MECHANICS OF FLUIDS (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University Dr. S.Ramachandran, M.E., Ph.D., Mr. K.Pandian, M.E., Mr. YVS. Karthick, M.E., Sathyabama University Jeppiaar Nagar, Chennai - 600 119 AIR WALK PUBLICATIONS (Near All India Radio) 80, Karneeshwarar Koil Street Mylapore, Chennai - 600 004. Ph.: 2466 1909, 94440 81904 Email: aishram2006@gmail.com, airwalk800@gmail.com www.airwalkpublications.com
First Edition: July 2016 Price : Rs. 300/- ISBN:978-93-84893-50-7 ISBN : 978-93-84893-50-7 and
ME203 MECHANICS OF FLUIDS Syllabus Chapter 1 : Introduction Introduction: Fluids and continuum, Physical properties of fluids, density, specific weight, vapour pressure, Newton s law of viscosity. Ideal and real fluids, Newtonian and non-newtonian fluids. Fluid Statics - Pressure-density-height relationship, manometers, pressure on plane and curved surfaces, center of pressure, buoyancy, stability of immersed and floating bodies, fluid masses subjected to uniform accelerations, measurement of pressure. Chapter 2: Kinematics of Fluid Flow Kinematics of fluid flow: Eulerian and Lagrangian approaches, classification of fluid flow, 1-D, 2-D and 3-D flow, steady, unsteady, uniform, non-uniform, laminar, turbulent, rotational, irrotational flows, stream lines, path lines, streak lines, stream tubes, velocity and acceleration in fluid, circulation and vorticity, stream function and potential function, Laplace equation, equipotential lines flow nets, uses and limitations FIRST INTERNAL EXAM Chapter 3: Dynamics of Fluid Flow Dynamics of Fluid flow: Fluid Dynamics: Energies in flowing fluid, head, pressure, dynamic, static and total head, Control volume analysis of mass, momentum and energy, Equations of fluid dynamics: Differential equations of mass, energy and momentum (Euler s equation), Navier-Stokes equations (without proof) in rectangular and cylindrical co-ordinates, Bernoulli s equation and its applications: Venturi and Orifice meters, Notches and Weirs (description only for notches and weirs). Hydraulic coefficients, Velocity measurements: Pitot tube and Pitot-static tube.
Chapter 4 : Pipe Flow Pipe Flow: Viscous flow: Reynolds experiment to classify laminar and turbulent flows, significance of Reynolds number, critical Reynolds number, shear stress and velocity distribution in a pipe, law of fluid friction, head loss due to friction, Hagen Poiseuille equation. Turbulent flow: Darcy- Weisbach equation, Chezy s equation Moody s chart, Major and minor energy losses, hydraulic gradient and total energy line, flow through long pipes, pipes in series, pipes in parallel, equivalent pipe, siphon, transmission of power through pipes, efficiency of transmission, Water hammer, Cavitation. SECOND INTERNAL EXAM Chapter 5 : Concept of Boundary Layer Concept of Boundary Layer: Growth of boundary layer over a flat plate and definition of boundary layer thickness, displacement thickness, momentum thickness and energy thickness, laminar and turbulent boundary layers, laminar sub layer, velocity profile, Von- Karman momentum integral equations for the boundary layers, calculation of drag, separation of boundary and methods of control. Chapter 6: Dimensional Analysis Dimensional Analysis and Hydraulic similitude: Dimensional analysis, Buckingham s theorem, important dimensional numbers and their significance, geometric, Kinematic and dynamic similarity, model studies. Froude, Reynold, Weber, Cauchy and Mach laws- Applications and limitations of model testing, simple problems only END SEMESTER EXAM
Contents C.3 Contents 1. Introduction 1.1 Introduction... 1.1 1.2 Fluids and Continuum... 1.2 1.2.1 Distinction between solid and fluid... 1.2 1.3 Properties of Fluids... 1.4 1.3.1 Gas and Liquid... 1.4 1.3.2 Density (or) mass Density... 1.5 1.3.3 Specific weight (or) Weight density... 1.5 1.3.4 Specific Volume v... 1.6 1.3.5 Specific gravity (or) Relative density s... 1.6 1.3.6 Temperature... 1.7 1.3.7 Viscosity... 1.7 1.3.7.1 Kinematic Viscosity ()... 1.11 1.3.8 Compressibility 1 K... 1.12 1.3.8.1 Relationship between Bulk Modulus K and pressure p of a Gas for Isothermal and Isentropic Process... 1.13 1.3.9 Vapour Pressure... 1.14 1.3.9.1 Cavitation... 1.14 1.3.9.2 Gas and Gas laws... 1.15 1.3.10 Surface Tension... 1.17 1.3.10.1 Surface Tension on Droplet... 1.19 1.3.10.2 Surface Tension on a Hollow Bubble... 1.21 1.3.10.3 Surface Tension on a Liquid Jet... 1.21 1.3.11 Capillarity... 1.22 1.3.11.1 Expression for Capillary Rise... 1.23 1.3.11.2 Expression for Capillary Fall... 1.24 1.3.12 Thermodynamic Properties... 1.25
C.4 Mechanics of Fluids 1.4 Newton s Law of Viscosity... 1.27 1.4.1 Types of Fluid... 1.29 1.5 Fluid Statics... 1.67 1.5.1 Concept of Fluid Static Pressure... 1.67 1.5.2 Pressure of Fluids P... 1.67 1.5.3 Atmospheric Pressure... 1.68 1.5.4 Absolute zero Pressure (or) Absolute pressure 1.69 1.5.5 Gauge Pressure... 1.69 1.5.6 Vacuum Pressure... 1.70 1.6 Pressure - Density - Height Relationship... 1.70 1.7 Manometry... 1.73 1.8 Pressure on Plane and Curved Surfaces... 1.81 Total pressure... 1.81 Centre of pressure... 1.82 1.8.1 Pressure on submerged horizontal plane surface... 1.82 1.8.2 Pressure on submerged vertical plane surface 1.83 1.8.3 The Hydrostatic paradox... 1.87 1.8.4 Pressure on immersed inclined plane surface 1.102 1.8.5 Pressure on immersed curved surface... 1.114 1.9 Buoyancy... 1.127 Centre of Buoyancy... 1.128 1.9.1 Stability of Immersed Body... 1.129 1.9.2 Stability of Floating body... 1.131 1.9.3 Metacentre M... 1.132 1.9.4 Oscillation (or) Rolling of a Floating body... 1.153 1.10 Fluid Masses Subjected to Uniform Accelerations.. 1.154 1.10.1 Fluid masses subjected to uniform horizontal acceleration... 1.154 1.10.2 Fluid mass subjected to uniform vertical acceleration... 1.160
Contents C.5 1.11 Measurement of Pressure... 1.175 Manometers... 1.175 Mechanical Gauges... 1.175 1.11.1 Pressure Measurement Methods... 1.176 1.11.1.1 Bourdon gauge (C-Type)... 1.176 1.11.1.2 Diaphragm-type pressure gauge... 1.178 1.11.1.3 Bellows... 1.180 1.11.1.4 Dead Weight Pressure Gauge... 1.182 1.11.1.5 Capacitive Pressure Transducer... 1.183 1.11.1.6 Strain Gauge Pressure Transducer... 1.184 1.11.2 Simple Manometers... 1.185 1.11.3 Differential Manometer... 1.189 Problems in Simple Manometer... 1.191 Problems In Differential Manometer... 1.197 2. Kinematics of Fluid Flow 2.1 Introduction... 2.1 2.1.1 Concept of System... 2.1 2.1.2 Control Volume... 2.1 2.2 Fluid Characteristics... 2.2 2.3 Types of Fluid Flow... 2.3 2.3.1 Steady Flow and Unsteady Flow... 2.3 2.3.2 Uniform and Non-Uniform Flows... 2.4 2.3.3 Laminar Flow and Turbulent Flow... 2.5 2.3.4 Incompressible and Compressible Flow... 2.6 2.3.5 Rotational Flow and Irrotational Flow... 2.7 2.3.6 One Dimensional Flow... 2.7 Two-dimensional Flow... 2.8 Three-dimensional Flow... 2.8 2.4 Flow Visualization - Lines of Flow... 2.8 2.4.1 Stream Line... 2.9
C.6 Mechanics of Fluids 2.4.2 Stream Tube... 2.10 2.4.3 Path Line... 2.10 2.4.4 Streak Line... 2.11 2.5 Velocity Field... 2.12 2.5.1 Velocity field and acceleration... 2.13 2.6 Mean Velocity of Flow... 2.18 2.7 Principles of Fluid Flow... 2.18 2.7.1 Principle of Conservation of mass... 2.18 2.7.1.1 Continuity Equation in one Dimension... 2.19 2.7.1.2 Continuity Equation In Cartesian Co-ordinates In Three Dimensions... 2.26 2.7.1.3 Equation of continuity in polar coordinates (Rotation and Circulation)... 2.29 2.8 Types of Motion or Deformation of Fluid Element.. 2.37 2.9 Circulation and Vorticity... 2.37 2.10 Velocity Potential Function... 2.42 2.10.1 Properties of Potential Function... 2.43 2.11 Stream Function... 2.43 2.11.1 Properties of Stream Function... 2.44 2.12. Relation between Stream Function and Velocity Potential Function... 2.45 2.13 Equipotential Line... 2.56 2.13.1 Line of constant stream function.... 2.57 2.14 Flow Net... 2.58 2.14.1 Condition of flow net to be a set of square. 2.58 2.14.2 Methods of drawing Flow Nets... 2.60 2.14.3 Uses of Flow nets... 2.65 2.14.4 Limitations of Flow Nets... 2.66 3. Dynamics of Fluid Flow 3.1 Introduction... 3.1
Contents C.7 3.2 Equations of Motion... 3.1 3.2.1 Principle of Conservation of Energy... 3.2 3.2.2 Euler s equation along a Stream Line... 3.4 3.2.3 Bernoulli s Equation... 3.6 3.2.4 Important points in Bernoulli s Equation... 3.6 3.2.5 Assumptions for derivation of Bernoulli s Equation... 3.7 3.2.6 Bernoulli s Equation for Real Fluid... 3.7 3.3 Navier-stokes Equations... 3.20 3.4. Bernoulli s Equation: Applications... 3.22 3.4.1 Venturi Meter... 3.22 3.4.2 Orifice Meter... 3.26 Problems in Venturimeter... 3.29 Problems in Orificemeter... 3.43 3.4.3 Principle of Conservation of Momentum... 3.47 3.4.4 Moment of Momentum Equation... 3.49 Problems in Momentum Equations... 3.49 3.4.5 Pitot-tube... 3.58 Problems in Pitot Tube... 3.60 3.4.6 Notches And Weirs... 3.62 3.4.6.1 Discharge over a rectangular notch... 3.66 3.4.6.1.1 Ventilation of Weirs... 3.69 3.4.6.2 Discharge over a Triangular Notch (or Weir) 3.70 3.4.6.3 Discharge over a Trapezoidal Notch (or Weir) 3.76 3.4.6.4 Discharge over a Stepped Notch... 3.80 3.4.6.5 Effect on discharge due to error in Measurement of Head... 3.83 3.4.6.6 Time required to empty a Reservoir or Tank 3.85 3.4.6.7 Velocity of approach... 3.90 3.4.6.8 End contractions... 3.91 3.4.6.9 Discharge over a Cippoletti weir (or Notch) 3.101
C.8 Mechanics of Fluids 3.4.6.10 Discharge over a Submerged or Drowned weir... 3.105 3.4.6.11 Discharge over a narrow crested weir... 3.109 3.4.6.12 Discharge over a Broad Crested Weir... 3.110 3.4.6.13 Discharge over a sharp-crested weir... 3.114 3.4.6.14 Discharge over an Ogee weir... 3.116 4. Pipe Flow 4.1 Introduction... 4.1 4.2 Reynolds Number... 4.2 4.2.1 Significance of Reynolds Number... 4.3 4.2.2 Critical Reynolds Number... 4.4 4.2.3 Reynolds Experiment... 4.4 4.3 Shear Stress Distribution and Velocity Distribution for Laminar Flow Through Circular Tubes... 4.6 4.3.1 Ratio of maximum velocity (U max to average velocity u... 4.10 4.3.2 Loss of head for a given length of pipe (Hagen poiseuille formula)... 4.11 4.4 Law of Fluid Friction... 4.13 4.4.1 Head loss due to friction (for laminar flow).. 4.15 4.4.2 Hagen Poiseuille Equation (in terms of Discharge)... 4.16 4.5 Turbulent Flow... 4.40 4.6 Frictional Loss in Pipe Flow... 4.41 4.6.1 Darcy- Weisbach s Equation - Expression For Loss Of Head Due To Friction In Pipes... 4.41 4.6.2 Chezy s Formula for Loss of head due to friction in pipes... 4.44 4.7 Shear Stress in Turbulent Flow... 4.45
Contents C.9 4.8 Velocity Distribution for Turbulent Flow in Pipes... 4.46 4.9 Hydrodynamical Smooth and Rough Boundary Pipe Roughness... 4.47 4.9.1 Velocity Distribution for Turbulent flow in Smooth pipes... 4.48 4.9.2 Velocity Distribution for Turbulent flow in Rough pipes... 4.48 4.10 Friction Factor - Resistance of Smooth and Rough Pipe in Darcy-weisbach Equation... 4.49 4.11 Moody s Chart... 4.49 4.12 Energy Losses in Pipes... 4.52 4.12.1 Major energy (Head) losses... 4.52 4.12.2 Minor energy losses... 4.54 1. Loss of head due to sudden enlargement h e... 4.54 2. Loss of head due to sudden contraction: h c... 4.55 3. Loss of head due to obstruction: h obs... 4.56 4. Loss of head due to entrance of pipe: h i... 4.57 5. Loss of head due to exit of pipe: h 0... 4.58 6. Loss of head due to bend in pipe h b... 4.58 7. Loss of head due to various pipe fittings h fitting 4.58 4.13 Hydraulic Gradient Line (H.G.L) and Total Energy Line (T.E.L)... 4.70 4.13.1 Total Energy Line (T.E.L) (or) Energy Gradient Line (E.G.L)... 4.70 4.13.2 Hydraulic Gradient Line (H.G.L)... 4.71 Salient points to draw the T.E.L and H.G.L... 4.71 4.14 Flow Through Long Pipes Under Constant Head H 4.85 4.15 Flow Through Pipes in Series (or) Flow Through Compound Pipes... 4.87 4.16 Equivalent Pipe... 4.98
C.10 Mechanics of Fluids 4.16.1 Total head loss in compound pipe neglecting minor losses... 4.98 4.16.2 Total head loss in equivalent pipe.... 4.99 4.17 Flow Through Parallel Pipes... 4.100 4.18 Flow Through Branched Pipes... 4.115 4.19 Siphon (Flow Through Pipeline with Negative Pressure)... 4.120 4.20 Power Transmission Through Pipes... 4.130 4.20.1 Efficiency of power transmission... 4.131 4.20.2 Condition for maximum power transmission 4.132 4.20.3 Maximum efficiency of Transmission of power max... 4.132 4.20.4 Important Note about Power... 4.138 4.21 Water Hammer... 4.140 4.22 Cavitation... 4.140 4.22.1 Nature of cavitation... 4.141 4.22.3 Effects of cavitation... 4.142 4.22.4 Precautions... 4.142 5. Concept of Boundary Layer 5.1 Concept of Boundary Layer... 5.1 5.1.1 Boundary layer theory... 5.1 5.2 Growth of Boundary Layer Over a Flat Plate... 5.2 5.2.1 Laminar boundary layer... 5.3 5.2.2 Turbulent Boundary layer... 5.4 5.2.3 Laminar Sub - layer... 5.4 5.2.4 Boundary layer thickness... 5.5 5.2.6 Displacement Thickness... 5.5 5.2.7 Momentum Thickness... 5.7 5.2.8 Energy thickness... 5.9
Contents C.11 5.3 Von Karman Momentum Integral Equation for Boundary Layer... 5.15 5.4 Drag Force F D on Plate of Length L... 5.15 5.4.1 Local Coefficient of Drag C D... 5.16 5.4.2 Average Coefficient of Drag C D... 5.16 5.5 Velocity Profiles... 5.26 5.6 Turbulent Boundary Layer on a Flat Plate... 5.29 5.7 Drag and Lift Coefficient... 5.31 5.7.1 Drag Force F D... 5.31 5.7.2 Lift force F L... 5.31 5.8 Total Drag on a Flat Plate... 5.34 5.9 Boundary Layer Separation... 5.41 5.9.1 Location of point of Separation... 5.42 5.9.2 Methods of controlling boundary layer separation... 5.44 6. Dimensional Analysis 6.1 Definition... 6.1 6.2 Introduction... 6.1 6.3 Dimensional Homogeneity... 6.4 6.4 Methods of Dimensional Analysis... 6.5 6.4.1. Rayleigh s Method... 6.5 6.4.2 Buckingham s - - Theorem... 6.8 6.5 Important Dimensional Numbers... 6.28 (a) Reynold s Number R e... 6.28 (b) Froude Number: F r... 6.29 (c) Euler Number E u... 6.30 (d) Weber Number W e... 6.30 (e) Mach Number (M)... 6.31 (f) Cauchy Number C a... 6.31
C.12 Mechanics of Fluids 6.6 Advantages of Dimensional Analysis... 6.32 6.7 Similarity Laws and Models... 6.33 6.7.1 Similitude... 6.34 6.7.1.1 Geometric Similarity... 6.34 6.7.1.2 Kinematic Similarity... 6.35 6.7.1.3 Dynamic Similarity... 6.35 6.7.2 Reynolds model law (Viscous forces are predominant in fluid flow)... 6.36 6.7.3 Froude model law: (Gravity force is predominant)... 6.38 6.7.4 Euler s model law: (Pressure forces are predominant)... 6.42 6.7.5 Weber model law: (Surface tension is a dominant force)... 6.43 6.7.6 Mach model law (Velocity of flow is comparable to the velocity of sound; compressible flow)... 6.44 6.8 Model Testing of Partially Submerged Bodies... 6.60 6.9 Classification of Hydraulic Models... 6.67 6.9.1 Undistorted models... 6.67 6.9.2 Distorted models... 6.68 6.9.3 Scale Ratios for Distorted models... 6.68 Other scale ratios... 6.69 6.10 Applications for Model Testing... 6.70 6.11 Limitations of Model Testing... 6.71 Supplementary - Chapter 3 Static, Dynamic and Total Head... S.1 Control Volume Analysis... S.2 Hydraulic Coefficients... S.3 Pitot-static tube... S.11
Chapter 1 Introduction Introduction: Fluids and continuum, Physical properties of fluids, density, specific weight, vapour pressure, Newtons law of viscosity. Ideal and real fluids, Newtonian and non-newtonian fluids. Fluid Statics-Pressure-density-height relationship, manometers, pressure on plane and curved surfaces, center of pressure, buoyancy, stability of immersed and floating bodies, fluid masses subjected to uniform accelerations, measurement of pressure. 1.1 INTRODUCTION Fluid mechanics is the science which deals with the mechanics of liquids and gases. Fluid is a substance capable of flowing. Fluid mechanics may be divided into three branches. 1. Fluid Statics 2. Fluid Kinematics 3. Fluid Dynamics rest. Fluid statics is the study of mechanics of fluids at Fluid kinematics is the study of mechanics of fluids in motion. Fluid Kinematics deals with velocity, acceleration and stream lines without considering the forces causing the motion. Fluid dynamics is concerned with the relations between velocities, accelerations and the forces exerted by (or) upon fluids in motion.
1.2 Mechanics of Fluids - www.airwalkpubilications.com 1.2 FLUIDS AND CONTINUUM A fluid is a substance that deforms continuously when subjected to even an infinitesimal shear stress. This continuous deformation under the application of shear stress constitutes a flow. Solids can resist tangential stress at static conditions undergoing a definite deformation while a fluid can do it only at dynamic conditions undergoing a continuous deformation as long as the shear stress is applied. 1.2.1 Distinction between solid and fluid Table 1.1 Solid Fluid 1. Solid is a substance which 1. A fluid is a substance undergoes a finite deformation which undergoes continuous depending upon elastic limit deformation under application on application of a force. of a shear force, no matter how small the force might be. 2. Atoms (molecules) are 2. Atoms are comparatively usually closer together in loosely packed in fluid. solid. 3. Intermolecular attractive 3. Inter molecular forces are forces between the molecules not so large enough to hold of a solid are large the various elements of the fluid together and hence fluid will flow under the action of slightest stress. 4. A solid has a definite shape. 4. A fluid has no definite shape of its own but it conforms to the shape of the container vessel.
Introduction 1.3 The concept of continuum assumes a continuous distribution of mass within the matter or system with no empty space. In the continuum approach, properties of a system such as density, viscosity, temperature, etc can be expressed as continuous functions of space and time. The continuum concept is basically an approximation, in the same way planets are approximated by point particles when dealing with celestial mechanics, and therefore results in approximate solutions. Consequently, assumption of the continuum concept can lead to results which are not of desired accuracy. However, under the right circumstances, the continuum concept produces extremely accurate results. A dimensionless parameter known as knudsen number K n /L, where is the mean free path and L is the characteristic length, aptly describes the degree of departure from continuum. The continuum concept usually holds good when k n 0.01. Fluid mechanics is a sub discipline of continuum mechanics as illustrated below. Continuum Mechanics (The study of the physics of continuous materials) Solid Mechanics (The study of the physics of continuous materials with a defined rest shape) Fluid Mechanics (The study of the physics of continuous materials which deform when subjected to a force)
1.4 Mechanics of Fluids - www.airwalkpubilications.com 1.3 PROPERTIES OF FLUIDS 1.3.1 Gas and Liquid A fluid may be either a gas or a liquid. The molecules of a gas are much farther apart than those of a liquid. Hence a gas is highly compressible and a liquid is relatively incompressible. A vapour is a gas whose temperature and pressure are very closely nearer to the liquid phase. So steam is considered as vapour. A gas may be defined as a highly super heated vapour, i.e its state is far away from the liquid phase. So air is considered as a gas. Fluids are having the following properties: Table 1.2 Quantity Symbol Unit 1. Density (or) mass density kg/m 3 2. Specific weight (or) weight density w N/m 3 3. Specific volume v m 3 /kg 4. Specific gravity s No unit 5. Compressibility 1 K 6. Vapour pressure p N/m 2 7. Cohesion and Adhesion 8. Surface tension N/m 9. Capillary rise (or) fall h m 10. Viscosity-Dynamic viscosity (or) Ns/m 2 viscosity 11. Kinematic viscosity m 2 /s The above properties are discussed in detail.
Introduction 1.5 1.3.2 Density (or) mass Density The density of a fluid is its mass per unit volume. Density mass volume kg m 3 So the unit of density is kg/m 3 and dimension is ML 3 The density of liquids is normally constant while that of gases changes with the variation of pressure and temperature. Density of water at 4C is 1000 kg/m 3 1.3.3 Specific weight (or) Weight density Specific weight is the weight per unit volume. Its symbol is w. Specific weight represents the force exerted by gravity on a unit volume of fluid. Specific weight w Weight of fluid Volume of fluid N m 3 Specific weight and density are related w Weight of fluid Mass of fluid g Volume of fluid Volume of fluid g... mass Volume So w g...(1.1) Where g acceleration due to gravity.
1.6 Mechanics of Fluids - www.airwalkpubilications.com The specific weight of water at 4 C is 9810 N/m 3. Density is absolute since it depends on mass and independent of location. Specific weight, on the other hand, is not absolute, since it depends on value of g which varies from place to place. Density and specific weight of fluids vary with temperature. 1.3.4 Specific Volume v Specific volume is the volume occupied by a unit mass of fluid. Its symbol is v. Its unit is m 3 /kg v Volume of fluid Mass of fluid V Mass m3 kg Specific volume is the reciprocal of density v 1...(1.2) Note: v Specific volume; V Volume; u, v and V Velocity of flow. 1.3.5 Specific gravity (or) Relative density s Specific gravity of a liquid is the ratio of its density to that of pure water at a standard temp. 4C. Its symbol is s. It has no unit. s Density of liquid Density of water w...(1.3) where w 1000 kg/m 3 Density of liquid s w
Introduction 1.7 Specific gravity can also be defined in terms of specific weight. s Specific weight of liquid specific weight of water w w w Specific weight of liquid w s w w...(1.4) where w w 9810 N/m 3 The specific gravity of gas is the ratio of its density to that of air. 1.3.6 Temperature It is intensive thermodynamic property which determines the hotness or the level of heat intensity of a body. A body is said to be hot if it is having high temperature indicating high level of heat intensity. Similarly a body is said to be cold if it is at low temperature indicating low level of heat intensity. The temperature of a body is measured by an instrument called thermometer. Temperatures are measured in well known two scales: (i) Centigrade scale C (ii) Fahrenheit scale F Both the scales are inter convertible as follows F 1.8C 32 C F 32 1.8 1.3.7 Viscosity Viscosity is the resistance offered to the movement of one layer of fluid by another adjacent layer of the fluid.
1.8 Mechanics of Fluids - www.airwalkpubilications.com u Velocity profile Y dy y u+du u du Fig. 1.1. Velocity Variation. Refer Fig. 1.1. Fluid is divided into different layers one over the other. Consider two layers of fluid. One is moving with velocity u. Another layer is moving with u du The distance between the layer is dy. The top layer causes a shear stress on the adjacent lower layer while the lower layer causes a shear stress on the adjacent top layer. This shear stress (denoted by ) is proportional to rate of change of velocity with respect to y. i.e du dy... (1.5) du dy... (1.6) The proportionality constant is and is known as coefficient of viscosity (or) absolute viscosity (or) dynamic viscosity (or) simply viscosity. du Velocity gradient (or) rate of shear strain (or) dy rate of shear deformation.
Introduction 1.9 The equation (1.6) can be rearranged as To find unit of : du/dy Shear stress Change in velocity Change of distance Force/Area Length Time 1 Length Force Time Length 2 N/m 2 m/s/m Ns m 2 In MKS Unit System, Force is measured in kgf So unit of kgfsec m 2 In CGS System, Force is Measured in dyne dyne sec So unit of cm 2 or poise [... dyne sec 1 cm 2 1 poise] In SI System Force is represented in Newton N So unit of N s m 2 Pa s [... N/m 2 Pascal Pa] Numerical Conversion From MKS unit to CGS unit.
1.10 Mechanics of Fluids - www.airwalkpubilications.com We know that 1 kgf 9.81 N So, 1 kgf Sec m 2 9.81 N S m 2 kgf Sec 1 m 2 2 9.81 1 kg 1 m/sec sec m 2 [... Force N m a] 9.81 103 g 10 2 cm/sec 2 sec 10 4 cm 2 9.81 105 g cm/sec 2 sec 10 4 cm 2 kgf Sec 1 m 2 98.1 dyne sec cm 2 [... 1 g cm sec 2 1 dyne] kgf Sec 1 98.1 poise m 2 In S.I Units [... dyne sec 1 cm 2 1 poise] kgf sec 1 m 2 98.1 poise kgf sec Also 1 m 2 9.81 N sec m 2 1 N Sec m 2 98.1 poise 10 poise 9.81 98.1 poise [... 1 kgf 9.81 N]
Introduction 1.11 So, 1 N sec m 2 10 poise [ in S.I unit sec is represented as S] So 1 Ns m 2 10 poise Sometimes unit of viscosity is given at centipoise 1 Centipoise CP 1 100 poise A widely known metric unit for viscosity is the poise (p) 1 poise 0.1 Ns/m 2 in S.I units 1 centipoise 0.01 poise 0.001 Ns/m 2 1.3.7.1 Kinematic Viscosity () Kinematic viscosity is the ratio of dynamic viscosity to the density of the fluid. The symbol for kinematic viscosity is...(1.7) Unit for kinematic viscosity N s/m2 kg/m 3 kg m s 2 s 1 m 2 m3 kg... N kg m s 2 m2 s In metric system, is in stoke 1 stoke 1 cm 2 /s 10 4 m 2 /s in S.I units 1 centistoke 10 6 m 2 /s
1.12 Mechanics of Fluids - www.airwalkpubilications.com Variation of Viscosity with temperature The viscosity of liquids decreases with the increase of temperature while the viscosity of gases increases with the increase of temperature. This is due to the reason that in liquids the cohesive forces predominates the molecular momentum transfer, due to closely packed molecules and with the increase in temperature, the cohesive forces decreases with the result of decreasing viscosity. But in case of gases the cohesive force are small and molecular momentum transfer predominates. With the increase in temperature, molecular momentum transfer increases and hence viscosity increases. The relationship between viscosity and temperature for (a) Liquids: 0 1 1 t t 2...(1.8) (b) Gases: Where 0 t t 2...(1.9) Viscosity of liquid/gas at tc, in poise 0 Viscosity of liquid/gas at 0C, in poise, are constants for liquid/gas. 1.3.8 Compressibility 1 K Compressibility of a liquid is inverse of its bulk modulus of elasticity.
Introduction 1.13 Bulk modulus K Compressive Stress Volumetric strain Consider a cylinder piston mechanism Let P 1 initial pressure inside the cylinder P 2 Final pressure inside the cylinder V 1 Initial volume V 2 Final volume K Increase in Pressure Volumetric strain C ylinder Piston dp dv V dp dv V...(1.10) Since rise in pressure reduces the volume by d V, the strain is indicated as dv V dv V Fig. 1.2 Compressibility 1 K 1.3.8.1 Relationship between Bulk Modulus K and pressure p of a Gas for Isothermal and Isentropic Process For Isothermal Process We know that for Isothermal process PV constant Partial diffirenting the above equation we get PdV VdP 0
1.14 Mechanics of Fluids - www.airwalkpubilications.com P VdP dv dp dv/v...(1.11) From equation 1.10 we know that K P K For Isothermal process. For Adiabatic (or) isentropic process dp dv/v We know that for Adiabatic process PV constant Ratio of Specific heat Partial differentiating the above equation we get P V 1 dv V dp 0 dividing above equation by V 1, we get P dv VdP 0 P VdP dv K We get P K, for adiabatic or Isentropic Process. 1.3.9 Vapour Pressure When the liquid is kept in a closed vessel, it evaporates into vapour and this vapour occupies the space between the free surface of the liquid and top of the vessel. This vapour exerts a partial pressure on the free surface of the liquid. This pressure is known as vapour pressure of liquid. 1.3.9.1 Cavitation Now consider a flow of liquid in a system. If the pressure at any point in this flowing liquid becomes equal to the vapour pressure, the vaporization of the liquid begins and bubbles are formed. When these bubbles are carried
Introduction 1.15 by flowing liquid into region of high pressure, these bubbles collapse creating very high pressure. The metallic surface above which this liquid is flowing is subjected to these high pressures causing pitting action on surface. This process is called cavitation. 1.3.9.2 Gas and Gas laws The gas is the term applied to the state of any substance of which the evaporation from the liquid state is complete. Substances like Oxygen, air, Nitrogen and hydrogen etc may be regarded as gases within the temperature limits. A vapour may be defined as a partially evaporated liquid and consists of pure gasesous state together with the particles of liquid in suspension. Examples of vapour are not steam, ammonia, SO 2, CO 2 etc. A perfect gas or an ideal gas is one which obeys all gas laws under all conditions of temperature and pressures. No gas known is perfect i.e., no gas strictly obeys the gas laws but within the temperature limits of applied thermodynamics many gases like H 2, O 2, N 2 and even air may be regarded as perfect gases. Gas laws: Three variables control the physical properties of a gas. The pressure exerted by gas P, the volume V occupied by it and its temperature T. If any of these two variable are known then the third can be calculated by gas laws. Gas laws does not apply to vapours.
1.16 Mechanics of Fluids - www.airwalkpubilications.com Boyle s law: Boyle s law states that The volume of a given mass of a gas varies inversely as its absolute pressure, provided the temperature remains constant. V 1 or PV constant P Charles laws: Charle s law states The volume of a given mass of a gas varies directly as its absolute temperature, provided the pressure is kept constant. The variation is same for all gases. V T or V T Constant law). Perfect gas law (combination of boyle s and charle s Let us assume we have a perfect gas at absolute pressure, volume and absolute temperature of P 1, V 1 and T 1 respectively. Suppose the gas expands or contracts at a constant temperature to its volume V 1 such that the corresponding volume of its new absolute pressure is P 2. According to boyle s law P 1 V 1 P 2 V 1 1...(1) Now let gas be expanded (or contracted) further such that the pressure remains constant and its volume and absolute temperature change from V 1 1 to V 2 and T 1 to T 2 respectively. According to charles law
Introduction 1.17 V 1 T 1 V 2 T 1 or V 1 V 2 T 1 T 2...(2) Substituting (2) in (1) we get i.e P 1 V 1 P 2 V 2 T 1 T 2 P V T constant. or P 1 V 1 T 1 P 2 V 2 T 2 If v is the Volume of unit mass of gas, then this constant is R (characteristic gas constant) i.e Pv R or Pv RT which is called the equation of T perfect gas or characteristic gas equation. If m is mass of gas under consideration, then the equation becomes PV mrt. 1.3.10 Surface Tension When a liquid is put inside a narrow tube, the free surface of the liquid displays either a rise (or) depression near the walls of the tube. This phenomena is attributed to a property of fluids known as surface tension. Soap bubbles, small droplets of water and dew on a dry solid surface also are attributed to surface tension. Surface Tension is also defined as the tensile force acting on the surface of a liquid in contact with a gas or on the surface between two immiscible liquids such that the contact surface behaves like a membrane under tension. The magnitude of this force per unit length of the free
1.18 Mechanics of Fluids - www.airwalkpubilications.com surface will have the same value as the surface energy per unit area. Surface tension in a liquid is caused by (i) Cohesive forces i.e forces of attraction between molecules of the same material or fluid and (ii) Adhesive forces i.e forces of attraction between molecules of different materials, say, the attraction between molecules of liquid and the molecules of container (or) air. Example for Cohesive Force When mercury is poured on the floor, it does not wet the surface of floor and forms sphere. When two spheres of mercury are brought close together, they combine together to form a bigger sphere. This means that the mercury molecules have cohesive tendency and have no tendency to adhere (adhesion) to the floor (solid surface). Example for Adhesive Force When water is poured on the floor, the water molecules wet the surface. This means that water molecules have adhesive tendency to adhere to the floor (solid surface). At the interface between a liquid and a gas ie at the liquid surface, and at the interface between two immiscible (not mixable) liquids, the out of balance attraction force between molecules forms an imaginary film capable of resisting tension. This liquid property is known as surface tension. Because the tension acts on a surface, we compare such forces by measuring the tension per unit length of surface. The surface tension is denoted by the symbol. The unit of surface tension is N/m.
Introduction 1.19 A B Consider a liquid in a vessel as shown in Fig. 1.3 Consider two molecules A and B of a liquid in a mass of liquid. The molecule A is attracted in all directions equally by the surrounding molecules of the liquid. Thus the resultant force acting on the molecule A is zero. But the molecule B is situated near the free surface. This molecule B is acted upon by upward and downward forces which are unbalanced. Thus net resultant force on molecule B is acting in a downward direction. Like molecule B, all the molecules near the free surface experience a downward force. So the free surface of the liquid acts like a very thin film under tension of the surface of the liquid. 1.3.10.1 Surface Tension on Droplet Let us consider a spherical droplet of liquid of radius r. The surface tension on the surface of droplet in N m P Pressue intensity inside the droplet r Radius of droplet Fig.1.3. Surface tension in excess of the outside pressure intensity
1.20 Mechanics of Fluids - www.airwalkpubilications.com P Water Droplet SurfaceTension Pressure Forces Fig.1.4 Forces on droplet Let us assume that the droplet is cut into two halves as shown in Fig. 1.4. The forces acting on one half (say left half) will be (i) Tensile force acting around the circumference of the cut portion (This tensile force is due to surface tension) and is given as circumference 2r (ii) Pressure force on the area P r 2 Under equibibrium conditions, these two forces will be equal and opposite in direction. Equate both forces, we get, Water droplet Fig.1.5.Pressure inside a water droplet d P r 2 2r P 2 r or 4 d...(1.12) where d dia of droplet
Introduction 1.21 The equation shows that with the decrease of radius of the droplet, pressure intensity inside the droplet increases. 1.3.10.2 Surface Tension on a Hollow Bubble Soap bubble d A hollow bubble (soap bubble) has two surfaces in contact with air, one inside and other outside. The above two surfaces are subjected to surface tension. In such case, Surface tension on both circumferences Fig.1.6. Pressure inside a soap bubble 2 2r we can equate two forces acting on bubble P r 2 2 2 r P 4 r or P 8 d...(1.13) 1.3.10.3 Surface Tension on a Liquid Jet Consider a liquid jet of diameter d and length L as shown in Fig. 1.7. P Pressure intensity inside the liquid jet above the outside pressure Surface tension of the liquid
1.22 Mechanics of Fluids - www.airwalkpubilications.com Consider the equilibrium of forces acting on the half of the liquid jet Force acting on jet P area of half of jet d (rectangle). P L d. Force due to surface tension 2L L P L d 2L P 2L L d 2 d r P r Fig.1.7 Force on liquid jet 1.3.11 Capillarity Capillarity is the property of exerting forces on fluids by fine tubes. It is due to cohesion and adhesion. Capillarity may be defined as a phenomenon of rise or fall of a liquid surface in a small tube relative to the adjacent general level of liquid when the tube is held vertically in the liquid. When a fine glass tube is partially immersed in water, the water will rise in the tube to a height of h m above the water level. This happens when cohesion is of less effect than adhesion. On the otherhand, if the same tube is partially immersed in mercury, the mercury level in tube will be lower than the adjacent mercury level. This happens because cohesion predominates than adhesion.
Introduction 1.23 The rise of the liquid surface is called capillary rise and the depression of the liquid surface is called capillary depression (or) capillary fall. Mercury h=capillary fall water (a) capillary Rise (b) Capillary Depression Fig.1.8. Capillary Rise (or) Fall 1.3.11.1 Expression for Capillary Rise Refer capillary rise in Fig 1.8 (a) Lifting force created by surface tension Vertical component of the surface tension force circumference cos d cos Weight of the liquid of height h in the tube mass g Volume g Area h g 4 d2 h g
1.24 Mechanics of Fluids - www.airwalkpubilications.com get, Under equilibrium condition, equate both forces, we d cos 4 d2 h g Capillary rise h 4 cos g d...(1.14) Where h: Height of liquid in the tube (or) capillary rise : Surface tension of liquid. : Angle of contact between liquid and glass tube. : Density of liquid Generally value is approximately equal to zero and hence cos 1 then Capillary rise h 4 gd...(1.15) 1.3.11.2 Expression for Capillary Fall Refer Capillary Fall in Fig. 1.8 (b) Hydrostatic force acting upward P 4 d2 g h 4 d2 Downward force making depression due to surface tension d cos Under equilibrium condition, g h 4 d2 d cos
Introduction 1.25 Capillary fall h 4 cos g d 1.3.12 Thermodynamic Properties The characteristic equation (or) equation of state for perfect gases is given as PV m RT... (i) Where V Volume of gas in m 3 m mass of gas in kg R Characteristic gas constant or simply gas constant R for air 0.287 kj/kg K T absolute temperature in K Kelvin T tc 273 The equation can be written Pv RT P R T... (ii)... (iii) Where v specific volume in m 3 /kg density in kg/m 3 Also, we can write the equation as P 1 V 1 P 2 V 2 P 3 V 3 constant T 1 T 2 T 3 For Isothermal process, temperature remains constant Pv constant or PV constant. For Isentropic process
1.26 Mechanics of Fluids - www.airwalkpubilications.com Pv constant or PV constant. For Polytropic process, Pv n constant [n index of compression (or) expansion] [n ranges from 0 to ] ratio of specific heats C p C v C p and C v specific heats of a gas at constant pressure and constant volume respectively According to Avagadro s hypothesis, all the pure gases at the same temperature and pressure have the same number of molecules per unit volume. For any gas, PVm MRT... (iv) Where Vm molar Volume. (molar volume is the volume occupied by the molecular mass of any gas at standard temperature and pressure). T Absolute temp in K; M Molecular weight So the equation (iv) can be written as PV m R T R MR Universal gas constant From Avagadro s law, When P 1.01325 10 5 N/m 2 and T 273.15 K Molar volume Vm 22.4 m3 /kg mol PV m R T R PV m T 1.01325 105 22.4 273.15
Introduction 1.27 8314 J/kg mol K 8.314 kj/kg mol K R 8.314 kj/kg mole K for any gas [Universal gas constant] R R for gas A M for gas A R R air M for air 8.314 0.287 kj/kg K 28.96 [... M for air 28.96 kg/kg mol] 1.4 NEWTON S LAW OF VISCOSITY A fluid is a substance that deforms continuously when subjected to a shear stress, no matter how small that shear stress may be. A shear force is the force component tangent to a surface, and this force divided by the area of the surface is the average shear stress over the area. Shear stress at a point is the limiting value of shear force to area as the area is reduced to the point. In the Fig 1.9, a substance is placed between two closely spaced parallel plates so large that conditions at their edges may be neglected. The lower plate is fixed and a force F is applied to the upper plate, which exerts a shear stress. If A is the area of plate, then shear stress is Y b b c c V u F t a d y x Fig.1.9. Deformation Due to Constant Shear Force
1.28 Mechanics of Fluids - www.airwalkpubilications.com F/A on any substance between plates. When the force F causes the upper plate to move with a steady (Non zero) velocity, no matter how small the magnitude of F, one may conclude that the substance between the two plates is a fluid. The fluid in the area abcd flows to the new position abcd, each fluid particle moving parallel to the plate and the velocity u varying uniformly from zero at the stationary plate to V at upper plate. Experimental results shows that F AV t...(1.16) where is the proportionality factor called coefficient of viscosity. If F/A then...(1.17) Substituting 1.17 in 1.16 we get Shear Stress V t V t is the angular velocity or rate of angular deformation. The angular velocity may also be written as du/dy as both V/t and du/dy express the velocity change divided by the distance over which the change occurs. Now in the differential form, du. This equation is called dy Newton s law of viscosity. Newton s law of viscosity states that the shear stress on a fluid layer is directly proportional to the rate of statics strain. The constant of proportionality is called the coefficient of viscosity Shear Stress du/dy
Introduction 1.29 1.4.1 Types of Fluid du dy...(1.18) All the fluids can be classified into the following types 1. Ideal fluid 2. Real fluid 3. Newtonian fluid 4. Non-newtonian fluid 5. Ideal plastic fluid 1. Ideal Fluid: A fluid with no viscosity is called ideal fluid. Practically, no fluid is ideal since all fluids have some viscosity. This fluid is represented by the horizontal axis in the following graph. 2. Real Fluid: A fluid having viscosity is called Real fluid 3. Newtonian Fluid: A fluid which obeys the Newton s law of viscosity is called Newtonian fluid. In Newtonian fluids there is a linear relation between and rate of deformation du/dy as shown in Fig.1.10 This means that regardless of the forces acting on a fluid, it continues to flow. For example, water is a Newtonian fluid as it continues to display fluid properties no matter how much it is stirred or mixed. 4. Non-Newtonian Fluid: The fluid which does not obey the newton s law of viscosity is called Non-newtonian fluid. Here there is a non-linear relationship between and du/dy. So, stirring a non-newtonian fluid can leave a gap behind which will gradually fill up over time, as such in materials such as pudding. Alternatively, stirring a
1.30 Mechanics of Fluids - www.airwalkpubilications.com non-newtonian fluid can cause the viscosity to decrease, so the fluid appears thinner as seen in non-drip paints. Elastic solid =shear stress ideal plastic Non - Newtonian fluid New tonian fluid ideal fluid (Velocity gradient) du/dy Fig. 1.10. Types of fluids. 5. Ideal Plastic Fluid: A fluid in which shear stress is more than the yield value and shear stress is proportional to the rate of shear strain (or velocity gradient), is called ideal plastic fluid. It is shown in Fig 1.10 by straightline intersecting the vertical axis at the yield stress. SOLVED PROBLEMS Problem 1.1 A mass of liquid weight 500 N, is exposed to standard earth s gravity g 9.806 m/s 2. (i) What is its mass? (ii) What will be its weight in a planet with acceleration due to gravity is 3.5 m/s 2?
Introduction 1.31 Solution Given: Weight W 500 N, g 9.806 m/s 2, g 1 3.5 m/s 2 Case (i) To find: Mass M and weight W 1 Weight W 500 N; g 9.806 m/s 2 W mg Case (ii) Mass m W g 500 9.806 Mass m 50.9892 kg 50.9892 kg Mass of liquid m 50.9892 kg (Mass remains constant) W 1 mg 1 where g 1 Acceleration due to gravity in a planet. W 1 50.9892 3.5 178.4622 N W 1 178.4622 N Problem 1.2 Determine the specific gravity of a fluid having viscosity of 0.07 poise and kinematic viscosity of 0.042 stokes. Given: 0.07 poise, 0.042 stokes. To Find: Specific Gravity s.
1.32 Mechanics of Fluids - www.airwalkpubilications.com Solution: s specific gravity w 0.07 poise 0.07 0.1 0.007 Ns m 2 0.042 stokes 0.042 10 4 m 2 s.. Ns. 1 poise 0.1 m 2 We know kinematic viscosity Dynamic viscosity Density of fluid... 1 stoke 10 4 m 2 Ns 0.042 10 4 0.007 ; 1666.7 kg/m 3 But s w So 1666.7 s 1000 [... w density of water 1000 kg/m 3 ] s 1.6667 Specific gravity of fluid s 1.6667
Introduction 1.33 Problem 1.3 Determine the viscosity of oil having kinematic viscosity of 6 stokes and specific gravity of 2. Given: 6 ; stokes; s 2 To Find: Solution: viscosity of oil? We know that s 2 ; 6 stokes 6 10 4 m 2 /s (FAQ) [... 1 stoke 10 4 m 2 /s] s w s w 2 1000 2000 kg/m 3 So 2000 6 10 4 1.2 Ns/m 2... w 1000 kg/m 3 Problem 1.4: The space between two parallel plates kept 3 mm apart is filled with an oil of dynamic viscosity of 0.2 Pa-s. What is the shear stress on the lower fixed plate, if the upper one is moved with a velocity of 1.5 m/s. Given: Solution: dy 3 mm, 0.2 Pa s 0.2 Ns m 2 du 1.5 m/s To Find: 0.2 Pa.s 0.2 Ns m 2 According to Newtons law of viscosity
1.34 Mechanics of Fluids - www.airwalkpubilications.com Shear Stress du dy where shear stress on the lower fixed plate. du Change of velocity u 0 1.5 0 1.5 m/s distance between two plates dy 3 mm Upper moving plate 3 10 3 m 1.5m/s du dy 1.5 0.2 3 10 3 100 N/m 2 Oil of = 0.2 Poise Lower fixed plate dy = 3 mm Shear stress on lower plate 100 N/m 2 Problem 1.5: A 2.5 cm wide gap between two large plane surfaces are filled with glycerine. What surface force is required to drag a very thin plate of 0.75 m 2 in area between the surfaces at a speed of 0.5 m/s, if it is at a distance of 1 cm from one of the surfaces? Take 0.785 Ns/m 2 Given: dy 2.5 cm dy 1 1 cm 1 10 2 m dy 2 1.5 10 2 m A 0.75 m 2 u 0.5 m/s 0.785 Ns/m 2
Introduction 1.35 Solution Note: Since it is very thin plate, the weight of the plate and buoyant force are neglected. 1cm dy 1 dy 2 Area of thin plate 0.75 m 2 du u 0 0.5 m/s dy 1 1 cm 1 10 2 m dy 2 1.5 cm 1.5 10 2 m 0.785 Ns/m 2 2.5cm To Find Drag Force to Lift the Plate: Drag force on both sides A Shear stress on both sides du dy 1 du dy 2 du dy 1 du dy 2 0.785 0.5 1 10 2 0.5 1.5 10 2 0.785 [50 33.333 ] 65.42 N/m 2 Drag force on both sides A 65.42 0.75 49.0625 N
1.36 Mechanics of Fluids - www.airwalkpubilications.com Problem 1.6: If the velocity distribution over a plate is given by u y y 2 in which u is the velocity in m/s at a distance of y meters above the plate, determine the shear stress at y 0.10 m when coefficient of viscosity is 0.86 Ns/m 2 Solution Given: u y y 2 ; 0.86 Ns/m 2 To Find: 0.1? y 0.1 m 0.86 Ns/m 2 u y y 2 Velocity gradient du dy 1 2y when y 0.1 m, du dy 1 2 0.1 0.8 s 1 0.1 To Find Shear stress at y 0.1 m 0.1 du dy 0.1 0.688 N/m 2 0.86 0.8 0.688 N/m 2 0.1 Problem 1.7: A fluid of specific gravity 0.9 flows along a surface with a velocity profile given by v 4y 8y 3 m/s, where y is in m. What is the velocity gradient at the boundary? If the kinematic viscosity is 0.36 S, What is the shear stress at the boundary? (FAQ)
Introduction 1.37 Given Solution: Specific gravity of fluid 0.9 Kinematic viscosity, 0.36 stokes Velocity V 4y 8y 3 m/s 0.36 10 4 m 2 /s (a) velocity gradient at the boundary =? (b) shear stress at boundary? V 4y 8y 3 dv dy 4 8 3y2 4 24 y 2 (a) Velocity gradient at the boundary, ie y 0, dv dy 4 24 0 2 4 Ans y 0 (b) Shear stress, dv dy Specific gravity S 0.9 Density of fluid Density of water Density of fluid 0.9 density of water 0.9 1000 900 kg/m 3 w.k.t 0.36 10 4 900 0.0324 NS/m 2
1.38 Mechanics of Fluids - www.airwalkpubilications.com Shear stress at boundary, i.e dv dy y 0 0.0324 4 0.1296 N/m 2 Ans Problem 1.8: A cubical block having 200 mm edge and mass of 25 kg slides down an inclined plane surface which makes an angle of 20 with the horizontal. On the plane, there is a thin film of oil of thickness 0.026 mm and viscosity 0.2 10 2 Ns/m 2. What terminal velocity will be attained by the block? Solution Given: l 200 mm 0.2 m Mass m 25 kg 20 t 0.026 mm 0.2 10 2 Ns/m 2 0.2 10 2 Ns/m 2 Area 0.2 0.2 0.04 m 2 W Wt. of block 25 9.81 245.25 N 20 o m =25kg dy=thickness=0.026m m w wsin20 o =20 o To Find Shear Stress: Component of W along their inclined plane i.e shear force F W sin
Introduction 1.39 245.25 sin 20 83.8804 N Shear stress on the F bottom surface of cube A 83.8804 0.04 To Find Terminal Velocity u We know, du dy 2097.01 N/m 2 2097.01 0.2 10 2 du 0.026 10 3 du 27.261 m/s du u 0 So, Terminal Velocity u 27.261 m/s Problem 1.9: A plate having an area of 0.8 m 2 is sliding down the inclined plane at 25 to the horizontal with a velocity of 0.36 m/s. There is a cushion of fluid, 1.8 mm thick between the plane and the plate. Find the viscosity of the fluid if the weight of the plate is 442 N. Given: A 0.8 m 2 25 u 0.36 m/s t dy 1.8 mm W 442 N Solution Area of plate 0.8 m 2 Weight of plate 442 N
1.40 Mechanics of Fluids - www.airwalkpubilications.com Velocity of plate u 0.36 m/s du u 0 0.36 m/s plate dy=1.8m m u=0.36m /s Thickness of film 25 o t dy 1.8 mm 25 o 1.8 10 3 m To Find Viscosity of Fluid w sin w = 442N Component of W along the plate Shear stress F A 187 0.8 We know, du dy shear force F 0.36 233.75 1.8 10 3 W sin 442 sin 25 187 N 233.75 N/m2 Viscosity 1.169 Ns/m 2 11.69 poise Problem 1.10: A metal plate of size 0.6 m 0.6 m and 1 mm thick and weighing 25 N is to be lifted up edgewise with uniform velocity of 0.2 m/s in the gap between two flat surfaces. The plate is in the middle of the gap of width 20 mm and the gap contains oil of relative density 0.85 and viscosity 16 poise. Calculate the vertical force required for this job.
Introduction 1.41 Solution: Given : Dimensions of the plate dy =dy 1 2 t=dy 1 1 t=dy 2 2 0.6 0.6 1 10 3 m Area of plate 0.6 0.6 0.36 m 2 Since the plate is in the middle of the gap, t 1 t 2 dy 1 dy 2 20 1 2 9.5 mm 9.5 10 3 m 20mm 1mm Relative density specific gravity s 0.85 Dynamic viscosity 16 poise 1.6 Ns/m 2 Velocity of the plate u 0.2 m/s; So du u 0 0.2 m/s Weight of the plate 25 N To Find Force Required to Lift the Plate Drag force (or viscous resistance) against the motion of the plate, F 1 A 2 A where 1 and 2 are the shear stresses on both sides of the plate. 1 du dy 1 and 2 du dy 2 F A [ 1 2 ] A du 1 1 dy 1 dy 2
1.42 Mechanics of Fluids - www.airwalkpubilications.com 0.36 1.6 0.2 1 9.5 10 3 1 9.5 10 3 Force F 24.253 N Upward thrust (or) buoyant force on the plate Specific weight Volume of oil displaced. 0.85 9810 0.6 0.6 1 10 3 3.00186 N [Note: When a body is immersed in a fluid, upward force (or) buoyant force acts on body to move the body upward. This buoyant force is equal to weight of fluid displaced by the body. So Weight of fluid displaced Sp. weight Volume of fluid displaced Now Weight of the body 25 N acts downward. Buoyant force is acting upward. Effective weight of the plate 25 3 22 N Total force required to lift the plate at a velocity of 0.2 m/s F effective weight of plate 24.253 22 46.253 N. Problem 1.11: A vertical gap of 30 mm width and infinitely long, contain oil of specific gravity 0.9 and viscosity 3.5 Ns/m 2. A metal plate 1.0 m 1.0 m 10 mm having a mass of 20 Kg is to be lifted through the gap at a constant speed of 0.15 m/s. Determine the force required. The plate is situated 5 mm from the left end.