This is the largest domain and the one that separates master-level candidates from journeymen. It is almost entirely calculation, and calculation is the good news: unlike a memory question, a sizing problem can be derived on the exam if you know the procedure. Four conversions carry most of the domain:
- WSFU to gpm (water supply),
- DFU to pipe size and slope to fall (drainage and venting),
- BTU/hr to cfh to pipe size (fuel gas), and
- roof area × rainfall to gpm (storm).
Master these four and you have earned the single biggest block of points on the exam.
1.1 Water Supply Sizing by Fixture Units
Teach. Water distribution is sized in three moves: total the load, convert the load to a flow, choose a pipe that keeps velocity in bounds.
The load is measured in water supply fixture units (WSFU), a probability-weighted number the code assigns to each fixture so that a lavatory and a flushometer water closet can be added on one scale. You total the WSFU that a given section of pipe must carry, then convert that single total to a design flow in gallons per minute (gpm) using a demand curve derived from Hunter's probability work. The curve's whole purpose is to discount the near-impossibility that every fixture runs simultaneously.
The rule that trips people: add fixture units first, convert to gpm once. Converting each branch to gpm and adding the flows badly overstates demand, because the curve has already discounted simultaneous use. Also, flush-valve (flushometer) fixtures ride a different, higher demand curve than flush-tank fixtures, because a flushometer draws a heavy momentary flow while a tank sips through a small refill valve.
Once you have design gpm, choose the pipe so velocity stays in bounds — generally about 8 ft/s for cold water and 5 ft/s for continuously circulated hot water — using:
v = 0.408 × Q / d² where v is velocity (ft/s), Q is flow (gpm), and d is the actual inside diameter (inches).
High velocity causes erosion-corrosion, noise, and water hammer; hot water is chemically more aggressive, so its limit is lower.
Worked Example 1A — total, convert, size.
A branch serves, using illustrative WSFU values (confirm against your adopted table):
- 3 lavatories at 1.0 WSFU each = 3.0
- 2 water closets, flush tank, at 3.0 WSFU each = 6.0
- 1 kitchen sink at 1.5 WSFU = 1.5
- 1 shower at 2.0 WSFU = 2.0
Total = 3.0 + 6.0 + 1.5 + 2.0 = 12.5 WSFU.
Convert once on a flush-tank demand curve. Using an illustrative demand relationship (again, verify against the adopted curve): 12.5 WSFU on the flush-tank curve corresponds to roughly 10 gpm of design flow.
Now size the pipe to keep cold-water velocity at or under 8 ft/s. Rearrange the velocity formula for the minimum diameter:
d = √(0.408 × Q / v) = √(0.408 × 10 / 8) = √(4.08 / 8) = √0.51 = 0.714 in
A nominal ¾-in tube has an inside diameter close to 0.785–0.811 in depending on type, which is larger than 0.714 in, so ¾-in copper holds the velocity below 8 ft/s. Check it: v = 0.408 × 10 / (0.785²) = 4.08 / 0.616 = 6.6 ft/s — under the limit. Good.
Worked Example 1B — why you convert only once.
Suppose two branches each carry 20 WSFU and meet at a main. On a flush-tank curve, 20 WSFU is about 16 gpm. The wrong method adds the flows: 16 + 16 = 32 gpm. The right method adds the fixture units first — 20 + 20 = 40 WSFU — then converts once: 40 WSFU is about 30 gpm, not 32. The single conversion is lower because the curve keeps discounting as the population grows. On a big system the error compounds badly, so the exam loves this trap.
Worked Example 1C — velocity limit sets a floor on size.
A hot-water recirculation return must carry 8 gpm and must stay at or under 5 ft/s (continuous circulation). Minimum diameter:
d = √(0.408 × 8 / 5) = √(3.264 / 5) = √0.653 = 0.808 in
So a nominal ¾-in line (ID ~0.785 in) is too small here — it would run slightly over 5 ft/s and invite erosion of the hot line. Step up to 1-in and recheck: v = 0.408 × 8 / (1.025²) = 3.264 / 1.05 = 3.1 ft/s. Comfortable.
Key-Code Callout — Water Supply Sizing - Demand/WSFU: UPC Table 610.3 / §610.0; IPC Appendix E (equivalent fixture-unit values and Hunter demand curve). Confirm your jurisdiction adopts Appendix E if you rely on it. - Velocity: hold cold ≈ 8 ft/s, hot ≈ 5 ft/s (UPC §610.12). Formula v = 0.408 Q / d². - A supply main never reduces in the direction of flow toward fixtures it must still feed. - Minimum fixture flow pressures (≈ 8 psi lavatory to ≈ 25 psi flushometer) feed the pressure budget (next section).
Common Traps - Adding gpm instead of WSFU. Always total units, then convert once. - Using the flush-tank curve for flushometers (or vice versa). Flushometers draw more. - Using nominal size as the diameter in the velocity formula. Use the actual inside diameter; type M, L, and K copper differ. - IPC vs UPC: the IPC houses the demand data in Appendix E, which is only enforceable if the jurisdiction adopts the appendix; the UPC carries it in the body of Chapter 6. Same physics, different citation.
1.2 The Pressure Budget
Teach. Every water system is a budget: the pressure available at the source must pay for everything between the source and the most remote, highest fixture. The four consumers are:
- Elevation (static lift): 0.433 psi per foot of rise. (Equivalently, 2.31 ft of water = 1 psi.)
- Fixed device losses: the water meter, softener, filter, and backflow assembly, each read from a manufacturer's chart at the design flow.
- Pipe friction: expressed as psi per 100 ft at the design flow.
- Minimum residual pressure at the fixture: commonly 8 to 25 psi.
The procedure is subtraction. Start from the minimum guaranteed street pressure, subtract elevation, subtract device losses, subtract the fixture's required residual, and whatever remains is the budget available for friction. Divide that remaining pressure by the developed length of the critical run and multiply by 100 to get the allowable friction loss per 100 ft. Carry that value across a friction chart at the design gpm to pick each pipe.
Developed length, not tape length: add the equivalent length of every fitting, valve, and the meter to the straight pipe, because a run full of elbows behaves as if it were much longer.
Worked Example 1D — the full budget.
Given:
- Minimum street (static) pressure: 60 psi
- Highest fixture is 30 ft above the meter
- Meter loss: 8 psi
- Backflow assembly loss: 9 psi
- Required residual at the critical fixture (flush tank): 15 psi
- Developed length of the critical run: 120 ft
Step 1 — elevation loss: 30 ft × 0.433 psi/ft = 12.99 ≈ 13.0 psi.
Step 2 — subtract all fixed consumers: 60 − 13.0 − 8 − 9 − 15 = 15.0 psi left for friction.
Step 3 — allowable friction loss per 100 ft: (15.0 psi ÷ 120 ft) × 100 = 12.5 psi per 100 ft.
Now enter the friction chart at the design gpm and pick the smallest pipe whose loss at that flow is ≤ 12.5 psi/100 ft. If the trial pipe exceeds 12.5, go up one size and recheck.
Worked Example 1E — how tall can gravity go?
A building has 55 psi minimum street pressure. The top fixture needs 15 psi residual, and meter plus backflow eat 15 psi. Ignoring friction for a first cut, the pressure available for elevation is 55 − 15 − 15 = 25 psi. Convert to height: 25 psi ÷ 0.433 psi/ft = 57.7 ft of rise — roughly five stories — before a booster pump is needed. Add friction and the real ceiling is lower, which is exactly why tall buildings pump.
Worked Example 1F — developed length matters.
A run measures 80 ft of straight pipe but includes fittings whose equivalent lengths total 40 ft. The developed length is 120 ft, not 80. If you had 15 psi for friction and used the tape length, you would compute 15/80 × 100 = 18.75 psi/100 ft and undersize the pipe; using the correct 120 ft gives 12.5 psi/100 ft and the right, larger pipe.
Key-Code Callout — Pressure Budget - Elevation: 0.433 psi/ft; 2.31 ft = 1 psi. - Subtract elevation + meter + device + residual from static supply; the remainder ÷ developed length × 100 = allowable psi/100 ft. - Use developed length (straight pipe + fitting equivalents), not measured length. - Both UPC and IPC accept the segmented friction-loss method; the IPC also permits a simplified prescriptive table for small systems.
Common Traps - Forgetting the residual. The fixture still needs 8–25 psi to work; that pressure is spent, not available. - Using measured instead of developed length, which undersizes the pipe. - Mixing units: psi vs feet of head. Convert with 2.31 ft/psi. - Assuming static = working pressure. Under flow, pressure drops; size to the minimum available.
1.3 Drainage and Vent Sizing by Fixture Units
Teach. Sanitary drainage is sized from drainage fixture units (DFU), a value that scales with a fixture's discharge rate and its potential to load the drain. The DFU tables are split into columns that must never be interchanged:
- horizontal fixture branch,
- vertical stack (often qualified by branch intervals), and
- building drain and sewer at a stated slope.
Total the DFU a section carries and pick the smallest pipe whose column capacity meets or exceeds the load. Two override rules bind no matter the count:
- A drain is never smaller than the largest trap or trap arm it serves.
- Any drain receiving a water closet is at least 3 inches.
Slope is the other axis. Both codes set a minimum of ¼ in per foot for pipe 2½ in and smaller and ⅛ in per foot for 3 in and larger (with 1/16 in/ft for very large sewers only by special approval). Total fall = slope × developed run.
Vents are sized in parallel: a vent is not less than half the diameter of the drain it serves, never smaller than 1¼ in, and is further limited by developed length. The trap arm — the distance from a trap to its vent — is capped by pipe size (roughly 6 ft for a 2-in arm; verify the table).
Worked Example 1G — size a building drain.
Using illustrative DFU values (verify against your table): 2 water closets at 4 DFU (8), 3 lavatories at 1 DFU (3), 2 showers at 2 DFU (4), 1 kitchen sink at 2 DFU (2), 1 clothes washer at 3 DFU (3).
Total = 8 + 3 + 4 + 2 + 3 = 20 DFU.
Because water closets discharge here, the drain must be at least 3 in regardless of count. Check a 3-in building drain at ¼ in/ft against an illustrative capacity of ~35 DFU: 20 ≤ 35, so 3-in works. If the same load were on a stack with several branch intervals, read the stack column, not the building-drain column — the capacities differ for the same nominal size.
Worked Example 1H — slope to fall.
A 4-in building drain runs 80 ft at the ⅛-in/ft minimum. Total fall = 0.125 in/ft × 80 ft = 10 inches of drop from start to end. If it ran at ¼ in/ft, fall = 0.25 × 80 = 20 inches — which may not fit under a slab, a real design constraint that pushes you to the flatter allowed slope on long runs.
Worked Example 1I — vent and trap arm.
A 3-in drain needs a vent of at least half its diameter: 3 ÷ 2 = 1.5, so a 1½-in vent minimum. A 4-in drain needs at least a 2-in vent. Separately, a 2-in trap arm (a 2-in fixture drain from trap to vent) is limited to roughly 6 ft of developed length; run it longer and the trap can self-siphon, so you must relocate the vent or upsize the arm.
Key-Code Callout — DWV Sizing - DFU tables: UPC Chapter 7 (§703); IPC Chapter 7 (§710 sizing, §704 slope). Read the correct column. - Water closet → 3-in minimum drain (IPC §710.1). - Slope minimums: ¼ in/ft ≤ 2½ in; ⅛ in/ft ≥ 3 in (IPC §704.1). Fall = slope × run. - Vent ≥ ½ drain diameter, ≥ 1¼ in (UPC §904).
Common Traps - Interchanging table columns (branch vs stack vs building drain). - Reducing size downstream as more load is added — never legal. - Ignoring the trap/trap-arm override — a 3-in trap forces at least a 3-in drain even at low DFU. - IPC vs UPC: the two assign slightly different DFU values to some fixtures and tabulate trap-arm lengths differently. Size to the adopted code, and never mix a UPC DFU value into an IPC table.