---
title: "Rectangular sedimentation tank, 9 million L peak demand"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-c75"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-c75
site: BCT Engineering
---

# Rectangular sedimentation tank, 9 million L peak demand

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2075 Chaitra, 8 marks
- Video: [Numerical on designing of sedimentation tank](https://www.youtube.com/watch?v=DsOIWRxw-lA) — Ritiksha Danu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-c75
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

A water treatment plant has to purify water for a town with daily peak demand of 9 million liters. Design a rectangular sedimentation tank assuming the velocity of flow as 20 cm/min and detention period of 4 hours.

## Short answer

Tank dimensions are 48 m long by 9 m wide with 3.5 m effective depth and total depth 4.5 m. Surface overflow rate is 20.8 m³/m²/day.

## Answer

#### Data
Peak demand $Q=9\times10^6$ l/day $=9000$ m³/day; velocity of flow $v=20$ cm/min $=0.2$ m/min; detention period $t=4$ h $=240$ min.

#### Step 1: Length of the tank
$$
L=v\,t=0.2\times240=48\ \text{m}
$$

#### Step 2: Cross-sectional area
$$
Q=\frac{9000}{24\times60}=6.250\ \text{m}^3/\text{min},\qquad A_x=\frac{Q}{v}=\frac{6.250}{0.2}=31.25\ \text{m}^2
$$

#### Step 3: Depth and width
Assume effective depth $H=3.5$ m (usual 3 to 4 m):
$$
B=\frac{A_x}{H}=\frac{31.25}{3.5}=8.93\ \text{m}\approx9\ \text{m}
$$

#### Step 4: Checks
$$
V=Q\,t=\frac{9000}{24}\times4=1500\ \text{m}^3\ (\text{tank volume}=48\times9\times3.5=1512\ \text{m}^3)
$$
$$
\text{SOR}=\frac{9000}{48\times9}=20.8\ \text{m}^3/\text{m}^2/\text{day}\ (\text{within }15\text{ to }30)
$$
$L:B=48/9=5.3:1$, slightly above the usual 3:1 to 5:1 range. This is acceptable for a long horizontal-flow tank; if a ratio within 5:1 is wanted, take $H=3.2$ m, giving $B=31.25/3.2=9.8$ m and $L:B=4.9:1$. Add 0.5 m free board and a 0.5 m sludge zone: total depth 4.5 m.

```text
 PLAN                                       ----> flow
  +-------+------------------------------+--------+
  | inlet |           L = 48 m           | outlet |  B = 9 m
  +-------+------------------------------+--------+
 SECTION: H = 3.5 m effective + 0.5 m free board + 0.5 m sludge zone = 4.5 m
```

**Answer: tank 48 m long $\times$ 9 m wide $\times$ 3.5 m effective depth (4.5 m total). SOR = 20.8 m³/m²/day.**

## Check yourself

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

1. How is the length of the tank found from velocity and detention time?
   A. L = v t
   B. L = Q/v
   C. L = v/t
   D. L = Q t
   Answer: A. L = 0.2 m/min x 240 min = 48 m. (Key point: Tank length)

2. How is the cross-sectional area found?
   A. Flow rate divided by velocity
   B. Flow rate times velocity
   C. Volume divided by length x depth
   D. Flow divided by area
   Answer: A. A_x = Q/v = 6.25/0.2 = 31.25 m2. (Key point: Cross-section)

3. With H = 3.5 m, what width results?
   A. About 9 m
   B. About 3.5 m
   C. About 18 m
   D. About 5 m
   Answer: A. B = 31.25/3.5 = 8.93 m, adopted as 9 m. (Key point: Width)

4. What is the SOR of the 48 m x 9 m tank for 9000 m3/day?
   A. 20.8 m3/m2/day
   B. 208 m3/m2/day
   C. 2.08 m3/m2/day
   D. 62.5 m3/m2/day
   Answer: A. SOR = 9000/432 = 20.8, within 15 to 30. (Key point: SOR check)

## Related in Water Supply Engineering

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## Related in other subjects

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