Tools › Building Systems › Fire Protection › Sprinkler Hydraulics › Example
Twelve-head light-hazard tree: calculation sheet from the remote head to the hydrant-test supply curve
| Tool | Sprinkler Hydraulics |
| Source | U.S. Department of Defense (2016). Fire Protection Engineering for Facilities. UFC 3-600-01 with Change 6, 2021, Tables 9-3 and 9-4 (design density, area, K-factor, hose allowance); U.S. Department of Agriculture, NRCS (2009). Engineering Field Handbook Part 650 Chapter 3, eqs 164-173 and Table 9 (Hazen-Williams formula, fitting loss coefficients). |
| Level | Professional |
Problem
A wet-pipe system protects a light-hazard occupancy: 0.10 gpm/ft² over the most remote 1,500 ft² with K 5.6 heads at 130 ft² each (12 flowing heads, three branch lines of four), a 7 psi minimum at the end head and a 250 gpm hose allowance. Branch lines are 13 ft of 1 in, 1 in and 1-1/4 in Schedule 40 steel then 7 ft of 1-1/2 in with a tee into the cross main (10 ft of 2 in, 10 ft of 2-1/2 in, 40 ft of 3 in with an elbow); the riser drops 20 ft through 3 in pipe with a gate valve, an elbow and an 8 psi double-check backflow preventer; a 100 ft 6 in main (C 140) with an elbow reaches the supply point, where a hydrant test gave 65 psi static and 50 psi residual at 1,000 gpm. Find the sprinkler demand, the total demand, the pressure required at the supply and the margin against the supply curve.
Inputs to Enter
- Density 0.10 gpm/ft², design area 1,500 ft², 130 ft² per head, spacing 13 ft, K 5.6, p_min 7 psi, hose 250 gpm at the supply node
- Supply: static 65 psi, residual 50 psi at 1,000 gpm; supply node SUP at elevation 0; heads and cross main at 20 ft
- Fitting K (NRCS Table 9): elbow 0.90, tee 1.80, gate valve 0.19; C 120 steel, C 140 feed main; Schedule 40
Answers
- Remote head S1: Q = 5.6 √7 = 14.82 gpm; first pipe p_f = 4.52 × 14.82^1.85 / (120^1.85 × 1.049^4.87) = 0.0747 psi/ft, 0.971 psi over 13 ft
- Branch line at the cross main: 70.80 gpm at 16.78 psi, K_eq 17.29 (tee charged as 9.00 ft of 1-1/2 in pipe)
- Sprinkler demand 216.05 gpm (12 heads); total with hose 466.05 gpm at 39.49 psi required at the supply (riser elevation 8.66 psi, device 8 psi)
- Supply at 466 gpm: 65 − 15 (466/1,000)^1.85 = 61.35 psi, margin 21.86 psi; 1,332.5 gpm available at 39.49 psi; exact nodal solution 466.13 gpm at 39.49 psi (+0.02 %)
Notes
The tool's default inputs are this system; press Submit. The network is illustrative (stated on the page): the density, area, K and hose values are UFC 3-600-01 Tables 9-3 and 9-4, the fitting coefficients NRCS Table 9, and every number is checked by the numpy / scipy run in References/Sprinkler-Hydraulics/checks/ through test_sprinkler_hydraulics.php (218 checks), which also reproduces the NRCS handbook's Hazen-Williams examples 62, 74 and 76.