1.3 Trade Math I: Pressure, Head, Fall, Volume & Expansion

Why trade math matters

Six questions are labeled Trade Math, and many Sizing and Design and System Installation questions also need a quick calculation. The exam allows only a simple four-function calculator, so learn the constants and set up each problem the same way every time. Mathematics for Plumbers and Pipefitters (8th edition), one of the approved references, uses the same methods.

Core constants

The pressure constant comes straight from the weight of water. One cubic foot weighs 62.4 lb and rests on 144 square inches, so:

$$\frac{62.4\ \text{lb}}{144\ \text{in}^2} = 0.433\ \text{psi per foot of height}$$

Pressure and head

Static pressure depends only on vertical height. The pipe's diameter and shape do not matter.

$$P\ (\text{psi}) = H\ (\text{ft}) \times 0.433 \qquad H\ (\text{ft}) = P\ (\text{psi}) \times 2.31$$

• A DWV section tested with the minimum 10-foot head of water in IPC 312.2 has $10 \times 0.433 \approx 4.3$ psi at its lowest point.
• A main at 60 psi static can lift water about $60 \times 2.31 = 138.6$ feet before the pressure drops to zero.
• Going up costs about 0.433 psi per foot, and going down gains the same amount.

Static vs. residual pressure

Static pressure is measured with no flow. Residual (flow) pressure is what remains at the fixture while water flows. It equals static pressure minus elevation loss, minus friction in pipe and fittings, minus losses through meters, backflow preventers and pressure-reducing valves. IPC Table 604.3 sets the minimum flow pressure each fixture needs. A siphonic flushometer-valve water closet, for example, needs 25 gpm at 35 psi.

Worked check. A meter shows 55 psi static. The top-floor flushometer valve is 42 feet above it, and peak friction loss is 12 psi.

$$\text{Elevation loss} = 42 \times 0.433 = 18.2\ \text{psi}$$ $$\text{Residual} = 55 - 18.2 - 12 = 24.8\ \text{psi}$$

24.8 psi is less than the 35 psi Table 604.3 requires, so the design fails. To pass with only 55 psi available, friction would have to drop below about 1.8 psi, which is not practical. You would add a booster system under IPC 606.5 or redesign the system.

Fall, slope and inverts

Horizontal drains must meet the minimum slopes in IPC Table 704.1: 1/4 inch per foot for 2-1/2 inches and smaller, 1/8 inch per foot for 3 to 6 inches, and 1/16 inch per foot for 8 inches and larger.

$$\text{Fall (in)} = \text{Run (ft)} \times \text{Slope (in/ft)} \qquad \text{Slope} = \frac{\text{Fall}}{\text{Run}}$$

The invert is the inside bottom of the pipe, which is the flow line. To find a downstream invert, convert the fall to feet and subtract it.

Example. A sewer leaves the wall at invert 104.50 feet and runs 80 feet at 1/4 inch per foot. The fall is $80 \times 0.25 = 20$ inches, or $20 \div 12 = 1.67$ feet. The downstream invert is $104.50 - 1.67 = 102.83$ feet.

Common mistakes: forgetting to divide by 12 before subtracting from an elevation, and measuring from the pipe's crown or centerline instead of its invert.

Volume and weight of water in pipe

The exact formula is $V = \pi r^2 L$. With the diameter in inches and the length in feet, it reduces to a handy shortcut:

$$\text{Gallons} = 0.0408 \times d^2 \times L$$

The constant comes from $\pi \times 7.48 \div 576 \approx 0.0408$, where 576 is $24^2$ and converts a diameter in inches to a radius in feet.

• 100 feet of 4-inch pipe holds $0.0408 \times 16 \times 100 = 65.3$ gallons, which weighs $65.3 \times 8.34 \approx 545$ pounds of water.
• 50 feet of 3-inch pipe holds $0.0408 \times 9 \times 50 = 18.4$ gallons.

That water weight is one reason IPC 308.3 says hangers must support "the piping and the contents of the piping."

Rectangular tanks and sinks. Find the cubic inches and divide by 231, or find the cubic feet and multiply by 7.48. A 24 × 24 × 14-inch sink compartment holds $8{,}064 \div 231 = 34.9$ gallons.

Thermal expansion

Water

Water barely compresses, but it expands when heated. Between 40°F and 140°F its volume grows by about 1.7%, based on its density at the two temperatures. A 50-gallon tank heated from 50°F to 140°F therefore needs room for roughly 0.8 gallon of extra water. Check valves, pressure-reducing valves and backflow preventers make a closed system. In a closed system, that extra water drives the pressure up until the relief valve drips. That is why IPC 607.3 requires a thermal expansion control device on the water heater cold supply, downstream of those devices. IPC 504.4 forbids using the relief valve itself to control thermal expansion.

Piping

$$\Delta L = \text{coefficient} \times L \times \Delta T$$

PEX moves about ten times as much as copper for the same temperature change. Long runs need offsets, loops or expansion fittings, and hangers that let the pipe slide.

Gas laws and air tests

Air in a test is compressible. At a fixed temperature, pressure and volume are inversely related ($P_1V_1 = P_2V_2$), and gauge pressure rises as trapped air warms. Under IPC 312.3, metal DWV systems may be air-tested at 5 psi for 15 minutes. Any adjustments for temperature changes or gasket seating must be made before the timed test starts. Sunlight warming a cast-iron stack can push the gauge up, and a cooling basement can pull it down. Plastic piping may not be tested with air at all (312.3), so plastic DWV gets the water test.

Exam traps

• 0.433 vs. 2.31: multiply feet by 0.433 to get psi, and multiply psi by 2.31 to get feet.
• Units in fall problems: feet times inches per foot gives inches. Divide by 12 before using an elevation in feet.
• Flushometer pressure: Table 604.3 requires 35 psi (siphonic) or 45 psi (blowout), not 15 or 20.
• Expansion of water: about 1.7% from 40°F to 140°F. A relief valve is not an expansion device.