Manitoba Wastewater Treatment Operator Level
II Exam 100 Verified Q&A ACTUAL EXAM
2026/2027 | MB Wastewater Level II | Verified
Q&A | Pass Guaranteed - A+ Graded
Section 1: Preliminary & Primary Treatment (Questions 1–12)
Q1: An operator at a Manitoba activated sludge plant notices excessive grit accumulation in the primary
clarifier influent channel. The aerated grit chamber's air flow is set to 0.8 m³/min. What is the most
appropriate immediate adjustment?
A. Decrease air flow to 0.4 m³/min [CORRECT]
B. Increase air flow to 1.2 m³/min
C. Bypass the grit chamber temporarily
D. Increase primary clarifier rake speed
Correct Answer: A
Rationale: Reducing air flow in an aerated grit chamber decreases horizontal velocity, allowing heavier
inorganic grit to settle while keeping lighter organics in suspension, which prevents grit carryover to
primary clarifiers.
Q2: A primary clarifier has a diameter of 20 m and a design flow of 4,000 m³/d. What is the surface
overflow rate (SOR) in m³/m²/d?
A. 8.5 m³/m²/d
B. 10.2 m³/m²/d
C. 12.7 m³/m²/d [CORRECT]
D. 15.9 m³/m²/d
Correct Answer: C
Rationale: Surface area equals πr² equals 3.14 times (10 m)² equals 314 m². SOR equals flow divided by
area equals 4,000 divided by 314 equals 12.74 m³/m²/d, which is within typical primary clarifier design
range for Manitoba municipal plants.
,Q3: A bar screen at a Winnipeg-area plant is clogged with debris, causing upstream flooding in the
influent channel. The automatic rake is operational but moving slowly. Which action should the operator
take first?
A. Increase the rake cycle frequency to continuous operation [CORRECT]
B. Shut down the influent pump station
C. Bypass the screen and send raw sewage directly to primary clarifiers
D. Add chlorine upstream of the screen
Correct Answer: A
Rationale: Increasing the rake cycle frequency to continuous operation addresses the immediate debris
accumulation without creating a bypass situation that violates Manitoba Environment Act discharge
requirements or disrupts downstream treatment processes.
Q4: A primary clarifier has a detention time of 2.5 hours at a flow of 3,500 m³/d. If the flow increases to
5,000 m³/d, what is the new detention time?
A. 1.25 hours
B. 1.5 hours
C. 1.75 hours [CORRECT]
D. 2.0 hours
Correct Answer: C
Rationale: Detention time is inversely proportional to flow rate. New detention time equals original
detention time times original flow divided by new flow, which equals 2.5 hours times 3,500 divided by
5,000 equals 1.75 hours.
Q5: An operator notices that primary clarifier effluent has elevated TSS of 180 mg/L when the typical
range is 80–120 mg/L. The sludge blanket level is 0.4 m. Which is the most likely cause?
A. Excessive primary sludge removal
B. Insufficient primary sludge removal causing sludge to rise [CORRECT]
C. Too much alum being added
D. Excessive aeration in the primary clarifier
Correct Answer: B
,Rationale: A sludge blanket level of 0.4 m indicates accumulated solids that can rise and wash out when
anaerobic conditions develop, causing elevated TSS in the effluent; increasing sludge removal frequency
will reduce the blanket level and improve clarifier performance.
Q6: A comminutor at a Brandon-area plant jams frequently during peak morning flows. The operator
notices large amounts of wipes and fibrous material in the screenings. Which operational adjustment is
most effective?
A. Increase the comminutor cutting speed
B. Install a fine screen upstream and implement public education about flushable wipes [CORRECT]
C. Reduce influent flow during morning peaks
D. Add chemical coagulant upstream
Correct Answer: B
Rationale: Wipes and fibrous materials do not break down effectively in comminutors and cause
repeated jamming; a fine screen removes these materials before they reach the comminutor, and public
education reduces future loading of non-flushable items.
Q7: A primary clarifier has a side water depth of 3.5 m and a diameter of 18 m. What is the approximate
volume in cubic metres?
A. 650 m³
B. 890 m³ [CORRECT]
C. 1,120 m³
D. 1,450 m³
Correct Answer: B
Rationale: Volume equals surface area times depth equals π times (9 m)² times 3.5 m equals 3.14 times
81 times 3.5 equals 890 m³, which is the operational volume available for sedimentation.
Q8: An operator measures the influent BOD₅ as 220 mg/L and the primary effluent BOD₅ as 150 mg/L.
What is the BOD removal efficiency of the primary clarifier?
A. 22%
B. 32% [CORRECT]
C. 42%
D. 52%
, Correct Answer: B
Rationale: Removal efficiency equals (influent minus effluent) divided by influent times 100, which
equals (220 minus 150) divided by 220 times 100 equals 31.8%, rounded to 32%, which is typical for
primary clarification.
Q9: A grit chamber is designed to remove particles with a specific gravity of 2.65 and a diameter of 0.2
mm. The operator notices that sand particles are settling but organic matter is also accumulating in the
grit hopper. Which adjustment is most appropriate?
A. Increase the horizontal velocity through the chamber
B. Decrease the detention time in the grit chamber [CORRECT]
C. Increase the channel depth
D. Add aeration to the existing chamber
Correct Answer: B
Rationale: Decreasing detention time reduces the opportunity for lighter organic particles to settle while
still allowing heavier inorganic grit to reach the hopper, which improves grit purity and reduces volatile
solids in the grit removed.
Q10: A primary clarifier weir is set at an elevation that creates a weir loading rate of 350 m³/m/d at
design flow. During wet weather events, the flow doubles. What happens to the weir loading rate?
A. It remains at 350 m³/m/d
B. It decreases to 175 m³/m/d
C. It increases to 700 m³/m/d [CORRECT]
D. It increases to 525 m³/m/d
Correct Answer: C
Rationale: Weir loading rate is directly proportional to flow rate when weir length remains constant;
doubling the flow doubles the weir loading rate to 700 m³/m/d, which may cause turbulence and
reduced settling efficiency during wet weather.
Q11: An operator is checking the velocity through a Parshall flume and records 0.45 m/s at a flow of 120
L/s. If the flow increases to 180 L/s, what is the expected velocity?
A. 0.55 m/s
B. 0.60 m/s
II Exam 100 Verified Q&A ACTUAL EXAM
2026/2027 | MB Wastewater Level II | Verified
Q&A | Pass Guaranteed - A+ Graded
Section 1: Preliminary & Primary Treatment (Questions 1–12)
Q1: An operator at a Manitoba activated sludge plant notices excessive grit accumulation in the primary
clarifier influent channel. The aerated grit chamber's air flow is set to 0.8 m³/min. What is the most
appropriate immediate adjustment?
A. Decrease air flow to 0.4 m³/min [CORRECT]
B. Increase air flow to 1.2 m³/min
C. Bypass the grit chamber temporarily
D. Increase primary clarifier rake speed
Correct Answer: A
Rationale: Reducing air flow in an aerated grit chamber decreases horizontal velocity, allowing heavier
inorganic grit to settle while keeping lighter organics in suspension, which prevents grit carryover to
primary clarifiers.
Q2: A primary clarifier has a diameter of 20 m and a design flow of 4,000 m³/d. What is the surface
overflow rate (SOR) in m³/m²/d?
A. 8.5 m³/m²/d
B. 10.2 m³/m²/d
C. 12.7 m³/m²/d [CORRECT]
D. 15.9 m³/m²/d
Correct Answer: C
Rationale: Surface area equals πr² equals 3.14 times (10 m)² equals 314 m². SOR equals flow divided by
area equals 4,000 divided by 314 equals 12.74 m³/m²/d, which is within typical primary clarifier design
range for Manitoba municipal plants.
,Q3: A bar screen at a Winnipeg-area plant is clogged with debris, causing upstream flooding in the
influent channel. The automatic rake is operational but moving slowly. Which action should the operator
take first?
A. Increase the rake cycle frequency to continuous operation [CORRECT]
B. Shut down the influent pump station
C. Bypass the screen and send raw sewage directly to primary clarifiers
D. Add chlorine upstream of the screen
Correct Answer: A
Rationale: Increasing the rake cycle frequency to continuous operation addresses the immediate debris
accumulation without creating a bypass situation that violates Manitoba Environment Act discharge
requirements or disrupts downstream treatment processes.
Q4: A primary clarifier has a detention time of 2.5 hours at a flow of 3,500 m³/d. If the flow increases to
5,000 m³/d, what is the new detention time?
A. 1.25 hours
B. 1.5 hours
C. 1.75 hours [CORRECT]
D. 2.0 hours
Correct Answer: C
Rationale: Detention time is inversely proportional to flow rate. New detention time equals original
detention time times original flow divided by new flow, which equals 2.5 hours times 3,500 divided by
5,000 equals 1.75 hours.
Q5: An operator notices that primary clarifier effluent has elevated TSS of 180 mg/L when the typical
range is 80–120 mg/L. The sludge blanket level is 0.4 m. Which is the most likely cause?
A. Excessive primary sludge removal
B. Insufficient primary sludge removal causing sludge to rise [CORRECT]
C. Too much alum being added
D. Excessive aeration in the primary clarifier
Correct Answer: B
,Rationale: A sludge blanket level of 0.4 m indicates accumulated solids that can rise and wash out when
anaerobic conditions develop, causing elevated TSS in the effluent; increasing sludge removal frequency
will reduce the blanket level and improve clarifier performance.
Q6: A comminutor at a Brandon-area plant jams frequently during peak morning flows. The operator
notices large amounts of wipes and fibrous material in the screenings. Which operational adjustment is
most effective?
A. Increase the comminutor cutting speed
B. Install a fine screen upstream and implement public education about flushable wipes [CORRECT]
C. Reduce influent flow during morning peaks
D. Add chemical coagulant upstream
Correct Answer: B
Rationale: Wipes and fibrous materials do not break down effectively in comminutors and cause
repeated jamming; a fine screen removes these materials before they reach the comminutor, and public
education reduces future loading of non-flushable items.
Q7: A primary clarifier has a side water depth of 3.5 m and a diameter of 18 m. What is the approximate
volume in cubic metres?
A. 650 m³
B. 890 m³ [CORRECT]
C. 1,120 m³
D. 1,450 m³
Correct Answer: B
Rationale: Volume equals surface area times depth equals π times (9 m)² times 3.5 m equals 3.14 times
81 times 3.5 equals 890 m³, which is the operational volume available for sedimentation.
Q8: An operator measures the influent BOD₅ as 220 mg/L and the primary effluent BOD₅ as 150 mg/L.
What is the BOD removal efficiency of the primary clarifier?
A. 22%
B. 32% [CORRECT]
C. 42%
D. 52%
, Correct Answer: B
Rationale: Removal efficiency equals (influent minus effluent) divided by influent times 100, which
equals (220 minus 150) divided by 220 times 100 equals 31.8%, rounded to 32%, which is typical for
primary clarification.
Q9: A grit chamber is designed to remove particles with a specific gravity of 2.65 and a diameter of 0.2
mm. The operator notices that sand particles are settling but organic matter is also accumulating in the
grit hopper. Which adjustment is most appropriate?
A. Increase the horizontal velocity through the chamber
B. Decrease the detention time in the grit chamber [CORRECT]
C. Increase the channel depth
D. Add aeration to the existing chamber
Correct Answer: B
Rationale: Decreasing detention time reduces the opportunity for lighter organic particles to settle while
still allowing heavier inorganic grit to reach the hopper, which improves grit purity and reduces volatile
solids in the grit removed.
Q10: A primary clarifier weir is set at an elevation that creates a weir loading rate of 350 m³/m/d at
design flow. During wet weather events, the flow doubles. What happens to the weir loading rate?
A. It remains at 350 m³/m/d
B. It decreases to 175 m³/m/d
C. It increases to 700 m³/m/d [CORRECT]
D. It increases to 525 m³/m/d
Correct Answer: C
Rationale: Weir loading rate is directly proportional to flow rate when weir length remains constant;
doubling the flow doubles the weir loading rate to 700 m³/m/d, which may cause turbulence and
reduced settling efficiency during wet weather.
Q11: An operator is checking the velocity through a Parshall flume and records 0.45 m/s at a flow of 120
L/s. If the flow increases to 180 L/s, what is the expected velocity?
A. 0.55 m/s
B. 0.60 m/s
