viscosity - ac

1

Viscosity

Department of Physics The Open University of Sri Lanka

Produced by The Open University of Sri Lanka

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.

Produced by The Open University of Sri Lanka 2014

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

Nawala, Nugegoda, Sri Lanka

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viscosity - ac

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