Wastewater Treatment Plant Operator Exam
Study Guide 2026/2027 – Complete Exam-
Style Questions with Correct Answers &
Detailed Rationales | Verified & Reliable
1. A wastewater treatment plant receives an average flow of
2.4 MGD. The influent BOD₅ concentration is 220 mg/L.
Approximately what is the daily BOD₅ loading?
A. 220 lb/day
B. 440 lb/day
C. 4,404 lb/day
D. 8,808 lb/day
The standard loading calculation is: lb/day = flow (MGD) ×
concentration (mg/L) × 8.34. Therefore, 2.4 × 220 × 8.34 =
approximately 4,404 lb/day. The factor 8.34 converts MGD and
mg/L to pounds per day.
2. An activated-sludge process has an aeration basin volume
of 0.75 million gallons and receives an average flow of 3.0
MGD. What is the approximate hydraulic retention time?
,A. 2 hours
B. 6 hours
C. 12 hours
D. 18 hours
HRT in days is calculated as basin volume divided by flow: 0.75
MG ÷ 3.0 MGD = 0.25 day. Multiplying 0.25 day by 24
hours/day gives 6 hours.
3. Which condition most directly indicates that nitrification is
occurring effectively in an activated-sludge process?
A. Increasing influent BOD₅
B. Increasing primary sludge production
C. Conversion of ammonia nitrogen to oxidized nitrogen forms
D. Increasing effluent total suspended solids
Nitrification is the biological oxidation of ammonia, generally
producing nitrite and then nitrate. It is performed primarily by
autotrophic microorganisms and requires adequate dissolved
oxygen, suitable temperature, sufficient solids retention time,
and appropriate environmental conditions.
4. A clarifier has a surface area of 10,000 ft² and receives 2.5
MGD. What is its approximate surface overflow rate?
,A. 25 gpd/ft²
B. 100 gpd/ft²
C. 250 gpd/ft²
D. 400 gpd/ft²
Surface overflow rate is calculated as flow in gallons per day
divided by surface area. Thus, 2,500,000 gal/day ÷ 10,000 ft² =
250 gpd/ft². This parameter is important because excessive
overflow rates can contribute to poor solids settling and
carryover.
5. A secondary clarifier begins experiencing a gradual increase
in effluent suspended solids while the sludge blanket remains
relatively low. Which condition should the operator
investigate first?
A. Excessively high chlorine residual
B. Poor floc formation or dispersed growth
C. Excessive digester gas production
D. Low influent screening efficiency
When effluent TSS increases without a high sludge blanket, the
problem may be related to the quality of the biological floc
rather than simply solids accumulation in the clarifier. Dispersed
growth, pin floc, or other poor-settling conditions can allow fine
solids to escape with the effluent.
, 6. A plant's influent flow is 4.0 MGD, and the influent TSS
concentration is 250 mg/L. What is the approximate influent
TSS loading?
A. 417 lb/day
B. 834 lb/day
C. 8,340 lb/day
D. 16,680 lb/day
Using the loading equation, 4.0 MGD × 250 mg/L × 8.34 = 8,340
lb/day. Correct loading calculations are fundamental for
evaluating treatment capacity, solids production, and process
performance.
7. Which activated-sludge parameter is most directly
associated with the amount of biomass maintained in the
aeration system relative to incoming biodegradable organic
matter?
A. SVI
B. DO
C. F/M ratio
D. Chlorine residual
The food-to-microorganism (F/M) ratio compares the organic
substrate available to the amount of microorganisms
Study Guide 2026/2027 – Complete Exam-
Style Questions with Correct Answers &
Detailed Rationales | Verified & Reliable
1. A wastewater treatment plant receives an average flow of
2.4 MGD. The influent BOD₅ concentration is 220 mg/L.
Approximately what is the daily BOD₅ loading?
A. 220 lb/day
B. 440 lb/day
C. 4,404 lb/day
D. 8,808 lb/day
The standard loading calculation is: lb/day = flow (MGD) ×
concentration (mg/L) × 8.34. Therefore, 2.4 × 220 × 8.34 =
approximately 4,404 lb/day. The factor 8.34 converts MGD and
mg/L to pounds per day.
2. An activated-sludge process has an aeration basin volume
of 0.75 million gallons and receives an average flow of 3.0
MGD. What is the approximate hydraulic retention time?
,A. 2 hours
B. 6 hours
C. 12 hours
D. 18 hours
HRT in days is calculated as basin volume divided by flow: 0.75
MG ÷ 3.0 MGD = 0.25 day. Multiplying 0.25 day by 24
hours/day gives 6 hours.
3. Which condition most directly indicates that nitrification is
occurring effectively in an activated-sludge process?
A. Increasing influent BOD₅
B. Increasing primary sludge production
C. Conversion of ammonia nitrogen to oxidized nitrogen forms
D. Increasing effluent total suspended solids
Nitrification is the biological oxidation of ammonia, generally
producing nitrite and then nitrate. It is performed primarily by
autotrophic microorganisms and requires adequate dissolved
oxygen, suitable temperature, sufficient solids retention time,
and appropriate environmental conditions.
4. A clarifier has a surface area of 10,000 ft² and receives 2.5
MGD. What is its approximate surface overflow rate?
,A. 25 gpd/ft²
B. 100 gpd/ft²
C. 250 gpd/ft²
D. 400 gpd/ft²
Surface overflow rate is calculated as flow in gallons per day
divided by surface area. Thus, 2,500,000 gal/day ÷ 10,000 ft² =
250 gpd/ft². This parameter is important because excessive
overflow rates can contribute to poor solids settling and
carryover.
5. A secondary clarifier begins experiencing a gradual increase
in effluent suspended solids while the sludge blanket remains
relatively low. Which condition should the operator
investigate first?
A. Excessively high chlorine residual
B. Poor floc formation or dispersed growth
C. Excessive digester gas production
D. Low influent screening efficiency
When effluent TSS increases without a high sludge blanket, the
problem may be related to the quality of the biological floc
rather than simply solids accumulation in the clarifier. Dispersed
growth, pin floc, or other poor-settling conditions can allow fine
solids to escape with the effluent.
, 6. A plant's influent flow is 4.0 MGD, and the influent TSS
concentration is 250 mg/L. What is the approximate influent
TSS loading?
A. 417 lb/day
B. 834 lb/day
C. 8,340 lb/day
D. 16,680 lb/day
Using the loading equation, 4.0 MGD × 250 mg/L × 8.34 = 8,340
lb/day. Correct loading calculations are fundamental for
evaluating treatment capacity, solids production, and process
performance.
7. Which activated-sludge parameter is most directly
associated with the amount of biomass maintained in the
aeration system relative to incoming biodegradable organic
matter?
A. SVI
B. DO
C. F/M ratio
D. Chlorine residual
The food-to-microorganism (F/M) ratio compares the organic
substrate available to the amount of microorganisms
