Master Plumber (UPC/IPC) — All Questions

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