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PHYS1004 Physics 1 (Environmental) – Semester 2, 2012
MODULE 2 OUTLINE: PROPERTIES OF MATTER
Unit Description
This module is one of 3 comprising PHYS 1004 Physics 1 (Environmental). This document describes details of
this module and should be read in conjunction with the more general unit of study outline for PHYS 1004
Physics 1 (Environmental).
General goals of this module
This module is concerned with the properties of fluids at rest and in motion, including the effects of viscosity.
We also look at the elastic behaviour of solids, surface tension in liquids, and adhesion between solids and
liquids.
•
To examine the properties of a fluid at rest and in motion, and to see the effects of a fluid on a body
completely or partially immersed in it.
•
To examine the additional effects of viscosity on a flowing fluid, and to appreciate the difference
between Newtonian and non-Newtonian fluids.
•
To examine the elastic behaviour of solid materials.
•
To examine the surface tension of liquids and the behaviour of liquid-surface interfaces.
•
To appreciate how the physical phenomena encountered are relevant to environmental and life
sciences
MODULE DEFINITION & OBJECTIVES – PROPERTIES OF MATTER
(specified as references from the text: College Physics: A Strategic Approach by R.D. Knight, B. Jones & S.
nd
Field, 2 edition published by Pearson/Addison Wesley, 2007.)
LEARNING OBJECTIVES
For each topic in this Module the Specific Objectives define what we expect you to learn and understand.
“Understanding” a term or concept means that you should be able to:
•
explain its meaning,
•
interpret it correctly when you read or hear it,
•
use it correctly in your own writing.
•
apply it correctly to examples and problems.
There is no easy road to learning. Your marks will depend on the work that you do. You should therefore read
through and understand the sections of the textbook specified below, and work through the specified
examples. You should then attempt as many as possible of the recommended questions, exercises and
problems. Problem solving skills can only be acquired by practice.
Knight et al: Chapter 13
FLUIDS
The physical properties of fluids at rest and in motion are discussed, and laws governing their behaviour are
explained and applied. Real fluids, that is, fluids where viscosity cannot be ignored, are included. These
principles are used to describe the movement of fluids in biological systems such as the human circulatory
system.
Text sections: 1-7
Supplementary notes: Turbulence, Power Dissipation in Fluids, Rheology
Examples: 13.1 - 13.14
*
*
Suggested questions: 5, 12, 19, 23, 27, 28, Q2 , Q3
*
*
*
Suggested problems: 5, 11, 12, 15, 24, 25, 27, 32, 33, 36, 38, P2 , P4 , P6
Specific Objectives - after studying this chapter you will be able to
•
describe the different states of matter in terms of atomic structure
•
define and use the concepts of mole and Avogadro’s number
•
define the term fluid and give examples
•
explain the concept of density
•
explain the concept of pressure
PHYS1004 Module 2 Outline
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Copyright © 2012, The University of Sydney
•
•
•
•
•
•
•
•
•
•
•
•
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calculate the pressure below the surface of a liquid
explain and use Pascal’s principle
explain and use Archimedes’ principle
explain the origin of the buoyancy force
solve problems involving stationary objects in a fluid
explain and use the equation of continuity
describe the difference between turbulent and laminar flow
explain and use Bernoulli’s equation
explain what is meant by the term ‘viscous’ when describing fluids and describe the velocity profile of a
viscous fluid flowing in a pipe.
explain why the pressure in a viscous fluid drops along a pipe and be able to apply Poiseuille’s equation
calculate the power dissipated in a tube by viscous forces
use the Reynolds number to determine whether flow is laminar or turbulent
describe the difference between Newtonian and non-Newtonian fluids. and give examples of the
behaviour of each.
Knight et al: Chapter 8
EQUILIBRIUM AND ELASTICITY
The concept of elasticity. The relationship between stretching, squeezing, shearing and twisting forces and the
particular types of deformation they cause in a body will be defined.
Text sections: 3,4
Supplementary notes: Elastic moduli
Examples: 8.6, 8.8, 8.9, 8.10
Suggested questions: 8, 9, 11, 12
Suggested problems: 19, 20, 26, 31, 33, 45, 57, 60,
Specific Objectives - after studying this section you will be able to
•
define the terms, stress, strain and modulus of elasticity
•
apply these terms to situations where there is tension, compression, shearing and hydraulic stress.
•
perform calculations involving stress, strain or elastic modulus for the different types of deformation
•
draw, and describe the common features of, the stress-strain curve for different materials
•
explain the terms ultimate strength and yield strength
Surface Tension
Forces between molecules at the surface of a liquid give rise to the phenomenon known as surface tension.
The adhesion force between a liquid and a solid allows us to explain, for example, how trees can grow to
great heights and how detergents work to dissolve grease. Laplace’s equation showing the relation between
tension in elastic membranes and pressure difference across the membrane can be applied to various
biological structures.
Supplementary Notes: Surface Tension, Capillarity, Laplace’s Law and Negative Pressure
Examples: see supplementary notes
Suggested questions: Q4
*
Suggested problems: P8-P14
∗
Specific Objectives - after studying this section you will be able to
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•
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•
•
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explain the concept of surface tension
explain the difference between cohesive and adhesive forces
explain the concept of contact angle between a solid and a free liquid surface
calculate the rise or fall of a liquid in a capillary tube.
explain Laplace’s Law and apply it appropriately to liquid drops, soap films, blood vessels and alveoli
describe the role of surfactant in the lungs
explain how sap can rise in tall trees and the concept of negative pressure
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Supplementary Questions
Q1
An ice cube floats in a glass of water. When the ice melts, will the water level in the glass rise, fall, or
remain unchanged? Justify your answer.
If the ice had been made from pure water, and the water in the glass was salt water (denser than pure
water) how would this change your answer, if at all? Justify your answer.
Q2
Poiseuille’s equation shows that for laminar flow the volume flow rate through a pipe is proportional to
the product of the pressure difference and the fourth power of the radius. The viscosity of water is 1.0
-3
 10 Pa.s
(a) Water in a pipe is flowing without turbulence under a certain pressure difference. If the radius of
the pipe is reduced by 20%, what percentage increase in pressure difference is required to
maintain the same flow rate?
-1
(b) In agricultural irrigation, typical values of flow velocity and pipe diameter are 1.0 m.s and 100
mm., respectively. Is a calculation such as in part (a) applicable? (i.e. is the flow in the pipe likely
to be laminar?)
Q3
For a Newtonian liquid flowing through a pipe the radial speed profile is parabolic. Qualitatively, explain
how the profile differs for (a) a pseudoplastic liquid; (b) a dilatant liquid.
Q4
Explain what is meant by the terms ‘cohesive’ and ‘adhesive’ when describing forces between
molecules. Use these concepts to explain why people put a coating of wax on their cars after washing
them to keep them clean.
Supplementary Problems
P1
At a bottling plant soft drink flows in a pipe at a rate that would fill two hundred 375 mL cans per
2
minute. At point 1 in the pipe the gauge pressure is 152 kPa and the cross-sectional area is 8.00 cm .
2
At point 2, 1.35 m above point 1, the cross-sectional area is 2.00 cm . Find
(a) the volume flow rate
(b) the flow speeds at points 1 and 2
(c) the gauge pressure at point 2
P2
What is the greatest average speed of blood flow at 37 °C in an artery of radius 2.0 mm (the viscosity
-3
of blood is 2.1 x 10 Pa.s at 37 °C) if the flow is to remain laminar? What is the corresponding flow
rate?
P3
An Echiuroid worm feeds by passing a current of water through a porous membrane 10 µm thick
-12
2
-3
2
containing cylindrical holes each of area 1.0  10 m . The area of the membrane is 2.0  10 m
-3
3
and half this area is holes. If 1.2  10 m of water flows through the membrane in an hour, calculate
-3
the pressure difference that exists across the membrane. The viscosity of water: 1.0  10 Pa.s
P4
Blood flows from the foreleg of an animal through a hypodermic needle inserted into an artery. The
needle has a length of 80 mm, and an internal radius of 0.30 mm. If it takes 400 s to obtain a 50 mL
-3
sample, calculate the gauge and absolute pressures in the artery. The viscosity of blood is 4.0  10
Pa.s
P5
In the human body, there are of the order of 10 capillaries. The average length of the capillaries is
-3
-6
about 2.0  10 m and the average radius about 4.0  10 m. These capillaries act in parallel to
-5
3
-1
convey blood through the body at a rate of 8.0  10 m . s . Calculate the total pressure drop along a
-3
capillary. The viscosity of blood is 4.0  10 Pa.s
P6
The pressure drop along a length of artery is 100 Pa. The radius of the artery is 10 mm, and the flow is
-1
laminar. The average speed of the blood is 15 mm s .
(a) What is the net force on the blood in this section of artery?
(b) What is the power expended maintaining the flow?
P7
The aorta is the principal blood vessel through which blood leaves the heart in order to circulate around
the body.
(a) Calculate the average speed of the blood in the aorta if the flow rate is 5.0 litres per minute, The
aorta has a radius of 10 mm.
(b) Blood also circulates through smaller blood vessels known as capillaries. When the rate of blood
flow in the aorta is 5.0 litres per minute, the speed of the blood in the capillaries is about 0.33
–1
mm.s . Given that the average diameter of a capillary is 8.0 mm, calculate the number of
capillaries in the blood circulatory system.
(c) Can changing the viscosity of the blood assist people who have blocked arteries? Briefly explain.
10
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P8
Bone, which has a Young’s modulus of 1.5 x 10 Pa, can undergo only a 1.2% decrease in length
before fracturing,
(a) What is the maximum compression force that can be applied to a bone whose minimum cross2
sectional area is 3.0 cm , without fracturing. (This is the approximate area of the tibia at its
narrowest.)
(b) What is the maximum height from which a 70 kg person could jump without fracturing the tibia.
Assume that stress is the same in both legs, and that the time between first touching the floor and
coming to rest is 0.035 s.
P9
You are given the following information: the surface tension of blood at 22 °C is 5.6 x 10 N.m , and its
3
-3
density is 1.1 x 10 kg.m .
(a) What vertical force is required to lift a fine wire ring of radius 15 mm from the surface of a
container of blood?.
(b) How high will blood rise in a capillary tube of radius 0.50 mm? (Such tubes are often used to
collect for analysis samples of blood from a drop produced by a pin prick.)
P10
Assume that each of the six legs of a water skater makes a depression 2.5 mm in radius on the water
surface, If at the point of contact, the water surface makes an angle of 45 °C with the vertical, calculate
-1
the mass of the insect. (The surface tension of water at 20 °C is 0.073 N.m .)
P11
To what height will water at 20 °C rise in a capillary tube of diameter 1.0 µm? The surface tension of
-1
water at 20 °C is 0.073 N.m . Assume a contact angle of zero degrees.
P12
When a capillary tube of radius 1.0 mm is immersed in a beaker of mercury, the level of mercury in the
tube is found to be 5.4 mm below the level in the beaker. Given that the surface tension of mercury is
-1
0.47 N.m , determine the contact angle between mercury and glass. Draw a diagram showing the
contact between the mercury surface and the glass, and label the contact angle. The density of
4
-3
mercury is 1.36 x 10 kg.m .
P13
Soap bubbles are blown from water to which detergent has been added so that its surface tension is
-1
reduced to 0.025 N.m . What is the difference in pressure between the inside and outside of the
bubble for a soap bubble of radius (a) 10 mm, (b) 100 mm? Which is higher: the pressure inside or
outside?
P14
The following data are taken from physiological measurements. They allow the change in the tension of
the membrane/surfactant system of alveoli during breathing to be determined.
(a) After breathing in the radius of the alveoli is 0.10 mm; the gauge pressures inside the lung and in
the pleural cavity are is 0.0 mm.Hg and –5.5 mm.Hg respectively. Determine the tension in the
membrane/surfactant system.
(b) After breathing out the radius of the alveoli is 0.050 mm; the gauge pressures inside the lung and
in the pleural cavity are is 0.0 mm.Hg and –2.5 mm.Hg respectively. Determine the tension in the
membrane/surfactant system.
-2
-1
Supplementary Notes
Turbulent Flow
Turbulent flow is mentioned briefly on p430 of Knight; here we provide some more detail. In contrast to
laminar flow, where streamlines are stationary, turbulent flow is characterised by non-steady streamlines. Flow
which starts out laminar can become turbulent: first the streamlines vary periodically and then randomly when
turbulence is fully developed. An example of this is the flow from a tap. The flow rate can be adjusted so that
the flow is laminar as the water leaves the tap and becomes turbulent after a short distance. We can
determine whether fluid flow is likely to be laminar or turbulent from the value of a dimensionless quantity
called the Reynolds number.
For flow in a pipe of diameter d, the Reynolds number NR is given by
NR =
ρdv
η
(1)
where r is the density of the fluid, v is its flow speed, d is the diameter of the tube, and h is the viscosity of the
liquid. The value of the Reynolds number provides an approximate indication of whether the flow is likely to be
laminar or turbulent. If
NR < 2000
2000 < NR < 3000
NR > 3000
€flow is laminar
flow is unstable - it may change from laminar to turbulent and vice versa
flow is turbulent
We can also apply the concept of Reynolds number to flow past an object (e.g. air flow past a ball, water flow
past a boat, air flow past an aerofoil). For this situation we can calculate Reynold's number using equation 1,
PHYS1004 Module 2 Outline
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with d replaced by the parameter L, which characterises the size of the object. For example, for a ball, L would
obviously be the diameter; for a boat we would take its length. The value of Reynolds number at which the
flow changes from laminar to turbulent will depend to some extent on the shape of the object. As a rough
guide, however, we can also say in this case that the transition from laminar to turbulent flow occurs around a
value of 2000. With increasing Reynolds number, before the flow becomes fully turbulent it usually becomes
oscillatory (for example the von Karman vortex street), and then unstable (changing randomly to and from
turbulent flow).
Power Dissipation in Viscous Flow
A pressure difference is needed to keep a viscous fluid flowing through a pipe at a constant speed (see
Poiseuille’s equation). This means that work is being done (which, because the flow is steady, will end up as
heat in the fluid and walls of the pipe). If the pressure difference across a section of pipe of length l and crosssectional area A is Dp, the net force across the section is ADp. The power required to maintain the steady
flow is equal to the net force times the average speed:
p = ΔpAv = ΔpQ
(2)
Exercise
The large artery in a dog has an inner radius of 4.0 mm. Blood flows through the artery at the rate of 1.0
3 -1
cm s . Find
(a) the average and maximum flow speeds of the blood;
(b) the pressure drop in a 0.10 m section of the artery;
(c) the power required to maintain the blood flow in this section of the artery.
Rheology
Rheology is the study of how bodies behave under the action of deforming forces. This term is normally used
in relation to all materials except those solids that are purely elastic and those liquids which obey Newton’s
law. Fluids which obey Newton’s law (shear stress is proportional to shear rate),
F
dv
=η x
A
dy
(3)
where the coefficient of viscosity h is a constant, are called Newtonian fluids. The velocity gradient
dv x /dy
describes the shearing the fluid experiences as a result of the application of shear stress F / A , and is thus
referred to as the shear rate. Examples of Newtonian fluids are water, alcohol, kerosene, and lubricating oil.
€
Non-Newtonian Fluids
Fluids that do not obey Newton’s law - i.e. for which shear stress is not proportional to the
€shear rate and thus
their viscosities depend on shear rate, are called non-Newtonian fluids. They fall
into
two
classes.
€
•
pseudoplastic fluids for which the coefficient of viscosity decreases as the shear rate increases.
This kind of behaviour is also called shear thinning. For such fluids flow rate through a tube will
increase at a rate greater than linearly with pressure difference, and they will appear to be less
viscous the more vigorously they are stirred. Examples include milks, paints, emulsions, and blood.
•
dilatant fluids for which the coefficient of viscosity increases as the shear rate increases. This kind of
behaviour is also called shear thickening. For such fluids flow rate through a tube will increase at a
rate less than linearly with pressure difference, and they will appear to be more viscous the more
vigorously they are stirred. Examples include syrups, corn flour paste (you can poke your finger
through it, but if you hit it with a hammer it becomes hard and virtually impenetrable), wet sand (it
stiffens when trodden on).
Apart from these two main categories there is one other type of non-Newtonian fluid which has the
complicating feature that it does not flow, i.e. it does not behave as a fluid, until a certain stress, the yield
stress, has been exceeded. Once this stress has been exceeded, the viscosity either remains constant or
decreases as the shear rate increases. Such materials are called plastic.
The flow characteristics of the different non-Newtonian fluids are summarised by the curves in Figure 1 which
show shear stress as a function of shear rate. On this diagram a Newtonian fluid is represented by a straight
line passing through the origin, for which the constant slope is equal to the coefficient of viscosity.
PHYS1004 Module 2 Outline
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Figure 1:
Variation of shear stress with shear rate for Newtonian and non-Newtonian fluids
These different fluid behaviours also reveal themselves if we plot flow rate as a function of pressure
difference, as shown in Figure 2. We know from Poiseuille’s equation that for a Newtonian fluid (h = constant)
the result will be a straight line. For a pseudo-plastic fluid, the result will be a curve which is concave upwards
(h decreases as flow rate increases); for a dilatant fluid, the result will be a curve which is convex upwards (h
increases as flow rate increases). For a plastic fluid there will be no flow until a threshold pressure difference
is exceeded.
Flow rate Q
Figure 2:
Flow rate as a function of pressure difference for Newtonian and non-Newtonian fluids
Another way in which the non-Newtonian nature of a fluid can show itself is in its radial velocity profile as it
flows through a narrow tube.
•
For Newtonian fluids the radial velocity profile is parabolic.
•
For non-Newtonian fluids the radial velocity profile is not parabolic - it is somewhat sharper for
dilatant fluids and for pseudoplastics it is blunter.
•
For plastic materials there is a completely flat region in the centre where the shearing stress is less
than the yield value.
Two other terms used to describe non-Newtonian fluids are thixotropic and rheopectic. These refer to the
way the fluids respond over time if the shear rate is kept constant. For thixotropic fluids the viscosity
decreases with time; for rheopectic fluids it increases with time. Rheopectic behaviour is rare, but thixotropic
behaviour is more common, examples including greases, heavy printing inks, paints, blood, and tomato sauce
(it will flow more easily – viscosity decreases - after being shaken vigorously).
Strictly speaking pseudoplastic refers to the way a fluid responds to changes in the magnitude of the shear
rate, while thixotropic refers to the way it responds at a function of time to a constant shear rate. Often
however the terms are used interchangeably, as a pseudoplastic fluid can be considered to be a thixotropic
fluid with a very short response time.
Viscoelastic Materials
The other major class of materials covered by the subject of rheology are the viscoelastic materials:
materials which are neither purely elastic nor purely viscous, materials which show the properties of both
solids and fluids. Whether these behave as solids or fluids depends on how long the stress is applied. They
behave as solids when stress is applied over a short time, but as fluids when stress is applied over long
periods.
A good example of a viscoelastic is the material known as “silly” putty. If stress is applied over a short time, it
behaves elastically - it will bounce. If, however, stress is applied over a long time, it will flow - it behaves as a
viscous liquid. Another example is egg white - it will flow but can be cut like a solid using scissors.
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Blood
Blood is a complex fluid, consisting of a plasma in which are suspended a variety of cells, the predominant
ones being the red cells. Measurement show that blood is pseudoplastic/thixotropic. At high shear rates,
normal blood has a viscosity of between 5 and 6 times that of water; but at low shear rates it may be several
hundred times that of water. The viscosity for people with certain diseases such as myocardial infarction and
thrombosis is much higher, particularly at low shear rates.
The data presented in Figure 3, is taken from many patients. The graphs show that the viscous properties of
blood can be used for diagnostic purposes.
Figure 3:
Blood flow data
Blood’s viscous behaviour is partly due to clusters of the red cells which form when the shear rate is low, but
at higher shear rates they break up, giving a lower viscosity. If allowed to rest the original viscosity returns as
the clusters reform. The fact that blood cells are not rigid also has important consequences. If the red cells
were rigid particles, when their concentration reached 65% (the fraction of the volume taken up by the cells),
blood would have the consistency of concrete. However, blood is still very fluid even at 99% red cell
concentration because the red cells are easily deformable. Thus blood can flow in the small capillaries
because the cells deform as they flow. Any condition which leads to more rigid red cells leads to a much
greater blood viscosity.
An important consequence of the rheological nature of blood is that when it is artificially pumped, when the
heart is by-passed during heart surgery, special roller pumps have to be used. They pump the blood so that
the red cells are not damaged by too high shear rates but at a rate sufficiently great so that aggregation does
not occur.
Elastic Moduli
Young's modulus for linear extension or compression is described in Knight et al.; but this is only one of three
Elastic Moduli of importance to us. The others are shear modulus and bulk modulus.
Figure 4 shows a body undergoing shear deformation. Originally a rectangular block, the object has been
subjected to equal and opposite forces along the top and bottom faces, as shown. There is no acceleration of
the block as a whole, because the forces are equal and opposite. But the forces do not act along the same
line, so they have the effect of shearing the object into the deformed shape shown. The shear stress is again
defined as force/area, but in this case the area is that of the faces over which the force acts (see Figure 4) and
the force is acting in the same plane as the area. The shear strain is defined as Δx/ℓ0.
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Figure 4:
An object under shear stress, which causes shear deformation (strain)
Shear deformation is relevant in many situations: for example scissors cut material by exerting a shear stress
which is greater than the shear strength of the material, leading to its fracture along the line of cutting. Rivets
preventing metal plates from sliding across each other are also under shear stress.
For an object under moderate shear stress, i.e. still within the elastic limit, the shear stress and shear strain
are related by the equation:
F
Δx
=S
A
l0
(4)
where S is the shear modulus, measured in Pa.
The last of the three elastic moduli applies to the situation where an object is subject to uniform compression
by an external pressure. Figure 5 shows an object with an initial volume of V0, subject to initial pressure p0.
After we apply additional pressure Δp, the volume is reduced (i.e. the object is compressed) by an amount ΔV.
Figure 5:
An object under bulk stress, causing uniform compression
Assuming that the compression remains within the elastic limit, the bulk stress and bulk strain are related by
the equation:
F
ΔV
= p = −B
A
V0
(5)
The minus sign is introduced because a positive stress (pressure) will lead to a decrease in volume, i.e. ΔV0
negative, hence the minus is needed for B, the bulk modulus, to be positive. The units of B are again Pa.
Surface Tension
A liquid is cohesive: it has a tendency to remain intact, for example liquids form drops. In fact the liquid
surface behaves like a membrane - it is under tension. This is due to the force of attraction between the
molecules of the liquid. Molecules near the surface experience a net inwards force; in contrast molecules in
the interior experience an essentially zero net force due to equal attraction in all directions.
In addition to attractive forces among themselves (cohesion), the molecules of a liquid experience attractive or
repulsive interactions with the molecules of other substances (adhesion). For example, water will rise in a
narrow glass tube inserted in a container of water (a phenomenon known as capillarity) while in the case of
mercury the level in the tube will be depressed below that of the level in the container.
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When the liquid adheres to a solid surface there will be a force on the solid due to the tension in the liquid
surface. The parameter surface tension g is defined as the force per unit length exerted by the liquid surface.
-1
The units of g are therefore N.m . For a typical liquid the value decreases with increasing temperature.
Example
If a U-shaped wire, with an attached sliding wire as shown in Figure 6 is dipped into a liquid then
withdrawn, a thin liquid film will fill the area enclosed by the wire. At equilibrium the force on the sliding
part of the wire due to the surface tension of the liquid is given by
F = 2γ l
(6)
where l is the length of the horizontal section of the wire. The factor 2 arises because the water film has two
surfaces.
Figure 6: Sliding wire supported by surface tension
Capillarity
The surface of a liquid in contact with a solid surface forms an angle with respect to the solid surface. The
contact angle q is a result of the competition between the cohesion and adhesion forces. Figure 7 shows
examples where (a) q < 90° (adhesion greater than cohesion: the surface attracts the liquid) and (b) q > 90°
(cohesion greater than adhesion: the surface repels the liquid).
Figure 7:
Liquid adhering to a solid surface for an angle of contact (a) less than 90° (e.g. water
and glass), and (b) greater than 90° (e.g. mercury and glass).
If a narrow glass tube is dipped into water, the liquid will rise up the tube, as shown in Figure 8, until the force
due to surface tension is equal in magnitude to the weight of the water column. The force on the glass due to
the liquid is equal in magnitude to the force of the glass on the liquid (Newton’s third law). Thus at equilibrium
the vertical component of the force on the water column due to adhesion with the glass surface is equal in
magnitude to the weight of the column of water.
Figure 8:
Capillarity: liquid rising in a tube due to adhesion between the liquid and the walls of
the tube.
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For a tube of radius r the length of contact is 2pr and so the vertical component of the force due to surface
tension is
2πrγ cosθ
(7)
If the water rises a distance h, the weight of water supported is
ρπ r2 hg
(8)
Equating these we find that the height of the column in the capillary tube is given by
h=
2γ cosθ
ρgr
(9)
Exercise
-5
The sap in trees, which consists mainly of water, rises in a system of capillaries of radius 2.5 x 10 m. The
contact angle is 0°. What is the maximum height to which the sap will rise? Since trees grow much higher
than this, capillary action cannot be the mechanism for sap rising in a tree.
Laplace’s Law
Laplace’s law relates the pressure difference across a closed membrane or liquid film to the tension in the
membrane or film. Consider a spherical drop of liquid of radius r. Consider the force balance on a hemisphere
as shown in Figure 9: the force due to surface tension along the contact length of the surface ‘membrane’ will
be balanced by the force due to the pressure difference between inside and outside, acting over the projected
area of the hemisphere. Thus
2 π r γ = ( pi − po ) π r 2
Figure 9:
(10)
Pressure inside a sphere of liquid is greater than the external pressure due to surface
tension.
Thus the pressure difference is
pi − po =
2γ
r
(11)
This is known as Laplace’s law. It implies that it requires a greater pressure difference to sustain a small
sphere than a larger one.
When a liquid is in equilibrium with its own vapour, the pressure of the gas phase is called the vapour
pressure. It may also be thought of as the pressure necessary to prevent more of the liquid from evaporating;
it balances the pressure difference across the liquid-vapour surface. Thus a liquid drop in equilibrium with its
vapour will have a pressure difference equal to the vapour pressure of the liquid at the ambient temperature.
Exercise
-2
-1
Given that the surface tension of water at 20 °C is 7.28 x 10 N.m , and the vapour pressure of water at
3
the same temperature is 2.35 x 10 Pa, what is the radius of the smallest spherical water droplet which can
form without evaporating?
Soap bubbles
A soap bubble differs from a drop in that there are two surfaces (an inside surface and an outside surface),
which introduces a factor of 2 into Laplace’s equation:
pi − p o =
PHYS1004 Module 2 Outline
4γ
r
(12)
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Copyright © 2012, The University of Sydney
Cylindrical tubes
When we apply the same arguments to a cylindrical membrane (e.g. a blood vessel) the result is
pi − p o =
γ
(13)
r
In Physiology, this pressure difference (across the wall of a blood vessel) is called the transmural pressure.
Note that this is quite different to the pressure difference that drives the fluid through the vessel.
Exercises
4
(a) An aorta has a transmural pressure of 1.3 x 10 Pa and a radius of 12 mm. What is the tension in its
wall?
3
-3
(b) A small capillary has a transmural pressure of 4.0 x 10 Pa and a radius of 6.0 x 10 mm. What is its
wall tension?
Alveoli
The small air sacs of the lungs, called alveoli, expand and contract every breath. Oxygen and carbon dioxide
are transported across the membrane of the alveoli. The tension in the walls of the alveoli is due to the
combined effect of the membrane and a liquid on the walls which contains long protein molecules called
surfactant. The effect of the surfactant is that the tension changes as the alveoli expand and contract.
Approximating alveoli as small spheres, and applying Laplace’s law:
r ( pi − po ) = rΔp = 2γ
(14)
Note that in order to keep the lungs inflated it is always the case that pi > po. During exhalation the pleural
pressure po increases, causing Dp ( = pi - po) to decrease, and due to muscle contraction the radii of the
alveoli decrease. If both r and Dp decrease, and g is constant the equilibrium condition cannot be satisfied. As
Dp is too small for equilibrium the alveoli will continue to contract until they are completely collapsed.
During inhalation pleural pressure po decreases, causing Dp to increase, and due to muscle relaxation the
radii of the alveoli increase. If both r and Dp increase, and g is constant the equilibrium condition cannot be
satisfied. As Dp is too large for equilibrium the alveoli will continue to expand until they rupture.
The surfactant solves this problem. As the radius of the alveoli increases the long molecules of the surfactant
are pulled apart and the tension increases. Correspondingly, as the radii decrease the molecules slide back
together and the tension decreases. In this way the molecular properties of the surfactant allow the tension to
vary so that equilibrium is maintained during the breathing cycle. Insufficient surfactant in the lungs is a cause
of death of newborn infants.
Negative Pressure
The tallest trees in the world are more than 60 m high. We have already seen that capillarity cannot explain
the rise of sap over such heights. Consider water in a cylinder with a piston in contact with the surface of the
water. Because of the adhesion between the water and the piston the water can be pulled up by the piston to
a height much greater than capillarity would allow and much greater than atmospheric pressure could support
(rgh = pA).
In trees sap moves up the tree through channels called xylem, which have radii of 25-250 µm. The xylem
channels are filled with water up to the leaves. As water evaporates from the leaf, the water column moves
upwards to keep itself intact. This phenomenon is known as negative pressure for the following reason. If the
pressure at the top of the xylem, at height h, is ph, and the pressure at the bottom is atmospheric pressure pA,
then the usual formula for variation of pressure with height gives
ph = p A − ρgh
(15)
If h is large enough, ph will be negative. For example if h = 60 m, ph = − 4.8 atmospheres. Thus the effect of
the adhesion of the water to the top of the xylem is equivalent to what would happen if we were able to create
a negative pressure at the top, hence the term negative pressure. This is of course not possible - pressure
can only be positive. In reality, the weight of the column of sap is supported by adhesion forces between the
sap and the top of xylem channel.
PHYS1004 Module 2 Outline
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Copyright © 2012, The University of Sydney