SI Units

Hydraulics · FE Reference Handbook section

Hydraulics

11 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 SI Units 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 si units describes physically and when it applies.
State every one of the 11 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. SI Units 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: si units.
Capstone Studio instructional photograph

Hydraulics — SI Units: 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 11 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

P	Quantity produced by "P = pressure loss (bars per meter of pipe)" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = flow (liters/minute)" — read its definition and unit from the handbook line directly above the equation.
D	Quantity produced by "D = pipe diameter (mm)" — read its definition and unit from the handbook line directly above the equation.
QR	Quantity produced by "QR = QF × (HR/HF)0.54" — read its definition and unit from the handbook line directly above the equation.
QF	Quantity produced by "QF = total test flow" — read its definition and unit from the handbook line directly above the equation.
HR	Quantity produced by "HR = PS - 20 psi" — read its definition and unit from the handbook line directly above the equation.
HF	Quantity produced by "HF = PS - PR" — read its definition and unit from the handbook line directly above the equation.
PS	Quantity produced by "PS = static pressure" — read its definition and unit from the handbook line directly above the equation.
PR	Quantity produced by "PR = residual pressure" — 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.

6.05 Q1.85 5
where
Values of Hazen-Williams Coefficient C
Pipe Material C
Ductile iron 140
Concrete (regardless of age) 130
Cast iron:
New 130
5 yr old 120
20 yr old 100
Welded steel, new 120
Wood stave (regardless of age) 120
Vitrified clay 110
Riveted steel, new 110
Brick sewers 100
Asbestos-cement 140
Plastic 150
Formula for Calculating Rated Capacity at 20 psi from Fire Hydrant
where
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

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units

A 150.0 mm main (C = 140) carries 0.140 m³/s over 1182 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 150.0 mm = 0.49 ft
Q = 0.140 m³/s = 2,219 gpm
L = 1182 m = 3,878 ft
C = 140

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.140/0.01767 = 7.92 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0375 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.30586 m / m$
Head loss
$h_{f} = S L = 0.30586 (1182) = 361.5 m$
Unit check
$361.5 m \times 3.281 = 1, 186 f t o v er 3, 878 f t o f p i p e$

Answer: h_f = 361.5 m (1,186 ft); Q = 2,219 gpm

Why the other options are there

561.2 m (U.S. constant used with SI data)
110.2 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 2

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (2)

A 150.0 mm main (C = 100) carries 0.030 m³/s over 1108 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 150.0 mm = 0.49 ft
Q = 0.030 m³/s = 475.5 gpm
L = 1108 m = 3,635 ft
C = 100

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.030/0.01767 = 1.70 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0375 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.03290 m / m$
Head loss
$h_{f} = S L = 0.03290 (1108) = 36.46 m$
Unit check
$36.46 m \times 3.281 = 119.6 f t o v er 3, 635 f t o f p i p e$

Answer: h_f = 36.46 m (119.6 ft); Q = 475.5 gpm

Why the other options are there

56.59 m (U.S. constant used with SI data)
11.11 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 3

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (3)

A 300.0 mm main (C = 100) carries 0.080 m³/s over 989 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 300.0 mm = 0.98 ft
Q = 0.080 m³/s = 1,268 gpm
L = 989 m = 3,245 ft
C = 100

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.080/0.07069 = 1.13 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0750 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.00692 m / m$
Head loss
$h_{f} = S L = 0.00692 (989) = 6.84 m$
Unit check
$6.84 m \times 3.281 = 22.4 f t o v er 3, 245 f t o f p i p e$

Answer: h_f = 6.84 m (22.4 ft); Q = 1,268 gpm

Why the other options are there

10.62 m (U.S. constant used with SI data)
2.09 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 4

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (4)

A 250.0 mm main (C = 120) carries 0.050 m³/s over 1380 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 250.0 mm = 0.82 ft
Q = 0.050 m³/s = 792.5 gpm
L = 1380 m = 4,528 ft
C = 120

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.050/0.04909 = 1.02 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0625 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.00502 m / m$
Head loss
$h_{f} = S L = 0.00502 (1380) = 6.93 m$
Unit check
$6.93 m \times 3.281 = 22.7 f t o v er 4, 528 f t o f p i p e$

Answer: h_f = 6.93 m (22.7 ft); Q = 792.5 gpm

Why the other options are there

10.76 m (U.S. constant used with SI data)
2.11 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 5

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (5)

A 250.0 mm main (C = 130) carries 0.140 m³/s over 1488 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 250.0 mm = 0.82 ft
Q = 0.140 m³/s = 2,219 gpm
L = 1488 m = 4,882 ft
C = 130

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.140/0.04909 = 2.85 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0625 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.02915 m / m$
Head loss
$h_{f} = S L = 0.02915 (1488) = 43.37 m$
Unit check
$43.37 m \times 3.281 = 142.3 f t o v er 4, 882 f t o f p i p e$

Answer: h_f = 43.37 m (142.3 ft); Q = 2,219 gpm

Why the other options are there

67.34 m (U.S. constant used with SI data)
13.22 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 6

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (6)

A 250.0 mm main (C = 140) carries 0.170 m³/s over 1088 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 250.0 mm = 0.82 ft
Q = 0.170 m³/s = 2,695 gpm
L = 1088 m = 3,570 ft
C = 140

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.170/0.04909 = 3.46 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0625 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.03641 m / m$
Head loss
$h_{f} = S L = 0.03641 (1088) = 39.61 m$
Unit check
$39.61 m \times 3.281 = 130.0 f t o v er 3, 570 f t o f p i p e$

Answer: h_f = 39.61 m (130.0 ft); Q = 2,695 gpm

Why the other options are there

61.49 m (U.S. constant used with SI data)
12.07 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 7

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (7)

A 200.0 mm main (C = 130) carries 0.030 m³/s over 1442 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 200.0 mm = 0.66 ft
Q = 0.030 m³/s = 475.5 gpm
L = 1442 m = 4,731 ft
C = 130

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.030/0.03142 = 0.95 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0500 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.00499 m / m$
Head loss
$h_{f} = S L = 0.00499 (1442) = 7.19 m$
Unit check
$7.19 m \times 3.281 = 23.6 f t o v er 4, 731 f t o f p i p e$

Answer: h_f = 7.19 m (23.6 ft); Q = 475.5 gpm

Why the other options are there

11.16 m (U.S. constant used with SI data)
2.19 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 8

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (8)

A 250.0 mm main (C = 140) carries 0.090 m³/s over 731 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 250.0 mm = 0.82 ft
Q = 0.090 m³/s = 1,427 gpm
L = 731 m = 2,398 ft
C = 140

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.090/0.04909 = 1.83 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0625 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.01121 m / m$
Head loss
$h_{f} = S L = 0.01121 (731) = 8.20 m$
Unit check
$8.20 m \times 3.281 = 26.9 f t o v er 2, 398 f t o f p i p e$

Answer: h_f = 8.20 m (26.9 ft); Q = 1,427 gpm

Why the other options are there

12.72 m (U.S. constant used with SI data)
2.50 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 9

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (9)

A 250.0 mm main (C = 130) carries 0.020 m³/s over 629 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 250.0 mm = 0.82 ft
Q = 0.020 m³/s = 317.0 gpm
L = 629 m = 2,064 ft
C = 130

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.020/0.04909 = 0.41 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0625 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.00079 m / m$
Head loss
$h_{f} = S L = 0.00079 (629) = 0.50 m$
Unit check
$0.50 m \times 3.281 = 1.6 f t o v er 2, 064 f t o f p i p e$

Answer: h_f = 0.50 m (1.6 ft); Q = 317.0 gpm

Why the other options are there

0.77 m (U.S. constant used with SI data)
0.15 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

Example 10

Hazen-Williams head loss computed in SI and in U.S. customary units — SI Units (10)

A 350.0 mm main (C = 130) carries 0.090 m³/s over 949 m. Compute the head loss with the SI Hazen-Williams form, then convert the data to U.S. customary units and confirm the loss in feet.

Given

D = 350.0 mm = 1.15 ft
Q = 0.090 m³/s = 1,427 gpm
L = 949 m = 3,114 ft
C = 130

Find

Head loss in metres and in feet

Start with the thinking

Hazen-Williams uses a different leading constant in each unit system: 0.849 in SI, 1.318 in U.S. customary — the exponents never change.
Convert the answer, not the constant, when you need the other unit system.

Step-by-step solution

Velocity
$V = Q / A = 0.090/0.09621 = 0.94 m / s$
Formula (SI)
$V = 0.849 C R^{0.63} S^{0.54}$
Hydraulic radius
$R = D /4 = 0.0875 m$
Solving for S
$S = [V / (0.849 C R^{0.63})]^{1/0.54} = 0.00250 m / m$
Head loss
$h_{f} = S L = 0.00250 (949) = 2.37 m$
Unit check
$2.37 m \times 3.281 = 7.8 f t o v er 3, 114 f t o f p i p e$

Answer: h_f = 2.37 m (7.8 ft); Q = 1,427 gpm

Why the other options are there

3.68 m (U.S. constant used with SI data)
0.72 ft (conversion inverted)

Reference: FE Reference Handbook — Hydraulics → SI Units

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

SI Units contains 11 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.

SI Units

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

P= 1.85 4.87 # 10Needs technical review

P=pressureloss(barspermeterofpipe)

Q=flow(liters/minute)

D=pipediameter(mm)

QR=QF×(HR/HF)0.54

QR=ratedcapacity(gpm)at20psi

QF=totaltestflow

HR=PS−20psi

HF=PS−PR

PS=staticpressure

PR=residualpressure

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section