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Chapter 1
Properties of Fluids

AL
RI
TE
1.1 Introduction MA
A fluid is a substance which deforms continuously, or flows, when subjected to shear
stresses. The term fluid embraces both gases and liquids; a given mass of liquid will occupy
a definite volume whereas a gas will fill its container. Gases are readily compressible; the
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low compressibility, or elastic volumetric deformation, of liquids is generally neglected in

computations except those relating to large depths in the oceans and in pressure transients
in pipelines.
HT

This text however deals exclusively with liquids and more particularly with Newtonian
liquids, i.e. those having a linear relationship between shear stress and rate of deformation.
Typical values of different properties are quoted in the text as needed for the various
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worked examples. For more comprehensive details of physical properties refer to tables
such as Kaye and Laby (1995) or internet versions of such information.
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1.2 Engineering units

The metre–kilogram–second (mks) system is the agreed version of the international system
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(SI) of units that is used in this text. The physical quantities in this text can be described
by a set of three primary dimensions (units): mass (kg), length (m) and time (s). Further
discussion is contained in Chapter 9 regarding dimensional analysis. The present chapter
refers to the relevant units that will be used.
The unit of force is called newton (N) and 1 N is the force which accelerates a mass of
1 kg at a rate of 1 m/s2 (1 N = 1 kg m/s2 ).
The unit of work is called joule (J) and it is the energy needed to move a force of 1 N
over a distance of 1 m. Power is the energy or work done per unit time and its unit is watt
(W) (1 W = 1 J/s = 1 N m/s).

1.3 Mass density and specific weight

Mass density (ρ) or density of a substance is defined as the mass of the substance per unit
volume (kg/m3 ) and is different from specific weight (γ ), which is the force exerted by
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2 Civil Engineering Hydraulics

the earth’s gravity (g) upon a unit volume of the substance (γ = ρg: N/m3 ). In a satellite
where there is no gravity, an object has no specific weight but possesses the same density
Chapter 1

that it has on the earth.

1.4 Relative density

Relative density (s) of a substance is the ratio of its mass density to that of water at a
standard temperature (4◦ C) and pressure (atmospheric) and is dimensionless.
For water, ρ = 103 kg/m3 , γ = 103 × 9.81  104 N/m3 and s = 1.

1.5 Viscosity of fluids

Viscosity is that property of a fluid which by virtue of cohesion and interaction between
fluid molecules offers resistance to shear deformation. Different fluids deform at different
rates under the action of the same shear stress. Fluids with high viscosity such as syrup
deform relatively more slowly than fluids with low viscosity such as water.
All fluids are viscous and ‘Newtonian fluids’ obey the linear relationship
du
τ =µ (Newton’s law of viscosity) [1.1]
dy
where τ is the shear stress (N/m2 ), du/dy the velocity gradient or the rate of deformation
(rad/s) and µ the coefficient of dynamic (or absolute) viscosity (N s/m2 or kg/(m s)).
Kinematic viscosity (ν) is the ratio of dynamic viscosity to mass density expressed in
metres squared per second.
Water is a Newtonian fluid having a dynamic viscosity of approximately 1.0 × 10−3
N s/m2 and kinematic viscosity of 1.0 × 10−6 m2 /s at 20◦ C.

1.6 Compressibility and elasticity of fluids

All fluids are compressible under the application of an external force and when the force
is removed they expand back to their original volume, exhibiting the property that stress
is proportional to volumetric strain.
pressure change
The bulk modulus of elasticity, K =
volumetric strain
dp
=− [1.2]
(dV/V)
The negative sign indicates that an increase in pressure causes a decrease in volume.
Water with a bulk modulus of 2.1 × 109 N/m2 at 20◦ C is 100 times more compressible
than steel, but it is ordinarily considered incompressible.

1.7 Vapour pressure of liquids

A liquid in a closed container is subjected to partial vapour pressure due to the escaping
molecules from the surface; it reaches a stage of equilibrium when this pressure reaches
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Properties of Fluids 3

saturated vapour pressure. Since this depends upon molecular activity, which is a function
of temperature, the vapour pressure of a fluid also depends upon its temperature and

Chapter 1
increases with it. If the pressure above a liquid reaches the vapour pressure of the liquid,
boiling occurs; for example if the pressure is reduced sufficiently, boiling may occur at
room temperature.
The saturated vapour pressure for water at 20◦ C is 2.45 × 103 N/m2 .

1.8 Surface tension and capillarity

Liquids possess the properties of cohesion and adhesion due to molecular attraction. Due
to the property of cohesion, liquids can resist small tensile forces at the interface between
the liquid and air, known as surface tension (σ : N/m). If the liquid molecules have greater
adhesion than cohesion, then the liquid sticks to the surface of the container with which
it is in contact, resulting in a capillary rise of the liquid surface; a predominating cohesion
on the other hand causes capillary depression. The surface tension of water is 73 × 10−3
N/m at 20◦ C.
The capillary rise or depression h of a liquid in a tube of diameter d can be written as
4σ cos θ
h= [1.3]
ρgd
where θ is the angle of contact between liquid and solid.
Surface tension increases the pressure within a droplet of liquid. The internal pressure
p balancing the surface tensional force of a small spherical droplet of radius r is given by
2σ
p= [1.4]
r

Worked examples

Example 1.1

The density of an oil at 20◦ C is 850 kg/m3 . Find its relative density and kinematic viscosity
if the dynamic viscosity is 5 × 10−3 kg/(m s).

Solution:
ρ of oil
Relative density, s =
ρ of water
850
=
103
= 0.85
µ
Kinematic viscosity, ν =
ρ
5 × 10−3
=
850
= 5.88 × 10−6 m2 /s
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4 Civil Engineering Hydraulics

Example 1.2
Chapter 1

If the velocity distribution of a viscous liquid (µ = 0.9 N s/m2 ) over a fixed boundary is
given by u = 0.68y − y2 , in which u is the velocity (in metres per second) at a distance y
(in metres) above the boundary surface, determine the shear stress at the surface and at
y = 0.34 m.

Solution:

u = 0.68y − y2
du
⇒ = 0.68 − 2y
dy
Hence, (du/dy) y=0 = 0.68 s−1 and (du/dy) y=0.34m = 0.
Dynamic viscosity of the fluid, µ = 0.9 N s/m2
From Equation 1.1,
shear stress (τ ) y=0 = 0.9 × 0.68
= 0.612 N/m2
and at y = 0.34 m, τ = 0.

Example 1.3

At a depth of 8.5 km in the ocean the pressure is 90 MN/m2 . The specific weight of the
sea water at the surface is 10.2 kN/m3 and its average bulk modulus is 2.4 × 106 kN/m2 .
Determine (a) the change in specific volume, (b) the specific volume and (c) the specific
weight of sea water at 8.5 km depth.

Solution:

Change in pressure at a depth of 8.5 km, d p = 90 MN/m2

2
= 9 × 104 kN/m
2
Bulk modulus, K = 2.4 × 106 kN/m
dp dV −9 × 104
From K = − =
(dV/V) V 2.4 × 106
= −3.75 × 10−2
Defining specific volume as 1/ γ (m3 /kN), the specific volume of sea water at the surface
= 1/10.2 = 9.8 × 10−2 m3 /kN.
Change in specific volume between that
at the surface and at 8.5 km depth, dV
= −3.75 × 10−2 × 9.8 × 10−2
= −36.75 × 10−4 m3 /kN
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Properties of Fluids 5

The specific volume of sea water at 8.5 km depth = 9.8 × 10−2 − 36.75 × 10−4

Chapter 1
= 9.44 × 10−2 m3 /kN

1
The specific weight of sea water at 8.5 km depth =
specific volume
1
=
9.44 × 10−2
3
= 10.6 kN/m

References and recommended reading

Kaye, G. W. C. and Laby, T. H. (1995) Tables of Physical and Chemical Constants, 16th edn,
Longman, London. http://www.kayelaby.npl.co.uk
Massey, B. S. and Ward-Smith, J. (2006) Mechanics of Fluids, 8th edn, Taylor & Francis,
Abingdon, UK.

Problems
1. (a) Explain why the viscosity of a liquid decreases while that of a gas increases with an
increase of temperature.
(b) The following data refer to a liquid under shearing action at a constant temperature.
Determine its dynamic viscosity.

du/dy (rad/s) 0 0.2 0.4 0.6 0.8

τ (N/m2 ) 0 0 1.9 3.1 4.0

2. A 300 mm wide shaft sleeve moves along a 100 mm diameter shaft at a speed of 0.5 m/s
under the application of a force of 250 N in the direction of its motion. If 1000 N of force
is applied, what speed will the sleeve attain? Assume the temperature of the sleeve to be
constant and determine the viscosity of the Newtonian fluid in the clearance between the
shaft and its sleeve if the radial clearance is estimated to be 0.075 mm.

3. A shaft of 100 mm diameter rotates at 120 rad/s in a bearing 150 mm long. If the radial
clearance is 0.2 mm and the absolute viscosity of the lubricant is 0.20 kg/(m s), find the
power loss in the bearing.

4. A block of dimensions 300 mm × 300 mm × 300 mm and mass 30 kg slides down a plane
inclined at 30◦ to the horizontal, on which there is a thin film of oil of viscosity 2.3 × 10−3
N s/m2 . Determine the speed of the block if the film thickness is estimated to be 0.03 mm.

5. Calculate the capillary effect (in millimetres) in a glass tube of 6 mm diameter when
immersed in (i) water and (ii) mercury, both liquids being at 20◦ C. Assume σ to be 73 ×
10−3 N/m for water and 0.5 N/m for mercury. The contact angles for water and mercury
are 0 and 130◦ , respectively.

6. Calculate the internal pressure of a 25 mm diameter soap bubble if the tension in the soap
film is 0.5 N/m.