11.1 Q = K√P and Standard K-Factors

Why Level I still has to own Q = K√P

Independent OpenExamPrep material in this chapter helps learners study NICET Water-Based Systems Layout Level I tasks 1.5.2 Solve basic hydraulic equations and 1.5.3 Assist with flow tests. It is not an NFPA or NICET handbook, and it does not claim official approval from either organization. The exam references are NFPA 13 (2022) and NFPA 291 (2022). The testing vendor supplies a built-in calculator. That calculator will multiply and take square roots; it will not remember which formula, which K-factor, or which C-value belongs in the stem. You still have to write Q = K√P from memory and keep the units straight.

Every listed sprinkler is identified by a K-factor, a discharge coefficient with U.S. units of gpm/psi^0.5. NFPA 13 hydraulic calculations use the same pair of equations at every head:

Q = K √P

P = (Q / K)²

Q is flow from that sprinkler in gallons per minute. P is the residual pressure at that sprinkler in pounds per square inch. √P is the square root of that pressure, not “P divided by two.” On the calculator, use the square-root key. Stems that look like simple arithmetic fail if you skip the root or square the wrong term.

Worked discharge: K-5.6 at 16 psi

The workhorse commercial spray sprinkler is K-5.6, the 1/2 in orifice in NFPA 13 Table 7.2.2.1. A remote head is calculated at 16 psi.

Step 1: √16 = 4, because 4 × 4 = 16.

Step 2: Q = 5.6 × 4 = 22.4 gpm.

That 22.4 gpm is the flow at that one sprinkler at 16 psi. It is not the remote-area demand, not a hose-stream allowance, and not a hydrant pitot flow. If the stem later asks for eight such heads at the same pressure, you still start here, then add; you do not invent a new formula.

Worked pressure: K-5.6 must deliver 20 gpm

A light-hazard sprinkler covering 200 ft² at 0.10 gpm/ft² needs 20 gpm, because 0.10 × 200 = 20. The head is K-5.6. Invert the discharge equation.

Step 1: Q / K = 20 / 5.6 = 3.5714 (keep extra decimals until you square).

Step 2: P = (3.5714)² = 12.755 ≈ 12.8 psi.

12.8 psi is above the 7 psi minimum in NFPA 13 28.2.4.11, so density—not the floor pressure—governs this head. If the sprinkler’s listing specifies a higher minimum for the intended use, that listing governs instead of 7 psi.

The 7 psi floor is not a design density

Section 28.2.4.11 sets 7 psi (0.5 bar) as the minimum operating pressure for any sprinkler unless the listing is higher. At 7 psi:

√7 ≈ 2.646

Q = 5.6 × 2.646 ≈ 14.8 gpm

Now change the occupancy, not the K-factor. A K-5.6 covering 130 ft² of ordinary hazard Group 2 at 0.20 gpm/ft² needs 0.20 × 130 = 26 gpm.

P = (26 / 5.6)² = (4.6429)² ≈ 21.6 psi

Here density governs. You cannot “take 7 psi” and call the head finished. Calculate the pressure that produces the required flow, then confirm it is at least 7 psi or the listed minimum, whichever is greater.

Identify the other catalog K-factors

Table 7.2.2.1 is an identification table. The percentages are relative to a K-5.6 at the same pressure. Memorize the row, not a story about “all 1/2 in heads.”

Run the same algebra on other catalog K-factors with original pressures so the 16 psi example does not become the only number you can compute:

• K-8.0 at 25 psi: √25 = 5; Q = 8.0 × 5 = 40 gpm.
• K-11.2 at 9 psi: √9 = 3; Q = 11.2 × 3 = 33.6 gpm.
• K-2.8 at 16 psi: Q = 2.8 × 4 = 11.2 gpm.
• K-4.2 at 9 psi: Q = 4.2 × 3 = 12.6 gpm.
• K-14.0 at 36 psi: √36 = 6; Q = 14.0 × 6 = 84 gpm.
• K-16.8 at 25 psi: Q = 16.8 × 5 = 84 gpm.
• K-5.6 must deliver 28 gpm: P = (28 / 5.6)² = 5² = 25 psi.

Because flow follows the square root of pressure, quadrupling pressure only doubles flow. K-5.6 at 16 psi is 22.4 gpm; at 64 psi (four times 16) it is 5.6 × 8 = 44.8 gpm. Doubling pressure from 16 psi to 32 psi multiplies flow only by √2 ≈ 1.41 (22.4 × 1.41 ≈ 31.6 gpm). If a stem claims that doubling pressure doubles flow, that choice is a trap.

NFPA 13 9.4.4 generally requires a minimum nominal K-5.6 unless a listed exception applies. K-2.8 and K-4.2 are permitted in light hazard when the system is hydraulically calculated and, for K-factors smaller than 5.6, on wet pipe—or under the exposure-protection and corrosion-resistant dry/preaction exceptions in 9.4.4.3 and 9.4.4.4. A listed strainer is required on the supply side of sprinklers smaller than K-2.8. CMSA and ESFR sprinklers start at K-11.2. New systems must not use a K-factor larger than 5.6 on 1/2 in NPT (9.4.5). Do not hang a K-8.0 head on a K-5.6 calculation without re-running Q = K√P.

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