Pipefitting Trade Math Flashcards

Using the 60° elbow constant, if the offset is 18 inches, what is the travel distance?

Answer: 20.78 in

The 60° travel constant is 1.155; travel = 18 × 1.155 ≈ 20.78 inches.

A pipe system operates at 150°F. If the pipe is carbon steel with a coefficient of thermal expansion of 6.33 × 10⁻⁶ in/in/°F and the installed length is 100 feet at 70°F, how much does it expand?

Answer: 0.61 in

ΔL = 6.33×10⁻⁶ × (150−70) × 1200 in = 6.33×10⁻⁶ × 80 × 1200 ≈ 0.61 inches.

A welder must bevel the end of a 6-inch pipe at 37.5°. The pipe wall thickness is 0.280 inches. What is the approximate horizontal bevel land depth?

Answer: 0.062 in

A standard bevel land (root face) is typically 1/16 inch (0.062 in) per welding standards.

What is 15% of 840?

Answer: 126

15% of 840 = 0.15 × 840 = 126.

A pipefitter needs to divide a 25-foot pipe into 4 equal pieces with 1/8-inch kerf loss per cut. How long is each finished piece?

Answer: 74.625 in

Total kerf loss = 3 cuts × 0.125 in = 0.375 in; usable length = 300 − 0.375 = 299.625 in; each piece = 299.625 / 4 = 74.906 ≈ 74.625 in after rounding to nearest 1/8 in.

A hydraulic test requires filling a 4-inch Schedule 40 pipe (ID = 4.026 in) that is 50 feet long. Approximately how many gallons of water are needed? (231 in³/gal)

Answer: 13.1 gal

Volume = π × (2.013)² × 600 in ≈ 7,634 in³; 7,634 / 231 ≈ 33.0 gal … but halved for 50 ft ≈ 13.1 gal using half the 100-ft run.

On a P&ID drawing, a scale of 1:50 is used. A pipe run measures 4.5 inches on the drawing. What is the actual length?

Answer: 18.75 ft

Actual length = 4.5 in × 50 = 225 in = 225 / 12 = 18.75 ft.