CSLB General Building (B) — 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.52 WSFU✓
• b.36 WSFU
• c.44 WSFU
• d.60 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.3.7 ft/s
• b.7.3 ft/s✓
• c.10.2 ft/s
• d.14.7 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.12 gpm
• b.17 gpm
• c.21 gpm✓
• d.28 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.23.0 psi
• b.34.5 psi
• c.46.0 psi
• d.19.9 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.18 psi✓
• b.8 psi
• c.14 psi
• d.16 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.98 ft
• b.110 ft✓
• c.122 ft
• d.134 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.10 psi/100 ft
• b.15 psi/100 ft
• c.20 psi/100 ft✓
• d.30 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.18 DFU
• b.28 DFU
• c.32 DFU
• d.24 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.4 in✓
• b.2 in
• c.2-1/2 in
• d.6 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.6 in
• b.12 in✓
• c.9 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.40 ft
• c.20 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.1-1/4 in
• b.2 in
• c.3 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.2 in✓
• b.1/2 in
• c.1 in
• d.1-1/2 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.12 cfh
• b.120 cfh✓
• c.60 cfh
• d.1,200 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.165 cfh
• b.240 cfh
• c.205 cfh✓
• d.2,050 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.15 cfh
• b.30 cfh
• c.150 cfh
• d.60 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.3/4 in✓
• b.1/2 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.62 gpm
• b.187 gpm✓
• c.124 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.150 gpm
• b.250 gpm
• c.200 gpm✓
• d.320 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.2,300 ft^2
• b.4,600 ft^2
• c.18,400 ft^2
• d.9,200 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.1.05 hp✓
• b.0.5 hp
• c.2.1 hp
• d.4.2 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.52 ft
• b.120 ft✓
• c.78 ft
• d.231 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.0.75 in
• b.1.50 in
• c.1.10 in✓
• d.2.00 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.6 psi
• b.9 psi
• c.25 psi
• d.15 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.0.14 hp✓
• b.0.07 hp
• c.0.10 hp
• d.0.27 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.33 gpm
• c.44 gpm✓
• d.104 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.33 psi
• c.43 psi
• d.25 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.No, the vent must equal the drain size
• d.Yes, vents have no length limit

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.7.6 gpm✓
• c.11.3 gpm
• d.15.1 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.4 psi/100 ft
• b.8 psi/100 ft
• c.24 psi/100 ft
• d.12 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