---
title: "Settling tank 20 m long, 3.5 m deep; 0.02 and 0.01 mm"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-b79"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-b79
site: BCT Engineering
---

# Settling tank 20 m long, 3.5 m deep; 0.02 and 0.01 mm

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2079 Bhadra, 8 marks
- 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-b79
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

In a continuous flow, settling tank 20 m long and 3.5 m deep, what detention time would you recommend for effective removal of 0.02 mm particles at 25°C? Assume specific gravity of particles = 2.65. Also determine the percentage of 0.01 mm particles removed in the same tank at 15°C.

## Short answer

Stokes' law vs for 0.02 mm at 25°C = 0.0403 cm/s. Detention time = depth/vs = 350/0.0403 = 2.41 h. Overflow rate v0 = 0.0403 cm/s. At 15°C, vs for 0.01 mm = 0.00790 cm/s; removal = vs/v0 = 19.6%.

## Answer

A continuous-flow ideal settling tank removes a particle completely if it settles through the full depth $H$ during the detention time $t$, i.e. $v_s \ge H/t$ (equivalently $v_s\ge$ overflow rate $v_0$). So the detention time for 100 % removal is $t = H/v_s$.

#### (a) Detention time for 0.02 mm at 25 °C
Take $S_s = 2.65$, $\nu_{25} = 0.00893$ cm²/s, $d=0.002$ cm, $H = 3.5$ m $=350$ cm.

$$
v_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.002^2}{18\times0.00893}=0.0403\ \text{cm/s}
$$
$Re = v_sd/\nu = 0.009<1$ (Stokes' law valid).

$$
t=\frac{H}{v_s}=\frac{350}{0.0403}=8689\ \text{s} = 2.41\ \text{h}
$$

Recommended detention time $\approx 2.4$ h. (Horizontal velocity $=L/t = 2000/8689 = 0.230$ cm/s, well below the scour velocity.)

#### (b) Removal of 0.01 mm particles at 15 °C in the same tank
The tank has overflow rate $v_0 = H/t=0.0403$ cm/s. At 15 °C, $\nu=0.01139$ cm²/s, $d=0.001$ cm:
$$
v_s=\frac{981\times1.65\times0.001^2}{18\times0.01139}=0.00790\ \text{cm/s}\quad (Re=0.0007<1)
$$
$$
\text{Removal}=\frac{v_s}{v_0}=\frac{0.00790}{0.0403}=0.196\Rightarrow 19.6\ \%
$$

**Answer: detention time = 2.41 h; removal of 0.01 mm particles at 15 °C = 19.6 %.**

## Check yourself

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

1. What is the detention time for complete removal of a particle in a continuous-flow ideal tank?
   A. t = H x vs
   B. t = H/vs
   C. t = L/vs
   D. t = vs/H
   Answer: B. The particle must settle the full depth H during t, so t = H/vs. (Key point: Detention time criterion)

2. With vs = 0.0403 cm/s and H = 350 cm, what is the detention time?
   A. About 2.4 h
   B. About 0.24 h
   C. About 24 h
   D. About 8689 min
   Answer: A. t = 350/0.0403 = 8689 s = 2.41 h. (Key point: Detention time result)

3. Why is the settling velocity of the 0.01 mm particles low at 15 C compared with 25 C?
   A. Kinematic viscosity is higher at 15 C
   B. Water density is lower at 15 C
   C. Particle size is smaller at 15 C
   D. Re exceeds 1
   Answer: A. vs is proportional to 1/nu, and nu is larger at 15 C (0.01139 vs 0.00893 cm2/s). (Key point: Temperature effect)

4. How is the removal of the 0.01 mm particles found?
   A. vs/v0 with v0 = H/t of the existing tank
   B. v0/vs
   C. vs x t
   D. H/vs
   Answer: A. The tank's overflow rate is fixed at 0.0403 cm/s, so removal = 0.00790/0.0403 = 19.6%. (Key point: Percentage removal)

## Close variants set in other exams

- [Settling tank 30 m x 3 m, 0.02 mm at 25°C; 0.01 mm at 10°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-as70) — Water Supply Engineering, chapter 6, asked 3×
- [Settling tank 30 m x 3 m, 0.03 mm at 25°C; 0.025 mm at 20°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-k72) — Water Supply Engineering, chapter 6, asked 3×

## Related in Water Supply Engineering

- [Settling tank 20 m x 2.5 m; velocity for 0.02 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b66) — Water Supply Engineering, chapter 6, asked 2×
- [Settling tank 30 m x 3 m, 0.02 mm, SG 2.50](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk80) — Water Supply Engineering, chapter 6
- [Settling tank 30 m x 3 m; velocity for 0.02 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk68) — Water Supply Engineering, chapter 6, asked 2×
- [Settling tank 15 x 4 x 2.5 m, 96% of 0.015 mm](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-k78) — Water Supply Engineering, chapter 6
- [Settling tank 15 m x 7 m x 3 m, 95% removal](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-b81) — Water Supply Engineering, chapter 6

## Related in other subjects

- [Continuous flow sedimentation tank particle size](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-continuous-4m-60m) — Sanitary Engineering, chapter 7