Pressure-flow table for a K280 sprinkler. Every value is calculated from Q = K√P, with Q in litres per minute and P in bar.
Metric K280 corresponds to K19.6 in US units (gpm/psi½): NFPA 13 Table 7.2.2.1 gives K280 as the metric nominal value for K19.6, with a permitted range of 272–301 L/min/bar½ (conversion factor 14.4). Typical application: ESFR and high-flow storage applications.
Pressure-flow table (K280)
| Pressure (bar) | Flow (L/min) | Flow (L/s) |
|---|---|---|
| 0.50 | 198.0 | 3.30 |
| 0.70 | 234.3 | 3.90 |
| 1.00 | 280.0 | 4.67 |
| 1.40 | 331.3 | 5.52 |
| 1.75 | 370.4 | 6.17 |
| 2.00 | 396.0 | 6.60 |
| 2.50 | 442.7 | 7.38 |
| 3.00 | 485.0 | 8.08 |
| 3.50 | 523.8 | 8.73 |
| 4.00 | 560.0 | 9.33 |
| 5.00 | 626.1 | 10.43 |
| 6.00 | 685.9 | 11.43 |
| 7.00 | 740.8 | 12.35 |
| 8.00 | 792.0 | 13.20 |
| 10.00 | 885.4 | 14.76 |
| 12.00 | 969.9 | 16.17 |
Required pressure at common design combinations
The table below shows the pressure a K280 sprinkler requires for a given density and coverage area. The calculation: required flow = density × coverage area; required pressure = (flow / K)².
| Density (mm/min) | Coverage area (m²) | Required flow (L/min) | Required pressure (bar) |
|---|---|---|---|
| 2.25 | 12.0 | 27.0 | 0.01 |
| 5.00 | 12.0 | 60.0 | 0.05 |
| 7.50 | 12.0 | 90.0 | 0.10 |
| 10.00 | 9.0 | 90.0 | 0.10 |
| 12.50 | 9.0 | 112.5 | 0.16 |
| 12.50 | 7.4 | 92.5 | 0.11 |
Note: K280 does not appear in EN 12845 Table 37a; sprinklers of this size are designed for storage protection to EN 12845-2 (ESFR and CMSA) or the NFPA 13 tables and the product listing, on a minimum-pressure-per-sprinkler basis. The density table above is an arithmetic example only. NFPA 13 requires at least 0.5 bar (7 psi) at any sprinkler (clause 28.2.4.11.1).
How to use it
- Take the design density (mm/min) from the hazard class.
- Find the coverage area (m²) from the sprinkler layout.
- Calculate the required flow by multiplying them.
- Read the required pressure from the table above, or calculate P = (Q/K)².
- Compare the result with the minimum sprinkler pressure the standard requires; if it is lower, use the minimum and recalculate the flow.
The table gives the pressure at the sprinkler inlet. In the hydraulic calculation, pipe friction loss, fitting equivalent lengths and elevation are added on top; the pressure at the system entry point is their sum.
Points to watch
- Confirm whether a catalogue K value is metric or US units; the two differ by a factor of 14.4.
- For ESFR sprinklers the K-factor comes with a minimum operating pressure defined in the listing; a table value does not mean you may go below that limit.
- A large K-factor works at low pressure but increases pipe diameter and water supply demand.
- A small K-factor produces finer droplets, weakening penetration into a stack in a storage fire.
Frequently Asked Questions
How much does a K280 sprinkler flow at 4 bar?
560.0 L/min from Q = K√P. That value corresponds to the pressure at the sprinkler inlet.
What is the US-unit equivalent of metric K280?
K19.6. NFPA 13 Table 7.2.2.1 gives K280 as the metric nominal value for K19.6. The conversion factor is 14.4, and which unit a catalogue uses must always be confirmed.
Can the sprinkler operate below the lowest pressure in the table?
No. NFPA 13 clause 28.2.4.11.1 sets a minimum operating pressure of 0.5 bar (7 psi) for any sprinkler; where the listing specifies a higher minimum, that governs (clause 28.2.4.11.2). EN 12845 clause 13.4.4 also sets minimum pressures by hazard class. For ESFR sprinklers the listed minimum operating pressure is especially binding.
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Download MEP Calc on the App StoreNFPA 13 (2025) Table 7.2.2.1, clause 28.2.4.11 · EN 12845:2015+A2:2026 Table 19, Table 37a, clause 13.4.4 · EN 12259-1. Table values are calculated from Q = K√P; the manufacturer's approval document governs the product-specific minimum and maximum operating pressure.