Question
Download Solution PDFThe maximum daily demand at a water purification plant has been estimated as 12 million litres per day.Find the cross-sectional area of the sedimentation tank required if a detention period of 6 hours and the velocity of flow of 0.2 m/minute is assumed.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
The Cross-sectional area required for the sedimentation tank is calculated as:
\(\text {Cross-Sectional Area} = \frac{\text{Design Flow}}{\text{Velocity of flow}}\)
Calculation:
The Design flow will be the maximum daily demand which is given as 12 million liters per day.
Q = 12 × 106 × 10-3 m3/day
Or
Q = 12000/(24× 60) = 8.33 cum/minute
Velocity of flow, V = 0.2 m/minute
Therefore, the cross-sectional area required for the sedimentation tank is calculated as:
A = 8.33/0.2
∴ A = 41.7 m2
Last updated on Jun 12, 2025
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