Combined Wastewater Treatment Plant Operator
MA Combined Wastewater Treatment Plant
Operator Grade 5 – Practice Examination
(Questions 1–40) Questions with Correct
DETAILED DETAILED ANSWERs and Explanations.
Domain 1: Preliminary & Primary Treatment
1. A grit chamber is designed to remove particles with a specific gravity of
approximately 2.65 and a diameter greater than 0.21 mm. If the flow through a
velocity-controlled grit channel is 9.0 MGD, and the design parameter is 1.0 ft of
length per 1.0 MGD of flow, what is the theoretical minimum required length of
the channel?
A) 3.0 ft
B) 6.0 ft
C) 9.0 ft
D) 18.0 ft
DETAILED ANSWER: C) 9.0 ft
Explanation: This is a direct application of a standard design rule of thumb for aerated
or velocity-controlled grit chambers: approximately 1 foot of channel length is needed
per MGD of peak flow to achieve adequate detention time for grit settling. With a flow
of 9.0 MGD, the required length is 9.0 MGD * 1 ft/MGD = 9.0 ft. This ensures sufficient
,time for heavier inorganic solids (sand, silt, eggshells) to settle out while organic solids
remain suspended.
2. A primary clarifier is 80 feet in diameter and has an average side water depth of
12 feet. The incoming flow is 4.0 MGD. What is the detention time in hours?
A) 1.5 hours
B) 2.7 hours
C) 3.6 hours
D) 4.5 hours
DETAILED ANSWER: B) 2.7 hours
Explanation:
Step 1: Calculate the volume of the clarifier in millions of gallons (MG).
Area = π * (diameter/2)² = 3.14 * (40 ft)² = 5,024 sq ft.
Volume = Area * Depth = 5,024 sq ft * 12 ft = 60,288 cubic feet.
Convert to gallons: 60,288 ft³ * 7.48 gal/ft³ = 450,954 gallons = 0.451 MG.
Step 2: Calculate Detention Time.
Detention Time (hours) = (Volume, MG) / (Flow, MGD) * 24 hours/day.
DT = (0.451 MG / 4.0 MGD) * 24 = 0.11275 * 24 = 2.706 hours ≈ 2.7 hours.
This is within the typical design range of 1.5–2.5 hours for primary clarification, allowing
adequate settling of settleable solids. The slightly higher value suggests a conservatively
designed unit.
3. In a comminutor, what is the primary mechanism of solids size reduction?
, A) High-pressure hydraulic shear
B) Biological degradation by attached growth
C) Rotating cutting teeth or screens that shred solids
D) Chemical oxidation with chlorine
DETAILED ANSWER: C) Rotating cutting teeth or screens that shred solids
Explanation: A comminutor is a mechanical device installed in the influent channel,
upstream of primary treatment. Its purpose is to shred or cut large suspended solids
(rags, plastics, debris) into smaller, more uniform particles without removing them from
the flow stream. This protects downstream equipment (pumps, sludge scrapers) from
clogging. It differs from a bar screen, which removes debris. The solids remain in the
wastewater after comminution.
4. A plant operator observes that the primary sludge pump is losing prime and
delivering erratic flow. The pump is a centrifugal type located above the wet well.
Which is the most likely cause?
A) The sludge is too thin (< 1% solids).
B) Air is leaking into the suction line.
C) The discharge valve is fully open.
D) The impeller is rotating too fast.
DETAILED ANSWER: B) Air is leaking into the suction line.
Explanation: A centrifugal pump located above the liquid source must create negative
pressure (suction) to lift the sludge. Any leak in the suction piping—through a failed
gasket, loose fitting, or crack—will allow atmospheric air to enter the line. Because air is
compressible, the pump loses its ability to create the necessary vacuum, leading to loss
MA Combined Wastewater Treatment Plant
Operator Grade 5 – Practice Examination
(Questions 1–40) Questions with Correct
DETAILED DETAILED ANSWERs and Explanations.
Domain 1: Preliminary & Primary Treatment
1. A grit chamber is designed to remove particles with a specific gravity of
approximately 2.65 and a diameter greater than 0.21 mm. If the flow through a
velocity-controlled grit channel is 9.0 MGD, and the design parameter is 1.0 ft of
length per 1.0 MGD of flow, what is the theoretical minimum required length of
the channel?
A) 3.0 ft
B) 6.0 ft
C) 9.0 ft
D) 18.0 ft
DETAILED ANSWER: C) 9.0 ft
Explanation: This is a direct application of a standard design rule of thumb for aerated
or velocity-controlled grit chambers: approximately 1 foot of channel length is needed
per MGD of peak flow to achieve adequate detention time for grit settling. With a flow
of 9.0 MGD, the required length is 9.0 MGD * 1 ft/MGD = 9.0 ft. This ensures sufficient
,time for heavier inorganic solids (sand, silt, eggshells) to settle out while organic solids
remain suspended.
2. A primary clarifier is 80 feet in diameter and has an average side water depth of
12 feet. The incoming flow is 4.0 MGD. What is the detention time in hours?
A) 1.5 hours
B) 2.7 hours
C) 3.6 hours
D) 4.5 hours
DETAILED ANSWER: B) 2.7 hours
Explanation:
Step 1: Calculate the volume of the clarifier in millions of gallons (MG).
Area = π * (diameter/2)² = 3.14 * (40 ft)² = 5,024 sq ft.
Volume = Area * Depth = 5,024 sq ft * 12 ft = 60,288 cubic feet.
Convert to gallons: 60,288 ft³ * 7.48 gal/ft³ = 450,954 gallons = 0.451 MG.
Step 2: Calculate Detention Time.
Detention Time (hours) = (Volume, MG) / (Flow, MGD) * 24 hours/day.
DT = (0.451 MG / 4.0 MGD) * 24 = 0.11275 * 24 = 2.706 hours ≈ 2.7 hours.
This is within the typical design range of 1.5–2.5 hours for primary clarification, allowing
adequate settling of settleable solids. The slightly higher value suggests a conservatively
designed unit.
3. In a comminutor, what is the primary mechanism of solids size reduction?
, A) High-pressure hydraulic shear
B) Biological degradation by attached growth
C) Rotating cutting teeth or screens that shred solids
D) Chemical oxidation with chlorine
DETAILED ANSWER: C) Rotating cutting teeth or screens that shred solids
Explanation: A comminutor is a mechanical device installed in the influent channel,
upstream of primary treatment. Its purpose is to shred or cut large suspended solids
(rags, plastics, debris) into smaller, more uniform particles without removing them from
the flow stream. This protects downstream equipment (pumps, sludge scrapers) from
clogging. It differs from a bar screen, which removes debris. The solids remain in the
wastewater after comminution.
4. A plant operator observes that the primary sludge pump is losing prime and
delivering erratic flow. The pump is a centrifugal type located above the wet well.
Which is the most likely cause?
A) The sludge is too thin (< 1% solids).
B) Air is leaking into the suction line.
C) The discharge valve is fully open.
D) The impeller is rotating too fast.
DETAILED ANSWER: B) Air is leaking into the suction line.
Explanation: A centrifugal pump located above the liquid source must create negative
pressure (suction) to lift the sludge. Any leak in the suction piping—through a failed
gasket, loose fitting, or crack—will allow atmospheric air to enter the line. Because air is
compressible, the pump loses its ability to create the necessary vacuum, leading to loss
