---
title: "Rapid sand filter design, 50,000 population"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:rsf-c72"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/rsf-c72
site: BCT Engineering
---

# Rapid sand filter design, 50,000 population

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2072 Chaitra, 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-c72
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

The population of a city is 50,000. Design a rapid sand filter including design of filter media, base material and underdrainage system.

## Short answer

Design flow 6885 m³/d. Four beds, each 3.6 m × 4.4 m, total depth 3.4 m. Filter media: 0.7 m sand (effective size 0.5 mm, UC ≤1.5) over 0.45 m graded gravel (3–40 mm). Underdrain: 500 mm manifold with 28 laterals of 75 mm, perforated with 421 holes (12 mm dia) per bed.

## Answer

Population 50,000 is taken as the design population. Per capita demand is assumed to be 135 lpcd (urban supply as per Nepal Water Supply Design Guidelines).

#### Step 1: Design flow
- Design population = 50,000 (given)
- Average daily demand $Q = 50,000 \times 135/1000 = 6,750$ m³/day
- Add 2% for wash water: $Q' = 1.02 \times 6,750 = 6,885$ m³/day, i.e. 286.9 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 3000–6000)
- Total area $A = 286.9/4.5 = 63.8$ m²
- Number of beds $N = 1.22\sqrt{Q_{MLD}} = 1.22\sqrt{6.75}$ = 3.17, adopt **4 beds** (minimum 4 so that one can be washed while the rest work)
- Area of one bed $= 63.8/4 = 15.94$ m²
- Take $L/B = 1.25$: $B = \sqrt{15.94/1.25} = 3.57$ m. Adopt **B = 3.6 m, L = 4.4 m** (area 15.84 m² each, total 63.36 m²)
- Actual rate with all beds working = 4.53 m/h; with one bed under wash = 6.04 m/h (about 6 m/h, acceptable for the short washing period)

#### 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 |

Gravel (base material) is 450 mm 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 wash flow per bed $= 0.01 \times 15.84 = 0.158$ m³/s
- Total perforation area = 0.3% of bed area $= 0.003 \times 15.84 = 0.048$ m²
- Perforation diameter 12 mm (area 1.131 cm²): number of holes $= 421$ per bed
- Manifold along the 4.4 m length, laterals at 0.3 m c/c on both sides: $4.4/0.3 \approx 14$ per side, i.e. **28 laterals**, about 15 holes each
- Area of laterals = 2 × perforation area $= 0.095$ m², i.e. $0.095/28 = 0.0034$ m² each, so $D = 66$ mm, adopt **75 mm**; length $\approx (3.6 - 0.5)/2 = 1.55$ m (L/D ≈ 21, below 60, OK)
- Area of manifold = 2 × area of laterals $= 0.190$ m², so $D = 492$ mm, adopt **500 mm**
- Wash velocity in manifold $= 0.158/0.190 = 0.83$ 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 3.6 m × 4.4 m, 3.4 m deep, filtration rate 4.5 m/h, sand 0.7 m on 0.45 m graded gravel, with a perforated underdrain of a 500 mm manifold and 28 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 flow found for 50 000 people at 135 lpcd?
   A. Q = 675 m^3/day
   B. Q = 6750 m^3/day, plus 2 % for wash water
   C. Q = 6750 m^3/day, minus 2 %
   D. Q = 50 000 × 135 m^3/day
   Answer: B. 50 000 × 135 L = 6750 m^3/day, and 2 % is added to get 6885. (Key point: Design flow)

2. With 4 beds and a total area 63.8 m^2, what is the area of one bed?
   A. 15.94 m^2
   B. 63.8 m^2
   C. 255 m^2
   D. 3.5 m^2
   Answer: A. Total area is divided by the number of beds. (Key point: Area per bed)

3. How is the total perforation area decided?
   A. 30 % of bed area
   B. Equal to manifold area
   C. 0.3 % of the bed area
   D. Equal to sand volume
   Answer: C. Perforations of 12 mm total 0.3 % of bed area, giving about 421 holes. (Key point: Underdrain)

4. How is the manifold area related to the laterals?
   A. Manifold area equals perforation area
   B. Manifold is half the lateral area
   C. Manifold area is 10 × lateral area
   D. Manifold area is 2 × lateral area, and lateral area is 2 × perforation area
   Answer: D. Each stage is made larger than the one it feeds so wash velocity stays under 2.5 m/s. (Key point: Manifold sizing)

## Related in Water Supply Engineering

- [Rapid sand filter, 45,000 population in 2021](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Arsf-a76) — Water Supply Engineering, chapter 6
- [Rapid sand filter for 60000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Arsf-c69) — Water Supply Engineering, chapter 6
- [Rapid sand filter construction and working](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Arsf-construction-working) — Water Supply Engineering, chapter 6, asked 5×
- [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
- [Rapid sand filter design for 25 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Arsf-b81) — 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

## Related in other subjects

- [Conventional ASP design, population 50,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-50000-80lpcd) — Sanitary Engineering, chapter 7, asked 2×
- [Conventional ASP design, population 1,00,000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Aasp-100000-150lpcd) — Sanitary Engineering, chapter 7
- [Time of critical DO deficit, 50 l/s city](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Acritical-do-time-city-50ls) — Sanitary Engineering, chapter 6, asked 2×