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Rapid sand filter design for 25 MLD

Short answer

6 filters (+1 standby), each 6.7 m × 5.4 m in plan, total area 217 m². Rate 5 m/h. Depth: freeboard 0.5 m, water 1.5 m, sand 0.7 m, gravel 0.45 m, underdrain 0.3 m; overall 3.45 m.

Design a rapid sand filter for treating 25 MLD of water. Also draw a proper sketch showing the dimensions of RSF.

Design data (assumed)

  • Flow Q=25Q=25 MLD =25 000=25\,000 m³/day
  • Rate of filtration 5 m/h =120=120 m³/m²/day (range 3 to 6 m/h)
  • 4 % of the filtered water is used for backwashing
  • Operation 24 h/day; L:B=1.25L:B=1.25 (1.25 to 1.33)

Step 1: Design flow and total filter area

Q′=25 000×1.04=26000 m3/day=1083.3 m3/hQ'=25\,000\times1.04=26000\ \text{m}^3/\text{day}=1083.3\ \text{m}^3/\text{h} Atotal=Q′rate=1083.35=216.7 m2A_{total}=\frac{Q'}{\text{rate}}=\frac{1083.3}{5}=216.7\ \text{m}^2

Step 2: Number and size of units

N=1.22QN=1.22\sqrt{Q} (Q in MLD) =1.2225=6.1=1.22\sqrt{25}=6.1. Adopt 6 units (plus 1 standby).

Aunit=216.76=36.11 m2,B=36.111.25=5.37 m,L=1.25BA_{unit}=\frac{216.7}{6}=36.11\ \text{m}^2,\qquad B=\sqrt{\frac{36.11}{1.25}}=5.37\ \text{m},\quad L=1.25B

Adopt B=5.4B=5.4 m, L=6.7L=6.7 m (A=36.18A=36.18 m²). Check: actual rate =1083.3/(6×36.18)=4.99=1083.3/(6\times36.18)=4.99 m/h; when one unit is being washed, the rate in the other 5 is 5.99 m/h (acceptable, close to 6 m/h).

Step 3: Depth of the unit

LayerDepth (m)
Free board0.5
Water above sand1.5
Sand (ES 0.45 to 0.7 mm, UC 1.3 to 1.7)0.7
Gravel (4 layers, 2 to 50 mm)0.45
Under-drain and manifold space0.3
Total3.45

Step 4: Under-drainage

  • Perforated area =0.3%=0.3\% of bed area =0.003×36.18=0.109=0.003\times36.18=0.109 m². With 12 mm orifices (area 1.13×10−41.13\times10^{-4} m²): 960 orifices.
  • Central manifold along the length, with laterals on both sides at 0.3 m c/c: 44 laterals, about 22 holes each.
  • Total lateral area ≈2×\approx2\times orifice area =0.217=0.217 m²; manifold area ≈2×\approx2\times lateral area =0.434=0.434 m² (diameter about 0.74 m).

Step 5: Backwash

Rate 0.6 m/min (36 m/h) for 10 min: flow =0.6×36.18/60=0.362=0.6\times36.18/60=0.362 m³/s; volume per wash =217=217 m³ per unit, which is about 5 % of a unit's daily output (26 000/6=433326\,000/6=4333 m³/day). Wash-water troughs are provided above the sand.

 SECTION (one unit)
  inlet                           wash-water trough
    |      _____________________________|__
    v     |  FB 0.5 m                       |
    ----> |~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~| water 1.5 m
          |.................................|
          |   sand 0.7 m                    |
          |---------------------------------|
          |   gravel 0.45 m                 |
          |---+----+----+----+----+----+----|
          |   laterals (0.3 m) / manifold   | -> filtered
          |_________________________________|    water out
              wash-water in ^   wash waste out ->

Answer: 6 rapid sand filters (+1 standby), each 6.7 m ×\times 5.4 m in plan and about 3.45 m deep.

✓ Answer checked by Claude Opus 5.5 (high effort)

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