1. JIF 104 PHYSICS II/PRACTICAL 1b
MODULE 2: PROPERTIES OF LIQUIDS
by
Sabar Bauk
Pusat Pengajian Pendidikan Jarak Jauh,
Universiti Sains Malaysia.
2008
1

2. Contents
CHAPTER 1 : LIQUID SURFACE EFFECTS
1.1 Introduction
1.2 Free surface
1.3 Free surface energy
1.4 Variations of surface energy with temperature
1.5 Surface tension
1.6 Relationship between the surface energy γ and the surface tension T
1.7 Angle of contact
1.8 Capillarity
1.9 Excess pressure theorem
1.10 Determination of the free surface energy γ for a liquid-vapour
interface
CHAPTER 2 : VISCOSITY OF LIQUID
2.1 Introduction
2.2 Viscosity
2.3 Origin of viscosity
2.4 Determination of viscosity
2.5 Relationship between viscosity and temperature
2.6 Newtonian and non-Newtonian liquids
CHAPTER 3 : LIQUID FLOW
3.1 Introduction
3.2 The continuity equation
3.3 The Euler’s equation
3.4 The Bernoulli’s equation
3.5 Applications of the Bernoulli’s equation
3.6 The Poiseuille’s equation
3.7 Motion of an object in a viscous liquid
2

3. 3.8 Boundary layer
3.9 Critical velocity and the Reynolds’ number
3.10 Comparison between laminar flow and turbulent flow
CHAPTER 4 : DIFFUSION
4.1 Introduction
4.2 Diffusion equation
4.3 Solution to the diffusion equation
4.4 Several processes involving diffusion process
REFERENCES
SOLUTIONS TO THE EXERCISES
3

4. CHAPTER 1
LIQUID SURFACE EFFECTS
LESSON OBJECTIVES
After completing this lesson, you should be able to:
• State some everyday examples of liquid surface effects.
• Define surface tension and free surface energy.
• Prove that surface tension is similar to free surface energy.
• Explain the relationship between free surface energy and temperature.
• Define angle of contact.
• Define capillarity based on the excess pressure theorem.
• Determination of the free surface energy for a liquid-vapour interface.
1.1 Introduction
As we have discussed in Module 1, the particles forming a solid are fixed at
specific positions, called the lattice points, in the bulk material. They do not move
translationally but they might vibrate about their positions due to thermal internal
energies.
The particles of some materials can move translationally from one place to
another. These materials have the ability to flow. A substance that flows readily and
tends to assume the shape of its container is called a fluid. Basically, liquids and gases
are fluids. Hence, many properties of liquids such as viscosity, flow and diffusion are
also applicable to gases.
For a solid, the molecular arrangement is ordered and periodical whilst for a
fluid it is disordered; with liquid it is partially ordered and completely disordered for a
gas. Figure 1.1 shows the distribution of molecules in a solid and a fluid.
4

5. (a) Solid
(b) Fluid
Figure 1.1: Molecular arrangement in a solid and a fluid.
In this lesson we will discuss some properties of the surface of liquids. The
surface of a liquid has some unique properties different from the bulk of the liquid.
We have seen some of the examples of liquid surface effects every day. Some insects
are able to walk across a water surface, a dry steel needle or razor blade may be made
to float on water and droplets of liquid tends to form spheres if sprayed in air. These
examples suggest that ‘the surface of a liquid acts like an elastic skin covering the
liquid’.
Figure 1.2: Forces acting on a needle floating on the surface of
water. W is the weight of the needle and T is the tension of the
liquid surface.
Particles in condensed matter exhibit short-range forces and these forces are
found to be disturbed at the surface of the matter. Some of the bondings of the surface
particles are broken as the surface particles are not completely surrounded by
neighbouring particles; that is, some parts of them are exposed to free space. Hence
the surface of a condensed matter can be considered as a defect and may have
completely different properties as compared to the interior of the matter.
5

6. Quiz
Can you give another two everyday examples of a liquid surface effect?
1.2 Free surface
Now, let us discuss the definition of a free surface. For molecules at depth, the
net resultant force acting on a molecule is zero since each molecule is completely
surrounded by its neighbouring molecules. However, for a molecule at or near the
surface of the liquid, the net resultant force acting on it is not zero as it is not
completely surrounded by its neighbours.
It is easier to visualise this effect by defining the sphere of molecular activity
as a sphere around a molecule where other molecules whose centres are within the
sphere will be mutually attracted to the molecule. As an example, a sphere of activity
around a molecule A is as shown in Figure 1.3. Other molecules whose centres are not
within the sphere of A are assumed not to influence the properties of A. Similar
spheres of activity may be drawn for other molecules in the liquid.
Liquid
surface
B
C
A
Figure 1.3: Spheres of activity for three liquid molecules A, B
and C. The spheres are incomplete for molecules B and C
causing the resultant force acting on the molecules to be directed
towards the interior of the liquid.
For a molecule A deep inside the liquid, the forces acting on it due to its
neighbouring molecules are equal in all directions. Hence the resultant force acting
on molecule A is zero.
For molecules B and C at or near the surface of the liquid, their spheres of
activity are incomplete. The resultant force acting on molecules B and C are toward
the inside of the liquid. Hence, the molecules at or near the surface of a liquid tend to
6

7. confine the rest of the molecules inside the liquid by applying a force towards the
interior of the liquid body. This tends to make the surface area of the liquid to become
smaller. In fact, for a given volume of a certain liquid, its surface is inclined to have
the smallest surface area. For example, a drop of liquid will form into a sphere as a
sphere is the minimum surface area for a given volume. For a minimum surface area,
the molecules are packed closer together. The process of minimising the surface area
stops when the surface force is balanced by the repelling forces between the closelypacked molecules.
Figure 1.4: (a) Resultant force FR acting on particles B and C are
directed towards the bulk of the liquid. The resultant force at
particle A is zero. (b) The resultant forces acting on molecules at
the surface of a liquid drop tend to minimise the surface area of
the drop. The minimum surface area for a given volume is a
sphere.
The horizontal broken line in Figure 1.3 is the limit of the surface effect. Any
molecule lying between this line and the surface of the liquid will contribute to the
surface effect. On this line, as represented by molecule A, and below towards the
interior of the liquid, the surface effect is nil as the resultant force acting on a
molecule in this region is zero.
The forces found in a liquid are a combination of cohesive forces between the
fluid molecules and the adhesive forces at the fluid’s surface. The adhesive forces can
also act when two surfaces meet to form an interface; even if they are from different
types of molecules. As an example, the adhesive force acting between water
molecules and the wall of a glass tube.
7

8. Quiz
Explain why the forces acting on a molecule deep inside a liquid is different from the
forces acting on a molecule at the surface of the liquid.
1.3 Free surface energy
The potential energy of the surface molecules is found to be higher than the
potential energy of molecules far from the surface. If the area of a surface is to be
increased, then molecules from the depth of the liquid have to be brought to the
surface. In the process, work has to be done as energy is needed to overcome the
attractive forces acting between the molecules. The work done is conserved as a
potential energy when the molecules reach the surface. This will increase the potential
energy of the surface.
If the work is carried out isothermally when the surface area is increased by
one unit, the work needed to do this is called the free surface energy. This quantity is
identical to the surface excess potential energy and is represented by the symbol γ.
The SI unit for γ is J m-2. Hence, the quantity γ is a special property of a liquid
surface.
Normally when the free surface energy γ of a liquid is mentioned, it is
assumed that the liquid surface is in contact with its vapour. Therefore, surface energy
can also be defined for any interface; that is, the surface of a liquid in contact with
different molecular materials. For example, at room temperature, the surface energy γ
for water (in contact with its vapour) is 72 × 10-3 J m-2 and for a water-benzene
interface γ is 35 × 10-3 J m-2.
Quiz
The potential energy of molecules at the surface of a liquid is smaller than the
potential energy of molecules at depth. Right or wrong?
Quiz
Give a definition for the term ‘free surface energy’.
8

9. 1.4 Variations of surface energy with temperature
Now, let us discuss the effects of temperature on the surface energy of liquids.
For a pure liquid in equilibrium with its vapour, the variation of surface energy γ with
temperature is given by


γ t = γ 0 1 −
T 

T′
n
(1.1)
where γt is the surface energy at temperature T, γ0 is the surface energy at 0°C, n is a
constant number which has a value between 1 and 2, and T′ is a temperature just
slightly below the critical temperature of the liquid. The value of γ decreases as the
temperature increases and γ is zero as the temperature approaches the critical
temperature. The critical temperature is the temperature of the matter at the critical
point in a phase diagram (Figure 1.5). Above the critical temperature, only gas and/or
vapour exist.
P
Liquid
Solid
Critical
point
Triple point
Vapour
T
Figure 1.5: A typical phase diagram of a material.
At a high temperature, the molecules of the liquid have higher kinetic energies
and generally the distances between them are bigger than when the liquid is at a lower
temperature. This situation lowers the mutual action among the molecules of the
liquid. At this point, the energy needed to move an interior molecule in the liquid to
the surface is lower. Therefore, the surface energy γ decreases when the temperature
increases.
9

10. Quiz
Explain why it is easier to move a molecule from the interior of the liquid to the
surface when the temperature of the liquid is higher?
1.5 Surface tension
Surface tension is a concept which explains why some insects can skate on the
surface of water and why a drop of water forms into a sphere. Surface tension makes
the surface of liquid looks like an elastic skin in a state of tension. Any line drawn on
the surface is acted on by two equal and opposite forces. As an analogy, if we cut the
skin of a blown-up balloon, the rubber draws away from the cut due to the action of
the two forces. Another example is the case of a loop of thread in a soap film (Figure
1.6. If the film in the loop is punctured, the surface tension acts on the loop radially
and the thread loop becomes a circle.
Figure 1.6: (a) A loop of thread in a soap film. (b) The thread
loop forms into a circle if the film inside the loop is punctured.
The surface tension T of a liquid is defined as numerically equal to the force in
the surface acting at right angles to one side of a unit length of a line drawn on the
surface. The unit of T is N m-1.
10

11. Figure 1.7: The surface force of the surfaces of a liquid film
balancing the weight of a sliding wire.
Figure 1.7 shows an inverted U-shaped wire with a sliding wire on the fourth
side. A film of a liquid is spread between the wires as shown. When the set-up is
allowed to stand vertically, the sliding wire is prevented from falling down by an
upward force due to the surface tension of the film. Since the film has two surfaces,
the total length along which the surface force acts is 2l.
When the sliding wire is in equilibrium
2Tl = W
W
T=
2l
(1.2)
where W is the weight of the sliding wire and l is the length of the sliding wire.
Hence, by using this simple set-up, we can determine the surface tension of a liquid.
1.6 Relationship between the free surface energy γ and the surface tension T
In this section, we will discuss the relationship between free surface energy γ
and the surface tension T of a liquid.
11

12. A
l
D
D′
2Tl
B
F
C
C′
x
Figure 1.8: A film of liquid stretched on a horizontal wire frame
to form a horizontal plane ABCD. Wire CD can be moved freely
on the wire frame.
Consider a film of a liquid stretched on a horizontal frame ABCD as shown in
Figure 1.8. Actually, the thin film has two surfaces; one facing out of the page and the
other facing into the page. If T is the surface tension, the force acting on CD is 2Tl.
The numerical 2 is due to the two surfaces acting on CD. If CD is moved a little to the
right with a constant speed to a new position C′D′, the force F needed to do this is
2Tl. If the distance traveled by CD is x and it is done isothermally, then the work done
for the process is
W = Force × Distance = 2Tlx
In the same process, the increase in the surface area of the film ∆A is 2lx for
the two surfaces facing out and into the page. Hence, the work done in increasing the
surface area
W = 2Tlx = T × 2lx = T × ∆A
W
∴T =
∆A
(1.3)
From the expression, surface tension can also be defined as the work done to increase
the surface area by one unit provided the process is carried out isothermally. The unit
for T, which was initially N m-1 as defined in Section 1.5, can also be written as J m-2,
which is the unit for γ as in Section 1.3. Hence, it can be concluded that surface
tension T is similar to free surface energy γ.
12

13. Quiz
Can you show that the surface tension is similar to the free surface energy?
Example
A circular wire of diameter 6 cm is immersed horizontally in a liquid sample. The
extra force (due to surface tension) needed to pull out the circular wire from the liquid
is 0.01 N. Calculate the surface tension of the liquid.
Solution
Given: d = 6 cm = 0.06 m; F = 0.01 N.
When the circular wire is pulled out, a cylindrical liquid film of radius r = 6 cm is
formed. The film has an internal surface and an external surface.
r
2T
2T
The lines of contact of the surfaces with the wire are two isocentric circles with
almost similar diameters. They are so close together that their radii can be considered
identical.
The surface tension acting on the wire is given by
F = 2 × 2πrT
F
∴T =
4πr
0.01
=
= 0.027 N m -1 #
4π × 0.03
1.7 Angle of contact
When a liquid surface meets with a solid surface, usually it does not spread
itself uniformly throughout the surface of the solid. Commonly, the two surfaces meet
at an angle α known as the angle of contact. The value of α depends on the properties
13

14. of the liquid as well as the type of the solid surface. The angle of contact is measured
within the liquid. Figure 1.9 shows the meeting point of the solid surface plane and
the contact line plane and the angle of contact.
γLV
Vapour
Liquid
α
γSV
γLS
Solid
Figure 1.9: The angle of contact α is measured within the liquid.
Let us determine the relationship between α and the free surface energy γ.
Consider the surface of an infinitely large drop of liquid on a solid surface. The
contact point between the liquid, solid and air (vapour) is as shown in Figure 1.9. The
angle of contact is represented by α which is measured within the liquid. Let γLS as the
surface energy of the liquid-solid surface, γLV as the surface energy of the liquid-air
surface and γSV as the surface energy for the solid-air surface.
Let the lateral distance of the liquid drop be increased by x as shown in Figure
1.10, then the surface energy of the solid-air surface decreases by γSV × x and the
surface energy of the solid-liquid surface increases by γLS × x.
Direction
of motion
x cosα
γLV
α
Liquid
γLS
γSV
x
Solid
Figure 1.10: The process of increasing the surface area of the
liquid.
14

15. The surface area of the liquid has also increased. The increase in liquid-air
surface is γLV × x cosα. Assuming x is small and the surface energy is conserved, then
γ LS x + γ LV x cos α = γ SV x
γ − γ LS
cos α = SV
γ LV
(1.4)
This is the Young’s definition of the angle of contact. If α is greater than 90°,
the liquid will not wet the solid surface as in the case of mercury on a glass plate. If α
is zero, the liquid will wet the solid surface; that is, the liquid will spread itself to the
whole surface of the solid.
Figure 1.11: The angle of contact for a drop of liquid on a
surface. The angle of contact would determine the wetting
property of the liquid with a solid surface.
Quiz
By using suitable diagrams, can you derive the relationship in equation 1.4?
1.8 Capillarity
One of the effects of a liquid surface is the capillarity phenomenon. It involves
the rise or depression of a column of a liquid in a capillary tube when the tube is
partly immersed in the liquid. This phenomenon contradicts the liquid level predicted
by hydrostatic principles. It can be shown that if the angle of contact is less than 90°,
the liquid level in the capillary tube will rise above the level of the liquid outside the
tube. If the angle of contact is more than 90°, the liquid in the capillary tube will be
15

16. depressed below the level of the liquid outside the capillary tube. The relevant
examples are water which would rise up and mercury which would be depressed in a
capillary tube.
Figure 1.12: The angle of contact and its effect on capillarity.
The rise of water in a glass capillary tube is due to the presence of adhesive
force between the molecules of water and the molecules of the tube. Cohesive force
between the water molecules causes the other molecules down the tube to follow
rising up the tube. The rise of the water level in the capillary tube stops when this
upward force is balanced by a downward gravitational force due to the weight of the
water column.
For mercury, the cohesive force between the mercury molecules is higher than
the adhesive force between the mercury and the tube. Mercury never wets the glass.
This causes the mercury level inside the glass capillary tube to be lower than the level
outside the tube. Hence, capillarity is due to the interplay between the adhesive force
acting between the liquid molecules and the tube wall molecules and the cohesive
force acting between the molecules of the liquid.
Quiz
Capillarity occurs only in tubes that are vertical. True or false? Explain.
16

17. 1.9 Excess pressure theorem
The pressure drop across a curved surface is responsible for the rise, or
depression, of a liquid surface in a capillary tube. This effect also leads to the
dependence of the vapour pressure of a liquid on the curvature (meniscus) of its
surface.
Consider a curved surface of a liquid; it might be the meniscus of the surface
of a liquid. Let an element of the surface is represented by ABCD as in Figure 1.13(a).
The sides of the elemental surface area is x and y. The radius of curvature of the sides
might be different and is represented by r1 and r2 respectively. The area of the surface
element is given by
A = xy
B
y
x
D
A
x
θ
D
r1
r2
r1
D′
δr
C
A
δx
A′
(a)
θ
(b)
Figure 1.13: (a) An element of the liquid surface ABCD and its
radii of curvature r1 and r2. (b) Increasing the surface side by δx
will increase the radius of curvature by δr.
Let the area A is slightly increased in an isothermal process so that its radius of
curvature is increased by δr as shown in Figure 1.13(b). Then the increase in the area
is
δA = δ (xy ) = xδy + yδx
17

18. From Figure 1.5(b) we found that
θ=
δx x
=
δr r1
∴ δx =
x
δr
r1
Similarly, on the y side we have
δy =
y
δr
r2
Hence, the change in the area is
δA = xy
δr
r2
+ xy
δr
r1
1 1
= xyδr  + 
r r 
2 
 1
Work done by the excess pressure ∆P to increase the surface area is ∆Pxyδr
(force × distance). This work is equal to the increase in the surface energy γδA where
γ is the free surface energy of the liquid. Hence
∆Pxyδr = γδA
1 1
= γxyδr  + 


 r1 r2 
1 1
∴ ∆P = γ  + 
r r 
2 
 1
(1.5)
The excess pressure ∆P is the difference in pressure between both sides of the liquid
surface. This equation is called excess pressure theorem. It can be used in any liquid
interface.
If a thin film of a liquid is considered, two excess pressures with identical sign
and magnitude will be found since a thin film has two surfaces. The excess pressure
between the first surface and the second surface is
1 1
∆P = 2γ  + 
r r 
2 
 1
(1.6)
If a spherical liquid drop is considered, r1 is equal to r2 and the equation
becomes
18

19. ∆P =
2γ
r
(1.7)
For a thin spherical bubble (with external and internal surfaces), the equation
becomes
∆P =
4γ
r
(1.8)
where r is the radius of the sphere. Note that the factor 2 in equation (1.8) is due to the
two surfaces (external and internal) of a thin film bubble.
Quiz
Write down the general equation for the theorem of excess pressure and explain the
meaning of each symbol used.
Example: Excess pressure in an air bubble in a liquid
Consider a bubble formed inside a liquid as shown in Figure 1.14. If we
consider the equilibrium in half of the bubble B, we have
Surface tension on B + Force due to external pressure P1
= Force due to internal pressure P2 inside the bubble
The surface tension acts round the circumference of the bubble which has a
length of 2πr; thus the force is 2πrγ. Pressure is basically force per unit area; hence
the force due to P1 is πr2 × P1 and the force due to P2 is πr2 × P2 where πr2 is the
effective area where the pressures act. The relationship above becomes
2πrγ + πr 2 P1 = πr 2 P2
P2 − P1 =
2γ
r
Now (P2 - P1) is the excess pressure ∆P. Therefore
∆P =
2γ
r
(1.9)
19

20. 2πγr
P1
P2
B
Figure 1.14: Excess pressure in an air bubble in a liquid.
Example: Excess pressure in a soap bubble
A soap bubble is actually a thin spherical film of soap solution. It has two
surfaces; one surface inside the bubble in contact with air and another surface is
outside the bubble in contact with the outside air. The force on half of the bubble B
due to surface tensions is γ × 2πr × 2 (Figure 1.15). The relationship between the
forces is
4πrγ + πr 2 P1 = πr 2 P2
Hence the excess pressure
∆P = P2 − P1 =
4γ
r
(1.10)
You might like to compare between this result and the result for an air bubble
inside a liquid in the previous example.
2πγr × 2
P1
P2
B
Figure 1.15: Excess pressure in a soap bubble.
20

21. 1.10 Determination of the free surface energy γ for a liquid-vapour interface
The free surface energy γ can be determined by using several methods.
However, only a handful of methods can give reliable results. In this section we
would discuss one of the reliable methods by using the capillary tube method.
When a capillary tube is vertically immersed in the liquid which its γ is to be
determined, the liquid level inside the tube rises as shown in Figure 1.16. Let h be the
height of the liquid column, and PA, PB and PC be the pressures at points A, B and C
respectively as shown.
A
B
h
C
Figure 1.16: The rise of a liquid in a capillary tube. Points A, B
and C are the positions where the pressures are being considered
in the text.
Point A is exposed to the atmosphere. Hence, the pressure at A is the
atmospheric pressure. Point C is at the same level as the surface of the liquid outside
the capillary tube. Therefore, according hydrostatic theory, point C should be at
atmospheric pressure too. The pressures at A and C is given by
PA = PC = Patm
(1.11)
where Patm is the atmospheric pressure.
In the case of a capillary tube dipping in water, the angle of contact is
practically zero. Hence, the excess pressure theorem gives
∆P = PA − PB =
2γ
r
(1.12)
21

22. where r is the radius of curvature of the liquid meniscus in the tube. At the meniscus
of the surface, the pressure is given by the hydrostatic equation
PA = Patm = PB + ρgh
(1.13)
where ρ is the density of the liquid. Therefore the difference in pressure between the
air and the liquid at the meniscus is
PA − PB = ρgh
(1.14)
But equation (1.14) can be substituted into equation (1.12) to give
2γ
= ρgh
r
ρghr
∴γ =
2
(1.15)
For a tube with a very small bore (diameter < 3 mm), the meniscus in the tube
can be assumed to be a part of a spherical surface as shown in Figure 1.17. From the
figure, the radius of curvature r is given by
R
r
R
∴r =
cos α
cos α =
(1.16)
where α is the angle of contact and R is the radius of the capillary tube. Substituting r
into equation (1.15) for the free surface energy
γ =
ρghR
2 cos α
(1.17)
The equation can be rearranged to indicate the rise, or depression, of the liquid
column in the capillary tube
h=
2γ cos α
ρgR
(1.18)
If α < 90°, cos α is positive and h is also positive. This means that the liquid
will rise in the tube. For a given type of liquid, h is inversely proportional to the radius
of the tube R. If the tube becomes smaller, the surface of the liquid inside the tube
becomes higher.
If α > 90°, h will have a negative value. This indicates that the surface of the
liquid in the capillary tube will be depressed below the surface level outside the tube.
22

23. In this case, for a given type of liquid, the depression becomes deeper (h becomes
more negative) as the radius of the tube R decreases.
r
α
R
α
Figure 1.17: The meniscus of a liquid surface in a capillary tube.
R is the radius of the tube, r is the radius of curvature of the
meniscus and α is the angle of contact.
Example
A capillary tube with an internal diameter a is fixed vertically with its lower end
immersed in a liquid with a density ρ and surface tension γ1. The upper end of the
tube is covered by a soap bubble with a diameter b and the surface energy of the soap
bubble is γ2. Determine the height of the liquid column inside the tube. Assume that
the liquid wets the tube.
Solution
Let the pressures at several points be represented by the points shown in the diagram.
Let h be the height of the liquid column and P3 = Patm be the atmospheric pressure.
Point P1 is not exposed to the atmosphere.
23

24. b
Patm
P1
P2
h
P3
a
For the soap bubble
P1 − Patm =
4γ 2
b
(i)
and for the meniscus
P1 − P2 =
2γ 1
a
(ii)
But, for the liquid inside the tube in equilibrium
Patm − P2 = ρgh
(iii)
Patm = P2 + ρgh
Substitute (iii) into (i)
P1 − (P2 + ρgh ) =
P1 − P2 =
4γ 2
b
4γ 2
+ ρgh
b
Substituting into (ii)
2γ 1 4γ 2
=
+ ρgh
a
b
2γ
4γ
ρgh = 1 − 2
a
b
2  γ 1 2γ 2 
∴h =

 −
ρg  a
b 
24

25. SUMMARY
1. The resultant force on a molecule at the a liquid surface is not zero. The resultant
force is directed towards the inside of the bulk liquid perpendicular to the plane of the
surface.
2. The free surface energy γ is similar to the surface tension T eventhough their units
are different.
3. The free surface energy varies with temperature according to the relationship


γ T = γ 0 1 −
T 

T′
where γ0 is the surface energy at 0°C and T′ is a temperature slightly below the critical
temperature.
4. The angle of contact α is measured within the liquid. If α < 90°, it is called a
wetting condition. If α > 90°, it is called a non-wetting condition.
5. The excess pressure theorem describes the pressure difference between both sides
of a liquid surface.
1 1
∆P = γ  + 
r r 
2 
 1
where r1 and r2 are the radii of curvature of the meniscus of the liquid surface.
6. The height or the depression of a liquid in a capillary tube is determined by
h=
2γ cos α
ρgR
where R is the radius of the capillary tube.
25

26. EXERCISE 1
1. Calculate the work needed to increase the diameter of a spherical soap bubble from
10 cm to 15 cm, if the surface tension is 0.04 N m-1.
[Ans: 3.14 × 10-3 J]
2. A loop of fine thread is placed on a horizontal plane soap film and the film inside
the loop is then punctured. Show that the loop will form into a circle and derive an
expression for the tension T of the thread in terms of the free surface energy γ of the
film as well as the radius r of the circle formed.
[Ans: T = 2γr]
3. Derive an expression for a column of mercury in a glass tube with a diameter of 0.3
mm which is fixed in a vertical position with one of its ends immersed in mercury.
The relative density of mercury is 13.6, the angle of contact is 130° and the surface
energy is 0.49 N m-1.
[Ans: -3.2 cm]
4. Two circular glass plates, each with a diameter of 0.06 m, are separated by a film of
water 1.0 × 10-5 m thick. Determine the minimum force required to pull the plates
apart. The free surface energy of water is 73 × 10-3 J m-2 and the angle of contact of
water with the glass is 0°.
[Ans: 165 N]
5. What is the pressure in a cylindrical water jet of diameter 5 mm with respect to the
atmospheric pressure around it if the surface tension of water is 0.073 N m-1?
[Ans: 29.2 N m-2]
26

27. CHAPTER 2
VISCOSITY OF LIQUIDS
LESSON OBJECTIVES
After completing this lesson, you should be able to:
• Define the viscosity of a liquid.
• Explain the relationship between viscosity and the velocity gradient of a moving
liquid.
• Describe the origin of viscosity.
• Describe a method to determine the viscosity of a liquid.
• Explain the effects of temperature on viscosity.
• Explain the viscosity of Newtonian and non-Newtonian liquids.
2.1 Introduction
In the previous module, we have discussed the behaviour of solids under
mechanical stress. In this chapter, we will discuss the behaviour of liquids under
dynamical or shear stress. One of the special properties of liquids is their ability to
flow. However, before we discuss the flow of liquids, we have to study a
characteristic of a liquid in motion called viscosity.
2.2 Viscosity
Viscosity may be thought as the internal friction of a fluid. It is common
knowledge that if an object is moving in a liquid, its motion is resisted; that is, the
liquid is opposing the motion of the object. For example, a man wading in a pool of
water feels that the water is resisting the movement of his legs. Different types of
liquids produce different amount of resistance to the moving object. The ability of
27

28. liquids to resist motion is similar to the friction of an object moving on a solid surface
but for a liquid it is called viscosity.
Similar resistance to motion can also be found when the liquid itself is flowing
in a channel. The behaviour of a flowing liquid will be discussed in the next chapter.
To define the concept of liquid viscosity, let us consider two parallel plates
initially at rest as shown in Figure 2.1. Each plate has an area A and a thin layer of
liquid of thickness l is in between the two plates. The problem is similar to that of the
shear stress and shear strain in a solid.
U
F
l
L
Figure 2.1: Two identical plates separated by a layer of liquid. A
shear force F is applied to the top plate.
The top plate U is then moved by a shear force F while the bottom plate L is
fixed. Since the liquid cannot oppose the shear stress, its form is directly changed at a
certain rate. The motion of U is resisted by the layer of liquid between the plates.
When the liquid resistance is in equilibrium with the applied force F, only then the
plate U will move a constant velocity v.
When the plate U moves, the layer of liquid directly in contact with its surface
will move at the same speed as that of the plate, while the liquid layers below will
move at lower speeds. The further down a layer of liquid is placed, the slower is its
speed. In fact the layer of liquid in direct contact with plate L will not move at all as
plate L is stationary. In this situation a velocity gradient v/l exists as shown in Figure
2.2.
U
L
velocity = v
velocity = 0
Figure 2.2: A velocity gradient exist when the top plate U is
moving at a velocity v relative to the bottom plate L.
28

29. As stated above, when a plate is in contact with a liquid, a layer of liquid
becomes attached to the surface of the plate. This situation can be explained as
follows. When a molecule of the liquid collides with a plate molecule, the resulting
bonding is easier to produce than that between two liquid molecules. Since initially
the plate is stationary, it can absorb more kinetic energy from the incident molecules.
Furthermore, the molecules at the surface of the plate have incomplete chemical
bonds as they are at the edges of a structured three-dimensional arrangement.
Consequently, a layer of the liquid molecules will stick to the surface of the plate. The
layer is called the adsorbed layer and is a monomolecular layer.
For most liquids, the velocity gradient existing in the liquid is proportional to
the shear stress F/A applied; that is
F v
∝
A l
F
v
∴ =η
A
l
(2.1)
where η is a proportional constant which can be rewritten as
η=
F
A = Fl
v
Av
l
(2.2)
This means that η which is constant at constant temperature and pressure is a
measure of stress per velocity gradient. It is a measure of the resistance of motion
because, if the velocity gradient is fixed, F/A increases as η increases and vice-versa.
The constant η is called the viscosity (or the coefficient of viscosity) of the liquid.
The SI unit for viscosity, as can be derived from equation 2.2, is Pascal second
(Pa s or N s m-2). However, its c.g.s. unit still widely used and is called the poise (p)
in honour of the French scientist Poiseuille. Thus
1 poise = 1 dyne s cm-2 = 0.1 N s m-2
The reciprocal of η is called fluidity and its unit is rhe.
Sometimes liquid viscosity is stated in terms of a quantity called kinematics
viscosity. Kinematics viscosity is η/ρ where ρ is the density of the liquid and, in c.g.s.
system, the unit used is stokes.
29

30. Quiz
Explain the followings:
a) fluidity,
b) kinematics viscosity.
2.3 Origin of viscosity
Figure 2.3 shows a parallel straight flow where layer SS is moving at a higher
speed than a neighbouring TT layer. Since the molecules in the liquid are in a random
thermal motion state, some molecules from the SS layer will move into the TT layer.
When this happens, momentum is also transferred from the SS layer to the TT layer.
Collisions with molecules in the TT layer will cause the higher momentum of the
molecules originally from the SS layer will be shared with the molecules in the TT
layer. Hence, the speed of the TT layer will increase. In the same way, the lower
speed molecules from the TT layer will be transferred to the SS layer and will
decrease the speed of the SS layer. Therefore, each molecule transfered will produce
an accelerating force or a decelerating force in the opposite direction to the difference
in velocity between the layers.
S
S
T
T
Figure 2.3: The transfer of molecules, and hence the momentum,
between two layers of flowing liquid due to random thermal
motion.
This transfer of momentum mechanism is one of the reasons for the existence
of viscosity in liquids and gases. For gases, this is the main mechanism for the
existence of viscosity η. Since the kinetic energy increases with an increase in
temperature, η will also increase with increasing temperature.
30

31. In liquids, another mechanism can also be found apart from the one discussed
above. In this case, the existence of η can also be assumed to be due to the forces
between particles in adjacent layers since the particles in a liquid is much closer than
those in a gas. These forces will be changed by the relative motion of the liquid
layers. When this happens, shear forces are produced and these will resist the relative
motion. This resistance also contributes to the viscosity in liquids.
Quiz
The existence of viscosity in a liquid is due to two mechanisms. Describe those
mechanisms.
2.4 Determination of viscosity
There are several methods to determine the viscosity of a liquid. Some of the
methods will directly produce viscosity values while the others produce relative
viscosity values. Figure 2.4 shows some of the methods to determine the coefficient of
viscosity: the rate of settling of a solid sphere in a liquid, determination of the rate of
flow through a capillary tube (Ostwald’s viscometer) and using the force required to
turn one of two concentric cylinders at a certain angular velocity. A method which
directly determines η will be described in the next chapter but here we will discuss
only one method which will produce relative viscosity.
Relative viscosity value may be determined by using the Ostwald’s viscometer
as shown in Figure 2.4(b). A certain amount of liquid is poured into the U-tube
through limb A so that the liquid in the tube reaches up to P and Q which are at the
same level. The liquid is then sucked into B so that its surface level is higher than L1.
Then the liquid is released to drop under gravity. The time taken t by the surface of
the liquid in B to fall from L1 to L2 is recorded. This time is dependent on the density
and viscosity of the liquid. For a given volume of liquid flowing in the capillary tube,
time t becomes higher if η value is higher, while the time decreases if the density
value is bigger. Hence, t is directly dependent to η/ρ which is the kinematics
viscosity. When the duration needed by different liquids to flow from L1 to L2 is
determined, a comparison of kinematics viscosity between the liquids can be
31

32. produced. If the densities of liquids involved are known, then the relative viscosity of
the liquids can be determined.
Suppose two liquids have densities ρ1 and ρ2 respectively. The excess
pressures for the two liquids along the capillary tube of height h are ρ1gh and ρ2gh
respectively. If the volume between L1 and L2 is V, then from Poiseuille’s equation
(will be discussed in the next chapter) we have
V π (ρ1 gh )a 4
=
t1
8η1l
(2.3)
where t1 is the time taken by the level of the first liquid to drop from L1 to L2, a is the
radius of the capillary tube, l is the length of the capillary tube and η1 is the
coefficient of viscosity of the first liquid.
Similarly, for the second liquid we have
V π (ρ 2 gh )a 4
=
t2
8η 2 l
(2.4)
Dividing equation (2.4) by equation (2.3)
t1 η 1 ρ 2
=
t 2 η 2 ρ1
t ρ
η
∴ 1 = 1⋅ 1
η2 t2 ρ2
(2.5)
The result shows that if we know the coefficient of viscosity of one of the
liquid, then the coefficient of viscosity of the other liquid can be determined.
32

33. A
L1
B
L2
P
Q
ω
Capillary
tube
vT
(c)
(a)
(b)
Figure 2.4: Methods to determine the coefficient of viscosity
using (a) the settling of a solid sphere in the liquid, (b) the rate of
flow of the liquid through a capillary tube (Ostwald’s
viscometer), and (c) the force required to turn one of two
concentric cylinders at a certain angular velocity ω.
Quiz
The Ostwald’s viscometer would directly give the viscosity value. Right or wrong?
Explain.
2.5 Relationship between viscosity and temperature
The relationship between viscosity and temperature can be derived by
considering the potential energy between adjacent liquid molecules. When any
molecule, such as a molecule M in Figure 2.5, tries to move, it has to overcome a
potential barrier. The molecule must have a suitable energy to overcome all forces
acting on it due to the presence of neighbouring molecules.
33

34. M
Figure 2.5: Molecule M with its neighbouring molecules in two
dimensions.
Let the height of the potential barrier be u. Since M is always vibrating due to
thermal or internal energy, let the frequency of the vibration be f0; which means, M is
trying to overcome the barrier f0 times in one second. The probability that M will have
a suitable energy to overcome the potential barrier in each attempt is e-u/kT where k is
the Boltzmann’s constant and T is the temperature in unit kelvin K.
Hence, the number of successful attempts f to overcome u in one second is
f = f0e
−
u
kT
(2.6)
f is also called the jumping frequency. If the liquid is more viscous, then the bondings
between the molecules are stronger. This means that it is more difficult for a molecule
to overcome the potential barrier. Thus we found that f ∝ 1/η or η ∝ 1/f. Therefore
η∝
1
f
u
∝ e kT
(2.7)
u
B
∴η = Ae kT = Ae T
where A is a proportional constant and B = u/k is a constant for a given liquid under
constant pressure.
Table 2.1 shows an example of the change in viscosity η with temperature for
castor oil, water and air. For liquids, its viscosity values decreases as the temperature
increases but for gases, the viscosity increases with increasing temperature.
For gas molecules, higher temperature means higher internal energy and more
energetic translational and vibrational motions. Consequently a molecules has higher
frequency of interactions with its neigbours causing higher viscosity.
34

35. Table 2.1: Typical values of viscosity
Temperature
Viscosity of
Viscosity of
Viscosity of
(°C)
castor oil
water
air
(poise)
(centipoise)
(micropoise)
0
53
1.792
171
20
9.86
1.005
181
40
2.31
0.656
190
60
0.80
0.469
200
80
0.30
0.357
209
100
0.17
0.284
218
We also found that when certain liquid is compressed at very high pressure, its
viscosity increases. This is due to the fact that the potential barrier increases as the
liquid is compressed.
2.6 Newtonian and non-Newtonian liquids
If the viscosity η is a constant number at a certain temperature and pressure,
the shear stress versus velocity gradient graph would be a straight line passing through
the origin of the graph. Such a liquid is called a Newtonian liquid. However, there are
liquids which do not follow a straight line behaviour as shown in Figure 2.6. Thus,
their viscosities η are not constants at constant temperature and pressure. Such liquids
are called non-Newtonian liquids.
Shear
stress
Newtonian
Non-Newtonian
Velocity gradient
(strain rate)
Figure 2.6: Newtonian and non-Newtonian liquids.
35

36. For non-Newtonian liquids, where the coefficient of viscosity depends on the
rate of shear stress, we can define an apparent viscosity. It can be determined from the
gradient of a straight line connecting the origin of the graph with a chosen point on
the non-linear curve. Hence, the apparent viscosity of the non-Newtonian liquid is
equal to the viscosity of a Newtonian liquid which has the same resistance at that
particular shear stress (or strain rate). Examples of non-Newtonian liquids are certain
colloidal suspensions and solutions of macro-molecules.
In most liquid engineering, whenever the changes in η is small, the liquids can
be considered as Newtonian liquids. Thin liquids commonly have Newtonian liquid
characteristics.
Quiz
Explain the meaning of the followings: (a) Newtonian liquid, (b) non-Newtonian
liquid.
Quiz
The apparent viscosity of a non-Newtonian liquid is determined from the gradient of a
straight line connecting the origin of the graph with a chosen point on the non-linear
curve. Can you sletch a graph to show this?
Example
Determine the shear stress needed to slide a plane plate at a rate of 0.08 m s-1 on a
plane surface which is separated from the plate by a film of oil 0.2 × 10-3 m and has a
viscosity of 0.45 N s m-2.
Solution
The shear stress
τ =η
v

= 0.45 ×
0.80
= 180 N m -2
0.2 × 10 −3
36

37. SUMMARY
1. Viscosity is a measure of resistance experienced when a layer of liquid attempts to
move over another neighbouring layer. The mechanisms causing the existence of
viscosity were also discussed.
2. Viscosity is defined as the ratio of the shear stress to the velocity gradient in the
liquid.
η=
F
A = Fl
v
Av
l
The unit for viscosity is N s m-2.
3. Viscosity is due to the forces acting between particles in adjacent layers of a
moving liquid. Furthermore, some particles (and hence, momentum) are transferred
between these layers due to the random thermal motion of the particles.
4. The coefficient of viscosity can be determined by several methods and one of them
is by using an Ostwald’s viscometer.
5. Viscosity depends on temperature as
η = Ae
u
kT
where A is a proportionality constant, u is the height of the potential barrier, k is the
Boltzmann’s constant and T is the temperature.
6. For a liquid with η constant at constant temperature and pressure, the liquid is
known as a Newtonian liquid.
37

38. EXERCISE 2
1. Explain why when the Ostwald’s viscometer is used, the time for the liquid to flow
from one level to the other is inversely proportional to the density of the liquid.
2. Two identical circular discs S and K of radii R are separated by a liquid with a
viscosity of η. The discs are parallel and are separated by a distance d and are
arranged in such a way that an axis passing through the centre of S will also pass
through the centre of K. Disc S is then rotated about the axis with a constant angular
velocity ω. Determine the couple q needed to maintain K at rest. Neglect edge effects.
[Ans: q =
ηωπR 4
2d
]
3. Since the viscosity of a liquid changes with temperature, what are the effects on a
car engine oil during winter and during summer?
38

39. CHAPTER 3
LIQUID FLOW
LESSON OBJECTIVES
After completing this lesson, you should be able to:
• Define the laminar and the turbulent flow.
• Apply the continuity and the Bernoulli’s equations.
• Derive the Bernoulli’s equation from the Euler’s equation.
• Describe and apply the Poiseuille’s equation.
• Describe the motion of an object in a viscous liquid and its usage to determine the
coefficient of viscosity of a liquid.
• Define the boundary layer.
• Define the Reynold’s number and its relationship with the critical velocity.
3.1 Introduction
In our secondary schools, we have come across several aspects of static liquids
such as Archimedes principle, buoyancy and others. In this chapter we will discuss
liquids in motion; or sometimes called hydrodynamics.
If we stand at the edge of a slow flowing river, we will realise that water in the
middle of the river is flowing more smoothly. In areas such as at the river bank where
the flow encounters obstacles or junctions, the smoothness of the flow is disturbed.
Thus we can conclude that in order to understand the flow of liquids, we first need to
know the types of flow involved. In physics, the flow of liquids basically can be
divided into two types; that is, laminar flow and turbulent flow. The path followed by
an element of a moving fluid is called a line of flow.
Laminar flow is a flow where the liquid is flowing smoothly by following an
orderly path (Figure 3.1(a)). This type of flow is usually found in situations where the
velocity of flow is low.
39

40. In turbulent flow, the paths of the liquid particles are no longer orderly paths
but are found to crisscross each other in a disorderly manner and in chaos (Figure
3.1(b)). Thus, in turbulent motion, secondary chaotic motions as well as fluctuating
velocity are found to be superimposed on the main or average flow of the liquid.
(a)
(b)
Figure 3.1: (a) Laminar flow. (b) Turbulent flow.
Other than the two types of flow, liquid flow can also be categorised as steady
flow, non-steady flow, uniform flow or non-uniform flow.
Steady flow is a flow where the flow parameters (such as velocity, pressure
and density) at a certain point in the liquid do not change with time. A flow where
changes occur with time is called non-steady flow.
If for a given time interval, the parameters of flow do not change from point to
point in a given region, the flow is called a uniform flow. Conversely if changes occur
from one point to another, then the flow is a non-uniform flow.
Three basic equations to describe the steady flow of liquids will be derived in
the coming sections. They are the continuity equation, the Euler’s equation and the
Bernoulli’s equation. These equations are analogous to the equations used to describe
the motion of solid particles.
Quiz
The flow of liquids can be divided into two types. What are they? Describe their
differences.
Quiz
What is meant by the secondary chaotic motion and the fluctuating velocity?
40

41. 3.2 The continuity equation
The equation describes the conservation of mass in liquid flow. Consider the
flow of a liquid through a tube as shown in Figure 3.2. Let A1 be the cross-sectional
area at the point where the liquid enters the tube (point 1), v1 is the average velocity
and ρ1 is the average density of the liquid at that point. Let the corresponding
parameters at the exit of the tube (point 2) be A2, v2 and ρ2. In a time interval of ∆t, the
volume of the liquid passing through point 1 is v1A1∆t and the mass of the liquid
passing through A1 is
M 1 = v1 A1 ρ1 ∆t
(3.1)
In the same time interval, the mass of the liquid passing through A2 is
M 2 = v 2 A2 ρ 2 ∆t
(3.2)
Since the flow is assumed to be a steady flow, then no liquid will be
accumulated anywhere in the tube. In this case, the mass of the liquid M1 passing
through A1 is equal to the mass of the liquid M2 passing through A2; that is
M1 = M 2
v1 A1 ρ1 ∆t = v 2 A2 ρ 2 ∆t
(3.3)
v1 A1 ρ1 = v1 A1 ρ1
This is the continuity equation. If the liquid is assumed to be incompressible,
then the density ρ1 = ρ2 and the continuity equation becomes
Q = v1 A1 = v 2 A2
(3.4)
where Q is the rate of flow of the liquid in m3 s-1. Q is also sometimes called the rate
of liquid discharge.
ρ2
v2
A2
A1
ρ1
v1
Figure 3.2: The flow of a liquid inside a tube.
41

42. Quiz
The continuity equation is similar to the conservation of mass equation. Please
comment on this statement.
3.3 The Euler’s equation
Let us consider a very small cylindrical element of a liquid. Let the length of
the cylinder be ds and the area at each end of the cylinder normal to the flow direction
be dA. The case is as shown in Figure 3.3 where v is the velocity and ρ is the density
of the liquid. Assume that, for a general case, the cylinder is tilted at angle θ to the
vertical represented by the z-axis.
dA
P
ds
z
θ
v (flow direction)
mg
P+
∂P
ds
∂s
Figure 3.3: An element of a flowing liquid for Euler’s equation
consideration.
The mass of the liquid inside the cylinder at any time is
m = ρ dA ds
(3.5)
The acceleration of m is dv/dt and the forces (P dA) as well as

∂P  
 P + ∂s ds dA are found to be acting on both ends of the cylinder where P is the
 

pressure. The weight of the liquid cylinder is mg. If the viscosity is assumed to be
zero, then the resultant force F acting on the liquid mass m is given by
42

43. ∂P 

F = P dA −  P +
ds  dA + ρg dA ds cos θ
∂s 

(3.6)
But, from Newton’s second law
F =m
dv
dv
= ρ dA ds
dt
dt
(3.7)
Hence, equation (3.6) equals equation (3.7)
dv
∂P 

P dA −  P +
ds  dA + ρg dA ds cos θ = ρ dA ds
dt
∂s 

dv
∂P
∴−
+ ρg cos θ = ρ
∂s
dt
(3.8)
If z is the vertical height from an arbitrary horizontal level chosen as reference, then
cos θ = −
∂z
∂s
θ
∂s
∂z
Also, for a very general case, v is a function of s and t; i.e. v ≡ v(s,t) and thus
dv
∂v ∂v
=v +
dt
∂s ∂t
Therefore, equation (3.8) becomes
−
∂P
∂z
 ∂v ∂v 
− ρg
= ρ v + 
∂s
∂s
 ∂s ∂t 
(3.9)
This is the Euler’s equation and is used to describe (in one dimension) the flow of a
non-viscous liquid. The flow involved may be a steady or non-steady flow.
If the flow is a steady flow type, then ∂v/∂t = 0 and the other derivatives
become full derivatives and the Euler’s equation becomes
ρv dv + dP + ρg dz = 0
(3.10)
Quiz
What are the conditions which should be satisfied before the above equation can be
used?
43

44. 3.4 The Bernoulli’s equation
Euler’s equation for the steady flow of a non-viscous liquid is given by
equation (3.10)
ρv dv + dP + ρg dz = 0
If ρ is a constant, i.e. the liquid is incompressible, then the above equation can be
integrated to become
ρv 2
2
+ P + ρgz = constant
(3.11)
This equation is called the Bernoulli’s equation where ρ is assumed to be constant, the
flow is assumed steady and the viscosity is zero. The Bernoulli’s equation can be used
in many normal flows where the assumptions are more or less obeyed, such as in a
low-velocity flow with low viscosity.
The constant in the Bernoulli’s equation is the total energy at any point in the
liquid. Generally, the constant is different for different streamline. A streamline is
defined as a curve whose tangent, at any point, is in the direction of the liquid velocity
at that point. In steady flow, the streamlines coincide with the lines of flow. This
means that the total energy is the same for any two points along a streamline if the
flow is steady, and the liquid is non-viscous as well as incompressible.
For a non-steady flow of a viscous liquid, the case is more complex. We will
not discuss the case here.
Quiz
What is the relationship between the Euler’s equation and the Bernoulli’s equation?
3.5 Applications of Bernoulli’s equation
First of all, we have to note that the equation of hydrostatic is actually a
special case of the Bernoulli’s equation. If the velocity is zero everywhere, i.e. v1 = v2
= 0, then the Bernoulli’s equation becomes
P1 − P2 = ρg ( y 2 − y1 ) = ρgh
which is the hydrostatic equation.
44

45. 3.5.1 The Venturi meter
Let us use the Bernoulli’s equation in an example of a Venturi meter. Consider
a liquid flow in a tube with varying cross-sections as shown in Figure 3.4. Let points 1
and 2 represent the two different cross-sectional areas of the tube. This type of tube is
called a Venturi meter. We would like to determine the volume rate of flow Q of the
liquid.
h
v1
v2
A2, P2
A1, P1
Figure 3.4: A Venturi meter.
Let P1 and v1 be the pressure and velocity of the liquid at point 1, and P2 and
v2 at point 2. Since points 1 and 2 are at the same level, then the term ρgz in equation
(3.11) can be cancelled out. Thus the Bernoulli’s equation becomes
1 2
1 2
ρv1 + P1 = ρv 2 + P2
2
2
1
2
P1 − P2 = ρ v 2 − v12
2
2(P1 − P2 ) = ρ (v 2 + v1 ) v 2 − v1
(
)
(
(3.12)
)
But according to the continuity equation, assuming steady and parallel velocities, the
rate of flow of the liquid is
Q = v1 A1 = v 2 A2
(3.13)
where A is the cross-sectional area of the tube. Thus
v1 =
Q
A1
and
v2 =
Q
A2
Then equation (3.12) becomes
 Q Q  Q Q 
2(P1 − P2 ) = ρ 


 A + A  A − A 
1  2
1 
 2
(3.14)
45

46. and it can be rewritten as
Q2 =
2(P1 − P2 )( A1 A2 )
ρ A12 − A22
(
2
)
(3.15)
Hence, if the pressure and the area of cross-section at any two points are known, the
volume rate of flow of the liquid can be determined.
The Venturi meter in Figure 3.4 can be used as a device to determine the rate
of flow of a liquid. Actually the Q value obtained by this method is bigger than the
real value as viscosity is neglected. Therefore, this device has to be calibrated first
before being put to use.
The reduced pressure at the constriction of a tube finds a number of technical
applications. Fuel vapour is drawn into the manifold of a car engine by the low
pressure produced in a Venturi throat to which the carburetor is connected.
Example
The diameter of a pipe is 12 cm at one end, 6 cm in the middle and 10 cm at the other
end as shown in the Figure below. If the average velocity of water at the 6 cm section
is 15 cm s-1, what are the average velocities at the other two sections? Assume that the
water is incompressible.
A1= 12 cm
A3 = 10 cm
A2 = 6 cm
v1
v2
v3
Solution
Let A1, A2 and A3 be the cross-sectional areas and v1, v2 and v3 be the velocity at the
sections as shown in the diagram above. Given that v2 = 15 cm s-1.
Since the water is incompressible, then the continuity equation
Q = v1 A1 = v 2 A2 = v3 A3
For v1A1 = v2A2 we found that
v1 =
A2 v 2 π × 3 2 × 15
=
= 3.7 cm s -1 #
A1
π × 62
For v3A3 = v2A2 we found that
46

47. A2 v 2 π × 3 2 × 15
=
= 5.4 cm s -1 #
2
A3
π ×5
v3 =
3.5.2 Speed of efflux
Figure 3.5 shows a large tank of cross-sectional area A1 filled to a depth h with
a liquid of density ρ. A small hole of cross-sectional area A2 is drilled near the base of
the tank. Both the surface of the liquid in the tank and the hole opening are exposed to
the atmospheric pressure Patm; i.e. P1 = P2 = P0.
Patm
1
•
h
•2
Patm
v2
Figure 3.5: A large tank holding a type of liquid with a small hole
at its base. The liquid is being discharged through the hole.
Let the surface of the liquid in the tank be represented by point 1 and the hole
be represented by point 2. The quantity v2 is called the speed of efflux. By using
Bernoulli’s equation
P1 +
(
1 2
1 2
ρv1 + ρgy1 = P2 + ρv 2 + ρgy 2
2
2
)
1
2
ρ v 2 − v12 = (P1 − P2 ) + ρg ( y1 − y 2 )
2
But since the cross-sectional area of the tank is very large compared to the crosssectional area of the hole, then A1 >>> A2 and v1 <<< v2. Hence, v1 is negligible
compared to v2 and can be neglected. We also have P1 = P2 = Patm and (y1 - y2) = h.
Then
47

48. 1 2
ρv 2 = ρgh
2
2
v 2 = 2 gh
∴ v 2 = 2 gh
This result is also called the Torricelli’s theorem. It is also similar to the speed
of a body after it fell freely through a distance of h under gravity.
3.5.3 Measurement of pressure in a moving fluid
(a) Open-tube manometer
Figure 3.6: The open-end manometer (a) channel wall opening
type, (b) probe type.
In Figure 3.6(a) shows one arm of the manometer is connected to an opening
in the channel wall. Figure 3.6(b) a probe is inserted in the stream of the flowing fluid.
The probe should be small enough so as not to disturb the flow and not to cause
turbulence. The pressure of the moving fluid is determined by
Patm − P = ρ m gh
P = Patm − ρ m gh
where ρm is the density of the manometer liquid.
48

49. (b) Pitot tube
P
P1
Patm
h
Figure 3.7: A Pitot tube.
The pitot tube is a probe with an opening at its upstream side (Figure 3.7). A
stagnation point forms at the opening where the pressure is P1 and the speed is zero.
Applying Bernoulli’s equation to the stagnation point and to another point at a large
distance from the probe where the pressure is P and the speed of the fluid is v
P1 = P +
1 2
ρv
2
where ρ is the density of the flowing fluid.
For the manometer part
P1 − Patm = ρ m gh
P1 = Patm + ρ m gh
where ρm is the density of liquid in the manometer.
Therefore, the pressure P of the moving fluid in the channel is given by
1 2
ρv
2
1
+ ρ m gh + ρv 2
2
Patm + ρ m gh = P +
∴ P = Patm
49

50. (c) Prandtl’s tube
2
1
h
Figure 3.8: A Prandtl’s tube.
This tube is actually a combination of the open probe and the pitot tube and is
as shown in Figure 3.8. The two ends of the manometer are not exposed to the
atmosphere. Hence
P1 − P2 = ρ m gh
As shown in the open probe diagram in the previous sub-section, P2 is the
actual pressure P of the flowing fluid in the channel. Hence
P1 − P = ρ m gh
But as shown in the pitot tube sub-section
P1 − P =
1 2
ρv
2
Hence
1 2
ρv = ρ m gh
2
2 ρ m gh
∴v =
ρ
This result indicates that if the Prandtl’s tube is held at rest, it can be used to
measure the speed of the flowing fluid by just reading h. This device is self-contained
and does not depend on the atmospheric pressure. If it is mounted on an aircraft, it
50

51. indicates the velocity of the aircraft relative to the surrounding air which is called the
airspeed.
Other examples of the uses of the Bernoulli’s equation are the lift on an
aircraft wing and the curved flight of a spinning ball. However, they will not be
discussed in this module.
Quiz
When two closely-spaced pieces of paper is blown in order to separate them, we
found that they become closer to each other. Explain this observation. You may carry
out this experiment.
3.6 The Poiseuille’s equation
In the previous sections, the viscosities of the liquids were ignored when
calculating the rate of flow. For liquids with a high and non-negligible viscosity,
Bernoulli’s equation cannot be used. For this type of liquid, another equation called
the Poiseuille’s equation should be used to calculate the rate of flow. It describes the
flow of a Newtonian liquid in a uniform and rigid cylindrical tube. Furthermore, the
flow in the tube should be a steady and laminar flow.
Consider a cylindrical tube of length L and radius R as shown in Figure 3.9.
Let P1 and P2 be the pressures at the ends of the tube.
P2
P1
v
r
R
dr
L
Figure 3.9: A cylindrical element of a viscous liquid with radius r
flowing in a cylindrical tube.
51

52. To derive the Poiseuille’s equation, let us consider the forces acting on a
cylindrical liquid element of radius r and a thickness dr flowing in a tube with a
constant velocity v as shown in Figure 3.9.
Viscosity force as shown in Chapter 2 is given by
F = −ηA
dv
dv
= −2πrLη
dr
dr
(3.16)
assuming that the liquid is a Newtonian liquid. The force acting on the liquid due to
the presence of the pressure difference ∆P is ∆Pπr2 where ∆P = P1 - P2. Since the
velocity v is constant, then the viscosity force should be balanced by the force due to
∆P. Hence
dv
= (P1 − P2 )πr 2
dr
∆P
dv = −
r dr
2 Lη
∆P 2
∴v = −
r +C
4 Lη
− 2πrLη
(3.17)
At the tube wall r = R, v = 0. Hence
C=
∆PR 2
4 Lη
(3.18)
Therefore, the general equation (3.17) becomes
v=−
∆Pr 2 ∆PR 2
∆P
+
=
R2 − r 2
4 Lη
4 Lη
4 Lη
(
)
(3.19)
From the last equation it could be seen that v is a function of r; i.e. the velocity of the
liquid is not uniform across the tube. The velocity is maximum at the centre of the
tube (r = 0) and zero at the wall of the tube (r = R).
52

53. v=0
R
dr
r
vmax
v=0
Figure 3.10: The velocity profile of a flowing viscous liquid
depends on r. vmax at the centre of the tube and v = 0 at the wall of
the tube.
The rate of flow Q in the rigid tube is
R
Q = ∫ 2πvrdr
0
R
= ∫ 2πr
0
=
∆P
R 2 − r 2 dr
4 Lη
(
)
2π∆P R 2
3
∫0 R r − r dr
4 Lη
(
)
R
2π∆P  R 2 r 2 r 4 
− 
=

4 Lη  2
4 0
=
π∆PR 4
8 Lη
(3.20)
This equation is called the Poiseuille’s equation and it gives the rate of flow in
terms of the pressure difference, the radius and the length of the tube as well as the
viscosity of the liquid. The volume flow rate decreases if the viscosity of the liquid
increases and the flow rate increases if the radius of the tube and/or the pressure
difference increases.
Another important property of equation (3.20) the pressure difference ∆P
between both ends of a uniform tube depends on the length L of the tube even when
the tube is placed horizontally. That is
∆P ∝ L
53

54. This property contradicts the case of non-viscous liquids where the Bernoulli’s
equation is valid. For a non-viscous liquid flowing in a uniform horizontal tube, there
will be no pressure difference along the tube. Figure 3.11 illustrates this point.
(a)
(b)
Figure 3.11: Pressure along the flow of a liquid in a uniform
horizontal tube. (a) For a non-viscous liquid, the Bernoulli’s
equation is obeyed. No pressure difference along the horizontal
tube. (b) For a viscous liquid, the Poiseuille’s equation is obeyed.
Pressure changes along the horizontal tube.
Quiz
State the conditions required in order to use the Poiseuille’s equation.
3.7 Motion of an object in a viscous liquid
So far we have discussed the flow of a liquid in stationary containers. Now let
us discuss the motion of solid objects in stationary liquids. We will limit our
discussions to a special case only; that is, the motion of a spherical object under
gravity in a stationary liquid. This is also a method to determine the coefficient of
viscosity of a liquid as stated in Chapter 2.
When an object falls under gravity in a liquid, initially it will be accelerated.
At this instance the force due to gravity mg is bigger than the summation of the
buoyancy force Fb and the viscous force (Stoke’s force) FS. The viscous force acting
on the object increases with increasing velocity, while the gravity force and the
buoyancy force are constants (Figure 3.12(a)). Thus the acceleration of the object
gradually decreases until a terminal velocity vT is attained. This happens when the
upward forces and the gravity force are in equilibrium (Figure 3.12(b)). The changes
54

55. in velocity from the time when the object is initially released until the terminal
velocity vT is attained is as shown in Figure 3.12(c).
v
Fb
Fb
FS
FS
vT
vT
a
mg
mg
(a)
0
(b)
x or t
(c)
Figure 3.12: Forces acting on a spherical object falling in a
viscous liquid. (a) Initially, the velocity increases. a is the
acceleration of the object. (b) The object attains a terminal
velocity vT. (c) Plot of velocity against time or distance traveled.
The viscous force depends on the shape and size of the object, nature of the
object’s surface, density of object and liquid, viscosity of liquid as well as the
characteristics of the liquid flow around the object as it falls. Thus the determination
of the viscous force is generally very complex. However, for a rigid spherical object
falling steadily and slowly in a Newtonian high viscosity liquid, Stokes had
determined that the viscous force acting on the object is given by FS =6πηrv where r
is the radius and v is the velocity of the spherical object.
Before vT is attained, the Newton’s second law gives
∑ F = ma
mg − Fb − FS = ma
mg − ml g − 6πηrv = ma
Hence, when the object attains the terminal velocity vT (i.e. a = 0) and the
forces are in equilibrium, the
55

56. mg − ml g − 6πηrvT = 0
4 3
4
πr ρ s g − πr 3 ρ l g − 6πηrvT = 0
3
3
4
4
6πηrvT + πr 3 ρ l g = πr 3 ρ s g
3
3
(3.21)
where ρs is the density of the object, ρl is the density of the liquid and g is the
gravitational acceleration.
From the equation above we have
η=
2r 2 g (ρ s − ρ l )
9v T
(3.22)
Thus by just observing the motion of a spherical object under gravity in a
viscous liquid, the viscosity coefficient of the liquid can be determined.
Example
If the viscosity drag acting on an object of volume V and density ρs falling through a
liquid of density ρc is proportional to v2 where v is the velocity of the object, derive an
expression for the terminal velocity vT of the object.
Solution
Let the viscous drag Fd as
Fd = kv 2
where k is a proportionality constant.
When the terminal velocity vT is attained
mg − Fb − Fd = 0
2
ρ sVg − ρ lVg − kvT = 0
2
kvT = (ρ s − ρ l )Vg
2
vT =
(ρ s − ρ l )Vg
∴ vT =
k
(ρ s − ρ l )Vg
k
#
56

57. 3.8 Boundary layer
In the study of low viscosity liquid flowing in contact with a solid surface, it is
useful to divide the flow field into two regions. One of the regions is the one close to
the solid surface such as the wall of a tube. In this region, the viscosity of the liquid
will affect the flow of the liquid.
The second region is the one far from the solid surface. In this region, it can be
assumed that the viscosity of the liquid does not have any significant effect. (The
division of the flow field into two regions is sometimes called the Prandtl’s boundary
layer hypothesis.) In many practical cases, the liquid in the second field region can be
assumed to be non-viscous. Hence, the equations for non-viscous liquids such as the
continuity and the Bernoulli’s equations can be used here.
The first region where liquid viscosity cannot be neglected is called the
boundary layer. We have discussed previously that when a liquid is flowing along a
stationary surface, a thin layer of liquid will stick to the solid surface and the velocity
of this layer is zero. The velocity of other layers increases as the layer is situated
further away from the surface. This condition continues until a limit is reached where
the velocity of the flow is equal to vmax (Figure 3.13). The region where the variation
of flow velocity occurs is called the boundary layer. Commonly, the thickness of the
boundary layer is taken as the distance from the solid surface where the velocity of
flow is 99% of the velocity of the main flow vmax. Boundary layer can be found in
laminar flow as well as in turbulent flow and the flow in boundary layer itself can be
laminar or turbulent.
57

58. y
v = vmax
99% vmax
Solid
surface
Boundary layer
v
Figure 3.13: Boundary layer in a liquid flowing in contact with a
solid surface. y is the distance from the solid surface and vmax is
the velocity of the main flow.
3.9 Critical velocity and the Reynolds’ number
When the velocity of liquid flow is low in a channel, then the flow is found to
be laminar. This state can be maintained even if the velocity of the flow is slightly
increased. However, this process cannot be continued indefinitely if we want to
maintain laminar flow, as in the process of increasing the flow velocity, we will reach
a certain value when the laminar flow turns turbulent. The velocity where turbulence
starts to occur is called the upper (or higher) critical velocity.
Conversely, if the velocity of a turbulent flow is reduced until a level is
reached where the flow becomes laminar, the velocity when laminar flow starts is
called the lower critical velocity. The critical velocity vc is the average of these
velocities, that is
vc =
Q
A
(3.23)
where A is the cross-sectional area of the channel and Q is the volume rate of flow of
the liquid.
Reynolds found that the critical velocity vc of a flow depends on the density ρ
as well as the viscosity η and the size d of the channel. For a cylindrical channel, the
size of the channel is the diameter of the cylinder. The relationship between vc with
those quantities is
vc =
Recη
ρd
(3.24)
58

59. where Rec is a number called Reynolds’ critical number and it is without a unit. Since
there are two vc values, then two Reynolds’ number can be defined. However, from
practicality point of view, it was found that the more important one is the lower
critical value. Thus, usually the lower Reynolds’ critical number is just called the
Reynolds’ critical number. Hence, for all values of v, the Reynolds’ number is
Re =
ρvd
η
(3.25)
As an example, for water flowing in a cylindrical pipe, the transition from
laminar flow to turbulent flow is found to occur at Reynolds’ critical number of about
2000; that is, the flow is laminar if critical NR < 2000 and it is turbulent if critical NR >
2000.
Hence, the properties of a flow can be characterised by NR. It can also be used
to describe dynamics similarities. This means that if the flow pattern is to be
maintained for any situation, the NR value should have the same value. For example, if
we want the flow pattern of two different liquids to be similar, the Re values of both
liquids should be the same. Hence, if any variable in NR expression changes, then the
other variables should be modified in order to maintain the type and pattern of the
flow. Hence NR values can be utilised by engineers to build large practical systems
using data derived from small model systems.
Quiz
Show that Reynold’s number has no unit.
3.10 Comparison between laminar flow and turbulent flow
For a laminar flow in a cylindrical tube, the velocity profile of the flow is
parabolic where the flow has been fully expanded. However, when the flow becomes
turbulent, the liquid particles have random cross motions. This state causes the
velocity of the flow at any point in the tube cross-section to be equalised. Hence, the
velocity profile of the turbulent flow tends to be more flattened. For a laminar flow,
the energy loss is proportional to the average flow velocity to the power of one, but
for a turbulent flow, the energy loss is found to be proportional to the average flow
velocity to the power of 1.85 up to 2.
59

60. In a turbulent flow, the liquid particles are always accelerated since their
motions are always changing; i.e. the velocity changes. Thus, the mass factor or the
liquid density will become more important. Higher quantity of energy is needed to
push the liquid particles around. This situation causes higher energy loss. If the
density of a liquid is higher, then the energy loss will be higher. Due to the presence
of higher energy loss, higher pressure is needed to produce a certain flow rate
compared to a laminar flow for the same flow rate. For a laminar flow, from the
Poiseuille’s equation, the pressure difference for a certain liquid flow is in the form
∆p = AQ
(3.26)
where A is a constant and Q is the rate of flow of the liquid in m3 s-1.
For a turbulent flow, the pressure difference is in the form
∆p = AQ + BQ 2
(3.27)
where B is a term as a function of viscosity and density of the flowing liquid.
Thus, the energy loss in a laminar flow is due to the presence of viscosity,
whilst in a turbulent flow the energy loss is due to viscosity as well as the
convectional motion of the liquid particles.
Quiz:
State the differences between a laminar flow and a turbulent flow.
SUMMARY
1. The continuity equation is
Q = A1v1 = A2 v 2 =  = An v n
where Q is the rate of flow in m3 s-1, A is the cross-sectional area of the channel and v
is the velocity of the liquid flow across A.
2. The Bernoulli’s equation is
P+
1 2
ρv + ρgy = constant
2
60

61. where P is the pressure at a point in the flowing fluid, ρ is the density of the fluid and
y is the height of the point against a reference level.
3. The hydrostatic equation (or gauge pressure equation) is
∆P = ρgh
4. The Poiseuille’s equation is
Q=
dV π∆PR 4
=
dt
8ηL
where Q is the rate of flow in m3 s-1, ∆P is the pressure difference at both ends of the
tube, R is the radius of the tube, η is the coefficient of viscosity and L is the length of
the tube.
5. The Stoke’s equation for a solid spherical object moving in a stationary liquid is
FS = 6πηrv
where η is the coefficient of viscosity, r is the radius of the sphere and v is the
velocity of the spherical object.
4. The boundary layer is taken as the distance from the solid surface (wall of channel)
where the velocity of flow is 99% of the maximum velocity vmax of the main current.
5. The Reynold’s number is given by
NR =
ρvd
η
6. The critical velocity vc occurs when NR ≈ 2000. If NR < 2000, the flow is laminar
and if NR > 2000, the flow is turbulent.
61

62. EXERCISE 3
1. Water is flowing steadily through a piping system as shown. The cross-sectional
area of pipes 1, 2 and 3 are identical at 1.4 cm2. The cross-sectional area of pipe 4 is
2.8 cm2. Water enters pipe 1 at a rate of 3 × 10-2 m3 s-1. What is the speed of flow in
each section of the pipe?
2
4
1
3
[Ans: v1 = 214 m s-1; v2 = v3 = v4 = 107 m s-1]
2. Water is flowing in a horizontal pipe from one section to another section which has
a cross-sectional area 1/3 of the first section. The velocity of water in the smaller
section is 8.0 m s-1. What is the pressure drop from the first section to the second
section?
[Ans: ∆P = 28.4 × 103 N m-2]
3. A glass tube is inserted into a hole at the centre of a piece of circular card B as
shown. Another piece of card C is then held so that it touches B. Air is blown steadily
into the tube and into the space between B and C. The support for C is then removed.
C is found to be sticking to B and it does not fall off. Explain this observation.
4. A pipe containing flowing water decreases in size from 0.4 m2 at A to 0.25 m2 at B.
Assuming a steady flow, the velocity at A is 1.8 m s-1 and the pressure is 105 N m-2. If
the loss due to viscosity is negligible, determine the pressure at B which is 5 m higher
than A.
[Ans: 4.85 × 104 N m-2]
5. The diagram below shows two identical pipes but with different diameters. Water
flows in the pipe with the bigger diameter whilst oil of density 0.8 g cm-3 flows in the
62

63. smaller pipe. If the rate of flow of water is 30 cm3 s-1, determine the rate of flow of the
oil if its viscosity is 0.012 poise. Assume that the flow pattern is similar in both cases.
Water
Oil
0.6 cm
0.4 cm
[Ans.: 30 cm3 s-1]
6. A liquid which has a kinematics viscosity of 0.38 × 10-3 N s m-2 flows through a
pipe of diameter 0.07 m at a rate of 0.012 m3 s-1. What is the type of flow obtained?
[Ans.: turbulent]
63

64. CHAPTER 4
DIFFUSION
LESSON OBJECTIVES
After completing this lesson, you should be able to:
• Define the term diffusion.
• Derive the Fick’s equation.
• Describe the solution to the Fick’s equation.
• List some applications of the Fick’s equation.
4.1 Introduction
In this chapter we will discuss an important process in physics. This process is
not only found in physical systems, but also in chemical and biological systems. The
process is known as the diffusion process.
We found that fluid molecules move randomly because they have thermal
energy. If for some reasons, the molecules are found to be located in one region, they
tend to move from that region to other regions with lower molecular concentration.
Think about what happens to a drop of ink in a glass of water. The net flow of
particles from higher concentration regions to lower concentration regions is called
diffusion. Diffusion process does not only present in liquids and gases, but also to
some degree in solids.
Diffusion process does not depend on any bulk motion of the materials such as
in a blowing wind or a convectional current. It also does not depend on the
disturbances due to pressure or temperature differences.
Diffusion process can also occur against the gravitational pull. For example, if
a higher density fluid layer (say in chamber A) is placed below a lower density fluid
layer (say in chamber B), after some time we will find that some fluid A molecules
can be found in B and vice-versa (Figure 4.1). The sliding partition is withdrawn for a
definite interval of time. When in contact with each other, they are found to diffuse
64

65. into each other. From the average composition of one chamber to the other, the
diffusion coefficient D may be calculated.
Figure 4.1: Apparatus for the measurement of the diffusion
coefficient D.
Quiz
Microscopically, describe what happen in a diffusion process.
Quiz
A heavy gas, such as nitrogen, is placed at the bottom of a container and the top of the
container is filled with a lighter gas, for example hydrogen. The container is then left
standing. After a time t, nitrogen can be found at the top and hydrogen can be found at
the bottom of the container. Explain this observation.
4.2 Diffusion equation
Consider the flow of particles, in one-dimension only, in a pipe from a higher
particle concentration region to a lower particle concentration region as shown in
Figure 4.2.
65

66. C1
C2
Flow
direction
High
concentration
x1
∆x
Low
A concentration
x2
Figure 4.2: One-dimensional diffusion process in a cylindrical
pipe.
Assume that the particle concentration is uniform in any cross-section of the
pipe at any specific time but the concentration is allowed to change with time.
However, the particle concentration is different from section to section of the pipe.
Let the concentration of particles at x1 and x2 planes be C1 and C2 respectively where
C1 > C2. From experiments, the mass of particles moved or transferred from x1 to x2 in
the time duration ∆t is proportional to C1 and the mass transferred in the opposite
direction is proportional to C2. Both are found to be proportional to the cross-sectional
area of the pipe A. Hence, the resultant net mass ∆m crossing x2 in time ∆t is
∆m
∝ AC1 − AC 2
∆t
∆m
= kA(C1 − C 2 )
∆t
∆m
∴
= −kA ∆C
∆t
(4.1)
where k is a proportional constant and the negative sign indicates that the particle
concentration decreases when ∆x increases. If the distance between x1 and x2 is large,
then the mass transfer rate from x1 to x2 is small and the k value is small. Similarly, if
the distance between x1 and x2 is small, then the k value is high. Thus, k is inversely
proportional to (x2 – x1) = ∆x. Hence k can be written as
k=
D
∆x
(4.2)
where D is a second constant called the Fick’s diffusion constant. Hence, equation 4.1
can be rewritten as
∆m
D
∆C
=−
A∆C = − DA
∆t
∆x
∆x
(4.3)
66

67. At its limit, when ∆t → 0 and ∆x → 0, we found
∂m
∂C
= − DA
∂t
∂x
(4.4)
These derivatives are partial derivatives as both variables C and m are time
and position functions. This final expression describing the diffusion process is called
the Fick’s first law of diffusion. The unit of D is m2 s-1.
Quiz
Why do the derivatives in equation (4.4) are partial derivatives?
The diffusion equation can also be written in another form. The mass per
second entering the right part of x1, from equation 4.4 is
 ∂C 
 ∂m 
 ∂t  = − DA ∂x 
  x1
  x1
(4.5)
and the mass per second exiting from the left part of x2 is
 ∂C 
 ∂m 
 ∂t  = − DA ∂x 
  x2
  x2
(4.6)
where the subscripts x1 and x2 indicate the positions where the derivatives are
calculated.
Hence, the rate of mass exchange in the region between x1 and x2 is
∂ ( AC∆x )  ∂m 
 ∂m 
=  − 
∂t
 ∂t  x1  ∂t  x2
∂C
 ∂C 
 ∂C 
A∆x
= − DA  + DA 
∂t
 ∂x  x2
 ∂x  x1
(4.7)
where AC∆x is the mass of particles in this region.
But
∂ (∂C ∂x )
 ∂C 
 ∂C 
∆x
=  +
 ∂x 
∂x
  x2  ∂x  x1
∂ 2C
 ∂C 
=   + 2 ∆x
 ∂x  x1 ∂x
(4.8)
Therefore equation 4.7 can be written as
67

68. A∆x
∂ 2C
∂C
 ∂C 
 ∂C 
= − DA  + DA  + DA 2 ∆x
∂t
∂x
 ∂x  x1
 ∂x  x1
∂ 2C
= DA 2 ∆x
∂x
(4.9)
Hence
∂C
∂ 2C
=D 2
∂t
∂x
(4.10)
This equation is also Fick’s law in one dimension written in another form. It is
called the Fick’s second law of diffusion. It can be used to determine the distance x a
particle may diffuse in a certain period of time.
The equation also contains the fact that C and the gradient of C will change
during the diffusion process. In three dimensions, equation 4.10 can be written in a
more general form as follows
 ∂ 2C ∂ 2C ∂ 2C 
∂C
= D 2 + 2 + 2 
∂t
∂y
∂z 
 ∂x
and in the vector form


∂C
= D∇ 2 C
∂t
(4.11)
(4.12)
4.3 Solution to the diffusion equation
The solution to the diffusion equation (4.10) has several forms, each
depending on the conditions applied.
Let the initial conditions is a situation where all particles are located at x = 0
when t = 0. Equation 4.10 is a second-order, linear and homogenous differential
equation. Its solution yields concentration as a function of time and distance
C=
β
t
e
 x2 

−
 4 Dt 


(4.13)
where β is a constant. It may be shown that at t = 0, C = 0 everywhere except at the
origin x = 0, where C → ∞. The constant β can be evaluated from the condition that at
any time all particles must be somewhere between x = -∞ and x = +∞. Therefore, the
number of particles initially at present at x = 0 is
68

69. +∞
−∞
−
+∞
1
2
 x2 

−
 4 Dt 


−∞
N = ∫ C dx = ∫ βt e
=
β
t
∫
 x2 


+ ∞ − 4 Dt 


−∞
e

dx
dx
 x2 
 x2 



+ ∞ − 4 Dt 


β  0 − 4 Dt 
e   dx + ∫ e   dx 
=
∫
t

−∞
0


But, the general solution for this definite integral is in the form of
∫
∞
0
1 π
2 a
e − ax dx =
2
(For example, see Fried et. al., 1977. Physical Chemistry. MacMillan Publishing Co.
Inc.)
Hence



β 1
π
π 
1

N=
+
t 2  1  2  1  





 4 Dt  
 4 Dt 

=
β
t
[
4πDt
]
= 2 β πD
or
β=
N
2 πD
Substituting β into equation (4.13), we have
C=
N
2 πDt
e
 x2 

−
 4 Dt 


(4.14)
where N is the number of particles initially present at x = 0, sometimes called the
strength of the diffusion source. C is the concentration expressed as the number of
particles per unit distance.
The probability that a particle will be found between x and x+dx is
69

70. P=
=
C (x )
dx
N
1
2 πDt
e
 x2 

−
 4 Dt 


(4.15)
dx
The solution is plotted in Figure 4.3 for several values of t. From the graph we
can see that the particles which are moving randomly will diffuse from x = 0 and
when the time approaching infinity, the concentration of the particles approaches zero
for all values of x as stated earlier.
Concentration C
t=0
t1
t3 > t2 > t1
t2
t3
0
x
Figure 4.3: Graph representing equation 4.13 for several t values.
Quiz
Equation (4.10) has one solution only. Right or wrong?
Since, microscopically, the diffusion process is actually a random motion
process, then it is impossible for us to predict accurately how far a particular particle
can travel after a certain time. The mean distance which a particle diffuses x is zero
because diffusion in +x and -x directions is equally probable. However, we can still
calculate, from the solution of the diffusion equation and the random motion process
equation, the mean square displacement x 2 of a particle after a time t. In one
dimension, the value of x 2 is given by
70

71. ∞
x 2 = ∫ x 2 f ( x )dx
−∞
∞
1
−∞
2 πDt
= ∫ x2
e
 x2 

−
 4 Dt 


dx
(4.16)
= 2 Dt
If we determine the square root of the expression we have
x 2 = 2 Dt
x rms ∝ t
(4.17)
The root-mean-square of the distance is proportional to the square root of the time t.
For a normal motion, we found that to cover a distance of 2l, we need a time of 2t if
the time taken to cover a distance l is t. However, for a random motion, this will not
happen. There are particles covering a distance of more than, and also less than, x 2 in
time t, but what is calculated here is the mean only.
In other words, to diffuse to a mean distance of x, the time needed is
proportional to x2. This characteristic is an important one for a diffusion process and
causes several unexpected effects. For example, diffusion coefficient D for small
molecules in water at normal temperature is about 10-3 cm2 s-1. For this D value, if a
tube of length 1 cm is used, the particle concentration in the tube will become uniform
in about 20 minutes. But if a tube of length 1 m is used, similar state may be attained
after a few months!
Quiz
1
Plot the equation x = t 2 and comment on the shape of the curve obtained.
4.4 Several processes involving diffusion process
The diffusion process is a manifestation of the macroscopic movement of the
microscopic particles existing in a fluid. This process can be found in many physical,
chemical as well as biological processes. Some examples of them are:
1. Diffusion process can control chemical reaction rates.
2. The movement of materials through a cell membrane is achieved by
diffusion.
3. Defects can diffuse in a crystal.
71

72. 4. Fragrance from one place will spread to another place by diffusion process
without convection.
5. Diffusion process can also be used to determine the volume of molecules.
SUMMARY
1. Diffusion is the transfer of particles from a region where its concentration is high to
a region of low concentration due to the thermal motion of the particles.
2. Diffusion does not depend on the bulk motion or flow of the material such as
convection or disturbances due to pressure or temperature differences.
3. Diffusion equation is defined by the Fick’s law as
∂m
∂C
= − DA
∂t
∂x
4. The solution to the Fick’s equation depends on the given conditions.
5. Diffusion process is not only found in physical systems but also in chemical and
biological systems.
EXERCISE 4
1. By substitution method, show that equation (4.13) is the solution to equation (4.10).
2. A small bit of a material is placed at the bottom of a water column 6 cm deep.
Estimate the time taken by the material to diffuse to the water surface. Given that the
Fick’s diffusion constant D for the material is 3.5 × 10-11 m2 s-1. Assume that there is
no agitation in the water column.
[Ans.: 1.6 years]
72

73. 3. The diffusion coefficient of a certain gas in a certain solid is 10-8 cm2 s-1. How far
would a gas molecule be expected to diffuse in the solid in a million years?
[Ans.: 7.94 m]
73

74. REFERENCES
Lim K.O. JIF104 Fizik II/Amali 1b. Pusat Pengajian Pendidikan Jarak Jauh,
Universiti Sains Malaysia.
Fried V., Blukis U. and Hameka H.F., 1977. Physical Chemistry. MacMillan
Publishing Co. Inc.
Nelkon M., 1969. Mechanis and Properties of Matter. Heinemann Educational Books
Ltd.
Sears F.W., Zemansky M.W. and Young H.D., 1980. College Physics.
Addison.Wesley Publishing Company.
Alberty R.A. and Daniels F., 1979. Physical Chemistry. John Wiley and sons.
Wilson J.D., 1994. College Physics. Prentice Hall.
Douglas J.F., Gasiorek J.M. and Swaffield J.A., 2001. Fluid Mechanics. Prentice Hall.
74

75. SOLUTIONS TO THE EXERCISES
SOLUTIONS TO EXERCISE 1
1. Given: γ = 0.04 N m-1; d1 = 10 cm = 0.1 m; d2 = 15 cm = 0.15 m.
The free surface energy is given by
W
2∆A
∴W = γ × 2∆A
γ =
(
)
= 0.04 × 8π (0.075 − 0.05 )
= γ × 2 × 4π r22 − r12
2
2
= 3.14 × 10 −3 J #
2. The free surface energy γ will stretch the soap film isotropically in two dimensions.
Hence, the forces acting on the thread due to the soap film are directed radially away
with equal magnitudes as shown figure (a) below. Therefore, the thread loop will form
a circle.
y
x
2γ
dl
T
θ
r
2γ
dF
θ
r
T
(a)
(b)
75

76. If we cut the thread loop into two identical halves, the radially outward force
due to free surface energy is balanced by the tension T of the thread as shown in
figure (b) above.
Consider the component of force dF directed to x-axis of figure (b) acting on
an infinitesimal length dl of the thread due to the free surface energy γ. The total
force in the x component is
Fx = ∫ dF = ∫ 2γ cos θ dl
But the arc dl is defined as
dl = r dθ
Then the total force in the x direction becomes
π
Fx = ∫ 2π 2γ cos θ ⋅ r dθ
−
2
π
= 2γr ∫ 2π cos θ dθ
−
2
π
= 2γr [sin θ ] 2π = 2γr × 2 = 4γr
−
2
This total force in the x direction is balanced by the two tensions in the -x direction;
that is
2T = 4γr
∴ T = 2γr
#
3. Given: d = 0.3 mm ⇒ r = 0.15 mm = 0.15 × 10-3 m; ρ = 13.6 × 103 kg m-3; α =
130°; γ = 0.49 N m-1.
The forces acting on the mercury column are:
Force due to surface effect acting downwards
= Force acting due to the displacement of mercury in the column
2πrγ cos α = πr 2 hρg
2γ cos α
2 × 0.49 cos130°
∴h =
=
= −3.15 × 10 − 2 m = −3.15 cm #
ρrg
13.6 × 10 3 × 0.15 × 10 −3 × 9.8
The negative sign indicates that the mercury level in the tube is below the
surface of the mercury outside the tube.
76

77. 4. Given: Thickness of water film x = 1.0 × 10-5 m; radius of plates R = 0.06 m; γ = 73
× 10-3 J m-2; α = 0°.
R
2R
γ
x
P
Po
x
γ
A
(b)
Projected effective cross-sectional
area acted by the pressures.
(a)
First, we have to determine the pressure difference between the water film and
the atmosphere.
Let us consider the equilibrium of one half of the plates. Figure (a) shows the
pressures acting on the exposed surface of the water film where P is the pressure
inside the water and Patm is the atmospheric pressure. Surface tensions are only acting
at the points where the water film touches the glass plates; i.e. only two horizontal γ in
Figure (a). The effective cross-sectional area A of the water film where the pressures
are being considered is as in Figure (b); i.e. the area of the rectangle 2Rx minus the
area of two half-circles πx2/4.
A = 2 Rx −
πx 2
4
=
8 Rx − πx 2
4
The forces acting on the one half of the plates
PA + 2 × 2 Rγ = P0 A
(Patm − P ) = 4 Rγ
A
(Patm − P ) =
4 Rγ
 8 Rx − πx


4

2




=
16 Rγ
16 Rγ
=
2
x(8 R − πx )
8 Rx − πx
The force needed to take the two plates apart
F = (Patm − P )A′
where A′ = πR 2 is the area of each plate.
77

78. F = (Patm − P )A′
=
16 Rγ
× πR 2
x(8 R − πx )
=
16 R 3γ
x(8 R − πx )
=
16π × 0.06 3 × 73 × 10 −3
= 165.1 N #
1 × 10 −5 8 × 0.06 − π × 1 × 10 −5
(
)
5. Given: d = 5 mm = 5 × 10-3 m; T = 0.073 N m-1.
Let us cut a section of the water cylinder along its axis to have a semi-cylinder
with length l and diameter d.
T
Patm
P
d
A
T
l
Projected effective cross-sectional
area of a semi-cylinder.
At equilibrium
PA = Patm A + 2lT
where A is the cross-sectional area of the cylinder along its axis.
(P − Patm )A = 2lT
(P − Patm )ld = 2lT
∴ (P − Patm ) =
2T 2 × 0.073
=
= 29.2 N m -2 #
d
5 × 10 −3
78

79. SOLUTIONS TO EXERCISE 2
1. Due to Poiseuille’s equation (2.3)
1
t∝
ρ
2.
ω
A liquid element of the liquid layer.
r
R
S
δl
d
(b)
K
(a)
The arrangement of the discs is as shown in Figure (a). Consider a circular
element of the liquid with radius r and thickness δl as shown in Figure (b). The shear
force acting on the plane surface of the element is
F = ηA
δv
δl
But A = πr 2 and v = rω. Hence
F = η ⋅ πr 2 ⋅
δ (rω )
δr
= η ⋅ πr 2 ⋅ ω
δl
δl
as ω is a constant. The couple is given by
q = F ⋅ D = F ⋅ 2r = 2ηπr 3ω
δr
δl
qδl = 2ηπω r 3δr
d
R
0
0
q ∫ δl = 2ηπω ∫ r 3δr
q[l ]
d
0
R
r4 
= 2ηπω  
 4 0
 R4
qd = 2ηπω 
 4

ηπωR 4
∴q =
2d




#
79

80. SOLUTIONS TO EXERCISE 3
1. Given: A1 = A2 = A3 = 1.4 cm2 = 1.4 × 10-4 m2; A4 = 2.8 cm2 = 2.8 × 10-4 m2; Q = 3
× 10-2 m3 s-1.
Using the continuity equation
Q = A1v1
∴ v1 =
3 × 10 − 2
Q
=
= 214 m s -1 #
−4
A1 1.4 × 10
Since A2 = A3, then the quantity of water passing through point 2 and point 3 is
Q/2 respectively. Hence
A2 v 2 = A3 v3 =
∴ v 2 = v3 =
Q
2
3 × 10 − 2
Q
=
= 107 m s -1 #
2 A2 2 × 1.4 × 10 − 4
For pipe 4
A4 v 4 = Q
∴ v4 =
2. Given: A2 =
3 × 10 − 2
Q
=
= 107 m s -1 #
−4
A4 2.8 × 10
1
A1; v2 = 8.0 m s-1.
3
A2 =
A1
1
A1
3
Using Bernoulli’s equation
1 2
1 2
ρv1 = P2 + ρv 2
2
2
(P1 − P2 ) = 1 ρ v22 − v12
2
P1 +
(
)
But from the continuity equation
80

81. A1v1 = A2 v 2
1
v1 = v 2
3
Hence, by substituting v1
1 2
v2 
2 
32 

1  1 2
= ρ  1 − v 2
2  9
1
 1
= × 10001 −  × 8 2 = 28.4 × 10 3 N m -2 #
2
 9

(P1 − P2 ) = 1 ρ  v 22 −

3. The cross-section of the set-up is as shown below.
Tube
1
B
•
•2
(atmosphere)
C
Let point 1 represents the space between card B and card C; and let point 2
represents the space outside the cards arrangement. Using Bernoulli’s equation
1 2
1 2
ρv1 = P2 + ρv 2
2
2
(P1 − P2 ) = 1 ρ v22 − v12
2
P1 +
(
)
Assuming that the air outside the card arrangement is not flowing; i.e. v2 = 0,
(P1 − P2 ) = − 1 ρv12
2
The negative sign indicates that P2 > P1.
Since the outside pressure is greater than the pressure in the space between the
cards, card C will stick to card B and it will not fall off.
81

82. 4.
B
h
A
Given: AA = 0.4 m2; AB = 0.25 m2; vA = 1.8 m s-1; PA = 1 × 105 N m-2; h = 5 m;
ρ = 1000 kg m-3.
Using the continuity equation
AA v A = AB v B
∴ vB =
AA v A 0.4 × 1.8
=
= 2.88 m s -1
AB
0.25
Using Bernoulli’s equation
1 2
1 2
ρv A + ρgy A = PB + ρv B + ρgy B
2
2
1
2
2
PB = PA + ρ v A − v B + ρg ( y A − y B )
2
1
2
2
= PA + ρ v A − v B + ρg (− h )
 y A − y B = −h
2
1
= 1 × 10 5 + × 1000 1.8 2 − 2.88 2 + 1000 × 9.8 × (− 5)
2
4
= 4.85 × 10 N m -2
#
PA +
(
)
(
)
(
)
5. Given:
For water:
Qw = 30 cm3 s-1 = 30 × 10-6 m3 s-1; dw = 0.6 cm = 0.6 × 10-2 m; ηw = 0.01 poise
= 0.01 × 10-1 N s m-2.
For oil:
82

83. ρo = 0.8 g cm-3 = 800 kg m-3; do = 0.4 cm = 0.4 × 10-2 m; ηo = 0.012 poise =
0.012 × 10-1 N s m-2.
Quantity of water flowing is given by
Qw = Aw v w =
2
πd w v w
4
4Qw
2
πd w
vw =
Reynold’s number is given by
N Rw =
ρ w v w d w 4 ρ w Qw
=
ηw
πd wη w
But the patterns of flow are similar, hence
N Rw = N Ro
4 ρ w Q w 4 ρ 0 Qo
=
πd wη w πd oη o
∴ Qo =
ρ w Qw d oη o
ρ o d wη w
1000 × 30 × 10 −6 × 0.4 × 10 − 2 × 0.012 × 10 −1
800 × 0.6 × 10 − 2 × 0.01 × 10 −1
= 3 × 10 −5 m 3 s -1
=
= 30 cm 3 s -1 #
6. Given: η = 0.38 × 10-3 N s m-2; d = 0.07 m; Q = 0.012 m3 s-1.
Continuity equation
Q = Av
Q
v=
A
Reynold’s number is given by
ρvd
η
ρQd
=
ηA
4 ρQ
=
πηd
NR =
=
4 × 1000 × 0.012
= 5.7 × 10 5
π × 0.38 × 10 −3 × 0.07
83

84. Since NR > 2000, the flow is turbulent.
SOLUTIONS TO EXERCISE 4
1. Fick’s law equation
∂C
∂ 2C
=D 2
∂t
∂x
To show that
C=
1
e
4πDt
2

− x
 4 Dt 


is a solution to the Fick’s law equation.
For the L.H.S.:
C=
=
1
e
4πDt
1
4πD
t
2

− x
 4 Dt 


−1
2
e
 x 2t −1 

−
 4D 


 x 2t −1 

4D 


∂C (− 1 )t 2 −
= 2
e 
∂t
4πD
−3
1
=
2 4πD × t
=
=
1
4πDt
e
1
2t 4πDt
3
2
e
 x2 

−
 4 Dt 


 x2 

−
 4 Dt 


e
+
1
4πD
t
−
1
2

x 2 t −2
 − (− 1)

4D

 x2

+
1 
4 Dt 2
2 
4πD × t
1
 x 2t −1 

4D 


 −
e 


 x2 


 − 4 Dt 
e  


 1
x2 
− +
2 
 2t 4 Dt 
 x2 

−
 4 Dt 



x2 
− 1 +

2 Dt 

For the R.H.S.:
84

85. C=
1
4πDt
e
 x2 

−
 4 Dt 


 x2 

−
 4 Dt 
∂C  2 x  1


e
= −

∂x  4 Dt  4πDt
 x2 

−
 4 Dt 
∂ 2C 
2  1
 2x 
e   − −
= −


2
∂x
 4 Dt 
 4 Dt  4πDt
=−
=
1
2 Dt 4πDt
1
2 Dt 4πDt
e
e
 x2 

−
 4 Dt 


 x2 

−
 4 Dt 


+
2
x2
4 D 2 t 2 4πDt
1
4πDt
e
e
 x2 

−
 4 Dt 


 x2 

−
 4 Dt 



x2 
−1+


2 Dt 

 x2 

−
 4 Dt  
∂ 2C
1
x2 
D 2 =
e   − 1 +

2 Dt 
∂x
2t 4πDt

⇒ L.H.S. = R.H.S.
⇒C=
2. Given:
1
4πDt
e
2

− x
 4 Dt 


is a solution of
∂C
∂ 2C
=D 2 .
∂t
∂x
x 2 = xrms = 6 cm = 6 × 10-2 m; D = 3.5 × 10-11 m2 s-1.
x rms = 2 Dt
2
x rms = 2 Dt
2
x rms
∴t =
2D
=
(6 × 10 )
−2 2
2 × 3.5 × 10 −11
= 1.6 years #
3. Given: D = 1 × 10-8 cm s-1; t = 1 × 106 years = 3.15 × 1013 s.
x rms = 2 Dt
= 2 × 1 × 10 −8 × 3.15 × 1013
= 793.7 cm
= 7.94 m #
85

86. 86