NFPA 13 Density/Area Method
NFPA 13 sizes wet/dry sprinkler systems by hazard classification. You pick a design density (mm/min or gpm/ft²) and an area of operation from the density/area curves for the hazard (light, ordinary, extra). The base demand is density × area. Each sprinkler in the design area must discharge at least density × its coverage, and the required pressure at each head comes from its nominal K-factor via Q = K√P. The full calculation then balances pipe friction and elevation back to the supply.
Frequently Asked Questions
Total design flow = design density (mm/min) × area of operation (m²). For example 5 mm/min over 140 m² = 700 L/min. Each sprinkler must deliver at least density × its coverage area, and the pressure needed follows Q = K√P.
A sprinkler discharges Q = K√P, where Q is flow (L/min), P is pressure at the head (bar) and K is the discharge coefficient. Rearranged, the pressure needed for a target flow is P = (Q ÷ K)². Common metric K-factors are 57, 80 and 115.
From the NFPA 13 density/area curves for the hazard: light hazard ≈ 4.1 mm/min over 140 m²; ordinary hazard 6.1–8.1 mm/min; extra hazard higher still. Choosing a smaller area requires a higher density and vice-versa along the curve.
No — this gives the base demand (flow and per-head pressure). A full hydraulic calculation adds Hazen-Williams pipe friction (C ≈ 120 for steel, 150 for plastic/copper) and elevation, balancing flows node-by-node back to the source to find the pump duty.
Approximately area of operation ÷ coverage per sprinkler. For 140 m² with 12 m² coverage that is about 12 heads in the most-remote design area used for the calculation.
Fire Sprinkler Hydraulic Design (NFPA 13)
NFPA 13 protects buildings by classifying the hazard and applying a water density over a design area. The density/area method sets how much water (mm/min) must fall over the most-remote operating area; the product is the base flow the system must deliver.
The discharge law
Each sprinkler is an orifice that follows Q = K√P. The K-factor is printed on the sprinkler; a higher K passes more water at the same pressure. To deliver the required per-head flow you solve for pressure: P = (Q ÷ K)². The most-remote head usually governs because it sees the lowest pressure.
From demand to pump duty
The full calculation starts at the most-remote head, then walks back to the riser adding Hazen-Williams pipe friction and elevation, summing flows at each node. The result is the flow and pressure the supply (town main, tank + pump) must provide — sized with the pipe sizing and pump tools.
Related: Pipe Sizing, Pump Head, Reynolds Number.