--- title: "Rapid sand filter design for 25 MLD" subject: "Water Supply Engineering" chapter: 6 key: "semester_5/water_supply_engineering:rsf-b81" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/rsf-b81 site: BCT Engineering --- # Rapid sand filter design for 25 MLD Water Supply Engineering · Chapter 6: Water Treatment - Asked in: 2081 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/rsf-b81 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6 ## Question Design a rapid sand filter for treating 25 MLD of water. Also draw a proper sketch showing the dimensions of RSF. ## 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. ## Answer #### Design data (assumed) - Flow $Q=25$ MLD $=25\,000$ m³/day - Rate of filtration 5 m/h $=120$ m³/m²/day (range 3 to 6 m/h) - 4 % of the filtered water is used for backwashing - Operation 24 h/day; $L:B=1.25$ (1.25 to 1.33) #### Step 1: Design flow and total filter area $$ Q'=25\,000\times1.04=26000\ \text{m}^3/\text{day}=1083.3\ \text{m}^3/\text{h} $$ $$ A_{total}=\frac{Q'}{\text{rate}}=\frac{1083.3}{5}=216.7\ \text{m}^2 $$ #### Step 2: Number and size of units $N=1.22\sqrt{Q}$ (Q in MLD) $=1.22\sqrt{25}=6.1$. Adopt **6 units** (plus 1 standby). $$ A_{unit}=\frac{216.7}{6}=36.11\ \text{m}^2,\qquad B=\sqrt{\frac{36.11}{1.25}}=5.37\ \text{m},\quad L=1.25B $$ Adopt $B=5.4$ m, $L=6.7$ m ($A=36.18$ m²). Check: actual rate $=1083.3/(6\times36.18)=4.99$ 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 | Layer | Depth (m) | |---|---| | Free board | 0.5 | | Water above sand | 1.5 | | 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 space | 0.3 | | **Total** | **3.45** | #### Step 4: Under-drainage - Perforated area $=0.3\%$ of bed area $=0.003\times36.18=0.109$ m². With 12 mm orifices (area $1.13\times10^{-4}$ 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 $\approx2\times$ orifice area $=0.217$ m²; manifold area $\approx2\times$ lateral area $=0.434$ m² (diameter about 0.74 m). #### Step 5: Backwash Rate 0.6 m/min (36 m/h) for 10 min: flow $=0.6\times36.18/60=0.362$ m³/s; volume per wash $=217$ m³ per unit, which is about 5 % of a unit's daily output ($26\,000/6=4333$ m³/day). Wash-water troughs are provided above the sand. ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the design flow taken as 1.04 × 25 000 m^3/day? A. 4 % of the filtered water is used for backwashing B. 4 % is lost to evaporation C. It allows for 4 % population growth D. It is a safety factor for filter rate Answer: A. The filters must produce the supply plus the wash water, so Q' = 1.04Q. (Key point: Design flow) 2. How is the total filter area obtained? A. A = Q'/ (wash rate) B. A = Q'/rate of filtration = 1083.3/5 = 216.7 m^2 C. A = Q' × rate D. A = Q/ (L:B) Answer: B. Area equals design flow per hour divided by the filtration rate in m/h. (Key point: Total area) 3. How many filter units are adopted and how is it estimated? A. 25 units from Q in MLD B. 2 units from N = Q/12 C. 6 units from N = 1.22 sqrt(Q in MLD), plus 1 standby D. 4 units from N = L/B Answer: C. N = 1.22 sqrt(25) = 6.1, so 6 units are adopted. (Key point: Number of units) 4. Why is the rate checked with one unit out of service? A. The rate falls to 4 m/h B. The wash water increases the area C. The rate must exceed 10 m/h D. The remaining 5 units carry the full flow at about 5.99 m/h, near the 6 m/h upper limit Answer: D. When one unit is being washed the flow is shared by N-1 units, so the rate rises; it must stay within the allowed range. 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