GATE CE 2020 Set 1 — Question 62

NAT+2 / -0HardPrimary SedimentationWastewater Treatment ProcessesEnvironmental Engineering

Environmental Engineering → Wastewater Treatment Processes → Primary Sedimentation

Last updated 5 September 2026

Question

20000 \text{ litre/m}^2 \text{ per day}

SOR = 20000 \text{ L/m}^2/\text{day} = \frac{20000 \times 10^{-3} \text{ m}^3/\text{day}}{1 \text{ m}^2} = 20 \text{ m/day}

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SOR = 20000 \text{ L/m}^2/\text{day} = \frac{20000 \times 10^{-3} \text{ m}^3/\text{day}}{1 \text{ m}^2} = 20 \text{ m/day}

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