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Annual bleaching powder 10 MLD; break point

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).

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.

Annual quantity of bleaching powder

Flow = 10 MLD =10×106= 10\times10^6 l/day; chlorine dose = 0.3 mg/l; available chlorine = 27 %.

Chlorine per day=0.3×10×106 mg=3×106 mg=3.00 kg/dayBleaching powder per day=3.000.27=11.11 kg/dayPer year=11.11×365=4056 kg≈4.06 tonnes\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.

 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 (Fe2+^{2+}, Mn2+^{2+}, H2_2S, 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 (NH2_2Cl, NHCl2_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 (N2_2O, NCl3_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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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