Page 52 - Water Engineering Hydraulics, Distribution and Treatment

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Chapter 2
Water Sources: Surface Water
EXAMPLE 2.3 CALCULATIONS OF MEAN ANNUAL RUNOFF AND DRAFT
A mean draft of 30.0 MGD (113.6 MLD) is to be developed from a catchment area of 40.0 mi (103.6 km ). First calculations
ask for a reservoir area of 1,500 acres (6.07 km ) at flow line. The mean annual rainfall is 47.0 in./year (119.4 cm/year), the
mean annual runoff is 27.0 in./year (68.6 cm/year), and the mean annual evaporation is 40.0 in./year (101.6 cm/year). Find the
following:
1. The revised mean annual runoff
2. The equivalent mean draft
3. The equivalent land area
4. The adjusted flow line
Solution 1 (US Customary System): 2 2 2
1. By Eq. (2.2), the revised annual runoff is
Q = Q − (Q + E − R)(fa/A)
r
Q = 27.0 − (27.0 + 40.0 − 47.0)[(0.9 × 1,500/640)/(40.0)] = 27.0 − 1.1 = 25.9 in./year.
r
2
Here, 1 mi = 640 acres.
2. By Eq. (2.3), the equivalent mean draft is
D = (Q + E − R)(fa/A)
e
D = (27.0 + 40.0 − 47.0)[(0.9 × 1,500/640)/(40.0)] = 1.1 in./year.
e
2
D = 1.1 in./year = 52,360 gpd/mi = 0.052 MGD/mi 2
e
and the effective draft D is
ed
2
2
D = D + D (A) = 30.0MGD + (0.052 MGD/mi )(40.0mi ) = 32.1MGD.
ed md e
3. By Eq. (2.4), the equivalent land area is
A = A − fa[1 − (R − E)/Q]
e
2
A = 40.0 − (0.9 × 1,500/640)[1 − (47.0 − 40.0)/27.0] = 40.0 − 1.6 = 38.4mi .
e
4. By Eq. (2.5), the adjusted flow line is
F = Q + E − R
F = 27.0 + 40.0 − 47.0 = 20 in., equaling 20 × 0.9 = 18 in. at spillway level.
Solution 2 (SI System):
1. By Eq. (2.2), the revised annual runoff is
Q = Q − (Q + E − R)(fa/A)
r
Q = 68.58 − (68.58 + 101.60 − 119.38)[(0.9 × 6.07)/(103.6)] = 68.58 − 2.68 = 65.9cm/year.
r
2. By Eq. (2.3), the equivalent mean draft is
D = (Q + E − R)(fa/A)
e
D = (68.58 + 101.60 − 119.38)[(0.9 × 6.07)/(103.6)] = 2.68 cm/year.
e
2
D = 2.68 cm/year = 0.07337 MLD/km and the effective draft D is
e
ed
2
2
D = D md + D (A) = 113.55 MLD + (0.07337 MLD/km )(103.6km ) = 121.2MLD.
e
ed
2
Here, 1 cm/year = 0.0273793 MLD/km .
3. By Eq. (2.4), the equivalent land area is
A = A − fa[1 − (R − E)/Q]
e
A = 103.6 − (0.9 × 6.07)[1 − (119.38 − 101.60)/68.58] = 103.6 − 4.47 = 99.13 km 2.
e
4. By Eq. (2.5), the adjusted flow line is
F = Q + E − R
F = 68.58 + 101.60 − 119.38 = 50.8 cm, equaling (50.8cm) × 0.9 = 45.7 cm at spillway level.