and Machineryairwalkbooks.com/images/pdf/pdf_73_1.pdf · 2018-06-13 · (Near All India Radio) 80, Karneeshwarar Koil Street, Mylapore, Chennai – 600 004. Ph.: 2466 1909, 94440

(Near All India Radio)

80, Karneeshwarar Koil Street,

Mylapore, Chennai – 600 004.

Ph.: 2466 1909, 94440 81904Email: [email protected],

[email protected]

www.airwalkbooks.com, www.srbooks.org

For III Semester B.E., Mechanical Engineering Students

Fluid Mechanicsand

MachineryAs per Latest Syllabus of Anna University - TN

New Regulations 2017

With Short Questions & Answers and University Solved Papers

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

Professor - MechSathyabama Institute of Science and Technology

Chennai - 119

Dr. R. VenkatasubramaniProfessor and Head - Department of Civil Engineering

Dr. Mahalingam College of Engineering and TechnologyPollachi, Coimbatore

300/-

Fifth Edition: June 2018

978-81-936958-9-0

www.srbooks.orgwww.airwalkbooks.com

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

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

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 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

Unit I

Fluid Properties and Flow

Characteristics

Units 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, energy equation and

momentum equation.

1.1 INTRODUCTIONFluid mechanics is the science which deals with the mechanics of

liquids and gases. Fluid is a substance capable of flowing. Fluid mechanics

may be divided into three branches.

1. Fluid Statics 2. Fluid Kinematics 3. Fluid Dynamics

Fluid statics is the study of mechanics of fluids at rest.

Fluid kinematics is the study of mechanics of fluids in motion. Fluid

Kinematics deals with velocity, acceleration and stream lines without

considering the forces causing the motion.

Fluid dynamics is concerned with the relations between velocities,

accelerations and the forces exerted by (or) upon fluids in motion.

1.1.1 Distinction between solid and fluid

Table 1.1

Solid Fluid

1. Solid is a substance whichundergoes a finite deformationdepending upon elastic limit onapplication of a force.

1. A fluid is a substance whichundergoes continuous deformationunder application of a shear force, nomatter how small the force might be.

2. Atoms (molecules) are usuallycloser together in solid.

2. Atoms are comparatively looselypacked in fluid.

Solid Fluid

3. Intermolecular attractive forcesbetween the molecules of a solid arelarge

3. Inter molecular forces are not solarge enough to hold the variouselements of the fluid together andhence fluid will flow under the actionof slightest stress.

4. A solid has a definite shape. 4. A fluid has no definite shape ofits own but it conforms to the shapeof the container vessel.

1.2 FLUID AND CONTINUUMA fluid is a substance that deforms continuously when subjected to

even an infinitesimal shear stress. This continuous deformation under the

application of shear stress constitutes a flow.

Solids can resist tangential stress at static conditions undergoing a

definite deformation while a fluid can do it only at dynamic conditions

undergoing a continuous deformation as long as the shear stress is applied.

The concept of continuum assumes a continuous distribution of mass

within the matter or system with no empty space. In the continuum approach,

properties of a system such as density, viscosity, temperature, etc can be

expressed as continuous functions of space and time.

The continuum concept is basically an approximation, in the same way

planets are approximated by point particles when dealing with celestial

mechanics, and therefore results in approximate solutions. Consequently,

assumption of the continuum concept can lead to results which are not of

desired accuracy. However, under the right circumstances, the continuum

concept produces extremely accurate results.

A dimensionless parameter known as knudsen number Kn /L,

where is the mean free path and L is the characteristic length, aptly

describes the degree of departure from continuum. The continuum concept

usually holds good when kn 0.01.

1.2 Fluid Mechanics and Machinery - www.airwalkbooks.com

Fluid mechanics is a sub discipline of continuum mechanics as

illustrated here.

1.3 UNITS AND DIMENSIONS IN FLUID MECHANICSGenerally all physical quantities are measured in certain units either

fundamental units or derived units. Physical quantities expressed in terms of

the Length L, mass (M) and time T are called fundamental quantities and

units are called fundamental units. Some units, derived from the above

fundamental units like pressure, velocity, acceleration etc. are called derived

units.

1.3.1 System of Units

There are four internationally accepted system of units namely

1. F.P.S unit system: In F.P.S. unit System Length is Measured in Foot,

mass is measured in pound and time is measured in seconds.

2. C.G.S unit system: In C.G.S unit system the Length, Mass and Time

are measured in centimeter, gram and seconds respectively.

3. M.K.S unit system: In M.K.S unit system the length, mass and time

are measured in Meter, Kilogram and second respectively.

C ontinuum M echan ics

So lid M echan ics Fluid M echa nics

(The study of the phys ics o f co ntinuous m ate ria ls)

(The study of the phys ics o f co ntinuous m ate ria ls w ith a de fined rest shape)

(The study of the phys ics o f continuo us m ateria ls w hich d efo rm w hen subjected to a force)

Fluid Properties and Flow Characteristics 1.3

4. S.I. unit system: It is also called as International System of units. It

has six basic units (Length - Meter, Mass - Kilogram, Time - Second,

Current - Ampere, Temperature - Kelvin, Luminous intensity - Candela),

two supplementary units (plane Angle - radians, Solid angle - steradian)

and twenty seven derived units (some are density - kg/m3, Force -

Newton, Power - Watts etc).

In fluid mechanics, the basic dimensions are

1. Length L

2. Mass M

3. Time T

4. Temperature K.

The above units ae fundamental units and physical quantities are

expressed in terms of above units.

Now According to Newton’s second law

Force F ma

Where m mass in kg

a acceleration in meter

s2 or ms2

L

T2 LT 2

[ here m meter, s second ]

So F ma MLT 2 [ here m mass ]

So it is a derived unit.

Unit of force is Newton N, So the dimension for force

N MLT 2. Similarly, a list of derived quantities used in fluid mechanics

and their dimensions in terms of L, M, T are tabulated.


Table 1.2

Quantity with symbol SI unit Dimension

Velocity V m/s LT 1

Acceleration a m/s2LT 2

Area A m2 L2

Density kg/m3 ML 3

Volume V m3 L3

Force F N MLT 2

Specific volume v m3/kg L3 M 1

Pressure P or p N/m2MLT 2

L2 ML 1 T 2

Flow rate (Discharge) Q) m3/sec L3T 1

Viscosity (Dynamic viscosity)

N s/m2 MLT 2 TL2

ML 1 T 1

Kinematic viscosity m2/s L2T 1

Frequency Hz hertz (Hz)

cycles

s

T 1

Energy, work (or)Quantity of heat

N m MLT 2 L ML2T 2

Power P N m/s ML2T 2

T ML2T 3

Specific weight w N/m3 MLT 2

L3 ML 2T 2


Example: Bernoulli’s equation for the flow of an ideal fluid is given as

follows

Pw

Z V2

2g constant

where P Pressure in N/m2

Z Elevation height in m

V Mean flow velocity in m/s

w Specific weight in N/m3

g Acceleration due to gravity 9.81 m/s2

Demonstrate that this equation is dimensionally homogeneous i.e all

terms have the same dimensions.

Term 1: Dimensions of Pw

N/m2

N/m3

ML 1 T 2

ML 2 T 2 L

Term 2: Dimensions of Z L

Term 3: Dimensions of V2

2g

m/s2

m/s2

m2/s2

m/s2

L2 T 2

LT 2 L

So all the terms have the same dimensions

1.4 PROPERTIES OF FLUIDS

1.4.1 Gas and Liquid

A fluid may be either a gas or a liquid. The molecules of a gas are

much farther apart than those of a liquid. Hence a gas is highly compressible

and a liquid is relatively incompressible.

A vapour is a gas whose temperature and pressure are very closely

nearer to the liquid phase. So steam is considered as vapour.

A gas may be defined as a highly super heated vapour, i.e its state is

far away from the liquid phase. So air is considered as a gas.

Fluids are having the following properties:


Table 1.3

Quantity Symbol Unit

1. Density (or) mass density kg/m3

2. Specific weight (or) weight density w N/m3

3. Specific volume v m3/kg

4. Specific gravity s No unit

5. Compressibility 1K

6. Vapour pressure P N/m2

7. Cohesion and Adhesion

8. Surface tension N/m

9. Capillary rise (or) fall h m

10. Viscosity-Dynamic viscosity (or) viscosity Ns/m2

11. Kinematic viscosity m2/s

The above properties are discussed in detail.

1.4.2 Density (or) mass Density

The density of a fluid is its mass per unit volume.

Density mass

volume

kg

m3

So the unit of density is kg/m3 and dimension is ML 3 [M for mass

in kg and L for length in m]

The density of liquids is normally constant while that of gases changes

with the variation of pressure and temperature.

Density of water at 4C is 1000 kg/m3


1.4.3 Specific weight (or) Weight density

Specific weight is the weight per unit volume. Its symbol is w.

Specific weight represents the force exerted by gravity on a unit

volume of fluid.

Specific weight w Weight of fluidVolume of fluid

N

m3

Specific weight and density are related

w Weight of fluidVolume of fluid

Mass of fluid gVolume of fluid

g ...

massVolume

So w g ...(1.1)

Where g Acceleration due to gravity.

The specific weight of water at 4 C is 9810 N/m3. Density is absolute

since it depends on mass and independent of location.

Specific weight, on the other hand, is not absolute, since it depends

on value of g which varies from place to place.

Density and specific weight of fluids vary with temperature.

1.4.4 Specific Volume vSpecific volume is the volume occupied by a unit mass of fluid. Its

symbol is v. Its unit is m3/kg

v Volume of fluid

Mass of fluid

VMass

m3

kg

Specific volume is the reciprocal of density

v 1 ...(1.2)

Note: v Specific volume; V Volume; u, v and V Velocity of flow.


1.4.5 Specific gravity (or) Relative density sSpecific gravity of a liquid is the ratio of its density to that of pure

water at a standard temp. 4C. Its symbol is s. It has no unit.

s Density of liquidDensity of water

w ...(1.3)

where w 1000 kg/m3

Density of liquid s w

Specific gravity can also be defined in terms of specific weight.

s Specific weight of liquidSpecific weight of water

w

ww

Specific weight of liquid w s ww ...(1.4)

where ww 9810 N/m3

The specific gravity of gas is the ratio of its density to that of air.

1.4.6 Temperature

It is intensive thermodynamic property which determines the hotness

or the level of heat intensity of a body. A body is said to be hot if it is

having high temperature indicating high level of heat intensity. Similarly a

body is said to be cold if it is at low temperature indicating low level of

heat intensity. The temperature of a body is measured by an instrument called

thermometer.

Temperatures are measured in well known two scales:

(i) Centigrade scale C

(ii) Fahrenheit scale F

Both the scales are inter convertible as follows

F 1.8C 32

C F 32 1.8


1.4.7 Viscosity (Dynamic Viscosity)Viscosity is the resistance offered to the movement of one layer of

fluid by another adjacent layer of the fluid.

Refer Fig. 1.1. Fluid is divided into different layers one over the other.

Consider two layers of fluid. One is moving with velocity u. Another layer

is moving with u du

The distance between the layer is dy. The top layer causes a shear

stress on the adjacent lower layer while the lower layer causes a shear stress

on the adjacent top layer. This shear stress (denoted by ) is proportional to

rate of change of velocity with respect to y.

i.e dudy ... (1.5)

dudy ... (1.6)

The proportionality constant is and is known as coefficient of viscosity

(or) absolute viscosity (or) dynamic viscosity (or) simply viscosity.

dudy

Velocity gradient (or) rate of shear strain (or) rate of shear

deformation.

The equation (1.6) can be rearranged as

du/dy

Y

dy

duuy

u+du

u Ve lo city p rofile

Fig. 1.1 Velocity Variation

Top layer

Lowe r layer


To find unit of :

Shear stress

Change in velocityChange of distance

Force/Area

LengthTime

1

Length

N/m2

m/s/m

Ns

m2

Force Time

Length2

In MKS Unit System, Force is measured in kgf

So unit of kgfsec

m2

In CGS System, Force is Measured in dyne

So unit of dyne sec

cm2 or poise

[. . . 1 dyne sec

cm2 1 poise]

In SI System Force is represented in Newton N

So unit of N s

m2 Pa s [... N/m2 Pascal Pa]

Numerical Conversion From MKS unit to CGS unit.

We know that 1 kgf 9.81 N

So, 1 kgf Sec

m2 9.81 N S

m2

1 kgf Sec

m2 9.81 1 kg 1 m/sec2 sec

m2 [. . . Force N m a]

9.81 103 g 102 cm/sec2 sec

104 cm2

9.81 105 g cm/sec2 sec

104 cm2

1 kgf Sec

m2

98.1 dyne sec

cm2

[. . . 1 g cm

sec2 1 dyne]


1 kgf Sec

m2 98.1 poise

[. . . 1 dyne sec

cm2 1 poise]

In S.I Units

1 kgf sec

m2 98.1 poise

Also 1 kgf sec

m2 9.81 N sec

m2 98.1 poise

[. . . 1 kgf 9.81 N]

1 N Sec

m2

98.19.81

poise 10 poise

So, 1 N sec

m2 10 poise [ in S.I unit sec is represented as S]

So 1 Ns

m2 10 poise

Sometimes unit of viscosity is given at centipoise

1 Centipoise CP 1

100 poise

A widely known metric unit for viscosity is the poise (p)

1 poise 0.1 Ns/m2 in S.I units

1 centipoise 0.01 poise 0.001 Ns/m2

1.4.7.1 Kinematic Viscosity ()

What is the importance of kinematic viscosity? (Nov/Dec 2014 - AU)Kinematic viscosity is the ratio of dynamic viscosity to the density

of the fluid. The symbol for kinematic viscosity is

...(1.7)

Unit for kinematic viscosity

N s/m2

kg/m3


kg m

s2 s 1

m2 m3

kg . . . N kg m

s2

m2

s

In metric system, is in stoke

1 stoke 1 cm2/s 10 4 m2/s in S.I units

1 centistoke 10 6 m2/s

1.4.7.2 Variation of Viscosity with temperature

Write down the effect of temperature on viscosity of liquids and gases.(Nov/Dec 2016, Nov/Dec 2015, Nov/Dec 2013 - AU)

The viscosity of liquids decreases with the increase of temperature

while the viscosity of gases increases with the increase of temperature. This

is due to the reason that in liquids the cohesive forces predominates the

molecular momentum transfer, due to closely packed molecules and with the

increase in temperature, the cohesive forces decreases with the result of

decreasing viscosity. But in case of gases, the cohesive forces are small and

molecular momentum transfer predominates. With the increase in temperature,

molecular momentum transfer increases and hence viscosity increases.

The relationship between viscosity and temperature for

(a) Liquids:

0

1

1 t t2 ...(1.8)

(b) Gases:

0 t t2 ...(1.9)

Where, Viscosity of liquid/Gas at tC, in poise

0 Viscosity of liquid/Gas at 0C, in poise

, are constants for liquid/Gas.


1.4.8 Compressibility 1K

Compressibility of a liquid is inverse of its bulk modulus of elasticity.

Bulk modulus K Compressive StressVolumetric strain

Consider a cylinder piston mechanism

Let P1 Initial pressure inside the cylinder; P2 Final pressure inside the cylinder

V1 Initial volume; V2 Final volume

K Increase in PressureVolumetric strain

dP

dV

V

dPdV

V

...(1.10)

Since rise in pressure reduces the volume by

d V, the strain is indicated as dV

V

Compressibility 1K

1.4.8.1 Relationship between Bulk Modulus K and pressure P of a Gas for Isothermal and Isentropic Process

For Isothermal Process

We know that for Isothermal process

PV constant

Partial diffirenting the above equation we get

PdV VdP 0

P VdPdV

dP

dV/V ...(1.11)

From equation 1.10 we know that K dP

dV/V

P K For Isothermal process.

Fig. 1.2

dv

V

Cylinder Piston


For Adiabatic (or) isentropic process

We know that for Adiabatic process PV constant

Ratio of Specific heat

Partial differentiating the above equation we get

P V 1 dV V dP 0

dividing above equation by V 1, we get

P dV VdP 0

P VdPdV

K

We get P K, for adiabatic or isentropic Process.

1.4.9 Vapour Pressure

When the liquid is kept in a closed vessel, it evaporates into vapour

and this vapour occupies the space between the free surface of the liquid and

top of the vessel. This vapour exerts a partial pressure on the free surface of

the liquid. This pressure is known as vapour pressure of liquid.

1.4.9.1 Cavitation

What is cavitation? What causes it? (Nov/Dec 2013 - AU)Now consider a flow of liquid in a system. If the pressure at any point

in this flowing liquid becomes equal to the vapour pressure, the vaporization

of the liquid begins and bubbles are formed. When these bubbles are carried

by flowing liquid into region of high pressure, these bubbles collapse creating

very high pressure. The metallic surface above which this liquid is flowing

is subjected to these high pressures causing pitting action on surface. This

process is called cavitation.


1.4.9.2 Gas and Gas lawsThe gas is the term applied to the state of any substance of which the

evaporation from the liquid state is complete. Substances like Oxygen, Air, Nitrogen

and Hydrogen etc may be regarded as gases within the temperature limits.

A vapour may be defined as a partially evaporated liquid and consists

of pure gasesous state together with the particles of liquid in suspension.

Examples of vapour are steam, ammonia, SO2, CO2 etc.

A perfect gas or an ideal gas is one which obeys all gas laws under all

conditions of temperature and pressures. No gas is perfect i.e., no gas strictly

obeys the gas laws but within the temperature limits of applied thermodynamics

many gases like H2, O2, N2 and even air may be regarded as perfect gases.

Gas laws: Three variables control the physical properties of a gas. The

pressure exerted by gas P, the volume V occupied by it and its temperature

T. If any of these two variable are known, then the third can be calculated

by gas laws. Gas law does not apply to vapours.

Boyle’s law: Boyle’s law states that “The volume of a given mass of a gas

varies inversely as its absolute pressure, provided the temperature remains constant”.

V 1P

or PV constant

Charles laws: Charle’s law states that “The volume of a given mass of a

gas varies directly as its absolute temperature, provided the pressure is kept

constant”. The variation is same for all gases.

V T or VT

Constant

Perfect gas law (combination of boyle’s and charle’s law).

Let us assume that we have a perfect gas at absolute pressure, volume

and absolute temperature of P1, V1 and T1 respectively. Suppose the gas

expands or contracts at a constant temperature to its volume V1 such that the

corresponding volume of its new absolute pressure is P2.

According to boyle’s law

P1 V1 P2 V2 ...(1)


Now let the gas be expanded (or contracted) further such that the

pressure remains constant and its volume and absolute temperature change

from V1 to V2 and T1 to T2 respectively.

According to charles law

V1

T1

V2

T2 ...(2)

According to Gay-Lussac’s Law

PT

constant

P1

T1

P2

T2 ... (3)

Combining these three gas laws, we can get a new equation,

P1 V1

T1

P2 V2

T2 constant

If, is the volume of unit mass of gas, then this constant is R

(characteristic gas constant)

i.e, PvT

R Pv RT

If m is the mass of gas under consideration, then the equation becomes

PV mRT, which is called the equation of perfect gas or characteristic gas equation.

Law Equation Constant variable

Boyle’s Law P1 V1

T1

P2 V2

T2 P1 V1 P2 V2

TemperatureT1 T2

Charle’s Law P1 V1

T1

P2 V2

T2

V1

T1

V2

T2

PressureP1 P2

Gay-Lussac’s Law P1 V1

T1

P2 V2

T2

P1

T1

P2

T2

VolumeV1 V2


1.4.10 Surface TensionWhen a liquid is put inside a narrow tube, the free surface of the

liquid displays either a rise (or) depression near the walls of the tube. This

phenomena is attributed to a property of fluids known as surface tension.

Soap bubbles, small droplets of water and dew on a dry solid surface also

are attributed to surface tension.

Surface Tension is also defined as the tensile force acting on the

surface of a liquid in contact with a gas or on the surface between two

immiscible liquids such that the contact surface behaves like a membrane

under tension. The magnitude of this force per unit length of the free surface

will have the same value as the surface energy per unit area.

Surface tension in a liquid is caused by (i) Cohesive forces i.e forces

of attraction between molecules of the same material or fluid and

(ii) Adhesive forces i.e forces of attraction between molecules of different

materials, say, the attraction between molecules of liquid and the molecules

of container (or) air.

Example for Cohesive ForceWhen mercury is poured on the floor, it does not wet the surface of

floor and forms sphere. When two spheres of mercury are brought close

together, they combine together to form a bigger sphere. This means that the

mercury molecules have cohesive tendency and have no tendency to adhere

(adhesion) to the floor (solid surface).

Example for Adhesive ForceWhen water is poured on the floor, the water molecules wet the

surface. This means that water molecules have adhesive tendency to adhere

to the floor (solid surface).

At the interface between a liquid and a gas ie at the liquid surface,

and at the interface between two immiscible (not mixable) liquids, the out of

balance attraction force between molecules forms an imaginary film capable

of resisting tension. This liquid property is known as surface tension. Because

the tension acts on a surface, we compare such forces by measuring the

tension per unit length of surface. The surface tension is denoted by the

symbol . The unit of surface tension is N/m.


Consider a liquid in a

vessel as shown in Fig. 1.3.

Consider two molecules

A and B of a liquid in a mass

of liquid. The molecule A is

attracted in all directions

equally by the surrounding

molecules of the liquid. Thus

the resultant force acting on the

molecule A is zero. But the

molecule B is situated near the free surface. This molecule B is acted upon

by upward and downward forces which are unbalanced. Thus net resultant

force on molecule B is acting in a downward direction. Like molecule B, all

the molecules near the free surface experience a downward force. So the free

surface of the liquid acts like a very thin film under tension of the surface

of the liquid.

1.4.10.1 Surface Tension on Droplet

Let us consider a spherical droplet of liquid of radius r.

The surface tension on the surface of droplet in Nm

P Pressue intensity inside the droplet

in excess of the outside pressure intensity

r Radius of droplet

Let us assume that the droplet is cut into two halves as shown in

Fig. 1.4. The forces acting on one half (say left half) will be

F ig. 1.3 Surface tension

A

B

W a te r D ro plet Su rfaceTen sion Pressu re Fo rces

P

F ig. 1.4 Forces on droplet


(i) Tensile force acting around thecircumference of the cutportion (This tensile force isdue to surface tension) and isgiven as circumference

2r

(ii) Pressure force on the area

P r2

Under equibibrium conditions, these two forces will be equal and

opposite in direction. Equate both forces, we get,

P r2 2r

P 2 r

or 4d ...(1.12)

Where d Dia of droplet

The equation shows that with the decrease of radius of the droplet,

pressure intensity inside the droplet increases.

The excess pressure P inside a bubble is known to be a function of thesurface tension and the radius. By dimensional reasoning determine howthe excess pressure will vary if we double the surface tension and the radius.

(Nov/Dec 2013 - AU)

1.4.10.2 Surface Tension on a Hollow Bubble

A hollow bubble (soap

bubble) has two surfaces in

contact with air, one inside

and other outside. The above

two surfaces are subjected to

surface tension. In such case,

Surface tension on both

circumferences

2 2r

We can equate two forces acting on bubble

d p

Soap bubble

Fig. 1 .7 Pressure inside a soap bubble

p

W ater d rop le t d

Fig. 1 .6 Pressure inside a water droplet


P r2 2 2 r

P 4r

or P 8 d ...(1.13)

1.4.10.3 Surface Tension on a Liquid Jet

Consider a liquid jet of diameter d

and length L as shown in Fig. 1.8.

P Pressure intensity inside the liquid jet

above the outside pressure

Surface tension of the liquid

Consider the equilibrium of forces

acting on the half of the liquid jet

Force acting on jet P area of

half of jet (rectangle).

P L d.

Force due to surface tension 2L

P L d 2L

P 2L

L d

2 d

r

P r

1.4.11 Capillarity

Capillarity is the property of exerting forces on fluids by fine tubes.

It is due to cohesion and adhesion.

Capillarity may be defined as a phenomenon of rise or fall of a liquid

surface in a small tube relative to the adjacent general level of liquid when

the tube is held vertically in the liquid.

When a fine glass tube is partially immersed in water, the water will

rise in the tube to a height of ‘h’ m above the water level. This happens

when cohesion is of less effect than adhesion.

Fig. 1.8 Force on liquid jet

d

L


On the otherhand, if the same tube is partially immersed in mercury,

the mercury level in tube will be lower than the adjacent mercury level. This

happens because cohesion predominates than adhesion.

The rise of the liquid surface is called capillary rise and the depression

of the liquid surface is called capillary depression (or) capillary fall.

1.4.11.1 Expression for Capillary Rise

Refer capillary rise in Fig 1.9 (a)

Lifting force created by surface tension

Vertical component of the surface tension force circumference cos

d cos

Weight of the liquid of height h in the tube mass g Volume g

Area h g 4

d2 h g

Under equilibrium condition, equate both forces, we get,

d cos 4

d2 h g

Capillary rise h 4 cos g d ...(1.14)

w ater

(a ) Capillary Rise

M ercuryh=cap illa ry

fa ll

(b) Capillary Depression

Fig. 1.9 Capillary Rise (or) Fall


Where h: Height of liquid in the tube (or) capillary rise

: Surface tension of liquid.

: Angle of contact between liquid and glass tube.

: Density of liquid

Generally value is approximately equal to zero and hence cos 1 then

Capillary rise h 4 gd ...(1.15)

1.4.11.2 Expression for Capillary FallRefer Capillary Fall in Fig. 1.9 (b)

Hydrostatic force acting upward

P 4

d2 g h 4

d2

Downward force making depression due to surface tension

d cos

Under equilibrium condition,

g h 4

d2 d cos

Capillary fall h 4 cos g d

1.4.12 Thermodynamic PropertiesThe characteristic equation (or) equation of state for perfect gases is given as

PV m RT ... (i)

Where

V Volume of gas in m3; M mass of gas in kg

R Characteristic gas constant or simply gas constant

R for air 0.287 kJ/kg K

T absolute temperature in K Kelvin

T tC 273


The equation can be written

Pv RT ... (ii)

P R T ... (iii)

Where v specific volume in m3/kg ;

density in kg/m3

Also, we can write the equation as

P1 V1

T1

P2 V2

T2

P3 V3

T3 constant

For Isothermal process, temperature remains constant

Pv Constant or PV Constant.

For Isentropic process

Pv Constant or PV Constant.

For Polytropic process, Pv n constant [n Index of compression (or)

expansion]

[n ranges from 0 to ]

Ratio of specific heats Cp

Cv

Cp and Cv Specific heats of a gas at constant pressure and constant

volume respectively

According to Avagadro’s hypothesis, all the pure gases at the same

temperature and pressure have the same number of molecules per unit volume.

For any gas, PV

m MRT ... (iv)

Where V

m Molar Volume. (Molar volume is the volume occupied

by the molecular mass of any gas at standard temperature and pressure).

T Absolute temp in K; M Molecular weight

So the equation (iv) can be written as

PV

m R

T


R

MR Universal gas constant

From Avagadro’s law,

When P 1.01325 105 N/m2 and T 273.15 K

Molar volume V

m 22.4 m3/kg mol

PV

m R

T

R

PV

m

T

1.01325 105 22.4273.15

8314 J/kg mol K 8.314 kJ/kg mol K

R

8.314 kJ/kg mole K for any gas [Universal gas constant]

R for gas A R

M for gas A

R air R

M for air

8.31428.96

0.287 kJ/kg K[. . . M for air 28.96 kg/kg mol]

1.5 NEWTON’S LAW OF VISCOSITY

Define Newton’s law of viscosity. (Nov/Dec 2012 - AU)A fluid is a substance that deforms continuously when subjected to a

shear stress, no matter how small that shear stress may be. A shear force is

the force component tangent to a surface, and this force divided by the area

of the surface is the average shear stress over the area. Shear stress at a point

is the limiting value of shear force to area as the area is reduced to the point.

Y

t

F

a d

y

x

b b � c c �

V

u

Fig. 1.12 Deformation Due to C onstant Shear Force


In the Fig 1.10, a substance is placed between two closely spaced

parallel plates so large that conditions at their edges may be neglected. The

lower plate is fixed and a force F is applied to the upper plate, which exerts

a shear stress . If ‘A’ is the area of plate, then shear stress is F/A on any

substance between plates. When the force F causes the upper plate to move

with a steady (Non zero) velocity, no matter how small the magnitude of

F, one may conclude that the substance between the two plates is a fluid.

The fluid in the area abcd flows to the new position abcd, each fluid particle

is moving parallel to the plate and the velocity u is varying uniformly from

zero at the stationary plate to V at upper plate.

Experimental results shows that

F AVt ...(1.16)

Where is the proportionality factor called coefficient of viscosity. If

F/A then ...(1.17)

Substituting 1.17 in 1.16 we get

Shear Stress Vt

Vt

is the angular velocity or rate of angular deformation.

The angular velocity may also be written as du/dy as both

V/t and du/dy express the velocity change divided by the distance over which

the change occurs. Now in the differential form, dudy

. This equation is

called Newton’s law of viscosity.

Newton’s law of viscosity states that the shear stress on a fluid

layer is directly proportional to the rate of statics strain. The constant of

proportionality is called the coefficient of viscosity

Shear Stress du/dy

dudy ...(1.18)


1.5.1 Types of Fluid

All the fluids can be classified into the following types

1. Ideal fluid

2. Real fluid

3. Newtonian fluid

4. Non-newtonian fluid

5. Ideal plastic fluid

1. Ideal Fluid: A fluid with no viscosity is called ideal fluid. Practically,

no fluid is ideal since all fluids have some viscosity. This fluid is represented

by the horizontal axis in the following graph.

2. Real Fluid: A fluid having viscosity is called Real fluid

3. Newtonian Fluid: A fluid which obeys the Newton’s law of viscosity is

called Newtonian fluid. In Newtonian fluids there is a linear relation between

and rate of deformation du/dy as shown in Fig. 1.11

du/d y

idea l flu idN ew ton ian fluid

- New tonia n flu

id

Non

idea l plastic

E las tic so lid

=

shea

r st

ress

(Ve loc ity gra dien t)

Fig. 1.11 Types of flu ids


This means that regardless of the forces acting on a fluid, it continues

to flow. For example, water is a Newtonian fluid as it continues to display

fluid properties no matter how much it is stirred or mixed.

4. Non-Newtonian Fluid: The fluid which does not obey the newton’s law

of viscosity is called Non-newtonian fluid. Here there is a non-linear

relationship between and du/dy. So, stirring a non-Newtonian fluid can leave

a gap behind which will gradually fill up over time, as such in materials such

as pudding. Alternatively, stirring a non-Newtonian fluid can cause the

viscosity to decrease, so the fluid appears thinner as seen in non-drip paints.

5. Ideal Plastic Fluid: A fluid in which shear stress is more than the yield

value and shear stress is proportional to the rate of shear strain (or velocity

gradient), is called ideal plastic fluid. It is shown in Fig. 1.11 by straightline

intersecting the vertical axis at the yield stress.

SOLVED PROBLEMSProblem 1.1: A mass of liquid weight 500 N, is exposed to standard earth’s

gravity g 9.806 m/s2. (i) What is its mass? (ii What will be its weight in a

planet with acceleration due to gravity is 3.5 m/s2?

Solution

Given: Weight W 500 N, g 9.806 m/s2, g1 3.5 m/s2

To find: Mass M and weight W1

Case (i)

Weight W 500 N; g 9.806 m/s2

W mg

Mass m Wg

500

9.806 50.9892 kg

Mass m 50.9892 kg

Case (ii)

Mass of liquid m 50.9892 kg (Mass remains constant)

W1 mg1


where g1 Acceleration due to gravity in a planet.

W1 50.9892 3.5 178.4622 N

W1 178.4622 N

Problem 1.2: 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

Problem 1.3: Determine the specific gravity of a fluid having viscosity of0.07 poise and kinematic viscosity of 0.042 stokes.

Given: 0.07 poise, 0.042 stokes.

To Find: Specific Gravity s.

Solution: s specific gravity w

0.07 poise 0.07 0.1 0.007 Nsm2

. . . 1 poise 0.1

Ns

m2

0.042 stokes 0.042 10 4 m2

s . . . 1 stoke 10 4

m2

Ns


We know kinematic viscosity

Dynamic viscosity

Density of fluid

0.042 10 4 0.007

; 1666.7 kg/m3

But s w

So 1666.7 s 1000 [ . . . w density of water 1000 kg/m3 ]

s 1.6667

Specific gravity of fluid s 1.6667

Problem 1.4: Determine the viscosity of oil having kinematic viscosity of 6stokes and specific gravity of 2. (Apr 2006, Nov/Dec 2008 - AU)(FAQ)

Given: 6 ; stokes; s 2

To Find:

Solution: viscosity of oil ?

We know that

s 2 ; 6 stokes 6 10 4 m2/s [. . . 1 stoke 10 4m2/s]

s w

s w 2 1000 2000 kg/m3 . . . w 1000 kg/m3

So 2000 6 10 4 1.2 Ns/m2

Problem 1.5: The space between two parallel plates kept 3 mm apart is filledwith an oil of dynamic viscosity of 0.2 Pa-s. What is the shear stress on thelower fixed plate, if the upper one is moved with a velocity of 1.5 m/s.

Given: dy 3 mm, 0.2 Pa s 0.2 Ns

m2 ; du 1.5 m/s

To Find:


Solution:

0.2 Pa.s 0.2 Ns

m2

According to Newtons law of viscosity

Shear Stress dudy

where Shear stress on the lower fixed plate.

du Change of velocity u 0 1.5 0 1.5 m/s

Distance between two

plates dy 3 mm

3 10 3 m

dudy

0.2 1.5

3 10 3

100 N/m2

Shear stress on lower plate 100 N/m2

Problem 1.6: A 2.5 cm wide gap between two large plane surfaces arefilled with glycerine. What surface force is required to drag a very thin plate

of 0.75 m2 in area between the surfaces at a speed of 0.5 m/s, if it is at a

distance of 1 cm from one of the surfaces? Take 0.785 Ns/m2

Given: dy 2.5 cm dy1 1 cm 1 10 2 m dy2 1.5 10 2 m

A 0.75 m2 u 0.5 m/s 0.785 Ns/m2

Solution

Note: Since it is very thin plate, the weight of the plate and buoyant force

are neglected.

Area of the thin plate 0.75 m2

du u 0 0.5 m/s

Upper m oving plate

Lower fixed plate

O il o f = 0.2 P oise

1.5m /s

dy = 3 m m


dy1 1 cm 1 10 2 m

dy2 1.5 cm 1.5 10 2 m

0.785 Ns/m2

To Find Drag Force to Lift the Plate:

Drag force on both sides A

Shear stress on both sides

dudy1

dudy2

dudy1

dudy2

0.785

0.5

1 10 2

0.5

1.5 10 2

0.785 [50 33.333 ]

65.42 N/m2

Drag force on both sides A

65.42 0.75 49.0625 N

Problem 1.7: If the velocity distribution over a plate is given by

u y y2 in which u is the velocity in m/s at a distance of y meters above

the plate, determine the shear stress at y 0.10 m when coefficient of viscosity

is 0.86 Ns/m2

Solution: Given: u y y2 ; 0.86 Ns/m2

To Find: 0.1 ?

y 0.1 m ; 0.86 Ns/m2 ; u y y2

Velocity gradient dudy

1 2y

dy1 dy2

2 .5cm

1cm


when y 0.1 m,

dudy

0.1

1 2 0.1 0.8 s 1

To Find Shear stress at y 0.1 m

0.1 dudy

0.1

0.86 0.8 0.688 N/m2

0.1 0.688 N/m2

Problem 1.8: A fluid of specific gravity 0.9 flows along a surface with a

velocity profile given by v 4y 8y3 m/s, where y is in m. What is the velocitygradient at the boundary? If the kinematic viscosity is 0.36 S, What is theshear stress at the boundary? (FAQ)

Given: Specific gravity of fluid 0.9

Kinematic viscosity, 0.36 stokes 0.36 10 4 m2/s

Velocity V 4y 8y3 m/s

Solution:

V 4y 8y3

dvdy

4 8 3y2 4 24 y2

(a) Velocity gradient at the boundary, ie y 0,

dvdy

y 0

4 24 02 4

(b) Shear stress, dvdy

Specific gravity S 0.9 Density of fluidDensity of water

Density of fluid 0.9 density of water 0.9 1000 900 kg/m3

w.k.t


0.36 10 4 900 0.0324 NS/m2

Shear stress at boundary, i.e dvdy

y 0

0.0324 4 0.1296 N/m2

Problem 1.9: A cubical block having 200 mm edge and mass of 25 kg slidesdown an inclined plane surface which makes an angle of 20 with thehorizontal. On the plane, there is a thin film of oil of thickness 0.026 mm

and viscosity 0.2 10 2 Ns/m2. What terminal velocity will be attained bythe block?

Solution: Given: l 200 mm 0.2 m ; Mass m 25 kg 20 ;

t 0.026 mm ; 0.2 10 2 Ns/m2 { 0.2 10 2 Ns/m2;

Area 0.2 0.2 0.04 m2; W Wt. of block 25 9.81 245.25 N

To Find Shear Stress:

Component of W along their inclined planei.e shear force F

W sin 245.25 sin 20

83.8804 N

Shear stress on the bottom surface of cube FA

83.8804

0.04 2097.01 N/m2

To Find Terminal Velocity u

We know, dudy

2097.01 0.2 10 2

du

0.026 10 3

du 27.261 m/s

du u 0

So, Terminal Velocity u 27.261 m/s

m =25kg

dy=th ickness=0.026mmwsin20 o =20o

w

20o


Problem 1.10: A plate having an area of 0.8 m2 is sliding down the inclinedplane at 25 to the horizontal with a velocity of 0.36 m/s. There is a cushionof fluid, 1.8 mm thick between the plane and the plate. Find the viscosity ofthe fluid if the weight of the plate is 442 N.

Given: A 0.8 m2; 25; u 0.36 m/s; t dy 1.8 mm; W 442 N

Solution

Area of plate 0.8 m2 ; Weight of plate 442 N; Velocity of plateu 0.36 m/s

du u 0 0.36 m/s

Thickness of film t dy 1.8 mm 1.8 10 3 m

To Find Viscosity of Fluid Component of W along the plate

shear force F

W sin 442 sin 25 187 N

Shear stress

FA

1870.8

233.75 N/m2

We know, dudy

233.75 0.36

1.8 10 3

Viscosity 1.169 Ns/m2 11.69 poise

Problem 1.11: Calculate the dynamic viscosity of oil which is used forlubrication between 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 330 Nand it slide down the inclined plane with a uniform velocity of 0.3 m/s. Thethickness 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

u=0.36m /s

dy=1.8m m

plate

25o

w = 442N

25o

w s in


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

Problem 1.12: A metal plate of size 0.6 m 0.6 m and 1 mm thick andweighing 25 N is to be lifted up edgewise with uniform velocity of 0.2 m/sin the gap between two flat surfaces. The plate is in the middle of the gapof width 20 mm and the gap contains oil of relative density 0.85 and viscosity16 poise. Calculate the vertical force required for this job.

Solution: Given: Dimensions of the plate 0.6 0.6 1 10 3 m

Area of plate 0.6 0.6 0.36 m2

Since the plate is in the middle of the gap,

t1 t2 dy1 dy2 20 1

2 9.5 mm 9.5 10 3 m

Relative density specific gravity s 0.85


Dynamic viscosity 16 poise 1.6 Ns/m2

Velocity of the plate u 0.2 m/s; So du u 0 0.2 m/s

Weight of the plate 25 N

To Find Force Required to Lift the Plate

Drag force (or viscous resistance) against the

motion of the plate, F 1 A 2 A where 1 and 2 are

the shear stresses on both sides of the plate.

1 dudy1

and 2 dudy2

F A [1 2 ] A du

1dy1

1

dy2

0.36 1.6 0.2

1

9.5 10 3

1

9.5 10 3

Force F 24.253 N

Upward thrust (or) buoyant force on the plate

Specific weight Volume of oil displaced.

0.85 9810 0.6 0.6 1 10 3 3.00186 N

[Note: When a body is immersed in a fluid, upward force (or) buoyant force

acts on body to move the body upward. This buoyant force is equal to weight

of fluid displaced by the body. So

Weight of fluid displaced Sp. weight Volume of fluid displaced

Now Weight of the body 25 N acts downward. Buoyant force is acting

upward.

Effective weight of the plate 25 3 22 N

Total force required to lift the plate ata velocity of 0.2 m/s

F effective weight of plate 24.253 22

46.253 N.

t = dy1 1 t = dy2 2

dy =dy1 2

1m m

20m m


Problem 1.13: A vertical gap of 30 mm width and infinitely long, contain

oil of specific gravity 0.9 and viscosity 3.5 Ns/m2. A metal plate

1.0 m 1.0 m 10 mm having a mass of 20 Kg is to be lifted through thegap at a constant speed of 0.15 m/s. Determine the force required. The plateis situated 5 mm from the left end.

Solution: Given: s 0.9 ; 3.5 Ns/m2

Area of plate 1 1 1 m2

Thickness t 10 mm 0.01 m

dy1 5 mm 5 10 3 m

dy2 30 [5 10] 15 mm 15 10 3 m

du u 0 0.15 0 0.15 m/s

Shear stress on both sides

1 2 dudy1

dudy2

3.5 0.15

5 10 3 3.5

0.15

15 10 3

105 35 140 N/m2

Drag force (or) shear force F A 140 1 140 N

Weight of plate 20 9.81 196.2 N

Buoyant force on the plateUpward thrust on the plate

Specific wt Volume of oil displaced

0.9 9810 1 1 0.01 88.29 N

Effective weight of plate 196.2 88.29 107.91 N

Total force required to lift the plate

Drag force Effective weight of plate

140 107.91 247.91 N

10m m

30m m

dy1 dy2


Problem 1.14: A skater weighing 800 N skates at 54 km/hr on ice. The

average skating area supporting him is 10 cm2 and the effective dynamiccoefficient of friction between the skates and the ice is 0.02. If there is actuallya thin film of water between the skates and ice, determine its averagethickness. Assume that the water has a viscosity of 1 cP at 0C.

Given: Weight of skater 800 N

u Velocity of skater

54 1000

3600 15 m/s

du u 0 15 m/s

Area 10 cm2 10 10 4 m2

Effective dynamic coefficient

of friction k 0.02

Viscosity of water

1 cP1 centipoise

0.001 Ns/m2

Solution

Friction force (or) Drag force

(or) shear force k RN

0.02 800 16 N

Shear stress FA

16

10 10 4 16000 N/m2

We know dudy

16000 0.001 15dy

dy 9.375 10 7 m

Thickness of water film dy 9.375 10 7 m.

W

Skater

Wate r film

dy Fric tion = Rk N

R =W =800NN


Problem 1.15: A flat plate 0.1 m2 area is pulled at 30 cm/s relative toanother plate located at a distance of 0.01 cm from it, the fluid separating

them is water of dynamic viscosity 0.001 Ns/m2. Find the force and powerrequired to maintain the velocity.

Solution: Given: A 0.1 m2 ; du 30 10 2 m/s 0.3 m/s;

dy 0.01 cm 0.01 10 2 m ; 0.001 Ns/m2

Shear Stress

dudy

0.001

0.3

0.01 10 2 3 N/m2

Shear Force F

F A 3 0.1 0.3 N/m2

Power Force Velocity 0.3 0.3 0.09 Watts

Power 0.09 Watts

Problem 1.16: The space between two square flat parallel plates is filledwith 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 force of 98.1N 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 gravity of theoil 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

dy=0.01x10 m-2

0 .3m /s


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]

Problem 1.17: A plate 0.025 mm distance from a fixed plate moves at 50

cm/s and requires a force of 1.5 N/m2 to maintain the speed. Find the fluidviscosity between the plates.

Solution: Given: dy 0.025 mm 0.025 10 3 m; u 50 cm/s 50 10 2 m/s

0.5 m/s du u 0 0.5 m/s; Shear stress 1.5 N/m2

Shear stress, dudy

1.5 0.5

0.025 10 3

7.5 10 5 Ns/m2

Viscosity of fluid 7.5 10 5 Ns/m2


Problem 1.18: In a stream of oil in motion, the velocity gradient is

0.25 m/s/m. The mass density of fluid is 1270 kg/m3 and kinematic viscosity

is 6.3 10 4 m2/s. Find the shear stress at that point.

Given: Velocity gradient dudy

0.25 m/s/m; Density, 1270 kg/m3

Kinematic viscosity 6.3 10 4 m2/s

Solution:

Kinematic viscosity,

Dynamic viscosity (or) viscosity 1270 6.3 10 4

0.8001 Ns/m2

Shear Stress dudy

0.8001 0.25

0.2 N/m2

Problem 1.19: A plate 1 m2 in area and weighing 150 N slides an inclined

plane of angle sin 1 5/13 with horizontal with a velocity of 5 cm/s. If thethickness of oil film over the plane is 1.5 mm, estimate the viscosity of oil. (April 2006 - AU)(FAQ)

Given: A 1 m2; W 150 N; sin 1 5/13 22.62

u 5 cm/s 0.05 m/s ;

du u 0 0.05 m/s

du thickness 1.5 10 3 m

Solution:

Drag force F W sin

150 sin 22.62

57.692 N

Shear stress F

Area

57.6921

57.692 N/m2

dy

w

=22 .62 o

ws in


dudy

57.692 0.05

1.5 10 3

1.731 Ns/m2

Viscosity of oil 1.731 Ns/m2

Problem 1.20: A 90 mm diameter shaft rotates at 1200 rpm in a 100 mmlong journal bearing of 90.5 mm internal diameter. The annular space in thebearing is filled with oil having a dynamic viscosity of 0.12 Pa-s. Estimatethe power dissipated (Nov/Dec 2008 - AU) (FAQ)

Given: d 90 mm 90 10 3 m; d0 90.5 mm 90.5 10 3 m

dy d0 d

2

90.5 10 3 90 10 3

2 0.25 10 3 m

N 1200 RPM ; l 100 mm 0.1 m

0.12 Ns/m2

Solution

Tangential Velocity of the Shaft u

u dN60

90 10 3 1200

60 5.65 m/s

To Find Shear Stress :

dudy

0.12 5.65

0.25 10 3

2712 N/m2

[ . . . du u 0 5.65 m/s ]

90m m 90.5m m

l=100m m

dy=0.25mm


Shear Force: F

Shear force Shear stress Shear area

F [ dl ] 2712 [ 90 10 3 0.1 ]

F 76.68 N

Viscous Torque, T

T Shear force radius of shaft

F d2

76.68 90 10 3

2 3.451 Nm

Power Dissipated: P

P 2NT

60

2 1200 3.45160

433.67 Watts

Problem 1.21: Lateral stability of a long shaft 150 mm in diameter isobtained by means of a 250 mm stationary bearing having an internaldiameter of 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 be required toovercome the visious resistance when the shaft is rotated at a constant rateof 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 lubricant, 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

Problem 1.22: Determine the torque and power required to turn a 10 cmlong, 5 cm diameter shaft at 500 rpm in a 5.1 cm diameter concentric bearingflooded with a lubricating oil of viscosity 100 centipoise.

Given: l 10 cm 0.1 m ; d 5 cm 0.05 m ; N 500 rpm d0 5.1 cm 0.051 m

dy d0 d

2

0.051 0.052

5 10 4 m 0.0005 m

Viscosity of oil

100 cP 100 1

1000 Ns/m2 0.1 Ns/m2

Solution:

To Find Tangential Velocity u

u dN60

0.05 500

60 1.309 m/s

To Find Shear Stress

du u 0 1.309 0

1.309 m/s

dudy

0.1 1.309

5

10 4 261.8 N/m2

l = 0 .1m

dy=

d=0.051md=

0.05m

0.05cm=0.0005m


Shear force F Shearing Area 261.8 [ dl ]

261.8 [ 0.05 0.1 ] 4.1123 N

To Find Torque T and Power P

Viscous Torque T F d2

4.1123 0.05

2 0.103 Nm

Power 2 NT

60

2 500 0.10360

5.393 watts

Problem 1.23: The dynamic viscosity of an oil used for lubrication betweena shaft and sleeve is 6 poise. The shaft is of diameter 0.4 m and rotates at190 rpm. 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

Problem 1.24: An oil of viscosity 5 poise is used for lubrication betweena shaft and sleeve. The diameter of shaft is 0.5 m and rotates at 250 rpm.Find the power lost in oil for a sleeve length of 100 mm. The thickness ofoil film is to be 1.0 mm. (Oct 2004 - AU) (FAQ)

Given: 5 poise 0.5 Ns/m2 [. . . 1 poise 0.1 Ns/m2]

d 0.5 m ; N 250 rpm ; l 100 mm 0.1 m

dy 1 mm 0.001 m; Tangential Velocity, u

u dN60

0.5 250

60 6.545 m/s

du u 0 6.545 m/s

Solution:

Shear stress dudy

0.5 6.5450.001

3272.5 N/m2

Shear force F A 3272.5 [ 0.5 0.1 ] [. . . Shear Area dl]

F 514.042 N

Viscous torque T F d2

514.042 0.52

128.51 N

Power dissipated P 2NT

60

2 250 128.5160

3364.39 Watts

3.364 kW


Problem 1.25: A 0.5 m shaft rotates in a sleeve under lubrication withviscosity 5 Poise at 200 rpm. Calculate the power lost for a length of 100mm if 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

Problem 1.26: Determine the viscous drag torque and power absorbed onone surface of a collar bearing of 0.2 m ID and 0.3 m OD with an oil filmthickness of 1 mm and a viscosity of 30 centi poise if it rotates at 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

0.03 Ns/m2

Drag Torque T

60t 2 N

R14 R2

4

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

Problem 1.27: A liquid of unknown viscosity is filled in the space betweentwo coaxial cylinders 120 mm and 125 mm dia and 300 mm high. When theouter cylinder is rotating at 100 rpm, a torque of 1.5 N-m is experienced bythe inner cylinder. What is the viscosity of the liquid?

Given:Diameter of inner cylinder d 120 mm 0.12 m

Diameter of outer rotating cylinder

D 125 mm 0.125 mm

N 100 rpm ; l 300 mm 0.3 m

Solution:

Torque T 1.5 Nm

Torque T F d2

1.5 F 0.12

2

F 25 N

Shear Stress FA

25dl

[ ... Shearing area dl]

25

0.12 0.3

221.05 N/m2

120mm

125mm

innercylinder

Outerrotatingcylinder

300 m m

2.5m m liquid


To Find Tangential Velocity udu u 0 u

u DN60

0.125 100

60 0.6545 m/s

du 0.6545 m/s [ . . . Outer cylinder is rotating ]

To Find Viscosity

dudy

; dy D d

2

0.125 0.122

0.0025 m

221.05 0.65450.0025

Viscosity, 0.8443 Ns/m2

Problem 1.28: A cylinder shaft of 90 mm rotates about a vertical axis insidea cylinderical tube of length 50 cm and 95 mm internal diameter. If the spacebetween them is filled with oil of viscosity of 2 poises, find the power lostin friction for a shaft speed of 240 rpm.

Given:

d 90 mm 0.09 m l 50 cm 0.5 m

D 95 mm 0.095 m

dy D d

2

0.095 0.092

2.5 10 3 m

2 poises 0.2 Ns/m2

N 240 rpm

Tangential Velocity

u dN60

0.09 240

60 1.131 m/s

du u 0 1.131 m/s

Solution

Shear stress dudy

0.2 1.131

2.5 10 3 90.5 N/m2

50cm

2 .5m m

90m m

95m m


Shear force shearing area

F dl 90.5 0.09 0.5

F 12.791 N

Power lost Force Velocity F u

P 12.791 1.131 14.467 Watts

Problem 1.29: A 15 cm diameter vertical cylinder rotates concentricallyinside another 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 11.77 N-mis required to rotate the inner cylinder of 100 rpm, determine the viscosityof 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

0 .05cm

15cm

15.1cm

25cm


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

Problem 1.30: 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

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


Problem 1.31: A liquid with a volume of 0.2 m3 at 3000 kPa is subjectedto a pressure of 3000 kPa and its volume is found to decrease by 0.2 %.Calculate the bulk modulus of elasticity of the liquid.

(April 2004 - AU)(FAQ)

Given: V 0.2 m3; P1 300 kPa ; P2 3000 kPa

Decrease of volume dVV

0.2100

0.002

Solution:

Increase in pressure dP 3000 300 2700 kPa

K dPdVV

27000.002

1.35 MN/m2

Bulk modulus K 1.35 MN/m2

Problem 1.32: Estimate the pressure inside a water droplet of 0.5 mmdiameter. Assume 0.073 N/m. (Oct 2006 - AU)

Given: d 0.5 mm 0.5 10 3 m; Surface tension 0.073 N/m

Solution:

Pressure inside the dropletabove atmospheric pressure

Excess pressure p

For Water droplet,

p 4 d

4 0.073

0.5 10 3 584 N/m2

Problem 1.33: A soap bubble 50 mm diameter has inside pressure of

20 N/m2 above atmosphere. Calculate the tension in soap film.

Given: Dia of bubble d 50 mm 0.05 m;

Inside pressure in excess of atmosphere 20 N/m2


Solution:

To Find Surface Tension

We know p 8 d

or

4 r

20 8 0.05

0.125 N/m

Problem 1.34: A soap bubble 25 mm in diameter has inside pressure of

20 N/m2 above atmosphere. Calculate the tension in soap film.

Given: d 25 mm 0.025 m; P 20 N/m2

Solution:

We know that pressure p 8 d

20 8 0.025

0.0625 N/m

Tension (surface tension) in the film 0.0625 N/m.

Problem 1.35: Calculate the gauge pressure and absolute pressure within(i) a droplet of water 0.4 cm in diameter and (ii) a jet of water 0.4 cm indiameter. Assume the surface tension of water as 0.03 N/m and the

atmospheric pressure as 101.3 kN/m2.

Given: Surface Tension 0.03 N/m ; Atmospheric pressure p 101.3 kN/m2

101.3 103 N/m2

Solution:

(i) Droplet of (water of) dia d 0.4 cm 0.4 10 2 m

Excess pressure P

[(or) pressure above atmosphere (or) gauge pressure].


[Note: The pressure gauge will show the excess pressure of the fluid

above the atmosphere. And it shows zero reading (zero pressure) when it is

exposed to atmosphere]

For droplet, P 4 d

4 0.03

0.4 10 2 30 N/m2

So Gauge pressure 30 N/m2

Absolute pressure inside droplet Gauge pressure Atmospheric pressure

30 101.3 103 101.33 103 N/m2

(ii) Jet of water with dia d 0.4 cm 0.4 10 2 m

For jet of water, P 2d

2 0.03

0.4 10 2 15 N/m2

Gauge pressure 15 N/m2

Absolute pressureinside jet of water

Gauge pressure Atmospheric pressure

15 101.3 103 101.315 103 N/m2

Problem 1.36: Find the excess pressure inside a soap bubble of 30 mmdiameter, if the surface tension of water is 0.074 N/m.

Given: d 30 mm 0.03 m; Surface tension 0.074 N/m;

d 0.03 m ; 0.074 N/m

Solution:

For soap bubble, Excess pressure P 8d

8 0.074

0.03 19.733 N/m2

Problem 1.37: The surface tension of water in contact with air is given as

0.07 N/m. The pressure inside a droplet of water is 0.015 N/cm2 greater thanoutside pressure. Find the diameter of the droplet of water.

Given:

Surface tension 0.07 N/m; Excess pressure P 0.015 N/cm2


0.015

10 4 150 N/m2

0.07 N/m

Solution:

For droplet of water, P 4d

150 4 0.07

d

Diameter of droplet d 1.867 10 3 m 1.867 mm

Problem 1.38: If the surface tension of water in contact with air is 0.075N/m, what correction need to be applied towards capillary rise in themanometric reading in tube of 3 mm diameter.

Given: Surface tension 0.075 N/m; Dia of tube d 3 mm 3 10 3 m.

Solution:

Height of capillary rise h 4 cos

wd

d Dia of capillary tube; angle of contact of the liquid surface

w Specific weight of liquid; For water and glass 0

So h 4wd

4 0.075

9810 3 10 3 0.0102 m 10.2 mm

There will be a capillary rise of 10.2 mm. So for any reading in the

manometer should be corrected by subtracting 10.2 mm from the manometer

reading value.

Problem 1.39: A Capillary tube having inside diameter 6 mm is dipped inCCl4 at 20C. Find the rise of CCl4 in the tube if surface tension is 2.67

N/m and Specific gravity is 1.594 and contact angle is 60 and specific

weight of water at 20C is 9981 N/m3 (Nov/Dec 2008 - AU) (FAQ)

Given:

Dia. of tube, d 6 mm 0.006 m; Surface tension, 2.67 N/m


Specific gravity of CCl4 1.594; Contact angle 60

Specific weight of water 9981 N/m3; Capillary rise of CCl4 ?

Solution:

Specific weight of CCl4 sp. gravity sp. wt., of water

1.594 9981 15909.714 N/m3

w 15909.714 N/m3

Capillary rise h 4 cos

wd

4 2.67 cos 6015909.714 0.006

0.05594 m

55.94 mm

Problem 1.40: A hollow cylinder of 150 mm OD with its weight equal tothe buoyant forces is to be kept floating vertically in a liquid with a surface

tension of 0.45 N/m2. The contact angle is 60. Determine the additional forcerequired 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

Problem 1.41: Calculate the capillary rise in a glass tube of 2.5 mm diameterwhen immersed vertically in (a) water and (b) mercury. Take surfacetension 0.0725 N/m for water and 0.52 N/m for mercury in contactwith air. The specific gravity for mercury is given as 13.6 and angle ofcontact 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:


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


wd

4 0.52 cos 13013.6 9810 2.5 10 3

h 4.008 10 3 m

Problem 1.42: A 1.9 mm diameter tube is inserted in to an unknown liquid

whose density is 960 kg/m3 and it is observed that the liquid rises 5 mm inthe tube, making a contact angle of 15. Determine the surface tension ofthe liquid. (April 2008 - AU)(FAQ)

Given: Dia. of tube, d 1.9 mm 0.0019 [m}, Density, 960 kg / m3 ;

Capillary rise, h 5 mm 0.005 m ; Contact angle 15 Surface tension, ?

Solution:

Capillary rise, h 4 cos gd


Surface tension, h gd

4cos

0.005 960 9.81 0.0019

4 cos 15

0.0232 N/m

Problem 1.43: Derive an expression for the capillary rise at a liquid in acapillary tube of radius r having surface tension and contact angle . Ifthe plates 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 Page1.22/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

Problem 1.44: Find the height through which water rises by capillary actionin a glass tube of 2 mm bore if the surface tension at the prevailing temp.is 0.075 g/cm.

Given: Diameter d 2 mm 2 10 3 m

Surface tension 0.075 g/cm 0.075 10 3

10 2 kg/m 0.0075 kg/m

To make it in S.I. Unit,1 kgf 9.81 N; So 0.0075 kg/m 9.81 0.0736 N/m

Solution:


wd

4wd

[ . . . 0 for water and glass]

h 4 0.0736

9810 2 10 3 0.015 m 15 mm


Problem 1.45: A U-tube is made of two capillaries of diameter 1.0 mm and1.5 mm respectively. The tube is kept vertically and partially filled with waterof surface tension 0.0736 N/m and zero contact angle. Calculate the differencein 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

Problem 1.46: Calculate the capillary rise in a glass tube of 4.0 mmdiameter when immersed in water. The surface tension of water in contactwith air is 0.075 N/m.

Given: Diameter d 4 mm 4 10 3 m; Surface tension 0.075 N/m

Solution: Capillary rise h 4 cos

wd [ 0 for water and glass]

So, cos 1

h 4 0.075

9810 4 10 3 7.645 10 3 m 7.645 mm


Problem 1.47: A gas weighs 16 N/m3 at 25C and at an absolute pressure

of 30 N/cm2. Determine the gas constant and density of the gas.

Given: Weight of gas w 16 N/m3 ; Temperature T 25 273 298 K

Pressure P 30 N/cm2 30 104 N/m2

Solution:

To Find Density

wg

16

9.81 1.631 kg/m3

Characteristic equation

P v RT

P

RT . . . V

1

Characteristic gas constant R PT

30 104

1.631 298

R 617.24 J/kg K

Problem 1.48: A cylinder of 0.6 m3 in volume contains air at 50C and 3

bar absolute pressure. The air is compressed to 0.3 m3. Find (i) pressureinside the cylinder assuming isothermal process and (ii) pressure and temp.assuming isentropic process (reversible adiabatic process).

Given: Initial Volume V1 0.6 m3 ; T1 50 273 323 K ; Take 1.4

P1 3 105 N/m2

. . . 1 bar 105 N / m2Final Volume V2 0.3 m3

Solution:

(i) Isothermal process: P V constant

P1 V1 P2 V2

P2 P1 V1

V2

3 105 0.6

0.3 6 105 N/m2 6 bar


(ii) Isentropic Process:

P V constant

P1V 1 P2V 2

P2 P1 V1

V2

3 105 0.60.3

1.4

7.92 105 N/m2

Final Pressure P2 7.92 bar

From thermodynamic relations,

T2

T1 P2

P1

1

T2 T1 P2

P1

1

323 7.92 105

3 105

0.41.4

426.25

Final temp. T2 426.25 K

Problem 1.49: Calculate the pressure exerted by 10 kg of nitrogen gas at

a temperature of 10C if the volume is 0.4 m3. Molecular weight of nitrogenis 28. Assume ideal gas laws are applicable.

Given: m 10 kg ; T 10 273 283 K ; V 0.4 m3 ; M 28

Solution:

By using ideal gas law,

P Vm m R T

Molar volume Vm V M 0.4 28 11.2 kgmole

Universal gas constant R

8.4132 kJ/kgmol K

P m R

TV

m

10 8.4132 28311.2

2125.84 kN/m2

Pressure P 21.26 bar


1.6 FLUID STATICSThe branch of fluid mechanics that studies incompressible fluids at rest

is known as fluid statics or hydrostatics.

When the fluid velocity is zero (no shear stress), the pressure variation

is only due to the weight of the fluid. Such a condition is termed as the

hydrostatic condition. So, the pressure of a static fluid does not vary in the

x or y direction but varies only in the z - direction.

1.6.1 Concept of Fluid Static PressureConsider a small area dA in large mass of fluid. If the fluid is

stationary, then the force exerted by surrounding fluid on the area dA will

always be normal to the surface. If dF is the force acting on the area dA in

the normal direction, then the ratio dF/dA is known as the intensity of

pressure. Hence mathematically the pressure at a point in a fluid at rest is

pressure P dFdA

If the force F is uniformly distributed over the area A, then pressure

at any point is given by

P FA

forceArea

in N/m2

So, pressure is defined as the average force per unit area. 1 N/m2 is

known as pascal and is represented by Pa

1.6.2 Pressure of Fluids PPressure is the normal force

exerted by fluid against unit area of the

bounding surface. Pascal’s law states that

the pressure P at a point in a

incompressible fluid in equilibrium is

same in all directions. In otherwords,

when a certain pressure is applied at any

point in a fluid at rest, the pressure is

transmitted in all directions and to every

other points in the fluid. Since this fact

was established by Blaise Pascal, it is known as Pascal’s law.

P

P

PP

P

P

PP

Fig. 1.12 Hydro Static Pressure Acting Equality from all directions

at a Point in a Fluid at Rest


Unit for pressure is in N/m2

1 N/m2 1 pascal

1 bar 1 105 N m2 1 105 Pascal

1.6.3 Atmospheric Pressure

The atmospheric air exerts a pressure upon all surfaces with which it

is in contact, and this pressure is known as atmospheric pressure. Depending

on the elevation (height) from sea level, the atmospheric pressure varies. Since

atmospheric pressure is measured by barometer, it is called barometricpressure.

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.

1. Absolute zero pressure (complete vacuum)

2. Atmospheric pressure.

1.6.4 Absolute zero Pressure (or) Absolute pressure

G augepressure

a t A

Locala tm ospheric

(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. 1.13 Relationship between absolute, gauge and vacuum pressures


Refer the Fig. 1.13

When pressure is measured above absolute zero, it is called as absolute

pressure.

1.6.5 Gauge Pressure

The pressure recorded by any pressure gauge is called gauge pressure.

Pressure gauge takes atmospheric pressure as datum and pressure is measured

either above (or) below atmospheric pressure. Practically any pressure gauge

reads zero when it is exposed to atmosphere and is used to measure the

difference between the fluid pressure and atmospheric pressure.

1.6.6 Vacuum Pressure

The pressure less than atmospheric pressure is called vacuum pressure

(or) negative pressure (or) suction pressure. This pressure is measured by

using pressure gauge called vacuum gauge.

Gauge pressures are positive if they are above atmospheric pressure,

and negative if they are below the atmospheric pressure.

Absolutepressure

Atmosphericpressure

Gauge pressure

...(1.19)

Absolutepressure

Atmosphericpressure

Vacuum pressure

..(1.20)


1.7 PRESSURE - DENSITY - HEIGHT RELATIONSHIPThe pressure at any given point

of a static fluid due to the force of

gravity is called hydrostatic pressure.

The hydrostatic law states that rate of

increase of pressure in a vertically

downward direction at a point in a static

fluid, is equal to the weight density of

the fluid at that point.

ie p

z g W

where

is the density of the fluid

g is the acceleration due to gravity

W is the weight density of the fluid

p

z rate of increase of pressure in vertical direction

Proof: Consider a small fluid element as shown in Fig. 1.14

Let A Cross-sectional area of element

Z Height of fluid element

p Pressure on face AB

Z Distance of fluid element from free surface.

The forces acting on the fluid element are:

1. Pressure force on AB p A and acting perpendicular to face AB in

the downward direction.

2.Pressure force on CD

p

p

Z Z

A, acting perpendicular to face

CD, in the upward direction.

3. Weight of fluid element = Density g Volume g A Z.

4. Pressure forces on surfaces BC and AD are equal and opposite.

Zpx A

A B

Z

CD

Z Ap

zp+

Fig. 1.14 Forces on a Flu id Element

Free su rfa ce o f f lu id


For equilibrium we have p A g A Z p

p

Z Z

A

p A p A p

Z Z A g A Z 0

p

z Z A g A Z 0

p

Z Z A g A Z or

p

Z g

p

Z g w

. . . g w ...(1.21)where w Weight density of fluid.

The above equation is called hydrostatic law.

By integrating the equation for liquids, we get

dp g d Z

or P1 P2 g Z1 Z2

Thus, the difference in pressure between two points in an

incompressible fluid at rest can be expressed in terms of the vertical distance

between the points. The pressure at free surface is the local atmospheric

pressure ie P2 P0 0, if we are measuring guage pressure,

P gh ...(1.22)

From the above equation it can be stated that the pressure, P at any

point on a fluid at rest, with a free surface exceeds that of the local

atmospheric pressure by an amount gh, where h is the vertical depth of the

point from the free surface, called the pressure head.

The above equation is a fundamental equation in Fluid Statics. It

defines the manner in which pressure varies with height or elevation and finds

many applications. Mainly it enables one to determine atmospheric pressures

at different elevations above the sea level and to determine pressure at various

depths of an ocean. The other application is in Manometry, which forms the

basis for pressure measuring instruments.


1.8 MANOMETRYManometry is the branch of fluid mechanics, which deals with the

measurement of pressures. We have seen that pressure is proportional to the

height of a column of fluid. Manometry exploits this condition for measuring

pressure using liquid columns in vertical or inclined tubes. The devices used

in this manner are called manometers. They work on the principle of

balancing the column of liquid whose pressure is to be determined by

same/another liquid column. Depending on the application, magnitude of

pressure and sensitivity requirement, the manometer can be selected. The most

commonly used manometric fluids are mercury, water, alcohol and kerosense.

Problem 1.50: A hydraulic press has a plunger of 5 cm diameter and aram of 40 cm. What is the weight lifted by the hydraulic press when theforce applied at the plunger is 600 N?

Solution: Given:

Dia of ram, D 40 cm 0.4 m;

Dia of plunger, d 5 cm 0.05 m

Force of plunger, F 600 N

Area of ram,

Ar 4

D2 4

0.42 0.1256 m2

Area of plunger Ap 4

0.052 0.00196 m2

Pressure intensity due to plunger Force on plungerArea of plunger

FAp

600

0.00196 N/m2

Due to Pascal’s law, the intensity of pressure will be equally

transmitted in all directions. Hence the pressure intensity of the ram.

600

0.00196 306122.4 N/m2

But pressure intensity at ram Weight

Area of ram

WAr

W

0.1256 N/m2

W0.1256

306122.4

Weight 38448.9 N 38.498 kN

RAM

PL

UN

GE

RW

F


Problem 1.51: A hydraulic press has ram of 30 cm diameter and a plungerof 4 cm diameter. It is used for lifting a weight of 40 kN. Find the forcerequired at the plunger.

Solution: Given: Dia of ram, D 30 cm 0.3 m

Area of ram, A 4

D2 4

.32 0.0706 m2

Dia. of plunger, d 4 cm 0.04 m

Area of plunger, a 4

.042 1.25 10 3 m2

Weight lifted, W 40 kN 40000 N

Pressure intensity developed due to plunger ForceArea

Fa

By Pascal’s Law, this pressure is transmitted equally in all directions

Hence pressure transmitted at the ram Fa

We know, Force acting on ram

Pressure intensity Area of ram

Fa

A F 0.0706

1.25 10 3 N

But force acting on ram Weight lifted 40000 N

40000 F 0.0706

1.25 10 3

Force required at the plunger,

F 40000 1.25 10 3

0.0706 708.2 N

Problem 1.52: Calculate the pressure due to a column of 0.4 m of (a)water, (b) an oil of sp. gr. 0.8 and (c) mercury of sp. gr. 13.6. Take density

of water, p 1000 kg/m3.


Solution: Given:

Height of liquid column, h 0.4 m

The pressure at any point in a liquid is given by the equation

P gh

(a) For water, 1000 kg/m3

P gh 1000 9.81 0.4 3924 N/m2

3924

104 N/cm2 0.3924 N/cm2

(b) For oil of sp. gr. 0.9,

We know that the density of a fluid is equal to specific gravity of

fluid multiplied by density of water.

Density of oil, 0 Sp. gr. of oil Density of water

0.9 1000 900 kg/m3

Now pressure, P 0 g h 900 9.81 0.4 3531.6 N/m2

0.3531 N/cm2

(c) For mercury, sp, gr. 13.6

Now, Density of mercury, s Specific gravity of mercury Density of water

13.6 1000 13600 kg/m3

P s g h 13600 9.81 0.4 53366.4 N/m2

5.3366 N/cm2

Problem 1.53: The pressure intensity at a point in a fluid is given

4.126 N/cm2. Find the corresponding height of fluid when the fluid is: (a)water, and (b) oil of sp. gr. 0.8

Solution: Given: Pressure intensity, P 4.126 N/cm2 4.126 104 N/m2


The corresponding height, h of the fluid is given by

h P

g

(a) For water, 1000 kg/m3

Height of column when water is used,

h P

g

4.126 104

1000 9.81 4.2 m of water.

(b) For oil, sp. gr. 0.8

Density of oil 0 0.8 1000 800 kg/m3

Height of column when oil is used,

h P

0 g

4.126 104

800 9.81 5.25 m of oil.

Problem 1.54: An oil of sp. gr. 0.8 is contained in a vessel. At a point theheight of oil is 50 m. Find the corresponding height of water at the point.

Solution: Given: Sp. gr. of oil, s 0.8; Height of oil, h0 50 m

Density of oil, 0 Sp. gr . of oil Density of water 0.8 1000 800 kg/m3

Intensity of pressure, P 0 g h0 800 9.81 50 N/m2

Corresponding height of water P

Density of water g

800 9.81 50

1000 9.81 40 m of water.

Problem 1.55: An open tank contains water upto a depth of 3 m and aboveit an oil of sp. gr. 0.8 for a depth of 1 m. Find the pressure intensity (i) atthe interface of the two liquids, and (ii) at the bottom of the tank.


Solution:

Given:

Height of water, h1 3 m

Height of oil, h2 1 m

Sp. gr. of oil, 0.8

Density of water, 1 1000 kg/m3

Density of oil,

2 Sp. gr. of oil Density of water 0.8 1000 800 kg/m3

Pressure intensity at any point is given by

P g h

(i) At interface, P 2 g 1.0 800 9.81 1.0

7848 N/m2 0.7848 N/cm2

(ii) At the bottom,

P 2 g h2 1 g h1 7848 1000 9.81 3.0

7848 29430 37278 N/m2 3.7278 N/cm2

Problem 1.56: The water level in a tank is 20 m above the ground. A hoseis connected to the bottom of the tank and the nozzle at the end of the hoseis pointed straight up. The tank is at sea level and the water surface is opento the atmosphere. In the line leading from the tank to the nozzle is a pump,which increases the pressure of water. If the water jet rises to a height of27 m from the ground, determine the minimum pressure rise supplied by thepump 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

1 .0

3 .0

O IL

WATE R


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

Problem 1.57: The diameters of a small piston and a large piston of ahydraulic jack are 4 cm and 10 cm respectively. A force of 100 N is appliedon the small piston. Find the load lifted by the large piston when:(a) the pistons are at the same level.(b) small piston is 30 cm above the large piston. Take water as the fluid.

Solution: Given: Dia of small piston, d 4 cm

Area of small piston, a 4

d2 4

42 12.566 cm2

Dia. of large piston, D 10 cm

Area of large piston, A 4

102 78.54 cm2

Force on small piston, F 100 N; Let the load lifted be W.

(a) When the pistons are at the same level

Pressure intensity on small piston

P 1

P2Pn

Nozzle

27m = h2

( P )p

Pum p

20m

= h

1


Fa

100

12.566 N/cm2

By pascal’s law, this pressure

intensity is transmitted equally on the

large piston.

Pressure intensity on the large

piston 100

12.566

Force on the large piston Pressure Area

100

12.566 78.54 N 625 N

(b) When the small piston is 30 cm above the large piston

Pressure intensity on the small piston

Fa

100

12.566 N/cm2

Pressure intensity at section A A

Fa

Pressure intensity due to

height of 30 cm of liquid.

But pressure intensity due to 30

cm of liquid

g h 1000 9.81 0.3 N/m2

. . .Density of water 1000 kg/m3

1000 9.81 0.3

104 N/m2 0.2943 N/m2

Pressure intensity at section A A 100

12.566 0.2943 8.25 N/m2

Pressure intensity transmitted to the large piston 8.25 N/m2

Force on the large piston = Pressure Area of the large piston

8.25 78.54 647.95 N

W LAR G EPIS TO N

SM

ALL

PIS

TO

N

F

0 .5

1 .5

W

F

30 cmA A

0.5

1 .5

1


1.9 MEASUREMENT OF PRESSUREThe pressure in a fluid is measured by the following devices.

1. Manometers 2. Mechanical Gauges

Manometers

Manometers are defined as the devices used for measuring the pressure

at a point in a fluid by balancing the column of fluid by the same or another

column of fluid.

Manometers are classified as:

(a) Simple Manometers; (b) Differential Manometers

Simple manometers are used to measure pressure at a point in a fluid flowing

through pipe (or) contained in vessel.

Differential manometers are used to measure the pressure differencebetween any two points in a fluid flowing through pipe or contained in a

vessel.

Mechanical Gauges

Mechanical gauges are devices used for measuring the pressure by

balancing the fluid column by the spring or dead weight. The commonly used

mechanical pressure gauges are

(a) Diaphragm pressure gauge

(b) Bourdon tube pressure gauge

(c) Dead-weight pressure gauge

(d) Bellows pressure gauge

1.9.1 Pressure Measurement Methods

Pressure can be measured by the following methods

1. Elastic pressure transducers: Bourdon tube pressure gauge (C-type,

Helical type, Spiral type), Diaphragm pressure transducers, Bellows.

2. Electric pressure transducers: Strain gauge type, potentiometer type

(resistance type), capacitance type etc.,

3. Manometer method


1.9.1.1 Bourdon gauge (C-Type)

Principle

The Bourdon pressure gauge uses the principle that a flattened tube

tends to change to a more circular cross-section when pressurized. Although

this change in cross-section may be hardly noticeable, the displacement of

the material of the tube is magnified by forming the tube into a C shape or

even a helix, such that the entire tube tends to straighten out or uncoil,

elastically, as it is pressurized as shown in the Fig. 1.15.

Mechanism and working

In practice, a flattened thin-wall, closed-end tube is connected at the

hollow end to a fixed pipe containing the fluid pressure to be measured. As

the pressure increases, the closed end moves in an arc, and this motion is

converted into the rotation of a (segment of a) gear by a connecting link

which is usually adjustable. A small diameter pinion gear is on the pointer

Section A.A

Geared sectorand pinion

Pressure

Socket

Bourdontube

AA

Poin ter

0

Scale

100

Tip(Sealed end)

Link

Fig. 1.15 M echanism of the Bourdon Gauge


shaft, so that the motion is magnified further by the gear ratio. The positioning

of the indicator card behind the pointer, the initial pointer shaft position, the

linkage length and initial position all provide means to calibrate the pointer

to indicate the desired range of pressure for variations in the behavior of the

Bourdon tube itself.

When the measured pressure is rapidly pulsing, such as when the gauge

is near a reciprocating pump, an orifice restriction in the connecting pipe is

frequently used to avoid unnecessary wear on the gears and provide an

average reading.

When the whole gauge is subject to mechanical vibration, the entire case

including the pointer and indicator card can be filled with an oil or glycerin.

Typical high-quality modern gauges provide an accuracy of 2% of span, and

a special high-precision gauge can be as accurate as 0.1% of full scale.

Other types of Bourdon gauges

The other types of Bourdon gauges are Spiral type (Fig. 1.16.(a)) and

Helical Type (Fig. 1.16.(b)). The principle of operation of spiral and Helical

type is same as that of the C-type Bourdon gauge. When the tube is

pressurized the tube gets deflected proportionally to the change in pressure

and the same is indicated by the pointer attached at the end.

(a)

P

T

P

(b)

T

Fig. 1.16 (a) Spiral type (b) Helical type


1.9.1.2 Diaphragm-type pressure gauge Principle

An elastic steel diaphragm usually is designed so that the

deflection-versus-pressure characteristics are as linear as possible over a

specified pressure range, and with a minimum of hysteresis and minimum

shift in the zero point.

When the diaphragm is subjected to pressure the diaphragm deflects

linearly and this deflection is magnified by mechanical linkages to indicate

the pressure as shown in Fig. 1.17(a) and (b).

M otion

Pressure

D iaphragm(corrugated

type)

M otion

Pressure

D iaphragm(Flat)

(a)

Fig 1.15 (a) flat diaphragm ; (b) corrugated diaphragm

PivotUnderrange

S top

D iaphragmcapsule

Overrangestop

Pressure orvacuum

Fig 1.15 (c) Use of capsule elem ent in pressure gauge

(b)

Fig 1.17

Fig 1.17

Fig. 1.17 (a) Flat diaphragam; (b) Corrugated diaphragm


Working

Fig. 1.17(c) shows the use of capsule element in pressure gauge. To

amplify the motion that a diaphragm capsule produces, several capsules are

connected end to end. Diaphragm type pressure gauges are used to measure

gauge, absolute, or differential pressure. They are normally used to measure

low pressures of 25 mm of Hg, but they can also be manufactured to measure

higher pressures in the range of 0 to 7 kPa. They can also be built for use

in vacuum service.

The material of the diaphragm has the following properties.

1. They are enough flexible to provide required sensitivity of elastic

transducer.

2. Physical properties of their material are comparable with load and their

natural frequency is high enough to provide good frequency response.

1.9.1.3 Bellows

Principle

The device consists of a precision potentiometer whose wiper alarm is

mechanically linked to bourdon tube or bellow. The movement of wiper alarm

across the potentiometer converts the mechanically detected sensor deflection

into a resistance measurement using a Wheatstone bridge circuit.

Features of Bellows

Made of Bronze, Stainless steel, Beryllium Copper, Monel etc.,

The movement is proportional to number of convolutions.

Sensitivity is proportional to size.

In general a bellows can detect a slightly lower pressure than a

diaphragm

The range is from 0-5 mmHg to 14 MPa

Accuracy in the range of 1% span.

Fig.1.18(b) shows a Bellows or a bourdon tube with a variable

resistor. Bellow expands or contract causing the attached slider to move along

the slidewire. This increase or decreases the resistance. Thus indicating an

increase or decrease in pressure.


1.9.1.4 Dead Weight Pressure Gauge

Fig shows schematic of dead weight pressure gauge. It is generally

used for calibrating pressure gauges and is also used for producing and

measuring pressures.

SpringLead w ires

Sliders lide w ire

Be llows

Pressure

Fig 1.18 (a) Simple bellow pressure gauge (b) Bellows resistance transducer

Needle

Be llows

Input p ressure

(a)

(b)

Pressu re gauge

W eigh ts

P lunger Screw

P iston

O ilValve

Fig 1.19 D ead weight pressure gauge


Working: Initially piston and dead weights are removed and plunger is at

the lower most end. Clean oil is poured through the opening of piston and

plunger is moved slowly upward and air gaps are removed. The piston is

fitted and pressure gauge to be calibrated is screwed. When the valve is

opened the pressure is transmitted to the gauge. The pressure is varied by

varying the weights as the plunger. The pressure exerted is calculated by

knowing the weights.

Pressure exerted P Weight on plunger

Area of plunger

W

D2

4

Where D Diameter of Plunger.

1.9.1.5 Capacitive Pressure Transducer

A capacitive pressure transducer consists of a pair of electrically

insulative elastic diaphragms disposed adjacent to each other and bonded

together in a spaced apart relationship to form a sealed cavity. A conductive

layer applied to the inside surface of each of the diaphragms and a small

absolute pressure provided in the cavity. This small absolute pressure cavity

essentially reduces the effects of the negative temperature coefficient of the

modulus of elasticity of the diaphragms. (Fig.1.20)

D iaphragmInsu la ted s tand offs

P ressu rebellow s Pressu re

port

C apa cito rp la tes

Fig. 1.20 Capacitor Type Pressure Transducer


The sensing diaphragm and capacitor form a differential variable

separation capacitor. When the two input pressures are equal the diaphragm

is positioned centrally and the capacitance are equal. A difference in the two

input pressure causes displacement of the sensing diaphragm and is sensed

as a difference between the two capacitances.

1.9.1.6 Strain Gauge Pressure TransducerA strain gauge is a passive type resistance pressure transducer whose

electrical resistance changes when it is stretched or compressed. The wire

filament is attached to a structure under strain and the resistance in the

strained wire is measured. A pressure transducer contains a diaphragm which

is deformed by the pressure which can cause a strain gauge to stretch or

compress. This deformation of the strain gauge causes the variation in length

and cross sectional area due to which its resistance changes.

Fixed resistor

Strain gaugee lements

Strain gaugee lements

Fixed poin ts

Pressure M ovableb lock

Diaphragm

Wire resistance strain gauge

Pressure

Strain gauges

Diaphragm

Double bonded strain gauge

Fig. 1.21 Strain Gauge Type Pressure Transducer


1.10 SIMPLE MANOMETERSA simple manometer consists of a glass tube having one of its ends

connected to a point where pressure is to be measured and other end remains

open to atmosphere. Common type of simple manometers are

(i) Piezometer (ii) U-tube Manometer (iii) Single column Manometer

(i) Piezometer:

It is the simplest form of manometer used

for measuring gauge pressure as shown in Fig.1.22.

If at a point A, the height of liquid say

water is h in piezometer tube, then pressure at

A is given by

PA g h

: Density of liquid in kg/m3

g : Acceleration due to gravity

(ii) Simple U tube manometer

A U tube manometer consists of a glass tube bent in U-shape as

shown in Fig 1.23. One end (left limb) is connected to the pipe and other

end (right limb) is open to atmosphere. This U tube is filled up with mercury,

since it is heavier than water and it will not mix with water.

The high pressure liquid in the pipe pushes the mercury down in the

left limb. So the mercury level in the right limb rises up.

The liquid in the pipe and mercury meets at point A.

The meeting point A is taken on datum line ZZ

Pressure head at A Pressure head of liquid in pipe pressure headdue to h1 of liquid above datum line.

hp h1 s1 in m of water

Where s1 Specific gravity of liquid in pipe

h

A

Piezometer

Fig. 1.22


Pressure head at B Atmosphere pressure head Pressure head dueto mercury head h2 above datum line

10.3 h2 sm in m of water

[ . . . Atmosphere 10.3 m of water and

sm specific gravity of mercury 13.6 ]

Under equilibrium at datum line,

Pressure head at A pressure head at B.

hp h1 s1 10.3 h2 sm

To find gauge pressure, we can neglect the atmosphere pressure (10.3).

So hp h1 s1 h2 sm

Pressure of liquid in pipe

hp h2 sm h1 s1 in m of water . ...(1.39)

Once we know hp we can convert the pressure head into pressure by

the following relation

P whp where w specific wt. of water 9810 N/m3

S 1

h ‘m ’ o f wa terp

z z

h1

h 2

A B

sm

F ig. 1.23 ‘U’ - tube Manom eter


(iii) Single column Manometer

Single column manometer is modified form of U tube Manometer

having a very large reservoir. There are two types of single column

manometer.

(a) Vertical single column Manometer

(b) Inclined single column Manometer.

(a) Vertical single column Manometer

Let X X be datum line in the reservoir and in right limb of the

manometer, when it is not connected to the pipe. When manometer is

connected to the pipe, due to pressure at A, the heavy liquid is paused

downward and will rise in right limb as shown in Fig.1.24.

If H : Fall of heavy liquid in reservoir

h2 : Rise of heavy liquid in right limb

h1 : Height of centre of pipe above X X

PA : Pressure at A

A : Cross sectional area of reservoir

a : Cross sectional area of right limb

s1 : Sp gr of liquid in pipe

s2 : Sp gr of heavy liquid.

1 : Density of liquid

2 : Density of heavy liquid

Then, PA a h2

A [2g 1g] h22g h11g

Here aA

is very small so

PA h2 2g h11g ...(1.40)

h

Ah2

Fig. 1.24 Vertical single colum n M anom eter

X X

Y Y


(b) Inclined Single Column Manometer

If the right limb is inclined at angle with horizontal instead of vertical as shown in

Fig.1.25.

L : Length of heavy liquid moved in rightlimb from X X

: Inclination of right limb

h2 L sin

Pressure at A

PA h22g h11g

PA L sin 2g h11g [. . . h2 L sin] ...(1.41)

1.10.1 Differential Manometer

Differential manometer consists of glass tube bent in U-shape. Two

ends of the U tube are connected to the two different points in a pipe between

which pressure difference is to be measured. Refer the Fig. 1.26.

F ig. 1.26 U - tube d ifferentia l m anom eters

h1

h1h3

h3

zz

x

A

A

B

B

z z

Two p ipes at different levels. Two p ipes at sam e levels.

hA h AhB

hB

sm sm

D

D

C

C

x

h1

h

A

Fig. 1.25 Inclined single column Manometer

X X

Y Y

X

Y

L h2


Assume that pressure of liquid in pipe A has more pressure than that

of pipe B. High pressure liquid in pipe A pushes the mercury in the left limb

downward. Hence, the mercury level in right side limb rises up.

Liquid in pipe A and mercury meet at point C. This meeting point is

taken as datum Z Z . x Difference of mercury level in U tube.

Pressure head at C

pressure head of liquid in

pipe A pressure head due to h1

of liquid in pipe A above datum

hA h1 s1 in m of water

Pressure head at D

Pressure head of liquid in pipeB Pressure head due to heightx of mercury pressure headdue to height h3 of liquid inpipe B above D

hB x sm h3 s3 in m of water

Under Equilibrium

Pressure head at C Pressure head at D.

hA h1 s1 hB x sm h3 s3

Pressure head difference hA hB x sm h3 s3 h1 s1 in m of water.

If both pipes A and B are at same level and contain same liquid, then

h1 x h3 and s1 s3

ie h3 h1 x


Now the equation (1) becomes,

Pressure headdifference

hA hB xsm h1 x s1 h1 s1

xsm h1 s1 xs1 h1 s1

x sm s1 in m of water.

So hA hB x sm s1 in m of water ...(1.42)

where x difference of mercury level in U tube and s1 specific

gravity of liquid in pipe.

If water is flowing in pipe line, then s1 1

Then hA hB x 13.6 1

hA hB 12.6 x ...(1.43)[sm specific gravity of mercury 13.6]

To Find Pressure,

We can use the equation P wh

Where w specific weight of water

PA whA and PB whB

PA PB w hA hB

PROBLEMS IN SIMPLE MANOMETER

Problem 1.58: The water pressure is measured by means of simplemanometer. The mercury level difference is 150 mm as shown in fig. Theheight of water in the left tube is 40 mm. Determine the static pressure inthe pipe.

Given: h1 40 mm 0.04 m; s1 1; h2 150 mm 0.15 m; s2 sm 13.6

Solution:

Pressure Head at A Pressure head in pipe h1 s1

hp h1 s1 hp 0.04 1 in m of water.


Pressure Head at B h2s2

Under equilibrium,Pressure head at A Pressure head at B

hp 0.04 1 h2 s2

hp 0.15 13.6 0.04 1

2 m of water head

Pressure in pipe whp 9810 2 19620 N/m2

Problem 1.59: A simple manometer is used to measure the pressure of oilof specific gravity 0.8 flowing in a pipe. The right limb is open to atmosphereand left limb is connected to pipe. The centre of the pipe is 9 cm below thelevel of mercury in the right limb. If the difference of mercury level in the twolimbs is 15 cm, find the pressure of oil in the pipe.

Given: Specific gravity of oil in pipe sp 0.8 ; Specific gravity of mercury

in tube sm 13.6

Height of oil in left limb h1 15 9 6 cm 0.06 m

Difference of mercury level x 0.15 m

h =40 m m1

h =15 0mm2

z

A B

z

h-’m ’ o f w ate r

s1

sm


Solution:

To find Gauge pressure (neglecting

atmospheric pressure)

Pressure head above xx in left limb =

Pressure head above xx in right limb.

hp Pressure head of oil in pipe

hp h1 sp x sm

hp x sm h1sp

hpipe oil 0.15 13.6 0.06 0.8

1.992 of water.

The pressure P is given by

P w hp 9810 1.992 19541.5 N/m2

19.542 kN/m2

Absolute pressure of oil in pipe.

Pabs Patm Pgauge 1.01325 105 19541.5

120866.5 N/m2 120.87 kN/m2

[atmospheric pressure 1.01325 105 N/m2 ]

Problem 1.60: Consider a double fluid manometer attached to an air pipeshown in fig. If the specific gravity of one fluid is 13.55, determine the specificgravity of the other fluid for the indicated absolute pressure of air. Take theatmospheric pressure to be 100 kpa. (April/May 2008 - AU)(FAQ)

Given: Air pressure in the air pipe 76 kPa (Absolute)

h1 22 cm 0.22 m

SG1 13.55

h2 40 cm 0.4 m

P 76 kPa 76 103 Pa

Patm 100 kPa 100 103 Pa

L iqu id (s =s =0 .8)1 p

P ipe

h1

h 2

90m

m

150

mm

x x

M ercu ry (s =s =13 .6 )2 m


Solution:

Starting with the pressure at

point “A” at the air liquid (1) interface

and reaching point “B” where

atmosphere make contact with liquid

(2).

i.e P

w g h1 SG1

Patm

w g h2 SG2

76 103

1000 9.81 0.22 13.55

100 103

1000 9.81 0.4 SG2

10.7282 10.1937 0.4 SG2

Specific gravity of fluid,

SG2 10.7282 10.1937

0.4 1.34 Ans

Problem 1.61: U-tube manometer containing mercury is used to find thenegative pressure in the pipe. Water is flowing through the pipe. The rightlimb is open to atmosphere. The left limb is connected to pipe. The differenceof mercury level in the two limbs is 80 mm and height of water in the leftlimb from the centre of the pipe is found to be 40 mm below. Find thevacuum pressure in the pipe.

Given: Specific gravity of water in pipe sp 1 ; Specific gravity of mercury

sm 13.6

Height of water in left limb h1 40 mm 0.04 m

Difference of mercury levelormercury height in left limb

x 0.08 m

h =40cm2

SG 2

SG =13 .551

A


Solution:

Neglecting atmospheric pressure, under

equilibrium,

Pressure head in left limb above xx Pressure head in right limb above xx

hp h1 sp x sm 0

hp 0.04 1 0.08 13.6

1.128 m of water

Gauge pressure

wh 9810 1.128

11.066 kN/m2 or

11.066 kN/m2 vacuum

Absolute Pressure:

Pabs Patm Pgauge 1.01325 105 11066

90259 N/m2

PROBLEMS IN DIFFERENTIAL MANOMETER

Problem 1.62: A differential manometer is connected at two points A andB in a horizontal pipe line containing oil of specific gravity 0.8 and thedifference in mercury level is 125 mm. Determine the difference of pressureat the two points.

Given: Specific gravity of oil s1 0.8 ; Difference in mercury level

x 125 mm 0.125 m

Solution:

We know pressure head difference between A and B is given by

hA hB x sm s1 0.125 13.6 0.8 1.6 m of water

Pressure difference PA PB w hA hB 9810 1.6 15696 N/m2

[ . . . w specific wt of water 9810 N/m3 ]

Pipe� �

M ercury(s =s =13.6)2 m

Water(s =s =1.0)1 p

h=

40 m

m1

x=80

mm


Problem 1.63: 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 . The centre ofpipe A is 3 meters above centre of pipe B. The centre of the pipe B is atthe level of mercury in the left limb connecting the pipe A. Find the differenceof 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

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 ]

sm

h =3m1

h =h3 2

P = 120kN/mA2

P = 220kN/mB2

Pipe ‘A’

Pipe ‘B’

zz C D

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

Problem 1.64: A pipe containing water at 170 kN/m2 pressure is connectedby a differential gauge to another pipe 1.5 m lower than the first pipe andcontaining water at high pressure. If the difference of heights of two mercurycolumn of gauge is equal to 7.5 cm, find the pressure in the lower pipe. Takespecific gravity of mercury as 13.6. (Oct 2004 AU)(FAQ)

Given: Take lower pipe as datum PA 170 103 N/m2 ;

Difference in heights of pipe h1 1.5 m; x 7.5 cm 0.075 m ;

s1 sp specific gravity of water 1; sm specific gravity of mercury 13.6

hA PA

w

170 103

9.81 103 17.33 m

hB ?

Solution:

Height of left limb above datum line (above c)

Pressure head at datum c hA h1 s1 xsm

Height of right limb above datum line (above d)

Pressure head at datum d hB xs1

Under equilibrium,

Pressure head at c Pressure head at d

hA h1 s1 xsm hB xs1


17.33 1.5 1 0.075 13.6 hB 0.075 1

hB 19.775 m

Pressure in pipe B, PB whB 9.81 103 19.775

PB 193.993 103 N/m2

Alternate Method

Since both pipes are containing same liquid,

Difference of pressure head

h x smsp

1 0.075

13.6

1 1

0.945 m

Also, h PB

w ZB

High Pressure

PAw

ZA

Low Pressure

0.945

PB

9.81 103 0 170 103

9.81 103 1.5

PB 193.985 103 N/m2

Problem 1.65: A ‘U’ tube manometer is used to measure water in a pipelinewhich is in excess of atmosphere pressure. The right limb of the manometercontains mercury and is open to atmosphere. The contact between water andmercury is in the left limb. Calculate the pressure of water in the mainlineif the difference in level of mercury in the limbs is 10.5 cm and the freesurface 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 thenew 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.

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

A

W ater

B C

D

M ercury

10.5 cmLeft L im b

R ightLim b

A

W ater 10.5-�

�

B

B* C *

C

D *

D

(10.5 -2 ) �

M ercury

�


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

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

Problem 1.66: A pipe containing water at 180 kN/m2 pressure is aconnected by a differential gauge to another pipe 1.6 m lower than the firstpipe and containing water at high pressure. If the difference in heights of 2mercury columns of the gauge is equal to 90 mm, What is the pressure inthe lower pipe? (Nov/Dec 2008 - AU)(FAQ)

A

W ater 10 .5 -�

�

B

B* C *

C

D *

D

(10.5-2 ) �

Mercury

�


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

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

Problem 1.67: A pipe contains oil of specific gravity 0.8. A differentialmanometer connected at the two points A and B of the pipe shows a differencein mercury levels of 20 cm. Calculate the difference of pressure at two points.

Given: Specific gravity of oil sP 0.8; Specific gravity of mercury

sm 13.6; Difference in mercury level x 0.2 m

x

hA+

hB+


Solution:

Difference of pressure heads between A and B h

h x smsp

1 0.2

13.60.8

1 3.2 m of water.

Difference of pressure between A and B wh

9.81 103 3.2 31.392 kN/m2

Problem 1.68: The maximum blood pressure in the upper arm of a healthyperson is about 120 mm Hg. If a vertical tube open to atmosphere is connectedto the vein in the arm of the person, determine how high the blood will rise in

the tube? Take the density of the blood to be 1050 kg/m3.(April/May 2008 - AU) (FAQ)

Given: Blood pressure, PB 120 mm Hg 0.12 m of Hg ; Density of blood,

B

1050 kg/m3

Blood pressure in terms of column of water

0.12 13.6 1.632 m of water

(. . . pressure in height of liquid sp. gravity of liquid pressure

in height of water)

(or) Blood pressure, PB wh in . . . P wh

9810 1.1632 16009.92 N/m2

Height of Rise of blood in the tube open to atmosphere,

h Pw

Pg

16009.92

1050 9.81 1.554 m Ans

Alternate Method

PB hm sm hB sB

where PB pressure of blood

hm mercury column height 0.12 m given


sm specific gravity of mecury 13.6

hB height of blood rise in the tube

sB specific gravity of blood 10501000

1.05

hB hm sm

sB

0.12 13.61.05

1.5543 m Ans

1.11 FLUID KINEMATICSThe kinematics of fluid motion deals only with space-time relationships

(Velocity and Acceleration) without taking into account the forces associated

with them. The fluid motion can be described completely by an expression

describing the location of a fluid particle in space at different times, thus

enabling determination of the magnitude and direction of velocity and

acceleration in the flow field at any instant of time.

1.11.1 Concept of System:

A system or fluid system is a certain mass of fluid within a closed

surface. When fluid flows through a constriction (or) when fluid is

compressed, it is deformed. So the size and shape of the system is changing

with respect to time, i.e the fluid system moves and deforms.

1.11.2 Control Volume

v in CV

v out C V

Control su rface o f contro l vo lume, fixed in space

boundary

Volum e o f moving fluid system

Fig. 1.27 Control Volum e


A control volume is a fixed region in space which does not move (or)

change shape. Refer Fig. 1.27 the fluid flows into and out of this fixed region.

Its closed boundaries are called control surface.

Actually, the control surface may be in motion though space relative

to an absolute frame of reference.

Concept of Control VolumeA fluid dynamic system can be analyzed using a control volume which

is an imaginary surface enclosing a volume of interest. The control volume

can be fixed or moving.

Control volume approach is widely applied in analysis. An arbitrary

fixed volume located at a certain place in the flow-field is identified and the

conservation equations (refer kinematics and dynamics chapters) are written.

The surface which bounds the control volume is called the control surface.

In the control volume approach, the control surface is first defined

relative to a coordinate system that may be fixed, moving or rotating. Mass,

heat and work can cross the control surface and mass and properties can

change with time within the control volume. Choice of location and shape

of control volume are important for mathematical formulation.

Kinematics of Fluid Flow1.12 CONTINUUM & FREE MOLECULAR FLOWS

1.12.1 Continuum Flow

It is a concept of idealization of the continuous description of matter

where the properties of the matter are considered as continuous functions of

space variable.

1.12.2 Free Molecular Flow

This is also termed as regime of high vacuum. It describes the fluid

dynamics of gas where the mean free path of the molecules is larger than

the size of the chamber.


1.13 FLOW CHARACTERISTICSFlow characteristics are studied by using two methods namely

(i) Lagrangian method (ii) Eulerian method

In Lagrangian method, a single fluid particle is taken into

consideration and it is analysed and followed to study its properties like

velocity, acceleration and density etc.

It is just like following one vehicle by another vehicle to study its motion.

In Eulerian method, a particular space (or) region may be selected

and fluid flows through this region. During this time, the fluid characteristics

are analysed. This is just like the traffic police stands on place and observes

all the vehicles passing through his region. In fluid mechanics, Eulerian

method is commonly used.

1.14 TYPES OF FLUID FLOWSThe fluid flows are classified as follows:

1. Steady flow and unsteady flow.

2. Uniform flow and non-uniform flow.

3. Laminar flow and turbulent flow.

4. Compressible flow and Incompressible flow.

5. Rotational flow and Irrotational flow.

6. One dimensional flow, two dimensional flow and three dimensional flow.

7. Subsonic flow.

8. Sonic flow.

9. Supersonic flow.

10. Subcritical flow.

11. Critical flow.

12. Supercritical flow.

1.14.1 Steady Flow and Unsteady Flow

In 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 change with respect to time.

Mathematically, for unsteady flow,

V

t 0;

P

t 0;

t

0

1.14.2 Uniform and Non-Uniform FlowsWhen velocity remains constant at all points in the moving fluid, then

the flow is called Uniform flow. In uniform flow, the velocity does not change

with respect to ‘the distance travelled by the fluid’ at any time.

H =C onstan t H =C onstan t

V=C onstan t V=C onstan t

Steady Flow Unsteady F low

Fig. 1.28 Steady and Unsteady F low

Y

X

Y

X

Un ifo rm Flow Mean tha t the Ve loc ity is Constan t a t Certa in tim e inD iffe rent Positions (doesn’t D epend

on any D im ension x o r y or z)

Non-U n iform F low M eans Veloc ity Changes at Certain tim e in D ifferent Positions ( D epend on D im ension

x or y or z)

Fig. 1.29 Uniform and Non - 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

1.14.3 Laminar Flow and Turbulent Flow

Laminar 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. In laminar

flow,the fluid particles of one layer will not disturb another layer or that layer

itself.

In this flow, the fluid particles moves in a well defined path and all

layers are parallel to each other and also parallel to the pipe wall.

The flow is highly influenced by viscosity. So the velocity of fluid

particle in any layer is low.

Turbulent Flow

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

flow is called turbulent flow.

Fig. 1.30 Laminar flow


In this turbulent flow, the fluid particles do not have any definite path.

The fluid particles cross each other and have full of turbulence. There is no

layers in this flow. The velocity fluctuations are violent and erratic.

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)

1.14.4 Incompressible and Compressible Flow

Define incompressible fluid (Nov/Dec 2014 - AU)In Incompressible flow, the density of fluid remains constant

constant

Though liquids are slightly compressible, it is assumed to be

incompressible.

In compressible flow, the density does not remain constant. It varies

with pressure and temperature.

Fig. 1.31 Turbulent flow


For ideal gas, P

RT; P

RT

Gases are compressible, since their densities vary with P and T as per

the characteristic equation.

1.14.5 Rotational Flow and Irrotational Flow

When fluid particles move along

stream lines, they rotate about their own

axis. This type of flow is rotationalflow. In case of irrotational flow, when

fluid particles move along stream lines,

they do not rotate about their own axis.

Mach number (M)

It is a non-dimensional number

and is defined as

M2 Inertia forceElastic force

P1 P 2

P2

P1

V1 V 2

V2

V1

Incompressible Fluid Compressible Flu id

Fig. 1.32 Compressible & Incompressible Flu id

(a) Rotational Flow

(b) Irrotational Flow

Fig. 1.33 Rotational & Irrotational F low


1.14.6 Subsonic FlowA fluid flow is called as subsonic,

when the fluid velocity is lower than the

acoustic speed (sound velocity) (M < 1).

When mach number is in the range of 0

- 0.99, the flow is considered as subsonic.

1.14.7 Sonic Flow

A fluid flow is called as sonic

flow, when the mach number is 1.

If the fluid velocity is equal to

sound velocity, then it is sonic flow.

1.14.8 Supersonic Flow

A fluid flow is called as

supersonic flow, when the mach number

is more than 1. M 1

Super critical flow, sub critical

flow and critical flow, these all are

categories of open channel flow.

1.14.9 Subcritical flow

This type of flow is deep, slow flow with a low energy state and has

a Froude number less than one. F 1

1.14.10 Critical flow

This type of flow occurs when Froude number is equal to one F 1and there is a proper balance between the gravitational and inertia forces.

1.14.11 Supercritical flow

This type of flow is shallow, fast flow with a high energy state and

has Froude number greater than one F 1

Fig. 1.35 Sonic Flow Sonic M = 1

. . .

Fig. 1 .36 Supersonic F low

Superson ic M > 1

Fig. 1 .34 Subsonic Flow

Subson ic 0 < M < 1


1.15 ONE DIMENSIONAL FLOWIn general all fluids flow three-dimensionally, with velocities and other

flow properties varying in all directions. But in many cases the greatest

changes occur only in two directions or even one. In these cases change in

other directions can be effectively ignored making analysis simpler.

When the velocity is a function of time and one space coordinate only

(say x direction only), then it is called one dimensional flow.

For one dimensional flow,

u f x, v 0 and w 0 where u, v and w are velocity components in

x, y and z direction respectively.

Two-dimensional FlowIn this flow, velocity is a function of time and two rectangular space

coordinates (say x and y direction).

For two-dimensional flow,u f1 x, y

v f2 x, y

w 0

Three-dimensional FlowIn this flow, velocity is a function of time and three mutually

perpendicular axes.

For three dimensional flow,u f1 x, y z

v f2 x, y, z

w f3 x, y, z

1.16 FLOW VISUALIZATION - LINES OF FLOWPatterns of flow can be visualized in several ways.

A flow pattern may be described by

1. Stream lines

2. Stream tube

3. Path line

4. Streak line


1.16.1 Stream Line

A stream line is an imaginary curve drawn through a fluid. The tangent

of this curve at any point gives the direction of velocity of flow at that point.

The pattern of fluid is represented by a series of stream lines. In the Fig 1.38,

at point P, velocity V is tangential to the stream line passing through P. P is

defined by the coordinates x and y.

If u and v are the horizontal

and vertical component of

velocity V, then

tan vu

dydx

From the above,

dxu

dyv

or

udy vdx 0. ...(2.1)

This is the differential

equation for stream line in two dimensional flow.

For three dimensional flow,

dxu

dyv

dzw ...(2.2)

1.16.2 Stream TubeGroup of stream lines passing through a small closed curve form a

stream tube as shown in Fig 1.38.

Y

X

stream lines

p(x,y)

v

u

Vv

v

Fig. 1.37 Streamlines

Stream lines

Fig. 1.38 Stream tube


1.16.3 Path LineA Path line is a line traced

by a single particle of fluid over

a period of time. A path line

shows the direction of velocity of

the same fluid particle at

successive instants of time.

On the otherhand, stream line shows the direction of velocity of a

number of fluid particles at the same instant of time.

In steady flow, stream lines and path lines are identical, since velocity

does not change with respect to time.

1.16.4 Streak LinePath Line and Streak line are

imaginary lines in a fluid flow helping

to better understand the flow. Path Line

is the path followed by a particle in the

flow. Streak line is the locus of all the

particles that have gone through a given

point in the flow.

So, a streak line is associated with a particular point in space which

has the fluid moving past it. All particles which pass through the given point

(at some instant of time) are said to form the streak line of that point.

A dye or some colour pigment is injected into the flow to trace the

motion of the fluid particles. If the flow is laminar, then a ribbon of colour

results. Main ribbon is called streak line (or) filament line.

The smoke rising from a cigarette gives appearance of a streak line.

[But this smoke does not properly represent the movement of atmospheric air

because it is less dense than the air and therefore it rises more rapidly. So

this is not actual streak line of air.]

It is important to use a dye (or) colour pigment having same density

as that of fluid being observed to study the streak line.

In a steady flow, stream lines, path lines and streak lines all coincide.

Path Lines

Streak Line

T

X

Fig. 1.40 Path Lines and Streak Lines

Pa th line

OD

B

Fig. 1.39 Path Line

O ne p artic le is re le ase d, the p ath line sho ws how the p artic le m o ves in the flu id.


1.17 MEAN VELOCITY OF FLOWThe velocity of fluid layer

nearer to the pipe wall is zero or very

low. On the otherhand the velocity of

fluid at the centre of the pipe is

maximum. This effect is because of

viscosity. So for real fluid, the velocity

will increase rapidly across the flow

from the pipe wall to centre of the

pipe and it produces the velocity profile as shown in Fig. 1.41. For solving

problem, we take the average velocity over the cross sectional area of the

pipe. This average velocity Vmin Vmax

2 Vavg

is known as Mean velocity.

1.18 PRINCIPLES OF FLUID FLOW1. Principle of conservation of mass

2. Principle of conservation of energy

3. Principle of conservation of momentum.

1.18.1 Principle of Conservation of massIt states that mass can neither be created nor be destroyed but it can

be transformed from one form to another. This is the basis for the continuityequation.

1.18.1.1 Continuity Equation in one Dimension

Write the continuity equation. (AU Nov/Dec 2012)

O �

O �

O

vm ax

vm in

Typical Velocity ProfilesFig. 1.41

Real fluid

Directionof flow

In le tExit

11 22

Fig. 1.42 Continutity Flow


Refer the Fig. 1.42

m mass of flowing fluid in kg/s.

Mass of fluid at inlet mass of fluid at exit.

m

1 m

2

m

1 1 A1 V1 and m

2 2 A2 V2

1 A1 V1 2 A2 V2 continuity equation

1, 2 density of fluid at inlet and outlet.

A1 and A2 Area of pipe at inlet and exit respectively.

V1 and V2 velocity of fluid at inlet and exit respectively.

For incompressible fluid, the density of fluid remains constant, ie 1 2

So the equation 1 A1 V1 2 A2 V2 becomes A1 V1 A2 V2 Q

where Q rate of flow (or) discharge in m3 s

Quantity of fluid flowing through pipe in m3 s

Q AV is known as theoretical discharge Qt. The actual discharge Qa is

less than the theoritcal discharge, since there will be losses in actual cases.

The Coefficient of discharge Cd is known as the ratio of actual discharge

to theoretical discharge

Cd Qa

Qt.

Problem 1.69: A pipe 15 cm in diameter carries an unsteady flow of anincompressible fluid. The pipe terminates in a nozzle 7.5 cm in diameter.Determine the velocity and acceleration of the parallel stream leaving thenozzle at the instant when the velocity and acceleration in the pipe are

2m/s and 0.5m/s2.


Given: d1 15 cm 0.15 m ; Unsteady flow ; d2 7.5 cm 0.075 m;

V1 2 m/s ; a1 0.5 m/s2 dudt

Solution:

The acceleration along a stream line is

dudt

u

t u

u

s

Since the streamlines are straight and parallel

u

s 0

The acceleration is dudt

u

t

In pipe dudt

0.5 m/s2

We know the continuity equation

Q A1V1 A2V2

Velocity in the jet

V2 A1

A2 V1

d1

d2

2

V1 157.5

2

2 8 m/s

Also from continuity A1 dV1 A2 dV2

Dividing with dt

A1 dV1

dt A2

dV2

dt

dV2

dt

A1

A2 dV1

dt 157.5

2

0.5 2 m/s2

Acceleration in the jet at given instant 2 m/s2


Problem 1.70: The diameters of the pipe at sections (1) and (2) are 15 cmand 20 cm respectively. Find the discharge through the pipe if the velocityof water at section (1) is 4 m/s. Determine also the velocity at section (2)

Given:

D1 15 cm 0.15 m

D2 20 cm 0.2 m

V1 4 m/s; V2 ?

At section (1),

A1 4

D12

4

0.152 0.0176 m2

At section (2)

A2 4

D22

4

0.22 0.0314 m2

Discharge through pipe, Q A1 V1 0.0176 4 0.0704 m3/s

By continuity equation, we have,

A1 V1 A2 V2

V2 A1 V1

A2

0.07040.0314

2.242 m/s

Velocity at section 2 2.242 m/s

Problem 1.71: A 40 cm diameter pipe conveying water branches into twopipes of diameters 30 cm and 20 cm respectively. If the average velocity inthe 40 cm diameter pipe is 3 m/s., find the discharge in this pipe. Also,determine the velocity in 20 cm diameter pipe if the average velocity in 30cm diameter pipe is 2 m/s.

Solution

Given:D1 40 cm 0.4 m

V1 3 m/s

1 1D =15cm1

D =20cm2

V =4m /s1

12


D2 30 cm 0.3 m

D3 20 cm 0.2 m

V2 2 m/s

Q3 ? V3 ?

A1 4

D12

4

0.42 0.12566 m2

A2 4

D22

4

0.32 0.07068 m2

A3 4

D32

4

0.22 0.0314 romam2

(i) To find discharge in the 40 cm pipe

Now, the discharge, Q1 in pipe 1 is

Q1 A1 V1 0.12566 3 0.3769 m3/s

(ii) To find V3

Q2 A2 V2 0.07068 2 0.414 m3/s

By continuity equation, we have

Q1 Q2 Q3

0.3769 0.414 Q3

Q3 0.2355 m3/s

Now Q3 A3 V3

ie V3 Q3

A3

0.23550.0314

7.5 m/s

Discharge in 40 cm pipe 0.3769 m3/s

Velocity in 20 cm in pipe 7.5 m/s

1

2

3


Problem 1.72: A 0.2 m diameter pipe carries oil of specific gravity 0.9 ata velocity of 4 m/s. At another section the diameter is 0.15 m. Find thevelocity at this section and mass rate of flow of oil.

Given:

D1 0.2 m; D2 0.15 m; V1 4 m/s; V2 ? Specific gravity of oil = 0.9

Solution

A1 4

D12

4

0.22 0.0314 m2

A2 4

D22

4

0.152 0.01767 m2

By continuity equation,

A1 V1 A2 V2

V2 A1 V1

A2

0.0314 40.01767

7.11 m/s

Mass rate of flow of oil = Mass density Discharge

Now, Density of oil Sp. gravity of oil Density of water

0.9 1000 900 kg/m3

Discharge, Q A1 V1 0.0314 4 0.1256 m3/s

Mass flow rate of oil 900 0.1256 113.04 kg/s

Velocity of section 2 7.11 m/s

Mass rate of flow of oil 113.04 kg/s


1.18.1.2 Continuity Equation In Cartesian Co-ordinates – In Three Dimensions

Consider a fluid element (control volume) - rectangular prism with

sides dx, dy and dz as shown in Fig 1.43.

Let Density of the fluid at a particular instant

u, v, w Components of velocity of flow entering the three faces of the prism

Rate of mass of fluid entering the face ABCD (i.e. fluid influx).

velocity in x direction area of ABCD udydz ... (i)

Rate of mass of fluid leaving the face EFGH (i.e. fluid efflux)

u dy dz x

udy dz dx u

x

u dx dydz

...(ii)

The gain in mass per unit time due to flow in the X-direction is given

by the difference between the fluid influx and efflux.

Mass accumulated per unit time due to flow in X-direction.

u dy dz u

x

u dx dy dz

x

u dx dy dz... (iii)

Similarly, the gain in fluid mass per unit time in the prism due to

flow in Y and Z-direction.

B

A

C

D

dx

E

F

G

H

dy

dz

Y

OX

zFig. 1.43 Fluid Element


y

v dx dy dz in Y direction... (iv)

z

w dx dy dz in Z direction... (v)

The total gain in fluid mass per unit time for fluid flow along three

co-ordinate axes

x

u y

v z

w dx dy dz

... (vi)

Rate of change of mass of the prism (control volume).

t

dx dy dz...(vii)

Equate (vi) and (vii), we get

x

u y

v z

w dx dy dz

t

dx dy dz

By simplifying the above expression, we get

x

u y

v z

w t

0...(2.9)

This equation (Eqn 2.9) is the general equation of continuity in

three-dimensions and is applicable to any type of flow and for any fluid

whether compressible or incompressible.

For steady flow dt

0 of incompressible fluids constant the

equation reduces to

u

x v y

w

z 0

... (ix)

For two dimensional flow, Eqn. (ix) reduces to

u

x v y

0 . . . w 0

For one dimensional flow, (in x-direction), Eqn (ix) reduces to


u

x 0 . . . v 0, w 0

Integrating with respect to x, we get

u constant ... (x)

If the area of flow is a then the rate of flow is

Q a u constant for steady flow.

If area of flow a is constant, the velocity of flow u will also be constant.

1.18.1.3 Equation of continuity in polar coordinates (Rotation and Circulation)

For compressible fluids:

1r

Vr r

Vr

r V 0

For incompressible fluids

Vr

r Vr

r

1r

V

0 . . . constant

Problem 1.73: A fluid flow field is given by

V x2 yi y2 zj 2xyz yz2 kProve that it is a case of possible steady incompressible fluid flow. Calculatethe velocity and acceleration at the point (3,1,4)

Given:

Flow field, V x2 yi y2 zj 2xyz yz2 k

Solution

Here

u x2 y u

x 2xy

v y2 z v

y 2yz

w 2xyz yz2 w

z 2xy 2yz


For a case of possible steady incompressible fluid flow, the continuity

equation should be statisfied.

i.e., u

x v

y w

z 0.

Substituting the values of u

x, v

y and

w

z, we get

u

x v

y w

z 2xy 2yz 2xy 2yz 0

Hence the velocity field V x2 yi y2 zj 2xyz yz2 k is a possible

case of fluid flow. Ans.

Velocity at (3, 1, 4)

Substituting the values x 3, y 1 and z 4 in velocity field, we get

V x2 yi y2 zj 2xyz yz2 k

32 li 12 4j 2 3 1 4 1 42 k

9i 4j 40k Ans.

and Resultant velocity

92 42 402 1697 41.194 units. Ans.

Acceleration at (3,1,4)

The acceleration components ax, ay and az for steady flow are

ax u u

x v

u

y w

u

z

ay u v

x v

v

y w

v

z

az u w

x v

w

y w

w

z

u x2 y, u

x 2xy,

u

y x2 and

u

z 0


v y2 z, v

x 0,

v

y 2yz,

v

z y2

w 2xyz yz2, w

x 2yz,

w

y 2xz z2,

w

z 2xy 2yz

Substituting these values in acceleration components, we get

acceleration at (3,1,4)

ax x2 y 2xy y2 z x2 2xyz yz2 0

2x3 y2 x2 y2 z

2 33 12 32 12 4 54 36

90 units

ay x2 y 0 y2 z 2yz 2xyz yz2 y2

2y3 z2 2xy3 z y3 z2

2 13 42 2 3 13 4 13 42

32 24 16 8 units

az x2 y 2yz y2 z 2xz z2 2xyz yz2 2xy 2yz

2x2 y2 z 2xy2 z2 y2 z3 [4x2 y2 z 2xy2 z2 4xy2 z2 2y2 z3]

2 32 12 4 2 3 12 42 12 43 [4 32

12 4 2 3 12 42 4 3 12 42 2 12 43]

72 96 64 144 96 192 128 328

Acceleration ax i ay j az k 90i 8j 328k Ans

or Resultant acceleration

902 82 3282 8100 64 107584 340.217 units


1.19 TYPES OF MOTION OR DEFORMATION OF FLUID ELEMENT

1.20 CIRCULATION AND VORTICITYCirculation and vorticity are two primary measures of rotation of fluid flow.

Circulation, which is a scalar integral quantity, is a macroscopic

measure of rotation for a finite area of the fluid. It is defined as the line

integral of the velocity field along a closed contour.

Vorticity however is a vector field that gives a microscopic measure

of the rotation at any point in the fluid. Hence it is a measure of the total

rotation of the fluid and it can be said that vorticity at a point is circulation

per unit area.

(a )

(b )

( c)

(d )

L inear Trans la tion

R o tat iona l Trans la tion

L inear D e form ation

Angu la r D e fo rm ation

Fig. 1.44 Fundamental Types of F luid E lem ent M otion or Deform ation : (a) Translation,

(b) Rotation, ( c) Linear Strain , and(d) Shear Stra in.


Consider a closed curve ABC within the fluid

medium as shown in Fig. 1.46. The circulation around

such a curve is defined as the summation of product

of velocity component along the element (such as

PQ) of the curve and elemental length ds. Let be

angle between the length elements ds and the velocity

vector V. The circulation is denoted by (Greek

Capital letter “gamma”) is given by

V cos ds

where denotes line integral taken around the closed curve in an

anticlockwise direction.

Vorticity is defined as the ratio of the differential circulation around an

infinitesimal closed curve enclosing the point to the area of the closed curve.

Vorticity is denoted as (Greek letter xi)

Vorticity z d

dxdy v

x

u

y

In case of a fluid rotating at a certain constant angular velocity , the

velocity V at a radial distance r is given by r.

Circulation d Vr d r2 d r2 2. [Here v r]

Circu la tionVorticity Fig. 1.45

P Qds

A

B

C

vFig. 1.46


Vorticity at a point

dr2

2r2

r2 2

So, vorticity is twice the angular velocity of a fluid particle.

For an arbitrary point in a flow field,

Any fluid element (particle) that occupies that point having a

non–zero vorticity, that point is called rotational.

Vice versa, Any fluid element (particle) that occupies that point

having a zero vorticity, that point is called irrotational, (particle

is not rotating).

Flow from P to Q is rotational (has voriticity) while flow from

P to R is irrotational (has no vorticity).

Flow through turbomachines is an example of rotational flow.

Rotation is defined as movement of a fluid element in such

a way that both of its axes (horizontal as well as vertical)

rotate in the same direction. It is equal to 1/2 v

x

u

y

for a two-dimensional element in X Y plane.

The rotational components are

z 1/2 vx

uy

P

RQ

P

Fig. 1.47��

dr

v=r

Fig. 1.48


x 1/2 y

v

z

y 1/2 uz

x

Problem 1.74: A fluid flow is given by V 8x3i 10x2yj State whether theflow is rotational or irrotational.

Given:

V 8x3i 10x2 yj

Solution:

The velocity components are u 8x3 ; v 10x2y

u

x 24x2 ;

v

x 20xy

u

y 0 ;

v

y 10x2

Rotation in X Y plane is given by

z 1/2 vx

u

y 1/2 20xy 0 10xy

As z 0. Hence flow is rotational

Problem 1.75: The velocity components in a two- dimensional flow are

u y3/3 2x x2y; v xy2 2y x3/3. Show that these functions represent apossible case of as irrotational flow.

Given:

u y3/3 2x x2y and V xy2 2y x3/3.

Solution:

The velocity components are

u y3/3 2x x2y v xy2 2y x3/3

u

x 2 2xy

v

x y2 x2

u

y y2 x2 v

y 2xy 2


For a 2 D flow, continuity equation is

u

x

v

y 0

u

x

v

y 2 2xy 2xy 2 0

It is a possible case of fluid flow.

For rotation, z is given by z 1/2 vx

u

y

z 1/2 [y2 x2 y2 x2] 0

Rotation is zero, which means that the flow is an irrotational flow.

1.21 STREAM FUNCTION

Stream function is denoted by Psi. It

is defined as follows. Mathematically for steady

flow, stream function is defined as f x, y.

Stream function is a scalar function of

space and time like potential function. But the

partial derivative of stream function with respect

to x direction gives the velocity components in

y direction. Generally, the partial derivative of

stream function with respect to any direction gives the velocity component at

right angles to that direction.

i.e x

v...(i)

y

u...(ii)

The continuity equation for two dimensional flow is

u

x v

y 0

... (iii)

(x ,y)=c1

P

(x ,y)=c2

Y

X

Fig. 1.49


Substitute u y

and v x

in equation (iii), we get

x

y

y

x

0

2x y

2 xy

0

So if the stream function exists, it is a possible case of fluid flow.

1.21.1 Properties of Stream Function

1. If the stream function satisfies Laplace equation 2x2

2 y2 0,

then the flow is irrotational flow.

2. If the stream function exists, it is a possible case of an

irrotational flow.

Summary

Potential Function

u x

and v y ...(2.12)

Stream Function

u y

and v x ...(2.13)

1.22 VELOCITY POTENTIAL FUNCTIONPotential function is defined as follows: Potential function is a scalar

function of space and time. It is denoted by Phi. Mathematically, the

velocity potential is defined as f x, y, z for steady flow.

The negative partial derivative of potential function with respect to

any direction gives the velocity of fluid in that direction.


i.e u x ... (i)

v y ... (ii)

w dz ... (iii)

where u, v and w are the velocity components in x, y and z direction

respectively. The negative sign indicates that the flow takes place in the

direction in which decreases.

For an incompressible steady flow, the continuity equation is

u

x v

y

wz

0... (iv)

Substitute u x

; v y

; w z

in the equation (iv), we get

x

x

y

y

z

z

0

i.e 2x2

2y2

2z2 0

...(2.10)

This equation is known as Laplace equation for three dimensional

flow.

Laplace equation for two dimensional flow is

2x2

2y2

0...(2.11)

1.22.1 Properties of Potential FunctionIf the potential function satisfies the Laplace equation, then there

is a possibility of fluid flow i.e. it represents the flow is possible and it is

incompressible steady and irrotational flow.


1.23 RELATION BETWEEN STREAM FUNCTION AND VELOCITY POTENTIAL FUNCTION

u x

y

v y

x ...(2.14)

These equations are called as Cauchy - Riemann equation.

Existsfor

Stream Function Velocity potential only 2D flow all flows

viscous or non-viscous flows Irrotational (i.e. Inviscid or zeroviscosity) flow

Incompressible flow (steady orunsteady)

Incompressible flow (steady orunsteady state)

compressible flow (steady stateonly)

compressible flow (steady orunsteady state)

Problem 1.76: The velocity potential function is given by the expression

xy3

3 x2

x3y3

y2

(i) Find the velocity components in x and y direction.(ii) Show that represents a possible case of flow.

Given: xy3

3 x2

x3y3

y2

Solution

To find the velocity components u and v

u x

y3

3 2x

3x2y3

0

u y3

3 2x x2y

v y

3xy2

3 0

x3

3 2y

v xy2 x3

3 2y


If it is a possible case of flow, then for given value of , it should

satisfy the Laplace equation.

i.e 2x2

2y2

0

x

y3

3 2x x2y

2x2

x

x

0 2 2xy

2x2 2 2xy

Since it satisfies the Laplace equation, it represents possible case of flow.

Problem 1.77: The velocity potential function is given by 8 x2 y2.Calculate the velocity components at point (6, 9).

Given:

8 x2 y2

x

8 2x 0 16x

y

8 0 2y 16y

Solution

To find velocity component u and v at points (6, 9)Here x 6 and y 9

u x

16x 16 6 96 units

v y

16y 16 9 144 units

y

xy2 x3

3 2y

2y2

y y

2xy 0 2

So

2x2

2y2 2 2xy 2xy 2 0


Problem 1.78: The velocity potential function is given by

2x3y 2xy3. Calculate the velocity components and the value of streamfunction at point (3, 2).

Given:

2x3y 2xy3

Solution

u x

[ 6x2y 2y3 ]

u 6x2y 2y3

v y

[ 2x3 6xy2 ]

v 2x3 6xy2

Magnitude of resultant velocity u2 v2 922 182 93.74 units

To find stream function

u y

;

So y

u 6x2y 2y3

y

6x2y 2y3

Integrating on both sides, we get

6x2y2

2

2y4

4

3x2y2 y4

2 ... (i)

Again v x

2x3 6xy2

i.e. x

2x3 6xy2

Substitute x 3 and y 2 in eqn. u and v,

we get

u 6x2y 2y3 6 32 2 2 23

108 16 92 units

v 2x3 6xy2 2 33 6 3 22

54 72 18 units


Integrating on both sides,

2 x4

4 6

x2

2 y2

x4

2 3x2y2

... (ii)

Combining these (i) and (ii), we get

[Common term 3x2y2 is taken only one time]

3x2y2 12

x4 y4

Substitute x 3 and y 2 in equation, we get

3 32 22 12

34 24

108 48.5 59.5 units.

Problem 1.79: Do the following velocity potential represents possible flow?

If so, determine the stream function, y x2 y2. (April/May 2007 AU)

Solution:

Given: y x2 y2

If a velocity potential represents a possible case of flow, then it should

satisfy the Laplace equation

2x2

2y2 0

x

0 2x 0 2x; 2x2

2

y

1 0 2y 1 2y; 2y2 2

2x2

2y2 2 2 0

Since it satisfies the laplace equation, it represents possible case of

flow.


To find stream function

u y

y

u x

2x

y

2x

Integrating both sides, 2xy ...(i)

v x

; x

v y

1 2y

x

2y 1

Integrating both sides, we get,

2xy x ...(ii)

Combining (i) & (ii),

2xy x Ans

Problem 1.80: A stream function is given by 4x 7y. Find the velocitycomponents and also magnitude and direction of the resultant velocity at anypoint.

Given: 4x 7y

Solution

To find u and v

u y

; v x

u [ 7 ] 7 units/sec

v [ 4 ] 4 units/sec

Resultant velocity u2 v2 72 42 8.062 units/sec


To find direction

tan vu

47

0.5714

tan 1 0.5714, 29.745

Problem 1.81: Determine the stream function, of the velocity components ofa two dimensional incompressible fluid flow are given by

u 13

y3 2x x2 y and v xy2 2y 13

x3.(April/May 2007 AU)

Solution:

u y3

3 2x x2 y

v x y2 2y x3

3

We know that, x

v xy2 2y x3

3 ...(i)

y

u y3

3 2x x2 y

y

y3

3 2x x2 y

...(ii)

Now x

xy2 2y x3

3

integrating both sides, we get

xy2 2y x3

3 dx

x2 y2

2 2xy

x4

4 3 C

...(iii)

Where C is a constant of integration which is independent of x but can be

a function of y

Differentiate (iii) w.r.t to y,


y

2x2y

2 2x 0

c

y x2y 2x

cy ...(iv)

comparing (iv) and (ii), we get

c

y

y3

3

Integrating this, we get

c y3

3 dy

y4

4 3 y4

12

Substituting this value in (iii), we get,

x2 y2

2 2xy

x4

12

y4

12 Ans.

Problem 1.82: The stream function for a two-dimensional flow is given by 2xy, calculate the velocity at the point P (2, 3). Find the velocity potential

function

Given:

2xy and Point P 2, 3

Solution:

2xy

The velocity components

u y

y

2xy 2x

v x

x

2xy 2y

At point P (2, 3) we get

u 2 2 4 units/sec

v 2 3 6 units/sec

Resultant velocity at point P u2 v2

42 62 7.21 units/sec


To find velocity potential function

We know x

u 2x 2x;...(i)

y

v 2y...(ii)

Intergrating (i) we get

d 2xdx

2 x2

2 C x2 C

...(iii)

where C is a constant independent of x but can be a function of y

Differentiating (iii) w.r.t. ‘y’ we get

y

C

y

from (ii) y

2y C

y 2y

...(iv)

Integrating (iv) we get

C 2ydy 2y2

2 y2

Substituting C in (iii) we get

x2 y2

Problem 1.83: Sketch the stream lines represented by x2 y2. Also findout the velocity and its direction at point (1, 2)

Given:

x2 y2 and Point P 1, 2

Solution:

x2 y2

The velocity components u and v are


u y

y

x2 y2 2y

v x

x

x2 y2 2x

At the point (1, 2)

u 2 2 4 units/sec

v 2 1 2 units/sec

Resultant velocity u2 v2 42 4.47 units/sec

Sketch of stream lines

x2 y2

Let 1, 2, 3, and so on.

Then we have 1 x2 y2

2 x2 y2

3 x2 y2

and so on.

Each equation is an equation of a circle. Thus we shall get concentric

circles of different diameters.

1.24 EQUIPOTENTIAL LINE

A line along which the velocity potential is constant, is called

equipotential line. i.e.

constant (for equipotential line)

d 0 ...(1)

We know fx,y for steady flow

d x

dx y

dy

d udx Vdy. . .

x

u, y

v

...(2)

x+y

=3

22

x+y

=2

22

x+y

=1

22

X

Y


Equating (1) and (2) we get,

0 udx vdy

dydx

uv ... (A)

called slope of equipotential line

1.24.1 Line of constant stream function.

A line along which remains constant is called line of constant stream.

constant

d 0 ...(3)

d x

dx y

dy

d vdx udy. . .

x

v ; y

u

...(4)Equating (3) and (4) we get

0 vdx udy

dydx

vu

Slope of stream line ...(B)

From equation (A) and (B) it is clear that the product of the slope of

the equipotential line and slope of stream line at the point of intersection is

equal to 1. Thus the equipotential lines are orthogonal to the stream lines

at all points of intersection.

Note: Equi-Potential line is an imaginary line in a field of flow such that

the total head is the same for all points on the line and therefore the direction

of flow is perpendicular to the line at all points.


1.25 FLOW NETA grid obtained by drawing a

series of streamlines and equipotentiallines is known as a Flow Net. It is an

important tool in analyzing two

dimensional irrotational flow problems,

when the mathematical calculation is

difficult and tedious.

Flow net is a graphical

representation of a family of streamlines

intersecting orthogonally a family of equipotential lines and in the process

forming small curvilinear squares. [Refer Fig. 1.50]. It is actually a graphical

solution of laplace’s equation in two dimensions.

1.26 DYNAMICS OF FLUID FLOWDynamics of fluid flow is the study of flow with the forces causing

flow. The fluid flow is analysed by the Newton’s second law of motion, to

relate the acceleration with the forces. Here the fluid is assumed to be

incompressible and non-viscous.

1.27 EQUATIONS OF MOTIONThe dynamic behaviour of fluid motion is governed by a set of equations,

known as equations of motion obtained by using Newton’s second law.

It may be written as Fx m ax ...(i)

where Fx is the net force acting in x direction upon a fluid element

of mass m producing an acceleration ax in the x direction.

The following forces are considered in fluid flow

(i) Fg gravity force (iv) Ft Turbulence force

(ii) FP pressure force (v) Fc Compressibility force

(iii) Fv Viscosity force

Thus net force in x direction is

Fx Fgx Fpx Fvx Ftx Fcx

+ 2

+

-

s90o

90o

S tream slines Equ ipo tentia l

lines

n

-2

Fig. 1.53


Now,

(a) If the force due to compressibility Fc is neglected, the resulting net

force

Fx Fgx Fpx Fvx Ftx which is called as Reynold’s Equation

of Motion.

(b) For flow, if Ft is negligible then the resulting equation,

Fx Fgx Fpx Fvx is called Navier-Stokes Equation.

(c) If the flow is ideal and viscous force Fv is zero, then the equation of

motion is known as Euler’s Equation of Motion and is given by

Fx Fgx Fpx

1.27.1 Euler’s equation along a Stream Line

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.

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


dm The mass of element dAds.

cos dZds

; dm cos dA ds dZds

dAdZ

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.

1.27.2 Principle of Conservation of Energy

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 calledkinetic 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

P2

w

This is known as Bernoullis equation.

1.28 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.

1.28.1 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.


1.28.2 Assumptions for derivation of Bernoullis Equation

State the assumptions used in the derivation of the Bernoulli’s equation.(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

1.28.3 Bernoulli’s Equation for Real Fluid

Real fluid has viscosity so some losses occur due to friction force.

This losses should be taken into account while writing the Bernoullis equation

for real fluid

P1

w Z1

V12

2g

P2

w Z2

V22

2g hL

Where hL loss of energy due to friction between inlet and outlet (or

between the two sections considered)

Problems in Bernoullis EquationProblem 1.84: A pipe slope down at 1 in 100 and tapers from 0.25 mdiameter to 0.15 m diameter at lower end. If the pipe carries 100 lps of oilof specific gravity 0.85, calculate the pressure at the lower end. The lengthof the pipe is 200 m and the upper end gauge reads 50 kPa

Given:Length of pipe 200 m; slope 1 in 100.

P2

D 2z1

D =0.6m1

Z =02

Da tum lineslope 1 in 100

200m

Flow d irectionp =50 kPa

1

1

2


Z1 1

100 200 2 m ; Specific gravity S 0.85

Dia of pipe at upper end D1 0.25 m;

A1 4

D12

4

0.252 0.05 m2

Dia of pipe at lower end D2 0.15 m;

P1 50 kPa 50 103 N/m2

A2 4

D22

4

0.152 0.0177 m2

Z2 0 because it is at datum line

Discharge Q 100 lits s 1001000

m3 s 0.1 m3 s

Continuity equation A1 V1 A2 V2 Q

where V1 and V2 are velocities at (1) and (2) respectively.

V1 QA1

0.10.05

2 m s; V2 QA2

0.1

0.0177 5.65 m s

To Find Pressure P2 at Lower End

Using Bernoulli’s equation,

P1

w

V12

2g Z1

P2

w

V22

2g Z2

50 103

0.85 9.81 103

22

2 9.81 2

P2

0.85 9.81 103

5.652

2 9.81 0

[w specific wt of oil specific gravity of oil specific wt of water

0.85 9.81 103 N m3]

8.2 P2

8338.5 1.627

P2 54809 N m2 54.809 kN m2

Pressure at lower end 54.81 kPa

. . .

1000 l 1 m3

1 l 1

1000 m3


Problem 1.85: A 5m long pipe is inclined at an angle of 15 with thehorizontal. The smaller section of the pipe which is at a lower level is 80mm diameter and the larger section of the pipe is 240 mm diameter.Determine the difference of pressures between the two sections, if the pipe isuniformly tapering and the velocity of water at the smaller section is 1 m/s.

Given:

Dia. of lower end of pipe 1

D1 80 mm 0.8 m

Dia. of upper end (2)

D2 240 mm 0.240 m

Solution:

At section 1

A1 4

D12

4

0.082 5.03 10 3 m2

V1 1 m s; Z1 0 m

At section 2

A2 4

D22

4

0.242 0.0452 m2; Z2 1.294 m

V2 ?

Length of Pipe l 5 m

Angle of Taper 15Velocity at Point 1 V1 1 m/s

Height of Point 2 Z2 5 sin 15 1.294 m

V =1m /s1

5m

15 =o

D =0.242

Z2

Datumline

1

2

D =0.08m ; Z =0m1 1


Continuity equation Q A1V1 A2V2

Discharge Q 5.03 10 3 1 5.03 10 3 m3 s

Velocity V2 A1V1A2

5.03 10 3

0.0452 0.1113 m s

Now, Using Bernoullis equation, we get

P1w

V1

2

2g Z1

P2

w

V22

2g Z2

Difference of Pressure head P1 P2

w

V22 V1

2

2g Z2 Z1

0.11132 12

2 9.81 1.294 0

0.0503 1.294

1.244 m of water

Difference of Pressure P1 P2 w 1.244 9.81 103 1.244

12203.64 N m2

Difference in pressure 12.204 kN m2

Problem 1.86: The water is flowing through a taper pipe of length 100 mhaving diameters 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 the pressure

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


P1

w

V12

2g Z1

P2

w

V22

2g Z2

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

Problem 1.87: A pipe 300 m long tapers from 1.2 m diameter to 0.6 mdiameter at its lower end and slopes downward at 1 in 100. The pressure at

the upper end is 69 kN/m2. Neglecting frictional losses, find the pressure at

the lower end when the rate of flow in 5.5 m3 min.

Given: l 300 m, D1 1.2 m; D2 0.6 m ; Slope 1 in 100 ; P1 69 kN m2 ;

Q 5.5 m3 min

Z 1 D =0.3m1

Z =02

S lope 1 in 30

Da tum L ine

100 m

0.6m =D1


Solution:

Height above datum of section, Z1 1

100 300 3 m

By Continuity Equation, . . . Q A1V1 A2V2

Discharge Q 5.560

0.0917 m3 s A1 V1

V1 0.0917

4

1.22

0.0811 m s

Velocity V2 QA2

0.0917

4

0.62

0.3243 m s


P1

w

V12

2g Z1

P2w

V2

2

2g Z2

69 103

9.81 103 0.08112

2 9.81 3

P2

9.81 103 0.32432

2 9.81 0

7.034 3.353 10 4 3 P2

9.81 103 5.36 10 3 0

P2 98380.7 N m2

Pressure at lower end, P2 98.381 kN m2

Z 1

1.2m =D1

D =0.6m2

Z =02

300 m

D atu rn line

S lope 1 in 100


Problem 1.88: Water is flowing through a pipe of diameter 30 cm and 20cm at sections 1 and 2 respectively. The rate of flow through pipe is 35 lps.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 intensity of pressure at section2. (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

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

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

P =44.5 x 10 N/m14 2

Z =8m1

Datum line

D =0.3m1

D =0.2m2

Z =16m2


P1w

V

1

2

2g Z1

P2

w

V2

2

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

Problem 1.89: Water is flowing through a pipe having diameters 20 cmand 10 cm at section 1 and 2 respectively. The rate of flow through the pipeis 35 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 the intensity ofpressure at section 2. (Nov/Dec 2008 AU)

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



Pressure

P1 39.24 N/cm2 39.24 104 N/m2

Pressure P2 ?


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

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

Problem 1.90: At a point in pipe line where the diameter is 25 cm, the

velocity of water is 3.5 m/s and the pressure is 35 N cm2. At a point 20 mdownstream, the diameter reduces to 15 cm. Calculate the pressure at thispoint, if the pipe line is (i) horizontal (ii) vertical with flow downward (iii)Vertical with flow upward.

Given: D1 25 cm 0.25 m ; V1 3.5 m/s; P1 35 N/cm2 35 104 N/m2 ;

length l 20 m; D2 15 cm 0.15 m

D = 0.2m1P =39.24N/cm1

2

Z =6m1

Datum line

D =0.1m2

Z = 4m2


Solution:

(i) If the pipe line is horizontal

Since it is horizontal pipe, there is no vertical height difference between

section (1) and (2)

So Z2 Z1 i.e Z2 Z1 0

A1 4

0.252 0.05 m2

A2 4

0.152 0.0177 m2

According to continuity equation

Q A1 V1 A2V2

0.05 3.5 0.0177 V2 V2 9.903 m s

Now Using Bernoullis equation,

P1

w

V12

2g Z1

P2

w

V22

2g Z2

35 104

9.81 103 3.52

2 9.81

P2

9.81 103 9.9032

2 9.81

35,678 0.6244 P2

9.81 103 4.998

[. . . Z2 Z1]

2

0 .25 m = D 1 D = 0 .15m2

V = 3.5 m /s1

1 20m

P 1 = 35X10 N /m 4 2


P2 307090 N m2

Pressure P2 307.1 kN m2

(ii) If the pipe line is vertical with flow downward

Take section 2 as datum.

P1w

V1

2

2g Z1

P2

w

V22

2g Z2

35 104

9.81 103 3.52

2 9.81 20

P2

9.81 103

9.9032

2 9.81 0

35.678 0.6244 20

P2

9.81 103 4.998 0

Pressure

P2 503276.5 N m2 503.28 kN m2

(iii) If the pipe line is vertical with flow upward

Take section (1) as datum

P1

w

V12

2g Z1

P2

w

V22

2g Z2

35 104

9.81 103

3.52

2 9.81 0

P2

9.810 103

9.9032

2 9.81 20 m

35.678 0.6244 P2

9.81 103 4.948 20

P2 110890.3 N m2 110.890 kN m2

1

2 da tum

Z =20m1

Z =0m2


Problem 1.91: Water flows at the rate of 200 litres per second upwardsthrough a tapered vertical tape. The diameter at the bottom is 240 mm andat the top 200 mm and the length is 5 m. The pressure at the bottom is 8bar and the pressure at the topside is 7.3 bar. Determine the head lossthrough the pipe. Express it as a function of exit velocity head. (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

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

V2 QA2

0.2

0.031415

V2 6.37 m/s

d2

LZ2

d1 P1

Z =01

P2

1

2

200 m m= 7.3 bar

= 5m

240 m m= 8 ba r


P1w

V1

2

2g Z1

P2

w

V22

2g Z2 HL

8 105

9810 0.69462 V2

2

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

Problem 1.92: Air flows at a rate of 200 litres/s through a pipe. The pipeconsists of two sections of diameters 20 cm and 10 cm with a smooth reducingsection that connects them. The pressure difference between the two pipesections is measured by a water manometer. Neglecting frictional effects,determine the differential height of water between the two pipe sections. Take

the air density to be 1.2 kg/m3. (April/May 2008 AU)

Given: Flow rate Q 200 litres/s 0.2 m3/s ; Dia. of section 1, D1 20 cm 0.2 m

Dia. of section (2), D2 10 cm 0.1 m; air 1.2 kg/m3

Solution:

Q A1 V1 A2 V2

20cm A ir 200 L/s

h

10cm


V1 QA1

0.2

4

0.22 6.37 m/s

V2 QA2

0.2

4

0.12 25.465 m/s

Applying Bernoulli’s equation between sections (1) & (2), we get,

P1

g

V12

2g Z1

P2

g

V22

2g z2

Since pipe is horizontal, then Z1 Z2

P1 P2

g

V22 V1

2

2g

But P1 P2

g difference in pressure heads between sections (1) & (2) h

h V2

2 V12

2g

25.4652 6.372

2 9.81 30.98 m

1.29 NAVIER-STOKES EQUATIONSFluid dynamics deals with the motion of fluids, which when studied

macroscopically, appear to be continuous in structure. All the variables are

considered to be continuous functions of the spatial coordinates and time. The

Navier-Stokes equations are a set of nonlinear partial differential equations

that describe the flow of fluids.

The Navier-Stokes equations describe how the velocity, pressure,

temperature, and density of a moving fluid are related. The equations were

derived independently by G.G. Stokes, in England, and M. Navier, in France.

The equations are extensions of the Euler Equations and include the effects

of viscosity on the flow.

The Navier-Stokes equations are the basic governing equations for a

viscous, heat conducting fluid. It is a vector equation obtained by applying

Newton’s Law of Motion to a fluid element and is also called the momentum


equation. It is supplemented by the mass conservation equation, also called

continuity equation and the energy equation. Usually, the term Navier-Stokes

equations is used to refer to all of these equations.

The Navier-Stokes equations for irrotational flow are

t

u 0,...Continuity Equation (1)

For an incompressible fluid the change rate of density is zero.

Therefore

u 0.

u

t u u

1

F

2 u,...Equations of Motion (2)

t

u KH T p u 0.

...Conservation of Energy (3)

Where,

u velocity vector field,

thermodynamic internal energy,

p pressure,

T temperature,

x

i y

j z

k

2 2

x2 2

y2 2

z2

There are four independent variables in the equation - the x, y, and z

spatial coordinates, and the time t; six dependent variables - the pressure p,

density , temperature T, and three components of the velocity vector u.

Together with the equation of state such as the ideal gas law V n R T,

the six equations are just enough to determine the six dependent variables.

density,

viscosity,

KH heat conduction cofficient,

F external force per unit mass,


1.30 BERNOULLI’S EQUATION: APPLICATIONS The Bernoulli’s equation can be applied to a great many situations.

The following devices are examples for the practical applications of

Bernoulli’s equation.

1. Venturi meter

2. Orifice meter

Bernoullis equation is also applied for measurement of flow through

pipes and open channels.

1.30.1 Venturi Meter

Venturimeter is used to measure the discharge (Q) of a fluid flowing

through a pipe. In venturimeter the cross-sectional area of the passage is

reduced to create a pressure difference. By measuring the pressure difference,

we can find the discharge through the pipe.

Venturimeter consists of three parts

1. A short converging part

2. Throat

3. A long diverging part

The venturimeter is fitted in a pipe as shown in Fig. 1.52. The U tube

manometer is fitted on one limb at the inlet pipe and other limb at the throat.

3. Flow nozzle

4. Pitot tube

d1

h

1

2

In le t Th roat

F ig. 1.52 Venturim eter


The included angle of convergent part is 15 to 20 while that of divergent

part is 5 to 7

A1 area of pipe at section (1) (inlet)

P1 Pressure at section (1) (inlet)

V1 velocity of fluid at section (1) (inlet) and A2, P2 and V2 are

corresponding values at section (2) (or) throat.

Apply Bernoullis equation between section (1) and (2) i.e between inlet

and throat, we get

P1

w

V12

2g Z1

P2

w

V22

2g Z2 ... (i)

When a pipe is horizontal, Z1 Z2

Then P1 P2

w

V22 V1

2

2g ... (ii)

Here P1 P2

w is the difference of pressure heads at section 1 and 2

and it is equal to h. So the equation becomes

h V2

2 V12

2g ... (iii)

V22 V1

2 2gh ... (iv)

Applying continuity equation at section 1 and 2

A1V1 A2V2 or V1 A2V2

A1

Substitute V1 A2V2

A1 in equation (iv), we get

V22

A2 V2

A1

2

2gh


V22

A22 V2

2

A12

2gh

V22 1

A22

A12 2gh

V2 2gh

A12

A12 A2

2

A1 2gh

A12 A2

2

Theoretical Discharge Qth A2V2 A1 A2 2gh

A12 A2

2

The above discharge is theoretical one. The actual discharge will be

less than the theoretical one.

So Qact Cd Qth Cd A1A2 2gh

A12 A2

2

where Cd coefficient of venturimeter and it is less than one (always).

Note: If the manometer contains liquid heavier than flowing liquid, then

h x sh

sf 1

where x manometer reading difference in levels of manometricliquid.

sh specific gravity of heavier liquid.

sf specific gravity of flowing fluid.

If the manometer contains liquid lighter than the flowing fluid, then

h x 1

sl

sf

where sl specific gravity of lighter liquid.

V22 A1

2 A22

A12

2gh

V22 2gh

A12

A12 A2

2


1.30.2 Orifice Meter

An orifice meter is a simple device used for measuring discharge of

fluid through a pipe. It works on basis of Bernoullis equation like venturi meter.

Ref. Fig. 1.53.

It consists of a flat circular plate having sharp edged hole (orifice)

concentric with a pipe. The diameter of the orifice varies from 0.4 to 0.8

times the pipe diameter (mostly 0.5 times)

Orifice meter is fitted on the pipe line to measure the discharge of

fluid. A differential manometer is connected so that one limb connects at

section 1 and other at section 2.

P1, V1 and A1 are pressure, velocity and area at the section 1

respectively

P2, V2, and A2 are pressure, velocity and area at the section 2

respectively.

Applying Bernoullis equation

P1

w

V12

2g Z1

P2

w

V22

2g Z2

1

1

2

2

P ipe O rif ice M e te r

D iffe rentialM anom ete r

x

F ig. 1.53 Orifice meter

D irec tion of flow

Vena con tra cta


P1

w Z1

P2

w Z2

V22

2g

V12

2g

But P1

w Z1

P2

w Z2

h Differential manometer head

So h V2

2 V12

2g

V22 V1

2 2gh ... (i)

The section (2) is at the venacontracta and A2 is the area of the

venacontrata

The coefficient of contraction Cc A2

A0

Where A0 area of the orifice. So A2 CcA0

[Vena contracta: When fluid is flowing

through orifice, the diameter of liquid jet will be

reduced (contracted) in front of the orifice. This

one, refers to vena contracta. When a fluid flows

through the orifice, it contracts and its dia

reduces at a distance d2

from the orifice. The

point where the flow contracts is called venacontracta. Beyond this vena contracta, the fluid jet diverges. At venacontracta,

the cross sectional area is less than the orifice. The stream lines of the flow

is parallel here.

The ratio of the area of venacontracta to the area of the orifice is

known as coefficient of contraction.

Coefficient of contraction

Cc Area of jet at vena contracta

Area of the orifice

According to continuity equation

A1V1 A2V2

Venacon tracta

C

C

Jet

2d

d/2

Fig. 1.61


V1 A2V2

A1

Cc A0 V2A1

Substitute V1 value in equation (i), we get

V22

Cc2A0

2V22

A12

2gh V22 1

Cc2A0

2

A12

2gh

V2 2gh

1

Cc2A0

2

A12

Theoretical discharge Qth V2A2 V2 A0 Cc . . . A2 A0 Cc

2gh

1

Cc2 A0

2

A12

A0Cc 2gh A0 Cc

A1

2 A02

A12

Cc

A1A0 2gh

A12 A0

2

Actual discharge Cd Qth

Qactual CdA1A0 2gh

A12 A0

2

where Cd coefficient of discharge for orifice meter.


PROBLEMS IN VENTURIMETER:

Problem 1.93: A 30 cm 15 cm venturimeter is provided in a verticalpipeline carrying oil of specific gravity 0.9, the flow being upwards. Thedifference of elevation of the throat section and entrance section of theventurimeter is 30 cm. The differential U tube mercury manometer shows agauge deflection of 25 cm. Calculate (a) the discharge of the oil (b) the pressure difference between the entrance and throat section. Thecoefficient of meter is 0.98. Specific gravity of mercury is 13.6.

(April 2004 - AU)

Given:

Dia of inlet

D1 30 cm 0.3 m ; Cd 0.98

Area

A1 4

0.32 0.071 m2

Dia of throat

D2 15 cm 0.15 m

Area of throat

A2 4

0.152 0.0177 m2

Specific gravity of mercury

sm 13.6

Specific gravity of fluid in pipe

sp 0.9

Solution:

Differential manometer reading x 25 cm 0.25 m

Pressure head difference

‘h’ P1

w Z1

P2

w Z2

x

smsp

1

0.25 13.60.9

1 3.53 m of oil

30cm

30cm

25cm

15cm

In let

Th roat


Discharge Q

We know Q Cd A1 A2

A12 A2

2 2gh

0.98 0.071 0.0177

0.0712 0.01772 2 9.81 3.53

Discharge Q 0.14905 m3/s

To find pressure head difference P1 P2

h P1

w Z1

P2

w Z2

3.53

P1 P2

w Z1 Z2 3.53

P1 P2 3.53 0.3 w . . . Z1 Z2 30 cm 0.3 m

P1 P2 3.83 9.81 103 0.9

33815.1 N/m2 33.8151 kN/m2

[Note : w specific wt of oil

specific weight of water specific gravity of oil

9.81 103 0.9 ]

Problem 1.94: A horizontal venturimeter of specification 200 mm 100 mmis used to measure the discharge of an oil of specific gravity 0.8. A mercurymanometer is used for the purpose. If the discharge is 100 litres per secondand the coefficient of discharge of meter is 0.98, find the manometerdeflection. (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 ?

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

Problem 1.95: A Venturimeter having inlet and throat diameters 30 cm and15 cm is fitted in a horizontal diesel pipe line Sp.Gr. 0.92 to measure thedischarge through the pipe. The venturimeter is connected to a mercurymanometer. It was found that the discharge is 8 liters/sec. Find the readingof mercury manometer head in cm. Take Cd 0.96 (Nov/Dec 2011 - AU)

Given: Horizontal Venturimeter

Inlet 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.e x 0.07713 cm

Problem 1.96: Crude oil (sp. gravity 0.85) flows upwards at a volumetric rateof 60 L/s through a vertical venturimeter with an inlet diameter of 200 mm and athroat diameter of 100 mm. The coefficient of discharge of venturimeter is 0.98. Thevertical distance between the pressure tappings is 300 mm.(a) Determine the difference in the readings of pressure gauges connectedto the inlet and throat section.(b) If a differential U-tube mercury manometer is used to connect the twotappings, determine the manometer reading.


Given:

Volumetric rate

Q 60 L/s 60

1000 m3/s

0.06 m3/sec

. . . 1 m3 1000 L

D1 200 mm 0.2 m

D2 100 mm 0.1 m

Cd 0.98; Z2 Z1 300 mm

0.3 m

Sp. gravity of oil flowing

through pipe sp 0.85

Area of inlet

4

0.22 0.03142 m2

Area of throat

4

0.12 7.854 10 3 m2

Solution:

To find pressure head difference h

We know that discharge

Q Cd A1 A2

A12 A2

2 2gh

0.06 0.98 0.03142 7.854 10 3

0.031422 7.854 10 32 2 9.81 h

0.06 0.03521 h

h 2.904 m of oil

300mm

x

1

2


(a) To find difference in readings of pressure gauges

For vertical venturimeter (also for inclined venturimeter)

h P1

w Z1

P2

w Z2

2.904

P1 P2

w 2.904 Z2 Z1

P1 P2

w 2.904 0.3 3.204 m

[ . .

. Z2 Z1 0.3 m ]

P1 P2 3.204 w 3.204 0.85 9.81 103

26715.1 N/m2 26.715 kN/m2

(Here specific weight of oil Sp. weight of water specific

gravity of oil)

(b) To find manometer reading x

h x sm

sp 1

; 2.904 x

13.60.85

1

x 0.1936 m

Difference in levels of mercury column x 0.1936 m

Problem 1.97: A mercury filled vertical U-tube manometer connected acrossa venturimeter records a difference of 30 mm. Diameters at inlet and throatof venturimeter are respectively 100 mm and 50 mm. If oil of specific gravity0.85 flows through a horizontal pipe. Calculate the discharge. Take Cd for

venturimeter is 0.9.

Given: Manometer difference x 30 mm 0.03 m

Dia of inlet D1 0.1 m ; A1 4

0.12 7.854 10 3 m2

Dia of throat D2 0.05 m ; A2 4

0.052 1.9635 10 3 m2

sp 0.85 ; sm 13.6 ; Cd 0.9


Solution:

Pressure head

h x smsp

1 0.03

13.60.85

1 0.45 m of oil

Discharge Q Cd A1 A2

A12 A2

2 2gh

0.9 7.854 10 3 1.9635 10 3

7.854 10 32 1.9635 10 32 2 9.81 0.45

Q 5.423 10 3 m3/s 5.423 litres/s . . . 1 m3/s 1000 litres/s

Problem 1.98: Calculate the flow rate through a venturimeter placed at30 to the horizontal carrying gasoline of specific gravity 0.8. The diametersat inlet and throat are 5 cm and 3 cm respectively. A mercury manometerreads a level of difference of 10 cm at these two points. Take Cd 0.98.

Given: D1 5 cm 0.05 m ; D2 3 cm 0.03 m ; x 10 cm 0.1 m ; Cd 0.98

Specific gravity of gasoline

in pipe sp 0.8

Specific gravity of mercury

sm 13.6

Solution:

For inclined venturimeter

h P1

w Z1

P2

w Z2

x sm

sp 1

x 0.1 m

x=0 .1m0 .81

2


Area of inlet A1 4

D12

4

0.052 1.9635 10 3 m2

Area of throat A2 4

D22

4

0.032 7.07 10 4 m2

Pressure head h x sm

sp 1

0.1

13.60.8

1 1.6 m

Discharge through venturimeter Q Cd A1 A2

A12 A2

2 2gh

0.98 1.9635 10 3 7.07 10 4

1.9635 10 32 7.07 10 42 2 9.81 1.6

1.36 10 6

1.832 10 3 5.603 4.1599 10 3 m3/s

Problem 1.99: A Venturimeter of size 50 mm 25 mm with coefficient ofdischarge 0.97 is to be replaced by an Orificemeter with a coefficient ofdischarge 0.62. If both the meters give the same difference of pressure for adischarge of 10.5 lps, determine the diameter of the orifice meter.[The inlet diameter for both venturimeter and orificemeter are same. (FAQ)

Given: For Venturimeter: D1 50 mm 0,05 m ; D2 25 mm 0.02 m

Cdv 0.97 Cdo 0.62, Qv Q0 10.5 LPS, Q 10.5 10 3 m3/s

Solution:

Discharge through venturimeter

Qventuri Cdv A1 A2 2gh

A12 A2

2

Discharge through orificemeter

Qorifice Cdo A1 A0 2gh

A12 A0

2

Both give same discharge, so Qventuri Qorifice


Cdv A1 A2 2gh

A12 A2

2

Cdo A0 A2 2gh

A12 A0

2

The above equation becomes, Cdv A2

A12 A2

2

Cdo A0

A12 A0

2... (i)

[ . . . h, A1 are common for both meters]

Coefficient of discharge for venturi Cdv 0.97

Coefficient of discharge for orifice Cdo 0.62

Area of inlet A1 4

D12

4

0.052 1.9635 10 3 m2

Area of throat A2 4

D22

4

0.0252 4.909 10 4 m2

Substitute these values in equation (i),

CdvA2

A12 A2

2

CdoA0

A12 A0

2, we get

0.97 4.909 10 4

1.9635 10 32 4.909 10 42

0.62 A0

1.9635 10 32 A02

0.2505 0.62 A0

3.86 10 6 A02

Squaring on bothsides, 0.063 0.3844 A0

2

3.86 10 6 A02

2.4318 10 7 0.063 A02 0.3844 A0

2

A0 7.373 10 4 m2

4

D02 7.373 10 4

D0 0.031 m

Dia of orificemeter 0.031 m 31 mm


Problem 1.100: A 150 mm 75 mm venturimeter with a coefficient ofdischarge 0.98 is to be replaced by an orificemeter having a coefficient ofdischarge 0.6. If both meters are to give the same differential manometer readingfor a discharge of 100 lit/s and the inlet dia is to remain 150 mm, what shouldbe the diameter of orifice. (Oct 2004 - AU)

Given: Coefficient of discharge for venturi Cdv 0.98

Coefficient of discharge for orifice Cdo 0.6

Both meter give same discharge Q 100 lit/s

Inlet dia for both metersD1 150 mm 0.15 m

Area of inlet for both meters

A1 4

0.152 0.0177 m2

Dia of throat D2 0.075 m

Area of throat A2 4

0.0752 4.42 10 3 m2

Solution:

Discharge is same Qventuri Qorifice

Cdv A1 A2

A12 A2

2 2gh

Cdo A1 A0

A12 A0

2 2gh

The above equation becomes

Cdv A2

A12 A2

2

Cdo A0

A12 A0

2

[ . . . A1 and h are common for both meters]

0.98 4.42 10 3

0.01772 4.42 10 32

0.6 A0

0.01772 A02


0.252713 0.6 A0

3.1329 10 4 A02

Squaring on both sides, we get

0.06386 0.62 A0

2

3.1329 10 4 A02

2 10 5 0.06386 A02 0.36 A0

2

A0 6.869 10 3 m2

A0 4

D02 6.869 10 3 m2

D0 0.09352 m

Dia of orifice D0 0.09352 m 93.52 mm

Problems in Orifice MeterProblem 1.101: An orifice meter consisting of 10 cm diameter orifice in a25 cm diameter pipe has coefficient 0.65. The pipe delivers oil of specificgravity 0.8. The pressure difference on the two sides of the orifice plate ismeasured by a mercury differential manometer. If the differential gauge reads80 cm of mercury, calculate the rate of flow in L/s.

Given: Dia of orifice D0 10 cm 0.10 m ; Area of orifice A0 4

0.12

7.854 10 3 m2

Dia of pipe D1 25 cm 0.25 m

Area of pipe A1 4

0.252 0.0491 m2

Coefficient of discharge Cd 0.65; Specific gravity of oil in pipe sp 0.8

Specific gravity of mercury in manometer sm 13.6

Reading of differential gauge x 80 cm 0.8m


Solution:

We know

Difference ofpressure headsin two sections

h x

smsp

1 0.8

13.60.8

1

12.8 m of oil

To find discharge QDischarge

Q Cd A0 A1 2gh

A12 A0

2

0.65 7.854 10 3 0.0491 2 9.81 12.8

0.04912 7.854 10 32

Q 0.08196 m3/sec 81.96 lit/s . . . 1 m3/sec 1000 L/SProblem 1.102: Determine the rate of oil of sp. gravity 0.88 through a pipeof 240 mm diameter fitted with an orifice meter of 120 mm diameter havinga coefficient of discharge as 0.65. Reading of the differential manometer fixedbetween the upstream and venacontracta is 400 mm of mercury.

Given: Specific gravity of oil in pipe sp 0.88

Specific gravity of mercury in manometer sm 13.6

Dia of pipe

D1 240 mm 0.24 m

Area of pipe

A1 4

0.242 0.04524 m2

Coefficient of discharge Cd 0.65

Reading of differential manometerfixed between upstream and venacontrata

x 400 mm 0.4 m

Dia of orifice D0 120 mm 0.12 m

Area of orifice A0 4

0.122 0.01131 m2


Solution

Difference of pressure headbetween upstream and vena contracta

h x

sm

sP 1

0.4 13.60.88

1

5.782 m of oil

To find discharge QDischarge

Q Cd A1 A0 2gh

A12 A0

2

0.65 0.04524 0.01131 2 9.81 5.782

0.045242 0.011312

Q 0.080867 m3/s 80.867 lit/s [1 m3/s 1000 L/s]

Problem 1.103: An orificemeter with orifice diameter 15 cm is inserted ina pipe of 30 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]

1.30.3 Principle of Conservation of Momentum

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

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

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 )


[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

1.30.4 Moment of Momentum Equation

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.

Problems in Momentum EquationsProblem 1.104: 250 lit/sec of water is flowing in a pipe having a diameterof 300 mm. 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 axis

Fx 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 45o

sin

45o

45 o

2

22

1

1

2

2

300mm

dia

A2cos 45oP2

A2

sin

45o

P2

135 o

300mm d ia

F=

20.6

23kN

y

F =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

Problem 1.105: A 30 cm diameter pipe carries water under a head of 20meters with a velocity of 3.5 m/s. If the axis of the pipe turns through 45.Calculate the magnitude of the resultant force at the bend.

Given: Dia of pipe at inlet and outlet D1 D2 30 cm 0.3 m

Area of pipe A1 A2 4

0.32 0.0707 m2

Pressure head of water in pipe h 20 m

Pressure of water in pipe P1 P2 wh

P1 P2 wh 9.81 103 20 196.2 103 N/m2

Velocity of water in pipe V1 V2 3.5 m/s

Angle of bend of pipe 45

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


Solution:

Force along x axis:

Fx Q V1 V2 cos p1 A1 p2 A2 cos

Here 45

Q A1 V1 0.0707 3.5 0.2474 m3/s

Fx 1000 0.2474 [3.5 3.5 0.707 ] 196.2 103 0.0707

196.2 103 0.0707 0.707

253.71 13871.34 9807.04 4318.01 N 4.318 kN

Fy Q V2 sin P2A2 sin

1000 0.2474 3.5 0.707 196.2 103 0.0707 0.707

612.19 9807.04 10419.2 N 10.419 kN

Resultant force

R Fx2 Fy

2 4.3182 10.4192 11.279 kN.

v1 P A1 1

P A2

2

45o

1

1

2

2

P A cos45o

2 2 V cos45o

2

V sin45o

2

P A s in452o

2

PA2

2

45o

V 2

FY

FX


Direction of resultant force with horizontal

tan FyFx

10.424.318

2.4132

67.49

Problem 1.106: A 30 reducing bend is connected in a pipe line; Thediameters at the inlet and outlet of the bend are 60 cm and 30 cmrespectively. Find the resultant force exerted by water on the bend, if thepressure at inlet to bend is 0.9 bar and the discharge of water is 600 lits/sec.

Given: Angle of bend 30; Dia of bend at inlet D1 60 cm 0.6 m

Area of bend at inlet A1 4

0.62 0.2827 m2

Dia of bend at outlet D2 30 cm 0.3 m

Area of bend at outlet A2 4

0.32 0.0707 m2

Pressure at inlet P1 0.9 bar 0.9 105 N/m2

Discharge Q 600 lit/s 0.6 m3

s [. . . 1000 L 1 m3]

Solution:

V1 QA1

0.6

0.2827 2.1224 m/s

V2 QA2

0.6

0.0707 8.49 m/s

Using Bernoullis equation in between (1) and (2)

P1

w

V12

2g Z1

P2

w

V22

g Z2

Here we assume Z1 Z2

So P1

w

V12

2g

P2

w

V22

2g


0.9 105

9.81 103 2.12242

2 9.81

P2

9.81 103 8.492

2 9.81

P2 56212.2 N/m2

Force along x direction

Fx Q [V1 V2 cos ] P1A1 P2A2 cos

1000 0.6 [2.1224 8.49 cos 30 ] 0.9 105 0.2827

56212.2 0.0707 cos 30

3138.1 25443 3442 18,863 N

Force along y direction:

Fy Q [ V2 sin ] P2A2 sin

9.81 103 0.6

9.81 [ 8.49 sin 30 ] 56212.2 0.0707 sin 30

2547 1987.1 4534.1 N

[ (negative) sign indicates that Fy is acting downward].

1

1

2

2

y

x

P A1 1

P A2

2

V 2

45o


Resultant force R

R Fx2 Fy

2 188632 4534.12 19400.28 N

Direction of Resultant force:

tan FyFx

4534.118863

0.2404

tan 1 0.2404 13.52

1.30.5 PITOT-TUBE

Pitot-tube is a device used for measuring the velocity of flow at any

point in a pipe or a channel. It is based on the principle that if the velocity

of flow at a point becomes zero, the pressure there is increased due to the

conversion of the kinetic energy into pressure energy. In the simplest form,

the pitot tube consist of a glass tube, bent at right angles as shown in

Fig. 1.56.

Consider two points (1) and (2) at same level in such a way that point

(2) is just at the inlet of pitot tube and point (1) is far away from the tube.

1 2

Pitot - tube

H

h

Fig. 1.56 Pitot Tube


Let P1 : intensity of pressure at point (1)

V1 : Velocity of flow at (1)

P2 : Intensity of pressure at point (2)

V2 : Velocity of flow at (2), which is zero

H : depth of tube in the liquid

h : rise of liquid in the tube above the free surface

Applying Benoullis equation at points (1) and (2) we get

P1

g

V12

2g Z1

P2

g

V22

2g Z2 ...(i)

But Z1 Z2 as points (1) and (2) are on same line Also V2 0

Pressure head at (1) P1

g H

Pressure head at (2) P2

g h H

Substituting these values we get in (i) we get

H V1

2

2g h H

h V1

2

2g or V1 2gh

This is theoretical velocity. Actual velocity is given by

V1act Cv 2gh Cv: Coefficient of pitot-tube

Velocity at any point V Cv 2gh

The various arrangement of pitot-tube adopted are

(i) Pitot-tube along with a vertical piezometer Fig. 1.57 (1)

(ii) Pitot-tube connected with piezometer. Fig. 1.57 (2)


(iii) Pitot-tube and vertical piezometer tube with differential V-tube

manometer. Fig. 1.57 (3)

(iv) Pitot-static tube, which consists of two circular concentric tubes one

inside the other as shown in Fig 1.57 (4). The outlet is connected to

the differential manometer where the difference of pressure head ‘h’ is

measured by knowing the difference of the level of manometer liquid

say x. Now h x sg

so 1

Problems in Pitot Tube

Problem 1.107: A pitot-static tube placed in the centre of a 300 mm pipe linehas one orifice pointing upstream and other perpendicular to it. The mean velocityin the pipe is 0.80 of the central velocity. Find the discharge through the pipe ifthe pressure difference between the two orifices is 60 mm of water. Take thecoefficient of pitot tube as Cv 0.98

Fig. 1.57 Arrangement of Pitot Tube


Given: Dia. of pipe d 300 mm 0.30 m; Cv 0.98

Difference of pressure head h 60 mm of water 0.06 m of water

Solution

Mean velocity V 0.80 central velocity

Central velocity

V Cv 2gh 0.98 2 9.81 0.06 1.063 m/s

V 0.80 1.063 0.850 m/s

Discharge through pipe

Q Area of pipe V

d2

4 V

4

0.302 0.850

Q 0.06 m3/s

Problem 1.108: A pitot-tube is inserted in a pipe of 300 mm diameter. The staticpressure in pipe is 100 mm of mercury (vacuum). The stagnation pressure at the

centre of the pipe, recorded by the pitot-tube is 0.981 N/cm2. Calculate the rateof flow of water through pipe, if the mean velocity of flow is 0.85 times thecentral velocity. Take Cv 0.98

Given: Diameter of pipe d 0.3m

Area of pipe d2

4

4

0.32 0.07068 m2

Static pressure head 100 mm of mercury (Vacuum)

1001000

13.6 1.36 m of water

Stagnation pressure 0.981 N/cm2 0.981 104 N/m2

Solution:

Stagnation pressure head

0.981 104

g

0.981 104

1000 9.81 1m


h stagnation pressure head static pressure head

1.0 1.36 2.36 m of water

Velocity at centre V Cv 2gh

V 0.98 2 9.81 2.36 6.668 m/s

Mean velocity V 0.85 6.668 5.6678 m/s

Rate of flow of water V Area of Pipe

Q 5.6678 0.07068

Q 0.40 m3/s

1.31 HEADIt is a concept that relates the energy in an incompressible fluid to the

height of an equivalent static column of that fluid.

The Bernoulli’s equation for incompressible flow is given by

P

gZ V2

2 constant

or

P g

V2

2g Z constant

Static Head

In above,

The term P g

is pressure energy per unit weight of fluid called the

pressure head or static head. It is the internal energy of a fluid due to the

pressure exerted on its container. For a pump, the static head is the maximum

height it can deliver.


Dynamic head

The term v2

2g is kinetic energy per unit weight called the kinetic head

or velocity head or dynamic head.

Potential head

The term Z is the potential energy per unit weight called the potentialhead or elevation head.

Hydraulic head (or) Piezometric head

It is composed of pressure head P

and elevation head Z.

Total head

It is the sum of the pressure head, kinetic head and potential head and

is constant.

1.32 CONCEPT OF CONTROL VOLUMEA fluid dynamic system can be analyzed using a control volume, which

is an imaginary surface enclosing a volume of interest. The control volume

can be fixed or moving.

Control volume approach is widely applied in analysis. An arbitrary

fixed volume located at a certain place in the flow-field is identified and the

conservation equations (refer kinematics and dynamics chapters) are written.

The surface which bounds the control volume is called the control surface.

In the control volume approach, the control surface is first defined

relative to a coordinate system that may be fixed, moving or rotating. Mass,

heat and work can cross the control surface and mass and properties can

change with time within the control volume. Choice of location and shape of

control volume are important for mathematical formulation.


1.33 HYDRAULIC CO-EFFICIENTSThe hydraulic coefficients are

(a) Coefficient of velocity, Cv

(b) Coefficient of contraction, Cc

(c) Coefficient of discharge, Cd

(d) Coefficient of resistance, Cr

(a) Coefficient of Velocity (Cv)

It is defined as the ratio between the actual velocity of a jet of liquid

at vena-contracta and the theoretical velocity of jet. The difference occurs

between the velocities due to friction of the orifice.

Cv V

2gH ...(1)

Here,

V actual velocity

2gh Theoretical velocity

where,

H is the height of liquid above the centre of the orifice

The value of Cv varies from 0.95 to 0.99, depending on the shape,

size of the orifice and the head under which flow takes place. Generally, for

sharp edged orifices, the value of Cv is taken as 0.98.

Also, when the coordinate (x, y) of a fluid particle at a time ‘t’ is

given, then V gx2

2y

where,

x horizontal distance travelled by particle in time ‘t’

y vertical distance between particle and centre of orifice


Substituting the above equation in (1), we have,

Cv gx2

2y

1

2gH

Cv x

4yH

(b) Coefficient of contraction Cc

It is defined as the ratio of the area of the jet at vena - contracta to

the area of the orifice.

Cc Ac

A

Where,

Ac area of jet at vena - contracta

A area of orifice

The Cc value varies from 0.61 to 0.69 depending on shape, size of

the orifice and the available head of liquid. Generally, the value of

Cc 0.64 is taken.

(c) Coefficient of discharge Cd

It is defined as the ratio of the actual discharge from an orifice to the

theoretical discharge.

Cd Q

Qth

Actual dischargeTheoritical discharge

Actual velocity Actual Area

Theoretical velocity Theoretical area

Cv Cc

So, the value of coefficient of discharge varies with the value of

coefficient of contraction and coefficient of velocity. It varies from 0.61 to

0.65 and generally, is taken as 0.62.


Also, Qth Area 2gH

Cd Q

A 2gH

(d) Coefficient of resistance CrIt is defined as the ratio of loss of head in the orifice to the head of

water available at the exit of the orifice.

Cd Loss of head in the orifice

Head of water

The loss of head takes place because the walls of the orifice offer

some resistance to the liquid as it comes out. The coefficient of resistance is

generally neglected while solving numerical problems.

PROBLEMS

Problem 1.109: The head of water over the centre of an orifice of diameter18 mm is 1.5 m. The actual discharge through the orifice is 0.9 litre/s. Findthe co-efficient of discharge.

Given: Diameter of orifice, d 18 mm 0.018 mm

Area, A 4

0.0182 2.54 10 4 m2

Head, H 1.5 m

Actual discharge, Q 0.9 litre/s 0.9 10 3 m3/s

Solution

Theoretical velocity, Vth 2gH 2 9.81 1.5 5.425 m/s

Theoretical discharge, Qth Area of orifice Vth

2.54 10 4 5.425

1.38 10 3 m3/s

Coefficient of discharge, Actual discharge

Theoretical discharge

0.9 10 3

1.38 10 3

0.65


Problem 1.110: A 150 mm diameter pipe reduces in diameter abruptly to100 mm diameter. If the pipe carries water at 30 liters per second, calculatethe pressure loss across the contraction. Take coefficient of contraction as0.6. (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

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

Problem 1.111: The head of water over an orifice of diameter 90 mm is 1m. The water coming out from orifice is collected in a circular tank of diameter1.5 m. The rise of water level in this tank is 1.0 m in 25 seconds. Also theco-ordinates of a point on the jet, measured from venta-contracts are 5.1 mhorizontal and 0.5 m vertical. Find the co-efficients Cd, Cv and Cc.


Solution: Given: Head, H 14 m;

Dia. of orifice, d 90 mm 0.09 m

Area of orifice, a 4

0.092 6.36 10 3 m2

Dia. of measuring tank, D 1.5 m

Area, A 4

1.52 1.767 m2

Rise of water, h 1 m; Time, t 25 seconds

Horizontal distance, x 5.1 m; y Vertical distance, 0.5 m

Now theoretical velocity, Vth 2gH 2 9.81 14 16.57 m/s

Theoretical discharge,

Qth Vth Area of orifice 16.57 6.36 10 3 0.1054 m3/s

Actual discharge, Q A h

t

1.767 1.025

0.07068

Cd Q

Qth

0.070680.1054

0.67

The value of Cv is given by equation,

Cv x

4yH

5.1

4 0.5 14

5.15.29

0.96

Cc is given by equation, Cc Cd

Cv

0.670.96

0.697

Problem 1.112: Water discharge at the rate of 85 litres/s through a 110mm diameter vertical sharp-edged orifice placed under a constant head of10 metres. A point on the jet, measured from the vena-contracta of the jethas co-ordinates 4.5 metres horizontal and 0.54 meters vertical. Find theco-efficient, Cv, Cc and Cd of the orifice.

Solution: Given: Discharge, Q 85 litres/s 0.085 m3/s

Dia. of orifice, d 110 mm 0.11 m


Area of orifice, a 4

0.112 0.0095 m2

Head, H 10 m

Horizontal distance of a point on the jeft from vena-contracta,

x 4.5 m

and vertical distance, y 0.54 m

Now theoretical velocity,

Vth 2g H 2 9.81 10 140 m/s

Theoretical discharge, Qth Vth Area of orifice

14.0 0.0095 0.133 m3/s

The value of Cd is given by,

Cd Actual discharge


QQth

0.0850.133

0.64

The value of Cv is given by equation

Cv x

4yH

4.5

4 0.54 10 0.968 Ans.

The value of Cc is given by equation

Cc Cd

Cv

0.640.968

0.66

Problem 1.113: A 25 mm diameter nozzle discharges 0.76 m3 of water perminute when the head is 60 m. The diameter of the jet is 22 mm. Determine(i) the values of co-efficients Cc, Cv and Cd and (ii) the loss of head due

to fluid resistance.

Solution

Given

Dia. of nozzle,

D 25 mm 0.025 m

Actual discharge,

Nozzle Jet of Water


Qact 0.76 m3/minute 0.7660

0.01267 m3/s

Head, H 60 m

Dia. of jet, d 22 mm 0.022 m

(i) Values of co-efficients:Co-efficient of contraction Cc is given by,

Cc Area of jet

Area of nozzle

4

d2

4

D2

d2

D2

0.0222

0.0252 0.774

Co-efficient of discharge Cd is given by,

Cd Actual discharge


0.01267

Theoretical velocity Area of nozzle

0.01267

2gH 4

D2

0.01267

2 9.81 60 4

0.0252

0.752

Co-efficient of velocity Cv is given by,

Cv CdCc

0.7520.774

0.97

(ii) Loss of head due to fluid resistance:

Applying Bernoulli’s equation at the outlet of nozzle and to the jet of

water, we get

p1

g

V12

2g z1

p2

g

V22

2g z2 Loss of head


But p1

g

p2

g Atomspheric presssure head

z1 z2, V1 2g H , V2 Actual velocity of jet Cv 2gH

2gH 2

2g

Cv 2gH 2

2g Loss of head

or H Cv2 H Loss of head

Loss of head H Cv2 H H

1 Cv

2

60 1 0.972 60 0.0591 3.546 m

1.34 PITOT-STATIC TUBE (OR PRANDTL TUBE)Many a time a Pitot static tube is made use of, as it senses both the

stagnation pressure and the static pressure in a single probe. A schematic of a

pitot static probe is shown in Fig. 1.58. The proportions of the probe are very

important in order to have an accurate measurement of the velocity. Typically

the probe diameter D would be 6 mm. The length of the probe would then

be around 10 cm or more.

>4D

V

8DD

Typically 8D

Statics H o les 4 to 8, 1mm

D iam eter H oles

R adius o f C u rva ture = 3D

Fig. 1.58 Pitot static tube (or Pradtl tube) showing typical proportions


The static pressure holes are positioned more than 4D from the

stagnation point. A suitable manometer may be connected between the inner

and outer tubes to measure the fluid velocity.

Pitot static tube measures the velocity at a given point in the pipe. The

instrument consists of two concentric tubes out of which the outer tube, which

contains perforations in the form of small holes drilled at right angles to its

periphery, measures the static head while the inner tube measures the total head.

The nose is designed so as to give least resistance to flow.

Important Formulae and Points

1. Mass density Mass

Volume

m

V in kg/m3

2. Specific weight w WeightVolume

W

V in N/m3

Also w g

3. Specific volume v

VolumeMass

V

m in m3/kg

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

5. Bulk modulus K dp

dV

V

; Compressibility 1K

6. Surface tension on droplet:

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

P 4d

; where Surface tension in N/m; d dia. of droplet in m


7. Surface tension on a Hollow bubble:

P 8d

8. Surface tension on a liquid jet:

P 2d

9. Capillary rise (or) fall h 4 cos gd

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

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

11. Kinematic viscosity:

in m2/s

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

12. Atmospheric pressure 760 mm of Hg 10.3 m of water 1.01325 bar

Absolute pressure Atmospheric pressure Gauge pressure

Absolute pressure Atmospheric pressure Gauge pressure Vacuum pressure

13. 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 molK

V

m Molar volume ; M Molecular weight

Also R for gas A R

M for gas A

14. Bernouli’s Equation

P1

g

V12

2g Z1

P2

g

V22

2g Z2 (Ideal Fluid)

(Pr.Head) (Kinetic head) (datun Head)

P1

g

V12

2g Z1

P2

g

V22

2g Z2 hL (Read Fluid)

15. Discharge through venturimeter.

Q Cd A1 A2

A12 A2

2 2gh

Cd: Coefficient of discharge

A1: Inlet Area

A2: Throut Area

h: Pressure head difference

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

h x Sh

S0 1

[Differental Manometer Contain Heavy Liquid]

h x 1

SL

S0

[Differental manometer Contain 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.

17. For pitot tube

Velocity V CV 2gh

CV Coefficient of pitot tube.

h x Sg

S0 1

rise of liquid in the tube.

18. 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

19. Moment of Momentum equation

Torque T Q [V2 r2 V1 r1]


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