Page 33 - Water Engineering Hydraulics, Distribution and Treatment

EXAMPLE 1.4 DETERMINATION OF THE CAPACITY OF TREATMENT PLANT UNITS
Estimate the capacity of the components of a rapid sand filtration plant (Fig. 1.5) that is to deliver 10 MGD or 6,940 gpm (37.85 MLD
or 26,268 L/min) of water to a city of 67,000 people.
Solution 1 (US Customary System):
1. Two mixing basins, H = 10 ft deep; number of mixing basins N = 2.
(a) Assumed detention period t = 2 min.
(b) Volume V = Qt∕N = 6,940 × 2∕2 = 6,940 gal = 928 ft each.
(c) Surface area A = V∕H = 928∕10 = 92.8ft = 0.785 D .
√ √ 2 2 3 1.7 Purification Works 11
A × 4 92.8 × 4
(d) Diameter D = = = 10.9ft.

2. Two flocculating basins, H = 10 ft deep.
(a) Assumed detention period t = 30 min; number of flocculating basins N = 2.
3
(b) Volume V = Qt∕N = 6,940 × 30∕2 = 104,000 gal = 13,900 ft .
2
3
(c) Surface area A = V∕H = (13,900 ft )∕(10 ft) = 1,390 ft each (such as 20 ft by 70 ft).
3. Two settling basins, H = 10 ft deep, but allow for 2 ft of sludge; number of settling basins N = 2.
(a) Assumed detention period t = 2h.
3
(b) Effective volume V = Qt∕N = 6,940 × 2 × 60∕2 = 416,000 gal = 55,700 ft .
2
(c) Surface area A = V∕H = 55,700∕(10 − 2) ft = 6,960 ft (such as 35 ft by 200 ft).
2
(d) Surface rating SR = Q∕A = 6,940∕6,960 = 1.0 gpm∕ft .
4. Six rapid sand filters.
2
(a) Assumed surface rating SR = Q∕A = 3 gpm∕ft ; number of filters N = 6.
2
(b) Area A = Q∕(N × SR) = 6,940∕(6 × 3) = 385 ft (suchas15ftby26ft).
Solution 2 (SI System):
1. Two mixing basins, H = 3.05 m deep; number of mixing basins N = 2.
(a) Assumed detention period t = 2min.
3
(b) Volume V = Qt∕N = (26,268 × 2)∕2 = 26,268 L = 26.27 m each.
2
(c) Surface area A = V∕H = (26.27∕3.048) = 8.62 m .
√ √
A × 4 8.62 × 4
(d) Diameter D = = = 3.31 m.

2. Two flocculating basins, H = 3.05 m deep.
(a) Assumed detention period t = 30 min; number of flocculating basins N = 2.
3
(b) Volume V = Qt∕N = (26,268 L∕min × 30 min)∕2 = 394,020 L = 394 m .
2
3
(c) Surface area A = V∕H = (394 m )∕(3.05 m) = 129.27 m each (such as 6.1m × 21.3m).
3. Two settling basins, H = 3.05 m deep, but allow for 0.61 m of sludge; number of settling basins N = 2.
(a) Assumed detention period t = 2h.
3
(b) Effective volume V = Qt∕N = (26,268 L∕min × 2 × 60 min)∕2 = 1,576,080 L = 1,576 m .
3
2
(c) Surface area A = V∕H = (1,576 m )∕(3.05 m − 0.61 m) = 646 m (such as 10.7 m by 61 m).
2 2
(d) Surface rating SR = Q∕A = (26,268 L∕min)∕(646 m ) = 40.7L∕min∕m .
4. Six rapid sand filters.
2
(a) Assumed surface rating SR = Q∕A = 122.1L∕min∕m ; number of filters N = 6.
2
(b) Area A = Q∕(N × SR) = (26,268 L∕min)∕(6 × 122.1) = 35.86 m (such as 4.6 m by 7.9 m).