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Question 1Irrigation hydraulics and system design fundamentals

In irrigation hydraulics, 1 psi of pressure is equivalent to approximately how many feet of water head?

  1. 8.33 ft
  2. 14.7 ft
  3. 0.433 ft
  4. 2.31 ft Correct answer.

Explanation

A column of water 1 ft high develops 0.433 psi, so the reciprocal relationship gives 1 psi = 1 / 0.433 = 2.31 ft of head. This conversion is used constantly when turning nozzle or pump pressures into elevation head.

Question 2Irrigation hydraulics and system design fundamentals

Which term describes the pressure reading at a hose bib when no water is moving through the system?

  1. Static pressure Correct answer.
  2. Operating pressure
  3. Friction loss
  4. Residual pressure

Explanation

Static pressure is measured with zero flow, so it reflects only the elevation and the supply pressure behind the system. Once water moves, friction and velocity changes reduce the reading to a dynamic or residual value.

Question 3Irrigation hydraulics and system design fundamentals

Which formula gives the precipitation rate in inches per hour for sprinklers spaced on a square grid, where S is the spacing in feet?

  1. PR = GPM ÷ 96.3 × S²
  2. PR = 96.3 × S² ÷ GPM
  3. PR = GPM × S² ÷ 96.3
  4. PR = 96.3 × GPM ÷ S² Correct answer.

Explanation

The constant 96.3 converts gallons per minute applied over square feet into inches per hour. Dividing the head flow by the area served by that head, S squared, gives the application rate.

Question 4Irrigation hydraulics and system design fundamentals

On a typical centrifugal pump performance curve, what happens to discharge head as flow rate increases?

  1. It rises then falls
  2. It rises steadily
  3. It stays constant
  4. It decreases Correct answer.

Explanation

A centrifugal pump curve shows head falling as flow increases, because more of the impeller energy is spent moving water rather than building pressure. The curve intersects the system curve at the operating point.

Question 5Irrigation hydraulics and system design fundamentals

Which backflow prevention device is typically required for a high-hazard irrigation connection to a potable supply?

  1. Double check valve assembly
  2. Atmospheric vacuum breaker
  3. RPZ assembly Correct answer.
  4. Pressure vacuum breaker

Explanation

A reduced pressure zone assembly protects against both backpressure and backsiphonage and vents to atmosphere when it fails, so it is used where contamination would be a serious health hazard. Devices such as double checks, pressure vacuum breakers and atmospheric vacuum breakers are limited to lower hazard situations.

Question 6Irrigation hydraulics and system design fundamentals

What is the commonly recommended maximum water velocity in irrigation laterals and mains to limit friction loss and water hammer?

  1. 5 ft/s Correct answer.
  2. 25 ft/s
  3. 12 ft/s
  4. 1 ft/s

Explanation

Design velocities are generally held to about 5 ft/s or less; friction loss rises sharply with velocity, and high velocities magnify surge when valves close. Keeping velocity at or below this value also limits wear and noise.

Question 7Irrigation hydraulics and system design fundamentals

Water hammer in an irrigation system is most commonly caused by which condition?

  1. A slowly closing valve
  2. Low static pressure
  3. Rapid valve closure Correct answer.
  4. An oversized mainline

Explanation

Stopping the flow abruptly converts velocity energy into a pressure surge that travels back through the piping. Slow-closing valves, air chambers and surge tanks are used to spread the deceleration over time.

Question 8Irrigation hydraulics and system design fundamentals

What is the purpose of a check valve built into a sprinkler head?

  1. To stop low-head drainage Correct answer.
  2. To regulate flow to a set rate
  3. To increase nozzle pressure
  4. To filter debris from the nozzle

Explanation

A built-in check valve closes when the zone shuts off and holds water in the lateral, preventing the lowest heads from draining the pipe. This avoids puddling and erosion at low points and reduces the water wasted on each cycle.

Question 9Irrigation hydraulics and system design fundamentals

A sprinkler sits 20 ft above the point where the pressure is 45 psi. Ignoring friction, what is the pressure at the sprinkler?

  1. 36.3 psi Correct answer.
  2. 45.4 psi
  3. 8.7 psi
  4. 53.7 psi

Explanation

Raising water 20 ft costs 20 × 0.433 = 8.7 psi of pressure, so the pressure at the higher point is 45 − 8.7 = 36.3 psi. Elevation gain always reduces pressure, while elevation drop adds to it.

Question 10Irrigation hydraulics and system design fundamentals

A sprinkler delivers 4 GPM and heads are spaced 30 ft by 30 ft on a square grid. What is the approximate precipitation rate?

  1. 0.13 in/hr
  2. 0.43 in/hr Correct answer.
  3. 1.28 in/hr
  4. 0.86 in/hr

Explanation

Each head covers 30 × 30 = 900 ft². Applying PR = 96.3 × GPM ÷ S² gives 96.3 × 4 ÷ 900 = 0.43 in/hr.

Question 11Irrigation hydraulics and system design fundamentals

A zone has nine sprinkler heads, each rated at 1.5 GPM at the design pressure. What is the required zone flow?

  1. 10.5 GPM
  2. 6.0 GPM
  3. 18.0 GPM
  4. 13.5 GPM Correct answer.

Explanation

Zone flow is the sum of the individual head flows at design pressure: 9 × 1.5 = 13.5 GPM. This total is then used to size the valve, lateral and mainline serving the zone.

Question 12Irrigation hydraulics and system design fundamentals

Which change most reduces friction loss in a long mainline carrying a fixed flow?

  1. Raising the operating pressure
  2. Adding more fittings
  3. Increasing the pipe diameter Correct answer.
  4. Increasing the water temperature

Explanation

In the Hazen-Williams relationship, friction loss varies roughly with the inverse of the diameter raised to about the 4.87 power, so a modest increase in pipe size produces a large drop in loss. At a fixed flow, system pressure does not change friction loss.

Question 13Irrigation hydraulics and system design fundamentals

A system applies 1.2 inches of water and 0.9 inch reaches the root zone. What is the application efficiency?

  1. 75% Correct answer.
  2. 90%
  3. 27%
  4. 133%

Explanation

Application efficiency is stored water divided by water applied: 0.9 ÷ 1.2 = 0.75, or 75%. The 25% difference is lost to evaporation, wind drift, runoff or deep percolation.

Question 14Irrigation hydraulics and system design fundamentals

To keep precipitation rates matched, a 360-degree head delivering 4 GPM should be paired with what flow for a 180-degree head on the same zone?

  1. 4 GPM
  2. 2 GPM Correct answer.
  3. 1 GPM
  4. 8 GPM

Explanation

Matched precipitation means flow is proportional to the arc covered. A half-circle head covers half the area of a full-circle head at the same spacing, so it should deliver half the flow, or 2 GPM.

Question 15Irrigation hydraulics and system design fundamentals

What does 'head-to-head' spacing mean in sprinkler layout?

  1. Spacing equal to the throw diameter Correct answer.
  2. Spacing equal to the riser height
  3. Spacing equal to the pipe diameter
  4. Spacing equal to the nozzle diameter

Explanation

Head-to-head spacing sets the distance between heads equal to the wetted diameter of the sprinkler, so each head throws to the adjacent head. This overlapping pattern gives the most uniform coverage at a given pressure.

Question 16Irrigation hydraulics and system design fundamentals

A catch-can test across an irrigation zone is used primarily to determine which value?

  1. Distribution uniformity Correct answer.
  2. Static pressure at the pump
  3. Total dynamic head of the pump
  4. Pipe friction loss coefficient

Explanation

Cans placed at intervals collect water for a set time; comparing the lowest quarter of the catches to the overall average gives the distribution uniformity of the zone. The same data also yield the actual precipitation rate.

Question 17Irrigation hydraulics and system design fundamentals

In pipe friction calculations, losses through valves and fittings are usually handled by which method?

  1. Equivalent length of pipe Correct answer.
  2. Ignoring them in short runs
  3. Doubling the pipe diameter
  4. Adding 10 psi per fitting

Explanation

Each fitting is assigned an equivalent length of straight pipe that produces the same pressure loss at the same velocity. These lengths are added to the actual pipe length before the friction calculation.

Question 18Irrigation hydraulics and system design fundamentals

Pump cavitation is most directly caused by which condition?

  1. A sticking foot valve
  2. An oversized suction line
  3. Excessively high discharge pressure
  4. Insufficient NPSH available Correct answer.

Explanation

Cavitation occurs when the pressure at the impeller eye falls below the water's vapor pressure, so vapor bubbles form and then collapse. This happens when net positive suction head available is less than the head required by the pump.

Question 19Irrigation hydraulics and system design fundamentals

A zone applies water at 0.4 in/hr and needs 0.6 in of irrigation. How long must it run, ignoring efficiency losses?

  1. 90 min Correct answer.
  2. 67 min
  3. 150 min
  4. 24 min

Explanation

Run time equals depth required divided by application rate: 0.6 ÷ 0.4 = 1.5 hours, or 90 minutes. Efficiency losses would be added by dividing the net requirement by the efficiency first.

Question 20Irrigation hydraulics and system design fundamentals

Total dynamic head for an irrigation pump is made up of which combination of terms?

  1. Lift, pressure head, friction Correct answer.
  2. Static pressure and elevation only
  3. Flow, velocity and pipe size
  4. Nozzle pressure and flow rate

Explanation

Total dynamic head is the sum of the elevation lift, the pressure head needed at the discharge point and all friction and fitting losses in the system. It represents the energy the pump must add per unit weight of water.

Question 21Irrigation hydraulics and system design fundamentals

Sprinklers delivering 4 GPM are spaced 30 ft apart on a triangular pattern. Using PR = 96.3 × GPM ÷ (0.866 × S²), what is the approximate precipitation rate?

  1. 0.43 in/hr
  2. 0.49 in/hr Correct answer.
  3. 0.99 in/hr
  4. 0.57 in/hr

Explanation

The area served by each head in a triangular pattern is 0.866 × 30² = 779 ft². Applying the formula gives 96.3 × 4 ÷ 779 = 0.49 in/hr, slightly higher than the same spacing on a square grid because less area is covered.

Question 22Irrigation hydraulics and system design fundamentals

Using the Hazen-Williams relationship, if flow through a given pipe doubles, friction loss increases by a factor of about:

  1. 1.85
  2. 3.6 Correct answer.
  3. 2.0
  4. 4.9

Explanation

Hazen-Williams relates head loss to flow raised to the 1.852 power. Doubling the flow gives 2^1.852, which is about 3.6 times the original loss.

Question 23Irrigation hydraulics and system design fundamentals

Water flows at 10 GPM through a pipe with an inside diameter of 1 inch. What is the approximate velocity?

  1. 1.0 ft/s
  2. 8.2 ft/s
  3. 4.1 ft/s Correct answer.
  4. 2.0 ft/s

Explanation

Velocity equals flow divided by cross-sectional area; using V = 0.4085 × GPM ÷ d² with diameter in inches gives 0.4085 × 10 ÷ 1 = 4.1 ft/s. This is near the upper end of typical irrigation design velocity.

Question 24Irrigation hydraulics and system design fundamentals

A pump delivers water at 50 psi at the outlet. The main rises 30 ft and has 15 psi of friction loss. What pressure remains at the top of the rise?

  1. 26 psi
  2. 48 psi
  3. 35 psi
  4. 22 psi Correct answer.

Explanation

The 30 ft rise costs 30 × 0.433 = 13 psi, and friction removes another 15 psi. Subtracting both from 50 psi leaves about 22 psi at the top of the rise.

Question 25Irrigation hydraulics and system design fundamentals

A pump running at 60 Hz delivers 60 GPM. Using the affinity laws, what flow will it deliver at 50 Hz?

  1. 50 GPM Correct answer.
  2. 72 GPM
  3. 36 GPM
  4. 60 GPM

Explanation

Flow changes in direct proportion to rotational speed, so the new flow is 60 × (50 ÷ 60) = 50 GPM. Head would change with the square of the speed ratio and power with the cube.

Question 26Irrigation hydraulics and system design fundamentals

Which equation correctly relates flow rate (Q), velocity (V), and pipe cross-sectional area (A)?

  1. Q = V × A Correct answer.
  2. Q = V × A²
  3. Q = V ÷ A
  4. Q = A ÷ V

Explanation

The continuity relationship states that discharge equals velocity multiplied by the cross-sectional area of the pipe. Rearranging it lets a designer solve for velocity when flow and pipe size are known. This is the basis for checking whether a chosen pipe diameter keeps velocity within acceptable limits.

Question 27Irrigation hydraulics and system design fundamentals

An air relief valve is installed at the highest point of an irrigation mainline primarily to do what?

  1. Prevent irrigation water from flowing backwards into the potable supply
  2. Relieve excess pressure created by large elevation changes in the mainline
  3. Vent trapped air that restricts flow and raises friction loss Correct answer.
  4. Drain the mainline automatically at the end of every irrigation cycle

Explanation

Air released from solution collects at high points and can throttle the pipe, cause erratic sprinkler operation, and aggravate water hammer. An air relief valve vents that accumulated air automatically. It is not a pressure control or backflow device.

Question 28Irrigation hydraulics and system design fundamentals

Distribution uniformity (DU) as determined in an irrigation audit describes which condition?

  1. The share of applied water that reaches the root zone
  2. The difference between static and dynamic pressure in a zone
  3. How evenly water is applied over the irrigated area Correct answer.
  4. How closely measured flow matches the nozzle's rated flow

Explanation

Distribution uniformity is calculated by comparing the lowest-quarter catch volumes with the average catch volume, so it measures evenness of application. A system can be uniform yet still inefficient if it applies too much water overall. Uniformity and efficiency are separate measures.

Question 29Irrigation hydraulics and system design fundamentals

On a typical centrifugal pump performance curve, what happens to total head as flow increases from shutoff toward the maximum end of the curve?

  1. It rises steadily as flow increases toward the maximum
  2. It stays the same at every point along the curve
  3. It decreases as flow increases Correct answer.
  4. It rises to a peak near mid-flow and then falls

Explanation

A centrifugal pump produces its highest head at shutoff and progressively less head as discharge increases. The curve slopes downward across the normal operating range. This downward slope is what allows the pump and system curves to intersect at a single stable operating point.

Question 30Irrigation hydraulics and system design fundamentals

Water flows through a reducer in which the pipe diameter decreases, with no change in elevation and negligible friction. What happens to the velocity and the static pressure?

  1. Velocity increases and pressure also increases
  2. Velocity decreases and pressure decreases
  3. Velocity increases and pressure decreases Correct answer.
  4. Velocity and pressure both stay the same

Explanation

Continuity requires the higher-velocity flow in the smaller section to pass the same discharge. Conservation of energy then converts part of the pressure head into velocity head, lowering the static pressure. This trade-off is the working principle of the Bernoulli relationship.

Question 31Irrigation hydraulics and system design fundamentals

The flow and head at which a pump actually operates in a particular irrigation system is determined by what?

  1. The rated flow stamped on the pump nameplate
  2. The setting of the pressure relief valve on the mainline
  3. Where the pump curve crosses the system curve Correct answer.
  4. The pump's shutoff head measured with no flow

Explanation

The system curve shows how much head the piping network demands at each flow, and the pump curve shows what head the pump can supply. The single point satisfying both is the operating point. Nameplate ratings are only a reference condition and do not fix the operating point.

Question 32Irrigation hydraulics and system design fundamentals

In the Hazen-Williams friction equation, what does a lower C value assigned to a pipe represent?

  1. A rougher pipe wall, so friction loss is higher Correct answer.
  2. A larger inside diameter at the same nominal size
  3. A lower velocity for the same rate of discharge
  4. A smoother pipe wall, so friction loss is lower

Explanation

The C factor is a roughness coefficient: smoother materials such as new PVC carry high values, while older or rougher pipe carries lower ones. A lower C increases the computed friction loss for the same flow and diameter. Diameter is a separate term in the equation.

Question 33Irrigation hydraulics and system design fundamentals

What advantage do pressure-compensating emitters provide on a long lateral that runs downhill?

  1. They hold a nearly constant discharge as inlet pressure varies Correct answer.
  2. They remove the need for a filter upstream of the zone
  3. They increase discharge at higher heads to even out the wetting pattern
  4. They lower the total flow that the zone requires from the pump

Explanation

A pressure-compensating emitter contains a flexible or moving element that throttles internally so discharge stays nearly flat over a stated pressure range. This allows longer laterals and tolerates elevation change without over-watering the low end. Filtration and zone flow requirements are unaffected by compensation.

Question 34Irrigation hydraulics and system design fundamentals

A sprinkler zone runs down a constant slope with no pressure regulation anywhere in the zone. What is the expected result?

  1. Discharge is unchanged but pump pressure rises
  2. All heads discharge equally because the pipe is full
  3. Lower heads discharge more than upper heads Correct answer.
  4. Upper heads discharge more because they sit higher

Explanation

Pressure rises about 0.433 psi for every foot of elevation drop, so the lowest heads see the highest pressure. Because fixed-nozzle discharge varies with the square root of pressure, those heads put out more water. The cure is pressure regulation or pressure-compensating devices.

Question 35Irrigation hydraulics and system design fundamentals

Which device is used to protect a potable water supply from contamination by an irrigation system?

  1. A pressure-reducing valve on the zone valve
  2. A reduced-pressure backflow preventer Correct answer.
  3. A wye strainer ahead of the valves
  4. A flow meter on the service line

Explanation

A reduced-pressure-zone assembly guards against both backsiphonage and backpressure by maintaining a lower pressure in the zone between its two check valves. Local codes generally require an approved backflow assembly on any irrigation connection to a drinking water line. The other devices perform unrelated hydraulic functions.

Question 36Irrigation hydraulics and system design fundamentals

Which statement correctly distinguishes static pressure from dynamic pressure in an irrigation system?

  1. Static pressure is read while water flows; dynamic pressure is read with the system off
  2. Static pressure applies to drip only; dynamic pressure applies to sprinklers
  3. Static pressure includes friction loss; dynamic pressure does not include it
  4. Static pressure is read with no flow; dynamic is read while water flows Correct answer.

Explanation

Static pressure exists whenever the system is full and shut off, and it reflects elevation and supply pressure only. Dynamic pressure is the lower value observed while water moves, because friction and velocity losses consume part of the static head. The difference between the two readings indicates the losses in the intervening piping.

Question 37Irrigation hydraulics and system design fundamentals

A designer finds that the peak demand of a single zone exceeds the available water supply. Which change directly reduces the flow required at any one time?

  1. Shorten the laterals by moving the valves closer to the heads
  2. Raise the operating pressure so the zone finishes faster
  3. Split the area into more zones run one at a time Correct answer.
  4. Increase the diameter of the mainline that feeds the zone

Explanation

Dividing the planted area into more zones reduces the number of heads operating simultaneously, so instantaneous demand falls even though total volume is unchanged. Pipe sizing and valve location affect losses and uniformity, not the flow demanded by the nozzles. Raising pressure would increase nozzle flow and worsen the problem.

Question 38Irrigation hydraulics and system design fundamentals

A pump delivers 200 GPM steadily to a 2.0-acre block of turf. Ignoring efficiency losses, what average application depth does this represent? (1 acre-inch = 27,154 gallons.)

  1. About 0.88 inch per hour
  2. About 0.22 inch per hour Correct answer.
  3. About 0.44 inch per hour
  4. About 0.11 inch per hour

Explanation

Convert flow to gallons per hour (200 × 60 = 12,000 gal/hr) and divide by 27,154 to get 0.44 acre-inch per hour. Spreading that volume over 2.0 acres halves the depth to about 0.22 inch per hour. This is the system capacity available before any efficiency losses are considered.

Question 39Irrigation hydraulics and system design fundamentals

A 5.0-acre field needs 0.25 inch of water per day, and the system is limited to 8 hours of run time per day. What pump discharge is required? (1 acre-inch = 27,154 gallons; ignore efficiency losses.)

  1. About 18 GPM
  2. About 141 GPM
  3. About 35 GPM
  4. About 71 GPM Correct answer.

Explanation

The daily volume is 5.0 acres × 0.25 inch = 1.25 acre-inches, which equals about 33,940 gallons. Dividing by 480 minutes of run time gives roughly 71 GPM. Less run time or higher daily depth would push the required discharge up proportionally.

Question 40Irrigation hydraulics and system design fundamentals

A fixed-orifice sprinkler nozzle delivers 3.0 GPM at 40 psi. What discharge should be expected at 55 psi?

  1. About 2.6 GPM
  2. About 4.1 GPM
  3. About 3.5 GPM Correct answer.
  4. About 3.0 GPM

Explanation

For a fixed orifice, discharge varies with the square root of the pressure ratio: 3.0 × √(55/40) = 3.0 × 1.17 ≈ 3.5 GPM. Because the relationship is a square root, a 37 percent pressure increase raises flow only about 17 percent. This is why pressure variation affects flow less than one might expect but still matters for uniformity.

Question 41Backflow prevention and cross-connection control

A cross-connection is best described as which of the following?

  1. A valve used to drain a pipeline after hydrostatic testing
  2. A fitting that joins two pipes made of different materials
  3. An actual or potential link between potable and nonpotable water Correct answer.
  4. A meter installed between the public main and a private service line

Explanation

A cross-connection is any actual or potential physical link between a potable water supply and a source of contamination or pollution. It is the pathway that makes backflow possible, so eliminating or protecting it is the purpose of cross-connection control.

Question 42Backflow prevention and cross-connection control

Backflow is defined as which of the following?

  1. The loss of pressure caused by friction in a pipe
  2. Reversal of flow opposite to its intended direction Correct answer.
  3. The draining of a piping system in preparation for winterization
  4. The normal forward flow of water through a meter

Explanation

Backflow is the reverse of the normal intended direction of flow in a piping system, which allows nonpotable water to move into the potable supply. It occurs as either backsiphonage or backpressure.

Question 43Backflow prevention and cross-connection control

A water main breaks and loses pressure while a submerged hose lies in a pesticide-filled tank. Which condition draws the tank contents into the piping?

  1. Backpressure created by a booster pump
  2. Backsiphonage caused by negative pressure in the main Correct answer.
  3. Thermal expansion of water in a closed tank
  4. A water hammer surge traveling through the service line

Explanation

Backsiphonage occurs when a negative pressure develops in the supply piping and atmospheric pressure pushes fluid from a cross-connection back into the system. A main break that drops system pressure below atmospheric is the classic cause.

Question 44Backflow prevention and cross-connection control

Which method provides the highest degree of protection against backflow?

  1. A pressure vacuum breaker
  2. A double check valve assembly
  3. An air gap Correct answer.
  4. A hose bibb vacuum breaker

Explanation

An air gap is a physical separation between the potable outlet and the flood level rim of the receiving vessel, so no backflow path exists at all. Codes recognize it as the highest level of protection and it is required for the most severe hazards.

Question 45Backflow prevention and cross-connection control

A reduced pressure principle assembly protects the potable supply primarily because it does which of the following?

  1. Relies on a single spring-loaded check valve and a strainer
  2. Uses an atmospheric vent to break a vacuum on the downstream side
  3. Holds the pressure between its two checks below inlet pressure Correct answer.
  4. Discharges supply water continuously to drain during normal flow

Explanation

The assembly has two independent check valves plus a differential relief valve that keeps the zone between the checks at least 2 psi lower than the inlet. If either check fails, the relief valve opens and discharges instead of allowing backflow.

Question 46Backflow prevention and cross-connection control

A double check valve assembly is an acceptable choice only when which condition applies?

  1. The hazard is non-health, and both backpressure and backsiphonage occur Correct answer.
  2. A continuous discharge to a drain is available at the point of installation
  3. The downstream hazard is a health hazard such as a sewage connection
  4. The assembly will be installed below grade in a vault subject to flooding

Explanation

Double check valve assemblies resist both backsiphonage and backpressure but are approved only for non-health hazards. They give no visible or mechanical warning if both checks fail, so health hazards require an air gap or reduced pressure assembly.

Question 47Backflow prevention and cross-connection control

Where must a pressure vacuum breaker assembly be installed?

  1. At least 12 inches below the highest downstream outlet
  2. In a vault below the water main
  3. At least 12 inches above the highest downstream outlet or use point Correct answer.
  4. In a location where it can be submerged by irrigation runoff

Explanation

A pressure vacuum breaker must sit at least 12 inches above the highest downstream outlet or point of use so its air inlet can open to atmosphere. It may be used under continuous pressure but must not be installed where backpressure can develop.

Question 48Backflow prevention and cross-connection control

Which limitation applies to an atmospheric vacuum breaker?

  1. It discharges to drain whenever the supply pressure fluctuates
  2. It may be used where backpressure exists downstream
  3. It cannot be under continuous pressure for more than 12 hours in 24 Correct answer.
  4. It may be installed at any elevation relative to downstream outlets

Explanation

An atmospheric vacuum breaker depends on a float that must be able to drop to atmosphere, so no shutoff valve or continuous pressure may exist downstream and service is limited to 12 hours in any 24-hour period. It also must not be installed where backpressure can occur.

Question 49Backflow prevention and cross-connection control

A hose bibb vacuum breaker protects the potable system by doing which of the following?

  1. Draining the hose after each use
  2. Raising downstream pressure above supply pressure
  3. Filtering debris from the hose before it enters the tap
  4. Admitting air to break a vacuum when supply pressure drops Correct answer.

Explanation

The device seals while water is flowing and opens to atmosphere when supply pressure falls, admitting air to break the vacuum. It must not be left under continuous pressure or have a shutoff valve downstream.

Question 50Backflow prevention and cross-connection control

The main advantage of a spill-resistant vacuum breaker over a standard atmospheric vacuum breaker is that it does which of the following?

  1. It may be installed at any elevation relative to downstream outlets
  2. It incorporates a check valve and may be used under continuous pressure Correct answer.
  3. It provides reliable protection against downstream backpressure
  4. It discharges the full supply flow to drain during normal operation

Explanation

A spill-resistant vacuum breaker combines a check valve with an air inlet, which allows it to remain under continuous pressure without spilling. It still must be installed above the highest downstream outlet and cannot be used where backpressure exists.