Fire Hydrant Discharging to Atmosphere

Hydraulics · FE Reference Handbook section

Hydraulics

5 formulas

10 exam-style examples

~55 min

All Hydraulics lectures

Learning objectives

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

This chapter section covers Fire Hydrant Discharging to Atmosphere 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 hydrant discharging to atmosphere describes physically and when it applies.
State every one of the 5 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 Hydrant Discharging to Atmosphere 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 hydrant discharging to atmosphere.
Capstone Studio instructional photograph

Hydraulics — Fire Hydrant Discharging to Atmosphere: 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 5 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 = 29.8 D2Cd P1/2" — read its definition and unit from the handbook line directly above the equation.
D	Quantity produced by "D = outlet diameter (in.)" — read its definition and unit from the handbook line directly above the equation.
P	Quantity produced by "P = pressure detected by pitot gauge (psi)" — read its definition and unit from the handbook line directly above the equation.
Cd	Quantity produced by "Cd = hydrant coefficient based on hydrant outlet geometry" — 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
OUTLET SMOOTH OUTLET SQUARE OUTLET SQUARE AND
AND WELL-ROUNDED AND SHARP PROJECTING INTO BARREL
COEF. 0.90 COEF. 0.80 COEF. 0.70
NFPA Standard 291, Recommended Practice for Fire Flow Testing and Marking of Hydrants, Section 4.10.1.2

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 hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere

A hydrant outlet 2.5 in. in diameter with a coefficient of 0.97 shows a pitot pressure of 35 psi. Static pressure is 71 psi and the residual pressure during the test is 25 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 2.5 in., C = 0.97
Pitot pressure = 35 psi
Static = 71 psi, residual = 25 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.97) (2.5)^{2} 35 = 1, 070 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 51 p s i; s t a t i c - r es i d u a l = 46 p s i$
Substituting
$Q_{20} = 1, 070 (51/46)^{0.54} = 1, 131 g p m$

Answer: Measured Q = 1,070 gpm; available at 20 psi ≈ 1,131 gpm

Why the other options are there

428.0 gpm (diameter not squared)
1,186 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 2

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (2)

A hydrant outlet 2.5 in. in diameter with a coefficient of 0.90 shows a pitot pressure of 42 psi. Static pressure is 67 psi and the residual pressure during the test is 33 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 2.5 in., C = 0.90
Pitot pressure = 42 psi
Static = 67 psi, residual = 33 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.90) (2.5)^{2} 42 = 1, 087 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 47 p s i; s t a t i c - r es i d u a l = 34 p s i$
Substituting
$Q_{20} = 1, 087 (47/34)^{0.54} = 1, 295 g p m$

Answer: Measured Q = 1,087 gpm; available at 20 psi ≈ 1,295 gpm

Why the other options are there

435.0 gpm (diameter not squared)
1,503 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 3

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (3)

A hydrant outlet 4.0 in. in diameter with a coefficient of 0.97 shows a pitot pressure of 39 psi. Static pressure is 55 psi and the residual pressure during the test is 27 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 4.0 in., C = 0.97
Pitot pressure = 39 psi
Static = 55 psi, residual = 27 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.97) (4.0)^{2} 39 = 2, 891 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 35 p s i; s t a t i c - r es i d u a l = 28 p s i$
Substituting
$Q_{20} = 2, 891 (35/28)^{0.54} = 3, 261 g p m$

Answer: Measured Q = 2,891 gpm; available at 20 psi ≈ 3,261 gpm

Why the other options are there

722.8 gpm (diameter not squared)
3,614 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 4

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (4)

A hydrant outlet 4.5 in. in diameter with a coefficient of 0.97 shows a pitot pressure of 32 psi. Static pressure is 43 psi and the residual pressure during the test is 21 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 4.5 in., C = 0.97
Pitot pressure = 32 psi
Static = 43 psi, residual = 21 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.97) (4.5)^{2} 32 = 3, 315 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 23 p s i; s t a t i c - r es i d u a l = 22 p s i$
Substituting
$Q_{20} = 3, 315 (23/22)^{0.54} = 3, 395 g p m$

Answer: Measured Q = 3,315 gpm; available at 20 psi ≈ 3,395 gpm

Why the other options are there

736.6 gpm (diameter not squared)
3,465 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 5

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (5)

A hydrant outlet 2.5 in. in diameter with a coefficient of 0.97 shows a pitot pressure of 34 psi. Static pressure is 68 psi and the residual pressure during the test is 30 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 2.5 in., C = 0.97
Pitot pressure = 34 psi
Static = 68 psi, residual = 30 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.97) (2.5)^{2} 34 = 1, 054 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 48 p s i; s t a t i c - r es i d u a l = 38 p s i$
Substituting
$Q_{20} = 1, 054 (48/38)^{0.54} = 1, 196 g p m$

Answer: Measured Q = 1,054 gpm; available at 20 psi ≈ 1,196 gpm

Why the other options are there

421.8 gpm (diameter not squared)
1,332 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 6

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (6)

A hydrant outlet 4.5 in. in diameter with a coefficient of 0.90 shows a pitot pressure of 61 psi. Static pressure is 99 psi and the residual pressure during the test is 33 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 4.5 in., C = 0.90
Pitot pressure = 61 psi
Static = 99 psi, residual = 33 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.90) (4.5)^{2} 61 = 4, 246 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 79 p s i; s t a t i c - r es i d u a l = 66 p s i$
Substituting
$Q_{20} = 4, 246 (79/66)^{0.54} = 4, 679 g p m$

Answer: Measured Q = 4,246 gpm; available at 20 psi ≈ 4,679 gpm

Why the other options are there

943.6 gpm (diameter not squared)
5,082 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 7

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (7)

A hydrant outlet 2.5 in. in diameter with a coefficient of 0.90 shows a pitot pressure of 39 psi. Static pressure is 79 psi and the residual pressure during the test is 35 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 2.5 in., C = 0.90
Pitot pressure = 39 psi
Static = 79 psi, residual = 35 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.90) (2.5)^{2} 39 = 1, 048 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 59 p s i; s t a t i c - r es i d u a l = 44 p s i$
Substituting
$Q_{20} = 1, 048 (59/44)^{0.54} = 1, 228 g p m$

Answer: Measured Q = 1,048 gpm; available at 20 psi ≈ 1,228 gpm

Why the other options are there

419.1 gpm (diameter not squared)
1,405 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 8

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (8)

A hydrant outlet 4.0 in. in diameter with a coefficient of 0.97 shows a pitot pressure of 34 psi. Static pressure is 59 psi and the residual pressure during the test is 25 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 4.0 in., C = 0.97
Pitot pressure = 34 psi
Static = 59 psi, residual = 25 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.97) (4.0)^{2} 34 = 2, 700 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 39 p s i; s t a t i c - r es i d u a l = 34 p s i$
Substituting
$Q_{20} = 2, 700 (39/34)^{0.54} = 2, 907 g p m$

Answer: Measured Q = 2,700 gpm; available at 20 psi ≈ 2,907 gpm

Why the other options are there

674.9 gpm (diameter not squared)
3,096 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 9

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (9)

A hydrant outlet 2.5 in. in diameter with a coefficient of 0.80 shows a pitot pressure of 55 psi. Static pressure is 88 psi and the residual pressure during the test is 44 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 2.5 in., C = 0.80
Pitot pressure = 55 psi
Static = 88 psi, residual = 44 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.80) (2.5)^{2} 55 = 1, 106 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 68 p s i; s t a t i c - r es i d u a l = 44 p s i$
Substituting
$Q_{20} = 1, 106 (68/44)^{0.54} = 1, 399 g p m$

Answer: Measured Q = 1,106 gpm; available at 20 psi ≈ 1,399 gpm

Why the other options are there

442.5 gpm (diameter not squared)
1,709 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

Example 10

Fire hydrant discharge and available fire flow at 20 psi residual — Fire Hydrant Discharging to Atmosphere (10)

A hydrant outlet 2.5 in. in diameter with a coefficient of 0.90 shows a pitot pressure of 41 psi. Static pressure is 75 psi and the residual pressure during the test is 42 psi. Compute the measured discharge and the flow available at a 20 psi residual.

Given

d = 2.5 in., C = 0.90
Pitot pressure = 41 psi
Static = 75 psi, residual = 42 psi

Find

Measured Q and Q available at 20 psi

Start with the thinking

Hydrant flow testing uses Q = 29.83 C d²√p with d in inches, p in psi and Q in gpm.
The 0.54 exponent projects the test to the standard 20 psi residual used for fire-flow ratings.

Step-by-step solution

Formula
$Q = 29.83 C d^{2} p$
Substituting
$Q = 29.83 (0.90) (2.5)^{2} 41 = 1, 074 g p m$
Formula
$Q_{20} = Q_{t es t} (\frac{p _{s} - 20}{p _{s} - p _{r}})^{0.54}$
Pressure drops
$s t a t i c - 20 = 55 p s i; s t a t i c - r es i d u a l = 33 p s i$
Substituting
$Q_{20} = 1, 074 (55/33)^{0.54} = 1, 416 g p m$

Answer: Measured Q = 1,074 gpm; available at 20 psi ≈ 1,416 gpm

Why the other options are there

429.8 gpm (diameter not squared)
1,791 gpm (0.54 exponent dropped)

Reference: FE Reference Handbook — Hydraulics → Fire Hydrant Discharging to Atmosphere

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 Hydrant Discharging to Atmosphere contains 5 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.