Definitions

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

16 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Definitions 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 definitions describes physically and when it applies.
State every one of the 16 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. Definitions 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: definitions.
Capstone Studio instructional photograph

Fluid Mechanics — Definitions: 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 16 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

t	Quantity produced by "t = limit Dm DV" — read its definition and unit from the handbook line directly above the equation.
c	Quantity produced by "c = limit DW DV" — read its definition and unit from the handbook line directly above the equation.
SG	Quantity produced by "SG = γ/γw = ρ/ρw" — read its definition and unit from the handbook line directly above the equation.
∆m	Quantity produced by "∆m = mass of infinitesimal volume" — read its definition and unit from the handbook line directly above the equation.
∆V	Quantity produced by "∆V = volume of infinitesimal object considered" — read its definition and unit from the handbook line directly above the equation.
γ	Quantity produced by "γ = specific weight" — read its definition and unit from the handbook line directly above the equation.
∆W	Quantity produced by "∆W = weight of an infinitesimal volume" — read its definition and unit from the handbook line directly above the equation.
tw	Quantity produced by "tw = density of water at standard conditions" — read its definition and unit from the handbook line directly above the equation.
γw	Quantity produced by "γw = specific weight of water at standard conditions" — 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.

Density, Specific Volume, Specific Weight, and Specific Gravity
The definitions of density, specific weight, and specific gravity follow:
DV " 0
DV " 0
DV " 0
also
where

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

Density, specific weight, specific gravity and viscosity — Definitions

A liquid has a specific gravity of 0.90 and a dynamic viscosity of 0.00150 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 2.5 m³.

Given

SG = 0.90
μ = 0.00150 Pa·s
Volume = 2.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 0.90 (1000) = 900.0 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 900.0 (9.81) = 8, 829 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/900.0 = 1.111 e - 3 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00150/900.0 = 1.667 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 900.0 (2.5) = 2, 250 k g$

Answer: ρ = 900.0 kg/m³, γ = 8,829 N/m³, ν = 1.67e-6 m²/s, m = 2,250 kg

Why the other options are there

ρ = 0.90 kg/m³ (SG reported as density)
ν = 1.350 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 2

Density, specific weight, specific gravity and viscosity — Definitions (2)

A liquid has a specific gravity of 0.97 and a dynamic viscosity of 0.00200 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 2.0 m³.

Given

SG = 0.97
μ = 0.00200 Pa·s
Volume = 2.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 0.97 (1000) = 970.0 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 970.0 (9.81) = 9, 516 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/970.0 = 1.031 e - 3 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00200/970.0 = 2.062 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 970.0 (2.0) = 1, 940 k g$

Answer: ρ = 970.0 kg/m³, γ = 9,516 N/m³, ν = 2.06e-6 m²/s, m = 1,940 kg

Why the other options are there

ρ = 0.97 kg/m³ (SG reported as density)
ν = 1.940 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 3

Density, specific weight, specific gravity and viscosity — Definitions (3)

A liquid has a specific gravity of 1.16 and a dynamic viscosity of 0.00180 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 4.0 m³.

Given

SG = 1.16
μ = 0.00180 Pa·s
Volume = 4.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.16 (1000) = 1, 160 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 160 (9.81) = 11, 380 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 160 = 8.621 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00180/1, 160 = 1.552 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 160 (4.0) = 4, 640 k g$

Answer: ρ = 1,160 kg/m³, γ = 11,380 N/m³, ν = 1.55e-6 m²/s, m = 4,640 kg

Why the other options are there

ρ = 1.16 kg/m³ (SG reported as density)
ν = 2.088 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 4

Density, specific weight, specific gravity and viscosity — Definitions (4)

A liquid has a specific gravity of 1.42 and a dynamic viscosity of 0.00110 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 3.0 m³.

Given

SG = 1.42
μ = 0.00110 Pa·s
Volume = 3.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.42 (1000) = 1, 420 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 420 (9.81) = 13, 930 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 420 = 7.042 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00110/1, 420 = 7.746 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 420 (3.0) = 4, 260 k g$

Answer: ρ = 1,420 kg/m³, γ = 13,930 N/m³, ν = 7.75e-7 m²/s, m = 4,260 kg

Why the other options are there

ρ = 1.42 kg/m³ (SG reported as density)
ν = 1.562 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 5

Density, specific weight, specific gravity and viscosity — Definitions (5)

A liquid has a specific gravity of 1.21 and a dynamic viscosity of 0.00100 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 4.5 m³.

Given

SG = 1.21
μ = 0.00100 Pa·s
Volume = 4.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.21 (1000) = 1, 210 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 210 (9.81) = 11, 870 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 210 = 8.264 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00100/1, 210 = 8.264 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 210 (4.5) = 5, 445 k g$

Answer: ρ = 1,210 kg/m³, γ = 11,870 N/m³, ν = 8.26e-7 m²/s, m = 5,445 kg

Why the other options are there

ρ = 1.21 kg/m³ (SG reported as density)
ν = 1.210 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 6

Density, specific weight, specific gravity and viscosity — Definitions (6)

A liquid has a specific gravity of 1.47 and a dynamic viscosity of 0.00140 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 1.0 m³.

Given

SG = 1.47
μ = 0.00140 Pa·s
Volume = 1.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.47 (1000) = 1, 470 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 470 (9.81) = 14, 421 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 470 = 6.803 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00140/1, 470 = 9.524 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 470 (1.0) = 1, 470 k g$

Answer: ρ = 1,470 kg/m³, γ = 14,421 N/m³, ν = 9.52e-7 m²/s, m = 1,470 kg

Why the other options are there

ρ = 1.47 kg/m³ (SG reported as density)
ν = 2.058 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 7

Density, specific weight, specific gravity and viscosity — Definitions (7)

A liquid has a specific gravity of 0.81 and a dynamic viscosity of 0.00060 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 4.0 m³.

Given

SG = 0.81
μ = 0.00060 Pa·s
Volume = 4.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 0.81 (1000) = 810.0 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 810.0 (9.81) = 7, 946 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/810.0 = 1.235 e - 3 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00060/810.0 = 7.407 e - 7 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 810.0 (4.0) = 3, 240 k g$

Answer: ρ = 810.0 kg/m³, γ = 7,946 N/m³, ν = 7.41e-7 m²/s, m = 3,240 kg

Why the other options are there

ρ = 0.81 kg/m³ (SG reported as density)
ν = 0.486 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 8

Density, specific weight, specific gravity and viscosity — Definitions (8)

A liquid has a specific gravity of 0.97 and a dynamic viscosity of 0.00190 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 0.5 m³.

Given

SG = 0.97
μ = 0.00190 Pa·s
Volume = 0.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 0.97 (1000) = 970.0 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 970.0 (9.81) = 9, 516 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/970.0 = 1.031 e - 3 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00190/970.0 = 1.959 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 970.0 (0.5) = 485.0 k g$

Answer: ρ = 970.0 kg/m³, γ = 9,516 N/m³, ν = 1.96e-6 m²/s, m = 485.0 kg

Why the other options are there

ρ = 0.97 kg/m³ (SG reported as density)
ν = 1.843 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 9

Density, specific weight, specific gravity and viscosity — Definitions (9)

A liquid has a specific gravity of 0.98 and a dynamic viscosity of 0.00100 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 5.0 m³.

Given

SG = 0.98
μ = 0.00100 Pa·s
Volume = 5.0 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 0.98 (1000) = 980.0 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 980.0 (9.81) = 9, 614 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/980.0 = 1.020 e - 3 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00100/980.0 = 1.020 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 980.0 (5.0) = 4, 900 k g$

Answer: ρ = 980.0 kg/m³, γ = 9,614 N/m³, ν = 1.02e-6 m²/s, m = 4,900 kg

Why the other options are there

ρ = 0.98 kg/m³ (SG reported as density)
ν = 0.980 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

Example 10

Density, specific weight, specific gravity and viscosity — Definitions (10)

A liquid has a specific gravity of 1.36 and a dynamic viscosity of 0.00200 Pa·s. Compute its density, specific weight, specific volume, kinematic viscosity, and the mass contained in 4.5 m³.

Given

SG = 1.36
μ = 0.00200 Pa·s
Volume = 4.5 m³
ρ_water = 1000 kg/m³

Find

ρ, γ, v, ν and the mass

Start with the thinking

Specific gravity is dimensionless — always multiply by the density of water to recover ρ.
Kinematic viscosity is dynamic viscosity divided by density, in m²/s.

Step-by-step solution

Formula
$ρ = S G \times ρ_{w} a t er$
Substituting
$ρ = 1.36 (1000) = 1, 360 k g / m^{3}$
Formula
$γ = ρ g$
Substituting
$γ = 1, 360 (9.81) = 13, 342 N / m^{3}$
Formula
$v = 1/ ρ$
Substituting
$v = 1/1, 360 = 7.353 e - 4 m^{3} / k g$
Formula
$ν = μ / ρ$
Substituting
$ν = 0.00200/1, 360 = 1.471 e - 6 m^{2} / s$
Formula
$m = ρ \times V o l u m e$
Substituting
$m = 1, 360 (4.5) = 6, 120 k g$

Answer: ρ = 1,360 kg/m³, γ = 13,342 N/m³, ν = 1.47e-6 m²/s, m = 6,120 kg

Why the other options are there

ρ = 1.36 kg/m³ (SG reported as density)
ν = 2.720 m²/s (multiplied instead of divided)

Reference: FE Reference Handbook — Fluid Mechanics → Definitions

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

Definitions contains 16 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.

Definitions

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

t=limitDmDV

c=limitDWDV

c=limitg:DmDV=tg

SG = γ/γw = ρ/ρwNeeds technical review

t=density(alsocalledmassdensity)

∆m=massofinfinitesimalvolume

∆V=volumeofinfinitesimalobjectconsidered

γ=specificweight

= ρgNeeds technical review

∆W=weightofaninfinitesimalvolume

SG=specificgravity

tw=densityofwateratstandardconditions

=1,000kg/m3​(62.4lbm/ft3​)

γw = specific weight of water at standard conditionsNeeds technical review

=9,810N/m3​(62.4lbf/ft3​)

=9,810kg/(m2​•s2​)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section