Page 63 - Water Engineering Hydraulics, Distribution and Treatment

EXAMPLE 2.5 SIZING OF CISTERN AND REQUIRED CATCHMENT AREA
A farmhouse for a family of four people has minimum drinking and culinary requirements of 120 gpd (454 L/d). The mean annual
rainfall is 45 in. (114 cm), and the effective period between rainy periods is 100 days. Determine
1. The size of the required rainfall collection cistern
2. The size of the required catchment area
Solution 1 (US Customary System):
1. The size of the required rainfall collection cistern:
The minimum volume of the cistern required will be
120 gpd × 100 days = 12,000 gal = 12,000∕7.48 = 1,604 ft 3. Problems/Questions 41
Say, a 10-ft-deep, 13-ft-square cistern.
2. The size of the required catchment area:
Because the mean annual rainfall is 45 in., the total design rainfall is 45 × 2∕3 = 30 in.
The total year’s requirement = 365 days × 120 gpd = 43,800 gal.
2
Referring to Fig. 2.18, the catchment area required to produce 43,800 gal from a design storm of 30 in. is 3,300 ft .
Say, a 100-ft × 33-ft catchment area.
Solution 2 (SI System):
1. The size of the required rainfall collection cistern:
The minimum volume of the cistern required will be
3
(454.2 L/d) × 100 days = 45,420 L = 45.42 m .
Say, a 3-m-deep, 4-m-square cistern (3m × 4m × 4m).
2. The size of the required catchment area:
Because the mean annual rainfall is 114 cm, then the total design rainfall is (114 cm) × 2∕3 = 76.2cm.
The total year’s requirement = (365 days)(454 L/d) = 165,800 L.
Referring to Fig. 2.18, the catchment area required to produce 165,800 L (43,800 gal) from a design storm of 76 cm (30 in.)
2
2
is 307 m (3,300 ft ).
Say, a 31-m × 10-m catchment area.

PROBLEMS/QUESTIONS
2.1 The response of a new water surface will establish new hydro-
Yield of catchment area (in thousand gallons) 110 Total precipitation (in.) 50 40 30 20 E = 40.5 in./year (102.87 cm/year); (c) areal rate of rainfall R =
150
logic equilibria. Determine the equivalent mean draft under the
following conditions: (a) areal rate of runoff of the original water-
130
shed Q = 27.5 in./year (69.85 cm/year); (b) areal rate of evaporation
60
Runoff = 0.75 total precipitation
47.5 in./year (120.65 cm/year); (d) reservoir area a = 1,500 acres
2
2
(6.07 km ); (e) catchment area A = 25,600 acres (103.60 km ); and
90
(f) mean annual water surface = 90% of reservoir area at spillway
level.
70
Determine the revised mean annual runoff, Q , assuming (a)
2.2
r
50
the original mean annual runoff Q = 27.5 in./year (69.85 cm/year);
and (b) the equivalent mean draft D = 1.08 in./year (2.74 cm/year).
e
10
30
Determine the effective catchment area assuming (a) orig-
2.3
2
inal catchment area A = 25,600 acres (103.60 km ); (b) reservoir
10
2
area a = 1,500 acres (6.07 km ); (c) mean annual water surface =
0
4000
2000
6000
90% of reservoir area at spillway level; (d) original water-
Horizontal area of catchment (in square feet)
shed’s runoff rate Q = 27.5 in./year (69.85 cm/year); (e) evapora-
Figure 2.18 Yield of impervious catchment area. Conversion tion rate E = 40.5 in./year (102.87 cm/year); and (f) rainfall rate
2
2
factors: 1 in. = 25.4 mm; 1 gal = 3.785 L; 1 ft = 0.0929 m . R = 47.5 in./year (120.65 cm/year).