--- title: "Rapid sand filter for 5000 persons" subject: "Water Supply Engineering" chapter: 6 key: "semester_5/water_supply_engineering:rsf-k78" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/rsf-k78 site: BCT Engineering --- # Rapid sand filter for 5000 persons Water Supply Engineering · Chapter 6: Water Treatment - Asked in: 2078 Kartik, 8 marks - Video: [Design of rapid sand filter](https://www.youtube.com/watch?v=i8TLhz1O17E) — Ambreesh - 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-k78 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6 ## Question Design and draw a neat sketch of a rapid sand filter for a community having 5000 number of persons. Assume necessary data with appropriate values. ## Short answer 2 rapid sand filters, each 2.6 m × 2.1 m, depth 2.85 m. Rate 4.76 m/h when one unit is washed. Sand 0.6 m, gravel 0.45 m, water 1.0 m, freeboard 0.5 m, underdrain 0.3 m. ## Answer #### 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 | Layer | Depth (m) | |---|---| | Free board | 0.5 | | Water above sand | 1.0 | | Sand (ES 0.5 mm, UC 1.5) | 0.6 | | Gravel (graded 2 to 50 mm) | 0.45 | | Under-drain space | 0.3 | | **Total** | **2.85** | 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. ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the flow multiplied by 1.04? A. To allow for population growth B. To provide 4 % wash water from the filtered output C. To allow for 4 % leakage in pipes D. To allow for sludge removal Answer: B. Filtered water is used for backwashing, so production must exceed the supply by the wash water. (Key point: Design flow) 2. Why are two units provided although the formula gives 0.95? A. A minimum of two units lets one be washed while the other works B. Two units cost less than one C. Two gives a better L:B ratio D. The code demands equal flows Answer: A. At least two units are needed so filtering continues during backwash. (Key point: Number of units) 3. How is the backwash volume found? A. Rate × time = 0.6 × 10 = 6 m^3 B. Area × depth = 3.2 × 2.85 C. Rate × area × time = 0.6 × 3.2 × 10 = 19.2 m^3 D. Q' × 24 Answer: C. A backwash rate of 0.6 m/min over 3.2 m^2 for 10 min uses 19.2 m^3. (Key point: Wash water) 4. What fraction of bed area is given to under-drain perforations? A. About 30 % B. About 3 % C. About 0.003 % D. About 0.3 % Answer: D. The total hole area is typically 0.3 % of the bed area, drawn from laterals joined to a manifold. 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