Rapid sand filter for 5000 persons

Assumed data

Population 5000; 120 lpcd; rate of filtration 5 m/h; 4 % wash water; two units (minimum); $L:B=1.25$ to 1.33.

Step 1: Flow

$$Q=5000\times120=600\,000\ \text{l/day}=600\ \text{m}^3/\text{day},\qquad Q'=600\times1.04=624\ \text{m}^3/\text{day}=26.00\ \text{m}^3/\text{h}$$

Step 2: Filter area

$$A=\frac{Q'}{\text{rate}}=\frac{26.00}{5}=5.20\ \text{m}^2$$

Number of units $N=1.22\sqrt{0.6}=0.95$, so provide 2 units, each designed for the full flow so that one can be washed while the other works: each unit area $A_u=5.20$ m² $\Rightarrow$ $B=\sqrt{5.20/1.25}=2.04$ m, $L=1.25B=2.55$ m. Adopt $B=2.1$ m, $L=2.6$ m ($A=5.46$ m² each). Rate with both units working $=26.00/(2\times5.46)=2.38$ m/h; with one unit working (the other being washed) $=26.00/5.46=4.76$ m/h, which is within 6 m/h.

Step 3: Depth of filter box

Backwash: 0.6 m/min for 10 minutes gives $0.6\times5.46\times10=32.8$ m³ of wash water per unit, supplied from an elevated wash-water tank.

 SECTION (one unit)
  inlet                           wash-water trough
    |      _____________________________|__
    v     |  FB 0.5 m                       |
    ----> |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~| water 1.0 m
          |.................................|
          |   sand 0.6 m                    |
          |---------------------------------|
          |   gravel 0.45 m                 |
          |---+----+----+----+----+----+----|
          |   laterals (0.3 m) / manifold   | -> filtered
          |_________________________________|    water out
              wash-water in ^   wash waste out ->

Under-drainage

Perforated lateral pipes (holes about 13 mm, total hole area 0.3 % of bed area) are connected to a manifold, which also carries the wash water.

Answer: 2 rapid sand filters, each 2.6 m $\times$ 2.1 m in plan and 2.85 m deep, with 0.6 m of sand over 0.45 m of gravel.