Environmental Engineering FOURTH EDITION

(backadmin) #1
118 ENVIRONMENTALENGINEERING

Exmm 6.5. Areservoir is needed to supply water demand for 9 out of 10 years. The
required reservoir capacities, which were determined by the method of Example 6.4,
are shown below:


Required reservoir Required reservoir
Year capacity (m3 x lo6) Year capacity (m3 x lo6)


1961
1962
1963
1964
1965
1966
1967
1968
1969
1970

60
40
85
30
67
46
60
42
90
51

1971
1972
1973
1974
1975
1976
1977
1978
1979
1980

53
62
73
80
50
38
34
28
40
45

These data must now be ranked, with the highest required capacity, or worst drought,
getting rank 1, the next highest 2, and so on. Data were collected for 20 years, so that
n = 20 and n + 1 = 21. Next, m/(n + 1) is calculated for each drought

Capacity Capacity
Rank (m3 x lo6) m/(n + 1) Rank (m3 x lo6) m/(n+ 1)

1 2 3 4 5 6 7 8 9

10

90
85
80
73
67
62
60
60
53
51

0.05
0.1
0.14
0.19
0.24
0.29
0.33
0.38
0.43
0.48

11
12
13
14
15
16
17
18
19
20

~
50
46
45
42
40
40
38
34
30
28

0.52
0.57
0.62
0.67
0.7 1
0.76
0.81
0.86
0.90
0.95

These data are plotted in Fig. 6-9. A semilog plot often yields an acceptable straight
line. If the reservoir capacity is required to be adequate 9 years out of 10, it may be
inadequate 1 year out of 10. Entering Fig. 6-9 at m/(n + 1) = 1/10 = 0.1, we find
that

m/n + 1 = 2/21 = 0.1.

Free download pdf