---
title: "Super chlorination; overflow rate 5000 L/m2/h"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:super-chlorination-settling-numerical"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/super-chlorination-settling-numerical
site: BCT Engineering
---

# Super chlorination; overflow rate 5000 L/m2/h

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2081 Baisakh, 8 marks; 2080 Bhadra, 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/super-chlorination-settling-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

When is super chlorination recommended in a water supply scheme? Write about the advantages of break point chlorination. A settling tank is designed for an overflow rate of 5000 liters per m$^2$ per hour. What percentage of particles of diameter 0.06 mm and 0.03 mm will be removed in this tank? Temperature of water is 20°C and specific gravity of particles is 2.65.

## Short answer

Super‑chlorination is used during epidemics, for new mains, or poor‑quality water; break‑point chlorination gives stable free residual, oxidizes ammonia, and removes tastes. For overflow rate 5000 L/m²/h (0.1389 cm/s) at 20°C: 0.06 mm vs=0.3215 cm/s, removed 100%; 0.03 mm vs=0.0804 cm/s, removed 57.9%.

## Answer

#### When super-chlorination is recommended
Super-chlorination (a dose far above the demand, followed by de-chlorination) is used:
- during **epidemics** of cholera, typhoid, hepatitis, etc.;
- when **a new main, pipe or tank** is commissioned or repaired;
- when the raw water is of **highly variable or very poor quality**, or the plant is overloaded;
- when the **contact time is short** (so a high dose compensates);
- for **emergency or camp supplies** and when pre-treatment is poor.
The dose is 3 to 15 mg/l and the excess is removed by sulphur dioxide, sodium thiosulphate or activated carbon.

#### Advantages of break-point chlorination
- Oxidises ammonia, organic matter and iron/manganese completely, which removes most tastes and odours.
- Kills pathogens and algae far more effectively than combined chlorine, including organisms that resist chloramines.
- Gives a stable **free residual** which protects the distribution system against recontamination.
- The chlorine demand is fully satisfied and known, so the dose can be controlled accurately.

#### Numerical: removal in the settling tank
Overflow rate $v_0 = 5000$ l/m²/h $= 5$ m³/m²/h:

$$
v_0 = \frac{5}{3600} = 1.389\times10^{-3}\ \text{m/s} = 0.1389\ \text{cm/s}
$$

At 20 °C, $\nu = 0.01007$ cm²/s, $S_s=2.65$, $g=981$ cm/s². By Stokes' law, $v_s = \dfrac{g(S_s-1)d^2}{18\nu}$.

**Particle 0.06 mm (0.006 cm):**
$$
v_s = \frac{981\times1.65\times0.006^2}{18\times0.01007} = 0.3215\ \text{cm/s}
$$
$Re = 0.19 < 1$ (Stokes' law holds). Since $v_s = 0.3215$ > $v_0=0.1389$ cm/s, the particles are removed **100 %**.

**Particle 0.03 mm (0.003 cm):**
$$
v_s = \frac{981\times1.65\times0.003^2}{18\times0.01007} = 0.0804\ \text{cm/s}
$$
$Re = 0.02<1$. As $v_s<v_0$, the removal is
$$
\frac{v_s}{v_0} = \frac{0.0804}{0.1389} = 0.579 \Rightarrow 57.9\ \%
$$

**Answer: 0.06 mm particles: 100 % removed; 0.03 mm particles: 57.9 % removed.**

## Check yourself

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

1. In which situation is super-chlorination recommended?
   A. Normal clean supply with long contact time
   B. Epidemics or commissioning and repair of a new main
   C. Only for removing hardness
   D. Only when the water has no organisms
   Answer: B. A very high dose with later de-chlorination suits epidemics, new or repaired mains, short contact time and poor water. (Key point: Use of super-chlorination)

2. An overflow rate of 5000 l/m2/h equals what v0?
   A. 0.1389 cm/s
   B. 1.389 cm/s
   C. 0.0139 cm/s
   D. 13.89 cm/s
   Answer: A. 5 m3/m2/h = 5/3600 m/s = 1.389e-3 m/s = 0.1389 cm/s. (Key point: Overflow velocity)

3. The 0.06 mm particle has vs = 0.3215 cm/s against v0 = 0.1389 cm/s. What is its removal?
   A. 43 %
   B. 57.9 %
   C. 100 %
   D. 24.8 %
   Answer: C. vs is greater than v0, so the particle is fully removed. (Key point: Removal of 0.06 mm)

4. How is the removal of the 0.03 mm particle found and what is it?
   A. vs/v0 = 0.0804/0.1389, about 57.9 %
   B. v0/vs, about 173 %
   C. 100 %, since it is below 0.1 mm
   D. vs x v0, about 1 %
   Answer: A. Because vs is less than v0, removal is the ratio vs/v0 = 0.579. (Key point: Removal of 0.03 mm)

## Related in Water Supply Engineering

- [Overflow rate 4000 L/m2/h at 10°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-overflow-4000) — Water Supply Engineering, chapter 6, asked 2×
- [Super/break point chlorination; bleaching 20 MLD](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ableaching-a76) — Water Supply Engineering, chapter 6
- [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×
- [Super vs break point chlorination; residual](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Asuper-vs-break-point-chlorination) — Water Supply Engineering, chapter 6
- [Break point chlorination; bleaching powder 500000 L](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ableaching-c71) — Water Supply Engineering, chapter 6
- [Break point chlorination; pH and temperature effects](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Abpc-ph-temperature-hocl) — Water Supply Engineering, chapter 6, asked 2×

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