Master Plumber (UPC/IPC) — All Questions

A residential branch serves the following fixtures with these water-supply fixture unit (WSFU) values: 8 water closets (tank) at 2.5 each, 8 lavatories at 1 each, and 8 showers at 2 each. What is the total demand load on the branch?

• a.60 WSFU
• b.36 WSFU
• c.52 WSFU✓
• d.44 WSFU

Multiply each fixture count by its WSFU value and add: (8 x 2.5) + (8 x 1) + (8 x 2) = 20 + 8 + 16 = 52 WSFU. This total is what you carry into a Hunter-curve demand chart to convert fixture units to gpm. Using the wrong per-fixture value is the usual cause of a low answer such as 44.UPC §610.0

Using the 0.408 velocity relationship v = 0.408 x Q / d^2 (v in ft/s, Q in gpm, d in inches inside diameter), what is the velocity of 18 gpm flowing in a pipe with a 1.00 in inside diameter?

• a.7.3 ft/s✓
• b.3.7 ft/s
• c.14.7 ft/s
• d.10.2 ft/s

v = 0.408 x 18 / (1.00)^2 = 7.34 / 1.00 = 7.3 ft/s. Because d^2 = 1, the velocity equals 0.408 x Q directly. This sits just under the 8 ft/s limit commonly set for cold-water piping to control erosion and water hammer.UPC §610.0

Cold-water piping is commonly limited to a maximum velocity of about 8 ft/s. Using v = 0.408 x Q / d^2, what is the approximate maximum flow a pipe with a 1.025 in inside diameter (nominal 1 in type L copper) can carry without exceeding 8 ft/s?

• a.28 gpm
• b.17 gpm
• c.12 gpm
• d.21 gpm✓

Rearrange to Q = v x d^2 / 0.408 = 8 x (1.025)^2 / 0.408 = 8 x 1.051 / 0.408 = 20.6 gpm, about 21 gpm. Above this flow the velocity exceeds 8 ft/s and erosion-corrosion and noise become a concern. Hot water is usually held to an even lower 5 ft/s.UPC §610.0

A fixture is located 46 ft above the water meter. Using 0.433 psi per foot of elevation, how much static pressure is lost to elevation between the meter and that fixture?

• a.46.0 psi
• b.19.9 psi✓
• c.34.5 psi
• d.23.0 psi

Static loss = height x 0.433 psi/ft = 46 x 0.433 = 19.9 psi. Every foot of rise costs 0.433 psi and must be subtracted from the available supply pressure before you can check the residual at the fixture. The 46 psi answer wrongly treats 1 ft as 1 psi.UPC §610.0

A system has 62 psi at the meter. It must overcome 22 psi of elevation, 14 psi of friction loss in the piping, and deliver a fixture that requires 8 psi minimum flow pressure. What residual pressure remains at the fixture?

• a.14 psi
• b.18 psi✓
• c.16 psi
• d.8 psi

Residual = 62 - 22 (elevation) - 14 (friction) = 26 psi available at the fixture, which exceeds the 8 psi required, leaving 26 - 8 = 18 psi of margin. The design works because the delivered 26 psi is greater than the 8 psi minimum. The pressure budget must always balance supply against elevation, friction, and fixture demand.UPC §610.0

A copper water line has 90 ft of straight pipe. Fittings add an equivalent length of 3 ft each for four elbows and 8 ft for one gate valve. What is the developed (equivalent) length used for friction-loss calculations?

• a.110 ft✓
• b.134 ft
• c.98 ft
• d.122 ft

Developed length = straight pipe + fitting equivalents = 90 + (4 x 3) + 8 = 90 + 12 + 8 = 110 ft. Friction loss is calculated on this equivalent length, not on the measured pipe alone, because fittings behave like extra pipe. Ignoring fittings understates the loss and can undersize the pipe.UPC §610.0

Available pressure for friction is 30 psi and the total developed length of the water line is 150 ft. What is the maximum allowable uniform friction loss per 100 ft of pipe (the value used to enter a sizing chart)?

• a.30 psi/100 ft
• b.15 psi/100 ft
• c.10 psi/100 ft
• d.20 psi/100 ft✓

Allowable loss per 100 ft = (available pressure / developed length) x 100 = (30 / 150) x 100 = 20 psi per 100 ft. This uniform-loss figure is the design value you carry across the sizing chart against the flow to pick a pipe size. Spreading the whole 30 psi over 150 ft gives the per-100-ft rate.UPC §610.0

A horizontal drainage branch carries the following drainage fixture units (DFU): 4 water closets at 4 DFU, 4 lavatories at 1 DFU, and 2 kitchen sinks at 2 DFU. What is the total DFU load carried by the branch?

• a.24 DFU✓
• b.28 DFU
• c.32 DFU
• d.18 DFU

Total = (4 x 4) + (4 x 1) + (2 x 2) = 16 + 4 + 4 = 24 DFU. This total is read against the horizontal fixture branch column of the drain sizing table to select the pipe. Because water closets are present, the branch can be no smaller than 3 in regardless of the DFU count.UPC §703.0

A horizontal fixture branch carries 30 DFU. The sizing table lists these branch capacities: 2 in = 6 DFU, 3 in = 20 DFU, 4 in = 160 DFU. Two of the fixtures are water closets. What is the minimum pipe size for the branch?

• a.2 in
• b.6 in
• c.4 in✓
• d.2-1/2 in

The 30 DFU load exceeds the 20 DFU capacity of a 3 in branch, so the table already forces the next size, 4 in (160 DFU). The presence of water closets independently forbids anything smaller than 3 in, but here the load alone requires 4 in. Always pick the smallest size whose capacity equals or exceeds the load.IPC §710.1

A 4 in building drain is run at the code minimum slope of 1/8 in per foot over a developed length of 96 ft. What is the total fall from the upstream end to the downstream end?

• a.9 in
• b.12 in✓
• c.6 in
• d.24 in

Fall = slope x length = 1/8 in/ft x 96 ft = 12 in, or 1 ft. Pipe 3 in and larger uses 1/8 in per foot as the minimum. Using 1/4 in per ft (which applies only to pipe 2-1/2 in and smaller) would wrongly double the answer to 24 in.IPC §704.1

A 2 in horizontal fixture drain must fall a total of 5 in over its run at the code minimum slope of 1/4 in per foot. How long is the run?

• a.24 ft
• b.20 ft✓
• c.40 ft
• d.60 ft

Length = fall / slope = 5 in / (1/4 in per ft) = 5 x 4 = 20 ft. Pipe 2-1/2 in and smaller uses the steeper 1/4 in per foot minimum. Dividing by 1/8 in per ft would incorrectly give 40 ft, the rate reserved for 3 in and larger pipe.IPC §704.1

A vent must be sized at not less than one-half the diameter of the drain it serves, and never smaller than 1-1/4 in. What is the minimum vent size for a 3 in drain?

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

Half of 3 in is 1-1/2 in, which is larger than the 1-1/4 in floor, so the minimum vent is 1-1/2 in. The half-diameter rule sets the size and the 1-1/4 in minimum only governs very small drains. A 4 in drain by the same rule would need at least a 2 in vent.UPC §904.1

A 2 in trap arm connects a fixture trap to its vent. The code limits the fall in a trap arm to no more than one pipe diameter between the trap weir and the vent. What is the maximum allowable fall in this trap arm?

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

The maximum fall equals one pipe diameter, and for a 2 in arm that is 2 in. If the arm falls more than one diameter, the vent opening drops below the crown weir and the trap can be self-siphoned. This limit is separate from and in addition to the maximum developed-length limit for the arm.UPC §906.1

A natural-gas furnace is rated at 120,000 BTU/hr. Natural gas has a heating value of about 1,000 BTU per cubic foot. What gas volume flow, in cubic feet per hour (cfh), must the piping deliver?

• a.1,200 cfh
• b.60 cfh
• c.12 cfh
• d.120 cfh✓

cfh = BTU/hr load / heating value = 120,000 / 1,000 = 120 cfh. Because natural gas is roughly 1,000 BTU per cubic foot, the cfh figure is simply the input in thousands of BTU. This cfh value is what you carry into the gas-pipe sizing table with the longest run.UPC §610.0

A house has three natural-gas appliances: a 100,000 BTU/hr furnace, a 40,000 BTU/hr water heater, and a 65,000 BTU/hr range. Using 1,000 BTU per cubic foot, what total cfh must the gas meter and main serve?

• a.2,050 cfh
• b.165 cfh
• c.205 cfh✓
• d.240 cfh

Total input = 100,000 + 40,000 + 65,000 = 205,000 BTU/hr, and at 1,000 BTU per cubic foot that is 205 cfh. The whole-house demand sizes the meter and the main from the meter to the first tee. Individual branches are then sized for the appliance each one serves.IFGC §402.4

A propane appliance is rated at 150,000 BTU/hr. Propane has a heating value of about 2,500 BTU per cubic foot. What is the required gas flow in cfh?

• a.150 cfh
• b.30 cfh
• c.60 cfh✓
• d.15 cfh

cfh = 150,000 / 2,500 = 60 cfh. Propane packs about 2.5 times the energy of natural gas per cubic foot, so the same BTU load needs far fewer cubic feet, which is why propane piping is often smaller than natural-gas piping for the same appliance.IFGC §402.4

A gas branch must deliver 55 cfh of natural gas over a 60 ft longest length. The sizing table for that length gives these capacities: 1/2 in = 42 cfh, 3/4 in = 88 cfh, 1 in = 165 cfh. What is the minimum pipe size?

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

The 55 cfh demand exceeds the 42 cfh capacity of 1/2 in pipe at 60 ft, so the next size up, 3/4 in (88 cfh), is required. Gas pipe is sized on the longest length from the meter to the most remote outlet, applied to every section. A 1/2 in pipe would be overloaded and starve the appliance.IFGC Table 402.4

Using the rational method Q = 0.0104 x A x i, where A is the projected roof area in square feet and i is the rainfall rate in in/hr, what is the design storm flow for a 6,000 ft^2 roof at a 3 in/hr rainfall rate?

• a.124 gpm
• b.187 gpm✓
• c.62 gpm
• d.312 gpm

Q = 0.0104 x 6,000 x 3 = 187 gpm. The 0.0104 factor converts one inch per hour of rain over one square foot into gpm. This design flow is then read against the vertical leader and horizontal storm-drain tables to size the conductors.UPC §1101.0

A roof drains 4,800 ft^2 at a design rainfall rate of 4 in/hr. Using Q = 0.0104 x A x i, what is the storm design flow?

• a.200 gpm✓
• b.320 gpm
• c.250 gpm
• d.150 gpm

Q = 0.0104 x 4,800 x 4 = 199.7 gpm, about 200 gpm. Doubling the rainfall rate doubles the flow for the same roof, which is why the local 100-year rainfall intensity is critical to storm sizing. This flow selects the leader and storm drain sizes.UPC §1101.0

A storm sizing table is published at 4 in/hr. A vertical leader lists a capacity of 4,600 ft^2 of roof at 4 in/hr. If the local design rate is only 2 in/hr, what roof area can that same leader serve?

• a.4,600 ft^2
• b.9,200 ft^2✓
• c.2,300 ft^2
• d.18,400 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: 4,600 x (4/2) = 9,200 ft^2. Conductor capacity is fixed in gpm, so a lighter storm lets the same pipe drain more roof. Always adjust table areas to the local rainfall rate.UPC §1101.0

A booster pump must deliver 40 gpm and raise pressure by 45 psi. Using approximate water horsepower WHP = (gpm x psi) / 1,714, what is the water horsepower required (before pump efficiency)?

• a.0.5 hp
• b.4.2 hp
• c.2.1 hp
• d.1.05 hp✓

WHP = (40 x 45) / 1,714 = 1,800 / 1,714 = 1.05 hp. This is the ideal hydraulic power; the actual motor must be larger to account for pump efficiency, often around 60 to 70 percent. Dividing the delivered fluid power by efficiency gives the brake horsepower to specify.UPC §610.0

A booster pump adds 52 psi. What is the equivalent head, in feet, that the pump develops? Use 1 psi = 2.31 ft of head.

• a.120 ft✓
• b.78 ft
• c.231 ft
• d.52 ft

Head = psi x 2.31 = 52 x 2.31 = 120 ft. The 2.31 factor is the reciprocal of the 0.433 psi-per-foot relationship and converts pressure to the equivalent vertical column of water. Pump curves are usually plotted in feet of head, so this conversion is needed to read them.UPC §610.0

A recirculation loop must deliver 6 gpm of hot water. Using v = 0.408 x Q / d^2, what inside diameter keeps the velocity at about 2 ft/s to limit erosion of the continuously circulated hot line?

• a.1.50 in
• b.1.10 in✓
• c.2.00 in
• d.0.75 in

Solve for d: d = sqrt(0.408 x Q / v) = sqrt(0.408 x 6 / 2) = sqrt(1.224) = 1.11 in. Hot recirculation lines are held to a low velocity, near 2 to 3 ft/s, because constant flow at high velocity erodes copper. A larger diameter lowers velocity for the same flow.UPC §610.0

A water main runs 250 ft of developed length and the friction chart shows a loss of 6 psi per 100 ft at the design flow. How much pressure is lost to friction over the full run?

• a.9 psi
• b.15 psi✓
• c.25 psi
• d.6 psi

Friction loss = (loss per 100 ft) x (length / 100) = 6 x (250 / 100) = 6 x 2.5 = 15 psi. Friction loss scales directly with developed length, so long runs consume a large share of the pressure budget. This 15 psi must be subtracted from supply along with elevation before checking residual.UPC §610.0

A sewage ejector pump must handle a building with a discharge of 30 gpm against 18 ft of total head. Using WHP = (gpm x head in ft) / 3,960, what is the water horsepower?

• a.0.07 hp
• b.0.10 hp
• c.0.27 hp
• d.0.14 hp✓

WHP = (30 x 18) / 3,960 = 540 / 3,960 = 0.136 hp, about 0.14 hp. The 3,960 constant is used when head is expressed in feet rather than psi. As with any pump, the motor is oversized above this by dividing by the pump efficiency.IPC §712.0

A sewage sump receiving a peak inflow of 45 gpm is served by a pump that discharges 90 gpm when running. If useful storage between pump-on and pump-off is 30 gallons, how long does one pump-down cycle (running time) last during peak inflow?

• a.30 s
• b.40 s✓
• c.60 s
• d.90 s

While running, the net removal rate is pump output minus inflow = 90 - 45 = 45 gpm. Draw-down time = storage / net rate = 30 / 45 = 0.667 min = 40 s. Sizing sump volume this way limits motor starts per hour to protect the pump; too small a volume causes short-cycling.IPC §712.4

Two demand loads combine on a main: branch A carries 40 WSFU and branch B carries 60 WSFU. If the Hunter demand chart converts 100 WSFU (predominantly flush tanks) to about 44 gpm, what design flow sizes the main?

• a.22 gpm
• b.104 gpm
• c.33 gpm
• d.44 gpm✓

You add the fixture units first, 40 + 60 = 100 WSFU, then convert the combined total to gpm on the demand curve, giving about 44 gpm. You must never convert each branch to gpm and add the flows, because the Hunter curve already accounts for the low probability of simultaneous use, and adding gpm overstates demand.UPC §610.0

A meter and service must supply 44 gpm. The static supply is 68 psi, elevation loss is 20 psi, the meter loss is 8 psi, and the most remote fixture needs 15 psi. How much pressure remains for pipe friction?

• a.18 psi
• b.25 psi✓
• c.33 psi
• d.43 psi

Available for friction = static - elevation - meter - fixture requirement = 68 - 20 - 8 - 15 = 25 psi. This leftover is spread over the developed length to set the allowable friction rate per 100 ft. If friction loss at the chosen pipe size exceeds 25 psi, the pipe must be enlarged.UPC §610.0

An individual vent serving a lavatory is 1-1/2 in. The vent sizing table limits a 1-1/2 in vent to a maximum developed length of 150 ft at the fixture-unit load carried. If the vent run is 90 ft, is the vent acceptable and why?

• a.Yes, 90 ft is within the 150 ft limit✓
• b.No, it exceeds the length limit
• c.Yes, vents have no length limit
• d.No, the vent must equal the drain size

A 90 ft developed vent length is within the 150 ft maximum allowed for a 1-1/2 in vent at that load, so the vent is acceptable. Vent sizing depends on both the fixture-unit load and the total developed length; exceeding either forces a larger vent. Vents do have length limits, so the no-limit answer is wrong.IPC §906.2

A hot-water line is limited to a maximum velocity of 5 ft/s. Using v = 0.408 x Q / d^2, what is the maximum flow through a pipe with a 0.785 in inside diameter (nominal 3/4 in type L copper)?

• a.4.9 gpm
• b.11.3 gpm
• c.15.1 gpm
• d.7.6 gpm✓

Q = v x d^2 / 0.408 = 5 x (0.785)^2 / 0.408 = 5 x 0.616 / 0.408 = 7.55 gpm, about 7.6 gpm. Hot water is held to a lower 5 ft/s than cold water because higher temperature accelerates erosion-corrosion of copper. Exceeding this flow risks pinholing the line over time.UPC §610.0

A drainage stack sizing table lists a 4 in stack at 500 DFU maximum and a 3 in stack at 48 DFU maximum for the total to a stack. A stack collects 240 DFU. What is the minimum stack size, and what secondary rule also applies?

• a.3 in, and a cleanout is required
• b.3 in, and no water closets allowed
• c.4 in, and no size reduction downward✓
• d.6 in, and a vent stack is required

240 DFU 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 size in the direction of flow as it descends. Downsizing a stack lower down would create a restriction that floods the branches above it. The stack must stay 4 in or larger to its base.IPC §710.1

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

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

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