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Wastewater Treatment Operator Class III study materials
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Wastewater Treatment Operator Class III practice questions

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Question 1Wastewater Treatment Processes and Microbiology

In the activated sludge process, what is the primary role of the bacterial population?

  1. Supply dissolved oxygen to the mixed liquor through photosynthesis
  2. Oxidize organic matter and synthesize new cell mass Correct answer.
  3. Filter suspended solids out of the flow
  4. Disinfect the waste stream before discharge

Explanation

Activated sludge bacteria use incoming organic matter as food, oxidizing part of it to carbon dioxide and water to obtain energy and converting the rest into new cell mass that is later wasted as excess sludge. This biological conversion is the core of secondary treatment.

Question 2Wastewater Treatment Processes and Microbiology

The 5-day BOD test measures which of the following?

  1. Dissolved oxygen concentration in the receiving stream
  2. Total suspended solids remaining after five days of settling
  3. Oxygen consumed by microbes as they oxidize organic matter Correct answer.
  4. Chlorine demand of the disinfected sample

Explanation

The BOD test measures the dissolved oxygen consumed by microorganisms while they stabilize biodegradable organic matter, typically over five days at 20 degrees C. The result reflects organic strength rather than solids content or chemical demand.

Question 3Wastewater Treatment Processes and Microbiology

What is the purpose of returning activated sludge (RAS) from the secondary clarifier to the aeration tank?

  1. Add nutrients that the microorganisms are lacking
  2. Control the pH of the aeration tank
  3. Reduce the organic loading on the primary clarifier
  4. Return settled microbes to maintain the solids inventory Correct answer.

Explanation

RAS carries settled microorganisms back to the aeration tank so that enough biomass remains in the system to treat the incoming waste. Without adequate return, solids wash out of the process and treatment efficiency collapses.

Question 4Wastewater Treatment Processes and Microbiology

Which organism is used as the primary indicator of fecal contamination in wastewater effluent?

  1. Sulfate-reducing bacteria
  2. Escherichia coli Correct answer.
  3. Zoogloea ramigera
  4. Nitrosomonas species

Explanation

E. coli is a normal inhabitant of the intestinal tract of warm-blooded animals and does not readily multiply outside the host, so its presence signals recent fecal contamination. Pathogen testing is slow and costly, so indicator organisms are monitored instead.

Question 5Wastewater Treatment Processes and Microbiology

The sludge volume index (SVI) is used to evaluate which of the following?

  1. The oxygen demand that the influent wastewater exerts
  2. The rate at which the digester produces gas
  3. The concentration of nitrogen in the effluent
  4. The settleability and compaction of activated sludge Correct answer.

Explanation

SVI is the volume in milliliters occupied by one gram of sludge after 30 minutes of settling, calculated from the settled volume and the MLSS. A low SVI indicates dense, well-compacting sludge, while a high SVI points to bulking.

Question 6Wastewater Treatment Processes and Microbiology

Which group of microorganisms produces methane in an anaerobic digester?

  1. Methanogenic archaea Correct answer.
  2. Filamentous bacteria
  3. Nitrifying bacteria
  4. Sulfur-oxidizing bacteria

Explanation

Methanogens are a specialized group of strictly anaerobic archaea that convert acetate and hydrogen with carbon dioxide into methane in the final step of digestion. They are sensitive to pH, temperature, and toxic substances, so digester stability depends on protecting them.

Question 7Wastewater Treatment Processes and Microbiology

What dissolved oxygen concentration is typically maintained in an activated sludge aeration tank to support aerobic treatment?

  1. 0.5 to 0.9 mg/L
  2. 0.1 to 0.3 mg/L
  3. 1.0 to 3.0 mg/L Correct answer.
  4. 6.0 to 8.0 mg/L

Explanation

Operators normally target roughly 1 to 3 mg/L of dissolved oxygen in the aeration tank, which supports both carbon oxidation and nitrification without wasting blower energy. Near-zero values create anaerobic conditions and encourage filamentous growth, while very high values are costly and unnecessary.

Question 8Wastewater Treatment Processes and Microbiology

Mean cell residence time (MCRT) is best described as which of the following?

  1. The average time microorganisms remain in the treatment system Correct answer.
  2. The time a drop of water takes to pass through the plant
  3. The time raw wastewater spends in the aeration tank
  4. The time required for the clarifier to form a sludge blanket

Explanation

MCRT, also called sludge age or SRT, is the mass of solids in the system divided by the mass of solids leaving each day. It is the key control parameter governing nitrification, sludge settling characteristics, and sludge production.

Question 9Wastewater Treatment Processes and Microbiology

Nitrification in an activated sludge plant requires which set of conditions?

  1. Anaerobic conditions and high phosphorus concentrations
  2. Aerobic conditions with adequate alkalinity and long SRT Correct answer.
  3. Anoxic conditions with a readily degradable carbon source
  4. Low dissolved oxygen and a very short solids retention time

Explanation

Nitrification is a two-step aerobic oxidation of ammonia to nitrite and then nitrate carried out by slow-growing autotrophic bacteria. Because those organisms grow slowly, the system needs a long SRT, sufficient dissolved oxygen, and adequate alkalinity.

Question 10Wastewater Treatment Processes and Microbiology

What is the purpose of the anoxic zone in a biological nutrient removal plant?

  1. To oxidize ammonia to nitrite and then to nitrate
  2. To aerate the mixed liquor before it enters the clarifier
  3. To convert nitrate to nitrogen gas using influent carbon Correct answer.
  4. To precipitate phosphorus as a metal salt

Explanation

In the anoxic zone, denitrifying bacteria use nitrate as an electron acceptor and influent organic carbon as an electron donor, releasing nitrogen gas to the atmosphere. This removes nitrogen from the liquid stream and recovers a portion of the alkalinity consumed during nitrification.

Question 11Wastewater Treatment Processes and Microbiology

A low food-to-microorganism ratio in an activated sludge system generally indicates which condition?

  1. An older sludge with a tendency toward pin floc Correct answer.
  2. A high oxygen demand and rapid cell growth
  3. Excessive organic loading being applied to the aeration tank
  4. A young sludge age with poor floc formation

Explanation

A low F/M means food is scarce relative to the biomass present, which corresponds to a long sludge age. Under extended aeration the flocs become compact and overoxidized and tend to shear into pinpoint floc that carries over in the effluent.

Question 12Wastewater Treatment Processes and Microbiology

Which condition most commonly promotes filamentous bulking in an activated sludge system?

  1. Sustained low dissolved oxygen in the aeration basin Correct answer.
  2. High dissolved oxygen in the aeration tank
  3. Addition of a supplemental carbon source
  4. Excessive wasting of mixed liquor solids from the system

Explanation

Filamentous organisms can extend into the bulk liquid and capture oxygen that floc formers cannot reach, so they outcompete the floc formers when dissolved oxygen is chronically low. The result is a high SVI, poor compaction, and solids loss in the clarifier overflow.

Question 13Wastewater Treatment Processes and Microbiology

Thick, stable brown foam on the aeration tank that persists through the winter is most often associated with which organism?

  1. Nitrosomonas, a nitrifying chemoautotroph
  2. Nocardia and related actinomycetes Correct answer.
  3. Paramecium, a free-swimming ciliate
  4. Bacillus, an aerobic spore-forming bacterium

Explanation

Nocardia and related actinomycetes are hydrophobic filamentous organisms that stabilize foams, especially in plants with long sludge ages, warm temperatures, and fats, oils, and grease in the influent. Their foam is brown, thick, and persistent on aeration basins and clarifiers.

Question 14Wastewater Treatment Processes and Microbiology

If the sludge blanket in the secondary clarifier rises above the effluent weir, what is the most likely result?

  1. Solids carryover into the effluent Correct answer.
  2. Reduced sludge production in the aeration tank
  3. Increased dissolved oxygen in the effluent
  4. Improved effluent clarity from increased settling

Explanation

When the blanket rises above the weir, solids escape with the overflow, raising effluent total suspended solids and BOD. The blanket is controlled by balancing RAS withdrawal, wasting rate, and hydraulic loading on the clarifier.

Question 15Wastewater Treatment Processes and Microbiology

Why do nitrifying bacteria recover slowly after a toxic discharge to an activated sludge plant?

  1. They can only grow in the absence of oxygen
  2. They produce a protective spore that must germinate
  3. They require chlorine as an essential nutrient
  4. They grow much more slowly than heterotrophic organisms Correct answer.

Explanation

Nitrifiers are autotrophs with a long generation time, so a population damaged by a toxic slug or pH swing takes weeks to rebuild even after conditions return to normal. Heterotrophic bacteria reproduce much faster and recover quickly by comparison.

Question 16Wastewater Treatment Processes and Microbiology

An aeration tank holds 0.8 million gallons with an MLSS of 2,500 mg/L. Solids are wasted at 5,000 lb/day and solids leaving in the clarifier effluent are negligible. What is the approximate solids retention time?

  1. 3.3 days Correct answer.
  2. 0.3 days
  3. 1.7 days
  4. 8.3 days

Explanation

The solids inventory is 0.8 MG times 8.34 lb/gal times 2,500 mg/L divided by 1,000,000, or about 16,700 lb. Dividing that inventory by the 5,000 lb/day leaving the system gives an SRT of roughly 3.3 days.

Question 17Wastewater Treatment Processes and Microbiology

An aeration tank receives 4,000 lb/day of BOD and contains 16,000 lb of MLVSS. What is the F/M ratio?

  1. 4.0 lb BOD per lb MLVSS per day
  2. 2.5 lb BOD per lb MLVSS per day
  3. 0.40 lb BOD per lb MLVSS per day
  4. 0.25 lb BOD per lb MLVSS per day Correct answer.

Explanation

F/M is the pounds of BOD applied per day divided by the pounds of volatile solids under aeration. Dividing 4,000 by 16,000 gives 0.25 per day, which corresponds to conventional or extended aeration operation.

Question 18Wastewater Treatment Processes and Microbiology

An oxygen uptake rate test performed on mixed liquor provides information about which of the following?

  1. The concentration of phosphorus in the influent
  2. The settling characteristics of the sludge in the clarifier
  3. The biological activity of the microorganisms Correct answer.
  4. The rate at which grit settles in the grit chamber

Explanation

The oxygen uptake rate measures how quickly the mixed liquor consumes dissolved oxygen, which reflects the metabolic activity of the biomass. Operators use it to detect toxicity, track load changes, and judge whether the solids inventory matches the organic load.

Question 19Wastewater Treatment Processes and Microbiology

The presence of large numbers of ciliates and rotifers in the mixed liquor generally indicates which condition?

  1. The dissolved oxygen level is below 0.5 mg/L
  2. A toxic discharge has just entered the plant
  3. The sludge age is very short and the sludge is young
  4. A stable, mature sludge with good settling properties Correct answer.

Explanation

Ciliates and rotifers are higher organisms that cannot survive a very short sludge age, low oxygen, or toxic conditions. Their abundance indicates a mature, stable biomass with a sludge age long enough to support a healthy food chain and good settling.

Question 20Wastewater Treatment Processes and Microbiology

What is a primary purpose of recirculating trickling filter effluent back to the filter influent?

  1. To cool the filter media during warm weather
  2. To dilute the influent strength and sustain flow Correct answer.
  3. To increase the organic loading applied to the filter
  4. To disinfect the wastewater as it passes through the media

Explanation

Recirculation dilutes the strength of the wastewater applied to the filter and keeps the media wetted at a steady hydraulic loading. It also helps flush sloughed biological growth out of the media and equalizes loading during low-flow periods.

Question 21Wastewater Treatment Processes and Microbiology

A 1,000 mL settleometer sample of mixed liquor settles to 240 mL in 30 minutes, and the MLSS is 3,000 mg/L. What is the SVI?

  1. 80 mL/g Correct answer.
  2. 800 mL/g
  3. 240 mL/g
  4. 125 mL/g

Explanation

SVI equals the settled volume in mL/L multiplied by 1,000 and divided by the MLSS in mg/L. Here, 240 times 1,000 divided by 3,000 equals 80 mL/g, which indicates a dense, well-settling sludge.

Question 22Wastewater Treatment Processes and Microbiology

A secondary clarifier shows sludge flocs rising to the surface while the effluent carries high suspended solids. The aeration tank dissolved oxygen is 2.5 mg/L, the sludge age is long, and the influent is warm. Which cause is most likely?

  1. A young sludge with poor floc formation
  2. A toxic shock that killed the grazing protozoa
  3. Filamentous bulking caused by low dissolved oxygen in the basin
  4. Denitrification in the clarifier releasing nitrogen gas Correct answer.

Explanation

A long sludge age, warm temperatures, and adequate oxygen favor complete nitrification, so nitrate-rich mixed liquor reaches the clarifier. Where oxygen runs out under the blanket, denitrifiers reduce that nitrate and release nitrogen gas bubbles that lift the sludge to the surface.

Question 23Wastewater Treatment Processes and Microbiology

Microscopic examination shows small, weak flocs with few filaments and no free-swimming bacteria. The SVI is low at 60 mL/g and the effluent is turbid with fine solids. What is the most likely cause?

  1. Excessive filamentous bacteria from a low F/M ratio
  2. A young sludge caused by excessive wasting
  3. An old, overoxidized sludge forming pinpoint floc Correct answer.
  4. A toxic discharge that dispersed the previously formed flocs

Explanation

Pinpoint floc develops when an old, overoxidized sludge has little filamentous backbone, so the flocs shear into tiny particles that settle slowly and carry over the weir. The low SVI together with small flocs and turbid effluent is characteristic, and the usual correction is to increase wasting and reduce the sludge age.

Question 24Wastewater Treatment Processes and Microbiology

Complete nitrification of 1 mg/L of ammonia nitrogen consumes approximately how much alkalinity, expressed as CaCO3?

  1. 1.0 mg/L
  2. 7.1 mg/L Correct answer.
  3. 4.3 mg/L
  4. 2.5 mg/L

Explanation

Oxidizing ammonia through nitrite to nitrate consumes about 7.1 mg/L of alkalinity as CaCO3 for each mg/L of ammonia nitrogen oxidized. If influent alkalinity is insufficient to cover this demand, pH drops and nitrification stalls.

Question 25Wastewater Treatment Processes and Microbiology

Wastewater that has gone septic in a long, slow force main arrives at the plant. Which set of conditions is most likely?

  1. Low BOD and a rise in pH above 9
  2. High dissolved oxygen and a growing nitrifier population
  3. Sulfide odors and an oxygen-depleted, septic flow Correct answer.
  4. High alkalinity and improved sludge settling

Explanation

Septic wastewater has been depleted of dissolved oxygen and nitrate, allowing sulfate reduction and fermentation that generate hydrogen sulfide and volatile organic acids. It arrives odorous, corrosive, and with a high immediate oxygen demand that can shock the biological system.

Question 26Wastewater Treatment Processes and Microbiology

What is the primary purpose of returning settled solids from the secondary clarifier to the aeration tank?

  1. To supply the alkalinity nitrifiers consume
  2. To maintain the biomass in the aeration tank Correct answer.
  3. To reduce the volume of waste sludge produced
  4. To raise the hydraulic loading on the clarifier

Explanation

Return activated sludge keeps the biomass inventory in the aeration tank at the level needed to consume BOD, so organisms are not washed out with the effluent. The return rate controls how solids are distributed between the aeration tank and the clarifier, while wasting controls the total mass of solids in the system.

Question 27Wastewater Treatment Processes and Microbiology

Which group of organisms is most commonly used as an indicator of recent fecal contamination in treated effluent?

  1. Nitrate-reducing bacteria
  2. Fecal coliform bacteria Correct answer.
  3. Sulfate-reducing bacteria
  4. Filamentous bacteria

Explanation

Fecal coliforms are present in very high numbers in the intestines of warm-blooded animals and are shed in feces, so their presence signals recent fecal contamination. They are inexpensive to culture and their survival characteristics roughly track those of many bacterial pathogens.

Question 28Wastewater Treatment Processes and Microbiology

In a facultative stabilization lagoon, which zone is essentially free of dissolved oxygen?

  1. The whole water column during daylight hours
  2. The upper sunlit layer near the surface
  3. The bottom sludge layer Correct answer.
  4. The layer just below the surface where algae grow

Explanation

A facultative lagoon is stratified into an aerobic surface zone where algae produce oxygen, an intermediate facultative zone, and an anaerobic bottom zone where settled solids are decomposed by anaerobic bacteria. Oxygen from photosynthesis and surface reaeration does not reach the bottom sludge layer.

Question 29Wastewater Treatment Processes and Microbiology

Mesophilic anaerobic digestion is typically operated in which temperature range?

  1. 45 to 55 degrees C (113 to 131 degrees F)
  2. 30 to 38 degrees C (86 to 100 degrees F) Correct answer.
  3. 10 to 20 degrees C (50 to 68 degrees F)
  4. 20 to 30 degrees C (68 to 86 degrees F)

Explanation

Mesophilic digesters are heated to roughly 35 degrees C (about 95 degrees F), a range in which methane-forming bacteria grow dependably. Thermophilic digestion operates near 50 to 57 degrees C and proceeds faster but is less stable.

Question 30Wastewater Treatment Processes and Microbiology

In a biological phosphorus removal plant, what occurs in the anaerobic zone?

  1. Phosphorus-accumulating organisms release phosphate Correct answer.
  2. Denitrifying bacteria reduce nitrate to nitrogen gas
  3. Nitrifying bacteria convert ammonia to nitrate
  4. Aerobic bacteria oxidize soluble BOD to cell mass

Explanation

In the anaerobic zone, phosphorus-accumulating organisms take up volatile fatty acids and store them, releasing previously stored phosphate in the process. Later, under aerobic conditions, they take up far more phosphate than they released, and the excess leaves the plant in waste sludge.

Question 31Wastewater Treatment Processes and Microbiology

What pH range is generally maintained in an activated sludge aeration tank?

  1. 9.0 to 10.0
  2. 4.5 to 5.5
  3. 6.5 to 8.5 Correct answer.
  4. 2.0 to 3.0

Explanation

Activated sludge organisms, including the nitrifiers, function best near neutral pH, with about 6.5 to 8.5 as the practical operating window. Nitrification consumes alkalinity and can push the pH below this range when influent alkalinity is low.

Question 32Wastewater Treatment Processes and Microbiology

Which sequence describes a typical operating cycle of a sequencing batch reactor?

  1. Decant, fill, idle, aerate, settle
  2. Fill, aerate, settle, decant, idle Correct answer.
  3. Fill, settle, aerate, decant, idle
  4. Aerate, fill, decant, settle, idle

Explanation

A sequencing batch reactor performs fill, react (aeration and mixing), settle, decant, and idle, all in the same basin. The cycle repeats continuously, which is why no separate secondary clarifier is needed.

Question 33Wastewater Treatment Processes and Microbiology

In a membrane bioreactor, what function does the membrane unit perform?

  1. It separates solids from water in place of a clarifier Correct answer.
  2. It thickens waste sludge before dewatering
  3. It transfers oxygen to the mixed liquor more efficiently
  4. It removes dissolved phosphorus by chemical precipitation

Explanation

In a membrane bioreactor, membranes replace gravity settling in the secondary clarifier, retaining biomass and allowing a much higher mixed liquor suspended solids concentration. Effluent quality is high because the membrane pores are smaller than bacteria and most suspended particles.

Question 34Wastewater Treatment Processes and Microbiology

A trickling filter suddenly discharges a large slug of accumulated solids in its effluent. What is this event called?

  1. Flow short-circuiting
  2. Filter unloading Correct answer.
  3. Media plugging
  4. Media ponding

Explanation

When flow, loading, or dissolved oxygen conditions change, biomass and trapped solids that accumulated in the media can detach in a surge, an event known as filter unloading. The resulting solids slug can shock the downstream process and produce high effluent suspended solids.

Question 35Wastewater Treatment Processes and Microbiology

In an anaerobic digester, a steadily rising volatile acid-to-alkalinity ratio indicates which of the following?

  1. The digester is receiving too little organic loading
  2. The digesting sludge is becoming sour and unstable Correct answer.
  3. Methane formers are consuming acids faster than they are produced
  4. Carbon dioxide in the digester gas has fallen sharply

Explanation

The volatile acid-to-alkalinity ratio compares acid production by acid-forming bacteria with the buffering capacity available to neutralize it. When methane formers are inhibited, acids accumulate, the ratio climbs, and pH falls, a condition commonly called a sour digester.

Question 36Wastewater Treatment Processes and Microbiology

A well-operated primary clarifier typically removes what percentage of influent suspended solids?

  1. 90 to 98 percent
  2. 50 to 70 percent Correct answer.
  3. 5 to 15 percent
  4. 20 to 30 percent

Explanation

A well-operated primary clarifier removes roughly 50 to 70 percent of suspended solids along with about 25 to 40 percent of BOD. Performance depends on detention time, flow distribution, and consistent sludge withdrawal.

Question 37Wastewater Treatment Processes and Microbiology

What is the main reason for removing grit early in the treatment train?

  1. It provides the alkalinity needed for nitrification
  2. It is the primary source of soluble phosphorus
  3. It causes abrasive wear on pumps and other equipment Correct answer.
  4. It contributes most of the BOD load to the aeration tank

Explanation

Grit consists of sand, gravel, and other dense inert material that abrades pumps, valves, and piping and accumulates in tanks and digesters. Removing it early protects mechanical equipment and reduces maintenance and cleanout costs.

Question 38Wastewater Treatment Processes and Microbiology

An oxidation-reduction potential reading of about -200 mV in a biological nutrient removal basin is most consistent with which condition?

  1. Anaerobic conditions that favor phosphate release Correct answer.
  2. Aerobic conditions that favor nitrification
  3. Anoxic conditions that favor denitrification
  4. Aerobic conditions that favor carbon oxidation

Explanation

Oxidation-reduction potential is reported in millivolts and indicates the oxidation state of the mixed liquor. Anaerobic zones typically read below about -200 mV, anoxic zones roughly -50 to -100 mV, and aerobic zones above about +50 mV, so the strongly negative reading identifies the anaerobic zone where phosphate is released.

Question 39Wastewater Treatment Processes and Microbiology

Using the common approximation that return sludge concentration in mg/L equals 1,000,000 divided by the SVI in mL/g, what return sludge concentration corresponds to an SVI of 250 mL/g?

  1. 4,000 mg/L Correct answer.
  2. 8,000 mg/L
  3. 2,000 mg/L
  4. 12,000 mg/L

Explanation

Dividing 1,000,000 by the SVI of 250 mL/g gives an expected return sludge concentration of about 4,000 mg/L. A sludge with that high an SVI settles poorly, so the underflow is thin and large return rates are needed to hold the desired mixed liquor solids.

Question 40Wastewater Treatment Processes and Microbiology

Approximately how many pounds of oxygen demand are credited per pound of nitrate nitrogen reduced to nitrogen gas during denitrification?

  1. 2.86 pounds Correct answer.
  2. 1.5 pounds
  3. 1.0 pound
  4. 4.6 pounds

Explanation

Nitrate can serve as an electron acceptor in place of oxygen, so reducing 1 pound of nitrate nitrogen to nitrogen gas provides the equivalent of about 2.86 pounds of oxygen demand. This credit lowers the oxygen and carbon demand that the aerobic portion of the process would otherwise carry.

Question 41Preliminary and Primary Treatment

What is the primary purpose of preliminary treatment in a wastewater treatment plant?

  1. Removing dissolved organic matter through biological action in an aerated basin
  2. Disinfecting the flow with chlorine before it reaches the primary clarifier
  3. Protecting downstream equipment and processes from damage and clogging Correct answer.
  4. Stabilizing sludge so that it can be dewatered and hauled to a landfill

Explanation

Preliminary treatment removes or reduces materials that would damage pumps, clog pipes, or interfere with downstream processes, and it includes screening, grinding, and grit removal. It does not rely on biological action, which occurs later, and it is not a disinfection or sludge stabilization step.

Question 42Preliminary and Primary Treatment

At what approach velocity should a mechanically cleaned bar screen typically be operated?

  1. About 10 ft/s (3 m/s)
  2. About 0.1 ft/s (0.03 m/s)
  3. About 1 ft/s (0.3 m/s) Correct answer.
  4. About 4 ft/s (1.2 m/s)

Explanation

Screen approach velocities are kept near 1 ft/s (0.3 m/s) so that solids are carried to the screen without being forced through the openings, while heavy grit is not held in suspension unnecessarily. Much lower velocities let grit and solids deposit in the channel, and much higher velocities push material through the bars.

Question 43Preliminary and Primary Treatment

Which statement best describes the material removed in a grit chamber?

  1. Floatable material such as grease, oil, and plastic beads that collects on the surface
  2. Settleable inorganic material such as sand, with a specific gravity near 2.65 Correct answer.
  3. Dissolved mineral salts that precipitate out of solution only after the flow is aerated
  4. Organic settleable material such as food waste and paper fibers that decompose in the clarifier

Explanation

Grit is the heavy, largely inorganic fraction of the solids, such as sand, gravel, and coffee grounds, with a specific gravity of roughly 2.5 to 2.65 and a settling velocity much higher than that of organic solids. It is removed early because it abrades pumps and accumulates in basins.

Question 44Preliminary and Primary Treatment

How is flow rate determined with a Parshall flume?

  1. It stays accurate even when the throat is deeply submerged by downstream water
  2. It needs a long, quiet upstream channel with a stilling well to produce accurate readings
  3. The head is measured at two points and the difference between them is used to compute flow
  4. Flow is computed from a single depth reading at a set point near the throat Correct answer.

Explanation

A Parshall flume develops critical flow at its throat, so the flow rate can be computed from a single head reading taken at a fixed point upstream of the throat. Free-flow conditions are required for the standard rating, and the flume has low head loss and is largely self-cleaning.

Question 45Preliminary and Primary Treatment

What surface overflow rate is typical for a primary clarifier at average design flow?

  1. 2,500 to 4,000 gpd/ft²
  2. 100 to 200 gpd/ft²
  3. 10,000 to 15,000 gpd/ft²
  4. 600 to 1,000 gpd/ft² Correct answer.

Explanation

Typical design surface overflow rates for primary clarifiers are about 600 to 1,000 gpd/ft² at average flow, with lower values where peak flows must be handled. The overflow rate, not detention time alone, governs how much settleable solids can be captured.

Question 46Preliminary and Primary Treatment

A primary clarifier has a volume of 0.35 MG and receives an average flow of 1.4 MGD. What is the hydraulic detention time?

  1. 10.0 hours
  2. 2.5 hours
  3. 6.0 hours Correct answer.
  4. 24.0 hours

Explanation

Detention time equals basin volume divided by flow: 0.35 MG ÷ 1.4 MGD = 0.25 day, which is 6 hours. This falls within the typical range for primary clarification at average flow.

Question 47Preliminary and Primary Treatment

What removal efficiency is expected from a well-operated primary clarifier?

  1. 50 to 65 percent of TSS and 25 to 40 percent of BOD Correct answer.
  2. 75 to 85 percent of TSS and 60 to 70 percent of BOD
  3. 90 to 95 percent of TSS and 80 to 90 percent of BOD
  4. 10 to 15 percent of TSS and 5 to 10 percent of BOD

Explanation

A well-operated primary clarifier typically removes about 50 to 65 percent of influent TSS and 25 to 40 percent of BOD. BOD removal is limited because much of the BOD is dissolved or colloidal and does not settle.

Question 48Preliminary and Primary Treatment

A primary clarifier treats 2.0 MGD and has 400 feet of effluent weir. What is the weir loading rate?

  1. 12,500 gpd/ft
  2. 5,000 gpd/ft Correct answer.
  3. 500 gpd/ft
  4. 2,000 gpd/ft

Explanation

Weir loading rate is the flow divided by the total effluent weir length: 2,000,000 gpd ÷ 400 ft = 5,000 gpd/ft. Values much above roughly 10,000 to 15,000 gpd/ft at peak flow can draw solids over the weir.

Question 49Preliminary and Primary Treatment

What does a comminutor do in the headworks of a treatment plant?

  1. It compacts and dewaters the screenings it removes so they can be landfilled
  2. It raises the flow velocity above 5 ft/s so that solids stay suspended
  3. It grinds large solids into smaller pieces that stay in the flow Correct answer.
  4. It removes grit from the flow by centrifugal action and deposits it in a hopper

Explanation

A comminutor is a rotating grinder that cuts large solids into small pieces that remain in the flow, avoiding the handling of screenings. Downstream equipment is protected, but the material is not removed and adds to the solids load on later processes.

Question 50Preliminary and Primary Treatment

Which practice best prevents septic conditions and hydrogen sulfide formation in a long preliminary and primary treatment train?

  1. Covering the channels so that sunlight cannot reach the wastewater
  2. Keeping the wastewater fresh by aeration or by adding nitrate Correct answer.
  3. Increasing the detention time in the grit chamber so that more solids settle out
  4. Recirculating primary sludge back to the head of the plant on a regular schedule

Explanation

Septic conditions develop when wastewater sits under oxygen-deficient conditions, allowing sulfate-reducing bacteria to produce hydrogen sulfide. Keeping the flow fresh with aeration or an oxidizing chemical such as nitrate prevents sulfide generation and the corrosion and odors that follow.