10 FE practice problems: clarifier overflow rate and detention time, with solutions
These ten fill-in problems cover the clarifier design checks in the Water and Wastewater area of the FE Environmental specification: surface overflow rate, tank area and dimensions, and detention time. Fill-in questions have no options to lean on, so carry units through every line and round only at the end. Each solution lists the common wrong answers and the mistake behind each one.
How to use these problems
Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.
The problems
Problem 1 · FE Environmental, Clarifier overflow rate and detention time
A rectangular clarifier is 55 ft long and 16 ft wide, with a side water depth of 14.0 ft. The design flow is 1.26 MGD. Find the detention time (hydraulic residence time) in hours.
Enter your answer in hours (h), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 1.76 h
- Given
- L = 55 ft, W = 16 ft, H = 14.0 ft, Q = 1.26 MGD, 7.481 gal/ft³ (handbook)
- Find
- detention time θ (h)
- Handbook
- Environmental Engineering, Clarifier (hydraulic residence time); Units and Conversion Factors (7.481 gal per ft³), pages 3 and 345
- Equation
θ = V/Q, with V = L·W·H- Substitute
V = L·W·H = (55 ft)(16 ft)(14.0 ft) = 12,320 ft³V = 12,320 ft³ × 7.481 gal/ft³ = 92,170 galθ = V/Q = (92,170 gal)/(1.26 × 10⁶ gal/d) = 0.07315 d × 24 h/d = 1.76 h
- Result
- 1.76 h, 3 significant figures
- Check
- the overflow rate is vo = Q/A = (1.26 × 10⁶ gal/d)/(55 ft × 16 ft) = 1,432 gpd/ft², and θ = H × 7.481/vo × 24 = 1.76 h, the same answer.
- Common wrong answers
- 0.235 h: divided the volume in ft³ by a flow in gallons per day without the 7.481 gal/ft³ conversion
- 0.0731 h: stopped at V/Q in days and did not multiply by 24 h/d
- 0.125 h: divided the surface area by Q and left out the depth
Problem 2 · FE Environmental, Clarifier overflow rate and detention time
A circular primary clarifier with a diameter of 12.5 m treats an average flow of 6,200 m³/d. Find the overflow rate (surface loading rate) in m³/(m²·d).
Enter your answer in m³/(m²·d), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 50.5 m³/(m²·d)
- Given
- D = 12.5 m, Q = 6,200 m³/d
- Find
- overflow rate vo (m³/(m²·d))
- Handbook
- Environmental Engineering, Clarifier (overflow rate), page 345
- Equation
vo = Q/A_surface, with A = πD²/4- Substitute
A = πD²/4 = π(12.5 m)²/4 = 122.7 m²vo = Q/A = (6,200 m³/d)/(122.7 m²) = 50.5 m³/(m²·d)
- Result
- 50.5 m³/(m²·d), 3 significant figures
- Check
- units (m³/d)/m² = m³/(m²·d), the same as m/d; the rate does not depend on the depth.
- Common wrong answers
- 12.6 m³/(m²·d): used πD² instead of πD²/4, so the area is 4 times too large and the rate 4 times too small
- 158 m³/(m²·d): divided Q by the circumference πD (that is the weir loading idea, per m of weir) instead of the surface area
- 202 m³/(m²·d): put the radius into πD²/4, so the area is 4 times too small and the rate 4 times too large
Problem 3 · FE Environmental, Clarifier overflow rate and detention time
A rectangular clarifier is 105 ft long and 25 ft wide, with a side water depth of 13.0 ft. The design flow is 2.83 MGD. Find the detention time (hydraulic residence time) in hours.
Enter your answer in hours (h), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 2.16 h
- Given
- L = 105 ft, W = 25 ft, H = 13.0 ft, Q = 2.83 MGD, 7.481 gal/ft³ (handbook)
- Find
- detention time θ (h)
- Handbook
- Environmental Engineering, Clarifier (hydraulic residence time); Units and Conversion Factors (7.481 gal per ft³), pages 3 and 345
- Equation
θ = V/Q, with V = L·W·H- Substitute
V = L·W·H = (105 ft)(25 ft)(13.0 ft) = 34,130 ft³V = 34,130 ft³ × 7.481 gal/ft³ = 255,300 galθ = V/Q = (255,300 gal)/(2.83 × 10⁶ gal/d) = 0.09021 d × 24 h/d = 2.16 h
- Result
- 2.16 h, 3 significant figures
- Check
- the overflow rate is vo = Q/A = (2.83 × 10⁶ gal/d)/(105 ft × 25 ft) = 1,078 gpd/ft², and θ = H × 7.481/vo × 24 = 2.16 h, the same answer.
- Common wrong answers
- 0.289 h: divided the volume in ft³ by a flow in gallons per day without the 7.481 gal/ft³ conversion
- 0.167 h: divided the surface area by Q and left out the depth
- 0.0902 h: stopped at V/Q in days and did not multiply by 24 h/d
Problem 4 · FE Environmental, Clarifier overflow rate and detention time
A circular primary clarifier has a diameter of 36.0 m and a side water depth of 3.7 m. It treats an average flow of 24,800 m³/d. Find the hydraulic residence time (detention time) in hours.
Enter your answer in hours (h), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 3.64 h
- Given
- D = 36.0 m, H = 3.7 m, Q = 24,800 m³/d
- Find
- hydraulic residence time θ (h)
- Handbook
- Environmental Engineering, Clarifier (hydraulic residence time and overflow rate), page 345
- Equation
θ = V/Q, with V = A·H and A = πD²/4- Substitute
A = πD²/4 = π(36.0 m)²/4 = 1,018 m²V = A·H = (1,018 m²)(3.7 m) = 3,766 m³θ = V/Q = (3,766 m³)/(24,800 m³/d) = 0.1519 d × 24 h/d = 3.64 h
- Result
- 3.64 h, 3 significant figures
- Check
- θ = H/vo gives the same time: the overflow rate is vo = Q/A = 24.36 m³/(m²·d), and 3.7 m/(24.36 m/d) × 24 h/d = 3.64 h.
- Common wrong answers
- 14.6 h: used πD² instead of πD²/4 (the diameter as if it were the radius), so the area and the time are 4 times too large
- 0.985 h: divided the surface area by Q and left out the depth
- 0.152 h: stopped at V/Q in days and did not multiply by 24 h/d
Problem 5 · FE Environmental, Clarifier overflow rate and detention time
A rectangular clarifier is 150 ft long and 37 ft wide, with a side water depth of 12.5 ft. The design flow is 6.77 MGD. Find the detention time (hydraulic residence time) in hours.
Enter your answer in hours (h), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 1.84 h
- Given
- L = 150 ft, W = 37 ft, H = 12.5 ft, Q = 6.77 MGD, 7.481 gal/ft³ (handbook)
- Find
- detention time θ (h)
- Handbook
- Environmental Engineering, Clarifier (hydraulic residence time); Units and Conversion Factors (7.481 gal per ft³), pages 3 and 345
- Equation
θ = V/Q, with V = L·W·H- Substitute
V = L·W·H = (150 ft)(37 ft)(12.5 ft) = 69,380 ft³V = 69,380 ft³ × 7.481 gal/ft³ = 519,000 galθ = V/Q = (519,000 gal)/(6.77 × 10⁶ gal/d) = 0.07666 d × 24 h/d = 1.84 h
- Result
- 1.84 h, 3 significant figures
- Check
- the overflow rate is vo = Q/A = (6.77 × 10⁶ gal/d)/(150 ft × 37 ft) = 1,220 gpd/ft², and θ = H × 7.481/vo × 24 = 1.84 h, the same answer.
- Common wrong answers
- 0.147 h: divided the surface area by Q and left out the depth
- 0.246 h: divided the volume in ft³ by a flow in gallons per day without the 7.481 gal/ft³ conversion
- 0.0767 h: stopped at V/Q in days and did not multiply by 24 h/d
Problem 6 · FE Environmental, Clarifier overflow rate and detention time
A circular primary clarifier with a diameter of 39.0 m treats an average flow of 66,100 m³/d. Find the overflow rate (surface loading rate) in m³/(m²·d).
Enter your answer in m³/(m²·d), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 55.3 m³/(m²·d)
- Given
- D = 39.0 m, Q = 66,100 m³/d
- Find
- overflow rate vo (m³/(m²·d))
- Handbook
- Environmental Engineering, Clarifier (overflow rate), page 345
- Equation
vo = Q/A_surface, with A = πD²/4- Substitute
A = πD²/4 = π(39.0 m)²/4 = 1,195 m²vo = Q/A = (66,100 m³/d)/(1,195 m²) = 55.3 m³/(m²·d)
- Result
- 55.3 m³/(m²·d), 3 significant figures
- Check
- units (m³/d)/m² = m³/(m²·d), the same as m/d; the rate does not depend on the depth.
- Common wrong answers
- 221 m³/(m²·d): put the radius into πD²/4, so the area is 4 times too small and the rate 4 times too large
- 13.8 m³/(m²·d): used πD² instead of πD²/4, so the area is 4 times too large and the rate 4 times too small
- 539 m³/(m²·d): divided Q by the circumference πD (that is the weir loading idea, per m of weir) instead of the surface area
Problem 7 · FE Environmental, Clarifier overflow rate and detention time
A circular primary clarifier with a diameter of 14.5 m treats an average flow of 4,300 m³/d. Find the overflow rate (surface loading rate) in m³/(m²·d).
Enter your answer in m³/(m²·d), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 26.0 m³/(m²·d)
- Given
- D = 14.5 m, Q = 4,300 m³/d
- Find
- overflow rate vo (m³/(m²·d))
- Handbook
- Environmental Engineering, Clarifier (overflow rate), page 345
- Equation
vo = Q/A_surface, with A = πD²/4- Substitute
A = πD²/4 = π(14.5 m)²/4 = 165.1 m²vo = Q/A = (4,300 m³/d)/(165.1 m²) = 26.0 m³/(m²·d)
- Result
- 26.0 m³/(m²·d), 3 significant figures
- Check
- units (m³/d)/m² = m³/(m²·d), the same as m/d; the rate does not depend on the depth.
- Common wrong answers
- 104 m³/(m²·d): put the radius into πD²/4, so the area is 4 times too small and the rate 4 times too large
- 6.51 m³/(m²·d): used πD² instead of πD²/4, so the area is 4 times too large and the rate 4 times too small
- 94.4 m³/(m²·d): divided Q by the circumference πD (that is the weir loading idea, per m of weir) instead of the surface area
Problem 8 · FE Environmental, Clarifier overflow rate and detention time
A circular clarifier 90 ft in diameter treats a flow of 5.30 MGD. Find the overflow rate in gpd/ft².
Enter your answer in gpd/ft², 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 833 gpd/ft²
- Given
- D = 90 ft, Q = 5.30 MGD
- Find
- overflow rate vo (gpd/ft²)
- Handbook
- Environmental Engineering, Clarifier (overflow rate), page 345
- Equation
vo = Q/A_surface, with A = πD²/4- Substitute
A = πD²/4 = π(90 ft)²/4 = 6,362 ft²Q = 5.30 MGD = 5,300,000 gal/dvo = Q/A = (5,300,000 gal/d)/(6,362 ft²) = 833 gpd/ft²
- Result
- 833 gpd/ft², 3 significant figures
- Check
- units (gal/d)/ft² = gpd/ft²; 1 MGD = 1,000,000 gal/d.
- Common wrong answers
- 111 gpd/ft²: converted the flow to ft³/d before dividing, which gives ft/d, not gpd/ft²
- 3,330 gpd/ft²: put the radius into πD²/4, so the area is 4 times too small and the rate 4 times too large
- 208 gpd/ft²: used πD² instead of πD²/4, so the area is 4 times too large and the rate 4 times too small
Problem 9 · FE Environmental, Clarifier overflow rate and detention time
A rectangular clarifier is 175 ft long and 39 ft wide, with a side water depth of 13.0 ft. The design flow is 8.85 MGD. Find the detention time (hydraulic residence time) in hours.
Enter your answer in hours (h), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 1.80 h
- Given
- L = 175 ft, W = 39 ft, H = 13.0 ft, Q = 8.85 MGD, 7.481 gal/ft³ (handbook)
- Find
- detention time θ (h)
- Handbook
- Environmental Engineering, Clarifier (hydraulic residence time); Units and Conversion Factors (7.481 gal per ft³), pages 3 and 345
- Equation
θ = V/Q, with V = L·W·H- Substitute
V = L·W·H = (175 ft)(39 ft)(13.0 ft) = 88,730 ft³V = 88,730 ft³ × 7.481 gal/ft³ = 663,800 galθ = V/Q = (663,800 gal)/(8.85 × 10⁶ gal/d) = 0.07500 d × 24 h/d = 1.80 h
- Result
- 1.80 h, 3 significant figures
- Check
- the overflow rate is vo = Q/A = (8.85 × 10⁶ gal/d)/(175 ft × 39 ft) = 1,297 gpd/ft², and θ = H × 7.481/vo × 24 = 1.80 h, the same answer.
- Common wrong answers
- 0.241 h: divided the volume in ft³ by a flow in gallons per day without the 7.481 gal/ft³ conversion
- 0.138 h: divided the surface area by Q and left out the depth
- 0.0750 h: stopped at V/Q in days and did not multiply by 24 h/d
Problem 10 · FE Environmental, Clarifier overflow rate and detention time
A circular primary clarifier with a diameter of 38.0 m treats an average flow of 28,300 m³/d. Find the overflow rate (surface loading rate) in m³/(m²·d).
Enter your answer in m³/(m²·d), 3 significant figures.
Handbook: Environmental Engineering, Clarifier, FE Reference Handbook 10.6
Show the worked solution
- Answer
- 25.0 m³/(m²·d)
- Given
- D = 38.0 m, Q = 28,300 m³/d
- Find
- overflow rate vo (m³/(m²·d))
- Handbook
- Environmental Engineering, Clarifier (overflow rate), page 345
- Equation
vo = Q/A_surface, with A = πD²/4- Substitute
A = πD²/4 = π(38.0 m)²/4 = 1,134 m²vo = Q/A = (28,300 m³/d)/(1,134 m²) = 25.0 m³/(m²·d)
- Result
- 25.0 m³/(m²·d), 3 significant figures
- Check
- units (m³/d)/m² = m³/(m²·d), the same as m/d; the rate does not depend on the depth.
- Common wrong answers
- 99.8 m³/(m²·d): put the radius into πD²/4, so the area is 4 times too small and the rate 4 times too large
- 237 m³/(m²·d): divided Q by the circumference πD (that is the weir loading idea, per m of weir) instead of the surface area
- 6.24 m³/(m²·d): used πD² instead of πD²/4, so the area is 4 times too large and the rate 4 times too small
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
What is a surface overflow rate?
It is the flow divided by the clarifier’s surface area, a velocity-like loading that sizes the tank. The handbook gives the relationships in the Environmental Engineering chapter, and each solution cites the page.
How precise should a fill-in answer be?
Enter the value in the units the question asks for. Our solutions round final answers to 3 significant figures and keep 4 in the steps.
Is this topic only on FE Environmental?
It is central to FE Environmental, and wastewater treatment also appears within the Water Resources and Environmental Engineering area of FE Civil.
Sources
- NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.