--- title: "Annual bleaching powder 10 MLD; break point" subject: "Water Supply Engineering" chapter: 6 key: "semester_5/water_supply_engineering:bleaching-s71" source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/bleaching-s71 site: BCT Engineering --- # Annual bleaching powder 10 MLD; break point Water Supply Engineering · Chapter 6: Water Treatment - Asked in: 2071 Shrawan, 8 marks - Video: [pressure filter! Disinfection! chlorination! factory effecting! Numerical! types of aeration!#ioe](https://www.youtube.com/watch?v=Pr8fCTxjjG0) — Sanjay64official - 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-s71 - Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6 ## Question Determine the amount of bleaching powder required annually in a water treatment plant treating 10 MLD of water if 0.3 ppm of chlorine dose is required. Available bleaching powder contains 27% of chlorine. Describe the break point chlorination in water treatment process. ## Short answer Break-point chlorination supplies chlorine to oxidize all demand compounds, leaving free residual; the break point is the dose at which residual dips to a minimum. For 10 MLD at 0.3 mg/L dose and27% chlorine, daily bleach=3.00/0.27=11.11 kg; annually 11.11×365=4056 kg(4.06 tonnes). ## Answer #### Annual quantity of bleaching powder Flow = 10 MLD $= 10\times10^6$ l/day; chlorine dose = 0.3 mg/l; available chlorine = 27 %. $$ \begin{aligned} \text{Chlorine per day} &= 0.3\times10\times10^6\ \text{mg} = 3\times10^6\ \text{mg} = 3.00\ \text{kg/day} \\ \text{Bleaching powder per day} &= \frac{3.00}{0.27} = 11.11\ \text{kg/day} \\ \text{Per year} &= 11.11\times365 = 4056\ \text{kg} \approx 4.06\ \text{tonnes} \end{aligned} $$ **Answer: about 4056 kg (4.06 tonnes) of bleaching powder per year.** #### Break-point chlorination in water treatment **Break-point chlorination** is the addition of chlorine to water in a quantity large enough to oxidise all the chlorine-demanding substances (reducing agents, ammonia, organic matter) so that a **free residual chlorine** is left. The dose at which the residual falls to its minimum and then starts to rise again as free chlorine is the **break point**. ```text Residual chlorine ^ | B | / \ / | / \ / | / \ / free residual | / \ / (HOCl + OCl-) | combined / \ / | residual / \ / | / \ / | / \/ +----------+-------+-------+-----------------> Chlorine dose O A B C | Zone 1 | Zone 2| Zone 3| Zone 4 ``` A = demand of reducing agents met, B = peak of combined residual (chloramines), C = break point 1. **Zone 1 (O to A)**: the chlorine first reacts with reducing compounds (Fe$^{2+}$, Mn$^{2+}$, H$_2$S, organic matter) and is used up. No residual appears. 2. **Zone 2 (A to B)**: once this demand is met, chlorine reacts with ammonia and amines to form **chloramines** (NH$_2$Cl, NHCl$_2$). The combined residual rises to a peak at B. 3. **Zone 3 (B to C)**: further chlorine oxidises the chloramines to nitrogen gas and other products (N$_2$O, NCl$_3$), so the residual **falls** to a minimum at C. Tastes and odours are most noticeable in this zone. 4. **Zone 4 (beyond C)**: all the ammonia and oxidisable organic matter have been destroyed; the added chlorine remains as **free chlorine** (HOCl and OCl$^-$), and the residual rises in direct proportion to the dose. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Chlorine per day for 10 MLD at 0.3 ppm? A. 30 kg/day B. 0.3 kg/day C. 11.11 kg/day D. 0.3 mg/l x 10^7 l = 3.0 kg/day Answer: D. 3x10^6 mg = 3 kg. (Key point: Chlorine per day) 2. Annual bleaching powder at 27 % chlorine? A. 3.0 / 0.27 = 11.11 kg/day, x 365 = about 4056 kg B. 3.0 x 365 = 1095 kg C. 3.0 x 0.27 x 365 = 296 kg D. 11.11 kg Answer: A. Divide by the available fraction, then multiply by 365. (Key point: Bleaching powder per year) 3. What exists beyond the break point? A. Chloramines peaking B. Free chlorine (HOCl and OCl-) rising with dose C. No residual D. Nitrate only Answer: B. All demand is oxidised so excess is free. (Key point: Zone 4) 4. What happens between a and b? A. Residual rises as free chlorine B. Chloramines form C. Chloramines are oxidised and residual falls D. No reaction occurs Answer: C. Tastes and odours are worst here. 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