Page 180 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 5
Water Hydraulics, Transmission, and Appurtenances
EXAMPLE 5.52 FREEZING INDEX AND FROST DEPTH DETERMINATION
The following are the average consecutive 2-week temperature records of a typical coldest month.
6
5
2
10
11
7
8
9
13
3
14
Day
1
4
12
◦
28
37
30
37
30
29
25
36
40
T ( F)
40
45
24
42
35
d
◦
T ( C)
1.67
2.78
−1.11
−1.66
d
Determine the freezing index and the required minimum frost depth.
Solution 1 (US Customary System):
◦
)
(
T − 32
Day 2.78 T ( F) −2.22 d −3.88 Σ T − 32 −1.11 −4.44 2.22 4.44 4.44 5.55 7.22
d
d
1 35 3 3
2 37 5 8
3 37 5 13 First maximum positive
4 30 −2 11
5 28 −4 7
6 25 −7 0
7 29 −3 −3
8 30 −2 −5
9 24 −8 −13 First maximum negative
10 36 4 −9
11 40 8 −1
12 40 8 7
13 42 10 17
14 45 13 30
Figure 5.24 can be plotted for determining F.
The freezing index = 13 + 13 = 26 (US customary unit).
The frost depth d = 1.65F 0.468 .
d = 1.65 (26) 0.468 = 7.58 in.
There shall be a minimum of 1 ft (12 in.) of fill material on the top of a sewer or water main for freezing protection. Certain US
states near Canada require a minimum of 3 ft of fill material.
Solution 2 (SI system):
◦
(
Day T ( C) T − 0 Σ T − 0 )
d
d
d
1 1.67 1.67 1.67
2 2.78 2.78 4.45
3 2.78 2.78 7.22 First maximum positive
4 −1.11 −1.11 6.11
5 −2.22 −2.22 3.9
6 −3.88 −3.88 0.02
7 −1.66 −1.66 −1.64
8 −1.11 −1.11 −2.75
9 −4.44 −4.44 −7.19 First maximum negative
10 2.22 2.22 −4.97
11 4.44 4.44 −0.53
12 4.44 4.44 3.91
13 5.55 5.55 9.46
14 7.22 7.22 26.68
The freezing index = 7.22 + 7.19 = 14.41 (SI unit).
The frost depth d = 55.18F 0.468 .
d = 55.18 (14.41) 0.468 = 192.33 mm.
There shall be a minimum of 305 mm of fill material on the top of a sewer or water main for freezing protection. Certain US
states near Canada require a minimum of 1 m (3 ft) of fill material.