Pressure-flow table for a K80 sprinkler. Every value is calculated from Q = K√P, with Q in litres per minute and P in bar.
Metric K80 corresponds to roughly K5.6 in US units (gpm/psi½), with a conversion factor of 14.4. Typical application: Used for the OH classes (K80 or K115) and for HHP/HHS ceiling sprinklers in EN 12845 Table 37a; in NFPA 13, K5.6 (K80) is the common standard choice for light and ordinary hazard.
Pressure-flow table (K80)
| Pressure (bar) | Flow (L/min) | Flow (L/s) |
|---|---|---|
| 0.50 | 56.6 | 0.94 |
| 0.70 | 66.9 | 1.12 |
| 1.00 | 80.0 | 1.33 |
| 1.40 | 94.7 | 1.58 |
| 1.75 | 105.8 | 1.76 |
| 2.00 | 113.1 | 1.89 |
| 2.50 | 126.5 | 2.11 |
| 3.00 | 138.6 | 2.31 |
| 3.50 | 149.7 | 2.49 |
| 4.00 | 160.0 | 2.67 |
| 5.00 | 178.9 | 2.98 |
| 6.00 | 196.0 | 3.27 |
| 7.00 | 211.7 | 3.53 |
| 8.00 | 226.3 | 3.77 |
| 10.00 | 253.0 | 4.22 |
| 12.00 | 277.1 | 4.62 |
Required pressure at common design combinations
The table below shows the pressure a K80 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 | 0.11 |
| 5.00 | 12.0 | 60.0 | 0.56 |
| 7.50 | 12.0 | 90.0 | 1.27 |
| 10.00 | 9.0 | 90.0 | 1.27 |
| 12.50 | 9.0 | 112.5 | 1.98 |
| 12.50 | 7.4 | 92.5 | 1.34 |
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: under EN 12845 13.4.4, 0.70 bar in LH, 0.35 bar in OH and 0.50 bar in HHP/HHS; under NFPA 13 28.2.4.11, 7 psi (0.5 bar) or the higher value in the listing. If the calculated pressure 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 heads 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 K80 sprinkler flow at 4 bar?
160 L/min from Q = K√P. That value corresponds to the pressure at the sprinkler inlet.
What is the US-unit equivalent of metric K80?
Roughly K5.6. The conversion factor is 14.4, and the unit in which a catalogue gives the K value must always be confirmed.
Can the sprinkler operate below a table value?
No. You cannot go below the standard's minimum sprinkler pressure (EN 12845 13.4.4, NFPA 13 28.2.4.11) or the minimum operating pressure in the product's listing or approval; for ESFR sprinklers that limit is especially binding.
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Download MEP Calc on the App StoreNFPA 13 (2025) · EN 12845:2015+A2:2026 · 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.