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Module 2 jif 104 liquid Module 2 jif 104 liquid Document Transcript

  • JIF 104 PHYSICS II/PRACTICAL 1b MODULE 2: PROPERTIES OF LIQUIDS by Sabar Bauk Pusat Pengajian Pendidikan Jarak Jauh, Universiti Sains Malaysia. 2008 1
  • 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
  • 3.8 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
  • 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
  • (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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • ∆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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • ∂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
  • 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
  • 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
  • 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
  • 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
  • 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
  • (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
  • (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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • +∞ −∞ − +∞ 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
  • 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
  • ∞ 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
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