Slow sand filter, 10000 population, base year 2068

The population of 10,000 at the base year is taken as the design population (no growth rate is given). A filtration rate of 0.2 m³/m²/h is assumed.

Step 1: Design flow

• Design population = 10,000 (given)
• Daily demand $Q = 10,000 \times 150/1000 = 1,500.0$ m³/day = 62.50 m³/h

Step 2: Area and number of units

• Filtration rate = 0.2 m³/m²/h (range for slow sand filters 0.1–0.4)
• Total area $A = Q/(\text{rate}) = 62.50/0.2 = 312.5$ m²
• Keep each unit below about 100 m² and use at least 2 working units: working units n = 4, plus 1 standby unit for cleaning, so 5 units in all
• Area of one unit $= 312.5/4 = 78.1$ m²
• Take $L = 2B$: $B = \sqrt{78.1/2} = 6.25$ m. Adopt B = 6.3 m, L = 12.6 m (area 79.4 m² each)
• Check: rate with 4 working units = 0.197 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: 4 working + 1 standby slow sand filters, each 6.3 m × 12.6 m (79.4 m²), total depth about 2.7 m.