Rapid sand filter, 45,000 population in 2021

A rapid sand filter is designed for the population forecast at the end of the design period. The design period is not stated, so 20 years is assumed (2021 to 2041).

Step 1: Design flow

• Population forecast (geometric increase): $P_n = P_0(1+r)^n = 45,000\times(1+0.021)^{20} = 68,191$ (design year 2041)
• Average daily demand $Q = 68,191 \times 110/1000 = 7,501$ m³/day
• Add 2 % for wash water: $Q' = 1.02 \times 7,501 = 7,651$ m³/day, i.e. 318.8 m³/h (plant works 24 h/day)

Step 2: Plan area and number of beds

• Filtration rate adopted = 4.5 m³/m²/h (4500 L/h/m², usual range 3000 to 6000)
• Total area $A = 318.8/4.5 = 70.8$ m²
• Number of beds $N = 1.22\sqrt{Q_{MLD}} = 1.22\sqrt{7.50} = 3.34$; adopt 4 beds, so that one can be washed while the other three work
• Area of one bed $= 70.8/4 = 17.71$ m²
• Take $L/B = 1.25$: $B = \sqrt{17.71/1.25} = 3.76$ m. Adopt B = 3.8 m, L = 4.7 m (area 17.86 m² each, total 71.44 m²)
• Actual rate with all beds working = 4.46 m/h; with one bed under wash = 5.95 m/h (within 3 to 6 m/h, OK)

Step 3: Filter media (sand and gravel)

The gravel (base material) is 450 mm deep in total. It supports the sand, stops it entering the laterals and spreads the wash water evenly.

Step 4: Underdrainage (perforated manifold and laterals)

• Backwash rate = 0.6 m³/min/m² (0.01 m/s) for 10 min, so the wash flow per bed $= 0.01 \times 17.86 = 0.179$ m³/s
• Total perforation area = 0.3 % of bed area $= 0.003 \times 17.86 = 0.054$ m²
• Perforation diameter 12 mm (area 1.131 cm²): number of holes $= 474$ per bed
• Manifold along the 4.7 m length, with laterals at 0.3 m c/c on both sides: $4.7/0.3 \approx 15$ per side, i.e. 30 laterals, about 16 holes each
• Total area of laterals = 2 × perforation area $= 0.107$ m², so each lateral $= 0.107/30 = 0.00357$ m², $D = 67$ mm, adopt 75 mm; length $\approx (3.8-0.55)/2 = 1.6$ m (L/D = 22, below 60, OK)
• Area of manifold = 2 × area of laterals $= 0.214$ m², so $D = 522$ mm, adopt 550 mm (area 0.238 m²)
• Wash velocity in manifold $= 0.179/0.238 = 0.75$ m/s (< 2.5 m/s, OK)

Step 5: Depth of the bed

Free board 0.5 m + water depth over sand 1.5 m + sand 0.7 m + gravel 0.45 m + underdrain space 0.25 m = 3.4 m.

 Inlet  ->  | Free board 0.5 m         |
            |~~~ water 1.5 m ~~~~~~~~~~|--> wash-water
            |__________________________|    trough
            | Sand 0.7 m               |
            | Gravel 0.45 m            |
            |___o_o_o_laterals_o_o_o___|
            |====== manifold ==========|--> filtered
            (3.4 m total depth)          water

Answer: 4 rapid sand filter beds, each 3.8 m × 4.7 m and 3.4 m deep, filtration rate 4.5 m/h, 0.7 m of sand on 0.45 m of graded gravel, with a perforated manifold–lateral underdrain (550 mm manifold, 30 laterals of 75 mm).