Fire Sprinkler Discharge

Hydraulics · FE Reference Handbook section

Hydraulics

4 formulas

10 exam-style examples

~53 min

All Hydraulics lectures

Learning objectives

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

This chapter section covers Fire Sprinkler Discharge 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 fire sprinkler discharge describes physically and when it applies.
State every one of the 4 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. Fire Sprinkler Discharge 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: fire sprinkler discharge.
Capstone Studio instructional photograph

Hydraulics — Fire Sprinkler Discharge: 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 4 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

Q	Quantity produced by "Q = KP1/2" — read its definition and unit from the handbook line directly above the equation.
K	Quantity produced by "K = measure of the ease of getting water out of the orifice, related to size and shape of the orifice in units of" — read its definition and unit from the handbook line directly above the equation.
P	Quantity produced by "P = pressure (psi)" — 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.

where
gpm per (psi)1/2

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 — Fire Sprinkler Discharge

Sprinkler heads with a K-factor of 8.0 operate at 19 psi. Compute the discharge of one head, the total flow if 10 heads in the design area operate simultaneously, and the resulting density over a 166 ft² per-head coverage.

Given

K = 8.0 gpm/psi^{0.5}
p = 19 psi
10 heads flowing
Coverage = 166 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
$Q = 8.0 19 = 34.9 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 34.9 (10) = 348.7 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 34.9/166 = 0.210 g p m / f t^{2}$
Check
$d es i g na r e a 10 (166) = 1, 660 f t^{2} a t 0.210 g p m / f t^{2} a v er a g e$

Answer: 34.9 gpm per head, 348.7 gpm total, density 0.210 gpm/ft²

Why the other options are there

152.0 gpm (pressure not square-rooted)
3.5 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 2

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

Sprinkler heads with a K-factor of 14.0 operate at 16 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 187 ft² per-head coverage.

Given

K = 14.0 gpm/psi^{0.5}
p = 16 psi
11 heads flowing
Coverage = 187 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
$Q = 14.0 16 = 56.0 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 56.0 (11) = 616.0 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 56.0/187 = 0.299 g p m / f t^{2}$
Check
$d es i g na r e a 11 (187) = 2, 057 f t^{2} a t 0.299 g p m / f t^{2} a v er a g e$

Answer: 56.0 gpm per head, 616.0 gpm total, density 0.299 gpm/ft²

Why the other options are there

224.0 gpm (pressure not square-rooted)
5.1 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 3

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

Sprinkler heads with a K-factor of 5.6 operate at 41 psi. Compute the discharge of one head, the total flow if 13 heads in the design area operate simultaneously, and the resulting density over a 129 ft² per-head coverage.

Given

K = 5.6 gpm/psi^{0.5}
p = 41 psi
13 heads flowing
Coverage = 129 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
$Q = 5.6 41 = 35.9 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 35.9 (13) = 466.1 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 35.9/129 = 0.278 g p m / f t^{2}$
Check
$d es i g na r e a 13 (129) = 1, 677 f t^{2} a t 0.278 g p m / f t^{2} a v er a g e$

Answer: 35.9 gpm per head, 466.1 gpm total, density 0.278 gpm/ft²

Why the other options are there

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

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 4

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

Sprinkler heads with a K-factor of 11.2 operate at 43 psi. Compute the discharge of one head, the total flow if 13 heads in the design area operate simultaneously, and the resulting density over a 165 ft² per-head coverage.

Given

K = 11.2 gpm/psi^{0.5}
p = 43 psi
13 heads flowing
Coverage = 165 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
$Q = 11.2 43 = 73.4 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 73.4 (13) = 954.8 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 73.4/165 = 0.445 g p m / f t^{2}$
Check
$d es i g na r e a 13 (165) = 2, 145 f t^{2} a t 0.445 g p m / f t^{2} a v er a g e$

Answer: 73.4 gpm per head, 954.8 gpm total, density 0.445 gpm/ft²

Why the other options are there

481.6 gpm (pressure not square-rooted)
5.6 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 5

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

Sprinkler heads with a K-factor of 14.0 operate at 60 psi. Compute the discharge of one head, the total flow if 13 heads in the design area operate simultaneously, and the resulting density over a 106 ft² per-head coverage.

Given

K = 14.0 gpm/psi^{0.5}
p = 60 psi
13 heads flowing
Coverage = 106 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
$Q = 14.0 60 = 108.4 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 108.4 (13) = 1, 410 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 108.4/106 = 1.023 g p m / f t^{2}$
Check
$d es i g na r e a 13 (106) = 1, 378 f t^{2} a t 1.023 g p m / f t^{2} a v er a g e$

Answer: 108.4 gpm per head, 1,410 gpm total, density 1.023 gpm/ft²

Why the other options are there

840.0 gpm (pressure not square-rooted)
8.3 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 6

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

Sprinkler heads with a K-factor of 5.6 operate at 54 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 157 ft² per-head coverage.

Given

K = 5.6 gpm/psi^{0.5}
p = 54 psi
9 heads flowing
Coverage = 157 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
$Q_{t} o t a l = 41.2 (9) = 370.4 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 41.2/157 = 0.262 g p m / f t^{2}$
Check
$d es i g na r e a 9 (157) = 1, 413 f t^{2} a t 0.262 g p m / f t^{2} a v er a g e$

Answer: 41.2 gpm per head, 370.4 gpm total, density 0.262 gpm/ft²

Why the other options are there

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

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 7

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

Sprinkler heads with a K-factor of 14.0 operate at 39 psi. Compute the discharge of one head, the total flow if 14 heads in the design area operate simultaneously, and the resulting density over a 100 ft² per-head coverage.

Given

K = 14.0 gpm/psi^{0.5}
p = 39 psi
14 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
$Q = 14.0 39 = 87.4 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 87.4 (14) = 1, 224 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 87.4/100 = 0.874 g p m / f t^{2}$
Check
$d es i g na r e a 14 (100) = 1, 400 f t^{2} a t 0.874 g p m / f t^{2} a v er a g e$

Answer: 87.4 gpm per head, 1,224 gpm total, density 0.874 gpm/ft²

Why the other options are there

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

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 8

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

Sprinkler heads with a K-factor of 5.6 operate at 49 psi. Compute the discharge of one head, the total flow if 10 heads in the design area operate simultaneously, and the resulting density over a 98 ft² per-head coverage.

Given

K = 5.6 gpm/psi^{0.5}
p = 49 psi
10 heads flowing
Coverage = 98 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
$Q = 5.6 49 = 39.2 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 39.2 (10) = 392.0 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 39.2/98 = 0.400 g p m / f t^{2}$
Check
$d es i g na r e a 10 (98) = 980.0 f t^{2} a t 0.400 g p m / f t^{2} a v er a g e$

Answer: 39.2 gpm per head, 392.0 gpm total, density 0.400 gpm/ft²

Why the other options are there

274.4 gpm (pressure not square-rooted)
3.9 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 9

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

Sprinkler heads with a K-factor of 11.2 operate at 43 psi. Compute the discharge of one head, the total flow if 12 heads in the design area operate simultaneously, and the resulting density over a 101 ft² per-head coverage.

Given

K = 11.2 gpm/psi^{0.5}
p = 43 psi
12 heads flowing
Coverage = 101 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
$Q = 11.2 43 = 73.4 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 73.4 (12) = 881.3 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 73.4/101 = 0.727 g p m / f t^{2}$
Check
$d es i g na r e a 12 (101) = 1, 212 f t^{2} a t 0.727 g p m / f t^{2} a v er a g e$

Answer: 73.4 gpm per head, 881.3 gpm total, density 0.727 gpm/ft²

Why the other options are there

481.6 gpm (pressure not square-rooted)
6.1 gpm total (divided instead of multiplied)

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

Example 10

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

Sprinkler heads with a K-factor of 8.0 operate at 44 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 195 ft² per-head coverage.

Given

K = 8.0 gpm/psi^{0.5}
p = 44 psi
11 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
$Q = 8.0 44 = 53.1 g p m p er h e a d$
Total system flow
$Q_{t} o t a l = 53.1 (11) = 583.7 g p m$
Formula
$density = \frac{Q}{A _{co v er a g e}}$
Substituting
$d e n s i t y = 53.1/195 = 0.272 g p m / f t^{2}$
Check
$d es i g na r e a 11 (195) = 2, 145 f t^{2} a t 0.272 g p m / f t^{2} a v er a g e$

Answer: 53.1 gpm per head, 583.7 gpm total, density 0.272 gpm/ft²

Why the other options are there

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

Reference: FE Reference Handbook — Hydraulics → Fire Sprinkler Discharge

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

Fire Sprinkler Discharge contains 4 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.