Master Plumber (UPC/IPC) — All Questions

A 2,000 lb cast-iron section is lifted by a two-leg sling with each leg at 60 degrees from horizontal. Each leg carries (load / 2) / sin(60 degrees), with sin(60) about 0.866. What is the tension in each sling leg?

• a.1,155 lb✓
• b.1,414 lb
• c.1,000 lb
• d.2,000 lb

Each leg tension = (2,000 / 2) / 0.866 = 1,000 / 0.866 = 1,155 lb. As the sling angle drops toward horizontal, the leg tension climbs above the simple share of the load, which is why slings must be de-rated for angle. At a straight vertical lift each leg would carry only 1,000 lb.29 CFR 1926.251

During sewer confined-space work, the attendant outside must maintain what capability?

• a.Enter to help immediately without equipment
• b.Perform the plumbing work
• c.Continuous communication with entrants and the ability to summon rescue without entering✓
• d.Leave to get tools as needed

The attendant remains outside, keeps continuous communication with the entrants, monitors conditions, and summons trained rescue if needed, but does not enter the space to attempt rescue alone. Untrained would-be rescuers are a leading cause of confined-space fatalities. Non-entry retrieval systems allow rescue from outside.29 CFR 1910.146

A trench 10 ft deep in Type B soil is sloped at 1 to 1 (horizontal to vertical) on both sides. If the trench bottom is 3 ft wide, how wide is the excavation at the top?

• a.13 ft
• b.20 ft
• c.30 ft
• d.23 ft✓

Each side slopes back 1 x depth = 10 ft, so the top width = bottom + 2 x 10 = 3 + 20 = 23 ft. Type B soil requires a 1-to-1 slope, while less stable Type C requires 1.5 to 1 and would open even wider. The soil classification set by a competent person drives the excavation width and spoil space needed.29 CFR 1926.652

A trench 90 ft long and 6 ft deep requires egress so no worker travels more than 25 ft laterally to a ladder. A ladder placed mid-run protects 25 ft in each direction, or 50 ft total. What is the minimum number of ladders required?

• a.2 ladders✓
• b.1 ladder
• c.3 ladders
• d.4 ladders

Each ladder covers 50 ft of trench (25 ft of lateral travel on each side), so 90 ft / 50 ft = 1.8, which rounds up to 2 ladders. The 25 ft maximum travel ensures rapid exit if a wall fails or the atmosphere changes. Egress is required in any trench 4 ft or deeper and must extend above the top edge.29 CFR 1926.651

Before servicing a sewage ejector pump, the plumber must control hazardous energy. What procedure prevents the pump from starting during service?

• a.Unplug only if convenient
• b.Lockout/tagout of the electrical disconnect, then verify zero energy✓
• c.Work quickly between cycles
• d.Post a verbal warning

Lockout/tagout requires de-energizing the pump at its disconnect, applying a personal lock and tag, and verifying zero energy by attempting to start it before hands go near moving parts. An automatic float could start the pump unexpectedly and cause severe injury. Each worker applies their own lock.29 CFR 1910.147

On a California jobsite, a trench 5 ft or deeper in which workers will enter also requires a specific state permit beyond federal rules. What is that requirement?

• a.A federal permit only
• b.A permit only above 20 ft
• c.No permit is ever required in California
• d.A Cal/OSHA excavation/trench permit for trenches 5 ft or deeper that workers enter✓

Cal/OSHA requires a project or annual excavation permit for trenches 5 ft or more deep into which a person must descend, in addition to the cave-in protection rules. California often has requirements at least as strict as, and sometimes stricter than, federal OSHA. The competent person and daily inspections still apply.Cal/OSHA T8 1541

Cutting and grinding cast iron or concrete for a plumbing penetration generates respirable crystalline silica. What is the primary control the plumber should use?

• a.Open a window only
• b.Work faster
• c.No control is needed for short tasks
• d.Wet cutting or on-tool dust collection (engineering controls) plus respiratory protection as needed✓

The primary defense against respirable silica is engineering controls, using water (wet cutting) or a vacuum dust-collection shroud on the tool to suppress dust at the source, supplemented by respiratory protection when needed. Silica dust causes silicosis and is regulated with a strict exposure limit. Dry cutting without controls quickly exceeds the permissible exposure limit.29 CFR 1926.55

A demand of 62 gpm must be delivered with an available friction pressure of 24 psi over a 200 ft developed length. What is the allowable friction loss per 100 ft used to size the pipe?

• a.12 psi/100 ft✓
• b.24 psi/100 ft
• c.8 psi/100 ft
• d.4 psi/100 ft

Allowable loss per 100 ft = (available pressure / developed length) x 100 = (24 / 200) x 100 = 12 psi per 100 ft. This uniform rate is read against the 62 gpm demand on the friction chart to choose the smallest adequate pipe. Spreading the full 24 psi over the whole run gives the per-100-ft design value.UPC §610.0

An air gap at a lavatory faucet must be maintained above the flood-level rim. If the effective opening of the faucet is 1/2 in and the outlet is near a single wall, the required gap is three times the opening. What is the minimum air gap?

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

Near a single wall the multiplier is three times the effective opening, so 3 x 1/2 in = 1-1/2 in. Away from walls the standard multiplier is two times the opening, but a nearby wall disrupts the free fall of air and requires the larger gap. The air gap is the most reliable backflow protection because it is a fixed physical break.UPC §603.4.6

A natural-gas dryer needs 35,000 BTU/hr and a range needs 65,000 BTU/hr on a shared branch. Using 1,000 BTU/ft^3, what is the combined cfh the common branch must carry?

• a.35 cfh
• b.1,000 cfh
• c.65 cfh
• d.100 cfh✓

Combined input = 35,000 + 65,000 = 100,000 BTU/hr, and at 1,000 BTU per cubic foot that is 100 cfh on the shared branch. The common section always carries the sum of the downstream appliance demands. Each appliance's own connector is then sized for only its individual load.IFGC §402.4

A commercial branch serves 6 flushometer-valve water closets at 10 WSFU each, 4 flushometer urinals at 5 WSFU each, and 6 lavatories at 1 WSFU each. What is the total water-supply fixture-unit load?

• a.76 WSFU
• b.86 WSFU✓
• c.80 WSFU
• d.96 WSFU

Multiply each fixture count by its WSFU value and add: (6 x 10) + (4 x 5) + (6 x 1) = 60 + 20 + 6 = 86 WSFU. Flushometer fixtures carry far higher WSFU than tank types because they draw a high instantaneous flow. This total is read on the flush-valve demand curve to convert to gpm. IPC Table 604.3.IPC Table 604.3

A branch serves 10 lavatories at 1 WSFU each, 4 service sinks at 3 WSFU each, and 2 hose bibbs at 2.5 WSFU each. What is the total WSFU demand?

• a.22 WSFU
• b.27 WSFU✓
• c.25 WSFU
• d.32 WSFU

Total = (10 x 1) + (4 x 3) + (2 x 2.5) = 10 + 12 + 5 = 27 WSFU. Each fixture type carries its own supply fixture-unit value, and only after summing them do you enter the demand curve. Forgetting the hose bibbs would give the too-low 22. IPC Table 604.3.IPC Table 604.3

A tank-type water closet is assigned 2.5 WSFU and a flushometer-valve water closet is assigned 10 WSFU. A remodel replaces 4 tank closets with 4 flushometer closets. By how much does the branch WSFU load increase?

• a.40 WSFU
• b.30 WSFU✓
• c.10 WSFU
• d.25 WSFU

New load = 4 x 10 = 40 WSFU; old load = 4 x 2.5 = 10 WSFU; increase = 40 - 10 = 30 WSFU. Switching to flushometer valves sharply raises the demand because they draw a high instantaneous flow, which often forces the branch to be resized. IPC Table 604.3.IPC Table 604.3

A 100 WSFU load converts to about 44 gpm on the flush-tank demand curve but about 65 gpm on the flush-valve demand curve. A branch whose water closets are flushometer valves carries 100 WSFU. Which design flow sizes the pipe?

• a.22 gpm
• b.44 gpm, the flush-tank value
• c.65 gpm, the flush-valve value✓
• d.50 gpm

Because the closets are flushometer valves, the flush-valve demand curve applies, giving about 65 gpm for 100 WSFU. Flush-valve systems demand more instantaneous flow than tank systems for the same fixture-unit count, so the correct curve must match the fixture type. Using the tank curve would undersize the supply. IPC §604.3.IPC §604.3

How many BTU are required to raise 50 gallons of water by 80 degrees F? Water weighs 8.33 lb/gal, and 1 BTU raises 1 lb by 1 degree F.

• a.16,660 BTU
• b.25,000 BTU
• c.41,650 BTU
• d.33,320 BTU✓

Energy = weight x delta-T = (50 x 8.33) x 80 = 416.5 lb x 80 = 33,320 BTU. The mass of water, not its volume, drives the heat load, so gallons are first converted to pounds. This figure underlies water-heater recovery and sizing calculations. IPC §501.0.IPC §501.0

A gas water heater has a 40,000 BTU/hr input and 80 percent recovery efficiency, heating water through a 90 degree F rise. Using gph = (input x efficiency) / (8.33 x delta-T), what is its recovery rate?

• a.38 gph
• b.30 gph
• c.53 gph
• d.43 gph✓

Useful output = 40,000 x 0.80 = 32,000 BTU/hr, and gph = 32,000 / (8.33 x 90) = 32,000 / 749.7 = 42.7, about 43 gph. Recovery is the gallons per hour the heater can raise through the design temperature rise. Ignoring efficiency would overstate the recovery. IPC §501.0.IPC §501.0

An electric water heater has a single 4.5 kW element (1 kW = 3,412 BTU/hr) heating water through a 90 degree F rise at about 100 percent efficiency. Using gph = BTU/hr / (8.33 x delta-T), what is the recovery rate?

• a.15 gph
• b.30 gph
• c.41 gph
• d.21 gph✓

Input = 4.5 x 3,412 = 15,354 BTU/hr, and gph = 15,354 / (8.33 x 90) = 15,354 / 749.7 = 20.5, about 21 gph. Electric heaters convert nearly all input to heat but recover far fewer gph than gas because their input is lower. This is why electric units rely on larger storage. IPC §501.0.IPC §501.0

A water heater's first-hour rating is approximately the usable storage plus one hour of recovery. A unit provides 35 gallons of usable storage and recovers 40 gph. What is its approximate first-hour rating?

• a.40 gal
• b.75 gal✓
• c.50 gal
• d.90 gal

First-hour rating = usable storage + one-hour recovery = 35 + 40 = 75 gallons. It measures how much hot water the heater can deliver during a busy hour, combining what is stored with what it can reheat. Sizing to the peak-hour demand prevents running out during showers. IPC §501.0.IPC §501.0

A water heater must continuously deliver 120 gal/hr heated from 60 degrees F to 120 degrees F. At 100 percent efficiency, what input is required? Water weighs 8.33 lb/gal.

• a.60,000 BTU/hr✓
• b.40,000 BTU/hr
• c.30,000 BTU/hr
• d.50,000 BTU/hr

Input = flow x weight x delta-T = 120 x 8.33 x (120 - 60) = 120 x 8.33 x 60 = 59,976, about 60,000 BTU/hr. Continuous demand ties the required input directly to flow and temperature rise. A larger rise or higher flow raises the input proportionally. IPC §501.0.IPC §501.0

Using Q = 0.0104 x A x i, what is the design storm flow for a 10,000 ft^2 roof at a rainfall rate of 2 in/hr?

• a.416 gpm
• b.150 gpm
• c.208 gpm✓
• d.104 gpm

Q = 0.0104 x 10,000 x 2 = 208 gpm. The 0.0104 factor converts one inch per hour over one square foot into gpm. This design flow is then read against the leader and horizontal storm-drain tables. IPC §1106.2.IPC §1106.2

A horizontal storm drain at 1/4 in/ft slope has these capacities: 4 in = 110 gpm, 5 in = 194 gpm, 6 in = 311 gpm. The design flow is 208 gpm. What is the minimum drain size?

• a.6 in✓
• b.8 in
• c.5 in
• d.4 in

The 208 gpm flow exceeds the 194 gpm capacity of a 5 in drain, so the next size, 6 in (311 gpm), is required. Always pick the smallest size whose capacity equals or exceeds the design flow. Steeper slope would raise each size's capacity. IPC §1106.3.IPC §1106.3

A flat roof measures 40 ft x 60 ft and the design rainfall rate is 3 in/hr. Using Q = 0.0104 x A x i, what is the storm design flow?

• a.37 gpm
• b.75 gpm✓
• c.100 gpm
• d.150 gpm

Area = 40 x 60 = 2,400 ft^2, and Q = 0.0104 x 2,400 x 3 = 74.9, about 75 gpm. The projected roof area and the local rainfall intensity together set the flow. This value then selects the leader and storm-drain sizes. IPC §1106.2.IPC §1106.2

A vertical roof leader has these capacities: 2 in = 30 gpm, 3 in = 92 gpm, 4 in = 192 gpm. A leader must carry 75 gpm. What is the minimum size?

• a.3 in✓
• b.4 in
• c.5 in
• d.2 in

The 75 gpm flow exceeds the 30 gpm capacity of a 2 in leader but fits within the 92 gpm capacity of a 3 in leader, so 3 in is the minimum. Leaders are vertical conductors sized on gpm, not roof area directly. Undersizing floods the roof during a design storm. IPC §1106.2.IPC §1106.2

At a rainfall rate of 1 in/hr, how many gpm does 1,000 ft^2 of roof produce? Use Q = 0.0104 x A x i.

• a.1.04 gpm
• b.10.4 gpm✓
• c.104 gpm
• d.6.9 gpm

Q = 0.0104 x 1,000 x 1 = 10.4 gpm. This unit figure, 10.4 gpm per 1,000 ft^2 per inch of rain, lets you scale flow quickly for any roof and rate. Doubling either the area or the rate doubles the flow. IPC §1106.2.IPC §1106.2

Using v = 0.408 x Q / d^2 (v in ft/s, Q in gpm, d in inches inside diameter), what is the velocity of 25 gpm in a pipe with a 1.25 in inside diameter?

• a.5.2 ft/s
• b.4.1 ft/s
• c.6.5 ft/s✓
• d.8.2 ft/s

v = 0.408 x 25 / (1.25)^2 = 10.2 / 1.5625 = 6.5 ft/s. This sits below the 8 ft/s cold-water limit, so the pipe is acceptable for erosion and noise. Velocity falls sharply as diameter grows because d is squared. IPC §604.0.IPC §604.0

Cold water is limited to 8 ft/s. Using v = 0.408 x Q / d^2, what is the maximum flow in a pipe with a 1.265 in inside diameter (nominal 1-1/4 in type L copper)?

• a.21 gpm
• b.25 gpm
• c.12 gpm
• d.31 gpm✓

Rearrange to Q = v x d^2 / 0.408 = 8 x (1.265)^2 / 0.408 = 8 x 1.600 / 0.408 = 31.4, about 31 gpm. Above this flow the velocity exceeds 8 ft/s and erosion-corrosion and water hammer become concerns. Larger pipe carries more flow at the same velocity limit. IPC §604.0.IPC §604.0

A fixture is 32 ft above the meter. Using 0.433 psi per foot, how much static pressure is lost to elevation?

• a.6.9 psi
• b.7.4 psi
• c.13.9 psi✓
• d.32.0 psi

Static loss = height x 0.433 = 32 x 0.433 = 13.9 psi. Every foot of rise costs 0.433 psi and must be subtracted from available pressure before checking residual at the fixture. Treating 1 ft as 1 psi wrongly gives 32. IPC §604.0.IPC §604.0

Using 1 psi = 2.31 ft of head, what head does 40 psi represent?

• a.92.4 ft✓
• b.17.3 ft
• c.120 ft
• d.40 ft

Head = psi x 2.31 = 40 x 2.31 = 92.4 ft. The 2.31 factor is the reciprocal of 0.433 psi per foot and converts pressure to the equivalent column of water. Pump curves in feet of head require this conversion. IPC §604.0.IPC §604.0

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