Water operator math tutorial
Surface Overflow Rate
Divide flow in gpd by a circular clarifier's surface area.
Surface overflow rate compares flow to the surface area of a clarifier. It is the rise rate particles must beat to settle, so lower is better for settling.
For a circular tank, find the area from the diameter first. Keep π and the square in the area calculation.
The formula
SOR (gpd/ft²) = flow (gpd) ÷ [0.785 × diameter² (ft²)], with flow (gpd) = MGD × 1,000,000
| Variable | Unit |
|---|---|
| Flow | MGD |
| Clarifier diameter | ft |
| Result: Surface overflow rate | gpd/ft² |
Conversion factors used: 1 MGD = 1,000,000 gpd; area of a circle = π × d² ÷ 4 (about 0.785 × d²).
Applies to: Water Treatment, Wastewater Treatment.
Common exam tip: Diameter is not radius. If the problem gives radius, double it first.
Calculator
Surface overflow rate
424.4 gpd/ft²
- 1. Area = π × 60² ÷ 4 = 2,827.4 ft²
- 2. 1.2 MGD × 1,000,000 = 1,200,000 gpd
- 3. 1,200,000 gpd ÷ 2,827.4 ft² = 424.4 gpd/ft²
For instructional use. Results assume the standard conversion factors shown on this page. Actual field calculations may need chemical strength, density, temperature, and other adjustments, so check your plant’s procedures.
Worked examples
Example 1
A 60 ft diameter clarifier treats 1.2 MGD.
- Area = π × 60² ÷ 4 = 2,827.4 ft²
- 1.2 MGD × 1,000,000 = 1,200,000 gpd
- 1,200,000 gpd ÷ 2,827.4 ft² = 424.4 gpd/ft²
Answer: 424.4 gpd/ft²
Example 2
An 80 ft diameter clarifier treats 2.0 MGD.
- Area = π × 80² ÷ 4 = 5,026.5 ft²
- 2 MGD × 1,000,000 = 2,000,000 gpd
- 2,000,000 gpd ÷ 5,026.5 ft² = 397.9 gpd/ft²
Answer: 397.9 gpd/ft²
Common mistakes
Wrong result: 106.1 gpd/ft²
using π × d² and forgetting to divide by 4
Wrong result: 6,366.2 gpd/ft²
forgetting to square the diameter
Practice questions
1. 0.8 MGD, 50 ft diameter. SOR in gpd/ft²?
- A101.9 gpd/ft²
- B5,093 gpd/ft²
- C4,074 gpd/ft²
- D407.4 gpd/ft²
Show answer and explanation
Correct answer: D. Area = π × 50² ÷ 4 = 1,963.5 ft² 0.8 MGD × 1,000,000 = 800,000 gpd 800,000 gpd ÷ 1,963.5 ft² = 407.4 gpd/ft² Answer: 407.4 gpd/ft². A common slip is using π × d² and forgetting to divide by 4, which gives 101.9 gpd/ft².
2. 3.5 MGD, 100 ft diameter. SOR in gpd/ft²?
- A445.6 gpd/ft²
- B111.4 gpd/ft²
- C11,140.8 gpd/ft²
- D4,456 gpd/ft²
Show answer and explanation
Correct answer: A. Area = π × 100² ÷ 4 = 7,854 ft² 3.5 MGD × 1,000,000 = 3,500,000 gpd 3,500,000 gpd ÷ 7,854 ft² = 445.6 gpd/ft² Answer: 445.6 gpd/ft². A common slip is using π × d² and forgetting to divide by 4, which gives 111.4 gpd/ft².
3. 1.6 MGD, 70 ft diameter. SOR in gpd/ft²?
- A103.9 gpd/ft²
- B415.8 gpd/ft²
- C7,275.7 gpd/ft²
- D4,158 gpd/ft²
Show answer and explanation
Correct answer: B. Area = π × 70² ÷ 4 = 3,848.5 ft² 1.6 MGD × 1,000,000 = 1,600,000 gpd 1,600,000 gpd ÷ 3,848.5 ft² = 415.8 gpd/ft² Answer: 415.8 gpd/ft². A common slip is using π × d² and forgetting to divide by 4, which gives 103.9 gpd/ft².
Where this formula applies
This calculation can appear in the disciplines below. Exact exam coverage depends on your state, certification level, and official exam outline.
Water Treatment
Wastewater Treatment
Study guides
Keep going with the full guide
Water Operator Math Made Simple walks through operator math step by step and includes printable references plus a full practice exam.
Water Operator Math Made Simple
A Step-by-Step Guide to the Math Every Water Operator Needs to Know