---
title: "Rectangular sedimentation tank, 8x10^6 L/day"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-as67"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-as67
site: BCT Engineering
---

# Rectangular sedimentation tank, 8x10^6 L/day

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2067 Asar (old course)
- Video: [Rectangular sedimentation tank design.](https://www.youtube.com/watch?v=cRtMpplW84I) — Engr Suliman Khan
- 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-as67
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Design a rectangular sedimentation tank for a town to purify the water at a rate of $8\times10^6$ litres per day. Assume velocity of flow as 15 cm/minute and detention period as 5 hrs.

## Short answer

Tank length = v × t = 45 m. Cross‑section area = Q/v = 37.04 m². With 3.5 m effective depth, width = 10.6 m. Provide tank 45 m × 10.6 m × 3.5 m (overall 4.5 m with freeboard and sludge zone).

## Answer

#### Data
$Q=8\times10^6$ l/day $=8000$ m³/day; velocity of flow $v=15$ cm/min $=0.15$ m/min; detention period $t=5$ h $=300$ min.

#### Step 1: Length
$$
L=v\,t=0.15\times300=45\ \text{m}
$$

#### Step 2: Cross-section
$$
Q=\frac{8000}{1440}=5.556\ \text{m}^3/\text{min},\qquad A_x=\frac{Q}{v}=\frac{5.556}{0.15}=37.04\ \text{m}^2
$$

#### Step 3: Depth and width
Assume effective depth $H=3.5$ m:
$$
B=\frac{37.04}{3.5}=10.58\ \text{m}\approx10.6\ \text{m}
$$

#### Step 4: Checks
$$
V=\frac{8000}{24}\times5=1667\ \text{m}^3,\qquad \text{SOR}=\frac{8000}{45\times10.6}=16.8\ \text{m}^3/\text{m}^2/\text{day}\ (\text{within }15\text{ to }30),\qquad L:B=4.2:1
$$
With a 0.5 m free board and 0.5 m sludge zone the overall depth is 4.5 m.

**Answer: tank 45 m long $\times$ 10.6 m wide $\times$ 3.5 m effective depth (4.5 m overall).**

## Check yourself

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

1. How is the length of the rectangular tank found?
   A. L = Q × t = 8000 × 5
   B. L = Q/v = 5.556/0.15
   C. L = velocity of flow × detention time = 0.15 × 300 = 45 m
   D. L = detention time/velocity
   Answer: C. Water travelling at v for time t covers L = vt, with t = 5 h = 300 min. (Key point: Tank length)

2. How is the cross-sectional area found?
   A. Ax = Q/v = 5.556/0.15 = 37.04 m^2
   B. Ax = Q × v
   C. Ax = L × t
   D. Ax = Q/t
   Answer: A. Flow equals area times velocity, so Ax = Q/v with Q in m^3/min and v in m/min. (Key point: Cross-section)

3. After assuming a depth of 3.5 m, the width is found by
   A. B = L/H
   B. B = Ax/H = 37.04/3.5 = 10.6 m
   C. B = Q/L
   D. B = H × Ax
   Answer: B. Width = cross-sectional area divided by depth. (Key point: Width)

4. What does the check SOR = Q/(L × B) = 16.8 m^3/m^2/day show?
   A. The tank is too short
   B. The velocity of flow is too high
   C. The detention time is 16.8 h
   D. The rate is within the accepted 15 to 30 range, so the design is acceptable
   Answer: D. SOR is the discharge per unit plan area and must lie in the usual limits for plain sedimentation. (Key point: Surface overflow rate)

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