Water operator math tutorial
Wet Well Pumping Rate
Find a pump's rate from how fast the wet well level drops.
Lift station operators estimate a pump's actual output by timing how long it takes to lower the wet well a measured distance with no inflow.
Volume removed is length times width times the drop in feet, converted to gallons, then divided by the pumping time.
The formula
pumping rate (gpm) = [length (ft) × width (ft) × drop (ft) × 7.48 gal/ft³] ÷ time (min)
| Variable | Unit |
|---|---|
| Well length | ft |
| Well width | ft |
| Level drop | ft |
| Pumping time | min |
| Result: Pumping rate | gpm |
Conversion factors used: 7.48 gal per cubic foot.
Applies to: Collection Systems, Wastewater Treatment.
Common exam tip: Units are the whole problem: feet, then ft³, then gallons, then divide by minutes.
Calculator
Pumping rate
119.68 gpm
- 1. 8 × 8 × 2.5 = 160 ft³
- 2. 160 ft³ × 7.48 = 1,196.8 gal
- 3. 1,196.8 gal ÷ 10 min = 119.68 gpm
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
An 8 ft by 8 ft well drops 2.5 ft in 10 minutes.
- 8 × 8 × 2.5 = 160 ft³
- 160 ft³ × 7.48 = 1,196.8 gal
- 1,196.8 gal ÷ 10 min = 119.68 gpm
Answer: 119.68 gpm
Example 2
A 10 ft by 12 ft well drops 3 ft in 15 minutes.
- 10 × 12 × 3 = 360 ft³
- 360 ft³ × 7.48 = 2,692.8 gal
- 2,692.8 gal ÷ 15 min = 179.52 gpm
Answer: 179.52 gpm
Common mistakes
Wrong result: 16 gpm
skipping the cubic feet to gallons conversion
Wrong result: 7,180.8 gpm
using hours instead of minutes
Practice questions
1. 6 ft by 6 ft well, 2 ft drop, 6 minutes. Pumping rate?
- A12 gpm
- B5,385.6 gpm
- C897.6 gpm
- D89.76 gpm
Show answer and explanation
Correct answer: D. 6 × 6 × 2 = 72 ft³ 72 ft³ × 7.48 = 538.6 gal 538.6 gal ÷ 6 min = 89.76 gpm Answer: 89.76 gpm. A common slip is skipping the cubic feet to gallons conversion, which gives 12 gpm.
2. 9 ft by 9 ft well, 1.5 ft drop, 5 minutes. Pumping rate?
- A181.76 gpm
- B24.3 gpm
- C10,905.84 gpm
- D1,817.6 gpm
Show answer and explanation
Correct answer: A. 9 × 9 × 1.5 = 121.5 ft³ 121.5 ft³ × 7.48 = 908.8 gal 908.8 gal ÷ 5 min = 181.76 gpm Answer: 181.76 gpm. A common slip is skipping the cubic feet to gallons conversion, which gives 24.3 gpm.
3. 12 ft by 10 ft well, 3.5 ft drop, 20 minutes. Pumping rate?
- A21 gpm
- B157.08 gpm
- C9,424.8 gpm
- D1,570.8 gpm
Show answer and explanation
Correct answer: B. 12 × 10 × 3.5 = 420 ft³ 420 ft³ × 7.48 = 3,141.6 gal 3,141.6 gal ÷ 20 min = 157.08 gpm Answer: 157.08 gpm. A common slip is skipping the cubic feet to gallons conversion, which gives 21 gpm.
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.
Collection Systems
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