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MI Water Operator practice questions

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Question 1Water Treatment

Which chemical is most commonly used as the primary coagulant in conventional surface water treatment plants?

  1. Sodium hypochlorite
  2. Sodium fluoride
  3. Potassium permanganate
  4. Aluminum sulfate (alum) Correct answer.

Explanation

Alum hydrolyzes in water to form positively charged hydrolysis products that neutralize the negative surface charge on clay and other particles, allowing them to aggregate. It is the standard primary coagulant for conventional plants, usually fed together with a polymer as a flocculant aid. Sodium hypochlorite and potassium permanganate are oxidants, and sodium fluoride is dosed only for caries prevention.

Question 2Water Treatment

A nephelometer is used to measure turbidity in a treatment plant. What does the instrument actually quantify?

  1. The total mass of dissolved solids per liter of sample
  2. Light scattered by suspended particles Correct answer.
  3. The electrical conductivity of the water sample
  4. The concentration of free available chlorine in the sample

Explanation

Nephelometric turbidity is measured by passing a light beam through the sample and detecting light scattered at roughly 90 degrees by suspended and colloidal material, with results reported in NTU. Because it responds to the particles that shield pathogens from disinfection and that indicate treatment performance, turbidity is a key regulatory measurement.

Question 3Water Treatment

What is the primary purpose of the flocculation stage that follows rapid mix?

  1. To raise the pH so that calcium hardness can precipitate
  2. To strip dissolved gases such as carbon dioxide
  3. To grow small particles into settleable floc Correct answer.
  4. To kill pathogenic bacteria with a strong oxidant

Explanation

Flocculation is gentle, prolonged mixing that encourages destabilized microfloc to collide and grow into larger floc that can be removed by sedimentation and filtration. It adds no chemicals; the coagulant has already been dispersed during rapid mix.

Question 4Water Treatment

In chlorination practice, chlorine demand is defined as which of the following?

  1. The maximum chlorine dose a clearwell is permitted to receive
  2. The free chlorine residual measured after 24 hours of contact
  3. The applied dose minus the residual remaining Correct answer.
  4. The amount of chlorine needed to reach breakpoint chlorination

Explanation

Chlorine demand is the portion of the applied dose consumed by oxidizable material such as natural organic matter, ammonia, iron, and manganese, so demand equals dose minus residual. It is a water quality characteristic, not a regulatory limit.

Question 5Water Treatment

Why is a granular media filter backwashed?

  1. To sterilize the media with chlorine gas between filter runs
  2. To flush captured solids out of the expanded bed Correct answer.
  3. To increase the media's capacity to adsorb taste and odor compounds
  4. To reduce the alkalinity of the finished water

Explanation

Backwash water is forced upward at a rate that expands the bed, breaks up the media, and carries captured particles to waste, restoring head loss and effluent quality. The filter is then returned to service, usually after a short filter-to-waste period.

Question 6Water Treatment

In a conventional surface water plant, what is the main function of the sedimentation basin?

  1. To let settleable floc settle out by gravity Correct answer.
  2. To disinfect the water with chlorine before filtration
  3. To remove dissolved hardness minerals by ion exchange
  4. To add coagulant chemicals to the raw water

Explanation

The sedimentation basin provides quiescent conditions and enough detention time for floc that has grown large and dense to settle to the bottom as sludge. Removing this solids load protects the filters and extends filter runs.

Question 7Water Treatment

Granular activated carbon in a contactor or filter bed is used mainly to do what?

  1. Provide disinfection contact time for the CT calculation
  2. Remove hardness by precipitating calcium carbonate
  3. Adsorb dissolved taste- and odor-causing organics Correct answer.
  4. Neutralize the negative surface charge on clay particles

Explanation

Activated carbon has a vast internal pore structure that adsorbs dissolved organic compounds responsible for taste, odor, and color, and it also removes some disinfection byproduct precursors. The carbon is gradually exhausted and must be replaced or regenerated.

Question 8Water Treatment

Hardness in a raw water supply is caused principally by which dissolved ions?

  1. Chloride and sulfate
  2. Iron and manganese
  3. Calcium and magnesium Correct answer.
  4. Sodium and potassium

Explanation

Hardness is conventionally expressed as the sum of calcium and magnesium, reported as milligrams per liter of calcium carbonate equivalent. These divalent ions react with soap and form scale in pipes and heaters.

Question 9Water Treatment

An operator runs a jar test on a new raw water source. What does the test determine?

  1. The length of filter run before terminal head loss
  2. The best coagulant dose and coagulation pH Correct answer.
  3. The backwash rate needed to expand the filter bed
  4. The chlorine dose required to reach a free residual

Explanation

A jar test simulates rapid mix, flocculation, and settling at several chemical doses and pH values so the operator can select the combination that produces the clearest settled water. It is repeated whenever raw water turbidity, alkalinity, pH, or temperature changes, because the optimum dose shifts with water quality.

Question 10Water Treatment

On a chlorination curve for a water containing ammonia, which residual dominates just past the breakpoint?

  1. Combined chlorine present as monochloramine
  2. Free chlorine as HOCl and OCl- Correct answer.
  3. Combined chlorine present as dichloramine
  4. Combined chlorine present as nitrogen trichloride

Explanation

Before breakpoint, applied chlorine is consumed by ammonia to form chloramines, so little free residual appears. Once the ammonia has been oxidized at the breakpoint, additional chlorine persists as free chlorine in the hypochlorous acid and hypochlorite forms.

Question 11Water Treatment

A well produces water containing 1.5 mg/L dissolved iron and 0.3 mg/L manganese. Which treatment approach is appropriate?

  1. Lime softening only, with no oxidation step
  2. Oxidation followed by filtration Correct answer.
  3. Chlorination alone, with no filtration step
  4. Aeration in the clearwell only, with no filtration

Explanation

Dissolved ferrous iron and manganous manganese must first be oxidized to insoluble ferric and manganic forms with aeration, chlorine, ozone, or permanganate. The resulting precipitates are then captured in a filter, often a manganese greensand or catalytic media bed.

Question 12Water Treatment

For a conventional filtration plant, what combined filter effluent turbidity must be met in at least 95 percent of the measurements each month?

  1. 0.3 NTU Correct answer.
  2. 0.1 NTU
  3. 5.0 NTU
  4. 1.0 NTU

Explanation

Combined filter effluent from a conventional or direct filtration plant must be 0.3 NTU or less in at least 95 percent of the measurements taken each month, and no single measurement may exceed 1 NTU. Individual filters also have their own separate effluent limits.

Question 13Water Treatment

The Lead and Copper Rule establishes a lead action level based on the 90th percentile of first-draw tap samples. What is that level?

  1. 0.05 mg/L
  2. 0.005 mg/L
  3. 0.15 mg/L
  4. 0.015 mg/L Correct answer.

Explanation

The lead action level is 0.015 mg/L and the copper action level is 1.3 mg/L, each evaluated at the 90th percentile of first-draw samples from high-risk sites. Exceeding an action level triggers corrosion control treatment, public education, and possibly lead service line replacement.

Question 14Water Treatment

The Stage 2 Disinfectants and Disinfection Byproducts Rule sets the maximum contaminant level for total trihalomethanes at what value?

  1. 80 µg/L Correct answer.
  2. 10 µg/L
  3. 100 µg/L
  4. 40 µg/L

Explanation

The TTHM maximum contaminant level is 0.080 mg/L, or 80 µg/L, while the HAA5 level is 0.060 mg/L. Compliance is judged on running annual averages calculated at each monitoring location.

Question 15Water Treatment

Compared with free chlorine, chloramines have which set of characteristics?

  1. Stronger disinfection and a longer-lasting residual
  2. Equal disinfection strength, shorter-lasting residual
  3. Stronger disinfection, shorter-lasting residual
  4. Weaker disinfection, longer-lasting residual Correct answer.

Explanation

Chloramines are weaker disinfectants than free chlorine, so larger CT values are required, particularly for virus inactivation. They persist much longer in the distribution system and form far fewer trihalomethanes and haloacetic acids, which is why many utilities use them as a secondary disinfectant.

Question 16Water Treatment

Why is UV disinfection used specifically for Cryptosporidium control?

  1. It inactivates Cryptosporidium without leaving a residual Correct answer.
  2. It raises the pH and precipitates calcium carbonate scale
  3. It leaves a long-lasting residual throughout the distribution system
  4. It removes dissolved organic carbon precursors from the water

Explanation

Cryptosporidium oocysts are highly resistant to chemical disinfection but are inactivated by ultraviolet light at practical doses. UV provides no disinfectant residual, so a chemical disinfectant such as chlorine or chloramine is still required to protect the distribution system.

Question 17Water Treatment

Under the Surface Water Treatment Rule, what minimum overall removal and inactivation of Giardia lamblia must a conventional filtration plant achieve?

  1. 1-log
  2. 4-log
  3. 6-log
  4. 3-log Correct answer.

Explanation

The rule requires 3-log, or 99.9 percent, removal and inactivation of Giardia and 4-log, or 99.99 percent, for viruses. Plants meet the requirement by combining filtration credit with disinfection credit earned through CT.

Question 18Water Treatment

Alum is fed to a low-alkalinity surface water and the coagulation pH drops below the target. Which statement best explains the drop?

  1. Alum removes hardness, which lowers the pH of the water
  2. Alum consumes carbon dioxide gas, which raises the pH
  3. Alum consumes alkalinity and releases hydrogen ions Correct answer.
  4. Alum adds sulfate, which acts as a strong base in water

Explanation

The alum hydrolysis reaction consumes bicarbonate alkalinity and releases hydrogen ions, so pH falls as the coagulant dose increases. When raw water alkalinity is low, the operator must feed lime, caustic soda, or soda ash to hold pH in the range where coagulation performs best.

Question 19Water Treatment

What fluoride concentration does the U.S. Public Health Service currently recommend for community water systems to prevent tooth decay?

  1. 4.0 mg/L
  2. 0.1 mg/L
  3. 2.0 mg/L
  4. 0.7 mg/L Correct answer.

Explanation

The 2015 Public Health Service recommendation is a single optimal level of 0.7 mg/L, intended to prevent dental caries while limiting the risk of dental fluorosis. The primary maximum contaminant level is 4.0 mg/L and the secondary standard is 2.0 mg/L.

Question 20Water Treatment

A filter has run for 30 hours, effluent turbidity has begun to climb, and head loss has reached the terminal setpoint. What is the appropriate action?

  1. Backwash the filter and return it to service Correct answer.
  2. Increase the coagulant dose to the filter influent
  3. Raise the filtration rate to push the solids through the bed
  4. Add chlorine ahead of the filter to oxidize the captured solids

Explanation

Rising effluent turbidity together with terminal head loss indicates the bed is loaded and beginning to pass particles. The filter should be backwashed according to the plant procedure and returned to service, and the cause should be reviewed if the run was unusually short.

Question 21Water Treatment

A clearwell holds water with a free chlorine residual of 1.5 mg/L. The theoretical detention time is 90 minutes and the baffling factor is 0.5. Using a required CT of 47 mg·min/L for 3-log Giardia inactivation, what is the calculated CT and does it satisfy the requirement?

  1. 135 mg·min/L; requirement met
  2. 45 mg·min/L; requirement not met
  3. 67.5 mg·min/L; requirement met Correct answer.
  4. 33.8 mg·min/L; requirement not met

Explanation

Effective contact time is theoretical detention time multiplied by the baffling factor: 90 min times 0.5 equals 45 min. CT is the residual multiplied by effective contact time: 1.5 mg/L times 45 min equals 67.5 mg·min/L, which exceeds the 47 mg·min/L requirement.

Question 22Water Treatment

A chlorinated groundwater contains 1.0 mg/L of free ammonia as N and essentially no other chlorine-demanding material. Approximately what chlorine dose is required to reach breakpoint?

  1. 15 mg/L
  2. 1.0 mg/L
  3. 7.6 mg/L Correct answer.
  4. 3.0 mg/L

Explanation

Reaching breakpoint requires about 7.6 mg/L of chlorine per mg/L of ammonia nitrogen by weight to oxidize the ammonia completely, so roughly 7.6 mg/L is needed here. Before that point most of the applied chlorine appears as chloramines rather than free residual.

Question 23Water Treatment

A conventional plant receives 2.5-log Giardia credit for filtration and its disinfection segment earns an additional 1.0-log credit, against a 3.0-log requirement. What is the compliance status?

  1. It fails, because the total credit is only 2.5-log
  2. It complies only if the combined total reaches 4.0-log
  3. It fails, because filtration and disinfection credits cannot be combined
  4. It complies, because the credits add to 3.5-log Correct answer.

Explanation

Under the Surface Water Treatment Rule, physical removal credit from filtration and inactivation credit from disinfection are additive. Here 2.5-log plus 1.0-log equals 3.5-log, which meets the 3.0-log Giardia requirement with a margin of 0.5-log that must be documented.

Question 24Water Treatment

A plant feeds 8 gallons per day of sodium hypochlorite solution that is 12.5 percent available chlorine by weight with a specific gravity of 1.20. The plant flow is 0.5 MGD. What is the applied chlorine dose?

  1. 0.3 mg/L
  2. 19.2 mg/L
  3. 2.4 mg/L Correct answer.
  4. 1.2 mg/L

Explanation

Pounds of chlorine per day equals gallons fed times 8.34 lb/gal times specific gravity times weight fraction: 8 times 8.34 times 1.20 times 0.125 equals 10.0 lb/day. Dividing by the flow expressed in million pounds per day, 0.5 MGD times 8.34 equals 4.17, gives a dose of 2.4 mg/L.

Question 25Water Treatment

Under the Stage 2 Disinfectants and Disinfection Byproducts Rule, how is compliance with the TTHM and HAA5 maximum contaminant levels determined?

  1. On a system-wide average of all samples collected during the year
  2. On a single sample collected at the plant effluent each quarter
  3. On a locational running annual average at each site Correct answer.
  4. On the highest single sample collected anywhere during the year

Explanation

Stage 2 requires each monitoring location to meet the maximum contaminant level when its results are averaged over a running 12-month period, so one problem location can put the system out of compliance. This replaced the earlier system-wide averaging approach and is paired with an initial distribution system evaluation used to select monitoring sites.

Question 26Water Treatment

In a conventional surface water treatment plant, what is the purpose of the rapid-mix step?

  1. To disperse coagulant rapidly and uniformly through the water Correct answer.
  2. To oxidize dissolved iron and manganese before filtration
  3. To strip volatile organic compounds from the water
  4. To grow large, settleable floc particles with gentle stirring

Explanation

Rapid mix supplies intense, brief turbulence (typically well under a minute) so the coagulant is dispersed before it hydrolyzes and destabilizes particles. Without fast, uniform mixing, coagulant can react locally and form poor floc. Gentle stirring for floc growth and oxidation or stripping are separate process steps.

Question 27Water Treatment

Which material is normally placed as the upper layer of a dual-media filter because its specific gravity is lower than that of sand?

  1. Gravel
  2. Anthracite coal Correct answer.
  3. Garnet
  4. Activated alumina

Explanation

Anthracite coal has a specific gravity of roughly 1.4 to 1.5, so after backwashing it restratifies above the denser sand and provides a coarse upper layer that holds more solids. This arrangement lets solids penetrate deeper, extending filter runs. Garnet and gravel are much denser and are used as support or bottom layers.

Question 28Water Treatment

What is the main function of the flocculation basin that follows rapid mix?

  1. To remove settled solids continuously from the bottom of the basin
  2. To provide gentle mixing that lets particles collide and grow into floc Correct answer.
  3. To add disinfectant and achieve the required CT for inactivation
  4. To provide intense shear that breaks up the coagulant precipitate

Explanation

Flocculation uses slow, tapering mixing so that destabilized particles contact one another and agglomerate into larger, denser floc that can be settled or filtered out. The mixing energy must be gentle enough to build floc without shearing it apart. Disinfection and sludge removal are performed in other units.

Question 29Water Treatment

Which disinfectant is most widely used to maintain a residual throughout a distribution system because it persists and is easily measured?

  1. Ultraviolet light
  2. Chlorine dioxide
  3. Free chlorine Correct answer.
  4. Ozone

Explanation

Free chlorine is relatively stable in treated water, can be measured quickly in the field, and leaves a residual that protects the distribution system against recontamination. Ozone and UV decay or are applied only at a single point and provide no lasting residual. Chlorine dioxide is used but is more costly and forms chlorite.

Question 30Water Treatment

A groundwater has a total hardness of 300 mg/L as CaCO3, made up of 200 mg/L calcium hardness and 100 mg/L magnesium hardness. Which treatment is best suited to reduce the hardness to about 85 mg/L?

  1. Excess-lime softening followed by soda ash addition Correct answer.
  2. Adsorption on granular activated carbon
  3. Aeration followed by filtration through manganese greensand
  4. Cartridge filtration followed by ultraviolet disinfection

Explanation

Lime raises pH and precipitates calcium as calcium carbonate, and excess lime is added to raise pH high enough to precipitate magnesium as magnesium hydroxide, which is needed to reach a low total hardness. Soda ash supplies carbonate to precipitate noncarbonate hardness. Aeration, carbon adsorption, and cartridge filtration do not remove hardness.

Question 31Water Treatment

For a fixed free chlorine residual and pH, what happens to the CT value required to inactivate Giardia lamblia as the water temperature drops from 20 degrees C to 5 degrees C?

  1. It depends only on the chlorine dose and not on temperature
  2. It decreases because cold water holds chlorine longer
  3. It remains essentially unchanged at all temperatures
  4. It increases, often by a factor of about two to three Correct answer.

Explanation

Disinfection reactions slow as water gets colder, so more contact time or a higher residual is needed to achieve the same log inactivation. Published CT tables show the required CT rising steeply as temperature falls, roughly doubling to tripling over the 20 to 5 degrees C range. Temperature is a required input whenever CT credit is calculated.

Question 32Water Treatment

A well sample contains 0.04 mg/L arsenic, and speciation shows it is mostly arsenite. What must be done before an anion exchange or coagulation process can remove it effectively?

  1. The arsenic must be reduced to arsine gas by a reducing agent
  2. The pH must be lowered below 5.0 to make the arsenic cationic
  3. Oxidize arsenite to arsenate with chlorine or permanganate Correct answer.
  4. The water must first be softened with excess lime and soda ash

Explanation

Arsenite (As(III)) is uncharged at normal water pH and passes through most treatment processes. Oxidizing it to arsenate (As(V)) produces a charged species that is removed by anion exchange, coagulation with alum or iron salts, or activated alumina. Oxidation is therefore the required first step.

Question 33Water Treatment

A clarifier receives 2.16 MGD and has a surface area of 2,400 square feet. What is its surface overflow rate?

  1. 1.11 gpm/ft2
  2. 0.63 gpm/ft2 Correct answer.
  3. 0.90 gpm/ft2
  4. 0.45 gpm/ft2

Explanation

Convert flow to gallons per minute: 2.16 MGD times 694.4 gpm per MGD equals about 1,500 gpm. Dividing by the 2,400 square feet of surface area gives 0.625, or about 0.63 gpm/ft2. Surface overflow rate is a key settling design parameter.

Question 34Water Treatment

A system supplied by low-alkalinity surface water has lead service lines. Which approach is the standard basis of a corrosion control program?

  1. Feed orthophosphate and adjust pH and alkalinity Correct answer.
  2. Raise the free chlorine residual to 4.0 mg/L throughout
  3. Add sodium chloride to raise the water's conductivity
  4. Lower the pH to 6.0 to keep the lead dissolved

Explanation

Optimal corrosion control adjusts pH and alkalinity and often adds an inhibitor such as orthophosphate so a low-solubility lead scale forms on pipe surfaces. This keeps lead from dissolving into the water at the tap. Higher chlorine, added salt, or lower pH do not form protective films.

Question 35Water Treatment

A community water system collects a routine sample that is total-coliform positive. Under the Revised Total Coliform Rule, what is the required next step?

  1. Issue a precautionary boil water notice for the whole system
  2. Flush the system and resample the same tap after one week
  3. Wait for the next scheduled monthly sample before taking any action
  4. Collect repeat samples within 24 hours at the site and nearby taps Correct answer.

Explanation

A routine positive triggers repeat sampling, normally within 24 hours, at the original tap plus upstream and downstream locations, and the system must also perform a level 1 assessment if repeats are positive. The repeat results determine whether the finding represents a system problem. Immediate boil water notices and week-long delays are not the prescribed response.

Question 36Water Treatment

After backwashing a filter, why is the initial filtrate often wasted for a period before being returned to the clearwell?

  1. The effluent contains excess fluoride for the first hour of the run
  2. The media is still coated with chlorine from the backwash water
  3. The first filtrate is turbid until the media resettles and ripens Correct answer.
  4. The backwash water must be dechlorinated before it can be reused

Explanation

Right after backwash the bed is disturbed and lacks a filter cake, so the first water through carries elevated turbidity and can pass particles and pathogens. Filter-to-waste diverts this water until turbidity falls and the bed ripens. This avoids a turbidity spike in the finished water.

Question 37Water Treatment

Which membrane process operates at the lowest pressure and removes particles, turbidity, and protozoa such as Giardia and Cryptosporidium while passing dissolved ions?

  1. Ultrafiltration Correct answer.
  2. Nanofiltration
  3. Reverse osmosis
  4. Electrodialysis reversal

Explanation

Ultrafiltration has pores in the range of roughly 0.01 to 0.1 micron and runs at low pressure, removing particles, turbidity, bacteria, and protozoa while dissolved minerals pass through. Reverse osmosis and nanofiltration operate at higher pressure and reject dissolved ions. Electrodialysis reversal uses an electric field to move ions.

Question 38Water Treatment

A treatment plant treats 4.0 MGD and feeds alum at 25 mg/L. The alum feed solution has a specific gravity of 1.02 and contains 5.0 percent alum by weight. What solution feed rate in gallons per day is required?

  1. 1,960 gpd Correct answer.
  2. 980 gpd
  3. 98 gpd
  4. 3,920 gpd

Explanation

Dry alum needed is 4.0 MGD times 8.34 times 25 mg/L, or 834 lb/day. Dividing by the 0.05 solution strength gives 16,680 lb/day of solution, and dividing by 8.34 times 1.02 lb/gal gives about 1,960 gpd.

Question 39Water Treatment

A clearwell holds 0.50 million gallons and receives a flow of 1.5 MGD. Tracer testing gives a baffling factor of 0.5, and the free chlorine residual at the outlet is 1.6 mg/L. What is the CT value in mg-min/L?

  1. 384 mg-min/L Correct answer.
  2. 48 mg-min/L
  3. 192 mg-min/L
  4. 768 mg-min/L

Explanation

Theoretical detention time is 0.50 divided by 1.5, times 1,440 minutes per day, or 480 minutes. Multiplying by the 0.5 baffling factor gives T10 of 240 minutes, and 1.6 mg/L times 240 minutes gives a CT of 384 mg-min/L.

Question 40Water Treatment

Finished water has a Langelier Saturation Index of minus 1.3. What does this indicate and what response is appropriate?

  1. The water is supersaturated; add a sequestering agent
  2. The water is at equilibrium; no corrective action is needed
  3. The water is corrosive; raise pH or alkalinity toward zero Correct answer.
  4. The water is scale-forming; lower the pH and alkalinity

Explanation

A negative Langelier index means the water is undersaturated with respect to calcium carbonate, so it tends to dissolve protective scale and attack metal and concrete surfaces. Operators respond by raising pH or alkalinity, or adding an inhibitor, until the index approaches zero. Positive values indicate scaling tendency and zero indicates equilibrium.

Question 41Water Distribution

What is the primary purpose of the cement-mortar lining applied to the interior of ductile iron water main?

  1. To protect the interior from corrosion and tuberculation Correct answer.
  2. To make the pipe lighter for handling and shipping
  3. To provide a bonding surface for solvent-cement joints
  4. To raise the pressure rating of the pipe so that thinner walls may be used

Explanation

Cement-mortar lining passivates the iron surface and creates an alkaline environment that discourages corrosion and tuberculation. Keeping the wall smooth preserves the pipe's carrying capacity (C-factor) over its service life. The lining does not change the pressure rating or weight of the pipe.

Question 42Water Distribution

After a gate valve is fully opened, the standard practice is to turn the handwheel back about a quarter turn so that the valve is:

  1. Left fully open with the gate wedged hard against the stem stop
  2. Held at roughly half open so that it can be used to throttle flow
  3. Slightly off its seat so the gate will not become stuck Correct answer.
  4. Cycled open and closed each time a water sample is collected

Explanation

Backing the gate off its seat keeps the wedge from binding so tightly that the valve cannot be reopened later, and it relieves stress on the stem and disc. Gate valves are designed for fully open or fully closed service, not throttling.

Question 43Water Distribution

When a fire hydrant is opened for flushing or fire flow, the operating nut should be turned:

  1. Clockwise, with the hydrant then closed as tightly as the wrench allows
  2. In whichever direction the bonnet arrows indicate, after the caps are left on
  3. One full turn only, so the barrel is not over-pressurized
  4. Counterclockwise and opened slowly to avoid a pressure surge Correct answer.

Explanation

Most hydrants open in the counterclockwise direction, and the valve should be opened gradually so the main is not subjected to a sudden velocity change. Opening slowly limits water hammer and keeps the hydrant from being damaged.

Question 44Water Distribution

Which meter type measures flow by counting the number of times a known volume of water fills and empties a measuring chamber?

  1. A positive displacement meter Correct answer.
  2. A propeller-driven turbine meter
  3. A magnetic induction flow meter
  4. An orifice plate differential meter

Explanation

Positive displacement meters, such as nutating disc and piston designs, trap and release a fixed volume with each cycle, so the count of cycles gives the total volume. They are accurate at the low flows typical of residential service.

Question 45Water Distribution

A reduced-pressure zone backflow prevention assembly protects a public water supply against:

  1. Both backsiphonage and backpressure from a health hazard Correct answer.
  2. Pollution hazards only, and never against health hazards
  3. Backpressure only, from a booster pump on the service line
  4. Backsiphonage only, when a vacuum forms in the main

Explanation

An RPZ assembly contains two independently acting check valves plus a relief valve, so it protects against both backsiphonage and backpressure and is rated for health-hazard connections. It discharges to atmosphere if either check fails.

Question 46Water Distribution

What is the purpose of the valve box installed over a buried gate valve?

  1. To carry the weight of the valve and the pipe on either side
  2. To shield the valve interior from corrosive water
  3. To provide access to the operating nut without excavating Correct answer.
  4. To let water drain out of the valve body between operations

Explanation

A valve box extends from the valve bonnet up to grade so crews can reach the operating nut with a valve key. The valve itself is supported by the pipe and bedding, and the box has no role in carrying water or loads.

Question 47Water Distribution

Which type of valve is best suited for throttling flow to control the rate of discharge?

  1. Double-disc gate valve
  2. Swing check valve
  3. Globe valve Correct answer.
  4. Resilient-seat gate valve

Explanation

The internal passage of a globe valve forces flow to change direction around the disc, which allows smooth and repeatable throttling. Gate and check valves are designed for isolation or one-way flow, not flow regulation.

Question 48Water Distribution

What is the function of a pressure-relief valve installed on a distribution system?

  1. It measures the volume of water passing a given point in the main
  2. It releases water to protect piping and fittings from excessive pressure Correct answer.
  3. It prevents backflow from the customer's service line into the main
  4. It maintains a constant chlorine residual downstream of the installation

Explanation

A relief valve opens when pressure exceeds its setting and discharges water, limiting pressure that could rupture pipe or damage appurtenances. It is a safety device against surges and over-pressure, not a flow or quality control device.

Question 49Water Distribution

A 12-inch PVC water main terminates at a 90-degree bend. What is normally installed to keep the bend from being pushed off by internal pressure?

  1. Additional pipe bedding compacted to 95 percent of maximum density
  2. A line valve placed within 5 feet upstream of the bend
  3. A concrete thrust block bearing against undisturbed soil Correct answer.
  4. A flexible expansion joint installed on the downstream side of the bend

Explanation

Unbalanced thrust at a bend is resisted by a thrust block, restrained joints, or a combination of the two. Thrust blocks transfer the force into undisturbed soil behind the fitting.

Question 50Water Distribution

A new water main has been chlorinated at 50 mg/L for 24 hours and then flushed. Before the main may be placed in service, the utility must:

  1. Complete the pressure test in the same operation as the sampling
  2. Collect bacteriological samples and confirm acceptable results Correct answer.
  3. Flush the main again with a 200 mg/L hypochlorite solution
  4. Apply a chlorine-resistant coating to the interior of the pipe

Explanation

After disinfection and flushing, the main is sampled for total coliform bacteria, and the results must be acceptable before the main is put into service. If samples fail, the main is rechlorinated and retested.