FRICTION FACTOR (f) *

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

6 formulas

10 exam-style examples

~57 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers FRICTION FACTOR (f) * within Fluid Mechanics. 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 friction factor (f) * describes physically and when it applies.
State every one of the 6 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: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. FRICTION FACTOR (f) * is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface 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 continuity plus energy, with one head-loss or force term. 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. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. 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.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: friction factor (f) *.
Capstone Studio instructional photograph

Fluid Mechanics — FRICTION FACTOR (f) *: 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 pipeline, jet or submerged surface. 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 6 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

f	Quantity produced by "f=64/Re" — read its definition and unit from the handbook line directly above the equation.
CD	Quantity produced by "CD = 24 , Re < 10" — 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.

.010
.008
.006
LAMINAR
FLOW CRITICAL Re .004
.002
.02 SM .0010
OO .0008
PIP .0006
.0004
.0002
.00010
.00006
.0000
2 .00004
.01 .000
6 8 103 2 4 6 8 104 2 4 6 8 105 2 4 6 8 106 2 3 01 6 8 107 2 4 6 8 108
* The Fanning Friction is this factor divided by 4.
ε (ft) ε (mm)
GLASS, DRAWN BRASS, COPPER, LEAD SMOOTH SMOOTH
COMMERCIAL STEEL, WROUGHT IRON 0.0001–0.0003 0.03–0.09
ASPHALTED CAST IRON 0.0002–0.0006 0.06–0.18
GALVANIZED IRON 0.0002–0.0008 0.06–0.24
CAST IRON 0.0006–0.003 0.18–0.91

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

Darcy-Weisbach head loss in a transmission main

A 300 mm main carries 0.10 m³/s over 1,200 m with f = 0.020. Compute the friction head loss.

Given

Q = 0.10 m³/s
D = 0.30 m
L = 1,200 m
f = 0.020

Find

h_f

Start with the thinking

Velocity from continuity first.
The velocity head is squared — a 10% velocity error is a 21% head error.

Step-by-step solution

Area
$A = π (0.30)^{2} /4 = 0.0707 m^{2}$
Velocity
$V = Q / A = 0.10/0.0707 = 1.41 m / s$
Velocity head
$V^{2} /2 g = 1.4 1^{2} /19.62 = 0.102 m$
Darcy-Weisbach
$h_{f} = f (L / D) (V^{2} /2 g)$
Substitute
$h_{f} = 0.020 (1, 200/0.30) (0.102) = 0.020 (4, 000) (0.102)$
Result
$h_{f} = 8.15 m$

Answer: h_f ≈ 8.15 m

Why the other options are there

0.82 m (L/D taken as 400)
16.3 m (velocity head doubled)

Reference: FE Reference Handbook — Fluid Mechanics — Darcy-Weisbach equation

Example 2

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) *

Water flows at 8.0 ft/s through 689 ft of 1.00 ft diameter pipe with f = 0.026. Find the friction head loss.

Given

L = 689 ft
D = 1.00 ft
f = 0.026
v = 8.0 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 689/1.00 = 689.0$
Velocity head
$v^{2} /2 g = (8.0)^{2} /64.4 = 0.9938 f t$
Substituting
$h_{f} = 0.026 (689.0) (0.9938)$
Evaluate
$h_{f} = 17.80 f t$

Answer: h_f ≈ 17.8 ft

Why the other options are there

35.6 ft (2g omitted)
143.3 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 3

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (2)

Water flows at 8.0 ft/s through 2270 ft of 1.25 ft diameter pipe with f = 0.024. Find the friction head loss.

Given

L = 2270 ft
D = 1.25 ft
f = 0.024
v = 8.0 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 2270/1.25 = 1, 816$
Velocity head
$v^{2} /2 g = (8.0)^{2} /64.4 = 0.9938 f t$
Substituting
$h_{f} = 0.024 (1, 816) (0.9938)$
Evaluate
$h_{f} = 43.31 f t$

Answer: h_f ≈ 43.3 ft

Why the other options are there

86.6 ft (2g omitted)
348.7 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 4

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (3)

Water flows at 3.5 ft/s through 1166 ft of 1.00 ft diameter pipe with f = 0.024. Find the friction head loss.

Given

L = 1166 ft
D = 1.00 ft
f = 0.024
v = 3.5 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 1166/1.00 = 1, 166$
Velocity head
$v^{2} /2 g = (3.5)^{2} /64.4 = 0.1902 f t$
Substituting
$h_{f} = 0.024 (1, 166) (0.1902)$
Evaluate
$h_{f} = 5.32 f t$

Answer: h_f ≈ 5.3 ft

Why the other options are there

10.6 ft (2g omitted)
97.9 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 5

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (4)

Water flows at 3.5 ft/s through 2998 ft of 0.50 ft diameter pipe with f = 0.027. Find the friction head loss.

Given

L = 2998 ft
D = 0.50 ft
f = 0.027
v = 3.5 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 2998/0.50 = 5, 996$
Velocity head
$v^{2} /2 g = (3.5)^{2} /64.4 = 0.1902 f t$
Substituting
$h_{f} = 0.027 (5, 996) (0.1902)$
Evaluate
$h_{f} = 30.79 f t$

Answer: h_f ≈ 30.8 ft

Why the other options are there

61.6 ft (2g omitted)
566.6 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 6

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (5)

Water flows at 5.0 ft/s through 3438 ft of 1.25 ft diameter pipe with f = 0.021. Find the friction head loss.

Given

L = 3438 ft
D = 1.25 ft
f = 0.021
v = 5.0 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 3438/1.25 = 2, 750$
Velocity head
$v^{2} /2 g = (5.0)^{2} /64.4 = 0.3882 f t$
Substituting
$h_{f} = 0.021 (2, 750) (0.3882)$
Evaluate
$h_{f} = 22.42 f t$

Answer: h_f ≈ 22.4 ft

Why the other options are there

44.8 ft (2g omitted)
288.8 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 7

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (6)

Water flows at 4.5 ft/s through 1026 ft of 1.25 ft diameter pipe with f = 0.021. Find the friction head loss.

Given

L = 1026 ft
D = 1.25 ft
f = 0.021
v = 4.5 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 1026/1.25 = 820.8$
Velocity head
$v^{2} /2 g = (4.5)^{2} /64.4 = 0.3144 f t$
Substituting
$h_{f} = 0.021 (820.8) (0.3144)$
Evaluate
$h_{f} = 5.42 f t$

Answer: h_f ≈ 5.4 ft

Why the other options are there

10.8 ft (2g omitted)
77.6 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 8

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (7)

Water flows at 5.5 ft/s through 844 ft of 1.25 ft diameter pipe with f = 0.024. Find the friction head loss.

Given

L = 844 ft
D = 1.25 ft
f = 0.024
v = 5.5 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 844/1.25 = 675.2$
Velocity head
$v^{2} /2 g = (5.5)^{2} /64.4 = 0.4697 f t$
Substituting
$h_{f} = 0.024 (675.2) (0.4697)$
Evaluate
$h_{f} = 7.61 f t$

Answer: h_f ≈ 7.6 ft

Why the other options are there

15.2 ft (2g omitted)
89.1 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 9

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (8)

Water flows at 4.5 ft/s through 1389 ft of 1.00 ft diameter pipe with f = 0.027. Find the friction head loss.

Given

L = 1389 ft
D = 1.00 ft
f = 0.027
v = 4.5 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 1389/1.00 = 1, 389$
Velocity head
$v^{2} /2 g = (4.5)^{2} /64.4 = 0.3144 f t$
Substituting
$h_{f} = 0.027 (1, 389) (0.3144)$
Evaluate
$h_{f} = 11.79 f t$

Answer: h_f ≈ 11.8 ft

Why the other options are there

23.6 ft (2g omitted)
168.8 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

Example 10

Darcy–Weisbach friction head loss — FRICTION FACTOR (f) * (9)

Water flows at 5.0 ft/s through 2026 ft of 0.75 ft diameter pipe with f = 0.030. Find the friction head loss.

Given

L = 2026 ft
D = 0.75 ft
f = 0.030
v = 5.0 ft/s

Find

h_f

Start with the thinking

Head loss varies with the square of velocity.
Diameter appears in the denominator — bigger pipe, far less loss.

Step-by-step solution

Darcy–Weisbach
$h_{f} = f (L / D) (v^{2} /2 g)$
Length ratio
$L / D = 2026/0.75 = 2, 701$
Velocity head
$v^{2} /2 g = (5.0)^{2} /64.4 = 0.3882 f t$
Substituting
$h_{f} = 0.030 (2, 701) (0.3882)$
Evaluate
$h_{f} = 31.46 f t$

Answer: h_f ≈ 31.5 ft

Why the other options are there

62.9 ft (2g omitted)
405.2 ft (velocity not squared)

Reference: FE Reference Handbook — Fluid Mechanics → FRICTION FACTOR (f) *

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 pipeline, jet or submerged surface, 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

FRICTION FACTOR (f) * contains 6 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
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.