---
title: "Settling tank for 41,600 population"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-c76"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-c76
site: BCT Engineering
---

# Settling tank for 41,600 population

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2076 Chaitra, 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-c76
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

Design a settling tank for a town having design year population of 41,600 numbers with a water supply rate of 120 lpcd. The detention period is expected as 4 hours, length width ratio as 4 and effective depth as 3.5 m. Also check for SOR and velocity. Sketch neat diagram with dimensions as designed.

## Short answer

Designed tank is 31.2 m long, 7.8 m wide, with 3.5 m effective depth and 4.3 m total depth. Surface overflow rate is 20.5 m³/m²/day and horizontal velocity is 0.127 m/min, both within limits.

## Answer

#### Design data and assumptions
Population = 41 600; rate = 120 lpcd; detention period $t = 4$ h; $L:B=4:1$; effective depth $H=3.5$ m.
The design flow is taken as the average daily demand (no peak factor is given).

#### Step 1: Flow
$$
Q=41\,600\times120=4\,992\,000\ \text{l/day}=4992\ \text{m}^3/\text{day}
$$

#### Step 2: Volume and plan area
$$
V=Q\,t=\frac{4992}{24}\times4=832.0\ \text{m}^3
$$
$$
A=\frac{V}{H}=\frac{832.0}{3.5}=237.71\ \text{m}^2
$$

#### Step 3: Length and width
$L=4B$, so $4B^2=237.71$:
$$
B=\sqrt{\frac{237.71}{4}}=7.71\ \text{m},\qquad L=4B=30.84\ \text{m}
$$
Adopt $B=7.8$ m, $L=31.2$ m (area 243.4 m²).

#### Step 4: Checks
$$
t_{actual}=\frac{243.4\times3.5}{4992}\times24=4.09\ \text{h}\ (\approx4\ \text{h, OK})
$$
$$
\text{SOR}=\frac{Q}{A}=\frac{4992}{243.4}=20.5\ \text{m}^3/\text{m}^2/\text{day}\ (15\text{ to }30,\ \text{OK})
$$
$$
\text{Velocity}=\frac{Q}{B\,H}=\frac{4992}{7.8\times3.5\times1440}=0.127\ \text{m/min}\ (<0.3,\ \text{OK})
$$
Add a free board of 0.3 m and a sludge zone of 0.5 m: total depth $=3.5+0.3+0.5=4.3$ m.

#### Sketch
```text
 PLAN (not to scale)
  +-------+---------------------------------+--------+
  | inlet |<-------- L = 31.2 m ----------->| outlet |
  | zone  |         settling zone           |  weir  |  B = 7.8 m
  +-------+---------------------------------+--------+
 SECTION
  inlet --> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ FB 0.3 m
  baffle    |           H = 3.5 m           |--> outlet
            |_______________________________|
             \_______ sludge 0.5 m ________/ -> drain
```

**Answer: tank 31.2 m long $\times$ 7.8 m wide $\times$ 3.5 m effective depth (4.3 m total); SOR = 20.5 m³/m²/day and velocity = 0.127 m/min, both within limits.**

## Check yourself

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

1. What flow is used for design?
   A. Population x lpcd = 4992 m3/day
   B. Half of it
   C. Peak factor 3 times it
   D. 4992 m3/h
   Answer: A. No peak factor is given, so average demand 41600 x 120 l/day is used. (Key point: Design flow)

2. How is the plan area obtained?
   A. A = Q t / H
   B. A = Q/t
   C. A = H/Qt
   D. A = Q x H
   Answer: A. V = Q t = 832 m3, and A = V/H = 237.7 m2. (Key point: Plan area)

3. With L = 4B and A = 237.71 m2, what are B and L?
   A. B = 7.71 m, L = 30.84 m
   B. B = 15.4 m, L = 15.4 m
   C. B = 5.9 m, L = 23.7 m
   D. B = 9.8 m, L = 24.3 m
   Answer: A. 4B^2 = 237.71, so B = 7.71 m and L = 30.84 m. (Key point: Length and width)

4. Which checks are made on the adopted tank?
   A. Detention time, SOR and horizontal velocity
   B. Only cost
   C. Only detention time
   D. Chlorine demand
   Answer: A. t = 4.09 h, SOR = 20.5 m3/m2/day (15 to 30) and velocity = 0.127 m/min (below 0.3). (Key point: Design checks)

## Related in Water Supply Engineering

- [Balancing reservoir: village 600 population, 65 lpcd](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ares-as70) — Water Supply Engineering, chapter 7
- [Rectangular sedimentation tank, 0.2 mm/s, 3 hours](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c71) — Water Supply Engineering, chapter 6, asked 2×
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- [Rectangular sedimentation tank, 20000 population, 0.4 mm/s](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s73) — Water Supply Engineering, chapter 6

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