---
title: "Overflow rate 4000 L/m2/h at 10°C"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:sed-overflow-4000"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/sed-overflow-4000
site: BCT Engineering
---

# Overflow rate 4000 L/m2/h at 10°C

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2074 Asoj, 8 marks; 2073 Chaitra, 8 marks (asked 2 times)
- Video: [Sedimentation tank Solved design dimension of tank | design of sedimentation tank | waste water engg](https://www.youtube.com/watch?v=P3phfraI2-g) — Civil Engineering Exam
- 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-overflow-4000
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

A settling tank is designed for an overflow rate of 4000 liters per m$^2$ per hour. What percentage of particles of diameter (a) 0.05 mm (b) 0.02 mm, will be removed in this tank at 10°C?

## Short answer

At 10°C, vo=0.1111 cm/s, Ss=2.65. By Stokes’ law, 0.05 mm vs=0.1721 cm/s (100% removed); 0.02 mm vs=0.0275 cm/s, removal = 24.8%.

## Answer

Specific gravity of the particles is not given, so the usual value for silt/sand, $S_s = 2.65$, is assumed. Particles whose settling velocity $v_s$ is less than the overflow rate $v_0$ are removed in the proportion $v_s/v_0$; those with $v_s \ge v_0$ are removed completely.

#### Data
- Overflow rate $v_0 = 4000$ l/m²/h $= 4$ m³/m²/h
- $v_0 = \dfrac{4\times100}{3600} = 0.1111$ cm/s
- Temperature 10 °C: $\nu = 0.01306$ cm²/s (kinematic viscosity of water at 10 °C), $g = 981$ cm/s²

#### Settling velocity (Stokes' law)
$$
v_s = \frac{g\,(S_s-1)\,d^2}{18\,\nu}
$$

**(a) d = 0.05 mm = 0.005 cm**
$$
v_s = \frac{981\times1.65\times0.005^2}{18\times0.01306} = 0.1721\ \text{cm/s}
$$
$Re = v_s d/\nu = 0.066 < 1$, so Stokes' law applies. Since $v_s = 0.1721$ > $v_0 = 0.1111$ cm/s, the removal is **100 %**.

**(b) d = 0.02 mm = 0.002 cm**
$$
v_s = \frac{981\times1.65\times0.002^2}{18\times0.01306} = 0.0275\ \text{cm/s}
$$
$Re = 0.0042 < 1$ (valid). Here $v_s < v_0$:
$$
\text{Removal} = \frac{v_s}{v_0} = \frac{0.0275}{0.1111} = 0.248 = 24.8\ \%
$$

**Answer: (a) 0.05 mm particles: 100 % removed; (b) 0.02 mm particles: 24.8 % removed.**

## Check yourself

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

1. What is v0 for an overflow rate of 4000 l/m2/h?
   A. 0.0111 cm/s
   B. 1.111 cm/s
   C. 0.1111 cm/s
   D. 0.4 cm/s
   Answer: C. 4 m3/m2/h x 100 cm/m / 3600 s/h = 0.1111 cm/s. (Key point: Overflow velocity conversion)

2. What decides whether a particle is fully removed in the ideal tank?
   A. Its settling velocity compared with v0: vs >= v0 gives 100 %
   B. Its colour
   C. Only the tank depth
   D. Its density only
   Answer: A. A particle with vs at least v0 reaches the floor before leaving the tank. (Key point: Removal rule)

3. At 10 C, vs for 0.05 mm is 0.1721 cm/s. What is the removal?
   A. 64.6 %
   B. 100 %
   C. 155 %
   D. 25 %
   Answer: B. 0.1721 is greater than v0 = 0.1111, so removal is 100 %. (Key point: Result for 0.05 mm)

4. For 0.02 mm at 10 C, vs = 0.0275 cm/s. What is the removal?
   A. 100 %
   B. 50 %
   C. about 24.8 %, from vs/v0
   D. about 4 %
   Answer: C. 0.0275/0.1111 = 0.248. (Key point: Result for 0.02 mm)

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