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Alum and chlorine quantity, 12 MLD

Short answer

With12 MLD flow and alum dose3 mg/L, alum required=36 kg/day. Assuming negligible chlorine demand, the dose equals desired residual0.2 mg/L, giving chlorine=2.4 kg/day.

Find the quantity of alum and chlorine required in a treatment plant of capacity 12 million lit/day. If optimum dose of alum is 3 mg/l and residual chlorine is expected in the distribution pipe is at the concentration of 0.2 mg/l.

Plant capacity = 12 MLD =12×106= 12\times10^6 l/day. Using 1 mg/l×1 MLD=1 kg/day1\ \text{mg/l}\times1\ \text{MLD} = 1\ \text{kg/day}:

Alum (optimum dose 3 mg/l):

Alum=3×12=36 kg/day\text{Alum} = 3\times12 = 36\ \text{kg/day}

Chlorine: the residual chlorine wanted in the distribution pipe is 0.2 mg/l. No chlorine demand is given, so the demand of water is assumed to be nil and the dose is taken as equal to the residual (if there were a demand, the dose = demand + residual):

Chlorine=0.2×12=2.4 kg/day\text{Chlorine} = 0.2\times12 = 2.4\ \text{kg/day}

If bleaching powder with 30 % available chlorine is used, the quantity would be 2.40.30=8.0\dfrac{2.4}{0.30} = 8.0 kg/day.

Answer: alum = 36 kg/day; chlorine = 2.4 kg/day.

✓ Answer checked by Claude Opus 5.5 (high effort)

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