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FAST READING

As per Latest Syllabus of Anna University - TN

For III Semester B.E., Mechanical Engineering Students

New Regulations 2017

With Short Questions & Answers and University Solved Papers

Dr. S. Ramachandran, M.E., Ph.D.,

Professor - Mech

Sathyabama Institute of Science and Technology

Chennai - 119

www.srbooks.orgwww.airwalkbooks.com

125/-

First Edition: June 2018

978-93-88084-02-4

UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS 12Units and dimensions- Properties of fluids- mass density, specific weight, specific volume,specific gravity, viscosity, compressibility, vapor pressure, surface tension and capillarity.Flow characteristics – concept of control volume - application of continuity equation, energyequation and momentum equation.UNIT II FLOW THROUGH CIRCULAR CONDUITS 12Hydraulic and energy gradient - Laminar flow through circular conduits and circular annuli-Boundary layer concepts – types of boundary layer thickness – Darcy Weisbach equation –friction factor- Moody diagram- commercial pipes- minor losses – Flow through pipes inseries and parallel.UNIT III DIMENSIONAL ANALYSIS 12Need for dimensional analysis – methods of dimensional analysis – Similitude –types ofsimilitude - Dimensionless parameters- application of dimensionless parameters – Modelanalysis.UNIT IV PUMPS 12Impact of jets - Euler’s equation - Theory of roto-dynamic machines – various efficiencies–velocity components at entry and exit of the rotor- velocity triangles - Centrifugal pumps–working principle - work done by the impeller - performance curves - Reciprocating pump-working principle – Rotary pumps –classification.UNIT V TURBINES 12Classification of turbines – heads and efficiencies – velocity triangles. Axial, radial and mixedflow turbines. Pelton wheel, Francis turbine and Kaplan turbines- working principles - workdone by water on the runner – draft tube. Specific speed - unit quantities – performancecurves for turbines – governing of turbines.TOTAL: 60 PERIODS

INDEX

AAbsolute zero Pressure, 1.64

Air Vessels, 4.178

Atmospheric Pressure, 1.64

Axial Flow Reaction Turbines, 5.76

BBellows, 1.79

Bernoulli’s Equation, 1.143

Boundary Layer Concepts, 2.129

Boundary Layer Theory, 2.129

Boundary Layer Separation, 2.162

Bourdon gauge, 1.76

CCapillarity, 1.21

Cauchy Number (Ca), 3.46

Cavitation, 1.15, 2.127, 4.107

Centrifugal Pump, 4.52

Chezy’s formula, 2.50

Circulation And Vorticity, 1.122

Co-efficients, 1.193

Coefficient of contraction (Cc), 1.194

Coefficient of resistance (Cr), 1.195

Coefficient of discharge (Cd), 1.194

Coefficient of Velocity (Cv), 1.193

Compressibility 1K

, 1.14

Continuum Flow, 1.101

Critical Reynolds Number, 2.3

Critical flow, 1.107

DDarcy - Weisbach formula, 2.49

Dimensional Analysis, 3.1

Dimensionless Numbers (NonDimensional Numbers), 3.43

Displacement Thickness , 2.132

Distorted models, 3.80

Draft Tube, 5.102

Drag And Lift Coefficient, 2.154

Drag Force (FD) on Plate of Length(L) , 2.141

Dynamic Similarity, 3.49

EElementary cascade theory, 4.38

Energy Losses In Pipes, 2.49

Energy thickness , 2.136

Equations of Motion, 1.139

Equipotential Line, 1.137

Equivalent Pipe, 2.94

Euler Number (Eu), 3.45

Euler’s model law, 3.58

Euler’s Equation, 4.44

FFlow Through Branched Pipes, 2.107

Flow Characteristics, 1.102

Flow Through Parallel Pipes, 2.96

Flow Net, 1.139

Fluid Kinematics, 1.100

Fluid Statics, 1.63

Francis Turbine, 5.41

Free Molecular Flow, 1.101

Froude model law., 3.55

Froude model law (gravity force ispredominant), 3.55

Froude Number (Fr), 3.44

GGauge Pressure, 1.65

Gear Pumps (Rotary Pumps), 4.201

Geometric Similarity, 3.48

Index I.1

Governing of Turbine, 5.113

Governing of Pelton Wheel, 5.29

HHead, 1.191

Hydraulic Gradient Line (H.G.L), 2.67

IImpact of Jets, 4.1

Indicator Diagram, 4.143

JJet Pump, 4.198

KKinematic Similarity, 3.48

Kinematic Viscosity (), 1.12

LLaminar Sub - layer, 2.131

Laminar boundary layer, 2.130

Law of Fluid Friction, 2.10

Lobe Pumps, 4.205

MMach Number (M), 3.46

Mach model law, 3.60

Main characteristic curves, 4.116

Major energy (Head) losses, 2.49

Manometry, 1.68

Measurement of Pressure, 1.75

Minor energy losses, 2.50

Model Analysis, 3.54

Momentum Thickness , 2.134

Moody’s Diagram, 2.41

Multi Stage Centrifugal Pumps, 4.111

NNavier-stokes Equations, 1.157

Net Positive Suction Head (NPSH),4.108

Newton’s Law of Viscosity, 1.25

NPSH Required (NPSHR) , 4.109

NPSH Available (NPSHA) , 4.110

OOne Dimensional Flow, 1.108

Operating Characteristics, 4.118

Orifice Meter, 1.162

PPath Line, 1.110

Pelton Turbine (or) Pelton Wheel, 5.5

Performance Curves, 4.116

Piezometer, 1.83

Piston Pumps, 4.208

Pitot-static Tube (or Prandtl Tube),1.200

Pitot-tube, 1.187

Positive Displacement Pump, 4.200

Power Transmission Through Pipes,2.119

Priming, 4.106

Properties of Fluids, 1.6

Pump Selection, 4.196

Pumps, 4.52

RReaction Turbines, 5.38

Reciprocating Pumps, 4.131

Reynold’s Number (Re), 2.2, 3.44

Reynolds Experiment, 2.3

Reynolds model law, 3.54

Roto Dynamic Machines, 4.38

SScale Ratios for Distorted models, 3.80

Selection of Turbines, 5.112

Separation, 4.163

Servomotor (or) Relay cylinder, 5.113

I.2 Fluid Mechanics and Machinery - www.airwalkbooks.com

Shear Stress In Turbulent Flow, 2.47

Similitude, 3.47

Simple Manometers, 1.83

Siphon, 2.110

Slip of Reciprocating Pump, 4.136

Sonic Flow, 1.107

Specific Speed of Turbine, 5.88

Specific weight (or) Weight density, 1.8

Specific Quantities, 3.50

Specific Speed, 3.51

Stoke’s Law, 2.37

Strain Gauge Pressure Transducer, 1.82

Streak Line, 1.110

Stream Tube, 1.109

Stream Line, 1.109

Stream Function, 1.126

Subcritical flow, 1.107

Subsonic Flow, 1.107

Supercritical flow, 1.107

Supersonic Flow, 1.107

Surface Tension, 1.18

Surge Tank, 5.116

TThermodynamic Properties, 1.23

Total Energy Line (T.E.L) (or) EnergyGradient Line (E.G.L), 2.66

Turbulent Bodary Layer on a FlatPlate, 2.153

Turbulent Flow, 2.38

Turbulent Boundary layer, 2.131

Type Number, 4.110

UUnit Discharge or Unit flow, 5.90

Unit Speed, 5.90

Unit Power, 5.91

Unit Quantities, 5.90

VVacuum Pressure, 1.65

Vane Pump, 4.207

Vapour Pressure, 1.15

Velocity Diagram For Kaplan Turbine,5.78

Velocity Profiles, 2.150

Velocity Potential Function, 1.127

Venturi Meter, 1.159

Viscosity (Dynamic Viscosity), 1.10

Von Karman Momentum IntegralEquation For Boundary Layer,2.140

WWater Hammer, 2.126

Weber Number (We), 3.45

Weber model law, 3.59

Index I.3

CONTENTS

Unit I: Fluid Properties and Flow Characteristics

1.3 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.2 Fluid and Continuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.21.3 Units and Dimensions in Fluid Mechanics . . . . . . . . . . . . . 1.31.4 Properties of Fluids. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6

1.4.1 Gas and Liquid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.61.4.2 Density (or) mass Density . . . . . . . . . . . . . . . . . . . . . . 1.71.4.3 Specific weight (or) Weight density . . . . . . . . . . . . . . 1.8

1.4.4 Specific Volume v . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8

1.4.5 Specific gravity (or) Relative density s . . . . . . . . . . 1.9

1.4.6 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.91.4.7 Viscosity (Dynamic Viscosity) . . . . . . . . . . . . . . . . . . 1.10

1.4.8 Compressibility 1K

. . . . . . . . . . . . . . . . . . . . . . . . . 1.14

1.4.9 Vapour Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.151.4.9.1 Cavitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.151.4.9.2 Gas and Gas laws. . . . . . . . . . . . . . . . . . . . . . . . . . . 1.16

1.4.10 Surface Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.181.4.10.1 Surface Tension on Droplet. . . . . . . . . . . . . . . . . . 1.191.4.10.2 Surface Tension on a Hollow Bubble . . . . . . . . . 1.201.4.10.3 Surface Tension on a Liquid Jet . . . . . . . . . . . . . 1.21

1.4.11 Capillarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.211.4.12 Thermodynamic Properties . . . . . . . . . . . . . . . . . . . 1.23

1.5 Newton’s Law of Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . 1.251.5.1 Types of Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.27

1.6 Fluid Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.631.6.1 Concept of Fluid Static Pressure . . . . . . . . . . . . . . . 1.63

1.6.2 Pressure of Fluids P . . . . . . . . . . . . . . . . . . . . . . . . 1.63

1.6.3 Atmospheric Pressure . . . . . . . . . . . . . . . . . . . . . . . . . 1.641.6.4 Absolute zero Pressure (or) Absolute pressure . . . . 1.641.6.5 Gauge Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.65

Contents C.1

1.6.6 Vacuum Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.651.7 Pressure - Density - Height Relationship . . . . . . . . . . . . . 1.661.8 Manometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.681.9 Measurement of Pressure. . . . . . . . . . . . . . . . . . . . . . . . . . . 1.751.10 Simple Manometers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.83

1.10.1 Differential Manometer . . . . . . . . . . . . . . . . . . . . . . 1.861.11 Fluid Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.100

1.11.1 Concept of System:. . . . . . . . . . . . . . . . . . . . . . . . . 1.1001.11.2 Control Volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.100

1.12 Continuum & Free Molecular Flows . . . . . . . . . . . . . . . 1.1011.13 Flow Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1021.14 Types of Fluid Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.102

1.14.1 Steady Flow and Unsteady Flow . . . . . . . . . . . . 1.1021.14.2 Uniform and Non-Uniform Flows . . . . . . . . . . . . 1.1031.14.3 Laminar Flow and Turbulent Flow . . . . . . . . . . 1.1041.14.4 Incompressible and Compressible Flow. . . . . . . . 1.1051.14.5 Rotational Flow and Irrotational Flow . . . . . . . 1.1061.14.6 Subsonic Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1071.14.7 Sonic Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1071.14.8 Supersonic Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1071.14.9 Subcritical flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1071.14.10 Critical flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1071.14.11 Supercritical flow . . . . . . . . . . . . . . . . . . . . . . . . . 1.107

1.15 One Dimensional Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1081.16 Flow Visualization - Lines of Flow . . . . . . . . . . . . . . . . 1.108

1.16.1 Stream Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1091.16.2 Stream Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1091.16.3 Path Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1101.16.4 Streak Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.110

1.17 Mean Velocity of Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1111.18 Principles of Fluid Flow. . . . . . . . . . . . . . . . . . . . . . . . . . 1.111

1.18.1 Principle of Conservation of mass . . . . . . . . . . . . 1.111

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1.19 Types of Motion Or Deformation of Fluid Element . . 1.1221.20 Circulation and Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . 1.1221.21 Stream Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1261.22 Velocity Potential Function . . . . . . . . . . . . . . . . . . . . . . . 1.1271.23 Relation Between Stream Function and Velocity PotentialFunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1291.24 Equipotential Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1371.25 Flow Net . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1391.27 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.139

1.27.1 Euler’s equation along a Stream Line . . . . . . . . 1.1401.27.2 Principle of Conservation of Energy . . . . . . . . . . 1.141

1.28 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1431.29 Navier-stokes Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1571.30 Bernoulli’s Equation: Applications . . . . . . . . . . . . . . . . . 1.159

1.30.1 Venturi Meter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1591.30.2 Orifice Meter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1621.30.3 Principle of Conservation of Momentum. . . . . . . 1.1781.30.4 Moment of Momentum Equation . . . . . . . . . . . . . 1.1801.30.5 Pitot-Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.187

1.31 Head . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1911.32 Concept of Control Volume . . . . . . . . . . . . . . . . . . . . . . . 1.1921.33 Hydraulic Co-efficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.193

(a) Coefficient of Velocity (Cv) . . . . . . . . . . . . . . . . . . . . . 1.193

(b) Coefficient of contraction Cc . . . . . . . . . . . . . . . . . . 1.194

(c) Coefficient of discharge Cd . . . . . . . . . . . . . . . . . . . . 1.194

(d) Coefficient of resistance Cr. . . . . . . . . . . . . . . . . . . . 1.195

1.34 Pitot-static Tube (or Prandtl Tube) . . . . . . . . . . . . . . . . 1.200

Unit II: Flow Through Circular Conduits

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.12.2 Reynolds Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.22.3 Laminar Flow Through Circular Tubes (Circular Conduitsand Circular Annuli) (Hagen Poiseullie’s Equation) . . . . . . . . . 2.5

Contents C.3

2.4 Law of Fluid Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.102.4.1 Head loss due to friction (for laminar flow) . . . . . 2.122.4.2 Hagen Poiseuille Equation (in terms of Discharge) 2.13

2.5 Stoke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.372.6 Turbulent Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.382.7 Hydrodynamical Smooth and Rough Surfaces – PipeRoughness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.392.8 Friction Factor - Resistance to Flow Through Smooth andRough Pipe - Darcy-weisbach Equation . . . . . . . . . . . . . . . . . . 2.402.9 Moody’s Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.412.10 Darcy- Weisbach’s Equation - Expression for Loss of HeadDue to Friction in Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.44

2.10.1 Chezy’s Formula for Loss of head due to friction inpipes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.46

2.11 Shear Stress in Turbulent Flow . . . . . . . . . . . . . . . . . . . . 2.472.12 Velocity Distribution For Turbulent Flow in Pipes . . . . 2.482.13 Energy Losses in Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.49

2.13.1 Major energy (Head) losses. . . . . . . . . . . . . . . . . . . 2.492.13.2 Minor energy losses . . . . . . . . . . . . . . . . . . . . . . . . . 2.50

2.14 Hydraulic Gradient Line (H.G.L) and Energy Gradient Line(EGL) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.66

2.14.1 Total Energy Line (T.E.L) (or) Energy Gradient Line(E.G.L) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.662.14.2 Hydraulic Gradient Line (H.G.L) . . . . . . . . . . . . . 2.67

2.15 Flow Through Long Pipes Under Constant Head H . . . 2.782.16 Flow Through Pipes in Series (or) Flow Through CompoundPipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.792.17 Equivalent Pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.942.18 Flow Through Parallel Pipes . . . . . . . . . . . . . . . . . . . . . . . 2.962.19 Flow Through Branched Pipes . . . . . . . . . . . . . . . . . . . . 2.1072.20 Siphon (Flow Through Pipeline With Negative Pressure) 2.1102.21 Power Transmission Through Pipes . . . . . . . . . . . . . . . . 2.119


2.22 Important Note About Power . . . . . . . . . . . . . . . . . . . . . 2.1252.23 Water Hammer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1262.24 Cavitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1272.25 Boundary Layer Concepts . . . . . . . . . . . . . . . . . . . . . . . . 2.1292.26 Boundary Layer Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 2.129

2.26.1 Laminar boundary layer . . . . . . . . . . . . . . . . . . . . 2.1302.26.2 Turbulent Boundary layer . . . . . . . . . . . . . . . . . . 2.1312.26.3 Laminar Sub - layer . . . . . . . . . . . . . . . . . . . . . . . 2.131

2.26.4 Boundary layer thickness . . . . . . . . . . . . . . . . 2.132

2.26.5 Displacement Thickness . . . . . . . . . . . . . . . . . 2.132

2.26.6 Momentum Thickness . . . . . . . . . . . . . . . . . . . 2.134

2.26.7 Energy thickness . . . . . . . . . . . . . . . . . . . . . . 2.136

2.27 Von Karman Momentum Integral Equation For BoundaryLayer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1402.28 Drag Force FD on Plate of Length L . . . . . . . . . . . . 2.1412.29 Velocity Profiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1502.30 Turbulent Boundary Layer on A Flat Plate . . . . . . . . . 2.1532.31 Drag and Lift Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . 2.1542.32 Total Drag on A Flat Plate. . . . . . . . . . . . . . . . . . . . . . . 2.1572.33 Boundary Layer Separation . . . . . . . . . . . . . . . . . . . . . . . 2.1622.34 High Lights and Important Formulae . . . . . . . . . . . . . . 2.165

Unit III Dimensional Analysis

3.1 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13.2 Need For Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . 3.13.3 Dimensional Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.43.4 Methods of Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . 3.43.5 Dimensionless Numbers (Non Dimensional Numbers) . . 3.433.6 Similitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.47

3.6.1 Types of Similitude . . . . . . . . . . . . . . . . . . . . . . . . . . 3.483.6.1.1 Geometric Similarity . . . . . . . . . . . . . . . . . . . . . . . . 3.483.6.1.2 Kinematic Similarity. . . . . . . . . . . . . . . . . . . . . . . . 3.48

Contents C.5

3.6.1.3 Dynamic Similarity. . . . . . . . . . . . . . . . . . . . . . . . . 3.493.6.2 Specific Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.50

3.7 Model Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.543.7.1 Reynolds model law. . . . . . . . . . . . . . . . . . . . . . . . . . 3.543.7.2 Froude model law . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.553.7.3 Euler’s model law . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.583.7.4 Weber model law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.593.7.5 Mach model law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.603.7.6 Problems in model laws . . . . . . . . . . . . . . . . . . . . . . 3.60

3.8 Model Testing of Partially Submerged Bodies . . . . . . . . . 3.753.9 Classification of Hydraulic Models . . . . . . . . . . . . . . . . . . . 3.80

3.9.1 Undistorted models . . . . . . . . . . . . . . . . . . . . . . . . . . 3.803.9.2 Distorted models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.803.9.3 Scale Ratios for Distorted models . . . . . . . . . . . . . . 3.80

3.10 Applications For Model Testing . . . . . . . . . . . . . . . . . . . . 3.823.11 Limitations of Model Testing. . . . . . . . . . . . . . . . . . . . . . . 3.83

Unit IV: Impact of Jets and Hydraulic Pumps

4.1 Impact of Jets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.14.2 Hydrodynamic Thrust of Jet on A Fixed Surfaces . . . . . . 4.14.3 Impact of Jet on A Hinged Plate . . . . . . . . . . . . . . . . . . . . . 4.84.4 Hydrodynamic Thrust of Jet on A Moving Surface (Flat andCurved Plates) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.134.5 Thrust of Jet of Water on Series of Vanes . . . . . . . . . . . 4.30

4.5.1 Workdone per second (or) Power of jet on a series of aradial curved vanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.324.5.2 Efficiency of the Radial curved vane . . . . . . . . . . . 4.32

4.6 Roto Dynamic Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.384.6.1 Elementary cascade theory . . . . . . . . . . . . . . . . . . . . 4.38

4.7 Theory of Rotodynamic (Turbo) Machines . . . . . . . . . . . . . 4.404.7.1 Roto dynamic machines classifications . . . . . . . . . . 4.41(i) Impulse and Reaction Turbines.. . . . . . . . . . . . . . . . . . 4.41(ii) Axial, Radial and Mixed flow machines . . . . . . . . . . 4.42


(iii) Backward, Radial and Forward Blade Impellers . . 4.434.8 Euler’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.44

4.8.1 Velocity components at the entry and exit of the rotor 4.454.8.2 Velocity triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.474.8.3 Degree of reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.50

4.9 Pumps:. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.524.10 Centrifugal Pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.524.11 H-Q Characteristics of A Centrifugal Pump . . . . . . . . . 4.1014.12 Typical Flow System Characteristics . . . . . . . . . . . . . . . 4.103

4.12.1 System characteristics Curve . . . . . . . . . . . . . . . . 4.1034.12.2 Pump characteristics curve . . . . . . . . . . . . . . . . . . 4.1044.12.3 Operating point . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.105

4.13 Priming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1064.14 Cavitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1074.15 Net Positive Suction Head (NPSH) . . . . . . . . . . . . . . . . 4.108

4.15.1 NPSH Required NPSHR . . . . . . . . . . . . . . . . . . . 4.109

4.15.2 NPSH Available NPSHA . . . . . . . . . . . . . . . . . . 4.110

4.16 Type Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1104.17 Multi Stage Centrifugal Pumps . . . . . . . . . . . . . . . . . . . 4.1114.18 Performance Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.116

4.18.1 Main characteristic curves . . . . . . . . . . . . . . . . . . 4.116(i) Q v/s H Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.116

(ii) Q v/s Curve. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.117

(iii) Q v/s P Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1184.18.2 Operating Characteristics . . . . . . . . . . . . . . . . . . . 4.118

4.19 Model Testing of Centrifugal Pumps . . . . . . . . . . . . . . . 4.1214.20 Specific Speed of Centrifugal Pump . . . . . . . . . . . . . . . . 4.122

4.20.1 Shape numbers Nq . . . . . . . . . . . . . . . . . . . . . . . 4.130

4.21 Reciprocating Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1314.22 Working Principle of A Reciprocating Pump . . . . . . . . 4.1334.23 Discharge, Workdone and Power Required to Drive ASingle Acting Reciprocating Pump. . . . . . . . . . . . . . . . . . . . . . 4.133

Contents C.7

4.24 Discharge, Work Done and Power Required to Drive ADouble Acting Pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1344.25 Slip of Reciprocating Pump . . . . . . . . . . . . . . . . . . . . . . . 4.136

4.25.1 Negative Slip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1374.26 Indicator Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.143

4.26.1 Effect of acceleration of piston in suction anddelivery pipes on indicator diagram . . . . . . . . . . . . . . . . 4.1454.26.2 Effect of acceleration in the suction pipe and deliverypipe. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1484.26.3 Effect of friction in the suction and delivery pipes onthe Indicator Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1514.26.4 Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.163

4.27 Maximum Speed of A Reciprocating Pump. . . . . . . . . . 4.1724.28 Air Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1784.29 Pump Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1964.30 Various Types of Pumps . . . . . . . . . . . . . . . . . . . . . . . . . 4.1974.31 Jet Pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1984.32 Positive Displacement Pump . . . . . . . . . . . . . . . . . . . . . . 4.2004.33 Gear Pumps (Rotary Pumps). . . . . . . . . . . . . . . . . . . . . . 4.201

4.33.1 Working Principle of External Gear Pump . . . . 4.2014.33.2 Working Principle of Internal Gear Pump. . . . . 4.2044.33.3 Lobe Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2054.33.4 Screw Pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.206

4.34 Vane Pump. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2074.35 Piston Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.208

Unit V: Hydraulic Turbines

5.1 Hydraulic Turbines - Introduction . . . . . . . . . . . . . . . . . . . . 5.15.2 Classification of Hydraulic Turbines . . . . . . . . . . . . . . . . . . 5.25.3 Heads and Efficiency of A Turbine . . . . . . . . . . . . . . . . . . . 5.35.4 Pelton Turbine (or) Pelton Wheel . . . . . . . . . . . . . . . . . . . . . 5.55.5 Governing of Pelton Wheel . . . . . . . . . . . . . . . . . . . . . . . . . 5.295.6 Reaction Turbines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.38


5.7 Francis Turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.415.7.1 Working of a Francis Turbine . . . . . . . . . . . . . . . . . 5.415.7.2 Velocity Triangles and Work done by water in FrancisTurbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.41

5.7.3 Hydraulic Efficiency h for Francis Turbine. . . . 5.43

5.7.4 Points to be remembered in Francis Turbine . . . . 5.435.7.5 Solved Problems on Francis Turbine . . . . . . . . . . . 5.44

5.8 Axial Flow Reaction Turbines . . . . . . . . . . . . . . . . . . . . . . . 5.765.8.1 Working Principle of a Kaplan Turbine. . . . . . . . . 5.775.8.2 Velocity Diagram For Kaplan Turbine . . . . . . . . . . 5.78

5.9 Specific Speed of Turbine. . . . . . . . . . . . . . . . . . . . . . . . . . . 5.885.9.1 Unit Quantities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.90

5.10 Draft Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1025.11 Cavitation in Reaction Turbines . . . . . . . . . . . . . . . . . . . 5.1075.12 Performance Curves of Turbines. . . . . . . . . . . . . . . . . . . 5.1085.13 Selection of Turbines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1125.14 Governing of Turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1135.15 Surge Tank. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.116

Contents C.9

Unit I

Fluid Properties and Flow

Characteristics

IMPORTANT FORMULAE AND POINTS

A fluid is a substance that deforms continuously when subjected to

even an infinitesimal shear stress.

PHYSICAL PROPERTIES OF FLUID

1.1. Mass density Mass

Volume

m

V in kg/m3

1.2. Specific weight w WeightVolume

W

V in N/m3

Also w g

1.3. Specific volume v

VolumeMass

V

m in m3/kg

1.4. Specific gravity s Density of fluidDensity of water

w

No unit

Also s Specific wt of fluidspecific wt of water

w

ww

w Density of water 1000 kg/m3

ww Specific weight of water 9810 N/m3

1.5. Bulk modulus K dp

dV

V

;

Compressibility 1K

1.6. Surface tension on a droplet:

P Pressure intensity inside the droplet (in excess of the outside pressure)

P 4d

Where, Surface tension in N/m;

d Diameter of droplet in ‘m’

1.7. Surface tension on a hollow bubble;

P 8d

1.8. Surface tension on a liquid jet;

P 2d

1.9. Capillary rise (or) fall, h 4 cos gd

1.10. Viscosity

dudy

(It is known as Newton’s law of viscosity)

Where, Shear stress in N/m2

Dynamic viscosity in Ns/m2

du Change in velocity

dy Distance between two layers or plates

dudy

Velocity gradient (or) rate of shear strain

Note: 1 poise 0.1 Ns/m2; 1 centipoise 0.001 Ns/m2

Shear force shearing area

Kinematic viscosity:

in m2/s

1 stoke 1 cm2/s 10 4 m2/s; 1 centistoke 10 6 m2/s

1.2 Fluid Mechanics and Machinery - www.airwalkbooks.com

1.11. Pascal’s Law

It states that the pressure P at a

point in a incompressible fluid in

equilibrium is same in all directions.

Unit for pressure is N/m2

1 N/m2 1 Pascal

1 bar 1 105 N m2 1 105 Pascal

1.12. Characteristic equation (or)Equation of state:

PV

m RT

R Characteristic gas constant;

R 0.287 for air

Also, for any gas,

PV

m MRT

(or) PV

m R

T

where R

Universal gas constant 8.314 kJ/kg mol K

V

m Molar volume ; M Molecular weight

Also R for gas A R

M for gas A

1.13. Pressure:

At sea level under normal conditions,

Atmospheric pressure 760 mm of mercury

10.3 m of water

1.01325 bar

1.01325 105 N/m2 or Pa

101.325 kPa

Normally pressure is measured with respect to two common datums.

P

P

PP

P

P

PP

Fig. 1.1 Hydro Static Pressure Acting Equality from all directions

at a Point in a Flu id at Rest

Fluid Properties and Flow Characteristics 1.3

1. Absolute zero pressure (complete vacuum)

2. Atmospheric pressure.

Absolutepressure

Atmosphericpressure

Gauge pressure

Absolutepressure

Atmosphericpressure

Vacuum pressure

1.14. Hydrostatic Law

P

Z g w

Where, w Weight density of fluid

P1 P2 gh

Where, h Pressure head

The pressure at free surface is the local atmospheric pressure.

P gh

P Gauge Pressure

Density of the fluid

g Acceleration due to gravity

G augepressure

a t A

Locala tm ospheric pressu re

(o r gauge zero)

PR

ES

SU

RE

Vacuum p ressure o r negative gauge p ressure a t B

B

Ab

solu

te p

ress

ure

at

A

Loc

al

bar

om

etri

c p

ress

ure

A

Abso lu te zero (or) C om plete vacuum

F ig. 1.2 Relationship between absolute, gauge and vacuum pressures


1.15. Types of Fluid Flow

(i) Steady and Unsteady flowIn steady flow, the fluid characteristics like velocity, pressure, density,

temperature etc at any point do not change with time. But these characteristics

may be different at different points.

For steady flow at any point, we can write as follows

V

t 0;

P

t 0;

t

0

In unsteady flow, the flow characteristics changes with respect to time.

Mathematically, for unsteady flow,

V

t 0;

P

t 0;

t

0

(ii) Uniform and Non-Uniform FlowWhen velocity remains constant at all points in the moving fluid, then

flow is called Uniform flow.

Mathematically, the uniform flow can be written as

V

s 0

V Change in velocity and (s distance travelled by fluid).

When velocity of fluid changes from point to point at any time, then

the flow is known as Non-uniform flow

Mathematically V

s 0

(iii) Laminar Flow and Turbulent FlowLaminar Flow

When the fluid particles

flow in well-defined ordered

layers in such a way that one

layer slides over another layer,

then it is called Laminar flow.

This flow is also called, stream

line flow and viscous flow. Fig. 1.3. Laminar flow


Turbulent Flow

When the fluid particles move in random order (not layer by layer), the

flow is called turbulent flow.

Note: Laminar flow or turbulent flow can be determined by a

non-dimensional number called Reynold Number Re

Re VD

Where, V Mean velocity of flow in pipe

D Diameter of pipe

Kinematic viscosity of fluid.

When, Re 2000, it is laminar flow

Re 4000, it a turbulent flow

When 4000 Re 2000 i.e when Reynold number lies between 2000 and

4000, then it may be laminar flow (or) turbulent flow. (Transient flow)

Incompressible and Compressible Flow

In incompressible flow, the density of fluid remains constant

Constant

(v) Rotational Flow and Irrotational Flow

When fluid particles move along stream lines, they rotate about their

own axis. This type of flow is rotational flow. In case of irrotational flow,

when fluid particles move along stream lines, they do not rotate about their

own axis.

Fig. 1.4 Turbulent flow


(vi) One Dimensional Flow

For one dimensional flow,

u f x, v 0 and w 0 ; where u, v and w are velocity component

in x, y and z direction respectively.

Two-dimensional Flow

For two-dimensional flow,

u f1 x, y

v f2 x, y

w 0

Three-dimensional Flow

For three dimensional flow,

u f1 x, y z

v f2 x, y, z

w f3 x, y, z

1.16. Velocity and acceleration in a fluid flow

The resultant velocity at any point in a fluid flow for the velocity

components u, v and w components in x, y, z direction is mathematically given

as if u f1 x, y, z, t, v f2 x, y, z, t and w t3 x, y, z, t

Then resultant velocity V ui vj wk

V u2 v2 w2

Let ax, ay and az be total acceleration in x, y and z directions

respectively.

Then,

ax u

t u

u

x v

u

y w

u

z


ay v

t u

v

x v

v

y w

v

z

az w

t u

w

x v

w

y w

wz

Resultant acceleration, A ax2 ay

2 az2

1.17. Principle of conservation of mass

m

1 m

2

1 A1 V1 2 A2 V2

A1 V1 A2 V2

[ Since density of fluid at inlet and outlet are same]

Where,

A1 and A2 Inlet and outlet cross section area in m2

V1 and V2 Inlet and outlet velocity in m/s

m Mass of flowing fluid in kg/s

Q A1 V1

Q Rate of flow (or) Discharge in m3/s

Note: Coefficient of discharge Cd Qa

Qt

Where, Qa Actual discharge

Qt Theoretical discharge

General equation of continuity in 3D for any kind of flow

x

u y

v z

w t

0


Rotational velocity components

z 12

v

x uy

x 12

w

y

v

z

y 12

u

z

w

x

Note: If z (or) y or z 0, then flow is irrotational

1.18. Euler’s Equation

dP

gdZ VdV 0

1.19. Bernoulli’s equation

Pw

Z V2

2g constant

(Pr.Head) (Kinetic head) (datum Head) = Constant

Some applications of Bernoulli’s equation are

Venturi meter

Orifice meter

Flow nozzle

Pitot tube

Bernoulli’s Equation for ideal and real fluid

P1

g

V12

2g Z1

P2

g

V22

2g Z2 (Ideal Fluid)

P1

g

V12

2g Z1

P2

g

V22

2g Z2 hL (Real Fluid)


1.20. Discharge through Venturimeter

Q Cd A1 A2

A12 A2

2 2gh

Cd: Coefficient of discharge

A1: Inlet Area

A2: Throat Area

h: Pressure head difference

1.21. For differential U-tube Manometer, the value of h is

h x Sh

S0 1

[Differental Manometer Containing Heavy Liquid]

h x 1

Sl

S0

[Differental manometer Containing Lighter Liquid]

h P1

g Z1

P2

g Z2

x

Sh

S0 1

[Ventimeter with differental Manometer heavy Liquid]

h P1

g Z1

P2

g Z1

x

1

SL

S0

[Ventimeter with differental Manometer lighter Liquid

x: Difference in the readings of differential manometer.

Sh: Sp. gravity of heavier liquid.

S0: Sp. gravity of fluid flowing.

Sl: Sp. gravity of lighter liquid.


1.22. For pitot tube

Velocity V CV 2gh

CV Coefficient of pitot tube.

h x Sg

S0 1

rise of liquid in the tube.

1.23. Force Exerted in bend tube

Fx Q V1 V2 cos P1 A1 P2 A2 cos

Fy Q V2 sin P2 A2 sin

Resultant force F Fx2 Fy

2 and tan FyFx

1.24. Moment of Momentum equation

Torque T Q [V2 r2 V1 r1]

F ig. 1.43 Forces on bend

V 1

FY

P A1 1

PA221

FX

2

V sin2

V s in2

V 2

(a )

P A sin2 2

P A cos2 2

P A2

2

O

(b )


UNIVERSITY SOLVED QUESTIONS 1.1. Calculate the specific weight and specific gravity of 1 litre of a liquid

with a density of 713.5 kg/m3 and which weighs 7 N.(Nov/Dec 2015 - AU)

Given: Volume 1 litre 10 3 m3

1. Specific weight w WeightVolume

7

10 3 7000 N/m3

2. wg

70009.81

kg/m3 713.5 kg/m3

3. Specific gravity Density of liquidDensity of water

713.51000

0.7135

1.2. Calculate the dynamic viscosity of oil which is used for lubricationbetween a square plate of size 0.8 m 0.8 m and an inclined plane

with angle of inclination 30. The weight of the square plate is 330N and it slide down the inclined plane with a uniform velocity of 0.3m/s. The thickness of the oil film is 1.5 mm. (Nov/Dec 2015 - AU)

Given: Area of plate, A 0.8 0.8 0.64 m2 ; Angle of plane, 30

Weight of plate, W 300 N ; Velocity of plate, u 0.3 m/s ; Thickness, t dy 1.5 mm

Solution: Consider viscosity as

Component of weight W, along the plane W cos 60 300 cos 60

150 N

Shear force on bottom surface of plate = 150 N

Shear stress FA

1500.64

N/m2

From equation,

dudy ...(1)

du Change in velocity

so,

du u 0 u 0.3 m/s

dy t 1.5 10 3 m


Applying values in equation (1)

1500.64

0.3

1.5 10 3

1.17 N s/m2

(or)

1.17 10 11.7 poise

1.3. The space between two square flat parallel plates is filled with oil.Each side of the plate is 600 mm. The thickness of the oil films is12.5 mm. The upper plate, which moves at 2.5 m/s requires a forceof 98.1 N to maintain the speed. Determine(i) The dynamic viscosity of the oil in poise and(ii) The kinematic viscosity of the oil in stokes if the specific gravityof the oil is 0.95. (Nov/Dec 2012 - AU)

Given

Side of plate 600 mm 0.6 m ; Oil film thickness, dy 12.5 mm 12.5 10 3 m

Velocity, u 2.5 m / s ; du u 0 2.5 m / s ; Force, F 9.81 N ; S 0.95

Solution

Area of square plate 0.6 0.6 0.36 m2

Shear force F A

FA

98.10.36

272.5 N/m2

dudy

Dynamic viscosity,

dudy

dydu

272.5 12.5 10 3

2.5

1.3625 N s/m2

0.13625 Poise (1 Poise 0.1 Ns/m2)


Specific gravity, S oil

water

Density of oil, oil S water 0.95 1000 950 kg/m3

Kinematic viscosity,

1.3625

950 1.4342 10 3 m2/s

1.4342 10 3

10 4 14.34 stokes

14.34 stokes [1 Stoke 1 cm2/s]

1.4. Lateral stability of a long shaft 150 mm in diameter is obtained bymeans of a 250 mm stationary bearing having an internal diameterof 150.25 mm. If the space between bearing and shaft is filled with

a lubricant having a viscosity 0.245 N s/m2, what power will berequired to overcome the visious resistance when the shaft is rotatedat a constant rate of 180 rpm? (Nov/Dec 2010 - AU)

Given: Dia. of shaft, D 150 mm 0.15 m ; Bearing length, L 250 mm 0.25 m

Bearing inner dia D1 150.25 mm 0.15025 m

Viscosity of lubr icant, 0.245 Ns/m2; Speed of shaft, N 180 rpm

Solution

Thickness of oil film, t D1 D

2

0.15025 0.152

t 1.25 10 4 m

Tangential speed of shaft,

V DN

60 0.15

18060

1.4137 m/s

Shear stress, dudy

Vt

0.245 1.4137

1.25 10 4 2770.85 N/m2

Shear force, F Area DL

2770.85 0.15 0.25 326.43 N


Resistance torque, T F D2

326.43 0.15

2 24.482 N.m

Power required to overcome the viscous resistance

2 NT

60

P 2 180 24.482

60

461.5 W or 0.462 kW

1.5. The dynamic viscosity of an oil used for lubrication between a shaft andsleeve is 6 poise. The shaft is of diameter 0.4 m and rotates at 190rpm. Calculate the power lost in the bearing for a sleeve length of 90mm. The thickness of oil film is 1.5 mm. (Nov/Dec 2016 - AU)

Solution: Given: Viscosity 6 poise 610

Ns

m2 0.6 Ns

m2

Dia. of shaft, D 0.4 m; Speed of shaft, N 190 r.p.m

Sleeve length, L 90 mm 0.09 m ; Thickness of oil film, t 1.5 mm 1.5 10 3 m

Tangential velocity of shaft, u D N

60

0.4 19060

3.98 m/s

Shear stress, dudy

where du Change of velocity u 0 u 3.98 m/s

dy Change of distance t 1.5 10 3 m

0 .4

1 .5 m m

9 0 m m

Sle eve

Sh a ft


0.6 3.98

1.5 10 3 1592 N/m2

Shear force on the shaft,

F Shear stress Area A

1592 D L 1592 0.4 0.09 180.05 N

Torque on the shaft, T Force D2

180.05 0.42

36.01 Nm

Power lost 2 NT

60

2 190 36.0160

716.48 W

1.6. A 0.5 m shaft rotates in a sleeve under lubrication with viscosity 5Poise at 200 rpm. Calculate the power lost for a length of 100 mmif the thickness of the oil is 1 mm. (Nov/Dec 2009 - AU) (FAQ)

Solution: Viscosity, 5 poise 0.5 Ns

m2 ; Dia. of shaft, D 0.5 m

Speed, N 200 rpm; Length, L 100 mm 0.1 m

Thickness of oil film, t 1 mm 0.001 m

Tangential velocity of shaft,

u DN60

0.5 200

60 5.24 m/s

w.k.t, dudy

du change in velocity u 0 u 5.24 m/s

dy change in distance t 0.001 m

0.5 5.24

0.001 2620 N/m2

Shear force on the shaft, F Area

2620 DL 2620 0.5 0.1 411.55 N

Torque on shaft F D/2 411.55 0.52

102.89 N.m

Power lost 2 NT

60

2 200 102.8960

2154.92 Watts


1.7. Determine the viscous drag torque and power absorbed on onesurface of a collar bearing of 0.2 m ID and 0.3 m OD} with an oilfilm thickness of 1 mm and a viscosity of 30 centi poise if it rotatesat 500 rpm. (Nov/Dec 2014 - AU)

Given data: Inner Diameter ID d2 0.2 m ; Outer Diameter OD d1 0.3 m

Oil from thickness t 1 mm ; Viscosity 30 centipoise ; Speed N 500 rpm

R1 d1

2 0.15 m ; R2

d2

2 0.1 m

t 1 10 3 m

30 centripoise 30 10 2 poise 30 10 2

10 Ns/m2 F 0.03 Ns/m2

Drag Torque T

60t 2 N

R1

4 R24

0.03 2 500 0.154 0.14

60 1 10 3 1.00238 Nm

Power P 2 NT

60

2 500 1.0023860

52.485 W

1.8. A 15 cm diameter vertical cylinder rotates concentrically insideanother cylinder of dia 15.1 cm. Both cylinders are 25 cm high. Thespace between the cylinder is filled with a liquid. If a torque of 11.77N-m is required to rotate the inner cylinder of 100 rpm, determinethe viscosity of the liquid

(April 2004 - AU)(Nov/Dec 2013 - AU)(FAQ)

Given: d 15 cm 0.15 m

D 15.1 cm 0.151 m

l 25 cm 0.25 m

N 100 rpm ; T 11.77 Nm

Tangental Velocity

u dN60

0.15 100

60

0.7854 m/s


du u 0 0.7854 m/s

dy D d

2

0.151 0.152

5 10 4 m

Solution:

Torque T F d2

11.77 F 0.15

2

F 156.93 N

Shear stress F

Shearing area

156.93 d l

156.93

0.15 0.25 1332.1 N/m2

But Shear stress dudy

1332.1 0.7854

5 10 4

0.848 Ns/m2

Viscosity of liquid 0.848 Ns/m2

1.9. A liquid is compressed in a cylinder having a volume of 0.012 m3 at

a pressure of 690 N/cm2. What should be the new pressure in order

to make its volume 0.0119 m3? Assume bulk modulus of elasticity (K)

for the liquid 6.9 104 N/cm2. (Nov/Dec 2013 - AU)

Given: P1 690 N/cm2 690 104 N/m2 6.9 106 N/m2 ; Bulk modulus

Kliquid 6.9 108 N/m2

% reduction in volume dVV

0.012 0.0119

0.012 8.3 10 3

0 .05cm

15cm

15.1cm

25cm


Solution:

We know K dP

dVV

6.9 108 dP

8.3 10 3

dP 5.73 106 N/m2 5.75 MN/m2

New pressure P2 6.9 106 5.73 106 12.63 106 N/m2

1.10. A hollow cylinder of 150 mm OD with its weight equal to the buoyantforces is to be kept floating vertically in a liquid with a surface

tension of 0.45 N/m2. The contact angle is 60. Determine theadditional force required due to surface tension.

(Nov/Dec 2014 - AU)

Given: Outer Diameter d1 150 mm 0.15 m ; Surface Tension 0.45 N/m2

Contact angle 60 Additional force ?

In this case a capillary rise occurs and this requires an additional force

to keep the cylinder floating.

Capillary rise, h 4 cos g d1

[ g w = specific weight]

Pi Po h g

h Pi Po

g

Pi Po

g

4 0.45 cos 60g0.15

6g

m

Pi Po 6 N/m2

Force A Pi Po 4

d12 Pi Po

4

0.152 6 0.106 N

As the immersion leads to additional buoyant force, the force required

to keep the cylinder floating will be double this value.

So additional force 2 0.106 0.212 N


1.11. Calculate the capillary rise in a glass tube of 2.5 mm diameter whenimmersed vertically in (a) water and (b) mercury. Take surfacetension 0.0725 N/m for water and 0.52 N/m for mercury incontact with air. The specific gravity for mercury is given as 13.6and angle of contact 130. (Nov/Dec 2016 - AU)

Given: Diameter 2.5 mm ; water 0.0725 N/m ; mercury 0.52 N/m

130; Specific gravity for mercury 13.6

Solution:

Capillary rise h 4 cos

wd [ 0 for water and glass]

For water w 9810 N/m3;

4 0.0725 cos 0

9810 2.5 10 3

h 0.01182 m 11.82 mm

Capillary rise h 4 cos

wd

4 0.52 cos 13013.6 9810 2.5 10 3

h 4.008 10 3 m

1.12. Derive an expression for the capillary rise at a liquid in a capillarytube of radius r having surface tension and contact angle . If theplates are of glass, what will be the capillary rise of water having 0.073 N/m, 0? Take r 1 mm. (Nov/Dec 2011 - AU)

Solution: Derivation in S.R BookPage 1.22 Section 1.4.11.1

Capillary rise, h 2 cos g r

4 cos gd

Given, r 1 mm, 0 and 0.073 N/m

Capillary rise,

h 2 0.073 cos 0

1000 9.81 1 10 3 0.015 m

15 mm


1.13. A U-tube is made of two capillaries of diameter 1.0 mm and 1.5 mmrespectively. The tube is kept vertically and partially filled with waterof surface tension 0.0736 N/m and zero contact angle. Calculate thedifference in the levels of the water caused by the capillary.

(Nov/Dec 2010 - Au)

Given: Dia of left capillary tube, d1 1.0 mm 1 10 3 m

Dia of right capillary tube, d2 1.5 mm 1.5 10 3 m

Surface tension, 0.0736 N/m ; Zero contact angle, i.e 0

Solution: h1 height of liquid rise in the left smaller capillary tube

h2 height of liquid rise in the right bigger capillary tube

water 1000 kg/m3

Capillary rise in left tube, h1 4g d1

Capillary rise in right tube, h2 4g d2

h1 h2 (. . . tube (1) is having less diameter)

Difference in levels,

h1 h2 4 g

1d1

1d2

4 0.0736

1000 9.81

1

1 10 3

1

1.5 10 3

0.0100034 m 10.003 mm

1.14. The water level in a tank is 20 m above the ground. A hose isconnected to the bottom of the tank and the nozzle at the end of thehose is pointed straight up. The tank is at sea level and the watersurface is open to the atmosphere. In the line leading from the tankto the nozzle is a pump, which increases the pressure of water. Ifthe water jet rises to a height of 27 m from the ground, determinethe minimum pressure rise supplied by the pump to the water line.

(Nov/Dec 2014 - AU)


P1 - Pressure at the top of the tank [Atmosphere pressure]

P2 - Pressure at bottom of the tank; Pn - Pressure at the nozzle inlet

Pp - Pump pressure

Pn P2 Pp [P1 1 bar]

P2 g h1 1000 9.81 20 1.96 105 N/m2 1.96 bar

Pn g h2 1000 9.81 27 2.648 105 N/m2

Pn P2 Pp

2.648 105 1.96 105 Pp

Pp 68870 N/m2 68.8 kN/m2

1.15. A U-tube manometer connects two pipes A and B. The pipe A contains

oil of specific gravity 1.6 and pressure 120 kN/m2 . The pipe B

contains oil of specific gravity 0.8 and pressure 220 kN/m2 . Thecentre of pipe A is 3 meters above centre of pipe B. The centre ofthe pipe B is at the level of mercury in the left limb connecting thepipe A. Find the difference of mercury levels.

Given: Pressure in Pipe A, PA 120 kN/m2 and in Pipe B, PB 220 kN/m2

Specific gravity of oil in pipe A s1 1.6 ; Specific gravity of oil in

pipe B s3 0.8

P 1

P2Pn

Nozzle

27m = h2

( P )p

Pum p

20m

= h

1


Solution:

Pressure of liquid in B is greater than that of A.

So hB hA PB

w

PA

w

where w specific weight of water 9810 N/m2

hB and hA Pressure heads of

liquids in pipe B and pipe A

respectively.

hB hA 220 103

9810

120 103

9810 10.194 m of water

Pressure head at

C Pressure head in pipe A

h1 s1 x sm

hA h1 s1 x sm

Pressure head at D Pressure head in pipe B h3 s3

hB x s3 [ . . . h3 x ]

Under equilibrium,

Pressure head at C Pressure head at D

hA h1 s1 xsm hB xs3

hB hA h1 s1 x sm x s3

10.194 3 1.6 x 13.6 0.8

x 10.194 3 1.6

13.6 0.8 0.4214 m

Difference in mercury level 0.4214 m

sm

h =3m1

h =h3 2

P = 120kN/mA2

P = 220kN/mB2

Pipe ‘A’

Pipe ‘B’

zz C D

x


1.16. A ‘U’ tube manometer is used to measure water in a pipeline which isin excess of atmosphere pressure. The right limb of the manometercontains mercury and is open to atmosphere. The contact between waterand mercury is in the left limb. Calculate the pressure of water in themainline if the difference in level of mercury in the limbs is 10.5 cmand the free surface of mercury is in level with centre of pipe. If the

pressure of water in the pipeline is reduced by (i) 10000 N/m2 and (ii)

12000 N/m2 find the new difference of level of mercury.(Nov/Dec 2017 AU)

Solution:

Difference at mercury = 10.5 cm = 0.105 m

PB PA Pressure due to 10.5 cm of water

PA 1 gh

PA 1000 9.81 0.105

PB PA 1030.05 N/m2

PC PD

Pressure due to 10.5cm mercury

0 2 gh2

PC 0 13.6 1000 9.81 0.105

PC 14008.7 N/m2

But PB PC

PA 1030.05 14008.7

PA 12978.65 N/m2

(i) If PA 10000 N/m2

And it is less than 12978.65 N/m2. Hence mercury in left limb will

rise and in right limb mercury will fall to equally.

A

W ater

B C

D

M ercury

10.5 cmLeft L im b

R ightLim b


Let us consider, rise of

mercury in left limb is ‘x’ cm.

Then fall of mercury in the

right limb is also ‘x’ cm.

We know, that PB PC

PA

Pressure due to10.5 x cm of water

PD

Pressure due to10 2x cm of mercury

PA 1 gh1 PD 2 gh2

10000 1000 9.81

10.5 x

100 0

13.6 1000 9.81

10 2x

100

10000 1030.05 98.1x 13341.6 2668.32x

x 0.90 cm

New difference of level of mercury 10 2x

8.20 cm

(ii) If PA 12000 N/m2,

We know that, PB PC

PA

Pressure due to10.5 x cm of water

PD

Pressure due to10 2x cm of mercury

12000 1000 9.81

10.5 x

100 0

13.6 1000 9.81

10 2x

100

A

Water 10.5-�

�

B

B* C*

C

D*

D

(10.5-2 ) �

M ercury

�


12000 1030.5 98.1x

13341.6 2668.32x

2668.32x 98.1x

13341.6 1030.5 12000

2570.22x 311.11

x 0.121 cm

New difference of level

of mercury

10 2x 10 2 0.121

9.75 cm

1.17. A pipe containing water at 180 kN/m2 pressure is a connected by adifferential gauge to another pipe 1.6 m lower than the first pipe andcontaining water at high pressure. If the difference in heights of 2mercury columns of the gauge is equal to 90 mm, What is thepressure in the lower pipe? (Nov/Dec 2008 - AU)(FAQ)

Given: Data given are shown in the figure, PA 180 kN/m2; h1 1.6 m

x 90 mm 0.09 m h2 h3;S1 S2 Sp. gravity of water 1

Sm Sp. gravity of mercury 13.6

Solution:

Let hA Pressure head in Pipe A, m of water

hB Pressure head in Pipe B, m of water

hA PA

w

180 103

9.81 103 18.35 m

Taking the lower level of mercury

(which is in the right limb of the

differential manometer) as datum, the

manometric equation is

hA h1 s1 h2 Sm hB h3 s2

A

W ater 10 .5 -�

�

B

B* C *

C

D *

D

(10.5-2 ) �

Mercury

�

x

hA+

hB+


hB hA h1 s1 h2 sm h3 s2 hA h1 s1 x sm x s2

hB hA h1 s1 x sm s2 . . . h2 h3 x

hB 18.35 1.6 1 0.09 13.6 1 21.084 m of water

PB hB 9.81 kN/m2

21.084 9.81 206.834 kN/m2 Ans

1.18. Derive the Euler’s equation of motion and deduce the expression toBernoulli’s equation.(Nov/Dec 2012 - AU) (Nov/Dec 2011 - AU) (Nov/Dec 2010 - AU)

Refer the Fig 1.51 Consider a small element of ideal fluid of length

ds along a stream line.

dm The mass of element dAds.

cos dZds

; dm cos dA ds dZds

dAdZ

P dA

dz

ds

(P+dP )dAs

Elem ent on stream line (ideal flu id)Fig. 1.51

Stream line

W eigh t co m ponent in the d ire ct ion o f m o tion dm g

= dm g cos


According to Newton’s second law, F ma

The forces tend to accelerate the fluid mass are the pressure forces on

the two ends of the element and weight component in the direction of motion.

Total forces PdA P dP dA dm g cos

dP dA dm g cos

dPdA gdAdZ

a Acceleration V dVds

Substitute all values in Newton’s second law.

F ma

dPdA g dA dZ dA ds V

dVds

Divide this equation by dA

dP

g dZ VdV

dP

g dZ VdV 0. This is Euler’s equation for one dimensional flow.

This Euler’s equation can be applied for both incompressible and

compressible flow.

PRINCIPLE OF CONSERVATION ENERGY

1.19. Derive the Bernoulli’s equation with the basic assumptions.(Nov/Dec 2016 - AU) (Nov/Dec 2015 - AU) (Nov/Dec 2013 - AU)

Energy can be neither created nor destroyed, but it can be converted

from one form to another form. This is the principle of conservation of

energy. This is the basis for Bernoullis theorem. According to this, total

energy at any point remains constant.

There are different forms of energy

1. Potential energy

2. Pressure energy (or) Flow energy

3. Kinetic energy (or) Velocity energy


Potential energy: The energy possessed by a fluid by virtue of its position

above or below the datum line is called potential energy.

Potential energy Wh mgh mgZ

where h Z height of fluid particle above datum

m mass of fluid in kg

g acceleration due to gravity; W weight of fluid in N

Epotential Potential energy per kg of fluid gZ in J/kg

Pressure Energy: The energy due to fluid’s pressure is called pressure energy

Pressure energy per kg of fluid Pv

where P pressure of fluid in N m2; v specific volume of fluid in m3

kg

But v 1

So pressure energy per kg of fluid P

Epressure P

in J kg

Kinetic Energy: The energy of the fluid by virtue of its velocity is called

kinetic energy.

Kinetic energy of fluid 12

mV2

K.E. for 1 kg of fluid V2

2 in J/kg

According to principle of conservation of energy, total energy of fluid

remains constant.

Total energy Epotential Ekinetic Epressure

gZ V2

2

P

in J kg

Total head Z V2

2g

P

w in m

[. . . w g]


Since total energy or head of fluid remains constant.

Z1 V1

2

2g

P1

w Z2

V22

2g

P2w

This is known as Bernoullis equation.

BERNOULLI’S EQUATION

In case of incompressible fluid, is constant. So we can integrate the

Euler’s equation and get Bernoulli’s equation.

Euler’s equation dP

g dZ VdV 0

Integrating, we get

dP

gdZ VdV constant

P

gZ V2

2 constant total head H

Pw

Z V2

2g constant

The above equation is called Bernoulli’s equation for a steady flow

of a frictionless incompressible fluid along a stream line.

Important points in Bernoulli’s Equation

1. In this equation, we assume that velocity is constant. But in actual

practice it is not so.

2. We assume that fluid is non-viscous. i.e frictional losses is neglected.

But no fluid is ideal in actual practice.

3. During turbulent flow, some energy will be lost (or) transformed from

kinetic energy to heat energy. This heat energy will be dissipated or

lost. This loss is neglected in Bernoulli’s theorem.

1.20. State the assumptions used in the derivation of the Bernoulli’sequation. (Nov/Dec 2014 - AU)


1. The fluid is ideal (or) non viscous. [there is no viscosity]

2. The fluid flow is steady flow.

3. The flow is incompressible

4. The velocity is uniform over the cross section of the passage.

5. The flow is irrotational

1.21. The water is flowing through a taper pipe of length 100 m havingdiameters 600 mm at the upper end and 300 mm at the lower end,at the rate of 50 litres/s. The pipe has a slope of 1 in 30. Find thepressure at the lower and if the pressure at the higher level is

19.62 N/cm2. (Nov/Dec 2013 - AU)

Given: l 100 m, D1 0.6 m; D2 0.3 m; Slope 1 in 30

P1 19.62 N cm2 19.62 104 N/m2; Q 50 lit/s 0.05 m3

s

Solution:

Height above datum of section, Z1 130

100 3.33 m

By Contrinuity Equation

. . . Q A1V1 A2V2

Discharge

Q 0.05 m3 s A1 V1

V1 0.05

4

0.62

0.1768 m s

Velocity V2 QA2

0.05

4

0.32

0.7074 m s

Using Bernoulli’s equation,

P1

w

V12

2g Z1

P2

w

V22

2g Z2

Z 1 D =0.3m1

Z =02

S lope 1 in 30

Da tum L ine

100 m

0.6m =D1


19.62 104

9.81 103 0.17682

2 9.81 3.33

P2

9.81 103 0.70742

2 9.81 0

20 1.593 10 3 3.33 P2

9.81 103 0.0255 0

P2 228632.7 N m2

Pressure at lower end,

P2 228.6327 kN m2

1.22. Water is flowing through a pipe of diameter 30 cm and 20 cm atsections 1 and 2 respectively. The rate of flow through pipe is 35lps. The section 1 is 8 m above datum and section 2 is 6 m above

datum. If the pressure at section 1 is 44.5 N/cm2, find the intensityof pressure at section 2. (Nov/Dec 2015 AU)

Given

Dia of section 1, D1 30 cm 0.3 m

Dia of section 2, D2 20 cm 0.2 m

Rate of flow,

Q 35 lit/sec 0.035 m3/sec

Solution

Area of section (1), A1 4

D12

4

0.32

A1 0.07065 m2


D22

4

0.22

A2 0.0314 m2

Datum head at sec (1), Z1 8 m

Datum head at sec (2) Z2 6 m

Pressure

P1 44.5 N/cm2 44.5 104 N/m2


We know that,

Q A1 V1 A2 V2

V1 QA1

0.035

0.07065

V1 0.4953 m/s

V2 QA2

0.035

0.0314

V2 1.1146 m/s

Applying Bernaullies equation between both sections

P1w

V

1

2

2g Z1

P2

w

V22

2g Z2

44.5 104

9810

0.49532

2 9.81 8

P2

9810

1.11462

2 9.81 6

45.36 0.0125 8 1.0913 10 4 P2 0.0633 6

53.372 1.0193 10 4 P2 6.0633

P2 53.372 6.0633

1.0193 10 4

P2 464129.30 N/m2

P2 464.13 kN/m2

1.23. Water is flowing through a pipe having diameters 20 cm and 10 cmat section 1 and 2 respectively. The rate of flow through the pipe is35 litres/sec. Section 1 is 6m above the datum and section 2 is 4m

above datum. If the pressure at section 1 is 39.24 N/cm2. Find theintensity of pressure at section 2. (Nov/Dec 2008 AU)

P =44.5 x 10 N/m14 2

Z =8m1

D atum line

D =0.3m1

D =0.2m2

Z =16m2


Given

Dia. of section 1, D1 20 cm 0.2 m

Dia. of section 2, D2 10 cm 0.1 m

Rate of flow,

Q 35 lit/sec 0.035 m3/sec


D12

4

0.22 0.03142 m2


D22

4

0.12 0.007854 m2

w.k.t Q A1 V1 A2 V2

Velocity at section (1), V1 QA1

0.035

0.03142 1.114 m/s

Similarly V2 QA2

0.035

0.007854 4.456 m/s

Applying Bernoulli’s equation between both sections, we get

P1

w

V12

2g Z1

P2

w

V22

zg Z2

39.24 104

9810

1.1142

2 9.81 6

P29810

4.4562

2 9.81 4



Pressure

P1 39.24 N/cm2 39.24 104 N/m2

Pressure P2 ?

D = 0.2m1P =39.24N/cm1

2

Z =6m1

Datum line

D =0.1m2

Z = 4m2


40 0.0633 6 P2

9810 1.012 4

46.0633 P2

9810 5.012

P2 46.0633 5.012 9810

402713.25 N/m2

40.27 N/cm2 Ans

1.24 Water flows at the rate of 200 litres per second upwards through atapered vertical tape. The diameter at the bottom is 240 mm and atthe top 200 mm and the length is 5 m. The pressure at the bottomis 8 bar and the pressure at the topside is 7.3 bar. Determine thehead loss through the pipe. Express it as a function of exit velocityhead. (Nov/Dec 2014 - AU)

Given data: Discharge Q 200 litres/sec 200 10 3 m3/sec

. . . 1 litre 1 10 3 m3Diameter (bottom) d1 240 mm; Diameter (top) d2 200 mm

Length of pipe L 5 m

Pressure P1 8 bar; Pressure P2 7.3 bar (Z - Datum Head)

Head loss HL ?

V1 - inlet velocity

V2 - Exit velocity

Z2 L 5 m

According to Bernoulli’s equation

P1

w

V12

2g Z1

P2

w

V22

2g Z2

d2

LZ2

d1 P1

Z =01

P2

1

2

200 m m= 7.3 bar

= 5m

240 m m= 8 ba r


A1 4

d12

4

240 10 32 0.04523 m2

A2 4

d22

4

200 10 32 0.031415 m2

Q A1 V1 A2 V2

V1 A2 V2

A1

0.031415 V2

0.04523

V1 0.6946 V2

w g 1000 9.81 9810 . . . water 1000 kg/m3

P1w

V1

2

2g Z1

P2

w

V22

2g Z2 HL

8 105

9810

0.69462 V22

2 9.81 0

7.3 105

9810

V22

2 9.81 5 HL

81.549 0.024589 V22 74.413 0.0509 V2

2 5 HL

2.127 0.026311 V22 HL

HL 2.127 0.026311 V22

HL 2.127 0.026311 6.372 1.059 m

1.25. A horizontal venturimeter of specification 200 mm 100 mm is used tomeasure the discharge of an oil of specific gravity 0.8. A mercurymanometer is used for the purpose. If the discharge is 100 litres persecond and the coefficient of discharge of meter is 0.98, find themanometer deflection. (April/May 2007 - AU) (FAQ)

Given: Horizontal venturimeter; Diameter of inlet of venturimeter,

d1 200 mm 0.2 m

Diameter of throat of venturimeter, d2 100 mm 0.1 m

Specific gravity of oil Sp 0.8; Discharge, Q 100 lit/sec 0.1 m3/s

Coefficient of discharge of meter, Cd 0.98; To find: Manometer

deflection, x ?

V2 QA2

0.2

0.031415

V2 6.37 m/s


Solution:

Area of inlet, A1 4

d12

4

0.22 0.03142 m2

Area of throat, A2 4

d22

4

0.12 0.007854 m2

Discharge Q Cd A1 A2

A12 A2

2 2gh

0.1 0.98 0.03142 0.007854

0.031422 0.0078542 2 9.81 h

0.03545 h

h 0.1

0.03545 2.821 m

h 7.958 m

But h x sm

sp 1

7.958 x 13.60.8

1 16x

Manometer deflection x 7.958

16 0.4974 m

49.74 cm Ans

1.25. A Venturimeter having inlet and throat diameters 30 cm and 15 cmis fitted in a horizontal diesel pipe line Sp.Gr . 0.92 to measurethe discharge through the pipe. The venturimeter is connected to amercury manometer. It was found that the discharge is 8 liters/sec.Find the reading of mercury manometer head in cm. TakeCd 0.96 (Nov/Dec 2011 - AU)

Given: Horizontal VenturimeterInlet diameter, D1 30 cm 0.3 m; Throat diameter, D2 15 cm 0.15 m

Sp. gravity of pipe liquid, S1 0.92; Discharge, Q 8 lit/sec 0.008 m3/s;

Cd 0.96


Solution:

Area of inlet of venturimeter, A, 4

D12

4

0.32 0.0707 m2

Area of throat, A2 4

D22

4

0.152 0.01767 m2

Discharge, Q Cd A1 A2

A12 A2

2 2 gh

h Q A1

2 A22

Cd A1 A2 2 g

0.008 0.08082 0.017672

0.96 0.0707 0.01767 2 9.91

0.1031 m

Venturi head, h 0.10312 0.01063 m

h x S2 S1

S1

Mercury reading x h

S1

S2 S1

. . . S2 Sp. gr.

of mercury 13.6

0.01063

0.9213.6 0.92

0.0007713 m

i.ex 0.07713 cm

1.27. An orificemeter with orifice diameter 15 cm is inserted in a pipe of30 cm dia. The pressure on the upstream and downstream of orifice

meter is 14.7 N/cm2 and 9.81 N/cm2}. Find the discharge. Cd 0.6.

(April 2006 AU)

Given: Dia of orifice, D0 15 cm 0.15 m;

P1 14.7 N/cm2 14.7 104 N/m2; P2 9.81 N/cm2 9.81 104 N/m2

Area of orifice A0 4

0.152 0.0177 m2

Dia of pipe D1 30 cm 0.3 m; Area of pipe A1 4

0.32 0.0707 m2


Coefficient of dischargefor orifice meter

Cd 0.6

Solution

Difference of pressurein between upstreamand down stream

P1 P2

14.7 104 9.81 104 N/m2

P1 P2 48900 N/m2

Difference of pressure headin between upstream anddown stream

h

P1 P2

w

[Here assume water is flowing through pipe.

So w specific wt of water 9.81 103 N/m3]

So h 48900

9.81 103 4.985 m of water

To find discharge Q:

Discharge Q Cd A1 A0 2gh

A12 A0

2

0.6 0.0707 0.0177 2 9.81 4.985

0.07072 0.01772

Q 0.1085 m3/s 108.5 lit/s[. . . 1 m3/s 1000 L/s]

1.28. PRINCIPLE OF CONSERVATION OF MOMENTUM (FAQ)

The impulse is equal to the change in momentum of the body.

Impulse force small interval of time Fdt

and Impulse change in momentum d mV

This is the basis for momentum equation

According to Newton’s second law, F ma


where a acceleration

F m dVdt

d mV

dt [ . . . m is constant ]

F dmV

dt This is momentum principle.

The above equation can be written as

F dt d mV

Impulse Change in momentum

The above equation is called Impulse-momentum equation.

This equation is used to determine the resultant force exerted by fluid

on the pipe bend.

Refer the Fig. 1.55

V1 velocity of flow at section 1

P1 pressure at section 1

A1 cross sectional area of pipe at section 1.

and V2, P2 and A2 are the corresponding values at section 2.

The component of forcesexerted by the flowing fluid on

the bend in x and y direction

Fx and Fy

F ig. 1.55 Forces on bend

V 1

FY

P A1 1

PA221

FX

2

V sin 2

V cos 2

V 2

(a )

P A sin2 2

P A cos2 2

PA2

2

O

(b )


Similarly the componentof forces exerted by the

bend on the flowing fluid inx and y direction

Fx and Fy but in opposite direction

The pressure force actingat section 1 and 2

P1 A1 and P2 A2

The momentum equation for x direction

P1A1 P2A2 cos Fx m dV

Q V2 cos V1

[where m mass of fluid flow in kg/s

and dV Final velocity - initial velocity in x direction] [ angle of bend ]

Fx Q V1 V2 cos P1A1 P2A2 cos

Similarly for y direction, the momentum can be given as

0 P2A2 sin Fy Q V2 sin 0

Fy Q V2 sin

P2A2 sin

The resultant force F is given by

F Fx2 Fy

2

The angle can be measured as follows

tan FyFx

1.29. MOMENT OF MOMENTUM EQUATION (FAQ)

Moment of Momentum principle states that the resulting torque acting

on a rotating fluid is equal to the rate of change of moment of momentum.

Let V1, V2 Velocity of Fluid at section 1 and Section 2

r1, r2 = radius of curvature at section 1 and section 2

Q = Rate of flow of fluid

= Mass density of fluid


Momentum of fluid at section 1 m v Q V1

Moment of momentum at section 1.

m V1 r1 Q V1 r1

Similarly moment of momentum at section 2.

m V2 r2 Q V2 r2

Rate of change of moment of momentum

Q V2r2 Q V1r1

Q [V2r2 V1r1]

According to the principle rate of change of moment of momentum is

equal to the resultant torque so

T Q [V2r2 V1r1]

The above equation is called moment of momentum.

Applications

1. Torque exerted by water on sprinkler.

2. Flow Analysis in turbines and centrifugal pumps.

1.30. 250 lit/sec of water is flowing in a pipe having a diameter of 300mm. If the pipe is bent by 135, find the magnitude and direction ofthe resultant force on the bend. The pressure of the water flowing is

400 kN/m2. Take specific weight of water as 9.81 kN/m3.(Nov/Dec 2011 AU)

Given: Diameter of the bend at inlet and exit D1 D2 300 mm 0.3 m

Area A1 A2 4

D12

4

0.32 0.0707 m2

Discharge Q 250 lit/s 0.25 m3/s[. .

. 1000 L 1 m3]

Pressure P1 P2 400 103 N/m2

Velocity at section 1–1 V1 QA1

0.25

0.0707 3.54 m/s

V2 V1 3.54 m/s


Force along x axisFx Q V1 V2 cos P1 A1 P2 A2 cos [ 135 ]Fx 1000 0.25 [ V1 V2 cos 135 ] P1 A1 P2 A2 cos 135

1000 0.25 [ 3.54 3.54 0.0707 400 103 0.0707

400 103 0.0707 cos 135

250 [ 3.54 0.2503 ] 28280 19997

49224.54 N 49.225 kN

Force along y axis:Fy Q V2 sin P2 A2 sin

Fy 1000 0.25 [ 3.54 sin 135 ] 400 103 0.0707 sin 135

625.7 19,997 20,623 N

20.623 kN

V 2

A2

V1

P 1

P 2

2

V 2

V 2

V2

cos 45 o

sin

45o

45 o

2

22

1

1

2

2

300mm

dia

A2cos 45oP2

A2

sin

45oP

2

135 o

300mm dia

F=

20.6

23kN

yF =49.225kNX

=22.731o

R=53.37kN


Magnitude of resultant force R Fx2 Fy

2 49.2252 20.6232 53.37 kN

The direction of R with horizontal x axis,

tan FyFx

20.62349.225

0.419

22.731

1.31. A 150 mm diameter pipe reduces in diameter abruptly to 100 mmdiameter. If the pipe carries water at 30 liters per second, calculatethe pressure loss across the contraction. Take coefficient ofcontraction as 0.6. (Nov/Dec 2012 - AU)

Given: Diameter before contraction, D1 150 mm = 0.15 m

Diameter after contraction, D2 100 mm = 0.1 m

Flow rate, Q 30 lit/sec 0.03 m3/s; Cc 0.6

To find: Pressure loss

Loss of head due to sudden contraction,

hc

1Cc

1 V2

2

2g

V2

V2V

2cos 45o

sin

45o

45o

2

2

A2

2

2

P2

A2cos 45oP 2

A2

sin

45o

P2

F=2

0.62

3kN

y

F =49.225kNx

=22.731 o

R=53.37kN


Q A2 V2

V2 QA2

0.03

4

0.12 3.82 m/s

hc

10.6

1

3.822

2 9.81 0.496 m of water

Pressure head loss P1

w

P2

w 0.496 m

Pressure loss, P1 P2 0.496 wwater . . . w 9.81

kN

m3

0.496 9.81 4.866 kN/m2


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