--- title: "Rapid sand filter, 45,000 population in 2021" subject: "Water Supply Engineering" chapter: 6 key: "semester_5/water_supply_engineering:rsf-a76" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/rsf-a76 site: BCT Engineering --- # Rapid sand filter, 45,000 population in 2021 Water Supply Engineering · Chapter 6: Water Treatment - Asked in: 2076 Asoj, 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-a76 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6 ## Question Design a rapid sand filter for the population of 45,000 in the year 2021. Water demand is 110 lpcd, annual population growth rate is 2.1%. ## Short answer 4 rapid sand filters, each 3.8 m × 4.7 m, 3.4 m deep. Design flow 7651 m³/day, rate 4.5 m/h. Sand 0.7 m, gravel 0.45 m; underdrain with 550 mm manifold and 30 laterals of 75 mm. ## Answer A rapid sand filter is designed for the population forecast at the end of the design period. The design period is not stated, so 20 years is assumed (2021 to 2041). #### Step 1: Design flow - Population forecast (geometric increase): $P_n = P_0(1+r)^n = 45,000\times(1+0.021)^{20} = 68,191$ (design year 2041) - Average daily demand $Q = 68,191 \times 110/1000 = 7,501$ m³/day - Add 2 % for wash water: $Q' = 1.02 \times 7,501 = 7,651$ m³/day, i.e. 318.8 m³/h (plant works 24 h/day) #### Step 2: Plan area and number of beds - Filtration rate adopted = 4.5 m³/m²/h (4500 L/h/m², usual range 3000 to 6000) - Total area $A = 318.8/4.5 = 70.8$ m² - Number of beds $N = 1.22\sqrt{Q_{MLD}} = 1.22\sqrt{7.50} = 3.34$; adopt **4 beds**, so that one can be washed while the other three work - Area of one bed $= 70.8/4 = 17.71$ m² - Take $L/B = 1.25$: $B = \sqrt{17.71/1.25} = 3.76$ m. Adopt **B = 3.8 m, L = 4.7 m** (area 17.86 m² each, total 71.44 m²) - Actual rate with all beds working = 4.46 m/h; with one bed under wash = 5.95 m/h (within 3 to 6 m/h, OK) #### Step 3: Filter media (sand and gravel) | Layer | Size | Depth | |---|---|---| | Filter sand | effective size 0.5 mm, UC ≤ 1.5 | 700 mm | | Gravel layer 4 (top) | 3–6 mm | 100 mm | | Gravel layer 3 | 6–12 mm | 100 mm | | Gravel layer 2 | 12–20 mm | 100 mm | | Gravel layer 1 (bottom) | 20–40 mm | 150 mm | The gravel (base material) is 450 mm deep in total. It supports the sand, stops it entering the laterals and spreads the wash water evenly. #### Step 4: Underdrainage (perforated manifold and laterals) - Backwash rate = 0.6 m³/min/m² (0.01 m/s) for 10 min, so the wash flow per bed $= 0.01 \times 17.86 = 0.179$ m³/s - Total perforation area = 0.3 % of bed area $= 0.003 \times 17.86 = 0.054$ m² - Perforation diameter 12 mm (area 1.131 cm²): number of holes $= 474$ per bed - Manifold along the 4.7 m length, with laterals at 0.3 m c/c on both sides: $4.7/0.3 \approx 15$ per side, i.e. **30 laterals**, about 16 holes each - Total area of laterals = 2 × perforation area $= 0.107$ m², so each lateral $= 0.107/30 = 0.00357$ m², $D = 67$ mm, adopt **75 mm**; length $\approx (3.8-0.55)/2 = 1.6$ m (L/D = 22, below 60, OK) - Area of manifold = 2 × area of laterals $= 0.214$ m², so $D = 522$ mm, adopt **550 mm** (area 0.238 m²) - Wash velocity in manifold $= 0.179/0.238 = 0.75$ m/s (< 2.5 m/s, OK) #### Step 5: Depth of the bed Free board 0.5 m + water depth over sand 1.5 m + sand 0.7 m + gravel 0.45 m + underdrain space 0.25 m = **3.4 m**. ```text Inlet -> | Free board 0.5 m | |~~~ water 1.5 m ~~~~~~~~~~|--> wash-water |__________________________| trough | Sand 0.7 m | | Gravel 0.45 m | |___o_o_o_laterals_o_o_o___| |====== manifold ==========|--> filtered (3.4 m total depth) water ``` **Answer: 4 rapid sand filter beds, each 3.8 m × 4.7 m and 3.4 m deep, filtration rate 4.5 m/h, 0.7 m of sand on 0.45 m of graded gravel, with a perforated manifold–lateral underdrain (550 mm manifold, 30 laterals of 75 mm).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the design population found when the design period is not given? A. Arithmetic increase with 10 years B. Present population only C. Geometric increase P0(1+r)^n with 20 years assumed D. Incremental increase with 5 years Answer: C. The filter is designed for the end of the design period, here 45 000 × 1.021^20 = 68 191. (Key point: Design population) 2. What addition is made to the average demand before sizing the beds? A. 2 % for wash water B. 25 % for fire C. 50 % for peak D. None Answer: A. Q' = 1.02 × Q so that backwash water is available. (Key point: Wash water) 3. Why are 4 beds adopted when the formula gives 3.34? A. Four makes L/B equal to 1 B. Four is the minimum so that one can be washed while the rest work C. Beds must be an even number D. 3.34 rounds to 4 by law only Answer: B. A minimum of four beds is used to keep the rate on the others within limits during wash. (Key point: Number of beds) 4. What is the function of the graded gravel layer? A. It removes bacteria B. It adds colour removal C. It reduces head loss only D. It supports the sand, stops sand entering laterals and spreads wash water evenly Answer: D. Gravel is a base material, not the filtering layer; it protects the under-drain and distributes backwash. 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