GATE 2024 CE (CE1) – Question 61
A water treatment plant treats 25 MLD water with a natural alkalinity of 4.0 mg/L (as CaCO$_3$). It is estimated that, during coagulation of this water, 450 kg/day of calcium bicarbonate (Ca(HCO$_3$)$_2$) is required based on the alum dosage.
Consider the atomic weights as: Ca-40, H-1, C-12, O-16.
The quantity of pure quick lime, CaO (in kg) required for this process per day is ___________ (rounded off to 2 decimal places).
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Correct answer: 99.06 to 100.06
Explanation
The molecular weight of Ca(HCO₃)₂ is 162, so 450 kg/day is equivalent to $450 \times \frac{100}{162} = 277.78$ kg/day of alkalinity as CaCO₃. The natural alkalinity of the water supplies $4.0 \text{ g/m}^3 \times 25000 \text{ m}^3/\text{d} = 100$ kg/day as CaCO₃. The shortfall of $277.78 - 100 = 177.78$ kg/day as CaCO₃ must be supplied by the lime. CaO (56) gives alkalinity equivalent to 100 per 56 as CaCO₃, so the lime needed is $177.78 \times \frac{56}{100} = 99.56$ kg/day.