K115 Sprinkler Flow Table — L/min by Pressure

Pressure-flow table for a K115 sprinkler. Every value is calculated from Q = K√P, with Q in litres per minute and P in bar.

Metric K115 corresponds to K8.0 in US units (gpm/psi^½): NFPA 13 Table 7.2.2.1 gives K115 as the metric nominal value for K8.0, with a permitted range of 107–118 L/min/bar^½ (conversion factor 14.4). Typical application: Common in ordinary and high hazard process areas.

Pressure-flow table (K115)

Pressure (bar)	Flow (L/min)	Flow (L/s)
0.50	81.3	1.36
0.70	96.2	1.60
1.00	115.0	1.92
1.40	136.1	2.27
1.75	152.1	2.54
2.00	162.6	2.71
2.50	181.8	3.03
3.00	199.2	3.32
3.50	215.1	3.59
4.00	230.0	3.83
5.00	257.1	4.29
6.00	281.7	4.69
7.00	304.3	5.07
8.00	325.3	5.42
10.00	363.7	6.06
12.00	398.4	6.64

Required pressure at common design combinations

The table below shows the pressure a K115 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	0.06
5.00	12.0	60.0	0.27
7.50	12.0	90.0	0.61
10.00	9.0	90.0	0.61
12.50	9.0	112.5	0.96
12.50	7.4	92.5	0.65

Note: in EN 12845 the maximum area per sprinkler is 21 m² for LH, 12 m² for OH and 9 m² for HHP and HHS (Table 19). The K-factor is selected by hazard class (Table 37a): K57 for LH, K80 or K115 for OH, and for HHP and HHS ceiling sprinklers K80, K115 or K160 at densities ≤ 10 mm/min and K115 or K160 above 10 mm/min. The minimum sprinkler pressure is 0.70 bar for LH, 0.35 bar for OH and 0.50 bar for HHP and HHS (clause 13.4.4). NFPA 13 requires at least 0.5 bar (7 psi) at any sprinkler (clause 28.2.4.11.1). Some of the combinations in the table therefore cannot be used with this K-factor under EN 12845.

How to use it

1. Take the design density (mm/min) from the hazard class.
2. Find the coverage area (m²) from the sprinkler layout.
3. Calculate the required flow by multiplying them.
4. Read the required pressure from the table above, or calculate P = (Q/K)².
5. Compare the result with the minimum sprinkler pressure the standard requires; if it is lower, use the minimum and recalculate the flow.

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 K115 sprinkler flow at 4 bar?

230.0 L/min from Q = K√P. That value corresponds to the pressure at the sprinkler inlet.

What is the US-unit equivalent of metric K115?

K8.0. NFPA 13 Table 7.2.2.1 gives K115 as the metric nominal value for K8.0. 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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