Surface Overflow Rate (SOR) for a Circular Clarifier
Calculate clarifier surface overflow rate from diameter and flow. Common trap: using diameter instead of radius. Full worked solution included.
Answer Choices
A) 409 gpd/ft² Correct
Correct. Convert flow: 2.6 MGD = 2,600,000 gpd. Area uses radius (90 ft ÷ 2 = 45 ft): A = 3.14 × (45 ft)² = 6,358.5 ft², so SOR = 2,600,000 gpd ÷ 6,358.5 ft² = 408.9 ≈ 409 gpd/ft². Sanity check: ~400 gpd/ft² is a reasonable secondary clarifier SOR at average flow.
B) 102 gpd/ft²
This comes from using the diameter as the radius (90 ft) in A = πr², which makes the area 4 times too large and the SOR 4 times too small.
C) 41 gpd/ft²
This results from a flow conversion error, such as using 260,000 gpd instead of 2,600,000 gpd (missing a zero), which makes the SOR 10 times too low.
D) 4,090 gpd/ft²
This results from a flow conversion error such as using 26,000,000 gpd instead of 2,600,000 gpd (adding an extra zero), which makes the SOR 10 times too high.
Step-by-Step Solution
- Given flow = 2.6 MGD = 2.6 × 1,000,000 gal/day = 2,600,000 gpd
- Given diameter = 90 ft → radius r = 90 ft ÷ 2 = 45 ft
- Surface area A = π × r² = 3.14 × (45 ft)² = 3.14 × 2,025 ft² = 6,358.5 ft²
- SOR = flow ÷ area = 2,600,000 gpd ÷ 6,358.5 ft² = 408.9 gpd/ft² ≈ 409 gpd/ft²
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