The K-factor formula
Sprinkler discharge is proportional to the square root of the pressure at the head: Q = K · √P. In SI units Q is L/min, P is bar and K is L/min/bar½. Rearranged, the pressure needed for a given flow is P = (Q / K)².
In US units K is gpm/psi½, and KSI = 14.4163 × KUS. That is why K80 ≈ K5.6, K115 ≈ K8.0 and K360 ≈ K25.2.
Minimum end-head pressure
NFPA 13 requires at least 7 psi (≈0.5 bar) for standard spray sprinklers. EN 12845 sets 0.35 bar for OH, 0.5 bar for HHP/HHS and 0.7 bar for LH. For specially listed sprinklers (ESFR, CMSA, extended coverage) the listed minimum pressure governs.
What the tool does not cover
It gives the behaviour of one sprinkler and the total flow of heads at equal pressure. In a real system every head sees a different pressure; the design-area demand needs a Hazen-Williams hydraulic calculation.
Frequently Asked Questions
Should I pick a larger K-factor for the same flow?
A larger K-factor delivers the same flow at lower pressure, reducing pump pressure and pipe sizes. Droplet size and coverage change too, so selection must follow the hazard class and the listing.
Are K80 and K5.6 the same?
In practice, yes. K5.6 × 14.4163 ≈ 80.7 L/min/bar½, which catalogues call K80.
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