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PSPICE FILTERS
FILTER DESIGN EXAMPLE
Dr. A. Saatci
USA -Rapid Sand Filter Design Inputs Design a rapid sand filter to treat Allowing filtered water for backwashing: Time used for bakwashing per day = Assume the rate of filtration = Number of Filters Length/ Width of Filters= Depth of Sand Media=
Q= Q_BW= t_BW=
20000 2% 0.50 10 2.00 1.30 1.00
m3/d hours m/h
m
Solution: Total filtered water = 20 000 x 1.02x 24/(24*(24-.5))
868 m3/h
=
2
Area of filter = 20 000 x 1.02 = 86.8 m 23.5 x 10 Each bed area 86.8/2 = 43.4 m2
86.8 m2 43.4 m2 2
Let the length of the filter as: L/B = 1.3; 1.3B = 43.4 m2 Width = Length = 5.78 x 1.3 = 7.51 m
5.78 m 7.51 m
Underdrainage system: Total area of holes = 0.2 to 0.5% of bed area. Assume 0.3% of bed area = 0.3 x 43.4 = 0.13 m 100 XSS Area of lateral = 2 (Area of holes of lateral) XSS Area of manifold = 2 (Area of laterals) 1/2
0.35% (0.2-0.5 % of bed area) 2
Diameter of manifold = (4 x 0.52 /pi) = 81.4 cm Choose a standard diameter = Assume c/c (center to center) of lateral = 30 cm.
0.152 m2 0.304 m2 0.608 m2 0.880 m 0.9 m 0.3 m
900
Holes Take dia of holes = 13 mm
10.00 mm 2
Number of holes: n pi (1.3) =Total Area of holes = 860 cm 4 n = 4 x Total Hole Area =
2
0.01 m
0.152 m2 1934
2
3.14(Hole Diam) Choose total number of holes
2000
Number of holes per lateral = No of holes/number of laterals = 13 2
40
2
Area of perforations per lateral = 13 x p (1.3) /4 = 17.24 cm Spacing of holes = L_lateral/no of holes per lateral = 19.5 cm. 2
XSS. area of lateral = 2 x area of perforations per lateral = 2 x 17.24 = 34.5 cm . 1/2
Diameter of lateral = (4 x 34.5/p) = 6.63 cm Choose a standard pipe diameter= Check:( Length of lateral < 60* Diam of Lateral) 60 x 6.63 = 3.98 m. l = 2.545 m.
0.00314 m2 0.061
31.4 cm2 6.13 cm
0.0063 m2
62.73 cm2
0.0894 m 90.00 mm 0.0900 m 5.40 OK > L_Lateral
BW Trough Design Number of BW troughs= Distance between BW troughs lengthwise at 5.75/3 = 1.9 m c/c.
3 1.93 m c/c
3
Discharge of each trough = Q/3 = 0.36/3 = 0.12 m /s.
0.12 m3/s
3/2
Q =1.71 x b x h Assume width of BW Trough= h= = 40 + (free board) 5 cm = 45 cm; slope 1 in 40
0.3 m 0.381 m
Clear water reservoir for backwashing Duration of BW
0.50 h
For 4 h filter capacity, Capacity of tank = 4 x 5000 x 7.5 x 5.75 x 2 = 1725 m 1000 Depth of BW Storage Tank= Surface area = 1725/5 = 345 m L/W =
2
2
L/B = 2; 2B = 345; B = 13 m & L = 26 m. L= Volume= Dia of inlet pipe coming from two filter = 50 cm. Velocity in filter effluent pipe during filtration= Diameter of the BW water pipe= BW Water flowrate= Velocity in the BW tank=
3
1302 m3 5m 260 m2 2 11.4 Roundup= 22.8 Roundup= 1380.0 m3 0.5 m 0.59 m/s 0.70 m 0.362 m3/s 0.94 m/s
12.0 23.0
700 OK
Air BW 2
Air Velocity 1000 l of air/ min/ m bed area (1 m/min = 60 m/h) Time of air BW= 3
Air required During BW = 1000 x 5 x 7.5 x 5.77 x 2 = 4.32 m of air. Blower Capacity=
