Master Plumber (UPC/IPC) — All Questions

A rooftop tank provides 60 ft of elevation head to the floor below. Using 1 ft = 0.433 psi, what static pressure does this produce?

• a.60 psi
• b.13 psi
• c.139 psi
• d.26 psi✓

Pressure = height x 0.433 = 60 x 0.433 = 25.98, about 26 psi. Gravity tanks convert their height into pressure at the fixtures below, so tank elevation must be high enough to meet fixture minimums. Confusing the 2.31 factor for 0.433 would give the wrong 139. IPC §604.0.IPC §604.0

A supply has 70 psi static at the meter. Elevation to the top fixture is 26 ft, the meter loses 9 psi, and pipe friction is 12 psi. Using 0.433 psi/ft, what pressure is delivered to the fixture?

• a.37.7 psi✓
• b.24.0 psi
• c.20.0 psi
• d.17.7 psi

Elevation loss = 26 x 0.433 = 11.3 psi, so residual = 70 - 11.3 - 9 - 12 = 37.7 psi delivered at the fixture. The pressure budget subtracts elevation, meter, and friction losses from the static supply. The delivered pressure must still exceed the fixture's minimum flow pressure. IPC §604.0.IPC §604.0

A water line has 120 ft of straight pipe, six elbows at 2.5 ft equivalent length each, and two tees at 4 ft each. What developed length is used for friction calculations?

• a.120 ft
• b.143 ft✓
• c.135 ft
• d.128 ft

Developed length = 120 + (6 x 2.5) + (2 x 4) = 120 + 15 + 8 = 143 ft. Fittings behave like extra pipe, so their equivalent lengths are added before computing friction. Ignoring fittings understates loss and can undersize the pipe. IPC §604.0.IPC §604.0

Available pressure for friction is 35 psi over a developed length of 175 ft. What is the allowable friction loss per 100 ft of pipe?

• a.35 psi/100 ft
• b.15 psi/100 ft
• c.25 psi/100 ft
• d.20 psi/100 ft✓

Allowable loss per 100 ft = (available / developed length) x 100 = (35 / 175) x 100 = 20 psi per 100 ft. This uniform rate is read against the design flow on the friction chart to pick the pipe. Spreading the whole 35 psi over 175 ft gives the per-100-ft value. IPC §604.0.IPC §604.0

A friction chart shows 8 psi per 100 ft at the design flow, and the line's developed length is 320 ft. What is the total friction loss?

• a.25.6 psi✓
• b.8.0 psi
• c.16.0 psi
• d.32.0 psi

Friction loss = (loss per 100 ft) x (length / 100) = 8 x (320 / 100) = 8 x 3.2 = 25.6 psi. Friction scales directly with developed length, so long runs consume much of the pressure budget. This loss is subtracted along with elevation before checking residual. IPC §604.0.IPC §604.0

Static supply is 80 psi, elevation loss is 15 psi, the meter loses 10 psi, and the most remote fixture needs 25 psi. How much pressure remains for pipe friction?

• a.25 psi
• b.45 psi
• c.30 psi✓
• d.20 psi

Available for friction = 80 - 15 - 10 - 25 = 30 psi. Whatever is left after elevation, meter, and the fixture's minimum flow pressure is the budget spread over the developed length as the allowable friction rate. If the chosen pipe's loss exceeds 30 psi, it must be enlarged. IPC §604.0.IPC §604.0

A demand chart lists: 40 WSFU converts to 30 gpm, 60 WSFU to 40 gpm, and 80 WSFU to 48 gpm (flush-tank). A branch carries 60 WSFU. What is the design flow?

• a.60 gpm
• b.40 gpm✓
• c.30 gpm
• d.48 gpm

Reading the curve, 60 WSFU converts to 40 gpm. Notice the gpm rises less than proportionally to fixture units because the Hunter curve accounts for the low probability of simultaneous use. You never simply equate WSFU to gpm one-for-one. IPC §604.3.IPC §604.3

A booster pump delivers 50 gpm at 40 psi. Water horsepower is (gpm x psi) / 1714, and the pump is 65 percent efficient. What brake horsepower must the motor provide?

• a.1.8 hp✓
• b.1.2 hp
• c.2.9 hp
• d.0.9 hp

Water horsepower = (50 x 40) / 1714 = 2,000 / 1714 = 1.17 hp; brake horsepower = 1.17 / 0.65 = 1.8 hp. The motor must exceed the ideal hydraulic power to cover pump inefficiency. Dividing by the efficiency gives the horsepower to specify. IPC §606.5.IPC §606.5

Using Q = v x d^2 / 0.408, what flow gives 6 ft/s in a pipe with a 2.067 in inside diameter (nominal 2 in type L copper)?

• a.50 gpm
• b.21 gpm
• c.63 gpm✓
• d.40 gpm

Q = 6 x (2.067)^2 / 0.408 = 6 x 4.272 / 0.408 = 25.6 / 0.408 = 62.8, about 63 gpm. Larger pipe carries much more flow at a given velocity because area grows with the square of the diameter. This is why mains upsize quickly with demand. IPC §604.0.IPC §604.0

A building has 96 psi static street pressure. Code limits fixture static pressure to 80 psi, requiring a pressure-reducing valve. If the PRV is set to 55 psi outlet, by how much does it reduce the pressure?

• a.80 psi
• b.41 psi✓
• c.16 psi
• d.55 psi

Reduction = inlet - outlet = 96 - 55 = 41 psi. A PRV is required where static exceeds 80 psi to protect fixtures and control water hammer, and it is set to a comfortable working pressure. A closed system downstream then needs thermal-expansion control. IPC §604.8.IPC §604.8

A recirculation return must carry 8 gpm at no more than 3 ft/s to limit erosion. Using Q = v x d^2 / 0.408, what minimum inside diameter is required?

• a.1.50 in
• b.0.75 in
• c.1.05 in✓
• d.2.00 in

Solve for d: d = sqrt(0.408 x Q / v) = sqrt(0.408 x 8 / 3) = sqrt(1.088) = 1.04, about 1.05 in. Continuously circulated hot lines are held to a low velocity because constant flow erodes copper. A larger diameter lowers velocity for the same flow. IPC §607.2.IPC §607.2

Hot water is limited to 5 ft/s. Using Q = v x d^2 / 0.408, what is the maximum flow in a pipe with a 1.025 in inside diameter (nominal 1 in type L copper)?

• a.20.6 gpm
• b.12.9 gpm✓
• c.5.1 gpm
• d.7.6 gpm

Q = 5 x (1.025)^2 / 0.408 = 5 x 1.051 / 0.408 = 5.255 / 0.408 = 12.9 gpm. Hot water uses a lower 5 ft/s limit than cold water's 8 ft/s because heat accelerates erosion-corrosion of copper. Exceeding this risks pinholing the line. IPC §604.0.IPC §604.0

A drainage branch serves 8 water closets at 4 DFU each, 8 lavatories at 1 DFU each, and 4 floor drains at 2 DFU each. What is the total drainage fixture-unit load?

• a.40 DFU
• b.44 DFU
• c.48 DFU✓
• d.56 DFU

Total = (8 x 4) + (8 x 1) + (4 x 2) = 32 + 8 + 8 = 48 DFU. This total is read against the horizontal branch or drain sizing table to pick the pipe. Because water closets are present, the branch can be no smaller than 3 in regardless of DFU. IPC Table 709.1.IPC Table 709.1

A horizontal fixture branch carries 42 DFU and includes water closets. The table lists 2 in = 6 DFU, 3 in = 20 DFU, 4 in = 160 DFU. What is the minimum branch size?

• a.3 in
• b.2-1/2 in
• c.2 in
• d.4 in✓

The 42 DFU load exceeds the 20 DFU capacity of a 3 in branch, so 4 in (160 DFU) is required. The presence of water closets independently forbids anything smaller than 3 in, but here the load alone forces 4 in. Pick the smallest size whose capacity meets the load. IPC §710.1.IPC §710.1

A building drain at 1/4 in/ft slope has capacities 3 in = 36 DFU, 4 in = 216 DFU, 5 in = 480 DFU. It carries 200 DFU. What is the minimum size?

• a.6 in
• b.5 in
• c.4 in✓
• d.3 in

The 200 DFU load exceeds the 36 DFU capacity of a 3 in drain but fits within the 216 DFU capacity of a 4 in drain, so 4 in is the minimum. Building-drain capacity rises steeply with size and slope. Always choose the smallest adequate size. IPC §710.1.IPC §710.1

A 4 in building drain is rated at 180 DFU at 1/8 in/ft slope but 216 DFU at 1/4 in/ft. The load is 200 DFU. What is the most economical compliant solution?

• a.Use a 6 in drain at any slope
• b.Reduce the load to 180 DFU
• c.Run the 4 in drain at 1/4 in per foot✓
• d.Use a 5 in drain at 1/8 in per foot

At 1/8 in/ft the 4 in drain only handles 180 DFU, below the 200 DFU load, but at 1/4 in/ft it handles 216 DFU, which covers it. Increasing slope raises capacity without upsizing pipe. Steeper slope keeps solids in suspension at the higher load. IPC §710.1.IPC §710.1

A soil stack collects 300 DFU. The table lists 3 in = 48 DFU and 4 in = 500 DFU for total load on a stack. What is the minimum stack size, and how may it change as it descends?

• a.4 in, and it may not be reduced in the direction of flow✓
• b.6 in, and it must enlarge at the base
• c.4 in, and it may reduce to 3 in below the lowest branch
• d.3 in, may reduce to 2 in at the base

The 300 DFU load exceeds the 48 DFU limit of a 3 in stack, so a 4 in stack is required, and a stack may never be reduced in the direction of flow. Downsizing lower down would restrict flow and flood upper branches. The stack stays 4 in to its base. IPC §710.1.IPC §710.1

A 4 in building drain runs at 1/8 in/ft over 120 ft. What is the total fall?

• a.24 in
• b.12 in
• c.30 in
• d.15 in✓

Fall = slope x length = 1/8 in/ft x 120 ft = 15 in. Pipe 3 in and larger uses 1/8 in per foot as the minimum slope. Using 1/4 in per ft would wrongly double the answer to 30 in. IPC §704.1.IPC §704.1

A 3 in drain at the minimum 1/8 in/ft slope runs 64 ft. What is the total fall?

• a.4 in
• b.8 in✓
• c.12 in
• d.16 in

Fall = 1/8 in/ft x 64 ft = 8 in. Pipe 3 in and larger takes the 1/8 in per foot minimum, so a 64 ft run drops 8 in. Applying the 1/4 in rate reserved for smaller pipe would overstate the fall. IPC §704.1.IPC §704.1

Expressed as a percent grade, what is a slope of 1/4 in per foot?

• a.0.5%
• b.2.08%✓
• c.4.0%
• d.1.0%

Percent grade = rise / run = 0.25 in / 12 in = 0.0208 = 2.08 percent. Converting the fractional-inch slope to a percent helps when laying pipe with a laser or level. The 1/8 in per foot minimum is half this, about 1.04 percent. IPC §704.1.IPC §704.1

Expressed as a percent grade, what is a slope of 1/8 in per foot?

• a.1.04%✓
• b.1.5%
• c.2.08%
• d.0.5%

Percent grade = 0.125 in / 12 in = 0.0104 = 1.04 percent. This is the minimum slope for pipe 3 in and larger, half the 2.08 percent of the 1/4 in rate. Too little slope lets solids settle and clog the drain. IPC §704.1.IPC §704.1

A vent must be at least one-half the diameter of the drain it serves and never less than 1-1/4 in. What is the minimum vent for a 4 in drain?

• a.1-1/4 in
• b.1-1/2 in
• c.3 in
• d.2 in✓

Half of 4 in is 2 in, which is larger than the 1-1/4 in floor, so the minimum vent is 2 in. The half-diameter rule sets the size and the 1-1/4 in minimum only governs very small drains. A 3 in drain by the same rule needs at least a 1-1/2 in vent. IPC §906.1.IPC §906.1

A vent-sizing table shows a 3 in vent stack serves up to 48 DFU for a developed length up to 212 ft. The vent carries 40 DFU over a 150 ft run. Is a 3 in vent acceptable?

• a.No, a vent must equal the drain size
• b.No, the length exceeds the limit
• c.Yes, both load and length are within limits✓
• d.No, the load exceeds the limit

The 40 DFU load is within the 48 DFU cap and the 150 ft run is within the 212 ft length limit, so a 3 in vent is acceptable. Vent sizing depends on both fixture-unit load and total developed length; exceeding either forces a larger vent. IPC §906.2.IPC §906.2

The maximum trap-to-vent (trap arm) developed length is 1-1/2 in = 6 ft, 2 in = 8 ft, 3 in = 12 ft. What is the maximum developed length for a 2 in trap arm?

• a.5 ft
• b.6 ft
• c.12 ft
• d.8 ft✓

For a 2 in trap arm the table caps the developed length at 8 ft. Beyond this distance the vent is too far to protect the seal and the trap can self-siphon. Larger trap arms are allowed to run farther because they drain more slowly. IPC §906.1.IPC §906.1

The total fall of a trap arm between the trap weir and the vent may not exceed one pipe diameter. What is the maximum fall for a 3 in trap arm?

• a.3 in✓
• b.6 in
• c.2 in
• d.1-1/2 in

The maximum fall equals one pipe diameter, so for a 3 in arm that is 3 in. If the arm falls more than one diameter, the vent opening drops below the crown weir and the trap can self-siphon. This limit is separate from the developed-length limit. IPC §906.1.IPC §906.1

A fixture trap must maintain a liquid seal. What is the minimum trap seal depth required by code?

• a.2 in✓
• b.4 in
• c.3 in
• d.1 in

The minimum trap seal depth is 2 in (and generally not more than 4 in unless a deep-seal trap is specified). The seal is the water held between the dip and the crown weir that blocks sewer gas. Too shallow a seal is easily broken by siphonage or evaporation. IPC §1002.4.IPC §1002.4

A building sewer at 1/8 in/ft slope has capacities 4 in = 180 DFU, 5 in = 390 DFU, 6 in = 700 DFU. It carries 350 DFU. What is the minimum size?

• a.6 in
• b.8 in
• c.5 in✓
• d.4 in

The 350 DFU load exceeds the 180 DFU capacity of a 4 in sewer but fits within the 390 DFU capacity of a 5 in sewer, so 5 in is the minimum. The building sewer is sized on the total DFU it carries at its installed slope. Choose the smallest adequate size. IPC §710.1.IPC §710.1

A fixture drain may not be smaller than the trap it serves. A lavatory has a 1-1/4 in trap. What is the minimum fixture-drain size?

• a.3 in
• b.1-1/2 in
• c.1-1/4 in✓
• d.2 in

The fixture drain must be at least as large as the trap, so a 1-1/4 in trap requires at least a 1-1/4 in drain. The drain may be larger but never smaller, which would restrict flow at the trap outlet. Lavatory traps are commonly 1-1/4 in. IPC §709.2.IPC §709.2

A branch serves two tank water closets totaling 8 DFU. The DFU table would allow a 2 in pipe, but what minimum size governs because water closets are present?

• a.4 in
• b.3 in✓
• c.2 in
• d.1-1/2 in

Any branch or drain that receives water-closet discharge must be at least 3 in, regardless of a low DFU count. The 8 DFU load alone would fit a 2 in pipe, but the water-closet rule overrides it. This prevents clogging from bulk waste. IPC §710.1.IPC §710.1

An individual vent must be at least half the drain diameter but never smaller than 1-1/4 in. What is the minimum vent for a 1-1/4 in fixture drain?

• a.3/4 in
• b.1-1/2 in
• c.1-1/4 in✓
• d.2 in

Half of 1-1/4 in is 0.625 in, which is below the 1-1/4 in floor, so the minimum vent is 1-1/4 in. For small drains the fixed 1-1/4 in minimum governs rather than the half-diameter rule. Vents smaller than this clog and lose venting capacity. IPC §906.1.IPC §906.1

A 45-degree offset must clear a 30 in vertical obstruction. The diagonal travel equals the offset times 1.414. What is the travel length of pipe between the two fittings?

• a.30.0 in
• b.42.4 in✓
• c.21.2 in
• d.60.0 in

Travel = offset x 1.414 = 30 x 1.414 = 42.4 in. For a 45-degree offset the rise, run, and diagonal form a right triangle where the diagonal is 1.414 (the square root of 2) times the offset. This constant lets a plumber lay out offsets from the required rise. IPC §704.2.IPC §704.2

For a 45-degree offset, the horizontal run (set) equals the vertical rise. If a stack offsets 18 in vertically at 45 degrees, what is the horizontal run?

• a.18 in✓
• b.25.5 in
• c.36 in
• d.12.7 in

At 45 degrees the horizontal run equals the vertical rise, so an 18 in rise gives an 18 in run. The equal legs are what make the diagonal 1.414 times either one. This symmetry is unique to the 45-degree fitting. IPC §704.2.IPC §704.2

A change order adds four lavatories at 1 DFU each and two floor drains at 2 DFU each to a branch already loaded to 30 DFU. What is the new total load?

• a.36 DFU
• b.42 DFU
• c.34 DFU
• d.38 DFU✓

Added load = (4 x 1) + (2 x 2) = 4 + 4 = 8 DFU, so the new total is 30 + 8 = 38 DFU. A change order that adds fixtures raises the drainage load, so the branch and stack must be rechecked against their capacity. If the total exceeds the pipe capacity, the branch must be enlarged. IPC Table 709.1.IPC Table 709.1

A horizontal branch is 3 in and the table caps a 3 in branch at 20 DFU. The branch already carries 20 DFU and a designer wants to add a 2 DFU sink. What is required?

• a.Add a separate relief vent to raise capacity
• b.Enlarge the branch to 4 in✓
• c.Reduce the slope to gain capacity
• d.Nothing further, because 22 DFU still fits a 3 in branch

Adding the sink brings the load to 22 DFU, which exceeds the 20 DFU cap of a 3 in branch, so it must be enlarged to 4 in (or a new branch run). Venting and slope do not raise a pipe's DFU capacity. Only a larger pipe carries more fixture units. IPC §710.1.IPC §710.1

A code table requires 1 water closet per 75 male occupants in a business occupancy. The male occupant load is 200. What is the minimum number of water closets?

• a.4
• b.2
• c.3✓
• d.5

Divide and round up: 200 / 75 = 2.67, which rounds up to 3 water closets. Fixture counts always round up because a fraction of a fixture cannot be installed and the table sets a minimum. The occupant load and ratio drive the count. IPC §403.1.IPC §403.1

Lavatories are required at 1 per 200 occupants. The occupant load is 480. What is the minimum number of lavatories?

• a.2
• b.3✓
• c.5
• d.4

480 / 200 = 2.4, which rounds up to 3 lavatories. Any fractional result rounds up to meet the minimum-fixture requirement. Lavatory ratios are typically lower than water-closet ratios. IPC §403.1.IPC §403.1

Drinking fountains are required at 1 per 100 occupants. The occupant load is 350. What is the minimum number required?

• a.4✓
• b.3
• c.2
• d.5

350 / 100 = 3.5, which rounds up to 4 drinking fountains. As with all fixture counts, a fractional result rounds up. Some codes allow bottle-filling stations to substitute for a portion of the fountains. IPC §410.1.IPC §410.1

For a male toilet room, code allows urinals to substitute for up to 50 percent of the required water closets. If 12 water closets are required, how many may be replaced by urinals?

• a.12
• b.6✓
• c.4
• d.8

Maximum urinal substitution = 50 percent of 12 = 6, so up to 6 water closets may be replaced by urinals. Substituting urinals saves water and space while keeping total fixtures adequate. At least half the required closets must remain water closets. IPC §424.2.IPC §424.2

The occupant load is 300, split 50/50 by sex. Female water closets are required at 1 per 40. How many female water closets are required?

• a.8
• b.4✓
• c.5
• d.3

Female occupants = 150, and 150 / 40 = 3.75, which rounds up to 4 water closets. Occupant loads are usually divided equally by sex before applying the ratio. Rounding up meets the minimum. IPC §403.1.IPC §403.1

An assembly occupancy requires 1 water closet per 40 occupants. The load is 260. What is the minimum number of water closets?

• a.6
• b.5
• c.8
• d.7✓

260 / 40 = 6.5, which rounds up to 7 water closets. Assembly occupancies use tighter ratios than business occupancies because of peak simultaneous use at intermission. Rounding up satisfies the minimum. IPC §403.1.IPC §403.1

Business occupancies must provide at least one service (mop) sink. What is the code purpose of that requirement?

• a.To provide at least one sink for janitorial and cleaning use✓
• b.To meet the building's drinking-water needs
• c.To serve as the required backflow-assembly test connection
• d.To provide an emergency eyewash station

The service sink requirement guarantees a dedicated fixture for janitorial and maintenance use so cleaning water is not drawn from food-prep or public fixtures. It is separate from lavatories, drinking fountains, and backflow test points. At least one is required per building or floor. IPC §403.2.IPC §403.2

A restaurant has 8 water closets total. Code requires at least 1 accessible (ADA) water closet per 6 fixtures or fraction thereof. How many must be accessible?

• a.3
• b.2✓
• c.1
• d.4

8 / 6 = 1.33, which rounds up to 2 accessible water closets. Accessibility counts round up like all fixture minimums. Larger fixture groups require proportionally more accessible fixtures. IPC §404.1.IPC §404.1

A sewage ejector discharges 40 gpm against 22 ft of total head. Using WHP = (gpm x head) / 3960, what is the water horsepower?

• a.0.44 hp
• b.0.11 hp
• c.0.33 hp
• d.0.22 hp✓

WHP = (40 x 22) / 3960 = 880 / 3960 = 0.22 hp. The 3960 constant is used when head is expressed in feet rather than psi. The motor is oversized above this by dividing by the pump efficiency. IPC §712.3.IPC §712.3

A sump receives 60 gpm peak inflow and its pump discharges 100 gpm. With 40 gallons of usable storage, how long is one pump-down (run time) at peak inflow?

• a.90 s
• b.40 s
• c.60 s✓
• d.24 s

Net removal while running = 100 - 60 = 40 gpm, so draw-down time = 40 / 40 = 1 min = 60 s. Sizing storage this way limits motor starts per hour and prevents short-cycling. Too small a volume overworks the pump. IPC §712.1.IPC §712.1

A booster pump must add 45 psi. What head, in feet, must it develop? Use 1 psi = 2.31 ft.

• a.195 ft
• b.104 ft✓
• c.45 ft
• d.19 ft

Head = psi x 2.31 = 45 x 2.31 = 103.95, about 104 ft. Pump curves are plotted in feet of head, so the required pressure boost must be converted before selecting a pump. The 2.31 factor is the reciprocal of 0.433 psi per foot. IPC §606.5.IPC §606.5

A water line has a developed length of 180 ft and the chart shows 5 psi per 100 ft at design flow. What is the total friction loss?

• a.25 psi
• b.18 psi
• c.5 psi
• d.9 psi✓

Friction loss = 5 x (180 / 100) = 5 x 1.8 = 9 psi. Friction scales directly with developed length at a fixed loss rate. This loss is subtracted from the pressure budget along with elevation and meter losses. IPC §604.0.IPC §604.0

The pressure available for friction is 27 psi over a 150 ft developed length. What allowable friction loss per 100 ft is used to size the pipe?

• a.18 psi/100 ft✓
• b.27 psi/100 ft
• c.9 psi/100 ft
• d.13.5 psi/100 ft

Allowable loss per 100 ft = (27 / 150) x 100 = 18 psi per 100 ft. This uniform rate is read against the design flow on the friction chart to select the smallest adequate pipe. Spreading all 27 psi over the run gives the per-100-ft value. IPC §604.0.IPC §604.0

Using v = 0.408 x Q / d^2, what is the velocity of 9 gpm in a pipe with a 0.785 in inside diameter (nominal 3/4 in type L copper)?

• a.8.1 ft/s
• b.3.7 ft/s
• c.4.9 ft/s
• d.6.0 ft/s✓

v = 0.408 x 9 / (0.785)^2 = 3.672 / 0.616 = 5.96, about 6.0 ft/s. For hot water this exceeds the 5 ft/s limit, so the flow or pipe size would need adjustment; for cold water it is acceptable. Velocity climbs quickly in small pipe. IPC §604.0.IPC §604.0

Hot water is held to 5 ft/s. Using Q = v x d^2 / 0.408, what is the maximum flow in a 0.785 in inside-diameter (3/4 in type L) pipe?

• a.7.6 gpm✓
• b.12.3 gpm
• c.15.1 gpm
• d.4.9 gpm

Q = 5 x (0.785)^2 / 0.408 = 5 x 0.616 / 0.408 = 3.08 / 0.408 = 7.55, about 7.6 gpm. Hot water uses the lower 5 ft/s limit because heat accelerates erosion-corrosion of copper. Exceeding this flow risks pinholing over time. IPC §604.0.IPC §604.0

Water expands about 2 percent by volume when heated from cold to hot. A 50-gallon heater is filled with cold water. Approximately how much expansion volume must a thermal-expansion tank accommodate?

• a.1 gal✓
• b.2.5 gal
• c.0.1 gal
• d.5 gal

Expansion volume = 2 percent x 50 = 0.02 x 50 = 1 gallon. On a closed system this expanded water cannot flow back to the main, so an expansion tank absorbs it to prevent a pressure spike. The tank is charged to system static pressure. IPC §607.3.IPC §607.3