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Water Distribution Operator Class III study materials
Water Distribution Operator Class III study resources

Water Distribution Operator Class III practice questions

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Question 1Distribution System Design, Layout, and Hydraulics

In a distribution system, the primary purpose of maintaining a minimum residual pressure at all service connections is to:

  1. Ensure meters register accurately at low flows
  2. Guarantee the water meets primary drinking water standards
  3. Prevent backflow and contamination from entering the main Correct answer.
  4. Reduce the need for disinfection at the treatment plant

Explanation

Positive pressure at every point in the system keeps the pipe full and prevents contaminated groundwater or cross-connections from being drawn into the main. Pressure is a barrier against intrusion, not a treatment or metering function.

Question 2Distribution System Design, Layout, and Hydraulics

A pipe carries 1,500 gallons per minute. Expressed in cubic feet per second, this flow is approximately:

  1. 5.6 cfs
  2. 2.2 cfs
  3. 3.3 cfs Correct answer.
  4. 0.45 cfs

Explanation

One cubic foot per second equals 448.8 gallons per minute. Dividing 1,500 gpm by 448.8 gpm per cfs gives about 3.34 cfs.

Question 3Distribution System Design, Layout, and Hydraulics

The hydraulic grade line in a pressurized distribution system represents the:

  1. Depth of cover over the pipeline
  2. Elevation of the pipe crown at each point
  3. Average velocity head in the main
  4. Height to which water would rise in a piezometer open to the atmosphere Correct answer.

Explanation

The hydraulic grade line is the elevation of the water surface that would stand in a riser open to atmosphere at a given point, combining pressure head and elevation head. It does not describe pipe geometry or velocity alone.

Question 4Distribution System Design, Layout, and Hydraulics

Which pipe material is most commonly specified for large-diameter transmission mains because of its strength and long service life?

  1. Polybutylene
  2. Galvanized steel
  3. Asbestos-cement
  4. Ductile iron Correct answer.

Explanation

Ductile iron is widely used for large transmission mains because of its high tensile strength, toughness, and long service life. Galvanized steel and polybutylene are used in small service lines, and asbestos-cement is no longer installed.

Question 5Distribution System Design, Layout, and Hydraulics

A water main must be laid with a minimum depth of cover primarily to:

  1. Permit easy access for tapping
  2. Protect the pipe from traffic loads and freezing Correct answer.
  3. Reduce the required pipe diameter
  4. Allow gravity drainage of the line

Explanation

Adequate cover distributes surface loads over the pipe and places the line below the frost line where freezing is a concern. Cover depth does not affect hydraulic capacity or tapping needs.

Question 6Distribution System Design, Layout, and Hydraulics

A distribution system has a total head loss of 40 feet over 8,000 feet of main. The hydraulic gradient in feet per 1,000 feet is:

  1. 8.0 ft/1,000 ft
  2. 5.0 ft/1,000 ft Correct answer.
  3. 2.0 ft/1,000 ft
  4. 0.5 ft/1,000 ft

Explanation

Divide total head loss by length and multiply by 1,000: (40 ÷ 8,000) × 1,000 = 5.0 ft per 1,000 ft. This gradient reflects the friction slope of the main.

Question 7Distribution System Design, Layout, and Hydraulics

The primary purpose of looping a distribution main network rather than using only dead-end branches is to:

  1. Lower the required disinfection residual
  2. Increase the number of meters required
  3. Provide alternate flow paths and reduce head loss Correct answer.
  4. Eliminate the need for valves

Explanation

Looped systems provide redundant paths so water can reach users from more than one direction, reducing head loss and stagnation and improving reliability. Looping does not change disinfection requirements or valve needs.

Question 8Distribution System Design, Layout, and Hydraulics

A pump delivers water at 500 gallons per minute against a total dynamic head of 120 feet. The water horsepower is approximately:

  1. 75.6 hp
  2. 120 hp
  3. 15.1 hp Correct answer.
  4. 1.5 hp

Explanation

Water horsepower equals (gpm × head in feet) ÷ 3,960. Here (500 × 120) ÷ 3,960 ≈ 15.1 hp. This is the power imparted to the water before efficiency losses.

Question 9Distribution System Design, Layout, and Hydraulics

A fire flow test records a static pressure of 70 psi and a residual pressure of 50 psi while flowing 1,000 gpm. The pressure drop of 20 psi indicates that the test point is:

  1. Isolated from the source by a closed valve
  2. Reasonably close to an adequate supply main Correct answer.
  3. On a main that is undersized for the flow
  4. At the end of a long dead-end main

Explanation

A modest pressure drop during a high flow indicates a relatively strong supply near the test point. Large drops at high flow point to restricted or remote mains.

Question 10Distribution System Design, Layout, and Hydraulics

Using the Hazen-Williams equation, head loss in a pipe varies inversely with which parameter raised to the 1.85 power?

  1. The flow rate
  2. The pipe length
  3. The C-factor Correct answer.
  4. The pipe diameter

Explanation

In the Hazen-Williams form, head loss is inversely proportional to C raised to about 1.85. Higher C values mean smoother pipe and less friction loss.

Question 11Distribution System Design, Layout, and Hydraulics

A 12-inch main carries 2,000 gpm. The flow velocity is closest to:

  1. 12 ft/s
  2. 2.8 ft/s
  3. 0.6 ft/s
  4. 5.7 ft/s Correct answer.

Explanation

Velocity equals flow divided by area. Converting 2,000 gpm to about 4.46 cfs and dividing by the 0.785 ft² area of a 12-inch pipe gives about 5.7 ft/s.

Question 12Distribution System Design, Layout, and Hydraulics

A distribution system serves a peak-hour demand of 4.0 mgd. If the peak-hour factor is 1.8 times the average day demand, the average day demand is approximately:

  1. 4.0 mgd
  2. 7.2 mgd
  3. 1.8 mgd
  4. 2.2 mgd Correct answer.

Explanation

Peak-hour demand equals average day demand times the peaking factor. Dividing 4.0 mgd by 1.8 gives about 2.22 mgd average day demand.

Question 13Distribution System Design, Layout, and Hydraulics

The primary function of an altitude valve in a distribution system is to:

  1. Prevent a storage tank from overflowing by closing at a set high level Correct answer.
  2. Throttle pump discharge to a constant rate
  3. Automatically flush dead-end mains
  4. Relieve excess pressure from water hammer surges

Explanation

An altitude valve senses tank level and closes the fill line when the tank reaches the set elevation, preventing overflow. Surge relief and flow control are handled by other valve types.

Question 14Distribution System Design, Layout, and Hydraulics

A community's system has a ground-level storage tank with a high-water elevation of 220 feet. The maximum static pressure at a service connection at elevation 100 feet is approximately:

  1. 52 psi Correct answer.
  2. 120 psi
  3. 220 psi
  4. 95 psi

Explanation

Static pressure in psi equals the head difference in feet divided by 2.31. The head difference is 220 − 100 = 120 feet, so 120 ÷ 2.31 ≈ 52 psi.

Question 15Distribution System Design, Layout, and Hydraulics

When designing a gridiron layout, engineers place larger transmission mains along major streets and smaller distribution mains on residential streets primarily because:

  1. Fire hydrants cannot be connected to large mains
  2. Flow capacity and reliability requirements are higher on the primary routes Correct answer.
  3. Residential customers require higher pressure than commercial areas
  4. Smaller mains are easier to tap under traffic

Explanation

Trunk mains carry the bulk of the flow to feed the grid, so they are sized larger and routed along primary corridors. Smaller branch mains then distribute water locally to customers and hydrants.

Question 16Distribution System Design, Layout, and Hydraulics

The term "minimum hour demand" is most relevant to distribution system design because it:

  1. Determines the maximum pipe diameter needed
  2. Sets the fire flow requirement
  3. Establishes the pump curve operating point
  4. Governs water age, storage turnover, and disinfection residual maintenance Correct answer.

Explanation

Low-flow periods cause long detention times in mains and tanks, which can degrade disinfectant residual and water quality. Designers consider minimum-hour conditions to manage storage turnover and stagnation.

Question 17Distribution System Design, Layout, and Hydraulics

A pressure-reducing valve is installed on a branch serving a low-elevation area. Its main design purpose is to:

  1. Prevent backflow into the transmission main
  2. Maintain disinfectant residual
  3. Increase flow capacity to the area
  4. Protect downstream piping and fixtures from excessive static pressure Correct answer.

Explanation

Areas at low elevation see higher static pressure from the same hydraulic grade line, so a pressure-reducing valve limits downstream pressure to protect pipes and fixtures. It does not add flow or residual.

Question 18Distribution System Design, Layout, and Hydraulics

A new subdivision will be served by a looped main with a design peak flow of 900 gpm. Using a target velocity of 3 ft/s, the minimum nominal pipe diameter is closest to:

  1. 10 inches Correct answer.
  2. 14 inches
  3. 6 inches
  4. 8 inches

Explanation

Flow of 900 gpm is about 2.0 cfs. For a target velocity of 3 ft/s, required area is 2.0 ÷ 3 ≈ 0.67 ft², giving a diameter of about 0.92 ft, or roughly 11 inches, so a 10-inch nominal main is the closest listed size that is close but slightly above target velocity.

Question 19Distribution System Design, Layout, and Hydraulics

A water system has a hydraulic grade line that falls below the elevation of a service connection during peak demand. This condition will most likely result in:

  1. Loss of pressure and possible negative pressure at that connection Correct answer.
  2. Increased disinfectant residual
  3. Automatic reduction in customer demand
  4. Higher flow velocity in the main

Explanation

If the hydraulic grade line drops below a service elevation, pressure at that point is zero or negative, risking outage and contamination intrusion. The system must be operated or sized to keep the grade line above all service elevations.

Question 20Distribution System Design, Layout, and Hydraulics

Water hammer in a distribution main is best mitigated by:

  1. Removing air release valves
  2. Increasing valve closure speed
  3. Reducing pipe wall thickness
  4. Slow-closing valves and surge protection devices Correct answer.

Explanation

Water hammer is caused by rapid changes in flow velocity, so slow-closing valves and surge control devices reduce the pressure surge. Air release and surge relief equipment also help protect the system.

Question 21Distribution System Design, Layout, and Hydraulics

A system has two tanks at the same ground elevation: tank A with a water surface at 180 feet and tank B at 160 feet, connected by an open main. Under static conditions, flow will:

  1. Cease immediately because both tanks are at the same ground elevation
  2. Move from tank B into tank A because it is lower
  3. Alternate direction depending on the pipe diameter
  4. Move from tank A into tank B until the surfaces equalize Correct answer.

Explanation

Static flow between connected tanks is driven by the difference in water surface elevation, not ground elevation. Water will flow from the higher surface to the lower until the levels equalize.

Question 22Distribution System Design, Layout, and Hydraulics

A distribution main must deliver 1,200 gpm through 2,000 feet of pipe with a maximum allowable head loss of 8 feet. Using the Hazen-Williams nomograph concept, the required diameter is closest to:

  1. 12 inches Correct answer.
  2. 20 inches
  3. 8 inches
  4. 16 inches

Explanation

For 1,200 gpm over 2,000 feet, the allowable gradient is 4 ft per 1,000 ft. A 12-inch main with a typical C-factor produces roughly that gradient near 1,200 gpm, making 12 inches the closest adequate size.

Question 23Distribution System Design, Layout, and Hydraulics

During a fire flow test, the observed residual pressure at 1,500 gpm is 18 psi with a static pressure of 65 psi. This result most strongly suggests that the test location is:

  1. Downstream of a fully open gate valve with no restrictions
  2. On a main with limited capacity relative to the flow demand Correct answer.
  3. At a point where the hydraulic grade line is well above the service elevation
  4. Near a large-diameter transmission main

Explanation

A drop of 47 psi while flowing 1,500 gpm indicates significant friction loss between the source and the test point, pointing to limited main capacity or a remote location. Strong supply near a large main would show a much smaller drop.

Question 24Distribution System Design, Layout, and Hydraulics

A distribution system is being evaluated for a new industrial customer with a large instantaneous demand. The most appropriate first step in the hydraulic evaluation is to:

  1. Increase the disinfection dose at the treatment plant
  2. Perform a network hydraulic model run under peak and fire flow conditions Correct answer.
  3. Reduce the storage tank high-water level
  4. Add additional meters at all residential connections

Explanation

A hydraulic model run under peak and fire flow conditions shows whether pressures, velocities, and storage can meet the new demand and where improvements are needed. Operational or treatment changes do not address capacity.

Question 25Distribution System Design, Layout, and Hydraulics

Two parallel mains of equal length and material connect the same two nodes: one is 8 inches in diameter and the other is 12 inches. Compared with the 8-inch main, the 12-inch main will carry approximately:

  1. Half as much flow
  2. Five times as much flow Correct answer.
  3. Twice as much flow
  4. The same flow

Explanation

For parallel pipes with the same head loss, capacity scales roughly with diameter to the 2.5 power under Hazen-Williams, so (12/8)^2.5 ≈ 5.2. The larger main therefore carries about five times the flow.

Question 26Distribution System Design, Layout, and Hydraulics

In a distribution system, the hydraulic grade line (HGL) at any point represents which of the following?

  1. The pressure in psi multiplied by the pipe diameter
  2. The elevation to which water would rise in a piezometer open to the main at that point Correct answer.
  3. The physical elevation of the pipe crown at that point
  4. The velocity head of the water moving through the main

Explanation

The HGL is the sum of the pressure head and the elevation head at a point, which equals the level water would rise to in an open standpipe. It is a measure of available head, not the pipe's physical elevation or velocity head alone.

Question 27Distribution System Design, Layout, and Hydraulics

Which pipe material characteristic most directly increases head loss for a given flow and diameter in the Hazen-Williams relationship?

  1. A shorter pipe length
  2. A higher roughness coefficient (C)
  3. A thicker pipe wall
  4. A lower roughness coefficient (C) Correct answer.

Explanation

In Hazen-Williams, head loss varies inversely with the roughness coefficient C raised to the 1.85 power, so a lower C yields greater head loss. Smooth materials such as cement-lined pipe carry higher C values and lose less head.

Question 28Distribution System Design, Layout, and Hydraulics

A storage tank has a low-water elevation of 150 feet and a high-water elevation of 190 feet. A pressure gauge at a service connection at elevation 120 feet reads the pressure when the tank is at its low-water level. What is the approximate static pressure?

  1. 17 psi
  2. 30 psi
  3. 13 psi Correct answer.
  4. 8 psi

Explanation

Static pressure equals the head difference between the water surface and the gauge elevation: 150 minus 120 equals 30 feet. Converting feet of head to psi by dividing by 2.31 gives about 13 psi.

Question 29Distribution System Design, Layout, and Hydraulics

Which layout pattern provides the greatest reliability because every main can be supplied from more than one direction?

  1. Radial spoke pattern
  2. Dead-end branching
  3. Single trunk line with laterals
  4. Gridiron or looped network Correct answer.

Explanation

A looped or gridiron network lets water reach any point by more than one path, so a break can be isolated without cutting off service. Dead-end and single-trunk layouts leave downstream customers without an alternate supply route.

Question 30Distribution System Design, Layout, and Hydraulics

A 10-inch diameter main carries a flow of 1,500 gpm. What is the flow velocity in feet per second? (1 cfs = 448.8 gpm)

  1. 6.8 ft/s
  2. 2.7 ft/s
  3. 4.1 ft/s Correct answer.
  4. 5.5 ft/s

Explanation

Convert flow to cfs: 1,500 divided by 448.8 equals about 3.34 cfs. The pipe area is pi times (0.4167 ft) squared, about 0.545 square feet, so velocity equals 3.34 divided by 0.545, or roughly 6.1 ft/s.

Question 31Distribution System Design, Layout, and Hydraulics

A distribution system has an average day demand of 3.0 mgd and a maximum day demand of 4.5 mgd. What is the maximum day peaking factor?

  1. 3.0
  2. 0.67
  3. 1.5 Correct answer.
  4. 2.25

Explanation

The peaking factor is the maximum day demand divided by the average day demand, or 4.5 divided by 3.0, which equals 1.5. Peaking factors are used to size storage and transmission capacity.

Question 32Distribution System Design, Layout, and Hydraulics

An engineer is sizing a transmission main to carry a design flow of 2,500 gpm at a target velocity of 4 ft/s. Which diameter is the minimum standard size that satisfies the target? (1 cfs = 448.8 gpm)

  1. 12-inch
  2. 14-inch
  3. 18-inch
  4. 16-inch Correct answer.

Explanation

Convert flow to cfs: 2,500 divided by 448.8 equals about 5.57 cfs. The required area is 5.57 divided by 4, or about 1.39 square feet, giving a diameter of about 1.33 feet, or 16 inches.

Question 33Distribution System Design, Layout, and Hydraulics

What is the primary purpose of a pressure-sustaining valve in a distribution system?

  1. To hold upstream pressure at or above a set minimum Correct answer.
  2. To automatically fill a storage tank to a set level
  3. To prevent backflow into the public supply
  4. To relieve excess pressure by discharging water to waste

Explanation

A pressure-sustaining valve throttles flow to keep upstream pressure from dropping below a setpoint, protecting higher-elevation customers. It differs from a pressure-relief valve, which discharges water to limit downstream overpressure.

Question 34Distribution System Design, Layout, and Hydraulics

A water system's hydraulic grade line drops below the elevation of a second-floor service connection during peak-hour demand. What is the most likely result for that customer?

  1. The customer receives water at excessive pressure
  2. The main develops negative pressure and cavitates
  3. The meter registers flow in reverse
  4. The customer experiences low or no pressure at the fixture Correct answer.

Explanation

When the HGL falls below the fixture elevation, there is insufficient head to deliver water, so the customer sees low pressure or no flow. This condition is a common reason for setting minimum service pressures in design standards.

Question 35Distribution System Design, Layout, and Hydraulics

Water hammer in a distribution main is most effectively reduced by which design or operational measure?

  1. Installing surge tanks or air chambers near the source of the pressure wave Correct answer.
  2. Increasing pump discharge velocity
  3. Closing valves as rapidly as possible
  4. Reducing the pipe diameter downstream of the pump

Explanation

Surge protection devices such as air chambers, surge tanks, and slow-closing valves absorb or dampen the pressure wave caused by rapid flow changes. Slower valve closure and controlled pump starts also limit surge magnitude.

Question 36Distribution System Design, Layout, and Hydraulics

Two reservoirs at different water surface elevations are connected by a single pipeline. Flow through the line is driven primarily by which quantity?

  1. The difference in water surface elevations between the reservoirs Correct answer.
  2. The absolute elevation of the lower reservoir
  3. The length of the pipeline alone
  4. The diameter of the pipe alone

Explanation

Flow between two reservoirs is driven by the head difference, which equals the difference in water surface elevations minus the head losses in the connecting pipe. Diameter and length affect the losses but do not create the driving head.

Question 37Distribution System Design, Layout, and Hydraulics

A distribution system must supply a fire flow of 1,500 gpm at 20 psi residual while maintaining a static pressure of 60 psi. Which factor most limits the ability to meet this requirement?

  1. The color of the pipe exterior coating
  2. The type of valve used at the service connection
  3. The number of meters installed in the area
  4. The elevation of the storage tank relative to the demand point Correct answer.

Explanation

Available fire flow depends on the head provided by storage and pumping relative to the elevation of the demand point, minus distribution losses. If the HGL at the fire location is too low, the residual pressure requirement cannot be met.

Question 38Distribution System Design, Layout, and Hydraulics

A 12-inch main and a 6-inch main of equal length and material are laid in parallel between the same two nodes. Approximately what fraction of total flow passes through the 12-inch main?

  1. About 50 percent
  2. About 94 percent Correct answer.
  3. About 75 percent
  4. About 60 percent

Explanation

For parallel pipes of equal length and material, flow splits in proportion to diameter raised to the 2.63 power under Hazen-Williams. The ratio of (12)^2.63 to (12)^2.63 plus (6)^2.63 is roughly 0.94.

Question 39Distribution System Design, Layout, and Hydraulics

A designer is evaluating a proposed looped subdivision main and must ensure adequate pressure during peak demand while limiting head loss. Which approach best reflects sound hydraulic design practice?

  1. Ignore elevation changes because pressure is set only at the source
  2. Size mains for the peak-hour flow at a velocity that keeps friction loss within allowable limits Correct answer.
  3. Size all mains to the smallest standard diameter to reduce material cost
  4. Use a single dead-end main to simplify construction

Explanation

Good design sizes mains so that peak-hour flows produce velocities and friction losses that keep the HGL above required service pressures at all elevations. Looping and adequate diameter both contribute to meeting that goal.

Question 40Distribution System Design, Layout, and Hydraulics

A pump delivers 800 gpm against a total dynamic head of 150 feet. What is the approximate water horsepower? (1 hp = 3960 for gpm and feet of head)

  1. 20 hp
  2. 40 hp
  3. 30 hp Correct answer.
  4. 25 hp

Explanation

Water horsepower equals flow in gpm times head in feet divided by 3960, so 800 times 150 divided by 3960 equals about 30 hp. This is the hydraulic power delivered to the water, before pump efficiency is applied.

Question 41Pumps, Motors, and Pumping Station Operation

A pump is discharging 1,200 gallons per minute against a total head of 150 feet. What is the approximate water horsepower delivered by the pump?

  1. About 45 hp Correct answer.
  2. About 90 hp
  3. About 60 hp
  4. About 32 hp

Explanation

Water horsepower equals flow in gpm times total head in feet divided by 3,960. Here 1,200 × 150 ÷ 3,960 ≈ 45.5 hp. This is the power actually transferred to the water, before pump and motor efficiency losses.

Question 42Pumps, Motors, and Pumping Station Operation

A centrifugal pump driven at constant speed is delivering water with a discharge pressure gauge reading steady, but the operator hears cavitation noise and the flow rate has dropped. Which condition is the most likely cause?

  1. The discharge valve has been throttled too far open
  2. The packing gland is too loose
  3. The suction lift has increased and net positive suction head available is insufficient Correct answer.
  4. The motor is running at a higher speed than nameplate

Explanation

Cavitation occurs when the absolute pressure at the pump suction falls to the vapor pressure of the water, so vapor bubbles form and collapse. Excessive suction lift, a clogged suction strainer, or a partly closed suction valve reduces net positive suction head available. The resulting noise and loss of capacity are classic signs of inadequate suction conditions.

Question 43Pumps, Motors, and Pumping Station Operation

In a centrifugal pump, what happens to the developed head and the power draw as flow increases from shutoff toward the best efficiency point on a typical rising-then-falling curve?

  1. Head falls and power rises Correct answer.
  2. Both head and power rise
  3. Both head and power fall
  4. Head rises and power falls

Explanation

On the usual centrifugal pump curve, head is highest at shutoff and declines as flow increases. Brake horsepower generally increases with flow over this range because the pump is moving more water. Operators use this relationship when setting throttling and when checking motor loading.

Question 44Pumps, Motors, and Pumping Station Operation

A pump station has two identical pumps operating in parallel into a common discharge header. Compared with one pump running alone, the parallel arrangement will generally:

  1. Double the head at the same flow
  2. Reduce the total flow because the pumps fight each other
  3. Increase total flow at a somewhat higher head than a single pump alone Correct answer.
  4. Keep total flow the same but halve the head

Explanation

Parallel pumps add their flows at a common head, so the combined curve is formed by adding capacities at each head. Because the system head rises with flow, the resulting operating point delivers more than one pump but less than twice the single-pump flow. Head is set by the system, not doubled.

Question 45Pumps, Motors, and Pumping Station Operation

Two identical centrifugal pumps are operated in series. The primary effect is that the combination:

  1. Adds the heads of both pumps at a common flow Correct answer.
  2. Adds the flows of both pumps at a common head
  3. Halves the required motor horsepower
  4. Squares the discharge pressure of one pump

Explanation

Series operation passes the same flow through both pumps, so each pump adds its head at that flow. The combined curve is the sum of the heads at each capacity. Series arrangements are used where a single pump cannot develop enough pressure.

Question 46Pumps, Motors, and Pumping Station Operation

A deep-well vertical turbine pump is installed with the bowl assembly set too high relative to the pumping water level. What is the most likely operating problem?

  1. Overheating of the motor windings from overloading
  2. Excessive discharge pressure at low flow
  3. Reverse rotation of the impellers
  4. Cavitation and loss of capacity from insufficient submergence Correct answer.

Explanation

Vertical turbine bowls require enough submergence below the pumping water level to provide adequate net positive suction head. If the setting is too shallow, the pump draws down to the bowl and vapor forms. The result is cavitation, noise, and reduced capacity.

Question 47Pumps, Motors, and Pumping Station Operation

An operator observes that a pump's discharge pressure gauge reads normal but the flow meter shows zero flow and the pump casing is warm. What is the most likely explanation?

  1. The suction line is plugged and the pump is running dry
  2. The pump is operating at shutoff with the discharge valve closed Correct answer.
  3. The check valve is stuck open
  4. The impeller has been removed

Explanation

A pump running against a closed discharge valve develops shutoff head, so the pressure gauge reads normal while flow is zero. The energy added to the water heats the casing, which is why extended shutoff operation can damage the pump. The warm casing with normal pressure and no flow is the signature of this condition.

Question 48Pumps, Motors, and Pumping Station Operation

Which device is installed on the discharge side of a pump to prevent water from flowing backward through the pump when it stops?

  1. Check valve Correct answer.
  2. Gate valve
  3. Air release valve
  4. Pressure relief valve

Explanation

A check valve allows flow in one direction only and closes when flow reverses. It protects the pump from backspin and prevents water hammer and reverse flow from the distribution system. Gate valves isolate but do not prevent backflow.

Question 49Pumps, Motors, and Pumping Station Operation

A pump motor nameplate lists a service factor of 1.15. What does this indicate to the operator?

  1. The motor may be operated continuously at 115 percent of nameplate horsepower under rated conditions Correct answer.
  2. The motor must be derated to 85 percent of nameplate horsepower for continuous service
  3. The motor can be started 1.15 times per hour
  4. The motor efficiency is 115 percent

Explanation

A service factor of 1.15 means the motor can carry 115 percent of its nameplate horsepower continuously without exceeding its rated temperature rise, provided voltage and ambient conditions are at nameplate values. It is a margin, not a target operating point. Relying on it continuously shortens motor life.

Question 50Pumps, Motors, and Pumping Station Operation

A three-phase induction motor driving a pump begins to hum loudly, draws high current, and does not come up to speed. Which cause is most consistent with these symptoms?

  1. The motor is running at synchronous speed
  2. The discharge pressure is too low
  3. One phase of the supply is lost Correct answer.
  4. The pump is operating at best efficiency point

Explanation

Loss of one phase on a three-phase motor leaves it single-phased, so it cannot develop starting torque and draws heavy current in the remaining windings. The motor hums and may trip overloads if not disconnected quickly. Single-phasing can overheat and damage the motor if left energized.