Slow sand filter, 10000 pop, 1.5% growth

The filtration rate is printed as 150 L/m²/day (0.15 m/day). That would need about 2,020/0.15 ≈ 13,500 m² of filter, which is unrealistic (slow sand filters work at about 0.1–0.2 m/h). It is therefore read as 150 L/m²/h (= 0.15 m³/m²/h).

Step 1: Design flow

• Population forecast (geometric increase): $P_n = P_0(1+r)^n = 10,000\times(1+0.015)^{20} = 13,469$ (base period 5 + design period 15 = 20 years from the survey year)
• Daily demand $Q = 13,469 \times 150/1000 = 2,020.3$ m³/day = 84.18 m³/h

Step 2: Area and number of units

• Filtration rate = 0.15 m³/m²/h (150 L/m²/h) (range for slow sand filters 0.1–0.4)
• Total area $A = Q/(\text{rate}) = 84.18/0.15 = 561.2$ m²
• Keep each unit below about 100 m² and use at least 2 working units: working units n = 6, plus 1 standby unit for cleaning, so 7 units in all
• Area of one unit $= 561.2/6 = 93.5$ m²
• Take $L = 2B$: $B = \sqrt{93.5/2} = 6.84$ m. Adopt B = 6.9 m, L = 13.8 m (area 95.2 m² each)
• Check: rate with 6 working units = 0.147 m/h (OK)

Step 3: Depth of unit

Sand is scraped 15–25 mm every 1–2 months; when its depth falls to 0.5 m it is topped up to 0.9 m. Inlet and outlet chambers, a filtered-water weir, a drain valve and a filter-to-waste connection complete each unit.

Step 4: Section

 inlet -> |  free board 0.3 m             |
          |~~~~ supernatant water 1.0 m ~~|
          |-------------------------------|
          |  filter sand 0.9 m            |
          |-------------------------------|
          |  gravel 0.3 m (3 layers)      |
          |==o==o== under-drain 0.2 m ==o=|
                       |
          outlet weir -> filtered water tank

Answer: 6 working + 1 standby slow sand filters, each 6.9 m × 13.8 m (95.2 m²), total depth about 2.7 m.