8.1
Under normal circumstances is the airflow though your trachea
(your windpipe) laminar or turbulent? List all assumptions
and show all calculations.
This is an open-ended student activity for which student responses will vary. No Solution is
available.
8.2
8.2
8.3
8.3
8.4
8.4
8.5
8.5
8.6
8.6
8.7
8.7
8.8
8.8
8.9
8.9
8.19
8.19
8.10
8.10
8.11
8.11
8.12
8.12
8.13
8.13
8.14
8.14
8.15
8.15
8.16
8.16
PROBLEM 8.17
T
Re
1876
1.21
1.21
2
2
8
8.18
8.18
8.18
8.19
8.19
8.20
8.20
8.21
8.21
8.22
8.22
PROBLEM 8.23
Re
PROBLEM 8.23
Re
The pipe size when choosing actual available pipe would be
Re
Re
PROBLEM 8.24
Re
PROBLEM 8.24
Re
When choosing actual available pipe, the new pipe size is
Re
PROBLEM 8.25
Q
2100
2100
Re
525
PROBLEM 8.25
Q
525
67200
67200
0.00190 m = 1.90 mm
PROBLEM 8.26
Re
PROBLEM 8.26
PROBLEM 8.26
Re
Re, tube
Re, needle
8.27
0.2
1.02
1.02
1.75 x 104
8.28
8.28
8.29
8.29
8.30
When soup is stirred in a bowl, there is considerable
turbulence in the resulting motion (see Video V8.7). From a very
simplistic standpoint, this turbulence consists of numerous intertwined
swirls, each involving a characteristic diameter and velocity.
As time goes by, the smaller swirls (the fine scale structure)
die out relatively quickly, leaving the large swirls that continue for
quite some time. Explain why this is to be expected.
This is an open-ended student activity for which student responses will vary. No Solution is
available.
8.31
8.31
8.32
8.32
8.33
8.33
8.34
8.34
8.35
8.35
8.35
8.35
8.36
8.36
8.37
8.37
head loss
8.38
8.38
8.39
8.39
8.40
8.40
8.41
8.41
8.42
8.42
8.43
8.43
8.44
8.44
PROBLEM 8.45
Since plastic is a smooth pipe, the given Blasius equation gives
Re
PROBLEM 8.45
Re
so the flow is turbulent and in the proper range for the given Blasius equation.
PROBLEM 8.46
wholly turbulent
PROBLEM 8.47
Re
Re
Re
Re
Re
PROBLEM 8.48
Re
Re
Re
Re
Re
PROBLEM 8.49
Re
Re
Re
Re
Re
Re
PROBLEM 8.49
Re
8.50
8.50
8.51
8.51
8.52
8.52
8.53
8.53
8.54
8.54
8.55
8.55
8.56
8.56
8.57
8.57
8.58
8.58
8.59
8.59
8.60
8.60
8.60
8.61
8.61
8.62
8.62
PROBLEM 8.63
Reh
1.0x10-6
so,
NOT IN 8E CHAPTER PROBLEMS
8.64
8.64
8.65
8.65
8.66
8.66
8.67
8.67
8.68
8.68
8.69
8.69
8.70
8.70
8.71
8.71
8.72
8.72
8.73
8.73
PROBLEM 8.74
8.1
Re
PROBLEM 8.74
PROBLEM 8.75
KL
KL
KL
KL
8.22
KL
interpolating Table B.2 gives
Re
8.4.2
KL
PROBLEM 8.75
PROBLEM 8.76
threaded
KLgate
KL
KL
KL
KL
threaded
8.22
KL
0.15
KL
1.5
KL
Table 8.1 and B.1 for
Re
1.21
5.35
KL
PROBLEM 8.76
0.018
0.018
15
100,900
105,515
-133,360
0.15
PROBLEM 8.77
Re
351 Q
1.21
PROBLEM 8.77
Re
55
8
51
65
7.02
7
7
7
8.78
8.78
PROBLEM 8.79
KL
KL
KL
For soldered (similar to threaded) pipe, Table 8.2 gives KLtee
Re
12
1
Re
40
1
Re
32
1
Table 8.1 gives
0.0129
0.0129
0.0132
0.0137
0.0132
0.0137
PROBLEM 8.79
1.84
2.34
3.29
1.34
36.6
3.29
2.34
8.80
8.80
8.81
8.81
8.82
8.82
8.83
8.83
PROBLEM 8.84
KL
KL
KL
KL
PROBLEM 8.84
KL
KL
KL
KL
Assuming standard air with v = 1.57 x 10-4 ft2/sec, the Reynolds numbers are (note the turbulent
flow as assumed):
Re
65
57
Re
77
8
0.0168
168
0.0385
285
0.107
8.85
8.85
8.86
8.86
8.87
8.87
8.88
8.88
8.89
8.89
PROBLEM 8.90
8.90
KL
KL
KL
KL
Fig. 8.22
KL
KL
10
KL
KL
KL
0.5, and Table 8.2 gives
the Reynolds number is
KL
KL
1.92
Re
8.31 x
8.1
KL
10-6
PROBLEM 8.90
0.5
+ 2(1.3) +10
0.0272
14.5
Re
92
0.0272
5.22
0.0262
0.0264
14.5
92
0.0262
0.0264
5.07
0.0262
0.0264
101.3
14.5
8.91
Gasoline is unloaded from the tanker truck shown in Fig.
P8.91 through a 4-in.-diameter rough-surfaced hose. This is
a “gravity dump” with no pump to enhance the flowrate. It is
claimed that the 8800-gallon capacity truck can be unloaded in
28 minutes. Do you agree with this claim? Support your answer
with appropriate calculations.
This is an open-ended student activity for which student responses will vary. No Solution is
available.
PROBLEM 8.92
8.92
Threaded
KL
KL
KL
KL
8.2
KL
0.15
8.22
1.5
(0.15)
1.5
20
63.1
8
Re
1.42
1.21
12.5
63.1
KL
KL
KL
KL
KL
= 20.
PROBLEM 8.92
1.775
12.5
1.42
0.0207
0.0207
1.42
12.3
63.1
1.75 x 105
12.3
0.0207
12.3
0.29
8.93
8.93
8.93
PROBLEM 8.94
P8.94
2
2
2
2
2
2
2
2
2
2
2
2
2
2
8.1
PROBLEM 8.94
wholly turbulent flow
2
Re
3750
2
1.052
2
53
Re
3436
2
1.052
0.054
Another iteration indicates convergence has been obtained,
2
8.95
8.95
8.96
8.96
8.96
PROBLEM 8.97
KL
8.22
0.8
KL
8.1
Re
Re
KL
PROBLEM 8.97
1.5
(4.9 x 10-6
0.6078
0.3199
5.374
Re
5.374
3.35
4.9
4000
5.374
373
13
1/4
PROBLEM 8.98
P8.98
KL
KL
KL
KL
KL
1.5
KL
KL
KL
KL
KL
KL
8.2
KL
= 10
1.5
19.3
Re
8.1
wholly turbulent flow
8.22
10 0.8
0.8
PROBLEM 8.98
0.000900
19.3
0.00667
0.000532
667
9.27
532
7.40
0.0256
Repeating the calculations for the average f25 = f5 =0.0258
0.0258 gives
19.3
863
640
863
511
640
511
0.0256
863
8.90
7.10
0.026
19.3
863
863
258)
(0.028)
PROBLEM 8.98
863
863
863
1.74
483
PROBLEM 8.99
P8.99
KL
8.22
KL
0.8
PROBLEM 8.99
KL
KL
KL
Q
KL
0.8
1.8
W
o
PROBLEM 8.99
1.8
1.8
1.8
Re
1
6
PROBLEM 8.99
Re
6
0.0172
0.0959
0.0959
1.21
0.0173
so with one more iteration convergence is obtained and
0.0957
957
4514
75.2
PROBLEM 8.100
PROBLEM 8.100
8.101
8.101
8.101
8.102
8.102
8.102
8.103
8.103
8.104
8.104
8.104
PROBLEM 8.105
Use next page in
place of this
deleted text.
PROBLEM 8.105
PROBLEM 8.105
Assuming the flow is turbulent, calculate the relative roughness. Using Table 8.1,
PROBLEM 8.105
2
1.94
2
Re
4
1
Re
9.90
4
8
0.349
2
8
0.345
Re
5
4
9.11
0.0211
0.346
2
1
PROBLEM 8.105
our preview guess
6
8.106
8.106
8.106
8.107
8.107
8.108
8.108
PROBLEM 8.109
P8.109
KL
KL
8.2
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
KL
8.22
KL
0.5
KL
KL
PROBLEM 8.109
KL
KL
KL
KL
4
Re
Re
KL
KL
1
4
6.32
8.2
KL
0.15
KL
KL
0.3
0.15
For D = 0.77 ft, the solution converges so
0.77
10
0.3
10
277
PROBLEM 8.110
KL
KL
KL
KL
KL
KL
KL
KL
Table 8.2 Fig. 8.26 (or the equation), and Fig. 8.25 give
f
PROBLEM 8.110
KL
KL
0.15
2
KL
KL
Re
Re
8
0.15
2
0.4
PROBLEM 8.110
1767
1767
8.111
8.111
8.112
8.112
8.112
8.113
Estimate the power that the human heart must impart to the
blood to pump it through the two carotid arteries from the heart to
the brain. List all assumptions and show all calculations.
This is an open-ended student activity for which student responses will vary. No Solution is
available.
PROBLEM 8.114
8.114
KL =
PROBLEM 8.114
KL
KL
KL
KL
KL
KL
8.28
KL
KL
KL
KL
KL
KL
With the given diameters, D2 = 2.067 in. and D3 = 3.068 in., and known numerical
values,
PROBLEM 8.114
Re
Re
0.041
0.041
38.6
8.115
8.115
PROBLEM 8.116
KL
Re =
Re
1
4
PROBLEM 8.116
Re
52
1
Re
2
1
0.0233
0.0174
0.0253
Assume the loss coefficient at the 90° at the standpipe inlet and outlet can be modeled by
90° standard, threaded elbows (KL = 1.5). The numerical values give
KL
2
KL
KL
2
0.0233
0.0174
1.5
41.2
271
117
1.5
0.0253
41.2
271
8.117
8.117
8.118
8.118
8.118
8.119
8.119
8.119
8.120
8.120
8.121
As shown in Fig. P8.121, cold water (T = 50 °F) flows
from the water meter to either the shower or the hot water heater.
In the hot water heater it is heated to a temperature of 150 °F.
Thus, with equal amounts of hot and cold water, the shower is at a
comfortable 100 °F. However, when the dishwasher is turned on,
the shower water becomes too cold. Indicate how you would predict
this new shower temperature (assume the shower faucet is not
adjusted). State any assumptions needed in your analysis.
This is an open-ended student activity for which student responses will vary. No Solution is
available.
8.122
8.122
8.123
8.123
8.124
8.124
8.125
8.125
8.126
8.126
8.127
8.127
8.128
8.128
8.129
8.129
8.130
8.130
8.131
8.131
8.132
8.132
8.133
8.133