---
title: "Bleaching powder for 5 MLD, chlorine demand 0.1"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:bleaching-a75"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/bleaching-a75
site: BCT Engineering
---

# Bleaching powder for 5 MLD, chlorine demand 0.1

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2075 Asoj, 4 marks
- Video: [Chlorine Dosing Calculation|How to calculate chlorine Dosing & Consumption|How to Prepare solution.](https://www.youtube.com/watch?v=COtK4Y62yzQ) — ETP, ZLD Plant Technical knowledge
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/bleaching-a75
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

How many kg/day of bleaching powder is required to treat 5 MLD of water if the chlorine demand of water is 0.1 mg/l and residual chlorine requirement is 0.4 mg/l? Assume bleaching powder contains 35% of available chlorine.

## Short answer

Total chlorine dose = demand + residual = 0.5 mg/L. Flow = 5 MLD gives chlorine mass 2.5 kg/day. Bleaching powder contains 35% available chlorine, so required = 2.5 / 0.35 = 7.14 kg/day.

## Answer

Total dose = chlorine demand + residual:
$$
\text{Dose} = 0.1 + 0.4 = 0.5\ \text{mg/l}
$$
Flow = 5 MLD $= 5\times10^6$ l/day.
$$
\begin{aligned} \text{Chlorine} &= 0.5\times5\times10^6\ \text{mg/day} = 2.5\ \text{kg/day} \\ \text{Bleaching powder} &= \frac{2.5}{0.35} = 7.14\ \text{kg/day} \end{aligned}
$$

**Answer: 7.14 kg/day of bleaching powder.**

## Check yourself

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

1. How is the total chlorine dose found?
   A. Demand minus residual
   B. Demand plus residual
   C. Residual only
   D. Demand multiplied by residual
   Answer: B. The dose must satisfy the demand and still leave the residual: 0.1 + 0.4 = 0.5 mg/l. (Key point: Total dose)

2. For 5 MLD at a dose of 0.5 mg/l, how much chlorine is needed per day?
   A. 0.5 kg
   B. 2.5 kg
   C. 5 kg
   D. 7.14 kg
   Answer: B. 0.5 x 5 = 2.5 kg/day. (Key point: Chlorine mass)

3. With 35% available chlorine, how much bleaching powder per day?
   A. 0.875 kg
   B. 2.5 kg
   C. 7.14 kg
   D. 8.75 kg
   Answer: C. 2.5/0.35 = 7.14 kg/day. (Key point: Bleaching powder)

## Related in Water Supply Engineering

- [Bleaching powder for 2 MLD, 30% chlorine](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ableaching-b79) — Water Supply Engineering, chapter 6
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