GATE CE 2024 Set 1 — Question 61
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Environmental Engineering → Drinking Water Treatment → Coagulation & Flocculation
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Question
A water treatment plant treats 25 MLD water with a natural alkalinity of 4.0 mg/L (as ). It is estimated that, during coagulation of this water, 450 kg/day of calcium bicarbonate () 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).
Correct answer
99 to 100
Solution
1.Calculate the total alkalinity required as equivalent:
The amount of calcium bicarbonate required is 450 kg/day.Equivalent weight of .
Equivalent weight of .
Alkalinity required (as ) = kg/day.
2.Calculate the natural alkalinity available in the water:
Flow rate = 25 MLD = L/day.Natural alkalinity = 4.0 mg/L as .
Alkalinity available = kg/day as .
3.Calculate the alkalinity to be added:
Alkalinity to be added = Total required - Natural available = kg/day as .4.Calculate the quantity of pure quick lime (CaO) required:
Equivalent weight of .CaO required = (Alkalinity to be added as )
CaO required = kg/day.Rounding to 2 decimal places, we get 99.56 kg/day.
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