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Bleaching powder for 5 MLD, chlorine demand 0.1

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.

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.

Total dose = chlorine demand + residual:

Dose=0.1+0.4=0.5 mg/l\text{Dose} = 0.1 + 0.4 = 0.5\ \text{mg/l}

Flow = 5 MLD =5×106= 5\times10^6 l/day.

Chlorine=0.5×5×106 mg/day=2.5 kg/dayBleaching powder=2.50.35=7.14 kg/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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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