22 Secondary Treatment & Activated Sludge Practice Questions & Answers

1. 1. Flow and load have been steady, but MLSS has climbed to 3,600 mg/L and MCRT is well above the plant's target. Which single adjustment will actually reduce the total solids inventory held in the system?

   • A.Increase the return activated sludge (RAS) rate
   • B.Increase the waste activated sludge (WAS) rate✓ Answer
   • C.Increase the air supply to the aeration basin
   • D.Decrease the return activated sludge (RAS) rate

   Only wasting physically removes solids from the plant. RAS simply moves solids back and forth between the final clarifier and the aeration basin, changing where the solids sit but not how many the system holds. Raising the WAS rate lowers MLSS and MCRT and raises F/M.

   Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1 (activated sludge process control); WPI 2025 Class II NTK, Treatment Process Evaluation and AdjustmentReport a problem with this question

2. 2. On a warm afternoon, clumps of sludge with gas bubbles clinging to them rise to the surface of the final clarifier and drift toward the weirs. A settleometer test on the mixed liquor shows a crisp interface, a clear supernatant and an SVI of 95. The most likely cause is:

   • A.Filamentous bulking caused by low dissolved oxygen
   • B.Denitrification inside the clarifier sludge blanket✓ Answer
   • C.Nutrient deficiency producing zoogleal slime growth
   • D.Hydraulic overload washing the blanket over the weirs

   Good settleability (SVI 95, crisp interface, clear supernatant) rules out bulking, so the sludge itself settles well. Nitrate carried into the clarifier is reduced to nitrogen gas in the oxygen-free blanket, and the gas bubbles float otherwise good sludge to the surface — classic rising sludge.

   Source: EPA Activated Sludge Process Control Troubleshooting (rising sludge); Maine DEP Activated Sludge Troubleshooting Guide (Schuyler/Jenkins conditions)Report a problem with this question

3. 3. Rising sludge from denitrification has been confirmed in a final clarifier. Which first operator response addresses the mechanism directly?

   • A.Increase the RAS rate to shorten blanket detention time✓ Answer
   • B.Raise basin DO to about 5 mg/L to add more oxygen
   • C.Decrease the RAS rate to let the blanket thicken more
   • D.Decrease wasting to build more mixed liquor solids

   Denitrification needs time in an oxygen-free blanket. Pulling sludge out faster with a higher RAS rate shortens that residence time so nitrogen gas has no chance to form and lift the floc. Adding air to the aeration basin does not oxygenate the clarifier blanket; the longer-term fix is an upstream anoxic zone.

   Source: EPA Activated Sludge Process Control Troubleshooting (rising sludge corrective actions); CSUS, Operation of Wastewater Treatment Plants, Vol. 1Report a problem with this question

4. 4. SVI has climbed to 240, the clarifier blanket is rising and effluent TSS is up. A diluted settleometer shows little improvement over the undiluted test, the microscope shows filaments bridging between floc particles, and mid-basin DO reads 0.4 mg/L. The diagnosis and first correction are:

   • A.Low-DO filamentous bulking; add air to hold DO at 2 mg/L✓ Answer
   • B.Rising sludge; raise the RAS rate to drop the blanket
   • C.Toxic shock; take the aeration basin out of service
   • D.Solids overload; increase wasting sharply to lower MLSS

   The diluted settleometer is the discriminator: if dilution had improved settling dramatically, the problem would be simple solids overload. Little improvement plus visible interfloc bridging means true filamentous bulking, and a mid-basin DO of 0.4 mg/L points to the low-DO filament group, so raising air to hold a working DO near 2 mg/L is the first correction.

   Source: Jenkins/Schuyler filament identification as reproduced in Maine DEP Activated Sludge Troubleshooting Guide (low-DO bulking: Sphaerotilus natans, type 1701); TCEQ RG-002 aeration basin DO 1–4 mg/LReport a problem with this question

5. 5. Thick, dark brown, greasy foam covers much of the aeration basin surface. MCRT has drifted to 22 days against a target of 10 days, F/M is low and SVI is 130. The appropriate response is:

   • A.Increase the RAS rate and lower the basin dissolved oxygen
   • B.Increase wasting and physically remove the foam✓ Answer
   • C.Dose a nutrient supplement and reduce the air supply
   • D.Decrease wasting and knock the foam down with sprays

   Dark brown, thick, leathery or greasy foam is the signature of an old sludge carrying Nocardia-type or Microthrix filaments favored by long MCRT and low F/M, so wasting must increase to shorten sludge age. The organisms concentrate in the foam itself and are not carried out with the WAS, so the foam has to be physically removed as well.

   Source: Maine DEP Activated Sludge Troubleshooting Guide (foam color/condition table); EPA Activated Sludge Troubleshooting (Nocardia/Microthrix foaming)Report a problem with this question

6. 6. Final clarifier effluent is turbid, with tiny dark pinpoint particles carrying over the weirs. The settleometer settles quickly to a small compact volume but leaves a cloudy supernatant, and MCRT has been running long. This is:

   • A.Young straggler floc; decrease wasting to raise MCRT
   • B.Filamentous bulking; chlorinate the return sludge line
   • C.Old sludge pin floc; increase wasting to cut MCRT✓ Answer
   • D.Rising sludge; increase the return sludge pumping rate

   Pin floc is the fingerprint of an over-oxidized, underloaded, old sludge: extended endogenous respiration shears the floc into small dense fragments that settle fast, while the fines that will not flocculate stay suspended and turn the effluent turbid. Increasing the wasting rate shortens MCRT and rebuilds younger, better-flocculating sludge.

   Source: Maine DEP Activated Sludge Troubleshooting Guide (pin floc conditions); CSUS, Operation of Wastewater Treatment Plants, Vol. 1Report a problem with this question

7. 7. Two weeks after a solids washout, MLSS is still low and large, light, fluffy floc particles drift over the clarifier weirs. The settleometer settles slowly with a fuzzy, poorly defined interface. The appropriate response is:

   • A.Increase RAS to move still more solids to the basin
   • B.Increase air to hold basin DO well above 4 mg/L
   • C.Increase wasting so a younger, lighter sludge forms
   • D.Reduce wasting so an older, denser sludge develops✓ Answer

   Straggler floc is the young-sludge symptom: at a high F/M the biomass grows fast, floc is large, light and buoyant, and the settling interface stays fuzzy. Cutting back the wasting rate lets sludge age and MLSS build so the floc becomes denser and settles cleanly; over-aeration would only shear it further.

   Source: Maine DEP Activated Sludge Troubleshooting Guide (straggler floc conditions); CSUS, Operation of Wastewater Treatment Plants, Vol. 1Report a problem with this question

8. 8. A settleometer filled with mixed liquor settles to 300 mL/L in 30 minutes, and the MLSS is 2,400 mg/L. Using SVI = (SSV30, mL/L × 1,000 mg/g) ÷ MLSS, mg/L, what is the sludge volume index?

   • A.8,000 mL/g
   • B.125 mL/g✓ Answer
   • C.125,000 mL/g
   • D.0.125 mL/g

   SVI = (300 × 1,000) ÷ 2,400 = 125 mL/g. The 1,000 mg/g conversion is what makes the units come out as millilitres occupied by one gram of sludge; leaving it out gives 0.125 and applying it to the wrong side or inverting the ratio gives the other choices, which is why this item is a units test rather than an algebra test.

   Source: WPI 2025 Formula/Conversion Table — SVI = (SSV30, mL/L × 1,000 mg/g) ÷ MLSS, mg/LReport a problem with this question

9. 9. An aeration basin holds 0.75 MG (2,840 m3) of mixed liquor at 2,800 mg/L MLSS and 2,100 mg/L MLVSS, and receives 2.0 MGD (7,570 m3/d) of primary effluent at 150 mg/L BOD5. What is the F/M ratio, in lb BOD5 per lb MLVSS per day (kg/kg per day)?

   • A.0.19✓ Answer
   • B.1.9
   • C.0.14
   • D.5.3

   Food = 2.0 × 150 × 8.34 = 2,502 lb BOD5/day; microorganisms = 0.75 × 2,100 × 8.34 = 13,135 lb MLVSS; 2,502 ÷ 13,135 = 0.19. F/M is defined on the volatile fraction, so using the 2,800 mg/L MLSS instead gives 0.14 — the single most common error on this calculation — and inverting the ratio gives 5.3.

   Source: WPI 2025 Formula/Conversion Table — F/M = lb BOD5/day ÷ lb MLVSS; lb/day = (MGD)(mg/L)(8.34)Report a problem with this question

10. 10. An operator doubles the waste activated sludge rate and holds flow, load, RAS and air unchanged. After several days at the new rate, what has happened to the process?

    • A.MLSS and MCRT rise while F/M falls
    • B.MLSS falls while MCRT and F/M both rise
    • C.MLSS rises while MCRT falls and F/M rises
    • D.MLSS and MCRT fall while F/M rises✓ Answer

    Wasting removes solids inventory, so MLSS drops and the solids in the system divided by the solids removed each day — the MCRT — gets shorter. Because the incoming BOD load is unchanged while the mass of organisms carrying it has fallen, each pound of MLVSS now sees more food, so F/M rises and the sludge gets younger.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1 (activated sludge process control levers); WPI 2025 Class II NTK Application exemplar on sludge wastingReport a problem with this question

11. 11. Compared with a conventional activated sludge plant treating the same wastewater, an extended aeration plant is characterized by:

    • A.Longer aeration time and less, better stabilized waste sludge✓ Answer
    • B.A higher F/M loading with a much shorter solids retention time
    • C.Greater primary sludge production with faster solids turnover
    • D.Shorter aeration time and more, less stabilized waste sludge

    Extended aeration deliberately runs at a low F/M and a long sludge age, so the biomass spends much of its time in endogenous respiration, consuming its own cell mass. That produces a smaller quantity of well-stabilized waste sludge and reliable nitrification, at the cost of more aeration energy and a risk of low effluent pH in low-alkalinity water.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1 (extended aeration); WPI 2025 Class II NTK, secondary suspended-growth processesReport a problem with this question

12. 12. Why does a complete-mix aeration basin generally tolerate a slug organic load better than a plug-flow basin of the same volume?

    • A.Its return sludge is introduced at the far end of the tank
    • B.It is designed to carry a much longer solids retention time
    • C.The incoming load is diluted at once throughout the tank✓ Answer
    • D.It holds a much higher dissolved oxygen at the inlet end

    In a complete-mix basin the influent is dispersed through the whole tank as it enters, so F/M and oxygen demand are uniform and a slug load is diluted by the entire mixed liquor volume instead of hitting an inlet zone. The same uniformly low substrate gradient is also why complete-mix basins are the variant most prone to filamentous bulking.

    Source: WEF, Wastewater Treatment Fundamentals 1 – Liquid Treatment (activated sludge configurations); CSUS, Operation of Wastewater Treatment Plants, Vol. 1Report a problem with this question

13. 13. In a biological nutrient removal train, an anoxic zone is deliberately maintained under which conditions, and for what purpose?

    • A.Dissolved oxygen near 2 mg/L, to oxidize ammonia nitrogen
    • B.No dissolved oxygen and no nitrate, to release phosphorus
    • C.No dissolved oxygen and no nitrate, to oxidize ammonia
    • D.No dissolved oxygen with nitrate present, to denitrify✓ Answer

    Anoxic means no dissolved oxygen but nitrate available, so heterotrophs strip oxygen from nitrate and release nitrogen gas, recovering roughly half the alkalinity nitrification consumed. Anaerobic means neither oxygen nor nitrate, and that is the condition that drives phosphorus release ahead of luxury uptake in the aerobic zone.

    Source: WPI 2025 Class II NTK, nutrient removal (anaerobic/anoxic/aerobic); WEF, Wastewater Treatment Fundamentals 1 – Liquid TreatmentReport a problem with this question

14. 14. Nitrification consumes about 7.14 mg/L of alkalinity as CaCO3 per mg/L of ammonia-N oxidized. A nitrifying plant oxidizing 25 mg/L NH3-N has an influent alkalinity of 120 mg/L; effluent pH has fallen to 6.5 and effluent ammonia is climbing. The best corrective action is:

    • A.Add alkalinity, such as soda ash or sodium bicarbonate✓ Answer
    • B.Reduce aeration so that less ammonia is oxidized daily
    • C.Increase wasting to shorten the solids retention time
    • D.Add ferric chloride to restore the buffering capacity

    The demand is 25 × 7.14 = 178.5 mg/L of alkalinity against only 120 mg/L available, so the buffer is exhausted, pH falls and the nitrifiers slow down — which is exactly why ammonia is rising. Adding alkalinity restores the buffer and the pH range nitrifiers need; ferric chloride would consume still more alkalinity, and shortening the sludge age would wash the slow-growing nitrifiers out.

    Source: EPA Process Design Manual for Nitrogen Control (7.14 mg alkalinity as CaCO3 per mg NH3-N); CSUS, Operation of Wastewater Treatment Plants, Vol. 1Report a problem with this question

15. 15. In a plant using biological phosphorus removal with an anaerobic selector ahead of the aerobic zone, phosphorus ultimately leaves the plant:

    • A.As a gas stripped off from the anaerobic zone water
    • B.In the return sludge sent back to the aeration basin
    • C.As a precipitate settled out inside the anoxic zone
    • D.In the waste activated sludge, held as cell material✓ Answer

    Phosphorus has no gaseous form in this process: the anaerobic zone makes the organisms release phosphorus, the aerobic zone makes them take up more than they released (luxury uptake), and the phosphorus stays inside the cells. It only leaves the plant when those cells are wasted, so wasting rate matters and a septic clarifier blanket or digester can release the phosphorus back.

    Source: WEF, Wastewater Treatment Fundamentals 1 – Liquid Treatment (enhanced biological phosphorus removal); WPI 2025 Class II NTK, nutrient removalReport a problem with this question

16. 16. Over several months, a positive-displacement blower's discharge pressure has risen from 7 psi to 9.5 psi (48 to 66 kPa) at the same delivered airflow, and aeration basin DO has drifted downward. The most likely cause is:

    • A.Fouled diffusers raising backpressure on the air system✓ Answer
    • B.A leaking air header between the blower and the basin
    • C.A drifting DO probe reading lower than the actual DO
    • D.A worn blower rotor developing less pressure than before

    A positive-displacement blower delivers a fixed volume and lets pressure follow the system, so a steady climb in discharge pressure at constant airflow means the system is resisting more — the classic signature of fouled or scaled diffusers. Fouling also lowers oxygen transfer, which explains the falling DO; a leak would reduce pressure, and a drifting probe would not move the pressure gauge at all.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1 (aeration equipment, diffuser fouling and blower operation); WPI 2025 Class II NTK, Equipment Evaluation, Maintenance, and/or OperationReport a problem with this question

17. 17. Compared with coarse-bubble diffusers in the same aeration basin, fine-bubble diffusers:

    • A.Mix the basin better but transfer less oxygen per scfm
    • B.Transfer oxygen less efficiently but resist fouling better
    • C.Transfer oxygen more efficiently and resist fouling better
    • D.Transfer oxygen more efficiently but foul more readily✓ Answer

    Fine bubbles give far more gas-liquid surface area per unit of air and rise more slowly, so oxygen transfer efficiency is the highest of the diffused-air options. The small membrane or ceramic openings that create those bubbles are also what plug with biofilm and mineral scale, so fine-bubble systems need cleaning, bumping and clean-air protection, while coarse-bubble units trade efficiency for fouling resistance and mixing.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1 (diffused aeration systems); WPI 2025 Class II NTK, aeration systems (blowers, surface aerators, diffusers)Report a problem with this question

18. 18. Blanket depth cored in all four quadrants of a 12-ft (3.7 m) deep final clarifier has risen from 2 ft (0.6 m) to 6 ft (1.8 m). Settleability is good (SVI 90) and there has been no hydraulic surge, but the RAS pump has run at reduced speed since a repair. The first action is:

    • A.Raise the effluent weirs to reduce the overflow rate
    • B.Lower the aeration basin DO to slow biological activity
    • C.Increase the RAS rate to match the solids reaching the unit✓ Answer
    • D.Increase wasting sharply to cut the basin MLSS in half

    With good settling and no hydraulic surge, the blanket is climbing simply because solids are entering the clarifier faster than the return system removes them. Restoring the RAS rate brings withdrawal back into balance; leaving a deep blanket in place invites denitrification, rising sludge and septicity, while a drastic wasting change would destroy the solids inventory the process needs.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1 (secondary clarifier operation and sludge blanket control)Report a problem with this question

19. 19. Water is standing on the surface of a rock-media trickling filter, odors around the unit have increased and the distributor arms have slowed. The most likely cause is:

    • A.Too little organic load reaching the filter media
    • B.Hydraulic loading far above the filter's design rate
    • C.Excess biological growth plugging the media voids✓ Answer
    • D.Loss of natural draft through the filter underdrains

    Ponding is the classic symptom of excess biological slime (or degraded media) filling the voids so water can no longer percolate and air can no longer move through by natural draft, which is why odors follow. Correction is to shear or remove the growth — raise hydraulic loading and recirculation, park the distributor arm over the pond, flood, flush or rake — and effluent TSS spikes briefly as the excess slime sloughs.

    Source: PA DEP Wastewater Treatment Plant Operator Training, Module 20 (Trickling Filters — ponding); CSUS, Operation of Wastewater Treatment Plants, Vol. 1Report a problem with this question

20. 20. Increasing recirculation around a trickling filter primarily:

    • A.Lowers hydraulic loading, letting the slime thicken evenly
    • B.Raises organic loading, feeding the slime growth more
    • C.Raises hydraulic loading, shearing slime and diluting slugs✓ Answer
    • D.Raises media temperature, speeding growth in cold weather

    Recirculated flow is already-treated water, so it adds hydraulic load without adding much BOD. That extra wetting keeps the media uniformly wet, shears excess slime before it can pond, dilutes slug and toxic loads reaching the biofilm, and suppresses filter flies and odors. On the WPI formula sheet the recirculation ratio is recirculated flow divided by primary effluent flow.

    Source: PA DEP Operator Training Module 20 (Trickling Filters — recirculation); WPI 2025 Formula/Conversion Table — Recirculation Ratio = recirculated flow ÷ primary effluent flowReport a problem with this question

21. 21. The biofilm on a first-stage RBC shaft is white and chalky instead of gray-brown and shaggy, and the influent is arriving septic with a strong sulfide odor. The appropriate response is:

    • A.Remove the last-stage shaft from service temporarily
    • B.Reduce first-stage loading by step feed or pre-aeration✓ Answer
    • C.Stop the shaft so the biofilm sloughs off completely
    • D.Increase shaft rotation speed to thicken the biofilm

    White, chalky growth is sulfur-oxidizing Beggiatoa, which takes over when the first stage is organically overloaded or receives septic, sulfide-bearing wastewater. The fix is to relieve that first stage and remove the sulfide driver — step feed to spread load across shafts, pre-aerate or add supplemental aeration — not to manipulate rotation, which mainly damages bearings or sheds healthy film.

    Source: CSUS, Operation of Wastewater Treatment Plants, Vol. 1 (rotating biological contactors — biofilm appearance and first-stage overloading)Report a problem with this question

22. 22. Organic loading on a trickling filter and organic loading on an RBC are conventionally expressed, respectively, as:

    • A.lb soluble BOD5/day per 1,000 ft2 and lb BOD5/day per 1,000 ft3
    • B.lb BOD5/day per 1,000 ft3 and lb soluble BOD5/day per 1,000 ft2✓ Answer
    • C.lb BOD5/day per 1,000 ft2 and lb soluble BOD5/day per 1,000 ft3
    • D.gpd per ft2 for both, since each is a fixed-film process

    A trickling filter's biomass fills a depth of media, so its organic load is volumetric — pounds of BOD5 per day per 1,000 cubic feet of media. An RBC's biomass lives on a defined disc area and responds to the readily degradable fraction, so its load is pounds of soluble BOD5 per day per 1,000 square feet of media. Gallons per day per square foot is hydraulic loading, not organic loading.

    Source: WPI 2025 Formula/Conversion Table — Organic Loading Rate–Trickling Filter (lb BOD5/day per 1,000 ft3) and Organic Loading Rate–RBC (lb soluble BOD5/day per 1,000 ft2)Report 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 →