---
title: "Number and size of RSF, 1,20,000 population"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:rsf-b78"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/rsf-b78
site: BCT Engineering
---

# Number and size of RSF, 1,20,000 population

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2078 Bhadra, 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-b78
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Propose the number and dimensions of a rapid sand filter for a town having population of 1,20,000 numbers. Assume average water supply rate as 120 lpcd with a filtration rate of 3000 lph/m$^2$ and length width ratio as 1.5.

## Short answer

5 rapid sand filters, each 7.8 m long × 5.2 m wide (40.6 m²). Filtration rate 3 m/h. Depth: freeboard 0.5 m, water 1.5 m, sand 0.7 m, gravel 0.45 m, underdrain 0.3 m; total 3.45 m.

## Answer

#### Data
Population 1 20 000; 120 lpcd; filtration rate 3000 l/h/m² $=3$ m/h; $L:B=1.5$. Filters work 24 h/day (wash water is neglected, since the average supply rate is given).

#### Step 1: Flow
$$
Q=120\,000\times120=14\,400\,000\ \text{l/day}=14400\ \text{m}^3/\text{day}=600\ \text{m}^3/\text{h}
$$

#### Step 2: Total filter area
$$
A=\frac{Q}{\text{rate}}=\frac{600}{3}=200\ \text{m}^2
$$

#### Step 3: Number of units
$N=1.22\sqrt{Q}$ (Q in MLD) $=1.22\sqrt{14.4}=4.63$, so adopt **5 units**.

#### Step 4: Size of each unit
$$
A_{unit}=\frac{200}{5}=40\ \text{m}^2,\qquad 1.5B^2=40\Rightarrow B=5.16\ \text{m},\ L=1.5B=7.75\ \text{m}
$$
Adopt $B=5.2$ m, $L=7.8$ m ($A=40.56$ m²).

#### Step 5: Check
Actual rate $=600/(5\times40.56)=2.96$ m/h. With one unit under backwash, the rate on the remaining 4 units is 3.70 m/h (acceptable).

Depth: free board 0.5 + water 1.5 + sand 0.7 + gravel 0.45 + under-drain 0.3 = 3.45 m.

**Answer: provide 5 rapid sand filters, each 7.8 m long $\times$ 5.2 m wide (area 40.6 m²), total area 203 m².**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the total filter area for 14 400 m^3/day at 3 m/h?
   A. 600 m^2
   B. 200 m^2 (600 m^3/h divided by 3)
   C. 50 m^2
   D. 1800 m^2
   Answer: B. Q = 14 400/24 = 600 m^3/h, and A = 600/3 = 200 m^2. (Key point: Flow and area)

2. How is the number of filter units chosen?
   A. N = 1.22 sqrt(14.4) = 4.63, adopt 5
   B. N = 14.4/3 = 5 exactly
   C. N = 1.5 × 4 = 6
   D. N = 200/1.5
   Answer: A. The empirical formula N = 1.22 sqrt(Q in MLD) gives 4.63, rounded up to 5. (Key point: Number of units)

3. With L:B = 1.5 and 40 m^2 per unit, the width is
   A. B = 40/1.5 = 26.7 m
   B. B = sqrt(40 × 1.5) = 7.7 m
   C. B = sqrt(40/1.5) = 5.16 m, L = 7.75 m
   D. B = 40/5 = 8 m
   Answer: C. A = 1.5B^2 gives B = 5.16 m and L = 1.5B. (Key point: Unit dimensions)

4. What is the rate on the remaining units while one is washed?
   A. About 2.96 m/h
   B. About 3.70 m/h on 4 units, which is acceptable
   C. About 6.0 m/h
   D. About 1.5 m/h
   Answer: B. 600/(4 × 40.56) = 3.70 m/h, still within the allowed rate range. (Key point: Check with one unit washed)

## Related in Water Supply Engineering

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- [Rapid sand filter design, 50,000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Arsf-c72) — Water Supply Engineering, chapter 6
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- [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, 10000 population, base year 2068](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Assf-bk68) — Water Supply Engineering, chapter 6

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