Rapid sand filter for 60000 population

Population 60,000 is taken as the design population for the new urban area. Per capita demand is assumed to be 135 lpcd (urban supply as per Nepal Water Supply Design Guidelines).

Step 1: Design flow

• Design population = 60,000 (given)
• Average daily demand $Q = 60,000 \times 135/1000 = 8,100$ m³/day
• Add 2% for wash water: $Q' = 1.02 \times 8,100 = 8,262$ m³/day, i.e. 344.2 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 3000–6000)
• Total area $A = 344.2/4.5 = 76.5$ m²
• Number of beds $N = 1.22\sqrt{Q_{MLD}} = 1.22\sqrt{8.10}$ = 3.47, adopt 4 beds (minimum 4 so that one can be washed while the rest work)
• Area of one bed $= 76.5/4 = 19.12$ m²
• Take $L/B = 1.25$: $B = \sqrt{19.12/1.25} = 3.91$ m. Adopt B = 3.9 m, L = 4.9 m (area 19.11 m² each, total 76.44 m²)
• Actual rate with all beds working = 4.50 m/h; with one bed under wash = 6.00 m/h (within 3–6 m/h, OK)

Step 3: Filter media (sand and gravel)

Gravel (base material) is 450 mm 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 wash flow per bed $= 0.01 \times 19.11 = 0.191$ m³/s
• Total perforation area = 0.3% of bed area $= 0.003 \times 19.11 = 0.057$ m²
• Perforation diameter 12 mm (area 1.131 cm²): number of holes $= 507$ per bed
• Manifold along the 4.9 m length, laterals at 0.3 m c/c on both sides: $4.9/0.3 \approx 16$ per side, i.e. 32 laterals, about 16 holes each
• Area of laterals = 2 × perforation area $= 0.115$ m², i.e. $0.115/32 = 0.0036$ m² each, so $D = 68$ mm, adopt 75 mm; length $\approx (3.9 - 0.55)/2 = 1.68$ m (L/D ≈ 22, below 60, OK)
• Area of manifold = 2 × area of laterals $= 0.229$ m², so $D = 540$ mm, adopt 550 mm
• Wash velocity in manifold $= 0.191/0.229 = 0.83$ 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 3.9 m × 4.9 m, 3.4 m deep, filtration rate 4.5 m/h, sand 0.7 m on 0.45 m graded gravel, with a perforated underdrain of a 550 mm manifold and 32 laterals of 75 mm.