Master Plumber (UPC/IPC) — All Questions

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

A hot-water recirculation loop loses 6,000 BTU/hr and the water may cool only 20 degrees F. Using Q (gpm) = BTU/hr / (500 x delta-T), what recirculation flow is required?

• a.0.6 gpm✓
• b.0.3 gpm
• c.2.0 gpm
• d.1.2 gpm

Q = 6,000 / (500 x 20) = 6,000 / 10,000 = 0.6 gpm. The 500 factor is the heat capacity of water in BTU per hour per gpm per degree F. A low recirculation flow offsets standby losses without eroding the piping. IPC §607.2.IPC §607.2

A thermostatic mixing valve blends 140 degree F hot with 50 degree F cold to deliver 110 degree F. What fraction of the blended flow is the 140 degree F hot water? Use 140f + 50(1 - f) = 110.

• a.50.0%
• b.33.3%
• c.66.7%✓
• d.75.0%

Set 140f + 50(1 - f) = 110, so 90f = 60 and f = 0.667, or 66.7 percent hot water. The valve stores water hot (140 F) for Legionella control while blending down to a safe 110 F delivery. The cold fraction is the remaining 33.3 percent. IPC §607.1.IPC §607.1

Using Q (gpm) = v x d^2 / 0.408, what flow produces 8 ft/s in a 1.025 in inside-diameter (1 in type L) pipe?

• a.12 gpm
• b.15 gpm
• c.21 gpm✓
• d.28 gpm

Q = 8 x (1.025)^2 / 0.408 = 8 x 1.051 / 0.408 = 8.408 / 0.408 = 20.6, about 21 gpm. This is the most a nominal 1 in type L line should carry on cold water before exceeding 8 ft/s. Beyond it, erosion and noise rise. IPC §604.0.IPC §604.0

A run has 200 ft of pipe, ten elbows at 2 ft each, one gate valve at 1 ft, and two tees at 3 ft each. What developed length is used for friction?

• a.227 ft✓
• b.207 ft
• c.220 ft
• d.230 ft

Developed length = 200 + (10 x 2) + 1 + (2 x 3) = 200 + 20 + 1 + 6 = 227 ft. Every fitting adds its equivalent length so friction is computed on the true hydraulic length. Omitting fittings understates the loss. IPC §604.0.IPC §604.0

A building needs 25 psi at the top fixture, which is 60 ft above a 40 psi service. Elevation costs 0.433 psi/ft. Ignoring friction, how much booster pressure is needed to satisfy elevation and the fixture requirement?

• a.25 psi
• b.40 psi
• c.26 psi
• d.11 psi✓

Elevation loss = 60 x 0.433 = 25.98 psi; the top fixture needs 25 psi, so total required is about 51 psi, and the deficit beyond the 40 psi service is 51 - 40 = 11 psi. A booster pump must supply that shortfall. Friction would add to the required boost. IPC §606.5.IPC §606.5

A meter/service must supply 44 gpm. Meter loss at 44 gpm is 8 psi. Static is 65 psi, elevation loss is 18 psi, and the fixture needs 15 psi. How much is left for friction?

• a.42 psi
• b.24 psi✓
• c.16 psi
• d.34 psi

Available for friction = static - meter - elevation - fixture = 65 - 8 - 18 - 15 = 24 psi. This leftover is spread over the developed length to set the allowable friction rate per 100 ft. If the chosen pipe's loss exceeds 24 psi, it must be enlarged. IPC §604.0.IPC §604.0

A storm table is published at 4 in/hr and lists a 6 in leader at 6,100 ft^2. If the local design rate is 2 in/hr, what roof area may that leader serve?

• a.12,200 ft^2✓
• b.6,100 ft^2
• c.3,050 ft^2
• d.24,400 ft^2

Capacity in area is inversely proportional to rainfall rate, so halving the rate from 4 to 2 in/hr doubles the allowable area: 6,100 x (4 / 2) = 12,200 ft^2. The conductor's gpm capacity is fixed, so a lighter storm lets it drain more roof. Always adjust table areas to the local rate. IPC §1106.2.IPC §1106.2

A roof's primary storm drainage handles 150 gpm. The separate secondary (overflow) system must be sized for what flow?

• a.75 gpm (one-half of the primary design flow)
• b.no separate overflow flow is ever required
• c.300 gpm (twice the primary design flow)
• d.150 gpm (at least the full primary flow)✓

The secondary (overflow) drainage must be sized for at least the full design storm, the same 150 gpm as the primary, because it must carry the entire flow if the primary is blocked. It is a required independent system with its own leaders. Undersizing it defeats its purpose. IPC §1107.1.IPC §1107.1

A leader table at 4 in/hr rates a 4 in leader at 4,600 ft^2. The local rate is 3 in/hr. What roof area may the 4 in leader serve?

• a.4,600 ft^2
• b.9,200 ft^2
• c.3,450 ft^2
• d.6,133 ft^2✓

Allowable area scales inversely with rate: 4,600 x (4 / 3) = 6,133 ft^2. Because the local 3 in/hr storm is lighter than the table's 4 in/hr, the same leader drains more roof. Adjusting to the local rainfall intensity is essential. IPC §1106.2.IPC §1106.2

How long will a 40,000 BTU/hr output heater take to raise 40 gallons by 90 degrees F? Energy = 40 x 8.33 x 90.

• a.60 min
• b.45 min✓
• c.30 min
• d.90 min

Energy = 40 x 8.33 x 90 = 29,988 BTU; time = energy / output = 29,988 / 40,000 = 0.75 hr = 45 min. Recovery time ties the heat load to the burner output. A larger burner or smaller rise shortens the time. IPC §501.0.IPC §501.0

Using v = 0.408 x Q / d^2, what is the velocity of 40 gpm in a 1.5 in inside-diameter pipe?

• a.10.9 ft/s
• b.3.6 ft/s
• c.7.3 ft/s✓
• d.5.4 ft/s

v = 0.408 x 40 / (1.5)^2 = 16.32 / 2.25 = 7.25, about 7.3 ft/s. This is under the 8 ft/s cold-water limit, so the 1.5 in line handles 40 gpm acceptably. For hot water it would exceed the 5 ft/s limit. IPC §604.0.IPC §604.0

A demand chart converts 20 WSFU (flush-tank) to about 16 gpm, which must flow at no more than 8 ft/s. Using Q = v x d^2 / 0.408, what is the smallest inside diameter that keeps velocity within 8 ft/s at 16 gpm?

• a.1.50 in
• b.2.00 in
• c.0.50 in
• d.0.90 in✓

Solve for d: d = sqrt(0.408 x Q / v) = sqrt(0.408 x 16 / 8) = sqrt(0.816) = 0.90 in. Any inside diameter of at least 0.90 in keeps the 16 gpm flow within the 8 ft/s limit. A nominal 3/4 in type L tube (0.785 in ID) would slightly exceed it, so 1 in is chosen. IPC §604.0.IPC §604.0

Which piping material is NOT approved for potable water distribution under the IPC and UPC?

• a.ABS plastic pipe✓
• b.Type L copper tube
• c.CPVC pipe
• d.PEX tubing

ABS is a DWV (drain, waste, vent) material and is not approved for pressurized potable water. Approved potable materials include copper, CPVC, PEX, and PVC (cold water only). Using ABS on a water line would fail inspection. IPC §605.4 / UPC §604.1.IPC §605.4 / UPC §604.1

Which material is approved for above-ground sanitary DWV piping?

• a.Schedule 80 PVC pressure pipe rated only for potable use
• b.Soft-annealed refrigeration-grade copper tubing
• c.Type M copper, for pressurized potable supply only
• d.ABS or PVC DWV pipe✓

ABS (ASTM D2661) and PVC (D2665) DWV pipe are the common plastic drainage materials, along with cast iron and DWV copper. The pressure-rated and refrigeration tubing listed are for supply or mechanical service, not gravity drainage. Match the pipe listing to the application. IPC §702.0 / UPC §701.0.IPC §702.0 / UPC §701.0

PEX tubing is joined by which method?

• a.Solvent cement, the same as used on PVC pipe
• b.Cold-expansion or crimp/clamp mechanical fittings✓
• c.Soldering with 95/5 tin-antimony solder
• d.Threaded couplings cut directly into the tubing wall by hand

PEX cannot be solvent-welded or soldered; it is joined with cold-expansion (ASTM F1960), crimp (F1807), or clamp fittings. The plastic is heat-set and will not accept cement or solder. Correct fittings and tools are essential to a reliable joint. IPC §605.5.IPC §605.5

For an underground water service in corrosive soil, which copper tube offers the thickest wall and greatest durability?

• a.Type K✓
• b.Type L, though thinner than the drainage grade
• c.Type M, the thin-wall grade used for drainage
• d.Type DWV, the drainage grade of tube

Wall thickness ranks K greater than L greater than M, so type K has the thickest wall and is preferred underground; type L is also acceptable, but M is often prohibited below grade. DWV copper is for drainage, not pressurized service. Thicker walls resist soil corrosion. IPC §605.4.IPC §605.4

Solder and flux used on potable water joints must meet what lead limit?

• a.Up to 50 percent lead is permitted for strength (50/50 solder)
• b.Up to 5 percent lead is allowed on cold-water lines only
• c.No lead limit applies to solder on domestic water lines
• d.Lead-free, not more than 0.2 percent lead✓

Since 1986, potable joints require lead-free solder containing no more than 0.2 percent lead, such as 95/5 tin-antimony. The older 50/50 tin-lead solder is banned on drinking-water lines. Lead exposure is a serious health hazard. IPC §605.14 / UPC §316.1.IPC §605.14 / UPC §316.1

A transition between a PVC drain and an ABS drain must be made how?

• a.By threading both pipe ends and using a steel union
• b.By applying ordinary PVC solvent cement, which bonds both plastics equally
• c.By heat-fusing the two pipes together
• d.With a listed transition coupling or approved transition cement✓

PVC and ABS cements are not interchangeable, so dissimilar plastics are joined with a listed mechanical (banded) transition coupling or a specifically listed transition cement where permitted. Ordinary PVC cement will not reliably bond ABS. The joint must match the listing. IPC §705.0 / UPC §705.0.IPC §705.0 / UPC §705.0

Hubless (no-hub) cast-iron soil pipe is joined by what method?

• a.A threaded ring twisted onto each pipe end
• b.Solvent cement applied to the spigot end
• c.A caulked joint hand-packed with oakum and poured molten lead
• d.A neoprene gasket inside a stainless shielded clamp✓

No-hub cast iron uses a neoprene sleeve enclosed in a stainless-steel shield and clamp band tightened to the specified torque. Lead-and-oakum is the old hub-and-spigot method, and cast iron is not solvent-cemented or threaded like plastic or steel. IPC §705.0.IPC §705.0

Where copper tube connects directly to galvanized steel pipe on a water line, what fitting prevents galvanic corrosion?

• a.A standard brass union tightened extra firmly
• b.A reducing bushing installed between the two metals
• c.A dielectric union or dielectric fitting✓
• d.A soldered copper coupling slipped over the steel thread

Copper and steel form a galvanic cell that corrodes the anodic steel, so a dielectric union or fitting isolates the metals electrically. A plain brass union or bushing still allows the dissimilar-metal contact. The dielectric barrier stops the corrosion current. IPC §605.14 / UPC §605.0.IPC §605.14 / UPC §605.0

Galvanized steel water pipe is generally prohibited in which location because of corrosion?

• a.In vertical risers within a heated building
• b.In short connectors at the water heater
• c.In exposed interior walls above the finished floor line
• d.Underground, in direct contact with soil✓

Galvanized steel corrodes rapidly in soil and is not permitted underground; where used at all it is limited to protected above-grade runs. Corrosion clogs the bore and causes leaks. Underground service is run in copper or approved plastic instead. UPC §604.0 / IPC §605.4.UPC §604.0 / IPC §605.4

Which material is prohibited for new potable water piping in all cases?

• a.Lead pipe✓
• b.Type L copper
• c.Cross-linked polyethylene (PEX)
• d.Chlorinated PVC (CPVC)

Lead pipe (and lead-based solder) is prohibited for potable water because it leaches lead into the drinking supply. Only lead-free materials such as copper, CPVC, and PEX are allowed. This ban is a core public-health protection. IPC §605.0 / UPC §604.0.IPC §605.0 / UPC §604.0