What is the pressure at a sprinkler head with a flow of 33 gpm (125 L/min) with a K-factor of 5.6 (80)?

The pressure required at a sprinkler head is calculated using the formula derived from the orifice equation: P = (Q/K)², where P is the pressure in psi, Q is the flow in gallons per minute (gpm), and K is the nominal K-factor.

Given the values:

- Q = 33 gpm

- K = 5.6

The calculation is as follows:

P = (33 / 5.6)²

P = (5.893)²

P = 34.73 psi

This value is rounded to 34.7 psi. The metric calculation confirms this result: P = (125 L/min / 80)² = 2.44 bar.

A. This pressure would result from a much lower flow rate (approximately 21.7 gpm) for the given K-factor, representing a significant miscalculation.

B. This pressure is an incorrect calculation and does not correspond to the relationship defined by the sprinkler discharge formula for the given parameters.

D. This pressure would require a significantly higher flow rate (approximately 41.7 gpm) through the sprinkler head with the specified K-factor.

1. NFPA 13, Standard for the Installation of Sprinkler Systems, 2022 Edition. Chapter 9, Section 9.1.2.1, "The quantity of water discharged from a sprinkler is determined by the formula q = K√p." (This formula is rearranged to P = (q/K)² to solve for pressure).

2. Society of Fire Protection Engineers (SFPE). SFPE Handbook of Fire Protection Engineering, 5th Edition. Chapter 21, "Hydraulic Calculations for Water-Based Fire Protection Systems," Section 21.3.1, "The K-Factor Formula," page 738. The text explicitly provides the formula Q = K√P.

3. Oklahoma State University, School of Fire Protection and Safety Engineering Technology. FSET 3153 - Sprinklers and Standpipes Courseware. Module on Hydraulic Calculations, which details the use of the formula P = (Q/K)² for determining required sprinkler pressure.