Hazen-Williams Equation

Hydraulics · FE Reference Handbook section

Hydraulics

15 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 Hazen-Williams Equation 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 hazen-williams equation describes physically and when it applies.
State every one of the 15 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. Hazen-Williams Equation 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: hazen-williams equation.
Capstone Studio instructional photograph

Hydraulics — Hazen-Williams Equation: 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 15 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

V	Quantity produced by "V = k1 CR H0.63 S0.54" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = k1 CAR H0.63 S 0.54" — read its definition and unit from the handbook line directly above the equation.
C	Quantity produced by "C = roughness coefficient" — read its definition and unit from the handbook line directly above the equation.
k1	Quantity produced by "k1 = 0.849 for SI units" — read its definition and unit from the handbook line directly above the equation.
RH	Quantity produced by "RH = hydraulic radius (ft or m)" — read its definition and unit from the handbook line directly above the equation.
S	Quantity produced by "S = slope of energy grade line (ft/ft or m/m) = L" — read its definition and unit from the handbook line directly above the equation.
hf	Quantity produced by "hf = Q1.852" — read its definition and unit from the handbook line directly above the equation.
L	Quantity produced by "L = pipe length (ft)" — read its definition and unit from the handbook line directly above the equation.
D	Quantity produced by "D = pipe diameter (ft)" — 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
where
Circular Pipe Head Loss Equation (Head Loss Expressed in Feet)
4.73 L
C1.852 D 4.87
where
Circular Pipe Head Loss Equation (Head Loss Expressed as Pressure)

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

Hazen-Williams head loss in a water main — Hazen-Williams Equation

A 8 in ductile iron main (C = 100) carries 1258 gpm over 1163 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 1258 gpm
D = 8 in
C = 100
L = 1163 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 125 8^{1} .85 = 542, 512$
Coefficient term
$C^{1} .85 = 10 0^{1} .85 = 5, 012$
Diameter term
$D^{4} .87 = 8^{4} .87 = 25, 006$
Substituting
$h_{f} = 10.44 \times 1163 \times 542, 512/ (5, 012 \times 25, 006)$
Evaluate
$h_{f} = 52.56 f t$
Gradient
$S = h_{f} / L = 0.04519 f t / f t$

Answer: h_f ≈ 52.6 ft over 1163 ft (S ≈ 4.519%)

Why the other options are there

52.6 ft (roughness coefficient mis-scaled)
6.11 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 2

Hazen-Williams head loss in a water main — Hazen-Williams Equation (2)

A 10 in ductile iron main (C = 140) carries 2835 gpm over 2201 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 2835 gpm
D = 10 in
C = 140
L = 2201 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 283 5^{1} .85 = 2, 439, 064$
Coefficient term
$C^{1} .85 = 14 0^{1} .85 = 9, 340$
Diameter term
$D^{4} .87 = 1 0^{4} .87 = 74, 131$
Substituting
$h_{f} = 10.44 \times 2201 \times 2, 439, 064/ (9, 340 \times 74, 131)$
Evaluate
$h_{f} = 80.95 f t$
Gradient
$S = h_{f} / L = 0.03678 f t / f t$

Answer: h_f ≈ 80.9 ft over 2201 ft (S ≈ 3.678%)

Why the other options are there

150.8 ft (roughness coefficient mis-scaled)
6.28 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 3

Hazen-Williams head loss in a water main — Hazen-Williams Equation (3)

A 8 in ductile iron main (C = 140) carries 561 gpm over 4390 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 561 gpm
D = 8 in
C = 140
L = 4390 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 56 1^{1} .85 = 121, 781$
Coefficient term
$C^{1} .85 = 14 0^{1} .85 = 9, 340$
Diameter term
$D^{4} .87 = 8^{4} .87 = 25, 006$
Substituting
$h_{f} = 10.44 \times 4390 \times 121, 781/ (9, 340 \times 25, 006)$
Evaluate
$h_{f} = 23.90 f t$
Gradient
$S = h_{f} / L = 0.00544 f t / f t$

Answer: h_f ≈ 23.9 ft over 4390 ft (S ≈ 0.544%)

Why the other options are there

44.5 ft (roughness coefficient mis-scaled)
7.34 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 4

Hazen-Williams head loss in a water main — Hazen-Williams Equation (4)

A 10 in ductile iron main (C = 100) carries 331 gpm over 1517 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 331 gpm
D = 10 in
C = 100
L = 1517 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 33 1^{1} .85 = 45, 886$
Coefficient term
$C^{1} .85 = 10 0^{1} .85 = 5, 012$
Diameter term
$D^{4} .87 = 1 0^{4} .87 = 74, 131$
Substituting
$h_{f} = 10.44 \times 1517 \times 45, 886/ (5, 012 \times 74, 131)$
Evaluate
$h_{f} = 1.96 f t$
Gradient
$S = h_{f} / L = 0.00129 f t / f t$

Answer: h_f ≈ 2.0 ft over 1517 ft (S ≈ 0.129%)

Why the other options are there

2.0 ft (roughness coefficient mis-scaled)
0.71 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 5

Hazen-Williams head loss in a water main — Hazen-Williams Equation (5)

A 6 in ductile iron main (C = 140) carries 522 gpm over 2285 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 522 gpm
D = 6 in
C = 140
L = 2285 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 52 2^{1} .85 = 106, 584$
Coefficient term
$C^{1} .85 = 14 0^{1} .85 = 9, 340$
Diameter term
$D^{4} .87 = 6^{4} .87 = 6, 160$
Substituting
$h_{f} = 10.44 \times 2285 \times 106, 584/ (9, 340 \times 6, 160)$
Evaluate
$h_{f} = 44.19 f t$
Gradient
$S = h_{f} / L = 0.01934 f t / f t$

Answer: h_f ≈ 44.2 ft over 2285 ft (S ≈ 1.934%)

Why the other options are there

82.4 ft (roughness coefficient mis-scaled)
14.44 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 6

Hazen-Williams head loss in a water main — Hazen-Williams Equation (6)

A 8 in ductile iron main (C = 100) carries 1194 gpm over 4741 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 1194 gpm
D = 8 in
C = 100
L = 4741 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 119 4^{1} .85 = 492, 559$
Coefficient term
$C^{1} .85 = 10 0^{1} .85 = 5, 012$
Diameter term
$D^{4} .87 = 8^{4} .87 = 25, 006$
Substituting
$h_{f} = 10.44 \times 4741 \times 492, 559/ (5, 012 \times 25, 006)$
Evaluate
$h_{f} = 194.5 f t$
Gradient
$S = h_{f} / L = 0.04103 f t / f t$

Answer: h_f ≈ 194.5 ft over 4741 ft (S ≈ 4.103%)

Why the other options are there

194.5 ft (roughness coefficient mis-scaled)
23.63 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 7

Hazen-Williams head loss in a water main — Hazen-Williams Equation (7)

A 10 in ductile iron main (C = 140) carries 1251 gpm over 4376 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 1251 gpm
D = 10 in
C = 140
L = 4376 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 125 1^{1} .85 = 536, 940$
Coefficient term
$C^{1} .85 = 14 0^{1} .85 = 9, 340$
Diameter term
$D^{4} .87 = 1 0^{4} .87 = 74, 131$
Substituting
$h_{f} = 10.44 \times 4376 \times 536, 940/ (9, 340 \times 74, 131)$
Evaluate
$h_{f} = 35.43 f t$
Gradient
$S = h_{f} / L = 0.00810 f t / f t$

Answer: h_f ≈ 35.4 ft over 4376 ft (S ≈ 0.810%)

Why the other options are there

66.0 ft (roughness coefficient mis-scaled)
5.51 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 8

Hazen-Williams head loss in a water main — Hazen-Williams Equation (8)

A 8 in ductile iron main (C = 100) carries 751 gpm over 4109 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 751 gpm
D = 8 in
C = 100
L = 4109 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 75 1^{1} .85 = 208, 898$
Coefficient term
$C^{1} .85 = 10 0^{1} .85 = 5, 012$
Diameter term
$D^{4} .87 = 8^{4} .87 = 25, 006$
Substituting
$h_{f} = 10.44 \times 4109 \times 208, 898/ (5, 012 \times 25, 006)$
Evaluate
$h_{f} = 71.50 f t$
Gradient
$S = h_{f} / L = 0.01740 f t / f t$

Answer: h_f ≈ 71.5 ft over 4109 ft (S ≈ 1.740%)

Why the other options are there

71.5 ft (roughness coefficient mis-scaled)
12.88 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 9

Hazen-Williams head loss in a water main — Hazen-Williams Equation (9)

A 12 in ductile iron main (C = 100) carries 596 gpm over 5388 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 596 gpm
D = 12 in
C = 100
L = 5388 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 59 6^{1} .85 = 136, 209$
Coefficient term
$C^{1} .85 = 10 0^{1} .85 = 5, 012$
Diameter term
$D^{4} .87 = 1 2^{4} .87 = 180, 141$
Substituting
$h_{f} = 10.44 \times 5388 \times 136, 209/ (5, 012 \times 180, 141)$
Evaluate
$h_{f} = 8.49 f t$
Gradient
$S = h_{f} / L = 0.00158 f t / f t$

Answer: h_f ≈ 8.5 ft over 5388 ft (S ≈ 0.158%)

Why the other options are there

8.5 ft (roughness coefficient mis-scaled)
1.86 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 10

Hazen-Williams head loss in a water main — Hazen-Williams Equation (10)

A 8 in ductile iron main (C = 140) carries 1421 gpm over 3728 ft. Find the friction head loss and the hydraulic gradient.

Given

Q = 1421 gpm
D = 8 in
C = 140
L = 3728 ft

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_{f} = 10.44 L Q^{1} .85/ (C^{1} .85 D^{4} .87)$
Flow term
$Q^{1} .85 = 142 1^{1} .85 = 679, 671$
Coefficient term
$C^{1} .85 = 14 0^{1} .85 = 9, 340$
Diameter term
$D^{4} .87 = 8^{4} .87 = 25, 006$
Substituting
$h_{f} = 10.44 \times 3728 \times 679, 671/ (9, 340 \times 25, 006)$
Evaluate
$h_{f} = 113.3 f t$
Gradient
$S = h_{f} / L = 0.03038 f t / f t$

Answer: h_f ≈ 113.3 ft over 3728 ft (S ≈ 3.038%)

Why the other options are there

211.1 ft (roughness coefficient mis-scaled)
15.80 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

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

Hazen-Williams Equation contains 15 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.