---
title: "Slow sand filter, 10000 pop, 1.5% growth"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:ssf-a74"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/ssf-a74
site: BCT Engineering
---

# Slow sand filter, 10000 pop, 1.5% growth

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2074 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/ssf-a74
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

A town with survey year population of 10000 and a growth rate of 1.5% per annum has a base period of 5 years, design period of 15 years and average water consumption rate of 150 lpcd. Taking length as twice of its width, propose number, length and width of a slow sand filter with filtration rate of 150 l/m$^2$/day to treat water in this average flow rate and sketch also.

## Short answer

Design population13,469 after20 years, flow2020 m³/d at0.15 m³/m²/h. Provide6 working plus1 standby units, each6.9 m ×13.8 m (95.2 m²), total depth2.7 m (free board0.3 m, water1.0 m, sand0.9 m, gravel0.3 m, underdrain0.2 m).

## Answer

The filtration rate is printed as 150 L/m²/day (0.15 m/day). That would need about 2,020/0.15 ≈ 13,500 m² of filter, which is unrealistic (slow sand filters work at about 0.1–0.2 m/h). It is therefore read as 150 L/m²/h (= 0.15 m³/m²/h).

#### Step 1: Design flow
- Population forecast (geometric increase): $P_n = P_0(1+r)^n = 10,000\times(1+0.015)^{20} = 13,469$ (base period 5 + design period 15 = 20 years from the survey year)
- Daily demand $Q = 13,469 \times 150/1000 = 2,020.3$ m³/day = 84.18 m³/h

#### Step 2: Area and number of units
- Filtration rate = 0.15 m³/m²/h (150 L/m²/h) (range for slow sand filters 0.1–0.4)
- Total area $A = Q/(\text{rate}) = 84.18/0.15 = 561.2$ m²
- Keep each unit below about 100 m² and use at least 2 working units: working units **n = 6**, plus **1 standby** unit for cleaning, so 7 units in all
- Area of one unit $= 561.2/6 = 93.5$ m²
- Take $L = 2B$: $B = \sqrt{93.5/2} = 6.84$ m. Adopt **B = 6.9 m, L = 13.8 m** (area 95.2 m² each)
- Check: rate with 6 working units = 0.147 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: 6 working + 1 standby slow sand filters, each 6.9 m × 13.8 m (95.2 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. How many years are used for the forecast?
   A. 20, base period 5 plus design period 15
   B. 15 only
   C. 5 only
   D. 10
   Answer: A. The population is projected from the survey year across both periods. (Key point: Design years)

2. What is the design demand?
   A. About 1500 m^3/day
   B. About 20 m^3/day
   C. About 2020 m^3/day (13 469 × 150 L)
   D. About 200 000 m^3/day
   Answer: C. Q = 13 469 × 150/1000 = 2020 m^3/day. (Key point: Demand)

3. How many units result?
   A. 1 unit of 561 m^2
   B. 2 working and none standby
   C. 20 working
   D. 6 working + 1 standby, each 6.9 m × 13.8 m
   Answer: D. Area 561.2 m^2/6 = 93.5 m^2 each with L = 2B. (Key point: Units)

## Related in Water Supply Engineering

- [Slow sand filter, 10000 population, base year 2068](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Assf-bk68) — Water Supply Engineering, chapter 6
- [Slow sand filter, 15,000 population, 150 L/h/m2](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Assf-a75) — Water Supply Engineering, chapter 6
- [Slow sand filter, 15000 present population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Assf-c76) — Water Supply Engineering, chapter 6
- [Population in survey year from design demand 800 m3](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apop-s73) — Water Supply Engineering, chapter 3, asked 2×
- [Slow sand filter for 3500 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Assf-c70o) — Water Supply Engineering, chapter 6
- [Average demand, village survey 2013, design 25 years](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ademand-c73) — Water Supply Engineering, chapter 3, asked 2×

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