Fundamentals of Fluid Mechanics PDF

FUNDAMENTALS OF FLUID MECHANICS BY M.O.IBIWOYE DEPARTMENT OF MECHANICAL ENGINEERING COLLEGE OF ENGINEERING AND TECHNOLOGY KWARA STATE UNIVERSITY,MALETE, P.M.B.1530, ILORIN, KWARA STATE  COURSE OUTLINE Characteristics of a Flui...

FUNDAMENTALS OF FLUID MECHANICS BY M.O.IBIWOYE DEPARTMENT OF MECHANICAL ENGINEERING COLLEGE OF ENGINEERING AND TECHNOLOGY KWARA STATE UNIVERSITY,MALETE, P.M.B.1530, ILORIN, KWARA STATE  COURSE OUTLINE Characteristics of a Fluid, Fluid Statics, Conservation Principles, Fluid Dynamics: uniform flow, steady flow, flow rate, Continuity equation, Bernoulli equation andMomentum equation REFERENCES 1. J.A. Robertson, and C.T. Crowe, Engineering Fluid Mechanics, Houghton Mifflin Comp., Boston, 1975. 2. J.F.Douglas, J.M. Gasiorek, J.A. Swaffield & Lynne B. Jack., Fluid Mechanics, Fifth Edition, 2005. 3. R.K. Rajput, A textbook of Fluid Mechanics and Hydraulic Machines in S.I. Units, 2014 4. Yunus A. Cengel and John M. Cimbala., Fluid Mechanics (Fundamentals and Applications),Fourth Edition, 2010. 5. P. Balanchandran, Fundamentals of Compressible Fluid Dynamics. 6. V.L. Streeter, K.W.Bedford and E.B. Wylie, Fluid Mechanics, Ninth Edition, McGraw Hill Book Company Inc., New York 1998. 7. Frank M. White, Fluid Mechanics, McGraw-Hill Fifth Edition, 2003. 8. Robert L. Mott Applied Fluid Mechanics, Sixth Edition, Pearson International Edition, 2006.  TABLE OF CONTENTS CHAPTER ONE - INTRODUCTION AND BASIC CONCEPTS 1.0 Introduction and basic concepts ….................................................................................6 1.1 Differences between Fluid and Solid ….........................................................................6 1.2 Commonalities between Liquids and gases …................................................................6 1.3 Differences between Liquids and Gasses........................................................................7 1.4 Scope of Fluid Mechanics................................................................................................8 CHAPTER TWO- PROPERTIES OF FLUIDS 2.0 Properties of Fluids...................................................................................................... 10 2.1 Density..........................................................................................................................10 2.1.1 Mass Density, …............................................................................................................10 2.1.2 Weight Density …..………………….…………………..............................................10 2.1.3 Specific Gravity…………………………….................................................................10 2.1.4 Property, State, Process and Cycle...............................................................................10 2.2 Viscosity.......................................................................................................................11 2.2.1 Coefficient of Dynamic Viscosity.................................................................................12 2.2.2 Kinematic Viscosity......................................................................................................13 2.2.3 Effect of Temperature on Viscosity...............................................................................13 2.2.4 Effect of Pressure on Viscosity …….............................................................................13 2.2.5 Newton’s Law of Viscosity..........................................................................................13 2.2.6 Newtonian / Non-Newtonian Fluids.............................................................................15 2.3 Thermodynamic Properties of Fluid..............................................................................17 2.3.1 Changes of state.............................................................................................................17 2.4 Surface Tension and Capillarity.....................................................................................18 2.4.1 Pressure Inside a Water Droplet, Soap Bubble and a Liquid Jet...................................18 CHAPTER THREE- FORCES IN STATIC FLUIDS 3.0 Forces in Static Fluids................................................................................................... 21 3.1 Fluids statics..................................................................................................................21 3.2 Pressure and Its Measurement.......................................................................................21 3.2.1 Pressure Head of a Liquid.............................................................................................22 3.2.2 Pascal’s Law for Pressure ata Point...............................................................................22 3.2.3 Variation of Pressure Vertically In a Fluid under Gravity.............................................24 3.2.4 Equality of Pressure at the Same Level in a Static Fluid..............................................25 3.2.5 General Equation for Variation of Pressure in a Static Fluid........................................26 3.3 Pressure and Head.........................................................................................................27 3.4 Pressure Measurement...................................................................................................28 3.4.1 Simple Manometers.......................................................................................................29 3.4.2 Choice of Manometer...................................................................................................37 3.4.3 Mechanical Gauges.......................................................................................................39 CHAPTER FOUR -Fluid Dynamics 4.0 Fluid Dynamics............................................................................................................ 41 4.1 Classification of Fluid flow..........................................................................................41 4.1.1 Uniform and non-uniform flow.....................................................................................41 4.1.2 Steady and unsteady flow..............................................................................................41 4.1.3 Compressible or Incompressible Flow.........................................................................41 4.1.4 One -, Two - and Three-dimensional flow.....................................................................42 4.1.4 Laminar and Turbulent flows........................................................................................43 4.2 Flow Rate......................................................................................................................43 4.2.1 Mass Flow Rate.............................................................................................................43 4.2.2 Volume Flow Rate - Discharge.....................................................................................43 4.2.3 Discharge and Mean Velocity.......................................................................................44 4.3 The Continuity Equation……………............................................................................45 4.4 The Bernoulli Equation..................................................................................................48 4.5 The Momentum Equation.............................................................................................51 CHAPTER FIVE – DIMENSIONAL ANALYSIS AND HYDRAULIC SIMILITUDE 5.0 Dimensional Analysis and Hydraulic Similitude........................................................ 55 5.1 Dimensional Analysis and its Applications...................................................................55 5.2 Dimensions and Units...................................................................................................55 5.3 Dimensional Homogeneity............................................................................................56 5.4 Methods of Dimensional Analysis...............................................................................57 5.4.1 Rayleigh’s Method........................................................................................................57 5.4.2 Buckingham’s p Method/Theorem................................................................................58 5.4.3 Limitations of Dimensional Analysis...........................................................................62 5.5 Model Analysis..............................................................................................................62 5.6 Similarity.......................................................................................................................62 CHAPTER SIX – Flow in pipes AND Ducts 6.0 Flow in pipes and ducts……………………………………………………………….63 6.1 Introduction…………………………………………………………………………...63 6.2 Classification of flow………………………………………………………………….63 6.3 Criterion of flow………………………………………………………………………64 6.3.1 Classification of flow based on Reynolds’ number…………………………………...65  CHAPTER ONE 1.0 INTRODUCTION AND BASIC CONCEPTS Fluid is any substance that can flow. In contrast, solid is any substance that cannot flow. We normally recognize the four states of matter as solid, liquid, gas and plasma. Liquids, gases, and plasmas are called fluid because of their shared characteristics which differentiate them from solid. Plasma which is the fourth state of matter and a fluid is beyond the scope of this course. Therefore, in this course, description of fluid will be restricted to only gases and liquids. The differences between fluid and solid, common characteristics of liquid and gas, and characteristic differences between liquid and gas are discussed in the following sub-sections. 1.1 DIFFERENCES BETWEEN FLUID AND SOLID Fluid Solid  The molecules of fluid have relatively  The molecules of solid are very very large spacing. closely spaced.  Fluid is capable of flowing.  Solid is typically not capable of flowing.  Fluid deforms continuously under the  Solid has the ability to resist shearing action of shearing force (shear stress). force (shear stress).  The applied shear stress on a fluid is  The applied stress on solid within its directly proportional to rate of strain. elastic limit is directly proportional (Newton’s law of viscosity). Here the strain (Hooke’s law). Here the constant of proportionality is known constant of proportionality is Young’s as coefficient of dynamic viscosity. modulus or modulus of elasticity.  It has no shape of its own, but  It typically has a shape of its own. conform to the shape of its container 1.2 COMMONALITIES BETWEEN LIQUIDS AND GASES The common characteristics between liquids and gases which qualify them as fluid are:  Fluid is capable of flowing.  Fluid deforms continuously under the action of shearing force (shear stress). Figure 1.1: Schematic of layers fluid element near a fixed solid surface 1.3 DIFFERENCES BETWEEN LIQUIDS AND GASES Although liquids and gasses behave in much the same way and share many similar characteristics, they also exhibit different characteristics as tabulated in the table below. Table 1.1: Differences between Liquids and Gasses S/N Liquids Gases  The spacing between different molecules The spacing between different of liquid is relatively large. molecules of gas is larger. 1 Liquids are difficult to compress and are Gases are easily compressed and are often classified as incompressible fluid. usually classified as compressible fluid. 2 It possesses a definite volume. It volume It possesses no definite volume. It is essentially constant with pressure. changes volume with pressure. 3 Its molecules are relatively free to change Its molecules are practically unrestricted their positions with respect to each other by cohesive forces. A gas has no fixed but restricted by cohesive forces so as to volume; it changes volume to expand to maintain a relatively fixed volume. Thus fill the containing vessel. It will it will occupy the container it is in and completely fill the vessel so no free form a free surface (if the container is of surface is formed. a larger volume). Figure 1.3 Figure 1.2 Now what is fluid mechanics? As its name suggests it is the study of fluids either in motion or at rest. Fluid mechanics is study of fluids in motion (fluid dynamics) or at rest (fluid statics). Fluid mechanics, like any other subject, has its own unique terminologies. Knowing the meaning of these terminologies will certainly leads to having a good grasp of the course. Thus, you are advice to review the fluid mechanics terminologies being defined in this lecture note with rapt attention. In this introductory section, the importance of fluid from engineering point of view will be stressed. This will make it fairly easy to see why engineers must have the knowledge of fluid behaviour to effectively analyze fluid related systems that will be encountered in a chosen field. Field of Fluid Mechanics can be divided into three (3) branches:  Fluid Statics: It is the study of the behavior of fluids at rest.   Kinematics: It is the study of velocity, acceleration and pattern of fluid particles in motion without considering the forces or energies that cause the motion.  Fluid Dynamics: It is the study of the behavior of fluid in motion and it is further divided into hydrodynamics and aerodynamics.  Hydrodynamics deal with flow of water.  Aerodynamics deals with flow of air. 1.4 SCOPE OF FLUID MECHANICS It takes pondering about life to recognize that fluid is one of the most important gifts of life. Life as we know it may not exist without fluids. The air we breathe, the water we drink, the blood that flow in our veins are all fluids. The flow of air keeps us comfortable in a warm room, and air provides the oxygen we need to sustain life. Similarly, our body fluids (liquid) are mostly water based. Any malfunction in the flow of fluids within our bodies, will essentially lead to poor health. Practically, fluids greatly influence our comfort; they are involved in our transportation systems in many ways; they have an effect on our recreation (e.g. basketballs and footballs are inflated with air) and entertainment (the sound from the speakers of a TV would not reach our ears in the absence of air). If fluid is this important, then we need to understand its concept. The knowledge of fluid mechanics is extremely important in many areas of engineering and science. Examples are: 1. Internal combustion engines 2. Turbojet, scramjet, rocket engines—aerospace propulsion systems 3. Waste disposal (a) Chemical treatment (b) Incineration (c) Sewage transport and treatment 4. Pollution dispersal—in the atmosphere (smog); in rivers and oceans 5. Steam, gas and wind turbines, and hydroelectric facilities for electric power generation 6. Pipelines (a) Transmission of crude oil and natural gas transfer (b) Irrigation facilities (c) Office building and household plumbing 7. Fluid/structure interaction (a) Design of tall buildings (b) Continental shelf oil-drilling rigs (c) Dams, bridges, etc. (d) Aircraft and launch vehicle airframes and control systems 8. Heating, ventilating and air-conditioning (HVAC) systems 9. Cooling systems for high-density electronic devices - digital computers from PCs to supercomputers 10. Solar heat and geothermal heat utilization 11. Artificial hearts, kidney dialysis machines, insulin pumps 12. Manufacturing processes (a) Spray painting automobiles, trucks, etc. (b) Filling of containers, e.g., cans of soup, cartons of milk, plastic bottles of soda (c) Operation of various hydraulic devices  (d) Chemical vapour deposition, drawing of synthetic fibres, wires, rods, etc. We conclude from the various preceding examples that there is essentially no part of our daily lives that is not influenced by fluids. As a consequence, it is extremely important that engineers be capable of predicting fluid motion. In particular, the majority of engineers who are not fluid dynamists still will need to interact, on a technical basis, with those who are quite frequently; and a basic competence in fluid dynamics will make such interactions more productive.  CHAPTER TWO 2.0 PROPERTIES OF FLUIDS Every fluid has its peculiar characteristic which helps to describe its condition and distinguish one fluid from another. Such characteristics are called fluid properties. The properties of fluid can be grouped into three classes: physical (general), thermodynamic and transport properties. We consider thermodynamic properties when fluid is been influenced by temperature and the equation of gas is use to describe the thermodynamic condition of a fluid. The transport properties of fluid are basically viscosity, thermal conductivity and mass diffusivity. However, only viscosity which is within the scope of this course will be considered in this group. This section provides basic knowledge of the main descriptors of fluid which will also include surface tension, capillarity and compressibility. 2.1 DENSITY The density of a fluid (or any other form of matter) is the amount of mass per unit volume. It can be expressed in there different ways. 2.1.1Mass Density,  It is defined as the mass of fluid per its unit volume at the standard temperature and pressure. It is commonly referred to as density. 3 m Its units are kg/m i.e.   (2.1) V Water has a density of 1000kg/m3 and air has a density of 1.23kg/m3 at P = 1.013×105N/m2, T = 15oC. N.B: Specific Volume is the reciprocal of mass density and thus, defined as volume per unit mass of a fluid. 3 V Its units are m /kg i.e.  (2.2) m 2.1.2 Weight Density (also known as Specific Weight),w It is defined as weight of a fluid per unit volume at the standard temperature and pressure. N.B: Recall that weight is a force, equal to mass times acceleration, then it is clear that the definition implies; w = ρg (2.3) where g is usually the gravitational acceleration. 2.1.3 Specific Gravity, SG (or Relative Density, σ) It is defined as the ratio of mass density a fluid to that of an equal volume of a standard reference fluid. The standard reference fluid for liquid is water (which usually occurs at 4oC) at atmospheric pressure. While the standard reference fluid for gas is air. Thus,  liquid  liquid Relative density for liquid, SGliquid   (2.4)  water 998kg / m 3   gas  gas Relative density for gas, SG gas   (2.5)  air 1.205kg / m 3 Because σis a ratio of densities, it is a dimensionless quantity and thus has no dimension or units. WORKED EXAMPLE 2-1 A 5-litre of liquid weighs30N. Calculate its specific weight, density, specific volume and specific gravity. Solution: Given Data: - Weight of the liquid = 30N Volume of the liquid = 5 litres = 5 ×10-3m3 Weight of liquid Specific weight, w = Volume of liquid 30 N w 3  6000 N / m 3  6 KN / m 3 5  10 m 3 Mass of liquid Density, ρ= Volume of liquid m w 6000 N / m 3    2  611.62kg / m 3 V g 9.81m / s 1 Specific volume, ѵ= Density 1 1    1.635  10 3 m 3 / kg  611.62kg / m 3 SGliquid Specific gravity, SGliquid= SGwater  liquid 611.62kg / m 3 SGliquid    0.61  water 1000 kg / m 3 2.2 VISCOSITY What is viscosity? Intuitively, viscosity is something like “stickiness or internal friction.”Viscous fluids tend to be sticky, indicating resistance to shear stresses. Viscosity is a measure of internal friction responsible for fluid resistance to flow. Since relative motion between layers requires the application of shearing forces, that is, forces parallel to the surfaces over which they act, the resisting forces must be in exactly the opposite direction to the applied shear forces and so they too are parallel to the surfaces. It is a matter of common experience that, under particular conditions, one fluid offers greater resistance to flow than another. Such liquids as tar, treacle and glycerine cannot be rapidly poured or easily stirred, and are commonly spoken of as thick liquid; on the other hand, so-called thin liquids such as water, petrol and paraffin flow much more readily. (Lubricating oils with small viscosity are sometimes referred to as light viscous fluid, and those with large viscosity are referred to as heavy fluid; but viscosity is not related to density.) Gases as well as liquids have viscosity, although the viscosity of liquids is larger than that of gases. For instant water is nearly 55 times as viscous as air. Viscosity can be described as the fluid property which determines its resistance to  shearing stresses. Viscosity of fluid arises on molecular scales due to two main physical effects: fluid intermolecular cohesion and interaction. It should be expected that the former would be important often dominant) in most liquids for which molecules are relatively densely packed, and the latter would be more important in gases in which the molecules are fairly far apart, but moving at high speed. Consider Figure 2.1 where two layers of fluid at a distance ‘dy’ apart, move over one and other at different velocities, say u and u+du. The viscosity together with relative velocity causes a shear stress acting between the fluid layers. The top layer of fluid causes a shear stress on the adjacent lower fluid layer while the lower layer causes a shear stress on the adjacent top layer. This shear stress (τ) is proportional to the rate of shear strain known as velocity gradient. Figure 2.1: Velocity gradient near a solid boundary du Mathematically,  dy du or   (2.6) dy where, μ is the constant of proportionality known as co-efficient of dynamic viscosity or simply called viscosity. du , is the rate of shear strain or rate of deformation or velocity gradient. dy 2.2.1 Coefficient of Dynamic Viscosity Coefficient of dynamic viscosity,  , can be defined as shear stress,  , (shear force per unit area) required to drag one layer of fluid with unit velocity past another layer a unit distance away from it in the fluid. From equation (2.6), we have F   Force      Area  =   =  A  = kgm1s 1 or Nsm2  du   du   Velocity   dy   dy   dis tan ce        The unit of coefficient of dynamic viscosity (or simply viscosity) is kgs -1m-1 or Nsm-2. However, the coefficient of dynamic viscosity is often measured in poise (P). 10 P = 1 kgs-1m-1 = 1 Nsm-2 The typical values of co-efficient of dynamic viscosity for water and air are respectively 1.14×10-3 kg/ms and 1.78×10-5 kg/ms. 1 Note: the viscosity of water at 20oC = poise = centipoises (CP) 100 2.2.2 Kinematic Viscosity The kinematic viscosity, , is defined as the ratio of dynamic viscosity to the density of the fluid.  / The unit of kinematic viscosity is m2/s. However, kinematic viscosity is also measured in stoke and 1 stoke = 10-4m2/s. 2.2.3 Effect of Temperature on Viscosity Temperature has effects on viscosity of fluid. In liquids, increases in temperature decreases viscosity. While in gases increases in temperature increases viscosity. In liquids, the shear stress is cause by the inter-molecular cohesion which decreases with increase in temperature. As the temperature of liquids increases, the internal friction or loosely saying stickiness will reduce and the liquid is then less viscous that it was at lower temperatures. In gases the intermolecular cohesion is negligible and the shear stress is due to exchange of momentum of the molecules, normal to the flow direction. The molecular activity increases with increase in temperature and so does the viscosity of gases.  For liquids:  T  Ae T 1 2 bT For gases: T  1 a T where, μT = Coefficient of dynamic viscosity at absolute temperature T, A, β = Constants of a given liquid, and a, b = constants for a given gas. 2.2.4 Effect of Pressure on Viscosity Under ordinary conditions, viscosity is significantly affected by the changes in pressure. However, increases in pressure have been found to appreciably increase the viscosity of some oil. 2.2.5 Newton’s Law of Viscosity Newton’s Law of Viscosity states that the shear stress exerted on the surface of a fluid element is directly proportional to the rate of shear strain. Let’s make use of Newton’s observations by considering a magnified 3Dof a layer of fluid element as shown in Figure 2.2.  Figure 2.2: Fluid element under a shear force The shearing force F, acts on the area on the top of the element. This area is given by A  z  x. We can thus calculate the shear stress which is equal to force per unit area i.e. F Shear stress,   A The deformation which this shear stress causes is measured by the size of the angle of deformation,  and is known as shear strain. In a fluid  increases for as long as  is applied - the fluid flows. In a solid shear strain,  , is constant for a fixed shear stress . It has been found experimentally that the shear stress is directly proportional to the rate of shear strain (shear strain per unit time,  /time) shear stress. If the particle at point E (in the above Figure 2.2) moves under the shear stress to point E’ and it takes time t toget there, it has moved the distance x. For small deformations we can write x Shear strain   y  x u Rate of shear strain =   t ty y x where, u t Using the experimental result that shear stress is proportional to rate of shear strain then u   cons tan t  y u The term is the change in velocity with y, or the velocity gradient, and may be written in y du differential form. The constant of proportionality is known as the coefficient of dynamic dy viscosity,  , of the fluid, giving du   dy This is known as Newton’s law of viscosity. The fluids which follow this law are known as Newtonian fluid. 2.2.6 Newtonian / Non-Newtonian Fluids Even among substances which are accepted as fluids there can be wide differences in behaviour under shear stress. Fluids obeying Newton’s law where the value of  is constant are known as Newtonian fluids. If  is constant the shear stress is linearly dependent on velocity gradient. This is true for most common fluids such as water, kerosene, air etc. Fluids in which the value of  is not constant are known as non-Newtonian fluids. Such fluids are generally complex mixtures and relatively uncommon. These fluids are studied under the science of deformation of flow called “rheology”. Examples of non-Newtonian fluids are slurries, mud flows, polymer solution, blood, etc. WORKED EXAMPLE 2.2 A moving plate spaced 0.05 mm apart from a stationary plate moves with a velocity of 1.2m/s as shown in the figure below. Find the viscosity of the fluid between the plates, if the force required to maintain the given speed of the plate is 2.2KN/m2. Figure 2.3: Two parallel spaced plates with fluid in-between. Solution: Given data: Velocity of the moving plate, u = 1.2m/s, Distance between plates, dy = 0.05mm = 0.05×10-3m, and Shearing force on the moving plate, F = 2.2KN/m2. To find: μ=? Assumption: the fluid in between the plate is a Newtonian fluid. du We know that,    dy where τ = shearing force (shear stress)per unit area = 2.2KN/m2. du = change of velocity = u – 0 = 1.2m/s dy = change of distance = 0.05×10-3m, and du 1.2m / s  =24×103s-1 dy 0.05  10 3 m   2.2 KN / m 2 Thus;    3 1 = 9.17×10-2Ns/m2 = 9.17×10-1poise  du  24  10 s    dy  ≈ 0.92poise Tutorials 1: 1. Prepare not more than three (3) pages of note on Newtonian and Non-Newtonian fluid. To be submitted within the next 72 hours. 2. A 0.3 square meter plate sliding down the 30o inclined plane to the horizontal with a velocity of 0.18m/s. There is a cushion of 0.9mm fluid thickness between the plane and the plate. If the weight of the plate is 140N, find the viscosity of the fluid. 3. The space between two square flat parallel plates is filled with oil. Each side of the plate is 720mm. The thickness of the oil film is 15mm. The plate, which moves at 3m/s, requires a force of 120N to maintain the speed. Determine: i. Dynamic viscosity of the oil. ii. Kinematic viscosity of the oil if the specific gravity of the oil is 0.95 4. A fluid with dynamic viscosity of 0.9Ns/m2flows over a flat plate with a velocity profile given by u = 2y – y2. Determine the velocity gradient and shear stress at the boundary and 0.15m from it. Note that u is the velocity in m/s at a distance y metre above the plate. 5. The velocity profile of flow over a flat plate is parabolic (u = ly2+ my +n) with vertex 30cm from the plate, where the velocity is 180cm/s as shown in figure 2.4. Plot the graph of shear stresses against rate of shear strains using a 5cm interval from the plate to the 30cm vertex. Figure 2.4: Flow over a flat plate. 6. A 400mm diameter shaft in Figure 2.5is rotating at 200 r.p.m. in a bearing of length 120mm. If the thickness of oil film is 1.5mm and the dynamic viscosity of the oil is 0.7Ns/m2. Assume linear velocity profile, determine: i. Torque required to overcome friction in bearing, and ii. Power utilised in overcoming viscous resistance.  Figure 2.5: Rotation shaft in a bearing 2.3 THERMODYNAMIC PROPERTIES OF FLUID The thermodynamic properties are considered using the equation of state of a perfect gas when a fluid is influenced by temperature changes. PV  mRT (2.7) or P  RT (2.8) Where, P = Absolute pressure, V = Volume of the fluid, T = Absolute temperature, m = mass of gas, and R = Characteristic gas constant Alternatively, the perfect gas equation can be derived in terms of kilogram-mole where m = nM. PV  nMRT (2.9) where, m = nM, n = number of moles, and M = molecular weight. For a given temperature and pressure, Equation (2.8) indicates that ρR= constant. By Avogadro’s hypothesis, all pure gases at the same temperature and pressure have the same number of molecules in a specified volume. The density is proportional to the mass of an individual molecule and so the product of R and the relative molecular mass M is constant for all perfect gases. This product MR is known as the universal gas constant, Ro. 2.3.1 Changes of state A change of density may be achieved both by a change of pressure and by a change of temperature. When the change in a state of the fluid system is affected at constant pressure the process is known to be isobaric process. When the change in a state of the fluid system is affected at constant temperature the process is known to be isothermal process. When no heat is transferred to or from the fluid during the change in the state of the fluid system, the process is known to be adiabatic process. P Here, P   = constant (2.10)  cp where,   , (2.11) cv cp=Specific heat capacity at constant pressure, and cv= Specific heat capacity at constant volume. 2.4 SURFACE TENSION AND CAPILLARITY Cohesion is the intermolecular attraction between molecules of the same liquid. It enables a liquid to resist small amount of tensile stresses. Also, cohesion is the tendency of the liquid to remain as one assemblage of particles rather than to behave like a gas and fill the entire space within which it is confined. Surface tension is caused by cohesion between particles at the free surface of a liquid. Adhesion is the attraction between the molecules of a liquid and the molecules of its solid boundary. Capillarity action is due to both cohesion and adhesion. Surface tension arises from the force of cohesion at the free surface. a liquid molecule in the interior of the liquid mass is surrounded by other molecules all around and is in equilibrium. It the free surface of liquid, there are no liquid molecules above the surface to balance the force of the molecules below it. Consequently, as shown in Figure 2.6, there is a net inward force on the molecule and it is normal to the liquid surface. At the free surface a thin layer of molecules is formed. It is because of this thin film that a thin small needle can float on the free surface of some liquid. Figure 2.6: Schematic molecules of liquid Surface tension is usually expressed in N/m. The value of surface tension depends upon the following factors:  i) Nature of the liquid ii) Nature of the surrounding matter (e.g., solid, liquid or gas), and iii) Kinetic energy (and hence the temperature of the liquid molecules). Water in contact with air has a surface tension of about 0.073 Nm−1 at usual ambient temperatures. Most organic liquids have values between 0.020 and 0.030 Nm−1 and mercury about 0.48 Nm−1, the liquid in each case being in contact with air. The surface tension of all liquids decreases with increase in temperature. The surface tension of water may be considerably reduced by the addition of small quantities of organic solutes such as soap and detergents. Salts such as sodium chloride in solution raise the surface tension of water. That tension which exists in the surface separating two immiscible liquids is usually known as interfacial tension. 2.4.1 Pressure inside a Water Droplet, Soap Bubble and a Liquid Jet Water droplet: Let, P = Pressure inside the droplet above outside atmospheric pressure, d = diameter of the droplet, and σ = surface tension of the liquid. Figure 2.7: Pressure inside a water droplet From free body diagram shown in Figure 2.7, we have:  i) Pressure force = P  d 2 , and 4 ii) Surface tension force acting around the circumference =   d. Pressure force and surface tension force are equal and opposite forces under equilibrium conditions.  i.e., P d 2 =   d 4   d 4  P  (2.12)  2 d d 4  Equation (2.12) shows that increases in the size droplet decreases the pressure intensity. Soap (or hollow) droplet: Soap bubbles have two surfaces on which surface tension (σ) acts. From the free body diagram shown in figure 2.8, we have:  P d 2 = 2 (   d ) 4 2(  d ) 8  P  (2.13)  2 d d 4 Since soap solution has a high value of surface tension, even with small pressure blowing, a soap bubble will tend to grow larger in diameter (hence formation of large soap bubbles). Figure 2.8: Pressure inside a soap bubble A Liquid jet: Let us consider a cylindrical liquid jet of diameter d and length l. Figure 2.9 shows a semi-jet. Pressure force = P  l  d Surface tension force =   2l Equating the two forces gives: P  l  d =   2l   2l 2  P  (2.14) ld d  Figure 2.9: Forces on liquid jet An interesting example of the effects of surface tension can be found in wet sand. It is common experience that we cannot walk on water. But it is also difficult to walk on dry sand-think beach volleyball. However, when we wet sand with water sufficiently to “activate” surface tension effects we can easily walk on the mixture. Another surface tension effect is capillarity. The manifestation of this is the curved shape of liquid surfaces near the walls if a container having sufficiently small radius to make surface tension forces non-negligible. Figure 2.10 displays this effect for two different cases. Details of the physics of capillarity, like that of surface tension, are rather complicated, and a rigorous treatment is beyond the scope of these lectures. But the basic idea is fairly simple. We see from Fig. 2.7 that surface tension acts in a direction parallel to the surface of the interface. Now the interaction of the gas-liquid interface is altered at the solid wall - one can imagine that this is caused by a combination of liquid molecular structure (including size of molecules) and details of the wall surface structure. This interaction causes the interface to make a generally nonzero angle φwith the solid boundary as shown in Fig. 2.10. This altered interface is called a meniscus, and the angle φ is termed the contact angle. The important point to note here is that unless φ = 90◦there will be a vertical component of force arising from surface tension. In Fig. 2.10(a) φ 90◦for which the surface tension force acts downward. Both situations occur in actual fluids. Figure 2.10: Capillarity for two different liquids  CHAPTER THREE 3.0 FORCES IN STATIC FLUIDS In this chapter, we will study the forces acting on or generated by fluids at rest. 3.1 FLUIDS STATICS The general rules of statics (as applied in solid mechanics) apply to fluids at rest. From earlier discussion we know that:  A static fluid can have no shearing force acting on it, and that  Any force between the fluid and the boundary must be acting at right angles to the boundary. Note that this statement is also true for curved surfaces, and in this case the force acting at any point is normal to the surface at that point. The statement is also true for any imaginary plane in a static fluid. We use this fact in our analysis by considering elements of fluid bounded by imaginary planes. Figure 3.1: Pressure force normal to the boundary We also know that:  For an element of fluid at rest– it applies that the sum of the components of forces in any direction will be zero.  The sum of the moments of forces on the element about any point must also be zero It is common to test equilibrium by resolving forces along three mutually perpendicular axes and also by taking moments in three mutually perpendicular planes and to equate these to zero. 3.2 PRESSURE AND ITS MEASUREMENT When a fluid is contained in a vessel, it will exert a normal force at all points on the boundary it is in contact with. Since these boundaries may be large and the force may differ from place to place it is convenient to work in terms of pressure, P, which is the force per unit area. Pressure intensity is the ratio of the normal force which exerts on the unit area of its boundary. Pressure is said to be uniform, if the force exerted on each unit area of a boundary is the same. F P (3.1) A where, P = pressure intensity, F = force, and A = area over which the force is applied. The unit of pressure is Newton’s per square metre, (N/m2) or Pascal (Pa), i.e. 1Pa = 1N/m2. Also frequently used is the alternative SI unit the bar. 5 5 Where, 1bar = 10 N/m2 =10 Pa 3.2.1 Pressure Head of a Liquid A liquid is subjected to pressure due to its own weight and this pressure increases with an increase in the depth of the liquid. Consider a liquid at rest contained by a vessel as illustrated in Figure 3.2. The liquid will exert pressure on sides and bottom of the vessel. Now let a cylinder be made to stand in the liquid as shown the figure. Figure 3.2: Schematic of Pressure head Let, h = height of the liquid in the cylinder, A = Area of the cylinder, w = Specific weight of the liquid, and P = Pressure intensity Pressure on the base of the cylinder = Weight of the liquid in the cylinder i.e. PA = wAh thus, P = wh [since w=ρg from equation (2.3)] P = ρgh (3.2) We could see from Equation (3.2) that the pressure intensity of liquid depends on the height which is the distance from the bottom of the vessel to the free surface. However, it is sometimes convenient to express the pressure intensity of a liquid in terms of the liquid height, h, known as the pressure head or static head. P P i.e. h  → Pressure head or static head g w 3.2.2 Pascal’s Law for Pressure ata Point Pascal’s law states that pressure intensity at any point in a liquid at rest is the same in all directions. (Proof that pressure acts equally in all directions.) By considering a small element of fluid in the form of a triangular prism which contains a point P, we can establish a relationship between the three pressures Px in the x direction, Py in the y direction and Ps in the direction normal to the sloping face. Figure 3.3: Triangular prismatic element of fluid The fluid is at rest, so we know there are no shearing forces on the faces of the element, and we know that all forces due to pressure are acting at rightangles to the surfaces.i.e. Pxis pressure acting perpendicular to surface ABFE, Pyis pressure acting perpendicular to surface FECD, and Ps is pressure acting perpendicular to surface ABCD. And, as the fluid is at rest, in equilibrium, the sum of the forces due to pressure in any direction is zero. The sum of forces in the x-direction is:  Fx = Px×AABFE+ ( - Ps ×AABCD)sinθ= Pxδyδz – Psδsδz(δy/δs) = Pxδyδz - Psδyδz For any fluid element at rest, the sum of the components of forces in any direction will be zero. i.e Pxδyδz – Psδyδz = 0 Px = Ps (3.3) Similarly, the sum of forces in the y-direction is given as;  Fy = Py×ACDEF+ ( - Ps ×AABCD)cosθ+ Wfluidelement = Pyδxδz – Psδsδz (δx/δs) + ρg×0.5δxδyδz Considering the fact thatδx, δy, and δz are all infinitesimal quantities, and that δxδyδz is negligible compared to δxδz coupled with the fact that F y = 0 for static fluid. Py = Ps (3.4) Thus, from Equations (3.3) and (3.4), Px = Py = Ps Considering the prismatic element again, Psis the pressure on a plane at any angle θ, the x, y and z-directions could be any orientation. The element is so small that it can be considered a point so thederived expression Px = Py = Ps indicates that pressure at any point is the same in all directions. WORKED EXAMPLE 3.1 The diameter of ram and plunger of an hydraulic press are 200mm and 30mm respectively. Find the weight lifted by the hydraulic press, if 400N force is applied at the plunger. Solution: Given data: Force on plunger, F = 400N Diameter of the plunger, d = 30mm = 0.03m Diameter of the ram, D = 200mm = 0.2m To find: Load lifted (W)? Schematic of the problem: Figure 3.4: Hydraulic press   Area of the plunger, A p  d2  0.03 2  7.07  10  4 m 2 4 4   Area of the ram, Ar  D2  0.2 2  0.03m 2 4 4 F 400 Pressure intensity on the plunger, Pp   4  5.66  10 5 N / m 2 Ap 7.07  10 m 2 We need to employ Pascal’s lawto determine the pressure intensity on the ram, Pr. Where, Pr  Pp  5.66  10 5 N / m 2 Load (W ) Pr   5.66  10 5 N / m 2 Ar Load(W )  Ar  5.66 105 N / m2  0.03m2  5.66 105 N / m2  17KN 3.2.3 Variation of Pressure Vertically In a Fluid under Gravity Shown in Figure 3.5 is a small element of fluid which is a vertical column of constant cross- sectional area, A, surrounded by the same fluid of mass density ρ. The pressure at the bottom of the cylinder is P1at level z1, and at the top isP2at level z2. The fluid is at rest and in equilibrium so all the forces in the vertical direction sum to zero. I.e. we have Force due to P1on A (upward) = P1A Force due toP2 on A (downward) = P2A Force due to weight of element (downward) =mg = mass density × volume = ρgA(z2 – z1) Taking upward forces as positive, in equilibrium we have P1A - P2A - ρgA(z2 – z1) = 0 P2 - P1= ρg(z2 – z1)=w(z2 – z1)(3.5) Thus in a fluid under gravity, pressure decreases with increase in height z = (z2 – z1).  Figure 3.5: Vertical elemental cylinder of fluid 3.2.4Equality of Pressure at the Same Level in a Static Fluid Consider the horizontally small cylindrical element of fluid shown in Figure 3.6, with cross- sectional area A, in a fluid of density ρ, pressure P1 at the left hand end and pressure at the right hand end. The fluid is at equilibrium so the sum of the forces acting in the x direction is zero. PlA=PrA Pl=Pr (3.6) Pressure in the horizontal direction is constant. Figure 3.6: Horizontal elemental cylinder of fluid 3.2.5 General Equation for Variation of Pressure in a Static Fluid Here we show how the above observations for vertical and horizontal elements of fluids can be generalized for an element of any orientation. Consider the cylindrical element of fluid in the figure above, inclined at an angle θ to the vertical, length δs, cross-sectional area A in a static fluid of mass density ρ. The pressure at the end with height z is P and at the end of height z+δz isP+δP.  Figure 3.7: A cylindrical element of fluid at an arbitrary orientation. The forces acting on the element are PA = force acting at right-angles to the end of the face at z (P+δP)A = force acting at right-angles to the end of the face at z+δz mg = ρ × Aδs × g There are also forces from the surrounding fluid acting normal to these sides of the element. For equilibrium of the element the resultant of forces in any direction is zero. Resolving the forces in the direction along the central axis gives PA− (P+δP)A−ρgAδscosθ = 0 δP = - ρgδscosθ δP/δs = - ρgcosθ Or in the differential form dP/ds = - ρgcosθ = - w cosθ If θ = 90 then s is in the x or y directions, (i.e. horizontal), so the weight of the element acting o vertically down  dP  dP dP     0  ds  90o dx dy Confirming that pressure on any horizontal plane is zero. If θ = 0o then s is in the z direction (vertical) so  dP  dP      g (3.7)  ds  0o dz The pressure variation with elevation is found by integrating Equation 3.7: dP    gdz z2 z2 P2  P1    gdz    wdz (3.8) z1 z1 3.3 PRESSURE AND HEAD dP In a static fluid of constant density we have the relationship   g   w , as shown above. dz This can be integrated to give  P  gz  cons tan t In a liquid, the pressure at any depth z measured from the free surface is given as: P  gh  cons tan t Where z = - h(seethe figure below) Figure 3.7:Fluid head measurement in a tank At the surface of fluids we are normally concerned with the atmospheric pressure,Patmoshpere. So P  gh  Patmoshpere As we live constantly under the pressure of the atmosphere, and everything else exists under this pressure, it is convenient (and often done) to take atmospheric pressure as the datum. So we quote pressure as above or below atmospheric. Pressure quoted in this way is known as gauge pressure i.e. Gauge pressure is Pguage  gh  wh Gauge pressure is the pressure measured with the help of pressure measuring instrument, in which the atmospheric pressure is taken as datum. The atmospheric pressure on the scale is marked as zero. Gauges record pressure above or below the local atmospheric pressure, since they measure the difference in pressure of the liquid to which they are connected and that of the surrounding air. If the pressure of the liquid is below the local atmospheric pressure, then the guage is designated vacuum gauge and the recoded value indicates the amount by which the pressure of the liquid is below local atmospheric pressure, i.e. negative pressure. Hence, vacuum pressure (or negative pressure) is defined as the pressure below the atmospheric pressure. Absolute pressure  Figure 3.8: Relationship between pressures The lower limit of any pressure is zero - that is the pressure in a perfect vacuum. Pressure measured above this datum is known as absolute pressure i.e. Absolute pressure is Pabsolute  gh  Patmoshpere (3.8) Absolute pressure = Gauge pressure + Atmospheric pressure As g is (approximately) constant, the gauge pressure can be given by stating the vertical height, h, of any fluid of density ρ which is equal to this pressure. P = ρgh This vertical height, h, is known as head of fluid. Note: If pressure is quoted in head, the density of the fluid must also be given. Example: −2 We can quote a pressure of 500KNm in terms of the height of a column of water of density,ρ = 1000kg/m3. Using P = ρgh, P 500  10 3 h   50.95m of water g 1000  9.81 And in terms of Mercury with density, ρ = 13.6×103kg/m3. P 500  10 3 h   3.75m of Mercury g 13.6  10 3  9.81 3.4 PRESSURE MEASUREMENT The pressure of a fluid is typically measured by engineers with the use of manometers or mechanical gauges. Manometers:  Manometers are defined as devices used for measuring the pressure at a point in a fluid by balancing the column of the fluid by the same or another column of liquid. Manometers are classified as follows; (a) Simple manometers; 1. Piezometer, 2. U-tube manometer, and 3. Single column manometer (Advance U-tube manometer). (b) Differential manometers Mechanical gauges: Mechanical gauges are devices in which pressure is measured by balancing the fluid column by spring (elastic element) or dead weight. Generally, these gauges are used for measuring high pressure and where high precision is not required. The commonly used mechanical gauges are: 1. Bourdon tube pressure gauge, 2. Diaphragm pressure gauge, 3. Bellow pressure gauge, and 4. Dead-weight pressure gauge. 3.4.1 Simple Manometers Simple manometer is one which consists of a glass tube whose one end is connected to a point where pressure is to be measured and the other end remains open to atmosphere. The relationship between pressure and head is used for pressure measurement in the simple manometer (also known as a liquid gauge). 3.4.1.1 The Piezometer The simplest manometer is a tube, open at the top, which is attached to the top of a vessel containing liquid at a pressure (higher than atmospheric) to be measured. An example is shown in the Figure 3.9. This simple device is known as a Piezometer tube. As the tube is open to the atmosphere the pressure measured is relative to atmospheric so is gauge pressure. Figure 3.9: A simple piezometer tube manometers Pressure at A = pressure due to column of liquid above A PA  gh1 Pressure at B = pressure due to column of liquid above B  PB  gh2 This method can only be used for liquids (i.e. not for gases) and only when the liquid height is convenient to measure. It must not be too small or too large and pressure changes must be detectable. Class Work 1: If the distance between the bottom and top of the liquid in the piezometer tube shown in Figure 3.9 is 4m. What is the maximum gauge pressure that can be measured if (a) the liquid is water and (b) the liquid is mercury? Remember that the densities of water and mercury are 1000kg/m3and 13.6×103kg/m3 respectively. 3.4.1.2The “U”-Tube Manometer The “U”-Tube manometer enables the pressure of both liquids and gases to be measured with the same instrument. The “U” is connected as shown in Figure 3.10 and filled with a fluid called the manometric-fluid. The fluid whose pressure is being measured should have a mass density less than that of the manometric fluid and the two fluids should not be able to mix readily - that is, they must be immiscible. Figure 3.10: U-tube manometer (i) For positive pressure – refer to Figure 3.10 (a). Pressure in a continuous static fluid is the same at any horizontal level so, Pressure at B = pressure at C PB  PC (i.e. Pressure above X-X in the left hand arm = Pressure above X-X in the right hand arm) For the left hand arm Pressure at B = pressure at A + pressure due to height h1 of fluid being measured PB  PA  gh1 For the right hand arm Pressure at C = pressure at D + pressure due to height h2 of manometric fluid PC  PAtmoshpere   man gh2 As we are measuring gauge pressure we can subtract PAtmoshperic giving  PB  PC PA   man gh2  gh1  wman h2  wh1 (3.9) If the fluid being measured is a gas, the density will probably be very low compared to the manometric fluid density i.e. ρman>>ρ (i.ewman>>w). In this case the term wh1can be neglected, and the gauge pressure is given by PA  wman h2 (3.10) (ii) For negative pressure – refer to Figure 3.10 (b). Pressure at B = pressure at C PB  PC (i.e. Pressure above X-X in the left hand arm = Pressure above X-X in the right hand arm) For the left hand arm Pressure at B = pressure at A + pressure due to height h1 of fluid being measured PB  PA  gh1   man gh2 For the right hand arm Pressure at C = 0 PB  PC PA  (  man gh2  gh1 )  (wman h2  wh1 ) (3.11) If ρman>>ρ (i.ewman>>w), then PA   man gh2  wman h2 (3.12) 3.4.1.3Advances to the “U” tube manometer (Single column manometer) The “U”-tube manometer has the disadvantage that the change in height of the liquid in both sides must be read. This can be avoided by making the diameter of one side very large compared to the other. In this case the side with the large area moves very little when the small area side move considerably more. The manometer in Figure 3.11 is arranged to measure the pressure difference in a gas of negligible density. If the datum line indicates the level of the manometric fluid when the pressure difference is zero and the height differences when pressure is applied is as shown in Figure 3.11, the volume of liquid transferred from the left side to the right = z 2  (d 2 / 4) And the fall in level of the left side is z1 = Volume moved /Area of left side z 2 (d 2 / 4) 2 d   z2   D / 4 2 D  Figure 3.11: U-tube with one leg enlarged We know from the theory of the “U” tube manometer that the height different in the two columns gives the pressure difference (see Equation 3.13) so  d  2 P1  P2  g  z 2  z 2      D     d 2   gz 2 1       D   2 Clearly if D is very much larger than d then (d/D) is very small so P1  P2  gz 2 So only one reading need be taken to measure the pressure difference. If the pressure to be measured is very small, then, tilting the arm provides a convenient way of obtaining a larger (more easily read) movement of the manometer. The above arrangement with a tilted arm is shown in the Figure 3.12. Figure 3.12: Tilted U-tube manometer The pressure difference is still given by the height change of the manometric fluid but by placing the scale along the line of the tilted arm and taking this reading large movements will be observed. The pressure difference is then given by  P1  P2  gz 2  gx sin The sensitivity to pressure change can be increased further by a greater inclination of the manometer arm, alternatively the density of the manometric fluid may be changed. 3.4.1.4 Differential Manometers A differential manometer is used to measure pressure differences between two points in a pipe or in two different pipes. Differential manometer in its simplest form consist of U-tube containing a heavy or lighter manometric fluid which its ends are connected to a pressurized vessel at two points and the pressure difference between these two points can be measured. The common types of differential manometers are: 1. U-tube differential manometer, and 2. Inverted U-tube (or n-tube) differential manometer U-tube differential manometer – Measurement of pressure difference between two points in a pipe. If the manometer is arranged as shown in Figure 3.11, then Pressure at C = pressure at D PC  PD PC  PA  ha PD  PB  w(hb  h)  wman h PA  wha  PB  w(hb  h)  wman h Giving the pressure difference PA  PB  w(hb  h)  wman h  wha PA  PB  w(hb  ha )  (wman  w)h (3.11) Again, if the fluid in the vessel shown in Figure 3.10 is gas, the pressure difference will be PA  PB  wman h (3.12) Figure 3.13: Measurement of pressure difference between two points with U-tube manometer Quiz 1: Briefly explain why Equation 3.11 is simplified as Equation 3.12. Assignment 2: A U-tube manometer is arranged to measure the pressure difference between point A and B in a pipeline conveying fluid as shown in Figure 3.13. The manometric liquid density is 13600kg/m3, h is half a meter, and point A is 200cm lower than point B. Calculate the pressure difference if the conveying fluid is (a) water with ρ = 1000kg/m3, and (b) methane with ρ = 0.72kg/m3. U-tube differential manometer connected to two pipes at the same level. If the manometer is arranged as shown in Figure 3.14, then Pressure at C = pressure at D PC  PD PC  PA  (h1  h)wA PD  PB  wB h1  wman h PA  (h1  h)wA  PB  wB h1  wman h The pressure difference between A and B is; PA  PB  wB h1  wman h  (h1  h)wA PA  PB  (wB  wA ) h1  wman h  hwA wB  w A but if , then; PA  PB  wman h  hwA PA  PB  (wman  wA )h  Figure 3.14: Two pipes at the same level where, h = difference in manometric fluid level h1 = distance from point A to the manometric fluid level in the right limb PA = gauge pressure at point A PB = gauge pressure at point B wA and wB = weight density of liquid A and B respectively wman = weight density of the manometric fluid U-tube differential manometer connected to two pipes at different levels. Pressure at C = pressure at D PC  PD PC  PA  (hA  h)wA PD  PB  wB hB  wman h PA  (hA  h)wA  PB  wB hB  wman h The pressure difference between A and B is; PA  PB  wB hB  wman h  (hA  h)wA PA  PB  (wman  wA )h  wB hB  wA hA  Figure 3.15: U-tube differential manometer Inverted U-tube ( or n-tube) differential manometer This type of manometer is used to measure pressure differences where accuracy is of major concern. Figure 3.16: Inverted U-tube manometer Pressure at C = pressure at D PC  PD PC  PA  (hA  h)wA PD  PB  wB hB  wman h PA  (hA  h)wA  PB  wB hB  wman h The pressure difference between A and B is; PA  PB  (hA  h)wA  wB hB  wman h PA  PB  wA hA  wB hB  (wman  wA )h 3.4.5 Choice of Manometer Care must be taken when attaching the manometer to vessel in the sense that no burrs must be present around this joint. The presence of burrs would alter the flow causing local pressure variations to affect the measurement. Some disadvantages of manometers:  Slow response - only really useful for very slowly varying pressures - no use at all for fluctuating pressures.  For the “U” tube manometer two measurements must be taken simultaneously to get the h value. This may be avoided by using a tube with a much larger cross-sectional area on one side of the manometer than the other.  It is often difficult to measure small variations in pressure - a different manometric fluid maybe required - alternatively a sloping manometer may be employed; It cannot be used for very large pressures unless several manometers are connected in series;  For very accurate work the temperature and relationship between temperature and ρ must be known. Some advantages of manometers:  They are very simple.  No calibration is required - the pressure can be calculated from first principles. Tutorials 2: 1. Assume the specific weight of the specific gravity of the diesel and liquid in the hydraulic jack shown mercury are 0.9 and 13.6 in Figure 3.17 is 9.8KN/m3; find the respectively. If the difference of load lifted by large piston when a mercury level in the two limbs is force of 0.4KN is exerted on the 0.16m, determine the absolute small piston. pressure of the oil in the pipe. Figure 3.17: Hydraulic jack Figure 3.18: Pipeline flow with U- tube manometer 2. The pressure of diesel flowing in a pipeline is measure using U-tube manometer as shown in figure 3.18. The left end of the manometer is connected to the pipeline and the right-limb is open to the atmosphere. The centre of the pipe is 100mm below the level of mercury in the right limb. The 3.. 5. ,, 4.. 6. , 2.5 Mechanical Gauges  CHAPTER FOUR 4.0 FLUID DYNAMICS In a static fluid (fluid at rest), specific weight is the only significant property of the fluid. Whereas, when fluid is in motion, several fluid properties become significant. Hence, the nature of fluid in motion is complex. The science that studies the behaviour of fluid in motion without consideration for the forces causing the motion is known as kinematics. Therefore kinematics merely involves the description of motion of fluid. The science that deals with the forces producing changes in fluid motion is known as kinetics. Therefore, detail study of fluid flow involves the consideration of both the kinematic and kinetic. Essentially, the knowledge of fluid flow description in addition to the kinetics is required to do justice to the study of fluid dynamics. The motion of fluids can be predicted in the same way as the motion of solids are predicted using the fundamental laws of physics together with the physical properties of the fluid. It is not difficult to envisage a very complex fluid flow. Spray behind a car; waves on beaches; hurricane sand tornadoes or any other atmospheric phenomenon are all example of highly complex fluid flows which can be analysed with varying degrees of success (in some cases hardly at all!). However, classification of fluid flow is required to facilitate its best analysis and this is discussed in the next subsection. 4.1 CLASSIFICATION OF FLUID FLOW The common types of fluid flow as follows:  Steady and unsteady flows  Uniform and non-uniform flows  One, two, and three dimensional flows  Compressible and incompressible flows  Laminar and turbulent flows 4.1.1 Uniform and non-uniform flow  Uniform flow: A flow is said to be uniform, if the flow velocity at any given instant is the same in magnitude and direction at every point in the fluid.  Non-uniform flow: A flow is said to be non-uniform, if at a given instant, the velocity is not the same at every points in the flow.(In practice, by this definition, every fluid that flows near a solid boundary will be non-uniform – as the fluid at the boundary must take the speed of the boundary, usually zero. However if the size and shape of the cross- section of the stream of fluid is constant the flow is considered uniform.) 4.1.2 Steady and unsteady flow  Steady flow: A flow is said to be steady if the conditions (velocity, pressure and cross- section)at any given instant DO NOT change with time.  Unsteady flow: A flow is said to be unsteady if at any point in the fluid, the conditions change with time. (In practice there are always slight variations in velocity and pressure, but if the average values are constant, the flow is considered steady. 4.1.3 Compressible or Incompressible Flow In reality all fluids are compressible - even water - their density will change as pressure changes. Under steady conditions, and provided that the changes in pressure are small, it is usually possible to simplify analysis of the flow by assuming it is incompressible and has constant density. As you will appreciate, liquid (e.g. water) is quite difficult to compress - so under most steady conditions they are treated as incompressible fluid. Gasses, on the contrary, are very easily compressed, it is essential in most cases to treat these as compressible, taking changes in pressure into account. Thus,  Compressible flow is the type of flow which is density of the fluid is not constant. i.e. ρ ≠ constant.  Incompressible flow is the type of flow with a constant fluid density. i.e ρ = constant. 4.1.4 One -, Two - and Three-dimensional flow Fluid flow is three-dimensional in nature. This means that the flow parameters like velocity, pressure and so on vary in all the three coordinate directions. In many cases the greatest changes only occur in two directions or even only in one. In these cases changes in the other direction can be effectively ignored making analysis much simpler. Flow is one dimensional if the flow parameters (such as velocity, pressure, depth etc.) at a given instant in time only vary in the direction of flow and not across the cross-section. The flow may be unsteady, in this case the parameter vary in time but still not across the cross- section. An example of one-dimensional flow is the flow in a pipe. Note that since flow must be zero at the pipe wall - yet non-zero in the centre – there is a difference of parameters across the cross-section. Should this be treated as two-dimensional flow? Possibly - but it is only necessary if very high accuracy is required. A correction factor is then usually applied. Figure 4.1: One dimensional flow in a pipe. Flow is two-dimensional if it can be assumed that the flow parameters vary in the direction of flow and in one direction at right angles to this direction. Streamlines in two-dimensional flow are curved lines on a plane and are the same on all parallel planes. An example is flow over a weir foe which typical streamlines can be seen in the figure below. Over the majority of the length of the weir the flow is the same - only at the two ends does it change slightly. Here correction factors may be applied.  Figure 4.2: Two-dimensional flow over a weir. 4.1.4 Laminar and Turbulent flows  Laminar flow occurs when a fluid layers flows in a parallel and undisrupted pattern.  Turbulent flow occurs when the particles of the fluid motion flows in a random, chaotic and undefined pattern. Figure 4.3: Laminar and turbulent flows In this course we will only be considering steady, incompressible one and two-dimensional flow. 4.2 FLOW RATE. 4.2.1 Mass Flow Rate If we want to measure the rate at which water is flowing along a pipe. A very simple way of doing this is to collect water coming out of the pipe in a bucket over a fixed time period. Measuring the weight of the water in the bucket and dividing this by the time taken to collect this water gives a rate of accumulation of mass. This is known as the mass flow rate. mass m V Al Note that mass flow rate, m      Au (4.1) time t t t WORKED EXAMPLE 4-1 An empty bucket weighs 2.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg. Find the mass flow rate. Solution: The mass flow rate is given by: m 82 m    0.857 kg / s t 7 WORKED EXAMPLE 4-2 If the mass flow rate of water is 1.7kg/s, how long will it take to fill a container with 8kg of water? Solution: The time taken to fill the container is m 8kg t   4.7 s m 1.7kg / s 4.2.2 Volume Flow Rate - Discharge More commonly we need to know the volume flow rate –most often referred to as discharge. The symbol normally used for discharge is Q. The discharge is the volume of fluid per unit time flowing through a section of pipe or channel. The mathematical expression is given in Equation 4.2. Volume flow rate should not be mistaken for mass flow rate, because volume flow rate multiplying by the density of the fluid gives us the mass flow rate. V m m Q    m  m 3 / s (4.2) t t  WORKED EXAMPLE 4-3 3 find its volumetric flow rate. If the density of water in Worked Example 4.1 is 1000kg/m Solution: V m m 0.857 kg / s Q     0.000857 m 3 / s  8.57  10  4 m 3 / s  0.857l / s t t  1000 kg / m 3 It should be noted that a consistent set of units should be applied to values in the equations in other to get calculations correctly. It would therefore make sense to always quote the values with consistent set of units. In some cases it is useful to use derived units, and in the case above, the useful derived unit is the litre (1 litre = 1.0 ×10-3m3). So the solution becomes 1008l / s. It is far easier to imagine 1 litre than1.0 ×10-3m3. Units must always be checked, and converted if necessary to a consistent set before using in an equation. 4.2.3 Discharge and Mean Velocity If we know the size of a pipe, and we know the discharge, we can deduce the mean velocity Figure 4.3: Discharge in a pipe If the cross sectional area of the pipe at point X is A, and the mean velocity here is um. During a time t, a cylinder of fluid will pass point X with a volume A×um×t. The volume per unit time (the discharge) will thus be Volume V A  u m  t Q    Au m time t t  WORKED EXAMPLE 4-4 −3 2 If a fluid flow in a pipe with a cross-section area of 1.2×10 m. Calculate the mean velocity if 24liters of fluid is discharged per second. Solution: Q 2.4l / s 2.4  10 3 m 3 / s um     2m / s A 1.2  10 3 m 2 1.2  10 3 m 2 Note how carefully we have called this the mean velocity. This is because the velocity in the pipe is not constant across the cross section. Crossing the centreline of the pipe, the velocity is zero at the walls increasing to a maximum at the centre then decreasing symmetrically to the other wall. This variation across the section is known as the velocity profile or distribution. A typical one is shown in the Figure 4.4.This idea, that mean velocity multiplied by the area gives the discharge, applies to all situations - not just pipe flow. Figure 4.4: A typical velocity profile across a pipe 4.3 THE CONTINUITY EQUATION Matter cannot be created or destroyed - (it is simply changed in to a different form of matter). This principle is known as the conservation of mass and we use it in the analysis of flowing fluids. The principle is applied to fixed volumes, known as control volumes (or surfaces) as shown in Figure 4.5. Figure 4.5: An arbitrarily shaped control volume. For any control volume the principle of conservation of mass says Mass entering = Mass leaving per unit + Increase of mass in the control volume per unit time time per unit time For steady flow there is no increase in the mass within the control volume, thus,  Mass entering per unit time = Mass leaving per unit time This can be applied to a tapered pipeline with mass of fluid entering and leaving through the two ends of the pipeline as shown in Figure 4.6. Figure 4.6: Tapered pipeline conveying mass of fluid We can then write Mass entering per unit = Mass leaving per unit + Increase in mass of time at the inlet (end 1) time at the outlet (end 2) the control volume 1  A1  u1   2  A2  u 2 + Increase in mass of the control volume Or for steady flow, 1  A1  u1   2  A2  u 2 = Constant This is the equation of continuity. The flow of fluid through a real pipe (or any other vessel) will vary due to the presence of a wall - in this case we can use the mean velocity and write 1  A1  um1   2  A2  um2  cons tan t  m When the fluid can be considered incompressible, the change in density is negligible, thus, 1   2   so(dropping the m subscript) A1u1  A2 u 2  Q This is the form of the continuity equation most often used. This equation is a very powerful tool in fluid mechanics and will be used repeatedly throughout the rest of this course. Example of the applications of continuity equation We can apply the principle of continuity to pipes with cross sections which change along their length. Consider the diagram shown in Figure 4.7.  Figure 4.7: Pipe with a contraction A liquid is flowing from left to right and the pipe is narrowing in the same direction. By the continuity principle, the mass flow rate must be the same at each section - the mass of fluid entering the pipe is equal to the mass of fluid leaving the pipe. So we can write: 1 A1u1   2 A2 u 2 where the sub-scripts 1 and 2 denote the values at the two sections. As we are considering a liquid, usually water, which is not very compressible (i.e incompressible), the density changes is negligible, so we can say 1   2  . This also says that the volume flow rate is constant or that Discharge at section 1 = Discharge at section 2 Q1  Q2 A1u1  A2 u 2 As the area of the circular pipe shown in Figure 4.7 is a function of the diameter we can present the downstream velocity as, 2 Au d 2 / 4 d  u 2  1 1  12 u1   1  u1 A2 d 2 / 4  d2  Another use of the continuity principle is to determine the velocities in pipes coming froma junction as shown in Figure 4.8. Figure 4.8: Diverging pipe Total mass flow into the junction = Total mass flow out of the junction 1Q1   2Q2  3Q3 When the flow is incompressible (e.g. if it is water) 1   2  3 Q1  Q2  Q3 A1u1  A2 u2  A3u3  WORKED EXAMPLE 4-5 2 2 If the contraction pipe shown in Figure 4.7 has an area A1= 0.01m and A2=0.03m and the upstream mean velocity,u1=2.1m/s, determine the downstream mean velocity Solution: The downstream velocity can be calculated using the equation of continuity as A1u1 = A2u2 0.01m x 2.1m/s = 0.03m2 x u2 2 Au u 2  1 1  7.0m / s A2 WORKED EXAMPLE 4-6 If Figure 4.8 has the following detail: pipe 1 diameter = 50mm, mean velocity 2m/s, pipe 2 diameter 40mm takes 30% of total discharge and pipe 3 diameter 60mm. What are the values of discharge and mean velocity in each pipe? Solution:  d 2  Q1  A1u1   u1  0.0039 m 3 / s  4  Q2  0.3Q1  0.00118m3 / s Q1  Q2  Q3 Q3  Q1  Q2  Q1  0.3Q1  0.7Q1  0.00275m 3 / s Q2  A2 u 2 u 2  0.936 m / s Q3  A3u3 u3  0.972m / s 4.4 THE BERNOULLI EQUATION Bernoulli postulated the most important and useful equations in fluid mechanics. He shows that the sum of pressure energy, kinetic energy and the potential energy is constant for a steady, continuous and ideal incompressible fluid flow. This equation has been named after the postulator and it is known as Bernoulli equation. Bernoulli’s equation can be expressed as; u12 u 22 P1   gh1  P2   gh2 2 2 or u 2 P   gh  constant 2 The assumptions made in the derivation of Bernoulli’s equation are; 1. The fluid is ideal and incompressible 2. The flow is continuous and steady, 3. The velocity is uniform over the section and is equal to mean velocity 4. The flow is along a streamline, thus it is one dimensional, and 5. Gravitational and pressure forces are the only two forces action on the fluid. Prove of Bernoulli’s Equation To prove Bernoulli equation, let us consider the small elemental fluid LMNO with a cross- section area dA and length ds as shown in Figure 4.1 Figure 4.1: Small elemental fluid element flowing in a pipe The normal pressure force acting on the face LN of the small fluid element is; PA The normal pressure force acting on the face MO of the small fluid element is; P  P A The body force (gravitational force acting on the small fluid element), W, is; W  mg cos  Vg cos  gAds cos dz but cos   ds thus, W  gAdz Assuming that the fluid is inviscid, the shearing force will be zero and the only forces acting on the fluid element will be the pressure and body forces. Newton’s second law of motion states that the sum of forces acting along the same direction is zero. Thus, F s  PA  P  P A  gAdz (i) Also, inertial force along streamline direction is; du du ma  V  Ads  Audu (ii) dt dt Equating equations (i) and (ii) gives Audu  PA  gAdz P  udu  gdz  0 2 2 2  P    udu  g  dz  0 1 1 1  P2  P1   u  u12 2   g z 2  z1   0 2 2 u 2 u 2 P1  1  gz1  P2  2  gz 2 2 2 or u 2 P   gh  constant (4.1) 2 Dividing equation (4.1) by weight density (w = ρg) gives: 2 P u   h  constant w 2g Where, P is called the pressure head or pressure energy, w u2 is called the velocity head or kinetic energy, and 2g h is called the potential head or potential energy. Density and acceleration due to gravity are constant in Equation (4.1). Thus, the three parameters required to analyze a continuous, steady and incompressible fluid motion are the pressure, velocity and potential head. The assumptions made in deriving Bernoulli’s equation has some restrictions in its applicability, they are:  Flow is steady,  Density is constant (which also means the fluid is incompressible),  Friction losses are negligible, and  The equation relates the states at two points along a single streamline, (not conditions on two different streamlines). All these conditions are impossible to satisfy at any instant in time! Fortunately for many real situations where the conditions are approximately satisfied, the equation gives very good results. WORKED EXAMPLE 4.7 The inlet pressure and velocity of the water flowing in a pipe shown in Figure 4.2 are 2atm and 2m/s respectively. Determine the pressure at the exit of the pipe which has the same inlet and exit cross-section area. Figure 4.2: Schematic of a pipe flow Solution: P1  2atm  2(1.013105 N / m2 ) u12 u 22 P1   gz1  P2   gz 2 2 2 For steady flow; Q1  Q2 A1u1  A2 u 2 A1u1 u2   u1  2m / s A2 Thus, P1  gz1  P2  gz 2 P2  P1  g z1  z 2   (2atm)  (1000kg / m3 )(9.81m / s 2 )(5m  10m) P2  2atm  49000N / m2  2atm  0.48atm  1.52atm WORKED EXAMPLE 4.8 The piping configuration in problem 4.2 above is modified to a new configuration as illustrated in Figure 4.3. Determine the exit pressure and velocity, if the inlet boundary conditions are unchanged, and the pipe elevation is zero. Figure 4.3: Schematic of a modified pipe flow configuration Solution: For steady flow; Q1  Q2 A1u1  A2 u 2 2 Au  d  2  5cm  1 u 2  1 1   1  u1    u1  u1  1m / s A2  d2   10cm  4 u12 u 22 P1   gz1  P2   gz 2 2 2 Since the pipe has a zero elevation, the Bernoulli’s equation is reduced to; u12 u 22 P1   P2  2 2 u u 2 2 P2  P1   21 2 2  P2  P1  (u12  u 22 ) 2 P2  (2atm)  ( 1000 kg / m 3 2  ) 4m / s   1m / s  2 2   P2  (2atm)  (500kg / m3 )15m2 / s 2  P2  2atm  7500N / m 2 P2  2atm  0.074atm P2  2.074 atm 4.5 THE MOMENTUM EQUATION We have all seen moving fluids exerting forces. The lift force on an aircraft is exerted by the air moving over the wing. A jet of water from a hose exerts a force on whatever it hits. In fluid mechanics the analysis of motion is performed in the same way as in solid mechanics - by use of Newton’s laws of motion. Account is also taken for the special properties of fluids when in motion. The momentum equation is a statement of Newton’s Second Law and relates the sum of the forces acting on an element of fluid to its acceleration or rate of change of momentum. You will probably recognize the equation F = ma which is used in the analysis of solid mechanics to relate applied force to acceleration. In fluid mechanics it is not clear what mass of moving fluid we should use so we use a different form of the equation. Newton’s 2nd Law can be written: The Rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force. To determine the rate of change of momentum for a fluid we will consider a streamtube as we did for the Bernoulli equation, We start by assuming that we have steady flow which is non-uniform flowing in a stream tube. A stream-tube in three and two-dimensions In time, δt, a volume of the fluid moves from the inlet a distance uδt, so the volume entering the stream-tube in the time δt is Volume entering the stream tube = Area × Distance = A1u1δt This has mass, Mass entering stream tube = Volume × Density = ρA1u1δt andthe momentum is Momentum of fluid entering stream tube = Mass × Velocity = ρA1u1δtu1 Similarly, at the exit, we can obtain an expression for the momentum leaving the steam-tube: Momentum of fluid leaving stream tube = Mass x Velocity = ρA2u2δtu2 We can now calculate the force exerted by the fluid using Newton’s 2nd Law. The force is equal to the rate of change of momentum. So Force = Rate of change of momentum A 2 u 2t  u 2  A1u1t  u1  F t We know from continuity that Q= A1u1 =A2u2, and if we have a fluid of constant density, i.e. ρ1= ρ2 = ρ, then we can write F = Qρ(u2 −u1) For an alternative derivation of the same expression, as we know from conservation of mass in a streamtube that mass into face 1 = mass out of face 2 We can write dm Rate of change of mass = m   1 A1u1   2 A2 u 2 dt The rate at which momentum leaves face 1 is 1 A1u1  u1  m u1 The rate at which momentum enters face 2 is  2 A2 u 2  u 2  m u 2 Thus the rate at which momentum changes across the stream tube is  2 A2 u 2  u 2  1 A1u1  u1  m u 2  m u1 i.e. Force = rate of change of momentum F  m (u 2  u1 ) F  Q (u 2  u1 ) This force is acting in the direction of the flow of the fluid. This analysis assumed that the inlet and outlet velocities were in the same direction - i.e. a one- dimensional system. What happens when this is not the case? Consider the two dimensional system in the figure below: Two-dimensional flow in a streamtube At the inlet the velocity vector, u1, makes an angle, θ1, with the x-axis, while at the outlet u2 make an angle θ2. In this case we consider the forces by resolving in the directions of the co- ordinate axes. The force in the x-direction Fx = Rate of change of momentum in x-direction = Rate of change of mass × change in velocity in x-direction   m (u 2 cos 2u1 cos 1)  Q(u 2 cos 2u1 cos 1)  Q(u 2 x  u1 ) x And the force in