Wastewater Class 4 Exam Virginia| 100
questions| with complete solutions
Course
Wastewater
1. A circular clarifier has a diameter of 50 ft and a depth of 12 ft. If the influent flow is 3.0
MGD, what is the hydraulic detention time (hours)?
Solution:
Area = 3.1416 × (25²) = 1,963 ft²
Volume = 1,963 × 12 = 23,556 ft³ = 176,146 gal
Flow = 3,000,000 gal/day
Detention time = 176,146 ÷ 3,000,000 days = 0.0587 days = 1.4 hr
2. MLSS = 4,500 mg/L. 30-min settleable solids = 300 mL/L. What is SVI?
Solution:
MLSS = 4.5 g/L
SVI = (300 ÷ 4.5) = 67 mL/g
3. Influent BOD = 250 mg/L. Flow = 5.0 MGD. How many pounds BOD enter per day?
Solution:
Pounds/day = 5 × 250 × 8.34 = 10,425 lbs/day
4. Secondary clarifier sludge blanket rises/billows. What adjustment is needed?
Solution: Increase WAS to remove more solids.
5. Ideal velocity in grit channels?
Solution: About 1 ft/sec.
6. Safe procedure: mixing acid and water.
Solution: Always add acid to water.
7. Most common dechlorination agent in chlorine gas disinfection?
Solution: Sulfur dioxide.
8. Typical suspended solids removal efficiency in a primary clarifier?
Solution: About 50%.
9. Where should flow be measured relative to a weir?
Solution: Upstream, at a point where the head is at least 4 times max head for accuracy.
10. What happens if pump packing is over-tightened?
Solution: Overheating, excessive wear, reduced efficiency. Packing should leak slightly (a drip).
11. A rectangular primary clarifier is 80 ft long, 20 ft wide, and 12 ft deep. Flow = 2.5 MGD.
What is the detention time?
, Solution:
Volume = 80 × 20 × 12 = 19,200 ft³
Gallons = 19,200 × 7.48 = 143,616 gal
Detention = 143,616 ÷ 2,500,000 = 0.0574 days × 24 = 1.38 hr
12. If the influent SS = 220 mg/L and effluent SS = 70 mg/L, what is the percent removal
efficiency?
Solution:
% Removal = (220 – 70) ÷ 220 × 100 = 150 ÷ 220 × 100 = 68%
13. Wastewater flow = 4 MGD. BOD influent = 300 mg/L. Effluent = 25 mg/L. What is plant
BOD removal efficiency?
Solution:
% Removal = (300 – 25) ÷ 300 × 100 = 275 ÷ 300 × 100 = 92%
14. An aeration tank has a volume of 1.2 million gallons. Flow = 4.0 MGD. What is the
hydraulic detention time?
Solution:
Detention = 1,200,000 ÷ 4,000,000 = 0.3 days × 24 = 7.2 hr
15. If a pump delivers 500 gpm, how many MGD is this?
Solution:
MGD = (500 × 1440) ÷ 1,000,000 = 720,000 ÷ 1,000,000 = 0.72 MGD
16. A digester has 400,000 gal volume. Sludge added = 20,000 gal/day. What is the detention
time?
Solution:
400,000 ÷ 20,000 = 20 days
17. Mixed liquor suspended solids = 3,800 mg/L. Aeration volume = 1.5 Mgal. What are the total
lbs of MLSS?
Solution:
lbs = 3,800 × 1.5 × 8.34 = 47,511 lbs
18. Plant flow = 3.0 MGD, influent BOD = 250 mg/L, aeration tank volume = 1.0 Mgal, MLSS
= 4,000 mg/L. Compute F/M ratio.
Solution:
Food = 3 × 250 × 8.34 = 6,255 lbs BOD/day
Mass = 4,000 × 1.0 × 8.34 = 33,360 lbs MLSS
F/M = 6,255 ÷ 33,360 = 0.19
19. What is the primary function of return activated sludge (RAS)?
Solution: Return concentrated biomass from clarifier to aeration basin to maintain MLSS.
questions| with complete solutions
Course
Wastewater
1. A circular clarifier has a diameter of 50 ft and a depth of 12 ft. If the influent flow is 3.0
MGD, what is the hydraulic detention time (hours)?
Solution:
Area = 3.1416 × (25²) = 1,963 ft²
Volume = 1,963 × 12 = 23,556 ft³ = 176,146 gal
Flow = 3,000,000 gal/day
Detention time = 176,146 ÷ 3,000,000 days = 0.0587 days = 1.4 hr
2. MLSS = 4,500 mg/L. 30-min settleable solids = 300 mL/L. What is SVI?
Solution:
MLSS = 4.5 g/L
SVI = (300 ÷ 4.5) = 67 mL/g
3. Influent BOD = 250 mg/L. Flow = 5.0 MGD. How many pounds BOD enter per day?
Solution:
Pounds/day = 5 × 250 × 8.34 = 10,425 lbs/day
4. Secondary clarifier sludge blanket rises/billows. What adjustment is needed?
Solution: Increase WAS to remove more solids.
5. Ideal velocity in grit channels?
Solution: About 1 ft/sec.
6. Safe procedure: mixing acid and water.
Solution: Always add acid to water.
7. Most common dechlorination agent in chlorine gas disinfection?
Solution: Sulfur dioxide.
8. Typical suspended solids removal efficiency in a primary clarifier?
Solution: About 50%.
9. Where should flow be measured relative to a weir?
Solution: Upstream, at a point where the head is at least 4 times max head for accuracy.
10. What happens if pump packing is over-tightened?
Solution: Overheating, excessive wear, reduced efficiency. Packing should leak slightly (a drip).
11. A rectangular primary clarifier is 80 ft long, 20 ft wide, and 12 ft deep. Flow = 2.5 MGD.
What is the detention time?
, Solution:
Volume = 80 × 20 × 12 = 19,200 ft³
Gallons = 19,200 × 7.48 = 143,616 gal
Detention = 143,616 ÷ 2,500,000 = 0.0574 days × 24 = 1.38 hr
12. If the influent SS = 220 mg/L and effluent SS = 70 mg/L, what is the percent removal
efficiency?
Solution:
% Removal = (220 – 70) ÷ 220 × 100 = 150 ÷ 220 × 100 = 68%
13. Wastewater flow = 4 MGD. BOD influent = 300 mg/L. Effluent = 25 mg/L. What is plant
BOD removal efficiency?
Solution:
% Removal = (300 – 25) ÷ 300 × 100 = 275 ÷ 300 × 100 = 92%
14. An aeration tank has a volume of 1.2 million gallons. Flow = 4.0 MGD. What is the
hydraulic detention time?
Solution:
Detention = 1,200,000 ÷ 4,000,000 = 0.3 days × 24 = 7.2 hr
15. If a pump delivers 500 gpm, how many MGD is this?
Solution:
MGD = (500 × 1440) ÷ 1,000,000 = 720,000 ÷ 1,000,000 = 0.72 MGD
16. A digester has 400,000 gal volume. Sludge added = 20,000 gal/day. What is the detention
time?
Solution:
400,000 ÷ 20,000 = 20 days
17. Mixed liquor suspended solids = 3,800 mg/L. Aeration volume = 1.5 Mgal. What are the total
lbs of MLSS?
Solution:
lbs = 3,800 × 1.5 × 8.34 = 47,511 lbs
18. Plant flow = 3.0 MGD, influent BOD = 250 mg/L, aeration tank volume = 1.0 Mgal, MLSS
= 4,000 mg/L. Compute F/M ratio.
Solution:
Food = 3 × 250 × 8.34 = 6,255 lbs BOD/day
Mass = 4,000 × 1.0 × 8.34 = 33,360 lbs MLSS
F/M = 6,255 ÷ 33,360 = 0.19
19. What is the primary function of return activated sludge (RAS)?
Solution: Return concentrated biomass from clarifier to aeration basin to maintain MLSS.
