---
title: "Sedimentation tank 6 m x 18 m x 3 m, 4 ML/day"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-a76"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-a76
site: BCT Engineering
---

# Sedimentation tank 6 m x 18 m x 3 m, 4 ML/day

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2076 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/sed-a76
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

In a sedimentation tank (dimensions: 6 m wide, 18 m long and 3 m depth), 4 million litres of water passes per day. Calculate (a) detention period and (b) surface overflow rate. Check the values with standard range.

## Short answer

Detention period is 1.94 h and surface overflow rate is 37.04 m³/m²/day. The tank is acceptable with coagulation, but for plain sedimentation the detention time is slightly low and the overflow rate is high.

## Answer

#### Data
Width $B=6$ m, length $L=18$ m, depth $H=3$ m, flow $Q=4$ ML/day $=4000$ m³/day.

#### (a) Detention period
$$
\begin{aligned} \text{Volume} &= 18\times6\times3 = 324\ \text{m}^3 \\ t &= \frac{V}{Q} = \frac{324}{4000} = 0.0810\ \text{day} = 1.94\ \text{h} \end{aligned}
$$

#### (b) Surface overflow rate
$$
\text{SOR} = \frac{Q}{A} = \frac{4000}{18\times6} = 37.04\ \text{m}^3/\text{m}^2/\text{day}
$$

#### Check with the standard range
Horizontal velocity $=\dfrac{Q}{B\,H}=\dfrac{4000}{6\times3}=222.2$ m/day $=0.154$ m/min (the limit is 0.3 m/min; OK).

| Item | Computed | Usual range (plain sedimentation) | Remark |
|---|---|---|---|
| Detention period | 1.94 h | 2 to 4 h (up to 6 h for plain tanks) | Just below 2 h, slightly low |
| SOR | 37.0 m³/m²/day | 15 to 30 m³/m²/day (plain); 30 to 40 (with coagulation) | Above the plain range, within that for coagulated water |
| Velocity | 0.154 m/min | below 0.3 m/min | Satisfactory |
| Depth | 3 m | 2.5 to 5 m | Satisfactory |
| L : B | 3 : 1 | 3:1 to 5:1 | Satisfactory |

**Answer: detention period = 1.94 h; SOR = 37.04 m³/m²/day. The tank is satisfactory only if coagulation is used; for plain sedimentation the tank is slightly small (detention a little low and SOR high), so the area should be increased.**

## Check yourself

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

1. How is the detention period found?
   A. Volume divided by flow
   B. Flow divided by volume
   C. Area divided by flow
   D. Length divided by velocity of settling
   Answer: A. t = V/Q = 324/4000 = 0.081 day = 1.94 h. (Key point: Detention period)

2. What is the SOR of the 18 m x 6 m tank for 4000 m3/day?
   A. 37.04 m3/m2/day
   B. 12.35 m3/m2/day
   C. 222 m3/m2/day
   D. 74 m3/m2/day
   Answer: A. SOR = Q/A = 4000/108 = 37.04. (Key point: Surface overflow rate)

3. How is the horizontal velocity checked against the 0.3 m/min limit?
   A. Q/(B H)
   B. Q/(L B)
   C. V/Q
   D. SOR x H
   Answer: A. Velocity = Q/(B H) = 0.154 m/min, which is below 0.3. (Key point: Horizontal velocity)

4. What is the conclusion about the tank?
   A. Satisfactory only with coagulation; for plain sedimentation it is slightly small
   B. Fully satisfactory for plain sedimentation
   C. Far too large
   D. Unsafe because velocity exceeds 0.3 m/min
   Answer: A. SOR 37 is above 15 to 30 for plain tanks but within 30 to 40 for coagulated water. (Key point: Conclusion on adequacy)

## Related in Water Supply Engineering

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- [Rectangular sedimentation tank, 9 million L peak demand](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c75) — Water Supply Engineering, chapter 6
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- [Storage tank capacity and water level, 20000 L/day](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Areservoir-level-river-20000) — Water Supply Engineering, chapter 7, asked 2×
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