Theory And Problems of Fluid Dynamics 1773613766, 9781773613765

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THEORY AND PROBLEMS OF FLUID DYNAMICS

THEORY AND PROBLEMS OF FLUID DYNAMICS

Edited by

Olga Moreira

ARCLER

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r

e

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s

www.arclerpress.com

Theory And Problems Of Fluid Dynamics Olga Moreira

Arcler Press 2010 Winston Park Drive, 2nd Floor Oakville, ON L6H 5R7 Canada www.arclerpress.com Tel: 001-289-291-7705         001-905-616-2116 Fax: 001-289-291-7601 Email: [email protected] e-book Edition 2019 ISBN: 978-1-77361-578-3 (e-book) This book contains information obtained from highly regarded resources. Reprinted material sources are indicated and copyright remains with the original owners. Copyright for images and other graphics remains with the original owners as indicated. A Wide variety of references are listed. Reasonable efforts have been made to publish reliable data. Authors or Editors or Publishers are not responsible for the accuracy of the information in the published chapters or consequences of their use. The publisher assumes no responsibility for any damage or grievance to the persons or property arising out of the use of any materials, instructions, methods or thoughts in the book. The authors or editors and the publisher have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission has not been obtained. If any copyright holder has not been acknowledged, please write to us so we may rectify.

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ABOUT THE EDITOR

Olga Moreira obtained her Ph.D. in Astrophysics from the University of Liege (Belgium) in 2010, her BSc. in Physics and Applied Mathematics from the University of Porto (Portugal). Her post-graduate travels and international collaborations with the European Space Agency (ESA) and European Southern Observatory (ESO) led to great personal and professional growth as a scientist. Currently, she is working as an independent researcher, technical writer, and editor in the fields of Mathematics, Physics, Astronomy and Astrophysics.

TABLE OF CONTENTS



List of Symbols.............................................................................................xiii

Peface...........................................................................................................xv Chapter 1

. Fluids.......................................................................................................... 1 1.1. Fluids Definition and Classification...................................................... 2 1.2. Density, Specific Weight, and Specific Gravity..................................... 3 1.3. Pressure and Specific Volume.............................................................. 4 1.4. Equations of State................................................................................ 5 1.5. Bulk Modulus...................................................................................... 6 1.6. Dynamic and Kinematic Viscosity........................................................ 7 1.7. Surface Tension.................................................................................. 11 1.8. Capillary Action................................................................................. 13 1.9. Vapor Pressure................................................................................... 16 1.10. Exercises.......................................................................................... 16

Chapter 2

Fluid Statics.............................................................................................. 21 2.1. Static Pressure and Variation Within a Fluid....................................... 22 2.2. Pressure Variation In Compressible Fluids.......................................... 25 2.3. Forces Acting on Surfaces by Static Fluids.......................................... 27 2.4. Pressure Center.................................................................................. 28 2.5. Force Analysis In Case of Curved Surface Submerged in a Liquid....... 30 2.6. How to Measure Pressure.................................................................. 31 2.7. Manometers....................................................................................... 32 2.8. Buoyancy, Law of Floatation and Floating Bodies............................... 35 2.9. Relative Motion in Fluids................................................................... 36 2.10. Metacenter Height........................................................................... 40 2.11. Exercises.......................................................................................... 41

Chapter 3

Kinematics of Flow................................................................................... 53 3.1. Flow Variables and Classifications...................................................... 54 3.2. Steady Flow....................................................................................... 54 3.3. Uniform Flow.................................................................................... 55 3.4. Elements of a Particle In Motion......................................................... 60 3.5. Equation of Continuity....................................................................... 62 3.6. Exercises............................................................................................ 65

Chapter 4

Dynamics of Flow.................................................................................... 69 4.1. Forces................................................................................................ 70 4.2. Equation of Motion in Streamline Coordinates .................................. 72 4.3. The Equation of Motion in an Inviscid Flow....................................... 73 4.4. Velocity Potential............................................................................... 75 4.5. Velocity Potential Versus Stream Function.......................................... 75 4.6. Flow Nets.......................................................................................... 76 4.7. Exercises............................................................................................ 79

Chapter 5

Bernoulli’s Theorem................................................................................. 83 5.1. Deduction of Bernoulli’s Theorem, Modifications, and Application.............................................................................. 84 5.2. Coefficient of Flow............................................................................ 84 5.3. Venturimeter and Orifice Meter......................................................... 87 5.4. Orifice Meter/Plate............................................................................ 88 5.5. Inflow and Outflow........................................................................... 89 5.6. Coefficient......................................................................................... 91 5.7. Time Relation in Discharging Liquids................................................. 92 5.8. Loses of Head In Flowing Fluids........................................................ 96 5.9. Weirs and Notches............................................................................ 99 5.10. Velocity of Approach..................................................................... 102 5.11. Siphon Spillway (Figure 5.20)........................................................ 102 5.12. Broad – Crested Weir..................................................................... 103 5.13. Submerged Weir (Figure 5.22)....................................................... 104 5.14. Sluice Gates (Figure 5.23).............................................................. 105 5.15. Borda’s Mouthpieces (Figure 5.24)................................................. 106

viii

Chapter 6

Force and Momentum............................................................................ 109 6.1. Introduction..................................................................................... 110 6.2. Impact of Jets................................................................................... 111 6.3. Forces on Bends............................................................................... 117 6.4. Rockets Motion................................................................................ 120 6.5. Jet Propulsion ................................................................................. 121 6.6. Loss of Energy in Pipe Expansion..................................................... 122 6.7. Lawn Sprinklers............................................................................... 124 6.8. Pelton Wheel................................................................................... 125 6.9. Reaction Turbines............................................................................ 127

Chapter 7

Irrotational Flow.................................................................................... 131 7.1. General Motion of a Fluid Element.................................................. 132 7.2. Uniform Flow.................................................................................. 136 7.3. Plane Potential Flow in Polar Coordinates........................................ 137

Chapter 8

Laminar Motion...................................................................................... 143 8.1. Introduction..................................................................................... 144 8.2. Shear Stresses.................................................................................. 144 8.3. Navier-Stokes Equations and Solutions............................................. 146 8.4. Flow Between Horizontal Parallel Plates.......................................... 151 8.5. Flow Between Two Plates With X-Axis Along the Lower Plate.......... 153 8.6. Couette Flow................................................................................... 154 8.7. Combined Hagen-Poiseuille and Couette Flows............................... 155 8.8. Hagen-Poiseuille Flow in Cylindrical Tubes..................................... 156 8.9. Laminar Flow in an Annulus Area.................................................... 158 8.10. Rotating Cylinders......................................................................... 159 8.11. Reynolds Number and Stability Parameters.................................... 160 8.12. Stability Parameter......................................................................... 161 8.13. Stability Curves.............................................................................. 163 8.14. Laminar Boundary Layer................................................................ 164 8.15. Boundary Layer Thickness.............................................................. 165 8.16. Displacement Thickness................................................................. 165

ix

Chapter 9

Turbulent Flow....................................................................................... 169 9.1. Characteristics of Turbulence and Classification............................... 170 9.2. Reynolds Equation........................................................................... 172 9.3. Derivation For Mean Turbulent Flow................................................ 173 9.4. Characteristics of Turbulence........................................................... 177 9.5. Turbulent Boundary Layer................................................................ 178 9.6. General Logarithm Formulation....................................................... 180 9.7. Momentum Equation....................................................................... 181 9.8. Turbulent Flow In Pipes.................................................................... 183 9.9. Exercises.......................................................................................... 188

Chapter 10 Flow Through Pipes................................................................................ 193 10.1. Friction In Pipes and Flow In Pipes................................................ 194 10.2. Reynolds Experiment..................................................................... 196 10.3. Distribution of Velocity in a Pipe................................................... 199 10.4. Steady Flow and Losses in Pipes.................................................... 200 10.5. Head Loss in Pipe and Fittings....................................................... 203 10.6. Hydraulic and Energy Gradients.................................................... 204 10.7. Problems in Pipe Flow................................................................... 207 10.8. Pipe Network................................................................................. 209 10.9. Transmission of Power In Pipes...................................................... 211 10.10. Unsteady Flow in Pipes............................................................... 213 10.11. Water Hammer Blow in Pipes...................................................... 215 10.12. Effect of Pipe Elasticity on Hammer Blow.................................... 218 10.13. Oscillations of Liquid Column in a U-Tube.................................. 219 10.14. Surge Tanks.................................................................................. 220 10.15. Exercises...................................................................................... 225 Chapter 11 Flow Through Channels.......................................................................... 239 11.1. Open Channels and Flow.............................................................. 240 11.2. Expression for The Flow in Open Channels.................................... 240 11.3. Manning’s Formula........................................................................ 242 11.4. Types of Channels and Discharge................................................... 242 11.5. Variation of Velocity Over The Cross Section of a Channel............. 245 11.6. Hydraulic Jump............................................................................. 249 11.7. Specific Energy of a Channel’s Cross Section................................. 251 x

11.8. Exercises........................................................................................ 253 Chapter 12 Compressible Flow................................................................................. 257 12.1. Introduction................................................................................... 258 12.2. Gas Laws And Work Done............................................................. 258 12.3. Application of Law of Conservation of Energy................................ 259 12.4. Enthalpy and Entropy..................................................................... 261 12.5. Velocity of Pressure Wave In Fluid................................................. 263 12.6. Variation of Atmospheric Pressure With Altitude............................ 265 12.7. Speed of Sound And Mach Number............................................... 267 12.8. Compressible Flow – One Dimensional Flow................................ 268 12.9. Normal Shock Wave...................................................................... 271 12.10. Compressible Flow In A Pipe With Friction.................................. 274 12.11. Mach Number Relations.............................................................. 276 12.12. Exercises...................................................................................... 277 Chapter 13 Dimensional Analysis............................................................................. 281 13.1. Introduction................................................................................... 282 13.2. Fundamental Units of Dimensions................................................. 282 13.3. Summation of Quantities............................................................... 283 Chapter 14 Fluid Power............................................................................................ 289 14.1. Introduction................................................................................... 290 14.2. Definitions, Coefficient, And Power Calculations........................... 291 14.3. Properties of Fluids In Hydraulics.................................................. 295 14.4. Elements of Power Hydraulics........................................................ 296 Index...................................................................................................... 333

xi

LIST OF SYMBOLS

(Note: These are default symbols, where not specified, it is assumed) 1 Kg = 2.2 Lbs 1 cubic Foot = 28.3168 L 1 Gallon = 3.785 liters M = Mass of an object V = Volume of an object ρ = Density of an object sg = specific gravity; p = pressure A = Total area F = Force N = Force in Newton’s Ma = Mach number (sometimes as M) w = Rotation in radians/sec r = Radius NTP = Normal temperature and pressure, defined as air at, pressure = 101 kPa Temperature = 20 Deg C; kPa = Kilo Pascal’s = 100 prascals. Deg C = Centigrade K = Bulk Modulus Ø = Angle of distortion µ = Coefficient of viscosity w = Specific weight of liquid p = pressure at a point in the fluid i, j, and k are units vectors in x, y, and z directions η = Kinematic viscosity Cd = Coefficient of discharge xiii

Cc = Coefficient of contraction Cv = Coefficient of velocity V = Del operator f0 = Shear stress Cf = Frictional drag coefficient Cp = Specific heat at constant pressure Cv = Specific heat at constant volume f = Coefficient of friction = 4xfd fd = Darcy frictional coefficient g = Standard gravity (9.81 m/s) = 32.2 Ft/sec2 m/s = Meters/second h = Pressure head (meters) Re OR R = Reynolds number hf = Head lost due to friction Vs = Specific volume of a gas d = Diameter rad = Radians dia = Diameter Q = discharge/sec T = Temperature Absolute g = Adiabatic constant K = Bulk modulus m = Hydraulic mean depth C = Chezy Constant bar = 1 Kg/cm2 (14.2 Psi) c = Velocity of sound at 1 bar v = Velocity, variable GGS = Centimeter-gram-second system of unit SI = International System of Units SQRT = Indicates Square root of a function

xiv

PREFACE

Fluid dynamics is a subdiscipline of fluid mechanics that explores and describes the movement and flow of fluids and the forces which act upon and influence them. It has a wide range of applications, with particular influence in the fields of mechanical and aerospace engineering. A solid knowledge of fluid dynamics is required for many areas of technology and design in present-day society, from an aircraft’s flight control and landing gear systems to the engineering of modern machines such as hydraulic turbines that can generate electricity and the hydraulic motors that can power road rollers, sugar mill crushers, stone crushers and much more. The main focus of this book is to analyze the behavior of various types of fluids when subject to normal and shear forces so we understand the changes in their physical properties and anticipate resultant effects. Knowledge of these factors will enable us to design hydraulic systems of aircraft, road rollers, crushing machines and many more. These machines, which are based on working fluid, are called fluid machines or hydraulic machines. Compared with any other machines, fluid machines have a very high power-to – weight ratio. For instance, a typical fluid pump generates 10 HP and weighs 0.5 kg, while a typical electric motor generates 10 HP and weighs as much as 20 kg. Thus, the power-to – weight ratio of the hydraulic pump is 10/0.5 = 20, whereas of that of the electric motor is 10/20 = 0.5. This represents a significant advantage, as high power-to – weight ratios are required in many mechanical engineering applications. It is a crucial factor for the development of aircraft, where minimizing the weight being carried while simultaneously optimizing the generation of higher power is the primary goal. Learning fluid dynamics is vital for aeronautical engineers as aircraft industries require as absolutely necessary both experimental data and a greater precision of fluid flow analysis. Considering the above aspects the scope of this work is focused on employing laws of Newtonian mechanisms to solve complex problems. The text also addresses the needs of modern engineers, advanced engineering students and occupational specialists working as hydraulic, mechanical, and civil engineers. The first chapter deals with properties of fluids extensively and their variation with environmental conditions. Following that we briefly explore and discuss

hydrostatics problems. Further on, flow kinematics of fluids is dealt in detail. This is necessary to understand the core foundations of fluid dynamics. In later chapters, a detailed treatment of the equations of motion and conservation of energy is presented to the reader. A practical example of the flow of fluids in pipes, channels, and other porous media are examined in detail. Two dimensional and three-dimensional fluid flows receive detailed mathematical treatments. All chapters are in brief and simplified as much as possible, although it is necessary that readers should have a working knowledge of calculus, vectors, and mathematics in particular. To assist the reader, the author has prepared a collection of fluid dynamics exercises. The author aims for readers to understand the subject fairly well and use the knowledge imparted in their own applications. Note: All base and derived units of measurement in this book are based on the modern metric system, the SI (International System of Units); and its variant, the GCS (Centimetre–gram–second system of units). For instance, we use either meter, kilogram, and second or gram centimeter and second as base units of length, mass, and time; Newton, Pascal, and Celsius as derived units of force, pressure, and temperature; Stokes, and Poise derived units of kinematic and dynamical viscosity. The final chapter of this book pertains to “Fluid Power”, it possesses significant importance to those who are involved in the design of hydraulic machines, and as such field engineers will find it of greater usefulness.

xvi

CHAPTER

1

FLUIDS

CONTENTS 1.1. Fluids Definition And Classification..................................................... 2 1.2. Density, Specific Weight, And Specific Gravity.................................... 3 1.3. Pressure And Specific Volume.............................................................. 4 1.4. Equations Of State............................................................................... 5 1.5. Bulk Modulus...................................................................................... 6 1.6. Dynamic And Kinematic Viscosity....................................................... 7 1.7. Surface Tension.................................................................................. 11 1.8. Capillary Action................................................................................. 13 1.9. Vapor Pressure................................................................................... 16 1.10. Exercises.......................................................................................... 16

2

Theory and Problems of Fluid Dynamics

1.1. FLUIDS DEFINITION AND CLASSIFICATION Fluids are a state of matter that can flow, i.e., the matter in fluids slides layer by layer. Fluids do not have a definite shape, they assumes the shape of the container. In physics, a fluid is usually defined as that which can deform continuously under minimum shear stress. Examples of fluids are water, petrol, oils, milk, air, steam, and water vapor. Based on their physical properties, fluids can be classified as liquids or gases. Liquids are far from being compressible fluids. Unlike the bonded molecules in solids, molecules move freely in liquids. The volume of liquids do not change significantly with the increase of temperature and pressure. Hence, for all practical purposes, liquids are considered be incompressible within a small range of temperature and pressure. Examples of liquids are water, oils, milk, and paints. Gases are highly compressible fluids. Molecules in gases move more freely, these are farther apart and have fewer interactions than those in liquids or solids. The volume of gases changes significantly with the change of pressure and temperature. Examples of gases are air, oxygen, nitrogen, steam, and water vapor. The following example illustrates the difference between liquids and gases. Suppose that by applying a pressure of 10 kilograms-force per square centimeter (kgf/cm2) on 10 liters (L) of liquid, it compresses it by 0.2 %. Thus, the liquid’s volume decreases to 9.98 L. Increasing, or decreasing, the temperature by 1°C changes the volume by 0.005%. In the case of gases, applying the same amount of pressure and same increase (or decrease) of temperature will compress the fluid by about 90%.

1.1.1. Classification of Liquids Liquids are classified as ideal or real. Ideal liquids cannot be compressed by applying any amount of pressure or temperature. When subjected to pressure, ideal liquids exert an equal pressure in all direction. In practice no liquid is ideal. On the contrary, real liquids are compressible. However, compressibility in real gases is usually very low, it is about 0.5% to 1% at low-pressure values. All liquids can be considered as ideal when working at low temperature and pressure. In conclusion, gases, and liquids behave significantly differently under the same variation of pressure and, or, temperature conditions.

Fluids

3

1.1.2. Classification of Gases Gases are classified as ideal, real, or vapors. Molecules in ideal gases occupy a negligible amount of space and have nearly no interactions. Ideal gases follow an equation of state known as the Perfect Gas Law (a.k.a Ideal Gas Law), PV = nRT

(1.1)

where P is the gas pressure, V the gas volume, n is the number of moles of the gas, R is the universal gas constant and T is the gas temperature. Real gases do not have the same properties of ideal gases. Their molecules occupy a small but non-negligible space. However, they do follow the gas laws with no significant deviations. Vapors have liquid molecules at all temperatures below a saturation point. As the temperature decreases more molecules fall back into the liquid state. Generally, vapors are a mixture of gases and liquids. Vapors that do not contain liquid molecules are called Supersaturated Vapors. For practical purposes, we can apply the gas laws to supersaturated vapors. In the case of unsaturated vapors, we can still apply the same gas laws but with a correction factor known as the Wetness Factor.

1.2. DENSITY, SPECIFIC WEIGHT, AND SPECIFIC GRAVITY It is important to know the properties of fluids. In general, the properties of gases vary with changes in pressure, temperature, and density. Density is defined as the mass of an object per unit volume. Mathematically it is expressed as,

M V (1.2) where ρ is density, M is the mass, and V is the volume. The SI unit for density is kilogram per cubic meter (kg/m3) and CGS unit is gram per cubic centimeter (g/cm3). Other commonly used units for density are kilogram per liter (kg/L) or gram per milliliter (g/mL). When the object has an irregular shape (i.e., when it does not have a definite shape), the density is determined using the following differential calculus expression:

ρ=

4

Theory and Problems of Fluid Dynamics

(1.3) where δm is an infinitesimal mass element and δv is an infinitesimal volume element. The weight of an object per unit volume is called Specific Weight. The weight of an object is defined as the force acting on a unit mass of the object due to gravity. Thus, (1.4) where is the specific weight and g is the standard acceleration due to gravity (standard gravity)? On Earth g = 9.80665 m/s2 measured at sea level, this value may vary with the altitude and latitude. The most common units for specific weight are expressed as weight (force) per unit volume, e.g., Newton per cubic meter (N/m3, where 1 N = 1 kg × 1 m/s2); pound per cubic foot (lb/ft3); kilograms-force per liter (kgf/L). Here, we will also define the Relative Density (RD) or Specific Gravity (SG) which an important quantity for the study of statistics and dynamics of liquids. It can be simply expressed as the ratio of weights (or densities) of a sample material to a reference material of equal volume, i.e.

wS ρ S = wR ρ R (1.5) where wS and ρS are the weight and density of a given material, respectively; while, wR and ρR are the weight and density of the reference material. Conventionally, the relative density of a sample material with respect to the reference material is expressed using the subscript “sample/reference”. For instance, RDH/air means the relative density of Hydrogen (H) with respect to dry air. The relative density of liquids is usually measured with respect to water at a temperature of 4°C and a pressure of 1 bar. It indicates how much time a liquid is heavier than water at 4°C and 1 bar pressure. Since it is the ratio of weights (or densities), the relative density is a dimensionless quantity. RD =

1.3. PRESSURE AND SPECIFIC VOLUME Pressure is a measure of force acting on fluid per unit of area. As aforementioned, a fluid assumes the volume of a container when this

Fluids

5

filled. One cannot determine a definite area occupied by a fluid. In order to calculate the total pressure of a fluid, we measure the amount of force applied perpendicularly to the surface of the fluid (normal force) and divide by the area of the surface on contact (contact area). Mathematically it is expressed as F P= − n A (1.6) where P is the total pressure, Fn is the magnitude of the normal force, and A is the contact surface area. Pressure may vary from point to point. The pressure at given point can be determined using the following differential calculus expression: δ Fn dF = − n p = l im δ A→ 0 δ A dA (1.7) where p is the pressure, 𝛿Fn is normal force acting on an infinitesimal surface element with 𝛿A area.

Pressure is a derived unit. The SI unit for pressure is the Pascal (Pa), named after the scientist Blaise Pascal and defined as the force of one Newton per square meter, i.e., 1Pa = 1N/m2. The standard atmosphere (atm) is one of most commonly used unit for pressure, it is defined as 1 atm = 101325 Pa, the bar defined 1 bar = 100000 Pa.

1.3.1. Specific Volume The ratio of the volume of a fluid to its mass is known as the Specific Volume (Vs). It is the reciprocal of the density, i.e.

V 1 = M ρ (1.8) The volume of a gas is significantly large compared to its weight, for instance, the volume of one cubic meter of air weighs only 1.225 kgs according to ISA. Hence, the specific volume is more convenient property to used than volume or density when analyzing the physical behavior of gases. V = s

1.4. EQUATIONS OF STATE At normal temperature and pressure (NTP), all matter exist in three basic states: solid, liquid, gas, and plasma. For example, at NTP, metals such as iron, silver, and copper exist in the solid state; oxygen, carbon dioxide and

6

Theory and Problems of Fluid Dynamics

nitrogen exist as gases; water, diesel, and hydraulic oils exist as liquids. These states of matter, are influenced by changes in temperature and pressure. Gases such as oxygen, ammonia, and carbon dioxide can exist as liquids at specific ranges of temperature and pressure. Similarly, water exists as a solid at and below a critical temperature and pressure. Metals such as iron, silver, and gold exist as liquids at high-temperature ranges. At each state, matter behaves and follows different laws. These state laws are mathematically expressed by an Equation Of State. For instance, as aforementioned, ideal gases follow the perfect gas law which is mathematically expressed by the equation state in Eq. (1.1). We shall further explain fluids equations of state in the later chapters.

1.5. BULK MODULUS Fluids do not have any definite shape, as aforementioned, they assume the shape of the container they occupy. It does not make sense to use length and width of a fluid as measures because these will vary with the container’s geometry. On contrary, the volume of a fluid will remain constant. Therefore, volume is a better physical property and more reliable measure to use for the study of fluids dynamics. In relation to this property, we define bulk modulus of fluids. First, let us define Volumetric Strain. Fluids do compress by applying pressure. Thus, let us increase the pressure by an infinitesimal amount, p’ = p + δp

(1.9)

where p is the initial pressure, 𝛿p is an infinitesimal increment of pressure and p’ is the final pressure. In this case, the volume decreases by an infinitesimal amount, V’ = V + δV

(1.10)

where V is the initial volume, 𝛿V is an infinitesimal increment of volume and V’ is the final volume. Thus, change in pressure is

p’ – p = (p + δp)-p,

(1.11)

Fluids

7

and the change in volume is V’ – V = (V + δV) – V = - δV (1.12) Volumetric strain (VS) is the ratio of the change in volume to initial volume when pressure is applied,

± '− −δ = V V (1.13) The ratio of the change in pressure to the volumetric strain is called Bulk Modulus (K), δp δp = −V K= δ V (1.14) −(δ V / V )

= VS

The minus sign indicates that the volume decreases as pressure increases and vice-versa. Bulk modulus has the same dimension as pressure, i.e., N/m2, or Pa. Normally it is constant for liquids within a small range of temperature. In case of gases, it is also dependent on the temperature. Since mass is conserved, K can also be expressed as

δp δp K ρ= γ = δρ

δγ (1.15)

where 𝛿ρ and 𝛿𝛾 are infinitesimal increments of density and specific weight, respectively.

1.6. DYNAMIC AND KINEMATIC VISCOSITY We have defined fluids as substances that flow. Let us consider, for instance, water flowing in a river. The liquid (water) flows over a flat surface (stream bed) and is composed of infinite stratified layers of infinitesimal thickness from bottom to top as shown in Figure 1.1. The lowermost layer of water is stationary, fixed to the base of the river, the next layer above it slides on it. One by one, every layer above slides on the layer below. The topmost layers have the maximum velocity and these are called the Flow. The stress due to the velocity difference between the adjustment layers is called Shear Stress. This phenomenon is called continuous deformation of layers.

Theory and Problems of Fluid Dynamics

8

where, v = velocity of the y layer; v + dv = velocity of the y + dy layer; y = distance from surface at rest; dy = distance between adjacent layers; fS = shear stress. Figure 1.1: Continuous deformation of layers.

Let us analyze this continuous deformation problem mathematically. Figure 1.2 shows the enlarged view of the AA area in Figure 1.1, the rectangle “abcd”. After one second, as fluid continuously shears from velocity “v” to “v + dv”, “abcd” is distorted to “abfe” with an angular shear strain θ. Angular shear strain is also known as the angular deformation rate. Experimentally, it has been found that fs =η θ (1.16) where fS is the shear stress; θ is the angular deformation rate; and η is a coefficient of viscosity1. Since θ is small, we can use the following approximation: 1. NOTE: It is interesting to note this is analogous to the case of a solid under shear force. The coefficient viscosity is analogous to the modulus of rigidity (G) which can be derived the same way. Imagine the rectangular “ABCD” as a solid and we arrive to an equivalent result G=fs/θ.

Fluids

9

CF dv = BC dy (1.17) where, as shown in Figure 1.2, represents the distance from point c to point f, represents the distance from point b to point c, dv is the velocity difference between adjacent layers and dy is the distance between adjacent layers.

θ ≈ tan θ =

where, dv = velocity difference between adjacent layers; θ = angular shear strain; dy = distance between adjacent layers; fS = shear stress. Figure 1.2: Enlarged view of AA (continuous deformation of layers).

By substituting Eq. (1.17) into Eq. (1.16), we obtain

 dv  fs = η    dy  (1.18) Therefore, the shear stress in fluids is proportional to the rate of change velocity from one layer to the adjacent layer. Dynamic Viscosity (also known as Absolute Viscosity) is the proportionality factor, i.e., the coefficient of viscosity η. Kinematic Viscosity is defined as the ratio of the absolute viscosity to the density,

v=

η ρ (1.19)

Theory and Problems of Fluid Dynamics

10

Viscosity varies with temperature. At a given temperature, fluids that display a linear relation between angular shear strain (dv/dy) and shear stress (fS) are called Newtonian Fluids. Fluids that do not manifest this linear relation at a given temperature are called Non-Newtonian fluids. Figure 1.3 shows graph of the relationship between Shear stress and Shear strain of several types of fluids. The slope of the curve at any point indicates the viscosity of the curve.

Figure 1.3: Fluid classification (Shear strain vs shear stress).

1.6.1. Dynamic and Kinematic Viscosity Units It is necessary to know the units of viscosities and their equal values. The dimensions of shear stress is that of pressure, i.e. (1.20) where [X] denotes the dimension of the physical quantity X. Since

(1.21)

where M, L, and T represent the three basic physical dimensions of mass, length, and time. And (1.22) Hence, Eq. (1.20) becomes

Fluids

11

(1.23) Furthermore, the dimensions of the angular deformation rate is (1.24) Finally, the dimensions of the dynamic viscosity (Eq. 1.18) is given by (1.25) Therefore, the dimensions of dynamic viscosity is that of mass per length and time. Using the SI system, where mass is given kilograms (kg); length is given in meters (m); and time is given in seconds (s), the unit of dynamic viscosity is kilogram per meter second (kg/m•s) which is equivalent to Newton second per square-meter (N•s/m2) or Pascal second or (Pa•s). Using the CGS system, where mass given in grams (g); length in centimeters (cm); and time in seconds (s), the unit of dynamic viscosity is gram per centimeter second which is equivalent to dyne second per square-centimeter (dyn•s/cm2) or the “Poise” (P): 1 P = 0.1 Pa·s = 0.1 N·s/m2 = 0.1 kg/m·s The Centipoise (cp) is another commonly used unit for kinematic viscosity: 1 cP = 0.01 P = 0.001 Pa·s = 0.001 N·s/m2 = 0.001 kg/m·s. The dimensions of kinematic viscosity (Eq. 1.19) is (1.26) which is square-length per time. Thus, the SI unit of kinematic viscosity is square-meters per second (m2/s) and CGS unit is square centimeter per second (cm2/s) also known as Stokes (St), 1 St = 1 cm2/s = 10−4 m2/s. Another commonly used unit for kinematic viscosity is square millimeter per second (mm2/s) also known as Centistokes (cSt), 1 cSt = 1 mm2s = 10−2 cm2/s = 10−6 m2/s.

1.7. SURFACE TENSION Let us consider two fluids with different physical properties, fluids A and B. These two fluids are in contact with each inside a container, A is at the

12

Theory and Problems of Fluid Dynamics

bottom and B is at the top. They are not miscible. We would assume that their contact surface is flat, but it is not true in most cases. The contact surface is usually curved. The question is why. These fluids will have an intermolecular attraction. Let us consider the partition of the surface tension. At the center of the container, the molecules of the top layer of the fluid A will be attracted by the molecules below it. Above it, there are no molecules of fluid A since this is the topmost layer. Hence, the force of attraction from the top layer is diminished. Yes, there is the force of attraction from the molecules of fluid B. However, the force is not the same, it may be weaker or stronger than that from the molecules of the fluid A. The same situation repeats in the bottom layer of Fluid B at the center of the container. Depending on the resultant force the molecules are pulled up or pulled down. Now consider the situation at the edge, i.e., wall of the container, and the molecular force exerted on the top layer molecules in contact with the wall surface. Here, the net molecular force behaves differently than at the top center because there is no force from the side nor from the top exerted on the molecules in contact with the vessel’s wall. This means that the force exerted on the molecules in contact with the wall of the vessel is lower than that exerted on top center molecules. Ultimately the fluids are in equilibrium under the action of varying molecular forces. This causes the fluids to form a curved surface. This curved surface is called elastic skin. This phenomenon is called surface tension. The curved surface at the border of two fluids is called meniscus (Figure 1.4).

Figure 1.4: Surface tension between fluids A and B.

Fluids

13

Now, let us analyze the situation mathematically. Let us consider a small rectangular box around a portion of the junction of two curves as in the sketch of Figure 1.4. Furthermore, consider the equilibrium condition. To balance surface tension forces = pδ s1δ s2 2α R1θ sin(φ / 2) + 2α R2θ sin(φ / 2) (1.27)

where p is the radial pressure; α is surface tension per unit length; R1 and θ are corresponding radius and arc length of the curve KPL; R2 and 𝜙 are the corresponding radius and arc length of the curve MPN. Since KPL and MPN arc lengths are very small,

(1.28) By substituting the small angle approximations in Eq. (1.28) into Eq. (1.27), we get (1.29) Assuming that the surface is spherically symmetric, both radii are the same and Eq.(1.29) becomes

p=

2α R

R=R1=R2 (1.30)

In the case of cylindrical surface, one of the radius is infinite and Eq.(1.29) becomes p=

α R

R = R1 , R2 = ∞ or R = R2 , R1 = ∞

(1.31)

We can conclude that the surface tension depends on the radius of the meniscus. In large vessels, the surface of the liquid is curved only near the perimeter.

1.8. CAPILLARY ACTION Let us consider the case of a narrow tube where the diameter is significantly smaller compared to the length. In this case, the liquid will rise above or fall below a certain height along the tube to balance surface tension force (see Figure 1.5 and 1.6). This phenomenon is as capillary action.

14

Theory and Problems of Fluid Dynamics

Let us analyze this situation mathematically as illustrated in Figure 1.5. The liquid inside a capillary tube is subject to two opposing vertical forces, the surface tension acting along the meniscus perimeter,

= Fc α= Lp cos(θ ) α 2r cos(θ )

(1.32)

where Lp is meniscus perimeter, r is the capillary tube radius, h the liquid height level inside the capillary tube, α is the liquid surface tension per unit length, and θ is the contact angle;

Figure 1.5: Capillary rise in a tube (concave meniscus).

and the body force acting throughout the liquid volume due to gravity,

Fg = −mg = − ρ gV = − − ρ gπ r 2 h = −γπ r 2 h

(1.33)

where m, ρ, and V are the mass, density, and volume of the liquid inside the capillary tube, respectively; and g is the standard gravity Considering equilibrium condition, the force balance is 2

d  d  γπ   h = 2π   α cos(θ ) 2 2 (1.34) where d is the capillary tube diameter. Simplifying Eq. (1.34), we get

Fluids

h =

15

4α cos(θ ) 2 cos(θ ) = γd g ρ r (1.35)

Eq. (1.35) is equivalent the Jurin’s law. A contact angle (θ) lower than 90° means that meniscus is concave and liquid will rise in a capillary tube due to the surface tension (Figure 1.5). When the contact angle is greater than 90°, the meniscus becomes convex and liquid will fall in a capillary tube due to the surface tension as shown in Figure 1.6.

Figure 1.6: Capillary fall in a tube (convex meniscus).

The occurrence of a convex meniscus depends on the type of liquids in contact. Table 1.1 shows the surface tension of common liquids. At room temperature (20°C), the surface tension of Mercury is significantly high compared to other liquids such as water. As a consequence, Mercury falls in a capillary tube and forms a convex meniscus while Water rises and forms a concave meniscus. Table 1.1: Surface Tensions of Common Liquids Liquid Benzene (20°C) Blood (37°C) Glycerin (20°C) Mercury (20°C) Water (20°C) Water (100°C)

Surface Tension (N/m) 0.029 0.058 0.063 0.47 0.073 0.059

Theory and Problems of Fluid Dynamics

16

1.9. VAPOR PRESSURE Let us consider a liquid is stored in an open vessel exposed to the atmosphere. The liquid starts evaporating as molecules that are weakly bound together move to the atmosphere. This will continue for some time. At the same time, some molecules may re-enter the vessel. Initially, the number of molecules leaving the liquid is higher than the number of molecules re-entering the vessel. After a certain time, the rate at which the molecules leave the liquid is equal to molecules re-entering the vessel. A certain amount of molecules remains in the atmosphere. The presence of molecules in the atmosphere creates vapor pressure of the liquid and the air above the liquid surface becomes saturated. The vapor pressure is dependent on temperature. When the pressure above the liquid is equal to vapor pressure then boiling occurs. When the pressure of the liquid decrease to less than the saturated vapor pressure at a critical temperature, the liquid state becomes unstable. Vapor pockets begin to form and spread throughout the liquid. These vapor pockets are called cavitations. When the cavitations come in contact with high-pressure areas, they collapse. Suppose these rise to the surface of the vessel, the force of their collapse will damage the vessel. When water is heated, before boiling, it bubbles start forming from the bottom to the top of the vessel. Cavities are also formed and these burst at the top. Similarly, vortices in the river are called cavities.

1.10. EXERCISES Problem 1: The density of kerosene is 810 kg/m3. Find the relative density, the specific weight and the specific volume of kerosene. Solution: • • • • •

Density of Water = 1000 kg/m3. Relative Density of Kerosene = Density of Kerosene/Density of Water = 810/1000 = 0.81. Standard Gravity = 9.81 m/s2. Kerosene Specific weight = Density of Kerosene × standard gravity = 810 9.81 = 7846 N/m2. Specific Volume = 1/Density of Kerosene = 1/810 = 1.23 L/kg (Note: 1 m3 = 1000 L.)

Fluids

17

Problem 2: The air density is 1.2 kg/m3 at 20 OC and 1 bar. Find the specific weight of air and its relative density with respect to hydrogen. Note: The hydrogen density is 0.09 kg/m3 at 20 OC and 1 bar Solution: •

Air Specific weight = Air Density × Standard Gravity = 1.2 × 9.81 = 11.77 N/m3. • Relative density of air with respect to hydrogen: Air Density/Hydrogen Density = 1.2/0.09 = 0.108 Problem 3: The velocity distribution between two parallel plates is given by

Find the shear stress and velocity at: • y = 0, a = 20 mm, η = 8.5 P, (dyn.s/cm2), dp/dx = – 1500 N/m; • y = 5 mm, a = 20 mm, η = 8.5 P (dyn.s/cm2), dp/dx = –1500 N/m. Note: Solve the problem in SI Units. Solution: • • •

Conversion to SI units: 1 P = 1 dyn.s/cm2 = 0.1 Ns/m2→ 8.5 P = 8.5 × 0.1 Ns/m2 = 0.85 Ns/m2 1 mm = 0.001 m → y = 5 mm = 0.005 m/s; a = 20 mm = 0.02 m Shear velocity (|u|): u = -[1/(2 × 0.85)] × (–1500) × (0.022- y2) = 882.35 × (0.0004 – y2) = 0.3594–882.35 × y2. At y = 0, |u| = 0.3594 m/s; At y = 0.005, |u| = | 0.3594–882.35 × 0.0052 | = | 0.3594–0.02205 | = 0.3373 m/s. Shear Stress = Shear Strain × Viscosity = |du/dy| × η: |du/dy| = |0–882.35 × (2 × y)| = 1764.7 × y. At y = 0, |du/dy| = 0, Shear stress = 0. At y = 0.005 mm, |du/dy| = 1764.7 × 0.005 = 8.8235, Shear stress = 8.8235 × 0.85 = 7.5000 N.s/m2.

Theory and Problems of Fluid Dynamics

18

Problem 4: At a distance of 0.25 cm from a fixed plate, a second plate moves at a velocity of 2 m/s. A force of 45 dyn/cm2 is required to maintain the moving plate’s velocity. Determine the coefficient of viscosity of the fluid between plates. Solution: •

Shear Strain = Velocity Gradient = |du/dy| = 200/0.25 = 800 cm/s = 8m/s • Coefficient of Viscosity = Shear Stress/Shear Strain = 45/800 = 0.0563 P = 0.0056 Ns/m2. Problem 5: A plate weighing 150 N, measures 80 cm × 80 cm. It slides down an inclined plane over oil film of thickness of 1.2 mm. For an inclination of 30o, the sliding velocity is 20 cm/s. Calculate shear stress and viscosity. Solution: •

Shear Stress = Force/Area = 150 × sin (30o)/(0.8 × 0.8) = 117.2 N/m2. • Shear strain = Rate of Deformation = |du/dy| = 20/0.12 = 167 cm/s = 1.67 m/s • Viscosity = Shear Stress/Shear strain = 117.2/1.66 = 0.706 N.s/ m2. Problem 6: A liquid is compressed inside a cylinder of volume 0.4 L3 (cubic liters) at pressure required to compress is 6.8 × 10 7 dyn/cm2. The same liquid is compressed to 0.396 L3 at pressure of 1.36 × 108 dyn/cm2. Find the bulk modulus of the liquid. Solution: •

Bulk Modulus = – Change Pressure/Volumetric Strain = – (V/ dV) × dp: dp = (1.36 × 10 8)- (6.8 × 10 7) = (13.6–6.8) × 10 7 = 6.8 × 10 7dyn/cm2; dv = 0.4–0.396 = 0.004L3 = 4 cm3; V = 0.4L3 = 400 cm3; V/dV = 400/ 4 = 100; Hence Bulk modulus = – 100 × 6.8 × 10 7 = – 6.8 × 10 8 dyn/cm2. Problem 7: A Newtonian fluid has an angular deformation of 1 rad/s. when acted upon a shear stress of 0.4 Pa. Find the viscosity.

Fluids

19

Solution: Newtonian fluids follow linear relation between shear stress and shear strain as expressed in Eq. (1.16) to Eq. (1.19) • Viscosity = Shear stress/Shear strain: 1 radian = 180o/𝜋 Shear Strain ≈ tan (180o/𝜋) = 1.557. Thus, viscosity = 0.4/1.557 = 0.257 N.s/m2. Problem 8: Determine the pressure within a droplet of water of 1 mm diameter, while, the atmospheric pressure = 101 kN/m2. Solution: •

For spherical surface, pressure difference (𝛥P) = 2 × Surface Tension/Radius = 4 × Surface Tension/Diameter. According to Tab. 1.1, the surface tension for water at 20°C is 0.073 N/m. 𝛥P = 4 × (0.073/0.0001) = 2920 N/m2 = 2.92 kN/m2 •

Pressure Difference (𝛥P) = Pressure within the water droplet (Pin) – Atmospheric Pressure (Pout). Hence, Pin = 𝛥P + Pout = 2.92 + 101 = 103.92 kN/m2 Problem 9: The water level in a steel tank is measured with a piezometer of 5 mm diameter. The reading of the water level, in the tube, is 90 cm. Find what will be the water level reading in the tank. Solution: The water in the tube raises above the water level in tank due to surface tension. The height which a liquid rises inside a tube is given by Eq. (1.33). We can consider cos (𝜃) = 1. Hence,

h=

4α gρd • • • • •

Density of Water = 1000 kg/m3, Standard Gravity = 9.81 m/s2, Water Surface Tension = 0.073 N/m h = 4 × 0.073/(1000 × 9.81 × 0.005) = 0.292/49 = 0.006 m = 0.6 cm. Hence, water level in tank = 90–0.6 = 89.4 cm.

Theory and Problems of Fluid Dynamics

20

1.10.1. Problems to Solve 1)

2)

3)

4)

5)

6)

What is the bulk modulus of a liquid which have been compressed from a volume 0.0125 m3 at 80 Ns/cm2 to a volume 0.0124 m3 at 150 N/ cm2? (Answer: 8.75 × 102 N/cm2). Find the height of a capillary in mm in a glass tube of diameter 4 mm when immersed in a container with: (a) water; (b) mercury. The surface tensions of water and mercury are 0.0735 N/m and 0.510 N/m, respectively. The contact angle of water with glass is 0o and that of mercury 130o. (Answer: Water h = 7.495 mm Mercury = –2.458). Consider a cylinder with a diameter of 100 mm, length of 200 mm, and weight of 20 N that has fallen at uniform velocity U into an outer cylinder with a diameter of 100.5 mm. The kinematic viscosity of oil between cylinders is 6.7 × 102 m2/s. Find U knowing that the density is ρ = 880 kg/m2. (Answer: 0.382 m/s). If in the problem (3) cylinder falls at a velocity of 0.5 m/s and decelerates at rate of 14 m/s2. Find the weight of body. (Answer: = 9.05). The water at a nozzle of a Pelton wheel is under pressure head of 600 m. Find mass density of water at Pelton wheel for water bulk modulus of 2.05 × 109 N/m2. What is the error in density if compressibility ignored? (Answer: ρ = 1002.97, error = 0.2927%). Find the mass density of seawater at a depth of 10,000 m. Density of seawater is 1026 kg/m3. Bulk modulus = 2.05 × 109 N/m2. (Answer: ρ = 1076 kg/m3).

CHAPTER

2

FLUID STATICS

CONTENTS 2.1. Static Pressure And Variation Within A Fluid...................................... 22 2.2. Pressure Variation In Compressible Fluids.......................................... 25 2.3. Forces Acting on Surfaces By Static Fluids.......................................... 27 2.4. Pressure Center.................................................................................. 28 2.5. Force Analysis In Case of Curved Surface Submerged In A Liquid...... 30 2.6. How To Measure Pressure.................................................................. 31 2.7. Manometers....................................................................................... 32 2.8. Buoyancy, Law Of Floatation And Floating Bodies............................. 35 2.9. Relative Motion In Fluids................................................................... 36 2.10. Metacenter Height........................................................................... 40 2.11. Exercises.......................................................................................... 41

22

Theory and Problems of Fluid Dynamics

2.1. STATIC PRESSURE AND VARIATION WITHIN A FLUID In this chapter, we will deal with the forces acting on the fluid at rest. As aforementioned in the first chapter (Eq. 1.7), the pressure at point within a fluid can be mathematically expressed as the negative of derivative of the normal force (F) with respect to the area (A). Therefore, d Fn=- pnd A (2.1)

where pn is the pressure at a point and Fn is the normal force acting on the area A of the surface element in contact within the fluid. The pressure at a point within a fluid at rest is called the Static Pressure. First, let us consider the surface area and forces acting upon a wedge-shaped fluid element “abcefg” as shown in Figure 2.1.

Figure 2.1: Wedge-shaped fluid element.

Let the small wedge dimensions be (2.2) and (2.3) Let us assume that the forces acting on the three faces of the wedge (“fecg”, “fabg” and “abce”) are due to the pressures px, pz, and pn, respectively.

In a fluid at rest, there is no shear stress. Thus, the balance of the horizontal and vertical forces acting on the wedge are respectively and

(2.4) (2.5)

Fluid Statics

23

where θ is the acute angle of the right triangle “fae” in Figure2.1, and δV is the wedge volume. Since (2.6) Eq. (2.4) reduces to px=pn (2.7)

and if we also take the limit as δV approaches zero, Eq. (2.5) reduces to py=pn (2.8)

Combining Eq.(2.7) and Eq.(2.8),

px= py=pn (2.9)

Therefore, the static pressure is equal in all directions. This is also known as the Pascal’s Law. Now, let us consider a differential element of fluid at rest as shown in Figure 2.2. Pressure increases in the X direction, Y direction and Z direction in equal proportions. Let the pressure on the faces [a1, a2, d2, d1] and [b1, b2, c2, c1] be (2.10) the pressure on the faces [a2, b2, c2, d2] and [a1, b1, c1, d1] be (2.11) the pressure on the faces [d1, d2, c2, c1] and [a1, a2, b2, b1] be (2.12)

Figure 2.2: Differential element of a fluid.

24

by

Theory and Problems of Fluid Dynamics

Thus, the balance of the surface forces acting in the X-direction is given pL dAL − pR dAR = 0,

δ d dx  δ d dx    dydz −  p − dydz = 0,  p− δ x 2  δ x 2    δd dydz = − 0, δx δp = 0. δx (2.13)

Similarly, in Z-direction, we have pB dAB − pF dAF = 0,

δ p dz  δ p dz    dydx −  p − dydx = 0,  p− δ x 2  δ x 2    δp dxdydz = − 0, δz δp = 0. δz (2.14)

Thus, the static pressure does not vary in X-direction (length) or Z-direction (width) as

δp δp = = 0 δx δz (2.15) In the Y-direction, besides surface forces, we also have the body force acting on the volume of the differential element of fluid due to gravity

= dFg g= ρ dV γ dxdydz

(2.16) Thus, the balance of forces in Y-direction is

pD dAD − pU dAU = 0,  δ p dy  δ p dy   dzdx − γ dxdydz = 0,  p−  dzdx −  p − δx 2  δ y 2    δp +γ = 0, δy δp = −γ . δy

(2.17)

The static pressure only varies in Y-direction. Furthermore, the pressure gradient is equal to specific volume as δp dp =−γ → =−γ δy dy (2.18)

Fluid Statics

25

Assuming the density remains a constant, this enables to write p y δp dp =−γ → =−γ ∫ dp =γ ∫ dp p y δy dy p1 − p0 = −γ ( y1 − y0 ) (2.19) 1

1

0

0

Thus, static pressure difference between two points in an incompressible fluid is proportional to the vertical distance between the two points. This is generally used to determine the pressure acting in the atmosphere (or in a gas volume) at given altitude, or height (h),

= p pa − γ h

(2.20)

and in a liquid volume (such as the ocean) at given depth (h),

= p pa + γ h (2.21) where pa is the pressure at the vertical datum.

In the case of a liquid volume, the commonly chosen datum is the free surface (h is measured from this point down); while, in case of the atmosphere, this is the sea level (h is measured from this point up). The positive sign in Eq. (2.21) indicates that the pressure increases with the depth and the negative sign in Eq. (2.20) indicates that the decreases in the altitude. The density of water is 1000 kg/m3 at ambient temperature, hence, the pressure difference (p -pa) is 1000 × h. When h = 10 m, the pressure difference is 10000 kg/m2 = 1 kg/cm2. This is approximately the atmospheric pressure in the metric system which is normally used in fluid dynamics calculations by field engineers. Thus, we may conclude that one atmospheric pressure (1 bar = 1 kg/cm2) is due to the linear pressure of a 10 meters water column. For example, a 1-meter water column exerts a pressure of 0.1 bar (0.1 kg/ cm2).

2.2. PRESSURE VARIATION IN COMPRESSIBLE FLUIDS As aforementioned, the compressibility of gases is negligible compared to that of liquids. Let us consider the following gas law: (2.22) where P is the gas pressure, Vs the specific volume, R is a gas constant and T0 is a fixed gas temperature.

Theory and Problems of Fluid Dynamics

26

Combining Eq.(1.4) and Eq.(1.8), we can express the specific volume as function of the specific weight: (2.23) From Eq. (2.22), we have (2.24) From Eq. (2.19), we know that (2.25) Therefore, (2.26) Integrating Eq. (2.22), we get

(2.27) Therefore, (2.28) If we choose the sea level as vertical datum, h0 = 0 and Eq. (2.24) is simply (2.29) It has been found that temperature decrease linearly with the altitude. At given altitude, temperature can be mathematically expressed as T= T0 − β h, dT = − β dh, dh = −

dT

β

.

(2.30)

where β is a constant. In this case, Eq. (2.22) is given by dp g dT = ± β

(2.31)

After integrating Eq. (2.27), we obtain

Fluid Statics

27

 p g T ln   = ln p R β T 0 0  

(2.32) where p0 and T0 are the temperature at sea level. As final result, we have

p  T0 − β h  =  p0  T 0 

g Rβ

(2.33)

Given the temperature and pressure at sea level, we can determine atmospheric pressure at an altitude h.

2.3. FORCES ACTING ON SURFACES BY STATIC FLUIDS We have proved that the pressure in a static fluid varies with depth. Now, let us analyze how and what amount of total force acts on a solid object submerged in a fluid. Here, we will use liquids as primary examples because the analysis of static fluids is generally applicable to any liquid (Figure 2.3).

where, D-D = liquid datum top surface; b = width of the lamina at a depth h; dh = elemental thickness of the lamina; h1 = depth of lamina upper edge; G = lamina center of gravity;

H = depth of the center of gravity of the lamina; p = liquid pressure acting on the lamina; h2 = depth of lamina lower edge; Figure 2.3: An irregular lamina fully submerged in a liquid.

Theory and Problems of Fluid Dynamics

28

Let us analyze the case of an irregular lamina (a two-dimensional planar closed surface) submerged in the liquid shown in Figure 2.3. The total force acting on the entire surface of the lamina is = ∫ dF A

pdA ∫ ∫= A

h2

h1

pbdh

(2.34)

From Eq. (2.21), we known how to determine the pressure at given depth, Eq. (2.34) reduces to h2

F= pa A + γ ∫ ( pa + γ h)bdh = h1

( ∫ bhdh)  h2

h1

first moment of surface area

(2.35)

The integral on the right side of the Eq. (2.35) is the first moment of surface area about vertical axis (h- direction), this is the depth of the lamina center of gravity (H) multiplied by the total area. Thus, the total force acting on the lamina is F =pa A + γ HA =( pa + γ H ) A =pc A

(2.36)

where p= pa + γ H c

(2.37)

Recalling Eq. (2.21), pc is the pressure acting at the center of gravity of the lamin. Furthermore, this central pressure equals the total pressure (i.e., P = F/A).

2.4. PRESSURE CENTER Figure 2.4 shows the vertical section of a river dam at the gate. The surface force due to water pressure acts on the gate that is suspended at the top by a rope. In order to close the gate, a force must be applied from the opposite side of dam, as indicated by a big arrow mark on the right side of the Figure 2.4., at a certain depth (C) from the datum. If force is applied at a point higher than C, the gate opens at the bottom, below C it opens at the top. P is the pressure acting from opposite side to close the gate. w = specific weight of water in the dam. Let us take the moment of force P on the datum = PxC. Moment of force on the strip = h x(whxdh). Thus, the moment of the whole area of gate under area on which force acts, = w ʃh1h2 (bh2)xdh;

Fluid Statics

29

= w x (Second moment of area Datum) = wx moment of inertia = wxI0. I0 = Ig + Axk2 where k is the distance of CG of the gate from the datum. A = area of gate. Ig = Moment of inertia at the center gravity of the gate on the horizontal axis. Moment of force to resist PxC; P = Equal to total force of liquid in opposite direction to hold the gate in position = wxAxk where k = distance of cg from the gate. Note point G in the sketch & k, cg of gate is different. With this, (wxAxk)xC = wxI0; C = I0/(Axk) C is called center of pressure, cp; Thus C = [(Ig + Axk2)/(Axk)];

where, C = hinge distance from the datum h1 = height at the top of the gate. h2 = height at the bottom of the gate. dh = gate’s element thickness h = depth of this strip. b = width of gate. A = area of the gate Figure 2.4: Vertical section of river dam at the gate.

30

Theory and Problems of Fluid Dynamics

2.5. FORCE ANALYSIS IN CASE OF CURVED SURFACE SUBMERGED IN A LIQUID

Figure 2.5: Curved surface submerged in a liquid (Force/pressure analysis).

We have analyzed the pressure acting on a plane surface when submerged in a liquid as in Figure (2.3) and we defined center pressure as in Figure (2.4). The analysis in case of a curved surface the analysis is to be done in a different method. We use the method used in statics in applied mechanics. As in Figure (2.4), we consider a curved surface submerged in the liquid. The normal force is acting at each points as shown f1,f2, and f3. Thus the curvature is different at surfaces. That is why we have to consider the normal forces at many points on the surface. These normal forces are resolved into two components f1h f1v, f2h f1v, f3h f3v, etc. as Horizontal components and vertical components. Then sum them all, We get Fh and Fv respectively. To get Fh and Fv we need to sum up all the horizontal and vertical components. Considering elementary strip and force on this as constant and

= f1h wh = dh, f1v wdv. (2.38) Since the surface is in equilibrium, it follows that

Fx = A ' y γ hg

(2.39)

where Fx is the resultant of the horizontal forces and A’y is projected area on vertical plane; Now we can treat the projected area as a plane lamina and calculate the center pressure and Force acting on the surface on account of by the liquid pressure

Fluid Statics

31

e is in equilibrium, it follows that

Fy = A 'x Fg

(2.40)

where Fy is the resultant of vertical forces, A’x is horizontal projected area and Fg the weight of liquid acting on it. This is how we have to analyze. We will be able to understand from the worked examples in the succeeding section.

2.6. HOW TO MEASURE PRESSURE There are different types of measuring pressure. Let us see how? We need to understand with sketches. Piezometer tube (Figure 2.6)

Figure 2.6: Piezometer tube submerged in a river.

The above sketch Figure (2.6) shows a section of river. The long narrow tube of L shape is called piezometer. Arrows show the direction of flow. The piezometer inserted vertically with L shape facing against the flow. Then h indicates total pressure including static head. The value vh indicates the pressure due to velocity. The same piezometer can be used to measure the pressure inside a lake, tank, or any static fluid. Once we know the head (h), then the pressure is by p = γ h (2.41)

32

Theory and Problems of Fluid Dynamics

2.7. MANOMETERS U-Tube manometer: Figure 2.7

Figure 2.7: U-tube manometer.

The above sketch is the schematic of U-tube manometer. One limb is normally shorter. The shorter is connected to the source whose pressure is to be measured. The tube is filled with a heavy liquid such as mercury and also has certain desired properties. This liquid should not mix with the source liquid and as well it should be chemically inert. The rise in mercury level, h is measured. The pressure p = wxh; where w is the specific weight of liquid filled in the tube. Thus we have discussed the method of measuring the pressure of a static liquid and also estimating pressure in a flowing liquid. There are many methods of measuring pressure of a liquid. One of them is inverted U-tube (Figure 2.8). The U-tube has equal limbs. Each limb is connected to two different sources where we need to find pressure difference. The sketch give below illustrated the method. The height h dive the pressure difference.

Fluid Statics

33

Figure 2.8: Inverted U-tube manometer.

These are the low-pressure measurement. When we want to measure high pressure in the order of 20 bars, 50 bars 100 bars, etc., separate pressure gauges are manufactured. These gages read the pressure directly. When we want to measure pressure which is greater than atmosphere, such gages is called pressure gauges. When we want to measure pressure below atmosphere, special gages are used. These games are called vacuum gages. The detail design and construction is not in the scope of this book. The readers may refer catalog of gage manufacturers. Inclined Manometer (Figure 2.9):

Figure 2.9: Inclined manometer.

Refer to the Figure 2.9. This is schematic diagram of an inclined manometer. When we wish to measure very small differences of pressure, we use this type of manometer. The limb of the manometer is inclined at an angle. As result of it, a larger reading or larger differences in the level

34

Theory and Problems of Fluid Dynamics

of manometer fluid in different limbs are produced for a small difference of pressure. The magnification can be readily seen as = {Δh/Sin θ}. Δh is the pressure head if the tube were to be vertical. The magnification can be adjusted to the desired level by setting the angle of inclination. Manometer with enlarged ends: The Figure 2.10(b) illustrates the principle of working of a manometer with enlarged ends. This is used when a very low pressure is to be measured with a high degree of accuracy. The magnification is higher than that of inclined manometers. The heavier gauge liquids fills the lower part of U-tube manometer. Let us say initial level at O-O. Lighter liquids are added to both sides of larger reservoirs say up to 1–1. Gas or liquid in the system under measuring fills above 1–1. Let us say the pressure at C is slightly greater than at D. The gauge fluid moves as shown in the Figure 2.10(b). The deflection is R. The volume of liquid moved in each reservoir is δx A = ax (R/2) where A is the section Area of reservoir and as is the section area of the tube. The equation for the movement of liquid and steady state is as below: Pc + w1, w2, w3, w4, and w5 the specific weight of liquids as indicated in the Figure 2.10. Let us say w2 = w4 = w1 = w5, Then we have (pc-pd) = Rx[w3 – w2x{1-(a/A)} – w1x(a/A]. (h1 + δ) xw1 + {h2 – δ + (R/2)} x w2 – w3 xR – (h2-R/2 + δ) w4 – (h1 – δ)xw5 = Pd

Figure 2.10: Manometer with an enlarged area.

The expression [w3 – w2x{1-(a/A)} – w1x(a/A] is gauge constant. Hence pressure difference is directly proportional to R. These types of manometers are called micro manometers. There are different brands in the market.

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35

2.8. BUOYANCY, LAW OF FLOATATION AND FLOATING BODIES When a solid object is floating or submerged in liquid and it is in rest, then the object is in the state of equilibrium. Then the object is acted upon by two forces, one vertical force due to liquid pressures. Besides the object also has its own weight. All these force keep the object at rest. Let us analyze these force mathematically. Look at the sketch below showing a floating object in a liquid. The small elemental vertical strip has an area – a, Let it be a rectangular prism, height perpendicular to paper space. The downward force = pxa; Upward force = pxa + wxaxH. If we apply this principle to the entire body instead of elemental strip, Then pxAwexHxA = pxAxw where A is the total area of the object/body perpendicular to pr. acting. The small elemental vertical strip has an area – a, Let it be a rectangular prism, height perpendicular to paper space. The downward force = pxa; Upward force = pxa + wxaxH. If we apply this principle to the entire body instead of elemental strip, Then pxA + wxHxA = pxAxw where A is the total area of the object/body perpendicular to pressure acting (Figure 2.11).

Figure 2.11: Floating object in a river.

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36

Suppose the object has weight, B, and it is floating and the object is at rest, Then pxA + wxHxA = pxA + B; It follows that B = wxHxA; It means the weight of floating body is equal to the weight of liquid displaced. This is called law of floatation. Equilibrium conditions of floating bodies: Suppose a body is floating in a liquid. It will have three status of equilibrium. Stable, Unstable, and Neutral If we give a small impulse to the floating body, then after that if the body oscillates and comes back to its original position, it is called stable equilibrium. If after giving a small impulse it heels further over, it is called unstable equilibrium. In case after it occupies a new position it is called neutral equilibrium.

2.9. RELATIVE MOTION IN FLUIDS a)

Fluid subjected to horizontal force: Let us consider a liquid as shown in the Figure 2.12. The liquid is in the vertical position initially. The liquid level is A-A. The force in vertical position is gravitational force. This gravitational is balanced by the pressure developed and thus the liquid is in equilibrium. Let the liquid is subjected to horizontal force Fx(fx is the acceleration). The liquid moves due to application of force. Then the liquid level takes a different position A1 – A1 and comes to rest. Now in this is the position the gravitational force and external horizontal force together with forces developed internally keeps the liquid in equilibrium. The entire liquid behaves as one solid. The analysis is similar to forces acting on solids. Under the action of external force, the liquid is in equilibrium. Hence all the forces should balance. Let us analyze mathematically.

Figure 2.12: Liquid under acceleration and subjected to an horizontal force.

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37

Consider a rectangular prism of liquid as shown. L = length. h1 and h2 are heights of liquid column from A1-A1 at the ends of liquid prism. Force acting on left = wxh1 x a and on the right = wxh2xa.(Note pressure p1 from left = wxh1 and pressure from right,p2 = wh2) The gravitational force is the vertical force and prism area/thickness is very small it balanced by weight of liquid. Now the horizontal force = ρxgxaxl x f ρxgx(h1 – h2)xa = ρxaxlxfx (w = ρg, hence substitution to w); With this we arrive at {(h1 – h2)/L} = (fx/g) = Tan θ Apply the analysis for the entire liquid, Then H/L = (fx/g) = Tan θ b)

Fluid subjected to Vertical force (Refer Figure 2.13):

Figure 2.13: Fluid subjected to a vertical force.

Let us say container is subjected to a vertical force (acceleration), fy as in the Figure 2.13 Horizontal force/ acceleration is zero. For equilibrium, we have

38

Theory and Problems of Fluid Dynamics

(pxdA-ρxhxdA) = [ρhx(dA/g)]xfy.------(1)

It indicates the mass (hxdAxρ), vertical column of liquid subjected vertical force, i.e., The force = Mass x Acceleration Force = (hxdAxρ) x fy. Hence we say the liquid column is subjected to acceleration. On simplifying (1) we get p = ρxhx[1 + (fx/g)]. Thus a vertical force will cause the pressure to increase by a factor = (fx/g). Acceleration along an inclined plane: Refer Figure 2.14

Figure 2.14: Liquid acceleration along an inclined plane.

The liquid surface is initially inclined at an angle α to the horizontal. Then a force acceleration is applied in the direction of α. Then the liquid takes the position as shown in the Figure 2.15. It is tilted by another angle θ. Then it can be proved that Tan (θ) = [{a × Cos(θ)}/{g + a × Sin (θ)}], where a = acceleration due to application of force. Rotation Cylinder: Another important practical problem is related to a fluid mass moving as a solid body under action of an external force is that of the uniform rotation of liquid contained in a container about the vertical axis. This is called forces vertex.

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Figure 2.15: Liquid column under rotation.

Study the Figure 2.15 which shows the liquid column is under rotation about a vertical axis. Let us analyze the forces acting on a liquid element PQRS. pxrxdθxdh-[ (p + (δp/δr)x(r + dr)x dθxdh = ρxdhx{r + (dr/2)}x dθx(-ω2xr); Note ω = angular velocity.(radians/s) (Note: We use “δ” for pressure and r because they are the partial derivatives. i.e., they are also function of other variables, such as “θ” etc. Neglecting small quantities and simplifying, We get dp/dr = (ρ/g)x (ω2xr); Remember p = ρgh; On integration, p = [(ρx ω2xr2)/2g] + Constant The value of constant is obtained in terms of pressure, if p = p0 r = 0

We get p = p0 + [(ρx ω2xr2)/2g]; In terms of head, h = h0; h = h0 + [(ω2xr2)/2g]

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Theory and Problems of Fluid Dynamics

2.10. METACENTER HEIGHT Refer Figure 2.16. Consider a section object PQRS submerged in a liquid. The object is at rest under the action of its own weight acting vertically downward against gravity and Buoyant force (Buoyancy). Let an impulse is applied to the object. The object will tilt and occupies a new position, P’Q’R’S’ as shown in the Figure 2.16.

Figure 2.16: Metacenter height.

Initially, G is the CG of the object. B is the center of buoyant force. The two are in the same line and hence object is stable. After the tilt, the position of CG and buoyant force are G’ and B’. The new position of buoyant force is located at B’. Now if the vertical line through B’ is drawn and extended upward and also vertical line through G’. These two line meet at a point M as seen in the Figure 2.16. This point is called metacenter. The position of M has an important significance. If the position M is above G (CG of object), the two forces, buoyant force and object weight constitute a couple which tries to restore the object in the original position. This is the stable position. If the position lies below G then the couple rotates the object further and thus make the object unstable. Ultimately the object would topple. When the position M and G are in the same line the body remains in the same position to which it is tilted. This is called neutral position. The position of metacenter plays an important role in shipbuilding. The design 1s ship is to be such that the position of M is always above the G. (Cg of the ship). The distance GM is called metacentric height.

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Analytical method of determining Metacentric height: The shift of the Buoyancy BB’ designated as x0 is found by considering two wedges shown as hatched. These two wedge forces are equal to weight of the object. Let the forces be ΔF. We have Wxx0 = ΔFx a (restoring couple) Total couple ΔFx a = ρxʃ(rxdθxdrxds)r = ρxʃ(r2xdA) = ρxθxI; drxds = dA and I is the moment of inertia of the area about the axis yy; Wx0 = ρxVxx0 = ρ θI V = Total volume of liquid displaced. From the geometry, we get MBxSin (θ) = MBx θ; since θ is small equal to Sin (θ); MB = (x0/ θ) = I/V.

2.11. EXERCISES Problem 1: Consider a triangular plate ABC (with AB = BC) immersed in water, as shown in the figure below. The base AB is horizontal. The plate makes an angle 60o with horizontal. All required distances are given in the figure. Determine the magnitude and the point of application of the force to water pressure on the triangular plate.

Solution:

Force on the plane = Pressure at the centroid x Area of the plate; Centroid = 2.5 + [(GD)] --Centroid = 2.5 + 0.866 = 3.366. m; Total force on the plate at centroid = (wh)x Area = 9806 x 3.66 x 3.0 Area = (0.5xACxBD); w = 9806 N/M3; Total force on the plate at centroid = [9806x3.366x(0.5x2x3.0)] = 99021 N. Moment inertia on G aixis = Ixg = (1/36) x2x 33 = 1.5

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42



hp (Vertical depth of P) = 3.366/Sin(60) = 3.887 m; Area 3 SqM; Center of pressure Depth = [1.5/(3.887x3)] + 3.887 = 4.015 m (Ans) Problem 2: An elliptical gate covers a circular pipe of 4-meter diameter. If the gate is hinged at the top, find the normal force required to open the gate when the water depth is 8 m. The gate is open to the atmosphere on the other side as shown in the figure below.

Solution:

Area elliptical gate = pixaxb = 3.14x2x2.5 = 15.7 M2 Force on the elliptical gate = wxhgx A = 9810x10x15.7 = 1540951 N Center pressure: hp = [I/(hgxA)] + hg = [(0.25 x 3.14x23x2.5)/(hg’x 3.14x2x2.5)] = [{(0.25x4)/hg’} + hg’] hg’ = Slant height = 12.5 M. Hence hp = 0.125 + 12.5 = 12.625 M From surface. But from the hinge, the center of pressure is located at a distance of 0.125 + 2.5 = 2.625. Taking moments Fx5 = 1540951x 2.625; F = 808999 = 809 KN (Ans). Problem 3: Calculate the magnitude and direction of the resultant force on the semicircular gate as shown in the figure below. The gate’s width is 3 meters.

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Solution:

The total horizontal force on the gate = Fx; This is the force on the vertical projection of the circular gate Area of projection 0.5x3 = 1.5 m2. CG of the area is at 0.75 + 0.25 = 1 metre. The magnitude of the horizontal component p Fx = wxhgxA, hg = Depth of CG. Fx = 9806x1x1.5 = 14715 N Fx acts at center of pressure,yp from the top surface of water. Yp = [{(bd3)/12}/(bxdxyg) ] + yg; Hence yp = [{(0.5)2/(12x1)} + 1] = 1.0283 m below the surface. Vertical components, Fy = Downward force on the Arc AC + Upward force acting of the Arc BC Vertical components, Fy = Weight of the liquid volume ECBAD + Weight of the liquid volume ADECA Vertical components, Fy = Weight of the liquid volume in the portion ACBA. Fy = 9806x3x3.14x(0.5)2/8 = 2888.1 N. This force acts at centroid which is 0.212d from the diameter. Resultant force = [Fx2 + Fy2]1/2 = [147152 + 2888.12]1/2 = 14990 N Angle of force with horizontal = Tan –1 (Fy/Fx) = 10.9 degrees. Problem 4: A rectangular box (dimensions: 2 m long, 1 m wide, and 3 m deep) is half filled with water and accelerating in the horizontal direction at 4.0 m/s2. Find the slope of the free surface and its location. Find the total pressure on the opposite sides of the wall and find its relation to the force

Theory and Problems of Fluid Dynamics

44

exerted on the fluid mass due to the acceleration (refer to Figure 2.12 if required). Solution: ×

Slope θ = Tan –1 (4/9.81) = 22.2 Deg. (fx = 4m/s; g = 9.81 m/s2) Let s = depth of water. at the front face. Then the depth at the rear face = s + 2xTan (22.2) s + 2xTan (22.2) = s + 0.41x2 = s + 0.82. The volume of water is the same before and after acceleration. We have (s + 0.82)x2x1 = 2x1x1.5 (Box is half filled, Hence 1.5); 2s + 1.64 = 3; s = (3–1.64)/2 = 0.68 m. Total pressure on the rear face = is wx(1.68)2/2 = 1.411xw. The total hydrostatic force on the front face = wx(0.68)2/2. = 0.23xw Force causing acceleration = Mass of the liquid in the boxx acceleration. = ρx2x1x1.5x4 = 12xρ Problem 5: A cylinder of 1 m diameter and 1.4 m height is filled with water up a depth of 0.75 m and is rotated around its central axis with a constant speed of 90 rpm. Determine: (a) (b)

depths of the liquid at the walls and at the center; the total pressures on the cylinder walls and on the bottom of the vessel; (c) the rpm required for the water level to only reach the top of the vessel; (d) the rpm required to attain a zero depth at the center as well as the amount of water spilled over the container’s brim. Refer to Figure 2.15 if necessary. Solution: •

(angular velocity rad/s), ω = 2xpixN/60 = 2x3.14x90/60 = 9.42 rad/s • hc = ω2xa2/4g = 9x3.142x0.52/(4x9.81) = 0.5665 m. Below the original level and depth at the wall is also above the original level at the wall by the same amount or the total depth = 0.75 + 0.5665 = 1.3165 m. Depth at the center = 0.75–0.5665 = 0.1835 m. The force on the outer wall of the cylinder is forced on the vessel projection of the curved surface equal to (ρxhw/2)xd = 0.866xρ.

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The total pressure on the bottom will remain the same as the one when the liquid is at rest at uniform depth of h = 0.70 m which is easily estimates as pixa2x(ρxh) = 0.55xρ; If the wall depth should reach the level of the top of the vessel, the rise in the level above the original depth is 1.4–0.75 = 0.65 which should be = ω2xa2/4g; Hence ω = [(0.65x4x9.81)/0.25]1/2. Simplifying ω = 10.096 or N = 96.40 RPM. The depth at the center is 0.75–0.65 = 0.10. Any further increase in speed of rotation will cause further decrease in the level at the center and overflow of liquid over top of the container. If the depth at the center has becomes zero, the depth at wall being 1.4 m, we equate ω2xa2/2g = 1.4 With this N = 100.05 RPM. The amount of overflow is pix(0.5)2x(0.75–0.7) = 0.0392 Cu m. Problem 6: A (30 cm diameter and 0.25 cm deep) closed cylinder is completely filled with water, rotates around the vertical axis at 240 RPMs. Calculate the total water pressure at each end. Solution: •



Consider a vertical thin hollow cylinder of water of radius, x and thickness dx • Centrifugal head on thin cylinder = [(ω2xx2)/2g]; • Hence intensity of pressure at radius x = px = wh = [wx(ω2xx2)/2g]; • Although this centrifugal pressure is horizontal it will also act vertically on top and bottom of cylinder, as pressure is transmitted equally in all direction. r = radius of cylinder in question. Total pressure on top and bottom of cylinder due to centrifugal pres-

sure, = ∫

r

0

px * 2* pi * x * dx

= ʃ[wx(ω2xx2)/2g] x2xpixxxdx (limit 0 to r) = [wx ω2 xPi)/g] xʃ x3 x dx

2xpixN/60 = ω; ω = 25 rad/s; r = 30/2 = 15 cm; g = 981 cm/s2 w = 1 gram/ cm2 = 9810 N/m3 = [wxpix ω2 xr4/(4xg)] = 27.3 Kg force = 260 N = total pressure on top of cylinder.

Theory and Problems of Fluid Dynamics

46

Total pressure on bottom = Centrifugal pressure + weight of water = 60.1 + (pixr2xdepthxw) = 262 N Problem 7: A 2 m diameter and 1.2 m long cylindrical buoy weighing 9806 N, floats in the water. Find the center of buoyancy and metacentric height. Refer to figure below.

Solution: × The CG is located at the center of the buoy = 0.6 m from each end. The submerged depth h. •

We have wx[pixd2/4]xh = 9806. By laws of floatation. w = specific weight of water., hence h = 0.318 m and V = [pixd2/4] xh = (3.14x22/4)x 0.318 = 0.998 cu. m Say 1 cu.m V = volume of liquid displaced. • Center buoyancy = 0.318/2 = 0.159 m below the surface of the water. • BM = I/V = [(3.14xd4/4)/V] = 0.785 m; where I = moment of inertial on yy axis of cylinder = ie. MI on axis plane as seen from top. BG = 0.6–0.159 = 0.441; GM = 0.785–0.441 = 0.344. M is above the G the body is stable. Problem 8: A, 1.5 m diameter and 2 m height, cylindrical buoy weighs 700 kg with its CG 0.8 m above the base S. Is the buoy stable when floating with its vertical axis in the seawater? Suppose that the buoy is anchored to the bottom by a chain attached to it. Find the chain tension required for the

Fluid Statics

47

buoy to float stably. The seawater density is 1020 kg/m3.

Solution: × Without anchor chain the depth of submersion of the buoy is = h = 700x4/(3.14x1.52x1020) = 0.388 M; Then BG = 0.8-(0.388/2) = 0.606 M; BM = I/V = [ (3.14x1.54/64)/ {3.14 x (1.52/4) }] = 0.362. Cg = 0.8 M Hence the buoy is unstable. When the anchor chain exerts a tension T on the body, the submerged depth has to be found. For equilibrium, 700x9.81 + = 1020x9.81x3.14x1.52/(4h1) = 1802.49x9.81xh1 T = (1802.49)xh1–700)x9.81. -------(1)

Let R be the location of the resultant of two forces W and T; Then we have 700x9.81 xRG = Tx(0.8-RG) ------(2) eliminating T from the relations We have T = 700x9.81 x[RG/(0.8-RG)] Simplifying, RG = 0.8–0.311/h1; --------(3)

Now m has to coincide with R, for a minimum stability condition. Then we have relation [0.8-RG-(h1/2)] = BM = I/V = [(3.14x1.54/64)/ (h1x3.14x 1.52/4)];

Theory and Problems of Fluid Dynamics

48

RG = 0.8 – (0.141/h1) – (h1/2); RG is also = 0.8–0.311/h1 --- from (3) Equating, we can solve for h1; h1 = 0.583

Now tension, T = (1802.49)xh1–700)x9.81 ---- From (1) Hence T = (1802.49)x0.583–700)x9.81 = 3440 N Problems to Solve 1.

2.

3.

4)

The air supplied to the gas engine is measured by drawing the air into a large chamber through a small orifice. If the pressure difference between the air outside and in the chamber is [45.5 cm of water], find the velocity at which the air flows through the orifice. The atmospheric temperature atmosphere is 18°C. A barometer reads 73.5 cm of mercury. The weight of 28 of air, at zero C and pressure of 76.0 cm of mercury, is 39 grams (Ans: 686 cm Ft/se) A cylindrical arm full of water rotates in the horizontal plane at 100 RPM at one end. The arm is 61 cm long and has a 5.0 cm diameter. Find the centrifugal head impressed on water and the total pressure on the outer end of arm. (Ans 2.07 M of water, 4.2 Kg) The U-Tube, shown in the figure below, rotates around the Y-axis. Find the difference between the level of the two limbs of the tube and the original level in the tubes. The distance from the vertical axis of the two tubes are as shown in the Figure 2.17 cm and 30 cm. (Ans = 6.44 cm as shown in the Figure 2.17.)

Determine the levels in the piezometers A and B. What are the pressures at C and D.

Fluid Statics

49

Shown in Figure 14 (l):

(Ans Level in A is 0.5 m above orifice. Level B 1.5 M above orifice. Pressure at C atmosphere. Head over the orifice is constant at 0.5 M, Pressure at D = –1M) 5)

A formwork holds slurry as shown in the Figure 14-(m). find the tensions in the top and bottom ties. (Ans Bottom 120 KN, Top 60 KN)

6)

A tank containing oil of specific gravity, 0.8 is accelerated horizontally at 3 M/s2. Find the surface slope and the pressure intensities at points A and B shown in Figure 14-(n). What is the acceleration is necessary to just expose point A? Find the depth at B when it happens (Ans. Depth at A = 1.5636 M and at B = 1.4364 M)

Theory and Problems of Fluid Dynamics

50

7)

A tank measuring 3 M long 2 M wide and 3 M deep is accelerated at 3 M/s2 up a slope of 15 Deg. Find the surface slope and the depths at the front and rear ends, if the water depth is initially 1 M (Ans. = 0.0125 M and 1.875 M; Angle 15.31 Deg) 8) A cylindrical block of wood 2 M in diameter and 1.75 M long has a specific weight of 8000 N/m3. (a) Will it float in stable condition with its axis vertical? (b) If the diameter is 1 m, is the block is stable? (Ans. BG = 0.16 1 M, BM = 0.175 M, Therefore stable (b) BM – 0.0766 M Not stable) 9) A hollow cylinder opens at both ends has an internal diameter of 25 cm and a wall thickness of 10 cm and a length of 100 cm. If the weight of cylinder is 600 N, find the stability of floating cylinder. (Ans.: Unstable) 10) Mercury fills a thin tube 0.5 cm in diameter and 0.5 M long. It is closed at the outer end. If the tube is rotated at a constant speed 125 RPM, in a vertical plane. What is the maximum force exerted on the closed end. (Ans force = 6.58 N) [Hint: Use equation δ(p + ρz)/δr = an/g = ω2xr/g] 11) A wooden cone of base diameter 1 M and height 1.5 M. The specific gravity is 0.6 floats in water. Is it stable? (ANS: Stable OG = 1.125; M OB = 1.054 M,; OM = 1.159 M). (Figure 14(p))

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CHAPTER

3

KINEMATICS OF FLOW

CONTENTS 3.1. Flow Variables And Classifications..................................................... 54 3.2. Steady Flow....................................................................................... 54 3.3. Uniform Flow.................................................................................... 55 3.4. Elements Of A Particle In Motion....................................................... 60 3.5. Equation Of Continuity...................................................................... 62 3.6. Exercises............................................................................................ 65

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54

3.1. FLOW VARIABLES AND CLASSIFICATIONS We have established in the previous chapters, the fluid parameters are direction, velocity, density, pressure, and temperature. At given instant of time, the behavior of fluids is characterized by these parameters. Furthermore, it is also known that these parameters change with time and their position in space. We can represent these parameters variation mathematically as follow. Let assume that at the coordinates (x,y,z) and at a time (t), the fluid’s pressure, density, and temperature are p, ρ, and T, respectively; and the components of the fluid’s flow velocity are u, v, and w. Then, u = F1 (x,y,z,t), v = F2 (x,y,z t) w = F3 (x,y,z,t); p = F4)x,y,z,t). ρ = F5(x,y,z,t), T = F6(x,y,z,t) (3.1) Mathematically, this states the fluid’s velocity, pressure, density, and temperature are multivariate functions dependent on both the spatial and time variables, x, y, z, and t. In terms of physics, this means that the fluid’s parameters (p, ρ, T, and V) change with its position in space and with time. In other words, these parameters will not remain constant over all instants of time nor at all positions in space. We shall classify flow systems as: • • • •

steady flow; unsteady flow; uniform flow; and non-uniform flow.

3.2. STEADY FLOW Let us consider that a flow of a fluid is a series of particles moving through time and space. Suppose that at any given instance of time, the particles occupy different positions (x,y,z) and have a different set of parameters ((u,v,w), p, ρ, T). Suppose the flow velocity is such that at a different instant of time, particles arriving at the same fixed position (x,y,z) have the same of ((u,v,w),ρ, T). Example: A flow of a fluid has the following parameters at time t = t0 are:

Kinematics of Flow

55

at a point with coordinates (x, y, z) = (120, 200, 300): (u, v, w) = (20, 10, 30), p = 30, ρ = 12 and T = 30; and at a point with coordinate (x, y, z) = (150, 250, 350): (u, v, w) = (25, 11, 33), p = 32, ρ = 14 and T = 35. After a few seconds, at time t = t1, fluid parameters are: at a point with coordinates (x, y, z) = (120, 200, 300), (u, v, w) = (20, 10, 30), p = 30, ρ = 12, T = 30; at a point with coordinates (x, y, z) = (150, 250, 350), (u, v, w) = (25, 11, 33), p = 32, ρ = 14 T = 35. Thus, the flow parameters at the positions (x, y, z) = (150, 250, 350) and (x, y, z) = (120, 200,300) are the same remain the for the time instants t0 and t1. Such a flow is called Steady Flow. We can conclude that in a steady flow, the fluid parameters may vary from point to point in space, but do not change over time. At the same point in space, the parameters remain the same. The parameters of a steady flow for are not time dependent. Mathematically, we can write down the following condition: A flow of fluid with parameters A = (u, v, w, p, ρ, T) is steady if (3.2) If the above condition not satisfied then the flow is called Unsteady Flow. A flow discharged through a taper-pipe from a fixed displacement pump running at a constant RPM speed is an example of a steady flow. As the flow velocities, pressure, density, and temperature may be different from position to position; but at the discharging point, for instance, these remain the same all the time.

3.3. UNIFORM FLOW Suppose the flow parameters (u, v, w, p, ρ, T) are the same at all points in space (x, y, z) whatever may be the point in time, then the flow is called uniform flow. Example: At time t = t0, the flow of a fluid as the following parameters: at a point with coordinates (x, y, z) = (120, 200, 300),

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(u, v, w) = (20, 10, 30), p = 30, ρ = 12, T = 30 At time t = t1, at same point of coordinates the parameters are: (u, v, w) = (20, 10, 30), p = 30, ρ = 12, T = 30, at positions (x = 120,y = 200,z–300). Hence, after some time has passed the parameters are remain at a point in space. Therefore, the parameters of a uniform flow do not vary over space. Once again Mathematically, we can write down the following condition: A flow of fluid with parameters A = (u, v, w, p, ρ, T) is uniform if (3.3) If the conditions are not satisfied the flow is called Non-uniform Flow. The flow of a fluid running through a parallel pipe or taper pipe is uniform. If the flow of fluid satisfies both conditions, it is called steady and uniform. Therefore, the flow parameters remain the same over time and also at all positions in space (x, y, z) (Figure 3.1);

Figure 3.1: Flow through a parallel pipe.

When the flow from a fixed displacement pump run through a parallel pipe is uniform and steady flow shown in Figure 3.1. V1, V3, and V2 are different because of the boundary effect, the flow near the pipe wall will have less velocity. The flow is still the same at a different position, P2, as it is not time-dependent The flow from a fixed displacement pump running through a taper-pipe (Figure 3.2) is steady and non-uniform flow as V1 ≠ V4; V2 ≠ V5 and V3 ≠ V6 however, V1, V2, V3, V4, V5, V6 remain constant within the pipe over time.

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Figure 3.2: Flow through a tapered pipe.

Streamline is line whose direction coincides with the velocity vector of the fluid particles and everywhere tangent to this vector at any given time. Figure 3.3 shows the velocities of the fluid particles have angles tangent to the flow curves (the locus of the fluid particles). This is not a streamline. In Figure 3.4 the velocities of fluid particles have the same direction as tangents to flow curve. Such a flow is called Streamline Flow. In short, the streamline flow exists when the velocities of fluid particles are tangential to the flow curve. The velocity normal to the flow curve is zero.

Figure 3.3.

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Figure 3.4.

Mathematically, using the vector method, let us say q is a velocity vector, ds is the length of the curve. The e is in equilibrium, it follows that

(3.4)

where ds=idx+jdy+kdz and q=iu+jv+kw where represents the cross product. Thu, we get (idx+jdy+kdz)×(iu+jv+kw)=0 (3.5) Therefore, (3.6) whereas, in the case that the flow is not a streamline, the velocities are not parallel with the tangent of flow’s curve. Hence, (3.7) Figure 3.5 shows a path line, t1, to t7 represent the time instances that a fluid particle is at that point in space.

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Figure 3.5: Path line.

Then, q = dr/dt; (3.8) where q is the velocity vector and r position vector. Decomposing Eq. (3.8) into the respective three dimensional components, this is

(3.9)

In a two dimensional steady flow, there are no changes in z components and parameters are time-independent at all points of the flow. Therefore, the equation for streamline becomes

uxdy-vxdx=0 (3.10) If, we choose u and v to be

then, we get (3.11) Hence is constant and a scalar. This is the stream function. The pathline varies with each fluid particle. It represents the velocity of a single particle at various instances of time. In a steady flow stream, it coincides with paths of the fluid particles; but not in an unsteady flow stream. Stream surface is the surface formed by the series streamlines as shown in Figure (3.6). Steam tube is obtained by drawing the streamline on the closed curve path as shown in Figure (3.7).

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Figure 3.6: Streamlines (A).

Figure 3.7: Streamlines (B).

3.4. ELEMENTS OF A PARTICLE IN MOTION There two main types of fluid deformations, linear, and angular. 1)

Linear deformation: In a linear deformation, the fluid particles in a rectangular lamina undergo a geometric translation to another position. Every particle shifts the same distance to a new position in a given direction as shown (Figure 3.8).

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Figure 3.8: Linear deformation.

This results in a linear shift of the position of the fluid particles along the coordinate axes. This means that fluid undergoes a rate of linear deformation, also known as linear strain, in the respective direction. The linear strain rate component in the X-direction (δu/δx), and the linear strain rate component in Y-direction (δv/δy), Similarly, the linear strain rate component in Z-direction (δw/δz) also exist. 2)

Angular deformation: In case, the fluid particles in a rectangular lamina undergo a continuous angular deformation along with linear deformations as shown in Figure 3.9.

Figure 3.9: Angular deformation.

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The rate of angular deformation is defined as the rate of change of angle between the linear segments AB and AD. The angular displacement of AB is (3.12) and that of the AD is (3.13) Since the direction of rotation is anti-clockwise, the average angular displacement is [(δα- δß)/δt] = (δv/δx) + (δu/δy) We can apply the same procedure is for the other blocks of fluid volume. Then, the rate of angular deformation in three planes are f = (δv/δx) + (δu/δy); g = (δu/δz) + (δw/δx): h = (δw/δy) + (δv/δz) Let us analyze the rotation of a fluid shown in Figure 3.10. If the fluid is to be irrotational, then, we have (δw/δy)-(δv – δz) = 0; (δu/δz)- (δw/δx) = 0; (δv/δx)-(δu/δy) = 0 These three are very important conditions.

Figure 3.10.

3.5. EQUATION OF CONTINUITY In this section, our goal is to determine when and under what conditions the fluid has continuous flow.

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Figure 3.11.

Figure 3.11 shows streamline flow. As already defined, the velocity normal to the streamline curve is zero. The velocity of the fluid particles is parallel with the tangent to the curve. There is only flow in the direction normal to the cross-section, in other words, flow is tangent to the stream curves. Because mass is conserved, the input flow at entry should be the same as output flow at the exit. The mass flow rate is the total mass flowing past a point in a given time interval, divided by that time interval. Δm/Δt = ρ ΔV/Δt = ρ dA Δx/Δt = ρAv (Δx/Δt = v flow is in direction normal to cross-sectional area A). Mass conservation implies that the mass flow rate is constant in any section of the flow. Therefore, ρdAv = constant or ρ1dA1v1 = ρ2dA2v2 = ρxdAxV The various flow parameters are: ρ = Density of the fluid. dA1, dA,dA2 are the cross sections of flow at the entrance, in between and at the exit.

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V1, V, and v2 are the corresponding velocities. In an incompressible fluid, the density is constant, dA1v1 = dA2v2 = dAv This is the equation of continuity. When we work with finite areas such as flow bounded by pipes and channels. The equation of continuity is ρ1A1v1 = ρ2A2v2 = ρAV When density is constant, the equation reduces to: A1V1 = AV = A2V2 = Constant. This means that in the narrow sections of the tube, the fluid flows faster.

Figure 3.12.

Differential equation mass flow in rectangular coordinates system is shown in Figure 3.12 Consider a 2D Plane Strain volume element based on 1 unit thickness, i.e., the volume in the Z-direction normal to XY-plane is 1. From the principle of mass conservation, and since the thickness of the elements is equal to unity, the volume stored in the element ABCD is equal to the difference between the volumes of the fluid inflow and outflow. [(ρ + (δρ/δx)x(dx/2)) x (u + (δu/δx)x(dx/2)) x dy.1] + [(ρ + (δρ/δy)x(dy/2)) x (v + (δv/δy)x(dy/2))x dx.1] – [{ρ-(δρ/δx)x(dx/2)}x{u-(δu/δx)x(dx/2)} xdy.1] -[{ρ-(δρ/δy)x(dy/2)} x{v-(δv/δy)x(dy/2)}xdx.1];

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After simplification, we have u(δρ/δx) + v(δρ/δy) + ρ(δu/δx) + ρ(δv/δy)] xdxxdy.1 = [{δ(ρu)/δx} + {δ(ρv)/ δy} ]xdx.dy = the difference of Flow = mass of the mass fluid stored in the element In a small time δt, the mass of the element is given by [ (ρdxdyx1) – (ρ + δρ/δt)xdxxdyx1] = -(δρ/δt)xdxxdyx1

(a)

By Equating the two (a) and (b) expression, after simplification, we get (δρ/δt) + δ(ρu)/dx + δ(ρv)/dy = 0; If the flow is steady, we have {δ(ρu)/δx + δ(ρv)/δy } = 0 and if the flow is incompressible as well (ρ = constant), we get {(δu/δx) + (δv/δy) } = 0 Similarly, if a three-dimensional fluid is an incompressible and steady flow, the continuity equation is {(δu/δx) + (δv/δy) + (δw/δz) } = 0; The polar coordinates form of the continuity equation is given by {(δu/δr) + (u/r) + (1/r) (δv/δθ) } where v = tangential velocity, u = radial velocity; θ = Polar angle r = radius The continuity equation can also be written in the vector form, Δ. V = 0 V = Vector = i.u + j.v + k.w; i,j,and k are position vectors, where Δ = [i (δ/δx) + j (δ/δy) + k (δ/δz)]; is the It states that the divergence of the velocity field of an incompressible and steady flow is zero everywhere. Thus, the continuity equation can also be derived by using vector methods, by using the cylindrical coordinates system, or by using a polar coordinates system. Deducing these alternative forms of the continuity equation will not be discussed here as it falls beyond the scope of this book.

3.6. EXERCISES Problem 1: In a two dimensional flow u = 3 xy and v = 3(x2-y2) Does the flow equation of continuity ‘ Solution:

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Equation of continuity, {(δu/dx) + (δv/dy) } = 0 in two dimensional flow. Here δu/dx = 3y; δv/dy = – 3x2y = – 6 y. Hence {(δu/δx) + (δv/δy)} = 3y–6y = –3y; Hence not satisfied. Problem 2: Does the following velocity field satisfy condition of equation of continuity u = mx2y; v = ny2z; w = -yz(2mx + nz) where m,n are constants. For continuity: {(δu/δx) + (δv/δy) + (δw/δz) } = 0; δu/δx = 2mxy; δv/δy = 2nyz; δw/δz = –2mxy-nyz (w = –2mxyz-nyz2); (δu/δx) + (δv/δy) + (δw/δz) = 0 Hence continuity is satisfied. Problem 3: Given u = x3yz/3 v = xy 3 z/3; Find w to satisfy continuity. Solution: δu/δx = x2yz; δv/δy = y2xz; For continuity; {(δu/δx) + (δv/δy) + (δw/δz) } = 0Hencewe have; x2yz + xy2z + (δw/δz) = 0; (δw/δz) = -xyz(x + y); Integrating, w = xy(x + y)xz2/2 + constant of integration., which could be a F(x,y) Problem 4: If u = -y/b2; v = x/a2 Show than (x2/a2 + y2/b2) = 1 is a stream line; Soultion: For stream line uxdy – vx dx = 0; Hence {(-y/b2) x dy} -{(x/a2)xdx Rearranging, (-y/b2) x dy = (x/a2)xdx Integrating both sides, we get (-y2/b2)x1/2 = (x2/a2)x1/2 Hence (x2/a2 + y2/ b2) = 1 Problem 5: Under what conditions following velocity fields satisfy Continuity: •

u = a1xx + b1xy + c1xz; v = a2xx-b2xy + c2xz, w = a3xx + b3xy + c3z; • u = ax2 + by2 + cz2; v = dxy-eyz-fzx Find w.Solution: For continuity, {(δu/δx) + (δv/δy) + (δw/δz) = 0 • (δu/δx) = a1; (δv/δy) = – b2; δw/δz = c3 It follows a1-b2 + 33 = 0; • (δu/δx) = 2ax; δv/δy = dx-ez; (δu/δx) + (δv/δy) + (δw/δz) = 2ax + dx-ez + (δw/δz) = 0 Hence δw/δz = –2ax-dx + ez, Integrating w = –2axz-dxz + exz2/2; Problems to Solve: •

Does he flow field u = V(x3 + xy2); and v = V(y3 + xy2) and w = 0 Satisfy continuity. (Ans No.)

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• • •

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Show that velocity distribution u = -x/(x2 + y2); v = -y/(x2 + y2) Satisfy continuity. Find the values of linear deformation. Angular deformation (ans (x2-y2)/ (x2 + y2)2; (y2-x2)/ (x2 + y2)2 u = C1y-C2x v = C2y-C2x Find Stream function. (ans = C1y2 + C3x2 – C2xy + constant) Test the following flow fields for continuity. ur = [UCos α + {q/ (2𝜋r)}]; vα = -USin α (And Satisfy) Test the following flow fields for continuity. ur = {UCos α – (µxCos α/r2)}; {vα = -UxSin α – (µx Sin α/r2)} (Ans: Satisfy).

CHAPTER

4

DYNAMICS OF FLOW

CONTENTS 4.1. Forces................................................................................................ 70 4.2. Equation Of Motion In Streamline Coordinates ................................. 72 4.3. The Equation Of Motion In An Inviscid Flow...................................... 73 4.4. Velocity Potential............................................................................... 75 4.5. Velocity Potential Versus Stream Function.......................................... 75 4.6. Flow Nets.......................................................................................... 76 4.7. Exercises............................................................................................ 79

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4.1. FORCES We know, from Newton’s laws of motion, that the rate of change of momentum is equal to the sum of external forces acting on the body/object. Mathematically this is F = dp/dt = mx dv/dt = m a m = mass of the body and dV = change in velocity; F = Mxf = m(u-v)/t; dF = mx(dV/dt) The force acts in the direction of motion. The total force acting F, is acceleration, both are vector quantities. The same laws are applicable to fluid particles in motion. Due to inertia, the fluid particles tend to move in a linear motion. In order to change the motion of the fluid to a circular motion, a force must be applied radially inwards towards the rotation center. This applied force acting in the direction normal to the motion is called centrifugal force. Let us analyze the motion of a fluid that is in a steady flow. We shall discuss how these applied forces behave and act on the fluid when it changes its direction of motion. As we discussed in the previous chapter, in a steady flow, the fluid velocities change with respect to the position in space but remain constant with respect to time. The mathematical analysis in Figure 4.1 demonstrates that although the velocity vector is tangent to the curve, as the fluid particles change their direction of motion to take the course of a curve, a rate of change of velocity gives rise to an acceleration in the direction normal to the curve, inwards, in the direction of the radius of the circular motion. This is called centripetal acceleration. As aforementioned, for the fluid to move in the curve, a centrifugal force must be acting in the same direction as the acceleration, in the direction normal to the curve.

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Figure 4.1.

The previous analysis can also be done by using the vector method. Readers should know vector calculus to be able to understand the calculations below. Let us say V velocity vector is V = (u.et + v.en) where e t n unit vectors in the direction of the tangent and normal to the streamline curvature on which fluid particles move. Acceleration vector is dV/dt = [ (du/dt)xet + (dv/dt) x en] + [ ux(det/dt) + v (den/dt)]; The differential coefficient of the unit vector det/dt = {(–1/ρ)xen x (ds/dt)}; en = {ρx(dθ/dt)xet). where (dθ/dt) is the rate of angular change of unit vector et with respect to time. Since the tangential velocity is constant, (dθ/dt) = 0; Hence dV/dt = [(du/dt)xet + dv/dt xen] + ux (–1/ρ)x (enxu) + vx {ρx(dθ/dt)xet)}; Here (du/dt) = 0, since tangential velocity is constant;

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vx {ρx(dθ/dt)xet)} = 0; Hence, we have dV/dt = (du/dt.et + dv/dt.en) = dv/dt xen + ux (–1/ρ)x (enxu); Thus, U = Constant; V = 0 That is normal velocity, this is also stream function. Hence it Follows dv/dt xen + ux (–1/ρ)x (enxu) = 0 Hence dv/dt = – (1/ρ) x u2;

4.2. EQUATION OF MOTION IN STREAMLINE COORDINATES The Euler’s Equation for fluid in one-dimensional flow in the differential form is derived in Figure 4.2. Now let us applied the same method but using different parameters.

Figure 4.2.

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Let’s assume that Fs is only gravity force which is constant at a place. Hence, ρ Fs δs δA = – ρFs(δs)(δA)gCos(θ); Fs = -g(δz/δs); δV/δt + V(δV/δs) = -g(δz/δs)-(1/ρ)(δp/δs); for steady flow, δV/δt = 0; Thus we have V(δV/δs) = -g(δz/δs)-(1/ρ)(δp/δs); Here V, z,p are all function of s. Hence, we have gδz + (1/ρ)δp + V δV = 0 If the Partial derivatives are replaced by total derivatives, we can obtain the equation in the differential form by integrating, ʃ (dp/ρ) + gz + (1/2) V2 = 0;

4.3. THE EQUATION OF MOTION IN AN INVISCID FLOW An Inviscid fluid is a fluid which has a very low viscosity, lower than the inertial forces. For all practical purposes, water may be considered an inviscid fluid. The resulting equations of motion for an inviscid fluid flow expressed in Cartesian coordinates can be derived using the technique from calculus known as Taylor series as demonstrated in Figure 4.3

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Figure 4.3: Taylor series method.

A simpler method based on the concept of the Law of conservation energy is described in Figure 4.4. This method consists in applying the conservation of energy principle to a fluid flowing through a non-uniform pipe. The only external force acting on the fluid element is the gravity. The Bernoulli’s and Euler’s equations are derived from the balance of gain in kinetic energy and loss of potential energy.

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Figure 4.4: Streamline method.

4.4. VELOCITY POTENTIAL The conditions for irrotationality has been established in chapter 3. (δu/δy)-(δv/dx) = 0; (δv/dz)-(δw/dy) = 0; (δw/δx)-(δu/δz) = 0; ----------(1) The condition of irrotationality implies the existence of a velocity potential such that u = (δφ/δx); v = (δφ/δy); w = (δφ/δz) If “φ” (“φ” Called velocity potential) is a function such that For any real values, all three equations in (1) are satisfied. This is potential can also be defined in normal and tangential coordinates, Suppose flow is tangential, u = (δφ/δt); v = (δφ/δn);

4.5. VELOCITY POTENTIAL VERSUS STREAM FUNCTION Suppose Ø(x,y) is a velocity potential function, Ψ(x,y) is a scalar and a stream function. u and v are velocities in two dimension.

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From Chapter 3, we have u = (δ Ψ/δy), v = – (δ Ψ/δx); ans since Ø is velocity potential, u = (δ Ø/δx), v = – (δ Ø/δy); Thus, (δ Ψ/δy) = (δ Ø/δx) = u (δ Ψ/δx) = (δ Ø/δy) = -v; δ Ψ = uxdy-vxdx = 0 and δ Ø = uxdx + vxdy dy/dx = v/u Tan(A); dy/dx = -(u/v) Tan (B) Thus Tan(A) and Tan B) are tangent of angles. Tan(A + B) = {Tan(A) + Tan (B)}/{1- (Tan(A)xTan(B)}; Tan(A)xTan(B) = (-u/v)(v/u). So, Tan(A)xTan(B) = –1 This means that the angles between A and B are Right Angle (90 Deg). Thus slope lines of Velocity potential and streamline intersect at right angles, i.e the slope lines are orthogonal to each other.

4.6. FLOW NETS Flow nets are the graphical representation of a fluid flow such as ground – water flows; seepage water from a dam as shown in Figure 4.5 or any fluid which can not be easily accessed. It has been widely used in soil mechanics applications. We can construct the flow nets by determining the velocity potential. Since the streamline is perpendicular at every point to velocity potential curve, we can interpolate the streamline curves. A Velocity potential curve is the locus of the points of constant velocity. Since velocity is proportional to the fluid head, we can estimate the velocity of the head knowing certain constants. we can plot a velocity curves for different pressures p1,p2,p3 (h1,h2,h3 etc) and from there we can draw streamline. Figure 4.5 shows a section of a water dam using this method. V1 to V7 are velocity curves. On each curve, there exists a point such as a1,a2 etc. where the velocities are the same. a1 is the velocity on the curves V1 to V7 and is equal. Similarly, a2 is the at some other different point and of different velocity, is equal and exists on all curves V1 to V7.

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Figure 4.5.

Because we know that the velocity potential curve and stream curve intersects orthogonally to each, each points a1,a2,a3 on curves V1 to V7 can be joined. This can be achieved as well by drawing velocity curve at an equal close interval (Angle) and also at equal close distance radial. The closer these points are, the better the accuracy is. Then Next, we draw squares by joining the equal velocity points as have done in trial methods. The other way of explaining to draw a flow net is: • draw the velocity potential curves; • select the points of constant velocity on each curve; • draw a tangent to the velocity potential curve at these points; • draw normal to the tangent to each tangent at the points; • draw an envelope to obtain the stream curve. The stream curve must be tangent to all the normal line drawn. Let us imagine a situation where we know the stream function, i.e., we know the streamline, for instance, a river flowing without turbulence and vortex or a channel. We can compute the velocity at a various point on the river/channel. Then, we can draw the line of velocity potential.

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Figure 4.6.

Laplace’s Equation: From the equation of continuity, if u, v, and w are velocity function x,y,z are space variables, then (du/dx) + (dv/dy) + (dw/dz) = 0; If flow potential is irrigational, then u = δ Ø/δx; v = δ Ø/δy; w = δ Ø/δz. In these two conditions, We get by substituting in the equation of continuity. [(δ 2Ø/δx2) + (δ 2Ø/δy2) + (δ 2Ø/δz2) ] = 0 -----------(1) Expression (1) is called Laplace’s equation. [(δ 2Ø/δx2) + (δ 2Ø/δy2) ] = 0 This is in two dimensions. The stream function Ψ satisfy the equation of continuity and velocity components are expressed as derivatives of stream function, u = (δΨ/δy); v = – (δΨ/δx). Flow is irrotational.

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Then δu/δy – δv/δx = 0 [(δ 2Ψ/δ x2) + (δ 2 Ψ/δy2) ] = 0 Laplace’s equation in polar coordinates: We are not proving. we just refer. ur = velocity in radial direction. vθ = Tangential velocity. Then ur = (δ Ø/δr); vθ = (1/r)x (δ Ø/δθ) This condition satisfy irrotational, (δ vθ/δr) + (vθ/r) -(1/r)x(δ ur/δr) = 0 Laplace’s Equation, [(δ 2Ø/δ r2) + (1/r)x(δ Ø/δr) + (1/r2)x(δ 2 Ø/δθ2) ] = 0

4.7. EXERCISES Problems 1: If u = 2x + 3y + z; v = 3x–3y + 2z; w = x + y + z find acceleration at x = y = z = 1 m; Solution: Let us put the equations: u,v,w are the functions of x,y,z,andt du/dt = (δu/δx)xu + (δu/δy)xv + (δu/δz) x w = 2x(2x + 3y + z) + (3)x(3x–3y + 2z) + 1x(x + y + z) = 2x(2 + 3 + 1) + 3x(3–3 + 1) + 1(1 + 1 + 1) = 12 + 3 + 3 = 18. dv/dt = (δv/δx)xu + (δv/δy)xv + (δv/δz) x w = 3x(2x + 3y + z) + (–3)x (3x–3y + 2z) + (2)x (x + y + z) = 3x(2 + 3 + 1) + (–3)x(3–3 + 2) + 2x(1 + 1 + 1) = 18–6 + 6 = 18. dw/dt = (δw/δx)xu + (δw/δy)xv + (δw/δz) x w = 1x(2x + 3y + z) + 1x(3x–3y + 2z) + 1x(x + y + z) = 1x(2x + 3y + z) + 1x(3x–3y + 2z) + 1(x + y + z) = 1(2 + 3 + 1) + 1(3–3 + 2) + 1(1 + 1 + 1) = 6 + 2 + 3 = 11. (Note: u,v,w are the function of x,y,z andt) Hence we have to consider partial derivatives. x,y,y are the function of t only hence dx/dt,dy/dt anddw/ dt are u,v,w respectively. Problem 2: In the above problem find pressure at x = y = z = 1 meter; consider one kg of mass and gravitational force g and fluid is water; Pressure, p = Weightx(V2/2g) = (u2 + v2 + w2)/2g = {(62 + 22 + 32)/2g }x1000 = (36 + 4 + 9)x1000/(2x9.81) = 49x1000/(2x9.81) = 2497 Kg/m2 = 24970 N/m2 Problem 3: Flow through a converging nozzle may be approximated to one dimensional flow. Parameter V = V(x), If the velocity varies linearly

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from V = V0 at the entrance to V = 3V0 at exit in a length L of nozzle. V(x) = V0[1 + (2x/L)] Find the acceleration at a distance x from the entrance. Now dv/dt = acceleration. dv/dt = (δV/δx)x(dx/dt) = V(x) x V0 [0 + 2/L) = V(x) xV0(2/L) = V0(1 + 2x/L) x V0x(2/L) = [ 2x(V0)2/L] x[ 1 + (2x/L)]

Problem 4: For the following values of Ø find the respective stream function. (a) Ø = y + x2-y2; (b) Ø = C/(r2xCosθ), (a) Solution: Find u and v, u = δ Ø/δx in case of velocity field (velocity Potential); u = 2x; Similarly v = δ Ø/δy = 1–2y; Equation of streamline at any instant, u.dy-v.dx (chapter 3): u = (δΨ/δy); v = (δΨ/δx); Ψ = stream function. dΨ = u.dy-v.dx = 2x.dy -(1–2y).dx.; Ψ = ʃ [2x.dy -(1–2y).dx] = 2xy-(1–2y)x. = 4xy–1 (b) Solution: Here the function is in the polar form. We have to apply the equation in polar form. ur = δ Ø/δr; v θ = (1/r) x (δ Ø/δ θ); ur = 2C/ (Cosθ) x (–1/r); v θ = (C/r2) (Tan θx Sec θ) Stream function in polar coordinate ur = (1/r)x(δΨ/δθ); v θ = – (δΨ/δr);

(δΨ/δθ) = urx r; (δΨ/δr) = – v θ

dΨ = (δΨ/δr)xdr + (δΨ/δθ)xdθ = [(- v θ)xdr + (urx r)xdθ] = {-(C/r2) (Tan θx Sec θ)}x dr + {2C/ (Cosθ) x (–1/r)x(r)}dθ

Ψ = [(Tan θx Sec θ)}x(-C) ʃ(1/r2)xdr] + [(–2C) xʃ1/cos θ x dθ] [{(Tan θx Sec θ)x(-C)} (–1/r)] + [(–2C) {In (Sec θ + Tan θ)}] + K2 + K2 (K1 and K2 are constant of integration) (Note: a lot of integration involved. Readers should understand the method than to correct answer) Problem 5: Find the value of Ø for the following values of Ψ(a) Ψ = Cr x Sin 2θ; 2

(b) Ψ = (r–2/r)x Sin θ Solution: (a) ur = (δΨ/δθ)(1/r) = 1/r[ Cr2x 2x Cos 2θ]; u θ = – (δΨ/δr) = –2Crx Sin 2θ; ur = δ Ø/δr; u θ = (1/r) x (δ Ø/δ θ); d Ø = (δ Ø/δr)xdr + (δ Ø/δθ)xdθ = urxdr + rx (u θ) xδθ

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81

2Crx Cos 2θxdr + rx(–2Crx Sin 2θ) xδθ Integrating { 2Cx Cos 2θxʃrdr} + {rx(–2Crx) ʃ(Sin 2θ) xδθ} Ø = Cr2 Cos 2θ + (–2Cr2 x (-Cos 2θ/2)) = 4Cr2 [Cos 2θ] Solution (b): ur = (δΨ/δθ)(1/r) = (1/r) (r–2/r) Cos θ;

u θ = – (δΨ/δr) = – Sin θx (1–2x(–1) (1/r2) = [{1 + (2/r2)}(– Sin θ)]; ur = δ Ø/δr; u θ = (1/r) x (δ Ø/δ θ)

d Ø = (δ Ø/δr)xdr + (δ Ø/δθ)xdθ = urxdr + r x u θ = (1/r) (r–2/r) Cos θxdr + rx [{1 + (2/r2)}(– Sin θ)] dθ Ø = Cos θʃ(1–2/r2)dr + (r + 1/r) ʃ(– Sin θ)) dθ = Cos θx(r + 2/r) + (1 + 2/r) Cos θ = 2(1 + 2/r)x Cos θ

CHAPTER

5

BERNOULLI’S THEOREM

CONTENTS 5.1. Deduction Of Bernoulli’s Theorem, Modifications, and Application.............................................................................. 84 5.2. Coefficient Of Flow........................................................................... 84 5.3. Venturimeter And Orifice Meter......................................................... 87 5.4. Orifice Meter/Plate............................................................................ 88 5.5. Inflow And Outflow........................................................................... 89 5.6. Coefficient......................................................................................... 91 5.7. Time Relation In Discharging Liquids................................................. 92 5.8. Loses Of Head In Flowing Fluids....................................................... 96 5.9. Weirs And Notches............................................................................ 99 5.10. Velocity Of Approach.................................................................... 102 5.11. Siphon Spillway (Figure 5.20)........................................................ 102 7.12. Broad – Crested Weir..................................................................... 103 5.12. Submerged Weir (Figure 5.22)....................................................... 104 5.13. Sluice Gates (Figure 5.23).............................................................. 105 5.14. Borda’s Mouthpieces (Figure 5.24)................................................. 106

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5.1. DEDUCTION OF BERNOULLI’S THEOREM, MODIFICATIONS, AND APPLICATION This theorem relates to the law of conservation energy. It is applicable to all type of flow including irrotational. Basically, energy is available in the various form, such as mechanical, electrical, chemical. Now in our discussion, we restrict to mechanical energy. There are mainly two type of mechanical energy (a) Potential energy (b) Kinetic energy. Bernoulli’ theorem, in short, says about the energy stored in fluids. The energy stored is conserved. We have already proved the theorem in Chapter 4. It is an extension of Euler’s theorem. By integrating the Euler’s equation we get Bernoulli’s theorem. With this preliminary introduction, we shall proceed. Bernoulli’s theorem finds its application in the various practical application. •

• •

It helps to find the fluid flow velocity, pressure, flow relations in pipes used to supply water in towns, supply water through channels in irrigation fields. Also to find out dam flow, the energy stored, etc. It also finds its application in flow around streamlined bodies, where little or no flow separation takes place. Where energy losses are negligible or where energy losses can be estimated to a fair degree of approximation.

5.2. COEFFICIENT OF FLOW Any flow from a pipe or orifice, nozzle or through a channel will not have a certain amount of energy losses due to various reasons. Here we analyze these losses and how it could be computed in our calculations by incorporating certain constant. These constants are (a) Coefficient of contraction Cc, (b) Coefficient of velocity Cv, (c) Coefficient of Discharge Cd. These coefficients are interrelated. If you know two of them, the other can be calculated. The coefficient of contraction: fluid energy is lost due to surface variation, surface roughness, and many other factors through which it flows. Figure 5.1 shows an arbitrary curve of streamline and velocity potential line for an efflux from a pipe a pipe orifice. The ratio of orifice dia and pipe diameter at the place of orifice fitting is 0.6. It is a streamlined pattern (curve). It has been found the coefficient of contraction which is d/D is pi/(pi + 2). This ratio also depends on the angle θ.

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The Figure 5.1 shows the flow through the nozzle shows the ratio d/D, besides angle θ is defined in Figure 5.2 shows the charts of Coefficient of contraction. When the fluid flows through such orifice or nozzle the actual discharge will not be calculated discharge, but it will be less than the actual. The actual discharge will be proportional to this coefficient discharge.

Figure 5.1.

Figure 5.2. Values of Cc for different values of d/D d/D

θ = 45

θ = 90

θ = 135

θ = 180

0.00

0.75

0.61

0.54

0.50

0.10

0.75

0.61

0.55

0.51

0.20

0.75

0.62

0.56

0.53

0.30

0.75

0.62

0.57

0.54

0.40

0.75

0.63

0.58

0.56

0.50

0.75

0.64

0.60

0.59

86

Theory and Problems of Fluid Dynamics 0.60

0.76

0.66

0.62

0.61

0.70

0.76

0.69

0.65

0.65

0.80

0.77

0.72

0.70

0.69

0.90

0.79

0.78

0.76

0.76

1.00

1.00

1.00

1.00

1.00

The coefficient contraction be also be measured in the hydraulic lab. This Cc will also depend on the fluid properties. Now let us apply energy the nozzle flow (Figure 5.3) and find the results. Fluid is incompressible.

Figure 5.3.

(p1-p2) = [(V22 -V12)/ρ] ------- (5.1) Under considering that the nozzle is horizontal. Apply the continuity theorem to the section at PP and section QQ. {ρV1x (0.785xD2)} = {

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87

ρV2xCcx0.785xd2}; Let as call; a1 = (0.785xD2)}; a1 = 0.785xd2. Then we have V1xa1 = V2 x Ccxa2. Substituting in (1) and simplifying; V2 = [1/[SQRT { 1-(1/Cc2x(a1/a2)2) ]] x [SQRT{2x(p1-p2)/ρ}]; say (p1-p2) = Δ p (Pressure drop) If we say q = Discharge; Then q = Cd x a2xSQRT[(Δ p/ρ)] Cd is called Coefficient of discharge. Cd = [1/[SQRT { 1-(1/Cc2x(a1/a2)2) ]]

5.3. VENTURIMETER AND ORIFICE METER The practical application of Bernoulli’s theorem is found in measuring the rate of flow, discharge of liquids. One of the instruments is called the venturimeter. In this instrument, the pressure head difference is used to measure the flow. Figure 5.4 briefly describes the function and construction of venturimeter. The meter has a converging portion to which pipe is connected and is the inflow of liquid. At the end of the converging portion is connected to a narrow tube. To the end of the narrow tube is connected the diverge cone through the liquid exit. On the OD of converge portion, a manometer is fitted. Similarly, another manometer tube is fitted to the narrow tube. The difference in the level of the liquid in the manometer tubes is the measure of the pressure difference between the pressure in the converging portion and narrow tube. Let us analyze the mathematical portion of the relation between the pressure difference of flow rate as under:

Figure 5.4.

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Apply Bernoulli’s theorem to section 1 and section 2. Let the meter is installed horizontally. (p1/w) + (v12/2g) = (p2/w) + (v22/2g); (p1-p2)/w = (v22 -v12)/2g; (p1-p2)/w = H; By equation of continuity, a1xv1 = a2xv2 = q (discharge) (w = constant); Substituting and simplifying we get q = {a1xa2/SQRT (a12-a22)}x {SQRT (2gxH)}; {a1xa2/SQRT (a12-a22)} = Constant C (For a given venturimeter) g = standard gravity. with these substitution we have q = Cx (SQRT H); Note when the liquid is discharge to atmosphere the pressure in the narrow tube is less than atmosphere, ie. vacuum pressure. In our calculations we not considered any pressure loss due to friction, due to restrictions of the narrow pipe. Even though this loss is small we need to take it into our calculations. The pressure loss is built in head H, The value H is theoretical without any loss. In normal circumstances the actual head measured head is h. This is higher than H; Thus our formula changes to; q = Ckx(SQRT h). hence k = SQRT (H/h). Normally k = 0.97. The venturimeter is not accurate for low velocities due to variation in k.

5.4. ORIFICE METER/PLATE Orifice meter is also a discharge measuring device. It is a symmetrical opening device fitted to a pipe. It is a plate having a small hole specially machined to have certain sizes. Figure 5.5 shows a typical section of an orifice meter. The principle used is the same as in venturimeter. On the same lines as that of venturimeter, you can deduce the formula for the manometer head discharge. The vena contracta area = Ccxa2 (a2 = orifice diameter) = a; Cc will be given by the manufacturers of orifice plate. It may vary from 0.85 to 0.9 q = [Ccxa//{SQRT (1-(Ccxa/a1)2}] x SQRT (2gxH) where H = (p1-p2)/w. If Cd = Discharge coefficient, Then q = Cdx SQRT (2gxH) Cd = [Ccxa// {SQRT (1-(Ccxa/a1)2}] is constant for an orifice plate. Refer the sketch Fig– 35 to understand more details. Cd is calibrated to each Orifice. It generally depends on area ratio(a2/a1) (Diameter ratio D/d). where d is the diameter of orifice and D is the diameter of pipe. The head loss can be calculated by Bernoulli’s theorem but can be estimated by the momentum theorem which will be discussed in succeeding chapters.

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Figure 5.5.

5.5. INFLOW AND OUTFLOW Discharge from Tank through outlets fittings: Refer Figures 5.6 and 5.7. Let us consider the liquid level in the following tanks are constant. The sketches show various types of orifices fitted. In places, it is also called mouthpieces. The theoretical relation based on Bernoulli’s theorem is Velocity V = SQRT(2gH) where v = velocity of liquid flowing out of orifice and H is the height of liquid column from a center line of the orifice. Actual velocity is not equal to theoretical velocity. This is due to friction in the pipe. Suppose Cv velocity co-efficient and Cc = area of orifice reduction due to vena contracta. Thus the actual discharge = (Cc)xaxCv(V) = CcxCvxaxSQRT (2gxH) this is further reduces to q = CdxSQRT(2gxH) Thus there the relation Discharge coefficient, Cd = Vene contracta, Ccx Velocity co-efficient Cv, Cd = CcxCv. The values of Cc,Cv shown below are indicative and flow to atmospheric pressure. It would defer depending on the flow conditions, such as flow is to a submerged tank and external pressure is not the atmosphere. Flow to mouthpieces depends on the angle of divergence/convergence, and length.

90

Theory and Problems of Fluid Dynamics Values of Cd, Cc, and Cv Type Sharp-edged orifice Rounded orifice Straight mouthpiece Borda Flowing full orifice Borda Flowing free

Cc 0.62

Cv 0.98

cd 0.61

1.00

0.98

0.98

1.00 1.00

0.80 0.75

0.80 0.75

0.50

1.00

0.50

Figure 5.6.

Figure 5.7.

Figure 5.8.

Large Vertical Orifice: Refer Figure 5.9: This is a large orifice. The orifice size is not small compared with the tank height and width. In such a situation, we cannot assume the flow along the thickness the orifice/diameter of the orifice is constant. The flow varies with the thickness of the orifice. We need to use differential/integral calculus to calculate the flow/velocity. Total flow across the orifice is calculated considering the flow through the elemental strip. Then integrating taking the limits the across the t strip of the

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91

orifice between the limits H1 and H2 which the depth to the upper lip and lower lip of the orifice. It is shown as under: dq = CdxBx{SQRT(2g)}xdh; q = CdxBx{SQRT(2g)}xʃdh (H2 – h2) Limits. Then, we have; q = CdxBx{SQRT(2g)}x(2/3)x[ H21.5 – H11.5]. If we consider the approach velocity U0 to the orifice, Then formula is modified and H’1 = [H1x(U0)2/2g] and H’2 = [H2x(U0)2/2g]; The formula is corrected to q = CdxBx{SQRT(2g)}x(2/3)x[ H’21.5 – H1’1.5].

Figure 5.9.

5.6. COEFFICIENT As we have seen earlier, the actual flow differ from the calculated flow. In order to estimate the actual flow, many laboratory experiments have been done. As per the results, it has been found the flow jet will contract to a smaller area/diameter after it comes out of the orifice. This is called as vena contracta. This we have defined in the previous section (5.2.). This is designated as Cc Cc = (minimum area of the jet as it comes out after orifice)/(area of the orifice); Hence Actual discharge = Ccxq Another factor is the velocity of the jet. Due to friction and other factors, the actual velocity is less than the calculated velocity. The call this coefficient as Cv

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Theory and Problems of Fluid Dynamics

Cv = (Calculated velocity/Actual velocity). Hence V = CvxVc where Vc is calculated. We know Discharge, q = Section are of flow x Actual velocity of flow. Thus actual discharge. q = (CcxCv) xqc where qc is the calculated discharge. The product (CcxCv) is called discharge coefficient Cd ie. Cd = (CcxCv); The values Cc and Cv vary from orifice to orifice and also liquids properties. The values are already provided in the previous section. In most of the cases, the value may be considered between 0.8 to 0.85. These default values.

5.7. TIME RELATION IN DISCHARGING LIQUIDS Time for emptying a tank fitted with an orifice (Refer Figure 5.10)

Figure 5.10.

Let the water level above the center line of orifice = H; Diameter of orifice = d Let at the time instant t we consider the level of water at height = h; let elemental strip of water = dh. Lrt the velocity at this instant = V; Then the flow for a short time dt be constant. Then we can write a small amount of discharge = dq; Now dq = Cd.av dt; where A = area of the tank. v = velocity of flow through at the instant under consideration. Hence we have v = SQRT(2gxh) Hence dq = Cdxax{SQRT(2gxh)}xdt where a is he area of orifice = 0.785xd2. v is constant for a small time dt. Total flow during a time dt; dq = Axdh and

Bernoulli’s Theorem

93

dq = = Cdxax{SQRT(2gxh)}xdt We have Axdh = Cdxax{SQRT(2gxh)}xdt Hence re-arranging, we have dt = (Axdh)/[ Cdxax{SQRT(2gxh)} Integrating, ʃdt = [A/ {CdxaxSQRT(2g)}] xʃh1/2 xdh On integrating, t = [2A/ {CdxaxSQRT(2g)}]x (h1/2); Now at start tank liquid level = H; we start reckoning time, t = 0; when H = 0 The t = time for emptying Hence time for emptying = say, T = [2A/ {CdxaxSQRT(2g)}]x (H1/2); Suppose we want to find the time to empty part of time, say level above orifice = H1 at start; we want to know time to bring level to H2; The T = [2A/ {CdxaxSQRT(2g)}]x [H11/2 – H21/2] Emptying a hemispherical vessel: (Figure 5.11)

Figure 5.11.

Let us find out the time required to empty a hemispherical vessel as shown in Figure 5.11. Now, x2 = R2 – (R – h)2; Hence area of section dh = Pixx2 = = {pix (R2 – (R – h)2}; dq = -pixx2 dh = Cdxaxv (v = velocity at the instant. a = area of the orifice.) dq = -pixx2 dh = Cdxaxv dt = CdxaxSQRT(2gh) dt; dt = -pixx2xdh/{ CdxaxSQRT(2gh)};

94

Theory and Problems of Fluid Dynamics

dt = -pix(2Rh – h2)xdh/{ CdxaxSQRT(2gh)} Hence T = [pi/ { CdxaxSQRT(2g)}]xʃ{(2Rh – h2)xh–1/2}xdh On integrating, T = – [pi/ { CdxaxSQRT(2g)}] x[ {(4/3)(xRxh3/2)} -{(2/5)x(h5/2)}] On substituting values T = [2pi/ { CdxaxSQRT(2g)}]x [ (2/3)xRx(H13/2 – H23/2)-(1/5)x(H15/2 – H25/2)]; Transfer time from one tank to another tank: (Refer Figure 5.12)

Figure 5.12.

Let us find out the time required to transfer liquid from one tank to another. The liquid flow from tank A1 to A2. The difference between the liquid level in the tanks = H1; Area = area of orifice. v = velocity of flow. let it is required to find the time taken to the difference in head H2. At certain instant, the difference in head = h. small quantity of liquid “dq” flow through the orifice in a time “dt”. Then the liquid level in tank A1 fall by “dH”. The level of liquid in tank A2 will rise by [dHx(A1/A2)] The flow volume from tank A1 = (dHxA1) = Flow to the tank A2 = dHxA1 But dHxA1 = dH1xA2; where “dH1” rise in liquid level in tank A2. Thus we have rise in level of liquid in Tank A2 = DH1 = [dH(A1/A2)]; The new difference in head after “dt” seconds = [h-dH- {dH(A1/A2)}] = [h-dH{1 + (A1/A2)}]; Therefore the change of head causing the flow, dh = dH{1 + (A1/A2)}; dH = [dh/{1 + (A1/A2)}];

Bernoulli’s Theorem

95

Small quantity of flow, dq = -A1xdH = Cdxaxvxdt v = SQRT(2gh); Thus we have dt = – A1xdH/{(CdxaxSQRT(2gh)}; Substituting for dH = [dh/{1 + (A1/A2)}]; We have dt = – (A1dh)/{Cdxax(1 + (A1/A2)}x{SQRT(2gh)}; dt = -[ (A1xh–1/2 x dh)/{Cdxax(1 + (A1/A2)}x{SQRT(2g)}]; T = – ʃ-[ (A1xh–1/2 x dh)/{Cdxax(1 + (A1/A2)}x{SQRT(2g)}]; T = – [A1/{Cdxax(1 + (A1/A2)}x{SQRT(2g)}]xʃ-[ (A1xh–1/2 x dh); T = – 2x[A1/{Cdxax(1 + (A1/A2)}x{SQRT(2g)}]x h1/2 apply limits H1 and H2; we Get T = –2A1x[H11/2 – H21/2]/ [{Cdxax(1 + (A1/A2)}x{SQRT(2g)}] when tank area is the same, then A1 = A2 T = -A1x[H11/2 – H21/2]/ [Cdxax{SQRT(2g)}]; Time required when there is in flow from other source and tank is emptying: (Figure 5.13)

Figure 5.13.

The liquid is flowing from one tank and liquid is discharged from the tank to which is flowing. Let the constant liquid flow from other sources = Q; q = discharge from the tank. Consider the tank at the instant when the liquid level in the tank = h above the orifice. Let the height be increased by dh in a small interval “dt”.

Theory and Problems of Fluid Dynamics

96

Amount of inflow from the source = “Qxdt”; Amount of discharge from tank = “qxdt”; qxdt = Cdxax{SQRT(2gh)}; a = area of orifice = 0.785xd2. Let k = Cdxax{SQRT(2g)} (Constant); Suppose the inflow > out flow, liquid level in the tank will increase. thus = Axdh; A = area of the tank. Adh = Qdt-kh1/2 dt; dt = [ Adh/{Q-(kxh1/2)}]; let p = {Q-(kxh1/2)}; p is a variable since h is a variable. h = (Q-p)2/k2); dh = –2[(Q-p)dp/k2]; substitute for h in the equation for dt = [ Adh/{Q-(kxh1/2)}]; We get dt = [Ax2(Q-p)xdp/(pxk2)]; Integrating we get T = (–2A/k2)x (QxIn p-p); Re-substitute vales for p and values for k and assign limits H1 and H2; (H1 – h2) is the level rise. Then finally, we get; Time taken to rise level H1 to H2 time T = –2A/k2x [ Qx[In {Q-K(H21/2)/ {Q-K(H11/2)}] + [k{H21/2 – h11/2}]]

5.8. LOSES OF HEAD IN FLOWING FLUIDS Here head means pressure. Let understand clearly. We know the pressure, p = wxh; h is referred as head, so many meters of the liquid column. w = weight of the fluid. p is pressure. Thus if we consider the weight as unity, then head represents pressure. Hence head means pressure. Fluid flowing along a straight uniform passage, would suffer certain loss. We consider water flowing through a pipe of about 10-meter length. and of uniform diameter 50 mm. Theoretically, velocity should be same Measure pressure at the beginning of pipe, say it is 1 Kg/cm2. Let us measure the pressure at the end of 10 meters. the pressure is not the same as 1 Kg/cm2, but is less and could be 0.9 Kg/cm2. Hence the pressure is down by 0.1 Kg/ cm2. This is called pressure loss. Since 10 meters of water column exerts 1 Kg/cm2, We say pressure loss is 1 Meter. There are many types for the pressure losses: •

Pressure loss due to the viscosity of fluid: This loss is the function of v2/2g and also depends on Reynolds number. We shall discuss the Reynolds number in the succeeding chapters. this depends on the length and diameter of the passage, besides roughness of pipe. We shall study this loss in pipes, channels, and various types of passages in the succeeding chapters.

Bernoulli’s Theorem

97

•

Pressure loss due to change in direction of the passage: It would be necessary to bend the pipes in which the fluid is flowing. It is normally expressed as a function of v2//g loss = kx(v2/g). k is found by experiments. The value depends on the angle of bend, as example 90 Deg bend, 45 Deg Bend, U-Bend etc. The head is lost in the friction of eddies formed. • Pressure loss due to change in section of the passage/pipe: Losses is due to sudden enlargement or contraction of passage section and a loss at the entrance of pipe. Besides, there are losses due to gradual enlargement of contraction of section. Head loss due to sudden enlargement = (v12 – v22)/2g, where v1 and v2 are velocities of flow at entrance and exit. Head loss due to sudden contraction = 0.2x(v2/2g) = Head loss at the entrance of the passage(pipe) •

Head loss due to obstruction: An obstruction to the passage to flow also cause loss. The obstruction will create a contraction of area of flow and then there would be sudden enlargement. This is the loss due to combination of enlargement and contraction. Head loss due to obstruction: [{A/(0.66(A-a)}–1] x (v2/2g) A and a area Proof for head loss due to a sudden enlargement: Refer Figure 5.14:

Figure 5.14.

Consider a fluid flowing in a pipe of area “a1”, with velocity “v1” and pressure “ p1”. Let the pipe be enlarged to an area “a2”. The pressure to

98

Theory and Problems of Fluid Dynamics

change to”p2” velocity to “a2”. The fluid flow as shown. Eddies formed at backwash. Shown as circles in the sketch. These will cause a pressure “p0”. The eddies press the annular area (a1-a2). It is found by experiment that p0 = p1(approximate). Consider the fluid between A-A and B-B. The resultant force acting = p2a2-p1a1-p0(a2-a1); assume p0 = p1; We have Total force = a2(p2-p1); Change of momentum per second of this mass of fluid is: [(wxa1xv12)/g ] – [(wxa2xv22)/g ]; a1v1 = a2v2; Hence Change of momentum = w/g{a2v2v1-a2v22} We know force = rate of change of momentum = change of momentum per second. Therefore, a2(p2-p1) = [(w/g)a2(v2v1 – v22)]; [(p2/w) – (p1/w) ] = [(v2v1v22)/g]; (5.2) hL = Head loss due to enlargement; Apply Bernoulli’s theorem: (p1/w) + (v12/2g) = (p2/w) + (v22/2g) + hL; [(p2/w) – (p1/w) ] = (V12/2g) – (V22/2g) – hL; Equate [(p2/w) – (p1/w) ] = (1); [(v2v1-v22)/g] = (V12/2g) – (V22/2g) – hL; finally we get hL = [(V1-V2)2/2g]; Head loss due to sudden contraction: (Refer Figure 5.15): The head loss by sudden contraction, but also due to enlargement as is clear from the sketch below:

Figure 5.15.

In this case fluid flow from enlarged portion of pipe to narrow area of pipe. we can see flow first contracts at vena contracta and then enlargement. It is this enlargement from vena contracts (CC) to narrow section (BB)

Bernoulli’s Theorem

99

causes the head loss. We have to find out a mathematical expression. ac = Ccxa where Cc is the coefficient of contraction. Loss head = [(vc-v)2/2g]; av = acxvc; vc = (v/Cc); Loss head = v2x [ (1/Cc–1)2/2g. Assuming Cc = 0.62 for circular orifice; We have Loss head = 0.375x(v2/2g); It has been found by experiment that the Value 0.375 is higher and is almost equal to 0.5. Head loss due to sudden contraction = 0.5x(v2/2g) Loss of head due to obstruction: Refer Figure 5.16. V = velocity of fluid in free section. Vc = velocity in the CC; BB is the area beyond the obstruction. LSS of head due to enlargement between CC and BB = {(Vc-V)2/2g}. Area at section CC = cc(A-a); Cc is coefficient of contraction. CcxVcx(A-a) = Va; Therefore Vc = [AxV{(Ccx(A-a)}] on simplifying loss of head = [{A/ Cc x(A-a)} – 1]2x (V2/2g); Assuming coefficient of contraction = 0.66; Loss Head = [{A/(0.66 x(A-a))} – 1]2x (V2/2g); The above principle made use of in measuring the flow of fluid in pipe. The instrument is called pipe orifice meter.

Figure 5.16.

5.9. WEIRS AND NOTCHES A notch may be regarded as an orifice with its fluid surface is below the upper edge. Notches are normally device used to measure the fluid flow

100

Theory and Problems of Fluid Dynamics

from a tank or reservoir. These notches are generally of rectangular or triangular shape. There is not much difference between a weirs and notches. Weir is the name given to a dam over which water flows. The top of the weir over which water flows is called crest. When weir is very wide it is divided into vertical partition. Rectangular notch: (Refer Figure 5.17)

Figure 5.17. Figure 5.17 shows the section of a rectangular notch. Let us consider the notch is used to measure the quantity of water. L = Breadth of notch; H = height level above the crest; Cd = discharge co-efficient; Discharge through the strip dh = Lx{SQRT(2gxh)}xdh; Total discharge = LxCd ʃ {SQRT(2g) xh1/2 xdh; (limit H to 0); Total discharge = Cdx(2/3)x SQRT (2g)xLxH3/2; To measure the water discharge, the notch is to be calibrated. Discharge, Q = k x H 3/2, where k = (2/3)x SQRT (2g); We have to find discharge for various values of H; From the graph we can estimate the discharge at any level H; Another way,; Log Q = Log k + 3/2xLog H; This is a straight line graph. Triangular or V-Notch: (Refer Figure 5.18)

Bernoulli’s Theorem

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Figure 5.18

Discharge of elemental strip, dQ = {CdxWidth of Stripxdh}; From the geometry, the width of strip: Width of strip = 2x(H – h)x Tan (θ/2); dQ = Cdx2x(H – h)x Tan (θ/2) xdh x SQRT(2gh) Q = Cdx2x Tan (θ/2)xʃ (H – h) x h1/2 dh = Cdx2x Tan (θ/2)x [ (2/3) x(Hxh3/2)-(2/5)x (h5/2)] Applying limits h = H h = 0 Cd = 0.6 Then we get 0.6x(8/15)x Tan (θ/2)xH5/2 = {2.56x Tan (θ/2)xH5/2}; Q = {2.56x Tan (θ/2) xH5/2}; when θ = 90 Deg, Then Q = {2.56x H5/2}; approximated to Q = {2.5x H5/2} Francis’ formula for rectangular weirs (Figure 5.19): An empirical formula for the discharge of a rectangular weir is given by Francis’: Q = 3.33x(L–0.1xnxH)x H3/2; n = no of end contraction. large weirs divided in to bays n depends on the number of bays the weir is divided. (Refer the Figure 5.19).

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Figure 5.19.

Bazin’s Formula for rectangular weirs: This is another type of equation used for obtaining the discharge over a rectangular: Q = (2/3Cd)xSQRT(2g) LxH3/2. Say m = 2/3xCd, Q = mx(SQRT(2g)) LxH3/2; The coefficient of m is to be found by experiments.

5.10. VELOCITY OF APPROACH If the water channel width is larger than the weir width, in case of rectangular, we have to consider the velocity approach. Suppose the discharge is Q. Then the area of channel = A, Then the velocity of approach v1 = Q/A. First, calculate the discharge by regular Weir. Then the formula is, Q = Cdx(2/3)x SQRT (2g)xLxH3/2; Knowing Q, find v1 = (Q/A); Then modify the formula Q = Cdx(2/3)x SQRT (2g)xLx[ H-(v1/2g)2 } – (v1/2g)2] 3/2

5.11. SIPHON SPILLWAY (FIGURE 5.20) The reservoirs need to have an automatic device that keeps the water level at constant height. The excess water at the time of flood should overflow. This can be achieved by having a weir and a sill at the same height. Then the excess water overflow to the channel below. No doubt this one of the

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methods, there are other methods to do the same function, more efficiently. The most popular method is siphon Spillway. Please refer Figure 5.20. The method of discharging the excess of water is called siphon spillway. This method employ the whole head between the reservoir level and the water level in the overflow channel. The sketch shows the cross section of an automatic siphon spillway. It consists of an ordinary weir still surrounded by an airtight rectangular cover as shown, thus converting the discharge face of the weir into a large rectangular sectioned pipe.

Figure 5.20.

The moment the water level rises in the reservoir to the level of sill, it strikes the cover, thus completely filling the portion below the rectangular airtight sheet. Now air is trapped in the section “B”. The water is sucked away by the stream of water flowing, without this rectangular sheet the velocity head would have been “ h” as in the normal weir. Now the head is “H” with a rectangular sheet. A1 = Area of section of spillway pipe (i.e., Section S-S) Discharge Q = CdxA1x{SQRT(2g)}

5.12. BROAD – CRESTED WEIR Figure 5.20. Sketch shows the water flowing on a broad crested weir, H = Head of water at upstream edge, h = Water at downstream edge. v = velocity of flow. b = breadth of weir.

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Figure 5.21.

Let us find out the relation between the head and Discharge Q. neglecting losses, H = [h + (v2/2g)]; Rearranging, v = SQRT(2gx(H – h); Discharge, Q = Cdxbxhx{SQRT(2g(H – h)}; Cd is to be found by experiment. Now Q = kx[SQRT{(Hh2 – h3)}]; k = CdxbxSQRT(2g); Now look at the equation Q = kx[SQRT{(Hh2 – h3)}]; Discharge is maximum when d[SQRT{(Hh2 – h3)}]/dh = 0 Thus Discharge is maximum when H = (3/2) x h; OR; h = 2/3(H) Maximum discharge Q = CdxSQRT(2g) x bx 2/3xHxSQRT(1/3) x H1/2; Simplifying Q = 3.09xCdxbxH3/2.

5.13. SUBMERGED WEIR (FIGURE 5.22) If the water depth downstream of the weir is above the top of the weir it is said to be submerged weir. See the Figure 5.22.

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Figure 5.22.

In this case, the weir is submerged in downstream water. H2 is the head on downstream and H1 is water head-on upstream side. We divide the solution into two parts. Top portion between the upstream depth H1 and the downstream depth H2. We consider H1 as free weir “H2” as an orifice. The discharge for free weir: Q1 = CdxBxSQRT(2g) xʃh1/2 x dh Limit (H1 – h2 to 0); Q1 = (2/3) xCdxBx{SQRT(2g)}x [H1 – h2]3/2 For the Orifice portion: Q2 = CdxBxH2[2gx(H1 – h2)]1/2 Hence Total discharge = Q = Q1 + Q2

5.14. SLUICE GATES (FIGURE 5.23) Sluice gates are used to as a control on canals, on rivers and on spillways of water flow. This is a gate which can be opened fully, partially or closed fully. In the fully closed condition, the flow is minimum as gate is vertical as shown. The gates may be swing types or vertical lifted type. This is how the area of flow under the gate is controlled. There are two types of flow under the gate. Free flow and submerged flow as shown in Figure 5.23. Let us consider the discharge per unit width of the gate.

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Figure 5.23.

In case of free flow: v1xH1 = CcxaxV2; or v1 = (v2xCcxa/H1); Now the energy equation: H1 + (v12/2g) = Ccxa + (v22/2g) Rearranging, v2 = SQRT[{ (Ccxa – h1)/(Ccxa)2 – H12)}] x SQRT(2gH1) v2 = SQRT [ { 1/(Ccxa + H1)}x SQRT(2g) Discharge q = (Ccxaxv2) = Ccxax[SQRT[{ 1/(Ccxa) + H1)}] x SQRT(2gH1)]; If Cv = 1 and Cd = [Cc/SQRT{1 + (Ccxa/H1)}] Then q = Cdxax(2gxH1)1/2 In case of sluice gate is submerged, then ΔH = H1 – h2; hence q = Cdxax(2gxΔH)1/2

5.15. BORDA’S MOUTHPIECES (FIGURE 5.24) The internal mouthpiece shown in Figure 5.24 is called Borda’s mouthpiece. Let us analyze both cases. If the jet after contraction, does not touch the sides of the mouthpiece, it is said to be running free. If the jet after contraction expands and fills the mouthpiece as shown in Figure (b) it is said to be running full. H = height of liquid level above the centerline of the mouthpiece. a = Area of the mouthpiece. v = velocity of flow through the mouthpiece. ac = contracted area of the jet. Force = Rate of change of momentum.

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Total pressure at entrance = Change of momentum per second. pxa = [(wx ac)/g ] xv; p = wxH; wxaxH = wx acx(v2/g); H = (v2/2g);

Therefore, (axv2/2g) = ac x(v2/g); ac = 0.5xa; Hence Coefficient of contraction, Cc = 0.5

Figure 5.24 (a) and (b).

Now consider the case (Figure 5.24b) when the liquid is filled the mouthpiece. This is similar to external mouthpiece. There will be vacuum at vena contracta, which will increase the velocity of jet. Loss of head due to enlargement = [(vc-v)2/(2g)] = [(1/Cc) – 1)]2x (v2/2g) = (v2/2g), as Cc = 0.5 Apply Bernoulli’s theorem to the liquid surface and to the end of the mouthpiece. Ha + H = Ha + (v2/2g) + (loss of head due to enlargement); v = {SQRT(gxH)} Discharge when running full = axv = ax{SQRT(gxH)} Discharge when running free = 0.5xax{SQRT(2gxH)}; Therefore, the discharge is increased by [1/(0.5xSQRT (2)] when running full. Coefficient of discharge when running full = (1/2)1/2 = 0.707. In practice, the Coefficient of discharge is slightly greater than this amount. Now, the pressure at vena contracta, may v\be found by applying Bernoulli’s theorem.

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Hc + (vc2/2g) H = Ha + (v2/g) + (v2/g); But vc = 2v;

Therefore Hc = (4v2/2g) = Ha + (v2/2g); Hc = Ha – h As H = (v2/g); Thus the pressure at vena contracta is less than atmospheric pressure by an amount equal to head of liquid in the vessel.

CHAPTER

6

FORCE AND MOMENTUM

CONTENTS 6.1. Introduction..................................................................................... 110 6.2. Impact Of Jets.................................................................................. 111 6.3. Forces On Bends.............................................................................. 117 6.4. Rockets Motion................................................................................ 120 6.5. Jet Propulsion ................................................................................. 121 6.6. Loss Of Energy In Pipe Expansion.................................................... 122 6.7. Lawn Sprinklers............................................................................... 124 6.8. Pelton Wheel................................................................................... 125 6.9. Reaction Turbines............................................................................ 127

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6.1. INTRODUCTION In the previous chapters, we studied about the forces acting by the fluid at rest and also its effects. We did many numerical examples. Then we studied about the conservation of mass, and also proved the condition in two and three-dimensional steady flow. We shall enumerate once again here these equations. {δ(ρu)/δx + δ(ρv)/δy } = 0 when flow is steady. Not equal to zero if flow is not steady. When flow is not steady we have to find the value of “{δ(ρu)/δx + δ(ρv)/δy }” considering boundary conditions. We studied the forces by fluids under static conditions. This we called as static force. Static force = F = axp = axwxh (p = wh and w = ρg)). h = gravitational head in case of fluid at rest with force acting is gravitational force. Besides when fluid is subjected to external force, as in by rotating a fluid container, that generates centrifugal force or as in case in case of fluid is subjected to linear acceleration generates external force. In such situations, the pressure will change and the head “ h” also change. Then we studied about the conservation of momentum and deduce the Euler’s equation as: (dp/ρ) + gxdz + vxdv = 0; This is conservation of momentum. (The is similar to one we study in dynamics, which states that the vector/algebraic sum of momentum before and after the collision is zero) On integration we get ʃ (dp/ρ) + gz + (1/2) V2 = 0: This is the Bernoulli’s equation related to conservation of energy. In this chapter, we learn about the fluid forces and momentum acting of other objects.

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6.2. IMPACT OF JETS

Figure 6.1.

Consider a plate as shown fixed. The fluid of density ρ impinge on the plate normal to it. The velocity of fluid = v. Let the discharge = ρxaxv, Weight of the fluid = wxaxv; where a = cross area of the jet impinging. After the fluid impinges on the plate leaves the plate tangentially. Neglect all losses. Then the velocity after it strikes the plate in the normal direction to the plate = 0. Force = rate of change of momentum. Change in momentum per second = Mass flowing per secondx velocity; Thus it follows Force = (ρxaxv) xv = (ρxaxv2) N = (wxaxv2/g) Kg where, g in m/sec2, v in m/s and ρ kg/m3; a = m2; Force on an inclined plate: All are data’s as above. The plate is inclined at angle θ to the horizontal.

Figure 6.2.

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Here the velocity after the leaves the plate is inclined at an angle θ. We consider the velocity normal to the plate = v Sin (θ) Hence Change in momentum = (ρxaxv) x v Sin (θ) = (ρxaxv2) Sin (θ) N. Force on the Curved surface:

Figure 6.3.

Consider a curved vane fixed as shown in Figure 6.3. θ1 and θ2 are at the start and end of vane. Let V be the constant velocity on the curved surface. The discharge (mass flow)is constant. Let the external pressure is the atmosphere. The area of the section of the jet is a Discharge = ρxaxV at the entry tangential to the vane. = Discharge at the exit. Horizontal momentum at entry = VxCos θ1and at the = VxCos θ2

Rate of change of momentum in X-direction = force in X direction Fx = Mass flow/sec x Velocity = ρxaxV2 x { VxCos θ2 – VxCos θ1}; Similarly Fy = ρxaxV2 x { VxSin θ2 – VxSin θ1};

Resultant Force = Fr = (Fx2 + Fy2)1/2 and Angle Tan –1 (Fx/Fy) Force on a moving flat plate: Refer the Figure 6.3.

Let us say the plate instead of fixed is moving with a velocity v. The velocity of jet in this case = V

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In such situation, we need to consider the relative velocity. The relative velocity to plate is (V-v) Hence flow per second = axρx(V-v) Hence the force = axρx(V-v) x (V-v) Force on the plate = axρx(V-v)2 N where V and v in meters/second; a = area of jet M2. ρ is density in Kg/m3. If w = unit Weight in N then force on plate = (w/g)x(V-v)2 N.

Figure 6.4.

Force on a moving plates (Vanes) fixed on a rotating wheel: Figure 6.4. Refer the figure as shown. This is a wheel on which plates are fixed radically. Such plates are called vanes. We shall call them as vanes. Let us analyze the force acting on such vanes and its cause. As shown in the figure, a = area of the jet, D = diameter of the wheel. That is the distance from the center of the wheel to the centerline of jet. The jet tangentially strikes the wheel. The effect is a torque due to which wheel start rotating. The angular velocity = ω rad/sec. Hence tangential velocity of wheel, v = (d/2)x ω. The weight of fluid striking the plates/sec = wxaxV. Here we are not considering relative velocity to find the flow/second, because the vanes are closely vanes are placed and one after the other the vanes come in front of the jet. Hence for the flow calculation, we have to consider actual velocity. Relative velocity = (V-v); here we have to consider the relative velocity Now force on the plates = (wxaxV/g) x (V-v) since it is the rate of change of momentum/second.

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Now we can find the work done per/second. Work done/sec = Force x Velocity of the object. Here the object is wheel and its velocity = v(Tangential velocity) Work done/second = [(wxaxV/g) x(V-v) x v] N-m/sec; Joules/sec (watts); This is also called power. Now kinetic energy stored in the jet = {wxaxVx(V2/2g)}; Note: wxaxV = weight of fluid/second Efficiency of work done = Use full work/Actual work = [{(wxaxV/g) x(V-v) x v}/ {wxaxVx(V2/2g)}]; Hence efficiency = [2x (V-v)xv/V2]; Here are two variable, V, and v. We get different values of efficiencies by varying the velocities V and v. There exit a condition where the value = [2x (V-v)xv/V2] is maximum, i.e., maximum efficiency. This can be obtained by differentiating the Value [2x (V-v)xv/V2] with respect to V and equating to zero. The we get V = 2v. substituting the value in the equation we have Maximum efficiency = 1/2 (i.e 50%) The problems related to impact of jet: The following are the various problems we solved: • Force on flat plates; • Force on fixed curved vanes; • Force on moving flat plates. We can solve these problems by pure mathematical methods: Force on flat plates: Here Force = Rate of change of momentum per second Momentum = mass x velocity = mxV: Here both are variable, m and V Force = d(mxV)/dt = Vx(dm/dt) + mx(dV/dt); Since there is no acceleration of the fluid, dV/dt = 0

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Force Vx(dm/dt): dm/dt = the mass flow per second = ρxaxV; Thus force = = ρxaxV2 = (w/g)xaxV2. When plate is inclined at an angle then V = VxSin θ Thus force = [(w/g) xaxV2x Sin2 2) Force on fixed curved vanes Force on fixed curved vanes: Let us consider V as vector, Ve = (VexCos θ1)xi + (VexSin θ1)xj

I and j are unit vectors. Similarly, Vx = (VxxCos θ2)xi + (VxxSin θ2)xj: Ve andVx are velocities vectors at entry and exit.

BY vector differentiation and re-arranging we get F = Fxxi + Fyxj; Note Vx = Ve = V m = ρxaxV2 (Constant)

Fx = [ ρxaxV2 x { VxCos θ2 – VxCos θ1}]; Fy = [ρxaxV2 x { VxSin θ2 – VxSin θ1}]; Force on moving flat plates:

Relative momentum = mx(V-v) where v = constant, wheel rotation. Hence Force = (dm/dt)x(V-v) + mx d(V-v)/dt = (dm/dt)x(V-v) + mx dV/ dt dV/dt = 0 Hence we have, Force = ρxaxVx(V-v) Work done/second = force x distance/second = [ρxaxVx(V-v)] x (ds/dt); where (ds/dt) is the distance moved by the object on which force acting. In our case it is vanes. ds/dt = v, velocity of vanes. Hence work done = ρxaxVx(V-v)xv. Force on Moving curved Vanes: Here we are doing vector algebra, since velocity is a vector and hte velocities we deal V and v are of different direction. Figure 6.5 consider a curved vane moving with velocity v in the direction AB. Let a fluid jet impinge on the vane at a velocity V (ab). The velocity of fluid vane is equal to the relative velocity of the jet to the vane = Vr (cb).

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Figure 6.5.

In the Figure 6.6 we have drawn ab = Velocity of jet. ac = Velocity of vane = v Hence by vectors Relative velocity V-v = ab-ac = cb (Vr). = Relative velocity at entrance. ed = Absolute velocity at exit = V1. Fluid pass over at exit with a velocity: df (Ve).

If the friction between the vane and fluid is zero or negligible there is no force. Hence no rate of change of velocity. Thus the velocity Ve = Vr (cb = df)

The absolute velocity of fluid at the exit is found by drawing triangle at exit. If there is no shock then the velocity is tangential to the vane at exit. Fe = magnitude of v, velocity of vane. Now θ is the angle between the relative velocity and direction of motion at inlet and Φ is the angle between the relative velocity and direction of motion at exit. If the fluid is to enter and leave the vane without a shock the angles of blades at inlet and outlet mode equal to θ and Φ respectively. Weight of fluid per second = (w) x a x V = W Force on the area = W/g x [cg-(-eh)] W/g = (W/g) x (Vw + Vw1):

where cg = Velocity of whirl at enteance = Vwand eh = Velocity of whirl at exit = Vw1

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There is no friction, We have Work done/sec = Force x velocity = (W/g) x (Vw + Vw1)xv In case direction of Vw1 is same as the direction of vanes, then Work done/sec = (W/g) x (Vw – Vw1)xv.

Now work done is also equal to change in kinetic energy = (WxV2/2g) – (WxV12/2g)

Efficiency = Work done/ Input energy = [{(WxV2/2g) – (WxV12/2g)}/ (WxV2/2g)] Efficiency = [1-(V12/V2)] Now let us see when we get maximum efficiency. The efficiency is maximum for a given angle of α If Φ is zero. Then Vw1 = (Vr -v) where Vr = relative velocity at the entrance which we have already defined. If α = 0; Then Vr = V-v and V1 = V–2v. therefore V1 = 0. When v = V/2. In this specific case efficiency = 1 if we consider there are no losses. The form vane is to be semicircular for unity Efficiency.

6.3. FORCES ON BENDS Fluid flows in pipes. In many circumstances, we have to change the direction of flow by bending the pipes that carries fluid. The fluid flows with certain velocity and it is a streamline flow. Whether the pipes are having different cross-section or uniform taper, there is always change in direction of velocity due to change in the direction of flow in pipes which are bent. This change in velocity and mass of the fluid flowing will cause force acting on the bend. The bend has to withstand this force. The bend are to be designed to withstand the force and also properly secured in position without movement or rotation of bend because force acting. Hence we need to know the relation between the force generated, fluid flow rate, and its velocity. Now we study this aspect. Now let us derive a general expression for relation among the forces acting and flow parameters. Figure 6.6 shows a reducing vertical bend. The discharge = Q. The weight of water in the bend = W. The area, velocity density, and pressure are (A1, V1, ρ1,p1)and (A2, V2, ρ2, p2), ρ1 = ρ1 = ρ(Constant)

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Figure 6.6.

The normal to entrance and exit the surfaces makes angles θ1 and θ2. The center of exit end at a height of y above the center of entrance. Now V1 = Q/ A1 = Q/A2. p1 + (V12/2g) + 0 = p2 + (V22/2g) + y;

Fx = ρxQx(V2x Cos θ2 – V1x Cos θ1) + p2xA2x Cos θ2 – p1xA1x Cos θ1

Fy = ρxQx(V2x Sin θ2 – V1x Sin θ1) + p2xA2x Sin θ2 – p1xA1x Sin θ1 + W Examples:

Parallel pipe right angle bend: Figure 6.7 The bend axis is horizontal. Hence; Area A1 = A2; θ1 = 0;

θ2 = 90; V1 = V2 = V,

Fx = ρxQx(V2x Cos θ2 – V1x Cos θ1) + p2xA2x Cos θ2 – p1xA1x Cos θ1

Fx = ρxQx(Vx 0 – Vx 1) + pxAx0 – pxAx1 = ρxAxVx(Vx 0 – Vx 1) – pxA = -(ρxAxV2 + pxA) Fy = ρxQx(V2x Sin θ2 – V1x Sin θ1) + p2xA2x Sin θ2 – p1xA1x Sin θ1 + W

Fy = ρxQx(Vx 1- V1x 0) + pxAx 1 – pxAx 0 + W; W = 0 Since Fy is not vertical. Fy = ρxAxVx(Vx 1 – Vx 0) + pxA = (ρxAxV2 + pxA)

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Figure 6.7.

Fy = ρxQx(V2x Sin θ2 – V1x Sin θ1) + p2xA2x Sin θ2 – p1xA1x Sin θ1 + W

Fy = ρxQx(Vx 1- V1x 0) + pxAx 1 – pxAx 0 + W; W = 0 Since Fy is not vertical. Fy = ρxAxVx(Vx 1 – Vx 0) + pxA = (ρxAxV2 + pxA) If it is a U-Bend as shown Figure 6.8:

Pipe diameters are equal in both limbs Area A1 = A2 = A; θ1 = 0; θ2 = 180; p1 = p2 = p; V1 = V2 = V

Figure 6.8.

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Fx = ρxQx(V2x Cos θ2 – V1x Cos θ1) + p2xA2x Cos θ2 – p1xA1x Cos θ1

Fx = ρxQx(Vx (–1) – Vx (1)) + pxAx (–1) – pxAx 1 = 2xρxV2 + 2xpxA = –2(ρxV2 + pxA) Similarly Fy = 2(ρxV2 + pxA);

In case the bend is vertical, then to Fy the weight of fluid in the bend is added.

6.4. ROCKETS MOTION The Figure 6.9 shows symbolic representation of rocket projected vertically up against gravity. It is in the direction of earth radius. Gravity g is acting. The rocket has a total mass that include the mass of all fuel, which being consumed continuously. This is substantial. Thus the mass of rocket is being reduced every moment. The loss of mass may be considered as uniform and proportional to time. Let the initial mass is M0. Let mass being reduced at a rate of α. Hence any instant loss of mass = αxt.

Figure 6.9.

The mass the rocket at an instant t = (M0 – αxt) Exit cross area = Ae, pressure pe, density of fuel = ρe. Velocity of exit fuel = ve, Then rate of brunt fuel ejected = α = ρexAexve. V = velocity of rocket at the instant considered.

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Now the force at any instant Mx Acceleration = force at the instant. (Inertia force + resultant force on the object = 0) Remember force is a vector. Hence we have, Mx(dV/dt) = (ρexAexve2) + pexAe – Mxg; M = mass at the instant = (M0 – αxt); Hence we have (dV/dt) = [{ρexAexve2) + pexAe – (M0 – αxt xg}/(M0 – αxt) ]; dV = [ { (αxve) + (pexAe)- (M0 – αxt)}/(M0 – αxt) ]xdt; dV = [{(αxve) + (pexAe)}/(M0 – αxt)} – g] xdt

On integrating we have V = -{ (αxve + pexAe)/α} x {In (M0 – αxt) – (In M0)} – (gxt)]; This further reduces to V = [ ve + (pexAe/ α)]x In [M0/ (M0 – αxt)] – gt;

6.5. JET PROPULSION

Figure 6.10.

In case of jet propulsions as in aircraft, the complete engine comprises of a compressor, combustion chamber, exhaust chamber and a jet nozzle. In the compressor chamber, atmospheric air is taken in, since we need oxygen for combustion. The compressor compresses the air to a high pressure. Then this air in the combustion chamber burns the fuel. The fuel expands into a very high volume at a constant pressure. Then the exhaust gas exits the nozzle with a very velocity. This high velocity provides a reaction force that propelled the aircraft. (It follows Newton’s third law.). Now let us analyze the relations of flow parameters, area, pressure, mass of gas, etc. Refer the Figure 6.10 for all fluid/flow parameters.

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Air enters at velocity V1 in the opposite direction to the aircraft velocity. V2 is the velocity of exhaust gases. The equation of motion is as under: p2 xA2 + pax(A – A2)- pa x (A-A1) -p1xA1 = (p2-pa)xA2 – (p1-pa)xA1 Now ma = ρxV1xA1; mf = mass rate of fuel = . Let us say fuel is mixed in the air in the air ratio = 1: N; Then, ρ1xV1xA1 + (1/N)xρ1 xV1xA1 = ρ2xV2xA2; Now ma x(V2-V1) + mfxV2 + (p2-pa)xA2- (p1-pa)xA1 = Thrust,(F) mf = (ρ2xA2xV2 – ρ1xA1xV1) Hence we have F + (p1-pa)xA1-(p2-pa)xA2 = (ρ2xA2xV2)xV2 – (ρ1xA1xV1) F = ρ1xA1xV1[{1 + (1/N)}xV2 -V1] + (p2-pa)xA2 – (p1-pa)xA1; We can make the following simplification after considering the following point/approximations 1. At subsonic speeds (Velocity of aircraft is less than that of air), the pressure at inlet is equal to pa p1 = pa; 2. In most of the cases the pressure at exhaust is very close to pa; Hence p2 = pa; 3. Fuel percentage compared with the air mass is very low. Hence N is very large compared with unity. Hence we can neglect (1/N) With the above modifications, we get Thrust, F = (ρ1xA1xV1)x(V2-V1).

6.6. LOSS OF ENERGY IN PIPE EXPANSION This has been dealt in the previous chapter. Angular Momentum: (Figure 6.11): We have seen fluid exerts force. We also saw in case of curved vanes, blades mounted on wheel, the fluid makes the object to rotate. That is fluid can also generate torque, moments that can be used in hydraulic turbines, rotating machinery and also in hydraulic motors to generate rotary motion. In order to study this phenomenon and also analyze the relations of the parameters such as pressure, velocity, Torques, moments, etc., we need to first define the angular momentum of the fluid and its relation with torques, pressure, velocity, etc.

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Figure 6.11.

Refer the Figure 6.11. We are analyzing torque, force, relation by vector method Consider a control volume (Boundary) Volume as shown. Let there be an elementary fluid particle at P of mass m having a mass δm. Let us define the position of the particle by a position vector R and angle θ, considering anti-clock – wise as positive. Let V be the velocity vector. R = erxR Then we have V = erxVr + e θx V θ; where er and eθ are unit vectors in the direction of R and in the direction of increasing angle of θ. The direction of unit vector is perpendicular to direction R. The momentum of the fluid particle due to its mass m and velocity V = (δm xV); Angular momentum is the product of momentum and perpendicular distance of the momentum of the particle. Let the angular momentum of this small fluid particle = δM (This is a vector) Rate of change of angular momentum is defined as Torque. (This is nothing but Force multiplied by the perpendicular distance from the point where we need to find Torque) Mathematically [R x (δm xV) ]; “x” represents cross products of vectors. Therefore δM = [R x (δm xV) ]; The angular momentum = (erxR) x[(erxVr + e θx V θ)x(dm);

erx e θ = k; k = unit vector perpendicular to paper

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Angular momentum = kxRxV θx(δm); k = unit vector perpendicular to paper. R = R x (Sin θ) This reduced to, δM = RxV θx(δm) x (Sin θ);

Let us say Torque due to the element of flow of fluid = δ T θ = δ[(δM)]/ δt = δ[Rx V θx(δm/δt)x (Sin θ)x(δm)]/δt; δm = ρxδv; δv = volume of flow. T θ = Rate of change of angular momentum is defined as Torque T θ = ʃ[Rx V θx(Sin θ)x(ρxVnx dA]

δm/δt = [ ρxδAx (δs/δt)] = (ρxVnx dA); δA = small quantity, While integrating, we represent dA Vn = Velocity in the normal direction.(ds/dt) Now let us consider a specific case as under. Suppose the turbine has R1and R2 are radius at inlet and exit. Vt1 and Vt2 are fluid velocities at entry and exit then the Torque = (R2xVt2 – R1 x Vt1) xʃ(Vnx dA); ʃ(ρxVnx dA) = Q, Discharge Hence Torque, T = (R2xVt2 – R1 x Vt1)xρxQ;

6.7. LAWN SPRINKLERS Refer to Figure 6.12. A,Q,Vt section area of nozzle, total discharge from both nozzles, Vt is the tangential velocity respectively. R is the radius of the center of rotation and center axis of nozzles. He sprinkler is symmetrical on the center axis. Now total discharge from each nozzle = Q/2. Refer to torque equation, T = (R2xVt2 – R1 x Vt1)xρxQ;

Here R2 = R1 = R; Vt1 = Vt2 = Vt.; Let ω = Rotation of sprinkler in radians/sec; Torque = [Vtx Rxρx (Q/2)] – [-Vtx Rxρx (Q/2)]; since half the discharge from each nozzle.

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Figure 6.12.

Torque = Vtx Rxρx Q; As you know Vt = absolute velocity.

The sprinkler is rotating with a velocity = ωxR; Relative velocity, V = Vt – (-ωxR); because the direction of rotation of sprinkler is opposite to the velocity of jet. Now V = Q/(2xA); Vt = (V- ωxR); Torque = Rxρx Qx [ {Q/(2xA)} – ωxR]; Suppose there is not friction of it is negligible, we can say T = 0 Hence we have ω = [Q/(2xAxR)]; The just limiting torque by which the sprinkler do not rotate, ω = 0 Hence we have Torque = Rxρx Qx [ {Q/(2xA)} – 0xR] = [Rxρx Q2/ (2xA)];

6.8. PELTON WHEEL This is one of the turbines most popular even today. This is discovered in the early stages. Figure 6.13 is a schematic diagram of a Pelton wheel. In its simplest form, it has a wheel keyed to a shaft. This shaft in turn coupled to a generator. On the periphery of the wheel are fitted a number of cups. The number of cups depends on the design. The number of cups is such it should not interfere with the jet. The typical section of the cup is shown in Figure 6.13. The fluid is water. The single jet enters at the center of the cup as shown. Then water exit on either side of cups at an angle. The jet is discharged to the atmosphere. The boundary (Control volume) encloses the wheel.

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Let us analyze force, Torque, and power: Vt = velocity of the jet. Tangential to the cup. It will impinge the cup directly. Ve is the exit velocity. Q = discharge from the nozzle and is constant till exit. Hence Momentum = Vtx Qxρ; at entry. Now exit velocity relative to cup = (Vt – ωxR);

Figure 6.13.

Hence tangential velocity, absolute (relative to ground) = Ve = [- (Vt – ωxR)x Cos β + ωxR]; This is because the jet actually exits at an angle β from either side of cup. The momentum of the water leaving the cups in the tangential direction = ρxQxVe; Substituting for Ve We have momentum, = ρxQx [- (Vt – ωxR)x Cos β + ωxR] Rate of change of momentum, = ρxVtxQ – [ρxQx [- (Vt – ωxR)x Cos β + ωxR]; Hence Force = ρxVtxQ – [ρxQx [- (Vt – ωxR)x Cos β + ωxR]; Simplifying, Force = ρxQx[Vt – ωxR] x (1 + Cos β);

Torque = Force x Radius of pelton wheel = ρxQxRx[Vt – ωxR] x (1 + Cos β)

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Power developed, P = Tx ω = ρxQxRx ω x[Vt – ωxR] x (1 + Cos β);

When β = 0 Torque is maximum = 2xρxQxRx[Vt – ωxR]. That is turns by 180 Deg. Let us find out maximum power when it occurs. P = ρxQxRx ω x[Vt – ωxR] x (1 + Cos β); P is max when dp/dω = 0; when we consider the RPM is a variable for a given configuration. We can have this as a variable. Hence we get max power occurs Vt = 2x ωxR Pmax = ρxQxR2x ω 2x (1 + Cos β);

6.9. REACTION TURBINES (Figure 6.14) This is the next popular turbine. Here the principle of working is slightly different. The principle is similar to sprinklers which we analyzed in the chapter. The water flows out with a velocity. The reaction of velocity makes the turbine to rotate. In this case of turbines, the water is not discharged to the atmosphere, instead, water is discharged above the tailrace. The turbine is totally is enclosed. The exit of the turbine is normally will have vacuum pressure, in order to make use of full energy of water. This, one will understand when they study the turbines in detail and the design. The portion, after the exit of the turbine that is connected to the tail end water level, is called a draft tube, this is like a diverging conical shape. The most popular turbine of this type is called Francis turbine. In this case, water enters on the external periphery of the turbine. Water flows on fixed guide vanes which decide the angle at which the water enters the runners. Runners are rotating vanes. The angle of fixed guide vanes is adjustable within the range. Thus we are able to change the angle of entry depending on the load on the turbines. The water discharges to the center and then flows out in the direction of the axis of the turbine. The designed can be designed with its axis either horizontal or vertical. Now let us find out the power and torque developed based on the discharge and velocity of flow.

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Figure 6.14.

Now, let us find the equations for power, torque, force etc. See the Figure 6.14 for nomenclatures for velocity, angles, etc. P = power developed. T = Torque developed. Q = discharges of water. We apply force angular momentum principle to find out the relationships among various parameters. Always the flow streamline. The turbine rotates on the vertical axis passing through center o. V1, V2, are absolute velocities at entry and exit respectively. Vt1,Vt2, are tangential velocities and Vr1, Vr2 radial velocities components. α β are guide vanes and runner vanes angles. From geometry, Vt1 = V1x Cos α; Vt2 = V2x Cos β; Torque developed is opposite to torque of the shaft. Apply T = (R2xVt2 – R1 x Vt1)xρxQ

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Thus, we have T = ρxQx [ R1xV1x Cosα – R2xV2x Cos β]; Power, P = ρxQx ω x [ R1xV1x Cosα – R2xV2x Cos β];

Q can be found from Vr1, Turbine width, and area occupied by the guide valves. Let us see how it can be calculated. The peripheral area through where water enters = 2xpi xR1x W, Velocity with which it enters, Vr1 = V1x Sin α. Hence Discharge = (V1x Sin α x pi xR1xW). But the entire area 2xpi xR1x W cannot be considered because a portion of the area is occupied by Guide Vanes. Say 2x k xpi xR1x W is the area occupied by the guide vanes, then we have area = 2xpi xR1x W(1-k) Hence discharge, Q = (V1x Sin α x pi xR1xW)x(1-k).

CHAPTER

7

IRROTATIONAL FLOW

CONTENTS 7.1. General Motion Of A Fluid Element................................................. 132 7.2. Uniform Flow.................................................................................. 136 7.3. Plane Potential Flow In Polar Coordinates........................................ 137

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7.1. GENERAL MOTION OF A FLUID ELEMENT In this chapter is focused on deriving the equations of motion of a fluid element, then, finding the conditions for which the flow is uniform, steady, non-uniform or unsteady. Let us consider a fluid element in the flow field plane. Let the space coordinates be (x,y,z) and let the time instant be t. After a time of δt, the fluid element is position at a new space coordinate [(x + δx),(y + δy),(z + δz)], Velocity at the instant are U, V, W Let the velocity functions are U(x,y,z,t), V(x,y,z,t) and W(x,y,z,t) These are the functions of x,y,z and t,in general. After δt seconds, Velocities are [U + δU), (V + δV), (W + δW)] (U + δU) = U[(x + δx),(y + δy), (z + δz), (t + δt)] (V + δV) = V[(x + δx),(y + δy), (z + δz), (t + δt)] (W + δW) = W[(x + δx),(y + δy), (z + δz), (t + δt)] Now we apply Taylor’s theorem to the above functions and expand. (Please refer advance mathematics to know about the theorem. Thus we have (U + δU) = U(x,y,z,t) + {(δU/δx)x Δx} + {(δU/δy)x Δy} + {(δU/δz)x Δz} + {(δU/δt)x Δt} + ---(Higher order of differentials which are neglected since they reduces to zero) (V + δV) = V(x,y,z,t) + {(δV/δx)x Δx} + {(δV/δy)x Δy} + {(δV/δz)x Δz} + {(δV/δt)x Δt} + --- ------(Higher order of differentials which are neglected since they reduces to zero) (W + δW) = W(x,y,z,t) + {(δW/δx)x Δx} + {(δW/δy)x Δy} + {(δW/δz)x Δz} + {(δW/δt)x Δt} + --- ------(Higher order of differentials which are neglected since they reduces to zero) Now note, Δx = Ux Δt; Δy = Vx Δt; Δz = Wx Δt Substitute the values: δU = {(δU/δx)x Ux Δt } + {(δU/δy)x Vx Δt } + {(δU/δz)x Wx Δt } + {(δU/ δt)x Δt} δV = {(δV/δx)x Ux Δt } + {(δV/δy)x Vx Δt } + {(δV/δz)x Wx Δt } + {(δV/ δt)x Δt} δW = {(δW/δx)x Ux Δt } + {(δW/δy)x Vx Δt } + {(δW/δz)x Wx Δt } + {(δW/δt)x Δt}

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δU/ Δt = {(δU/δx)x Ux } + {(δU/δy)x V } + {(δU/δz)x W } + {(δU/δt)} δV/Δt = {(δV/δx)x U } + {(δV/δy)x V } + {(δV/δz)x W } + {(δV/δt)} δW/ Δt = {(δW/δx)x U } + {(δW/δy)x V } + {(δW/δz)x W } + {(δW/δt)} Hence we have by limiting Δt -- 0; We have: (dU/dt) = {(δU/δx)x U } + {(δU/δy)x V } + {(δU/δz)x W } + {(δU/δt)} (dV/dt) = {(δV/δx)x U } + {(δV/δy)x V } + {(δV/δz)x W } + {(δV/δt)} (dW/dt) = {(δW/δx)x U } + {(δW/δy)x V } + {(δW/δz)x W } + {(δW/δt)} From the above we can say (Total acceleration) = (Acceleration due to position) + (Acceleration due to time) In case of steady flow the flow, do not vary with time: Hence we have: (δU/δt) = 0; (δV/δt) = 0; (δW/δt) = 0 In case of uniform flow: Flow is constant at any position. Not the function of co-ordinates. {(δU/δx)x U } + {(δU/δy)x V } + {(δU/δz)x W } = 0 {(δV/δx)x U } + {(δV/δy)x V } + {(δV/δz)x W } = 0 {(δW/δx)x U } + {(δW/δy)x V } + {(δW/δz)x W } = 0 In case of steady and uniform flow, both are zero, no acceleration. In case Non-uniform and non-steady none them are zero. Let us understand the difference between Steady flow and uniform flow with a practical example.

Steady flow example: Refer the Figure 7.1a. Here the flow is out of a tank through a taper pipe. The fluid level is maintained by pumping fluid from outside source. Thus fluid level, H is constant. We know in this case we know velocity is proportional to fluid level. Hence there is no change in velocity with respect to time. But the flow velocity at sections A, B and C are not same and they varying at any instant of time. This because the discharge is constant and area of crosssection of pipe varies with x-axis. This type of flow is called steady flow. In this particular case, flow is in one-dimensional. Same thing applies for three dimensional of flow.

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Figure 7.1a.

Uniform flow example: Consider the case of flow as in Figure 7.1-b.

Figure 7.1-b.

In this case, the fluid level in the tank is not constant and varies with time. As the water flows with time the fluid level reduces. Hence the velocity also reduces. It means the velocity changes with time. The fluid velocity at section A, B and C are same, since the pipe is parallel and area of cross section of flow is constant. Hence the flow does not vary with x -coordinates. It varies only with time. This flow is uniform flow.

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Uniform and steady flow: Refer to the Figure 7.1c. Here the flow does not vary with time since H head constant. It also do not very at sections A, B and C. Hence flow may be called uniform and steady.

Figure 7.1-c.

Non-Uniform and Unsteady example: Refer Figure 7.1-d:

Figure 7.1d.

Here the flow is not steady because the fluid level is not constant, it varies with the time. The pipe through which it flows also varies with diameter.

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Hence, the velocity at A, B, and C are not the same. This means the flow is not uniform. This is a non-uniform and unsteady flow.

7.2. UNIFORM FLOW Let us consider an uniform flow of constant velocity, U parallel to X-Axis. δΦ/δx = u = U = δΨ/δy: v = 0 = δΦ/δy = -(δΨ/δx) (Refer velocity potential and stream function, Chapter 4); Keep in mind Φ and Ψ are orthogonal By integrating we have Φ = Ux and Ψ = Uy; Integrating constant are Φ = 0 and Ψ = 0 at the origin. Suppose the flow is uniform, constant velocity, V parallel to Y-axis, Then we have u = δΦ/δx = δΨ/δy = 0; and δΦ/δy = v = V = -(δΨ/δx); We have Φ = Vy; Ψ = Vx. Let see the graphs from both conditions: Figure 7.2: U = constant; In case of V = constant, Similar to Figure 7.2, but Φ, Ψ, are interchanged. Suppose the flow is inclined to X and Y axis, let us see. U = V0x Cos θ; V = V0x Sin θ; Hence we have

δΦ/δx = δΨ/δy = V0x Cos θ AND δΦ/δy = – δΨ/δx = V0x Sin θ; Om integration We get Φ = Uxx + Vxy; Ψ = -Vxx + Uxy; The flow net is as shown in the Figure 7.3.

Figure 7.2

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137

Figure 7.3

7.3. PLANE POTENTIAL FLOW IN POLAR COORDINATES Consider the flow in natural springs. The flow is from center and distributing radial. In such cases, we call the flow originating point as source. The discharge is defined. The velocity is very high, cannot be measured. When flow is streamline the velocity at radial points are defined and can be measured. This is considered as flow from a center of a sphere. This is a three-dimensional flow. Q = Discharge constant. Velocity at the center is infinite (Very high). But velocity at different can be measured o calculated. U = Velocity = dΦ/dR; where R = radius of sphere; U = (Q/4xpixR2) M = (Q/4xPi) is called strength of source. This is similar to the light emitting from Sun, in the ideal case. A light rays emitting from a point source like light bulb. In case of 2 -Dimensional flow, as in the case of flow from the center and diverging between two narrow plates as shown in sketch. Let us say q = Discharge, ur = dΦ/dr = (q/2xpixr);

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ON integration we get Φ = [q/(2xPi)xIn r] = mxIn r where m = q/(2xpi); ur = (1/r)x dΨ/d θ = q/(2xpixr); On simplifying we get Ψ = mxθ; The equations, Φ = [q/(2xPi)xIn r] and Ψ = mxθ indicates that lines of constant Φ are circles and lines of constant Ψ are radial stream lines.

Figure 7.4

Sink If the flow is inward it is called sink. Similar to flow in wash basin. Then Φ = -[mx In r] and Ψ = mxθ; m = q/(2xpi); Source and Sink: Let us consider a case where there are two situations, a source, located at a distance of a sink. (Refer Figure 7.5)

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139

Figure 7.5.

Velocity potential: Φp = Φsource + Φsink = (mxIn r2 – mxIn r1) = mx (In r2/In r1); Stream function:

Ψp = m(θ2 – θ1);

The magnitude and direction of the velocity at P may be obtained by adding the two quantities by vectos addition. Flow around Corners: Consider now family of general solution to the Laplace’s equation: Φ = Axrnx(Cos nθ); nθ)

By substitution, we have ur = dΦ/dr = (1/r) x(dΨ/dθ) = nxAxrn–1x (Cos

uθ = (1/r) x (d Φ/dθ) = – (dΨ/dr) = – nxAxrn–1 x (Sin nθ) We can evaluate dΨ as under:

dΨ = (δΨ/δr) x dr + (δΨ/δθ) x dθ; We get dΨ = nxAxrn–1 x (Sin nθ) x dr + Axrnx(Cos nθ) xdθ; Ψ = Axrnx Sin nθ (By integrating) between limits (r,0 and 0,0) and (r,θ and r,0); For every values of n there are two values of θ for which Ψ = 0 is satisfied. Applying boundary conditions, we are in position ot obtain solutions for flow around corners.

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A table here below gives atypical values of n and real values of A. Flow pattern around corners n Ψ=0 1/2 1 2 4

 

θ = 0 and θ = 2pi θ = 0and θ = (Pi) θ = 0and θ = (Pi/2) θ = 0and θ = (Pi/4)  

flow Around plate Past a plate Around Right angle bend Around 45 Deg bend

 

           

Irrotational Flow

Flow around corner 270°.

Flow at corner 120°.

Flow around Circular cylinder.

141

CHAPTER

8

LAMINAR MOTION

CONTENTS 8.1. Introduction..................................................................................... 144 8.2. Shear Stresses.................................................................................. 144 8.3. Navier-Stokes Equations And Solutions............................................ 146 8.4. Flow Between Horizontal Parallel Plates.......................................... 151 8.5. Flow Between Two Plates With X-Axis Along The Lower Plate.......... 153 8.6. Couette Flow................................................................................... 154 8.7. Combined Hagen-Poiseuille And Couette Flows.............................. 155 8.8. Hagen-Poiseuille Flow In Cylindrical Tubes..................................... 156 8.9. Laminar Flow In An Annulus Area.................................................... 158 8.10. Rotating Cylinders......................................................................... 159 8.11. Reynolds Number And Stability Parameters................................... 160 8.12. Stability Parameter......................................................................... 161 8.13. Stability Curves.............................................................................. 163 8.14. Laminar Boundary Layer................................................................ 164 8.15. Boundary Layer Thickness.............................................................. 165 8.16. Displacement Thickness................................................................. 165

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8.1. INTRODUCTION We have studied the viscosity in chapter one. We defined that the fluid flows layer by layer. The bottom-most layer sticking to the vessel has zero velocity and increases with increasing distance from bottom. It reaches maximum at the center of pipe/ vessel in which it is flowing. We all seen the oil flowing from a barrel when it is to be drawn out by a tap, the flow is as if there is a transparent tube coming out and it looks still. In such conditions, the velocity has certain value. If velocity increases it may looks still and transparent. There is a limiting value of velocity. The flow up to certain velocity is called laminar flow. This is just a commercial explanation. Now let us analyze under what conditions the flow is laminar and also let us find the fluid parameters related to this type of flow. In the laminar flow, the fluid particles do not cross over and particles do not move at random. Any small external disturbance will change the conditions. Then the flow becomes turbulent. You note in such case there are two forces acting, Inertia forces, and viscous forces. So long the inertia force is less than viscous force, flow tending to be laminar. If inertia exceeds viscous force then the flow becomes turbulent. The factor that decide the flow laminar or not, is the ratio of inertia force to viscous force. Hence we define Re = [Inertia Force/Viscous Force]; This ratio is called Ronald’s number, Re Let us find its value of Re: Inertia force = Mass x Acceleration.

Viscous force = {Dynamic viscosity x Velocity gradient}x Area of cross section. Re = [{ρx L3x v/t} x/ {μx (v/L)x L2}] = [ρxL2x(1/t)/ μ] = [ρxL x (L/t)/ μ] Re = [ρxL x v/ μ] = vL/η; η = μ/ρ; η = Kinematic viscosity.

8.2. SHEAR STRESSES We have learned about viscosity in a earlier chapter. In this chapter, we will study stability in great detail. Viscosity plays a very important role in laminar flows. We need to know complete details about shear stresses and its relation with velocity, pressure, and other fluid parameters. Let us derive an expression that relates shear stress with velocity of flow and the distance from a datum. The viscosity is the property of fluid that

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145

develops a resistance against flow under action of shear forces (see Figure 8.1). Let us find the relation between shear stress and velocity gradient. Shear stress is the force per unit area. Velocity is the flow/unit time. The ratio of change in velocity per unit length in the direction perpendicular to shear stress acting. ρ is the density, dy is the infinitely small layer of fluid element. The velocity increases from the bottom to top. Now let us change in momentum from bottom to top of “dy” thickness.

Figure 8.1

Change in momentum = [(ρ + δρ)x(u + δu) – ρu] = δ(ρu); The momentum across the depth = [δ(ρu)/δy] This change momentum proportional to force acting. The only force acting is hear force. Shear stress = shear force/unit area = say = fs;

We have fs = η x[ρx(δu/δy)]; Let us say; η xρ = μ. Then we have fs = [μx((δu/δy)]; In the differential form; fs = μx(du/dy); This is one dimensional flow. Let us fine in three dimensional flow as under:

Let us say fxy = shear stress in the xy plane. fyz = Shear stress in the yzplane; fzx = Shear stress in zx plane.

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u = velocity in x direction, v = velocity in y -direction; w = velocity in zdirection. Now is the function of x,y,z, u = u(x,y,z); v = v(x,y,z); w = w(x,y,z); Please refer applied mechanics, where it is shown, fxy = fyx; fyz = fzy; fzx = fzx; You consider elemental thickness, δx, δy, δz; Velocities u,v,w

The we get fxy = μx[ (δv/δx) + (δu/δy)]; fyz = μx[ (δw/δy) + (δv/δz)]; fzx = μx[ (δu/δz) + (δw/δx)] -- (1) The equations as (1) are very important and be used in proving NavierStokes equations.

8.3. NAVIER-STOKES EQUATIONS AND SOLUTIONS Consider the parallelepiped the front view is as shown. This is a XY-plane. ZX and Zy planes are perpendicular to paper space. , fyx, fzy, fxx are shear stresses in planes XY, YZ, and ZX planes. The stresses on the opposite faces of these planes are opposite and of equal magnitude., fxx = Normal stress in the X-direction. ,[fxx + (δfxx/δx)x(dx/2)] = Stress after a length of dx. (δfxx/δx) = Rate of increase in normal stress.

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147

Figure 8.2

Mass of the element = ρx(dxxdyxdy); u,v,w are velocities of fluid elemental in X, Y and Z directions , dx,dy, and dz are elemental infinitely small length/thicknesses. δfzx/δy, δfzy/δx, δfyx/δx, δfxy/δz, are rate of shear in XZ, YZ, XY, planes fzx, fzy, fyx are shear stresses. fxx is the normal stresses.

Figure 8.2 shows XY and XZ planes, Once again apply Newton,s second law: We have to consider all the forces acting on the fluid element. The forces are: Inertia force = Mass of element x acceleration; Weight of the element (External forces acting on the element. In our case only gravitational force is acting); • Shear forces acting on the faces; • Normal forces acting parallel to axis. For equilibrium of the fluids element: • •

Inertia force = Weight of the fluid element + Shear forces + Normal forces. The components of all these forces should be considered in the respective directions, i.e., X, Y and Z direction and XY, YZ and ZX planes. The components should be zero individually. Now let us consider forces in X-direction: (datum center of fluid element) ρx(dxxdyxdz)x (du/dt) = X + [{fxx + ((δfxx/δx)xdx)}x(dyxdz) – fxxx(dyxdz)] + [{fxzxδxxδz) + (δfxz/δyx(dy/2))x(δxxδz)}-{ (fzxxδxxδz)-(δfzx/δy)x(dy/2) x(δxxδz}] +

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; [{fyxxδxxδy) + (δfyx/δzx(dz/2))x(δxxδy)}-{ (fxyxδxxδy)-(δfxy/δz)x(dz/2) x(δxxδy}]; ,ρx(dxxdyxdz)x (du/dt) = X + (δfxx/δx)x(dxxdyxdz) + [ δfxz/δyx(dyxδxxδz] + [δfxy/δzx(dyxδxxδz)] ρx (du/dt) = X + (δfxx/δx) + (δfxz/δy) + (δfxy/δz); ------(A);

We have fxy = μx{ (δv/δx) + (δu/δy); fzx = μx{ (δu/δz) + (δw/δx) (refer 7.1 of this chapter) ----(B) QUOTE Refer to chapter Seven.7.0 General motion of fluid: We have: (dU/dt) = {(δU/δx)x Ux } + {(δU/δy)x V } + {(δU/δz)x W } + {(δU/δt)} (8.1) (dV/dt) = {(δU/δx)x U } + {(δU/δy)x V } + {(δU/δz)x W } + {(δU/δt)} (8.2) (dW/dt) = {(δW/δx)x U } + {(δW/δy)x V } + {(δW/δz)x W } + {(δW/δt)} (8.3) QUOTE END. Now du/dt = du/dt = {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w } + {(δuδt)}; It can be shown for in-compressible fluid flows, (The proof is quite long and it is not present here). Take it granted proved. fxx = -p + 2 μ (δu/δx);

(8.4)

fyy = -p + 2 μ (δv/δx); fzz = -p + 2 μ (δw/δx); where, -p = [(fxx + fyy + fzz)/3]; ------(5) Substituting B, 1,4 and 5 in A, η xρ = μ RHS(X) = {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w } + {(δu/δt)} (8.5) RHS(Y) = {(δv/δx)x u } + {(δv/δy)x v } + {(δv/δz)x w } + {(δv/δt)} (8.6) RHS(Z) = {(δw/δx)x u } + {(δw/δy)x v } + {(δw/δz)x w } + {(δw/δt)} (8.7) {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w } + {(δu/δt)} -----RHS(X) = X – (1/ρ) x(δρ/δx) + 2x ηxδ2u/δx2 + ηx {(δ2u/δy2) + (δ2v/δx δy) } + ηx{ (δ2u/δz2) + (δ2w/δx δz))

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RHS(X) = X – (1/ρ) x(δρ/δx) + ηx {(δ2u/δx2) + (δ2v/δy2) + (δ2u/δz2)} + ηx[δ/ δx {(δu/δx) + (δv/δy) + (δw/δz)}] Because of equation of continuity, ηx[δ/δx {(δu/δx) + (δv/δy) + (δw/δz)}] = 0; RHS(X) = [X – (1/ρ) x(δρ/δx) + ηx {(δ2u/δx2) + (δ2v/δy2) + (δ2u/δz2)}]; RHS(Y) = {(δv/δx)x u } + {(δv/δy)x v } + {(δv/δz)x w } + {(δv/δt)} RHS(Z) = {(δw/δx)x u } + {(δw/δy)x v } + {(δw/δz)x w } + {(δw/δt)} RHS(Y) = [Y – (1/ρ) x(δρ/δy) + ηx {(δ2v/δx2) + (δ2v/δy2) + (δ2v/δz2)}]; RHS(Z) = [Z – (1/ρ) x(δρ/δz) + ηx {(δ2w/δx2) + (δ2w/δy2) + (δ2w/δz2)}]; Now {(δ2/δx2) + (δ2/δy2) + (δ2/δz2)} is termed as V 2. Hence We have RHS(X) = [X – (1/ρ) x(δρ/δx) + ηx V2u];

(X)

RHS(Y) = [Y- (1/ρ) x(δρ/δy) + ηxV v];

(Y)

RHS(Z) = [Z-(1/ρ) x(δρ/δy) + ηxV w];

(Z)

2

2

Please note external forces are X,Y, Z; X = 0,Z = 0 and Y = -g (Gravity force). Vector form of Navier’ Stokes equation: It is easier to remember. The readers should know about vectors to understand the equation. Let us say V = Velocity vector = iu + jv + kw, where I, J, and K are unit vectors. Then RHS(X) + RHS(Y) + RHS(Z) = DV/Dt; V = (δ/δx) + (δ/δy) + (δ/ δz)]; We have ρ(DV/Dt) = – (Δp) + ρ g + μΔ2V Vector method and Equate i,j, k vectors components. , μ = Dynamic viscosity. We get the values. ρ(DV/Dt) = ρ{D (iu + jv + kw)/Dt} = ρ[ ix(du/dt) + jx(dv/dt) + kx(dw/dt); ix[du/dt] = ix [{(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w } + {(δu/δt)}]; jx[dv/dt] = jx[{(δv/δx)x u } + {(δv/δy)x v } + {(δv/δz)x w } + {(δv/δt)}]; kx[du/dt] = kx[{(δw/δx)x u } + {(δw/δy)x v } + {(δw/δz)x w } + {(δw/δt)}] This is because u = u(x,y,z,t); v = v(x,y,x,z); w = w(x,y,z,t); ρ g = ρx(ixgx + jxgy + kxgz); gx,gy,gz are external forces,X,Y,Z.

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μΔ2V = μΔ2[iu + jv + kw] = μ[i V 2u + V 2v + V 2w] Δ2u = [ (δ2u/δx2) + (δ2u/δy2) + (δ2u/δz2)] Δ2v = [ (δ2v/δx2) + (δ2v/δy2) + (δ2v/δz2)] Δ2w = [ (δ2w/δx2) + (δ2w/δy2) + (δ2w/δz2)] Substitute values and separate i,j, and k components. We get Naviers stokes equation in X, Y, z direction. The solution for the Navier’s equation depends on boundary conditions and the flow type. For steady flow, RHS(X) = {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w } + {(δu/δt)}; (δu/ δt) = 0 RHS(X) = {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w }; Similarly it follows, RHS(Y) andRHS(Z) For uniform flow {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w} = 0; For uniform and steady flow RHS(X) = 0; RHS(Y) = 0 RHS(Z) = 0 Hence equate 0 to LHS; for uniform and steady flow. Navier-Stokes Equations in Cylindrical form: The derivation is purely of mathematical in nature. Hence the equations are produced in polar form. Interested reader may refer book on advanced mathematics. Nomenclature: u, v, w are velocities in radial, tangential, and axial direction. Do not be confused with that used in coordinates system. They are also the same nomenclatures. r = radius, θ = angle, tangential direction R = Radial force, external, Z = external force in axial direction. T = external force in tangential direction. p = Pressure. Radial form of equation: (δu/δt) + u(δu/δr) + (v/r)(δu/δθ) + w(δu/δz) -(v/r2) = R-(1/ρ)(δp/δr) + η[(δ2u/δr2) + (1/r) x(δu/δr) + (1/r2)x(δ2u/δθ2) + (δu2/δz2) -(u/r2)-(2/r2)x(δv/δθ)] Tangential form of equation: (δv/δt) + u(δv/δr) + (v/r)(δv/δθ) + w(δv/δz) + (uv/r) = θ -(1/ρr)(δp/δθ) + η[(δ2v/δr2) + (1/r) x(δv/δr) + (1/r2)x(δ2v/δθ2) + (δ2v/δz2) -(v/r2) + (2/r2)x(δv/δθ)];

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Axial Form of equation: (δw/δt) + u(δw/δr) + (v/r)(δw/δθ) + w(δw/δz) = Z -(1/ρ)(δp/δz) + η[(δ2w/δr2) + (1/r) x(δw/δr) + (1/r2)x(δ2w/δθ2) + (δ2w/ δz2)];

8.4. FLOW BETWEEN HORIZONTAL PARALLEL PLATES (Hagen-Poiseuille flow) Refer Figure 8.3. Fluid is flowing between two parallel plates, as shown. Plates are separated by a distance of 2a. Let us analyze the forces acting and keeping the fluid in dynamic equilibrium.

Figure 8.3

Consider fluid element as shown in the Figure 8.3. Pressure = p; increase in pressure = [p + (δp/δx)xdx]; Shear stress = fx; Increase shear stress on the elemental thickness = [fx + (δfx/δy)xdy];

These two forces pressure force and viscous force, cause motion and in dynamic equilibrium. There is no acceleration. Let us consider force acting at an infinitely small distance y from center line. p x dy – [p + (δp/δx)xdx]xdy – fxx dx + [fx + (δfx/δy)xdy]xdx = 0;

On simplification, we get (δp/δx) = (δfx/δy); Note that p is the function of x only and fx is the function of y only, the partial derivative is the total derivative. Hence we have, (dp/dx) = (dfx/dy); By Newton’s law, fx = μx (du/dy);

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We have (dp/dx) = d{μx (du/dy)}/dy; This further simplified to, (dp/dx) = μx(d2u/dy2); (d2u/dy2) = (1/ μ)x(dp/dx); (AA) Integrating once, we have (du/dy) = (1/ μ)x(dp/dx)xy + C1 (constant of integration.) du/dy = 0 at y = 0, we have C1 = 0; Integrating, du/dy, We have, u = (1/ μ)x(dp/dx) x (y2/2) + C2; C2 can be evaluated by boundary conditions. That is, u = 0, y = -a or + a Substituting, We have u = -{1/ (2μ)}x (dp/dx)x (a2 – y2); The is the parabolic equation in y and dp/dx); The velocity is maximum at center, where y = 0; umax = -{1/ (2μ)}x (dp/ dx)x (a2); Negitive sing indicates that pressure decreases ni the direction of flow. The shear stress, fx = ±(dp/dx)xa at y = 0, fx = (dp/dx)xy Indicates it is the linear variation. ow let us say dq = elemental discharge. = uxdy;

We have total discharge between plates q = ʃ uxdy = -ʃ{1/ (2μ)}x (dp/dx)x (a2 – y2)dy; Limits are -a and + a between plates thickness = 2a, q = {2/(3 μ)} x (-dp/dx)xa3; Average velocity, uav = q/2a = {1/(3 μ)} x (-dp/dx)xa2}; The force on one plate length L due to shear is:

ʃfxdx = (Δp/L)xaxL = Δpxa; where Δp indicates pressure loss. For both plates combined the force is 2x(Δpxa);

Power required, P = force x average velocity = {2x(Δpxa)}x {1/(3 μ)} x (-dp/dx)xa2}; Power, P = {2/(3 μ)}x (Δp/L)2xa3xL.

Now let see how the above relation can be derived from Navier’s-Stokes equations: N-S equation in the x-direction: RHS(X) = [X – (1/ρ) x(δρ/δx) + ηx V2u]; Here the flow is uniform and stady. Hence RHS = 0 [X – (1/ρ) x(δρ/δx) + ηx V2u] = 0; There is no change in z-direction. Hence we have, for uniform flow,

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RHS is {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w } + {(δu/δt)}; (δu/δt) = 0 since flow is stady. {(δu/δx)x u } + {(δu/δy)xv } + {(δu/δz)x w } = 0 For uniform flow. No change in z-direction, (δu/δz) = 0; (δ2u/δz2) = 0; (δw/δz) = 0; Equation of continuity, if (δu/δx) = (δw/δxz) = 0 Hence we have (δv/ δy) = 0 This shows by integration v = constant, v = 0 at boundary and everywhere. Hence we have, [X – (1/ρ) x(δρ/δx) + ηx V2u] = [- (1/ρ) x(δρ/δx) + ηxV 2 u] = 0 - (1/ρ) x(δρ/δx) + μx[ δ2u/δy2] = 0; d2u/dy2 = (1/μ) x (dp/dx) (note ρ is the function of pressure) Note μ = dynamic viscosity and η = kinematic viscosity. Hence (1/ρ) x(δρ/δx) = dp/dx; compare the result with that of (AA). Both are the same.

8.5. FLOW BETWEEN TWO PLATES WITH X-AXIS ALONG THE LOWER PLATE Earlier we consider x-axis at middle of plates. Suppose we shift it to lower plat then the equation becomes: u = (1/μ)x(dp/dx)x{(y2/2) – ay} Inclined parallel plates: Let both be inclined as shown in Figure 8.4. Angle of inclination = a0 Gravitational component = gxSin(a) per unit mass.

Figure 8.4

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Equation of equilibrium: p x dy – [p + dp]xdy – fxx dx + [fx + (δfx/δy)]xdx + wxdxxdyx Sin(a) = 0; We have dxx Sin(a) = dh. On substitution and simplification we get,

dfx = d(p + wh)/dx You can say, d2u/dy2 = (1/ μ)x [d(p + wh)/dx];

Solving the differential for u,We have u = (1/ μ)x [d(p + wh)/dx] x {(y2/2)ay}; p + wh is called piezometric pressure. w = specific weight.

8.6. COUETTE FLOW Here one plate is fixed and other plate is moving at a uniform velocity. Nomenclature is as per Figure 8.5.

Figure 8.5

Lower plate is stationary. Upper plate is moving with a uniform velocity. You can visualize that a narrow space between plates is occupied with a viscous fluid. Top plate moves with a uniform velocity. This occurs in case of sliding guideways in Lathes or slides in Boring machines or any such machines. U = Uniform velocity. Here the pressure gradient is zero. There is a uniform pressure in between plates. The only forces acting are shear forces in opposite directions. We [fx + (δfx/δy)] – fx = 0; This reduces to (dfx/dy) = μ(d2u/dy2) = 0;

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Integrating, du/dy = C1, Further u = C1y + C2, u = 0 on stationery plate y = 0 and C2 = 0 And u = U at y = 2a (top plate moving with uniform velocity U); C1 = (U/2a); Thus we have u = (u/2a)xy; This indicates that velocity distribution is linear. Shear stress f0 = (μx U/2a) Average velocity = (U/2)x2a = Uxa;

The force required to move top plate of unit width and length, L = f0xL = [μx UxL/2a]; Now power required, P = Force x Average velocity = = [μx UxL/2a] x U = = [μx U2xL/2a];

8.7. COMBINED HAGEN-POISEUILLE AND COUETTE FLOWS We will consider the lower plate is stationary and to plate moving with a uniform velocity U. We consider a velocity gradient as in case of (dp/dx) in the direction of flow. The solution is obtained by superpose the two linear equations: 1) Flow between two plates with X-axis at lower plate. This is, u = (1/μ)x(dp/dx)x{(y2/2) – ay} 2) Couette flow; – This is, u = (u/2a)xy Then we have: , u = = (1/μ)x(dp/dx)x{(y2/2) – ay} + (u/2a)xy; q = discharge, q = ʃuxdy On integrating and assigning limits as 0, 2a,we get q = [Ua – {(dp/dx)x (2a)3}]/(12x μ); Velocity distribution is: (u/U) = (y/2a) + [Kx(y/2a)x{1-(y/2a)}] where K = {(2a)2/(2 μU)}x(-dp/dx); The velocity pattern for different values of K are shown in the graph Figure 8.6. When you have pressure gradient is positive, it indicates flow due to gradient opposes the Couette flow. The average velocity, Vav = (q/2a)xʃuxdy limits 0 to 2a, V = {(1/2) + (K/6)}xU and q = 2xaxVav.

In a specific case when {(1/2) + (K/6)} = 0, the average velocity and the discharge will become zero. There is no net flow n the negative direction.

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When {(1/2) + (K/6)} = 0; K = {(2a)2/(2 μU)}x(-dp/dx) = –3, We have dp/dx = 6μU/(2a)2; , du/dy = (1/μ)x(dp/dx)x(y-a) + (U/2a); u is maximum when du/dy = 0; When y = a[1 + (1/K)]; Shear stress = fx = μ(du/dy);

Figure 8.6

In the graph P = K of our description. (y/b) = (y/h); (u/V) = (u/U);

8.8. HAGEN-POISEUILLE FLOW IN CYLINDRICAL TUBES The flow is steady and uniform. Let us find the solution by applying N-S equation. N-S equation in cylindrical coordinates, in our problem there is not tangential and radial flow, only axial. We need to consider the axial flow equations. Besides, in axial direction the flow is uniform and steady. Also velocity will not change with angle θ. On simplification we get, [Note: (δw/δt) + u(δw/δr) + (v/r)(δw/δθ) + w(δw/δz)

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= Z -(1/ρ)(δp/δz) + η[(δ2w/δr2) + (1/r) x(δw/δr) + (1/r2)x(δ2w/δθ2) + (δ2w/ δz2)]; } (AAA)

Figure 8.7

The equation (AAA) reduces to: [(δ w/δr2) + (1/r) x(δw/δr)] = (1/μ)[ δ(p + Wh)/δz], 2

(BBB)

Since w changes with r and piezometric pressure (p + wh) We write: [(δ2w/δr2) + (1/r) x(δw/δr)] = (1/r)[d(r(dw/dr)/dr] = (1/μ)[ d(p + Wh)/dz], On successive integration, and substitutions of boundary conditions, ,w = 0; r = a and dw/dr = 0 at r = 0; we get w = (1/4 μ)x(r2-a2)x[ d(p + Wh)/ dz],; The discharge, Q = wx2xpixr dr = (1/4 μ)xʃ (r2-a2)x[ d(p + Wh)/dz] x2xpixrxdr; limit 0 to a On integrating we get Q = {pixa4/(8 μ)} x[- d(p + Wh)/dz]; The average velocity = Vav = Q/(Pixa2) = (a2/8 μ) x[- d(p + Wh)/dz]

Let us pipe loss for a length L = hf;

,hf = D(p + Wh)/W = {8 μVav/(Wa2)}x L; W = ρxg; a = (d/2)

Multiply and dividing RH by factor 2xVav; hf = [(64xV2xL)/{(ρxVxd/μ) x dx2g}]; Vav = V The factor (ρxVxd/μ) = R, Renold’s number

Hence we have Loss = hf = [(64/R)x{V2xL/(2gxd)}]; Normally f is called friction factor = (64/R)

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8.9. LAMINAR FLOW IN AN ANNULUS AREA Let us say r = a (external radius) r = b (internal radius) The boundary conditions are w = 0 at r = a and also r = b; First integration of (BBB) yields: rx(dw/dr) = (1/ μ))x[ d(p + Wh)/dz] x(r /2) + C1 2

Integrate once again, we get, w = – (1/4 μ)x(r2)x[ d(p + Wh)/dz] + C1xIn r + C2; Applying boundar conditions r = a and r = b w = 0; Finally we get, w = -(1/4 μ)x [ d(p + Wh)/dz]x [ a2-r2 + (a2-b2)x {In (a/r)/In b/a)}x In (a/r3)].

Co-axial Couette flow:

Figure 8.8

Let us consider another type of Axi-symmetric flow. Here smaller diameter cylinder dragged inside a larger cylinder. Nomenclature are given. The cylinders are Co- axial. Refer the Figure 8.8. Let the common axis of the cylinder is axis, z.The equation of motion in the direction of flow is: d2w/dr2 + (1/r)x(dw/dr) = 0; (1/r)d[r(dw/dr)]/dr = 0 Integrating successively, we Have, w = C1In r + C2 Boundary conditions are w = 0 atr = a and w = U at r = . After evaluating we have w = [U/{(In (a/b)}x In (a/r) ];

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8.10. ROTATING CYLINDERS Refer Figure 8.9. Let us consider that one cylinder rotating in the other cylinder. This is the case of shaft bearing, where shaft is rotating inside a bush and the bush is lubricated. There is fluid in between clearance shaft OD and bush ID. We are studying the flow pattern when shaft rotates. In our case, both cylinders are rotating. Let w1 and w2 are angular velocities of cylinder of radius a and cylinder of radius b. Here the radial and axial components of N-S equations are absent. We need to consider only tangential components. This is because there is no motion in the radial and axial direction. Let v = tangential component of velocity. We have, ρxv2 = dp/dr; The pressure gradient in the radial direction is dependent upon the centrifugal force. The equation of motion in the tangential direction: d2v/dr2 + d(vr)/dr = 0; ʃ d2v/dr2 + d(vr)/dr = C1 We have dv/dr + (v/r) = (1/r)d(vr)/dr;

Figure 8.9

d2v/dr2 + d(vr)/dr = 0; ʃ d2v/dr2 + d(vr)/dr = C1 We have dv/dr + (v/r) = (1/r)d(vr)/dr; = C1; Second integration yields us, v = (C1/2)xr + (C2/r); We have to find constant of integration. We have v = w1xa at r = a and v = w2xb at r = b. On simplification the tangential velocity, v = {1/(a2-b2)}x[{ (a2x w1) – (b2x w2) }xr – {(ab)2 x(w1- w2)/r}];

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The angular deformation in the r w – plane is h = {(1/r)x (δu/δw)} + {r δ(v/r)/dr}; If v = C/r and u = 0 then h = –2C/r2; , fs (shear stress) = μxh = –2C/r2; Torque, T = –4xpix μxC

The power, P required to overcome this torque is Txw = –4xpixμxC2xr2; The difference in power required the inner and outer cylinders is 4xpixμxC2x{(1/b2) – 1/a2)}

8.11. REYNOLDS NUMBER AND STABILITY PARAMETERS We have defined laminar flow. We have extensively derived expressions and equations for laminar flow. We also defined the Ronald’s number which is the ratio of inertia force to viscous force. In order, the flow to be laminar the viscous force is to be more than the inertia force. The laminar flow will change its status if there is an external disturbances, such as an impulse on the media in which fluid is flowing, a sort of shock or vibration etc. These things create a disturbances and flow become turbulent. If the disturbance is small then it will be absorbed after a while the flow becomes laminar. This is like damping. However, if the disturbance is more than the force is beyond damping. The flow becomes turbulent. There are many researchers in this field and many theories have put forward to define the stability of laminar flow. However, as of today the main factor that affects the stability of laminar flow is Refolds’ number, designated as R OR Re. This is a dimensionless number. We have already seen the value of Reynolds number = [ρxVxL/ μ] OR R = VxL/ η; where η = Kinematic viscosity, μ = Dynamic viscosity. Note in this expression, if the numerator is greater, viscous force is more. i.e., if the viscosity is the factor that influence the laminar flow. Hence we can arrive at a conclusion that it depends on the fluid because different fluids will have different viscosities. When the viscosity is more we say fluid in commercial term, the fluid is thicker. Thicker the fluid it will have laminar characteristics for a larger range of velocity and length. Once again density is also a factor. Now for a given fluid its laminar conditions largely depend on velocity of flow, length. Here we have to understand the meaning of length. It is the surface of media in which fluid is in contact. This is also termed as wetted perimeter. As an example, if the flow is through pipe of certain diameter and

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full, then diameter perimeter) is the value of L. In case of rectangular it is the perimeter. Hence the value which is the pure number decides whether the flow is laminar or not. No doubt there are many theories have been put forward, but Reynolds’s number is the main factor. In general, the fluid in motion, it is laminar if the Refold’s number is less than 2000. It is turbulent if this number is greater than 2500. Between the values, the flow will have mixed properties and it is called transition stage.

8.12. STABILITY PARAMETER Refer Figure 8.10

Figure 8.10

Figure 8.10 shows the Helmholtz model of two superposed ideal fluid. A small disturbance occurs at the interface of fluids. This is shown schematically y in Figure 8.10. You can see how the parameters are changed. The disturbance is further increased it leads to formation of eddies/vortex.

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Figure 8.11

Instability of laminar flow. (Figure 8.11))

Figure 8.12

Transitional Flow (Figure 8.12)

Figure 8.13

Turbulent flow in large scale (Figure 8.13) The Figure 8.11 to Figure 8.13 are the images of flow variation due to external disturbances. Figure 94 the flow will have lowest Ronald’s number. Figure 8.13 has highest Ronald’s number.

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8.13. STABILITY CURVES The flow of a fluid is stable only when there are no external or internal forces disturbing the flow. When there is a disturbance then the flow is changed from its laminar status. It has been found that such disturbances are characterized by the magnitude and frequency or wavelength. We have shown this property in the Figure 8.10 and there on description. It is also known that the laminar status very much influenced by Refold’s number. It has been investigated that neutral stability curve linking disturbance wavelength (l = 2pi/a) and characteristic curves Ronald’s number of the flow. Such a stability curve for a laminar boundary layer is shown in Figure 8.14. We shall discuss about the boundary layer in the next chapter. These stability curves plays important role is shipbuilding. The stability of ships is determined by these curves and their characteristics. Refer the Figure 8.15, which is the stability curve for plane.

Figure 8.14

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Figure 8.15

8.14. LAMINAR BOUNDARY LAYER Boundary layer an important concept in fluid dynamics. It is the layer of the fluid in the immediate vicinity of boundary surface where fluid flows. Here the effects of viscosity are significant. The concept of such flow was first given by Prandtl in 1904. Boundary layer is a thin layer of viscous fluid close to the solid surface of a wall in contact with the moving stream in which the flow velocity varies from zero at the wall (velocity of fluid that sticks to the wall surface) up to U velocity just after the boundary with an error of 1%.

Figure 8.16

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Refer to the Figure 8.16. Let us consider the fluid flows with a constant velocity of U and of uniform flow. The fluid enters a plate as shown in the figure. The entry side of the plate is called leading edge and exit is called trailing edge. The fluid enters the plate with a velocity of U. When the fluid enters, the particle in the vicinity of leading edge will be at rest. As the fluid flow all the particles at the lowest point, sticks to plate are at rest. The particles just above the surface of the plate will have certain velocity, and just above it a little higher have increased velocity. This is how velocity increases with the distance from the plate surface. The velocity thus reaches almost the entry velocity U (99% of U). The reason will be explained later in this description). However, we consider that velocity has reached an entry value, even though it is 99%. Thus there exist a velocity gradient. Besides, as the particles away the leading edge the velocity gradient varies and increases with distance from the leading edge. The velocity curves are shown at points A, B, and C. (Velocity gradients are a1, a2 and a3.a3>a2 > a1) The velocities at A, B, and C ultimately reaches the velocity of U with 1% error at different distances from the surface of the plate. At A, velocity, U is reached at a distance of δ1, At B velocity U is reached at a distance of δ2 and at C velocity, U is reached at a distance of δ3. The boundary thicknesses at A, B, and C are δ1, δ2, and δ3. The velocity at C where δ3 is the boundary thickness reaches U. Beyond the point C there will not be any change in velocity gradient and it remains the same. This occurs at a distance x3 from the entry of fluid.

8.15. BOUNDARY LAYER THICKNESS The velocity defect in the boundary layer disappears a asymptotically as the distance from the edge of the plate increase, i.e., value X increases. Hence it is difficult to a practical limit for boundary thickness corresponding to zero defect. However, with an error within 1% in velocity of the local stream, we can define the boundary layer thickness as δ.

8.16. DISPLACEMENT THICKNESS We know due to loss of velocity the discharge within the area is reduced. If width is constant, then to maintain the equal discharge as per equation of continuity, the thickness of flow is to increase. The displacement thickness, say δ1 can be computed as below: Please refer the Figure 8.17.

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Figure 8.17

Refer to nomenclature stipulated in Figure 8.17. The displacement thickness can be obtained by equating the discharge of an ideal fluid in the boundary layer to the discharge of the ideal fluid in the thickness (δ – δ1) Hence we have ʃ udy = U(δ – δ1) limit 0 to δ OR δ1 = ʃ(1-u/U) dy limit 0 to δ. 13.17. Momentum thickness of Boundary layer: Refer Figure 8.18. The boundary shear forces causes loss of momentum and also energy. Let us consider flow on a flat plate as shown in Figure 8.18. Consider the momentum of a small strip of fluid dy. Momentum = Mass x velocity = δxux(u) x dy. Velocity distribution after striking plate in the boundary layer is shown in Figure 8.18. Initially fluid has uniform velocity before the striking plate. Initial momentum = Discharge x velocity = (ρxuxdy)xU; Momentum after striking plate in the boundary layer = (ρxuxdy)xu. Loss of momentum = (ρxuxdy)x(U-u) for a small thickness. Hence for the entire boundary thickness, δ: ,ʃ(u/U)x[1-(u/U)] dy limits are 0, δ. Hence momentum thickness, δ2 = ʃ(u/U) x[1-(u/U)] dy

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Figure 8.18

Now let us find out the energy loss: Now let us find out the energy loss: The energy of small strip after striking in the boundary = ρxu3xdy. We get energy loss = δ3 = ʃ(u/U)x[1-(u2/U2)] dy; limits 0 to δ.

CHAPTER

9

TURBULENT FLOW

CONTENTS 9.1. Characteristics Of Turbulence And Classification............................. 170 9.2. Reynolds Equation........................................................................... 172 9.3. Derivation For Mean Turbulent Flow................................................ 173 9.4. Characteristics Of Turbulence.......................................................... 177 9.5. Turbulent Boundary Layer................................................................ 178 9.6. General Logarithm Formulation....................................................... 180 9.7. Momentum Equation....................................................................... 181 9.8. Turbulent Flow In Pipes.................................................................... 183 9.9. Exercises.......................................................................................... 188

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9.1. CHARACTERISTICS OF TURBULENCE AND CLASSIFICATION The flow which is smooth and particles moves parallel without interfering the neighboring fluid particle is defined as laminar flow. On the contrary in a flow if the fluid particles instead of traveling in layers, interfering neighboring fluid particles, then the velocity of fluid particles keep on changing with momentary time and in the system coordinates. But the average flow velocity may be same. Do not get confused with unsteady flow where flow velocities getting changed, that is there could be acceleration of the fluid. In the steady flow, the velocities are momentarily getting fluctuating, because of external forces or damping effects. The image shown in Figure 9.1 shows clearly the laminar flow, transition stage, and turbulent flow.

Figure 9.1

Let us analyze this phenomena mathematically. Refer Figure 9.2. This shows a variation of velocity with discrete time. The average velocity line is in green. Blue curve shows variation.

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Figure 9.2

Here the velocity variation with average velocity is discrete. Theoretically, it is to be straight line in case laminar. But in practice, it is difficult to achieve this condition unless it is in simulated in a laboratory. However, the velocity variation is very small with the average velocity. We can consider as laminar flow. The main factor that differentiate the laminar and turbulent flow is Reynolds’s number. If the Reynolds number is less than 2000 it is consider as laminar. If the number is between 2000 to 2400 it is in the transitional stage. It means partly behaves as laminar and partly turbulent.

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Refer the Figure 9.3. This is a condition of turbulent flow. Here the momentary variation of velocity with respect to time is very large. The variation is at microscopic level. Turbulent flow will not have a well-defined velocity vector. As long as the average velocity is constant we can consider as steady flow. Now let us say Vm = Mean velocity. (green line in the Figure 9.3) V1, V2, V3, -------, Vn are velocities at different times dt,2dt,3dt.ndt. Now mean velocity is defined as: Vm = (V1 + V2 + V3 + V4 + ------ + Vn)/ n, where is the number of counts. Now let the total time is T = nxdt; I.e., Time from V1 to Vn. Then we have n = T/dt;

Hence we have (V1 + V2 + V3 + V4 + ------ + Vn)/ n = [(V1 + V2 + V3 + V4 + ------ + Vn)xdt/T]; Now let us consider V = velocity at any instant. this represents V1, V2, V3 etc.

We have Vm = [Σ Vxdt]/T This is the sum of discrete numbers and can be integrated. We have Vm = (1/T)[ʃT0Vxdt] as T tends to infinity.

Suppose Vrms = Root mean square velocity, [(V12 + V22 + V32 + V42 + ----- + Vn2)] are the sum of the velocity squared at points 1,2,3, etc. In the same way, we can show that SQRT(Vrms) = [(1/T){ʃT0Vxdt}]1/2as T tends to infinity.

9.2. REYNOLDS EQUATION Let us restrict our analysis to the two two-dimensional incompressible flow. The equation of continuity is: {(δu/δx) + (δv/δy) + (δw/δz) } = 0; (Refer Chapter 3, 3.5) In case of two dimensional flow it is: {(δu/δx) + (δv/δy) } = 0 --------(A) This is applicable for instant values of flow parameters. Now we will prove it is true for mean values. Let The values of instant value of velocities in case of turbulent flow is: Vm + V’, where we define the values V’ as the velocity increase or decrease at an instant under consideration. We have instant velocity, V = (Vm + V’); V = V(u,v,w)

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Now we apply the continuity equation: [{δ(um + u’)/δx} + {δ(vm + v’)/ δy} + {δ(wm + w’)/δz} ];

Differentiating, we get [{δum/δx + δu’/δx} + {δvm/δx + δv’/δy} + {δwm/ δz + δw’/δz} ]; Rearranging, we have [{(δum/δx) + (δvm/δx) + (δwm/δz)} + { (δu’/δx) + (δu’/δx) + (δv’/δy) + (δw’/δz)} ] = 0; (δwm/δz) = 0; because no flow in w direction, but,(δw’/δz) need not be zero. We have [{(δum/δx) + (δvm/δx)} + { (δu’/δx) + (δv’/δy) + (δw’/δz)} ] = 0; -----(XX) Now we note the following mathematical truth: •

The average differential coefficient is same as differential coefficient of the average. • The average of the average is the average itself. • The average of fluctuating quantities is zero. Apply these conditions to the equation (XX); We get – {(δum/δx) + (δvm/δx)} = 0

9.3. DERIVATION FOR MEAN TURBULENT FLOW It is a steady flow, no variation of velocity with time } (δu/δt) = 0 The general N-S equation in two dimension: RHS(X) = {(δu/δx)x u } + {(δu/δy)x v } + {(δu/δz)x w } + {(δu/δt)} RHS (X) = [X – (1/ρ) x(δρ/δx) + ηx {(δ2u/δx2) + (δ2v/δy2) + (δ2u/δz2)}]; {(δu/δx)x u } + {(δu/δy)x v } = [X – (1/ρ) x(δρ/δx) + ηx {(δ2u/δx2) + (δ v/δy2)]; ------(YY) 2

(Refer Chapter 8, 8.2, it is a three dimensional equation. For two dimensional some of the terms disappear.); Now we have u = (um + u’); v = (vm + v’); p = pm + p’ where pm = mean pressure and p’ = momentary pressure with respect to mean pressure,pm Substitute these values in (YY);

(um + u’) x [ δ(um + u’)/δx] + (vm + v’) x [ δ(vm + v’)/δy] = X – [(1/ρ) { δ(pm + p’)/δx}] + ηx[{ δ2(um + u’)/δx2 } + { δ2(vm + v’)/δy2 }];

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Expanding we get, [{umx (δum/δx) } + {umx (δu’/δx)} + {u’x (δum/δx) } + {u’ x (δu’/δx)}] +

[{vmx (δvm/δy) } + {vmx (δv’/δy)} + {v’x (δvm/δy) } + {v’ x (δv’/δy)}] = X -(1/ρ)[{ (δpm/δx)} + { (δp’/δx)}] + ηx [{ (δ2um/δx2) } + { (δ2u’/δx2) } + { (δ2vm/δy2) } + { (δ2v’/δy2) }];

Now u’m = v’m = p’m = 0; It means average of fluctuating velocities andpressures from datum = 0; All – signs and + signs get cancelled. But products of fluctuating values are not zero because the -x- = + ; Hence we have u’xv’ ≠ 0 (u’)x(δu’/δx) ≠ 0 and (v’)x(δu’/δy) ≠ 0; um = (um)m, vm = (vm)m; (Average of average is Average itself) {umx (δum/δx)}m = {umx (δum/δx)}; {vmx (δum/δx)}m = {vmx (δum/δx)}; (Average of average is Average itself); (δu’/δx)m = (δu’m/δx) = 0; (δv’/δx)m = (δv’m/δx) = 0 (Average of differential is differential average and mean of average is zero); {umx (δu’/δx)}m = {umx (δu’/δx)m} = {umx (δu’m/δx)} = 0; {umx (δu’/δx)}m = 0; (While averaging, a product of mean quantity and a turbulent one, the mean quantity can be taken out of averaging process); Now apply the averaging laws and simplify, We get {umx (δum/δx)} + {vmx (δum/δy)} + {u’x (δu’/δx)}m + {v’x (δu’/δy)}m = X-[(1/ρ)x(δpm/δx) ] + [ η x{(δ2um/δx2) + (δ2um/δy2) }]; ------(ZZ)

We add the following term, since they are zero and no change in the values to LHS. [(u’)x(δu’/δx) + (u’)x(δv’/δy)]m = [(u’)x{(δu’/δx) + (δv’/δy)}]m, to equation (ZZ); Note equation of continuity. Group the terms appropriately. We have; umx (δum/δx) + vmx (δum/δy) + [(u’)x(δu’/δx) + (u’)x(δu’/δx)]m + [(u’)x(δv’/ δy) + (v’)x(δu’/δy)]m = X-[(1/ρ)x(δpm/δx) ] + [ η x{(δ2um/δx2) + (δ2um/δy2) }];

Hence we get [ umx (δum/δx) + vmx (δum/δy)] + [ {δ(u’m)2/δx)] + [ {δ(u’xv’) /δy)] m

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= X-[(1/ρ)x(δpm/δx) ] + [ η x{(δ2um/δx2) + (δ2um/δy2) }]; Re-arranging we get:

[ umx (δum/δx) + vmx (δum/δy)] =

X-[(1/ρ)x(δpm/δx) + {δ(u’m)2/δx} + {δ(u’xv’)m/δy} ] + [ η x{(δ2um/δx2) + (δ2um/δy2) }] The terms ρx(u’m)2, ρx(u’xv’)m are called as Turbulent shear stress OR Apparent shear stress OR Reynolds Shear stress.

[ umx (δum/δx) + vmx (δum/δy)] = X-[(1/ρ)x(δpm/δx) + {δ(u’m)2/δx} + {δ(u’xv’) /δy} ] + ηxV2um m [ umx (δvm/δx) + umx (δvm/δy)] = X-[(1/ρ)x(δpm/δy) + {δ(v’m)2/δy} + {δ(u’xv’) /δx} ] + ηxV2vm m And we have -[{δ(u’xv’)m/δx} + {δ(v’xw’)m/δy}] = 0;

In three dimensional when we analyze we get following Reynolds shear stresses: ρx(u’m)2, ρx(v’m)2, ρx(w’m)2, ρx(u’v’)m, ρx(v’w’)m, ρx(w’u’)m, Thus there are six Reynolds stresses in three dimensional analysis. The terms ρx(u’m)2, ρx(v’m)2, ρx(w’m)2 are called normal turbulent stresses. The physical significance of these normal stresses[-ρx(u’m)2, etc. ] may be interpreted as additional momentum leaving the respective directions. per unit surface in unit time due to the fluctuating components in the respective direction. ρx(u’v’)m, ρx(v’w’)m, ρx(w’u’)m These are called as shear stresses. We can see the significance by an example as below: Exchange of momentum: The analogy of Reynolds shear stress well explained by the principle of exchange of momentum. Two open trucks are loaded with a material are moving in the same direction parallel to each other. the velocities are Va and Vb of trucks P and Q respectively. Some material of mass m per unit time is thrown continuously from the truck Q with a velocity V1 (m = ρV1) to the truck P. If the difference velocities of two trucks are U1 = Vb – Va. Then the change of momentum of truck P in the direction of motion is mU1 This is the tangential force acting on truck P in the direction of motion.

Shearing stress = Tangential force/Surface area = ρx(Vb-Va)xU1 = ρxV1xU1

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Figure 9.3

9.4.

We can apply the above principle to random motion. Refer the Figure

Figure 9.4

The velocity at level B = (u-u’) At level C = (u + u’); The fluctuating velocity component v’ will lead to mean rate of momentum transfer = -ρx(u’xv’)m; This mean rate of momentum per unit area is the turbulent shear stress.

fyx = -ρx(u’xv’)m.

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9.4. CHARACTERISTICS OF TURBULENCE We have seen in them while deriving Reynolds equation the number of variables are increased. It is important that in many cases there is a difficulty of getting analytical solution for velocities and other fluid parameters. In order to solve the turbulence problems, we may need to adapt statistical methods. Eddy Scales: In the fluid flow, eddies are the swirling of flow and reverse current created. This happens generally in turbulent flow. Eddies are created when the fluid in the turbulent status, flow past an obstacles. When the fluid flows past a obstacles, reverse current is created. This reverse current is interfering with the regular flow. The flow is further distorted. In turbulent flow eddies of different sizes in length and diameters are generated. The eddies sizes may vary from a few s in diameter/length to a few kilometers. Large eddies are generated in rivers and Ocean. The eddies may be shortlived only for a few seconds. But in ocean/rivers, they live for hours/days and even months. Each of the characteristics of eddies cannot be find by analytical methods. We need to adapt statistical methods to find their characteristics and their relation with other fluid parameters such as velocity, discharge, pressure etc. Vortex in fluid is also termed as eddy. But the formation of vortex, there is a low-pressure area. and no void. The formation of eddies, there is a turbulent flow and external energy is injected into fluid. Then eddies are formed.

Figure 9.5

9.5.

One can see eddy formation flow past a sphere as in the picture Figure Refer Picture Figure 9.6, the photograph of eddy formation in fluid.

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Figure 9.6

9.5. TURBULENT BOUNDARY LAYER We have studied in the laminar flow boundary layer and layer thickness. Figure 9.7 depicts the flow past an rectangular plate. The uniform flow before entering the edge of plate is shown by parallel lines. As soon as it enters the velocity will not remain constant as before but it is changed and varies from zero velus at the leading edge of the plate to a original value at an infinite distance. However, we consider 99% of the velocity is reached is equal to original value after a certain distance from leading edge. We studied the mathematical portion in the previous chapter.

Figure 9.7

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Figure 9.8 (a) and (b)

Comparison of formation of velocity profile in laminar and turbulent flow. Refer the Figure 9.8 (a) shows the velocity profile in case of a laminar at the boundary. Figure 9.8(b) shown the variation of velocity profile at the same boundary layer, in case of Turbulent flow. You can see in case of turbulent flow the velocity gradient is steep compared with that of velocity gradient in case of laminar flow. The curved green line indicates the eddies, those have reverse swirling flow. Now, refer to Figure 9.9. The inner layer next to the boundary is called the laminar sub-layer. In the laminar sub-layer mean velocity, u at any height y above the boundary, u is dependent on the boundary shear stress, We are not deriving any expressions relating the various parameters. We may assume close to wall, the shear stress is equal to boundary shear stress. fm = mean shear stress, f0 = boundary shear stress, The we have: fm = f0 = μx(du/dy) ~ (μxum/y); where μ = dynamic viscosity, um = mean velocity. Further, (um/ux) = {ρxyxux/ μ}; ux = {f0/ρ}1/2 = This is called shear stress velocity. Graphs are there showing the relation between (um/ux) and {ρxyxux/ μ}. Now let us define the law of the wall. The law states that the average velocity of turbulent flow(um) at certain point is proportional to the logarithm distance from that point to the “Wall” or boundary of the fluid region. This

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law was published by Theodore von Karman in 1930. This is applicable only to the

Figure 9.9.

Parts of the flow that are close to “Wall” (boundary region.< 20% of the height of the flow). This is a good approximation of velocity profile for natural streams.

9.6. GENERAL LOGARITHM FORMULATION (Figure 9.10) The logarithm law of the wall is a solution for the mean Velocity parallel to the wall. It is accurate for flows at high Reynolds numbers in an overlap region with approximately constant shear stress. The viscous effects at this distance is negligible. Let u1 = dimensionless velocity. The velocity u parallel to the wall as a function of y(distance from wall) divided by the friction velocity ut

y1 = The wall coordinate, the distance y to wall, made dimensionless with friction velocity ut and kinematic viscosity η.

Tw = Wall shear stress. ρ = fluid density. ut = Friction velocity or shear velocity. k = Von Karman constant; C1 = Constant. In = Natural logarithm. Now u1 = (1/k)x In (y1) + C1. y1 = (yxut/ η); ut = (Tw/ρ)1/2 and u1 = (u/ut)

From the experiment Van Karman found: k = 0.41; C1 = 5 for a smooth wall. with dimension, the logarithm law of the wall can be written as: u = (ut/k)xIn (y/y0). y0 = Distance from the boundary at which the idealized velocity given by the law of the wall goes zero. This is necessarily non-zero

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because the turbulent velocity profile defined by the law of the wall does not apply to the laminar sub-layer. The distance from the wall at which it reaches zero is determined by comparing the thickness of laminar sub-layer with the roughness surface over which it is flowing. For a near – wall laminar sublayer of thickness δx and a characteristics roughness length -scale ks

Figure 9.10

Graph Law of wall, Horizontal velocity near the wall with mixing length Model.

9.7. MOMENTUM EQUATION Karman’s momentum integral equation can be applied to a turbulent layer in the same manner as obtained for a laminar boundary layer. The equation: (dδ2/dx) = f0/(ρxU2) + (1/U2) x [d/dx {ʃδ0(u’2)m – (v’2)mdy }].

The equation differs from the laminar flow momentum equation by the presence of the last integral. The boundary layer shear stress f0 differs in magnitude in the two types of boundary layer, laminar, and Turbulent. A scientist, Blasius able to get an empirical formula for boundary shear stress for the smooth plates. That is, f0 = 0.0225xρxU2x[µ/(ρxUxδ]1/4;

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Finally we get for the boundary shear stress,(δ/x) = 0.376x(Rx)–1/5

f0 = 0.0286x(Rx)–1/5x(ρxU2); This is from Blasius formula and Prandtles power distribution of velocity in turbulent boundary layer. Force per unit area = 0.036(Rx)–1/5x(ρxU2); This is also called turbulent drag. (Note to Readers: There are many empirical formulae and mixture of derived theories. It may be difficult to follow in details for the readers. Maybe it is interesting for the researchers to know in detail. Readers may remember the end formula to solve problems related to boundary conditions) Frictional drag coefficient for flow past Flat plates: The frictional resistance drag, or total drag for flow past smooth flat plates surface is given by the equation Df = CfxρxAxV2, where Df = Drag in Newtons. ρ = density, A = wetted area in m2, V = Relative velocity of flow between surface and fluid in m/s. Cf = Coefficient of drag. The curves showing complete results of tests on smooth plates for large range values R is shown in Figure 9.11.

Figure 9.11

Empirically the drag coefficient by: Cf = 2.654 x(1/R)1/2 For laminar flow.

Cf = 0.144(1/R)1/5 for low values of R. Known as 1/5 Power law Cf = (1/R)1/7 for high values of R. Known as 1/7 Power law.

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9.8. TURBULENT FLOW IN PIPES (Note: refer Reynolds equation for symbols abbreviation used here) Smooth pipes: The velocity distribution in circular pipes may calculated by Prandtl’s mixing length hypothesis. fm = ρxl2x (dum/dy)2;

The distribution of shear stress is not known. As such further assumptions are necessary to obtain the velocity distribution. Prandtl’s assumed that shear stress is regarded as nearly constant over the entire cross-section of the fully established turbulent flow. in the pipe. Suppose the shear stress at any point across the boundary layer is equal to the local boundary shear stress, Then (f0/ρ) = (ux)2 = l2x(du/dy)2; Mixing length is defined, l = kx[ (δum/ δy)/ (δ2u/δy2)]; Close to boundary we may approximate, l = kx(u/y)x(u/y2) = kxy. Hence the expression for velocity gradient = (dum/dy) = {ux/(kxy)}: ux = (f0/ρ)1/2 = This is called shear stress velocity.(Same as defined ut in General logarithm formulation (Figure 9.10). On integration, we get um = (ux/k)xIn (y) + C1

Suppose we write C1 = ux + (ux/k)xIn (ρxux/μ). Then velocity distribution is given by (um/ux) = (1/k)x In{(ρxuxxy)/μ} + C; In the neighbourhood of the boundary, the value of k = 0.4. and the values of C = 5.5 Hence We have (um/ux) = 2.5x In{(ρxuxxy)/μ} + 5.5 = 5.75 x Log10 {(ρxuxxy)/μ} + 5.5; There are many changes and modifications on this equation. However the equation very well agrees with experimental results. Rough Pipes: The equation derived as above is valid for smooth pipes. When the pipe is too rough, then the roughness will protrude into laminar sub-layer. The causes the formation of eddies and leads to breaking of laminar sub-layer. Figure 9.12 shows such condition.

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Figure 9.12

It is estimated that velocity reaches a zero value at a distance equal to k/30 from the boundary. k is the average height of roughness. Boundary condition equation, (dum/dy) = (2.5x ux/y); We obtain the rough pipe equation as under:

(um/ux) = 5.75x Log 10 (y/k) + 8.5. The boundary is said to be hydraulically smooth if k/δl 6. The range between these two values is transition stage. Mean velocity and maximum velocity: The discharge in a pipe is given by the equation: ʃa02xpixrxdrx(um) where mean velocity is given by (um/ux) = 5.75 x Log10 {(ρxuxxy)/μ} + 5.5 for smooth pipes and (um/ux) = 5.75x Log 10 (y/k) + 8.5 for rough pipes. The average velocity in pipes, smooth or rough = , V = Q/(Pixa2); Q = discharge and a = Radius of the pipe. The values in case of smooth pipes is given by V/ux = 5.75 x Log10 {(ρxuxxa)/μ} + 1.75 The value in case of rough pipes is given by V/ux = 5.75x Log 10 (a/k) + 4.75. Now we have (um/ux) – (V/ux) = [5.75 x Log10 {(ρxuxxa)/μ} + 5.5 ] – [5.75 x Log10 {(ρxuxxa)/μ} + 1.75] Hence [(um -V)/ux] = [5.75 xLog10 (y/a) + 3.75] for smooth pipes

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[(um -V)/ux] = [5.75 xLog10 (y/a) + 3.75 ] for rough pipes. The location where the mean velocity is equal to average velocity in pipes can be obtained by Left side equal to zero ie [(um -V)/ux] = 0;

We get {Log10 (y/a) } = -(3.75/5.75) = –0.652 With this we get (y/a) = 0.223. The maximum velocity umax occurs where y = a and Log10 (y/a) = 0;

We have (umax-V)/ux = 3.75; ux = (f0/ρ)1/2 = Vx(f/8)1/2. f is called pipe resistance factor. f0 = Shear stress. (You know about f in the next section.).

We can write umax = 3.75 x ux + V = 3.75xVx(f/8)1/2 + V; umax = Vx[1.326x(f)1/2 + 1] The velocity at any distance from the boundary can be expressed as below: (umax – um)/ux = [5.75 x Log10 {a/(a-r)} ]. Pipe resistance factor:

Figure 9.13

Refer the Figure 9.13. Darcy found in pipes where there is a uniform flow, head losses (pressure drop) considering the equilibrium of flow. Consider the control volume as shown in the figure. Apply momentum theorem. Then (p + dp)xpixa2 -pxpixa2 + f0x(2xpixaxdl) – wxa2xdlx Sin(θ) = 0;

Which on simplification with all dlx Sin(θ) = dz; d{(p/w) + z} = dh; (w = specific heat of fluid) h = piezometric head.

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dh = [2xf0xdl/(axw)] = 4xf0xdl/(dxw)]

Now the boundary shear stress may be expressed in terms of the drag;

f0 = cfxρx(V2/2) where cf = co-efficient of resistance.

On integration we get, h = 4xf0xl/(dxw)] = [4x{cfx ρx(V2/2)}xl/ (ρxgx2xd) ] Note w = ρxg; h = fxlxV2/2xgxd where f = 4cf [ f is called the pipe resistance factor] Now we have Shear stress, f0 = (f/4)x(ρxV2/2); Shear velocity, ux = (f0/ ρ)1/2 = Vx(f/8)1/2;

Substituting of this value for ux in the expression for the average velocity in smooth and rough pipes, We get (1/f)1/2 = 2.03x Log10 { ρxVxdxf1/2/ µ} –0.91

(1/f)1/2 = 2.03x Log10 { a/k} + 1.68;

The constants are adjusted to suit experimental data to give the following expression for smooth and rough pipes. (1/f)1/2 = 2x Log10 {Rxf1/2} –0.8; remembering,Reynolds number, R = ρxVxd/ µ; (1/f)1/2 = 2x Log10 {a/k} + 1.74; for rough pipes. k = Average rate of roughness projections. The pipe resistance factor is dependent on the value of Reynolds number for smooth pipes. Laminar sub-layer thickness, δl is proved as, δl = [11.6x η/ux]; Kinematic viscosity = η; The value of ux = Vx(f/8)1/2; On simplifying, δl = [11.6x η/{ Vx(f/8)1/2 }]; On further adjusting and simplification, We have; (δl/a) = [65.6/(Rxf1/2)]; From this we can conclude the relative thickness of laminar sub-layer dependent of radius of the pipe. Blasius obtained an empirical formula, f = [0.316/R1/4] This is applicable for resistance factor for smooth pipes. This formula is valid for 2000 V2. How the velocity increases, i.e kinetic energy?

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Figure 11.12

Now total energy at, E = D + (V2/2g) = Depth of water in Chanel + Velocity head. Thus it is the sum of static head and velocity head. Hence the water utilizes the static head for increasing velocity, at lower depth and also when velocity decreases the energy is stored as static head. The energy, E is called specific energy. Now, look at the graph, Figure 11.13. Here the potential energy, in X-axis and depth D in Y-axis, is drawn in green line. This is a straight line. Then v, for various values in Y-axis draw corresponding values of (v2/2g) in X-axis. This is a curve, red. Then combine the straight line and curve in red by adding corresponding values of X-axis and Y-axis. Draw another curve. This is total energy curve and in blue. The total energy curve is called specific energy curve. Now consider vertical line ABC. The total energy conditions are favorable for a hydraulic jump to occur at this section. Before the jump, flow is rapid, and depth is BC. One the jump occurs the flow become slower and the depth increases to BA. But the total energy, E remains constant except certain losses. At the point M the energy is minimum and it is the depth at minimum energy. You can call depth as critical depth and velocity as critical velocity.

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Figure 11.13

Now let us derive an expression for critical velocity/critical depth. E = D + (v2/2g); Q is the discharge and is constant. Let us consider unit length of channel. Then, v = Q/D, because velocity is uniform throughout, as assumed. The Substituting, we have E = D + (Q2xD–2/2g); for the minimum value of E, the condition is: dE/(dD) = 0,Differentiating we have, 1-(2(Q2xD–3)/2g} = 0; Substitute v = (Q/D); We have v2 = gD; Hence depth for minimum specific energy = (v2/g); It may be noticed that: (v/gD)1/2 = 1; The term (v/gD)1/2 is called Froude number. It follows for minimum specific energy, Froude number is unity.

11.8. EXERCISES Solved problems:1

Problem:- A trapezoidal channel , having sides of smooth stone , has a base of 180 cm and side of 2 vertical to 1 horizontal. The depth of water in the channel is 120 cm. Find the quantity of water flowing if the slope of channel is 2 m per Km. Take C = 71 Solution:- Section of channel as shown in the sketch problem 1.

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Sketch Problem 1 Area = 1.8x1.2 + 0.6x1.2 = 2.88 m2 . Wetted 1.8+2x(1.22+0.62)1/2 = 4.48 m .

perimeter =

m = A/P = 2.88/4.48 = 0.643 m slope = 2/1000 C =71 v = 71 {0.643x(2/1000)}1/2 = 71 x{0.6431/2 x0.045)} = 71x(0.802x0.045) = 2.566 m/s ; Discharge = 2.88x2.566 = 7.39 m3 . 2.

Problem :- A trapezoidal Channel is to be designed for conveying 283 m3/s water per minute. Determine the cross sectional dimensions of the channel from the following data Here we need to use the economical section of channel (Refer Fig 11.14) Slope 1in 1600 ; Sides inclined 450; Cross section is to be a minimum V = 49.5x(mi)1/2 Ref Fig 11.14 . Let BF = nd = GE ; DB = dx(n2+)1/2 = CG ; QO = d ; Let us say n =1 ; Then Wetted perimeter , (b+2d)/2 = dx21/2 ; (refer the condition derived for economical section) b = 0.828 ; Area = (b+d)xd = 1.828d2 . Wetted perimeter = b+2dxSQRT(2) Hence P = 3.656d ; m = 0.5 d Quantity per second = AxV 0.5d/1600] ;

;

(283/60 ) = 1.828d2x49.5xSQRT[

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;

255

4.72 = 90.486 x d2x d1/2 [0.5/1600]1/2 = = 90.486xd5/2 x 0.0177 = 1.602xd5/2 We get d5/2 = 4.72/1.602 = 2.95 ; d = 1.54 m ; b = 0.828x1.54 = 1.275 m .

Figure 11.14

3.

Problem :-Waters enters a horizontal channel of uniform width , with a velocity of 6.1 m/s . The depth of the water at entrance is 61 cm . Calculate the critical depth of the water. We have proved under specific energy , condition for minimum depth , V = SQRT(gD) Hence D = V2/g ; Q = AxV = 6.1x0.61 = vxD = VxV2/g m.

V3 = 9.81x6.1x0.61 = 3.32 m/s ; Critical depth , Dc = 3.322/9.81 = 1.12 4.

Problem ;- Water is flowing along a channel of uniform width , the quantity of flow being 3.71 m3 /s per meter of width of channel, causing a standing wave to occur. If the depth of the water on the upstream side of the standing wave is 90 cm find the height of the wave. Let us say 1 meter of channel . q = 3.71 m3/s D1 = 0.9 ; Hence we have:D2 = -(0.9/2)+ SQRT[ {2x3.712/(9.81x0.9)}+(0.92/4)] = -0.45 + SQRT[3.11+0.20] = -0.45+1.77 ; D2 = 1.32 m ; Height of the wave = D2-0.9 =0.0.42 m .

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Problems to Solve:1. Problem A channel 10 ‘ wide at the bottom and with sides sloping 1 to 1 has a slope 56 cm per Km. What would be the discharge if water is 1.22 m deep in the channel and C =49.5 , v = CxSQRT(mi)? (Ans =5.81 m3) 2. Problem :-Find the maximum discharge for the least excavation of a rectangular channel 10 ‘ wide C= 58 . (Ans 7.43 m3 .) 3. Problem:-Find the depth of maximum discharge in a circular brick sewer 1.23 m in diameter. (Ans 1.16 m) 4. Problem:- A channel with side slope 450 is to have a cross section of 11.15 m2. Determine the dimension of the best section.( Depth = 247 cm ; Base = 204 cm) 5. Problem:-A rectangular channel is 5’ deep and 10’ wide. If the value of C = 55, determine the discharge if the gradient is 1:1000. (Ans 7.1 m3/s) 6. Problem :- The depth of water in a circular brick-lined conduit , 1.83 m in diameter, is to be 1.5 m and its capacity 225000 m3/day .Water surface subtends an angle of 960 20’ at the axis of the conduit. What must be the gradient . Assume C = 68 .(Ans = 1/2030) 7. Problem:- A concrete lined channel has a bottom width of 3 m side slopes of 1 horizontal to 3 vertical , and a gradient of 1 in 800 .When flowing 92 cm deep, it found to have a capacity of 6.3 m3/s . What is the value of C . (Ans = 73). 8. Problem:- An irrigation channel has a gradient of 1 in 2000 , a bottom width 4.87 m and side slopes of 1 vertical to 2 horizontal. If the depth of water is 1.23 m and the value of C=49.5, What is the mean velocity and capacity of the channel. (Ans 1.03 m/s ; 9170 LPM). 9. Problem:-Find the downstream height of a hydraulic jump occurring on a level bed when the upstream depth is 91 cm and velocity 10.7 m/s. (Ans = 4.5 m) 10. Problem:- A brick lined sewer has a semicircular bottom and vertical side walls 60 cm apart. If the slope is 1 to 1000 determine the discharge when the maximum depth of water is 91 cm. C= 49.5. (Ans = 398 LPM) . 11. Problem:- Water flows in a horizontal conduit of rectangular section with a velocity of 3 m/s, where the depth is 0.61 m, at which section a jump occurs. Calculate the depth after the jump and the energy lost per Kg of water. (81 cm. 194Nm) .

CHAPTER

12

COMPRESSIBLE FLOW

CONTENTS 12.1. Introduction................................................................................... 258 12.2. Gas Laws And Work Done............................................................. 258 12.3. Application Of Law Of Conservation Of Energy............................. 259 12.4. Enthalpy And Entropy.................................................................... 261 12.5. Velocity Of Pressure Wave In Fluid................................................ 263 12.6. Variation Of Atmospheric Pressure With Altitude........................... 265 12.7. Speed Of Sound And Mach Number.............................................. 267 12.8. Compressible Flow – One Dimensional Flow................................ 268 12.9. Normal Shock Wave...................................................................... 271 12.10. Compressible Flow In A Pipe With Friction.................................. 274 12.11. Mach Number Relations.............................................................. 276 12.12. Exercises...................................................................................... 277

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12.1. INTRODUCTION We studied the flow of fluid, assuming density as constant. Because the fluids we studied so far are basically, their compressibility is negligible. Hence we consider the density as constant. In this chapter, we study the flow of fluid when density is not constant and it is variable like pressure and volume. In such cases, the problems are more complicated. Besides we need to consider the variation in temperature since it is not constant. This is precisely study of gas flow. In this section, we do not study the flow properties and the relationship between various fluid parameters in detail. The study of flow is dealt with in detail as a major section in Heat transfer/Thermodynamics. However, we need to know certain laws, related to thermodynamics. Let us look into those laws here below: In this chapter, the fluids are called as gases also. Let p1,p2 are pressures of in the initial conditions and final conditions. v1,v2 are velocities. ρ1,ρ2 are densities. V1, V2 are volumes. Let us consider 1 Kg of fluid. w = weight per one m3 volume = ρxg

Cp, Cv specific heats of gases at constant pressure and constant volume respectively. U = Internal energy of 1 Kg of fluid/gas. J = Joule’s equivalent = 4186 joules/Kg. 1 Calorie = 4.18 Joules. T = Absolute temperature in Kelvin = (t + 273)0 K.

12.2. GAS LAWS AND WORK DONE Let us know about gas laws: Boyle’s Law: If a gas is expanded or compressed at a constant temperature, the product of pressure and volume is constant at any instant. pxV = Constant, p1xV1 = p2xV2 = p3xV3 = etc.: provided temperature is constant. Charle’s Law: If a gas is expanded or compressed at constant pressure, Then V/T = Constant V1/T1 = V2/T2 = V3/T3 = etc.

If the change takes place at constant volume then p/T = constant; p1/T1 = p2/ T2 = p3/T3 = etc. Characteristic Equation of a Gas: By combining the Boyle’s law x Charel’s law we get another relation:

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(pv/T) = RxT; where R is called the gas constant. Value of R is different for different gases. Its value depends on units. In metric unit (SI units) R = 287 J/Kg/Deg F for air. Note: All these gas laws have been experimentally verified. The internal energy of a Gas: This is the energy stored in a gas in the form of heat. This is the energy due to the vibration of molecules. It has been proved by experiment that the internal energy of 1 Kg gas at absolute temperature T is given by U = Cvx(T–273) CHU; The internal energy at any temperature T°K Suppose the temperature of the gas change from T1 to T2, then the change in the internal energy is given by (U2-U1) = Cvx(T2-T1) CHU heat units.

12.3. APPLICATION OF LAW OF CONSERVATION OF ENERGY When a given mass of gas is expanded so that it does some form of work, then the total energy stored before and after expansion remains the same; Let us say heat absorbed by a gas = dQ; Work is done during expansion = dW Increase in internal energy = dU; Then; – dQ = dW + dU; in other words, [Q = W + (U2-U1)]

Heat absorbed by 1 Kg of Gas = work done by 1 Kg of Gas + Increase energy per Kg of Gas; when gas rejects the heat, then dQ is negative.; If external work was done by compression then dW is negative. If dU is negative it represents a decrease in internal energy. Isothermal process: When a gas is expanded or compressed at a constant temperature the process is called isothermal. Then the gas does some external work. See the graph Figure 12.1. The graph represents the variation of pressure with respect to volume when it is expanded with temperature is constant. pxv = Constant, p1xV1 = p2xV2; Apply law of conservation of energy:

Q = W + (U2-U1); The temperature is constant. Internal energy is the function of temperature.

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Figure 12.1

T2-T1 = 0; Hence we have Q = W (heat units); It follows that heat absorbed by the gas is utilized in doing external work. Work done can be calculated: Work done = Area under the curve in the graph. W = ʃpxdV Limit V2 to V1; pV = p1V1; p = [p1V1/V];

W = p1V1xʃpxdV/V limits V2 to V1; We get W = p1V1x[In (V2/V1]; we have p1V1 = RT;

Substituting we have: (RxT)x[In r] where, (V2/V1) = r; This explains when heat is absorbed. When the process is compression then we have r > 1; It indicates heat is rejected. Adiabatic process: Refer the graph in Figure 12.2.

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261

Figure 12.2

The process is called adiabatic process. In this case, the gas is expanded/ compressed in such a manner such that no interchange of heat takes place. But some work is done. It is clear that gas utilizes the internal energy to the work. Thus, the internal energy is decreased. Consider the conservation of energy, Q = W + (U2-U1) = 0; Q = 0; W = + (U1-U2); It can be proved that in such an expansion/compression; pxVg; p1xV1g = p2xV2 g; where g = (Cp/Cv)

Work done is the area under the graph Figure 12.2. W = ʃpxdV Limit V2 to V1; pVg = p1V1g; p = [p1V1gxV-g];

W = p1V1gxʃpxV-gx dV; limits V2and V 1On integration we get:

W = [{p1xV1 – p2xV2}/(g–1)]; It is further simplified to; W = Rx[(T2-T1)/ (g–1)] per Kg of gas; In adiabatic process it is worth to know following relations: (p1/p2) = (V2/V1) g; (T1/T2) = (V2/V1) g–1; (T2/T1) = (p2/p1) [(g–1)/g];

12.4. ENTHALPY AND ENTROPY Total Heat of a Gas: The total heat of a gas at a given temperature is called enthalpy of gas. enthalpy is defined as the amount of heat required to raise the

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temperature of unit of the gas from 00 C to the given temperature. Enthalpy, H = Cpx(T–272); T is the absolute temperature in degrees centigrade. The enthalpy of gas is independent of pressure. But it does dependent at a very high temperature. The total heat can be found by knowing internal energy and external energy work done. When total is reckoned from the absolute zero temperature and assuming the gas occupies no volume at that temperature, H = [U + (pxV/J)]; U = CvxT, If it is reckoned from 00C, H = [U + {px(V-V0)/J}];

Entropy of Gases: The term entropy is defined as the amount of heat energy available for conversion into work. We do not describe this thing in more details. It is dealt in detail in the books related to Heat transfer or thermodynamics. The increase in temperature is accompanied by a reduction in the rate of availability of heat energy for conversion into work. The entropy of a substance of a unit weight, at a given temperature and pressure, is a physical property of the substance and can be stated in units of entropy measured from datum 00 C. Let us say dQ = a small amount of heat absorbed by a unit weight of a substance during a small interval of time. T = absolute temperature at that instant. S = entropy of the substance. Then dS = dQ/T; By integrating, we have ʃ s1dS = ʃT2T1 (dQ/T); s2

Hence we have (S2-S1) = = ʃT2T1 (dQ/T).

If heating takes place adiabatically, dQ = 0; dS = 0; No heat transfer takes place. This called constant entropy. Suppose heating takes place isothermally, T = constant; (S2-S1) = Q/T; Q = p1xV1xIn (r); dS = dQ/T; We write; dQ = TxdS hence we have Q = ʃTxdS;

Let us draw a curve, with S (entropy) as X-axis and T as Y axis as shown Figure 12.3. This is one Kg of substance.

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263

Figure 12.3

The area under the curve represents heat absorbed. Such curve called T-S diagram.

12.5. VELOCITY OF PRESSURE WAVE IN FLUID Bulk Modulus of a fluid: First let us understand bulk modulus of a fluid. The bulk elastic modulus of a fluid is the ratio between the increase of pressure and volumetric strain caused by this increase pressure. Volumetric strain = – (dV/V); V = Volume; Bulk modulus = Increase in pressure/Volumetric strain = dp/(-dV/V); K = -Vx(dp/dV); The bulk modulus of the liquids is very large compared with that of gases, because of very low compressibility of liquids. Thus for a gas, the value of K is relatively small as dV is large.

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Figure 12.4

Now let us find the pressure wave in a fluid. Refer the Figure 12.4 for all our explanations. Consider a tube of fluid of unit sectional area through which pressure wave being transmitted from right to left having a velocity of v. Imagine for a moment the wave is brought to rest, fluid having velocity in the opposite direction. Let intensity of pressure at C = p and velocity v. This section (C) has a least pressure. Consider an adjacent layer D has a maximum pressure. Let the increase in pressure between C and D = dp. This is due to local circulation caused by wave. Figure 12.4 shows the pressuredistance diagram. We cannotice the section C trough of pressure wave and section D to the crest. We have, Pressure of fluid at D = p + dp; Velocity at D = v + dv; Let V = Volume of fluid at A compresses per second by wave and ρ = Density of fluid at C. Let the volume compresses per second increase by dV between C and D. The at section C, V = axv = v, because a = 1 as assumed. At section D, volume compressed per second = V + dV = ax(v + dv) = v + dv; V = v; Hence dV = dv. Now force on fluid between sections C and D = Change of momentum per second = mass per second xChange in velocity. Thus we have,

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265

[p-(p + dp)]xa = ρxaxVxdv; We have dp = -ρxVxdv; but dV = dv; We have, (dp/dV) = – ρxV (12.1); We have already proved, K = -Vx(dp/dV); (dp/dV) = -K/V

(12.2);

Hence we have, ρxV = K/V; But as V = v, we have v = (K/ρ)1/2. (12.3) Thus the velocity in the fluid depends on the bulk modulus and density. It may be noted that sound is propagated by means of a pressure wave transmitted in a fluid, the equation (3) gives the velocity of sound in a fluid or as matter in a media. Wave velocity for isothermal process: The velocity of a pressure wave in a media depends on whether transmission is Isothermal OR Adiabatic. Now let us find out the expression, for pressure wave transmission by isothermal process. We have for isothermal, pxV = C; Differentiating and re-arranging, we have (dp/dV) = -(p/V) p;

We have: (dp/dV) = – (K/V); Equating K/V = p/V; Hence we have K = Hence velocity, v = (p/ρ)1/2; (12.4)

The velocity by the method does not give accurate values. This is because the temperature do not remain constant. Hence if we assume that the fluid expands adiabatically we get better results and very close to experimental values. However, the equation (12.4) give good results for liquids, because the temperature variation in liquids is very small. Wave velocity for adiabatic process: The transmission is very close to adiabatic, because the change in pressure is sudden. There is not time to exchange of heat from external source. It uses internal energy. Adiabatic process, pVg = C; C = constant. Differentiating, We have gxpxVg–1xdV = Vgxdp; Hence we have, (dp/dV) = -gxp/V; We also have (dp/dV) = -(K/V); Thus we have (K/V) = (gxp/V); Hence K = gxp; But, v = (K/ρ)1/2 We have Velocity of wave = v = (gxp/ρ)1/2.

12.6. VARIATION OF ATMOSPHERIC PRESSURE WITH ALTITUDE Now let us study the variation of atmospheric pressure with altitude. We

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know that the atmospheric pressure on the earth surface is due to air column weight. The pressure is the weight of air column per unit area. This vary with altitude because the weight of column varies with the length/height of liquid column. But this is not directly proportional to the height, as the in case of liquids. This is because the density of air is not constant throughout the altitude. Our intention is to study how it is dependent on altitude, density, pressure etc. If the temperature of the column of air is constant, the variation in pressure follow an isothermal process and obeys Boyle’s law. But this is not the case. It is found that atmosphere gets colder as the altitude increases. Once it is more accurate to consider variation to follow adiabatic law. Then we need to assume there is no interchange of heat between layers. The ratio between the temperature drop and the altitude is known as temperature gradient. In the solution, we assume the atmosphere is quiet and contain a constant amount of moisture. It is found from measurements and experiments that the true condition is neither adiabatic not adiabatic. It appears to follow the law pxVn = Constant. The value of n = 1.235 assumed. The pressure variation can also known, if we know temperature gradient. Now refer to Figure 12.5. Consider a column of air of unit cross section of area. The pressure difference between two sections: dp = w(dhx1); dp = – wxdh. V is the volume of unit weight of air. w = (1/V); We have,dh = -Vxdp; From gas equation, we have pxV = RxT;

Figure 12.5

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267

We have dh = -(RT/p)x dp. We can obtain the variation of atmospheric pressure with altitude can be obtained by four methods. Suppose we assume isothermal process, ʃ dh limits (h2 to h1) = -RTxʃdp/p; limits p2 to p1; h2 – h1 = RTx In (p2/p1)We assume h1 = 0, ground level, we have h2 = -RTx [In p2/105] N/m2; h2 in meters.

Assuming the variation follow adiabatic process: In the same way we proceed, we get: h2 – h1 = {g/(1-g)}xRT1x [(p2/p1){(g–1)/g}–1]; h1 = 0, T1 = temperature at h1; h2 = {g/(1-g)}xRT1x [(p2/105){(g–1)/g}–1] N/m2; h2 in meters. Assuming the variation follow pxVn = constant; Then we have h2 = {n/(1-n)}xRT1x [(p2/105){(n–1)/n}–1] N/m2; Suppose we know temperature gradient: T0 = absolute temperature at ground level. p0 = atmosphere pressure at ground level. T = Temperature at any altitude h; p = pressure at any altitude. p1 = pressure at a given height h1 t = temperature drop per meter increase of altitude. We have T = (T0 – txh); We have dh = -RTxdp/p = -Rx (T0-th)xdp/p;

We have (dp/p) = -dh/{Rx(T0-th)}; ʃ (dp/p) = ʃ -dh/{Rx(T0-th)} On integrating and substituting limits we have In{p1/p0} = (1/Rt) x In [1-(th1/ T0)]; This equation gives the pressure at any height h1from ground level.

12.7. SPEED OF SOUND AND MACH NUMBER We defined the speed of sound in 12.3. It review, it is the pressure wave traveling in a fluid or a media. We also proved that v = (K/ρ)1/2; where K is the bulk modulus and ρ is the density. Let us call v = c velocity of sound c = (K/ρ)1/2 Mach number is defined as the ratio of velocity fluid flow to the local velocity of sound in the media. It is designated as M; M = Velocity of fluid/ Velocity of sound in the media.

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M = V/c; Mach number is also measure of compressibility of fluid, because it is a function of K. In case incompressible fluid, K is infinity. Hence M = 0; We proved in 12.3 for Adiabatic compression (Called Isentropic), K = gxρ; The compressible flow can be classified based on Mach number as below: •

• • • •

If each number is below 0.3, the fluid may be classified as incompressible flow for all practical purposes. Density variation with pressure is very low. Most of the liquids come under this category. Subsonic flow; Mach number lies between 0.3 to 0.8. (< 980 Kmph) Transonic flow; Mach number lies between 0.8 to 1.3 (980 Kmph –1470 Kmph) Supersonic flow: Mach number lies between 1.3 to 5 (1470Kmph to 6120 Kmph) Hypersonic flow: Mach number lies between 5 to 10 (6120 Kmph to 12250 Kmph).

12.8. COMPRESSIBLE FLOW – ONE DIMENSIONAL FLOW Refer Figure 12.6, the stream tube.

Figure 12.6

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269

For steady flow, ρxAxV = constant. We have A(x)x V(x) x ρ(x) = constant. Let us differentiate with respect to x; We have (1/A)x(dA/dx) + (1/V)x(dV/dx) + (1/ρ)x(dρ/dx) = 0; We get (dA/A) + (dV/V) + (dρ/ρ) = 0; let us say m1` m2` is the rate of mass change. Apply energy equation to section A-A and B-B in the Figure 12.6: We have: -m1` [h1 + (V12/2) + (gxz1)] + [m2` [h2 + (V22/2) + (gxz2)] = Q`; The area of stream tube A(x) at section x-x as shown in Figure 12.6, of pressure, density, velocity etc. across the tube are ignored. The pressure, velocity, density along x-direction are p(x), V(x) and ρ(x) respectively. h = enthalpy which we have already defined. pressure force pxvs; where vs = (1/ρ) Specific volume. For continuity equation mass flow rates are to be equal, m1` = m2`; Besides the value z1 andz2 gravity force is negligibly small hence ignored. With this We have: [h1 + (V12/2)] = [h2 + (V22/2)] = constant.

We can conclude [h + (V2/2)] = Constant along the streamline in adiabatic flow. adiabatic because there is no change in temperature. The expression [h + (V2/2)] is called Total enthalpy. Suppose the velocity is zero, Then the enthalpy = h0 = total energy. This is called stagnation enthalpy. Enthalpy with Temperature: The enthalpy for an ideal gas is CpxT. We know. Just for verification let us do an experiment. Suppose we heat a fluid, in an underneath airtight cylinder, There is a weight above say W. Then there is pressure on the gas trapped in the cylinder. This pressure is constant, since weight is constant. Then Heat added per unit mass = Specific heatxDifference in temperature = Cpx(T2-T1) thus we have h = Cpx(T), it the initial temperature is absolute zero. We have [h + (V2/2)] = h0; We have CpxT + (V2/2) = CpxT0; 1 + {V2/ (2xCp8T)} = T0/T; T0 is called stagnation temperature. (When flow brought by Adiabatically); Cp = [gxR/(g–1)] and c2 = (gxRxT) already proved.

On simplifying and remembering M = c/V; We have T0 = Tx [1 + {(g–1) xM2/2}]; Enthalpy with pressure; We know in case of adiabatic (p1/p2) = (T1/T2) ; Substituting for M we have (p1/p2) = {[{1 + (g–1)xM22/2)}]/[{1 + (g–1) 2 xM1 /2)}]}g/(g–1); p0 = p x [ 1 + {(g–1)xM2/2}] g/(g–1) g/(g–1)

Enthalpy with density: We have (p1/p2) = (ρ1/ρ2)1/2 the total density at

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the point where the velocity becomes zero is: ρ0 = ρ x [ 1 + {(g–1)xM2/2}] 1/(g–1);

Area Velocity relation in one dimensional isentropic flow (Adiabatic flow): when the flow is in viscid Euler equation foe steady flow is written as: ρxVx(dV/dx) + (dp/dx) = 0; c2 = (dp/dρ); dp = c2xdρ; Substitute values. Hence we have: -(V/c2)x(dV/dx) = (1/ρ)x (dρ/dx); Refer equation: get (dA/A) + (dV/V) + (dρ/ρ) = 0; Now We get (1/V) x (dV/dx) = [(V/c2)–1] = (1/A)x(dA/dx); We have (1/V)x(dV/dx) = [(1/A)x(DA/dx)/(M2–1)]; This is an important equation in analysing the flow in ducts and Nozzles. For subsonic flow M 1; This indicates decrease area of contraction leads to decrease in velocity and an increase in area or expansion leads to increase in velocity. The velocity is minimum when contraction is minimum. This principle is used in jet engines. First, the area of section is reduced to decrease the velocity so that sufficient time is available for combustion chamber. The divergence of nozzle causes the velocity to increase providing sufficient thrust for the aircraft. A`, ρ` V` are the sections at sonic section of tunnel. A,ρ,V at any other place. By continuity we have: A`xρ`xV` = AxρxV; V = Mxc = M(gxRxT)1/2; c = (gxRxT`)1/2 We get; (A/A`) = (ρ`/ρ)x(T`/T)1/2 x(1/M); We have: We have T0 = [1 + {(g–1)xM2/2}; Flow is isentropic. ρ` = ρ0, T` = T0and p` = p0. Substitute values and simplifying, We get: A/A` = (1/M)x [[ 1 + {(g–1)/2}xM2]/[(g + 1)/2] ](g + 1)/{2x(g–1)} Mass flow Rate at throat: Let us designate ρ`, V` A` p` and T` as density, velocity, area,presuure, temperature at the sonic section. that is the section for which M = 1; We have (dm/dt) = ρ` x V` xA` We get T` = T0[2/(g + 1)]; ρ` = ρ0x[2/(g + 1)]1/(g–1)

Hence mass flow rate dm/dt = {ρ0xA`/(RT0)1/2 }xg1/2x{2/(g + 1)}{(g + ; Note,g = 1.4 On substituting we get Mass flow rate (dm/dt) = [0.685xρ0xA`/(RT0)1/2 ]; 1)/2x(g–1)}

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12.9. NORMAL SHOCK WAVE This is also called shock front. This is a type of propagating disturbance. It occurs suddenly and abrupt. This wave als0 carries energy. During shock wave pressure, density, temperature raises to a very high values within a short time. This is the very thing region of flow where supersonic flow is decelerated to subsonic flow. It is analogous to hydraulic jump in open channels. We can derive the relationship with flow parameters. Refer the Figure 12.7

Figure 12.7

Control volume chosen has two sections, A and B; 1) Continuity equation: ρ1xV1xA1 = ρ2xV2xA2; 2) Momentum equation: (p1-p2)xA = A(ρ22xV22-ρ12xV12); ρxV2 = ρxc2xM2 = gxpxM2; We can write: [p1 + gxp1xM12] = [p2 + gxp2xM22]: We have (p1/p2) = [ (1 + gxM12)/ (1 + gxM22)] 3) Energy equation: (h1 + V12/2) = (h2 + V22/2); But,h = CpxT; We get, h = [(gxp)/(g–1)xρ]; Final form {g/(g–1)}x(p1/ρ1) + (V12/2) = {g/(g–1)}x(p2/ρ2) + (V22/2); Eliminating ρ2, V2 from the three equations We get:

(p1/p2) = 1 + [2gx(M 12–1)/(g + 1) ];

Finally we get three equations as under: 1) M22 = {(g–1)xM12 + 2}/ { 2gxM12 + 1-g}; 2) (ρ2/ ρ1) = (V1/V2) = [{(g + 1)xM12}/ {(g–1)xM12 + 2}];

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3) (T1/T2) = 1 + [{2x(g–1)xgxM12 + 1}/{(g + 1)2xM12}]x[M12–1]. Nozzles Flow as function of back pressure: 1)

Consider the flow through convergent nozzle. The Figure 12.8 shows the section of a convergent Nozzle and pressure distribution along the Nozzle.

Figure 12.8

When the flow is subsonic throughout nozzle, the pressure along the nozzle increases and pressure decreases. This continues till sonic condition is reached at the minimum section of the nozzle. Pressure drop curve is b. Further reducing of back pressure does not have any effect, as sonic barrier is crossed. The signal for the demand cannot flow. At this stage, nozzle is choked. If the back pressure is further reduced supersonic flow is generated outside the nozzle. The jet pressure adjusts itself to the back pressure by a series of expansion waves as shown in graph c. 2)

Convergent-divergent Nozzle: The curve a indicates the flow throughout the nozzle. Refer the Figure 12.9. The fig indicates the section of a convergent-divergent nozzle and pressure distribution. The curve indicates subsonic flow throughout. The lowering of back pressure, sonic conditions are reached at the throat and maximum flow occurs for a given p0. Further reduction of pressure has no effect on the flow upstream of throat. This is the chocked condition. Beyond the throat, on the divergent side of the nozzle causes supersonic flow. Depending on the values of

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back pressure, a normal shock wave occur in the diffuser section. (inside the nozzle) as shown curve c.

Figure 12.9

3)

Flow through a Truncated Nozzle: Figure 12.10 shows the truncated nozzle. In this case, Nozzle is cut off at the throat section. Diffuser section is absent. Flow at the section can never be supersonic as divergent section is not there. The critical pressure ratio is: (pc/p0) = [2/(g + 1)] {g/(g–1)}; and for air g = 1.4; (pc/p0) = 0.528; pc = critical pressure, (sonic pressure); Me = [{ 2/(g–1)}x[ (p0/ρ0){(g–1)/g} – 1]1/2; Me = Mach number at exit.

Figure 12.10

The static temperature and speed of sound are found by the following expression:

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Te = T0/ [1 + {(g–1)/2}xMe2 ]; ce = [gxRxTe]1/2; fluid density at exit is given by ρe = pb/(RTe). Mass flow = [ρxAxMxc]e (All values at exit);

12.10. COMPRESSIBLE FLOW IN A PIPE WITH FRICTION Refer the Figure 12.11. Now we shall consider the flow of compressible fluids in pipe.

Figure 12.11

We consider two conditions. 1) when pipe is insulated and flow is adiabatic. 2)Another case when pipe is not insulated. It will be isothermal approximately long pipes. 1)

Adiabatic flow: As usual equation of continuity, ρxV = constant, when section is constant. We have (dV/V) + (dρ/ρ) = 0. h + (V2/2) = constant since it is adiabatic. (Note: in all cases the external force due to gravity is negligible) h = CpxT; Cp = [(gxR)/(g–1)]; we have energy equation in differential form: [(gxRxdT)/ (g–1)] + VxdV = 0; Momentum equation written as: A{p-(p + dp)} – f0xPx(∆x) = ρxVxA x[(V + dV)-V)]; (1)P = circumference and a = area of section of pipe. (f0/ ρ) = fxV2/8; (A/P) = R = D/4; P = 4xA/D; Momentum equation(1) in the differential form:

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ρxVxdV + dp + [(fx ρxV2xdx)/2D] = 0; Dividing throughout by p, We have (ρxVxdV/p) + (dp/p) + [(fx ρxV2xdx)/(2 pxD)]; Each term of this equation can be represented in Mach number. V2 = (M2xgxp)/ρ; OR {ρxV2/p} = gxM2; [(fx ρxV2xdx)/2D] = fxgxM2xdx/(2D); (p/ρ) = RT = (c2/g); This renders first term, ρxVxdV/p = gxVxdV/c2 = gxM2x(dV/V); Consider the equation of state: (dp/p) = (dρ/ρ) + (dT/T); Now from the equation of continuity: (dV/V) = (dρ/ρ); From the energy equation: (dT/T) = [-(g–1)xVxdV/(gxRxT)] = -(g–1) xM2x(dV/V); We have: (dp/p) = -(dV/V)-[(g–1)xM2x(dV/V)]; Substitution these expression in the momentum equation, we get (M2–1)x(dV/V) + [gxfxM2xdx/(2xD)] = 0; M = (V/c) = V/(gxRxT)1/2; This in differential form: (dM/M) = (dV/V)(dT/2T); Eliminate T, from above: (dT/T) = -(g–1)xM2x(dV/V) Again use momentum equation: (dM/M) = (dV/V)[ 1 + {(g–1)xM2/2}]; Then eliminating dV/V, We get [ {(1-M2)xdM}]/ [M3x[ 1 + {(g–1)/2}xM2] = {gxfx(dx)/(2D)}; Integrate the equation after re-arranging. Then integrating constant, C can be evaluated by defining L as distance corresponding to M = 1 and Evaluating substituting We get the following expression: [{(1-M2)/(gxM2) }] + [{(1 + g)/(2xg)}] x In [{(g + 1)xM2}/ { 2 + (g–1) xM2}] = [fx(L-x)/D]; x = Distance from the origin corresponding to a mach number M Origin M = 1; The expression indicates that when the flow is subsonic (dM/dx) >0; and mach number increases along the pipe length. If (dM/dx) < 0, Mach number decreases along the pipe. The effect of wall friction is to cause the mach number to approach unity. Suppose M0 = mach number at the entrance at x = 0, the length of pipe L, Sonic velocity is achieved m = 1 is for Air g = 1.4 We have (fxL/D) = (5/7)8[(1/M02)–1] + (6/7) x In [ (6xM02)/(M02 + 5)]; Isothermal Flow: Here t = Constant c = (gxRxT)1/2, (dp/p) = (dρ/ρ); Momentum derived earlier:

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gxM2x(dV/V) + (dp/p) + {gxfxM2xdx/(2d)} = 0 This is the momentum equation [ Same as (ρxVxdV/p) + (dp/p) + [(fx ρxV2xdx)/(2 pxD)]; (dp/p) = – (dV/V) and (dV/V) = (dM/M); We get (dM/dx) = (f/2D)x[(gxM2xdx)/{ 1- gxM2}]; Re-arranging and substituting We get finally {fxL/D} = [In(gxM2}] + [{ 1- gxM2}/(gxM2)]; For length greater than L,choking occurs and mass rate flow decreases.

12.11. MACH NUMBER RELATIONS

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12.12. EXERCISES 1.

Problem:- Calculate the velocity of sound in water of Weight of 9.807KN/m3 , if the value of its bulk modulus is 21090 Kg/cm2. ρ = 9.807x1000/9.81 = 999.7 Kg/m3 . v = SQRT (K/ρ) = (21090x104 /9.807)1/2 = 21.1321/2 x 100 = 4.597 x100 = 459.7 m/s . 2. Problem:- Calculate the velocity of sound in air at a pressure of 1Kg/cm2 and at a temperature of 00 C, g = 1.4 pV = RT ; V=1/w , R = 130 N-m and T = 273 0 C ; p = 1 Kg/cm2 = 100000 N/m2 . 100000x(1/w) = 130x273 ; w= 100000/(130x273) = 2.81 Kg/m3 . ρ = 2.91/9.81 = 0.296 kg/m2 . velocity of sound in air ,c = SQRT(gx100000/0.296) = (1.4x100000/0.296)1/2 = 100(1.4x10/0.298) = 100x6.87 = 687 m/s . 3. Problem:- If the atmospheric air at sea level has a pressure of 1 kg/cm2 and a temperature of 160C, Find the pressure at a height of 3050 metre. 1) Assuming isothermal 2) Adiabatic change (g=1.4). R = 130 N-m . T1 = 273+16 = 2890 C . 1) Isothermal:- h2 = -RT1xIn (p2/1 ) ; Since it is the ratio we can take unit of pressures as 1Kg/cm2. 3050 = -(130)x 289xIn (p2/1) ; In (p2) = -3050/(130x289) = 0.0812 ; p2 = 0.695 Kg/cm2 . Adiabatic : h2 = {g/(1- g)}xRxT1x [ (p2/1)(g-1)/g - 1] ; 3050 = -3.49x130x289[p20.286 -1] ; p2 = 0.68 Kg/cm2 . 4. Problem:- A supersonic air tunnel is required to operate at a mach number of 3. what is the test area if the throat area is 12 cm2. g=1.4 . We have A/A` = (1/M)x [[ 1+{(g-1)/2}xM2] / [(g+1)/2] ](g+1)/{2x(g-1)} As in Area Velocity relation. This has been proved. A/A` = (1/3) x [ 1+(0.4/2)x32]/[ 1.4/2] 3 = 4.23 ; Hence A = 4.23x12 = 51.06 cm2 ;

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5.

Problem:- An air tunnel has a test -to throat area of 3.5. The absolute total pressure and temperature are 5 MPa & 4000K. Find the Mach number, pressure, temperature and velocity at the rest of the test section. A/A` = 3.5 = (1/M)x[{1+0.2M2}/1.2]3. We can use tables for solving problems. M has two values. one less than one & other more than 1 , Take =2.8 T/T0 = 0.389 ; T = 0.389x400 =155.60K. p/p0 = 0.037 ; p = 0.037x5 = 0.185MPa. 6. Problem:- A rocket motor generates a thrust of 30000N at an altitude of 20 Km, Where the exit pressure is 5467 Pa . The pressure inside the combustion chamber (Throat) is 1000 kpa. The temperature is 25000K. Take R= 280 m2/s2 K and g=1 ,Assuming flow is isentropic the throat and exit area. ρe = 5467 Pa , p0 = 1000 Kpa . T0 = 2500 K . (ρe/ p0) = 5467/ (1000x1000) = 5.467x10-3 Thrust (Force) = massx velocity = (pexAexVe)xVe ; (p0/p) = [1+ {(g-1)/2}xM2]g/(g-1) . Substitute for pe,p0 ; We get M =4.15 ; (Ae/ A`) = (1/M)x[{1+0.2M2}/1.2]3 = 12.2 ; T0 = Tx [1+{(g-1)xM2/2}]; (Te/T) = 0.225 ; Te = 0.225x2500 = 562 K. Exit density , ρe = pe/(RxTe) = 5467/(280x562) = 0.0347 Kg/m3 ;

Exit velocity ,Ve = Mex(gxRxTe)1/2 = 4.15(1.4x280x562)1/2 = 1948 m/s; Exit Area Ae = Thrust/( ρexVe2) = 30000/(0.0347x19482) = 0.228 m2. Throat area = 0.228/12.2 = 0.0187 m2 . Problems to Solve 1) Problem:- Find the pressure and density of the atmosphere at a height of 3658 m and the pressure & temperature & density at the ground level are 1Kg/cm2 , 15 Deg C and 1.22 Kg/m3 respectively . Assume that the temperature quiescent atmosphere diminishes with the height at a uniform rate 20 C per 305 Metre Assume for air pV=130T. (Ans 0.638 Kg/cm2 , 0.854 Kg/m3.) 2) Problem: Calculate the velocity of a pressure wave transmitted through a liquid having a specific gravity of 0.85 and a bulk modulus of 19320 Kg/cm2 (Ans 1519 m/s). 3) Problem: Calculate velocity of sound in air having a pressure of 9.38 PSI. temperature -90C . a)Assuming isothermal process . b) Assum-

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ing Adiabatic process . R = 130 Kg-m g = 1.4 . (Ans a = 275 m/s , b = 324 m/s). 4) Problem: Find the pressure and density of atmosphere at an altitude of 3962 m Assuming Isothermal atmosphere at sea level pressure = 1 Kg/cm2 T = 150 C. R = 130 kg-m . (Ans 0.626 Kg/cm2 , 0.768 Kg/ m3 ) 5) Problem: If the pressure & temperature of the atmosphere at ground level are 14.7 Psi & 150C . Calculate the pressure & density at an altitude of 4877 m assuming adiabatic atmosphere. Find also mean temperature gradient up to this altitude , R = 130 g = 1.4. (0.532 Kg/ cm2 , 0.049 lb/ft3, 10/100 m). 6) Problem- calculate the speed of sound in helium & Hydrogen at 500 C . For helium ρ = 0.168 Kg/3 . R = 2077m2/s2 K. Cp = 5187J/Kg K, g = 1.66 = Cp/Cv . For Hydrogen ρ =0.0851 Kg/m3 R = 4127 .Cp = 14223. g = 1.41. 7) Problem:- What is the temperature at nose of a supersonic fighter flying at M = 2 at 273 K. 8) Problem:- An airplane travels at 850Km/Hr at sea level. T = 150C How far would be the airplane be flying at the same Mach number at an altitude where temperature is -500 C ;R=287m2/s2 K . g = 1.4. 9) Problem:- an object is placed in an air flow with a static pressure of 250 Kpa absolute at a static temperature 250 C and a velocity 300 m/s . What are the pressure & temperature at the stagnation point. 10) Problem:- A truncated nozzle is used for measuring the mass rate of flow of methane. The data : area of nozzle 2.5 cm2 . Area of pipe 10 cm2 . Upstream total pressure & temperature are 150 kpa and 400 C. Back pressure 100 KPa . Calculate mass rate of flow of methane. Also assume that Bernoulli’s theorem is valid calculate the mass rate of flow. The density being the density at exit. g=1.31 ; R = 518 m2/ s2 K.

CHAPTER

13

DIMENSIONAL ANALYSIS

CONTENTS 13.1. Introduction................................................................................... 282 13.2. Fundamental Units Of Dimensions................................................ 282 13.3. Summation Of Quantities.............................................................. 283

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13.1. INTRODUCTION Dimensional is a mathematical method of obtaining the equations governing many physical phenomena. It is also checking correctness the phenomena, by finding the left side of the equations matches right side. Here we mean matching that they have to be dimensional satisfied with certain constant as a multiplying factor. Reader will understand the meaning in the latter part of chapter. This is the analysis of the relationship between different physical quantities, mass, length and time. Any physical meaningful equation will have the same dimensions on left and right sides. This property is known as dimension homogeneity. Checking of dimensional homogeneity is the common application of dimensional analysis. Dimensional analysis is also known as factor-label method of or unit factor method is a method of converting one different unit to other. To convert one unit to another we use a converting factor, which will be multiplied or divide to the number or quantity we want to convert.

13.2. FUNDAMENTAL UNITS OF DIMENSIONS First, let us understand the fundamental dimensions. The fundamental dimensions are 1) Mass 2) Length and Time. These dimensions are abbreviated as M(Mass), L Length and T(Time). Now let us understand how it is applied to certain units. Suppose we say volume of a physical quantity. We know the volume = LengthxWidthx Height. It is the products of length thrice. Dimensions of Volume = LxLxL (Length, width, and height are all units of lengths) Volume = kxL3. Example volume of a sphere = 4xpixR3 = constant xL3; the constant = 4xpi. This does not have any dimensions. L represents radius, R Another example: Density = Mass/Volume = Constantx M/L3 = constantxMxL–3. Suppose density is represented by ρ. Density of Sphere of a material = M/(4xpixR–3) = 1/(4xPi)x{MxL–3} = ρ; Analysis for force: We know Force = F = Mass and Accleration = MxLenghs/sec2 = MxLxT–2. Physical quantity and dimensions: Every physical quantity can be expressed as products of fundamental dimensions. Suppose let us say a physical quantity designated by W and it has dimensions M, L, T, and Ø,

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where Ø = an angle. (Ø is not a fundamental dimensions.). This mean, W = F(M,L,T and Ø). W is a function of M.L, T, and Ø. This is a rule. The proof is beyond the scope of this book. The proof can be found in advanced textbook on dimensions. As a rule W = F(M,L,T,Ø) = MaxLbxTcxØd. Suppose the physical quantity is Force, Then we have W = F = M 1xL1xT–2xØ0; It follows, a = 1,b = 1,c = –2 and d = 0 W = Density, ρ = MxL–3, Here we have a = 1, b = –3, c, and d do not exist. it means not the function of density.

13.3. SUMMATION OF QUANTITIES Let us say there exits a physical quantity, V such that, Dimensionally We have V = MaxLbxLcxØd V = F(V1,V2, V3, V4.), Then We have V1 = F(Ma1,L,b1,Tc1,Ød1); V2 = F(Ma2,L,b2,Tc2,Ød2); V3 = F(Ma3,L,b3,Tc3,Ød3); V4 = F(Ma4,L,b4,Tc4,Ød4); etc.

The dimension homogeneity exits. If the relation between function is of type, V = V1 + V2 + V3 + V4 + ----- Then we have a = a1 + a2 + a3 + a4 + ------; b = b1 + b2 + b3 + b4 + ------; c = c1 + c2 + c3 + c4 + ------; d = d1 + d2 + d3 + d4 + ------; This is called dimensional homogeneity. Products of Quantity: We have U = U1m1U2m2 U3m3, ---- and we have

MaxLbxTcxØd = Ma1m1x Lb1m1Tc1m1 Ød1m1 + Ma2m2x Lb2m2Tc2m2 Ød2m2 + Ma3m3x Lb3m3Tc3m3 Ød3m3 + ----; Then we have; a = a1m1 + a2m2 + a3m3 + a4m4 + ----- = M b = b1m1 + b2m2 + b3m3 + b4m4 + ----- = L c = c1m1 + c2m2 + c3m3 + c4m4 + ----- = T

d = d1m1 + d2m2 + d3m3 + d4m4 + ----- = Ø Non-Dimensional parameters:

In this case U is the non-dimensional value. then we have

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0 = a1m1 + a2m2 + a3m3 + a4m4 + ----- = M0 0 = b1m1 + b2m2 + b3m3 + b4m4 + ----- = L0 0 = c1m1 + c2m2 + c3m3 + c4m4 + ----- = T0

0 = d1m1 + d2m2 + d3m3 + d4m4 + ----- = Ø0;

The above expression can be expressed in the form Matrix. Readers may refer books on linear algebra.

13.4. Limitations of Dimensional Analysis Although the method is more useful, it cannot lead us too far, because of the following reasons: • • • •

Dimensional analysis cannot be used to derive equations involving additions and subtractions. Numerical constants having no dimensions, cannot be obtained by this method. Equations using trigonometric, logarithmic, and exponential functions cannot be derived by this method. If dimensions are given, physical quantity may not be unique as many physical quantities have same dimensions.

13.5. Frictional or Viscous Resistance We know that the governing factor of this problem is the coefficient of viscosity. Let µ = Coefficient of viscosity. Let us obtain equations of all unknown indices in terms of the index µ. We know viscous resistance of a fluid depends on the wetted perimeter, l, Velocity, v, the density, ρ, and coefficient of viscosity, µ. Let us assume the law is unknown. Then equation for viscous resistance, R = k1xρax µbx lcx vd. In other words, R ∝ ρax µbx lcx vd Here a,b,c, and d are indices and not known. k1 = constant

We know resistance is the force and its dimensions = MLT–2. We have Dimensional equations, MLT–2 = k[(ML–3)a x(ML–1T–1)b x(L1)cx (LT–1)d. µ is the governing variable. Its index is b. Equate index of M on both sides, 1 = a + b; a = 1-b (13.1); Indices of L on each sides are equated. We have 1 = –3a-b + c + d (13.2).

Dimensional Analysis

The indices of T; –2 = -b-d; d = 2-b

285

(13.3);

Substitute values of (13.1) and (13.3) in (13.2); We have, 1 = –3(1-b)-b + c + (2-b) on simplifying we get c = 2-b (13.4); R ∝ [ρ (1-b)x µbx l(2-b)x v(2-b) ]; Rearranging we Have R ∝ ρxl2xv2x{ µ/ (ρlv)}b.x{ µ/(ρlv)} = (1/Re);

where Re = Reynolds number. This is dimensionless. Hence we write expression as R = k1xρxlxv2; k1 is a frictional coefficient and normally represented by kd.

The solution has proved that frictional coefficient kd is a function of the non-dimensional constant, (ρlv) which is the Reynolds number.

13.6. Surface Wave Resistance of Ships The resistance is due to formation of surface when ship sails. The pressure in the water due to ship’s motion cannot be maintained at water surface. Because the pressure on the surface of water is atmospheric. Hence the water is lifted against gravity by the water pressure. Thus governing factor is gravity, g. The total resistance depend on the density of the liquid, the value of g, the wetted perimeter (linear dimension) and the velocity. Now resistance is the force, R ∝ [ρaxgbxlcxvd ]; g is the governing variable, obtain all indices in terms of the index of g, that is b. MLT–2 = (ML–3)ax(LT–2)bxLcx T–2bxLcxLdxT-d; Equating indeces: Equating M 1 = a

(LT–1)d;

MLT–2 = MaxL–3axLbx

(13.5);

Equating T, –2 = –2b-d ie. d = 2–2b Equating L 1 = –3a + b + c + d

(13.6);

(13.7);

Substituting for a and d from equations (13.5) and (13.6); We have c = 2 + b; We have R ∝ [ρaxgbxlcxvd ]; Re-arranging substituting the values putting terms containing inside bracket, We have R ∝ [ρxl2xv2x (gl/v2)b]; The term (gl/v2) = (1/Fr)2;

Fr = Froude number. This proves that the non-dimensional constant is Froude number. R = k2x[ρxl2xv2 ]; It also can be written in the form, D = [CDxρxAxv2/2]; (Cd∝2k2);

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A is a function of l2 and immersed cross-sectional area amidships. D = Drag due to surface wave formation. 23.7. Non-dimensional factor for small orifices: Refer Figure 13.1. Consider a pipe of Diameter D and an orifice it of Diameter d; p = pressure difference between the two sides of orifice, due to heat loss. Our problem is to produce no-dimensional factors which can be found by applying of dimensional similarity. Let us assume Discharge through orifice, Q ∝ paxdbxDcx ρdxµe. Substituting, Dimensions, We have L3xT–1 = k[ (MaxL-ax T–2a) xL xLcx(MdxL–3d)x(MexL-exT-e); b

(MaxL-ax T–2a) = Dimension for p; LbxLc = Dimension for D and d; x(MdxL–3d) = Dimension for ρ; (MexL-exT-e) = dimension for µ; Let us obtain all indices in terms of c and e; 0 = a + d + e; d = -a-e (13.8);

Figure 13.1

Now equate indices of T, –1 = –2a-e; a = -(1/2)-(1/2)e; Substitute values of a in (13.8); d = -(1/2)-(1/2)e; substitute for a and d, 3 = [-(1/2) + (1/2)e] + b + c + [(3/2) + (3/2)e]-e; b = 2-e-c; Substituting values; Q∝ (p1/2xd2xρ–1/2) x[ µ/(dxp1/2xρ1/2) ]ex (D/d) c ; (13.9)

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287

Hence the constant of proportionality, say k is the function of [ µ/ (dxp1/2xρ1/2) ], which corresponds to Reynolds number, and is also a function of (D/d); Thus the equation (13.9), Q = kxd2x(p/ρ)1/2; i.e x[ µ/(dxp1/2xρ1/2) ]ex (D/d)c = k.

CHAPTER

14

FLUID POWER

CONTENTS 14.1. Introduction................................................................................... 290 14.2. Definitions, Coefficient, And Power Calculations........................... 291 14.3. Properties Of Fluids In Hydraulics................................................. 295 14.4. Elements Of Power Hydraulics....................................................... 296

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14.1. INTRODUCTION In this chapter, we study the utilization of fluid to generate, control, and transmit power. Many regard this to be a separate branch of fluid mechanics. Some categorize it as hydraulics and pneumatics. Ultimately we study the utilization of fluids to make them more useful towards achieving our own purposes. Some terms, such as density, pressure, discharge, etc., are redefined here for the purpose of refreshing the memory. The definitions should universally remain the same. We deal with Hydraulic- defined as properties, effects, and the power derived from liquids- and Pneumaticmeaning that which encompasses the same traits but from gases; for example, compressible fluids utilized for power conversion. Utilization of fluid for purposes such as transferring energy, lifting objects or similar efforts goes back to 200 BC. The first person known to the world who utilized fluid power is Archimedes. Other historical figures known to have explored and employed the tools and principles of hydraulics are Pascal, Bernoulli, and Euler to name just a few. Definitions: Let us define a fluid. No doubt there are many ways of defining fluid, but in simple terms, it is defined as a mass/matter that flows continuously. A few examples of fluids are water (called as universal fluid), oil, milk, etc. any of such kind. However, air is also a fluid. Is it not different from water or oil? Yes, it is true- fluids are further classified as liquids and gases. This classification is made on their differences in unique properties at ambient temperature. A liquid like water, for all practical purposes, is incompressible. whereas air is compressible. It is based on this property that we determine how fluids are classified as either liquids or gases. It is common knowledge that many solids, under different conditions, exist in fluid form. The fluid is essentially only a state of matter. Fluid power: This is the use of fluid under certain conditions to generate, transmit, and control power. Suppose we use liquid for this purpose, then we call it hydraulics/ hydraulic power. In case we use gas for the purpose it is called pneumatics/pneumatic power. The method of transmitting power, controlling power is different in case of liquids and gases. In the next Para, we restrict our discussion to Hydraulics and then move on to Pneumatics. Hydraulics: The unique advantage in this method of transmitting power, controlling power etc; is that its power to weight ratio is very high compared

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with any other methods. What this means can be demonstrated with the following example. In any method of transmitting generated power, we need to design machines or equipment to perform said function- to transmit electrical power we need design an electric motor, as with hydraulic power we design hydraulic motors. Ultimately we find that in comparing the two motors the weight of hydraulic motor is very low yet it can transmit a very high amount of power. For example, an electric motor of 10 HP. The motor weight would be in the order of 20 KG or more. For a hydraulic motor to transmit the power of 10 HP, its weight may be as low as 1 KG. This is the reason why in many places where weight is a crucial consideration, hydraulic power transmission is universally preferred. In aeronautical engineering, for example, hydraulic machines are used extensively because it is commonly accepted that aircraft should have minimum weight and produce maximum power, as they work directly against gravity while taking off- requiring power to move both forward and upward simultaneously. With trains, cars, and other motor land vehicles less efficient motors are acceptable as they are not moving vertically, only horizontally- unlike airplanes.

14.2. DEFINITIONS, COEFFICIENT, AND POWER CALCULATIONS Pressure: Now suppose the area cross section where the liquid is filled is “A” Sq cm, Then the ratio “W/A” is called the pressure and expressed in kg/ cm2. Thus we arrive at a definition that pressure is the force per unit area of section perpendicular to the direction of force acting. Generally, it abbreviated as “p”. [“p = W/A” W is the load/force or weight in Kegs and “A’ area in Sq cm] Discharge: Now let us consider the case when an object moves up due to pressure. Suppose it moves with a velocity of “V” cm/sec. Then per second, the area swept is [ = Axv.] Because the entirety of the liquid moves per second “V” cm/sec. The value “AxV “is called discharge and expressed as volume/sec. That is cubic cm/sec. It is normally expressed as liters per minute, LPM. It is only conversion. (One litre = 1000 cc). Discharge, Q [Q = AxV] Suppose the weight of liquid is “w”, grams/cc. Then the value [“ wxAxV”]is called mass flow/sec. Mass flow M

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[M = wxAxV] In hydraulics, we deal with discharge and we assume the specific weight remains constant, since liquid weight will not change appreciably with temperature. Work done by liquid and Power: By definition, Work done = [ForcexDistance] moved in the direction of force application. Suppose the stroke of cylinder is ‘L” then for the object to move from lower level to top level, by definition, work done = WxL = pressurex AreaxL = pxAxL. Suppose the object takes “t” seconds to move up, then what is power? Power is defined as work done/second. Hence power, P = pxAxL/t. Now let us consider the value “L/t”. This is the distance traveled by the object/sec and the liquid in the cylinder also travels same distance, since the liquid is pushing the object. The distance traveled by the liquid per second is nothing but velocity that we have defined as ‘V Hence, Power, P = pxAxV, This with suitable units expressed as Horse Power, HP or Kilowatts, KW. Liquid Head, H•(Figure 14.1) In hydraulics we more frequently speak of “Head”, abbreviated as “H” and it is expressed in meters, cm or Feet. This value enters into many of our calculations. Let us find out what is it.

Figure 14.1

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The sketch is the symbolic representation of tank filled with liquid, say water or oil. Let the weight of liquid per unit volume is “w”. As we have defined this is called as specific weight. Let us say pressure is “p”. The total weight of liquid = wxAxH. where “A” is surface area of tank and is uniform. The total force at the bottom of tank is due to the entire weight of liquid. Therefore, total force = pressure and area, pxA this value is also equal to wxAxH, Thus pxA = wxAxH, p = wxH. This how we can relate the pressure to the Head, “H” In other words: p/w = H Now once again going back to expression to power, P = pxAxV, replace P with who, We have Power, P = wxHxAxV, We have defined, Discharge, Q = AxV Therefore, P = wxQxH The expression in this form for Power, P is very important and used extensively in our calculations. Velocity Head: This term also used more frequently in hydraulic power and calculations. Let us define it as the following: suppose in the Figure 14.2 you may see an orifice (Hole) at the bottom. Suppose this is open, then liquid flows out. It flows with a certain velocity. We maintain “H” liquid level at a constant. That means we replenish liquid that flows out. Then it can be proved from Newton’s law and mathematically H = (VxV)/ (2xg), where g = standard gravity. The value related to H, i.e., (VxV/ (2xg) is called velocity heat. Certain Constants and Coefficient: Discharge Coefficient: The theoretical values of velocity, discharge, and size of orifice are not equal to actual values. Due to frictions, viscosity, and various other factors, these values are less than the calculated values. In order to calculate actual values, we use coefficient. These are determined experimental values. These coefficient multiplied by the theoretical values give actual values. There are three co-efficient. liquids.

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Figure 14.2

In the sketch Figure, 14.2 shown liquid is flowing from a tank through an orifice (a hole) at the bottom of tank. The actual size of whole diameter is “d” and area of flow ‘a”. When liquid flows through this hole the flow diameter is not ‘d” or area of flow is not “a”. In other words, the liquid will not flow through the full area of section and it less than “a”. The ratio, “ Cc = Actual flow area/ area of orifice “ is called coefficient of contraction, abbreviated “Cc” The value Cc depends on various factors, such as area profile, viscosity, and also depends on the type of liquids. Normally these values are given in standard table It may vary from 0.65 to 0.85. When liquid flows the actual velocity of flow is not equal to calculated value, i.e., not equal to V = Sqrt(2xH). But it is less than this calculated value. The ratio “Cv = Actual velocity of flow/theoretical calculated flow” is called coefficient and abbreviated as Cv The value of Cv also depend environmental conditions, such as type of liquid, velocity range etc. its values vary with type of liquid and also temperature of the liquid. This value can also determine experimentally. Standard values are given in the form of tables. Its value data not available could be assumed 0.85

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Now we know Q = AxV. When “A” theoretical and also calculated. We know A (actual) = Ccx A; V (Actual) = Cvx V. Thus it follows that Q(actual) = A(Actual)xV(Actual) = CcxAxCvxV = (CcxCv)xAxV. Thus actual discharge = (CcxCv)x Q(Theoretical) The product of “CcxCv “abbreviated as “Cd” This is called as discharge Co-efficient. Cd = Q(Actual)/Q(Theoretical) To sum up A(actual) = Ccx A(Theoretical) V(Actual) = Cvx V(Theoretical) Q(Actual) = Cdx Q(Theoretical) When Cc, and Cv is not known besides not known, it may be assumed 0.85 to 0.9.

14.3. PROPERTIES OF FLUIDS IN HYDRAULICS Properties of liquids used in power transmission: By now we should know the properties of liquids. They are: Viscosity, Density, Specific volume. Viscosity is the ability to flow easily in layers. Its units are Stokes and poise in the metric system. We are not going to be discussing them in detail as the properties are given in manufacturers manuals and textbooks. Regarding density, specific volume, it is assumed readers know about it. Normally specific volume is used in gases. When transmitting power in hydraulics we must select specific media to potentiate results. We know they are liquids, but they must also act efficiently and without reacting adversely to the materials used to construct the containers or pipes they will contact. Ultimately only a very few liquids are selected and can be used. In the past, water was used for power transmission. Water has several good properties such as its heat transmission, inexpensiveness due to abundance, it is both noncombustible and non-toxic. However, it reacts with almost all materials of which it comes into contact and its lubricating properties are very low and not ideal. This causes issues such as leaking at high pressures. Examples of successful water applications are abundant in the hydraulic presses in steel mills and forge shops. The other liquid used in hydraulics for transmitting power is mineral oils. Now in majority of industries, machinery, and constructions equipments use petroleum based oils for power transmission. Oils have very good properties such as good lubrication, being noncorrosive, (it does not react

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with material that it contacts) and its boiling point is higher than that of water. Compressibility is lessened with a rise in oil temperature, and it is also non-conductive. However, once again in modern times water is being considered as a more suitable for use in hydraulic machines. The main reasons for this are cost factors and rising environmental consciousness. We give here below the trend in the form of a chart. This is only a brief review of fluids used in hydraulics and their properties. More details are available in books and relevant manufacturers’ manuals.

Figure 14.3

[The graph shows the application trend of water Vs oil in fluid power. (Figure 14.3)]

14.4. ELEMENTS OF POWER HYDRAULICS Hydraulics Machines: Now let us talk about hydraulic machines. For the purposes of this text, we shall consider a machine to be that which is composed of certain material elements arranged in a definite manner to transmit power. A few examples of hydraulic machines are the hydraulic pump, hydraulic motor, hydraulic cylinder etc. These are the machines that are mainly responsible for transferring power. They do nothing but transmit it as it is. These types of hydraulic elements are called active elements. As we know we need more power, sometimes we do not need or at some time we need less power depending on our needs. That is the power developed to be used as much as we need and also to be used. This technique is generally called as controlling of power produced. In order to control power, we need certain other equipments like that of pumps, motors. These

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are not called machine. But these can be called as mechanisms for all general purposes. These mechanisms are called as passive elements. These items do not take any main role in the transforming the power. These elements are divided in three classifications depending on their functions. 1) Pressure controls; 2) flow controls; 3) Direction controls. Besides, the above said passive elements; there are other special hydraulic elements. They are: 4) Filers; 5) Suction strainers; 6) Hose pipes; 7) Pipes and fittings; 8) Hydraulic intensifiers; 9) pressure switches. Let us discuss these things later in following pages. Now let us list important fluid elements which are frequently used for power transmission: Active elements: As we have discussed these elements transmit/ transform power and are directly involved in the process. Hydraulic Cylinders • Hydraulic cylinder; • Hydraulic pump; • Hydraulic motor. These are the main active elements. These are machine elements, are further classified into subcategories depending on their construction, function, and method of transmitting power. Hydraulic Cylinder: These are also called linear actuators. Let us discuss these items with reference to a sketch Figure 14.4:

Figure 14.4

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The above sketch (Figure 14.4) is the schematic of a hydraulic cylinder. Its function is to transmit load “W” from one level to other level. The movement of the load is linear. (In Straight line), hence the name is linear actuator. It is comprised of a cylindrical barrel normally made of steel, inside which a plate slides in and out. The plate called Piston. It’s fixing is such that liquid does not surpass, but the plate can slide up and down easily inside cylinder. To this piston is connected a rod called piston rod. It moves inside the cylinder, such that no liquid would surpass it. To the other end of piston a plate is fixed and on which load may be placed. It need be of any shape which can hold object. There are two holes one at the top and the other at the bottom. These are called ports. Depending on how liquid enters one is port called inlet port and other is outlet port. The end of cylinder where rod comes out is called rod end and the other end where piston is nearer is called dead end. The function of cylinder is to lift an object from one level to other level. Here the object placed on the piston rod is called load, normally symbolized as “W” Referring to the sketch, the load initially at a lower position as in the first sketch the load is at a lower level. The liquid enters the cylinder at the dead-end port. The liquid is entrapped between cylinder and piston. As more and more liquid enters the liquid, forces the piston up. The piston, rod, and object placed on it moves up. When the cylinder is vertical the object moves against gravitational force. Types: Cylinders: Cylinders are classified based on their functions and Mountings Based on functions: classified as Single acting and Double acting Cylinder. The figure–14.5 shows double acting cylinder. In this case, liquid can enter from both sides, through ports. The cylinder has two ports. In case of single acting, liquid can enter from one side only. It has only one port. After it is extended to do certain work, the cylinder will retract on its own depending on its mounting. Suppose it is vertically then by selfweight it retracts. If it is mounted horizontally, it retracts by spring force. There would be a spring or air in a compressed condition housed inside on the rod end of the cylinder. Cylinder and Ram: The single acting cylinder when mounted vertically the liquid enters inside bore for lifting. In this case, piston need not fit exactly in the bore. It would allow liquid on the other side. Still, the Piston rod (Ram) can move. This is by displacement. When the liquid enters with pressure, it displaces equal volume of Ram. (In this case, piston rod is called

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Ram). This Ram moves. When the liquid comes out from the same port, the Ram retracts by its own weight due to gravity. Thus in case of Cylinder and Ram, Cylinder with Rod at both ends, In this case, the piston rod would project on both ends in either direction the force acting is same. Cylinder with cushions: In order to avoid the jerk, a sort of cushion is provided at the ends. It has to be mounted vertically. The cost of manufacturing this type of cylinder and Ram is lesser than other types of cylinders. Telescopic cylinders. Normally for a cylinder, single acting or double acting, the extended length at full stroke is = 2x length of barrel + extra of around 500 to 700 (Extra depends on stroke and size of cylinder) When sufficient space is not available to accommodate the cylinder in the retracted condition, then this type of cylinder are used. The rod is in the form of a telescope in another. For detail construction, please refer the books or manufacturers guide. Specification of cylinders: Cylinders are specified to the requirement on the following parameters: Stroke in mm– Length it comes out of the barrel Pressure: Maximum operating pressure in Kg/cm2 Pushing force in Kgs. What is the maximum load it can lift? Rod diameter in mm: Piston rod diameter Barrel diameter in mm: This is related with pushing force and pressure. (Only two of three parameters, 2,3 and 5 are required.) Mounting dimensions in length units: How the cylinder is fixed/fitted. Hydraulic power: Power = [Load(Kg s)x speed of movement in (meter/ sec)/102] KW OR Power = (Pressure kg/cm2/(100 x 100) x (0.785xBarrel Dia in meter) x (Barrel Dia in meter) x Speed of movement of Piston rod meter/sec/102 KW. The hydraulic symbols for various types of cylinders are shown in Figure 14.5.

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Figure 14.5

Hydraulic pumps and Hydraulic Motors: Generally people use simple hydraulic pumps. Hydraulic pumps have two types: the first is employed to lift liquid from lower to higher level. Here the purpose is to transfer liquid from one level to higher level. Example: centrifugal pump. The main purpose is to transfer liquid. Let us differentiate from Hydraulic power pumps, that which produces high pressure in liquids. In this case, pumps produce very high pressure, in the order of 200 kg/cm2–500 Kg/cm2. Depending on application it would in special circumstances is more? Now let us classify these pumps based on construction and function. Hand pump

2) Power driven pumps

Here classification depends on how it driven. As name indicates, a hand pump is operated manually. These types of pumps develop very high power, in the order of 200 kg/cm2–500 kg/cm2. But the flow of liquid at that pressure is very low. It would be generally 25 cc per stroke. Here we call stroke since it is operated, manually by hand or leg. One upward or downward thrusting movement of handle of hand pump is one stroke. Normally these pumps are reciprocating type. In principle, it has a barrel inside which a plunger moving up and down for a short distance. There would be a spring and steel ball assembly that does not allow oil to flow back to sump once oil is sucked out.

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Figure 14.6

Figure 14.6 shows the hydraulic symbol of the pump Power driven pumps (Figure 148) In this case pump will have a shaft that is coupled to a prime mover such as an electric motor or engine or any such prime movers. The shaft rotates at an RPM and sucks liquid from a sump/tank above which pump is mounted. The pump has two ports. Inlet and Outlet. Liquid is sucked from inlet port and discharge under pressure in outlet port.

Figure 14.7

The hydraulic symbol of pump is shown (Figure 14.7). The power driven pumps are classified based on their construction. 1) Piston pump 2) Gear pumps 3) Vane pump 4) Pressure compensated pumps Piston pumps: As the name indicates, there will be a piston, a barrel and a check valve similar to pump in principle. Instead of handle, it has a shaft driven by a prime mover. Piston pumps

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The piston pump further classified, based on how they discharge the liquid. •

Fixed displacement: The pump discharges a fixed quantity of oil for every rotation of pump shaft. • Variable displacement: In this case, the quantity of liquid displaced per rotation can be changed. Within the design limits. That is, the pump flow can be adjustable, (Ex: say we can adjust 2cc/rev or 3 cc/rev etc. up to the maximum value of say 5cc/rev) The maximum pressure developed in piston pumps could be as high as 700 kg/cm2. Piston normally used in high-pressure power packs used in hydraulic presses, rolling mills, etc. For detail application, we have to refer manufacturer’s catalogs. Compared with rotary pumps (Gear and Vane pumps- called as rotary pumps), these pumps give less discharge and pressure is higher than that of rotary pumps. The volumetric efficiency is lesser compared to rotary pumps. The pistons pumps are also classified based on their internal design as 1) Radial piston pumps 2) Axial piston pumps. Variable discharge is possible only in axial discharge pumps. Gear Pumps: This is a simple design concept. Two gears meshed and one of the gears has an extended shaft and comes out of the Housing. The gears are housed inside enclosure and supported on bearings. Oil comes out from the clearance between the gears and inner wall of the housing. For detail construction and function please refer websites or manufacturers guide. These pumps are fixed displacement pumps and discharge per revolution is fixed. These pumps are available for pressure up to 300 kg/cm2. Vane Pumps These pumps have a rotor (Cylindrical solid shaft), housed inside a barrel. The rotor has radial slots at equal intervals. In these slots, rectangular plates called vanes will slide in and out radially within the design limits (Say not more than5 mm). The barrel in which rotor housed is oval. As the rotor rotates the vanes inside slots moves in and out due to centrifugal force. Thus it is always buts with wall of the barrel. The rotor further extension/fixed another shaft that is coupled to a prime mover. The inlet and outlet are diametrically tapped holes on barrel. These are sucking and discharge ports. For more details of constructions, viewers may refer manufacturers guide or Hydraulic books.

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Pressure compensated pumps: These are normally available as piston pumps. It is a special feature added to the pump. The pump has a relief valve built-in. It is integral part of pump. When the system pressure attains the set value the pump relief valve, pump discharge becomes zero. Thus extra power being wasted is avoided. This would be explained under hydraulic circuit design. Variable displacement pumps: In these type of pumps, the pump flow can be varied within maximum and minimum limits. Normally these types of pumps are axial piston pumps. Here, 3 or 5 or 7 pistons barrels are arranged parallel to the axis of rotation. Pistons reciprocate inside the barrel and pumps the oil. There is a plate inclined to the axis of pumps, called swash plate. The inclination is varied and stroke of each pistons are increased or reduced, within design limits. Thus when stroke varies pump flow also changes. It is difficult to explain with sketches or drawing. Readers are requested to refer to the manufacturer’s guide or any books on hydraulic pumps.

Figure 14.8

Screw pumps: These pumps are based on the principle of a threaded screw moving inside a hole. The barrel acts as a hole and a pump rotor is the screw. As rotor rotates, oil is entrapped in the clearance between the barrel and screw thread. Oil drifts as the screw rotates. Its usage is lesser when compared with other pumps. For specific details refer to manufacturers guide. Double pumps/ Tandem pumps: Two pumps of similar design and construction are assembled together. The pumps have one drive shaft. Inside there is coupling to connect the rear pump. This is called tandem pump. When one gear pump and one piston are connected it is called as double pumps. There are no hard rules defining tandem or double pumps.

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Such pumps have two inlets and two outlets. They function as a separate pumps, but driven by one drive shaft. They produce two different pressures. As such the pressures can be adjusted independently. One pump may be at 100 kg/cm2 and other at 50 kg/cm2. This helps to operate two hydraulic power circuits. (This you will know in our chapter on Hydraulic circuit design) This is analogous with two different voltages in electricity, like 220 V Supply and 410 Volt supply. Efficiency related to Hydraulic pumps: There are two types of efficiency when referring to Hydraulic pumps: 1) Volumetric efficiency 2) Hydraulic efficiency Volumetric efficiency: Theoretical swept volume in CC per revolution of shaft/ Actual volume of liquid discharged in CC per revolution of shaft. CC: Cubic centimeters Generally, the value for gear pumps would be in the order of 92%–96%. Generally, the value for Vane pumps would be in the order of 88%–92% Generally, value in case of Piston pumps would be in the range of 80% – 88% Generally, value for Screw pumps would be in the range of 75%–60% Note: Volumetric efficiency varies with oil properties. Hydraulic Efficiency: This is the ratio of input power to input power Efficiency = Output power/ Input power. This value varies from type of pumps, Manufacturers design and properties of oil used. However, generally for hydraulic pimps, it would be in range 80%–90% Performance curves of hydraulic pumps: (Figure 14.9, Figure 14.10) Manufacturers’ test hydraulic pumps after it is assembled and it is tested for its performance. The values are plotted in relation to discharge, pressure, volumetric efficiency, RPM, Power transmitted. These are called performance curves. Performance curve shows the relation of flow in LPM (Litres per minute) to RPM (Shaft revolution per minute) for pumps P–1, P–2, P–3 and P–4. The black lines shows theoretical flow and red lines are actual flow obtained during test. The graph (Figure 14.10) shows the relation between pressures, Flow in LPM and Input HP (KW). For pumps P1, P2, P3 andP4. The Power, at different RPM, can be read out from a graph. However theoretical Power

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can also calculated if we know Pressure and Flow. Actual power can be calculated if we know the efficiency of the pumps. Many manufacturers give performance curves in relation with HP or pressure.

Figure 14.9

Figure 14.10

Theoretical power required in case hydraulic pumps: N = RPM. v = oil discharges cc/revolution, k = Volumetric efficiency, Qa = (Nxvxk/1000); k =

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Volumetric efficiency. Qa = Actual flow in LPM, pressure = p kg/cm2. m = mechanical efficiency k and m are less than 1. Power, P required to drive pump = [Qaxp/(mx450)] in HP HP.

In terms of RPM Power = [(Nxv/1000)xp/(mx450) = [2.22xNxvxp/(m)]

There are different ways of calculating the HP. By any means, the end result is the same. Hydraulic Motors: Hydraulic motors are analogous to electric motors. The amount of discharge input in relation to the current pressure at which it is operating is similar to current and voltage of electric motor. The construction of Hydraulic motor is similar to that of Hydraulic pumps. Types of hydraulic motors are also similar to hydraulic pumps, such as Gear motors, Vane motors, and Piston motors. There are no screw motors. The hydraulic motor receives Oil of discharge Q at input pressure p and its shaft rotates. We can couple to it a Gearbox or a winch, etc. the manufacturers of hydraulic motors are the same as that of hydraulic pumps. Performance curves are similar to that of hydraulic pumps. For details, you have to refer the manufacturers’ guide. Power calculations are similar to that of pumps. In hydraulic motors, it is important to know the mathematical relation between the Torque, RMP, Discharge, and pressure. Let us say volumetric efficiency = k, Hydraulic efficiency = m (The mechanical efficiency may be called as hydraulic efficiency in Motors) Qa = Actual flow in LPM. Qa = (Nxvxk/1000) p = pressure kg/cm2, In put power = Qaxp/450; Output power = Qaxpxm/450. Hydraulic Motor power = 2x3.14xNxT(Torque)/4500; Now Hydraulic motor power = Out power, NT = 0.065xQaxpxm T = Torque in Kg-Meter. Motor manufacturers normally furnish the performance curves showing the relation between N, T, Qa, and p. Hydraulic Valves: 01) Pressure control valves, Flow control valves and Direction control valves. Now let us know the function of these parameters. We are not going to discuss any construction features, as that depends on manufacturers. The

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hydraulic circuit explains the functions of all these hydraulic elements. It is the graphical representation of various hydraulic elements assembled together to form a hydraulic system that transform hydraulic power into any other form of power, such as mechanical/electrical. The elements that connected these elements are pipes/tubes. In the Figure 14.11 Hydraulic system does mechanical work by lifting a load. Thus hydraulic energy is transformed to mechanical energy ([Eq: Power = Energy x time]).

Figure 14.11

The Figure 14.11 shows how various hydraulic elements are connected (assembled) to lift a load. Hydraulic pump driven is by prime movers discharge hydraulic oil at pressure. The oil comes through pipe, Valve 02, 04 and applies pressure at the bottom of 01. Then the load is lifted. The pressure is always developed by the load acting on it. We have already defined the pressure. Now let us understand the function of each element: . Besides we also discuss various types of hydraulic valves and the sub-categories. 01) Hydraulic Cylinders: We have already discussed 02) Flow control valve (Figure 14.11) In principle this valve has an orifice through oil flows. The orifice area of flow could be adjustable by screwing/unscrewing a screw/nut. Arrow mark is graphical symbol for adjustment. For detail construction, refer to the manufacturers guide/catalog. By adjusting the quantity of flow, speed is controlled. Thus speed of movement of “W”

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is controlled. By the side of this valve, there is another valve, this called check valve. This valve is categorized under direction control valve. The function of this valve is not to allow oil to flow into the cylinder. It allows oil to flow out of cylinder. While oil is flowing inside its flow is controlled by orifice adjustment. While oil is going it flows through the check valve that has a larger area of section. By this function, the cylinder moves freely and faster in a downward direction. There are several types of flow control valves depending on their construction and function which we discuss separately. 03) Pressure Control valve: In this application, we call it a relief valve. This maintains the pressure in the hydraulic system. Its function is similar to the function of the heart in the human body that maintains blood pressure. The graphical symbol is shown in the circuit. It is connected to line but of line. It measures the pressure and does not allow extension beyond the set value. When the pressure exceeds the limit, then it allows excess oil that causes the pressure increase to flow to the tank directly. The pressure can be set to different values such as 100 kg/cm2, 120 kg/cm2, 200 kg/cm2, etc. This also acts as a safety valve for the pump, just like electrical fuse. In principle, it has a spool and spring. The spring controls pressure. Pressure control valves have various functions related to their application. We shall discuss later in this chapter. Depending on its controls it is called in different names, such as relief valve, safety valve, sequence valve, reducing valve, etc. 04) Direction control valve (Figure 14.11) As the name indicates it changes the direction of flow of liquid. This is a spool valve. The graphic is shown in the circuit. It can have two ports, three ports, four ports, and multiports, depending on the function it has. Ports are simply inlet and outlet holes. In the circuit shown the valve has four ports. Two are incoming ports and two are outgoing ports. The valve has a spool (cylindrical shaft) inside of its housing. The spool can be actuated (moving back and forth) by an electrical solenoid. An electrical solenoid is a pin that pushes the valve back and forth powered by electricity. When electrical power is switched off to both solenoids the spool remains in the neutral position. In the neutral position (circuit shows valve in

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05) 06)

07) 08)

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neutral position) the ports on one side are connected. This makes the oil flows to tank. There are two solenoid one on each side of valve. Let us say bottom solenoid pushes the spool. Then the spool moves to parallel arrow mark position. In this condition, the farther port is connected to the dead end of the cylinder and near the arrow mark connects the rod end of the cylinder to the tank. This causes oil to flow to the dead end of the cylinder and oil from the rod end flows to the tank. The load starts rising. Let us say the load has risen to its’ full length, and we want to bring it down. Then the top solenoid is actuated (When current is applied to this solenoid). Then the ports are cross-connected (i.e., crossed arrows). This makes the oil from dead-end flow out and oil from the valve flow to the rod end of the cylinder. This is the function of the direction control valve. In essence, it changes the direction of flow. We shall discuss in detail various types of valve later on. Hydraulic Pump: This we have dealt with. Return line filter: This we can call auxiliary element in the hydraulic system. Its function is to clean the oil. Oil is circulated in valves, pipes, cylinders etc. and should continue to repeat the operations. When oil returns to tank, it is contaminated. In order to remove this a contamination filter is used. The various types filter and their classifications will be discussed later in the chapter. Suction strainer This allows clean oil into system. Its function is almost similar to the filter that we have already dealt with. We shall deal with its size and types etc. discussed later in the chapter. Tank/Hydraulic reservoir This is a fabricated container that holds oil required for the system. There are other parts in the tank. Sizes of tanks required for a particular operation will be discussed later in the chapter. Prime mover: This is the main source of power. This can be an electrical motor, Engine etc. Electrical controls for the hydraulic system: In the hydraulic, we have seen electrical solenoid; it is powered by electricity in a controlled manner. Depending on the complexity of the hydraulic system, there may be many more hydraulic elements, such as pressure switches, limit switches etc. which may need electrical power in a controlled manner. Hence there would be an

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electrical controls system. It is necessary for a hydraulic engineer to know these aspects if not in detail, an overview. The electrical control system is called control panel and arranging the electrical elements such as switch gears, fuses, relays, solenoids etc. as per the requirements is called an electric control circuit. Now let us discuss in detail the hydraulic valves in-detail. 02) Flow Control Valves: Depending on how flow is controlled, these valves are categorized as: • 02.1 Throttle valve – fixed; • 02.2 Throttle – variable; • 02.3 Flow control valve variable with pressure compensation; • 02.4 Flow control valve with pressure and temperature compensation; • 02.5 All the flow control valve s is available with built-in check valve; • 02.6 Way flow control valve. Further, all these flow control valves are classified based on how their mounting. They are: • Line mounted flow control valve; • Sub-plate mounted flow control valve; • Modular type flow control valve. Let us analyze each type of flow control valve: •

02.1 Throttle valve – Fixed: With this type the flow through these valves remains constant once the pressure is constant. We cannot adjust the flow and it gives fixed flow. Graphical symbol is here below. The flow changes in accordance with changes in pressure from the upstream side:

Figure 14.12 (Graphical symbol)

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02.2 Throttle valve – Variable: In this case, there is a knob/ screw with which you can adjust the flow. However, the flow changes with pressure on upstream side. (Fig 14.13)

Figure 14.13

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02.3 Flow control valve with pressure compensation- In this case, the pressure is adjustable, besides the pressure will not change with change with pressure on upstream side. (Refer Figure 14.14)

Figure 14.14

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02.4 Flow control valve with pressure and temperature compensation: As we know liquid properties change with temperature. If the liquid temperature changes this causes change in flow quantity even when all other parameters remain constant. In order to keep the flow un-altered, the valve would be specially designed so that even there is temperature variations, flow will not change.

Figure 14.15. Flow control valve with pressure compensation.

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02.5 Flow control valve with check valve: In these types of valves there is a check built in the valve so that when oil returns it flows without any restriction. The graphic symbol all is the same except a check valve symbol is added thus here below. Flow control valve with check valve – graphical representation. Oil flow freely in opposite direction to arrow.

Figure 14.16 (Flow control valve with check)

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Line mounted Flow control valves: Flow control valves can be directly mounted online. It means incoming flow line is directly coupled to the valve “IN” and outgoing directly connected to “OUT”. These valves are provided with threaded knob for flow adjustment. Sub-plate mounted flow control valves: these valves have rectangular valve housing. They are mounted on another plate called sub-plate. To connect the valve with the flow line, piping work is required. Modular type flow control valves: These are special design and will be discussed on the chapter under “Modular Valves” Pressure and flow ranges: The pressure between “IN” and “OUT” of the valve is called differential pressure.” ∆p”. This pressure is proportional to the flow through the valve. Higher ∆p, larger is the flow. Also, flow depends on the inlet pressure. Flow control valves are manufactured to the pressure of 300 kg/cm2 and even more. The flow in LPM is from 0.5 LPM to 100 LPM or even more. The actual flow and actual differential pressure can be calculated, mathematically if we know inlet pressure, differential pressure, and temperature. These can also be found in manufacturer’s manuals and in some books on hydraulics.

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Manufacturers furnish performance curves showing relations between differential pressures in kg/cm2 flow in LPM for their different models of valves. Just for reference, a typical curve is depicted. For detailed performance curves, readers have to refer to the relevant manufacturers’ guide/manual.

Figure 14.17. Performance curve- flow control valves models P1, P2, P3, and P4.

•

03) Pressure Control Valves: The pressure controls are categorized based on their functions in specific applications. Different types of pressure control valves have totally varied construction features and design features. These valves are always connected offline. The graphic symbols also differs from one type to another. Various types of pressure control valves: (03.1) Safety Valve; (03.2)Relief Valve direct acting; (03.3) Relief valve pilot operated cum Un-loading valve; (03.4) Sequence Valve; (03.5) Pressure reducing valve.; (03.6) Brake valve.; (03.7) De-compression valve The basic function is the same in that all of them control pressure. Now let us evaluate one by one.

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(03.1) Safety Valve: As the name indicates, its purpose is to protect the system. Every system has to operate within the limit of pressure. If the pressure increases beyond the design value, the system could be damaged. In construction, the safety valve has a hole in the housing, a conical valve pin with spring, preset compression force acting on the valve pin. The hole through which oil flows to the tank is closed by this spring force. The valve has two ports, one connected to the system pressure and other to the tank. When the system pressure increases beyond design value, Oil pushes the valve pin against the spring and it opens to the tank. Thus oil flows to tank, until the pressure acting on the valve pin s equal to the system pressure. This way the system is protected from rising pressure. The graphic symbol is given here below. It may be designed to match any pressure.

Figure 14.18. Safety relief valve -Graphical symbol)

Figure 14.19. Section of a safety valve. • 03.2) Relief valve direct acting: This valve is used after the safety valve and its maximum pressure setting less than that of safety. In this valve the pressure setting ranges. Example range

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10 kg/cm2 to 100 kg/cm2. This means that it can be set to any pressure between 10 to 100. Suppose the pressure is set to 50 kg/ cm2, then the system pressure will not increase beyond this limit. Direct acting refers to the fact that it is directly fitted in line like a safety valve. Its construction is same as safety valve, but there is a screw that adjusts the tension/ compression of the spring. By adjusting the spring force you are altering the set pressure. For detail design and construction, you may refer manufacturers’ guide. Graphic symbol is given here below

Figure 14.20. Relief valve Adjustable (Graphical symbol)

•

03.3) Relief Valve – pilot operated cum on-loading: This is the combination of a pressure relief valve and a pilot operated pr. relief valve. This has two functions. It controls the pressure and also unloads, i.e., the pressure line is connected to tank line. Thus pressure in the system is atmospheric. Thus when there is no work, there won’t be power loss. The graphic symbol of the valve is shown in Figure 14.20. The main valve has another port ‘x”. Oil at pressure enters the port apply pressure. When system pressure exceeds this pressure is applied. Because of this pressure, the pilot spool (Small spool) in the valve opens and allows the oil to spring housing/main housing. This pressure pushes the main spool. Thus the oil from the system flows to tank until pressure is neutralized with system pressure. The oil that is flowing from external line through ‘x” combines with the main tank line. Hence it is called internal drain.

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Figure 14.21. Functional diagram of pilot operated relief valve.

Figure 14.22

Figure 14.23. (Relief Valve with un-Loading Electrically control (Graphical symbol)]

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This is another case of pilot control valve relief valve cum unloading valve, where the pilot line “y” is separate. Here the pressure is controlled electrically. When system is not working, the pilot line is connected to tank (y). and the oil goes to tank. This way system pressure attains atmospheric. When system should be put on pressure, the solenoid valve (two position and two – way valve electrically operated is energized. Thus the pilot pressure acts on the valve and hence main pressure. Note: It may be possible to people who are new to hydraulic power to understand these functions. Therefore they need to read books on hydraulics or read manufacturers’ guide. •

03.4) Sequence Valves: (Figure 14.24) This is a pressure control valve, that has definite function. Consider a machine clamp, operated hydraulically. In this case, first job should hold in the clamp with full force. Then any operation such as drilling or milling or pressing can be done. Without the job being held firmly drilling/milling cannot be done, otherwise the job would fly off due to drilling/milling force. It means the operation is sequenced. 1) Job to be held; 2)The operation drilling/Milling In such application, the sequence valve is used. The valve allows oil first to the holding fixture and pressure to build up to the set value. Only after set value pressure is reached is oil then allowed to the drilling/milling/pressing spindle to rotate/move for further operation on the job. In the event of any pressure drop on the first stage (i.e., clamp becomes loose due low pressure), the second stage stops.

Figure 14.24. Graphical symbol.

The graphic representation is shown above. (Figure 14.24). It is similar to the relief valve except the line is connected to the system, whereas in

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the case of the relief valve it is connected to the tank. Its application with reference to a system is dealt with in hydraulic circuit design. •

03.5) Pressure Reducing Valve: This is also called as pressure regulator. Let us say in a hydraulic system, we want to limit the pressure to a particular value. Oil should flow at the pressure and pressure should not increase. In such cases, a pressure reducing valve is used. As an example, if overall system pressure is 120 kg/cm2. In the system, there is one cylinder where we do not want to increase the pressure beyond 40 kg/cm2. In such a case the pressure reducing valve is used just before connection to the cylinder. Even though the incoming pressure is 120 kg/cm2, the cylinder pressure will not increase beyond 40 kg/cm2. In short, it is called as PRV

Figure 14.25. Graphical symbol.

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03.6) Brake Valve: As the name indicates its role in the hydraulic system is to stop the system function at the time of emergency. It functions similar to vehicle brakes. It will be discussed with an example in hydraulic circuit design. 03.7) De-compression Valve: In hydraulic systems let us consider a situation when a cylinder/container are under full pressure, say 200 kg/cm2. We need to bring it to atmospheric pressure or low pressure say 10 kg/cm2. When the cylinder/container is of huge volume it contains high volume of oil at high pressure. Thus a huge amount of energy stored. In such situation, we cannot suddenly open cylinder/container to atmosphere/ low-pressure area. If it is

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done it may damage system or an accident could occur. In order to avoid this situation, we use a decompression valve. It opens the pressurized oil passage in the cylinder/container gradually through an orifice and pressure is reduced gradually. Its function and application are discussed with an example in the section on hydraulic circuit design. 03.8) Counterbalance valve: The function of this valve is to balance a load that is accelerating/decelerating. To better understand the purpose of the counterbalance valve, consider an electric lift. The other end of lift is fixed with some dead weight. Its role is to balance the load and lower the lift uniformly and gradually. Without the deadweight, there would be too much strain on the lift motor to control its speed/velocity uniform. This the exact function the counterbalance valve does in hydraulic system.

Figure 14.26. Graphical symbol

The application of counterbalance valves with examples will be reviewed in hydraulic circuit design. 04) Direction Control Valves: These are the valves used in hydraulic system to change the direction of liquid. Here we deal various types of valves, in detail. Depending on number position the valve can operate: 1) One position and two way: The example: Check valve;

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Figure 14.27. Graphical symbol

Check valve is a two-valve single position. It allows oil in one direction and blocks the oil in the opposite direction. Depending on the function it can be categorized into three types. Check valve without external assistance allows oil in one direction and steel ball in the valve closes by fluid pressure in case oil tries to flow in the opposite direction. Check valve with pilot operation, oil can flow freely in one direction. Oil can flow in opposite direction, in case there is pressure acting on the ball to unseat. Check valve with spring, a little pressure is required for the oil to flow in the forward direction. In the opposite direction, oil flow is blocked. There are many types of check valves which will operate from pneumatic pressure, electrical solenoid valve etc. Check valve with definite amount of spring force (maybe some time adjustable) will operate only when the oil pressure exceeds spring force. For the flow in the forward direction, there should be certain amount of oil pressure. This pressure is called cracking pressure. Flow in the reverse direction is blocked. Two position three – way valves. Graphics shown here.

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This type valve is used when the load return on its own, due to spring force or gravity force or pneumatic pressure. Oil pressure lifts the load. After work is done the load returns gradually to its original position. This type of valve used in Upstroking hydraulic presses, Scissors platform etc. The valve is available as spring solenoid operated, detent type, pilot operated etc. Two position four – way valve

This valve has two positions and paths for the oil flow to travel. In one position the ports are connected to the system. The graphic symbol is selfexplanatory. It is available in various configurations and may be spring solenoid actuated, detent solenoid actuated or pilot or lever operated as well. Three position four – way valve

This type of valve has three positions. Left, right, and center position. In the center position, the valve is not connected to the actuator. By operating the valve through electrical solenoid/manual lever, in one position the left side of valve is aligned. That is if the cylinder is the actuator, dead end of cylinder is connected to pressure port and rod end to tank port. Similarly, if the other solenoid is energized then the connection is reversed. This type of valve is commonly available in many forms or configurations.

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Multi-position valves: . These are mobile valves. Normally used in construction machines, such as Dumpers, Loaders, Riggers etc. The valves in this grouping are considered special types. Depending on the type of actuation the valve is classified as: •

Electrical Solenoid operated and spring controlled. Spring controlled solenoid operated. In this type of valves actuation force for the valve spool to move is supplied by electrical solenoid coil. Restoring force would be from compression spring. For more detail, construction readers may refer manufacturers’ guide. • Electrical solenoid operated and detent controlled: In this case, there would be detent to hold in position; electrical solenoid would supply force sufficient to push the valve spools. • Pilot air pressure operated. As the name indicates, air pressure operates the valve and is restored in position either by detent or spring force. • Manual lever operated: Here a hand lever is provided to operate the valve spool. Once it is pushed it holds in position by spring/ detent. • Electro – hydraulically operated. These are heavy duty valves that handle high volume of Oil. Here there are two valves. Main valve carrying high flow. The main valve spool is operated by hydraulic oil pressure. The oil is directed by another small direction control valve that is solenoid operated. Depending on inlet and outlet port configuration: Before, we go to port configuration, let us know about ports. Normally there are two inlets and two outlets to the direction controls valves. These are called ports. Two inlets ports are “P” and T”. One of them is entry of pressure oil to the valve, is called “P” port. The other port is for the exit of the expelled oil with no/ less pressure. This called as port “T” Outlet is, one through oil enters cylinder/actuator is called “A” port, the other through oil exit is called port “B”.

H-Port

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All ports blocked.

Pressure port blocked & others connected.

P & T connected, A & B blocked.

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A&B connected & P&T Blocked

There are many ways depending on the application where the valves are to be used. One has to refer the manufacturers guide. The above-described configurations are more frequently in hydraulic circuit design. Direction valve construction and its function: The readers may refer to books on hydraulics or manufacturers guide for details construction and functioning. In short, the valve assembly has three parts: • Housing; • Valve Spool assembly; and • Actuator assembly. Housing is normally of steel/cast iron or machined out of steel block and precision machined to close tolerances Spool assembly: This is cylindrical shaft of alloy steel precision machined. The spool slides inside housing bore. The clearance between is lowest and the spools can move and no oils can leak through. Actuator assembly: This could be electrical solenoid and spring assembly, detent, and solenoid assembly or a Hand lever to operate the spools. Hydraulic Schematic/Hydraulic circuit design: Now with the details furnished we are in a position to prepare/design a hydraulic design. Hence it is absolutely necessary to design the hydraulic circuit before machine is build. Hydraulic circuit design is based on the total power required to do a certain function in hydraulic machines. Hydraulic circuit is an arrangement and assembly of various hydraulic elements in an orderly manner to do required function as quickly as possible and with maximum efficiency, besides with no hazards. The hydraulic system is absolutely necessary to be accompanied with the machine sold to any client. At the time of any fault in the machine, it helps to detect it. It is absolutely necessary for the

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maintenance engineer in case one deals with hydraulic machines, to have a thorough knowledge of hydraulic circuit and also able design at the time it I needed. On the client’s side, as everyone knows marketing department sell the machines, the marketing engineer should have a thorough knowledge of the hydraulic circuit, besides he should be able to design one. This is more required in case the client is a manufacturer of Hydraulic power pack/ Hydraulic system, for customers’ application. That is why many hydraulic company hires engineer as “sales and application engineer” who are specially trained in hydraulic system and hydraulic circuit design. In short, the following persons should have a thorough knowledge of hydraulic circuit • Designer; • Assembly person; • Sales and application engineer; • Service engineer; • Maintenance engineer. Hydraulic circuit reading: In order to read and understand hydraulic circuit, the person should have a thorough knowledge of basic fluid mechanics and about some technical terms, their relations and calculations. Basic Fluid parameters: • Pressure; • Discharge; • Power; • Discharge coefficients; • Fluid velocity; • Liquid; • Head; • Load; • Hydraulic loss. One should also have a knowledge of Hydraulic elements, their functions and in brief their constructions.

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Hydraulic Elements are: • Hydraulic pumps; • Hydraulic cylinders; • Pressure control valves; • Flow control valves; • Direction control valves; • Accumulators; • Intensifiers; • Flow dividers; • Hydraulic motors; • Rotary actuators; • Pipes and Fittings; • Bell housing and Couplings; • Filters; • Suction Strainers; • Coolers. The function of these elements and brief construction features are already explained in this article at the beginning. The other supporting hydraulic elements are: • Pressure Switches; • Temperature Gages; • Pressure sensors; • Temperature sensors; • Limit switches; • Flow meters; • Manometers. Hydraulic Circuit design: Let us deal with hydraulic circuit design a little more detail. Hydraulic circuit based on type of hydraulic elements used, the way they connected and arranged could be classified into three type: • • •

Conventional Hydraulic circuit; Manifold built Hydraulic circuit; Sand witch type/Stacked/Hydraulic circuit

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• Cartridge valves hydraulic circuit. There are no hard and fast rules in the classifications, but for easy understanding, we classified as above. Conventional hydraulic circuit: In this case, the various hydraulic elements are connected by pipes and fittings only in an orderly way to perform the function as designed/required. Simple example is here below.

Figure 14.28

Look at the above hydraulic circuit. Here all the hydraulic elements 01 to 09 are connected by pipes and fittings. In this type of occupies, more space and a lot of pipes are used. Same hydraulic circuit may be made very compact by using manifold. On cost considerations in most of cases, the circuit with pipes and fittings is expensive compared with that of Manifold construction. Manifold built hydraulic Circuit: (Figure 14.29) Hydraulic circuit is shown below. We can visualize the difference between the conventional and manifold built hydraulic circuit. In this case, pipes are bare minimum. The only pipes connected are P to Pump, T to Cooler, A, and B to cylinder dead end and Rod end. Internal connections between valves are all holes drilled in

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the Steel/Al block called Manifold (10). The manifold is manufactured out of block, machined all faces and sides on a milling machine/CNC machine. Holes are drilled for oil to flow as required. The hydraulic elements are mounted on face of block and fastened bolts. Besides in between the face of manifold and valve surface O-rings/rubber seal are interfaced to avoid leakage from the contact surface. This type of construction is compact and occupies less space, besides, man hours spent on bending pipes are totally avoided. When the unit is required in batches, it is comparably cheaper. The only problem, in this case, is that the manifold is to be designed, machined, and drilled with at most care. If any internal holes are interconnected the entire manifold would be rejected, since it is of no use besides, repairing is not economical. However, nowadays many 3D modeling software are available and also CNC machines are available it is easier to manufacture such manifold block. Another advantage is the connections from P and T, A and B to external system could be hoses. This makes the system more flexible. As an example the cylinder is working in hazardous area, then the pump, manifold can be placed on remote safer area.

Figure 14.29

Sand witch/Stacked Hydraulic Circuit•(Figure 14.30); – In this case, all the valves except direction control valves are of plated construction. The valve will be designed like a rectangular type. All the holes have holes

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through for P, T, A and B ports. The plates are assembled one above the other depending on the design. The top one is always a directional control valve. In the circuit shown the green line shows the valves. There are three valves, Relief valve, flow control and top DC Valve. These valves are tied together with studs. There is an ISO standard for these types of valves. These type of valves are known as CETOP Valves. CETOP – Valves: All the valves, pressure controls, flow controls direction control valves are available in the CETOP range of valves. We are not going to discuss this standard in detail. Interested readers may in website http://www.cetop.org/. Then you can search for cetop standards. In general, three Cetop standard are popular. (1) CETOP–3 (2) CETOP –5 (3) CETOP–7.

Figure 14.30

Cartridge Hydraulic Circuit: All valves are available in the form of cartridges. They can be inserted inside housing that user can manufacture from steel/aluminum or any material. Basically, the valve sits in the bore machined for the purpose. The bore will be circular and cartridge will be cylindrical. Besides, there will be a cap, the design of which varies depending on the function valve has to do, i.e., pressure control, flow control or direction control. Readers may refer manufacturers guide for detail construction and functions one can get from these valve. These valves are also called Logic Valves. These cartridge valves are also manufactured by the same manufacturers who manufacturers CETOP valves. We are not showing the hydraulic circuit from these valves. These circuits are designed when a very large quantity of fluid is to be handled and many more function is required. To name a few industries, Steel plant industries hydraulic controls, Heavy

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duty hydraulic presses Rolling Mills etc. There is no definite rule that it should be used for a specific application. It can be used, theoretically in any hydraulic function. The advantage is interchangeability. Cartridge valves are interchangeable from pressure control to flow control etc. Hence we can build up any type of valves with cartridges. We come across in circuit design; more frequently, a condition where we have operate at low pressure very high volume of oil and at the same time operate at very high pressure and at low volume of oil. As an example let us consider a hydraulic press, where cylinder has to move certain distance, up – word (against gravity) at fairly high speed and after moving certain distance it has to compress material into the mold at a very pressure but slowly. In such situation, to optimize pressure we use a typical circuit and called as High-Low Hydraulic circuit. Let us analyze how it works. High-Low Hydraulic Circuit: (Hi-Lo) (Figure 14.31) Here below we have designed a hydraulic circuit, to demonstrate the use of Hi-Lo Circuit, use of two Pos –3 Way valve and Cylinder Ram assembly P1 and P2 are 5 LPM and 100 LPM discharge pumps. P1 develops pressure up to 150 kg/ cm2 and P2 can develop only 25 kg/cm2 pressure. Relief valve 04 set at 200 kg/cm2 and Relief valve 05 set at 20 kg/cm2. Cylinder ram assembly, 10 is single acting. It has only one port. Two Pos Three – way valve used 11. In the system, the ram has to move at a fast rate, 100 mm/sec to a distance of 500 mm. At this position, it is very close to mold and has to apply very high force to press material (15) into the mold. The speed at which it presses is very low, say 5 mm/sec. Let analyze, how it happens.

Figure 14.31

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Oil from pumps P2 and P1 flows into the bottom of cylinder ram. This makes ram moves out of cylinder. Thus it moves at high speed. Now the load to be lifted is only the dead weight of the ram and accessories, hence less force. Accordingly, pressure developed is only 10 kg/cm2. Once material (15) comes in contact the punch surface for pressing, the resistance increases. Force developed to resist also increases and hence the pressure increases. The pump P2 cannot increase its pressure and hence discharge oil to tank through relief valves 05. Now the check valve 06 does not allow oil to flow into the tank from the high-pressure pump P1. Thus the flow from pump P1 flows only to the cylinder via DC valve 11. Relief valve, 04 is set high pressure. Thus oil flows at high pressure, but at low speed. When you want the ram moves downward, the valve 11 is actuated. The ram moves down by gravity. The downward speed is controlled by the flow control valve (12). In this circuit, we have a learned how to use cylinder and ram assembly and the Two pos–3Way valve besides the application of Hi-Lo circuit. Whenever there is an up- stroking press/ system, we can design cheaper hydraulic circuit Regeneration hydraulic circuit: We shall discuss about regeneration in hydraulic circuit. It is basically using the leftover energy from the hydraulic oil that goes to tank. This is possible in case of double acting cylinder and only one side piston rod. Here below the hydraulic circuit shown. Let us discuss the circuit. Here as usual oil enters to the dead of cylinder and load is pushed/ actuated. Pump has certain flow from fixed displacement pump, say Q. Let us say p1 kg/cm2 on dead end side of cylinder. Oil from the rod end will return to tank in normal case. In this case, what is done, oil is pumped to the input line. (Pressure line). Return oil from rod end joins the main flow at “x”. This is clear from the following argument. Let us say cylinder is moving uniform velocity/speed pushing the load. Suppose we offer a resistance to the rod end side of cylinder by having a sequence valve set at 100 Kg/cm2. The pressure developed on rod end is p1 (A/A-b). That is greater than p1. The sequence valve does not allow oil to flow to tank. Hence it flows via check valve 10 to the point x. Since the pressure, [p1(A/A-b)] is more than input pressure (p1), oil joins the main flow. Hence the load is pushed faster than it would have been without oil returned to rod end side. In the return direction, when DC valve crossconnected oil flows to the rod end of cylinder. Oil tries to flow through check

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valve 10, but it cannot, as it is at the same pressure. Hence it is forced to flow to the rod end of cylinder and pushing the piston in return direction. Oil in the dead end would return to tank.

Figure 14.32

INDEX

A acceleration 4, 70, 79, 80, 110, 114 Accumulators 326 active elements 296, 297 Adiabatic compression 268 adiabatic flow 269 adiabatic process 261, 265, 267 aeronautical engineering 291 altitude 25, 26, 27 ambient temperature 25 Angle of inclination 153 angles of blades 116 Angular deformation 61 angular displacement 62 Angular Momentum 122 angular velocities 159 applied mechanics 30 atmosphere 5, 16 atmospheric pressure 25, 27, 89, 108, 240 average velocity 170, 171, 172,

179, 184, 185, 186, 191, 192 axial flow equations 156 Axi-symmetric flow 158 B Bazin’s Formula 102 Bernoulli’s equation 110, 200 Bernoulli’ theorem 84 Blasius formula 182 Borda’s mouthpiece 106 Boring machines 154 boundary conditions 139 Boundary layer 164, 166 bulk modulus 263, 265, 267, 277, 278 Buoyant force 40 C capillary tube 14, 15 Cartesian coordinates 73 cavitations 16

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centripetal acceleration 70 Chezy formula 241 Chezy’s law 249 circular channel 240 coefficient of contraction 84, 99 Coefficient of discharge 107 Coefficient of Discharge 84 Coefficient of velocity 84 coefficient of viscosity 284 combustion 121 compressibility 2, 20, 258, 263, 268 compressible fluids 2 compressor 121 conservation of energy 74, 217 conservation of momentum 110 constant discharge 251 constant velocity 76, 77 continuity equation 65 continuous deformation of layers 7, 9 continuous flow 62 contracta 88, 89, 91, 98, 107, 108 contraction 200, 205, 206, 207, 226, 232 coordinates 54, 55, 56, 64, 65 Couette flow 155, 158 crest 100 critical velocity 194, 198, 199 curvature 30 D Darcy’s constant 206 Darcy’s equation 200 datum 25, 26, 27, 28, 29 dead end 298, 309, 321, 327, 331, 332 density 3, 4, 5, 7, 9, 14, 16, 17, 20,

54, 55, 64 dimensional homogeneity 282, 283 Discharge Coefficient 293 Discharge constant 137 downstream 103, 104, 105 drilling 317 Dynamic Viscosity 9 E Eddy Scales 177 elastic string 219 elementary fluid particle 123 Enthalpy 262, 269 entropy 262 equilibrium 30, 31, 35, 36, 37, 47, 147, 151, 154 Euler’s Equation 72 Euler’s theorem 84 F Filters 326 Flow dividers 326 flow fluid passage 240 Flow nets 76 flow parameters 55, 56, 63, 117, 121 fluid 22, 23, 24, 25, 27, 31, 34, 38, 44 fluid element 132 fluid flows 144, 148, 164, 165 fluid jet impinge 115 fluid of density 111 fluid parameters 54, 55, 144 fluid particles 57, 59, 60, 61, 63, 70, 71, 170 Fluid power 290 Fluids 1, 2, 6, 10 Fluid velocity 325

Index

forces 22, 24, 30, 31, 35, 36, 38, 39, 40, 41, 47 Francis turbine 127 friction 88, 89, 91, 97 frictional resistances 194 friction factor 195, 206, 208, 231 G Gases 2, 3, 6 gas laws 258, 259 gauge constant 34 geometric translation 60 gravitational force 36, 37 gravity force 73

335

Inertia forces 144 initial mass 120 inlet port 298, 301 inlet pressure 312 input flow 63 Integrating constant 136 Intensifiers 326 intermolecular attraction 12 Internal energy 258, 259 inviscid fluid flow 73 Isothermal process 259 J jet propulsions 121

H

K

Hagen-Poiseuille flow 195 Helmholtz model 161 Horizontal components 30 Hydraulic circuit 304, 324, 325, 326, 327, 330 Hydraulic cylinder 297 Hydraulic gradient 196 hydraulic jump 249, 250, 251, 252, 256 hydraulic machines 291, 296, 324 hydraulic mean depth 195 Hydraulic motor 297, 306 hydraulic power 290, 291, 293, 304, 307, 317 hydraulic power plants 220 Hydraulic pump 297, 307 hydraulics 290, 292, 295, 296, 312, 317, 324 hydraulic turbines 122

Karman’s momentum integral equation 181 Kinematic Viscosity 9, 10 kinetic energy 74

I inertia 70

L lamina 27, 28, 30 laminar flow 144, 160, 162, 170, 171, 178, 179, 181, 182 Law of conservation energy 74 Linear deformation 60, 61 linear equations 155 liquid column 37, 38, 39 liquid molecules 3 Liquids 2, 15 liquid surface 14, 16 M magnitude 146, 163 Manometer 33, 34 mass 3, 4, 5, 7, 10, 11, 14, 20 mass conservation 64 Mass oscillation 224

336

Theory And Problems Of Fluid Dynamics

Maximum efficiency 114 mean velocity 240, 242, 245, 247, 256 metacenter 40 Metacentric height 41 micro manometers 34 momentary time 170 momentum 110, 111, 112, 113, 114, 115, 122, 123, 124, 126, 128, 145, 166 momentum theorem 88 motion 70, 73 N Navier’s equation 150 Newtonian fluid 18 Newton’s laws of motion 70 nomenclatures 150 Non-Newtonian fluids 10 Non-uniform Flow 56 non-uniform pipe 74 normal temperature and pressure (NTP) 5 notches 100 nozzle 84, 85, 86 O open channel 240 orifice 84, 85, 88, 89, 90, 91, 92, 93, 94, 95, 96, 99, 105 P Pascal 5, 11 Pascal’s Law 23 passive elements 297 Pelton wheel 125 Perfect Gas Law 3 physical quantity 282, 283, 284 piezometric head 208

pipe resistance factor 185, 186, 187 Pipes and Fittings 326 pitot tube 247, 248 pneumatics 290 potential energy 74 pressure 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 48, 49, 54, 55 Pressure drop curve 272 pressure gauges 33 pressure gradient 24, 206 pressure head 205, 216, 231, 237 pressure regulator 318 R Rectangular Channel 242 rectangular lamina 60, 61 rectangular prism 35, 37 Refolds’ number 160 Relative Density (RD 4 relative velocity 113, 115, 116, 117 reservoirs 102 resistance 284, 285 resistance factor 195, 206 retardation 216, 231 Ronald’s number 144, 160, 162, 163 Rotary actuators 326 S saturation point 3 shear stress 2, 8, 9, 10, 17, 18, 19, 144, 145, 151, 152, 160 shock front 271 siphon Spillway 103 space coordinates 132 specific energy 252, 253, 255

Index

Specific Gravity (SG 4 spillways 105 stability curve 163 stable equilibrium 36 Static Pressure 22 steady flow 54, 55, 56, 59, 65, 133, 135 straight line graph 100 strain energy 217 streamline curvature 71 streamline curves 76 Streamline Flow 57 Stream surface 59 stream velocity 188 subsonic flow 270, 271, 272 Suction Strainers 326 Supersaturated Vapors 3 supersonic flow 270, 271, 272 surface area 22, 28 surface element 22 surface tension 12, 13, 14, 15, 19 surge tank 221, 222, 223, 224, 237 T Taylor series 73, 74 thermodynamics 258, 262 torque equation 124 Torques 122 trailing edge 165 transition stage 170, 184 Trapezoidal Channel 243 turbulent 144, 160, 161 turbulent flow 170, 171, 172, 177, 179, 183 U uniform diameter 96

337

uniform velocity 154, 155, 166 unit vectors 115, 123 unstable equilibrium 36 unsteady flow 136 U tube manometer 196 V vacuum gages 33 valve 203, 204, 213, 215, 216, 217, 221, 223, 224, 231, 232, 233, 236, 237, 238 velocity 54, 56, 57, 58, 59, 63, 65, 66, 67, 285 velocity distribution 246 velocity functions 132 velocity gradient 145, 155, 165 Velocity head 205, 212, 215, 225 velocity variation 246 velocity vector 57, 58, 59, 70, 71 vena contracta 202 venturimeter 87, 88 vertical components 30 viscosity 144, 149, 153, 160, 164 viscous fluid 154, 164 viscous forces 144 volume 2, 3, 4, 5, 6, 7, 14, 16, 18, 20 Volumetric efficiency 304, 305 Volumetric Strain 6, 18 W water flows 127 wedge 22, 23, 41 weirs 100, 101, 102 Wetness Factor 3 wetted perimeter 284, 285