Page 53 - Water Engineering Hydraulics, Distribution and Treatment

EXAMPLE 2.4 VOLUME OF SILTING
Determine the volume of silt accumulations for a drainage area having the following characteristics:
2
2
2
8
2
Area = 100 mi = 259 km = 2.59 × 10 m = 259,000,000 m
Average deposition of silt: c = 1.7
Area located in the southwestern United States: n = 0.77
Solution 1 (US Customary System):
(2.7)
s
where
V = volume of silt deposited annually, acre-ft V = cA n 2.6 Area and Volume of Reservoirs 31
s
A = the size of the drainage area, mi 2
V = 1.7 × (100) 0.77
s
= 1.7 × 34.7
= 59 acre-ft.
Solution 2 (SI System):
6
V = 1,233.5c[A∕(2.59 × 10 )] n (2.7a)
s
where
V = volume of silt deposited annually, m 3
s
A = the size of the drainage area in m 2
6
V = 1,233.5c[A∕(2.59 × 10 )] n
s
8
6
V = 1,233.5(1.7)[(2.59 × 10 )/(2.59 × 10 )] 0.77
s
= 1,233.5(1.7)(34.6737)
3
= 72, 708.98 m .
2
2
ML/km ) because an acre 3 ft deep (4,047 m area by 0.9144 For general use, surface areas and volumes are com-
m deep) is about 1 MG (3.785 ML). monly plotted against contour elevations as in Fig. 2.10.
Note that volumes must be determined from the surface area
2.6 AREA AND VOLUME OF RESERVOIRS
Surface area
The surface areas and volumes of water at given horizons
are found from a contour map of the reservoir site. Areas Surface area a
enclosed by each contour line are planimetered, and volumes
between contour lines are calculated. The average-end-area
method is generally good enough for the attainable precision a dH
of measurements. Elevation of water surface above reservoir bottom or above datum
For uniform contour intervals h (ft or m) and successive
2
contour areas a , a , … , a (acre or m ), the volume V of
0 1 n Volume of water stored, V
3
water (acre-ft or m )storeduptothe nth contour is V = ∫ a dH
V = 1/2 h[(a + a ) + (a + a ) + ⋯ + (a n−1 + a )]
1
2
1
n
0
Storage volume
n−1
( )
∑
V = 1/2 h a + a + 2 a (2.8) Figure 2.10 Surface area of a reservoir and volume of water
0
n
1 stored.