16 Preliminary & Primary Treatment Practice Questions & Answers

1. 1. A mechanically cleaned bar screen is taken out of service and raw wastewater is routed through the open bypass channel for two days. Which downstream problem should the operator expect first?

   • A.Heavy grit deposits building up on the aeration basin floor
   • B.Rags and debris binding the raw wastewater pump impellers✓ Answer
   • C.A sharp rise in dissolved BOD reaching the primary clarifiers
   • D.A rapid drop in influent pH caused by decaying screenings

   A bar screen's function is to intercept rags, sticks and other large debris before they reach pumps, valves and mechanical equipment, so bypassing it shows up first as rag fouling and binding of the raw wastewater pumps. Screens remove essentially no dissolved BOD and are not designed to capture grit, which is handled in the grit chamber downstream.

   Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., "Racks, Screens, Comminutors and Grit Removal" (purpose of bar screens); EPA 832-F-03-011, Screening and Grit RemovalReport a problem with this question

2. 2. Why is grit removed at the headworks instead of being allowed to settle out in later treatment units?

   • A.Grit exerts a high oxygen demand that depresses aeration basin DO
   • B.Grit dissolves at neutral pH and releases sulfide into the plant flow
   • C.Grit is inert and abrasive; it wears pumps and consumes tank volume✓ Answer
   • D.Grit blinds the bar screen and drives the headloss across it upward

   Grit is inert, dense mineral material (sand, gravel, eggshells, coffee grounds) that abrades pump impellers and piping and, if it passes the headworks, accumulates in channels, clarifiers and digesters where it steals working volume and forces expensive cleanouts. It is not readily biodegradable, so it exerts little oxygen demand, and it is removed after screening, not before it.

   Source: EPA 832-F-03-011, Screening and Grit Removal; CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., grit removal purposeReport a problem with this question

3. 3. The measured velocity through a channel-type grit chamber has fallen to about 0.5 ft/s (0.15 m/s). What should the operator expect to find in the grit hopper?

   • A.Grease and floatable material collecting in place of settled grit
   • B.Putrescible organic solids settling with the grit and causing odors✓ Answer
   • C.Clean, washed grit with almost no organic material remaining
   • D.Very little grit, because the flow scours the channel floor clean

   A channel grit chamber is controlled to hold a velocity near 1.0 ft/s (0.3 m/s), which is fast enough to keep lighter organic solids in suspension while dense grit still settles. At roughly half that velocity the organics settle too, so the hopper fills with putrescible, odorous material that must be washed or handled as sludge; at excessive velocity the opposite occurs and grit is carried downstream.

   Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., grit channel control velocity 1.0 ft/s (0.3 m/s); WEF MOP 8, grit removalReport a problem with this question

4. 4. Grit pulled from an aerated grit chamber has become noticeably lighter and the volume removed has dropped, while influent flow and character are unchanged. What adjustment should the operator make first?

   • A.Raise the water level in the chamber to lengthen detention time
   • B.Increase the air rate to strengthen the roll across the chamber
   • C.Increase grit pump run time so the hopper is emptied more often
   • D.Reduce the air rate so grit can settle out of the spiral roll✓ Answer

   In an aerated grit chamber the diffused air sets up a spiral roll sized to keep light organics suspended while dense grit falls out; too much air keeps the grit itself rolling in suspension so it passes downstream and the little that is captured is light and organic. Air rate is the operating control for that unit, so trimming the air back is the first move, whereas too little air would give more grit but fouled with settled organics.

   Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., aerated grit chamber air rate control; EPA 832-F-03-011Report a problem with this question

5. 5. How does a vortex-type grit unit separate grit from the wastewater flow?

   • A.Rising air bubbles float the grit to the surface where it is skimmed off
   • B.An added coagulant binds the grit into floc that settles in the basin
   • C.Fine screen media strain the grit particles out as the flow passes through
   • D.Rotating flow drives dense grit to the floor while organics stay suspended✓ Answer

   A vortex unit admits flow tangentially into a cylindrical chamber and a paddle or turbine maintains a controlled rotation; the resulting centrifugal and gravitational forces move the denser grit particles to the outside and down into a center hopper while lighter organic solids remain suspended and leave with the flow. It is a gravity/inertial separation, not straining, flotation or chemical treatment.

   Source: EPA 832-F-03-011, Screening and Grit Removal (vortex grit chambers); WEF Wastewater Treatment Fundamentals I – Liquid Treatment, preliminary treatmentReport a problem with this question

6. 6. A plant runs a grinder (comminutor) in the influent channel instead of a screen with a screenings washer and hopper. What should the operator expect as a result?

   • A.The shredded material is destroyed, lowering the plant's total solids load
   • B.Grit is cut fine enough to pass through the plant without settling anywhere
   • C.Influent BOD drops sharply because the organic material is broken apart
   • D.The shredded material stays in the flow and reports to the primary sludge✓ Answer

   A comminutor or grinder only reduces particle size; it removes nothing from the wastewater, so the shredded rags and solids stay in the flow and settle out later, mostly in the primary sludge, where they can rope together and plug sludge pumps and lines. Because nothing leaves the plant at the headworks, neither the solids load nor the influent BOD is reduced.

   Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., "Racks, Screens, Comminutors and Grit Removal" (comminution vs. removal)Report a problem with this question

7. 7. A flow equalization basin ahead of the primary clarifiers is drained and taken offline for cleaning. What effect should the operator anticipate during the daily peak flow period?

   • A.Steadier organic loading on the secondary process throughout the day
   • B.Higher clarifier overflow rates and more solids carried over the weirs✓ Answer
   • C.Lower chlorine demand because the load is spread out over the day
   • D.Longer clarifier detention time and thicker primary sludge withdrawn

   Equalization stores the diurnal peak and releases it at a steadier rate, so removing it lets the raw diurnal curve pass straight through: peak-hour flow raises the clarifier surface overflow rate, shortens detention time and lifts settled and light solids over the weirs, while chemical dose and downstream organic loading swing with the flow. Steadier loading and longer detention are what the basin provides, not what its loss produces.

   Source: WEF, Wastewater Treatment Fundamentals I – Liquid Treatment, flow equalization; CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., hydraulic loading effects on sedimentationReport a problem with this question

8. 8. Primary effluent quality has deteriorated and the operator observes a strong sheet of water pouring over one section of the effluent weir while other sections are barely wetted. What is the most likely cause and first correction?

   • A.The blanket is too deep; raise the sludge withdrawal rate at once
   • B.The weir is out of level; check it and adjust it to equalize the flow✓ Answer
   • C.The influent is septic; begin chemical addition at the plant headworks
   • D.The collector is running too fast; slow the sludge collector drive down

   Effluent weirs must be level so flow leaves the tank uniformly around the full weir length; when a weir settles or is set out of level, most of the flow concentrates on the low section, and the resulting high local weir loading creates approach velocities that lift settling and light solids over the plate. Leveling the weir restores uniform withdrawal, which is a different problem from a deep blanket, septicity, or collector speed.

   Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., "Sedimentation and Flotation" (weir loading and weir leveling)Report a problem with this question

9. 9. Dye added to the influent of a primary clarifier appears at the effluent weir well before one detention time has elapsed, and the influent is several degrees colder than the tank contents. What is happening?

   • A.The scum baffle is submerged and is passing floatables over the weir
   • B.The sludge collector is dragging settled solids toward the effluent end
   • C.A density current is carrying influent along the floor toward the outlet✓ Answer
   • D.Nitrogen gas from denitrification is lifting the settled sludge blanket

   Colder or more concentrated influent is denser than the water already in the tank, so it plunges and travels as a distinct layer along the floor to the outlet instead of mixing and using the full basin volume; the dye's early arrival is the classic signature of this short-circuiting. Because part of the tank is bypassed, the effective detention time is far less than the calculated value and removal efficiency falls.

   Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., "Sedimentation and Flotation" (short-circuiting and density currents)Report a problem with this question

10. 10. Gas bubbles and clumps of black sludge are rising to the surface of a primary clarifier and the area smells strongly of hydrogen sulfide. What is the most likely cause?

    • A.Sludge is being held too long in the tank and has turned septic✓ Answer
    • B.Sludge is being pumped too often and is drawing water, not solids
    • C.Scum has been skimmed so often that the baffle no longer holds it
    • D.The surface overflow rate is far below the clarifier's design value

    Primary sludge is highly degradable, and if it is not withdrawn often enough it goes anaerobic in the hopper, where sulfate-reducing and acid-forming bacteria generate hydrogen sulfide and other gases whose bubbles float clumps of black septic sludge to the surface. The correction is to pump the sludge more frequently or for longer cycles, and to keep the scum and sludge collectors in service so nothing sits in the tank.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., "Sedimentation and Flotation" (septic sludge, gasification and sludge withdrawal frequency)Report a problem with this question

11. 11. Primary sludge concentration has fallen from about 4% solids to about 1.5% solids and the volume pumped to solids handling has climbed sharply. What is the most likely cause?

    • A.The tank is short-circuiting and solids are leaving over the weirs
    • B.The sludge blanket is too deep for the collector to move it along
    • C.The influent has turned septic and gasified inside the sludge hopper
    • D.The pump runs too long and draws liquid down through the blanket✓ Answer

    When a primary sludge pump runs past the point where the blanket in the hopper is drawn down, it cones through the sludge and pulls supernatant, so the concentration drops while the volume pumped rises. Thin sludge is a real operating penalty because it hydraulically overloads downstream thickening, digestion and dewatering, and the fix is to shorten the pump cycle or control pumping from blanket depth or sludge density.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., "Sedimentation and Flotation" (primary sludge pumping cycles and coning)Report a problem with this question

12. 12. Influent BOD averages 210 mg/L and primary effluent BOD averages 145 mg/L, with the primary clarifiers otherwise operating normally. How should the operator interpret this?

    • A.Normal; the remaining BOD was already destroyed in the grit chamber
    • B.Abnormal; the sludge blanket must be too deep and releasing BOD back
    • C.Normal; primaries remove settleable solids and the BOD attached to them✓ Answer
    • D.Abnormal; a well-run primary clarifier should remove most of the BOD

    That is about 31% BOD removal, which sits inside the 25-40% range expected of a plain primary clarifier, because sedimentation can only capture settleable and some colloidal material and the BOD attached to it. Dissolved BOD passes straight through and must be removed biologically in the secondary process, which is why heavier primary removal lowers the organic load and F/M carried by the secondary system.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1, 8th ed., "Sedimentation and Flotation" (typical primary removals: 25-40% BOD, 50-70% TSS)Report a problem with this question

13. 13. Raw influent arrives at the headworks black in color and smelling strongly of hydrogen sulfide after a long, flat force main. Which operator response addresses the cause of the odor?

    • A.Add oxygen or an oxidizing chemical to the flow to control sulfide✓ Answer
    • B.Shorten the primary sludge pumping cycles to thin the settled sludge
    • C.Increase the bar screen cleaning frequency to carry the odor away
    • D.Lower the grit chamber velocity so the sulfide-bearing solids settle

    Septicity develops when wastewater sits oxygen-free for a long detention, as in a full force main, and sulfate-reducing bacteria convert sulfate to dissolved sulfide that releases hydrogen sulfide, blackens the flow and corrodes concrete and metal. Restoring an oxygen or oxidant source, by aeration or by feeding a chemical such as hydrogen peroxide, a nitrate salt or an iron salt, attacks the sulfide itself, while screen cleaning, grit velocity and sludge pumping do not.

    Source: EPA/625/1-85/018, Design Manual: Odor and Corrosion Control in Sanitary Sewerage Systems and Treatment Plants (sulfide generation and control)Report a problem with this question

14. 14. A circular primary clarifier is 60 ft (18.3 m) in diameter and receives a flow of 1.2 MGD (4,540 m3/day). What is the surface overflow rate?

    • A.848 gpd/ft2 (34.6 m3/m2/day)
    • B.106 gpd/ft2 (4.3 m3/m2/day)
    • C.424 gpd/ft2 (17.3 m3/m2/day)✓ Answer
    • D.212 gpd/ft2 (8.6 m3/m2/day)

    Surface overflow rate is flow divided by surface area, and the area of a circular tank is 0.785 x diameter squared: 0.785 x (60 ft)^2 = 2,827 ft2, so 1,200,000 gpd / 2,827 ft2 = 424 gpd/ft2. In metric, 0.785 x (18.3 m)^2 = 263 m2 and 4,540 m3/day / 263 m2 = 17.3 m3/m2/day; the common error is using the diameter in place of the radius in pi times r squared, which gives the 106 answer.

    Source: WPI Wastewater Formula/Conversion Table: Surface Overflow Rate = Flow / Surface Area; area of circle = 0.785 x D squaredReport a problem with this question

15. 15. A rectangular primary clarifier measures 80 ft long by 20 ft wide with a 10 ft side water depth (24.4 m by 6.1 m by 3.05 m) and receives 1.0 MGD (3,785 m3/day). What is the detention time?

    • A.2.9 hours✓ Answer
    • B.0.4 hours
    • C.5.7 hours
    • D.1.4 hours

    Detention time is tank volume divided by flow, so the volume must be expressed in the same units as the flow: 80 x 20 x 10 = 16,000 ft3 x 7.48 gal/ft3 = 119,680 gal, and 119,680 gal / 1,000,000 gpd = 0.12 day, or about 2.9 hours. In metric, 24.4 x 6.1 x 3.05 = 454 m3 and 454 / 3,785 m3/day = 0.12 day = 2.9 hours; forgetting the 7.48 gal/ft3 conversion produces the 0.4 hour answer.

    Source: WPI Wastewater Formula/Conversion Table: Detention Time = Volume / Flow; 7.48 gal/ft3Report a problem with this question

16. 16. Primary influent TSS is 240 mg/L and primary effluent TSS is 90 mg/L at a flow of 2.0 MGD (7,570 m3/day). How much TSS is being removed by the primary clarifiers?

    • A.2,502 lb/day (1,135 kg/day)✓ Answer
    • B.4,004 lb/day (1,816 kg/day)
    • C.1,251 lb/day (568 kg/day)
    • D.300 lb/day (136 kg/day)

    Mass removed uses the concentration difference, not the influent value: (240 - 90) = 150 mg/L, and lb/day = mg/L x MGD x 8.34 = 150 x 2.0 x 8.34 = 2,502 lb/day. In metric, kg/day = mg/L x m3/day / 1,000 = 150 x 7,570 / 1,000 = 1,135 kg/day; using 240 mg/L instead of the difference gives 4,004 lb/day and dropping the 8.34 factor gives 300 lb/day.

    Source: WPI Wastewater Formula/Conversion Table: lbs/day = mg/L x MGD x 8.34; kg/day = mg/L x m3/day / 1,000Report a problem with this question

Practice questions written against the 2025 standardized Wastewater Treatment Operator Need-to-Know Criteria published by Water Professionals International (formerly the Association of Boards of Certification) and standard references from the CSUS Office of Water Programs and the Water Environment Federation. This site is not affiliated with or endorsed by WPI/ABC, WEF, or the US EPA. Operator certification is issued by your state's certifying authority, which sets plant classification tiers, operator grades, eligibility, and the passing standard — confirm those with your state before testing, and confirm which edition of the exam your program has adopted. Effluent limits, monitoring frequencies and reporting requirements come from an individual discharge permit and are revised over time, so no answer here should be relied on as a current regulatory or permit value; consult the permit in force for your facility. This bank covers the wastewater treatment exam only — drinking-water treatment, water distribution, and wastewater collection are separate certifications. About the Need-to-Know Criteria →