find: [2] If it satisfies BC’s; Evaluate δ*/δ and θ/δ Solution:
3 4 The boundary layer equation is u
y
⎛ y⎞ ⎛ y⎞
= 2 − 2⎜ ⎟ + ⎜ ⎟ for which u = U at y = δ
U
δ
⎝δ ⎠ ⎝δ ⎠ The BC’s are u (0) = 0 du
dy =0
y =δ u
3
4
= 2(0) − 2(0) + (0) = 0
U
⎛1
⎛1
du
y2
y3 ⎞
δ2
δ3 ⎞
=U⎜2 − 6 3 + 4 4 ⎟
=U⎜2 − 6 3 + 4 4 ⎟ = 0
⎜δ
⎜δ
dy
δ
δ ⎟ y =δ
δ
δ⎟
⎝
⎠
⎝
⎠ At y = 0
At y = δ ∞ ⎛
0⎝ δ u⎞
u⎞
⎛
⎟dy = ∫ ⎜1 − ⎟dy
U⎠
U⎠
0⎝ For δ*: δ * = ∫ ⎜1 − Then 1
1
u ⎞ ⎛ y⎞
δ* 1δ⎛ u⎞
u⎞
⎛
⎛
= ∫ ⎜1 − ⎟dy = ∫ ⎜1 − ⎟d ⎜ ⎟ = ∫ ⎜1 − ⎟dη
U ⎠ ⎝δ ⎠ 0 ⎝ U ⎠
δ δ 0⎝ U⎠
0⎝ with u
= 2η − 2η 3 + η 4
U δ*
u⎞
1
1⎤
3
⎛
⎡
= ∫ ⎜1 − ⎟dη = ∫ (1 − 2η + 2η 3 − η 4 )dη = ⎢η − η 2 + η 4 − η 5 ⎥ =
= 0.3
δ
U⎠
2
5 ⎦ 0 10
⎣
0⎝
0
1 Hence 1 1 ∞ δ u⎛
u⎞
u⎛
u⎞
⎜1 − ⎟dy = ∫ ⎜1 − ⎟dy
U⎝ U⎠
U⎝ U⎠
0
0 For θ: θ =∫ Then 1
1
u⎛
u ⎞ ⎛ y⎞
θ 1δ u⎛ u⎞
u⎛
u⎞
= ∫ ⎜1 − ⎟dy = ∫ ⎜1 − ⎟d ⎜ ⎟ = ∫ ⎜1 − ⎟dη
U ⎝ U ⎠ ⎝δ ⎠ 0 U ⎝ U ⎠
δ δ 0U ⎝ U⎠
0 Hence θ
u⎛
u⎞
= ∫ ⎜1 − ⎟dη = ∫ (2η − η 3 + η 4 )(1 − 2η + η 3 − η 4 )dη = ∫ (2η − 4η 2 − 2η 3 + 9η 4 − 4η 5 − 4η 6 + 4η 7 − η 8 )dη
δ 0U ⎝ U ⎠
0
0
1 1 1 θ ⎡ 2 4 3 1 4 9 5 4 7 1 8 1 9⎤
37
= ⎢η − η − η + η − η + η − η ⎥ =
= 0.117
δ⎣
3
2
5
7
2
9 ⎦ 0 315
1 Problem 9.13 Given: Laminar boundary layer profile Find: [3] If it satisfies BC’s; Evaluate δ*/δ and θ/δ Solution: u
y
=2
U
δ
u
y
= 2− 2 +
U
δ The boundary layer equation is ( u (0) = 0 The BC’s are du
dy )( δ 0< y< ) 2 −1 2 δ 2 < y < δ for which u = U at y = δ =0
y =δ u
= 2 (0) = 0
U
du
1⎤
⎡
=U⎢ 2 − 2 ⎥
≠ 0 so it fails the outer BC.
dy
δ ⎦ y =δ
⎣ At y = 0 ( At y = δ ) This simplistic distribution is a piecewise linear profile: The first half of the layer has velocity gradient ( second half has velocity gradient 2 − 2 2 U δ = 1.414 U δ , and the )U = 0.586 U . At y = δ, we make another transition to zero velocity gradient.
δ
δ
δ ∞ For δ*: u⎞
u⎞
⎛
⎛
δ * = ∫ ⎜1 − ⎟dy = ∫ ⎜1 − ⎟dy
U⎠
U⎠
0⎝
0⎝ Then 1
1
δ* 1δ⎛ u⎞
u⎞
u ⎞ ⎛ y⎞
⎛
⎛
= ∫ ⎜1 − ⎟dy = ∫ ⎜1 − ⎟d ⎜ ⎟ = ∫ ⎜1 − ⎟dη
δ δ 0⎝ U⎠
U ⎠ ⎝δ ⎠ 0 ⎝ U ⎠
0⎝ u
1
= 2η
0 <η <
U
2
u
= 2 − 2 η + 2 −1
U with ( )( ) 1
<η <1
2 Hence δ*
u⎞
⎛
= ∫ ⎜1 − ⎟dη =
δ
U⎠
0⎝
1 ∫ (1 − 12 0 ) 2η dη + ∫ [1 − (2 − 2 )η − (
1 12 )] ⎡1
2 − 1 dη = ⎢
⎣2 2 ( ) 12 ⎤
⎡1
2
2η − 1 ⎥ + ⎢ (η − 1)
2
⎦0 ⎣
2 ( ) 1 ⎤
2 −2⎥
⎦1 2 δ * ⎡1
2 ⎤ ⎡1
2⎤ 3
2
= 0.396
=⎢ −
⎥= −
⎥+⎢ −
δ
2 8 ⎦ ⎣4 8 ⎦ 4 4
⎣
δ ∞ For θ: u⎛
u⎞
u⎛
u⎞
θ = ∫ ⎜1 − ⎟dy = ∫ ⎜1 − ⎟dy
U⎝ U⎠
U⎝ U⎠
0
0 Then θ 1δ u
=
δ δ ∫U
0 1 u⎞
u
⎛
⎜1 − ⎟dy = ∫
U
⎝ U⎠
0 1 u ⎞ ⎛ y⎞
u
⎛
⎜1 ⎟d ⎜ ⎟ = ∫
⎝ U ⎠ ⎝δ ⎠ 0 U u⎞
⎛
⎜ 1 − ⎟ dη
⎝ U⎠ Hence, after a LOT of work θ
u⎛
u⎞
= ∫ ⎜ 1 − ⎟ dη =
δ 0U ⎝ U ⎠
1 12 ∫ ( ) 2η 1 − 2η dη + 0 12 ⎛ 2η 1 ⎞⎤
θ⎡
⎡⎛ 1
⎟
= ⎢ 2η 2 ⎜
⎜ 3 − 2 ⎟ ⎥ + ⎢⎜ 3
δ⎢
⎠ ⎥ 0 ⎣⎝
⎝
⎣
⎦ ∫ [((2 − 2 )η + ( ))( ( 1 )( 2 −1 1− 2 − 2 η − ))] 2 − 1 dη 12 ( ) ( ) 1 1⎞
21
2
21
2⎤
2 − 2 (η − 1) − ⎟ 2 − 2 (η − 1) ⎥ =
−+
=
− = 0.152
2⎠
8 12 24
6 12
⎦1 2 Problem 9.14 [2] Problem 9.15 [3] Problem 9.16 [2] Problem 9.17 [2] Problem 9.18 Given: Data on fluid and boundary layer geometry Find: [3] Mass flow rate across ab; Drag Solution:
ρ = 800⋅ The given data is kg U = 3⋅ 3 m m
s L = 3⋅ m δ = 25⋅ mm b = 1⋅ m Governing equations:
Mass
Momentum
Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a Applying these to the CV abcd
δ Mass ⌠
( −ρ⋅ U⋅ b⋅ δ) + ⎮ ρ⋅ u⋅ b dy + mab = 0
⌡0 For the boundary layer u
y
=
=η
U
δ Hence ⌠
1
mab = ρ⋅ U⋅ b⋅ δ − ⎮ ρ⋅ U⋅ η⋅ δ dy = ρ⋅ U⋅ b⋅ δ − ⋅ ρ⋅ U⋅ b⋅ δ
⌡0
2 dy
= dη
δ
1 1
mab = ⋅ ρ⋅ U⋅ b⋅ δ
2 kg
mab = 30
s
δ Momentum ⌠
Rx = U⋅ ( −ρ⋅ U⋅ δ) + mab⋅ uab + ⎮ u⋅ ρ⋅ u⋅ b dy
⌡0
uab = U Note that and δ 1 ⌠
⌠
2
2
⎮ u⋅ ρ⋅ u⋅ b dy = ⎮ ρ⋅ U ⋅ b⋅ δ⋅ η dη
⌡0
⌡0 1 ⌠
2
2
Rx = −ρ⋅ U ⋅ b⋅ δ + ⋅ ρ⋅ U⋅ b⋅ δ⋅ U + ⎮ ρ⋅ U ⋅ b⋅ δ⋅ η dy
⌡0
2
2 1 2 1 R x = − ρ ⋅ U ⋅ b⋅ δ + 2 2 ⋅ ρ⋅ U ⋅ δ + 1
3 2 ⋅ ρ⋅ U ⋅ δ 1
2
R x = − ⋅ ρ ⋅ U ⋅ b⋅ δ
6 Rx = −30 N We are able to compute the boundary layer drag even though we do not know the viscosity because it is the viscosity that creates the
boundary layer in the first place Problem 9.19 [3] Given: Data on fluid and boundary layer geometry Find: Mass flow rate across ab; Drag; Compare to Problem 9.18 Solution:
The given data is ρ = 800⋅ kg U = 3⋅ 3 m m
s L = 1⋅ m δ = 14⋅ mm b = 3⋅ m Governing equations:
Mass
Momentum Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a
Applying these to the CV abcd
δ Mass ⌠
( −ρ⋅ U⋅ b⋅ δ) + ⎮ ρ⋅ u⋅ b dy + mab = 0
⌡0 For the boundary layer u
y
=
=η
U
δ Hence ⌠
1
mab = ρ⋅ U⋅ b⋅ δ − ⎮ ρ⋅ U⋅ η⋅ δ dy = ρ⋅ U⋅ b⋅ δ − ⋅ ρ⋅ U⋅ b⋅ δ
⌡0
2 dy
= dη
δ
1 1
mab = ⋅ ρ⋅ U⋅ b⋅ δ
2 kg
mab = 50.4
s
δ Momentum ⌠
Rx = U⋅ ( −ρ⋅ U⋅ δ) + mab⋅ uab + ⎮ u⋅ ρ⋅ u⋅ b dy
⌡0
uab = U Note that and
1 1 2 1 R x = − ρ ⋅ U ⋅ b⋅ δ + 1 ⌠
2
2
⋅ ρ⋅ U⋅ b⋅ δ⋅ U + ⎮ ρ⋅ U ⋅ b⋅ δ⋅ η dy
⌡0
2 2 R x = − ρ ⋅ U ⋅ b⋅ δ + δ ⌠
⌠
2
2
⎮ u⋅ ρ⋅ u⋅ b dy = ⎮ ρ⋅ U ⋅ b⋅ δ⋅ η dη
⌡0
⌡0 2 1
2
R x = − ⋅ ρ ⋅ U ⋅ b⋅ δ
6 2 ⋅ ρ⋅ U ⋅ δ + 1
3 2 ⋅ ρ⋅ U ⋅ δ Rx = −50.4 N We should expect the drag to be larger than for Problem 9.18 because the viscous friction is mostly
concentrated near the leading edge (which is only 1 m wide in Problem 9.18 but 3 m here). The reason viscous
stress is highest at the front region is that the boundary layer is very small (δ <<) so τ = μdu/dy ~ μU/δ >> Problem 9.20 [3] Problem 9.21 [2] Given: Data on wind tunnel and boundary layers Find: Displacement thickness at exit; Percent change in uniform velocity through test section Solution:
The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it
reduces the size of the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in
the boundary layer; an easier approach is to simply use the displacement thickness!
δ Basic equations ⌠
δdisp = ⎮
⎮
⌡0 (4.12) ⎛ 1 − u ⎞ dy
⎜
⎟
⎝ U⎠ Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal
1
7 and u
y⎞
=⎛ ⎟
⎜
U ⎝ δ⎠ w = 20⋅ cm h = 20⋅ cm For this flow ρ⋅ U⋅ A = const The design data is Udesign = 50⋅ The volume flow rate is Q = Udesign⋅ Adesign Q=2 We also have δin = 10⋅ mm δexit = 25⋅ mm m
s 2 Adesign = w⋅ h Adesign = 0.04 m 3 δ ⌠
⎮
δ
⎮
⌠⎛
⎮ 1 − u ⎞ dy = ⎮
δdisp =
⎟
⎮⎜
⎮
U⎠
⌡0
⌡0 ⎝ Hence m
s 1
1⎤
⎡
⌠
1⎞
⎮⎛
⎢
7⎥
⎛ y ⎞ ⎥ dy = δ⋅ ⎮ ⎜ 1 − η 7 ⎟ dη
⎢1 − ⎜ ⎟
⎝
⎠
⌡0
⎣ ⎝ δ⎠ ⎦ where η= y
δ Hence at the inlet and exit
δdispin = δin
8 δdispin = 1.25⋅ mm δdispexit = δexit
8 δdispexit = 3.125⋅ mm δdisp = δ
8 Hence the areas are ( )( ) 2 Ain = w − 2⋅ δdispin ⋅ h − 2⋅ δdispin ( )( Ain = 0.0390⋅ m ) Aexit = w − 2⋅ δdispexit ⋅ h − 2⋅ δdispexit 2 Aexit = 0.0375⋅ m Applying mass conservation between "design" conditions and the inlet (−ρ⋅ Udesign⋅ Adesign) + (ρ⋅ Uin⋅ Ain) = 0
Adesign or Uin = Udesign⋅ Also Adesign
Uexit = Udesign⋅
Aexit Ain The percent change in uniform velocity is then Uin = 51.3 m
s Uexit = 53.3
Uexit − Uin
Uin m
s
= 3.91⋅ % The exit displacement thickness is δdispexit = 3.125⋅ mm Problem 9.22 [2] Given: Data on wind tunnel and boundary layers Find: Uniform velocity at exit; Change in static pressure through the test section Solution:
δ Basic equations ⌠
u
δdisp = ⎮ ⎛ 1 − ⎞ dy
⎟
⎮⎜
U⎠
⌡0 ⎝ (4.12) 2 pV
+
+ g⋅ z = const
ρ
2 Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal For this flow ρ⋅ U⋅ A = const and u
y
=⎛ ⎞
⎜⎟
U ⎝δ⎠ The given data is U1 = 25⋅ m
s
δ1 = 20⋅ mm h = 25⋅ cm A=h We also have 2 2 A = 625⋅ cm δ2 = 30⋅ mm
δ Hence 1
7 ⌠
⎮
δ
⎮
⌠
u
δdisp = ⎮ ⎛ 1 − ⎞ dy = ⎮
⎟
⎮⎜
⎮
U⎠
⌡0
⌡0 ⎝ 1
1⎤
⎡
⌠⎛
1⎞
⎢
⎮⎜
7⎥
⎟
⎢1 − ⎛ y ⎞ ⎥ dy = δ⋅ ⎮ ⎝ 1 − η 7 ⎠ dη
⎜⎟
⌡0
⎣ ⎝δ⎠ ⎦ η= where y
δ δdisp = δ
8 Hence at the inlet and exit
δdisp1 = δ1
8 Hence the areas are δdisp1 = 2.5⋅ mm δdisp2 = δ2 δdisp2 = 3.75⋅ mm 8 (
)
2
A2 = (h − 2⋅ δdisp2)
A1 = h − 2⋅ δdisp1 2 2 A1 = 600⋅ cm 2 A2 = 588⋅ cm Applying mass conservation between Points 1 and 2 (−ρ⋅ U1⋅ A1) + (ρ⋅ U2⋅ A2) = 0
The pressure change is found from Bernoulli
Hence Δp = ρ⎛ 2
2
⋅ U1 − U2 ⎞
⎝
⎠
2 A1
U2 = U1⋅
A2 or
p1
ρ 2 + U1
2 = p2
ρ Δp = −15.8 Pa m U2 = 25.52 s 2 + U2
2 with ρ = 1.21⋅ kg
3 m The pressure drops slightly through the test section Problem 9.23 [2] Given: Data on boundary layer in a cylindrical duct Find: Velocity U2 in the inviscid core at location 2; Pressure drop Solution:
The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it reduces the size
the core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the boundary layer; an easier approa
is to simply use the displacement thickness!
The given or available data (from Appendix A) is
ρ = 1.23⋅ kg U1 = 12.5⋅ 3 m m
s D = 100⋅ mm δ1 = 5.25⋅ mm δ2 = 24⋅ mm Governing equations:
Mass
2 pV
+
+ g⋅ z = constant
ρ
2 Bernoulli (4.24) The displacement thicknesses can be computed from boundary layer thicknesses using Eq. 9.1
1 ⌠⎛
1⎞
⎮⎜
⌠⎛
δ
u⎞
7⎟
δdisp = ⎮ ⎜ 1 − ⎟ dy = δ⋅ ⎮ ⎝ 1 − η ⎠ dη =
⎮⎝
⌡0
8
U⎠
⌡0
δ Hence at locations 1 and 2 δdisp1 = δ1
8 Applying mass conservation at locations 1 and 2 δdisp1 = 0.656⋅ mm δ2 δdisp2 = δdisp2 = 3⋅ mm 8 (−ρ⋅ U1⋅ A1) + (ρ⋅ U2⋅ A2) = 0 A1
U 2 = U 1⋅
A2 or The two areas are given by the duct cross section area minus the displacement boundary layer
A1 = Hence π
4 ( ⋅ D − 2⋅ δdisp1 ) 2 A1 = 7.65 × 10 −3 2 m A2 = π
4 ( ⋅ D − 2⋅ δdisp2 ) 2 A1
U2 = U1⋅
A2 For the pressure drop we can apply Bernoulli to locations 1 and 2 to find A2 = 6.94 × 10 U2 = 13.8 p1 − p2 = Δp = ρ⎛ 2
2
⋅ U − U1 ⎞
⎠
2⎝ 2 m
s Δp = 20.6 Pa −3 2 m Problem 9.24 [2] Given: Data on wind tunnel and boundary layers Find: Uniform velocity at Point 2; Change in static pressure through the test section Solution:
δ Basic equations (4.12) ⌠
u
δdisp = ⎮ ⎛ 1 − ⎞ dy
⎟
⎮⎜
U⎠
⌡0 ⎝ 2 pV
+
+ g⋅ z = const
ρ
2 Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal For this flow ρ⋅ U⋅ A = const and u
y⎞
=⎛ ⎟
⎜
U ⎝ δ⎠ The given data is U1 = 60⋅ ft
s
δ1 = 0.4⋅ in W = 12⋅ in 1
7 A=W We also have 2 A = 144⋅ in δ2 = 0.5⋅ in
δ Hence 2 ⌠
⎮
δ
⎮
⌠
u
δdisp = ⎮ ⎛ 1 − ⎞ dy = ⎮
⎟
⎮⎜
⎮
U⎠
⌡0
⌡0 ⎝ 1
1⎤
⎡
⌠⎛
1⎞
⎢
⎮⎜
7⎥
⎟
⎞
⎢1 − ⎛ y ⎟ ⎥ dy = δ⋅ ⎮ ⎝ 1 − η 7 ⎠ dη
⎜
⌡0
⎣ ⎝ δ⎠ ⎦ η= where y
δ δdisp = δ
8 Hence at the inlet and exit
δdisp1 = δ1
8 Hence the areas are δdisp1 = 0.050⋅ in δdisp2 = δ2
8 (
)
2
A2 = (W − 2⋅ δdisp2)
A1 = W − 2⋅ δdisp1 2 δdisp2 = 0.0625⋅ in
2 A1 = 142⋅ in 2 A2 = 141⋅ in Applying mass conservation between Points 1 and 2 (−ρ⋅ U1⋅ A1) + (ρ⋅ U2⋅ A2) = 0
The pressure change is found from Bernoulli Hence Δp = In terms of inches of water p1
ρ 2 + A1
U2 = U1⋅
A2 or U1
2 = p2
ρ U2 = 60.25⋅ with ρ = 0.00234⋅ ft
s 2 + U2
2 ft
−4 ρ⎛ 2
2
⋅ U1 − U2 ⎞
⎝
⎠
2 slug Δp = −2.47 × 10 ⋅ psi Δp = −0.0356⋅ 3 lbf
ft ρH2O = 1.94⋅ slug
ft 3 Δh = Δp
ρH2O⋅ g 2 Δh = −0.00684⋅ in Problem 9.25 Given: Data on wind tunnel and boundary layers Find: [2] Pressure change between points 1 and 2 Solution:
2 Basic equations (4.12) pV
+
+ g⋅ z = const
ρ
2 Assumptions: 1) Steady flow 2) Incompressible 3) No friction outside boundary layer 4) Flow along streamline 5) Horizontal
For this flow ρ⋅ U⋅ A = const The given data is U0 = 100⋅ We also have δdisp2 = 0.035⋅ in Hence at the Point 2 A2 = h − 2⋅ δdisp2 ft
s ( U1 = U0 ) h = 3⋅ in 2 2 2 A1 = h A1 = 9⋅ in 2 A2 = 8.58⋅ in Applying mass conservation between Points 1 and 2 ( −ρ⋅ U1⋅ A1 ) + ( ρ⋅ U2⋅ A2 ) = 0
The pressure change is found from Bernoulli Hence Δp = p1
ρ 2 + A1
U2 = U1⋅
A2 or U1
2 ρ⎛ 2
2
⋅ U − U2 ⎞
⎠
2⎝ 1 The pressure drops by a small amount as the air accelerates = p2
ρ U2 = 105⋅ with ρ = 0.00234⋅ ft
s 2 + U2
2 slug
ft −3 Δp = −8.05 × 10 ⋅ psi Δp = −1.16⋅ lbf
ft 2 3 Problem 9.26 [3] Problem 9.27 [3] Problem 9.28 Given: Data on fluid and boundary layer geometry Find: [3] Gage pressure at location 2; average wall stress Solution:
The solution involves using mass conservation in the inviscid core, allowing for the fact that as the boundary layer grows it reduces the size
core. One approach would be to integrate the 1/7 law velocity profile to compute the mass flow in the boundary layer; an easier approach i
simply use the displacement thickness!
The average wall stress can be estimated using the momentum equation for a CV
The given and available (from Appendix A) data is
ρ = 1.23⋅ kg
3 m
s U1 = 15⋅ m L = 6⋅ m D = 400⋅ mm δ2 = 100⋅ mm Governing equations:
Mass Momentum
2 Bernoulli pV
+
+ g⋅ z = constant
ρ
2 (4.24) Assumptions: (1) Steady flow (2) No pressure force (3) No body force in x direction
The displacement thickness at location 2 can be computed from boundary layer thickness using Eq. 9.1
1 δ2 ⌠
δdisp2 = ⎮
⎮
⌡0
Hence δdisp2 = δ2 π
4 δdisp2 = 12.5 mm 8 Applying mass conservation at locations 1 and 2 A1 = ⌠
1⎞
⎮
⎛ 1 − u ⎞ dy = δ ⋅ ⎮ ⎜ 1 − η 7 ⎟ dη = δ2
⎜
⎟
⎠
2⌡ ⎝
8
⎝ U⎠
0 ⋅D (−ρ⋅ U1⋅ A1) + (ρ⋅ U2⋅ A2) = 0 2 A1
U 2 = U 1⋅
A2 or
2 A1 = 0.126 m The area at location 2 is given by the duct cross section area minus the displacement boundary layer
A2 = π
4 ( ⋅ D − 2⋅ δdisp2 ) 2 2 A2 = 0.11 m Hence A1
U2 = U1⋅
A2 U2 = 17.1 m
s For the pressure change we can apply Bernoulli to locations 1 and 2 to find
p1 − p2 = Δp =
Hence ρ⎛ 2
2
⋅ U − U1 ⎞
⎠
2⎝ 2 Δp = 40.8 Pa p2 ( gage) = p1 ( gage) − Δp p2 = −Δp p2 = −40.8 Pa For the average wall shear stress we use the momentum equation, simplified for this problem
D ⌠2
2
2π
2
⎮
2
Δp⋅ A1 − τ⋅ π⋅ D⋅ L = −ρ⋅ U1 ⋅ A1 + ρ⋅ U2 ⋅ ⋅ D − 2⋅ δ2 + ρ⋅ ⎮
2⋅ π⋅ r⋅ u dr
4
⌡D −δ ( ) 2 2 1
7 where y⎞
u ( r) = U2⋅ ⎛ ⎟
⎜δ
⎝ 2⎠ The integral is ⌠
2
⎮
7
y⎞
D
⎮
2
2⎮
ρ⋅ ⎮
2⋅ π⋅ r⋅ u dr = −2⋅ π⋅ ρ⋅ U2 ⋅ ⎮ ⎛ − y⎞ ⋅ ⎛ ⎟ dy
⎜
⎟⎜
⎠ ⎝ δ2 ⎠
⎮ ⎝2
⌡D −δ
⌡δ
2 r= and D
−y
2 dr = −dy 0 D
⌠2 2 2 D ⌠2
⎛ D δ2 ⎞
⎮
2
2
ρ⋅ ⎮
2⋅ π⋅ r⋅ u dr = 7⋅ π⋅ ρ⋅ U2 ⋅ δ2⋅ ⎜ − ⎟
⎝9 8⎠
⌡D −δ
2 2 2 Hence τ= 2π Δp⋅ A1 + ρ⋅ U1 ⋅ A1 − ρ⋅ U2 ⋅ τ = 0.461 Pa 4 ( ⋅ D − 2⋅ δ 2
π⋅ D ⋅ L ) 2 2 ⎛D − 7⋅ π⋅ ρ⋅ U2 ⋅ δ2⋅ ⎜ ⎝9 − δ2 ⎞ ⎟ 8⎠ Problem 9.29 [5] Part 1/2 Problem 9.29 [5] Part 1/2 Problem *9.30 [2] Problem *9.31 [3] Problem *9.32 [3] Problem *9.33 [3] Problem *9.34 [4] Problem 9.35 [4] Given: Blasius solution for laminar boundary layer Find: Point at which u = 0.95U; Slope of streamline; expression for skin friction coefficient and total drag; Momentum thicknes Solution:
Basic equation: Use results of Blasius solution (Table 9.1 on the web), and η = y⋅
f' = u
= 0.9130
U at η = 3.5 f' = u
= 0.9555
U at ν⋅ x
U η = 4.0 Hence by linear interpolation, when f' = 0.95 η = 3.5 +
− 4 ft From Table A.9 at 68oF ν = 1.62 × 10 Hence y = η⋅ The streamline slope is given by ⋅ 2 dy
v
=
dx
u U = 15⋅ and s ν⋅ x
U ( 4 − 3.5)
⋅ ( 0.95 − 0.9310)
( 0.9555 − 0.9310) ft
s η = 3.89 x = 7.5⋅ in y = 0.121 in u = U⋅ f' where and v= 1 ν⋅ U
⋅
⋅ ( η⋅ f' − f )
x
2 dy
1 ν⋅ U
ν ( η⋅ f' − f )
( η⋅ f' − f )
1
1
1
=⋅
⋅ ( η⋅ f' − f ) ⋅
=⋅
⋅
=
⋅
dx
2
x
f'
U⋅ f'
2 U⋅ x
f'
2⋅ Rex
We have Rex = U⋅ x Rex = 5.79 × 10 ν 4 From the Blasius solution (Table 9.1 on the web)
f = 1.8377 η = 3.5 f = 2.3057
Hence by linear interpolation at
at η = 4.0 f = 1.8377 + ( 2.3057 − 1.8377)
( 4.0 − 3.5) ⋅ ( 3.89 − 3.5) f = 2.2027 dy
1
( η⋅ f' − f )
=
⋅
= 0.00326
dx
f'
2⋅ Rex
The shear stress is ⎛∂ τw = μ⋅ ⎜ ⎝ ∂y u+ ∂ ⎞ v ⎟ = μ⋅ ∂x ⎠ ∂
∂y u at y = 0 (v = 0 at the wall for all x, so the derivative is zero there) 2 τw = μ⋅ U ⋅ U df
⋅
ν⋅ x dη2 2 and at η = 0 df
2 dη = 0.3321 (from Table 9.1) τw = 0.3321⋅ U⋅ The friction drag is ρ⋅ U ⋅ μ
x 2 τw = 0.3321⋅ ρ⋅ U ⋅ L
⌠
⌠
F D = ⎮ τ w dA = ⎮ τ w ⋅ b dx
⎮
⌡0
⌡ where b is the plate width L L
⌠
2
⎮
ρ⋅ U
2 ν⌠ 1
⎮
⋅
dx
FD = ⎮ 0.3321⋅
⋅ b dx = 0.3321⋅ ρ⋅ U ⋅
1
U⎮
Rex
⎮
⎮
⌡0
2
⎮x
⌡0
2 FD = 0.3321⋅ ρ⋅ U ⋅ For the momentum integral τw
ρ⋅ U 2 θL =
We have dθ
dx = or 2 F D = ρ ⋅ U ⋅ b⋅ L ⋅ dθ = τw
ρ⋅ U 2 0.6642
ReL ⋅ dx L ⌠
1 FD
0.6642⋅ L
⋅ ⎮ τ w dx =
⋅
=
2⌡
2b
ReL
ρ⋅ U 0
ρ⋅ U
1 L = 3⋅ ft θL = ν
⋅ b⋅ 2⋅ L
U 1
2 ReL = 0.6642⋅ L
ReL U⋅ L
ν ReL = 2.78 × 10
θL = 0.0454 in 5 2 μ
ρ⋅ U
= 0.3321⋅
ρ⋅ U ⋅ x
Rex Problem *9.36 Given: Blasius nonlinear equation Find: [5] Blasius solution using Excel Solution:
The equation to be solved is 2 d3 f
dη 3 +f d2 f
dη 2 =0 (9.11) The boundary conditions are
f = 0 and df
= 0 at η = 0
dη df
= 1 at η → ∞
dη
Recall that these somewhat abstract variables are related to physically meaningful variables:
f′= (9.12) u
= f′
U and η=y U νx ∝ y δ Using Euler’s numerical method f n+1 ≈ f n + Δη f n′ (1) f n′+1 ≈ f n′ + Δη f n′′ (2) f n′′+1 ≈ f n′′ + Δη f n′′′
th In these equations, the subscripts refer to the nth discrete value of the variables, and Δη = 10/N is the step
size for η (N is the total number of steps).
But from Eq. 9.11
f ′′′ = − 1
f f ′′
2 so the last of the three equations is
⎛1
⎞
f n′′+1 ≈ f n′′ + Δη ⎜ − f n f n′′ ⎟
⎝2
⎠ (3) Equations 1 through 3 form a complete set for computing f , f ′, f ′′ . All we need is the starting condition
for each. From Eqs. 9.12
f 0 = 0 and f 0′ = 0 We do NOT have a starting condition for f ′′ ! Instead we must choose (using Solver) f 0′′ so that the last
condition of Eqs. 9.12 is met:
′
fN =1 Computations (only the first few lines of 1000 are shown):
Δη = 0.01 Make a guess for the first f ''; use Solver to vary it until f 'N = 1
Count
0
1
2
3
4
5
6
7
8
10
9
10
8
11
12
6
13
η
14
4
15
16
2
17
18
0
19
0.0 20
21
22 η
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.20
0.21
0.22 f
f'
f''
0.0000
0.0000
0.3303
0.0000
0.0033
0.3303
0.0000
0.0066
0.3303
0.0001
0.0099
0.3303
0.0002
0.0132
0.3303
0.0003
0.0165
0.3303
0.0005
0.0198
0.3303
Blasius Velocity Profile
0.0007
0.0231
0.3303
0.0009
0.0264
0.3303
0.0012
0.0297
0.3303
0.0015
0.0330
0.3303
0.0018
0.0363
0.3303
0.0022
0.0396
0.3303
0.0026
0.0429
0.3303
0.0030
0.0462
0.3303
0.0035
0.0495
0.3303
0.0040
0.0528
0.3303
0.0045
0.0562
0.3303
0.0051
0.0595
0.3303
0.0056
0.0628
0.3303
0.4 0.0661
0.6
0.0063
0.3302
0.0069
0.0694
u/U = f '0.3302
0.0076
0.0727
0.3302 0.8 1.0 Problem 9.37 Given: Data on flow over flat plate Find: [2] Plot of laminar thickness at various speeds Solution:
Governing equations: Tabulated or graphical data:
ν= 2 m /s
1.50E05
(Table A.10, 20oC) Computed results:
U (m/s)
x crit (m) 1
7.5 2
3.8 3
2.5 4
1.9 5
1.5 10
0.75 x (m) δ (mm) δ (mm) δ (mm) δ (mm) δ (mm) δ (mm) 0.000
0.025
0.050
0.075
0.100
0.2
0.5
1.5
1.9
2.5
3.8
5.0 0.00
3.36
4.75
5.81
6.71
9.49
15.01
25.99
29.26
33.56
41.37
47.46 0.00
2.37
3.36
4.11
4.75
6.71
10.61
18.38
20.69
23.73
29.26 0.00
1.94
2.74
3.36
3.87
5.48
8.66
15.01
16.89
19.37 0.00
1.68
2.37
2.91
3.36
4.75
7.50
13.00
14.63 0.00
1.50
2.12
2.60
3.00
4.24
6.71
11.62 0.00
1.06
1.50
1.84 6.0
7.5 51.99
58.12 Laminar Boundary Layer Profiles
70 U = 1 m/s
U = 2 m/s
U = 3 m/s
U = 4 m/s
U = 5 m/s
U = 10 m/s 60
50
δ (mm) 40
30
20
10
0
0 2 4
x (m) 6 8 δ δ
1⎤
u
u ⎞ ⎛ y⎞
⎡
δ * = ∫ ⎛1 − ⎞dy = δ ∫ ⎛1 − ⎟d ⎜ ⎟ = δ ∫ (1 − 2 + η 2 )dη = δ ⎢η − η 2 + η 3 ⎥ =
⎜
⎟
⎜
3 ⎦0 3
U⎠
U ⎠ ⎝δ ⎠
⎣
0⎝
0⎝
0 Hence 1 1 1 Tabulated or graphical data: Given data:
ν = 1.01E06 m2/s
(Table A.8, 20 oC) L= 0.25 m U= 1.75 m/s Computed results: 0.0000
0.0125
0.0250
0.0375
0.0500
0.0625
0.0750
0.0875
0.1000
0.1125
0.1250
0.1375
0.1500
0.1625
0.1750
0.1875
0.2000
0.2125
0.2250
0.2375
0.2500 0.00.E+00
2.17.E+04
4.33.E+04
6.50.E+04
8.66.E+04
1.08.E+05
1.30.E+05
1.52.E+05
1.73.E+05
1.95.E+05
2.17.E+05
2.38.E+05
2.60.E+05
2.82.E+05
3.03.E+05
3.25.E+05
3.47.E+05
3.68.E+05
3.90.E+05
4.12.E+05
4.33.E+05 Laminar Boundary Layer Profiles δ (mm) δ* (mm) τw (Pa)
0.000
0.465
0.658
0.806
0.931
1.041
1.140
1.231
1.317
1.396
1.472
1.544
1.612
1.678
1.742
1.803
1.862
1.919
1.975
2.029
2.082 0.000
0.155
0.219
0.269
0.310
0.347
0.380
0.410
0.439
0.465
0.491
0.515
0.537
0.559
0.581
0.601
0.621
0.640
0.658
0.676
0.694 2.5 10.40
7.36
6.01
5.20
4.65
4.25
3.93
3.68
3.47
3.29
3.14
3.00
2.89
2.78
2.69
2.60
2.52
2.45
2.39
2.33 10 2.0 8 δ 1.5 6 τw (Pa) Re x δ and δ* (mm) x (m ) τw 1.0 4 δ* 0.5 2 0.0
0.00 0
0.05 0.10
x (m) 0.15 0.20 0.25 Problem 9.39 Given: Parabolic solution for laminar boundary layer Find: [2] Derivation of FD; Evaluate FD and θL Solution:
u
⎛ y⎞ ⎛ y⎞
= 2⋅ ⎜ ⎟ − ⎜ ⎟
U
⎝ δ⎠ ⎝ δ⎠ Basic equations: L = 0.25⋅ m Assumptions: 1) Flat plate so ∂
∂x 2 δ
5.48
=
x
Rex b = 1⋅ m U = 1.75⋅ m
s ρ = 1000⋅ kg
3 m p = 0, and U = const 2) δ is a function of x only 3) Incompressible ( τw δ ) d
2
=
U ⋅θ
ρ
dx The momentum integral equation then simplifies to ⌠ u⎛
u
θ=⎮
⋅ 1 − ⎞ dy
⎟
⎮ U⎜
U⎠
⎝
⌡0 where 2 dθ For U = const τ w = ρ⋅ U ⋅ The drag force is then L
L
⌠
⌠
⌠
2 dθ
2⌠
FD = ⎮ τw dA = ⎮ τw⋅ b dx = ⎮ ρ⋅ U ⋅ ⋅ b dx = ρ⋅ U ⋅ b⋅ ⎮ 1 dθ
⎮
⎮
⌡0
dx
⌡0
⌡
⌡0 For the given profile 1
1
⌠
⌠
θ ⌠ u⎛
2
u⎞
2
2
2
3
4
⎮
=
⋅ 1 − ⎟ dη = ⎮ 2⋅ η − η ⋅ 1 − 2⋅ η + η dη = ⎮ 2⋅ η − 5⋅ η + 4⋅ η − η dη =
⎮ U⎜
⌡0
⌡0
δ
15
⎝ U⎠
⌡0 dx
θ L 1 θ=
From Table A.8 at 20oC ( )( ) 2 FD = ρ⋅ U ⋅ b⋅ θL ( ) 2
⋅δ
15
2
−6 m ν = 1.01 × 10
δL = L⋅ ⋅ 5.48
ReL 2
θL =
⋅δ
15 L
2 FD = ρ⋅ U ⋅ b⋅ θL s ReL = U⋅ L
ν δL = 2.08 mm
θL = 0.278 mm
FD = 0.850 N ReL = 4.332 × 10 5 Problem 9.40 [2] Problem 9.41 [2] Given: Data on fluid and plate geometry Find: Drag at both orientations using boundary layer equation Solution:
The given data is ρ = 800⋅ kg μ = 0.02⋅ 3 m N ⋅s
m ReL = First determine the nature of the boundary layer 2 ρ⋅ U⋅ L
μ U = 3⋅ m
s L = 3⋅ m b = 1⋅ m 5 ReL = 3.6 × 10 The maximum Reynolds number is less than the critical value of 5 x 105
Hence:
Governing equations: τw
cf =
1
2
⋅ ρ⋅ U
2 cf = (9.22) 0.730 (9.23) Rex L The drag (one side) is ⌠
F D = ⎮ τ w ⋅ b dx
⌡0
L Using Eqs. 9.22 and 9.23 1
2⌠
F D = ⋅ ρ ⋅ U ⋅ b⋅ ⎮
2
⎮
⎮
⌡0 0.73
ρ⋅ U ⋅ x
μ FD = 0.73⋅ b⋅ μ⋅ L⋅ ρ⋅ U
Repeating for 3 L = 1⋅ m FD = 0.73⋅ b⋅ μ⋅ L⋅ ρ⋅ U dx FD = 26.3 N (Compare to 30 N for Problem 9.18) b = 3⋅ m
3 FD = 45.5 N (Compare to 50.4 N for Problem 9.19) Problem 9.42 Given: Triangular plate Find: [3] Drag Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 cf =
3
2 L = 0.50⋅ cm⋅
From Table A.10 at 20oC ⋅ Rex L = 0.433⋅ cm −5 m ν = 1.50 × 10 0.730 2 ρ = 1.21⋅ s ReL = U = 5⋅ ReL = 1443 so definitely laminar kg
m First determine the nature of the boundary layer m
s W = 50⋅ cm U⋅ L
ν 3 L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.730
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
2
2
Rex ⌠
FD = ⎮ τw⋅ w( x) dx
⌡0 L Hence FD = 1
2W⌠
⋅ ρ⋅ U ⋅ ⋅ ⎮
2
L⎮
⎮
⌡0 w ( x) = W⋅ x
L L ⌠
1
⎮
0.730
0.730⋅ x
2
dx =
⋅ ρ⋅ U ⋅ ⋅ ν ⋅ ⎮ x dx
⌡0
2
L
U⋅ x
ν
3
2W L The integral is ⌠
1
3
⎮
22
2
⎮ x dx = ⋅ L
⌡0
3 so 3 −4 FD = 0.243⋅ ρ⋅ W⋅ ν ⋅ L⋅ U FD = 4.19 × 10 Note: For twosided solution 2⋅ FD = 8.38 × 10 N −4 N Problem 9.43 [3] Plate is reversed from this! Given: Triangular plate Find: Drag Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 cf =
3
2 L = 0.50⋅ cm⋅
From Table A.10 at 20oC ⋅ Rex L = 0.433⋅ cm −5 m ν = 1.50 × 10 0.730 2 ρ = 1.21⋅ s ReL = U = 5⋅ ReL = 1443 so definitely laminar kg
m First determine the nature of the boundary layer 3 U⋅ L
ν
L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.730
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
2
2
Rex Hence L
⌠
⌠
⎮
3
⎮ 0.730⋅ ⎛ 1 − x ⎞
⎮
⎜
⎟
⎮
0.730
1
L⎠
2
2
⎝
dx =
FD = ⋅ ρ⋅ U ⋅ W⋅ ⎮
⋅ ρ⋅ U ⋅ W⋅ ν ⋅ ⎮
2
⎮
2
U⋅ x
⎮
⌡0
⎮
ν
⌡0
L The integral is ⌠
⎮
⎮
⎮
⎮
⌡0 m
s W = 50⋅ cm ⌠
FD = ⎮ τw⋅ w( x) dx
⌡0 w( x) = W⋅ ⎛ 1 −
⎜ ⎝ L x⎞
⎟
L⎠ 1⎞
⎛1
⎜−
2⎟
⎜ x 2 − x ⎟ dx
L
⎝ 1⎞
3
⎛1
1
⎜−
2⎟
2
⎜ x 2 − x ⎟ dx = 2⋅ L 2 − 2 ⋅ L = 4 ⋅ L
L⎠
3L
3
⎝
3 FD = 0.487⋅ ρ⋅ W⋅ ν ⋅ L⋅ U FD = 8.40 × 10
Note: For twosided solution The drag is much higher (twice as much) compared to Problem 9.42. This is because τw is largest near the
leading edge and falls off rapidly; in this problem the widest area is also at the front −4 N −3 2⋅ FD = 1.68 × 10 N Problem 9.44 Given: Parabolic plate Find: [3] Drag Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 cf = 0.730
Rex ⎛ W⎞
⎜⎟
2
L= ⎝ ⎠ W = 25⋅ cm 2 L = 6.25⋅ cm 25⋅ cm U = 7.5⋅ m
s Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L
From Table A.10 at 20oC −5 m ν = 1.50 × 10 ⋅ 2 ρ = 1.21⋅ s kg
m ReL = First determine the nature of the boundary layer 3 U⋅ L
ν ReL = 3.12 × 10 4 so just laminar L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.730
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
2
2
Rex Hence ⌠
3
⎮ 0.730⋅ x
L
⎮
0.730
1
L
ν⌠
2
2
dx =
FD = ⋅ ρ⋅ U ⋅ W⋅ ⎮
⋅ ρ⋅ U ⋅ W⋅
⋅ ⎮ 1 dx
2
2
L ⌡0
U⋅ x
⎮
⎮
ν
⌡0 ⌠
FD = ⎮ τw⋅ w( x) dx
⌡0 w ( x) = W⋅ x
L L 3 FD = 0.365⋅ ρ⋅ W⋅ ν ⋅ L⋅ U FD = 2.20 × 10
Note: For twosided solution −3 N −3 2⋅ FD = 4.39 × 10 N Problem 9.45 [4] Note: Plate is now reversed! Given: Parabolic plate Find: Drag Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 0.730 cf = Rex ⎛ W⎞
⎜⎟
2
L=⎝ ⎠ W = 25⋅ cm 2 L = 6.25⋅ cm 25⋅ cm U = 7.5⋅ m
s Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = 0
From Table A.10 at 20oC −5 m ν = 1.50 × 10 ⋅ 2 ρ = 1.21⋅ s kg
m U⋅ L
ν ReL = First determine the nature of the boundary layer 3
4 ReL = 3.12 × 10 so just laminar L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.730
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
2
2
Rex Hence ⌠
3
⎮ 0.730⋅ 1 − x
L
⌠
⎮
0.730
1
L
2
2
dx =
FD = ⋅ ρ⋅ U ⋅ W⋅ ⎮
⋅ ρ⋅ U ⋅ W⋅ ν ⋅ ⎮
⎮
2
2
U⋅ x
⎮
⌡0
⎮
ν
⌡0 The tricky integral is (this
might be easier to do
numerically!) ⌠
⎮
⎮
⎮
⌡ ⌠
FD = ⎮ τw⋅ w( x) dx
⌡0 w( x) = W⋅ 1 − x
L L FD = 2 i
⎛ −L − x −
x−
− ⋅ L⋅ ln ⎜
L
2
⎝ −L − x +
x 1
− dx =
xL
1 0.730
2 3
2 x⎞ ⎟
x⎠ 1
x − 1
dx
L L so ⌠
⎮
⎮
⌡0 1
x − 1
dx = 0.393⋅ m
L L ⌠
⋅ ρ⋅ U ⋅ W⋅ ν ⋅ ⎮
⎮
⌡0 1
x − 1
dx
L Note: For twosided solution The drag is much higher compared to Problem 9.44. This is because τw is largest near the leading edge
and falls off rapidly; in this problem the widest area is also at the front FD = 3.45 × 10 −3 N −3 2⋅ FD = 6.9 × 10 N Problem 9.46 Given: Pattern of flat plates Find: [3] Drag on separate and composite plates Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 0.730 cf = For separate plates L = 7.5⋅ cm From Table A.8 at 20oC ν = 1.01 × 10 Rex W = 7.5⋅ cm
−6 m ⋅ 2 ρ = 998⋅ s ReL = m
s kg
m First determine the nature of the boundary layer U = 1⋅ 3 U⋅ L
ν 4 ReL = 7.43 × 10 so definitely laminar L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.730
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
2
2
Rex ⌠
F D = ⎮ τ w ⋅ W dx
⌡0 L Hence FD = ⌠
1
2
⋅ ρ⋅ U ⋅ W⋅ ⎮
2
⎮
⎮
⌡0 L ⌠
1
⎮−
0.730
0.730
2
dx =
dx
⋅ ρ⋅ U ⋅ W⋅ ν⋅ ⎮ x
⌡0
2
U⋅ x
ν
3
2 L The integral is ⌠
1
1
⎮−
2
2
⎮x
dx = 2⋅ L
⌡0 This is the drag on one plate. The total drag is then so FD = 0.730⋅ ρ⋅ W⋅ ν⋅ L⋅ U 3 FTotal = 4⋅ FD FTotal = 0.0602 N
For both sides: For the composite plate L = 4 × 7.5⋅ cm FD = 0.0150 N 2⋅ FTotal = 0.120 N L = 0.30 m FComposite = 0.730⋅ ρ⋅ W⋅ ν⋅ L⋅ U 3 FComposite = 0.0301 N
For both sides: 2⋅ FComposite = 0.0602 N The drag is much lower on the composite compared to the separate plates. This is because τw is largest near the
leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times! Problem 9.47 [2] Problem 9.48 [3] Problem 9.49 [3] Problem 9.50 [3] Problem 9.51 Given: Water flow over flat plate Find: [3] Drag on plate for linear boundary layer Solution: From Table A.8 at 10oC du
τw = μ⋅
dy at y = 0, and also L = 0.35⋅ m Basic equations: ⌠
FD = 2⋅ ⎮ τw dA
⎮
⌡ W = 1⋅ m U = 0.8⋅ 2
−6 m ν = 1.30 × 10 ⋅ ρ = 1000⋅ s The velocity profile is
Hence
We also have y
= U⋅ η
δ
du
U
τw = μ⋅
= μ⋅
dy
δ τ w = ρ⋅ U ⋅ 2 dδ ⌠
⎮ τ w = ρ⋅ U ⋅
1 ( 1 ⋅ dx ⎮
⌡0 (1) m
s 3
5 ReL = 2.15 × 10 so laminar 1 ) Comparing Eqs 1 and 2 τw = μ⋅ Separating variables δ⋅ dδ = Hence δ= but we need δ(x) u⎛
u
2 dδ ⌠
⋅ ⎜ 1 − ⎞ dη = ρ⋅ U ⋅ ⋅ ⎮ η⋅ ( 1 − η) dη
⎟
U⎝
U⎠
dx ⌡0 ⌠
1
2
⎮ η − η dx =
⌡0
6 so 2 dδ τ w = ρ⋅ U ⋅ dx 1
2 dδ
⋅ ρ⋅ U ⋅ (2)
6
dx = U
1
2 dδ
= ⋅ ρ⋅ U ⋅
δ
6
dx
2 ⋅ dx or δ
6⋅ μ
=
⋅x + c
2
ρ⋅ U ⋅x or δ
=
x 6⋅ μ
ρ⋅ U 12⋅ μ
ρ⋅ U but δ(0) = 0 so c = 0 12
3.46
=
Rex
Rex
L L ⌠
⌠
FD = 2⋅ ⎮ τw dA = 2⋅ W⋅ ⎮
⎮
⎮
⌡
⌡ L 0 ⌠
⌠
1
1
⎮
⎮−
−
μ⋅ W⋅ U U ⎮
U
ρ⋅ U
2
2
dx =
⋅
dx
μ⋅ dx = 2⋅ W⋅ ⎮ μ⋅ U⋅
⋅x
⋅
x
⎮
ν ⌡0
δ
12⋅ μ
3
⌡0 L The integral is u⎛
u
⋅ 1 − ⎞ dη
⎟
⎮ U⎜
dx
U⎠
⎝
⌡0
⋅ u = U⋅ The integral is Then 1 kg m
U⋅ L
ReL =
ν First determine the nature of the boundary layer 2 dδ ⌠
⎮ ⌠
1
⎮−
⎮ x 2 dx = 2⋅ L
⌡0
FD = 2
3 ⋅ ρ⋅ W⋅ ν⋅ L⋅ U so
3 FD = 2⋅ μ⋅ W⋅ U
3 FD = 0.557 N ⋅ U⋅ L
ν Problem 9.52 Given: Data on flow in a channel Find: [3] Static pressures; plot of stagnation pressure Solution:
The given data is h = 30⋅ mm Appendix A ρ = 1.23⋅ δ2 = 10⋅ mm U2 = 22.5⋅ m
s w = 1⋅ m (Arbitrary) kg
3 m Governing equations
Mass
Before entering the duct, and in the the inviscid core, the Bernoulli equation holds
2 pV
+
+ g⋅ z = constant
2
ρ (4.24) Assumptions: (1) Steady flow (2) No body force in x direction
For a linear velocity profile, from Table 9.2 the displacement thickness at location 2 is
δ2 δdisp2 = δdisp2 = 5 mm 2 From the definition of the displacement thickness, to compute the flow rate, the uniform flow at location 2 is assumed to
take place in the entire duct, minus the displacement thicknesses at top and bottom ( A2 = w⋅ h − 2⋅ δdisp2 ) 2 A2 = 0.02 m 3 Then Q = A2⋅ U2 Q = 0.45 m
s Mass conservation (Eq. 4.12) leads to U2
U1⋅ A1 = U2⋅ A2
U1 = A2
A1 where A1 = w⋅ h ⋅ U2 2 A1 = 0.03 m
U1 = 15 m
s The Bernoull equation applied between atmosphere and location 1 is
patm
ρ = 2 p1 + ρ U1
2 or, working in gage pressures
1
2
p1 = − ⋅ ρ⋅ U1
2 p1 = −138 Pa
(Static pressure) Similarly, between atmosphere and location 2 (gage pressures)
1
2
p2 = − ⋅ ρ⋅ U2
2 p2 = −311 Pa
(Static pressure) The static pressure falls continuously in the entrance region as the fluid in the central core accelerates into a
decreasing core
The stagnation pressure at location 2 (measured, e.g., with a Pitot tube as in Eq. 6.12), is indicated by an application
of the Bernoulli equation at a point
pt
ρ 2 = pu
+
ρ
2 where pt is the total or stagnation pressure, p = p2 is the static pressure, and u is the local velocity, given by
u
y
=
U2
δ2 y ≤ δ2 u = U2 δ2 < y ≤ h
2 (Flow and pressure distibutions are symmetric about centerline)
Hence 1
2
pt = p2 + ⋅ ρ⋅ u
2 The plot of stagnation pressure is shown in the associated Excel workbook Problem 9.52 (In Excel) [3] Given: Data on flow in a channel
Find: Static pressures; plot of stagnation pressure
Solution:
Given data: The relevant equations are: h=
U2 =
δ2 = 30
22.5
10 mm
m/s
mm
ρ=
p2 = 1.23
311 kg/m3
Pa Stagnation Pressure Distibution in a Duct y (mm) u (m/s) p t (Pa)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0 0.00
2.25
4.50
6.75
9.00
11.25
13.50
15.75
18.00
20.25
22.50
22.50
22.50
22.50
22.50
22.50 311.00
307.89
298.55
282.98
261.19
233.16
198.92
158.44
111.74
58.81
0.34
0.34
0.34
0.34
0.34
0.34 15 10
y (mm)
5 0
400 300 200 100 p t (Pa gage) The stagnation pressure indicates total mechanical energy  the curve indicates significant loss close to the walls
and no loss of energy in the central core. 0 Problem 9.53 Given: Data on flow over a flat plate
Find: Plot of laminar and turbulent boundary layer; Speeds for transition at trailing edge
Solution: Given data:
U=
L= 10
5 m/s
m Tabulated data:
ν = 1.45E05 m2/s
(Table A.10) [3] Computed results:
x (m) Re x 0.00
0.125
0.250
0.375
0.500
0.700
0.75
1.00
1.50
2.00
3.00
4.00 0.00E+00
8.62E+04
1.72E+05
2.59E+05
3.45E+05
4.83E+05
5.17E+05
6.90E+05
1.03E+06
1.38E+06
2.07E+06
2.76E+06 5.00 3.45E+06 (a) Laminar (b) Turbulent (c) Transition
δ (mm)
δ (mm)
δ (mm)
0.00
0.00
0.00
2.33
4.92
2.33
3.30
8.56
3.30
4.04
11.8
4.04
4.67
14.9
4.67
5.52
19.5
5.5
5.71
20.6
20.6
6.60
26.0
26.0
8.08
35.9
35.9
9.3
45.2
45.2
11.4
62.5
62.5
13.2
78.7
78.7
14.8 94.1 94.1 Boundary Layer Profiles on a Flat Plate
100
75
δ (mm) Laminar
Turbulent
Transitional 50
25
0
0 1 2 3
x (m) The speeds U at which transition occurs at specific points are shown below
x trans
(m)
5
4
3
2
1 U (m/s)
1.45
1.81
2.42
3.63
7.25 4 5 Problem 9.54 [3] Note: Figure data applies to problem 9.18 only Given: Data on fluid and turbulent boundary layer Find: Mass flow rate across ab; Momentum flux across bc; Distance at which turbulence occurs Solution:
Basic equations: Mass Momentum
Assumptions: 1) Steady flow 2) No pressure force 3) No body force in x direction 4) Uniform flow at ab
The given or available data (Table A.9) is
U = 165⋅ ft
s Consider CV abcd δ = 0.75⋅ in b = 10⋅ ft ρ = 0.00234⋅ slug
ft 3 slug
mad = −0.241
s mad = −ρ⋅ U⋅ b⋅ δ
δ Mass ⌠
mad + ⎮ ρ⋅ u⋅ b dy + mab = 0
⌡0 and in the boundary layer − 4 ft ν = 1.62 × 10 ⋅ 2 s (Note: Software cannot render a dot) u
y⎞
=⎛ ⎟
⎜
U ⎝ δ⎠ 1
7 =η 1
7 dy = dη⋅ δ 1 Hence ⌠
1
⎮
7
7
m.ab = ρ⋅ U⋅ b⋅ δ − ⎮ ρ⋅ U⋅ η ⋅ δ dη = ρ⋅ U⋅ b⋅ δ − ⋅ ρ⋅ U⋅ b⋅ δ
⌡0
8 1
mab = ⋅ ρ⋅ U⋅ b⋅ δ
8 slug
mab = 0.0302
s 1 δ The momentum flux
across bc is ⌠
mfbc = ⎮
⌡0
mfbc = 7
9 ⌠
2
⎮
→ → ⌠δ
7
2
2
7
u⋅ ρ⋅ V dA = ⎮ u⋅ ρ⋅ u⋅ b dy = ⎮ ρ⋅ U ⋅ b⋅ δ⋅ η dη = ρ⋅ U ⋅ b⋅ δ⋅
⌡0
⌡0
9
2 ⋅ ρ ⋅ U ⋅ b⋅ δ mfbc = 31 slug⋅ ft
2 s From momentum −Rx = U⋅ ( −ρ⋅ U⋅ δ) + mab⋅ uab + mfbc Transition occurs at Rex = 5 × 10 5 and 2 Rx = ρ⋅ U ⋅ b⋅ δ − mab⋅ U − mfbc
Rex = U⋅ x
ν xtrans = Rx = 3.87 lbf
Rex⋅ ν
U xtrans = 0.491 ft Problem 9.55 [3] Problem 9.56 [3] Problem 9.57 Given: Triangular plate Find: [3] Drag Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 cf = Rex
3
2 L = 0.50⋅ cm⋅
From Table A.10 at 20oC 0.0594 L = 0.433⋅ cm −5 m ν = 1.50 × 10 ⋅ 1
5 2 ρ = 1.21⋅ s ReL = U = 25⋅ ReL = 7217 so definitely still laminar, but we are
told to assume turbulent! kg
m First determine the nature of the boundary layer m
s W = 50⋅ cm U⋅ L
ν 3 L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.0594
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
1
2
2 ⌠
FD = ⎮ τw⋅ w( x) dx
⌡0 Rex
L Hence 1
2W⌠
F D = ⋅ ρ⋅ U ⋅ ⋅ ⎮
2
L⎮
⎮
⎮
⎮
⌡0 0.0594⋅ x w( x) = W⋅ 5
9
5W 1⌠
⎮
5⎮ The integral is L 4 0.0594
5
dx =
⋅ ρ⋅ U ⋅ ⋅ ν ⋅
x dx
1
⌡0
2
L ⎛ U⋅ x ⎞
⎜
⎟
⎝ν⎠ 5 L ⌠
4
9
⎮
55
⎮ x 5 dx = ⋅ L
⌡0
9 x
L so ( 4 FD = 0.0165⋅ ρ⋅ W⋅ L ⋅ ν⋅ U ) 9 Note: For twosided solution 1
5 FD = 4.57 × 10 −3 N −3 2⋅ FD = 9.14 × 10 N Problem 9.58 Given: Parabolic plate Find: [3] Drag Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 cf = 0.0594
Rex 1
5 ⎛ W⎞
⎜⎟
2
L= ⎝ ⎠ W = 25⋅ cm 2 L = 6.25⋅ cm 25⋅ cm U = 25⋅ m
s Note: "y" is the equation of the upper and lower surfaces, so y = W/2 at x = L
From Table A.10 at 20oC −5 m ν = 1.50 × 10 ⋅ 2 ρ = 1.21⋅ s kg
m ReL = First determine the nature of the boundary layer 3 U⋅ L
ν 5 ReL = 1.04 × 10 so still laminar, but we are
told to assume turbulent! L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.0594
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
1
2
2 ⌠
FD = ⎮ τw⋅ w ( x) dx
⌡0 Rex w ( x) = W⋅ x
L 5 L Hence L
⌠
⎮
x
9
1 1⌠
3
⎮
−
⎮ 0.0594⋅ L
0.0594
1
2
5
25
10
dx =
FD = ⋅ ρ⋅ U ⋅ W⋅ ⎮
⋅ ρ⋅ U ⋅ W⋅ L ⋅ ν ⋅ ⎮ x dx
1
⌡0
2
2
⎮
⎮
5
⎛ U⋅ x ⎞
⎮
⎜
⎟
⎮
ν⎠
⌡0 ⎝ ( 4 FD = 0.0228⋅ ρ⋅ W⋅ ν⋅ L ⋅ U ) 9 1
5 FD = 0.0267 N
Note: For twosided solution 2⋅ FD = 0.0534 N Problem 9.59 Given: Pattern of flat plates Find: [3] Drag on separate and composite plates Solution:
Basic equations: τw
cf =
1
2
⋅ ρ⋅ U
2 cf = 0.0594
Rex For separate plates L = 7.5⋅ cm From Table A.8 at 20oC ν = 1.01 × 10 1
5 W = 7.5⋅ cm
2
−6 m ⋅ ρ = 998⋅ s U = 10⋅ m
s kg
3 m
ReL = First determine the nature of the boundary layer U⋅ L
ν ReL = 7.43 × 10 5 so turbulent L The drag (one side) is ⌠
F D = ⎮ τ w dA
⎮
⌡ We also have 1
1
2
2 0.0594
τw = cf ⋅ ⋅ ρ⋅ U = ⋅ ρ⋅ U ⋅
1
2
2 ⌠
F D = ⎮ τ w ⋅ W dx
⌡0 Rex 5 Hence 1⌠
⎮
5⎮ 9
5 L 1 −
⌠ 0.0594
0.0594
1
2
5
dx =
dx
FD = ⋅ ρ⋅ U ⋅ W⋅ ⎮
⋅ ρ⋅ U ⋅ W⋅ ν ⋅
x
1
⌡0
2
⎮
2
⎮
5
⎮ ⎛ U⋅ x ⎞
⎮ ⎜ν⎟
⎠
⌡0 ⎝
L The integral is L ⌠
1
4
⎮−
55
⎮ x 5 dx = ⋅ L
⌡0
4 This is the drag on one plate. The total drag is then so ( 4 FD = 0.371⋅ ρ⋅ W⋅ ν⋅ L ⋅ U ) 9 1
5 FTotal = 4⋅ FD FD = 13.9 N FTotal = 55.8 N
For both sides: 2⋅ FTotal = 112 N For the composite plate L = 4 × 7.5⋅ cm L = 0.30 m ( 4 FComposite = 0.371⋅ ρ⋅ W⋅ ν⋅ L ⋅ U ) 9 1
5 FComposite = 42.3 N
For both sides: 2⋅ FComposite = 84.6 N The drag is much lower on the composite compared to the separate plates. This is because τw is largest near the
leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times! Problem 9.60 [3] Problem 9.61 [3] Problem 9.62 [3] Problem 9.63 [3] Problem 9.64 [3] Problem 9.65 [3] Problem 9.66 [3] Problem 9.67 [3] Problem 9.68 [3] Given: Data on flow in a duct Find: Velocity at location 2; pressure drop; length of duct; position at which boundary layer is 20 mm Solution:
The given data is D = 150⋅ mm Table A.10 ρ = 1.23⋅ δ1 = 10⋅ mm δ2 = 30⋅ mm U1 = 25⋅ m
s 2
−5 m kg ν = 1.45 × 10 3 ⋅ m s Governing equations
Mass In the boundary layer δ
0.382
=
1
x
Rex (9.26) 5 In the the inviscid core, the Bernoulli equation holds
2 pV
+
+ g⋅ z = constant
ρ
2 (4.24) Assumptions: (1) Steady flow (2) No body force (gravity) in x direction
For a 1/7power law profile, from Example 9.4 the displacement thickness is
Hence δdisp1 =
δdisp2 = δ1 δdisp = δ
8 δdisp1 = 1.25 mm 8
δ2 δdisp2 = 3.75 mm 8 From the definition of the displacement thickness, to compute the flow rate, the uniform flow at locations 1 and 2 is
assumed to take place in the entire duct, minus the displacement thicknesses
A1 = π
4 ( ⋅ D − 2⋅ δdisp1 ) 2 2 A1 = 0.0171 m A2 = ( ) π
2
⋅ D − 2⋅ δdisp2
4 2 A2 = 0.0159 m Mass conservation (Eq. 4.12) leads to U2 (−ρ⋅ U1⋅ A1) + (ρ⋅ U2⋅ A2) = 0 or A1
U2 = U1⋅
A2 U2 = 26.8 m
s The Bernoulli equation applied between locations 1 and 2 is
p1
ρ
or the pressure drop is 2 U1 + 2 2 p2 U2
=
+
2
ρ p1 − p2 = Δp =
ρ⎛ 2
2
⋅ U − U1 ⎞
⎠
2⎝ 2 Δp = 56.9 Pa (Depends on ρ value selected) The static pressure falls continuously in the entrance region as the fluid in the central core accelerates into a decreasing core.
If we assume the stagnation pressure is atmospheric, a change in pressure of about 60 Pa is not significant; in addition, the velocity changes
about 5%, again not a large change to within engineering accuracy
To compute distances corresponding to boundary layer thicknesses, rearrange Eq.9.26 δ
0.382
ν⎞
=
= 0.382⋅ ⎛
⎜
⎟
1
x
⎝ U⋅ x ⎠
Rex 1
5 so 5
4 x=⎛
⎜ δ ⎞ ⎛U⎞
⎟ ⋅⎜ ⎟
⎝ 0.382 ⎠ ⎝ ν ⎠ 5 Applying this equation to locations 1 and 2 (using U = U1 or U2 as approximations)
5
4 ⎛ δ1 ⎞ ⎛ U 1 ⎞
⎟ ⋅⎜ ⎟
⎝ 0.382 ⎠ ⎝ ν ⎠ 1
4 x1 = ⎜ 5
4 ⎛ δ2 ⎞ ⎛ U 2 ⎞
⎟ ⋅⎜ ⎟
⎝ 0.382 ⎠ ⎝ ν ⎠ x1 = 0.382 m
1
4 x2 = ⎜ x2 = 1.533 m x2 − x1 = 1.15 m
For location 3 (Depends on ν value selected) δ3 = 20⋅ mm δdisp3 = A3 = π
4 ( ⋅ D − 2⋅ δdisp3 ) 2 A1
U3 = U1⋅
A3 δ3
8 δdisp3 = 2.5 mm
2 A3 = 0.017 m U3 = 25.9
5
4 ⎛ δ3 ⎞ ⎛ U 2 ⎞
⎟ ⋅⎜ ⎟
⎝ 0.382 ⎠ ⎝ ν ⎠ x3 = ⎜ x3 − x1 = 0.542 m m
s 1
4 x3 = 0.923 m
(Depends on ν value selected) 1
4 Problem 9.69 [3] Given: Data on a large tanker Find: Cost effectiveness of tanker; compare to Alaska pipeline Solution:
The given data is L = 360⋅ m B = 70⋅ m D = 25⋅ m kg ρ = 1020⋅ m
s U = 6.69⋅ 3 m
P = 9.7⋅ MW 4 P = 1.30 × 10 hp
P
70⋅ % The power to the propeller is Pprop = The shaft power is Ps = 120%⋅ Pprop The efficiency of the engines is Q= The journey time is t= The total energy consumed is Qtotal = Q⋅ t (Power consumed by drag)
4 η = 40⋅ % Hence the heat supplied to the engines is x = 2000⋅ mi Ps Pprop = 1.86 × 10 hp
4 Ps = 2.23 × 10 hp Q = 1.42 × 10 η
x
U 8 BTU hr t = 134 hr
10 Qtotal = 1.9 × 10 BTU From buoyancy the total ship weight equals the displaced seawater volume
Mship⋅ g = ρ⋅ g⋅ L⋅ B⋅ D 9 Mship = ρ⋅ L⋅ B⋅ D Mship = 1.42 × 10 lb Hence the mass of oil is Moil = 75%⋅ Mship Moil = 1.06 × 10 lb The chemical energy stored in the petroleum is q = 20000⋅ The total chemical energy is E = q⋅ Moil BTU
lb The equivalent percentage of petroleum cargo used is then The Alaska pipeline uses BTU
but for the ship
epipeline = 120⋅
ton⋅ mi The ship uses only about 15% of the energy of the pipeline! 9 13 E = 2.13 × 10
Qtotal
E BTU = 0.089 % Qtotal
eship =
Moil⋅ x BTU
eship = 17.8
ton⋅ mi Problem 9.70 Given: Linear, sinusoidal and parabolic velocity profiles Find: [3] Momentum fluxes Solution:
δ The momentum flux is given by ⌠
2
mf = ⎮ ρ⋅ u ⋅ w dy
⌡0 where w is the width of the boundary layer
For a linear velocity profile u
y
=
=η
U
δ (1) For a sinusoidal velocity profile u
π y⎞
π⎞
= sin⎛ ⋅ ⎟ = sin⎛ ⋅ η⎟
⎜
⎜
U
⎝ 2 δ⎠
⎝2 ⎠ (2) For a parabolic velocity profile u
y
y
2
= 2⋅ ⎛ ⎞ − ⎛ ⎞ = 2⋅ η − ( η)
⎜⎟ ⎜⎟
U
δ⎠ ⎝δ⎠
⎝ For each of these u = U⋅ f ( η) Using these in the momentum flux equation ⌠
2
2
mf = ρ⋅ U ⋅ δ⋅ w⋅ ⎮ f ( η) dη
⌡0 For the linear profile Eqs. 1 and 4 give ⌠2
mf = ρ⋅ U ⋅ δ⋅ w⋅ ⎮ η dη
⌡0 For the sinusoidal profile Eqs. 2 and 4 give ⌠
2
⎮
⎛ π ⋅ η⎞ dη
mf = ρ⋅ U ⋅ δ⋅ w⋅ ⎮ sin ⎜
⎟
⎝2 ⎠
⌡0 For the parabolic profile Eqs. 3 and 4 give ⌠
2
⎮
2
mf = ρ⋅ U ⋅ δ⋅ w⋅ ⎮ ⎡2⋅ η − ( η) ⎤ dη
⎣
⎦
⌡0 2 (3) y = δ⋅ η
1 2 (4) 1 mf = 1
2
⋅ ρ⋅ U ⋅ δ ⋅ w
3 mf = 1
2
⋅ ρ⋅ U ⋅ δ ⋅ w
2 1 2 1 2 The linear profile has the smallest momentum, so would be most likely to separate mf = 8
15 2 ⋅ ρ⋅ U ⋅ δ ⋅ w Problem *9.71 Given: Laminar (Blasius) and turbulent (1/7  power) velocity distributions
Find: Plot of distributions; momentum fluxes Solution: [4] Computed results: (Table 9.1) (Simpsons Rule)
η
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0 Laminar Weight Weight x
u/U
w
(u/U )2
0.000
1
0.00
0.166
4
0.11
0.330
2
0.22
0.487
4
0.95
0.630
2
0.79
0.751
4
2.26
0.846
2
1.43
0.913
4
3.33
0.956
2
1.83
0.980
4
3.84
0.992
1
0.98
Simpsons': 0.525 y /δ = η
0.0
0.0125
0.025
0.050
0.10
0.15
0.2
0.4
0.6
0.8
1.0 Turbulent
u/U
0.00
0.53
0.59
0.65
0.72
0.76
0.79
0.88
0.93
0.97
1.00 Laminar and Turbulent Boundary Layer
Velocity Profiles
1.0
0.8
y /δ Laminar 0.5 Turbulent
0.3
0.0
0 0.25 0.5 0.75
u/U 1 Problem 9.72 [2] Problem 9.73 [3] Part 1/2 Problem 9.73 [3] Part 2/2 Problem 9.74 [4] Problem 9.75 [5] Given: Channel flow with laminar boundary layers Find: Maximum inlet speed for laminar exit; Pressure drop for parabolic velocity in boundary layers Solution:
Basic equations: 2 δ
5.48
=
x
Rex 5 Retrans = 5 × 10 pV
+
+ g⋅ z = const
ρ
2 Assumptions: 1) Steady flow 2) Incompressible 3) z = constant
From Table A.10 at 20oC 2
−5 m ν = 1.50 × 10 ⋅ ρ = 1.21⋅ s Umax⋅ L Then Retrans = For Retrans = 5 × 10 For a parabolic profile ν
5 δdisp
δ δ2 = L⋅ 1 L = 3⋅ m 3 m
Retrans⋅ ν
L
5.48
Retrans Umax = 2.50 ( 1
⋅δ
32 h = 15⋅ cm
m
s U1 = Umax U1 = 2.50 ) where δtrans is the displacement thickness δdisp2 = 0.00775 m ( U1⋅ w⋅ h = U2⋅ w⋅ h − 2⋅ δdisp2 ) h
U2 = U1⋅
h − 2⋅ δdisp2 U2 = 2.79 m
s Since the boundary layers do not meet Bernoulli applies in the core
p1
ρ 2 + Δp =
From hydrostatics U1
2 = p2
ρ 2 + U2
2 ρ⎛ 2
2
⋅ U − U1 ⎞
⎠
2⎝ 2 Δp = ρH2O⋅ g⋅ Δh with ρ
2
2
Δp = p1 − p2 = ⋅ ⎛ U2 − U1 ⎞
⎝
⎠
2
Δp = 0.922 Pa
ρH2O = 1000⋅ kg
3 m
Δh = Δp
ρH2O⋅ g Δh = 0.0940 mm m
s δ2 = 0.0232 m 1
⌠⎛
⎮ 1 − u ⎞ dλ = ⌠ 1 − 2⋅ λ + λ2 dλ = 1
⎮
=
⎟
⎮⎜
⌡0
3
U⎠
⌡0 ⎝ δdisp2 =
From continuity Umax = kg Δh = 0.00370 in Problem 9.76 [5] Part 1/2 Problem 9.76 [5] Part 2/2 Problem 9.77 [5] Part 1/2 Problem 9.77 [5] Part 2/2 Problem 9.78 [2] Problem 9.79 Given: Pattern of flat plates Find: [3] Drag on separate and composite plates Solution:
FD Basic equations: CD = For separate plates L = 7.5⋅ cm From Table A.8 at 20oC ν = 1.01 × 10 1
2
⋅ ρ⋅ V ⋅ A
2
W = 7.5⋅ cm
−6 m ⋅ 2 s 0.0742
ReL 1
5 A = 5.625 × 10 −3 2 kg
m ReL = First determine the Reynolds number
CD = ρ = 998⋅ A = W⋅ L 3 V⋅ L
ν 5 ReL = 7.43 × 10 so use Eq. 9.34 CD = 0.00497 1
2
The drag (one side) is then FD = CD⋅ ⋅ ρ⋅ V ⋅ A
2 FD = 1.39 N This is the drag on one plate. The total drag is then FTotal = 4⋅ FD FTotal = 5.58 N
For both sides: For the composite plate L = 4 × 7.5⋅ cm ReL = First determine the Reylolds number
CD = 0.0742
ReL L = 0.300 m 1
5 1
2
The drag (one side) is then FD = CD⋅ ⋅ ρ⋅ V ⋅ A
2 V⋅ L
ν 2⋅ FTotal = 11.2 N A = W⋅ L A = 0.0225 m ReL = 2.97 × 10 2 6 so use Eq. 9.34 CD = 0.00377 FD = 4.23 N For both sides: 2⋅ FD = 8.46 N The drag is much lower on the composite compared to the separate plates. This is because τw is largest near the
leading edges and falls off rapidly; in this problem the separate plates experience leading edges four times! m V = 10⋅ m
s Problem 9.80 [3] Problem 9.81 Given: Aircraft cruising at 40,000 ft Find: [3] Skin friction drag force; Power required Solution:
Basic equations: FD CD = 1
2
⋅ ρ⋅ V ⋅ A
2
We "unwrap" the cylinder to obtain an equivalent flat plate
L = 125⋅ ft
From Table A.3, with
For D = 12⋅ ft z = 40000⋅ ft 2 z = 12192 m
ρ z = 12000⋅ m ρSL A = L ⋅ π⋅ D with = 0.2546 A = 4712⋅ ft V = 500⋅ mph ρSL = 0.002377⋅ slug
ft z = 13000⋅ m ρ
= 0.2176
ρSL z = 12192 m ρ
( 0.2176 − 0.2546)
= 0.2546 +
⋅ ( 12192 − 12000) = 0.255
ρSL
( 1300 − 12000) ρ = 0.255⋅ ρSL Hence at ρ = 0.000606⋅ slug
ft From Appendix A3 Hence 3 μ= b⋅ T 1
2 with S
1+
T b⋅ T μ= 1+ −6 kg ⋅ m⋅ s⋅ K 1
2 μ = 1.42 × 10 S S = 110.4⋅ K 1
2 − 5 N⋅ s
⋅
2 μ = 2.97 × 10 m − 7 lbf ⋅ s
⋅
2 ft T
ReL = Next we need the Reynolds number
CD = b = 1.458 × 10 T = 216.7⋅ K and also 3 0.455 ρ⋅ V ⋅ L ReL = 1.87 × 10 μ 2.58 1
2
F D = C D ⋅ ⋅ ρ⋅ V ⋅ A
2 FD = 1500⋅ lbf The power consumed is P = FD ⋅ V P = 1.100 × 10 ⋅ so use Eq. 9.35 CD = 0.00195 The drag is then 8 ( log ReL ) 6 ft⋅ lbf s P = 1999⋅ hp Problem 9.82 [3] Given data:
L=
B=
D= 80
35
5 ν= 1.21E05 ft
ft
ft
ft2/s (Table A.7) ρ= 1.94 slug/ft A= 3600 ft2 3 (Table A.7) Computed results: U (mph) Re L CD P (hp) 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15 9.70E+06
1.94E+07
2.91E+07
3.88E+07
4.85E+07
5.82E+07
6.79E+07
7.76E+07
8.73E+07
9.70E+07
1.07E+08
1.16E+08
1.26E+08
1.36E+08
1.45E+08 0.00285
0.00262
0.00249
0.00240
0.00233
0.00227
0.00222
0.00219
0.00215
0.00212
0.00209
0.00207
0.00205
0.00203
0.00201 0.0571
0.421
1.35
3.1
5.8
9.8
15
22
31
42
56
72
90
111
136 Power Consumed by Friction on a Barge
150
120
P (hp) 90
60
30
0
0 3 6 9
U (mph) 12 15 Problem 9.84 [3] Problem 9.85 [4] Problem 9.86 Given: Plastic sheet falling in water Find: [3] Terminal speed both ways Solution:
Basic equations: ΣFy = 0 for terminal speed h = 10⋅ mm
From Table A.8 at 20oC
Hence Also Solving for V 2
−6 m ν = 1.01 × 10 ⋅ ρ = 998⋅ s 0.0742 CD = ReL
L = 0.5⋅ m kg (9.34) (assuming 5 x 105 < ReL < 107) 1
5 A = W⋅ L SG = 1.5 for water 3 m FD + Fbuoyancy − W = 0 FD = W − Fbuoyancy = ρ⋅ g⋅ h⋅ A⋅ ( SG − 1)
4
5 1
51 1
0.0742
1
0.0742
2
2
2
FD = 2⋅ CD⋅ A⋅ ⋅ ρ⋅ V = 2⋅
⋅ A ⋅ ⋅ ρ⋅ V =
⋅ W⋅ L⋅ ρ⋅ V = 0.0742⋅ W⋅ L ⋅ ν ⋅ ⋅ ρ⋅ V
1
1
2
2
2
5 ⎛ V⋅ L ⎞
⎜
⎟
⎝ν⎠
4
5 1
51 ρH2O⋅ g⋅ h⋅ W⋅ L⋅ ( SG − 1) = 0.0742⋅ W⋅ L ⋅ ν ⋅ ⋅ ρ⋅ V
2
1⎤
⎡
⎢
5⎥
g⋅ h⋅ ( SG − 1) ⎛ L ⎞
V=⎢
⋅⎜ ⎟ ⎥
⎣ 0.0742 ⎝ ν ⎠ ⎦ Check the Reynolds numberReL = Repeating for 1
2
⋅ ρ⋅ V ⋅ A
2 W = 1⋅ m ReL Hence FD CD = Check the Reynolds numberReL = ν
1⎤
⎡
⎢
5⎥
g⋅ h⋅ ( SG − 1) ⎛ L ⎞
V=⎢
⋅⎜ ⎟ ⎥
⎣ 0.0742 ⎝ ν ⎠ ⎦ V⋅ L
ν 5 Note that we double FD because
we have two sides! 9
5 5
9 V = 3.41 V⋅ L L = 1⋅ m 9
5 5
9 m
s ReL = 1.69 × 10 V = 3.68 6 Hence Eq. 9.34 is reasonable m
s ReL = 3.65 × 10 6 Eq. 9.34 is still reasonable The short side vertical orientation falls more slowly because the largest friction is at the region of the leading edge (τ tails off as the
boundary layer progresses); its leading edge area is larger. Note that neither orientation is likely  the plate will flip around in a chaotic
manner Problem 9.87 [4] Problem 9.88 [4] Problem 9.89 Given: "Resistance" data on a ship
Find: Plot of wave, viscous and total drag (protoype and model); Power required by prototype
Solution: [4] Given data:
Lp = 409 Ap =
Lm = 19500
5.11 Am =
SG =
μ=
ρ= 3.05
1.025
2.26E05
1023 ft
ft2
ft (1/80 scale)
ft2
(Table A.2)
(Table A.2) lbf.s/ft2
slug/ft3 Computed results:
Model
Fr
0.10
0.20
0.30
0.35
0.40
0.45
0.50
0.60 Wave
"Resistance"
0.00050
0.00075
0.00120
0.00150
0.00200
0.00300
0.00350
0.00320 Viscous
Total
Wave
U (ft/s)
"Resistance" "Resistance"
Drag (lbf)
0.0052
0.0057
1.28
0.641
0.0045
0.0053
2.57
3.85
0.0040
0.0052
3.85
13.9
0.0038
0.0053
4.49
23.6
0.0038
0.0058
5.13
41.0
0.0036
0.0066
5.77
77.9
0.0035
0.0070
6.42
112
0.0035
0.0067
7.70
148 Viscous
Drag (lbf)
6.67
23.1
46.2
59.7
78.0
93.5
112
162 Total
Drag (lbf)
7.31
26.9
60.0
83.3
119
171
224
309 Drag on a Model Ship
350
300 Total
Wave
Viscous 250
F (lbf) 200
150
100
50
0
0 1 2 3 4 5 6
U (ft/s) 7 8 9 Prototype
Fr Wave
"Resistance" 0.10
0.20
0.30
0.35
0.40
0.45
0.50
0.60 0.00050
0.00075
0.00120
0.00150
0.00200
0.00300
0.00350
0.00320 Wave
Viscous
Total
Viscous
Total
Drag (lbf Drag (lbf x
Drag
U (ft/s)
"Resistance" "Resistance"
6
6
x 10 )
(lbf x 106)
10 )
0.0017
0.0022
11.5
0.328
1.12
1.44
0.0016
0.0024
23.0
1.97
4.20
6.17
0.0015
0.0027
34.4
7.09
8.87
16.0
0.0015
0.0030
40.2
12.1
12.1
24.1
0.0013
0.0033
45.9
21.0
13.7
34.7
0.0013
0.0043
51.6
39.9
17.3
57.2
0.0013
0.0048
57.4
57.5
21.3
78.8
0.0013
0.0045
68.9
75.7
30.7
106 Drag on a Prototype Ship
120
100 Total
Wave 80
F (lbf x 10 )
60
6 Viscous 40
20
0
0 10 20 30 40 50
U (ft/s) For the prototype wave resistance is a much more significant factor at high speeds! 60 70 80 Problem 9.90 [1] Problem 9.91 Given: Fishing net Find: [3] Drag; Power to maintain motion Solution:
Basic equations: FD CD = 1
2
⋅ ρ⋅ V ⋅ A
2
We convert the net into an equivalent cylinder (we assume each segment does not interfere with its neighbors)
d= 1
⋅ in
32 Total number of threads of length L is n1 = Total number of threads of length W is n2 = L = 40⋅ ft Total length of thread W = 5⋅ ft LT = L1 + L2 The frontal area is then A = LT⋅ d
From Table A.7 ρ = 1.94⋅ V = 7⋅ knot V = 11.8 W
D n1 = 160 Total length L1 = n1⋅ L L1 = 6400 ft L
D n2 = 1280 Total length L2 = n2⋅ W L2 = 6400 ft LT = 12800 ft LT = 2.42 mile A lot! A = 33.3 ft slug
ft The Reynolds number is Red = 3
⋅ in
8 Spacing: D = 3 V⋅ d
ν 2 Note that L⋅ W = 200 ft
− 5 ft ν = 1.21 × 10 ⋅ 2 s Red = 2543 For a cylinder in a crossflow at this Reynolds number, from Fig. 9.13, approximately
Hence 1
2
F D = C D ⋅ ⋅ ρ⋅ V ⋅ A
2 FD = 3611 lbf The power required is P = FD ⋅ V P = 42658 CD = 0.8 ft⋅ lbf
s P = 77.6 hp 2 ft
s Problem 9.92 [2] Problem 9.93 Given: Data on a rotary mixer Find: [3] New design dimensions Solution:
The given data or available data is
R = 0.6⋅ m P = 350⋅ W ω = 60⋅ rpm ρ = 1099⋅ kg
3 m
For a ring, from Table 9.3 CD = 1.2 The torque at the specified power and speed is
T= P
ω T = 55.7 N ⋅ m 1T
⋅
2R The drag on each ring is then FD = The linear velocity of each ring is V = R⋅ ω FD = 46.4 N
V = 3.77 m
s The drag and velocity of each ring are related using the definition of drag coefficient
CD = FD
1
2
⋅ ρ⋅ A ⋅ V
2
FD Solving for the ring area A= But A= The outer diameter is do = 125⋅ mm Hence the inner diameter is di = 1
2
⋅ ρ⋅ V ⋅ C D
2 A = 4.95 × 10 π⎛ 2
2
⋅ d − di ⎞
⎠
4⎝o 2 4⋅ A
do −
π di = 96.5 mm −3 2 m Problem 9.94 [2] Problem 9.95 Given: Data on airplane and parachute
Find: Time and distance to slow down; plot speed against distance and time; maximum "g"'s
Solution: Given data:
M=
Vi =
Vf =
CD = 8500
400
100
1.42 kg
km/hr
km/hr
(Table 9.3) ρ = 1.23 kg/m
Single: D = 6 m 3 Triple: D = 3.75 m [3] Computed results:
A = 28.3 m2 A = 11.0 m2 t (s) x (m) V (km/hr) t (s) x (m) V (km/hr) 0.0 0.0
1.0 96.3
2.0 171
3.0 233
4.0 285
5.0 331
6.0 371
7.0 407
8.0 439
9.0 469
9.29 477 0.0 0.0
1.0 94.2
2.0 165
3.0 223
4.0 271
5.0 312
6.0 348
7.0 380
7.93 407
9.0 436
9.3 443 400
302
243
203
175
153
136
123
112
102
100 400
290
228
187
159
138
122
110
100
91
89 "g "'s = 3.66 Max Aircraft Velocity versus Time V (km/hr) 400
350
300
250
200
150
100
50
0 One Parachute
Three Parachutes 0 2 4 6
t (s) 8 10 Aircraft Velocity versus Distance
400
350 One Parachute
Three Parachutes 300
V (km/hr) 250
200
150
100
50
0
0 100 200 300
x (m) 400 500 Problem 9.96 [3] Given: Data on airplane landing
Find: Single and threeparachute sizes; plot speed against distance and time; maximum "g''s
Solution: Given data:
M = 9500 kg
V i = 350 km/hr
V f = 100 km/hr
x f = 1200 m
C D = 1.42 (Table 9.3)
ρ = 1.23 kg/m 3 Computed results:
Single:
A = 11.4 m2
D = 3.80 m Triple:
A = 3.8 m2
D = 2.20 m "g "'s = 1.01 Max
t (s) x (m) V (km/hr)
0.00
2.50
5.00
7.50
10.0
12.5
15.0
17.5
20.0
22.5
24.6 0.0
216.6
393.2
542.2
671.1
784.7
886.3
978.1
1061.9
1138.9
1200.0 350
279
232
199
174
154
139
126
116
107
100
Aircraft Velocity versus Time 350
300
250
V (km/hr) 200
150
100
50
0
0 5 10 15
t (s) 20 25 Aircraft Velocity versus Distance
350
300
250
V (km/hr) 200
150
100
50
0
0 200 400 600 800
x (m) 1000 1200 Problem 9.97 [2] Problem 9.98 [2] Problem 9.99 [2] Problem 9.100 Given: Data on cyclist performance on a calm day Find: [2] Performance hindered and aided by wind Solution:
The given data or available data is
FR = 7.5⋅ N
CD = 1.2 ρ = 1.23⋅ 2 M = 65⋅ kg A = 0.25⋅ m kg
m The governing equation is FD = V = 30⋅ 3 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 km
hr FD = 12.8 N The power steady power generated by the cyclist is ( ) P = FD + FR ⋅ V
Now, with a headwind we have Vw = 10⋅ P = 169 W km
hr V = 24⋅ km
hr The aerodynamic drag is greater because of the greater effective wind speed
FD = ( ) 1
2
⋅ ρ⋅ A ⋅ V + V w ⋅ C D
2 FD = 16.5 N The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed ( P = V ⋅ FD + FR ) P = 160 W This is less than the power she can generate She wins the bet! With the wind supporting her the effective wind speed is substantially lower
VW = 10⋅
FD = 1
2 km V = 40⋅ hr ( ) 2 ⋅ ρ⋅ A ⋅ V − V W ⋅ C D km
hr FD = 12.8 N The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed ( P = V ⋅ FD + FR
This is more than the power she can generate ) P = 226 W
She loses the bet P = 0.227 hp Problem 9.101 [3] Given: Data on cyclist performance on a calm day Find: Performance hindered and aided by wind; repeat with hightech tires; with fairing Solution:
The given data or available data is
FR = 7.5⋅ N
CD = 1.2 ρ = 1.23⋅ 2 M = 65⋅ kg A = 0.25⋅ m kg
m V = 30⋅ 3 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 The governing equation is FD = Power steady power generated by the cyclist is P = FD + FR ⋅ V Now, with a headwind we have Vw = 10⋅ ( km
hr FD = 12.8 N ) P = 169 W P = 0.227⋅ hp km
hr The aerodynamic drag is greater because of the greater effective wind speed
FD = ( ) 1
2
⋅ ρ⋅ A ⋅ V + V w ⋅ C D
2 (1) The power required is that needed to overcome the total force FD + FR, moving at the cyclist's speed is ( P = V ⋅ FD + FR ) (2) Combining Eqs 1 and 2 we obtain an expression for the cyclist's maximum speed V cycling into a
headwind (where P = 169 W is the cyclist's power)
Cycling into the wind: P = ⎡FR +
⎢ ⎣ 1
2 ⎤
⋅ ρ⋅ A⋅ V + Vw ⋅ CD⎥ ⋅ V ( ) 2 ⎦ (3) This is a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use Excel's Goal Seek (or
Solver). From the associated Excel workbook
V = 24.7⋅ From Solver km
hr By a similar reasoning:
Cycling with the wind: P = ⎡FR +
⎢ ⎣ 1
2 ⎤
⋅ ρ⋅ A⋅ V − Vw ⋅ CD⎥ ⋅ V ( ) 2 ⎦ (4) km
hr From Solver V = 35.8⋅ With improved tires FR = 3.5⋅ N Maximum speed on a calm day is obtained from 1
2
⎛
⎞
P = ⎜ F R + ⋅ ρ⋅ A ⋅ V ⋅ C D ⎟ ⋅ V
2
⎝
⎠ This is a again a cubic equation for V; it can be solved analytically, or by iterating. It is convenient to use Excel's Goal
Seek (or Solver). From the associated Excel workbook
V = 32.6⋅ From Solver km
hr Equations 3 and 4 are repeated for the case of improved tires
From Solver Against the wind V = 26.8⋅ km
hr With the wind V = 39.1⋅ km
hr V = 29.8⋅ km
hr With the wind V = 42.1⋅ km
hr For improved tires and fairing, from Solver
V = 35.7⋅ km
hr Against the wind Problem 9.101 (In Excel) Given: Data on cyclist performance on a calm day
Find: Performance hindered and aided by wind; repeat with hightech tires; with fairing
Solution:
Given data:
FR =
M=
A=
CD = 7.5
65
0.25
1.2 N
kg
m2 ρ=
V=
Vw = 1.23
30
10 kg/m3
km/hr
km/hr Computed results:
F D = 12.8 N
P = 169 W Using Solver : Left (W) Right (W) Error V (km/hr)
169
169
0%
24.7 Left (W) Right (W) Error V (km/hr)
Using Solver : 169
169
0%
35.8 [3] With improved tires:
FR = 3.5 N Left (W) Right (W) Error V (km/hr)
Using Solver : 169
169
0%
32.6 Left (W) Right (W) Error V (km/hr)
Using Solver : 169
169
0%
26.8 Using Solver : Left (W) Right (W) Error V (km/hr)
169
169
0%
39.1 With improved tires and fairing:
FR =
CD = 3.5
0.9 N Left (W) Right (W) Error V (km/hr)
Using Solver : 169
169
0%
35.7 Left (W) Right (W) Error V (km/hr)
Using Solver : 169
169
0%
29.8 Left (W) Right (W) Error V (km/hr)
Using Solver : 169
169
0%
42.1 Problem 9.102 Given: Data on cyclist performance on a calm day Find: [3] Performance on a hill with and without wind Solution:
The given data or available data is
FR = 7.5⋅ N M = 65⋅ kg CD = 1.2 ρ = 1.23⋅ A = 0.25⋅ m kg
m V = 30⋅ 3 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 The governing equation is FD = Power steady power generated by the cyclist is P = FD + FR ⋅ V Riding up the hill (no wind) 2 km
hr θ = 5⋅ deg ( FD = 12.8 N ) P = 169 W P = 0.227 hp For steady speed the cyclist's power is consumed by working against the net force (rolling resistance, darg, and gravity)
Cycling up the hill: P = ⎛ FR +
⎜ ⎝ 1
2
⎞
⋅ ρ⋅ A⋅ V ⋅ CD + M⋅ g⋅ sin( θ) ⎟ ⋅ V
2
⎠ This is a cubic equation for the speed which can be solved analytically, or by iteration, or using Excel's Goal Seek
or Solver. The solution is obtained from the associated Excel workbook
From Solver V = 9.47⋅ km
hr Now, with a headwind we have Vw = 10⋅ km
hr The aerodynamic drag is greater because of the greater effective wind speed
FD = 1
2 ( ) 2 ⋅ ρ⋅ A ⋅ V + V w ⋅ C D The power required is that needed to overcome the total force (rolling resistance, drag, and gravity) moving at the cyclist's speed is
Uphill against the wind: P = ⎡FR +
⎢ 1 ⋅ ρ⋅ A⋅ V + Vw ⋅ CD + M⋅ g⋅ sin ( θ)⎤ ⋅ V
⎥ ⎣ 2 V = 8.94⋅ km This is again a cubic equation for V
From Solver hr ( ) 2 ⎦ Pedalling downhill (no wind) gravity helps increase the speed; the maximum speed is obtained from
Cycling down the hill: ⎛
⎝ 1
2
⋅ ρ⋅ A⋅ V ⋅ CD − M⋅ g⋅ sin ( θ)⎞ ⋅ V
⎟
2
⎠ P = ⎜ FR + This cubic equation for V is solved in the associated Excel workbook
V = 63.6⋅ From Solver km
hr Pedalling downhill (wind assisted) gravity helps increase the speed; the maximum speed is obtained from
Windassisted downhill: 1
2
⎡
P = ⎢FR + ⋅ ρ⋅ A⋅ V − Vw ⋅ CD − M⋅ g⋅ sin ( θ)⎤ ⋅ V
⎥
2
⎣
⎦ ( ) This cubic equation for V is solved in the associated Excel workbook
V = 73.0⋅ From Solver km
hr Freewheeling downhill, the maximum speed is obtained from the fact that the net force is zero
Freewheeling downhill: FR + V= Wind assisted: FR + 1
2
⋅ ρ⋅ A⋅ V ⋅ CD − M⋅ g⋅ sin ( θ) = 0
2
M⋅ g⋅ sin ( θ) − FR
1
⋅ ρ⋅ A ⋅ C D
2 ( V = 58.1 km
hr V = 68.1 km
hr ) 1
2
⋅ ρ⋅ A⋅ V − Vw ⋅ CD − M⋅ g⋅ sin ( θ) = 0
2 V = Vw + M⋅ g⋅ sin ( θ) − FR
1
⋅ ρ⋅ A ⋅ C D
2 Problem 9.102 (In Excel) [3] Given: Data on cyclist performance on a calm day
Find: Performance on a hill with and without wind
Solution:
Given data:
FR =
M=
A=
CD = 7.5
65
0.25
1.2 N
kg
m2 ρ=
V=
Vw =
θ= 1.23
30
10
5 kg/m
km/hr
km/hr
deg 3 Computed results:
1
2
FD = ⋅ρ ⋅A ⋅V ⋅CD
2 ( FD = Cycling up the hill: Using Solver :
Uphill against the wind: Using Solver : Cycling down the hill: Using Solver : Windassisted downhill: Using Solver : ⎛
⎝ P = ⎜ FR + Left (W)
169 1
2 N P= ) P = FD + FR ⋅V 12.8
169 W ⎞
⎠ ⋅ρ ⋅A⋅V ⋅CD + M ⋅g⋅sin(θ ) ⎟ ⋅V
2 Right (W)
169 Error
0% V (km/hr)
9.47 1
⎡
2
⎤
P = ⎢FR + ⋅ρ ⋅A⋅ V + Vw ⋅CD + M ⋅g⋅sin( θ )⎥ ⋅V
2
⎣
⎦ ( Left (W)
169 ) Right (W)
169 Error
0% V (km/hr)
8.94 1
⎛
2
⎞
P = ⎜ FR + ⋅ρ ⋅A ⋅V ⋅CD − M ⋅g⋅sin(θ ) ⎟ ⋅V
2
⎝
⎠ Left (W)
169 Right (W)
169 Error
0% V (km/hr)
63.6 1
⎡
2
⎤
P = ⎢FR + ⋅ρ ⋅A⋅ V − Vw ⋅CD − M ⋅g⋅sin(θ )⎥ ⋅V
2
⎣
⎦ ( Left (W)
169 Right (W)
169 ) Error
0% V (km/hr)
73.0 Problem *9.103 [3] FBnet V FD
y
x θ
T Given: Series of party balloons Find: Wind velocity profile; Plot Wlatex Note: Flagpole is actually 27 ft tall, not 63 ft! Solution:
Basic equations: FD CD = FB = ρair⋅ g⋅ Vol 1
2
⋅ ρ⋅ V ⋅ A
2
The above figure applies to each balloon
For the horizontal forces FD − T⋅ sin ( θ) = 0 (1) For the vertical forces −T⋅ cos ( θ) + FBnet − Wlatex = 0 Here π⋅ D
FBnet = FB − W = ρair − ρHe ⋅ g⋅
6 ( (2) ) 3 1
Mlatex =
⋅ oz
10 Wlatex = Mlatex⋅ g ft⋅ lbf
lbm⋅ R pHe = 16.2⋅ psi THe = 530⋅ R ft⋅ lbf
lbm⋅ R pair = 14.7⋅ psi D = 9⋅ in
We have (Table A.6) RHe = 386.1⋅
Rair = 53.33⋅ ( ) FBnet = ρair − ρHe ⋅ g⋅ π⋅ D
6 FD = T⋅ sin ( θ) = ) FD = FBnet − Wlatex ⋅ tan( θ) But we have 1
1
2
2 π⋅ D
FD = CD⋅ ⋅ ρair⋅ V ⋅ A = CD⋅ ⋅ ρair⋅ V ⋅
4
2
2 CD⋅ ρair⋅ π⋅ D
From Table A.9 − 4 ft ν = 1.63 × 10 ⋅ Rair⋅ Tair slug
ft ρair = 0.00233 ft V = 9.00 2 CD = 0.4 with from Fig. 9.11 (we will
check Re later) ft
s The Reynolds number is Red = V⋅ D
ν Red = 4.14 × 10 4 We are okay! 3 slug FD = 0.0167 lbf 2 s pair ρHe = 0.000354 ⋅ sin ( θ) Hence 2 RHe⋅ THe FBnet = 0.0140 lbf cos ( θ) 8⋅ FD ρair = pHe 3 FBnet − Wlatex ( Tair = 530⋅ R ρHe = Wlatex = 0.00625 lbf θ = 65⋅ deg Applying Eqs 1 and 2 to the top balloon, for which V= →
ΣF = 0 3 For the next balloon ( θ = 60⋅ deg
8⋅ FD V= CD⋅ ρair⋅ π⋅ D
The Reynolds number is Red =
For the next balloon The Reynolds number is Red = V⋅ D
ν The Reynolds number is Red = V⋅ D
ν V⋅ D
ν V⋅ D 8⋅ FD
2 V⋅ D V = 2.58 4 CD⋅ ρair⋅ π⋅ D 2 V = 1.82 FD = 0.00449 lbf with CD = 0.4 FD = 0.00283 lbf with CD = 0.4 FD = 0.00137 lbf with CD = 0.4 with CD = 0.4 ft
s
4 We are okay! ) FD = FBnet − Wlatex ⋅ tan( θ)
8⋅ FD CD = 0.4 We are okay! ) ( θ = 5⋅ deg with ft
s Red = 1.19 × 10 ν FD = 0.00544 lbf We are okay! FD = FBnet − Wlatex ⋅ tan( θ) CD⋅ ρair⋅ π⋅ D V= V = 3.71 ( θ = 10⋅ deg The Reynolds number is Red = 4 ) Red = 1.71 × 10 ν V= For the next balloon 2 CD = 0.4 ft
s ( 8⋅ FD with We are okay! FD = FBnet − Wlatex ⋅ tan( θ) CD⋅ ρair⋅ π⋅ D
The Reynolds number is Red = 4 ) Red = 2.15 × 10 θ = 20⋅ deg
V= For the next balloon 2 V = 4.67 FD = 0.00777 lbf ft
s ( 8⋅ FD CD = 0.4 We are okay! FD = FBnet − Wlatex ⋅ tan( θ) CD⋅ ρair⋅ π⋅ D For the next balloon 4 ) Red = 2.37 × 10 θ = 30⋅ deg The Reynolds number is Red = V = 5.14 with ft
s ( 2 FD = 0.00927 lbf We are okay! FD = FBnet − Wlatex ⋅ tan( θ)
8⋅ FD V= 4 ) Red = 2.83 × 10 θ = 35⋅ deg CD⋅ ρair⋅ π⋅ D For the next balloon 2 V = 6.15 CD = 0.4 ft
s ( V⋅ D
ν V= ) Red = 3.09 × 10 8⋅ FD with We are okay! FD = FBnet − Wlatex ⋅ tan( θ) CD⋅ ρair⋅ π⋅ D For the next balloon 2 θ = 45⋅ deg The Reynolds number is Red = 4 ( V = 6.71 FD = 0.0135 lbf ft
s FD = FBnet − Wlatex ⋅ tan( θ)
8⋅ FD V= V = 8.09 Red = 3.72 × 10 θ = 50⋅ deg CD⋅ ρair⋅ π⋅ D For the next balloon 2 V⋅ D
ν V= ) FD = FBnet − Wlatex ⋅ tan( θ) ft
s FD = 0.000680 lbf The Reynolds number is Red = V⋅ D
ν Red = 8367.80 We are okay! V = ( 1.82 2.58 3.71 4.67 5.14 6.15 6.71 8.09 9.00 ) ⋅ In summary we have ft
s h = ( 3 6 9 12 15 18 21 24 27 ) ⋅ ft 30 h (ft) 20 10 0 2 4 6 V (ft/s) This problem is ideal for computing and plotting in Excel 8 10 Problem 9.104 [2] FB V FD y θ T W x Given: Sphere dragged through river Find: Relative velocity of sphere Solution:
Basic equations: FD CD = →
ΣF = 0 FB = ρ⋅ g⋅ Vol 1
2
⋅ ρ⋅ V ⋅ A
2
The above figure applies to the sphere
For the horizontal forces FD − T⋅ sin ( θ) = 0 (1) For the vertical forces −T⋅ cos ( θ) + FB − W = 0 Here D = 1⋅ ft (2)
− 5 ft SG = 0.25 ⋅ 2 s ρ = 1.94⋅ θ = 45⋅ deg Applying Eqs 1 and 2 to the sphere, for which
FD = T⋅ sin ( θ) = and from Table A.7 ν = 1.41 × 10 FB − W
cos ( θ) slug
ft 3 ⋅ sin ( θ) = ρ⋅ g⋅ Vol⋅ ( 1 − SG) ⋅ tan( θ) 3 π⋅ D
⋅ ( 1 − SG) ⋅ tan( θ)
6 Hence F D = ρ ⋅ g⋅ But we have 1
1
2
2 π⋅ D
F D = C D ⋅ ⋅ ρ⋅ V ⋅ A = C D ⋅ ⋅ ρ⋅ V ⋅
4
2
2
8⋅ FD V= C D ⋅ ρ⋅ π ⋅ D
The Reynolds number is Red =
Try V⋅ D
ν CD = 0.15 The Reynolds number is Red = V⋅ D
ν 2 V = 8.97⋅ FD = 24.5⋅ lbf
with CD = 0.4 from Fig. 9.11 (we will
check Re later) ft
s Red = 6.36 × 10
V= 2 5 A bit off from Fig 9.11 8⋅ FD
C D ⋅ ρ⋅ π ⋅ D Red = 1.04 × 10 6 2 V = 14.65⋅ ft
s A good fit with Fig 9.11 (extreme right of graph) Problem 9.105 [2] Fn
W Given: Circular disk in wind Find: Mass of disk; Plot α versus V Solution:
CD = Basic equations: →
ΣM = 0 FD
1
2
⋅ ρ⋅ V ⋅ A
2 Summing moments at the pivot W⋅ L⋅ sin ( α) − Fn⋅ L = 0 Fn = and 1
2
⋅ ρ⋅ V n ⋅ A ⋅ C D
2 2 Hence
The data is ρ = 1.225⋅ 1
2 π⋅ D
⋅ ρ⋅ ( V⋅ cos ( α) ) ⋅
⋅ CD
4
2 M⋅ g⋅ sin ( α) =
kg V = 15⋅ 3 m 2 2 m
s D = 25⋅ mm 8⋅ M⋅ g V= ⋅ 2 π ⋅ ρ⋅ D ⋅ C D CD = 1.17 2 π⋅ ρ⋅ V ⋅ cos ( α) ⋅ D ⋅ CD
M=
8⋅ g⋅ sin ( α)
Rearranging α = 10⋅ deg M = 0.0451 kg tan( α)
cos ( α) V = 35.5⋅ m tan( α)
⋅
s cos ( α) We can plot this by choosing α and computing V
80 V (m/s) 60 40 20 0 10 20 30 40 Angle (deg) This graph can be easily plotted in Excel 50 60 70 (Table 9.3) Problem 9.106 Given: Data on dimensions of anemometer Find: [3] Calibration constant; compare to actual with friction Solution:
The given data or available data is D = 50⋅ mm R = 80⋅ mm ρ = 1.23⋅ kg
3 m The drag coefficients for a cup with open end facing the airflow and a cup with open end facing downstream are, respectively, from Table 9
CDopen = 1.42 CDnotopen = 0.38 1
2
The equation for computing drag is FD = ⋅ ρ⋅ A⋅ V ⋅ CD
2
A= where π⋅ D
4 (1) 2 A = 1.96 × 10 −3 2 m Assuming steady speed ω at steady wind speed V the sum of moments will be zero. The two cups that are momentarily parallel to the flow
will exert no moment; the two cups with open end facing and not facing the flow will exert a moment beacuse of their drag forces. For eac
the drag is based on Eq. 1 (with the relative velocity used!). In addition, friction of the anemometer is neglected
1
1
2
2
⎤
ΣM = 0 = ⎡ ⋅ ρ⋅ A⋅ ( V − R⋅ ω) ⋅ CDopen⎥ ⋅ R − ⎡ ⋅ ρ⋅ A⋅ ( V + R⋅ ω) ⋅ CDnotopen⎤ ⋅ R
⎢
⎢
⎥
2
2
⎣
⎦
⎣
⎦
or 2 2 ( V − R⋅ ω) ⋅ CDopen = ( V + R⋅ ω) ⋅ CDnotopen This indicates that the anemometer reaches a steady speed even in the abscence of friction because it is the relative
velocity on each cup that matters: the cup that has a higher drag coefficient has a lower relative velocity Rearranging for k= V
ω 2 2 ⎛ V − R⎞ ⋅ C
⎛V
⎞
⎜
⎟ Dopen = ⎜ + R⎟ ⋅ CDnotopen
⎝ω
⎠
⎝ω
⎠ Hence ⎞
⎛
⎜ 1 + CDnotopen ⎟
⎜
CDopen ⎟
⎝
⎠ ⋅R
k=
⎞
⎛
⎜ 1 − CDnotopen ⎟
⎜
CDopen ⎟
⎝
⎠ k = 0.251 m k = 0.0948 km
hr rpm For the actual anemometer (with friction), we first need to determine the torque produced when the anemometer is stationary but
about to rotate
Minimum wind for rotation is Vmin = 1⋅ km
hr The torque produced at this wind speed is
2
2
⎛1
⎛1
⎞
Tf = ⎜ ⋅ ρ⋅ A⋅ Vmin ⋅ CDopen⎟ ⋅ R − ⎜ ⋅ ρ⋅ A⋅ Vmin ⋅ CDnotopen⎞ ⋅ R
⎟
2
2
⎝
⎠
⎝
⎠ Tf = 7.75 × 10 −6 N⋅ m A moment balance at wind speed V, including this friction, is
1
1
2
2
⎤
ΣM = 0 = ⎡ ⋅ ρ⋅ A⋅ ( V − R⋅ ω) ⋅ CDopen⎥ ⋅ R − ⎡ ⋅ ρ⋅ A⋅ ( V + R⋅ ω) ⋅ CDnotopen⎤ ⋅ R − Tf
⎢
⎢
⎥
2
2
⎣
⎦
⎣
⎦
or 2⋅ Tf
2
2
( V − R⋅ ω) ⋅ CDopen − ( V + R⋅ ω) ⋅ CDnotopen =
R ⋅ ρ⋅ A
km
hr This quadratic equation is to be solved for ω when V = 10⋅ After considerable calculations ω = 104 rpm This must be compared to the rotation for a frictionless model, given by
V
ωfrictionless =
k The error in neglecting friction is ωfrictionless = 105 rpm
ω − ωfrictionless
ω = 1.12 % Problem 9.107 [2] Problem 9.108 [2] Problem 9.109 [3] Problem 9.110 [3] W=
CD =
A=
ρ= 1970's Sedan
4500
lbf
0.5
24
0.00234 ft2
slug/ft3 Current Sedan
3500
lbf
0.3
20
(Table A.9) ft2 Computed results: V (mph) F D (lbf) 1970's Sedan
F T (lbf) P (hp) F D (lbf) 20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100 12.1
18.9
27.2
37.0
48.3
61.2
75.5
91.4
109
128
148
170
193
218
245
273
302 79.6
86.4
94.7
104
116
129
143
159
176
195
215
237
261
286
312
340
370 4.24
5.76
7.57
9.75
12.4
15.4
19.1
23.3
28.2
33.8
40.2
47.5
55.6
64.8
74.9
86.2
98.5 6.04
9.44
13.6
18.5
24.2
30.6
37.8
45.7
54.4
63.8
74.0
84.9
96.6
109
122
136
151 Current Sedan
F T (lbf)
58.5
61.9
66.1
71.0
76.7
83.1
90.3
98.2
107
116
126
137
149
162
175
189
204 P (hp)
3.12
4.13
5.29
6.63
8.18
10.0
12.0
14.4
17.1
20.2
23.6
27.5
31.8
36.6
42.0
47.8
54.3 V (mph) F D (lbf) F R (lbf) V (mph) F D (lbf) F R (lbf) 47.3 67.5 67.5 59.0 52.5 52.5 The two speeds above were obtained using Solver Power Consumed by Old and New Sedans
150
1970's Sedan
Current Sedan 120
P (hp) 90
60
30
0
20 30 40 50 60
V (mph) 70 80 90 100 Problem 9.112 [3] Given: Data on a bus Find: Power to overcome drag; Maximum speed; Recompute with new fairing; Time for fairing to pay for itself Solution:
1
2
Basic equation: FD = ⋅ ρ⋅ A⋅ V ⋅ CD
2 P = FD ⋅ V The given data or available data is V = 50⋅ mph FD =
The power available is 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 V = 73.3 FD = 478 lbf ft
s A = 80⋅ ft 2 CD = 0.95 P = 3.51 × 10 slug
ft 4 ft⋅ lbf P = FD ⋅ V ρ = 0.00234⋅ 3 P = 63.8 hp s Pmax = 450⋅ hp The maximum speed corresponding to this maximum power is obtained from 1
2
Pmax = ⎛ ⋅ ρ⋅ A⋅ Vmax ⋅ CD⎞ ⋅ Vmax
⎜
⎟
⎝2
⎠ ⎛ ⎞
⎟
⎜ 1 ⋅ ρ⋅ A ⋅ C D ⎟
⎝2
⎠ Vmax = ⎜ or We repeat these calculations with the new fairing, for which
FD = 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 FD = 428 lbf ⎛ Pmax ⎞
⎟
⎜ 1 ⋅ ρ⋅ A ⋅ C D ⎟
⎝2
⎠
Pmax Vmax = ⎜ The initial cost of the fairing is Vmax = 141 ft
s Pnew = 3.14 × 10 4 ft⋅ lbf s The cost per day is reduced by improvement in the bus performance at 50 mph Vmax = 146 ft Vmax = 99.5 mph s dollars
Costday = 200⋅
day The fuel cost is
Gain = Pnew Gain = 89.5 % P The new cost per day is then Costdaynew = Gain⋅ Costday dollars
Costdaynew = 179
day Hence the savings per day is Saving = Costday − Costdaynew Saving = 21.1 The initial cost will be paid for in τ= Saving Pnew = 57.0 hp 1
3 Cost = 4500⋅ dollars Cost Vmax = 95.9 mph CD = 0.85 Pnew = FD⋅ V The maximum speed is now 1
3 dollars τ = 7.02 month day Problem 9.113 [3] Problem 9.114 [4] Given: Data on a sports car Find: Speed for aerodynamic drag to exceed rolling resistance; maximum speed & acceleration at 55 mph;
Redesign change that has greatest effect Solution:
1
2
Basic equation: FD = ⋅ ρ⋅ A⋅ V ⋅ CD
2 P = FD ⋅ V The given data or available data is M = 2750⋅ lbm A = 18.5⋅ ft Pengine = 165⋅ hp FR = 0.01 × M⋅ g 2 CD = 0.32
ρ = 0.00234⋅ slug
ft The rolling resistance is then FR = 27.5 lbf To find the speed at which aerodynamic drag first equals rolling resistance, set the two forces equal
Hence V= 2⋅ FR
ρ⋅ A ⋅ C D V = 63.0 ft
s FD = η= V = 80.7 1
2
⋅ ρ⋅ V ⋅ A ⋅ C D
2 The power consumed by drag and rolling resistance at this speed is
Hence the drive train efficiency is 1
2
⋅ ρ⋅ V ⋅ A ⋅ C D = F R
2 V = 43.0 mph To find the drive train efficiency we use the data at a speed of 55 mph V = 55⋅ mph
The aerodynamic drag at this speed is 3 ft
s Pengine = 12⋅ hp FD = 45.1 lbf ( ) Pused = FD + FR ⋅ V Pused Pused = 10.6 hp
η = 88.7 % Pengine The acceleration is obtained from Newton's second lawM⋅ a = ΣF = T − FR − FD
T= where T is the thrust produced by the engine, given by P
V The maximum acceleration at 55 mph is when we have maximum thrust, when full engine power is used.
Because of drive train inefficiencies the maximum power at the wheels is max = η⋅ Pengine
P
Hence the maximum thrust is Tmax = The maximum acceleration at 55 mph is then amax = Pmax Pmax = 146 hp Tmax = 998 lbf V Tmax − FD − FR
M ft
amax = 10.8
2
s Pengine = 165⋅ hp The maximum speed is obtained when the maximum engine power is just balanced by power consumed by drag and rolling
resistance
2
⎛1
⎞
Pmax = ⎜ ⋅ ρ⋅ Vmax ⋅ A⋅ CD + FR⎟ ⋅ Vmax
⎝2
⎠ For maximum speed: This is a cubic equation that can be solved by iteration or by using Excel's Goal Seek or Solver Vmax = 150 mph We are to evaluate several possible improvements:
For improved drive train η = η + 5⋅ % η = 93.7 % Pmax = η⋅ Pengine Pmax = 155 hp 2
⎛1
⎞
Pmax = ⎜ ⋅ ρ⋅ Vmax ⋅ A⋅ CD + FR⎟ ⋅ Vmax
2
⎝
⎠ Solving the cubic (using Solver) Vmax = 153 mph Improved drag coefficient: CDnew = 0.29
1
2
Pmax = ⎛ ⋅ ρ⋅ Vmax ⋅ A⋅ CDnew + FR⎞ ⋅ Vmax
⎜
⎟
2
⎝
⎠ Solving the cubic (using Solver) Vmax = 158 mph This is the best option! Reduced rolling resistance: FRnew = 0.93⋅ %⋅ M⋅ g FRnew = 25.6 lbf 1
2
Pmax = ⎛ ⋅ ρ⋅ Vmax ⋅ A⋅ CD + FRnew⎞ ⋅ Vmax
⎜
⎟
⎝2
⎠
Solving the cubic (using Solver) Vmax = 154 mph Problem 9.115 [4] Problem 9.116 Given: Data on dimensions of anemometer Find: [5] Calibration constant Solution:
The given data or available data is D = 50⋅ mm R = 80⋅ mm ρ = 1.23⋅ kg
3 m The drag coefficients for a cup with open end facing the airflow and a cup with open end facing downstream are, respectively, from Table 9
CDopen = 1.42 CDnotopen = 0.38 Assume the anemometer achieves steady speed ω due to steady wind speed V
k= The goal is to find the calibration constant k, defined by V
ω We will analyse each cup separately, with the following assumptions
1) Drag is based on the instantaneous normal component of velocity (we ignore possible effects on drag coefficient of
velocity component parallel to the cup)
2) Each cup is assumed unaffected by the others  as if it were the only object present
3) Swirl is neglected
4) Effects of struts is neglected ωR
Relative velocity
= Vcos θ  ωR θ V
Vcosθ In this more sophisticated analysis we need to compute the instantaneous normal relative velocity.
From the sketch, when a cup is at angle θ, the normal component of relative velocity is
Vn = V⋅ cos ( θ) − ω⋅ R (1) The relative velocity is sometimes positive sometimes negatiive. From Eq. 1, this is determined by ⎛ ω⋅ R ⎞
θc = acos ⎜
⎟
⎝V⎠
0 < θ < θc Vn > 0 θc < θ < 2⋅ π − θc Vn < 0 θc < θ < 2⋅ π For (2) Vn > 0 0 90 180 270 360 Vn ( θ) θ The equation for computing drag is FD = where A= 1
2
⋅ ρ⋅ A ⋅ V n ⋅ C D
2
π⋅ D
4 (3) 2 A = 1.96 × 10 −3 2 m In Eq. 3, the drag coefficient, and whether the drag is postive or negative, depend on the sign of the relative velocity
0 < θ < θc CD = CDopen FD > 0 θc < θ < 2⋅ π − θc CD = CDnotopen FD < 0 θc < θ < 2⋅ π For CD = CDopen FD > 0 The torque is 1
2
T = F D ⋅ R = ⋅ ρ⋅ A ⋅ V n ⋅ C D ⋅ R
2 The average torque is 1⌠
Tav =
⋅⎮
2⋅ π ⌡θ 2⋅ π π 1⌠
T dθ = ⋅ ⎮ T dθ
π ⌡θ where we have taken advantage of symmetry
θc Evaluating this, allowing for changes when θ = θc 1⌠
Tav = ⋅ ⎮
π⎮
⌡θ π 1⌠ 1
2
⋅ ρ⋅ A⋅ Vn ⋅ CDopen⋅ R dθ − ⋅ ⎮
⋅ ρ⋅ A⋅ Vn ⋅ CDnotopen⋅ R dθ
2
π⎮ 2
⌡θ
1 2 c ⎡ Using Eq. 1 θc π ⎤ ⌠
⌠
ρ⋅ A ⋅ R ⎢
2
2⎥
Tav =
⋅ CDopen⋅ ⎮ ( V⋅ cos ( θ) − ω⋅ R) dθ − CDnotopen⋅ ⎮ ( V⋅ cos ( θ) − ω⋅ R) dθ
⎢
⎥
⌡θ
2⋅ π
⌡θ ⎣ ⎦
θc
π
⎡
⎤
⌠
2
⌠
2
2
⎮
ρ⋅ A ⋅ R ⋅ ω ⎢
⎮ ⎛V
⎛ V ⋅ cos ( θ) − R⎞ dθ⎥
⎞ dθ − C
Tav =
⋅ ⎢CDopen⋅
⋅ cos ( θ) − R⎟
⎟⎥
Dnotopen⋅ ⎮ ⎜ ω
⎮ ⎜ω
2⋅ π
⎠
⎠⎥
⎮⎝
⌡θ ⎝
⎢
⌡θ
c
⎣
⎦
c and note that V
=k
ω The integral is ⌠
1⎞
2
2 ⎛1
2
⎮
⎮ ( k⋅ cos ( θ) − R) dθ = k ⋅ ⎜ 2 ⋅ cos ( θ) ⋅ sin ( θ) + 2 ⋅ θ⎟ − 2⋅ k⋅ R⋅ sin ( θ) + R ⋅ θ
⎝
⎠
⌡ For convenience define 1
2 ⎛1
2
f ( θ) = k ⋅ ⎜ ⋅ cos ( θ) ⋅ sin ( θ) + ⋅ θ⎞ − 2⋅ k⋅ R⋅ sin ( θ) + R ⋅ θ
⎟
2⎠
⎝2 Hence Tav = ρ⋅ A ⋅ R
⋅ ⎡C
⋅ f θc − CDnotopen⋅ f ( π) − f θc ⎤
⎦
2⋅ π ⎣ Dopen () ( ( )) For steady state conditions the torque (of each cup, and of all the cups) is zero. Hence () ( ( )) = 0 CDopen⋅ f θc − CDnotopen⋅ f ( π) − f θc
or CDnotopen
f θc =
⋅ f ( π)
CDopen + CDnotopen Hence CDnotopen
1⎞
21
2
2π
2⎞
k ⋅ ⎛ ⋅ cos θc ⋅ sin θc + ⋅ θc⎟ − 2⋅ k⋅ R⋅ sin θc + R ⋅ θc =
⋅ ⎛ k ⋅ + R ⋅ π⎟
⎜
⎜
2
2⎠
CDopen + CDnotopen ⎝ 2
⎝
⎠ Recall from Eq 2 that ω⋅ R ⎞
θc = acos ⎛
⎜
⎟
⎝V⎠ Hence () () () () or R⎞
θc = acos ⎛ ⎟
⎜
⎝k⎠ CDnotopen
R⎞
1
R ⎞⎞
R⎞
R⎞
2 1R
2
2π
2⎞
⎛
⎛
k ⋅ ⎛ ⋅ ⋅ sin ⎜ acos ⎛ ⎟ ⎞ + ⋅ acos ⎛ ⎟ ⎟ − 2⋅ k⋅ R⋅ sin ⎜ acos ⎛ ⎟ ⎞ + R ⋅ acos ⎛ ⎟ =
⋅ ⎛ k ⋅ + R ⋅ π⎟
⎜
⎜⎟
⎜
⎜⎟
⎜
⎜
⎝2 k
⎝
⎝ k ⎠⎠ 2
⎝ k ⎠⎠
⎝
⎝ k ⎠⎠
⎝ k ⎠ CDopen + CDnotopen ⎝ 2
⎠ This equation is to be solved for the coefficient k. The equation is highly nonlinear; it can be solved by iteration or
using Excel's Goal Seek or Solver
From the associated Excel workbook
k = 0.316⋅ m k = 0.119⋅ km
hr rpm Problem 9.116 (In Excel) Given: Data on dimensions of anemometer
Find: Calibration constant
Solution:
Given data:
D=
R=
C Dopen = 50
80
1.42 C Dnotopen = 0.38 Use Solver to find k to make the error zero!
k (mm)
Left
Right
315.85
37325.8
37326
k=
k= 0.316
0.119 m
km/hr/rpm mm
mm Error
0% [5] Problem 9.117 [4] Problem 9.118 [4] Problem 9.119 [5] Problem 9.120 [2] Given: Data on advertising banner Find: Power to tow banner; Compare to flat plate; Explain discrepancy Solution:
Basic equation: FD = 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 P = FD⋅ V
V = 55⋅ mph The given data or available data is V = 80.7⋅ ft
s A = L⋅ h
FD = FD = 771⋅ lbf 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 − 4 ft CD = V⋅ L
ν ⋅ 7 h = 4⋅ ft CD = 0.05⋅ L
h CD = 0.563 4 ft⋅ lbf P = 6.22 × 10 ⋅ s 2 (Table A.9, 69oF) s ) 2.58 log ReL
1
2
F D = ⋅ ρ⋅ A ⋅ V ⋅ C D
2 − ρ = 0.00234⋅ slug
ft ReL = 2.241 × 10 0.455 ( 2 P = FD⋅ V ν = 1.62 × 10 For a flate plate, check Re
ReL = A = 180⋅ ft L = 45⋅ ft so flow is fully turbulent. Hence use Eq 9.37b 1610
ReL CD = 0.00258 FD = 3.53⋅ lbf This is the drag on one side. The total drag is then 2⋅ FD = 7.06⋅ lbf . This is VERY much less than the banner
drag. The banner drag allows for banner flutter and other secondary motion which induces significant form drag. P = 113⋅ hp 3 Problem 9.121 Given: Data on car antenna Find: [1] Bending moment Solution:
1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 Basic equation: FD = The given or available data is V = 120⋅ km
hr V = 33.3⋅ A = L⋅ D
ρ = 1.225⋅ m
s A = 0.018 m
kg
3 L = 1.8⋅ m
2
−5 m ν = 1.50 × 10 ⋅ m
V⋅ D
ν D = 10⋅ mm 2 s (Table A.10, 20oC) 4 For a cylinder, check Re Re = Re = 2.22 × 10 From Fig. 9.13 CD = 1.0 FD = The bending moment is then L
M = FD ⋅
2 M = 11.0⋅ N ⋅ m 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 FD = 12.3 N Problem 9.122 Given: Data on wind turbine blade Find: [1] Bending moment Solution:
1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 Basic equation: FD = The given or available data is V = 45⋅ m
s L = 0.45⋅ m
2 A = L⋅ W
ρ = 1.225⋅ W = 35⋅ m A = 15.75 m 2
−5 m kg ν = 1.50 × 10 ⋅ ReL = 1.35 × 10 3 6 m
For a flat plate, check Re ReL =
CD = V⋅ L
ν
0.0742
ReL FD =
The bending moment is then 1
5 − 1740
ReL 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 W
M = FD ⋅
2 CD = 0.00312 FD = 61.0 N
M = 1067⋅ N⋅ m s (Table A.10, 20oC) so use Eq. 9.37a Problem 9.123 Given: Data on wind turbine blade Find: [4] Power required to maintain operating speed Solution:
1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 Basic equation: FD = The given or available data is ω = 20⋅ rpm
ρ = 1.225⋅ L = 0.45⋅ m
−5 m kg
m w = 35⋅ m ν = 1.50 × 10 3 ⋅ 2 (Table A.10, 20oC) s The velocity is a function of radial position, V ( r) = r⋅ ω, so Re varies from 0 to Remax = V ( w) ⋅ L
ν Remax = 2.20 × 10 6 The transition Reynolds number is 500,000 which therefore occurs at about 1/4 of the maximum radial distance; the boundary layer is
laminar for the first quarter of the blade. We approximate the entire blade as turbulent  the first 1/4 of the blade will not exert much
moment in any event
L
L⋅ ω
⋅ V( r ) =
⋅r
ν
ν Re( r ) = Hence CD = Using Eq. 9.37a 0.0742
ReL The drag on a differential area is dFD = 1
5 − 1740
=
ReL 0.0742 ⎛ L⋅ ω ⋅ r ⎞
⎜
⎟
⎝ν ⎠ 1 1
5 − 1
1
5−
ν⎞
5 1740
= 0.0742⋅ ⎛
⎜
⎟ ⋅r
L⋅ ω
⎝ L⋅ ω ⎠
⋅r
ν 1
2
2
⋅ ρ⋅ dA⋅ V ⋅ CD = ⋅ ρ⋅ L⋅ V ⋅ CD⋅ dr
2
2 − 1740⋅ ⎛
⎜ The bending moment is then ν ⎞ −1
⎟⋅r
⎝ L⋅ ω ⎠ dM = dFD⋅ r w w Hence ⌠
⌠1
2
⎮
M=
1 dM = ⎮
⋅ ρ⋅ L⋅ V ⋅ CD⋅ r dr
⎮
⎮2
⌡
⌡0
w ⌠
⎮
⎮
1
2
M = ⋅ ρ⋅ L⋅ ω ⋅ ⎮
⎮
2
⌡0
M = 1.43⋅ kN⋅ m ⌠
1
⎮
⎡
⎤
1
⎮
⎢
⎥
5−
1
ν⎞
23
5
⎛ ν ⎞ ⋅ r − 1⎥ dr
M=⎮
⋅ ρ⋅ L⋅ ω ⋅ r ⋅ ⎢0.0742⋅ ⎛
⎜
⎟ ⋅ r − 1740⋅ ⎜
⎟
⎮2
⎣
⎝ L⋅ ω ⎠
⎝ L⋅ ω ⎠
⎦
⌡0 1
⎡
⎤
14
⎢
⎥
5
⎞
⎞
⎢0.0742⋅ ⎛ ν ⎟ ⋅ r 5 − 1740⋅ ⎛ ν ⎟ ⋅ r 2⎥ dr
⎜
⎜
L⋅ ω ⎠
L⋅ ω ⎠ ⎦
⎣
⎝
⎝ Hence the power is P = M⋅ ω 1
⎡
⎤
19
⎢
⎥
5
1
1740 ⎛ ν ⎞ 3
2 5⋅ 0.0742 ⎛ ν ⎞
5
M = ⋅ ρ⋅ L ⋅ ω ⋅ ⎢
⋅⎜
⋅⎜
⎟ ⋅w −
⎟⋅w ⎥
2
3 ⎝ L⋅ ω ⎠ ⎦
⎣ 19 ⎝ L⋅ ω ⎠ P = 3.00 kW Problem 9.124 [2] Problem 9.125 [2] Problem 9.126 [2] Problem 9.127 [2] Problem 9.128 Given: 3 mm raindrop Find: [2] Terminal speed Solution:
1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 Basic equation: FD = Given or available data is D = 3⋅ mm ρH2O = 1000⋅ ΣF = 0 kg
3 ρair = 1.225⋅ m
Summing vertical forces M⋅ g − FD = M⋅ g −
M = ρH2O⋅ π⋅ D
6 m 1
2
⋅ ρ ⋅ A ⋅ V ⋅ CD = 0
2 air
M = 1.41 × 10 Check Re Re = V⋅ D
ν V = 8.95 ⋅ (Table A.10, 20oC) s Buoyancy is negligible 3 2⋅ M⋅ g
CD⋅ ρair⋅ A ν = 1.50 × 10 3 −5 kg A= π⋅ D
4 2 A = 7.07 × 10 −6 2 m CD = 0.4 Assume the drag coefficient is in the flat region of Fig. 9.11 and verify Re later
V= 2
−5 m kg m
s
3 Re = 1.79 × 10 which does place us in the flat region of the curve Actual raindrops are not quite spherical, so their speed will only be approximated by this result Problem 9.129 [3] Problem 9.130 [3] Problem 9.131 [3] F n2 Fn 1
W Given: Circular disk in wind Find: Mass of disk; Plot α versus V Solution:
Basic equations: CD = →
ΣM = 0 FD
1
2
⋅ ρ⋅ V ⋅ A
2 1
D
Summing moments at the pivot W⋅ L⋅ sin( α) − Fn1⋅ L − ⋅ ⎛ L − ⎞ ⋅ Fn2 = 0
⎜
⎟
2⎝
2⎠ (1) and for each normal drag Fn = 1
2
⋅ ρ⋅ V n ⋅ A ⋅ C D
2 Assume 1) No pivot friction 2) CD is valid for Vn = Vcos(α)
The data is ρ = 1.225⋅ − 5 N ⋅s
⋅
2 kg μ = 1.8 × 10 3 m m D = 25⋅ mm d = 3⋅ mm CD1 = 1.17 (Table 9.3) Red = V = 15⋅ m
s L = 40⋅ mm
ρ⋅ V ⋅ d
μ α = 10⋅ deg Red = 3063 so from Fig. 9.13 CD2 = 0.9 2 Hence Fn1 =
Fn2 = 1
2 π⋅ D
⋅ ρ⋅ ( V⋅ cos ( α) ) ⋅
⋅ CD1
4
2
1
2 Fn1 = 0.077 N D
2
⋅ ρ⋅ ( V⋅ cos ( α) ) ⋅ ⎛ L − ⎞ ⋅ d⋅ CD2
⎜
⎟
2⎠
⎝ Fn2 = 0.00992 N The drag on the support is much less than on the disk (and moment even less), so results will not be much different from those of Problem 9
2 Hence Eq. 1 becomes 1
1
D1
D
2 π⋅ D
2
M⋅ L⋅ g⋅ sin ( α) = L⋅ ⋅ ρ⋅ ( V⋅ cos ( α) ) ⋅
CD1 + ⋅ ⎛ L − ⎞ ⋅ ⎡ ⋅ ρ⋅ ( V⋅ cos ( α) ) ⋅ ⎛ L − ⎞ ⋅ d⋅ CD2⎤
⎜
⎟⎢
⎜
⎟
⎥
4
2
2⎝
2 ⎠ ⎣2
2⎠
⎝
⎦
2 2 ρ⋅ V ⋅ cos ( α) ⎡ 1
D ⎞⎛
D⎞
2
⎤
M=
⋅ ⎢ ⋅ π⋅ D ⋅ CD1 + ⎛ 1 −
⎜
⎟ ⋅ ⎜ L − ⎟ ⋅ d⋅ CD2⎥
4⋅ g⋅ sin ( α) ⎣ 2
2⎠
⎝ 2⋅ L ⎠ ⎝
⎦ M = 0.0471 kg 4⋅ M⋅ g tan( α)
1
⋅
⋅
cos ( α) ⎡ 1
ρ
D ⎞⎛
D⎞
2
⎛
⎤
⎢ ⋅ π⋅ D ⋅ CD1 + ⎜ 1 −
⎟ ⋅ ⎜ L − ⎟ ⋅ d⋅ CD2⎥
2
2⋅ L ⎠ ⎝
2⎠
⎣
⎝
⎦
We can plot this by choosing α and computing V
V= Rearranging V = 35.5⋅ 80 V (m/s) 60 40 20 0 10 20 30 40 Angle (deg) This graph can be easily plotted in Excel 50 60 70 m tan( α)
⋅
s cos ( α) Problem 9.132 [3] Given: Data on a tennis ball Find: Terminal speed time and distance to reach 95% of terminal speed Solution:
The given data or available data is M = 57⋅ gm Then A= π⋅ D
4 D = 64⋅ mm ν = 1.45⋅ 10 2 FD = Hence the terminal speed is Vt = Check the Reynolds number Re = s kg
3 m FD = M⋅ g The drag at speed V is given by ρ = 1.23⋅ m CD = 0.5 At terminal speed drag equals weight ⋅ −3 2 A = 3.22 × 10 Assuming high Reynolds number 2
−5 m (from Fig. 9.11) 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2
M⋅ g
1
⋅ ρ⋅ A ⋅ C D
2
Vt⋅ D Vt = 23.8 m
s 5 Re = 1.05 × 10 ν Check! For motion before terminal speed Newton's second law applies
M⋅ a = M⋅ dV
1
2
= M⋅ g ⋅ − ⋅ ρ⋅ V ⋅ A⋅ CD
dt
2 2
d
V = g − k⋅ V
dt or V Separating variables ⌠
1
⎮
dV = t
⎮ g − k⋅ V2
⌡0
g Hence V ( t) = Evaluating at V = 0.95Vt 0.95⋅ Vt = For distance x versus time, integrate dx
=
dt k k= where ρ⋅ A ⋅ C D k = 0.0174 2⋅ M ⌠
1
⎮
dV =
⎮
2
g − k⋅ V
⎮
⌡ 1
g⋅ k ⎛k⎞
⋅ V⎟
⎝g⎠ ⋅ atanh ⎜ ⋅ tanh( g⋅ k⋅ t) g
k ⋅ tanh( g⋅ k⋅ t) g
⋅ tanh( g⋅ k⋅ t)
k t= 1
g⋅ k ⎛
⎝ ⋅ atanh ⎜ 0.95⋅ Vt⋅ t ⌠
x=⎮
⎮
⌡0 g
k ⋅ tanh( g⋅ k⋅ t) dt k⎞ ⎟ g⎠ t = 4.44 s 1
m Note that ⌠
1
⎮
⎮ tanh( a⋅ t) dt = a ⋅ ln ( cosh ( a⋅ t) )
⌡ Hence x ( t) = Evaluating at V = 0.95Vt t = 4.44 s 1
⋅ ln (cosh ( g⋅ k⋅ t))
k so x ( t) = 67.1 m Problem 9.133 Given: Data on model airfoil Find: [3] Lift and drag coefficients Solution:
Basic equation: FD CD = 1
2
⋅ ρ⋅ A ⋅ V
2
Given or available data is D = 2⋅ cm
V = 30⋅ m
s CL = FL where A is plan area for airfoil, frontal area for rod 1
2
⋅ ρ⋅ A ⋅ V
2
(Rod)
L = 25⋅ cm
FL = 50⋅ N b = 60⋅ cm c = 15⋅ cm FH = 6⋅ N Note that the horizontal force FH is due to drag on the airfoil AND on the rod
ρ = 1.225⋅ kg
3 2
−5 m ν = 1.50 × 10 ⋅ m
For the rod Rerod = V⋅ D
ν Arod = L⋅ D Rerod = 4 × 10
Arod = 5 × 10 1
2
FDrod = CDrod⋅ ⋅ ρ⋅ Arod⋅ V
2
Hence for the airfoil A = b⋅ c CD = (Table A.10, 20oC) s 4 so from Fig. 9.13 −3 2 m FDrod = 2.76 N FD = FH − FDrod
FD 1
2
⋅ ρ⋅ A ⋅ V
2 CD = 0.0654 CDrod = 1.0 CL = FD = 3.24 N
FL
1
2
⋅ ρ⋅ A ⋅ V
2 CL = 1.01 CL
CD = 15.4 (Airfoil) Problem 9.134 [3] Problem 9.135 [4] Problem 9.136 [3] Problem 9.137 [4] Problem 9.138 [4] Problem 9.139 [4] Given: Data on a tennis ball Find: Terminal speed time and distance to reach 95% of terminal speed Solution:
The given data or available data is M = 57⋅ gm Then A= π⋅ D
4 From Problem 9.130 CD = D = 64⋅ mm 2 24
Re A = 3.22 × 10 Re −3 2 m Re ≤ 1
24 CD = ν = 1.45⋅ 10 1 < Re ≤ 400 0.646 CD = 0.5 400 < Re ≤ 3 × 10 CD = 0.000366⋅ Re 0.4275 5 6 3 × 10 < Re ≤ 2 × 10
6 CD = 0.18 Re > 2 × 10 At terminal speed drag equals weight FD = M⋅ g
1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 The drag at speed V is given by FD = Assume CD = 0.5 Hence the terminal speed is Vt = Check the Reynolds number Re = M⋅ g
1
⋅ ρ⋅ A ⋅ C D
2
Vt⋅ D This is consistent with the tabulated CD values! ν 5 Vt = 23.8 m
s 5 Re = 1.05 × 10 2
−5 m ⋅ s ρ = 1.23⋅ kg
3 m For motion before terminal speed, Newton's second law is M⋅ a = M⋅ dV
1
2
= M⋅ g ⋅ − ⋅ ρ⋅ V ⋅ A⋅ CD
dt
2 Hence the time to reach 95% of terminal speed is obtained by separating variables and integrating
0.95⋅ Vt ⌠
t=⎮
⎮
⎮
⎮
⌡0 1
g− ρ⋅ A ⋅ C D
2⋅ M dV
⋅V 2 For the distance to reach terminal speed Newton's second law is written in the form
M⋅ a = M⋅ V⋅ dV
1
2
= M⋅ g ⋅ − ⋅ ρ⋅ V ⋅ A⋅ CD
dx
2 Hence the distance to reach 95% of terminal speed is obtained by separating variables and integrating
0.95⋅ Vt ⌠
x=⎮
⎮
⎮
⎮
⌡0 V
g− ρ⋅ A ⋅ C D
2⋅ M dV
⋅V 2 These integrals are quite difficult because the drag coefficient varies with Reynolds number, which varies with
speed. They are best evaluated numerically. A form of Simpson's Rule is
⌠
ΔV
⎮
⎮ f ( V) dV = 3 ⋅ f V0 + 4⋅ f V1 + 2⋅ f V2 + 4⋅ f V3 + f VN
⌡ (( ) () () ( ) ( )) where ΔV is the step size, and V0, V1 etc., are the velocities at points 0, 1, ... N.
Here V0 = 0 From the associated Excel workbook 0.95⋅ Vt VN = 0.95⋅ Vt ΔV = t = 4.69⋅ s x = 70.9⋅ m N These results compare to 4.44 s and 67.1 m from Problem 9.132, which assumed the drag coefficient was constant and analytically
integrated. Note that the drag coefficient IS essentially constant, so numerical integration was not really necessary! Problem 9.139 (In Excel) [4] Given: Data on a tennis ball
Find: Terminal speed time and distance to reach 95% of terminal speed
Solution: Given data:
M=
ρ=
D=
CD = 57
1.23
64
0.5 gm
kg/m3
mm
(Fig. 9.11) ν = 1.45E05 m2/s Computed results:
A = 0.00322 m2
V t = 23.8 m/s
N=
20
ΔV = 1.19 m/s
For the time:
V (m/s)
Re
0
1.13
2.26
3.39
4.52
5.65
6.78
7.91
9.03
10.2
11.3
12.4 0
4985
9969
14954
19938
24923
29908
34892
39877
44861
49846
54831 CD
5438
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500 13.6
14.7
15.8
16.9
18.1
19.2
20.3
21.5
22.6 59815
64800
69784
74769
79754
84738
89723
94707
99692 0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500 W
1
4
2
4
2
4
2
4
2
4
2
4 f (V ) W xf (V ) For the distance:
f (V ) W xf (V ) 0.102
0.102
0.103
0.104
0.106
0.108
0.111
0.115
0.119
0.125
0.132
0.140 0.102
0.409
0.206
0.416
0.212
0.432
0.222
0.458
0.238
0.499
0.263
0.561 0.00
0.115
0.232
0.353
0.478
0.610
0.752
0.906
1.08
1.27
1.49
1.74 0.000
0.462
0.465
1.41
0.955
2.44
1.50
3.62
2.15
5.07
2.97
6.97 2
4
2
4
2
4
2
4
1 0.151
0.165
0.183
0.207
0.241
0.293
0.379
0.550
1.05 0.302
0.659
0.366
0.828
0.483
1.17
0.758
2.20
1.05 2.05
2.42
2.89
3.51
4.36
5.62
7.70
11.8
23.6 4.09
9.68
5.78
14.03
8.72
22.5
15.4
47.2
23.6 Total time: 4.69 s
Total distance: 70.9 m
(This compares to 4.44s for the exact result)
(This compares to 67.1 m for the exact result)
Note that C D is basically constant, so analytical result of Problem 9.132 is accurate! Problem 9.140 [4] Given: Data on an air bubble Find: Time to reach surface; plot depth as function of time; repeat for different sizes Solution:
The given data or available data is d0 = 0.3⋅ in h = 100⋅ ft ρw = 1000⋅ kg SG = 1.025 3 (Table A.2) m
2
−7 m ρ = SG⋅ ρw ν = 1.05 × 8.03 × 10 ⋅ s (Tables A.2 & A.8) patm = 101⋅ kPa The density of air is negligible compared to that of water, so Newton's second law is applicable with negligible MdV/dt
M⋅ dV
= 0 = ΣF = F B − F D
dt or FB = FD (1) where FB is the buoyancy force and FD is the drag (upwards is positive x)
FB = ρ⋅ Vol⋅ g FD = 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 (2) For a sphere, assuming high Reynolds number, from Fig. 9.11 CD = 0.5
The volume of the sphere increases as the bubble rises and experiences decreased pressure. Assuming the air is an isothermal
idea gas
p0⋅ Vol0 = p⋅ Vol
where p0 and Vol0 are the initial pressure and volume (at depth h), and p and Vol are the pressure and
volume at any depth
p0 = patm + ρ⋅ g⋅ h p = patm + ρ⋅ g⋅ ( h − x)
Hence (patm + ρ⋅ g⋅ h)⋅ 6 ⋅ d0
π π3
= ⎡patm + ρ⋅ g⋅ ( h − x)⎤ ⋅ ⋅ d
⎣
⎦6 (patm + ρ⋅ g⋅ h) 3 d = d0⋅ 3 ⎡patm + ρ⋅ g⋅ ( h − x)⎤
⎣
⎦ (3) For example, at the free surface (x = h) d = 12.1 mm
Combining Eqs. 1, 2 and 3 ρ⋅ π3 1 π22
⋅ d = ⋅ ρ⋅ ⋅ d ⋅ V ⋅ C D
6
24 V= 4⋅ g⋅ d
3⋅ CD V= ( ) patm + ρ⋅ g⋅ h
4⋅ g⋅ d0 ⎡
⎤
⋅⎢
⎥
3⋅ CD ⎡patm + ρ⋅ g⋅ ( h − x)⎤
⎣⎣
⎦⎦ Strictly speaking, to obtain x as a function of t we would have to integrate this expression (V = dx/dt). 1
6 However, evaluating V at depth h (x = 0) and at the free surface (x = h) x=0 V0 = 0.446 x=h V = 0.563 m
s m
s we see that the velocity varies slightly. Hence, instead of integrating we use the approximation dx = Vdt where dx is an increment of
displacement and dt is an increment of time. (This amounts to numerically integrating)
Note that the Reynolds number at the initial depth (the smallest Re) is Re0 = V0⋅ d0 so our use of CD = 0.5 from Fig. 9.11 is reasonable
The plots of depth versus time are shown in the associated Excel workbook
The results are d0 = 0.3⋅ in t = 63.4⋅ s d0 = 5⋅ mm t = 77.8⋅ s d0 = 15⋅ mm t = 45.1⋅ s ν Re0 = 4034 Problem 9.140 (In Excel) [4] Given: Data on an air bubble
Find: Time to reach surface; plot depth as function of time; repeat for different sizes
Solution:
The equation is where Given data:
h = 100 ft
h = 30.5 m
3
ρw = 1000 kg/m
SG = 1.025 Table A.2)
C D = 0.5 (Fig. 9.11)
ρ = 1025 kg/m3
p atm = 101 kPa
Computed results:
d 0 = 0.3 in
d 0 = 7.62 mm d0 = 5 d0 = mm 15 mm t (s) x (m) V (m/s) t (s) x (m) V (m/s) t (s) x (m) V (m/s) 0
5
10
15
20
25
30
35
40
45
50
63.4 0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
77.8 0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.1 0
2.23
4.49
6.76
9.1
11.4
13.8
16.1
18.6
21.0
23.6
30.5 0.446
0.451
0.455
0.460
0.466
0.472
0.478
0.486
0.494
0.504
0.516
0.563 0
1.81
3.63
5.47
7.32
9.19
11.1
13.0
14.9
16.9
18.8
20.8
22.9
25.0
27.1
29.3
30.5 0.362
0.364
0.367
0.371
0.374
0.377
0.381
0.386
0.390
0.396
0.401
0.408
0.415
0.424
0.435
0.448
0.456 0
3.13
6.31
9.53
12.8
16.1
19.5
23.0
26.6
30.5 0.626
0.635
0.644
0.655
0.667
0.682
0.699
0.721
0.749
0.790 Use Goal Seek for the last time step to make x = h ! Depth of Air Bubbles versus Time 30
25
20
x (m) 15
10
Initial Diameter = 5 mm
Initial Diameter = 0.3 in 5 Initial Diameter = 15 mm 0
0 10 20 30 40 50
t (s) 60 70 80 Problem 9.141 Given: Data on a tennis ball Find: [4] Maximum height Solution:
The given data or available data is M = 57⋅ gm Then A= π⋅ D
4 From Problem 9.130 CD = D = 64⋅ mm m
s ν = 1.45⋅ 10 2 24
Re CD = Vi = 50⋅ A = 3.22 × 10 −5 m ⋅ 2 s −3 2 m Re ≤ 1 24 1 < Re ≤ 400 0.646 Re 5 CD = 0.5 400 < Re ≤ 3 × 10
0.4275 5 6 CD = 0.000366⋅ Re
CD = 0.18
The drag at speed V is given by 3 × 10 < Re ≤ 2 × 10
Re > 2 × 10 FD = 6 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2
dV
1
2
= − ⋅ ρ⋅ V ⋅ A⋅ CD − M⋅ g
dt
2 For motion before terminal speed, Newton's second law (x upwards) is M⋅ a = M⋅ For the maximum height Newton's second law is written in the form M⋅ a = M⋅ V⋅ Hence the maximum height is dV
1
2
= − ⋅ ρ⋅ V ⋅ A⋅ CD − M⋅ g
dx
2 Vi
0
⌠
⌠
V
V
⎮
⎮
dV
dV =
xmax =
⎮
⎮
ρ⋅ A ⋅ C D 2
ρ⋅ A ⋅ C D 2
⎮
⎮−
⋅V − g
⋅V + g
2⋅ M
2⋅ M
⎮
⎮
⌡0
⌡V
i This integral is quite difficult because the drag coefficient varies with Reynolds number, which varies with
speed. It is best evaluated numerically. A form of Simpson's Rule is
⌠
ΔV
⎮
f ( V) dV =
⋅ f V0 + 4⋅ f V1 + 2⋅ f V2 + 4⋅ f V3 + f VN
⎮
3
⌡ (( ) () where ΔV is the step size, and V0, V1 etc., are the velocities at points 0, 1, ... N. () ( ) ( )) ρ = 1.23⋅ kg
m 3 Here V0 = 0 VN = Vi becomes N CD = 0.5
Vi the integral Vi xmax = 48.7⋅ m From the associated Excel workbook
If we assume ΔV = − ⌠
xmax = ⎮
⎮
⎮
⎮
⌡0
xmax = V
dV
ρ⋅ A ⋅ C D 2
⋅V + g
2⋅ M ⎛ ρ⋅ A ⋅ C D 2 ⎞
M
⋅ ln ⎜
⋅ Vi + 1⎟
ρ⋅ A⋅ CD ⎝ 2⋅ M⋅ g
⎠ xmax = 48.7 m The two results agree very closely! This is because the integrand does not vary much after the first few steps so the numerical
integral is accurate, and the analytic solution assumes CD = 0.5, which it essentially does! Problem 9.141 (In Excel) Given: Data on a tennis ball
Find: Maximum height
Solution:
The equation is Given data:
M=
V0 = 57
50.0 ρ = 1.23
D=
64
CD =
0.5 gm
m/s
kg/m3
mm
(Fig. 9.11) ν = 1.45E05 m2/s Computed results:
A = 0.00322 m2
N=
20
ΔV = 2.50 m/s
V (m/s)
0.0
2.5
5.0
7.5
10.0
12.5
15.0
17.5
20.0
22.5
25.0
27.5
30.0
32.5
35.0
37.5
40.0
42.5
45.0
47.5
50.0 Re
0
11034
22069
33103
44138
55172
66207
77241
88276
99310
110345
121379
132414
143448
154483
165517
176552
187586
198621
209655
220690 CD
0.000
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500
0.500 W
1
4
2
4
2
4
2
4
2
4
2
4
2
4
2
4
2
4
2
4
1 f (V ) W xf (V )
0.000
0.252
0.488
0.695
0.866
1.00
1.09
1.16
1.19
1.21
1.21
1.20
1.18
1.15
1.13
1.10
1.06
1.03
1.00
0.970
0.940 0.000
1.01
0.976
2.78
1.73
3.99
2.19
4.63
2.39
4.84
2.42
4.80
2.36
4.62
2.25
4.38
2.13
4.13
2.00
3.88
0.940 Maximum height: 48.7 m
(This is the same as the exact result)
Note that C D is basically constant, so analytical result of Problem 9.132 is accurate! [4] Problem 9.142 [3] Given: Data on rooftop carrier Find: Drag on carrier; Additional fuel used; Effect on economy; Effect of "cheaper" carrier Solution:
Basic equation: Given or available data is FD CD = 1
2
⋅ ρ⋅ A ⋅ V
2
w = 1⋅ m
V = 100⋅ h = 50⋅ cm km
hr V = 27.8 ρH2O = 1000⋅ kg
3 r = 10⋅ cm m
s ηd = 85⋅ % FE = 12.75⋅ A = w⋅ h A = 0.5 m km
L FE = 30.0 2 mi
gal BSFC = 0.3⋅ m
ρ = 1.225⋅ −5 m kg ν = 1.50 × 10 3 ⋅ m
From the diagram
Additional power is
Additional fuel is r
= 0.2
h
FD ⋅ V
ΔP =
ηd so kg
kW ⋅ hr 2 (Table A.10, 20oF) s
1
2
FD = CD⋅ ⋅ ρ⋅ A⋅ V
2 CD = 0.25 FD = 59.1 N ΔP = 1.93 kW ΔFC = BSFC⋅ ΔP ΔFC = 1.61 × 10 − 4 kg ΔFC = 0.00965 s kg
min Fuel consumption of the car only is (with SGgas = 0.72 from Table A.2)
FC = V
FE ⋅ SGgas⋅ ρH2O The total fuel consumption is then FCT = FC + ΔFC Fuel economy with the carrier is FE = For the squareedged: r
=0
h Additional power is ΔP = so
FD ⋅ V
ηd V
FCT ⋅ SGgas⋅ ρH2O CD = 0.9
ΔP = 6.95 kW FC = 1.57 × 10 − 3 kg FCT = 1.73 × 10
FE = 11.6 s
− 3 kg s km
L 1
2
FD = CD⋅ ⋅ ρ⋅ A⋅ V
2 FC = 0.0941
FCT = 0.104
FE = 27.2 kg
min
kg
min mi
gal FD = 213 N Additional fuel is ΔFC = BSFC⋅ ΔP ΔFC = 5.79 × 10 − 4 kg The total fuel consumption is then FE = s FCT = FC + ΔFC Fuel economy withy the carrier is now V
⋅ SGgas⋅ ρH2O
FCT The cost of the trip of distance d = 750⋅ km for fuel costing p = Cost =
The cost of the trip of with the rounded carrier ( FE = 11.6⋅ ΔFC = 0.0348 FCT = 2.148 × 10
FE = 9.3 km
L − 3 kg s FCT = 0.129
FE = 21.9 kg
min mi
gal $⋅ 3.50
with a rental discount = $⋅ 5 less than the rounded carrier is then
gal d
⋅ p − discount
FE Cost = 69.47 $ plus the rental fee Cost = 59.78 $ plus the rental fee km
) is then
L Cost = kg
min d
⋅p
FE Hence the "cheaper" carrier is more expensive (AND the environment is significantly more damaged!) Problem 9.143 [4] Problem 9.144 [4] Problem 9.145 [4] Problem 9.146 [4] Given: Data on a rocket
Find: Plot of rocket speed with and without drag
Solution:
From Example 4.12, with the addition of drag the momentum equation becomes
FB y + FS y − ∫ CV a rf y ρ dV = ∂
∂t ∫ CV v xyz ρ dV + ∫ CV r
r
v xyz ρV xyz ⋅ dA where the surface force is
FS y = − 1
ρAV 2 C D
2 Following the analysis of the example problem, we end up with
2
&
dVCV Ve me − 1 ρAVCV C D
2
=
−g
&
dt
M 0 − me t This can be written (dropping the subscript for convenience)
dV
= f (V , t )
dt (1) where
f (V , t ) = &
Ve me − 1 ρAV 2 C D
2
&
M 0 − me t −g (2) Equation 1 is a differential equation for speed V.
It can be solved using Euler’s numerical method
Vn +1 ≈ Vn + Δt f n where Vn+1 and Vn are the n + 1th and nth values of V, fn is the function given by Eq. 2 evaluated at the nth
step, and Δt is the time step.
The initial condition is V0 = 0 at t = 0 Given or available data:
M 0 = 400 kg
m e = 5 kg/s
V e = 3500 m/s
ρ = 1.23 kg/m3
D = 700 mm
C D = 0.3
Computed results:
A = 0.385 m2
N = 20
Δt = 0.50 s
With drag:
n t n (s) V n (m/s) f n V n+1 (m/s)
0 0.0
1 0.5
2 1.0
3 1.5
4 2.0
5 2.5
6 3.0
7 3.5
8 4.0
9 4.5
10 5.0
11 5.5
12 6.0
13 6.5
14 7.0
15 7.5
16 8.0
17 8.5
18 9.0
19 9.5
20 10.0 0.0
17.0
34.1
51.2
68.3
85.5
102
119
136
152
168
184
200
214
229
243
256
269
282
293
305 33.9
34.2
34.3
34.3
34.2
34.0
33.7
33.3
32.8
32.2
31.5
30.7
29.8
28.9
27.9
26.9
25.8
24.7
23.6
22.5
21.4 17.0
34.1
51.2
68.3
85.5
102
119
136
152
168
184
200
214
229
243
256
269
282
293
305
315 Without drag:
V n (m/s) f n V n+1 (m/s)
0.0
17.0
34.1
51.3
68.7
86.2
104
122
140
158
176
195
213
232
251
270
289
308
328
348
368 33.9
34.2
34.5
34.8
35.1
35.4
35.6
35.9
36.2
36.5
36.9
37.2
37.5
37.8
38.1
38.5
38.8
39.1
39.5
39.8
40.2 17.0
34.1
51.3
68.7
86.2
104
122
140
158
176
195
213
232
251
270
289
308
328
348
368
388 Trajectory of a Rocket
400
300
V (m/s) 200
Without Drag 100 With Drag 0
0 2 4 6 8
t (s) 10 12 Problem 9.147 [5] Problem 9.148 [5] Problem 9.149 [5] Part 1/2 Problem 9.149 [5] Part 2/2 Problem 9.150 [5] Part 1/2 Problem 9.150 [5] Part 2/2 Problem 9.151 [2] Given: Antique airplane guy wires Find: Maximum power saving using optimum streamlining Solution:
Basic equation: Given or available data is CD = FD 1
2
⋅ ρ⋅ A ⋅ V
2
L = 50⋅ m
A = L⋅ D ρ = 1.21⋅ P = FD⋅ V
D = 5⋅ mm
A = 0.25 m kg
3 V = 175⋅ −5 m The Reynolds number is Hence V = 48.6 m
s 2 ν = 1.50 × 10 m
V⋅ D
Re =
ν km
hr ⋅ 2 s 1
2
P = ⎛ CD⋅ ⋅ ρ⋅ A⋅ V ⎞ ⋅ V
⎜
⎟
2
⎝
⎠ 4 Re = 1.62 × 10 (Table A.10, 20oC)
so from Fig. 9.13 CD = 1.0 P = 17.4⋅ kW with standard wires Figure 9.19 suggests we could reduce the drag coefficient to CD = 0.06
Hence 1
2
Pfaired = ⎛ CD⋅ ⋅ ρ⋅ A⋅ V ⎞ ⋅ V
⎜
⎟
2
⎝
⎠ Pfaired = 1.04⋅ kW The maximum power saving is then ΔP = P − Pfaired ΔP = 16.3⋅ kW Thus ΔP
= 94⋅ %
P which is a HUGE savings! It's amazing the antique planes flew! Problem 9.152 [4] Problem 9.153 [4] Problem 9.154 [1] Problem 9.155 [5] Problem 9.156 Given: Aircraft in level flight Find: [1] Effective lift area; Engine thrust and power Solution:
Basic equation: For level, constant speed
Given or available data is FD CD = 1
2
⋅ ρ⋅ A ⋅ V
2
FD = T
km
V = 225⋅
hr
ρ = 1.21⋅ kg
3 CL = FL 1
2
⋅ ρ⋅ A ⋅ V
2
FL = W
m
V = 62.5
s P = T⋅ V CL = 0.45 CD = 0.065 M = 900⋅ kg (Table A.10, 20oC) m
Hence Also 1
2
FL = CL⋅ ⋅ ρ⋅ A⋅ V = M⋅ g
2
FL
FD = CL
CD FL = M⋅ g T = FD
The power required is then T = 1275 N P = T⋅ V P = 79.7 kW A= 2⋅ M⋅ g
C L ⋅ ρ⋅ V FL = 8826 N 2 2 A = 8.30 m CD
FD = FL ⋅
CL FD = 1275 N Problem 9.157 [2] Problem 9.158 Given: Data on an airfoil Find: [2] Maximum payload; power required Solution:
The given data or available data is ρ = 1.23⋅ kg
m L = 1.5⋅ m 3 w = 2⋅ m V = 12⋅ m
s CL = 0.72 CD = 0.17 2 Then A = w⋅ L A = 3m The governing equations for steady flight are W = FL and T = FD FL = 191 N FL = 43⋅ lbf where W is the model total weight and T is the thrust
The lift is given by FL = 1
2
⋅ ρ⋅ A ⋅ V ⋅ C L
2 W = M⋅ g = FL The payload is then given by
FL or M= The drag is given by FD = Engine thrust required T = FD T = 45.2 N The power required is P = T⋅ V P = 542 W g
1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 M = 19.5 kg M = 43⋅ lb FD = 45.2 N FD = 10.2⋅ lbf P = 0.727⋅ hp The model ultralight is just feasible: it is possible to find an engine that can produce about 1 hp that weighs less than about 45 lb Problem 9.159 [3] Problem 9.160 [3] Given: Data on a light airplane Find: Angle of attack of wing; power required; maximum "g" force Solution:
The given data or available data is ρ = 1.23⋅ kg m
m
V = 63⋅
s
The governing equations for steady flight are 3 2 M = 1000⋅ kg A = 10⋅ m CL = 0.72 CD = 0.17 W = M⋅ g = FL T = FD where W is the weight T is the engine thrust
1
2
⋅ ρ⋅ A ⋅ V ⋅ C d
2 The lift coeffcient is given by FL = Hence the required lift coefficient is CL = From Fig 9.17, for at this lift coefficient α = 3⋅ deg and the drag coefficient at this angle of attack is CD = 0.0065 M⋅ g CL = 0.402 1
2
⋅ ρ⋅ A ⋅ V
2 (Note that this does NOT allow for aspect ratio effects on lift and drag!)
1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 Hence the drag is FD = FD = 159 N and T = FD T = 159 N The power required is then P = T⋅ V P = 10 kW The maximum "g"'s occur when the angle of attack is suddenly increased to produce the maximum lift
From Fig. 9.17 CL.max = 1.72
FLmax = 1 The maximum "g"s are given by application of Newton's second law
M⋅ aperp = FLmax
where aperp is the acceleration perpendicular to the flight direction 2 ⋅ ρ⋅ A⋅ V ⋅ CL.max
2 FLmax = 42 kN FLmax Hence aperp = In terms of "g"s m
aperp = 42
2
s aperp
= 4.28
g M Note that this result occurs when the airplane is banking at 90o, i.e, when the airplane is flying momentarily in a circular
flight path in the horizontal plane. For a straight horizontal flight path Newton's second law is
M⋅ aperp = FLmax − M⋅ g
FLmax Hence aperp = In terms of "g"s aperp
= 3.28
g M −g m
aperp = 32.2
2
s Problem 9.161 Given: Data on an airfoil Find: [3] Maximum payload; power required Solution:
m
s The given data or available data is V = 12⋅ Then the area is ar = kg
m A = b⋅ c and the aspect ratio is ρ = 1.23⋅ A = 3m b
c c = 1.5⋅ m 3 b = 2⋅ m 2 ar = 1.33 The governing equations for steady flight are
W = FL and T = FD where W is the model total weight and T is the thrust
CL = 1.4 At a 12o angle of attack, from Fig. 9.17 CDi = 0.012 where CDi is the section drag coefficient
CL CD = CDi +
π⋅ ar The wing drag coefficient is given by Eq. 9.42 The lift is given by FL = 2 1
2
⋅ ρ⋅ A ⋅ V ⋅ C L
2 CD = 0.48 FL = 372 N FL = 83.6 lbf W = M⋅ g = FL The payload is then given by
FL or M= M = 37.9 kg The drag is given by FD = Engine thrust required T = FD T = 127.5 N The power required is P = T⋅ V P = 1.53 kW g
1
2 2 ⋅ ρ⋅ A ⋅ V ⋅ C D M = 83.6 lb FD = 127.5 N FD = 28.7 lbf P = 2.05 hp NOTE: Strictly speaking we have TWO extremely stubby wings, so a recalculation of drag effects (lift is unaffected) gives
b = 1⋅ m c = 1.5 m and A = b⋅ c 2 A = 1.5 m ar = b
c ar = 0.667
CL 2 CD = CDi +
π⋅ ar so the wing drag coefficient is CD = 0.948
FD = 56.6 lbf The drag is 1
2
F D = 2⋅ ⋅ ρ ⋅ A ⋅ V ⋅ C D
2 FD = 252 N Engine thrust is T = FD T = 252 N The power required is P = T⋅ V P = 3.02 kW P = 4.05 hp Problem 9.162 [3] Problem 9.163 [3] Problem 9.164 Given: Data on an airfoil Find: [3] Maximum payload; power required Solution:
The given data or available data is Vold = 225⋅ m
s ρ = 1.23⋅ kg
m 3 Assuming the old airfoil operates at close to design lift, from Fig. 9.19 A = 180⋅ m 10
arold =
1.8 CL = 0.3 arold = 5.56 CDi = 0.0062 2 (CDi is the old airfoil's
section drag coefficient) 2 CL Then CDold = CDi +
π⋅ arold The new wing aspect ratio is CDold = 0.0114 arnew = 8
2 CL Hence CDnew = CDi +
π⋅ arnew The power required is CDnew = 0.00978 1
2
P = T⋅ V = FD⋅ V = ⋅ ρ⋅ A⋅ V ⋅ CD⋅ V
2 If the old and new designs have the same available power, then
1
1
2
2
⋅ ρ⋅ A⋅ Vnew ⋅ CDnew⋅ Vnew = ⋅ ρ⋅ A⋅ Vold ⋅ CDold⋅ Vold
2
2
3 or CDold
Vnew = Vold⋅
CDnew Vnew = 236 m
s Problem 9.165 Given: Aircraft in circular flight Find: [3] Drag and power Solution:
Basic equations: FD CD = CL = 1
2
⋅ ρ⋅ A ⋅ V
2
The given data or available data are
ρ = 0.002377⋅ slug
ft P = FD ⋅ V 1
2
⋅ ρ⋅ A ⋅ V
2 R = 3250⋅ ft 3 V = 150⋅ mph V = 220⋅ ft
s →
→
Σ⋅ F = M⋅ a M = 10000⋅ lbm FL M = 311⋅ slug A = 225⋅ ft 2 ar = 7 Assuming the aircraft is flying banked at angle β, the vertical force balance is
or FL⋅ cos ( β) − M⋅ g = 0 1
2
⋅ ρ⋅ A⋅ V ⋅ CL⋅ cos ( β) = M⋅ g
2 or 1
M⋅ V
2
⋅ ρ⋅ A⋅ V ⋅ CL⋅ sin ( β) =
2
R (1) The horizontal force balance is
M⋅ V
−FL⋅ sin ( β) = M⋅ ar = −
R 2 Then from Eq 1 FL = Hence CL = M⋅ g
cos ( β)
FL
1
2
⋅ ρ⋅ A ⋅ V
2 tan( β) = V
R⋅ g β = atan ⎜ P = FD ⋅ V β = 24.8⋅ deg 4 FL = 1.10 × 10 ⋅ lbf
CL = 0.851
CL For the section, CDinf = 0.0075 at CL = 0.851 (from Fig. 9.19),
so
CD
Hence
FD = FL ⋅
FD = 524⋅ lbf
CL
The power is (2) ⎛ V2 ⎞
⎟
⎝ R⋅ g ⎠ 2 Equations 1 and 2 enable the bank angle β to be found 2 2 CD = CDinf +
π⋅ ar 5 ft⋅ lbf P = 1.15 × 10 ⋅ s P = 209⋅ hp CD = 0.040 Problem 9.166 [4] Given: Aircraft in circular flight Find: Maximum and minimum speeds; Drag and power at these extremes Solution:
Basic equations: FD CD = 1
2
⋅ ρ⋅ A ⋅ V
2
The given data or available data are
ρ = 0.002377⋅ slug
ft A = 225⋅ ft P = FD ⋅ V 1
2
⋅ ρ⋅ A ⋅ V
2 R = 3250⋅ ft 3 2 →
→
Σ⋅ F = M⋅ a M = 10000⋅ lbm FL CL = M = 311⋅ slug ar = 7 The minimum velocity will be when the wing is at its maximum lift condition. From Fig . 9. 17 or Fig. 9.19
CL = 1.72 CDinf = 0.02 where CDinf is the section drag coefficient
2 CL CD = CDinf +
π⋅ ar The wing drag coefficient is then CD = 0.155 Assuming the aircraft is flying banked at angle β, the vertical force balance is
or 1
2
⋅ ρ⋅ A⋅ V ⋅ CL⋅ cos ( β) = M⋅ g
2 or FL⋅ cos ( β) − M⋅ g = 0 1
M⋅ V
2
⋅ ρ⋅ A⋅ V ⋅ CL⋅ sin ( β) =
2
R (1) The horizontal force balance is
M⋅ V
−FL⋅ sin ( β) = M⋅ ar = −
R 2 2 (2) Equations 1 and 2 enable the bank angle β and the velocity V to be determined
2 2
⎛
⎞
M⋅ V
⎜
⎟
2
M⋅ g
R
2
2⎜
⎟ +⎛
⎞ =1
sin ( β) + cos ( β) =
⎟
⎜1
⎟ ⎜1
2
2
⋅ ρ⋅ A ⋅ V ⋅ C L ⎟
⎜ ⋅ ρ⋅ A ⋅ V ⋅ C L ⎟
⎜2
⎝
⎠ ⎝2
⎠ 2 or M ⋅V
R 4 2 2 2 4 ρ ⋅ A ⋅ V ⋅ CL
+ M ⋅g =
4 2 22 4 22 M ⋅g V= 2 2 ρ ⋅ A ⋅ CL
4
2 tan( β) = V
R⋅ g V = 149 2 2 − M
R ft
s V = 102 mph 2 ⎛ V2 ⎞
⎟
⎝ R⋅ g ⎠ β = atan ⎜ β = 12.0 deg The drag is then FD = 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2
P = FD ⋅ V The power required to overcome drag is FD = 918 lbf
5 ft⋅ lbf P = 1.37 × 10 P = 249 hp s The analysis is repeated for the maximum speed case, when the lift/drag coefficient is at its minimum
value. From Fig. 9.19, reasonable values are
CL = 0.3 CDinf = corresponding to α = 2o (Fig. 9.17) 47.6
CL 2 CD = CDinf +
π⋅ ar The wing drag coefficient is then
4 From Eqs. 1 and 2 CL 22 M ⋅g V= 2 2 2 V = ( 309.9 + 309.9i) 2
ρ ⋅ A ⋅ CL
M
−
2
4
R CD = 0.0104
ft
s Obviously unrealistic (lift is just too low,
and angle of attack is too low to
generate sufficient lift) We try instead a larger, more reasonable, angle of attack
CL = 0.55 CDinf = 0.0065
CL 4
2 2 2 2 tan( β) =
FD = CD = 0.0203 m
s V = 204 mph ⎛ V2 ⎞
⎟
⎝ R⋅ g ⎠ β = 40.6 deg 22 M ⋅g V= 2
ρ ⋅ A ⋅ CL
M
−
2
4
R The drag is then 2 CD = CDinf +
π⋅ ar The wing drag coefficient is then From Eqs. 1 and 2 corresponding to α = 4o (Fig. 9.17) V
R⋅ g 1
2
⋅ ρ⋅ A ⋅ V ⋅ C D
2 The power required to overcome drag is V = 91.2 β = atan ⎜ FD = 485 lbf
P = FD ⋅ V 5 ft⋅ lbf P = 1.45 × 10 s P = 264 hp Problem 9.167 [4] Problem 9.168 [3] Problem 9.169 Given: Car spoiler Find: [4] Whether they are effective Solution:
To perform the investigation, consider some typical data
For the spoiler, assume b = 4⋅ ft c = 6⋅ in kg ρ = 1.23⋅ 3 A = b⋅ c m
From Fig. 9.17 a reasonable lift coefficient for a conventional airfoil section is
Assume the car speed is V = 55⋅ mph Hence the "negative lift" is FL = CL = 1.4 1
2
⋅ ρ⋅ A ⋅ V ⋅ C L
2 FL = 21.7 lbf This is a relatively minor negative lift force (about four bags of sugar); it is not likely to produce a noticeable
difference in car traction
The picture gets worse at 30 mph: FL = 6.5 lbf For a race car, such as that shown on the cover of the text, typical data might be
b = 5⋅ ft
In this case: c = 18⋅ in A = b⋅ c FL = 1078 lbf Hence, for a race car, a spoiler can generate very significant negative lift! A = 7.5 ft 2 V = 200⋅ mph A = 2 ft 2 Problem 9.170 [5] Part 1/2 Problem 9.170 [5] Part 2/2 Problem 9.171 [5] Problem 9.172 [5] Problem 9.173 [5] Problem 9.174 [2] Problem 9.175 [2] Problem 9.176 [2] Problem 9.177 [2] Problem 9.178 [3] Problem 9.179 [4] x R L θ Given: Baseball pitch Find: Spin on the ball Solution:
Basic equations: The given or available data is M = 5⋅ oz
Compute the Reynolds number →
→
Σ⋅ F = M⋅ a FL CL = 1
2
⋅ ρ⋅ A ⋅ V
2
slug
ρ = 0.00234⋅
3
ft
C = 9⋅ in D= − 4 ft ν = 1.62 × 10
C
π V⋅ D
ν Re = ⋅ 2 L = 60⋅ ft s
2 D = 2.86 in A= π⋅ D
4 2 A = 6.45 in V = 80⋅ mph 5 Re = 1.73 × 10 This Reynolds number is slightly beyond the range of Fig. 9.27; we use Fig. 9.27 as a rough estimate
The ball follows a trajectory defined by Newton's second law. In the horizontal plane (x coordinate)
2 V
FL = M⋅ aR = M⋅ ax = M⋅
R FL = and 1
2
⋅ ρ⋅ A ⋅ V ⋅ C L
2 where R is the instantaneous radius of curvature of the trajectory
From Eq 1 we see the ball trajectory has the smallest radius (i.e. it curves the most) when CL is as large as possible.
From Fig. 9.27 we see this is when CL = 0.4
2⋅ M Solving for R R= (1) Also, from Fig. 9.27 ω⋅ D
= 1.5
2⋅ V Hence ω = 1.5⋅ From the trajectory geometry x + R⋅ cos ( θ) = R ω = 1.8⋅ where D ω⋅ D
= 1.8
2⋅ V ω = 14080 rpm 2⋅ V R = 463.6 ft to C L⋅ A ⋅ ρ sin ( θ) = 2 Hence L⎞
x + R⋅ 1 − ⎛ ⎟ = R
⎜
⎝ R⎠ Solving for x L⎞
x = R − R⋅ 1 − ⎛ ⎟
⎜
⎝ R⎠ 2 x = 3.90 ft 2⋅ V
D
L
R defines the best range ω = 16896 rpm Problem 9.180 [4] x R L θ Given: Soccer free kick Find: Spin on the ball Solution:
1
2
⋅ ρ⋅ A ⋅ V
2
kg
ρ = 1.21⋅
3
m The given or available data is M = 420⋅ gm →
→
Σ⋅ F = M⋅ a FL CL = Basic equations: C = 70⋅ cm Compute the Reynolds number D= C
π ν = 1.50⋅ 10 ⋅ D = 22.3 cm V⋅ D
ν Re = 2
−5 m L = 10⋅ m s A= π⋅ D
4 2 5 This Reynolds number is beyond the range of Fig. 9.27; however, we use Fig. 9.27 as a rough estimate
The ball follows a trajectory defined by Newton's second law. In the horizontal plane (x coordinate)
2 and FL = where R is the instantaneous radius of curvature of the trajectory
2⋅ M
C L⋅ A ⋅ ρ Hence, solving for R R= From the trajectory geometry x + R⋅ cos ( θ) = R (1)
where sin ( θ) = 2 Hence L⎞
x + R⋅ 1 − ⎛ ⎟ = R
⎜
⎝ R⎠ Solving for R R= Hence, from Eq 1 CL = For this lift coefficient, from Fig. 9.27 ω⋅ D
= 1.2
2⋅ V Hence ω = 1.2⋅ (And of course, Beckham still kind of rules!) (L2 + x2)
2⋅ x
2⋅ M
R⋅ A ⋅ ρ 2⋅ V
D R = 50.5 m
CL = 0.353 ω = 3086 rpm L
R 2 A = 0.0390 m Re = 4.46 × 10 V
FL = M⋅ aR = M⋅ ax = M⋅
R x = 1⋅ m 1
2
⋅ ρ⋅ A ⋅ V ⋅ C L
2 V = 30⋅ m
s ...
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This note was uploaded on 07/10/2011 for the course CHE 144 taught by Professor Tuzla during the Spring '11 term at Lehigh University .
 Spring '11
 TUZLA

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