--- title: "Slow sand filter, 15000 present population" subject: "Water Supply Engineering" chapter: 6 key: "semester_5/water_supply_engineering:ssf-c76" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/ssf-c76 site: BCT Engineering --- # Slow sand filter, 15000 present population Water Supply Engineering · Chapter 6: Water Treatment - Asked in: 2076 Chaitra, 5+3 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/ssf-c76 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6 ## Question Determine the size of slow sand filter for a present population of 15000 nos, design period = 20 yrs, annual population growth rate = 2.5%, water consumption = 45 lpcd, and also draw the section of slow sand filter. ## Short answer Design population 24,579, flow 1106 m³/d at 0.2 m³/m²/h. Use 3 working plus 1 standby beds, each 6.2 m × 12.4 m (76.9 m²), total depth 2.7 m (free board 0.3 m, water 1.0 m, sand 0.9 m, gravel 0.3 m, underdrain 0.2 m). ## Answer The design population is found for the end of the 20-year design period. A filtration rate of 0.2 m³/m²/h is assumed. #### Step 1: Design flow - Population forecast (geometric increase): $P_n = P_0(1+r)^n = 15,000\times(1+0.025)^{20} = 24,579$ - Daily demand $Q = 24,579 \times 45/1000 = 1,106.1$ m³/day = 46.09 m³/h #### Step 2: Area and number of units - Filtration rate = 0.2 m³/m²/h (range for slow sand filters 0.1–0.4) - Total area $A = Q/(\text{rate}) = 46.09/0.2 = 230.4$ m² - Keep each unit below about 100 m² and use at least 2 working units: working units **n = 3**, plus **1 standby** unit for cleaning, so 4 units in all - Area of one unit $= 230.4/3 = 76.8$ m² - Take $L = 2B$: $B = \sqrt{76.8/2} = 6.20$ m. Adopt **B = 6.2 m, L = 12.4 m** (area 76.9 m² each) - Check: rate with 3 working units = 0.200 m/h (OK) #### Step 3: Depth of unit | Component | Depth | |---|---| | Free board | 0.3 m | | Supernatant water | 1.0 m | | Filter sand (effective size 0.2–0.3 mm, UC 3–5) | 0.9 m | | Gravel in 3 layers (3–6, 6–12, 12–40 mm) | 0.3 m | | Under-drain space (central drain and laterals) | 0.2 m | | **Total** | **2.7 m** | Sand is scraped 15–25 mm every 1–2 months; when its depth falls to 0.5 m it is topped up to 0.9 m. Inlet and outlet chambers, a filtered-water weir, a drain valve and a filter-to-waste connection complete each unit. #### Step 4: Section ```text inlet -> | free board 0.3 m | |~~~~ supernatant water 1.0 m ~~| |-------------------------------| | filter sand 0.9 m | |-------------------------------| | gravel 0.3 m (3 layers) | |==o==o== under-drain 0.2 m ==o=| | outlet weir -> filtered water tank ``` **Answer: 3 working + 1 standby slow sand filters, each 6.2 m × 12.4 m (76.9 m²), total depth about 2.7 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which population is used for the filter? A. The forecast at the end of the 20-year design period, 24 579 B. The present 15 000 C. The population after 5 years D. The population after 40 years Answer: A. The design is for P0(1+r)^20. (Key point: Design population) 2. Slow sand filters work at what range of rate? A. 3 to 6 m^3/m^2/h B. 0.1 to 0.4 m^3/m^2/h C. 10 to 15 m^3/m^2/h D. 0.001 m^3/m^2/h Answer: B. They are slow filters, with 0.2 m/h adopted here. (Key point: Filtration rate) 3. Why is one standby unit provided? A. To increase rate B. To reduce sand depth C. So a unit can be out for cleaning while working units continue D. To serve as sedimentation Answer: C. Slow sand filters need scraping, so a spare unit keeps supply continuous. (Key point: Standby unit) 4. How is clogged slow sand cleaned? A. Backwash at 0.6 m/min B. Add alum C. Replace all sand weekly D. Scrape 15 to 25 mm of top sand and top up when depth falls to 0.5 m Answer: D. Slow sand filters are cleaned by scraping the surface layer, not backwashing. 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