Sprinkler K Factors

Hydraulics · FE Reference Handbook section

Hydraulics

29 formulas

10 exam-style examples

~60 min

All Hydraulics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Sprinkler K Factors within Hydraulics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what sprinkler k factors describes physically and when it applies.
State every one of the 29 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI).

Lecture

Why this section exists. Sprinkler K Factors is the part of Hydraulics that lets you connect a channel, culvert or control structure to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as open-channel normal depth, specific energy, or a weir discharge. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI). Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: sprinkler k factors.
Capstone Studio instructional photograph

Hydraulics — Sprinkler K Factors: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a channel, culvert or control structure. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 29 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

r	Quantity produced by "r= v" — read its definition and unit from the handbook line directly above the equation.
Gp	Quantity produced by "Gp = 3.2 + S + 0.27Nped" — read its definition and unit from the handbook line directly above the equation.
t	Quantity produced by "t = driver reaction time (sec)" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v = vehicle approach speed (ft/sec)" — read its definition and unit from the handbook line directly above the equation.
W	Quantity produced by "W = width of intersection, curb-to-curb (ft)" — read its definition and unit from the handbook line directly above the equation.
l	Quantity produced by "l = length of vehicle (ft)" — read its definition and unit from the handbook line directly above the equation.
y	Quantity produced by "y = length of yellow interval to nearest 0.1 sec (sec)" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = crosswalk length (ft)" — read its definition and unit from the handbook line directly above the equation.
Sp	Quantity produced by "Sp = pedestrian speed (ft/sec), default 3.5 ft/sec" — read its definition and unit from the handbook line directly above the equation.
Nped	Quantity produced by "Nped = number of pedestrian in interval" — read its definition and unit from the handbook line directly above the equation.
a	Quantity produced by "a = deceleration rate (ft/sec2)" — read its definition and unit from the handbook line directly above the equation.
± G	Quantity produced by "± G = percent grade divided by 100 (uphill grade "+")" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Orifice Size Name K Factor
1/2" Standard 5.6
17/32" Large 8.0
5/8" Extra large 11.2
Transportation
Queueing models are found in the Industrial Engineering section.
Traffic Signal Timing
where
Stopping Sight Distance
30 dc 32.2 m ! G n
where
Peak Hour Factor
Hourly Volume V
PHF Hourly Flow Rate 4 * V15
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors

Sprinkler heads with a K-factor of 8.0 operate at 48 psi. Compute the discharge of one head, the total flow if 9 heads in the design area operate simultaneously, and the resulting density over a 169 ft² per-head coverage.

Given

K = 8.0 gpm/psi^{0.5}
p = 48 psi
9 heads flowing
Coverage = 169 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 8.0 48 = 55.4 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 55.4 (9) = 498.8 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 55.4/169 = 0.328 g p m / f t^{2}$
Check
$d es i g na r e a 9 (169) = 1, 521 f t^{2} a t 0.328 g p m / f t^{2} a v er a g e$

Answer: 55.4 gpm per head, 498.8 gpm total, density 0.328 gpm/ft²

Why the other options are there

384.0 gpm (pressure not square-rooted)
6.2 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 2

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (2)

Sprinkler heads with a K-factor of 14.0 operate at 26 psi. Compute the discharge of one head, the total flow if 15 heads in the design area operate simultaneously, and the resulting density over a 133 ft² per-head coverage.

Given

K = 14.0 gpm/psi^{0.5}
p = 26 psi
15 heads flowing
Coverage = 133 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 14.0 26 = 71.4 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 71.4 (15) = 1, 071 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 71.4/133 = 0.537 g p m / f t^{2}$
Check
$d es i g na r e a 15 (133) = 1, 995 f t^{2} a t 0.537 g p m / f t^{2} a v er a g e$

Answer: 71.4 gpm per head, 1,071 gpm total, density 0.537 gpm/ft²

Why the other options are there

364.0 gpm (pressure not square-rooted)
4.8 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 3

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (3)

Sprinkler heads with a K-factor of 14.0 operate at 12 psi. Compute the discharge of one head, the total flow if 7 heads in the design area operate simultaneously, and the resulting density over a 195 ft² per-head coverage.

Given

K = 14.0 gpm/psi^{0.5}
p = 12 psi
7 heads flowing
Coverage = 195 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 14.0 12 = 48.5 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 48.5 (7) = 339.5 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 48.5/195 = 0.249 g p m / f t^{2}$
Check
$d es i g na r e a 7 (195) = 1, 365 f t^{2} a t 0.249 g p m / f t^{2} a v er a g e$

Answer: 48.5 gpm per head, 339.5 gpm total, density 0.249 gpm/ft²

Why the other options are there

168.0 gpm (pressure not square-rooted)
6.9 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 4

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (4)

Sprinkler heads with a K-factor of 14.0 operate at 30 psi. Compute the discharge of one head, the total flow if 8 heads in the design area operate simultaneously, and the resulting density over a 197 ft² per-head coverage.

Given

K = 14.0 gpm/psi^{0.5}
p = 30 psi
8 heads flowing
Coverage = 197 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 14.0 30 = 76.7 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 76.7 (8) = 613.4 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 76.7/197 = 0.389 g p m / f t^{2}$
Check
$d es i g na r e a 8 (197) = 1, 576 f t^{2} a t 0.389 g p m / f t^{2} a v er a g e$

Answer: 76.7 gpm per head, 613.4 gpm total, density 0.389 gpm/ft²

Why the other options are there

420.0 gpm (pressure not square-rooted)
9.6 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 5

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (5)

Sprinkler heads with a K-factor of 5.6 operate at 55 psi. Compute the discharge of one head, the total flow if 11 heads in the design area operate simultaneously, and the resulting density over a 100 ft² per-head coverage.

Given

K = 5.6 gpm/psi^{0.5}
p = 55 psi
11 heads flowing
Coverage = 100 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 5.6 55 = 41.5 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 41.5 (11) = 456.8 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 41.5/100 = 0.415 g p m / f t^{2}$
Check
$d es i g na r e a 11 (100) = 1, 100 f t^{2} a t 0.415 g p m / f t^{2} a v er a g e$

Answer: 41.5 gpm per head, 456.8 gpm total, density 0.415 gpm/ft²

Why the other options are there

308.0 gpm (pressure not square-rooted)
3.8 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 6

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (6)

Sprinkler heads with a K-factor of 5.6 operate at 54 psi. Compute the discharge of one head, the total flow if 16 heads in the design area operate simultaneously, and the resulting density over a 189 ft² per-head coverage.

Given

K = 5.6 gpm/psi^{0.5}
p = 54 psi
16 heads flowing
Coverage = 189 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 5.6 54 = 41.2 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 41.2 (16) = 658.4 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 41.2/189 = 0.218 g p m / f t^{2}$
Check
$d es i g na r e a 16 (189) = 3, 024 f t^{2} a t 0.218 g p m / f t^{2} a v er a g e$

Answer: 41.2 gpm per head, 658.4 gpm total, density 0.218 gpm/ft²

Why the other options are there

302.4 gpm (pressure not square-rooted)
2.6 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 7

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (7)

Sprinkler heads with a K-factor of 11.2 operate at 33 psi. Compute the discharge of one head, the total flow if 8 heads in the design area operate simultaneously, and the resulting density over a 138 ft² per-head coverage.

Given

K = 11.2 gpm/psi^{0.5}
p = 33 psi
8 heads flowing
Coverage = 138 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 11.2 33 = 64.3 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 64.3 (8) = 514.7 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 64.3/138 = 0.466 g p m / f t^{2}$
Check
$d es i g na r e a 8 (138) = 1, 104 f t^{2} a t 0.466 g p m / f t^{2} a v er a g e$

Answer: 64.3 gpm per head, 514.7 gpm total, density 0.466 gpm/ft²

Why the other options are there

369.6 gpm (pressure not square-rooted)
8.0 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 8

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (8)

Sprinkler heads with a K-factor of 11.2 operate at 13 psi. Compute the discharge of one head, the total flow if 15 heads in the design area operate simultaneously, and the resulting density over a 119 ft² per-head coverage.

Given

K = 11.2 gpm/psi^{0.5}
p = 13 psi
15 heads flowing
Coverage = 119 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 11.2 13 = 40.4 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 40.4 (15) = 605.7 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 40.4/119 = 0.339 g p m / f t^{2}$
Check
$d es i g na r e a 15 (119) = 1, 785 f t^{2} a t 0.339 g p m / f t^{2} a v er a g e$

Answer: 40.4 gpm per head, 605.7 gpm total, density 0.339 gpm/ft²

Why the other options are there

145.6 gpm (pressure not square-rooted)
2.7 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 9

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (9)

Sprinkler heads with a K-factor of 8.0 operate at 34 psi. Compute the discharge of one head, the total flow if 6 heads in the design area operate simultaneously, and the resulting density over a 167 ft² per-head coverage.

Given

K = 8.0 gpm/psi^{0.5}
p = 34 psi
6 heads flowing
Coverage = 167 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 8.0 34 = 46.6 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 46.6 (6) = 279.9 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 46.6/167 = 0.279 g p m / f t^{2}$
Check
$d es i g na r e a 6 (167) = 1, 002 f t^{2} a t 0.279 g p m / f t^{2} a v er a g e$

Answer: 46.6 gpm per head, 279.9 gpm total, density 0.279 gpm/ft²

Why the other options are there

272.0 gpm (pressure not square-rooted)
7.8 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Example 10

Fire sprinkler discharge from the K-factor and required system flow — Sprinkler K Factors (10)

Sprinkler heads with a K-factor of 5.6 operate at 30 psi. Compute the discharge of one head, the total flow if 11 heads in the design area operate simultaneously, and the resulting density over a 100 ft² per-head coverage.

Given

K = 5.6 gpm/psi^{0.5}
p = 30 psi
11 heads flowing
Coverage = 100 ft² per head

Find

Q per head, total flow and the design density

Start with the thinking

The K-factor already contains the orifice area and discharge coefficient, so Q = K√p is complete.
Density in gpm/ft² is what the hazard classification actually specifies.

Step-by-step solution

Formula
$Q = K p$
Substituting
$Q = 5.6 30 = 30.7 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 30.7 (11) = 337.4 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 30.7/100 = 0.307 g p m / f t^{2}$
Check
$d es i g na r e a 11 (100) = 1, 100 f t^{2} a t 0.307 g p m / f t^{2} a v er a g e$

Answer: 30.7 gpm per head, 337.4 gpm total, density 0.307 gpm/ft²

Why the other options are there

168.0 gpm (pressure not square-rooted)
2.8 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Sprinkler K Factors

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a channel, culvert or control structure, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Sprinkler K Factors contains 29 relations; you must be able to find this page in under 15 seconds.
Exam style: open-channel normal depth, specific energy, or a weir discharge.
Unit rule: Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI).
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

Manning's n is unitless but the constant is 1.486 (US) or 1.0 (SI)
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.