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Viscosity
Department of Physics The Open University of Sri Lanka
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2014
Introduction
The gases and liquids, because they are able to flow are known as fluids. We see that some
liquids flow very rapidly on a surface while others flow very slowly. For instance consider
water and Tar flowing on a surface. Water moves much faster than the tar. If a swimmer
moves through the water in a pool he experiences a resistance to his motion. He has to do
work against this resistance. Resistance to motion is different from liquid to liquid.
Now let us consider two identical small ball bearings one falling through water and the other
through Glycerine. The ball in the Glycerine falls much more slowly, than the ball in the
water. This shows that force on the ball resisting its motion is smaller in water than in
Glycerine.
It is seen that a fluid exerts a resistance force on a body moving through it. Such forces of a
fluid and their effects are said to arise from its viscosity. Consider again two liquids flowing
through a glass tube. The liquid which flows more slowly is said to be more “viscous” than
the other.
Viscosity of a fluid has considerable practical importance. The viscosity of lubricating oil is
one of the major factors which decide its suitability for use in an engine.
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Fluid flow
Consider a liquid flowing over a solid surface. It can be seen that the molecules in contact
with the surface tend to be at rest while the molecules which are far from the surface flow
fast, the speed increasing as the distance from the surface increases.
If we divide the liquid in to a set of parallel layers as shown in the figure 01 the bottom layer
is considered to be stationary. Starting from the bottom layer the velocity of each layer
increases uniformly from one layer to the other as shown in the diagram, V2 > V1 > V0=0
Therefore there is a relative velocity between any two layers of the liquid.
Figure: 01
In the study of friction between solid surfaces, you have learnt that when a surface is drawn
over another surface, it is subjected to a force which acts opposite to the direction of motion
of the body. Similarly, when a layer of liquid moves over another layer it is also subjected to
a force opposing the motion as in the figure 02.
Figure: 02
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F is the resistive force which acts on the layers when V1 is greater than V2.
Now consider a liquid flowing uniformly over a flat surface. If the velocities of the liquid
layers are represented by arrows, they would be as shown in figure 03
Where V5 > V4 > V3 > V2 > V1
Figure: 03
Liquid flow in a cylindrical tube
When a liquid flows through a tube the layer A of the liquid in contact with the tube is
practically stationary. But the central part D of the liquid is moving relatively fast. (See the
figure 04). At other layers between A and D, such as B, C the velocity is less than at D. The
magnitude of the velocity of a liquid layer is represented by the length of the arrow. When a
liquid flows in a cylindrical tube, the central part of the liquid has the greatest velocity.
Figure: 04
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Steady flow and turbulent flow
Consider a liquid in motion. The liquid moves as the constituent particles or elements of the
fluid move.
The path followed by an element of the moving fluid is called a line flow, in general the
velocity of the element changes in both magnitude and direction along its line of flow.
Figure: 05
If every element passing through a given point follows the same line of the flow, the flow is
said to be steady, stream line or laminar.
In steady flow the velocity at each point in space remain constant in time. But the velocity of
particular particles of the fluid may change as it moves from one point to the other.
A stream line is defined as a curve whose tangent at any point is in the direction of the fluid
velocity at that point. In steady flow the stream line coincide with the line of flow.
If we construct all of the stream lines passing through the periphery of an element of area “a”
such as in Fig: 06 these lines enclose the tube which is called a tube of flow. By the definition
of a stream line no fluid can cross the side wall of a tube of flow. In steady flow there can be
no mixing of the fluid from different tubes of flow.
Figure: 06
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Fig 07 illustrate nature of the streamlines in fluid flow around a number of obstacles (a,b,c)
and in a channel of varying cross section (d).
Figure: 7
It will be noted that each obstacle is completely surrounded by a tube of flow.
Turbulent flaw
Consider the motion of a fluid in which the elements move irregularly and in a much more
complex manner such flow is said to be turbulent flow. In turbulent flow, the velocity at any
point changes with time in both direction and magnitude.
In figure 08 that follow you are shown examples taken from everyday life.
Consider a cigarette burning in calm air. As shown in the figure 09 the smoke rises steadily
for a while and then becomes turbulent. Water flowing from a tap is also another good
example.
Figure: 08
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Figure: 09
A method to examine whether a fluid flow is steady
By introducing a jet of potassium permanganate solution into a tube through which water is
flowing, various types of flow may be studied. See figure 10.
Figure: 10
The potassium permanganate solution is fed into the water stream through a hypodermic
needle and the rate of flow is controlled by a screw clip. If the water and potassium
permanganate are flowing slowly through the tube a fine purple thread can be obtained. This
is streamline flow. If the flow rate is increased, the thread begins to break up shortly after
leaving the needle and the coloured liquid whirls around. This is turbulent flow.
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Newton’s Formula, coefficient of viscosity
The basic expression for the frictional force in a liquid was first suggested by “Newton”.
He stated that the frictional force or the viscous force is directly proportional to the “velocity
gradient” in the part of the liquid considered, and that larger the area of the liquid surface
greater the viscous force.
Figure: 11
Consider two liquid layers X and Y moving with velocities V1 and V2 respectively. Their
distance apart is “d”:
Then the velocity gradient between these layers, X and Y is defined as
Therefore the unit of velocity gradient is or s-1
If “A” is the surface area of the liquid layer considered and F, the viscous force on the layer,
Then,
01.........21
−=
d
VVAF η
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Definition: units and dimensions of viscosity
According to the formula (01) “ “ is given by
Therefore, is defined as follows:-
“The frictional force per unit area of a liquid when it is in a region of unit velocity gradient”
Units of “ ”
In the Formula for
The unit of F is the Newton (N), of A is “m2” and of is “s
-1”
Therefore unit of “ “is given by,
But, from Newton’s second law of motion,
and
The dimensions of are therefore MT-1
L-1
The units of the coefficient of viscosity may be given as “Nm-2
s” or as “kgs-1
m-1
”
The variation of viscosity with temperature is illustrated by the values given in the table
below:
Note that the viscosity increases with temperature for gases while it decreases for liquids.
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Temperature Viscosity “kgs-1
m-1
”
Castor oil Water Air
0 5.3 1.792 x 10-3
171 x 10-7
20 0.986 1.005 x 10-3
181 x x10-7
40 0.231 0.656 x 10-3
190 x 10-7
60 0.08 0.469 x 10-3
200 x x10-7
80 0.03 0.357 x 10-3
209 x 10-7
100 0.017 0.284 x 10-3
218 x 10-7
Fluids for which the equation (01) holds are called “Newtonian fluids”. This description is an
idealized model which not all fluids obey. In general fluids which are suspensions or
dispersions are often Non-Newtonian in their viscous behaviour.
Example 01
A flat plate of area 0.2 m2 is placed on a flat surface and is separated from it by a film of oil
2x10-5
m thick whose coefficient of viscosity is 1.5 Nsm-2
. Calculate the force required to
cause the plate to slide on the surface at a constant speed of 1 mms-1
(Assume that the flow is laminar and that the oil adjacent to each surface moves with that
surface.)
Not all fluids behave according to the direct proportionality between force and
velocity gradient as expected by equation (01). An interesting example is
“Blood”, for which the velocity increases more rapidly than the force.
Doubling the force, produces more than a two fold increase in velocity. This
behaviour can be understood on a microscopic scale.Blood is not a
homogeneous fluid, but rather a suspension of solid particles in a liquid. These
particles have characteristic shapes: for example red cells are roughly disk-
shaped. At small velocities their orientations are random, but as velocity
increases they tend to become oriented so as to facilitate flow.
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Solution
Figure: 12
Since the oil surface adjacent to the surface of plate, moves with the velocity of the plate.
The velocity gradient
Where V, is the velocity of the plate, is the velocity of the liquid layer in the bottom.
But =1 x 10-3
ms-1
Velocity gradient
From Newton’s Formula;
Viscous drag
Since the plate is moved with uniform velocity the force required is equal to the viscous
force.
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Steady flow of liquid through a horizontal uniform tube;
Poiseuille’s formula
The steady flow of a liquid through a tube was first investigated by Poiseuille who derived an
expression for the volume of liquid flowing per second through the tube.
Here, we shall consider only the flow of liquid through a horizontal tube with sectional area.
Consider a horizontal cylindrical tube with uniform cross sectional area.
Figure: 13
When the pressures at the two ends are p1 and p2 (p1 > p2) the pressure difference between the
ends p1-p2
If this pressure difference is constant, the flow of liquid will be steady. Assume that the
length and the radius of the tube are l and r respectively, and the coefficient of viscosity of
liquid is . Then, the volume of liquid issuing per second the tube depends on the following.
(a) The coefficient of viscosity .
(b) The radius “r” of the tube.
(c) The pressure gradient
If V is the volume of liquid flowing in a time t, the relation between V/t and , r, can be
derived by the method of dimensions.
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Figure: 14
Assume that V/t is proportional to
Therefore,
)02..(..........21
Z
yx
l
PPrk
t
V
−= η
Where is a dimensionless constant and its value are can be found experimentally.
Now substitute the dimensions for the quantities in the equation (02)
Then
By equalizing the indices of both sides
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Solving these we get
Putting these values in the equation (02) and we get
Therefore the equation becomes as
)03.(..........
)( 214
η
l
PPkr
t
V
−
=
By experiment, it has been shown that the value of k is equal to . Therefore, Poiseuille’s
formula can be written as
( ))04.....(..........
8
21
4
η
π
l
Ppr
t
V −=
This equation can be used only for streamline flow through horizontal tubes with uniform
cross sectional area.
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Reynolds Number
When the velocity of a fluid flowing in a tube exceeds a certain critical value, (which
depends on the viscosity of the fluid and the diameter of the tube) the nature of the flow
becomes very complicated.
Within a very thin layer adjacent to the tube walls, called the boundary layer, the flow is still
laminar. The flow velocity in the boundary layer is zero at the tube walls and increases
uniformly through the layer.
Beyond the boundary layer, the motion is irregular; random local circular currents called
vortices develop within the fluid, with a large increase in the resistance to flow and the flow
becomes turbulent.
Experiments show that combinations of four factors determine whether the flow is laminar or
turbulent. This combination is called the Reynold’s number and is given by
Where, is the density, the average velocity, the viscosity of the liquid and D the
diameter of the tube. The Reynold’s number is a dimensionless quantity.
All experiments show that when the Reynold’s number is less than about 2000 the flow is
laminar (steady) whereas above about 3000 the flow is turbulent. In the transition region
between 2000 and 3000 the flow is unstable and may change from one type to the other.
The Reynold’s number of a system forms the basis for the study of the behaviour of real
systems through the use of small scale models. A common example is the wind tunnel, in
which one measures the aerodynamic forces on a scale model of an aircraft wing. The forces
on a full size wing are then, deduced from these measurements.
Two systems are said to be dynamically similar if the Reynold number is the same for
both. The symbol D may refer in general, to any dimension of a system, such as the span or
chord of an aircraft wing.
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The variation of the rate of flow of liquid through a cylindrical tube,
with the pressure difference between the ends of the tube
Poiseuille’s equation is
If are constant
The equation is of the type
This is valid only for steady or streamlines flow.
We can increase in steps the pressure difference between the ends of the tube, and measure
the volume of liquid flowing through. If we plot these values we get a graph similar to the
one shown in figure 15. However, the steady flow of liquid is maintained only up to a certain
value of the pressure.
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Figure: 15
If we increase the pressure above this, the volume of liquid issuing per second is constant.
But this is “turbulent flaw” and Poisseuille’s equation does not hold in this region
Example 02
The diagram shows a capillary tube of inner radius
0.7 mm and length 22 cm, connected to a constant pressure apparatus, which provides a
steady head of 30 cm of water. Find the volume of water issuing from the free end of the
capillary tube in 1 minute. The density of water is 1000 kgm-3
, the coefficient of viscosity for
water is 1 x 10-3
kgs-1
m-1
and the acceleration due to gravity is 10 ms-2
Solution
Let the height of the liquid column from the capillary tube to the top surface of the liquid be
h, and the length of the capillary tube l and the pressures at the ends of the capillary tube p1
and p2 respectively. If V is the volume discharged in time t, from Poiseuille’s equation.
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where a is the radius of the tube.
Pressure
- Atmospheric pressure
From (1)
Given that,
Volume discharged in 1 minute
Example 03
Water flows steadily through a horizontal tube which consists of two parts joined end to end.
One part is 20 cm long and has a diameter of 0.2 cm and other is 5 cm long and has a
diameter of 0.05 cm. If the pressure difference between the ends of the tube is 10 cm of water
find the pressure difference between the ends of each part.
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Figure: 16
Solution
Let the tubes be as shown in the diagram, if pressures at the end points are p1, p2, and p3
The pressure difference between the ends of the first tube=
And pressure difference between the ends of the second tube =
Given that,
If V/t is the volume of water issuing per second, through first tube, the same quantity passes
through the second, since the flow is steady.
Applying Poiseuille’s equation to the first tube,
Second tube
(2)⇒
(3)⇒
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But
From (6)
( )
+
−=
4
2
2
4
1
1
318
a
I
a
I
PP
t
V
π
η
Substituting this value in (4) and (5) pressure differences between the ends of each tube can
be calculated.
Example 04
An incompressible liquid flows along pipes of varying diameter as shown in the diagram.
Figure: 17
The ratio of the speeds is equal to
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Solution
Since the liquid is incompressible we may assume that, there is a steady flow in the tube. So,
the mass of fluid passing per second through the first tube also passes through the second
tube. Let the density of the liquid be
Then mass flowing per second through the first tube
Mass flowing per second through 2nd tube
Since the masses are equal
Correct answer (iii)
Example 05
The dimensions of the coefficient of viscosity
(i) MLT-2
(ii) ML-2
T-3
(iii) ML-1
T-3
(iv) MLT-1
(v) ML-1
T-1
Solution
Newton’s Formula,
Substitute the dimensions.
Correct answer is (v)
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Example 06
Water flows in a streamline manner through a capillary tube of radius a, the pressure
difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure
increased to 2P the rate of flow becomes.
(i) 4Q (ii) Q (iii) Q/2 (iv) Q/4 (v) Q/8
Solution
In the first tube
Rate of flow = Q
Radius = a
Pressure difference = P
If the length of the tube is l and the coefficient of viscosity is
From Poiseuille’s equation
Now, let the new rate of flow be Q1,
When the radius = a/2
And the pressure difference = 2p
Poiseulle’s equation
Correct answer is (v)
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Example 07
A capillary tube x is connected to a vessel of water. The height h of the water column is kept
constant. Consider the statements given below about the flow of water through the tube;
Figure: 18
(A) Rate of flow of water is increased very much more by an increase of tube diameter than
by the same percentage increase of pressure.
(B) The rate of flow of water through the tube is increased, when the diameter of the vessel is
increased.
(C) The rate of flow of water is decreased when the length of tube is increased.
Of the above statements,
(i) Only A and B are correct
(ii) Only B and C are correct
(iii) Only A and C are correct
(iv) All A, B and C are correct
(v) All statements A, B and C are wrong.
Solution
According to the Poiseulle’s formula
The flow through tube depends only on pressure difference, radius, length and viscosity.
When “a” is increased, rate of flow is increased by a factor four times than by an increase of
pressure. So, A is correct.
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When the diameter of the vessel is changed the pressure at the end of the tube does not
change as it depends only on ‘h’.
So, B is wrong.
When l is increased, the rate of flow is decreased.
So, C is correct.
∴A and C are correct
(iii) is correct.
Example 08
Figure: 19
The capillary tubes A, B and C of the same diameter, are connected to a constant pressure
tank in which there is a viscous liquid. The lengths of A, B and C are l, 2l and 3l respectively.
The volume of water passing through A, B and C per second are VA ,VB and VC respectively.
Then,
(i) VA = VB = VC (ii) VA = 2VB = 3VC
(iii) VA = 4VB = 9VC (iv) 3VA = 2VB = VC
(v) 9VA = 4VB = VC
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Solution
Consider tube A, Poiseulle’s equation
To B
To C
Therefore VA = VB = VC
Correct answer is (i)
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Determination of the viscosity of a liquid by the capillary tube
method
A clean capillary tube of uniform cross section must be chosen. To clean the capillary tube it
should be first washed in Sodium hydroxide then in dilute Nitric acid and finally well
rinsed in distilled water.
The cleaned and washed capillary tube is connected horizontally to a constant pressure head
as shown in the figure 22.
Figure: 20
The constant pressure apparatus enables the pressure difference between the ends, of the tube
to be kept constant. A small piece of thread may be connected to the open end of the tube to
allow smooth flow. This reduces the possibility of turbulence. Any excess liquid flowing in
from the tap into the constant pressure apparatus can flow out through the tap of the tube
fixed through the bottom of the vessel. By measuring the volume of liquid flowing in a given
time, the volume flowing per second through the tube can be calculated. Let the capillary
tube be XY and its length l. The height of the liquid column above the capillary tube in the
constant pressure apparatus is h.
The pressure at the end y of the tube =P0,
where P0 is atmospheric pressure.
The pressure at the end x= P0 + h g
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Where is the density of the liquid and g is the acceleration of gravity.
Pressure difference between x and y = h g
Let V be the volume of liquid discharged at in time t,
Then,
Where a is the radius of the tube.
The quantities V, t, l, h, and a are measured and the coefficient of viscosity calculated.
Note: - In this case, the radius “a” of the tube must be measured very accurately. If there is a
small error in measuring the radius, the error in the final results will be four times as much,
since “a” occurs to the fourth power in the equation. So, the radius of the tube may be
measured by weighing a measured length of mercury thread drawn into the capillary tube.
Errors that may arise
(i) It is very difficult to measure “h” accurately.
(ii) The coefficient of viscosity changes with the temperature.
Hence the temperature of the liquid must be kept constant.
To minimize these errors, graphical method is used to determine the coefficient of viscosity.
From Poisseulle’s equation
The constant pressure head h is varied in steps and the volume of liquid issuing per second is
measured for each h. Plot V/t Vs h. Then a straight line graph passing through the origin is
obtained.
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Figure: 21
The gradient of this graph;
Comparison of viscosity by Ostwald viscometer
Ostwald viscometer shown in figure 24 is widely used for comparing the viscosities of two
liquids and to examine the variation of viscosity with temperature.
In this viscometer T is a capillary tube, while P, Q and S are bulbs of equal volume. The
liquid is introduced at S and drawn by suction above P, and the time taken for the liquid level
to fall from the fixed mark P to the mark Q is observed. Let this time be t1 . If the viscosity of
the liquid is and the density , then guided by Poiseuille’s formula,
1
1
1 η
ρα
t
V------------- (1)
Where is the volume of liquid that has flowed? In this case, since the capillary is uniform
the pressure difference is proportional to the density .
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Similarly, consider another liquid of density and coefficient of viscosity , If the time
taken for the liquid level to drop from P to Q is ,
2
2
2 η
ρα
t
V----------- (2)
From equation (1) and (2)
If we measure , and t2 then the viscosities of the liquids can be compared.
Figure: 22
Now suppose that we want to study the variation of viscosity of a liquid with temperature, for
this purpose the viscometer is placed inside a beaker of water which is heated to a particular
temperature.
As before let the time taken by the liquid to flow from P to Q is . If this temperature is .
Then,
Changing the temperature to , measure the time again, then
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If the viscosity at one temperature is known, that at the other temperature can be calculated.
Note: - The densities at the temperature and may be read from tables or
determined separately.
Self-assessment questions
1. Write down Polseuille’s formula for steady flow of a fluid through a horizontal tube,
Calculate the mass of water flowing in 10 minutes through a tube 0.1 cm in diameter, 40
cm long, if there is a constant pressure head of 20 cm of water. The coefficient of
viscosity of water is and the density of water
2. Write down Poiseuille’s formula for the rate of flow of a liquid through a capillary tube.
Hence show that if two capillaries of radii and having lengths and respectively
are connected in series, the rate of flow is given by
where p is the Pressure across the arrangement and η the coefficient
of viscosity of the liquid.
3. Three capillaries of lengths 8L, 0.2L and 2L with their radii r, 0.2r and 0.5r respectively,
are connected in series. If the total pressure across the system is p, deduce the pressure
across the shortest capillary.
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Solutions
(1)
Figure: 23
Poiseuille’s equation is
Where is the volume of water flowing in time t. Given that
Where
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Where V is the volume of water flowing in . V = 0.613 x 10-4
m3
Mass of water = volume x density
Mass of water =
(2)
Figure: 24
Let be pressure across the first and across the second, capillary. So that
Obviously, the rate of flow through either capillary will be the same, say Q. Therefore from
Poiseuille’s formula,
For the first tube,
For the second tube
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and,
From (2)
Therefore
(3)
Figure: 25
Let three tubes be A, B and C the shortest tube being B
.Let the pressure differences across A, B and C be and so that
The rate of flow of liquid across each capillary is the same.
Let this rate be Q then,
For the first tube
For the second tube
For the third tube
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Equating (1), (2), and (3)
This could also be easily obtained from the solution to question 2 as follows. Extending the
argument to three tubes,
And
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Learning out comes
After reading this session, you should be able to discuss,
� the difference between streamline flow and turbulent flow of a liquid.
� the definition of the coefficient of viscosity of a liquid.
� the factors governing the rate of flow of liquid in a horizontal tube and Poiseuille’s
formula for it.
� the meaning of the Reynolds number of a liquid.
� the experiment to determine the coefficient of viscosity of a liquid.
� a method for the comparison of viscosities of liquids using Ostwald’s viscometer.
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Course Team
Course Team Chairperson Content Editors
Mr. D. L. N. Jayathilake Prof. E. M. Jayasinghe
Mr. D. L. N. Jayathilake
Authors Editorial Assistants
Mr. R. M. Gunasinghe Ms. H. G. Chandrani
Mr. L. S. G. Liyanage Ms. W. D. M. Srikanthi
Mr. M. M.S.G.K. Tennakoon Ms. M. D. P. Alahakoon
Mr. J. D. Vithanage Ms. P. D. Sumanawathi
Mr. G. U. Sumanasekara
Mr. D. M. Nanda
Ms. R. D. Hettiarachchi
Ms. D. R. Abeydeera
Ms. I. R. Wickramasinghe
Graphic Artist Desktop Publishing
Mrs. M. R. P. Perera Mr. W.C. Deshapriya
Mrs. N. W.C. Kularatne
Cover Design Web Content Developer
Mrs. N. W. C. Kularatne Ms. B. K. S. Perera
The Open University of Sri Lanka
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