Advanced Fluid Mechanics
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Advanced Fluid
Mechanics
W. P. Graebel
Professor Emeritus, The University of Michigan
AMSTERDAM • BOSTON • HEIDELBERG • LONDON
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Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiv
Chapter 1
Fundamentals
1.1
1.2
1.3
1.4
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Velocity, Acceleration, and the Material Derivative . . . . . . . . . . . . . . . . . . . . . . .
The Local Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Path Lines, Streamlines, and Stream Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 Lagrange’s Stream Function for Two-Dimensional Flows . . . . . . . . . . .
1.4.2 Stream Functions for Three-Dimensional Flows, Including
Stokes Stream Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Newton’s Momentum Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Rates of Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 Constitutive Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9 Equations for Newtonian Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.10 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.11 Vorticity and Circulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.12 The Vorticity Equation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.13 The Work-Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.14 The First Law of Thermodynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.15 Dimensionless Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.16 Non-Newtonian Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.17 Moving Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
4
5
7
7
11
13
14
21
24
27
28
29
34
36
37
39
40
41
43
vii
viii
Contents
Chapter 2
Inviscid Irrotational Flows
2.1 Inviscid Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Irrotational Flows and the Velocity Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Intersection of Velocity Potential Lines and Streamlines
in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 Basic Two-Dimensional Irrotational Flows . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.3 Hele-Shaw Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.4 Basic Three-Dimensional Irrotational Flows . . . . . . . . . . . . . . . . . . . . . . . .
2.2.5 Superposition and the Method of Images . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.6 Vortices Near Walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.7 Rankine Half-Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.8 Rankine Oval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.9 Circular Cylinder or Sphere in a Uniform Stream . . . . . . . . . . . . . . . . . . .
2.3 Singularity Distribution Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Two- and Three-Dimensional Slender Body Theory . . . . . . . . . . . . . . . . .
2.3.2 Panel Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Forces Acting on a Translating Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Added Mass and the Lagally Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Theorems for Irrotational Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.1 Mean Value and Maximum Modulus Theorems . . . . . . . . . . . . . . . . . . . . .
2.6.2 Maximum-Minimum Potential Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.3 Maximum-Minimum Speed Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.4 Kelvin’s Minimum Kinetic Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . .
2.6.5 Maximum Kinetic Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.6 Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.7 Kelvin’s Persistence of Circulation Theorem . . . . . . . . . . . . . . . . . . . . . . . .
2.6.8 Weiss and Butler Sphere Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
47
49
51
57
58
59
61
65
67
68
69
69
71
77
79
81
81
81
82
82
83
84
84
84
85
Chapter 3
Irrotational Two-Dimensional Flows
3.1
3.2
3.3
3.4
3.5
3.6
3.7
Complex Variable Theory Applied to Two-Dimensional
Irrotational Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
Flow Past a Circular Cylinder with Circulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
Flow Past an Elliptical Cylinder with Circulation . . . . . . . . . . . . . . . . . . . . . . . . . . 93
The Joukowski Airfoil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
Kármán-Trefftz and Jones-McWilliams Airfoils . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
NACA Airfoils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
Lifting Line Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .101
Contents
3.8 Kármán Vortex Street . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
3.9 Conformal Mapping and the Schwarz-Christoffel Transformation . . . . . . . . . . 108
3.10 Cavity Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
3.11 Added Mass and Forces and Moments for
Two-Dimensional Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
Problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
Chapter 4
Surface and Interfacial Waves
4.1
Linearized Free Surface Wave Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.1.1 Infinitely Long Channel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.1.2 Waves in a Container of Finite Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
4.2 Group Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
4.3 Waves at the Interface of Two Dissimilar Fluids . . . . . . . . . . . . . . . . . . . . . . . . . 125
4.4 Waves in an Accelerating Container . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
4.5 Stability of a Round Jet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
4.6 Local Surface Disturbance on a Large Body of
Fluid—Kelvin’s Ship Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
4.7 Shallow-Depth Free Surface Waves—Cnoidal
and Solitary Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
4.8 Ray Theory of Gravity Waves for Nonuniform Depths . . . . . . . . . . . . . . . . . . . 136
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
Chapter 5
Exact Solutions of the Navier-Stokes Equations
5.1
Solutions to the Steady-State Navier-Stokes Equations
When Convective Acceleration Is Absent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
5.1.1 Two-Dimensional Flow Between Parallel Plates . . . . . . . . . . . . . . . . . . . 141
5.1.2 Poiseuille Flow in a Rectangular Conduit . . . . . . . . . . . . . . . . . . . . . . . . . 142
5.1.3 Poiseuille Flow in a Round Conduit or Annulus . . . . . . . . . . . . . . . . . . . 144
5.1.4 Poiseuille Flow in Conduits of Arbitrarily
Shaped Cross-Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
5.1.5 Couette Flow Between Concentric Circular Cylinders . . . . . . . . . . . . . . 147
5.2 Unsteady Flows When Convective Acceleration Is Absent . . . . . . . . . . . . . . . . 147
5.2.1 Impulsive Motion of a Plate—Stokes’s First Problem . . . . . . . . . . . . . . 147
5.2.2 Oscillation of a Plate—Stokes’s Second Problem . . . . . . . . . . . . . . . . . . 149
5.3 Other Unsteady Flows When Convective Acceleration
Is Absent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
5.3.1 Impulsive Plane Poiseuille and Couette Flows . . . . . . . . . . . . . . . . . . . . . 152
5.3.2 Impulsive Circular Couette Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
ix
x
Contents
5.4
Steady Flows When Convective Acceleration Is Present . . . . . . . . . . . . . . . . . . . 154
5.4.1 Plane Stagnation Line Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
5.4.2 Three-Dimensional Axisymmetric Stagnation Point Flow . . . . . . . . . . . . 158
5.4.3 Flow into Convergent or Divergent Channels . . . . . . . . . . . . . . . . . . . . . . 158
5.4.4 Flow in a Spiral Channel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
5.4.5 Flow Due to a Round Laminar Jet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
5.4.6 Flow Due to a Rotating Disk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
Chapter 6
The Boundary Layer Approximation
6.1 Introduction to Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
6.2 The Boundary Layer Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
6.3 Boundary Layer Thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
6.4 Falkner-Skan Solutions for Flow Past a Wedge . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
6.4.1 Boundary Layer on a Flat Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
6.4.2 Stagnation Point Boundary Layer Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
6.4.3 General Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
6.5 The Integral Form of the Boundary Layer Equations . . . . . . . . . . . . . . . . . . . . . . . 179
6.6 Axisymmetric Laminar Jet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
6.7 Flow Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
6.8 Transformations for Nonsimilar Boundary Layer Solutions . . . . . . . . . . . . . . . . . 184
6.8.1 Falkner Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
6.8.2 von Mises Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
6.8.3 Combined Mises-Falkner Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . 187
6.8.4 Crocco’s Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
6.8.5 Mangler’s Transformation for Bodies of Revolution . . . . . . . . . . . . . . . . 188
6.9 Boundary Layers in Rotating Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
Chapter 7
Thermal Effects
7.1 Thermal Boundary Layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
7.2 Forced Convection on a Horizontal Flat Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
7.2.1 Falkner-Skan Wedge Thermal Boundary Layer . . . . . . . . . . . . . . . . . . . . . 195
7.2.2 Isothermal Flat Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
7.2.3 Flat Plate with Constant Heat Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
7.3 The Integral Method for Thermal Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
7.3.1 Flat Plate with a Constant Temperature Region . . . . . . . . . . . . . . . . . . . . . 198
7.3.2 Flat Plate with a Constant Heat Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
Contents
7.4 Heat Transfer Near the Stagnation Point of an Isothermal Cylinder . . . . . . . . 200
7.5 Natural Convection on an Isothermal Vertical Plate . . . . . . . . . . . . . . . . . . . . . . . 201
7.6 Natural Convection on a Vertical Plate with Uniform Heat Flux . . . . . . . . . . . 202
7.7 Thermal Boundary Layer on Inclined Flat Plates . . . . . . . . . . . . . . . . . . . . . . . . . 203
7.8 Integral Method for Natural Convection on an
Isothermal Vertical Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
7.9 Temperature Distribution in an Axisymmetric Jet . . . . . . . . . . . . . . . . . . . . . . . . . 204
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
Chapter 8
Low Reynolds Number Flows
8.1 Stokes Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
8.2 Slow Steady Flow Past a Solid Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
8.3 Slow Steady Flow Past a Liquid Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
8.4 Flow Due to a Sphere Undergoing Simple Harmonic Translation . . . . . . . . . . 212
8.5 General Translational Motion of a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
8.6 Oseen’s Approximation for Slow Viscous Flow . . . . . . . . . . . . . . . . . . . . . . . . . . 214
8.7 Resolution of the Stokes/Whitehead Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
Chapter 9
Flow Stability
9.1
9.2
Linear Stability Theory of Fluid Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
Thermal Instability in a Viscous Fluid—Rayleigh-Bénard
Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
9.3 Stability of Flow Between Rotating Circular Cylinders—
Couette-Taylor Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
9.4 Stability of Plane Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
Chapter 10
Turbulent Flows
10.1
10.2
10.3
10.4
10.5
10.6
10.7
The Why and How of Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
Statistical Approach—One-Point Averaging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
Zero-Equation Turbulent Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
One-Equation Turbulent Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
Two-Equation Turbulent Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
Stress-Equation Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
Equations of Motion in Fourier Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
xi
xii
Contents
10.8 Quantum Theory Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
10.9 Large Eddy Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
10.10 Phenomenological Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
10.11 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
Chapter 11
Computational Methods—Ordinary
Differential Equations
11.1
11.2
11.3
11.4
11.5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
Numerical Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
Numerical Integration of Ordinary Differential Equations . . . . . . . . . . . . . . . . 267
The Finite Element Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
Linear Stability Problems—Invariant Imbedding and
Riccati Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
11.6 Errors, Accuracy, and Stiff Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
Chapter 12
Multidimensional Computational Methods
12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
12.2 Relaxation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
12.3 Surface Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
12.4 One-Step Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
12.4.1 Forward Time, Centered Space—Explicit . . . . . . . . . . . . . . . . . . . . . . . 297
12.4.2 Dufort-Frankel Method—Explicit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
12.4.3 Crank-Nicholson Method—Implicit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
12.4.4 Boundary Layer Equations—Crank-Nicholson . . . . . . . . . . . . . . . . . . 299
12.4.5 Boundary Layer Equation—Hybrid Method . . . . . . . . . . . . . . . . . . . . . 303
12.4.6 Richardson Extrapolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
12.4.7 Further Choices for Dealing with Nonlinearities . . . . . . . . . . . . . . . . . 304
12.4.8 Upwind Differencing for Convective
Acceleration Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
12.5 Multistep, or Alternating Direction, Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
12.5.1 Alternating Direction Explicit (ADE) Method . . . . . . . . . . . . . . . . . . . 305
12.5.2 Alternating Direction Implicit (ADI) Method. . . . . . . . . . . . . . . . . . . . 305
12.6 Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
12.7 Leapfrog Method—Explicit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
12.8 Lax-Wendroff Method—Explicit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310
12.9 MacCormack’s Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
Contents
12.9.1 MacCormack’s Explicit Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312
12.9.2 MacCormack’s Implicit Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312
12.10 Discrete Vortex Methods (DVM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
12.11 Cloud in Cell Method (CIC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
Appendix
A.1 Vector Differential Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318
A.2 Vector Integral Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
A.3 Fourier Series and Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323
A.4 Solution of Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
A.4.1 Method of Frobenius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
A.4.2 Mathieu Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326
A.4.3 Finding Eigenvalues—The Riccati Method . . . . . . . . . . . . . . . . . . . . . . 327
A.5 Index Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
A.6 Tensors in Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333
A.7 Tensors in Orthogonal Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 337
A.7.1 Cylindrical Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
A.7.2 Spherical Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
A.8 Tensors in General Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .356
xiii
Preface
This book covers material for second fluid dynamics courses at the senior/graduate
level. Students are introduced to three-dimensional fluid mechanics and classical theory,
with an introduction to modern computational methods. Problems discussed in the text
are accompanied by examples and computer programs illustrating how classical theory
can be applied to solve practical problems with techniques that are well within the
capabilities of present-day personal computers.
Modern fluid dynamics covers a wide range of subject areas and facets—far too
many to include in a single book. Therefore, this book concentrates on incompressible
fluid dynamics. Because it is an introduction to basic computational fluid dynamics,
it does not go into great depth on the various methods that exist today. Rather, it
focuses on how theory and computation can be combined and applied to problems to
demonstrate and give insight into how various describing parameters affect the behavior
of the flow. Many large and expensive computer programs are used in industry today
that serve as major tools in industrial design. In many cases the user does not have any
information about the program developers’ assumptions. This book shows students how
to test various methods and ask the right questions when evaluating such programs.
The references in this book are quite extensive—for three reasons. First, the originator of the work deserves due credit. Many of the originators’ names have become
associated with their work, so referring to an equation as the Orr-Sommerfeld equation
is common shorthand.
A more subversive reason for the number of references is to entice students to
explore the history of the subject and how the world has been affected by the growth of
science. Isaac Newton (1643–1747) is credited with providing the first solid footings
of fluid dynamics. Newton, who applied algebra to geometry and established the fields
of analytical geometry and the calculus, combined mathematical proof with physical
observation. His treatise Philosophiae Naturalis Principia Mathematica not only firmly
established the concept of the scientific method, but it led to what is called the Age of
Enlightenment, which became the intellectual framework for the American and French
Revolutions and led to the birth of the Industrial Revolution.
The Industrial Revolution, which started in Great Britain, produced a revolution in
science (in those days called “natural philosophy” in reference to Newton’s treatise)
of gigantic magnitude. In just a few decades, theories of dynamics, solid mechanics,
fluid dynamics, thermodynamics, electricity, magnetism, mathematics, medical science,
and many other sciences were born, grew, and thrived with an intellectual verve never
before found in the history of mankind. As a result, the world saw the invention of steam
engines and locomotives, electric motors and light, automobiles, the telephone, manned
flight, and other advances that had only existed in dreams before then. A chronologic
and geographic study of the references would show how ideas jumped from country to
xiv
Preface
country and how the time interval between the advances shortened dramatically in time.
Truly, Newton’s work was directly responsible for bringing civilization from the dark
ages to the founding of democracy and the downfall of tyranny.
This book is the product of material covered in many classes over a period of
five decades, mostly at The University of Michigan. I arrived there as a student at the
same time as Professor Chia-Shun Yih, who over the years I was fortunate to have as
a teacher, colleague, and good friend. His lively presentations lured many of us to the
excitement of fluid dynamics. I can only hope that this book has a similar effect on its
readers.
I give much credit for this book to my wife, June, who encouraged me greatly
during this work—in fact, during all of our 50+ years of marriage! Her proofreading
removed some of the most egregious errors. I take full credit for any that remain.
xv
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Chapter 1
Fundamentals
1.1 Introduction
1.2 Velocity, Acceleration, and the
Material Derivative
1.3 The Local Continuity Equation
1.4 Path Lines, Streamlines, and
Stream Functions
1.4.1 Lagrange’s Stream Function
for Two-Dimensional
Flows
1.4.2 Stream Functions for
Three-Dimensional Flows,
Including Stokes Stream
Function
1.5 Newton’s Momentum Equation
1
4
5
7
7
11
13
1.6 Stress
1.7 Rates of Deformation
1.8 Constitutive Relations
1.9 Equations for Newtonian Fluids
1.10 Boundary Conditions
1.11 Vorticity and Circulation
1.12 The Vorticity Equation
1.13 The Work-Energy Equation
1.14 The First Law of Thermodynamics
1.15 Dimensionless Parameters
1.16 Non-Newtonian Fluids
1.17 Moving Coordinate Systems
Problems—Chapter 1
14
21
24
27
28
29
34
36
37
39
40
41
43
1.1 Introduction
A few basic laws are fundamental to the subject of fluid mechanics: the law of conservation of mass, Newton’s laws, and the laws of thermodynamics. These laws bear
a remarkable similarity to one another in their structure. They all state that if a given
volume of the fluid is investigated, quantities such as mass, momentum, and energy
will change due to internal causes, net change in that quantity entering and leaving
the volume, and action on the surface of the volume due to external agents. In fluid
mechanics, these laws are best expressed in rate form.
Since these laws have to do with some quantities entering and leaving the volume
and other quantities changing inside the volume, in applying these fundamental laws to
a finite-size volume it can be expected that both terms involving surface and volume
integrals would result. In some cases a global description is satisfactory for carrying
out further analysis, but often a local statement of the laws in the form of differential
equations is preferred to obtain more detailed information on the behavior of the quantity
under investigation.
1
2
Fundamentals
There is a way to convert certain types of surface integrals to volume integrals
that is extremely useful for developing the derivations. This is the divergence theorem,
expressed in the form
· UdV
(1.1.1)
n · UdS =
S
V
1
In this expression U is an arbitrary vector and n is a unit normal to the surface S. The
closed surface completely surrounding the volume V is S. The unit normal is positive
if drawn outward from the volume.
The theorem assumes that the scalar and vector quantities are finite and continuous
within V and on S. It is sometimes useful to add appropriate singularities in these functions, generating additional terms that can be simply evaluated. This will be discussed
in later chapters in connection with inviscid flows.
In studying fluid mechanics three of the preceding laws lead to differential
equations—namely, the law of conservation of mass, Newton’s law of momentum,
and the first law of thermodynamics. They can be expressed in the following descriptive form:
⎡
⎤
mass
Rate at which ⎣ momentum ⎦ accumulates within the volume
energy
⎡
⎤
mass
+ rate at which ⎣ momentum ⎦ enters the volume
energy
⎡
⎤
mass
− rate at which ⎣ momentum ⎦ leaves the volume
energy
⎡
⎤
0
⎦
net force acting on volume
=⎣
rate of heat addition + rate of work done on volume
These can be looked at as a balance sheet for mass, momentum, and energy, accounting
for rate changes on the left side of the equation by describing on the right side the
external events that cause the changes. As simple as these laws may appear to us today,
it took many centuries to arrive at these fundamental results.
Some of the quantities we will see in the following pages, like pressure and
temperature, have magnitude but zero directional property. Others have various degrees
of directional properties. Quantities like velocity have magnitude and one direction
associated with them, whereas others, like stress, have magnitude and two directions
associated with them: the direction of the force and the direction of the area on which
it acts.
The general term used to classify these quantities is tensor. The order of a tensor
refers to the number of directions associated with them. Thus, pressure and temperature
are tensors of order zero (also referred to as scalars), velocity is a tensor of order one
(also referred to as a vector), and stress is a tensor of order two. Tensors of order higher
1
Vectors are denoted by boldface.
1.1
Introduction
than two usually are derivatives or products of lower-order tensors. A famous one is the
fourth-order Einstein curvature tensor of relativity theory.
To qualify as a tensor, a quantity must have more than just magnitude and directionality. When the components of the tensor are compared in two coordinate systems that
have origins at the same point, the components must relate to one another in a specific
manner. In the case of a tensor of order zero, the transformation law is simply that the
magnitudes are the same in both coordinate systems. Components of tensors of order one
must transform according to the parallelogram law, which is another way of stating that
the components in one coordinate system are the sum of products of direction cosines
of the angles between the two sets of axes and the components in the second system.
For components of second-order tensors, the transformation law involves the sum of
products of two of the direction cosines between the axes and the components in the
second system. (You may already be familiar with Mohr’s circle, which is a graphical
representation of this law in two dimensions.) In general, the transformation law for a
tensor of order N then will involve the sum of N products of direction cosines and the
components in the second system. More detail on this is given in the Appendix.
One example of a quantity that has both directionality and magnitude but is not
a tensor is finite angle rotations. A branch of mathematics called quaternions was
invented by the Irish mathematician Sir William Rowan Hamilton in 1843 to deal with
these and other problems in spherical trigonometry and body rotations. Information
about quaternions can be found on the Internet.
In dealing with the general equations of fluid mechanics, the equations are easiest
to understand when written in their most compact form—that is, in vector form. This
makes it easy to see the grouping of terms, the physical interpretation of them, and
subsequent manipulation of the equations to obtain other interpretations. This general
form, however, is usually not the form best suited to solving particular problems. For
such applications the component form is better. This, of course, involves the selection
of an appropriate coordinate system, which is dictated by the geometry of the problem.
When dealing with flows that involve flat surfaces, the proper choice of a coordinate
system is Cartesian coordinates. Boundary conditions are most easily satisfied, manipulations are easiest, and equations generally have the fewest number of terms when
expressed in these coordinates. Trigonometric, exponential, and logarithmic functions
are often encountered. The conventions used to represent the components of a vector,
for example, are typically vx vy vz v1 v2 v3 , and u v w. The first of these
conventions use x y z to refer to the coordinate system, while the second convention
uses x1 x2 x3 . This is referred to either as index notation or as indicial notation, and
it is used extensively in tensor analysis, matrix theory, and computer programming. It
frequently is more compact than the x y z notation.
For geometries that involve either circular cylinders, ellipses, spheres, or ellipsoids,
cylindrical polar, spherical polar, or ellipsoidal coordinates are the appropriate choice,
since they make satisfaction of boundary conditions easiest. The mathematical functions
and the length and complexity of equations become more complicated than in Cartesian
coordinates.
Beyond these systems, general tensor analysis must be used to obtain governing
equations, particularly if nonorthogonal coordinates are used. While it is easy to write
the general equations in tensor form, breaking down these equations into component
form in a specific non-Cartesian coordinate frame frequently involves a fair amount of
work. This is discussed in more detail in the Appendix.
3
4
Fundamentals
1.2 Velocity, Acceleration, and the Material Derivative
A fluid is defined as a material that will undergo sustained motion when shearing forces
are applied, the motion continuing as long as the shearing forces are maintained. The
general study of fluid mechanics considers a fluid to be a continuum. That is, the fact
that the fluid is made up of molecules is ignored but rather the fluid is taken to be a
continuous media.
In solid and rigid body mechanics, it is convenient to start the geometric discussion
of motion and deformation by considering the continuum to be made up of a collection
of particles and consider their subsequent displacement. This is called a Lagrangian,
or material, description, named after Joseph Louis Lagrange (1736–1836). To illustrate
its usage, let XX0 Y0 Z0 t YX0 Y0 Z0 t ZX0 Y0 Z0 t be the position at time
t of a particle initially at the point X0 Y0 Z0 . Then the velocity and acceleration of
that particle is given by
XX0 Y0 Z0 t
t
XX0 Y0 Z0 t
vy X0 Y0 Z0 t =
t
XX0 Y0 Z0 t
vz X0 Y0 Z0 t =
t
vx X0 Y0 Z0 t =
(1.2.1)
and
2
vx X0 Y0 Z0 t XX0 Y0 Z0 t
=
t
t2
vy X0 Y0 Z0 t 2 YX0 Y0 Z0 t
=
ay X0 Y0 Z0 t =
t
t2
v X Y Z t 2 ZX0 Y0 Z0 t
=
az X0 Y0 Z0 t = z 0 0 0
t
t2
The partial derivatives signify that differentiation is performed holding X0 Y0 and
Z0 fixed.
This description works well for particle dynamics, but since fluids consist of an
infinite number of flowing particles in the continuum hypothesis, it is not convenient to
label the various fluid particles and then follow each particle as it moves. Experimental
techniques certainly would be hard pressed to perform measurements that are suited to
such a description. Also, since displacement itself does not enter into stress-geometric
relations for fluids, there is seldom a need to consider using this descriptive method.
Instead, an Eularian, or spatial, description, named after Leonard Euler
(1707–1783), is used. This description starts with velocity, written as v = vx t, where
x refers to the position of a fixed point in space, as the basic descriptor rather than
displacement. To find acceleration, recognize that acceleration means the rate of change
of the velocity of a particular fluid particle at a position while noting that the particle
is in the process of moving from that position at the time it is being studied. Thus, for
instance, the acceleration component in the x direction is defined as
ax X0 Y0 Z0 t =
vx x + vx t y + vy t z + vz t t + t − vx x y z t
t→0
t
vx
vx
vx
vx
+ vx
+ vy
+ vz
=
t
x
y
z
ax = lim
1.3
5
The Local Continuity Equation
since the rate at which the particle leaves this position is vdt. Similar results can be
obtained in the y and z direction, leading to the general vector form of the acceleration as
v
+ v · v
(1.2.2)
t
The first term in equation (1.2.2) is referred to as the temporal acceleration, and the
second as the convective, or occasionally advective, acceleration.
Note that the convective acceleration terms are quadratic in the velocity components
and hence mathematically nonlinear. This introduces a major difficulty in the solution
of the governing equations of fluid flow. At this point it might be thought that since the
Lagrangian approach has no nonlinearities in the acceleration expression, it could be
more convenient. Such, however, is not the case, as the various force terms introduced
by Newton’s laws all become nonlinear in the Lagrangian approach. In fact, these
nonlinearities are even worse than those found using the Eularian approach.
The convective acceleration term v · v can also be written as
a=
1
v · v = v · v + v × × v
(1.2.3)
2
This can be shown to be true by writing out the left- and right-hand sides.
The operator t + v · , which appears in equation (1.2.2), is often seen in fluid
mechanics. It has been variously called the material, or substantial, derivative, and
represents differentiation as a fluid particle is followed. It is often written as
D
= + v ·
(1.2.4)
Dt t
Note that the operator v · is not a strictly correct vector operator, as it does not
obey the commutative rule. That is, v · = · v. This operator is sometimes referred
to as a pseudo-vector. Nevertheless, when it is used to operate on a scalar like mass
density or a vector such as velocity, the result is a proper vector as long as no attempt
is made to commute it.
1.3 The Local Continuity Equation
To derive local equations that hold true at any point in our fluid, a volume of arbitrary
shape is constructed and referred to as a control volume. A control volume is a device
used in analyzing fluid flows to account for mass, momentum, and energy balances. It
is usually a volume of fixed size, attached to a specified coordinate system. A control
surface is the bounding surface of the control volume. Fluid enters and leaves the
control volume through the control surface. The density and velocity inside and on the
surface of the control volume are represented by and v. These quantities may vary
throughout the control volume and so are generally functions of the spatial coordinates
as well as time.
The mass of the fluid inside our control volume is
dV . For a control volume
V
fixed in space, the rate of change of mass inside of our control volume is
d
dV =
dV
(1.3.1)
dt
V t
V
The rate at which mass enters the control volume through its surface is
v · n dS
(1.3.2)
S
6
Fundamentals
where v · n dS is the mass rate of flow out of the small area dS. The quantity v · n is
the normal component of the velocity to the surface. Therefore, a positive value of v · n
means the v · n flow locally is out of the volume, whereas a negative value means that
it is into the volume.
The net rate of change of mass inside and entering the control volume is then found
by adding together equations (1.3.1) and (1.3.2) and setting the sum to zero. This gives
v · n dS = 0
(1.3.3)
dV +
S
V t
To obtain the local continuity equation in differential equation form, use (1.1.1) to
transform the surface integral to a volume integral. This gives
· vdV
v · n dS =
S
V
Making this replacement in (1.3.3) gives
· vdV = 0
dV +
V t
V
Rearranging this to put everything under the same integral sign gives
+ · v dV = 0
t
V
(1.3.4)
Since the choice of the control volume was arbitrary and since the integral must vanish
no matter what choice of control volume was made, the only way this integral can
vanish is for the integrand to vanish. Thus,
+ · v = 0
t
(1.3.5)
Equation (1.3.5) is the local form of the continuity equation. An alternate expression
of it can be obtained by expanding the divergence term to obtain
+ v · + · v = 0
t
The first two terms represent the material derivative of . It is the change of mass density
D
≡ t + v · ,
as we follow an individual fluid particle for an infinitesimal time. Writing Dt
equation (1.3.5) becomes
D
+ · v = 0
Dt
(1.3.6)
The first term in equation (1.3.6) represents the change in density following a particle,
and the second represents (as shall be shown later) mass density times the change in
volume per unit volume as a mass particle is followed.
An incompressible flow is defined as one where the mass density of a fluid particle
does not change as the particle is followed. This can be expressed as
D
= 0
Dt
(1.3.7)
Thus, the continuity equation for an incompressible flow is
· v = 0
(1.3.8)
1.4
7
Path Lines, Streamlines, and Stream Functions
1.4 Path Lines, Streamlines, and Stream Functions
A path line is a line along which a fluid particle actually travels. Since it is the time
history of the position of a fluid particle, it is best described using the Lagrangian
description. Since the particle incrementally moves in the direction of the velocity
vector, the equation of a path line is given by
dt =
dX
dY
dZ
=
=
Vx
Vy
Vz
(1.4.1)
the integration being performed with X0 Y0 , and Z0 held fixed.
A streamline is defined as a line drawn in the flow at a given instant of time such
that the fluid velocity vector at any point on the streamline is tangent to the line at that
point. The requirement of tangency means that the streamlines are given by the equation
dx dy dz
=
=
vx
vy
vz
(1.4.2)
While in principle the streamlines can be found from equation (1.4.2), it is usually easier
to pursue a method utilizing the continuity equation and stream functions described in
the following.
A stream surface (or stream sheet) is a collection of adjacent streamlines, providing
a surface through which there is no flow. A stream tube is a tube made up of adjoining
streamlines.
For steady flows (time-independent), path lines and streamlines coincide. For
unsteady flows (time-dependent), path lines and streamlines may differ. Generally path
lines are more difficult to find analytically than are streamlines, and they are of less use
in practical applications.
The continuity equation imposes a restriction on the velocity components. It is a
relation between the various velocity components and mass density. It is not possible
to directly integrate the continuity equation for one of the velocity components in terms
of the others. Rather, this is handled through the use of an intermediary scalar function
called a stream function.
Stream functions are used principally in connection with incompressible flows—
that is, flows where the density of individual fluid particles does not change as the
particle moves in the flow. In such a flow, equation (1.3.8) showed that the continuity
equation reduces to
vx vy vz
+
+
= 0
x
y
z
(1.4.3)
Because the equations that relate stream functions to velocity differ in two and three
dimensions, the two cases will be considered separately.
1.4.1
Lagrange’s Stream Function for Two-Dimensional Flows
For two-dimensional flows, equation (1.4.3) reduces to
vx vy
+
= 0
x
y
(1.4.4)
8
Fundamentals
indicating that one of the velocity components can be expressed in terms of the other.
To attempt to do this directly by integration, the result would be
v
x
vy = −
dy
x
Unfortunately, this cannot be integrated in any straightforward manner. Instead, it is
more convenient to introduce an intermediary, a scalar function called Lagrange’s
stream function, that allows the integration to be carried out explicitly.
Let
(1.4.5)
vx =
y
Then equation (1.4.4) becomes
vy
2
+
= 0
y xy
which can be integrated with respect to y to give
(1.4.6)
x
Therefore, expressing the two velocity components in terms of
in the manner of
equations (1.4.5) and (1.4.6) guarantees that continuity is satisfied for an incompressible flow.
In two dimensions, the tangency requirement equation (1.4.2) reduces to dx/vx =
dy/vy on the streamline. Using equations (1.4.5) and (1.4.6) reduces this to dx
= −dy ,
y
x
or, upon rearrangement,
vy = −
0=
dx +
dy = d
x
y
(1.4.7)
Equation (1.4.7) states that along a streamline d vanishes. In other words, on a
streamline, is constant. This is the motivation for the name stream function for .
Sometimes it is more convenient in a given problem to use cylindrical polar coordinates rather than Cartesian coordinates. The relation of cylindrical polar coordinates
to a Cartesian frame is shown in Figure 1.4.1, with the appropriate tangent unit vectors
shown at point P. The suitable expressions for the velocity components in cylindrical
polar coordinates using Lagrange’s stream function are
and v = −
(1.4.8)
r
r
Note from either equation (1.4.5), (1.4.6), or (1.4.8) that the dimension of is length
squared per unit time.
A physical understanding of can also be found by looking at the discharge through
a curve C as seen in Figure 1.4.2. The discharge through the curve C per unit distance
into the paper is
B
Q=
vx dy − vy dx
(1.4.9)
vr =
A
When the velocity components are replaced by their expressions in terms of , equation (1.4.9) becomes
B
B
Q=
dy − −
dx =
d = B − A
(1.4.10)
y
x
A
A
1.4
9
Path Lines, Streamlines, and Stream Functions
z
P
z
y
r
θ
x
Figure 1.4.1 Cylindrical coordinate conventions
B
ds
dy
Curve C
vx
dx
vy
A
Figure 1.4.2 Discharge through a curve
Thus, the discharge through the curve C is equal to the difference of the value of
the endpoints of C.
at
Example 1.4.1 Lagrange stream function
Find the stream function associated with the two-dimensional incompressible flow:
vr = U 1 −
a2
cos
r2
v = −U 1 +
a2
sin
r2
Solution. Since vr = /r , by integration of vr with respect to , find
=
U 1−
a2
a2
cos
dr
=
U
r
−
sin + fr
r2
r
The “constant of integration” f here possibly depends on r, since we have integrated a
partial derivative that had been taken with respect to , the derivative being taken with
r held constant in the process.
10
Fundamentals
Differentiating this
with respect to r gives
a2
dfr
= −v = U 1 + 2 sin +
r
r
dr
Comparing the two expressions for v , we see that df/dr must vanish, so f must be a
constant. We can set this constant to any convenient value without affecting the velocity
components or the discharge. Here, for simplicity, we set it to zero. This gives
=U r−
a2
sin
r
Example 1.4.2 Path lines
Find the path lines for the flow of Example 1.4.1. Find also equations for the position
of a fluid particle along the path line as a function of time.
Solution. For steady flows, path lines and streamlines coincide. Therefore, on
a path line,
is constant. Using the stream function from the previous example,
we have
a2
sin = constant = B (say).
r
=U r−
The equations for a path line in cylindrical polar coordinates are
dt =
Since on the path line sin =
vr = U 1 −
0
2
r− ar
d
dr
=r
vr
v
, then
a2
r2
1 − sin2 =
U
r2
r 2 − a2 2 −
0 r
2
If for convenience we introduce new constants
F 2 = a2 +
2
0
2
+
0
a2 +
2
0
4
G 2 = a2 +
2
0
2
−
0
and a new variable s = r/F . Then, since a4 = FG2 and 2a2 +
vr =
U
F 2 s2
F 4 s4 − F 2 + G2 F 2 s2 + F 2 G2 =
UG
F 2 s2
2
0
a2 +
4
0
2
k=
F
G
= F 2 + G2 ,
1 − s1 − k2 s2
Consequently, we have
dt =
F 2 s2
ds
UG1 − s1 − k2 s2
Upon integration this gives
F 2 r/F
s2
ds where r0 is the value of r at t0
t − t0 =
UG r0 /F 1 − s1 − k2 s2
1.4
11
Path Lines, Streamlines, and Stream Functions
The integral in the previous expression is related to what are called elliptic integrals.
Its values can be found tabulated in many handbooks or by numerical integration. Once
r is found as a function of t on a path line (albeit in an inverse manner, since we have
t as a function of r), the angle is found from = sin−1 0a2 . From this we see that
r− r
a2
the constant 0 can also be interpreted as 0 = r0 − r sin 0 . Note that calculation of
0
path lines usually is much more difficult than are calculations for streamlines.
1.4.2
Stream Functions for Three-Dimensional Flows, Including
Stokes Stream Function
For three dimensions, equation (1.4.3) states that there is one relationship between the
three velocity components, so it is expected that the velocity can be expressed in terms
of two scalar functions. There are at least two ways to do this. The one that retains the
interpretation of stream function as introduced in two dimensions is
v = ×
(1.4.11)
where and are each constant on stream surfaces. The intersection of these stream
surfaces is a streamline. Equations (1.4.4) and (1.4.5) are in fact a special case of this
result with = z.
Although equation (1.4.11) preserves the advantage of having the stream functions
constant on a streamline, it has the disadvantage that v is a nonlinear function of
and . Introducing further nonlinearities into an already highly nonlinear problem is not
usually helpful!
Another possibility for solving the continuity equation is to write
v = × A
(1.4.12)
which guarantees ·v = 0 for any vector A. For the two-dimensional case, A = 0 0
corresponds to our Lagrange stream function. Since only two scalars are needed, in three
dimensions one component of A can be arbitrarily set to zero. (Some thought must be
used in doing this. Obviously, in the two-dimensional case, difficulty would be encountered if one of the components of A that we set to zero was the z component.) The form
of equation (1.4.12), while guaranteeing satisfaction of continuity, has not been much
used, since the appropriate boundary conditions to be imposed on A can be awkward.
A particular three-dimensional case in which a stream function is useful is that of
axisymmetric flow. Taking the z-axis as the axis of symmetry, either spherical polar
coordinates,
z
y
R = x2 + y2 + z2 = cos−1
= tan−1
(1.4.13)
R
z
(see Figure 1.4.3, with the appropriate tangent unit vectors shown at point P), or
cylindrical polar coordinates with
y
(1.4.14)
= tan−1 z = z
r = x2 + y 2
x
(see Figure 1.4.1) can be used. The term axisymmetry means that the flow appears the
same in any = constant plane, and the velocity component normal to that plane is
zero. (There could in fact be a swirl velocity component v without changing anything
12
Fundamentals
z
P
β
R
y
θ
x
Figure 1.4.3 Spherical coordinates conventions
we have said. It would not be related to or determined by .) Since any plane given by
equal to a constant is therefore a stream surface, we can use equation (1.4.11) with
= , giving
vR =
in spherical coordinates with
1
R2 sin
v =
−1
R sin R
(1.4.15)
vz =
1
r r
(1.4.16)
= R , or
vr = −
1
r z
in cylindrical coordinates, with = r z. This stream function in either a cylindrical
or spherical coordinate system is called the Stokes stream function. Note that the
dimensions of the Stokes stream function is length cubed per unit time, thus differing
from Lagrange’s stream function by a length dimension.
The volumetric discharge through an annular region is given in terms of the Stokes
stream function by
B
B
−1
1
dr −
dz
2r
2rvz dr − vr dz =
Q=
r r
r z
A
A
(1.4.17)
B
B
= 2
dr +
dz = 2
d = 2 B − A
r
z
A
A
Example 1.4.3 Stokes stream function
A flow field in cylindrical polar coordinates is given by
vr = −
15Ua3 rz
r 2 + z2 5/2
vz =
Find the Stokes stream function for this flow.
Solution. Since vr =
1
,
r z
= −15Ua3
Ua3 r 2 − 2z2
2r 2 + z2 5/2
integration of vr with respect to z gives
rz
r2
3
rdz
=
−Ua
+ fr
r 2 + z2 5/2
2r 2 + z2 3/2
1.5
13
Newtons Momentum Equation
Differentiating this with respect to r, we have
rr 2 − 2z2 dfr
= rvz = 05Ua3 2
+
r
r + z2 5/2
dr
Comparing this with the preceding expression for vz , we see that dfr/dr = rU , and
so df /dr = 05Ur 2 . Thus, finally
=−
Ur 2
a3
Ua3 r 2
2
+
05Ur
=
1
−
2r 2 + z2 3/2
2
r 2 + z2 3/2
The preceding derivations were all for incompressible flow, whether the flow is steady or
unsteady. The derivations can be extended to steady compressible flow by recognizing
that since for these flows the continuity equation can be written as · v = 0, the
previous stream functions can be used simply by replacing v by v.
A further extension, to unsteady compressible flows, is possible by regarding
time as a fourth dimension and using the extended four-dimensional vector V =
vx vy vz together with the augmented del operator a = x y z t . The
continuity equation (1.3.5) can then be written as
a · V = 0
(1.4.18)
The previous results can be extended to this general case in a straightforward manner.
1.5 Newtons Momentum Equation
Next apply
the interior
the control
momentum
Newton’s momentum equation
to the control volume. The momentum in
of the control volume is
vdV . The rate at which momentum enters
V
volume through its surface is S v v · n dS. The net rate of change of
is then
v
d
v v · n dS =
v dV +
dV +
· vvdV
dt
t
V
S
V
V
v
=
+ v · v + v · v dV
(1.5.1)
t
V
Dv
v
=
v
+ · v +
+ v · v dV =
dV
t
t
Dt
V
V
The first bracket in the third line of equation (1.5.1) vanishes by virtue of equation
(1.3.5), the continuity equation. The second bracket represents the material derivative
of the velocity, which is the acceleration.
Note that in converting the surface integral to a volume integral using equation (1.1.1) the “vector” was vv, which is a product of two vectors but is neither the dot
nor cross product. It is sometimes referred to as the indefinite product. This usage is in
fact a slight generalization of equation (1.1.1), which can be verified by writing out the
term in component form.
The forces applied to the surface of the control volume are due to pressure and
viscous forces on the surface and gravitational force distributed throughout the volume.
The pressure force is normal to the surface and points toward the volume. On an
14
Fundamentals
infinitesimal element its value is thus −pn dS—the minus sign because the pressure
force acts toward the area dS and thus is opposite to the unit normal.
The viscous force in general will have both normal and tangential components. For
now, simply write it as n dS acting on a small portion of the surface, where n is
called the stress vector. It is the force per unit area acting on the surface dS. The n
superscript reminds us that the stress vector is applied to a surface with normal pointing
in the n direction.
The gravity force per unit volume is written as g dV , where the magnitude of g is 9.80
or 32.17, depending on whether the units used are SI or British. The net force is then
g dV
−pn + n dS +
V
S
Equating this to the net change in momentum gives
Dv
g dV
−pn + n dS +
dV =
Dt
V
S
V
Use of the divergence theorem allows us to write
−p dV
−pn dS =
S
(1.5.2)
(1.5.3)
V
This reduces equation (1.5.2) to
Dv
n dS
−p + gdV +
dV =
Dt
S
V
V
(1.5.4)
At this point a similar simplification for the viscous term is not possible. This term is
investigated further in the next section.
1.6 Stress
Stress is defined as a force applied to an area divided by that area. Thus, two directions
are associated with stress: the direction of the force and the direction (orientation) of the
area. Therefore, stress has a more complicated mathematical structure than does either
a scalar or a vector. To put this into its simplest form, three special stress vectors will
be introduced that act on mutually orthogonal surfaces whose faces are orientated with
normals along our coordinate axes.
When a material is treated as a continuum, a force must be applied as a quantity
distributed over an area. (In analysis, a concentrated force or load can sometimes be a
convenient idealization. In a real material, any concentrated force would provide very
large changes—in fact, infinite changes—both in deformation and in the material.) The
previously introduced stress vector n , for example, is defined as
n = lim
S→0
F
S
(1.6.1)
where S is the magnitude of the infinitesimal area. In the limit as S approaches zero
the direction of the normal to S is held fixed.
It appears that at a given point in the fluid there can be an infinity of different
stress vectors, corresponding to the infinitely many orientations of n that are possible.
To bring order out of such confusion, we consider three very special orientations of n
and then show that all other orientations of n produce stress vectors that are simply
related to the first three.
1.6
15
Stress
x
τ (–y )
n (–y ) n
n (–z )
τ
z
τ (–z )
y
n (–x )
τ (–x )
Figure 1.6.1 Stress vector
First consider the three special stress vectors x y z , corresponding to forces
acting on areas with unit normals pointing in the x, y, and z directions, respectively.
These act on the small tetrahedron shown in Figure 1.6.1. For a force F acting on a
surface Sx with unit normal pointing in the x direction (thus, n = i), write the stress
on this face of our tetrahedron as
F
x = lim
= xx i + xy j + xz k
(1.6.2)
Ax →0 Ax
where xx is the limit of the x component of the force acting on this face, xy is the
limit of the y component of the force acting on this face, and xz is the limit of the z
component of the force acting on this face.
Similarly, for normals pointing in the y n = j and z n = k directions we have
y = lim
F
= yx i + yy j + yz k
Ay
(1.6.3)
z = lim
F
= zx i + zy j + zz k
Az
(1.6.4)
Ay →0
and
Az →0
As in equation (1.6.2), the first subscript on the components tells the direction that
the area faces, and the second subscript gives the direction of the force component on
that face.
Upon examination of the three stress vectors shown in equations (1.6.2), (1.6.3),
and (1.6.4), we see that they can be summarized in the matrix form
⎛ x ⎞ ⎛
⎞⎛ ⎞
xx xy xz
i
= ⎝ y ⎠ = ⎝ yx yy yz ⎠ ⎝ j ⎠
z
zx zy zz
k
The nine quantities in this 3 by 3 matrix are the components of the second-order stress
tensor.
In the limit, as the areas are taken smaller and smaller, the forces acting on the four
faces of the tetrahedron are − x dS x − y dS y − z dS z , and n dS. The first three
of these forces act on faces whose normals are in the −x −y −z directions. In writing
them we have used Newton’s third law, which tells us that −x = − x −y = − y ,
16
Fundamentals
and −z = − z . The fourth of these forces acts on the slant face with area dS and
normal n.
Summing the surface forces and using dSx = nx dS dSy = ny dS dSz = nz dS,
the result is
FS = − x dS x − y dS y − z dS z + n dS
= − x nx − y ny − z nz + n dS
(1.6.5)
Considering the summation of forces as the tetrahedron shrinks to zero, we have
FS + FV =
Dv
dV
Dt
Our surface forces depend on dS, and thus are second order in the tetrahedron dimensions. The body forces and the mass times acceleration terms depend on dV, which is
of the third order in the tetrahedron dimensions. If in our analysis the tetrahedron is
decreased in size, the dV terms approach zero faster than do the dS terms. Thus, the
body force and acceleration terms are of higher (third) order in the dimensions of the
tetrahedron than the surface force terms (second). In the limit as the tetrahedron goes
to zero, we are thus left with only the second-order terms, resulting in
n = x nx + y ny + z nz
(1.6.6)
Therefore, from knowledge of the three special stress vectors x y z , we can
find the stress vector in the direction of any n at the same point.
Next, combine equations (1.6.2), (1.6.3), and (1.6.4) with (1.6.6). This gives
n = nx xx i + xy j + xz k + ny (yx i + yy j + yz k
+ nz (zx i + zy j + zz k
= inx xx + ny yx + nz zx + jnx xy + ny yy + nz zy )
(1.6.7)
+ knx xz + ny yz + nz zz
the expression after the second equals sign being a rearrangement of the preceding.
Taking the dot product of the three unit vectors i, j, k with (1.6.7) gives
xx = i · x
xy = j · x
xz = k · x
yx = i · y
yy = j · y
yz = k · y
z
zx = i ·
z
zy = j ·
(1.6.8)
z
zz = k ·
This is in agreement with our definition of xx (and so on) being the components of the
various stress vectors.
Stress components associated with force components that act in the same direction as their normal (i.e., xx yy zz ) are referred to as normal stresses. Stress
components associated with force components that act perpendicular to their normal
(i.e., yx yz zx xy zx xz ) are referred to as shear stresses. Note that the first
subscript on the components tells us the direction in which the area faces, and the
second subscript gives us the direction of the force component on that face. Positive
sign conventions for viscous stress components are given in Figure 1.6.2. The nine components xx xy xz yx yy yz zx zy zzj of our three special stress vectors
x y z can be shown to be components of a second-order tensor called the
1.6
17
Stress
τ yx
x
τ yy
τ yz
τ xx
τ zy
τ xz
τ xy
τ zz
τ zx
τ yx
τ zx
τ zz
τ yz
τ yy
τ zy
z
y
τ xy
τ xz
τ xx
Figure 1.6.2 Stress tensor sign convention
stress tensor. The stress vectors x y , and z are thus expressible in terms of these
nine components, as shown above. As seen from the manner in which the components
of the three stress vectors were introduced while carrying out the development, the
nine-component stress tensor is the collection of the three x, y, z components of each of
these three stress vectors. Thus, the stress tensor is the collection of nine components
of the three special stress vectors x y z .
Besides its use in formulating the basic equations of fluid dynamics, the stress
vector is also used to apply conditions at the boundary of the fluid, as will be seen
when we consider boundary conditions. The stress tensor is used to describe the state
of stress in the interior of the fluid.
Return now to equation (1.5.4), and change the surface integral to a volume integral.
From equation (1.6.7) we have
nx xx i + xy j + xz k + ny (yx i + yy j + yz k + nz (zx i + zy j + zz kdS
n dS =
S
s
=
xx i + xy j + xz k +
yx i + yy j + yz k
y
V x
+
( i + zy j + zz k dV
z zx
(1.6.9)
18
Fundamentals
=
i
V
+k
xy yy zy
xx yx zx
+
+
+
+
+j
x
y
z
x
y
z
xz yz zz
+
+
x
y
z
dV
Inserting this into equation (1.5.4) gives
V
yx zx
Dv
xx
dV =
+
+
−p + gdV +
i
Dt
x
y
z
V
V
+j
xy yy zy
+
+
x
y
z
+k
xz yz zz
+
+
x
y
z
(1.6.10)
dV
As was the case for the continuity equation, equation (1.6.10) must be valid no matter
what volume we choose. Therefore, it must be that
xx yx zx
Dv
= −p + g + i
+
+
Dt
x
y
z
xy yy zy
+j
+
+
x
y
z
xz yz zz
+
+
+k
x
y
z
(1.6.11)
Before leaving these equations, note that the continuity equation (1.3.5) and the
momentum equation (1.6.11) can be written in a combined matrix form as
⎛ ⎞
⎛
⎛
⎞
⎞
vx
vy
⎜
⎜
⎟
⎟
⎟
⎜
⎜ vx ⎟ + ⎜p − x − xx ⎟ + ⎜−yx
⎟
⎝
⎝
⎠
⎠
⎝
vy
t
x −xy
y p − y − yy ⎠
−xz
−yz
vz
(1.6.12)
⎛
⎞
vz
⎟
⎜−zx
⎟ = 0
+ ⎜
⎝
⎠
−
z
zy
p − z − zx
This form of the equations is referred to as the conservative form and is frequently used
in computational fluid dynamics.
Moments can be balanced in the same manner as forces. Using a finite control
mass and taking R as a position vector drawn from the point about which moments are
being taken to a general fluid particle, equating moments to the time rate of change of
moment of momentum gives
d
R × vv · ndS
R × v dV +
dt
S
V
R × n dS
R × g dV +
R × −pndS +
=
S
But
V
(1.6.13)
S
D
d
DR × v
R×v+
dV
R × vdV =
dt
Dt
Dt
V
V
(1.6.14)
1.6
19
Stress
and using the product rule and the fact that a vector crossed with itself is zero,
DR × v DR
Dv
Dv
Dv
=
×v+R×
= v×v+R×
= R×
Dt
Dt
Dt
Dt
Dt
(1.6.15)
Also, using equation (1.6.7), R × n can be written as
R × n = R × inx xx + ny yx + nz zx + jnx xy + ny yy + nz zy
+ knx xz + ny yz + nz zz
(1.6.16)
= nx iyxz − zxy + jzxx − xxz + kxxy − yxx
+ ny iyyz − zyy + jzyx − xyz + kxyy − yyx
+ nz iyzz − zzy + jzzx − xzz + kxzy − yzx
The surface integral involving n then becomes
S
R × n dS =
S
nx iyxz − zxy + jzxx − xxz + kxxy − yxx
+ ny iyyz − zyy + jzyx − xyz + kxyy − yyx
+ nz iyzz − zzy + jzzx − xzz + kxzy − yzx dS
=
iyxz − zxy + jzxx − xxz + kxxy − yxx
x
V
iyyz − zyy + jzyx − xyz + kxyy − yyx
y
+ iyzz − zzy + jzzx − xzz + kxzy − yzx dV
z
=
iyz − zy + jzx − xz + kxy − yx
+
(1.6.17)
V
+i y
xy yy zy
xz yz zz
+
+
+
+
−z
x
y
z
x
y
z
xx yx zx
xz yz zz
+
+
+
+
+j z
−x
x
y
z
x
y
z
xy yy zy
xx yx zx
+
+
+
+
−y
+k x
x
y
z
x
y
z
=
iyz − zy + jzx − xz + kxy − yx
V
xy yy zy
xx yx zx
+R× i
+
+
+
+
+j
x
y
z
x
y
z
xz yz zz
+
+
+k
dV
x
y
z
dV
20
Fundamentals
Thus, using equations (1.6.14), (1.6.15), (1.6.16), (1.6.17), equation (1.6.13) becomes
Dv
R × −pdV
· vR × vdV =
R × v + R×
dV +
Dt
V
V
V t
R× g dV +
iyz − zy + jzx − xz + kxy − yx
+
V
V
xy yy zy
xx yx zx
+
+
+
+
+j
x
y
z
x
y
z
xz yz zz
+k
+
+
dV
x
y
z
+R× i
Collecting terms and rearranging, the result is
Dv
R×v
R×
+ · v dV +
dV
t
Dt
V
V
xx yx zx
+
+
R × − p + g + i
=
x
y
z
V
(1.6.18)
xy yy zy
xz yz zz
+
+
+
+
+k
dV
x
y
z
x
y
z
iy z − zy + jzx − xz + kxy − yx dV
+
+j
V
Examining the various terms, it is seen that the integrand in the first integral is
zero by virtue of the continuity equation. The integrand in the last integral on the left
side of the equals sign is R crossed with the mass density times the acceleration, which
is the left side of equation (1.6.11). The integrand in the first integral on the right of
the integral sign is R crossed with the right side of equation (1.6.11). Thus, the two
integrals cancel and we are left with
0=
iyz − zy + jzx − xz + kxy − yx dV
V
Again, for this to be true for any volume it is the integrand that must vanish. This yields
0 = iyz − zy + jzx − xz + kxy − yx
or
yz = zy
zx = xz
xy = yx
(1.6.19)
Thus, equation (1.6.19), the result of summing moments, tells us that the order of writing
the subscripts is immaterial, since the stress due to the y force component acting on the
face with normal in the x direction is equal to the stress due to the x force component
acting on the face with normal in the y direction, and so on for the other two faces. This
interchangeability of the indices tells us that the stress tensor is a symmetric tensor,
and there are only six rather than nine numerically unique values for its components
at a point. (It has been proposed that there is the possibility of a magnetic material
to have an antisymmetric stress tensor. Such a material would indeed have a complex
mathematical description.)
1.7
21
Rates of Deformation
1.7 Rates of Deformation
In the equations developed so far, we have not identified the nature of the material we
are studying. In fact, our equations are so general at this stage that they apply to solid
materials as well as to fluids. To narrow the subject to fluids, it is necessary to show
how the fluid behaves under applied stresses. The important geometric quantity that
describes the fluids’ behavior under stress is the rate of deformation.
To define the deformation of a fluid under acting stresses, first consider motion
in the two-dimensional xy plane. Choose three neighboring points ABC (Figure 1.7.1)
selected to make up a right angle at an initial time t. A is a distance x along the x-axis
from B, and C is a distance y above B along the y-axis.
As the flow evolves through a short time interval t, the angle ABC will change,
as will the distance between the three points. At this later time t + t, these points will
have moved to A′ B′ , and C′ , as shown. Using Taylor series expansions to the first
order, the fluid that initially was at point A will have x and y velocity components
v
given by vx + vxx x and vy + xy x. The fluid initially
at point A will move a distance
v
vx
vx + x x t to the right of A, and a distance vy + xy x t above A. Similarly,
the fluid initially occupying point B has moved to point B′ —that is, vx t to the right
of B and vyt above B, and the fluid initially
at point C has moved to C′ , which is
v
vy + yy y t above C and vx + vyx y t to the right of C.
Looking first at the time rate of change of lengths, it is seen that after a time interval
t the rate of change of length along the x-axis per unit length, which will be denoted
by dxx , is the final length minus the original length divided by the original length, all
divided by t. In the limit this is
v
x + vx + x x t − x + vx t
v
x
= x
(1.7.1)
dxx =
xt
x
(vx +
∂vx
dy )dt
∂y
C′
d θ2
vx dt
vy +
∂vy
A′
dy
∂y
C
vx +
∂vx
dy
vy +
dy v
y
d θ1
B′
∂y
∂vy
dx
∂x
vy dt
y
B
vx
A
dx
vx +
x
Figure 1.7.1 Rates of deformation—two dimensions
∂vx
∂x
dx
(vy +
∂vy
∂x
dx )dt
22
Fundamentals
A similar analysis along the y-axis would give the rate of change of length per unit
length as measured along the y-axis, dyy , as
vy
y t − y + vy t
y + vy +
vy
y
dyy = lim
=
(1.7.2)
t→0
yt
y
and similarly in the z direction
v
z + vz + z z t − z + vz t
v
x
= z
(1.7.3)
dzz = lim
t→0
zt
z
The dxx dyy , and dzz are the normal rates of deformation and can loosely be thought
of as rates of normal, or extensional, strain. The term “loosely” is used, since the
definitions of strain you might be familiar with from the study of solid mechanics are
for infinitesimal strains. For finite strain, many different definitions are used in solid
mechanics for rates of strain. Since in fluid mechanics strains are always finite—and
also usually very large—using the term “rates of deformation” avoids confusion.
Note that the rate of change of volume per unit volume is the volume at t + t
minus the original volume at t divided by the original volume times t, or
lim
xyzt→0
x + dxx xt + dyy ytz + dzz zt − xyz
xyzt
(1.7.4)
= dxx + dyy + dzz = · v
This is the dilatational strain rate.
Besides changes of length, changes of angles are involved in the deformation.
Looking at the angles that AB and BC have rotated through, it is seen that
−1
˙ 1 ≈ tan
t
1
≈
and similarly
−1
˙ 2 ≈ tan
t
2
≈
vy
xt
x
xt1 + vxx t
vx
yt
y
yt1 +
vy
t
y
≈
vy
x
(1.7.5)
≈
vx
y
(1.7.6)
The sum of the two angular rates, ˙ 1 + ˙ 2 , represents the rate of change of the angle
ABC. Let
1 vy vx
dxy = dyx = 05 ˙ 1 + ˙ 2 =
+
(1.7.7)
2 x
y
be defined as the rate of shear deformation as measured in the xy plane. Similarly
define
1 vy vz
dyz = dzy =
+
(1.7.8)
2 z
y
and
dzx = dxz =
1
2
vz vx
+
x
z
(1.7.9)
as the rates of shear deformation as measured in the yx and xz planes, respectively.
1.7
23
Rates of Deformation
The one-half factor in the definition of the rate of deformation components is
introduced so that the components transform independent of our selection of axes. Since
the definition of angular rate of deformation is to some degree our option, any choice
that gives a measure of the deformation is suitable. In this case we wish to relate the
deformation rate to stress, and we would like to do it in such a manner that if we change
to another coordinate system, all quantities change correctly.
The rate of deformation components that have been arrived at can be shown to transform
as a second-order tensor. Note that if we interchange the order in which the subscripts
are written in the definitions, there is no change in the various components. That is,
dxy = dyx
dxz = dzx
and dyz = dzy
(1.7.10)
As in the case of the stress tensor, such a tensor is said to be a symmetric tensor.
It may be helpful to your physical understanding of rate of deformation to look at
what is happening from a slightly different viewpoint. Consider any two neighboring
fluid particles a distance dr apart, where the distance dr changes with time but must
remain small because the particles were initially close together. To find the rate at which
the particles separate, take the time derivative of dr, obtaining
Ddr
Dr
=d
= dv
(1.7.11)
Dt
Dt
where dv is the difference in velocity between the two points, as shown in Figure 1.7.2.
Since the magnitude of the distance between the two points, or more conveniently its
square, is dr2 = dr · dr, we have
Ddr2
Ddr
= 2dr ·
= 2dr · dv
Dt
Dt
vy
vy
vy
vx
v
v
= 2dx
dx + x dy + x dz + 2dy
dx +
dy +
dz
x
y
z
x
y
z
+ 2dz
=2
+
vz
v
v
dx + z dy + z dz
x
y
z
(1.7.12)
vx 2 vy 2 vz 2
vx vy
vx vz
dx +
dy +
dz +
+
+
dxdy +
dxdz
x
y
z
y
x
z
x
vz vy
+
dzdy
y
z
= 2 dxx dx2 + dyy dy2 + dzz dz2 + 2dxy dxdy + 2dxz dxdz + 2dyz dydz
v + dv
v
dr
v
Figure 1.7.2 Velocity changes
dv
24
Fundamentals
Thus, after choosing the two neighboring points that we wish to describe (i.e., selecting dr) to find the rate at which the distance between the points change, we need
to know the local values of the six components of the rate of deformation—that is,
dxx dyy dzz dxy dyz dzx .
Example 1.7.1 Rigid-body rotation
Find the rate of deformation for rigid-body rotation as given by the velocity field
−y x 0.
Solution. From the definition of rate of deformation, dxx = dyy = dzz = 0 dxy =
dyz = dxz = 0. The absence of rates of deformation confirms that the fluid is behaving
as a rotating rigid body.
Example 1.7.2 Vortex motion
Find the rate of deformation for a line vortex with velocity v =
xG
−yG
0 .
x2 + y 2 x2 + y 2
Solution. Again from the definition,
dxx = −dyy =
2xyG
x2 + y2 2
dzz = 0
dxy =
xyG
x2 + y2 2
dxz = dyz = 0
1.8 Constitutive Relations
Although the physical laws presented so far are of general validity, very little can be said
concerning the behavior of any substance, as can be ascertained by noting that at this
point there are many more unknown quantities than there are equations. The “missing”
relations are those that describe the connection between how a material is made up,
or constituted, and the relation between stress and the geometric and thermodynamic
variables. Hooke’s law and the state equations of an ideal gas are two familiar examples
of constitutive equations. Whereas in a few cases constitutive relations can be derived
from statistical mechanics considerations using special mathematical models for the
molecular structure, the usual procedure is to decide, based on experiments, which
quantities must go into the constitutive equation and then formulate from these a set
of equations that agree with fundamental ideas, such as invariance with respect to the
observer and the like.
There is much inventiveness in proceeding along the path of determining constitutive
equations, and much attention has been given to the subject in past decades. The mental
process of generating a description of a particular fluid involves a continuous interchange
between theory and practice. Once a constitutive model is proposed, mathematical
predictions can be made that then, it is hoped, can be compared with the experiment.
Such a procedure can show a model to be wrong, but it cannot guarantee that it will
always be correct, since many models can predict the same velocity field for the very
simple flows used in viscometry (tests used in measuring the coefficient of viscosity)
and rheogoniometry (measuring the properties of complicated molecular structures such
as polymers). As an example, if a fluid is contained between two large plane sheets
placed parallel to each other, and the sheets are allowed to move parallel to one another
at different velocities, many constitutive models will predict a fluid velocity that varies
1.8
25
Constitutive Relations
linearly with the distance from one plate, with all models giving the same shear stress.
The various models, however, usually predict quite different normal stresses.
The familiar model presented by Newton and elaborated on by Navier and Stokes
has withstood many of these tests for fluids of relatively simple molecular structure and
holds for many of the fluids that one normally encounters. This model is not valid for
fluids such as polymers, suspensions, or many of the fluids encountered in the kitchen,
such as cake batter, catsup, and the like. A good rule of thumb is that if the molecular
weight of the fluid is less than a half million or so, and if the distance between molecules
is not too great (as in rarefied gases), a Newtonian model is very likely to be valid.
We will not delve deeper into a rigorous justification of a particular constitutive
equation here. We simply put down the minimum requirements that we expect of our
constitutive law and give a partial justification of the results.
Considering a fluid of simple molecular structure, such as water or air, experience
and many experiments suggest the following:
1. Stress will depend explicitly only on pressure and the rate of deformation.
Temperature can enter only implicitly through coefficients such as viscosity.
2. When the rate of deformation is identically zero, all shear stresses vanish, and
the normal stresses are each equal to the negative of the pressure.
3. The fluid is isotropic. That is, the material properties of a fluid at any given
point are the same in all directions.
4. The stress must depend on rate of deformation in a linear manner, according to
the original concepts of Newton.
The most general constitutive relation satisfying all of the above requirements is
xx = −p + ′ · v + 2dxx yy = −p + ′ · v + 2dy y zz = −p + ′ · v + 2dzz
xy = yx = 2dxy xy = yx = 2dxz yz = zy = 2dyz
(1.8.1)
where we have used the abbreviation
dxx + dyy + dzz =
vx vy vz
+
+
= · v
x
y
z
Here, is the viscosity and ′ is the second viscosity coefficient. Both of these viscosities
can depend on temperature and even pressure. The fluid described by equation (1.8.1)
is called a Newtonian fluid, although the term Navier-Stokes fluid is also used.
Up until now, pressure has deliberately been left undefined. The definition of
pressure varies in different instances. For instance, in elementary thermodynamics texts,
the term pressure is commonly used for the negative of mean normal stress. Summing
our constitutive equations gives
Mean normal stress =
xx + yy + zz
sum of the total normal stress components
= −p +
3
3
From equation (1.8.1), however, we see that
xx + yy + zz
2
= ′ +
· v
3
3
(1.8.2)
is called the bulk viscosity, or volume viscosity, since it
The coefficient ′ + 2
3
represents the amount of normal stress change needed to get a unit specific volume rate
26
Fundamentals
change. The second law of thermodynamics can be used to show that ′ + 2/3 ≥ 0.
If we are to have the mean normal stress equal to the negative of pressure, it follows
that the bulk viscosity must be zero. At one time Stokes suggested that this might in
general be true but later wrote that he never put much faith in this relationship. Since
for most flows the term is numerically much smaller than p, Stokes’s assumption is still
widely used.
Usually one thinks of p as being the thermodynamic pressure, given by an equation
of state (e.g., p = RT for an ideal gas). In such a case, the bulk viscosity is not
necessarily zero, and so the thermodynamic pressure generally must differ from the
mean normal stress. Values of ′ for various fluids have been determined experimentally
in flows involving very high-speed sound waves,2 but the data is quite sparse. Statistical
mechanics tells us that for a monatomic gas, the bulk viscosity is zero. In any case, in
flows where both the dilatation and the bulk viscosity tend to be very small compared
to pressure, the effects of the bulk viscosity can usually be neglected.
Some elaboration of the four postulates for determining our constitutive equation
are in order. We have said that the only kinematic quantity appearing in the stress is
the rate of deformation. What about strain or rate of rotation of fluid elements?
The type of fluid we are considering is completely without a sense of history.
(One class of fluids with a sense of history of their straining is the viscoelastic fluids.
For these fluids strain and strain rates appear explicitly in their constitutive equations.)
A Newtonian fluid is only aware of the present. It cannot remember the past—even the
immediate past—so strain cannot enter the model. Although such a model predicts most
of the basic features of flows, it does have its disturbing aspects, such as infinite speed
of propagation of information if compressibility is ignored. For most flows, however, it
seems a reasonable assumption.
Our assumption that when the rate of deformation is zero the stresses reduce to the
pressure is simply a reaffirmation of the principles of hydrostatics and is a basic law
used for practically all materials.
The isotropy of a fluid is a realistic assumption for a fluid of simple molecular
structure. If we had in mind materials made up of small rods, ellipsoids, or complicated
molecular chains, all of which have directional properties, other models would be called
for, and this constraint would have to be relaxed.
The linearity assumption can be justified only by experiment. The remarkable thing
is that it quite often works! If we relax this point but retain all the other assumptions,
the effect is to add only one term to the right side of equation (1.8.1), making stress
quadratic in the rate of deformation. Also, the various viscosities could also depend on
invariant combinations of the rate of strain, as well as on the thermodynamic variables.
While this adds to the mathematical generality, no fluids are presently known to behave
according to this second-order law.
We have already remarked that a state equation is a necessary addition to the
constitutive description of our fluids. Examples frequently used are incompressibility
D /Dt = 0 and the ideal gas law p = RT. Additionally, information on the heat flux
and internal energy must be added to the list. Familiar laws are Fourier’s law of heat conduction, where the heat flux is proportional to the negative of the temperature gradient, or
q = −kT
2
See, for example, Lieberman (1955).
(1.8.3)
1.9
27
Equations for Newtonian Fluids
and the ideal gas relation for internal energy,
U = UT
(1.8.4)
The latter is frequently simplified further by the assumption of the internal energy
depending linearly on temperature so that
U = U0 + c T − T0
(1.8.5)
c being the specific heat at constant density or volume.
Knowledge of the constitutive behavior of non-Newtonian fluids is unfortunately
much sparser than our knowledge of Newtonian fluids. Particularly with the everincreasing use of plastics in our modern society, the ability to predict the behavior of
such fluids is of great economic importance in manufacturing processes. Although many
theoretical models have been put forth over the last century, the situation in general is
far from satisfactory. In principle, from a few simple experiments the parameters in a
given constitutive model can be found. Then predictions of other flow geometries can
be made from this model and compared with further experiments. The result more often
than not is that the predictions may be valid only for a very few simple flows whose
nature is closely related to the flows from which the parameters in the constitutive model
were determined. There are many gaps in our fundamental understanding of these fluids.
1.9 Equations for Newtonian Fluids
Substitution of equation (1.8.1) into equation (1.6.5) gives the result
p − ′ · v
vx
vx vy
Dvx
=−
+ gx +
+
2
+
Dt
x
x
x
y
y
x
vx vz
+
+
z
x
z
Dvy
vy vx
vy
p − ′ · v
=−
+ gy +
+
+
2
Dt
y
x
x
y
y
y
v
v
y
+ z
+
z
z
y
Dvz
p − ′ · v
vz vx
vz vy
=−
+ gz +
+
+
+
Dt
z
x
x
z
y
y
z
vz
+
2
(1.9.1)
z
z
When and are constant, and for incompressible flows, this simplifies greatly
with the help of the continuity condition · v = 0 to the vector form
Dv
= −p + g + 2 v
Dt
where
2 ≡
is called the Laplace operator.
2
2
2
+ 2+ 2
2
x
y
z
(1.9.2)
28
Fundamentals
Equation (1.9.2) can be written in component notation as
v
v
v
vx
+ vx x + vy x + vz x
t
x
y
z
=−
p
+ gx + 2 vx
x
vy
vy
vy
vy
+ vx
+ vy
+ vz
t
x
y
z
=−
p
+ gy + 2 vy
y
vz
v
v
v
+ vx z + vy z + vz z
t
x
y
z
=−
(1.9.3)
p
+ gz + 2 vz
z
Either form (1.9.2) or (1.9.3) is referred to as the Navier-Stokes equation.
The last form (1.9.3) is the form most useful for problem solving. Form (1.9.2)
is useful for physical understanding and other manipulations of our equations. The
Navier-Stokes equations in non-Cartesian coordinate systems are given in the Appendix.
1.10 Boundary Conditions
To obtain a solution of the Navier-Stokes equations that suits a particular problem, it is
necessary to add conditions that need to be satisfied on the boundaries of the region of
interest. The conditions that are most commonly encountered are the following:
1. The fluid velocity component normal to an impenetrable boundary is always
equal to the normal velocity of the boundary. If n is the unit normal to the boundary, then
n · vfluid − vboundary = 0
(1.10.1)
on the boundary. If this condition were not true, fluid would pass through the boundary.
This condition must hold true even in the case of vanishing viscosity (“inviscid flows”).
If the boundary is moving, as in the case of a flow with a free surface or moving body,
then, with Fx t = 0 as the equation of the bounding surface, (1.10.1) is satisfied if
DF
= 0 on the surface F = 0
Dt
(1.10.2)
This condition is necessary to establish that F = 0 is a material surface—that is, a
surface moving with the fluid that always contains the same fluid particles. An important
special case of the material surface is the free surface, a constant pressure surface,
typically the interface between a liquid and a gas.
2. Stress must be continuous everywhere within the fluid. If stress were not
continuous, an infinitesimal layer of fluid with an infinitesimal mass would be acted
upon by a finite force, giving rise to infinite acceleration of that layer.
At interfaces where fluid properties such as density are discontinuous, however, a
discontinuity in stress can exist. This stress difference is related to the surface tension of
the interface. Write the stress in the direction normal to the interface as n and denote
the surface tension by (a force per unit length). By summing forces on an area of the
interface, if the surface tensile force acts outwardly along the edge of S in a direction
locally tangent to both S and C, the result is
n
n
dS
=
t × n ds
lower
−
fluid
upper fluid
S
where t and n are the unit tangent and normal vectors, respectively.
1.11
29
Vorticity and Circulation
A variation of the curl theorem allows us to write the line integral as a surface integral.
Letting
n
n = lower
fluid
n
− upper
fluid
we have
s
n dS =
S
− + nn · + · ndS = 0
(1.10.3)
But n · is zero, since
is defined only on the surface S and n · is perpendicular
to S. Also, · n = − R1 + R1 , where R1 and R2 are the principal radii of curvature of
1
2
the surfaces. (For information on this, see McConnell (1957), pp. 202–204.) By taking
components of this equation in directions locally normal and tangential to the surface,
equation (1.10.3) can be conveniently split into
n · n =
t ·
n
1
1
+
R1 R2
(1.10.4)
= −t ·
where n is the unit normal and t is a unit tangent to the surface. In words, if surface
tension is present, the difference in normal stress is proportional to the local surface
curvature. If gradients in the surface tension can exist, shear stress discontinuities can
also be present across an interface.
3. Velocity must be continuous everywhere. That is, in the interior of a fluid,
there can be no discrete changes in v. If there were such changes, it would give rise
to discontinuous deformation gradients and, from the constitutive equations, result in
discontinuous stresses.
The velocity of most fluids at a solid boundary must have the same velocity tangential
to the boundary as the boundary itself. This is the “no-slip” condition that has been
observed over and over experimentally. The molecular forces required to peel away fluid
from a boundary are quite large, due to molecular attraction of dissimilar molecules.
The only exceptions observed to this are in extreme cases of rarified gas flow, when
the continuum concept is no longer completely valid.3
1.11 Vorticity and Circulation
Any motion of a small region of a fluid can be thought of as a combination of translation,
rotation, and deformation. Translation is described by velocity of a point. Deformation
is described by rates of deformation, as in Section 1.7. Here, we consider the rotation
of a fluid element.
In rigid-body mechanics, the concept of angular rotation is an extremely important
one—and rather intuitive. Along with translational velocity, it is one of the basic
descriptors of the motion. In fluid mechanics we can introduce a similar concept in the
following manner.
3
Those interested in the history of this once-controversial condition should read the note at the end of
Goldstein’s Modern Developments in Fluid Dynamics (1965) for an interesting account.
30
Fundamentals
Consider again the two-dimensional picture shown in Figure 1.7.1. In Section 1.7,
we saw that the instantaneous rates of rotation of lines AB and BC were ˙ 1 and ˙ 2 ,
given by equations (1.7.5) and (1.7.6).
Since equations (1.7.5) and (1.7.6) generally will differ, it is seen that the angular
velocity of a line depends on the initial orientation of that line. To develop our analogy
of angular velocity, we want a definition that is independent of orientation and direction
at a point, and depends only on local conditions at the point itself. In considering the
transformation of ABC into A′ B′ C′ , it is seen that two things have happened: The angle
has changed, or deformed, by an amount d 1 + d 2 , and the bisector of the angle ABC
has rotated an amount 05d 1 − d 2 . Considering only this rotation, the rate of rotation
of the bisector is seen to be
1 vy vx
−
(1.11.1)
2 x
y
Writing the curl of v in Cartesian coordinates, we have
curl v = × v = i
vz vy
−
y
z
+j
vy vx
vx vz
−
−
+k
z
x
x
y
We see by comparison that equation (1.11.1) is one-half the z component of the curl
of v.
If we were to consider similar neighboring points in the yz and xz planes, we
would find that similar arguments would yield the x and y components of one-half the
curl of the velocity. We therefore define the vorticity vector as being the curl of the
velocity—that is,
= curl v = × v
(1.11.2)
and note that the vorticity is twice the local angular rotation of the fluid. (Omitting the
one-half from the definition saves us some unimportant arithmetic and does not obscure
the physical interpretation of the concept.) Since vorticity is a vector, it is independent
of the coordinate frame used. Our definition agrees with the usual “right-hand-rule” sign
convention of angular rotation.
Vorticity also can be represented as a second-order tensor. Writing
xx = yy = zz = 0
xy =
1
2
vx vy
−
y
x
(1.11.3)
1 vx vz
1 vy vz
xz =
−
−
yz =
2 z
x
2 z
y
note that × v = 2 zy i + xz j + yx k . Thus, the vector and the second-order tensor
contain the same information.
Since interchanging the indices in the definitions in equation (1.11.3) only changes
the sign, the vorticity (second-order) tensor is said to be skew-symmetric.
If we represent the rate of deformation and vorticity in index notation, we can
summarize some of our findings as follows:
dij =
1
2
vi vj
+
xj xi
v
dij + ij = i
xj
= dji
ij =
1
2
vi vj
−
xj xi
= −ji
(1.11.4)
1.11
31
Vorticity and Circulation
Flows with vorticity are said to be rotational flows; flows without vorticity are said
to be irrotational flows. Note for later use that, from a well-known vector identity, it
follows that
div = · × v = 0
(1.11.5)
You might wonder whether vorticity should enter the constitutive equation along
with the rates of deformation. What has been called the “principle of material objectivity”
or “principle of isotropy of space” or “material frame indifference,” among other things,
states that all observers, regardless of their frame of reference (inertial or otherwise),
must observe the same material behavior. Therefore, an observer stationed on a rotating
platform, for example, sees the same fluid behavior as an observer standing on the floor
of the laboratory. As we have seen in the last section, vorticity is not satisfactory in this
regard in that it is sensitive to rigid rotations. (If you find the idea of material frame
indifference unsettling, see Truesdell (1966), page 6. Truesdell was one of the first
expounders of this concept, but he apparently had many doubts on the same question
initially.) Many constitutive equations that have been proposed violate this principle
(both intentionally and unintentionally). Present work in constitutive equations tends to
obey the principle religiously, although doubts are still sometimes expressed.
Example 1.11.1 Rigid body rotation.
Find the vorticity associated with the velocity field v = −y x 0.
Solution. This is the case of rigid body rotation, with the speed being given by
times the distance of the point from the origin and the stream lines being concentric
circles. Taking the curl of the velocity we have
= × −y i + x j = 2 k
Thus, the flow everywhere has the same vorticity.
Example 1.11.2 Vortex motion
Find the vorticity associated with the velocity field v =
−yG
xG
0 .
x2 + y 2 x2 + y 2
Solution. This is a velocity field again with streamlines that are concentric circles
but with the speed now being proportional to the reciprocal distance from the origin.
This flow is called a line vortex. Taking the curl of this velocity, we have
=×
xG
−yG
i+ 2
j = 0.
x2 + y 2
x + y2
We see that the vorticity is zero, except perhaps at the origin, where the derivatives
become infinite, and a more careful examination must be made.
The preceding two examples point out what vorticity is and what it is not. In both
flows, the streamlines are concentric circles, and a fluid particle travels around the origin
of the coordinate system. In the rigid-body rotation case, particles on two neighboring
streamlines travel at slightly different velocities, the particle on the streamline with the
greater radius having the greater velocity. An arrow connecting the two particles on
different streamlines will travel around the origin, as shown in Figure 1.11.1.
32
Fundamentals
Direction of
Circulation
Figure 1.11.1 Rigid body rotation
Direction of
Circulation
Figure 1.11.2 Line vortex
Considering a similar pair of neighboring streamlines in the line vortex case, the
outer streamline has a slower velocity and the outer particle will lag behind the inner
one (Figure 1.11.2). An arrow connecting the two particles in the limit of zero distance
will always point in the same direction, much as the needle of a compass would. It is
this rotation that vorticity deals with and not the rotation of a point about some arbitrary
reference point, such as the origin.
Since vorticity is a vector, many of the concepts encountered with velocity and
stream functions can be carried over. Thus, vortex lines can be defined as being lines
instantaneously tangent to the vorticity vector, satisfying the equations
dx dy
dz
=
=
x
y
z
(1.11.6)
Vortex sheets are surfaces of vortex lines lying side by side. Vortex tubes are closed
vortex sheets with vorticity entering and leaving through the ends
of the tube.
Analogous to the concept of volume flow through an area, S v · dA, the vorticity
flow through an area, termed circulation, is defined as
· dS
(1.11.7)
circulation = = v · ds =
C
S
1.11
33
Vorticity and Circulation
y
(1, 1, 0)
(–1, 1, 0)
x
(1, –1, 0)
(–1, –1, 0)
Figure 1.11.3 Calculation of circulation about a square
The relation between the line and surface integral forms follow from Stokes theorem,
where C is a closed path bounding the area S.
Example 1.11.3 Circulation for a rigid rotation
Find the circulation through the square with corners at +1 +1 0 −1 +1 0
−1 −1 0 +1 −1 0 for rigid-body rotation flow shown in Figure 1.11.3.
Solution. Starting first with the line integral form of the definition in equation
(1.11.7), we see that
+1
+1
−1
−1
= v · ds =
vy x=1 dy
vx y=−1 dx +
vy x=−1 dx +
vx y=1 dx +
C
=
+1
−dx +
+1
+1
+1
−1
−1
+1
−dx +
+1
+1
dx +
+1
+1
dy
+1
= −−2 − −2 + 2 + 2 = 8
Of course, we could have obtained this result much faster by using the area integral
form of the definition (1.11.7),
· dA
=
S
Since the integrand is constant 2 and the area is 4, the result for the circulation
follows from a simple arithmetic multiplication.
Example 1.11.4 Circulation for a vortex motion
Using the square given in Example
1.11.3, findthe circulation for the vortex flow given
xG
in Example 1.5.2—that is, v = x−yG
2 +y 2 x2 +y 2 0 .
34
Fundamentals
Solution. In this case, we cannot easily use the area form of the definition, since
the vorticity is not defined at the origin. The line integral form of equation (1.11.7)
gives us
1 G
−1 −G
1 G
−1 −G
dy
+
dy
dx
+
dx
+
= v · ds =
2
2
x 2 +1
−1 x +1
1
−1 1 + y
1
C
1 + y2
= −G
−1 1
1 1
−1 1
−
−
+
−G
+G
4
4
4
4
4 4
1 1
+
4 4
+G
= 2G
If we take any path that does not include the origin, we could in fact use the area
form of the definition, and we would conclude that the circulation about that path was
zero. Any path containing the origin would have circulation 2G. Therefore, we say
that the vorticity at the origin is infinite for the line vortex, being infinite in such a
manner that the infinite vorticity times the zero area gives a finite, nonzero, value for the
circulation.
1.12 The Vorticity Equation
Differential equations governing the change of vorticity can be formed from the NavierStokes equations. Dividing equation (1.9.2) by the mass density and then taking the curl
of the equation the result after some manipulation and use of the continuity equation is
D
1
= · v −
× p + 2
Dt
(1.12.1)
The right-hand side of equation (1.12.1) tells us that as we follow a fluid particle, there
are three mechanisms by which its vorticity can change. The first term, · v, is
vorticity change due to vortex line stretching. The operator · is the magnitude of
the vorticity times the derivative in the direction of the vortex line. Consequently, if the
velocity vector changes along the vortex line (thus “stretching” the vortex line), there
will be a contribution to the change of vorticity. The second term says that unless the
pressure gradient and the density gradient are aligned so that they are parallel to one
another, the local vorticity will be changed by the density gradient. The third term says
that vorticity will be diffused by viscosity.
Some further insights into the nature of vorticity and circulation can be gained by
considering several theorems introduced first by Helmholtz. The first states the following:
The circulation taken over any cross-sectional area of a vortex tube is a constant.
The proof is simple. Use a vortex tube segment with ends S1 and S2 , and side surface
S0 as shown in Figure 1.11.4. By one of Green’s theorems,
· dV = 0
· dA =
· dA +
· dA +
S0
S1
S2
V
by virtue of equation (1.11.2). On S0 , vorticity is normal to the surface, so the integral
is zero. Then
· dA = 0
· dA +
S1
S2
1.12
35
The Vorticity Equation
ω∼2
S1
S3
S2
ω∼2
Figure 1.11.4 Vortex tube
from which it follows that
S2
· dA = −
S1
· dA
With the proper interpretation of signs of the outward normal, this proves the theorem.
A very important corollary of this theorem follows:
Vortex lines can neither originate nor terminate in the interior of the flow. Either
they are closed curves (e.g., smoke rings) or they originate at the boundary.
Notice in this proof we did not use any dynamical information. Only kinematics and
definitions were used.
Another useful theorem has to do with the rate of change of circulations. Starting
with the line integral definition of circulation and noting that Dds/Dt = dv, then
Dv v · v
Dv
D
D
v · ds =
=
+
· ds + v · dv =
· ds
Dt
Dt C
Dt
Dt
2
C
C
The second term in the last integral is an exact differential, so integrating it around
a closed path gives zero. Thus, the rate of change of circulation is given by
D Dv
=
· ds
(1.12.2a)
Dt
C Dt
or, with the help of the Navier-Stokes equations,
D
1
=
− p + 2 v · ds
Dt
C
(1.12.2b)
36
Fundamentals
Here, the body force terms vanish because they are an exact differential, so an integration around a closed path gives zero. Therefore, as we follow a curve C drawn
in the flow as it is carried along with a flow, the pressure gradient can change
the circulation associated with the curve only if 1 p · ds is not an exact differential. Thus, pressure gradients must be accompanied by density variations to affect the
circulation.
The understanding of the nature of vorticity is extremely important to the study of
fluid mechanics. Vorticity plays a central role in the transition from laminar to turbulent
flow. It can affect the performance of pumps and the lift of airfoils. Being able to control
vorticity, either by causing it or by eliminating it, is at the core of many engineering
problems.
1.13 The Work-Energy Equation
Another useful equation derived from the Navier-Stokes equations is a work-energy
statement. If we take the dot product of equation (1.9.1) with the velocity, with the help
of the product rule of calculus we have
D v·v
xx yx zx
= v·g+v· i
+
+
Dt 2
x
y
z
xy yy zy
xz yz zz
+
+
+
+
+k
x
y
z
x
y
z
vx xx vx yx vx zx
+
+
= v·g+
x
y
z
vy xy vy yy vy zy
vz xz vz yz vz zz
+
+
+
+
+
+
x
y
z
x
y
z
+ j
− xx
vy
vy
vy
vx
v
v
+ yx x + zx x − xy
+ yy
+ zy
x
y
z
x
y
z
vz
v
v
+ yz z + zz z
x
y
z
= v·g+
v + vy xy + vz xz
x x xx
+
vx yx + vy yy + vz yz +
vx zx + vy zy + vz zz
y
z
v
vx vy
vx vz
+
+
+ zx
− xx x + yx
x
y
x
z
x
− xz
vy vz
v
+
+ zz z
z
y
z
= p − ′ · vv + v · g +
vx xx + vy xy + vz xz
x
vx yx + vy yy + vz yz +
vx zx + vy zy + vz zz −
+
y
z
+ yy
vy
+ zy
y
(1.13.1)
1.14
37
The First Law of Thermodynamics
where
= xx
vx
vx vy
+ yx
+
x
y
x
+ zx
vy
vx vz
+
+ yy
z
x
y
vy vz
v
+
+ zz z + p − ′ · v · v
(1.13.2)
z
y
z
= xx dxx + yy dyy + zz dzz + 2 yx dyx + zx dzx + zy dzy + − ′ · v · v
2
2
2
+ dyz
+ dzx
= 2 dxx + dyy + dzz + 2 dxy
+ zy
The function represents the rate of dissipation of energy by viscosity and is called
the dissipation function. Since the viscosity is always positive, the quantity is
positive definite. That is, no matter what the velocity field is, is greater or equal to
zero.
Note that in deriving (1.13.1) we used only the Navier-Stokes equations—that is,
Newton’s law. We next combine (1.13.1) with the first law of thermodynamics.4
1.14 The First Law of Thermodynamics
The conservation of energy principal (first law of thermodynamics) in its rate form
states that the rate of change of energy of the system is equal to the rate of heat addition
to the system due to conduction from the surroundings, radiation, and internal reactions
plus the rate at which work is done on the system. In equation form, this is
dE
dQ dW
=
+
dt
dt
dt
For the rate of energy change we have
dE e
ev · dS
=
dV +
dt
t
V
S
(1.14.1)
(1.14.2)
where e is the specific energy (energy per unit mass), given by
e=
v·v
+ u
2
with u being the specific internal energy. Also, write
dr
dQ
q · dS +
=−
dV
dt
V dt
S
(1.14.3)
where q is the heat flux vector represent heat transfer from the surroundings. The body
term dr/dt represents heat generated either internally or transferred by radiation. Often
Fourier’s law of conductivity is used to relate the heat flux vector to the temperature.
This law states that q = −kT T being the temperature and k the coefficient of thermal
conduction.
4
The idea of conservation of energy was first published by Émilie du Châtelet (1706–1749), a physicist
and mathematician who made the first translation into French (along with her commentary) of Newton’s
Principia Mathematica. Her book Lessons in Physics was written for her 13-year-old son. It elaborated on
and advanced the leading scientific ideas of the time.
38
Fundamentals
The rate at which work is being done by the various forces can be written as
dW
v · n dS
(1.14.4)
g · v dV +
=
dt
S
V
Putting these expressions into equation (1.14.1), we have
e
dV +
ev · dS
t
S
V
dr
q · dS +
dV +
v · n dS
=−
g · vdV +
V dt
S
S
V
With the help of the divergence theorem, this becomes
e
+ · ev dV
t
V
dr
=
− ·q+
+ g · v + vx xx + vy yx + vz zx
dt
x
V
+
v + vy yy + vz zy + vx xz + vy yz + vz zz dV
y x xy
z
Again, the choice of the control volume is arbitrary, so the integrand on the right-hand
side of the equation must equal the integrand on the left. The result is
e
De
D
+ · ev =
+e
t
Dt
Dt
dr
+ g · v + vx xx + vy yx + vz zx
=− ·q+
dt
x
+ vx xy + vy yy + vz zy + vx xz + vy yz + vz zz
y
z
(1.14.5)
Since many of the viscous terms on the right-hand side of equation (1.14.5) are also
in equation (1.13.1), that equation can be used to eliminate terms here. The result after
subtracting equation (1.13.1) from equation (1.14.5) is
D v·v
De
D
dr
2
+e
−
= − · q +
− p − ′ · v · v +
Dt
Dt
Dt
dt
The kinetic energies cancel and we are left with
Du
dr
= − · q +
+ −p + e + ′ · v · v
(1.14.6)
Dt
dt
In words, this equation states that the internal energy of the fluid will be changed by
the addition of heat transfer from the surroundings (first term), heat generated internally
(second term), viscosity (third term), and compressibility effects (fourth term). For
incompressible flow, the fourth term on the right side is zero by virtue of the continuity
equation. Equation (1.14.6) can then be rewritten as
Du
dr
= − · q +
+
(1.14.7)
Dt
dt
These represent the thermodynamic portion of the first law. The thermodynamic
field quantities are thus seen to be coupled to the mechanical portion through the
convective change of the internal energy, the viscous dissipation, and the pressure.
1.15
Dimensionless Parameters
1.15 Dimensionless Parameters
The development of fluid mechanics over the years has relied on both experimentation
and analysis, with the former leading the way in many cases. To express data in the most
useful form, dimensional analysis was used extensively. It is also extremely useful in
analysis. The number of such parameters introduced from time to time in the literature is
certainly in the hundreds, if not thousands. There are a few, however, that predominate in
general usage. Most of them come from the Navier-Stokes equations and their boundary
conditions.
The Reynolds number, named after Osborne Reynolds (1842–1912), a British
scientist/mathematician, represents the relative importance of the convective acceleration
terms in the Navier-Stokes equations to the viscous terms. Reynolds used the term in
1883 in a paper presenting his results on the transition from laminar to turbulent flow
of liquids in round pipes. It is typically found in the form VD/.
The Froude number was named after William Froude (1810–1879), a British
mathematics professor who became interested in ship construction. He started his studies
of ship resistance by building scale models and then towing them in long, narrow basins
he had constructed himself. Towing basins are now used extensively in the design
of ships to determine the proportion of ship resistance to waves. They also are used
to study wave forces on offshore drilling cables and pipelines. The Froude number
(actually never used by him) represents the ratio of the convective acceleration terms in
the Navier-Stokes equations to √
the wave forces as represented by the gravity terms. It
is typically used in the form V/ gh, although the square of this is also used. Froude’s
ideas on model studies were initially ridiculed by his peers, but his perseverance led to
their acceptance.
The Richardson number V/ gh/ , named after colonel A. R. Richardson,
a faculty member of the University of London, is a variant of the Froude number.
It is used in studying waves in flows with density stratification. He introduced it
in 1920.
The Strouhal number is named after C. Strouhal (1850–1922), a German physicist
who studied the aeolian sounds generated by wind blowing through trees. It is an
important parameter in studying the shedding of vortices and is written as D/V or
fD/V , where f is the frequency and the circular frequency.
The pressure coefficient p/ 21 V 2 , sometimes called the Euler coefficient
(Leonard Euler, 1707–1783, a Swiss mathematician) is a form suited for the
presentation of pressure data. It is the ratio of pressure forces to the convective
acceleration.
The drag coefficient FD / 21 V 2 A is used for presenting the drag force (the force in
the direction of motion), where A is the projected area.
The lift coefficient FL / 21 V 2 A is similar to the drag force but perpendicular to the
direction of motion.
The moment coefficient M/ 21 V 2 AL is convenient for measuring the moment on a
wing or rudder.
The Weber number V 2 D/ was named after Moritz Weber (1871–1951), a professor of naval mechanics at the Polytechnic Institute of Berlin. He introduced the name
similitude to describe model studies that were scaled both geometrically and using
dimensionless parameter for forces, and introduced a capillary parameter, including
surface tension.
39
40
Fundamentals
1.16 Non-Newtonian Fluids
Fluids such as large molecular weight polymers that do not obey the Newtonian constitutive equation are encountered frequently in the chemical and plastics industry. Paints,
slurries, toothpastes, blood, drilling mud, lubricants, nylon, and colloids all exhibit nonNewton behavior. Many such fluids also can be found in the kitchen: catsup, suspensions
of corn or rice starch in water, melted chocolate, eggwhites, mayonnaise, milk, and gelatine all are examples! They exhibit such effects as die swell when exiting a tube, climbing
of a rotating rod in an otherwise still container of fluid, self-siphoning, drag reduction,
and transformation into a semisolid when an electric or magnetic field is applied.
As will be seen, finding solutions to the Navier-Stokes equations is more than a
sufficient challenge. For these fluids of greater constitutive complexity, we can hope
to find solutions for only the very simplest flows. Such flows are called viscometric
flows and are the flows found in simple viscometers. They are the flows involving
pipe flow, flow between rotating cylinders, and flows between a rotating cone and a
plate.
Section 1.8 listed the four considerations that a constitutive equation for a fluid
must satisfy. The last one, the requirement that the stress-rate of deformation equation
be linear in the rate of deformation, can be broadened to include quadratic terms in the
rate of deformation. (There is no need to go to higher powers: The Cayley-Hamilton
theorem states that powers higher than the second can be expressed in lower powers.)
Thus, the most general form of constitutive equation we could propose for flows that
are described by only stress and rate of deformation is given by
ij = −p + ′ · vij + dij + ′′ dik dkj
(1.16.1)
where ′′ is an additional viscosity coefficient. Note that its dimensions differ from the
standard viscosity by an additional unit of time. This constitutive law describes what are
called Stokesian fluids, after George Stokes, or sometimes Reiner-Rivlin fluids, after
Markus Reiner of Israel and Ronald Rivlin of the United States. Because of the added
difficulty that this second power term introduces into problem solution, it is fortunate
that no fluids have so far been found that obey this law better than they do the first
power law!
Suspensions such as paint, clays, and wood pulp solutions5 appear to behave as if
they must have a certain level of stress applied before the fluid deforms. Such fluids have
been referred to as Bingham fluids, or sometimes visco-plastic fluids. The constitutive
equation that has been proposed for them is
ij = Tij
if
1
T T ≤ T 2 or
2 ij jk
ij = Tij + −p + · vij + dij if
2T
d .
2Tmn Tmn ij
In the latter case, Tij = √
1
T T > T 2
2 mn mn
(1.16.2)
Here, Tij are components of the yield stress tensor
and T is the yield stress according to the von Mises yield criterion.
5
There is some doubt as to whether such fluids are indeed Bingham fluids. It has been claimed that the
only true Bingham fluid that has been discovered is an aluminum soap dispersed in a petroleum fraction
(McMillan, 1948). Controversy is not unheard of in the discussion of non-Newtonian fluids.
1.17
41
Moving Coordinate Systems
The non-Newtonian fluids that are more interesting in practical use are polymers.
Their long chain molecules act like springs and have often been modeled as springs
or dumbbells where the weights are connected by springs. Such fluids can also have
memory effects, where they do not respond immediately to changes in the applied force.
In viscometric flows they typically exhibit stress effects normal to the direction of flow,
and precise measurement of such effects is difficult.
A substantial number of constitutive models have been introduced for such fluids
(Macosko, 1994). They involve either time derivatives or integrals of the stress and/or
rate of deformation. The time derivative used is a convective one, a derivative that
is concerned with the relative rate of motion between two molecules, and thus is
more complicated than the time derivative involved with computing acceleration from
velocity. Since a number of different convected time derivatives have been introduced
over the years, deciding on the “correct” one is not a simple task.
Much of the terminology used in the field is associated with simple experiments,
where the substance is placed in either simple extension or simple shear. Based on such
experiments, classification can be carried out as shown in Table 1.16.1.
Table 1.16.1 shows that between the states of elastic (Hookean) solids and viscous
(Newtonian) fluids, there is a continuum of effects exhibited that take on many weird
and wonderful forms. Applications including drug delivery, drag reduction, body armor,
and automobile brakes and suspensions have been proposed and investigated. Heat and
stress over time tend to degrade long chain polymers, which have made their practical
application a nontrivial task.
It is fair to say that the problems found in fitting constitutive equations to such
fluids is an extremely difficult one, and a one-size-fits-all solution is unlikely to be
found. The best hope appears to be to fit a given constitutive relationship to a specialized
set of circumstances and applications.
1.17 Moving Coordinate Systems
Occasionally, it is necessary to use moving coordinate systems to understand a given
flow. For instance, the flow past an aircraft appears different to a person on the ground
than it does to a person in the aircraft. The changes that are introduced when moving
coordinates are used is primarily in the velocity and acceleration.
Consider first a coordinate system rotating with an angular velocity 0 and whose
origin translates with a velocity v0 . Then the velocity is given by
vabs = vrel + v0 + 0 × r
(1.17.1)
Here vabs is the velocity with respect to a nonmoving axes system, vrel is the velocity
measured in the moving system, and r is the position in the moving system. The
acceleration is given by
Dvabs
Dvrel dv0
d0
=
+
+ 0 × v0 +
× r + 0 × 0 × r + 20 × vrel
Dt
Dt
dt
dt
(1.17.2)
These follow from well-known results in dynamics. The last two terms on the right-hand
side represent the centripetal and Coriolis accelerations, while the second, third and
fourth represent the acceleration due to the coordinate system.
42
TABLE 1.16.1 Classification of non-Newtonian fluids
Type of fluid
Classification
Stress/rate of deformation behavior
Examples
Elastic solids
Hookean
Linear stress-strain relation
Most solids below the yield stress
Plastic solids
Perfectly plastic
Strain continues with no additional stress
Bingham plastic
Behaves Newtonian when threshold shear is
exceeded
Yield like the Bingham plastic, but the relation
between stress and rate of deformation is not
linear
Dilatant when threshold shear is exceeded
Exhibits both viscous and elastic effects
Ductile metals stressed above the yield
point
Iron oxide suspensions
Visco-plastic
Yield dilatant∗
Visco-elastic
Power-law
fluids
Time-dependent
viscosity
Apparent viscosity reduces as shear rate
increases
Dilatant or shear
thickening
Apparent viscosity increases as shear rate
increases
Rheopectic
Apparent viscosity increases the longer stress
is applied
Apparent viscosity decreases the longer stress
is applied
Some lubricants
Electrorheologic
Becomes dilatant when an electric field is
applied
Magnetorheologic
Becomes dilatant when a magnetic field is
applied
Melted chocolate bars, single- or
polycrystalline suspensions in insulating
fluids
Colloids with nanosize silica particles
suspended in polyethylene glycol
Linear stress-rate of deformation relationship
Water, air
Dilatant here refers to shear thickening as stress increases.
Some colloids, clay, milk, gelatine, blood,
liquid cement, molten polystyrene,
polyethylene oxide in water
Concentrated solutions of sugar in water,
suspensions of rice or cornstarch, solutions
of certain surfactants
Nondrip paints, tomato catsup
Fundamentals
Newtonian
fluids
∗
Eggwhite, polymer melts, and solutions
Shear thinning
Thixotropic
Electromagnetic
Drilling mud, nuclear fuel slurries,
mayonnaise, toothpaste, blood
43
Problems—Chapter 1
This result for the acceleration can be put in a more useful form by a bit of
rearranging. Consider the following:
v
dv
d0
Dvabs
= rel + vrel · vrel + 0 + 0 × v0 +
× r + 0 × 0 × r + 20 × vrel
Dt
t
dt
dt
= vrel + v0 + 0 × r + 0 × vrel + v0 + 0 × r + vrel · vrel + v0 + 0 × r
t
(1.17.3)
+ 0 × +vrel · vabs
=
t
since
vrel · v0 = 0 and vrel · 0 × r = 0 × vrel
This can also be written as
Dvabs
v
= abs + 0 × vabs + vabs − vrel − 0 × r · vabs
Dt
t
(1.17.4)
Problems—Chapter 1
1.1 For the two-dimensional flow field defined by the velocity components vx =
vy = 1 vz = 0, find the Lagrangian representation of the paths taken by the fluid
particles.
1
1+t
1.2 Find the acceleration at point (1, 1, 1) for the velocity v = yz + t xz − t xy.
1.3 a. Find the relationship between velocity components in cylindrical polar coordinates in terms of components in Cartesian coordinates, as well as the
inverse relations. Use Figure 1.4.1.
b. Find the relationships between velocity components in spherical polar coordinates in terms of components in Cartesian coordinates, as well as the
inverse relations. Use Figure 1.4.3.
1.4 Derive the continuity equation in cylindrical coordinates by examining a control
volume bounded by the following: two cylinders perpendicular to the x-y plane, the first
of radius r, the second of radius r + dr; two planes perpendicular to the x-y plane, the
first making an angle with the x-axis, the second an angle + d ; two planes parallel
to the x-y plane, the first above it an amount z, the second an amount z + dz.
Top View
Side View
y
z
dr
dr
dz
r
dθ
θ
x
Figure P1.4 Problem 1.4—Cylindrical coordinates
44
Fundamentals
1.5 a. Find the stream function for the two-dimensional incompressible flow with
velocity components v = x2 − 2xy cos y2 −2xy + sin y2 0.
b. Find the discharge per unit between the points 1 and (0, 0).
1.6 For the following flows, find the missing velocity component needed for the
flow to satisfy the incompressible continuity equation.
a. vx = x2 + y2 + a2
b. vx = ln y2 + z2
c. vx = ?
vy = −xy − yz − xz
vy = sin x2 + z2
vy =
y
x2 + y2 + z2 3/2
vz = ?
vz = ?
vz =
z
x2 + y2 + z2 3/2
.
1.7 For the flow field given by = A lnx2 + y2 + yS, find the discharge per unit
width in the z direction between the points (1, 1, 0) and −1 −1 0.
1.8 Find the stream function for the two-dimensional incompressible flow with
a radial velocity (cylindrical polar coordinates) given by vr = √Ar cos . Also find the
missing velocity component.
1.9 Find the stream function forthe two-dimensional
incompressible
flow with
2
2
velocity components given by vr = U 1 − ar 2 cos v = −U 1 + ar 2 sin .
1.10 Find the stream function for the two-dimensional incompressible flow
with radial velocity component vr = 23 Ar 3/2 cos 32 . Also find the missing velocity
component.
1.11 Find the stream function for the velocity field v = x2 − 2x + 1
−2xy + 2y − x 0.
1.12 For steady, incompressible inviscid flows with body forces neglected, it was
shown by Yih (1958) that the flow can be reduced to that of a constant-density flow by
the transformation
vx∗ =
where
0
0
vx vy∗ =
0
vy vz∗ =
0
vz
is a constant reference density. Verify this result.
1.13 Find the rates of deformation and vorticity in Cartesian coordinates for the
velocity field
v=
Bx
x2 + y2 + z2 3/2
By
x2 + y2 + z2 3/2
Bz
x2 + y2 + z2 3/2
1.14 a. Compute the unit normal for the inclined surface shown.
b. For the stress vector with components b = 10 3 7, calculate the normal
and tangential components of the stress vector on the inclined surface.
1.15 Given the velocity field v = 5yi, find the circulation about a rectangle 6 units
long and 4 units high, centered at the origin. Compute it two ways, once by the line
integral and again using the area formula.
45
Problems—Chapter 1
z
y
4
5
3
x
Figure P1.14 Problem1.14—Stress tetrahedron
1.16 For two dimensional incompressible flow, insert the stream function
vx =
y
vy = −
x
into the Navier-Stokes equation, and eliminate the pressure to find the governing equation for the stream function.
Chapter 2
Inviscid Irrotational Flows
2.1 Inviscid Flows
2.2 Irrotational Flows and the
Velocity Potential
2.2.1 Intersection of Velocity
Potential Lines and Streamlines in Two Dimensions
2.2.2 Basic Two-Dimensional
Irrotational Flows
2.2.3 Hele-Shaw Flows
2.2.4 Basic Three-Dimensional
Irrotational Flows
2.2.5 Superposition and the
Method of Images
2.2.6 Vortices Near Walls
2.2.7 Rankine Half-Body
2.2.8 Rankine Oval
2.2.9 Circular Cylinder or Sphere
in a Uniform Stream
2.3 Singularity Distribution Methods
2.3.1 Two- and Three-Dimensional
Slender Body Theory
2.3.2 Panel Methods
46
47
49
51
57
58
59
61
65
67
68
69
69
71
2.4 Forces Acting on a
Translating Sphere
2.5 Added Mass and the Lagally
Theorem
2.6 Theorems for Irrotational Flow
2.6.1 Mean Value and Maximum
Modulus Theorems
2.6.2 Maximum-Minimum
Potential Theorem
2.6.3 Maximum-Minimum
Speed Theorem
2.6.4 Kelvin’s Minimum Kinetic
Energy Theorem
2.6.5 Maximum Kinetic Energy
Theorem
2.6.6 Uniqueness Theorem
2.6.7 Kelvin’s Persistence of
Circulation Theorem
2.6.8 Weiss and Butler Sphere
Theorems
Problems—Chapter 2
77
79
81
81
81
82
82
83
84
84
84
85
2.1 Inviscid Flows
Finding solutions with the Navier-Stokes equations that were introduced in Chapter 1 is
a formidable challenge, particularly for flows where convective acceleration is present.
When, however, the Reynolds number is sufficiently high, of the order of 105 or more,
viscosity effects usually are of importance to the flow only in the boundary layer near a
body or a wall or possibly in confined regions in the wake of a body. In many problems,
such as the case of waves on a free surface, viscosity effects many be of secondary
importance in most of the flow field.
46
2.2
47
Irrotational Flows and the Velocity Potential
In solving such flows, it is convenient—and useful—to first omit viscosity terms
completely. Since this reduces the order of the differential equations, this means that
fewer boundary conditions can be applied. The zero normal velocity condition generally
is the most important condition and so is retained, whereas the no-slip velocity condition
is ignored. For many flows, viscous effects can be included later by considering the
boundary layer flow using the slip velocity found from the inviscid flow at the outer
edge of the boundary layer.
Most 19th-century fluid mechanics was concerned with the study of inviscid flows.1
There was no clear understanding of the effects of the Reynolds number on the flow, and
the study of turbulence was left largely untouched. This was, however, a time of great
ferment in the fields of fluid mechanics, electricity and magnetism, and thermodynamics.
Particularly in the first two areas, scientists discovered a great similarity in their fields,
and it was not unusual for a researcher to make contributions in both. As a result, today
both fields share many terms such as source, sink, potential, and current, among others.
Workers in one field often use analogies in the other, perhaps feeling that the other field
can give a clearer understanding. In this chapter, the electrical and magnetic analogies
are set aside, but keep in mind that despite this, a minor change in terminology is all
that is needed to change the topic to electrostatics, electrodynamics, and magnetostatics.
If viscosity terms are omitted in the Navier-Stokes equations, they reduce to the form
Dv
= −p + g
Dt
(2.1.1)
This is called the Euler equation.
2.2 Irrotational Flows and the Velocity Potential
From the circulation theorem (equation 1.12.2b),
if mass density is constant and viscous
1
effects can be neglected, D/Dt = C − p · ds. The integrand on the right-hand
side of the equation is an exact differential, so the integral around the path vanishes.
Consequently, for a flow with no upstream circulation, as the flow moves downstream,
it must continue to be vorticity-free, or irrotational. This is called the persistence of
irrotationality. (In a real fluid, viscosity effects will introduce vorticity at a boundary,
but at high Reynolds numbers this vorticity will be convected downstream and chiefly
confined to the vicinity of the boundary and the wake.)
By the definition of irrotational flows,
= × v = 0
(2.2.1)
This suggests that two of the velocity components can be solved for in terms of the
third component or, alternatively, that, as in the case of the continuity equation, scalar
functions can be introduced that have the effect of accomplishing this.
An easier approach is to realize that since for irrotational flows = C v · ds = 0
for any C, it follows that the integrand v · ds must be an exact differential, and therefore
for an irrotational flow field with velocity v, it must be expressible as the gradient of a
scalar. This allows us to write
v =
(2.2.2)
1
Inviscid fluids do not exist, but fluids can flow in such a manner that viscosity effects are negligible in
most of the region of flow.
48
Inviscid Irrotational Flows
where is called the velocity potential. For any velocity field written as the gradient
of a scalar as in equation (2.2.2), it is guaranteed that for any scalar function , v will
automatically be an irrotational velocity field. Because of equation (2.2.2), irrotational
flows are often also called potential flows.
From equation (2.2.2) it follows that surfaces of constant are locally orthogonal to
the velocity vector. Thus, surfaces of constant must be orthogonal to stream surfaces.
The introduction of a velocity potential guarantees irrotationality, but it is still
required that the flow field satisfy the basic dynamical equations. To simplify the
discussion, we will consider only incompressible flows. Then continuity requires that
the divergence of the velocity field vanish. Therefore, the continuity equation for an
irrotational incompressible flow is
0 = · v = 2
(2.2.3)
This equation, called the Laplace equation, is used to determine for a given flow
situation.
What then of the dynamics of the flow? Our flow field at this point seems to
be completely determined from irrotationality and continuity, yet we have not considered Euler’s equation. From equations (1.2.3) and (2.1.1), since × v = 0, Euler’s
equation (2.1.1) can be written in the form
v v2
+
= −p + g
t
2
For irrotational flows v = , and g can be written as g = −gh, where h is the
elevation of the point in the direction in which gravity acts.Euler’s equation can then
be
1 2
p
rearranged and, after dividing by , becomes the form
+ v + gh +
= 0.
t
2
Upon integration this gives
1 2
p
+ v + gh + = f t
t
2
(2.2.4)
where f t is a constant or at most a function of time and is determined from either
conditions at a reference point or a point far upstream. Equation (2.2.4) is known as the
Bernoulli equation for irrotational flows.
The Bernoulli equation could also have been derived for the case where mass density
depends only on pressure. In that case, the p/ term in equation (2.2.4) would be replaced
by dp/.
For most incompressible flows, then, the velocity field is found using only the
conditions of irrotationality (usually by the introduction of the velocity potential )
and the continuity equation in the form of equation (2.2.3), along with the imposition
of conditions on the normal velocity at boundaries. Once is known, pressure can be
found from the Bernoulli equation (2.2.4). Note that all of the mathematical nonlinearities appear only in the Bernoulli equation. For interface problems, however, further
nonlinearities can be introduced by boundary conditions.
The linearity of equation (2.2.3) allows for the superposition of velocity fields. The
nonlinearity of equation (2.2.4) means that pressure fields cannot be superposed in a
linear manner.
2.2
49
Irrotational Flows and the Velocity Potential
Note that for irrotational flows equation (2.2.1) can be written as
=
C
· ds = 2 − 1 =
(2.2.5)
where 1 and 2 are the values of the velocity potential at the start and end points of
the traverse of C. Thus, if is a single-valued function (i.e., if we go around a closed
loop and the value of has not changed), since the curve C is closed, will be zero.
It is, however, possible that can be multivalued if there exist points or isolated
regions where is either singular or not uniquely defined. A line vortex, described
following, has one such nonuniquely defined potential function. For such functions
the circulation will usually be different from zero. Multivalued velocity potentials are
associated with the presence of vorticity, corresponding to multivalued stream functions
that are associated with discharge.
Since we will be looking at some methods for solving Laplace’s equation, you may
wonder whether any solution you might have obtained for a given flow is unique. That
is, if you and your neighbor both solve the same problem but use different methods,
will you end up with the same velocity field? The answer is yes, provided that you
both stay with the same set of rules (and, of course, both do your work accurately). The
irrotational flow field around a body is unique for a given set of boundary conditions,
provided we also specify the circulation and do not allow the development of cavities
that do not contain fluid. Always remember that the primary interest is in finding a
flow field that models a real physical phenomenon to some degree of accuracy. Since
vorticity, and hence circulation, is present in the boundary layer, it may need to be
included in a model to give a realistic model of the flow field. Cavities might possibly
be a reasonable model for wake flows. As long as the circulation is prescribed, together
with rules concerning whether or not cavities are present, the flow field will be unique.
It is possible that the methods used to solve Laplace’s equation will introduce a
“mathematical” flow inside a body as well. In that region the flow is not unique. Any
flows that methods generate inside bodies lie outside our domain of interest and are
artifices of the mathematics with little if any physical meaning.
2.2.1
Intersection of Velocity Potential Lines and Streamlines
in Two Dimensions
Surfaces of constant velocity potential and of constant stream function intersect one
another throughout the flow. Since by definition the velocity is always normal to a
constant potential surface, it follows that the constant stream surfaces are generally
perpendicular to the constant potential surfaces. There are some exceptions to this, such
as near stagnation points. We next investigate this for two-dimensional flows to see at
what angles these intersections can take place.
To see the relationship between and lines, start with
d =
dx +
dy = vx dx + vy dy
x
y
and
d =
dx +
dy = −vy dx + vx dy
x
y
50
Inviscid Irrotational Flows
From these it follows that the slope of a = constant line d = 0 is given by
dy
v
=− x
dx
vy
and the slope of a
= constant line d = 0 is given by
vy
dy
=+
dx
vx
Multiplying the two slopes gives
dy
dx
dy
dx
= −1
which leads to the conclusion that constant and lines are orthogonal to one another
except possibly at places where the velocity is either zero (stagnation points) or infinite
(singularities).
In either of these two special cases we can investigate the situation further by
considering the second-order terms to see what occurs. For example, at a stagnation
point a Taylor series expansion gives to second-order terms in dx and dy
d =
2 dx2
2
2 dy2
dx +
dy + 2
+
dxdy + 2
x
y
x 2
xy
y 2
Since the first derivatives vanish at the stagnation point, on a line of constant passing
through a stagnation point this expression becomes
d =
2 dx2
2
2 dy2
+
dxdy
+
x2 2
xy
y2 2
First-order terms vanish at the stagnation point, so this becomes
2
2 dy 2 dy 2 dx2
0=
+2
+
x2
xy dx y2 dx
2
Solving this quadratic equation for the slope at the stagnation point, we have
2 2
2 2
2
±
− 2 2
−
xy
xy
x y
dy
=
2
dx
y2
A similar expression can be found involving by the same process.
Note that the term underneath the square root sign must always be positive, since
by Laplace’s equation
2
2
=
−
x2
y2
and so this term is the sum of two squares.
The conclusion then is that there will be two values for the slope at the stagnation
point, hence the line divides, or bifurcates, at that point. For details of the angle between
and , individual examples must be considered. Since the Laplace equation is a great
averager of things (for instance, it can be shown from the Gauss-Green theorems that the
value of at the center of a circle is the average of all the values it takes on the circle), the
constant lines can be expected to fall midway between the constant lines.
2.2
2.2.2
51
Irrotational Flows and the Velocity Potential
Basic Two-Dimensional Irrotational Flows
Next, consider several basic simple flows that are the building blocks of potential flow
theory from which all other potential flows can be constructed. These basic flows have
their counterparts in electrostatics and electromagnetics (point charges, dipoles), beam
deflection theory (concentrated loads), and many other branches of engineering physics,
and they are special cases of what are termed Green’s functions.
In two-dimensional flows the basic solutions must satisfy the equations
=
x
y
(2.2.6a)
=−
y
x
(2.2.6b)
vx =
and
vy =
Also, the equation
2 = 0
(2.2.7)
is the incompressible continuity equation for irrotational flows, corresponding to
· v = 0, while
2 = 0
(2.2.8)
is the irrotationality condition for two-dimensional incompressible flows satisfying continuity, corresponding to z = 0. These must be satisfied as well.
Uniform stream
A uniform stream is a flow whose velocity is the same at every point in space. Therefore,
the velocity components are
vx = Ux =
x
(2.2.9a)
vy = Uy =
y
(2.2.9b)
and
Integrating equations (2.2.9a) and (2.2.9b), we find that
uniform
stream
= xUx + yUy
(2.2.10)
where the constant of integration has arbitrarily been set to zero, since it does not
contribute to the velocity field in any way. Lines of constant and (both are straight
lines and mutually orthogonal) are shown in Figure 2.2.1. The stream function is found
from use of equations (2.2.6a) and (2.2.6b) to be
uniform stream
= yUx − xUy
(2.2.11)
52
Inviscid Irrotational Flows
U
θ
Streamlines
Potential Lines
Figure 2.2.1 Uniform stream streamlines and iso-potential lines
Line source or sink (monopole)
The velocity potential
m
m
ln r − r0 =
ln x − x0 2 + y − y0
=
2
2
2
(2.2.12)
is called a line source (if m is positive) or line sink (if m is negative) of strength m,
located at the point x0 y0 . It extends from − to + in the z direction. The velocity
components are given by
vx =
m x − x0
=
x
2 r − r0 2
vy =
m y − y0
=
y
2 r − r0 2
and
and
(2.2.13a)
(2.2.13b)
Differentiating the velocity components gives
m
2
2
2
−
r
−
2
x
−
x
=
r
0
0
x2
2 r − r0 4
2
m
2
2
2
=
−
−
r
−
2
y
−
y
=
r
0
0
y2
x2
2 r − r0 4
Therefore, continuity is satisfied everywhere except possibly at the point x0 y0 .
To investigate what is happening at the location of the source itself, integrate the
flow normal velocity about a 1 by 1 square centered at x0 y0 , as is seen in Figure 2.2.2.
(The size and shape of the square is actually arbitrary, since the same result would be
2.2
53
Irrotational Flows and the Velocity Potential
y
(–1, 1, 0)
(1, 1, 0)
x
(–1, –1, 0)
(1, –1, 0)
Figure 2.2.2 Source/sink discharge calculation
obtained for a contour of any size or shape that encloses the source.) Starting from the
definition of discharge, we have
y0 +1
x0 +1
Q=
vy x 1 − vy x −1 dx
vx 1 y − vx −1 y dy +
y0 −1
=
y0 +1
y0 −1
+
x0 −1
−1
1
dy
−
1 + y − y0 2 1 + y − y0 2
−1
1
m
dx = m
−
2 1 + x − x0 2 1 + x − x0 2
m
2
x0 +1
x0 −1
Thus, m represents the flow rate per unit length in the z direction being emitted from
the source at x0 y0 , and is called the strength of the source. Lines of constant
(concentric circles) and (radial lines) are shown in Figure 2.2.3.
The issue of rotationality at x0 y0 should also be considered. If the circulation
around a square similar to the one we just used is computed, the result is zero. Therefore, there is a concentrated source of mass at x0 y0 but no concentrated source of
vorticity there.
From equations (2.2.6a) and (2.2.6b) the stream function for a source is seen to be
=
y − y0
m
tan−1
2
x − x0
(2.2.14)
The arctangent is a multivalued function, changing by 2 as we go around a contour
enclosing from equations (2.2.6a) and (2.2.6b). Therefore, the change in as we go
around the contour is m, consistent with its representation of rate of flow.
From equation (2.2.12) we see that constant lines are concentric circles centered
at x0 y0 . From equation (2.2.14), we see that the constant lines are radial lines
emanating from x0 y0 . Thus, the flow will be along the radial streamlines emanating
from x0 y0 . To conserve mass, the velocity must decrease inversely with the distance
from x0 y0 .
54
Inviscid Irrotational Flows
Streamlines
Potential lines
Figure 2.2.3 Source/sink streamlines and iso-potential lines
Line doublet (dipole)
Consider a source and sink pair of equal strengths, and let them approach one another
along a connecting line in such a manner that their strengths increase inversely as the
distance between them increases. The pair then has the combined potential
source r + a + sink r − a
2 a
and since the sink is negative with respect to the source, in the limit as a goes to zero
this becomes a derivative. Therefore, define the potential of a line doublet as
doublet = B · source
of strength 2
=
Bx x − x0 + By y − y0
(2.2.15)
r − r0 2
where B gives the strength and direction of the doublet. Often a doublet is denoted by
a half-filled circle, the filled part representing the “source end” and the unfilled part the
“sink end,” to show the source-sink nature and the directionality. From its relation to
the source, we expect that there is no net discharge or vorticity at x0 y0 . This can
be verified by taking appropriate integrations around that point. To find what lines of
constant are like, rearrange equation (2.2.15) as
x − x0 2 + y − y0
2
=
Bx x − x0 + By y − y0
or equivalently
B x − x0
x − x0 − x
2
By y − y0
+ y − y0 −
2
=
Bx x − x0
2
By y − y0
+
2
For constant values of this is the equation of a circle whose radius is B/2 centered
at x0 + Bx /2 y0 + By /2 . Therefore, it is on a line through x0 y0 in the direction
of B. Constant and lines are shown in Figure 2.2.4. The constant lines are nested
circles centered at Bx /2 By /2 and with radius Bx /2 2 + By /2 2 1/2 . Constant
2.2
55
Irrotational Flows and the Velocity Potential
B
Potential Lines
Streamlines
Figure 2.2.4 Doublet streamlines and iso-potential lines
lines are a similar family of circles, but rotated 90 degrees with respect to the circles.
They are centered at By /2 −Bx /2 and have radius Bx /2 2 + By /2 2 1/2 .
The velocity field associated with the doublet is
vx =
Bx y − y0 2 − x − x0 2 − 2By x − x0 y − y0
=
x
x − x0 2 + y − y0 2
By x − x0 2 − y − y0 2 − 2Bx x − x0 y − y0
=
vy =
y
x − x0 2 + y − y0 2
(2.2.16)
Along a line in the direction of B, we see from equation (2.2.16) that the velocity
dies out as the square of the distance. Since source satisfies Laplace’s equation, and
since 2 source = 0 = 0, therefore
0 = B · 2 source = B · 2 source = 2 B · source = 2 doublet
(2.2.17)
The velocity potential for a line doublet therefore satisfies Laplace’s equation.
From equation (2.2.6) the stream function for a line doublet is
=
By x − x0 − Bx y − y0
r − r0 2
(2.2.18)
Since we have also labeled the source and doublet as monopole and dipole, you
might wonder whether taking even higher derivatives would be useful. The derivative
of the dipole is the quadrapole, which is mainly of interest in acoustic problems. The
dipole is usually sufficient for fluid mechanics use.
Line vortex
The vortex is a “reverse analog” of the source, in that it has concentrated vorticity rather
than concentrated discharge and also because the constant lines are radial lines while
the lines of constant are concentric circles. Its velocity potential is
vortex =
y − y0
tan−1
2
x − x0
(2.2.19)
56
Inviscid Irrotational Flows
Streamlines
Potential Lines
Figure 2.2.5 Vortex streamlines and iso-potential lines
with velocity components
vx =
− y − y0
=
x
2 r − r0 2
(2.2.20)
x − x0
=
vy =
y
2 r − r0 2
Thus, the velocity decreases inversely with the distance from x0 y0 . Constant (radial
lines) and (concentric circles) lines are shown in Figure 2.2.5.
Since
y − y0 x − x0
2
=
2
x
x − x0 2 + y − y0 2 2
and
2
y − y0 x − x0
2
=
=− 2
2
2
2
2
y
x − x0 + y − y0
x
Laplace’s equation is seen to be satisfied. The vortex is counterclockwise if is positive,
clockwise if is negative.
Since vortex is multivalued (in traversing a path around the point r0 vortex changes
by ), we anticipate that there is circulation associated with vortex flows. Checking this
by calculating the circulation around a 2 by 2 square centered at x0 y0 , we have
y0 −1
x0 −1
y0 +1
Circulation =
vy x0 − 1 y dy
vx x y0 + 1 dx +
vy x0 − 1 y dy +
+
x0 +1
x0 −1
vx x y0 − 1 dx
x0 −1
−dx
dy
+
2
x0 +1 1 + x − x0
y0 −1 1 + y − y0
x0 +1
dx
=
2
x0 −1 1 + x − x0
=
2
+
y0 +1
x0 +1
y0 −1
y0 +1
2
+
y0 −1
y0 +1
−dy
1 + y − y0
2
a result that depends only on the fact that the path encircles x0 y0 , not on any other
details of the path such as shape or size. Any path not enclosing the vortex has zero
2.2
57
Irrotational Flows and the Velocity Potential
TABLE 2.2.1 Velocity potentials and stream functions for irrotational flows
Flow Element
Two-dimensional
Uniform stream
Source or sink
Doublet
Line vortex
xUx + yUy′
m
ln x − x0
2
2+
yUx − xUy′
y − y0
Bx x − x0 + By y − y0
r − r0 2
By x − x0 − Bx y − y0
r − r0 2
−
y − y0
tan−1
2
x − x0
Three-dimensional
Uniform stream
xUx + yUy + zUz
Source or sink
−m
4 r − r0
Line vortex
2
y − y0
tan−1
2
x − x0
Flow Element
Doublet
m
y − y0
tan−1
2
x − x0
B · r − r0
r − r0 3
None
(axisymmetric)
05Uz r 2
−m z − z0
4 r 2 + z − z0
2
2
−Bz r
r 2 + z − z0 2
None
circulation, as can be verified by Stokes’s theorem. Therefore, a vortex has concentrated
vorticity but no concentrated mass discharge.
From equation (2.2.6) the stream function for a vortex is
(2.2.21)
ln r − r0 = −
ln x − x0 2 + y − y0 2
vortex = −
2
2
The velocity potentials and stream functions for all of these flows plus their threedimensional counterparts are summarized in Table 2.2.1.
2.2.3
Hele-Shaw Flows
The four solutions we have just seen—uniform stream, source/sink, doublet, and
vortex—are the fundamental solutions for two-dimensional potential flow. Before considering combinations of these to obtain physically interesting flows, it is useful to see
how these basic flows can be produced in a laboratory.
One way of producing them is by means of a Hele-Shaw table, which consists of
a flat horizontal floor with a trough at one end for introducing a uniform stream and
another trough at the other end for removing it. Holes in the bottom of the floor are
connected to bottles that may be raised or lowered. These provide the sources and sinks,
with the elevation of the bottle controlling their strength. Doublets can be produced
by putting a source and sink pair almost together, having the source height above the
table being equal to the sink height below. Sometimes a top transparent plate is added
to ensure that the flow is of the same depth everywhere. The vorticity component
perpendicular to the floor will be essentially zero; components in the plane of the floor
will be nonzero.
58
Inviscid Irrotational Flows
Vortices are more difficult to produce in Hele-Shaw flows than are sources or sinks.
One way of producing them is to use a vertical circular rod driven by an electric motor.
Hele-Shaw flows are slow flows, and because the velocity will vary roughly parabolically with the coordinate in the direction normal to the floor inherent to these flows,
there is a great deal of vorticity. However, this vorticity is largely parallel to the bed,
the vorticity component normal to the floor being virtually zero. Thus, Hele-Shaw flows
viewed perpendicular to the floor are good models of two-dimensional irrotational flows.
Streamlines can be traced by inserting dye into the flow. A permanent record of
these streamlines can be made by photography. Alternatively, one method that has been
used is to cast the bed of the table out of a hard plaster, with inlets for the source and
sink permanently cast into the plaster. If the bed is painted with a white latex paint,
streamlines can be recorded by carefully placing potassium permanganate crystals on
the bed. A record of the streamlines remains as dark stains on the paint.
2.2.4
Basic Three-Dimensional Irrotational Flows
Except for the vortex, all our two-dimensional irrotational flows have three-dimensional
counterparts that qualitatively are much like their two-dimensional counterparts. (An
example of a three-dimensional vortex is a smoke ring. Mathematical representation of
such a three-dimensional phenomenon is more complicated than in the case of twodimensional flows.) Here, we simply list these counterparts. The analysis proceeds as
in the two-dimensional case.
Uniform stream
The velocity potential for a uniform stream is
uniform
stream
(2.2.22)
= xUx + yUy + zUz = r · U
with a velocity field
v = = U
Surfaces of constant are planes perpendicular to U.
When U has only a component in the z direction, a Stokes stream function can be
found in the form
uniform stream
= 05Uz R2 sin2 = 05Uz r 2
Point source or sink (point monopole)
The velocity potential for a point source of strength m at r0 is
−m
−m
=
source =
4 r − r0 4 x − x0 2 + y − y0 2 + z − z0
(2.2.23)
2
(2.2.24)
Here, m is the volume discharge from the source, with continuity satisfied everywhere
except at r0 . If m is positive, represents a source. If negative, it represents a sink.
Irrotationality is satisfied everywhere.
Surfaces of constant are concentric spheres centered at r0 . The velocity is directed
along the radius of these spheres and dies out like the reciprocal of the distance squared
to satisfy continuity. The velocity is given by
v = =
m r − r0
4 r − r0 3
(2.2.25)
2.2
59
Irrotational Flows and the Velocity Potential
When the source lies on the z-axis, a Stokes stream function can be found in the form
source
=
−m z − z0
4 r 2 + z − z0 2
(2.2.26)
Point doublet (point dipole)
The velocity potential for a doublet can again be found by differentiating the potential
for a source, giving
doublet = B · source
of strength 4
=
with velocity components
v=
B
3
r − r0
−
B · r − r0
r − r0 3
3 r − r0 B · r − r0
r − r0 5
(2.2.27)
(2.2.28)
The constant surfaces are no longer simple geometries like circles and spheres.
When the doublet lies on the z-axis and additionally B points parallel to the z-axis,
a Stokes stream function for a doublet can be found in the form
doublet
=
r 2 +
−Bz r 2
z − z0 2 3/2
(2.2.29)
These results are summarized in Table 2.2.1.
2.2.5
Superposition and the Method of Images
In a number of simple cases, the solution for flow past a given body shape can be
obtained by recognizing an analogy between potential flow and geometrical optics, since
the Laplace equation also governs the passage of light waves. Boundaries can be thought
of as mirrors, with images of the fundamental solutions appearing at appropriate points
to generate the “mirror” boundary.
Line source near a plane wall
Suppose we have a source of strength m a distance b from a plane wall. According
to the method of images, the plane wall can be regarded as a mirror. As the source
“looks” into the mirror, it sees an image source of the same strength at distance b
behind the mirror (see Figure 2.2.6). A mirror interchanges right and left. This does not
affect the source, but will affect signs of vortices and doublets. The potential for the
two-dimensional case is, with the x-axis acting as the wall,
m 2
=
(2.2.30)
ln x + y − b 2 + ln x2 + y + b 2
2
with velocity components
m
x
x
vx =
+
2 x2 + y − b 2 x2 + y + b 2
(2.2.31)
y+b
y−b
m
+
vy =
2 x2 + y − b 2 x2 + y + b 2
Note that vy vanishes on y = 0, satisfying the necessary boundary condition that the
velocity normal to the stationary wall be zero.
60
Inviscid Irrotational Flows
Source at (1, 0) near a wall at (x, 0)
5
3
1
–1
–3
–5
–6
–4
–2
0
2
4
6
Figure 2.2.6 Source and its image by a wall—streamlines
This solution can be used as a simple model of physical problems such as water
intakes or pollution sources near a straight coastline. Additional sources for more
complicated flows can be easily added.
Note that if the source is in a corner, both walls act as mirrors, and there will
be three image sources plus the original source. The third image comes about because
the mirrors extend to plus and minus infinity, so the two images that you might think
would be sufficient in turn have their own image. The preceding procedure can be easily
extended to vortices and doublets near a wall and to three-dimensional examples. It
can also be used for acoustic sources. Some loudspeaker enclosures are designed to be
placed in corners or near walls to increase their apparent power output by the image
speakers.
Example 2.2.1 Point source near walls
A point source of strength m is located at the point (1, 2, 3). There is a wall at x = 0
and a second wall at y = 0. What is the velocity potential for this flow?
Solution. From equation (2.2.24), the given source has a velocity potential
0 = −
m
where r0 = i + 2j + 3k
4 r − r0
To generate the first wall, an image source of strength m at (−1, 2, 3) is needed. This
m
, where r1 = −i + 2j + 3k. When the second
will have a velocity potential 1 = − 4r−r
1
wall is added, both the original source and the image at (−1, 2, 3) will have images,
with a combined velocity potential
2 + 3 = −
m
m
−
where r2 = i − 2j + 3k and r3 = −i − 2j + 3k
4 r − r2 4 r − r3
Thus, our velocity potential is the sum of the potentials of four sources, each of strength
m, and located at the points (1, 2, 3), (−1, 2, 3), (1, −2, 3), and (−1 −2, 3). The total
potential is
2.2
61
Irrotational Flows and the Velocity Potential
total = 0 + 1 + 2 + 3 = −
m
m
m
m
−
−
−
4 r − r0 4 r − r1 4 r − r2 4 r − r3
Notice that if there were a third wall at z = 0, eight sources would be needed.
2.2.6
Vortices Near Walls
An interesting variation of the source near a wall is that of a vortex near a wall. The
image will have a circulation in the reverse direction of the original vortex because left
and right are interchanged in a reflection (see Figure 2.2.7). For a wall at x = 0 the
velocity potential for the original vortex plus its image is
−1 y + b
−1 y − b
=
− tan
tan
(2.2.32)
2
x
x
with velocity components
vx =
=
x
2
vy =
=
y
2
− y−b
2
x + y−b
x
x2 + y − b
y+b
2
x + y+b
2
+
2
x
+ 2
x + y+b
2
2
(2.2.33)
The velocity at (0, b) due to the image vortex is /4b 0 . This is called the induced
velocity. The vortex at (0, b) will tend to travel with the induced velocity, carrying its
image with it, since the induced velocity at the image will be the same value.
Wall
Figure 2.2.7 Vortex and its image by a wall—streamlines
62
Inviscid Irrotational Flows
The stream function to accompany equation (2.2.31) is
2
ln x + y − b 2 + ln x2 + y + b 2
2
x2 + y − b 2
ln 2
=
4 x + y + b 2
=
Therefore, lines of constant
(2.2.34)
are given by
x2 + y − b 2 = Cx2 + y + b 2
(2.2.35)
where C is a constant related to the value of the stream function by = /4 ℓn C.
Expanding and rearranging, equation (4.3.33) becomes
b 1+C 2
4Cb2
x2 + y −
=
1−C
1−C 2
Thus, the streamlines are circles of radius r = 2b C/ 1 − C centered at (0, d), where
1+C
.
d = b 1−C
Since the streamlines in this example are circles, a physical realization of this flow
is a single vortex in a cup of radius r, the vortex being a distance b from the center of
the cup. The vortex will travel around the cup on a circular path of radius r at a speed
of /4b.
Still another realization of a vortex flow is a vortex pair near a wall, the vortices
being of equal but opposite circulation. Take them to be a distance a from the wall and
separated by a distance 2b (Figure 2.2.8). To generate the wall, it is necessary to add a
vortex image pair as well. The total velocity potential is then
−1 y − a
−1 y − a
−1 y + a
−1 y + a
− tan
− tan
+ tan
tan
(2.2.36)
=
2
x−b
x+b
x−b
x+b
wall
b
a
Figure 2.2.8 Trajectories of a vortex pair and its images by a wall
2.2
63
Irrotational Flows and the Velocity Potential
with stream function
=
ln x − b 2 + y − a 2 − ln x + b 2 + y − a 2
4
− ln x − b 2 + y + a 2 + ln x + b 2 + y + a 2
and with velocity components
− y−a
vx =
=
x
2 x − b 2 + y − a
y−a
x+b 2 + y−a 2
y+a
−
x+b 2 + y+a 2
y+a
x−b 2 + y+a 2
x−b
vy =
=
y
2 x − b 2 + y − a
+
−
x−b
x−b 2 + y+a
2
2
(2.2.37)
+
x+b
2
x+b 2 + y−a 2
x+b
+
x+b 2 + y+a 2
(2.2.38)
−
The induced velocity at b a is
a
b
1
−1
−
+ 2
i+
j
Vinduced =
4
a a2 + b 2
b
a + b2
Therefore, the equations of motion for this vortex are
1
a
db
=
− 2
dt
4 a a + b2
da
−1
b
=
+ 2
dt
4 b
a + b2
(2.2.39)
With appropriate sign changes, similar equations hold for the other three vortices.
The path the vortices travel can be found by rearranging (2.2.39) in the form
a
b
dt
= 2 db = − 2 da
2
2
4 a + b
b
a
The variables a and b can be separated and the resulting equation integrated, giving as
the path the vortex travels
1
1
1
1
= 2 + 2 − 2
2
b
b 0 a0 a
(2.2.40)
a0 b0 being the initial position of the vortex.
This model of a traveling vortex pair is useful in describing the spreading of the
vortex pair left by the wing tips of an aircraft on takeoff. In this case, the wall represents
the ground. It is also responsible for the ground effect—the increased lift an airplane
experiences on takeoff due to the proximity of the ground.
Example 2.2.2 A vortex near a plane wall
A wall is located at x = 0. A vortex with circulation 20 meter2 /second is placed 1 meter
above the wall. What is the velocity potential, and at what speed does the vortex move?
Solution. For convenience, take the instantaneous position of the vortex to be
.
(0, 1, 0). Then the velocity potential for the original vortex is orig = 10 tan−1 y−1
x
64
Inviscid Irrotational Flows
Remembering that the circulation of the image vortex is reversed, from equation (2.2.19)
the velocity potential for the original vortex plus its image is
−1 y + 1
−1 y − 1
− tan
total = orig + image = 10 tan
x
x
Taking the gradient of the velocity potential, the velocity components are found to be
− y−1
y+1
vx = 10 2
+
x + y − 1 2 x2 + y + 1 2
x
x
vy = 10 2
−
x + y − 1 2 x2 + y + 1 2
The induced velocity at (0, 1, 0) is the velocity at that point due to the image vortex.
This gives a velocity at (0, 1, 0) of vx = 10 · 2/22 = 5 m/s vy = 0. The vortex thus
moves parallel to the wall at a speed of 5 m/s.
Example 2.2.3 A vortex pair in a cup
A vortex pair is generated in a cup of coffee of radius c by brushing the tip of a
spoon lightly across the surface of the coffee. The pair so generated will have opposite
circulations (try it!). If the vortex with positive circulation is at a b , and the vortex
with negative circulation at a −b , verify that the flow with the cup is generated by
an image vortex with positive circulation at ac2 / a2 + b2 −bc2 / a2 + b2 , plus an
image vortex with negative circulation at ac2 / a2 + b2 bc2 / a2 + b2 . These image
points are located at what are called the inverse points of our cup.
Solution. What is needed is a pair of opposite-rotating vortices plus the images
needed to generate the cup. Each vortex moves because of the induced velocity generated
by the images and the other vortex. The given vortex pair has a stream function
ln x − a 2 + y − b 2 − ln x − a 2 + y + b 2
4
x−a 2 + y−b 2
=−
ln
4
x−a 2 + y+b 2
=−
The proposed stream function consists of the original stream function plus the stream
function due to a pair of vortices at ak bk and ak −bk , where k = c2 / a2 + b2 .
The combined stream function is then
ln x − a 2 + y − b 2 − ln x − a 2 + y + b 2
4
+ ln x − ak 2 + y − bk 2 − ln x − ak 2 + y + bk 2
=−
As a check of the result, on the circle of radius c
x = c cos
and
y = c sin and so
ln c cos − a 2 + c sin − b 2 + ln c cos − ak 2 + c sin + bk 2
4
− ln c cos − a 2 + c sin + b 2 − ln c cos − ak 2 + c sin − bk 2 .
=−
2.2
Irrotational Flows and the Velocity Potential
But since cos2 + sin2 = 1 and using the definition of k
c cos − ak 2 + c sin − bk 2 = c2 − 2k a cos + b sin + a2 + b2 k2
= kc2 /k − 2 a cos + b sin + a2 + b2 k
= k a2 + b2 − 2 a cos + b sin + c2
= k c cos − a 2 + c sin − b 2
and
c cos − ak 2 + c sin + bk 2 = c2 − 2k a cos − b sin + a2 + b2 k2
= kc2 /k − 2 a cos − b sin + a2 + b2 k
= k a2 + b2 − 2 a cos − b sin + c2
= k c cos − a 2 + c sin + b 2
Substituting these into the expression of the stream function, we have
=−
ln c cos − a 2 + c sin − b 2 + ln k c cos − a 2 + c sin + b 2
4
− ln c cos − a 2 + c sin + b 2 − ln k c cos − a 2 + c sin − b 2
= 0
Thus, the cup of radius c is a streamline with the stream function equal to zero.
To find the equations that govern how the vortex at a b moves, the induced
velocity components are computed by taking the derivatives of , omitting the term
from the vortex at (a, b), and then letting x = a y = b. The result is
b
a 2 + b 2 a2 + b 2 + c 2
1
1
a=
−
+
,
2 2b2 a2 a2 + b2 − c2 2 + b2 a2 + b2 + c2 2 a2 + b2 − c2
1
a 2 + b 2 a2 + b 2 + c 2
a
−
.
b=
2 a2 a 2 + b 2 − c 2 2 + b 2 a 2 + b 2 + c 2 2 a2 + b 2 − c 2
2.2.7
Rankine Half-Body
A source located at the origin in a uniform stream (Figure 2.2.9) has the velocity
potential and a stream function
m 2
= xU +
ln x + y2
2
(2.2.41)
y
m
tan−1
= yU +
2
x
in a two-dimensional flow, and
m
= zU −
√
4 r 2 + z2
(2.2.42)
1
m
= r 2U −
√
2
4 r 2 + z2
in a three-dimensional flow.
65
66
Inviscid Irrotational Flows
2.0
1.0
Y
0.0
–1.0
–2.0
–6
–4
–2
0
X
2
4
6
Figure 2.2.9 Rankine half-body—two dimensions
Examining the three-dimensional case in further detail, we find the velocity components to be
vr =
mr
=
r
4 r 2 + z2
3/2
vz =
mz
=U+
x
4 r 2 + z2
3/2
(2.2.43)
It is seen from equation
√ (2.2.43) that there is a stagnation point vstagnation point = 0
at the point r = 0 z = − m/4U . On r = 0, the stream function takes on values
= −m/4 for z positive and = m/4 for z negative. The streamline = m/4
goes from the source to minus infinity. At the stagnation point, however, it bifurcates
and goes along the curve z2 = b2 − 2r 2 2 /4 b2 − r 2 z2 = b2 − 2r 2 2 /4 b2 − r 2 with
b2 = m/U . This follows by putting = m/4 into equation (2.2.42) and solving for
z. The radius of the body goes from 0 at the stagnation point to r = b far downstream
from the source. This last result can be obtained either from searching for the value
of r needed to make z become infinite in equation (2.2.43) or by realizing that far
downstream from the source the velocity must be U and all the discharge from the source
must be contained within
the body. In either case, the result is that far downstream the
√
body radius is b = m/U .
This flow could be considered as a model for flow past a pitot tube of a slightly
unusual shape. A pitot tube determines velocity by measuring the pressure at the stagnation point and another point far enough down the body so that the speed is essentially U .
The difference in pressure between these two points is proportional to U 2 . This pressure
difference can be found from our analytical results by writing the Bernoulli equation
between the stagnation point and infinity.
In the two-dimensional counterpart of this, the velocity is
vx =
mx
=U+
x
2 x2 + y2
vy =
my
=
y
2 x2 + y2
with the stagnation point located at −m/2U 0 . On y = 0 = 0 for x positive and
m/2 for x negative. The streamline = m/2 starts at − 0 , going along the x-axis to
the source. At the stagnation point the streamline bifurcates, having the shape given by
2Uy/m = − tan−1 y/x
The asymptotic half-width of the body is y = m/2U .
(2.2.44)
2.2
67
Irrotational Flows and the Velocity Potential
It is instructive to examine the behavior of the and lines at the stagnation point.
For the two-dimensional case, first expand
about the stagnation point in a Taylor
series about −m/2U 0 , giving to second order in x + m/2U and y
=
m 4U 2
m
x+
y+···
−
2
m
2U
This tells us that at the bifurcation point, the = m/2 line is either along y = 0 or
perpendicular to it (i.e., locally either on y = 0 or on x = −m/2U .
Expanding about −m/2U 0 to second order in x + m/2U and y, we have
m
m 2U 2
m 2
=
−1 + ln
+
+ y2 + · · ·
− x+
2
2U
m
2U
m
m
This tells us that = 2
−1 + ln 2U
along lines with slope ±45 y = ± x + m/2U ,
thereby bisecting the lines as we had earlier expected would happen.
Notice that if we wish to model this flow on a Hele-Shaw table, one way of
accomplishing this without the necessity of drilling any holes in our table would be to
cut out a solid obstacle of the shape given by setting = m/2 and then placing it on
the table aligned with the flow. The flow exterior to the obstacle is the same as if we
had drilled a hole and inserted the source.
2.2.8
Rankine Oval
The previous Rankine half-body was not closed because there was a net unbalance in
mass discharge. By putting an aligned source and sink pair in a uniform stream with
the source upstream of the sink, a closed oval shape is obtained. The velocity potential
and stream function then become
m
= xU +
ln x + a 2 + y2 − ln x − a 2 + y2
2
(2.2.45)
m
y
y
−1
−1
= yU +
− tan
tan
2
x+a
x−a
for a two-dimensional body, and
m
1
1
= zU +
−
4
r2 + z + a 2
r2 + z − a 2
1 2
m
z+a
z−a
= r U+
−
2
4
r2 + z + a 2
r2 + z − a 2
for a three-dimensional body.
The velocity for the two-dimensional body is
m
x−a
x+a
vx =
=U+
−
x
2
x + a 2 + y2
x − a 2 + y2
m
y
y
vy =
=
−
y
2
x + a 2 + y2
x − a 2 + y2
The stagnation points are therefore at ± a2 + ma/U 0 .
(2.2.46)
68
Inviscid Irrotational Flows
For the two-dimensional case, according to equation (2.2.45) the streamline that
makes up the body is given by = 0. (Notice that along y = 0 = 0, except in the
range −a < x < a, where = −m/2.) Therefore, the equation giving the body shape is
m
y
y
0 = yU +
− tan−1
tan−1
(2.2.47)
2
x+a
x−a
From symmetry, the maximum height of the body will be at x = 0. This height is given
from equation (2.2.47) as a solution of the equation
y
m
ymax =
− 2 tan−1 max
(2.2.48)
2U
a
Similar results hold for the three-dimensional case. The parameter governing shape
is m/Ua in the two-dimensional case and m/Ua2 in the three-dimensional case. If m/Ua
is large, the body is long and slender. If m/Ua is small, the body is short and rounded.
More complicated Rankine ovals can be formed by putting more sources and sinks
in a uniform stream. For the body to close, it is necessary that the sum of the source
and sink strengths be zero. This, however, is not a sufficient condition. Notice that in
our simple example that if the source and sink were interchanged (just change the sign
of m), there will be no closed streamlines about the source and sink pair.
2.2.9
Circular Cylinder or Sphere in a Uniform Stream
The Rankine oval is an unfamiliar geometrical shape, but if you plot its shape for various
values of the separation a and the shape parameter m/Ua, you can see that as the
source-sink pair get closer together while the shape parameter is held constant, the oval
shape becomes more and more circular. This suggests that in the limit as the source-sink
pair becomes a doublet, a circular shape would be achieved. The source portion of the
doublet should be facing upstream and the sink portion facing downstream in order to
generate a closed stream surface.
The velocity potential and stream function for a uniform stream plus a doublet is
= xU +
in a two-dimensional flow, and
= zU +
xBx
2
x + y2
zBz
r 2 + z2 3/2
= yU −
yBy
2
x + y2
1
r 2 Bz
= r 2U − 2
2
r + z2
(2.2.49)
3/2
(2.2.50)
in a three-dimensional flow.
If we let Bx = Ub2 in the two-dimensional case, equation (2.2.50) shows that = 0
both on y = 0 and on x2 + y2 = b2 . Therefore, we have flow past a circular cylinder of
radius b. Letting Bz = Ub3 /2 in three dimensions gives = 0 on a sphere of radius b.
How does this solution relate to the method of images? The interpretation is complicated by having to consider a curved mirror, but the flow can be thought of as the
body that focuses the uniform stream upstream (a very large distributed source) into the
source part of the doublet. The downstream part of the uniform stream (a very large
distributed sink) is focused into the sink part of the doublet.
More orderly ways of distributing sources to generate flows about given body
shapes are known. For thin bodies such as wings or airplane fuselages, sources are
distributed on the centerline, the strength of the source distribution per unit length being
2.3
69
Singularity Distribution Methods
proportional to the rate that the cross-sectional area changes. For more complicated
shapes, sources are distributed on the surface of the body. Vortices are included if lift
forces are needed, as indicated in the following section.
2.3 Singularity Distribution Methods
When generating flows past bodies by placing sources, sinks, and other basic flows
inside bodies, as was just done for Rankine bodies, we are solving inverse problems.
That is, we first place the basic flow elements, and then find the location of the closed
streamline corresponding to this distribution of mathematical singularities. We would
prefer to solve the more difficult direct problem, where we first specify a body shape
and then find where we must put the singularities. Since generally we do not want the
flow to be singular within the flow domain (i.e., to have infinite velocities outside of the
body), that leaves two possibilities: singularities distributed either within the interior of
the body or on its surface. We shall discuss both possibilities.
2.3.1
Two- and Three-Dimensional Slender Body Theory
First consider the possibility of putting singularities within the body. For simplicity,
restrict attention to two-dimensional, symmetric, slender bodies. (The symmetry restriction can easily be removed but with some complication in the mathematics.) By slender,
we mean that the thickness of the body is small compared to its length and that the
rate at which the thickness changes is also small. To emphasize this, write the body
thickness as
for 0 ≤ x ≤ L.
y = ±T x
(2.3.1)
where , the slenderness ratio, is defined as the maximum thickness divided by the
length L. For slender body theory to be valid, the slenderness ratio must be a small
parameter. Saying that the rate of thickness change is small implies that dy/dx is
everywhere small or that dT/dx is everywhere of order one or less.
To generate the flow past this body, assume that this can be accomplished by using
a uniform stream together with a continuous distribution of sources and sinks along the
center line of the body. Thus, write
1 L
= xU +
(2.3.2)
m ln x − 2 + y2 d
2 0
where m is the source strength per unit length along the body. We know that the velocity
derived from equation (2.3.2) automatically satisfies Laplace’s equation (continuity),
and, since it is a velocity potential, it will also be an irrotational flow. If we also satisfy
the boundary condition of zero normal velocity on the boundary, or
dx
x
=
dy
y
on
y = ±T x 0 ≤ x ≤ L
(2.3.3)
we have solved the problem.
Rearranging equation (2.3.3), and recognizing that on the body dy/dx = ±T ′ ,
where the prime denotes differentiation with respect to x, equation (2.3.3) becomes
= ± T ′
y
x
on
y = ±T
(2.3.4)
70
Inviscid Irrotational Flows
Using equation (2.3.2), this results in
L
T x
m
x− m
1 L
′
=
d = T x U +
d (2.3.5)
2
x − 2 + T 2
2 0 x − 2 + T 2
0
for 0 ≤ x ≤ L.
The task now is to solve equation (2.3.5) for m, given a thickness distribution T .
This is by no means a trivial job. This type of equation is called a Fredholm integral
equation of the first kind, and its solution by either analytic or numerical means requires
care and expertise.
To simplify the mathematics and obtain an approximate solution valid when is
small, notice that the effect of the source distribution is to give a y velocity component
(the left-hand side of equation (2.3.5)) that is small compared with U and also an
accompanying small x velocity component (the second term in the right-hand side of
equation (2.3.5)). The small y component cannot be neglected, since this term controls
the slope of the streamline in the boundary, but it might be reasonable to neglect the
x component compared to U . (This may sound a little ad hoc, but bear with me). The
validity of the assumptions can be checked once the solution has been found.
Making this approximation, and again with a little rearranging, equation (2.3.5)
becomes
1 L−x /T m x + T
d = UT ′ x
(2.3.6)
2 −x/T
1 + 2
where the substitution = − x /T has been made. Taking the limit as becomes
very small, equation (2.3.6) reduces to
m x d
≃ UT ′ x
(2.3.7)
2 − 1 + 2
or
m x ≃ U
d2T x
dx
(2.3.8)
Thus, the local source strength is the stream velocity times the slenderness ratio times
the local rate of change of total thickness of the body 2T . The appearance of in
equation (2.3.8) justifies the neglect of the x velocity component due to the source
distribution.
It might be argued that we were a little cavalier in bringing m outside the integral
sign in equation (2.3.6) because even though is small, becomes large in part of the
range of integration, and, in any case, is finite. Recall, however, that m is multiplied
by 1/1 + 2 and that in these regions this factor guarantees that the integrand is small.
Another question that arises is whether the body shape that we have generated
closes. To check this, in order for the body to close, the total source/sink strength must
be zero. Thus, we must have
L
m d = 0
(2.3.9)
0
From equation (2.3.8), we see that this condition is met, providing T 0 = T L . In that
part of the body where the thickness increases, from equation (2.3.8) it is seen that m
is positive, so it is a source. Where the body thickness decreases, m is negative, so it is
a sink. Where the body thickness does not change, we have neither a source nor a sink.
2.3
71
Singularity Distribution Methods
This analysis can be easily extended to curved two-dimensional slender bodies and
to three-dimensional bodies. For the three-dimensional case, a similar analysis shows
that the local source strength is again proportional to the rate of change of the body’s
area. For a uniform stream parallel to the z-axis and the body along the z-axis between
a and b, the result is
dA
1
1 b
d
(2.3.10)
r z = W z −
4 a r 2 + z − 2 3/2 d
Cross-flow can be included in the three-dimensional case with a little more effort,
the result being (since close to the slender body the flow appears to see what looks
like a long cylinder of circular cross-section) a doublet facing the cross-flow and with
strength equal to the cross-flow speed times the radius squared. For the cross-flow in
the x direction, the result is
A
1 b
d
r z = Ur cos 1 +
2 a r 2 + z − 2 3/2
Slender body theory has also been extended to curved, nonsymmetric bodies with
circulation, and the procedure for obtaining accuracy to any order of has been
explained. See van Dyke (1964), Cole (1968), and Moran (1984).
2.3.2
Panel Methods
The previous development is unfortunately only valid for slender bodies and, it turns
out, bodies of relatively simple shape. Even in such cases, it is likely that you will
not be able to carry out the integration of the source distribution potential by analytic
means and must resort to numerical integration to be able to proceed with applying the
results.
For more general shapes, surface distributions of sources and doublets allow us
to find solutions for any body shape—in three as well as two dimensions. Computer
solution of the resulting algebraic equations is a necessity, and the availability of
high-speed computers has revolutionized our approach to these problems. All of the
mathematical theory needed to carry out this approach was well known in the middle
of the nineteenth century. However, use of this mathematics was impractical before
computers.
To see the theory behind this, consider Green’s theorem from Appendix A. If x is a
point within the flow field, and xs is a generic point on the surrounding surface S, then
the potential corresponding to the distribution is
g r rS n · rS − rS n · g r rS dS
(2.3.11)
r =
S
where g r rS is the potential for a source of unit strength located at xs on the surface
and n is a unit normal pointing into the flow. Thus, the Green’s function in the two
cases is given by
g r rS =
−1
for three-dimensional flows,
4 r − rs
1
ln r − rs for two-dimensional flows.
=
2
(2.3.12)
72
Inviscid Irrotational Flows
The operation n · is equivalent to /n, the derivative in the direction locally normal
to the surface dS. If r is the position vector on the interior of S, the left side of equation
(2.3.11) is replaced by zero. Since g is the potential for a source, the first term in the
integral can be interpreted as a source distribution of strength /n and the second as
an outward-facing doublet distribution (recall that the derivative of the source potential
is a doublet potential) of strength .
We could now proceed to set up the boundary conditions as was done for the slender
body case, knowing that our solution already satisfies irrotationality and continuity
exactly, since we are working with velocity potentials that have already been shown to
satisfy the Laplace equation. Our effort, then, need only be directed toward satisfying
the boundary conditions.
If we were to proceed directly, however, we would find ourselves with an embarrassment of riches. We would have at any point on the surface two unknowns: the
source strength and the doublet strength. Roughly speaking, we would end up with
twice as many unknowns as we have conditions. Presumably we could discard half of
our unknowns—but which half is “best” discarded? And what are the consequences of
our actions and their physical interpretation?
To clarify this, realize that as long as the body surface is a stream surface, we really
don’t care about any “flow” that our velocity potential may give inside the body. Thus,
think of S as being a double surface—that is, a surface made up of our original surface
plus a very slightly smaller surface of the same shape inside it. On this inner surface, we
will construct a second source distribution that will have a velocity potential ′ inside
of S and zero outside of S. Then, with rS referring to a point outside of S, we have
0=
g r rs n · ′ rs − ′ rs n · g r rs dS
S
Adding this to equation (2.3.12), and recognizing that the preceding normal vector is
the negative of the one in equation (2.3.11), the result is
g r rs n · ′ rs − ′ rs n · g r rs dS
(2.3.13)
rS =
S
where represents the discontinuity in the velocity potential across the surface S, and
r lies outside of S.
There are then two distinct choices that can be made concerning the flow inside the
body:
l.
= 0 on S. In this case, the velocity potential is continuous across S. Then
the second term in equation (2.3.13) vanishes, and only a source distribution
remains.
2. n · = 0 on S. In this case, the normal velocity is continuous across S. Then the
first term in equation (2.3.13) vanishes, and only a doublet distribution remains.
If we are interested in flows with no lift forces, in principle either choice is correct,
although the numerical implementation in one case may be easier and/or have greater
accuracy than the other. If, however, we are interested in lifting flows, doublets must
be included.
You may wonder why the only doublets that appear in equation (2.3.13) are those
oriented perpendicular to the surface S. Why are there no doublets tangent to the surface?
The reason for this is that if we were to include a tangent doublet distribution of the form
Bt · s gdS
(2.3.14)
td =
S
2.3
73
Singularity Distribution Methods
where Bt is tangent to S and s involves derivatives with respect to xs , then with td as
the potential due to these tangent doublets, since
Bt · s g = s · Bt g − gs · Bt
s · Bt g − gs · Bt dS
td =
S
gs · Bt dS
= − gBt · n × ds −
(2.3.15)
S
C
The first integral on the right-hand side of equation (2.3.15) represents a ring of sources
bounding the surface S. If the surface is closed, the integral disappears. The second
integral is a surface source distribution of strength s · Bt . Thus, the tangent doublet
distribution is already realized in equation (2.3.13) through the source distribution.
The normal doublet distribution is equivalent to a ring vortex bounding the surface
S plus a surface vortex distribution on S. This should not be surprising, as the velocity
potential jump across S is typical of a vortex sheet, which itself is nothing more than a
velocity potential discontinuity.
The proof of this equivalence is quite involved in three dimensions but elementary
in two dimensions. Suppose we have a normal doublet distribution along a curve C
extending from A to B. Then the velocity potential is
B
doublet =
A
B
=
n · s ln
A
x − xs 2 + y − ys 2 ds
x − xs nx + y − ys ny
ds
x − xs 2 + y − ys 2
Defining the tangent vector t by t = n × k, then the components of the unit tangent are
related to those of the unit normal by tx = ny ty = −nx . Note also that
− y − ys
x − xs 2 + y − ys
x − xs
x − xs 2 + y − ys
2
2
y − ys
tan−1
x
x − xs
−1 y − ys
=
tan
y
x − xs
=
Then the velocity potential for the doublet can be written as
B
y − ys
ds
x − xs
A
B
t · s vortex ds = −
=−
doublet = −
t · s tan−1
A
= vortex BA +
A
B
A
t · s vortex − vortex t · s ds
B
vortex t · s ds
The first term is the two-dimensional version of the ring vortex; the second is the vortex
distribution.
74
Inviscid Irrotational Flows
To illustrate the use of equation (2.3.13), we next demonstrate how it can be
approximated in two dimensions to give the flow past a body of arbitrary shape. We
first elect to use the option of a source distribution. The potential for flow of a uniform
stream past the body represented by a source distribution is then, from equation (2.3.16),
given by
1
r = xU +
(2.3.16)
m rs ln x − xs 2 + y − ys 2 dS
2 S
To approximate the integration, the body will be covered with a series of M flat
panels (in two dimensions, panels become straight lines). On each of these panels the
source strength will be taken to be constant. Then equation (2.3.16) is replaced by the
approximation
r = xU + U
M
(2.3.17)
j Ij r
j=1
where j = mj /2U is evaluated on the jth panel and
Lj /2
Ij r =
ln x − xs 2 + y − ys 2 dsj
−Lj /2
Let Xj Yj and Xj+1 Yj+1 denote the endpoints of the jth panel (sometimes called
nodes), xcj ycj denote the center of the jth panel, and, referring to Figures 2.3.1
and 2.3.2,
xs = xcj + sj sin j ys = ycj − sj cos j Lj = Xj+1 − Xj 2 + Yj+1 − Yj 2
and xcj = 05 Xj + Xj+1 ycj = 05 Yj + Yj+1 .
To solve for the j , require that at the center point of the ith panel with coordinates
xci yci the normal velocity must be zero. Since
= cos
+ sin
n
x
y
differentiating equation (2.3.17) and forming the normal derivative gives, after carrying
out the integration,
0 = cos i + i +
M
j Iij i = 1 2 M
j =i
ni
Panel 1
Panel M
U
nel
Control Point
(xj , yj )
Figure 2.3.1 Paneled body
ne
x
Pa
j
βi
Pa
y
li
(2.3.18)
2.3
75
Singularity Distribution Methods
ni
(Xi , Yi )
θi
βi
(xi , yi )
(Xj + 1, Yj + 1)
ti
(Xi + 1, Yi + 1)
βj
nj
rij
i ′th panel
dsj
(xj , yj )
sj
θj
(Xj , Yj )
j ′th panel
Figure 2.3.2 Panels i and j detail
where
Iij =
025L2j + Aij Lj + Cij
1
sin j − i ln
2
025L2j − Aij Lj + Cij
−05Lj + Aij
05Lj + Aij
− tan−1
+ cos j − i tan−1
Bij
Bij
Aij = − xci − xcj sin j + yci − ycj cos j
Bij = xci − xcj cos j + yci − ycj sin j
and
Cij = xci − xcj 2 + yci − ycj 2
Equation (2.3.18) represents M algebraic equations in the M unknowns Iij and can
readily be solved by most any algebraic solver.
Once the j are known, the velocity at any point can be found. In particular, the
tangential velocity on the ith body panel is given by
Vi =
where
M
= cos i
+ sin i
= cos i + j Jij i = 1 2 M (2.3.19)
tangent
x
x
j=1
025L2j + Aij Lj + Cij
1
Jij = − cos j − i ln
2
025L2j − Aij Lj + Cij
−05Lj + Aij
05Lj + Aij
−1
−1
− tan
+ sin j − i tan
Bij
Bij
and
Dij = xci − xcj cos j + yci − ycj sin j
The tangential velocity is of use in computing the pressure at any point on the body
through use of Bernoulli’s theorem.
76
Inviscid Irrotational Flows
Several arbitrary decisions were made in setting up this approximation. For instance,
we took the nodes to be on the body, so our “panelized body” is the polygon inscribed by
the body. Alternately, we could have taken the center of the panel on the body, in which
case the nodes are outside of the body and the panelized body is a superscribing polygon.
Further, our choice of where to put the nodes, as well as the number of nodes, is our
engineering decision. Certainly we want the nodes to be placed closer together where
the curvature of the body is greatest. Generally, we would expect that the more nodes
there are, the better, but factors other than the number of nodes can be more important.
For instance, to find the flow past a symmetric body such as a circular cylinder, a panel
scheme that represents all symmetries of the problem will generally fare better than a
scheme with perhaps more nodes, but that doesn’t reflect the symmetries.
Clearly, curved body panels with the source strength varying along the panel
in a prescribed fashion is also an option. The result should be more accurate, with
necessarily more analysis. For example, with the previous approximation, the velocities
at the endpoints of the linear panels are infinite. Curved panels could eliminate these
singularities and make the velocity continuous. Thus, there is a certain amount of “art”
in actually carrying out the modeling. The present simple approach is sufficient to
illustrate the method.
The aerospace industry has developed these programs to a high degree of sophistication, modeling such flows as a Boeing 747 carrying piggyback a space vehicle such
as the space shuttle, even including details such as flow into the engines and the like.
Panel methods are now an important design tool, particularly in external flows.
We could also have elected to use the option of a normal doublet distribution so
that it would be possible to simulate flow past lifting bodies. The potential for flow of
a uniform stream past the body would then be given by
1
(2.3.20)
rs n · ln x − xs 2 + y − ys 2 dS
r = xU −
2 S
Again, to carry out the integration, the body is covered with a series of M straight
line panels. On these panels the doublet strength is taken to be constant. Then equation
(2.3.20) is replaced by the approximation
r = xU + U
where
j
M
j Kj r
(2.3.21)
j=1
= /2U on the jth panel, and, referring to Figures 2.3.1 and 2.3.2,
Kj r =
Lj /2
−Lj /2
x − xs cos j + y − ys sin j
dsj
x − xs 2 + y − ys 2
where the notation is as before. For this case the integration can be carried out explicitly,
giving
Kj r = tan−1
−05Lj − Pj
05Lj − Pj
− tan−1
Rj
Rj
where
Pj r = x − Xj sin j − y − Yj cos j
Rj r = x − Xj cos j + y − Yj sin j
(2.3.22)
2.4
77
Forces Acting on a Translating Sphere
The physical interpretation of this solution can be seen by noting that Pj = R − rc ·
tj Rj = R − rc · nj 05Lj − Pj = Rj+1 − r · tj , and −05Lj − Pj = Rj − r · tj , where
nj and tj are the unit normal and tangent to the jth panel. Then Kj r is the potential
of a pair of opposite-signed vortices at the nodes at the panel ends (located at Rj and
Rj+1 .
Solving for the j , on the ith panel, again requires that the normal velocity at the
center point must be zero. Differentiating equation (2.3.22) and then dividing by U , the
result is
0 = cos i + i +
where
Nij = − Rji sin j − i −
−
M
j=1
j Nij i = 1 2 M
(2.3.23)
05Lj − Pji cos j − i
+ Rji sin j − i
Lj − Pj 2 + R2ji
05Lj − Pji cos j − i
05Lj − Pji 2 + R2ji + Pji2
The second subscript i in the P and R indicates evaluation of the particular quantity
at ri . That is, we have Pji = Pj ri Rji = Rj ri .
Equation (2.3.23) represents M algebraic equations in the M unknowns j . The
solution of this system of equations is not usually as simple as in the case of the source
distribution, since the system is poorly conditioned and very sensitive to roundoff errors
and the like. In algebraic terms, this means that there is a great disparity in magnitude
among the various eigenvalues of the system. Systems that behave in this manner are
called stiff systems. (Similar problems can arise in systems of differential equations.
In the case of differential equations, they are characterized by having solutions that
contain terms that are negligible except in specific small regions.) Stiff systems can
render standard algebraic and differential equation methods useless. Thus, much greater
care must be used in finding its solution than in the case of the source distribution, and
special methods suited to dealing with such cases must be used.
Since a pure doublet distribution is difficult to work with, it is sometimes preferable
to use a source distribution supplemented by a doublet distribution, where all of the
doublet strengths have the same value. This adds one unknown to the set (the doublet
strength), and another condition must be added to have a determinate system. One
possibility is the Kutta condition, discussed in Chapter 3.
2.4 Forces Acting on a Translating Sphere
For a sphere translating in a real fluid, energy will be dissipated by viscous stresses
and must be replenished, even for the sphere to maintain a constant velocity. In inviscid
flows, a translating sphere will leave the energy unchanged unless the sphere is to
be accelerated with respect to the flow, if the sphere sees boundary changes such as
an uneven wall or a free surface, or if a cavity is allowed to form in the fluid. In
these cases the fluid kinetic energy can change, meaning that the sphere must exert a
force on the fluid so the fluid exerts an equal and opposite force on the sphere. These
forces are an inviscid drag force on the sphere, exerted by the pressure acting on the
sphere.
78
Inviscid Irrotational Flows
First consider a sphere of radius b moving at velocity U along the z-axis. In equation
(2.2.42) we found that for a stationary sphere in a moving stream, the velocity potential
3
was given by = zU + 2 r 2zb+zU2 3/2 . Thus, for a sphere moving in an otherwise stationary
fluid the potential is given by subtracting the uniform velocity, giving
=
zb3 U
2
r 2 + z2 3/2
=
b3 U cos
2R2
(2.4.1)
The fluid velocity as seen from the perspective of the sphere is then
vR =
b3 U cos
=−
R
R3
v =
1
b3 U sin
=−
R
2R3
(2.4.2)
We shall find the force on the sphere by two methods. The first is a global consideration
of the rate of work and change of energy. If the sphere is to be accelerated, the fluid kinetic
energy has to be changed. Thus, the first task is to compute the kinetic energy of the fluid.
Throughout the region R ≥ b, the square of the speed is given by equation vR2 +
6 2
2
v = U 2 Rb
cos + 41 sin2 . Accordingly, by equation (2.4.2) the kinetic energy is
given by
1 2 b 6
b3 2
1
T= U
U
cos2 + sin2 2R2 sin dRd =
2
R
4
3
0
b
The rate of work done by the force moving the sphere is F · U = Fz U = dT /dt; therefore,
d b3 U 2
dU
2
Fz U =
= b3 U
dt
3
3
dt
giving finally
dU
2
Fz = b3
3
dt
(2.4.3)
Note that it was necessary to compute the kinetic energy of a body moving in a quiescent
fluid. If the uniform stream had been left in the picture, the kinetic energy would have
been infinite.
The force can also be found by direct integration of the pressure over the surface of the
sphere. Since the sphere is moving and coordinates relative to the sphere are being used,
the pressure is found from the Bernoulli equation in translating coordinates to be given by
p
1 2
p
+
−U
+ vR + v2 = or
t
z 2
1 2
p − p
+U
− v + v2
=−
t
z 2 R
b3 1 − 3 cos2
1 2 b 6
1 2
dU b3 cos
2
cos
+
U
−
U
sin
+
=−
dt 2R2
2R3
2
R
4
On the surface of the sphere this becomes
1 − 3 cos2
p − p
1
1
dU b cos
=−
+U
− U 2 cos2 + sin2
dt
2
2
2
4
To compute the pressure force on any body due to inviscid effects, it is necessary to
carry out the integration
2.5
79
Added Mass and the Lagally Theorem
p cos sin d
2
1
dU
dU
dU
3
2
k
k − cos = b
k
cos sin d = b
= b
dt
dt
3
3
dt
0
0
F=−
pn · dA = −2k
0
(2.4.4)
where n is the unit outward normal on the body surface and the integration is taken over
the entire surface of the body. This second result is, of course, the same as equation
(2.4.3), found by energy considerations.
If the sphere is moving at a constant velocity, the drag force due to pressure vanishes.
When it is accelerating, the drag force on a sphere is proportional to the acceleration,
so the constant of proportionality must have the dimension of mass. Thus, we write
F = madded dU /dt, where madded is termed the added mass. In the case of the sphere,
madded = 2/3b3 , which is one-half the mass of the displaced fluid. Thus, the total force
needed to move the sphere is, by virtue of Newton’s law, F = mbody + madded dU
. The
dt
combination mbody + madded is sometimes called the virtual mass.
Although both approaches are capable of finding the added mass, the energy method
often is the simplest.
2.5 Added Mass and the Lagally Theorem
For inviscid flows, drag forces for translating bodies will always be of the form acceleration times added mass, where added mass will be some fraction of the mass of the
fluid displaced by the body. In general, if the body in a quiescent fluid is translating
with a velocity U and rotating with an angular velocity , the velocity potential can be
written in terms of axes fixed to the body in the form
=
6
V! !
1
where V! = Ux Uy Uz "x "y "z
Then the kinetic energy will be of the form
T=
6
6
1
A V V
2 !=1 =1 ! !
A! =
fluid volume
where
! · dV =
(2.5.1)
body
dS
!
n
surface
The A! are the components of what is referred to as the added mass tensor. Of the
36 possible A! , only 21 are distinct, since A! = A! . A11 A22 A33 represent pure
translation and have the dimension of mass. A44 A55 A66 are for pure rotations and have
the dimension of mass times length squared. The remaining terms represent interactions
between translations and rotations and have the dimension of mass times length.
The name added mass is clarified if we consider the kinetic energy of the solid
body without the fluid being present. That is given by
1
TB =
body V1 +V5 x3 −V6 x2 2 + V2 +V6 x1 −V4 x3 2
2 body volume
6
6
1
+ V3 +V4 x2 −V5 x1 2 dV =
B V V
(2.5.2)
2 !=1 =1 ! !
80
Inviscid Irrotational Flows
where
B11 = B22 = B33 = M
B44 = I11 B55 = I22 B33 = I33 B12 = −I12 B13 = −I13 B23 = −I23
(2.5.3)
B26 = 2Mx B35 = −2Mx B34 = 2My B16 = −2My B15 = 2Mz B24 = −2Mz
Here M is the mass of the body, the Is are components of the mass moment of inertia
of the body, and the barred coordinates are the position of the body center of mass. The
total kinetic energy of the body is then
T + Tbody =
6
6
1
A! + B! V! V
2 !=1 =1
(2.5.4)
To move the body through the fluid, then, the energy change required is the same as if
its inertia was increased by the added mass.
For bodies made up of source, sink, and doublet distributions, the calculations
for M! have been variously evaluated by Taylor (1928), Birkhoff (1953), and
Landweber (1956). Their results do not include the possibility of vortex lines or
distributed singularities, so using their results are limited to non-lifting cases.
As just mentioned, flows can be unsteady even for bodies that move at constant
velocities if their position relative to boundaries varies. The general calculation for
such cases was carried out originally by Lagally (1922) and have come to be known
as Lagally forces and moments. Lagally’s results were later generalized by Cummins
(1953) and by Landweber and Yih (1956). The derivations and formulae are lengthy
and somewhat complicated, so they will not be repeated here. Again, vorticity is not
included in their results.
We can see from simple global momentum analysis that for steady flows that involve
translating bodies generated solely by source-sink distributions inside or on the body
surface, no drag forces are found. It is, however, possible for such bodies to generate
forces perpendicular to the direction of translation. These forces are the lift forces that
normally one would expect to find even with the neglect of viscous effects. This absence
of lift in the formulation can be corrected by including vorticity in any model where lift
forces are desired. For instance, for the cylinder in the previous example, including a
vortex at the center of the cylinder would give the velocity potential and stream function
y
2
yB
xB
tan−1
= yU − 2 x 2 +
ln x + y2
(2.5.5)
= xU + 2 x 2 +
x +y
2
x
x +y
2
We can see that the stream function is constant on the cylinder x2 + y2 = a2 ; therefore,
the boundary condition on the body is still satisfied. Evaluation of the pressure force now,
however, gives a lift force proportional to U, called the Magnus effect after its discoverer.
Where would this vorticity come from in a physical situation? The cylinder could
be caused to rotate, and the effects of viscosity then provide the tangential velocity that
is provided in our mathematical model by the vortex. This has been attempted in ships
(the Flöettner rotor ship, Cousteau’s Alcone) and experimental airplanes, but it requires
an additional power source and is not generally practical. The effect of this rotation is
instead provided by having a sharp trailing edge for the lifting surface, or by providing
a “flap” on a blunter body. This is done to force the velocity on the body to appear the
same as in our model and thereby generate the desired force. The relationship between
lift force and vorticity is called the Joukowski theory of lift.
2.6
81
Theorems for Irrotational Flow
For the flow given in equation (2.5.2), the potential can be expressed in cylindrical
polar coordinates as
b2
(2.5.6)
= Ur cos 1 + 2 +
r
2
giving velocity components
b2
= U cos 1 − 2
vr =
r
r
b2
1
= −U sin 1 + 2 +
v =
r
r
2r
(2.5.7)
On the boundary r = b, the radial component vanishes and the tangential component
becomes v = −2U sin + /2b. Stagnation points are thus located at sin = /4Ub.
The techniques for lift generation mentioned previously amount to control of the locations of the stagnation points, thus generating the circulation necessary for lift. Chapter 3
discusses this in more detail.
2.6 Theorems for Irrotational Flow
Many theorems for irrotational flows exist that are helpful both in analysis and computational methods. They are also helpful for understanding the properties of such flows.
2.6.1
Mean Value and Maximum Modulus Theorems
Theorem: The mean value of over any spherical surface throughout whose interior
2 = 0 is equal to the value of at the center of the sphere.
Proof: Let c be the value of at the center of a sphere of radius R, and m be
the mean value of over the surface of that sphere. Then
1
1
dS
=
d#
m =
4R2 S
4
where d# = dS/R2 is the solid angle subtended by dS. From Green’s theorem, it
follows that
d
n · dS =
· dV =
2 dV = 0 =
dS = R2
d#
dR
S R
S
V
V
Therefore, the integral d# must be independent of the radius R. In particular, if we let
R approach zero, 4m = d# → 4c , proving the theorem. While we have proven
this theorem in three dimensions, a similar theorem can be proved in two dimensions.
The important concept to be learned from this is that the Laplace equation is a great
averager. At any point the solution is the average of its value at all neighboring points.
This is the basis for several numerical methods used in the solution of Laplace’s equation.
2.6.2
Maximum-Minimum Potential Theorem
Theorem: The potential cannot take on a maximum or minimum in the interior of any
region throughout which 2 = 0.
Proof: Again, taking a sphere around a point in a region where 2 = 0, if c were
a maximum, from the previous theorem it would be greater than the value of at all
points on the surface of the sphere, which contradicts the previous theorem.
82
Inviscid Irrotational Flows
2.6.3
Maximum-Minimum Speed Theorem
Theorem: For steady irrotational flows both the maximum and minimum speeds must
occur on the boundaries.
Proof: Consider two points P and Q that are close together. Choose axes at P
such that vP 2 = /x 2P and vQ 2 = 2Q . Since 2 /x = 0, from the previous theorem we can find a point Q such that /x 2P < /x 2Q , and therefore
vP 2 < vQ 2 . Thus, vP 2 cannot be a maximum in the interior. It also follows by the
same argument that for steady irrotational flows the speed must also take on a minimum
on the boundaries. The previous theorems hold only if 2 = 0 throughout the region.
Generally, if there are interior boundaries,2 modifications of the above result may be
necessary.
2.6.4
Kelvin’s Minimum Kinetic Energy Theorem
Theorem: The irrotational motion of a liquid that occupies a simply connected region
(that is, a region where all closed surfaces in the region can be shrunk to a point
without crossing boundaries) has less kinetic energy than any other motion with the
same normal velocity at the boundary. This theorem is credited to Lord Kelvin (born
William Thomson, 1824–1907) and published in 1849.
Proof: Let vQ represent the velocity field of any motion satisfying · vQ = 0 and
having kinetic energy TQ . Let be the velocity potential of an irrotational flow having
kinetic energy T . Let the boundary have a normal velocity component vn so that
/n = vQ · n = vn on the boundary (i.e., we are requiring that both velocity fields
satisfy the same boundary condition).
Define vdifference = vQ − , and note that vdifference = 0 on the boundary. Then
1
1
TQ =
vQ · vQ dV =
vdifference + · vdifference + dV
2
2
V
V
1
1
=
· vdifference dV
vdifference · vdifference dV +
· dV +
2
2
V
V
V
· vdifference dV
= Tdifference + T +
V
But vdifference · = · vdifference − · vdifference = · vdifference because its two
components satisfy the continuity equation, the difference velocity also must satisfy the
continuity equation. Thus,
V
· vdifference dV =
It follows that
V
· vdifference dV =
TQ = Tdifference + T ≥ T since Tdifference ≥ 0
2
Such regions are said to be periphractic.
S
vdifference · ndS = 0
(2.6.1)
2.6
83
Theorems for Irrotational Flow
The equality holds only if vdifference is zero everywhere in the region.
Note also for future use that
1
1
· dV =
· − 2 dV
T =
2
2
V
V
1
1
· dV =
dS
=
2
2
n
V
S
all
(2.6.2)
boundaries
2.6.5
Maximum Kinetic Energy Theorem
Theorem: The irrotational motion of a liquid occupying a simply connected region (that
is, a region where all closed surfaces in the region can be shrunk to a point without
crossing boundaries) has more kinetic energy than any other irrotational motion that
satisfies the boundary conditions only in the average.
Proof: Let Q represent the velocity field of any motion satisfying 2 Q = 0 and
having kinetic energy TQ . Let be the velocity potential of an irrotational flow having
kinetic energy T . Let the boundary have a normal velocity component vn so that
/n = vn on the boundary. The potential Q satisfies the averaged boundary condition
Q
Q vn −
dS ≥ 0
n
S
Define difference = − Q . Then
1
1
T =
· dV =
Q + difference · Q + difference dV
2
2
V
V
1
1
Q · Q dV +
difference · difference dV
=
2
2
V
V
(2.6.3)
Q · difference dV
+
V
Q · difference dV
= TQ + Tdifference +
V
But because its two potentials satisfy the Laplace equation, it follows that
difference · Q = · Q difference − Q · difference = · Q difference
and so
Q
Q vn −
· Q difference dV =
Q · difference dV =
dS ≥ 0
n
S
V
V
Thus
(2.6.4)
T = TQ + Tdifference ≥ TQ
since
Tdifference ≥ 0
The equality holds only if vdifference is zero everywhere in the region.
This and the previous theorem give us upper and lower bounds on the kinetic
energy. This also is useful for estimating the added mass coefficients and thus the drag
forces in accelerating flows.
84
Inviscid Irrotational Flows
2.6.6
Uniqueness Theorem
Theorem: There cannot be two different forms of acyclic irrotational motion3 of a
confined mass of a fluid in which the boundaries have prescribed values.
Proof by contradiction: Suppose we have two velocity potentials 1 and 2 , both
satisfying the Laplace equation and the same prescribed boundary conditions. Let 3 =
1 − 2 . Then the normal derivative of 3 vanishes on the boundary, and it follows that
the kinetic energy is zero. Thus, the velocity associated with 3 is zero everywhere,
and 3 must be a function of at most time. Notice that this breaks down if the velocity
potential is multivalued, which we have already seen is true for flows with concentrated
vortices. It also breaks down if there are cavities in the flow.
2.6.7
Kelvin’s Persistence of Circulation Theorem
Theorem: For an irrotational flow whose density is either constant or a function of
pressure, as we follow a given closed circuit the circulation does not change. This
theorem is credited to Lord Kelvin, and was published in 1869.
Proof: In equation (1.11.7) circulation was defined by = C v · ds. If we take the
time derivative following the same fluid particles as they move downstream, we have
Dv
D
Dds
1
D
− p + g · ds + v · dv
v · ds =
=
· ds + v ·
=
Dt
Dt C
Dt
Dt
C
C
If g is a conservative body force such as gravity and the conditions on density stated
above are met, then the integrand is an exact differential. Since the path is closed, this
means that
D
= 0
Dt
2.6.8
(2.6.5)
Weiss and Butler Sphere Theorems
Weiss sphere theorem: For a velocity potential ′ having no rigid boundaries and with
all singularities at least a distance a from the origin, the flow due to a rigid sphere of
radius a introduced into the flow at the origin is characterized by the velocity potential
a ′ a2 R
2 a ′ 2 R
′
−
d
(2.6.6)
R = R +
R
a R 0
R2
R2
Butler sphere theorem: If the preceding flow is also axisymmetric and characterized by
a stream function ′ R , where R are spherical polar coordinates, then the flow
due to introduction of the sphere is given by
R ′ a2
′
(2.6.7)
R −
R =
a
R
The proof of the Weiss theorem is fairly long and will be omitted here.
The proof of the Butler sphere theorem follows from the Weiss theorem quite
simply. Substitution of a for R makes the right-hand side of equation (2.6.7) vanish.
3
An acyclic irrotational motion is one for which the velocity potential is a single-valued function.
Problems—Chapter 2
Problems—Chapter 2
2.1 Examine the following functions to see if they could represent inviscid irrotational flows.
a. = x + y + z.
b. = x + xy + xyz.
c. = x2 + y2 + z2 .
d. = zx2 − x2 − z2 .
e. = sin x + y + z .
f. = ln x.
2.2 Show that if 2 1 = 0 and 2 2 = 0, then 1 + 2 C1 , and C + 1 also
satisfy the Laplace equation. Here C is a function of time.
2.3 Given a flow field written in terms of a scalar , with the velocity defined by
v = , indicate under what conditions the velocity field satisfies each of the following.
a. irrotationality.
b. conservation of mass.
c. the Euler equation.
d. the Navier-Stokes equation.
2.4 The potential , the velocity potential, is defined by v = , stating that the
velocity is the gradient of this potential. This means that the velocity is orthogonal to
equipotential surfaces. For inviscid flows, find a corresponding potential whose gradient
is the acceleration.
2.5 A shear flow has the velocity field v = U + yS 0 0 , where U and S are
constants.
a. Find the circulation about a unit square with the bottom side on the x-axis.
b. Find the circulation about the unit circle centered at the origin.
c. Find the circulation about the ellipse with major axis of length 2 and minor
axis of length 1. The ellipse is centered at the origin.
2.6 Select a proper combination of uniform stream and source strength to generate
an axisymmetric Rankine body 1.4 meters in diameter. The pressure difference between
ambient and stagnation is 5 kilo-Pascals. The fluid is water = 1000 kg/meter 3
2.7 a. Determine the stream function for a simple axisymmetric Rankine body that
is 3 meters long and 2 meters maximum in diameter. It is in a uniform
stream of strength 3 meters/second. The fluid is water.
b. Find the pressures at the stagnation points and at the place of largest diameter.
2.8 Find the length and breadth of the closed three-dimensional axisymmetric body
that is formed by a distributed line sink of strength m per unit length extending from
−1 0 to (0, 0), a distributed line source of the same strength from (0, 0) to (1, 0),
and a uniform flow along the z-axis. Use m = 20 and U = 5.
2.9 Find the drag force on an accelerating sphere of radius a by integrating the
pressure over the body.
85
86
Inviscid Irrotational Flows
2.10 A solid sphere 30 mm in diameter and having a specific gravity of 7 is released
in still water. Find its initial acceleration.
2.11 A point doublet of strength is located at r = 0 z = b. Find the stream
function if a sphere of radius a < b is inserted at the origin.
2.12 a. Find the equation that the Stokes stream function must satisfy to represent
an irrotational flow in three dimensions. Use spherical polar coordinates.
b. Use separation of variables in the form
equations that F and T must satisfy.
R = F R T , and find the
Chapter 3
Irrotational Two-Dimensional Flows
3.1
3.2
3.3
3.4
3.5
Complex Variable Theory Applied
to Two-Dimensional Irrotational
Flow
Flow Past a Circular Cylinder
with Circulation
Flow Past an Elliptical Cylinder
with Circulation
The Joukowski Airfoil
Kármán-Trefftz and
Jones-McWilliams Airfoils
87
91
93
95
98
NACA Airfoils 99
Lifting Line Theory 101
Kármán Vortex Street 103
Conformal Mapping and the SchwarzChristoffel Transformation 108
3.10 Cavity Flows 110
3.11 Added Mass and Forces and Moments
for Two-Dimensional Bodies 112
Problems—Chapter 3 114
3.6
3.7
3.8
3.9
3.1 Complex Variable Theory Applied to Two-Dimensional
Irrotational Flow
The theory of complex variables is ideally suited to solving problems involving twodimensional flow. The term complex variable means that a quantity consists of the sum
of a real and an imaginary number.
An imaginary number is a real number multiplied
√
by the imaginary number i = −1. (The terms imaginary and complex distinguish these
numbers from real numbers.) A complex number is in fact the sum of two real numbers,
the second one being multiplied by the square root of minus one. In many ways complex
variable theory is simpler than real variable theory and much more powerful.
Briefly, a complex function F that depends on the coordinates x and y is written in
the form
Fx y = fx y + igx y
(3.1.1)
where f and g are real functions. This type of representation has some of the properties of
a two-dimensional vector, with the real part standing for the x component and the imaginary part the y component. Thus, the complex number represented by equation (3.1.1)
has the directionality properties of the unit vector representation F = f i + gj, (here i
and j are Cartesian unit vectors), at least as far as representation and transformation of
87
88
Irrotational Two-Dimensional Flows
F plane
g
f
Figure 3.1.1 Complex variable—general complex plane
z plane
(x,y )
y
x
Figure 3.1.2 Complex variable—z plane
coordinates is concerned. The two forms of representation differ considerably, however,
in operations like multiplication and division.
A complex function F can be represented in graphical form as in Figure 3.1.1, and
a spatial position can be represented as in Figure 3.1.2, where the horizontal axis is the
x-axis, and the vertical axis is the y-axis. To denote the position vector, it is traditional
to write z = x + iy, where z is a complex number (z here is not the third-dimensional
coordinate) representing the position of a point in space. The plane containing the
x- and y-axes is called the complex z plane. Similarly, the complex F plane is shown
in Figure 3.1.2, with f and g measured along the horizontal and vertical axes.
The principal interest of complex function theory is in that subclass of complex
functions of the form in equation (3.1.1) that have a unique derivative at a point (x, y).
Unique derivative means that, if F is differentiated in the x direction z = x,
obtaining
F
f
g
=
+i
x
x
x
or if F is differentiated in the y direction z = iy, obtaining
f
g
f g
F
=
+i
= −i +
iy iy
iy
y y
(3.1.2a)
(3.1.2b)
the results at a given point are the same. For this to be true, the real parts and the
imaginary parts of equations (3.1.2a) and (3.1.2b) must be equal to each other. Thus, it
must be that
f
g
g
f
=
=−
(3.1.3)
x
y
x
y
3.1
89
Complex Variable Theory Applied to Two-Dimensional Irrotational Flow
The equations in (3.1.3) are called Cauchy-Riemann conditions, and functions
whose real and imaginary parts satisfy them are called analytic functions. Most functions of the complex variable z that involve multiplication, division, exponentiation,
trigonometric functions, hyperbolic functions, exponentials, logarithms, and the like are
analytic functions. Functions that can be expressed in terms of x only or y only, or
involving operations such as magnitude, arguments, or complex conjugations are the
few commonly used functions that are not analytic. (Recall that if F = f + ig is a complex number, its complex conjugate is F ∗ = f − ig. If F satisfies the Cauchy-Riemann
conditions, F ∗ will not.) Analytic functions have many useful properties, such as the
ability to be expanded in power series, the fact that an analytic function of an analytic
function is analytic, and that a transformation of the form z1 = Fz2 is angle-preserving.
Angle-preserving transformations are said to be conformal.
The preceding discussion was phrased in terms of derivatives in x and y. Since the
choice of a coordinate system is arbitrary, it should be clear that in fact, at any point in
the complex space, derivatives taken in any arbitrary orthogonal directions must satisfy
the Cauchy-Riemann conditions.
Comparison of the equations in (3.1.3) with equations (2.2.6a) and (2.2.6b) shows
that the complex function
(3.1.4)
w = +i
with as the velocity potential and
as Lagrange’s stream function is an analytic
function, since we have already seen from the stream function and velocity potential that
vx =
=
x
y
vy =
=−
y
x
which in fact are the Cauchy-Reimann conditions. The complex function w is termed
the complex velocity potential, or just the complex potential.
From differentiation of w find that
dw
=
+i
= vx − ivy
dz
x
y
(3.1.5)
That is, the derivative of the complex velocity potential is the complex conjugate of the
velocity, which is thus an analytic function of z.
Example 3.1.1 Complex variables—analytic functions
For Fz = az3 with a real, find the real and imaginary parts of F , show that F is an
analytic function, and decide whether the mapping from z to F is conformal.
Solution. Putting z = x + iy into F F = ax + iy3 = ax3 + 3x2 iy + 3xi2 y2 + i3 y3 .
Since i2 = −1, this reduces to F = ax3 + 3x2 iy − 3xy2 − iy3 = a x3 − 3xy2 + i3x2 y −
y3 . Separation into real and imaginary parts gives f = ax3 − 3xy2 g = a3x2 y − y3 .
To study the analyticity of F , form the partial derivatives of f and g, giving then
g
f
= 3ax2 − y2 =
x
y
f
g
= −6axy = −
y
x
Thus, F satisfies the Cauchy-Riemann equations, and therefore F is an analytic function
of z.
Since dF/dz = 3az2 has no singularities for finite z, and is zero only at z = 0, the
mapping from the z plane to the F plane is angle preserving except at z = 0.
90
Irrotational Two-Dimensional Flows
Example 3.1.2 Complex variables—analytic functions
Repeat the previous example, but with Fz = a ez with a real.
Solution. Putting z = x + iy into F and using DeMoive’s theorem, which states
that eiy = cos y + i sin y, find that F = aex+iy = aex cos y + i sin y. Thus, f = aex cos y
g = aex sin y. Taking the partial derivatives of f and g,
f
g
= aex cos y =
x
y
f
g
= −aex sin y = −
y
x
and so F is an analytic function of z in the entire finite z plane. Since dF/dz = aez is
finite and nonzero for all finite z, the mapping from the z plane to the F plane is angle
preserving.
Comparison of equation (3.1.4) with our basic flows in Chapter 2 gives the following
representations for these flows:
uniform stream w = U ∗ z
m
source/sink at z0 w =
lnz − z0
2
doublet at z0 w = B/z − z0
(3.1.6)
(3.1.7)
(3.1.8)
−i
(3.1.9)
lnz − z0
2
where again an asterisk denotes a complex conjugate. The vortex in equation (3.1.9) is
counterclockwise if > 0, and clockwise if < 0. The preceding expressions are much
more compact and easier to remember and work with than the forms given in Chapter 2,
although separating the complex velocity potential into real and imaginary parts can be
tedious.
If compactness and ease of use were the only advantages gained from the introduction of complex variables, there would be little justification for introducing them.
The power of complex variable theory is that, since an analytic function of an analytic
function is analytic, we can solve a flow involving a simple geometry and then use an
analytic function to transform, or map, that geometry into a much more complicated one.
In the process we concern ourselves only with the geometry, realizing that because we
are dealing with analytic functions, the fluid mechanics of the flow (that is, continuity
and irrotationality) is automatically satisfied.
Another important use of complex variables is in the integration of functions.
Perhaps surprisingly, it is often easier to integrate analytic functions than it is to integrate
real functions. The reason is Cauchy’s integral theorem. It states that if a function f (z)
is analytic and single-valued inside and on a closed contour C, then
fzdz = 0
(3.1.10)
vortex at z0 w =
C
Further, for a point z0 inside C,
f n z0 =
n!
fz
dz
2 i C z − z0 n+1
(3.1.11)
Here, f n z0 is the nth derivative of f evaluated at z0 . This theorem is useful in
determining forces on bodies.
3.2
91
Flow Past a Circular Cylinder with Circulation
Note that the theorem is restricted to single-valued functions. By that we mean
for a given z, there is only one possible value for f(z). An example of a single-valued
function is fz = sin z. An example of a multivalued function is fz = sin−1 z, which is
arbitrary to a multiple of 2 . The stream function for sources and the velocity potential
for vortexes are examples of multiple-valued functions. They must be treated with more
care than single-valued functions.
3.2 Flow Past a Circular Cylinder with Circulation
We illustrate the mapping process by showing how the flow past a circular cylinder can
be transformed into the flow past either an ellipse or an airfoil shape. Recalling from
Chapter 2 that the flow of a uniform stream past a circular cylinder is a doublet facing
upstream in a uniform stream, write
w = zU ∗ +
i
z
a2
U+
ln
z
2
a
(3.2.1)
where, for purposes that will become clearer as we proceed, a vortex has been added
at the center of the circle. To assure ourselves that this is indeed flow past a circular cylinder, note that, since on the cylinder z = aei , the complex potential on the
cylinder is
i
a2
i
ae
ln
= 2a U cos − −
waei = aei U ∗ + i U +
ae
2
a
2
which is real—hence
= 0 on z = a. Also, since
dw
a2
i
= U∗ − 2 U +
dz
z
2 z
(3.2.2)
far away from the cylinder the flow approaches a uniform stream. (Note: In writing the
vortex, a constant a was introduced as dividing z in the logarithm. This has the virtue
of making zero on the circle rather than a more complicated expression. It has the
additional advantage that we are now taking the logarithm of a dimensionless quantity,
which is, after all, a necessity. It has no other effect on the flow whatsoever and was
only inserted for convenience.)
From equation (3.2.2) we can easily find the stagnation points in the flow. Setting
dw/dz = 0 and then multiplying by z2 , the result is the quadratic (in z) equation
2
−i
± 4a2 UU ∗ − 2
i
2
2
2 ∗
z − a U = 0 with roots zseparation =
(3.2.3)
zU +
2
2U ∗
This can be put into a more understandable form by writing U = U e−i G =
/4 a U , giving
√
(3.2.4)
zstagnation = ae−i −iG ± 1 − G2
When G lies in the range −1 ≤ G ≤ 1, the stagnation points lie on the cylinder, and
equation (3.2.4) can be simplified further by letting G = sin . Then the two stagnation
points are at
zDSP = ae−i+
and zUSP = −ae−i− = ae−i−−
(3.2.5)
92
Irrotational Two-Dimensional Flows
y
a
x
α
U
USP
DSP
Figure 3.2.1 Cylinder in a uniform stream with circulation
as shown in Figure 3.2.1. In the figure, DSP stands for downstream stagnation point
and USP for upstream stagnation point. The stagnation points on the cylinder are thus
oriented at angles ± with the uniform stream. For the case where G is greater than
one, equation (3.2.4) shows that the stagnation points move off the circle, with one of
them lying inside and the other outside the circle. Both lie on a line that is perpendicular
to the uniform stream. Streamlines that show the flow past the cylinder are given in
Figure 3.2.2.
4
2
0
–2
–4
–6
–4
–2
0
Figure 3.2.2 Cylinder in a uniform stream—streamlines
2
4
3.3
93
Flow Past an Elliptical Cylinder with Circulation
Before going on, consider the force on the cylinder. From equation (3.2.2) the
velocity on the cylinder is given by
dw
i −i
e
= U ∗ − Ue−2i +
dz @z=aei
2 a
By Bernoulli’s equation the pressure on the cylinder is then
1
sin
p = p0 − Ux 1 − cos 2 + Uy sin 2 +
2
2 a
+ Ux sin 2 + Uy 1 + cos 2 +
2
2
2 a
cos
where p0 is the pressure at the stagnation point. The hydrostatic term has been omitted
in the pressure, for it can more easily be included later using Archimede’s principle.
The force on the cylinder per unit distance into the paper is then
2
F = Fx + iFy =
2
=
0
0
p− cos − i sin ad
1
acos + i sin
2
Ux 1 − cos 2 + Uy sin 2 +
+ Ux sin 2 + Uy 1 + cos 2 +
cos
2 a
2 a
2
sin
2
d
= Uy + iUx = iU ∗
(3.2.6)
Recall from the definition of G and that
= 4 a U sin
(3.2.7)
The stagnation pressure has been omitted in the preceding calculation because its integral
is zero.
Note three things about equation (3.2.5):
1. The force is independent of the cylinder size, being simply the fluid density times
the circulation times the stream speed. (Later it will be seen that the circulation
can be affected by the geometry.)
2. The force is always perpendicular to the uniform stream, so it is a lift force.
3. To have this lift force, circulation must be present.
3.3 Flow Past an Elliptical Cylinder with Circulation
To find the flow past an ellipse, introduce the Joukowski transformation:
z′ = z +
b2
z
(3.3.1)
This transformation adds to a circle of radius b its inverse point to the position z. (An
inverse point to a circle is the point zinverse such that z · zinverse = b2 .)
94
Irrotational Two-Dimensional Flows
a
d
c
z ′ plane
z plane
Figure 3.3.1 Ellipse in a uniform stream—transformations
To see what the Joukowski transformation does, let z = aei in equation (3.3.1) and
then divide the resulting expression into real and imaginary parts. We then have
b2
b2
′
′
cos y = a −
sin
(3.3.2)
x = a+
a
a
and the circle of radius a in the z plane is seen to have been transformed into an ellipse
in the z′ plane with semimajor axis c = a + b2 /a and semiminor axis d = a − b2 /a
(Figure 3.3.1). Upon elimination of the equation of the ellipse is found to be given by
2
2
y′
x′
+
= 1
(3.3.3)
a + b2 /a
a − b2 /a
With this knowledge in hand, the flow past an ellipse can be computed.
The direct approach would be to solve equation (3.3.1) for z, obtaining
1 ′ ′ 2
z + z − 4b2
z=
2
(3.3.4)
and then substituting this into equation (3.2.1), obtaining
√
2Ua2
z′ + z′2 − 4b2
i
1 ∗ ′ ′2
2
ln
+
w = U z + z − 4b +
2
2a
z′ + z′ 2 − 4b2 2
(3.3.5)
The velocity is then found by forming dw/dz′ from equation (3.3.5).
A slightly less direct approach—but with some advantages—would be to compute
the z coordinate corresponding to a point z′ by use of equation (3.3.4). The velocity can
then be found using this with equation (3.2.2) and the chain rule of calculus to be
1
dw
dw dw dz
=
=
dz′
dz dz′
1 − b2 /z2 dz
(3.3.6)
From equation (3.3.6) it is easy to see that the velocity is infinite at the points z = ±b.
Using the expressions for the semimajor and -minor axes, the parameter b in the
transformation is given by
(3.3.7)
b = a c − d/c + d
The points of infinite velocity in equation (3.3.6) lie inside the ellipse unless d = 0,
in which case they lie on the ellipse. For d = 0 the ellipse degenerates to a flat plate,
3.4
95
The Joukowski Airfoil
and the velocity is infinite at both endpoints of the plate, unless of course the plate is
aligned parallel to the uniform stream.
Notice that the possibility exists in the preceding analysis that, at a point where
the denominator in equation (3.3.6) becomes zero, dw/dz also is zero. We will use this
when we discuss airfoils.
3.4 The Joukowski Airfoil
The flow past an ellipse provides the clue needed to generate an airfoil shape. The
principle characteristic of an airfoil is that it has a sharp trailing edge. At that point, the
airfoil has a discontinuity in slope, and the mapping will not be conformal. To obtain
such a shape, Joukowski proposed the following sequence of transformations.
The circle in the z plane is first translated in a new coordinate system with origin
at z′c (Figure 3.4.1) as given by
z′ = z + z′c
where z′c = xc′ + iyc′ and xc′2 + yc′2 ≤ a2
(3.4.1)
Under this translation the circle z = aei in the z plane will still be a circle in the z′
plane, where it now satisfies the equation x′ − xc′ 2 + y′ − yc′ 2 = a2 and has its center
at z′c . The stagnation points originally at zDSP = aei− and zUSP = aei++ in the z
plane are now at
z′DSP = z′c + aei−
and z′USP = z′c + aei++
(3.4.2)
in the z′ plane, as shown in Figure 3.4.2.
The z′ transformation is next followed by the Joukowski transformation
z′′ = z′ + b2 /z′
(3.4.3)
dw
dw dz dz′
dw z′2
=
=
′′
′
′′
dz
dz dz dz
dz z′2 − b2
(3.4.4)
Here b is a real number. Since
it is seen that the velocity in the z′′ plane will be infinite at the points z′ = ±b unless
one of these points coincides with a stagnation point in the z plane. Therefore, to make
the transformed circle into an airfoil shape, the first stagnation point in equation (3.4.2)
(x,y )
y′
y
yc′
x
xc′
z and z ′ planes
Figure 3.4.1 Joukowsky transformation—z and z′ planes
x′
96
Irrotational Two-Dimensional Flows
y
y′
a
x
α
b
yc′
x′
U
DSP
USP
xc′
Figure 3.4.2 Joukowsky airfoil—z and z′ planes
must lie on the circle and be where z′ = +b. This will give a sharp trailing edge, and the
mapping of the cylinder to the airfoil will not be conformal at that point. This results in
yc′ = −a sin − and b = xc′ + a cos − = xc′ + a2 − yc′2
(3.4.5)
Further, require for later purposes that the point z′ = −b 0 does not lie outside
the circle, this condition being satisfied provided that
2
b + xc′ + yc′2 < a2
(3.4.6)
This means that, except inside the body, the only possible place where the velocity can
be singular is at z′ = b. Satisfaction of both parts of equation (3.4.5) means that xc′ ≤ 0.
Investigating the transformation further, note that on the circle in the z′′ plane, at
the trailing edge/stagnation point the angle the surface turns through is 2 (or zero),
whereas on the airfoil the corresponding angle is . The reason that the requirement
was made that z′ = −b must not lie outside the circle is that we wanted to ensure that
no singularities were introduced into the flow, only inside the circle.
The shape of the airfoils that this family of transformations has generated can now
be found. The following outlines the approach:
1. Specify U , the attack angle and the coordinates of the center of the translated
circle xc′ and yc′ . The latter is to be done such that xc′ < 0 and xc′2 + yc′2 ≤ a2 . (The
choice of a is in fact arbitrary, representing a scaling of the coordinate system,
and could just as well be taken as unity. The same is true for U , which scales
the velocity.) The parameter xc′ controls the airfoil thickness, while yc′ controls
its camber.
2.
Compute b , and .: From equation (3.4.5), = + sin−1 y ′c /a b = xc′ +
a2 − y ′2
c . From equation (3.2.7)
= 4 a U sin
3.4
97
The Joukowski Airfoil
y″
x″
Figure 3.4.3 Joukowsky airfoil shape
3. Choose an arbitrary point z = aei on the circle by selecting a value for .
Compute the corresponding transformed position of this point on the airfoil using
equations (3.4.1) and (3.4.3).
4. To find the velocity and pressure on the airfoil, use equation (3.4.4) for the veloc
ity and then compute the pressure coefficient according to Cp = 21 1 − U/U 2 .
5. Repeat steps 3 and 4 for a number of values of to generate the airfoil shape
and the pressure distribution.
6. Repeat steps 1 through 5 for various values of x ′c and y ′c to see the effect of
these parameters on airfoil shape.
Figure 3.4.3 shows a typical airfoil generated by this procedure. Here
U = 1
= 0
A = 1
xc′ = −01
yc′ = 01
In the beginning of the analysis, the vortex was somewhat arbitrarily added to the
flow. Its presence is an absolute necessity if a lift force is to be present. How, then,
does the flow introduce circulation?
In the case of the circular cylinder, circulation would have to be added artificially,
either by rotating the cylinder (the Magnus effect) and letting viscosity and the no-slip
condition introduce the swirl or by forcing the position of the stagnation points either
by the addition of a small flap where the trailing edge would be (Thwaite’s flap) or by
some other means to control the pressure on the boundary. In the case of the airfoil,
note that our theory predicts an infinite velocity at the trailing edge unless dw/dz = 0
at that point. In fact, unless we impose the condition dw/dz = 0 at the trailing edge,
the velocity there is not only infinite, but it turns through 180 degrees in zero distance.
A real fluid cannot do this because it implies infinite acceleration requiring infinite force.
Kutta proposed that the circulation will adjust itself so that both the infinite velocity
and the sharp turn in the flow at the trailing edge are avoided. At this stage both
numerator and denominator in equation (3.4.4) are zero, meaning that dw/dz′′ is of the
form zero over zero. To resolve the value of the true speed at the trailing edge, use
L’Hospital’s rule, in the form
⎡
⎤
d2 w
⎢ ′2
⎥ b2 2a2 U
dw
i
dz2
⎢z
⎥=
= lim
−
′
⎣
⎦ 2b
′′
3
dz d ′2
z′ →b
dz
z
2
z2 z=ae−i+
2
(3.4.7)
z
−
b
dz dz′
b
= U e2i+ cos
a
This tells us that the body slope at the trailing edge is turned an angle of 2 from the
attack angle .
98
Irrotational Two-Dimensional Flows
Kutta’s hypothesis (usually referred to as a Kutta condition) gives a very good
value for the circulation on a Joukowski airfoil, as has been verified in a series of
tests (National Advisory Committee for Aeronautics Technical Report number 391 and
Technical Memorandums 422 and 768). Joukowski airfoils do have very good stalling
behavior and can achieve angles of attack as high as 30 degrees with only a slight
drop in lift. They have the disadvantage of having a rather high drag coefficient, and
the very thin, sharp trailing edge makes their construction very difficult.
3.5 Kármán-Trefftz and Jones-McWilliams Airfoils
There have been several variations of the Joukowski airfoil that add several helpful
features. One, proposed by von Kármán and Trefftz, eliminates the disadvantage of the
thin trailing edge.
Transformation equation (3.4.1) can be written in the equivalent forms
z′′ + 2b = z′ + b2 /z′
z′′ − 2b = z′ − b2 /z′
Taking the ratio of these, the Joukowski transformation can therefore be written in
the form
′
z′′ + 2b
z +b 2
=
(3.5.1)
z′′ − 2b
z′ − b
Von Kármán and Trefftz suggested replacing the Joukowski transformation with the
alternate transformation
′
z +b n
z′′ + 2b
=
(3.5.2)
z′′ − 2b
z′ − b
The trailing edge then has an inside angle of 2 − n , rather than zero. The details of
the shape can be carried out in a manner similar to the Joukowski airfoil, with one more
variable n available to the designer.
A second variation of the Joukowski airfoil was proposed by R. T. Jones and
R. McWilliams in a pamphlet distributed at an Oshkosh Air Show. They also start with
equation (3.4.1) but then follow it with the two transformations
z′′ = z′ − ′
(3.5.3)
z −
and
b2
(3.5.4)
z′′
where is a complex number and is real. Carrying through an analysis similar to
what we did for the Joukowski profile, it can be shown using the same general analysis
as for the Joukowski transformation that
z′′′ = z′′ +
= xT′ − bxT′ − − yT′ 2 + iyT′ 2xT′ − − b
(3.5.5)
where xT′ yT′ is the location of the trailing edge in the z′ plane.
It is convenient to set b, a scale factor, arbitrarily to 1. The parameters
xc′ yc′ xT′ yT′ are then set by the designer. The parameter is determined by equation (3.5.5) subject to the inequality
− b ± + b2 + 4 − 2xc′ + iyc′ < 2a
(3.5.6)
3.6
99
NACA Airfoils
and the parameter a is determined by
a = xT′ − xc′ 2 + yT′ − yc′ 2
(3.5.7)
The circulation is then found from
= 4 aU B
(3.5.8a)
with B given by
B=
xT′ − xc′ sin − yT′ − y′ cos
.
a
(3.5.8b)
With a careful selection of the parameters, Jones and McWilliams have generated airfoils
with the properties of the NACA 6 series, the 747 series, the Clark Y, and the G-387.
3.6 NACA Airfoils
Early airfoil designs by Eiffel, the Wright brothers, and many others were largely
the result of intuition and trial and error, although the Wright brothers did utilize a
crude wind tunnel and were well read as to the aerodynamic theory of their day.
During World War I, Prandtl at Gottingen, Germany, started to put these designs
on a firmer basis, and efficient airfoils such as the Gottingen 398 and the Clark Y
were developed. These airfoil shapes, or extensions of them, were utilized almost
exclusively during the 1920s and 1930s. In the 1930s, the NACA (National Advisory
Committee for Aeronautics, which later became NASA, the National Aeronautics and
Space Administration) started a comprehensive and systematic development program
that used both theory and experiment. They were able to separate out the effects of
camber and thickness distribution and also carried out experiments at higher Reynolds
numbers than had previously been possible.
Airfoil shapes are generally given by specifying many coordinate points on the
surface of the foil. The effects of camber and thickness can be separated by writing
xU = xc − yt sin
yU = yc + yt cos
(3.6.1a)
(3.6.1b)
on the upper surface of the foil, and
xL = xc + yt sin
yL = yc − yt cos
(3.6.1c)
(3.6.1d)
on the lower surface. Here xc yc are the coordinates of a point on the mean line, tan
is the slope of the mean line, and yt is the half-thickness. Mean lines are frequently
designed to have specific load distributions. For example, the following mean line
equation has constant loading over the chord (Abbott and von Doenhoff, pages 74 and
75, 1959):
c
x x x
x
y = − Li 1 −
ln 1 −
+ ln
(3.6.2)
c
c
c c
4
where cLi = 2/U is the circulation per unit chord length, and U is the free stream
velocity.
100
Irrotational Two-Dimensional Flows
For two famous series of airfoils, the NACA four-digit and five-digit series, the
shapes are defined completely by formulae. In the four-digit series, the thickness is
given by
x
x 3
x 4
x
x 2
yt = 5ymax 02969
+ 028431
− 01015
− 0126 + 03537
c
c
c
c
c
(3.6.3)
where ymax is the maximum value of the half-thickness, and c is the chord length. Near
the leading edge, neglecting all terms on the right-hand side of equation (3.6.3) beyond
the first, the foil is approximately given by the parabola yt /ymax 2 ≈ 220374x/c. The
local radius of curvature at the leading edge is thus
2
RLE ≈ 110187 ymax
/c
(3.6.4)
The mean lines for the airfoils of this series are given by
yc =
=
m2px − x2
p2
for 0 ≤ x ≤ p
mc − xx − 2p + c
c − p2
(3.6.5)
for p ≤ x ≤ c
where m is the maximum camber (distance of the mean line from the chord line), and
p is the chordwise location of the maximum camber point.
The numbering system for this series is based on the foil shape. The first integer
is 100 m/c, the second integer is 10 p/c, and the third and fourth are 100 ymax /c. Thus,
for an NACA 2415 foil, we have
m/c = 002
p/c = 04
ymax /c = 015
(3.6.6)
Foils with numbers NACA 00xx are symmetrical—that is, their mean and chord lines
coincide.
For the four-digit series, tests show that the maximum lift coefficient increases as
the position of maximum camber is shifted forward from midchord. The mean lines
for the four-digit series were not suitable for forward positions of maximum camber,
so the five-digit series was introduced. For the five-digit series, the thickness is given
by equation (3.6.3) and the mean lines by
yc = k x3 − 3mx2 + m2 x3 − m/c
=
km3 c − x
c
for m ≤ x ≤ c.
for 0 ≤ x ≤ m
(3.6.7)
where k originally was calculated to give a design lift coefficient of 0.3, and m determines
where the maximum camber points occur. Differentiating equation (3.6.7), the maximum
camber points are found to occur at
xMC = m1 − m/3c
(3.6.8)
The numbering system for the five-digit series is based on a combination of foil shape
and aerodynamic characteristics. The first integer is 666667 · CL , where CL is the design
lift coefficient. The second and third integers are 200 · xMC /c, and the fourth and fifth
are 100 · ymax /c, where ymax is the maximum thickness. Thus, an NACA 23012 wing
3.7
101
Lifting Line Theory
section has a design lift coefficient of 0.3, a maximum camber at 0.15 of the chord
length, and ymax /c = 012.
Details of these series and others are given in Abbott and van Doenhoff (1959).
This book includes a good discussion of airfoil theory along with data on lift, drag,
and moment coefficients. Modern airfoils generally have more complicated shapes than
given by the NACA four- and five-digit series. Their shapes are usually given in tabular
form rather than by formulae.
3.7 Lifting Line Theory
At this point in the discussion there is a good bit of theory on foils of infinite span and
also information on test data on a large number of foil shapes. All good things come
to an end, however, including airplane wings, turbine blades, and vorticity lines (recall
that the last, however, do not end in the interior of the fluid). How, then, does the body
of two-dimensional knowledge fit into a three-dimensional world?
Suppose for example that a wing is to be designed and that the lift coefficient
distribution on the wing is given. This is a good starting point in the design process, as
the local value of the lift coefficient gives the load on the wing at that point, and this
load is intimately connected to the structural requirements of the wing. From the lift
coefficient, previous results for the Joukowski airfoil give the circulation distribution.
If a very simple wing design is used where the circulation is constant along the
span, in the interior of the wing, the portion of the vortex line along the span would
simply be a straight line. This portion of the vortex is called the bound vortex, since
it is bound to the wing. When this vortex line reaches the wing tips, it leaves the
wing and is convected rearward with the flow. Considering the situation where the
flow starts from rest, the vortex line for this case will be (approximately) a flat, closed
rectangle, consisting of the bound vortex, the tip vortices, and the starting vortices, as
shown in Figure 3.7.1. The length of this rectangle increases with time, as the starting
vortices move away from the wing. (The tip vortices can frequently be seen from an
airplane window. Because of the high rotational speeds at their core, the pressure there
is lowered. Water vapor can then condense, and flow visualization occurs.) The vortex
line is thus closed, so the requirement that a vortex line can neither originate nor end in
the fluid is met. When the starting vortex is very far away (at infinity), the pattern we
have described is called a horseshoe vortex.
Tip vorte
St
a
rti
ng
vo
rte
x
Bo
un
d
vo
rte
x
x
Tip v
ortex
Figure 3.7.1 Wingtip, boundary, and starting vortices
102
Irrotational Two-Dimensional Flows
Recalling that a doublet distribution is equivalent to a vortex sheet, with the local
vorticity being proportional to the change of the doublet strength, the velocity potential
for the vorticity associated with a finite wing can be found readily. Since we want
vorticity only on the boundaries of our rectangle, a constant-strength doublet facing
normal to the x-y plane is sufficient. This results in
=
4
=
4
=
4
L
s
d
−s
s
−s
xa
tan
zd
x − 2 + y − 2 + z2 3/2
z − x
2
2
− y + z
− x2 + y − 2 + z2
−1
− yx − xa
− tan
z x − xa 2 + − y2 + z2
−1
L
d
xa
− yx − L
z x − L2 + − y2 + z2
s
−s
(3.7.1)
From Figure 3.7.1, and with the help of the right-hand rule of vectors, it is seen that the
effect of this vortex rectangle is to induce a downward velocity in the interior, called the
downwash. The downwash in turn contributes a drag to the wing, called the induced drag.
If the circulation distribution is not constant, such as for tapered wings, this pattern
is repeated, with each point in change in the circulation acting (weakly) as a wing tip.
Thus, the vorticity pattern might be as in Figure 3.7.2. The velocity potential for such
a pattern could be generated by including the effects of variable inside the original
integral in equation (3.7.1). The integration involved in such a procedure would very
quickly overtax our calculus skills!
The preceding procedure reduces the wing to a line. It was originated by Prandtl
(1921) and is called lifting-line theory. The vortex-lattice method is a more detailed
procedure than a lifting line theory. It gives some geometry to the wing and is similar
to the panel method introduced earlier. The constant-strength doublet panels with which
we might expect to cover our wing (with some mathematical difficulties) are instead
replaced by a series of horseshoe vortices as we just saw. Experience suggests that
the best results are found by placing the bound vortex portion of the horseshoe at the
quarter-chord point. Computer programs to calculate the resulting velocities are closely
related to our panel programs. (See, for instance, Moran (1984) for more details.)
Figure 3.7.2 Vortex pattern for a tapered wing
3.8
Kármán Vortex Street
It was recognized as early as the 1890s that to increase wing lift and decrease
drag, it was important to keep the flow over the wings as two-dimensional as possible.
Frederick W. Lanchester1 in England patented the first concept of winglets, vertical
surfaces at the tips of a wing, in 1897. His version was essentially an endplate that, by
making the flow more two-dimensional, did reduce induced drag but had the unfortunate
consequence of increasing viscous drag, negating the benefits. In the 1970s, Richard T.
Whitcomb at NASA Langley Research Laboratory reexamined the concept and found
that the airflow above the wing tip of a typical airfoil is directed inward, while below
the wing tip it is directed outward. By careful design of the cant (upward angle) and
toe (inward angle) of the winglet the small lift force developed by the winglet could
be directed forward so that the trailing vortex contributes to thrust rather than drag.
This has strongly influenced design procedures used by general aviation and business
jet manufacturers and led to the introduction of winglets on many new planes and to
the retrofit of many older ones.
3.8 Kármán Vortex Street
For flows past two-dimensional bodies, wake vortices are found to alternately shed from
the two sides of the body. Their pattern tends to be initially symmetrical (providing the
body is symmetrical) but soon changes to an alternating pattern, as seen in Figure 3.8.1.
Von Kármán questioned whether this alternating pattern (called a vortex street) was
necessarily the only pattern that could exist; in other words, was the pattern stable?
To idealize the problem, he omitted the body and considered two rows of vortices,
each consisting of vortices extending to infinity for both positive and negative x. The
upper row consists of counterclockwise vortices at x = na n = 1 2 y = b/2,
with circulation , and the lower row of clockwise vortices at x = n + ca n =
1 2 −1 < c < 1 y = −b/2 with circulation −. The complex velocity potential
for this flow (the calculations become unmanageable if we do not use complex variables
here) is then
ib
ib
−i
ln z − na −
− ln z − na − ca +
w=
2 n=−
2
2
⎤
⎤⎫
⎧ ⎡
⎡
ib
ib
⎪
⎪
(3.8.1)
⎪
⎪
z−
z − ca +
⎢
⎥
⎥⎬
−i ⎨ ⎢
2
2
⎥ − ln ⎢sin
⎥
ln ⎢
=
⎣sin
⎣
⎦
⎦⎪
2 ⎪
a
a
⎪
⎪
⎭
⎩
Figure 3.8.1 Kármán vortex street
1
Frederick William Lanchester (1868–1946) made many contributions to aerodynamics and automotive
engineering. His work was influential in the study of aircraft stability, and his 1919 book Aviation in Warfare:
The Dawn of the Fourth Arm eventually led to the development of the study of logistics and the field of
operations research.
103
104
Irrotational Two-Dimensional Flows
The second form follows from
n=−
lnp − na = ln p +
n=1
lnp2 − n2
p2
p2
= ln p1 − p2 1 − 2 · · · 1 − 2 · · · + constant
2
n
=
sin p
where the constant is unimportant for our purposes.
In the analysis that follows, we will be using several formulas that come from
Fourier series. Three particularly helpful expansions are the following:
# cos m
k −
= 1 + 2k
Expansion #1. coshsinh
n=1 n2 +k2 . Evaluating this at = 0 gives
k
coth k =
Expansion #2.
sinh k −
cosh k
=
2
1
k cos n
2
+
2
2
k
n=1 n + k
k cosn+1/2
n=0 n+1/22 +k2 .
#
tanh k =
Evaluating this at = 0 gives
2
k
2
2
n=1 n + 1/2 + k
Differentiating this with respect to k gives
2
n + 1/22 − k2
dtanh k
=
=
2
2 2
dk
cosh2 k
n=0 n + 1/2 + k
# 1−cos n
.
Expansion #3. 4 2 − =
n=1
n2
The complex conjugate velocity corresponding to the complex potential (equation (3.8.1)) is
⎤
⎡
ib
ib
z−
z − ca +
dw −i ⎢
2
2 ⎥
⎢cot
⎥
u − iv =
=
−
cot
(3.8.2)
⎣
⎦
dz
2a
a
a
The first cotangent represents the velocity due to vortices in the upper row, and the
second the velocities due to the vortices in the lower row.
Considering the velocity induced at an arbitrary vortex in the upper row, notice that
the other vortices in the same row induce no net velocities, as the induced velocities
due to vortices at equal distances right and left from our chosen vortex will cancel
out each another. Thus, the net induced velocity at any vortex in the upper row (for
computational convenience consider the vortex to be at z = ib/2) is found from the
vortices in the lower row to be
ib
i
cot
−c
(3.8.3)
Vu∗ =
2a
a
the asterisk denoting the complex conjugate induced velocity. A similar computation
for the lower vortices gives Vl∗ = Vu∗ , as should be expected.
To a good approximation, the vortices in a vortex street behind a bluff body do not
drift either up or down. Equation (3.8.3), on the other hand, states that there will be
3.8
105
Kármán Vortex Street
a vertical component to the induced velocity unless either c = 0 (when the two vortex
rows are aligned), for which
V∗ =
b
coth
2a
a
(3.8.4)
or when c = 1/2 (the vortices in the lower row are placed below the midpoints of the
vortices in the upper row), when
V∗ =
b
tanh
2a
a
(3.8.5)
Thus, the two cases we will consider are for c either 0 or 1/2, and the vortices will
move with velocity V ∗ given by either equation (3.8.4) or (3.8.5).
To examine the stability of this arrangement, notice that at time t the vortices in
the upper row will be at zn0 = na + V ∗ t + ib/2 − < n < , and the vortices in the
lower row will be at z′n0 = n + ca + V ∗ t − ib/2 − < n < .
For the flow disturbance, displace the vortex originally at zn0 in the upper row to the
location zn0 + zn and displace the vortex in the lower row originally at z′n0 to the location
z′n0 + z′n . Further assume that our disturbance displacement is of the form zn = cos n
and z′n = ′ cosn + c, where and ′ are small time-dependent complex numbers,
so small that equations can be linearized. This corresponds to a wavy displacement of
each vortex row. Of course, this is not the most general displacement that is possible.
If, however, we use this displacement and find that and ′ grow with time, then,
because the flow is unstable to at least one disturbance, we can conclude that the flow
is unstable. If, however, we find that for this displacement and ′ do not grow with
time, since our choice of disturbance was not a general one, we cannot draw any definite
conclusions other than to say that it is a possible configuration.
The new velocity potential is now
w=
−i
lnz − zn0 − zn − lnz − z′n0 − z′n
2 n=−
(3.8.6)
with a complex velocity
dw −i
1
1
=
−
dz
2 n=− z − zn0 − zn z − z′n0 − z′n
(3.8.7)
If we consider a vortex in the upper row—say, the one corresponding to n = 0—
then it moves with a velocity V ∗ + d ∗ /dt and is at z0 + Vt + ib/2. Setting this equal
to equation (3.8.7) and linearizing the right-hand side, we have with the help of the
binomial theorem
⎞
⎛
∗
1
−i ⎝
1
d
⎠
=
−
V∗ +
′
′
dz
2
z
−
z
−
z
z
−
z
n=− 0
n0
n
n0 − zn
n=− 0
n=0
=
−i
1
1
1
i
+
+
2 n=1 z0 − zn0 − zn z0 − z−n0 − z−n
2 n=− z0 − z′n0 − z′n
=
1
1
−i
+
2 n=1 −na + − cos n na + − cos n
106
Irrotational Two-Dimensional Flows
+
i
1
2 n=− + ib − n + ca − ′ cosn + c
−i
−1
− cos n
1
− cos n
≃
1+
+
1−
2 n=1 na
na
na
na
+
di
2 n=0
1−
− ′ cosn + c
− ′ cos−n + c
1
−
i
ib − n + ca
ib − −n + ca
+
ib − n + ca
2 n=1
ib − −n + ca
1 − cos n i
1
− ′ cosn + c
i
+
1−
=
n2 a2
2 n=0 ib − n + ca
ib − n + ca
n=1
− ′ cosn + c
1
1−
ib + n + ca
ib + n + ca
1 − cos n
2ib − ca
2
i
i
+
+
=
2
2
2
2
ib − ca n=1 ib − ca − n a
n 2 a2
n=1
+
i
− ′ cosn + c − ′ cosn + c
−
+
2 n=0
ib − n + ca 2
ib + n + ca 2
(3.8.8)
Canceling V ∗ from both sides of the equation, we are left with
1 − cos n i
− ′ cosn + c − ′ cosn + c
i
d ∗
=
−
+
dt
n2 a2
2 n=0
ib − n + ca 2
ib + n + ca 2
n=1
=
i
A + C ′
a2
(3.8.9)
where
A=
=
1 − cos n
1
1
1
−
+
2 n=0 n + c − ib/a2 n + c + ib/a2
1 − cos n
−
n=1
n=1
n2
n2
n=0
n + c2 − b/a2
n + c2 + b/a2 + 2
and
C=
n + c2 − b/a2
cosn + c
2
2 2
n=0 n + c + b/a
are both real.
A similar calculation for the lower row gives
−i
d ′∗
=
A ′ + C
dt
a2
(3.8.10)
3.8
107
Kármán Vortex Street
To solve the set of equations for and ′ , differentiate the complex conjugate of
equation (3.8.9) with respect to time and use equation (3.8.10) to eliminate d ′ /dt. The
result is
2 2
d2
+
C − A2 = 0
(3.8.11)
dt2
a2
Consider first the case where c = 0—that is, the vortices are not staggered. Then
2
1 − cos n
n2 − b/a2
A=
−
=
−
−
2
2 2
b
n2
2
2
n=1
n=0 n + b/a
2 sinh2
a
n2 − b/a2
C=
cos n = −
2
2 2
n=0 n + b/a
Note that for =
2
A=−
4
2
2
−
2 sinh2
b
a
=−
b
b
sinh −
a −
a
b
2 b
sinh
sinh
a
a
2
cosh
b
cosh2
+1
a
b
4 sinh2
a
2
and
C=−
cosh
sinh2
b
a
b
a
so that
4
C 2 − A2 = −
4 b
2 b
cosh
+ cosh
+1
a
a
< 0
4 b
16 sinh
a
Therefore, there is at least one value of for which this arrangement is unstable.
For the staggered configuration c = 1/2,
2
n + 1/22 − b/a2
1 − cos n
−
=
−
+
A=
2
2
2
2
n
2
2 b
n=0 n + 1/2 + b/a
n=1
cosh
a
C=
n + 1/22 − b/a2
cosn + 1/2
2
2 2
n=0 n + 1/2 + b/a
When = C = 0 and A = 2 1/2 − 1/ cosh2 b /a. Thus, a necessary condition for
stability is that A = 0, giving cosh2 b /a = 2, or b/a = 0281. For that value of
V∗ =
03536
tanh 2 = √ =
2a
a
2a 2
(3.8.12)
In practice, the c = 1/2 case is usually found to be a good model several vortices
downstream from the body. The spacing parameter b is determined by the body shape
and size, as is the circulation.
Vortex shedding can be frequently seen in everyday life, from the waving of tree
branches in the wind to the torsional oscillations of stop signs on windy corners. Ancient
Greeks called the sounds made by wind passing tree branches Aeolian tones. Open
automobile windows can cause low-frequency oscillations in the automobile interior,
108
Irrotational Two-Dimensional Flows
with subsequent discomfort to passengers. In open areas with frequent crosswinds,
power and telephone lines can be seen to “gallop” on windy days. (Simple weights hung
at the quarter-wave points can alter the natural frequency of the line and thus prevent
catastrophic motion.) Perhaps the most famous case of motion due to vortex shedding
is the Tacoma Narrows Bridge, dubbed “galloping Gertie,” that self-destructed months
after it was built. (Von Kármán was on the committee that was formed to investigate the
collapse.) The final word on this bridge has not yet been written, as scientific articles
are still appearing with additions and further explanations for the collapse. Pictures can
be found by searching the Internet.
Our model for a vortex street here was the most simple one, where the vortices
are concentrated. In three dimensions no such simple models are found, and the analysis becomes more complicated. More details are given in the chapter on numerical
calculations and in the book by Saffman (1992).
3.9 Conformal Mapping and the Schwarz-Christoffel
Transformation
In performing the transformation of the circle to the ellipse by means of the Joukowski
transform, we say that we have mapped the circle in the z plane into an ellipse and
that the mapping is conformal, or angle-preserving. This means that if two lines cross
at some point in the z plane with an angle between them, at the corresponding point
in the z′ plane the angle between the mapped lines is still . The relation between w
and z can be thought of as a mapping of one plane into another.
There are a few mappings that have been found to be generally useful, particularly
for dealing with free streamline flows.2 The simplest one is the logarithm function
used in mapping the hodograph plane dw/dz. Letting dw/dz = u − iv = qe−i , we see
that ln dw/dz = ln q − i. Thus, regions of constant speed are vertical line segments in
the hodograph plane, whereas regions of constant direction of velocity are vertical line
segments.
The Schwarz-Christoffel transformation is used to map the interior of a closed
polygon into either the upper or lower half plane. The definition of polygon in this case
includes cases where the length of one or more sides can be infinite.
To illustrate the Schwarz-Christoffel transformation, consider for the sake of argument a five-sided polygon in the z plane with corners A, B, C, D, and E. We wish to
transform this into a half-plane in the plane. The angles one turns through in passing a
corner of the polygon are defined in Figure 3.9.1. By virtue of the polygon being closed
the angles must satisfy
++ ++ = 2
(3.9.1)
This mapping can be carried out by the transformation
dz
= R − a−/ − b−/ − c−/ − d−/ − e−/
d
(3.9.2)
where a, b, c, d and e are real numbers. Usually three of these numbers may be chosen
arbitrarily. Notice that it is necessary to traverse the polygon in a counterclockwise
2
A free streamline or free surface is a line or surface of constant pressure.
3.9
109
Conformal Mapping and the Schwarz-Christoffel Transformation
D
δ
γ
C
E
ε
β
C′
D′
B
A
α
E′
t plane
A′
B′
z plane
Figure 3.9.1 Schwarz-Christoffel transformation
direction to map to the upper-half t plane. If the polygon has n corners rather than five,
all of the preceding statements hold with the obvious addition (or subtraction if n < 5)
of terms to equations (3.9.1) and (3.9.2).
A variation of the Schwarz-Christoffel transformation allows the polygon to be made
up of circular arcs instead of straight lines. To simplify matters, it often is convenient
to map one point into the point at infinity, making one of the constants a b, and so on
be infinite. To accomplish this, simply drop the corresponding term in equations (3.9.1)
and (3.9.2).
As an example, consider a rectangle in the z plane of semi-infinite length with
thepoints c and d at infinity in the t plane. Since = = /2, we have dz/d =
R/ − a − b. If we exercise our freedom to choose, let a = 1 and b = −1. Then
z=R
dt
= R cosh−1 + S
√
2
−1
Exercising our freedom of choice once more, let S = 0. Then
z
= cosh
R
(3.9.3)
Then z = 0 maps to = +1, and z = i R maps to = −1. In other words, our semiinfinite rectangle is bounded by the y-axis and horizontal lines starting at z = 0 and
z = i R. See Figure 3.9.2.
πR
A
D
B
C
D
–1
1
A
B
t plane
z plane
Figure 3.9.2 Schwarz-Christoffel transformation for a semi-infinite rectangle
C
110
Irrotational Two-Dimensional Flows
3.10 Cavity Flows
To illustrate further the application of the Schwarz-Christoffel transformation, we consider the two-dimensional case of a uniform stream impinging on a flat plate. We could
use the result for an ellipse and let one of the axes be of zero length, collapsing the
ellipse into a plate perpendicular to the uniform stream. We would find that at the corners of the plate the speed would be infinite due to the flow turning through 180 degrees
at those points. This is not a physically realizable situation.
Instead, allow the fluid to leave the plate tangentially and have a constant pressure
region, or cavity, develop behind the plate. Neglecting gravity, the condition that the
pressure in the cavity be constant means that the Bernoulli equation says the fluid speed
must be constant on the free surface.
Notice that for steady flows this theory requires that the cavity never close. For it
to do so, one of two things would have to happen: Either the closure point where the
top and bottom flows meet would have be a stagnation point (ruled out by the constant
speed condition), or the two streams would have to form a cusp at closure (not easily
realizable).
The following presentation is given in Lamb (1932, pp. 99–102). The original
solution is credited to Kirchhoff. Since for symmetric flow the flows above and below
the streamline going to the stagnation point are the same, we need to consider only
the upper flow. For convenience let the just described streamline have the value zero,
and let the velocity potential be zero at the stagnation point. We will use the following
planes in our discussion. They are shown in Figure 3.10.1.
• z plane: This is the physical plane. The plate lies on the line segment DAB, A
being the stagnation point, D and B the edges of the plate. BC and DC are the free
streamline parts of the streamline = 0 going from the plate edges to infinity,
while AC is the portion of that streamline going from the stagnation point to
C
D
A
B
A
B(1, 0)
D(–1, 0)
A
C
Cavity
ζ plane
z plane
C
C
A
B
C
D
A
(–1, 0) (0, 0) (1, 0)
C
D
t plane
A
B
(0, –π/2)
(0, –π)
A
D
A
B
A
Ln ζ plane
w ′ plane
Figure 3.10.1 Cavity flow transformations
ψ
C
φ
3.10
111
Cavity Flows
far upstream at infinity. Generally, in mapping theory infinity is regarded as a
single point.
• t = dw/dz−1 plane: This is the inverse of the hodograph plane.3 The stagnation
point is set at infinity. Since the free streamlines require pressure to be constant,
in the absence of gravity this means that the speed is constant. Thus, the free
streamlines make up a semicircle BCD, while the flow along the plate are horizontal lines going from B and D to A. The flow lies in the region below the curve
ADCBA. If we let dw/dz = qe−i where q is the speed (real), then t = q1 ei .
• ln t plane: This transformation maps the flow region into a rectangle with corners
at A, B, and D. Notice that ln t = − ln q + i.
• plane: The Schwarz-Christoffel transformation is used to transfer the rectangle
in the ln t plane into the upper-half plane. The boundary is ABCDA.
• w plane: This is the complex potential plane used to relate the velocity potential
and stream function to the above planes. It consists of a slit ABCDA in the
entire plane.
• w′ plane: Here w′ = 1/w. This inverse w plane looks much like the w plane but
with some rearranging. The slit is now at CDABC.
The analysis proceeds much as that used in transforming the rectangle in the
previous section. Using the Schwarz-Christoffel transformation between the and ln t
planes, we have
1
1
+
(3.10.1)
ln t = cosh−1 − i or t = coshln + i = −
2
The latter form of the expression follows from the definition of the hyperbolic function.
Using the Schwarz-Christoffel transformation between the and w′ planes, we have
dw′ /d = −R, so w′ = −1/2R2 + S. At C we choose to have = 0 w−1 = 0, which
makes S = 0, leaving
1
w′ = − R2
2
(3.10.2)
To determine R, notice that so far the length L of the plate has not been used.
Along the plate t is real, so
1
1
q+
=−
2
q
√
giving q = − − 2 − 1. Here, the sign of the radical was chosen so as to make q
approach zero as approaches minus infinity.
Along the plate /x = q so that dx/d = 1/q. Also, w = 1/w′ = −2/R2 =
+ i , so that d/d = 4/R3 .
Then
1
−1 1 1
−1
−1 dx d
2 −1
L=2
−
+
d = 4R
d
=
2R
dv (3.10.3)
3
3
− q
−
− d d
3
In fluid mechanics the hodograph plane usually refers to the complex velocity plane dw/dz.
112
Irrotational Two-Dimensional Flows
Letting ′ = −1/, this becomes
8 1
2
L=
1 + 1 − ′2 d′ = 4 +
R 0
R
giving R =
2 + 4
L
(3.10.4)
Along the free boundary BC, ln t = i. Then = − cos and = 2R sec2 , giving
the intrinsic equation of the curve in the form s = L+4 sec2 with varying between
zero and − 2 . In terms of x and y, the coordinates of the free surface are then
2L
L
x=
sec +
y=
+
sec tan − ln
(3.10.5)
+4
4
+4
4 2
The origin is taken at the center of the plate.
To find the force on the plate
notice that the difference in pressure between front
and back is given by 21 1 − q 2 , so the net force is given
−1
−1
−1
1
d
d 1
dx
−q
= −2R
R
2 − 1 3 =
1 − q 2 d = −R
F =
3
d
q
2
−
−
−
Using equation (3.10.4), the net force is found to be
F=
+4
U 2 L = 0440U 2 L
(3.10.6)
3.11 Added Mass and Forces and Moments
for Two-Dimensional Bodies
For two-dimensional flows, the 21 independent added masses for three-dimensional
flows discussed in Chapter 2 reduce to six: A11 A12 A22 A61 A62 A66 . The remaining
A12 A16 A26 follow from these by symmetry. Of these six, a simplifying formula can
be found for all but A66 . While it is possible to derive the two-dimensional results from
the three-dimensional, it is easier to start afresh.
The added masses are going to(be generated by the p = −
term. Thus, the force
t
on the body is given by F = dtd nds, where the integration is around the body.
#
Letting = =126 V as before, then
A1 + iA2 = n1 ds + i n2 ds = i dz
= i w dz+
(3.11.1)
dz
(
since n1 ds + in2 ds = dy − idx = −idz. Also, s = n , and so
(
(
− z s ds = − zn ds. Let
B1 + iB2 = x1 n ds + i x2 n ds
dz
(
= − z z dz =
(3.11.2)
so equation (3.11.1) becomes
A1 + B1 + i A2 + B2 =
w dz
(3.11.3)
Note from the definitions in equation (3.11.2), B11 = B22 = B B12 = B21 = 0 B61 =
−Bȳ B62 = Bx̄, where B is the mass of the fluid displaced by the body and x̄ and ȳ
are the coordinates of the mass center of the displaced fluid. For a body made up of
3.11
113
Added Mass and Forces and Moments for Two-Dimensional Bodies
distributed and isolated sources and sinks as well as doublets, application of Cauchy’s
theorem gives
A1 + B1 + i A2 + B2 = 2
(3.11.4)
zdA + m zs +
where is the distributed source/sink strength, m the isolated source/sink strength
and zs its location, and is the doublet strength. For a circular cylinder, the doublet
strength is Ub2 , so A11 + B11 = A11 + b2 = 2 b2 . Thus the added mass of a cylinder
is A11 = madded = b2 .
For computation of other forces and moments on the body, again recourse could be
made to the general three-dimensional Lagally theorem. However, an approach introduced
by Blasius (1910) preceded the Lagally work and is in fact easier to use and understand.
The force on an infinitesimal piece of the body surface is
dF = dX + idY = ipei ds = ipdz
where is the inclination of the body surface. The complex conjugate of dF is then
dF ∗ = dX − idY = −ipe−i ds = −ipdz∗ = −ipe−2i dz
(3.11.5)
Similarly, the moment of dF about the origin is
dM = −ydX + xdY = pds x sin + y cos
(3.11.6)
the right-hand side being the real part of izdF ∗ . Thus, add the imaginary part of izdF ∗
to equation (3.11.6), obtaining
dM + idN = izdF ∗ = pzdze−i2
(3.11.7)
where idN is introduced strictly for mathematical convenience and has no particular
physical meaning.
2 2i
e , then integraSince for steady irrotational flows p = p0 − 21 q 2 = p0 − 21 dw
dz
tion of equations (3.11.5) and (3.11.7) gives us Blasius’s results—namely,
i dw 2
X − iY =
dz
(3.11.8)
2
dz
and
2
dw
1
dz
M + iN = − z
2
dz
(3.11.9)
Recalling the Cauchy integral theorems from earlier in this chapter, particularly
equations (3.1.10) and (3.1.11), the following infinite series representation of analytic
functions can be made.
Taylor series: If fz is analytic on and inside a simple closed contour C, and if z0
is a point inside C, then
fz = fz0 + z − z0 f ′ z0 + · · · +
=
z − z0 n
n=0
n!
f n z0
is convergent everywhere inside C.
z − z0 n n
f z0 + · · ·
n!
(3.11.10)
114
Irrotational Two-Dimensional Flows
Laurent series: If fz is analytic on and between two concentric circles C and C′
with center at z0 is a point inside C, then fz can be expanded in positive and negative
powers of z − z0 . It is convergent everywhere inside the ring-shaped region between C
and C′ .
an z − z0 n
(3.11.11)
fz =
n=−N
Here, an = 21 i
If N = , the function f is said to have an essential singularity at z0 . If N is a
positive nonzero number, f is said to have a singularity called a pole of order N at z0 .
An example of an essential singularity is e1/z . Examples of poles of order one and two
are 1/z and 1/z2 , respectively. Essential singularities are seldom encountered in flow
problems, while poles of orders one to four are frequently encountered in computing
forces and moments due to sources, sinks, vortices, and doublets.
You have no doubt already encountered the Taylor/Laurent series for real functions
in calculus classes. You may have wondered at that time why the expansion of 1/1 − x
about the origin failed at both x = +1 and x = −1. The circle domains required by these
theorems affords an explanation of that question.
(
fzdz
.
z−z0 n+1
Example 3.11.1 Find the force exerted on a circular cylinder, center at the origin, and
with a radius a, in a uniform stream with circulation
Solution. From our earlier work, the complex potential is
i
a2
ln z
+
w = U z+
z
2
Thus,
dw
dz
2
i
a2
= U 1− 2 +
z
2 z
2
a2 a4
iU
a2
2
= U2 1−2 2 + 4 +
1− 2 −
z
z
z
z
2 z
When this is inserted into equations (3.11.8) and (3.11.9), Cauchy’s theorems tell us
that the only contribution is from the first order pole. Thus,
X = 0
Y = U
M = 0
Problems—Chapter 3
3.1 a. Find the equation that the stream function must satisfy in order to represent
an irrotational flow in two dimensions. Use cylindrical polar coordinates.
b. Use separation of variables in the form
equations that R and T must satisfy.
r = RrT, and find the
3.2 Show that for steady
two-dimensional incompressible flow the acceleration can
dw ∗ d2 w ∗
be written as dz dz2 .
Problems—Chapter 3
3.3 It is sometimes convenient to regard a uniform flow as a source at minus
infinity and a sink at plus infinity, To demonstrate this, consider the potential for a
source of strength m at −a 0 and a sink of equal strength at a 0. Fix z, and expand
the complex potential in a Taylor series of z/a. Show that as a and m each go to infinity
such that their ratio remains constant, the result is a uniform stream of speed U = m/ a.
3.4 Show that, if w1 z represents the complex potential of a two-dimensional
irrotational flow and has no singularities in z < a, then wz = w1 z + w1 ∗ a2 /z
represents the same flow with a circle of radius a introduced at the origin. The
star represents that the complex conjugate of w1 z is to be taken. This is known
as the circle theorem. It is a two-dimensional counterpart of the Weiss sphere
theorem.
3.5 Use the circle theorem to find the complex potential for a circular cylinder of
radius a in a flow field produced by a counterclockwise vortex of strength located
at the point z = b, where b > a. Also find the location of the stagnation points on the
circle.
3.6 a. Write the complex potential for a uniform stream, a source of strength 2m
at −a 0, and sinks of strength m at (0, 0) and (a, 0).
b. Find the equation of the closed streamline.
c. What is the length of the closed body when m/2 aU = 2/3 a = 2?
3.7 A line vortex with circulation 10 is placed in a corner at (2, 3). There are walls
at x = 0 and at y = 0.
a. Write the complex potential for the vortex and two corners.
b. Find the induced velocity at the vortex.
3.8 Find the complex velocity potential for a line vortex at the origin and between
two parallel walls at x = 0 y ± a. Hint: Multiple mirrors are frequently encountered in
clothing stores, generating multiple reflections.
3.9 Show that the flow past a flat plate on the real z-axis is given by wz =
Ua cosh − i + 2i , where z = a cosh and U represents a uniform stream. The angle
is the angle between the plate and the uniform stream. The plate is at 0 ≤ Imag ≤ 1
in the plane.
3.10 Show that the flow of fluid inside a rotating elliptical cylinder with semimajor
axes a and b is given by wz = iAz2 . Find A in terms of the geometry
and angular rate
of rotating. Hint: The boundary condition is = 21 x2 + y2 + constant.
3.11 Show that the flow of fluid outside a rotating elliptical cylinder with semimajor axes a and b is given by wz = iAe−2 , where z = c cosh . Find A in terms
of the
and angular rate of rotating. Hint: The boundary condition is =
2 geometry
1
2
x
+
y
+
constant.
2
3.12 a. Write the complex potential for a line vortex of circulation located at
z = ih and a line vortex of circulation − located at z = −ih. Show that the
x-axis is a streamline and so can be considered an infinite flat plate.
b. Calculate the pressure on the surface of the plate from the Bernoulli equation,
letting the pressure at infinity be zero. Integrate this pressure over the entire
length of the plate in order to find the force acting on the plate due to the
vortex.
115
116
Irrotational Two-Dimensional Flows
c. If a pair of sources, each of strength m, had been used, what would the
force be?
3.13 Using the method of images, find the complex potential for a vortex located
along the centerline of a channel of width a. The vortex is at point (b, 0), and the
channel is in the region x > 0 −a/2 ≤ y ≤ a/2.
3.14 Show that the flow about a circular arc can be found using the Joukowski
transformation with an intermediate translation of axes that puts the origin on the vertical
axes. This means that both the leading and trailing edges lie on the surface of the arc.
3.15 Show that the transformation that maps the interior of the sector 0 ≤ ≤ /n, n
an integer, in the z plane onto the upper half of the plane is = zn . Then if w = A A
real, represents a uniform stream in the plane, find the corresponding flow in the
z plane.
3.16 The infinite strip shown can be regarded as a two-sided polygon and hence the
Schwarz-Christoffel transformation can be used to transform it into the upper-half plane.
Find the transformation that puts A and D at infinity and B and C at the origin in the upperhalf plane. Take the width of the strip to be h.
B
A
–∞
∞
C
D
A
B, C
D
∞
–∞
z plane
ζ plane
Figure P3.16 Problem 3.16—Infinitely long strip
3.17 The semi-infinite strip shown can be regarded as a three-sided polygon, hence
the Schwarz-Christoffel transformation can be used to transform it into the upper-half
plane. Find the transformation that puts A and D at infinity and B and C at plus and
minus one in the upper-half plane. Take the width of the strip to be h.
B
A
A
B
C
D
–∞
–1
1
∞
∞
C
D
z plane
Figure P3.17 Problem 3.17—Semi-infinitely long strip
ζ plane
117
Problems—Chapter 3
3.18 The Schwarz-Christoffel transformation can be used to transform the rightangle 90-degree bend shown into the upper-half plane. The bend can be thought of
as a four-sided polygon. Find the transformation that places the corner A at 1, B at the
origin, C at −1, and D at infinity in the plane.
B
–∞
C
A
–1
B
1
ζ plane
D
U
C
D
A
∞
z plane
Figure P3.18 Problem 3.18—Elbow
Chapter 4
Surface and Interfacial Waves
4.1 Linearized Free Surface Wave
Theory 118
4.1.1 Infinitely Long Channel 118
4.1.2 Waves in a Container of
Finite Size 122
4.2 Group Velocity 123
4.3 Waves at the Interface of Two
Dissimilar Fluids 125
4.4 Waves in an Accelerating
Container 127
4.5 Stability of a Round Jet 128
4.6 Local Surface Disturbance on a Large
Body of Fluid—Kelvin’s Ship
Wave 130
4.7 Shallow-Depth Free Surface Waves—
Cnoidal and Solitary Waves 132
4.8 Ray Theory of Gravity Waves for
Nonuniform Depths 136
Problems—Chapter 4 139
Many different mechanisms for wave propagation in fluids exist. Compressibility of both
liquids and gases allows for compression waves such as sound waves and shock waves.
These generally have quite large velocities of propagation. Normally, they cannot be
seen by the human eye without suitable instrumentation or lighting. Waves on surfaces
and interfaces, on the other hand, are generally slower and can be observed easily.
Gravity, inertial forces, surface tension, and viscosity are also important mechanisms in
generating waves.
Much of the work presented in this chapter was originated by three English scientists: Lord Kelvin (born William Thomson, 1824–1907), Lord Rayleigh (born William
Strutt, 1842–1919), and Sir Horace Lamb (1848–1934). The names in the literature for
the first two can be confusing, because, for example, both Kelvin and Thomson are
used, depending on whether the date of publication preceded or followed the bestowing
of honors. Much of the work is summarized in the books by Rayleigh (1945) and Lamb
(1932), and in the lengthy review in Wehausen and Laitone (1960).
4.1 Linearized Free Surface Wave Theory
4.1.1
Infinitely Long Channel
Consider the case of small amplitude waves in a two-dimensional channel of constant
depth h, an infinite length in the x direction, and having a uniform stream of velocity
118
4.1
119
Linearized Free Surface Wave Theory
U . It is convenient to choose a coordinate system with an origin on the undisturbed
free surface. When dealing with an unsteady flow problem with a free surface, it is
necessary to find the pressure by using the Bernoulli equation. Therefore, working with
the velocity potential, rather than the stream function, is indicated.
Write the total velocity potential in the form = Ux + ′ , where ′ represents the
velocity field due to the waves. The equations to be solved are then
2 = 0
(4.1.1)
such that
= vy = 0
y
on y = −h
(4.1.2)
Also, if the free surface is elevated an amount xt from the static position, then
p
1
2
1
=−
+
+ g +
+
= constant on y =
(4.1.3)
t
2
R1 R2
and
D
=
= vy
Dt
y
on y =
(4.1.4)
(see Figure 4.1.1). Here, is the surface tension and R1 and R2 are the principal radii
of curvature. Since the pressure above the free surface is constant and can be taken as
gauge pressure, it can usually be set to zero.
Equation (4.1.4) states that the free surface moves up and down with a velocity
equal to the vertical component of the fluid velocity. If this problem were to be solved
with all nonlinearities included, it would be necessary to work instead with the vector
, which is the total vector displacement of a fluid particle on the free surface, and to
replace equation (4.1.4) by
D 1
= 2 − gy
Dt
2
D
=
Dt
on the free surface D/Dt = v. Here, is the vertical component of .
Even though equation (4.1.1) is linear, the boundary conditions equations (4.1.3)
and (4.1.4), together with the fact that the free surface position is a priori unknown
make the problem highly nonlinear. Only in a few very special cases can a closed form
expression be found. Instead, consider the case of waves whose amplitude is small
y
η
Static position
h
Figure 4.1.1 Wave definitions—two dimensions
x
120
Surface and Interfacial Waves
compared to their wavelength, and replace the exact free surface conditions in equations
(4.1.3) and (4.1.4) by the linearized conditions
′
′
+U
+ g =
t
x
′
+U
=
t
x
y
2
x2
on y = 0
on y = 0
(4.1.5)
(4.1.6)
Here, the problem has been simplified still further by imposing equations (4.1.5) and
(4.1.6) at the static position y = 0 rather than on the actual surface. The problem is now
amenable to analysis by the method of separation of variables.
Start by assuming that can be written as the superposition of terms like
′ xyt = Tt Xx Yy
(4.1.7)
and insert this form into equation (4.2.1). The result is
TXxx Y + TXYyy = 0
Divided by , this becomes
Xxx /X + Yyy /Y = 0
Since the first term depends only on x and the second depends only on y, this relationship
can hold throughout the flow region only if each term is itself a constant. Thus,
Yyy /Y = −Xxx /X = constant = k2 say
The form of the constant k2 has been selected by “looking ahead” at the solution. It
is expected that the disturbance will be oscillatory in the x direction and also that it
will die out away from the free surface. Making this decision now is not necessary,
but it does save a lot of cumbersome and confusing notation later on. Experience with
separation of variables problems, and a little foresight, can be very useful in avoiding a
morass of symbols.
The X solutions are then trigonometric functions of kx, and the Y solutions are
hyperbolic functions of ky. Taking equation (4.1.2) into account, using DeMoive’s
theorem that ei = cos + i sin , write
X = A eikx + B e−ikx
Y = D cosh ky + h
(4.1.8)
(4.1.9)
Since both D and T multiply the still unknown constants A and B in X, at this
point D and T can be combined with A and B, letting
′ = cosh ky + h At eikx + Bt e−ikx
Guided by this form for the velocity potential, choose
′ = Ft eikx + Gt e−ikx
Substituting these expressions into equations (4.1.5) and (4.1.6), it is found that for
these boundary conditions to be satisfied for all x it is necessary that
1
1
dF
dG
A=
+ ikUF B =
− ikUG
(4.1.10)
k sinh kh dt
k sinh kh dt
4.1
121
Linearized Free Surface Wave Theory
and
F =−
cosh kh
g + k2
dA
+ ikUA
dt
and G = −
cosh kh
g + k2
dB
− ikUB
dt
(4.1.11)
Combining these equations and letting
2
= k tanh kh g + k2
(4.1.12)
results in
A=−
1
2
dA
d2 A
+ 2ik
− k2 U 2 A
dt2
dt
and
B=−
1
2
dB
d2 B
2 2
−
2ik
−
k
U
B
dt2
dt
These have solutions like A = e−ikUt k ei t + k e−i t , B = eikUt k ei t +
k e−i t .
Combining these results gives finally
′ = cosh ky + h eikx−Ut k ei t + k e−i t + e−ikx−Ut k ei t + k e−i t
(4.1.13)
′ = i
sinh kh
eikx−Ut k ei t − k e−i t + e−ikx+Ut k ei t − k e−i t
(4.1.14)
These represent a wave traveling upstream with a velocity U ± /k and are called
traveling waves. The parameters k and
are called the wavenumber and circular
frequency, respectively. The wavelength of the wave is given by = 2/k and the
wave speed relative to the uniform speed by
1
2
g+k
tanh kh
(4.1.15)
c= =
k
k
For depths large compared to the wavelength (the long wave case) tanh kh ≈ 1 and
c ≈ g + k2 2
. For shallow depths, tanh kh ≈ kh and c = k = hg + k2 . If
√
surface tension effects are not important, the wave speed reduces to c ≈ gh and could
have been found by elementary momentum considerations.
The fact that the wave speed depends on the wave number means that the wave
is dispersive. That is, if a wave consists of two or more wave numbers, as the wave
propagates, each component moves at a different speed, so the wave disperses, or
changes shape, as it travels.
Since waves on a very large body of water are being considered, with x extending
to infinity in both directions, there are no boundary conditions left to determine k.
Thus, there can be a continuous spectrum of k. The linearity of the problem allows for
superposition of wave numbers, leading to a solution in the form of a Fourier integral.
This is conveniently written as
′′ xyt =
−
′ xytk dk
′′ xyt =
−
′ xytk dk
(4.1.16)
122
Surface and Interfacial Waves
where is given by equation (4.1.12). Information on the initial shape and velocity of
the disturbance is necessary to determine , and , and to be able to carry out the
integrations.
Undersea earthquakes generate waves of very long wavelengths, giving rise to
waves called tsunamis, which can travel at very high speeds across the ocean. In deep
water these waves are of small amplitude, and ships encountering such waves at sea may
not experience much if any of a disturbance as the waves pass. However, as the waves
approach shallow water near the shore, the previous results for wave speed show that
they slow down and, by exchanging their kinetic energy for potential energy, steepen
drastically, frequently causing catastrophic damage and loss of life. This explanation
also can be used to understand wave amplification and breaking of waves on beaches.
Waves caused by high winds such as hurricanes can be of large amplitude near the
driving force but do not tend to steepen as much.
4.1.2
Waves in a Container of Finite Size
If there are walls at the ends of our channel, reflections will occur and the traveling
wave form of the solution is no longer convenient for describing the behavior of the
flow. Instead of the e±ikx in equation (4.1.8) a better selection is to use cos kx and sin
kx. If the channel is bounded at x = 0 and at x = L, then /x = 0 at those locations.
This rules out the use of sin kx and requires that sin kL = 0. This gives k = n/L,
where n is any positive integer greater than zero. Our solution now becomes
′ =
An t cos
n=0
nx
cosh ky + h
L
(4.1.17)
The boundary conditions still to be satisfied are
′
+ g =
t
2
x2
on y = 0
(4.1.18)
and
′
=
t
y
on y = 0
(4.1.19)
The U term has been dropped because the boundaries do not allow a uniform stream.
Taking the displacement of the form =
n = 0 Fn t cos nx/L and applying
equations (4.1.18) and (4.1.19) results in
n2 2
nx
nx
dAn
cos
cosh kh = −
g+ 2
Fn cos
L
L
L
n=0
n = 0 dt
nx
nx
dFn
cos
=
kAn sinh kh cos
L
L
n=0
n = 0 dt
Using the orthogonality properties of the cosines by multiplying each side of the above
by cos mx/L and then integrating over the channel length gives
n2 2
dFn
dAn
cosh kh = − g + 2
= kAn sinh kh
Fn
dt
L
dt
4.2
123
Group Velocity
Solving these for An and Fn results in
′ =
n=0
=−
an sin t + bn cos t cos
nx
cosh ky + h
L
nx
cosh kh
an cos t − bn sin t cos
2 2
L
n
n=0
g+ 2
L
(4.1.20)
(4.1.21)
with
n =
g+
n2 2
L2
nh
n
tanh
L
L
(4.1.22)
The solutions in equations (4.1.21) and (4.1.22) are in the form of Fourier series, whose
coefficients can be found by imposing initial conditions on and /t. The solution
given by equations (4.1.20) and (4.1.21) is referred to as a standing wave.
It is important to understand how the physics and mathematics have interacted.
Laplace’s equation is said to be of elliptic type. This type of equation governs phenomena such as steady-state heat transfer, deformation of membranes, and flow in porous
media. Normally, waves are not anticipated when dealing with such equations. The
free surface conditions have altered the physics of the situation, giving solutions that
in mathematics are more usual to equations of hyperbolic type—equations that permit
waves.
Wave behavior was clearly seen in the case where the channel extended to infinity
in the x direction. When the x extent was finite, this wave nature was concealed by the
form of the solution. Consider, for instance, a wave that starts out traveling to the right.
It will have a solution of the form eimx − t . When it encounters a wall, it is reflected
and becomes a left-traveling wave with the form e−imx + t . This wave is reflected
at the left wall, and the process repeats endlessly. The presence of the wall selects
waves—that is, it allows only certain wave numbers to be present. The form of the
solution in turn breaks down the exponential eimx ± t into sin mx cos mx sin t, and
cos t—thus disguising the traveling wave nature of the solution. The infinite domain
case with no walls present and solutions like eimx ± t results in traveling waves with the
general solution of the form fx − ct + gx + ct . In the finite domain case, because of
reflections from walls, solutions like sin mx sin t or cos mx cos t result in standing
waves after startup conditions have occurred.
4.2 Group Velocity
In the previous section it was stated that because the wave speed depended on the wave
number, the waves were dispersive. That means that if a wave shape is broken down into
Fourier components and more than one wave number is present, the wave shape would
change as the wave proceeds. If one observes an isolated group of waves traveling with
nearly the same wavelength, some of the waves are observed to proceed through the
group until it approaches the front of the group. At that point they appear to die out.
The process is then repeated by other waves.
124
Surface and Interfacial Waves
To analyze this situation, consider the sum of two sine waves of equal amplitude
and nearly equal wave length, and observe the change in the envelope of the resulting
wave. Let
= A sinkx − t + A sin k + k x − +
t
= 2A cos 05kx − t sin k + 05k x − + 05
(4.2.1)
t
This has the appearance of an envelope that looks much like a slowly traveling wave
of wavelength = /k that encloses a much more rapidly traveling wave of wavelength = 2/k. In the limit, as the wave numbers approach each other, the speed of
propagation of the envelope is then
cg =
d kc
dc
dc
d
=
= c+k
= c−
dk
dk
dk
d
(4.2.2)
The quantity cg is called the group velocity.
In the previous section we saw that the general formula for c as a function of wave
number was given by equation (4.1.15) as
1
g + k2
tanh kh
c= =
k
k
Thus, for these waves the group velocity is given by
2
2
g + 3k
tanh kh + kh g + k
sech2 kh
cg =
2
tanh kh
k g + k2
(4.2.3)
For large wavelengths compared to the depth this is approximated by
g + 3k2
(4.2.4)
cg ≈
2
k g + k2
whereas for long waves, where surface tension effects are small ≫ 2
cg ≈
1
2
g
1
≃ c
k 2
√
/g ,
(4.2.5)
There are at least two ways for explaining the concept of group velocity. Since
the group velocity is smaller than the wave velocity, the waves advance within their
envelopes until they approach the neck, or nodal point, at the right of the envelope.
As they approach that point, they are gradually extinguished and replaced by their
successors that are formed on the right side of the following node. The waves are thus
“grouped” within the necks of their envelope.
Another explanation of group velocity has to do with the speed at which the
energy
0 of the wave travels. The rate at which work is done on a section of the fluid is
p dy. The only part of the Bernoulli equation that contributes to the work in the
− x
4.3
125
Waves at the Interface of Two Dissimilar Fluids
linearized theory is
. Thus, taking the velocity potential from equation (4.1.15) in the
t
abbreviated form = eky + ikx − ct (no uniform stream) results in
0
p
−
0
dy =
x
−
1
dy =
ck2 cos2 kx − ct
t x
2
(4.2.6)
where the real part of the integrand has been used. Averaging this over a wavelength
gives 21 ck2 2 = 21 c2 . The kinetic energy passing through the section is
1
2
0
−
x
2
dy =
1
2
2
1
k2 cos2 kx − ct
ikeky + ikx−ct dy =
4
−
0
(4.2.7)
which when averaged over the same period gives 41 k2 2 = 41 2 . Thus, the kinetic
energy is moved through the fluid at only one-half the wave speed c, which corresponds
to the group velocity.
4.3 Waves at the Interface of Two Dissimilar Fluids
Following the procedure described in the previous section, the theory of waves at an
interface can also be developed. This time consider two fluids: the upper one y > 0
moving with a velocity U1 and the lower one y ≤ 0 moving with a velocity U2 . The
investigation will be to determine whether a small disturbance introduced at the interface
will either grow or decay. To accomplish this, choose a flow plus a disturbance in the
form of a progressive wave such that
1 = xU1 + C1 e−ky + ikx − ct for y > 0
=
(4.3.1)
2 = xU2 + C2 eky + ikx − ct
for y ≤ 0
Notice that both of these velocity potentials satisfy the Laplace equation. They represent
waves that are traveling in the positive x direction whose amplitude dies out away from
the interface. The (real) parameter k is the wave number and is twice pi divided by
the wave length. The real part of c is the wave speed, and the imaginary part of kc
is the growth rate. If Imag (kc) is positive, the wave will grow, whereas if it is negative
the wave decays. Neutral stability is obtained if c is real.
The form of the interface disturbance that is suited to these potentials is =
Aeikx − ct and the appropriate boundary conditions to be imposed at y = 0 are
D
=
Dt
n
and
p2 − p1 =
R
(4.3.2)
where is the surface tension and R is the radius of curvature of the interface.
As we saw in Section 4.1, imposing nonlinear boundary conditions on an unknown
boundary is a very difficult task. Instead, again assume that the disturbance is very
small—so small, in fact, that nonlinear terms can be neglected—and also that these
boundary conditions can be applied at the undisturbed interface. This will tell us whether
or not a small disturbance will grow. It does not say what happens if growth occurs. In
fact, it says the growth will be unbounded. Obviously, the nonlinear terms will become
important long before that, and in that situation a nonlinear analysis must be used.
126
Surface and Interfacial Waves
Our linearized boundary conditions to be applied then on y = 0 are
1
2
+ U1
=
+ U2
=
and
t
x
y t
x
y
2
1
1
2
+ U2
+ g − 1
+ U1
+ g =
p2 − p1 = 2
t
x
t
x
(4.3.3)
2
x2
The radius of curvature has been linearized here according to
d2 y
1
d2 y
dx2
=
≈
2 3/2
R
dx2
dy
1+
dx
Substitution of our velocity potentials into this and removing multiplying terms gives
ikU1 − c A = −kC1
2 ik U2 − c
ikU2 − c A = kC2
C2 + gA −
1 ik U1 − c
(4.3.4)
C1 + gA = −k2 A
Eliminating the C1 C2 , and A and solving for c gives
2
+ g − 1 k U1 − c 2 + g = −k2 or finally
2 1/2
g 2 − 1 + k2
1 2 U1 − U2
1 U1 + 2 U2
±
c=
−
k 1 + 2
1+ 2 2
1+ 2
2 −k U2 − c
(4.3.5a)
We can also write this as
where c0 =
1 U1 +
1+
2 U2
2
c = c0 ± c1
g 2 − 1 + k2
and c1 =
−
k 1 + 2
1 2 U1 − U2
1+ 2 2
2
1/2
(4.3.5b)
To decide on the stability of this flow note that c0 is always real and positive. If
the flow is to be unstable, it must be c1 that is imaginary. That would occur provided
g 22 − 21 + k2 1 + 2
2
U1 − U2 >
(4.3.6)
1 2k
We can see that surface tension is always a stabilizing influence, but for long waves
(small values of k) its effect loses significance. Having a denser fluid on top of a lighter
fluid is of course always destabilizing.
Helmholtz pointed out that where U1 = U2 1 = 2 can be used to explain the
flapping of sails and flags. In this case c1 vanishes, and our solution procedure fails
because of the double root. Repeating the analysis for this special case would be the
same except for the expression for , which now would have to include a time to the
power one multiplier. A disturbance introduced at the edge of a sail or flag would then
grow as it travels along the flag, a frequently observed phenomenon.
The difference in velocities on either side of the interface means that the interface
can be considered as a concentrated vortex sheet. To see this, suppose that rather than
having a discrete jump, the undisturbed flow was given by a velocity profile such as
4.4
127
Waves in an Accelerating Container
U = U1 +2 U2 + U1 −2 U2 tanh y . In the limit as gets small this approaches the discontinuous
1 − U2
= 2Ucosh
profile. It has the vorticity component dU
2 y . Vortices have a tendency to
dy
roll up, as exhibited in the nature of the displacement of the interface.
4.4 Waves in an Accelerating Container
Transporting a liquid in a moving container is a very effective way of causing surface waves,
as anyone who has carried a cup of coffee across a room has experienced. In transporting
oil across the oceans in large tankers, the breaking of these waves against the bulkheads
can easily create substantial damage. Even in airplanes, as fuel tanks empty the sloshing of
the fuel can create problems. One solution is to break up the surface into smaller regions by
floating dividers, but this leads to difficulties in cleaning the tanks.
Different types of container motion lead to different mechanisms of wave generation.
One of these—waves due to an acceleration in the vertical direction—is examined next.
Consider a rectangular tank of liquid subjected to a translational acceleration in the
vertical direction. Neglect surface tension effects. The liquid will be taken to originally
be in the rectangular region 0 ≤ x ≤ a 0 ≤ y ≤ b −h ≤ z ≤ 0, where the coordinate
system is fixed to the tank. The velocity potential then satisfies the Laplace equation
together with the boundary conditions
=
= 0
=
= 0
=0
x x = 0
x x = a
y y = 0
y y = b
z z = −h
(4.4.1)
dVz
=
= 0
+ g+
t
z z = 0
t z=0
dt
The solution of the Laplace equations that satisfies the boundary conditions at the walls
is of the form
=
dAmn
mx
my
cos
cos
cosh rz + h
a
b
m = 0 n = 0 dt
(4.4.2)
and
= r
2
Amn cos
m=0 n=0
my
mx
cos
sinh rh
a
b
(4.4.3)
2
where r = ma2 + nb2 . This solution also satisfies the kinematic condition at the free
surface.
The pressure condition at the free surface requires that
d2 Amn
dVz
+ r g +
Amn tanh rh = 0
(4.4.4)
dt2
dt
Next introduce the circular frequency 2 = rg tanh rh and take Vz = −V0 sin t.
Make time dimensionless by choosing T = 21 t and introduce dimensionless parameters
a=
2
2
V
2q = 0
g
2
2
128
Surface and Interfacial Waves
30
a
25
20 unstable
unstable
15
10
unstable
unstable
5
unstable
unstable
0
–5
–10
–15
unstable
unstable
unstable
–10
–5
0
q
5
10
15
Figure 4.4.1 Strutt diagram
Equation (4.4.4) then becomes
d2 Amn
+ a − 2q cos 2T Amn = 0
dT 2
(4.4.5)
This is the standard form of the Mathieu equation,1 which has been well studied,
particularly by Rayleigh. He showed that regions of stability were governed by combinations of the parameters p and as shown in Figure 4.4.1. This figure is called
the Strutt diagram. If the parameter pair (a, q) lie in a region not labeled, the solution
is oscillatory but not periodic. This nonperiodicity of the wave is an example of the
“weakness” of this type of wave. If the pair p lie in a region labeled unstable, at
least one of the solutions will increase without limit with increasing time.
The lines separating the stable-unstable regions in Figure 4.4.1 are lines along
which one of the solutions is periodic with period 2/, where is the largest integer
with p as an upper bound. The other solution increases in time without bound. Thus,
these dividing lines are considered part of the unstable region. Further information on
the computation of these lines is given in the Appendix.
4.5 Stability of a Round Jet
Liquid jets have a number of industrial applications, including the production of a stream
of droplets. In the early days of the development of the inkjet printer two technologies
were considered. The first one used jet breakup to produce a steady stream of electrically charged droplets, which then were steered by electric fields to either write on a
1
This equation is also encountered in analyzing certain mechanical vibrations, including the inverted
pendulum and the pendulum of varying length.
4.5
129
Stability of a Round Jet
page or to be collected and subsequently recirculated. The second approach, the “dropon-demand,” produced individual droplets all of which struck the page. It is the technology most frequently used today for home printers.
Modern “airless” paint sprayers used for painting automobiles depend on a rotating
cone to throw off a series of jets, each of which rapidly break up into droplets. The
water-based paint is electrically charged as it enters the cone, and an electric potential
between the cone and the item to be painted steers the paint to the target.
The breakup of a liquid jet was first studied by Raleigh in 1876. For a jet of
radius a traveling at a speed U , the velocity potential for the disturbed flow is taken
as = 1 r eikz + s + t . This is in the form of a wave traveling along the jet moving
at a speed c. The dependence has been selected so that traveling around the jet
circumference returns us to the starting point. The z dependence gives a wavelike shape
in the z direction.
Inserting this form into the Laplace equation gives
2
2
d
1 2
1 2
2
1 d2
n2
ikz + n + t
2
2 =
=
e
+
+
+
+
−
−
k
1 = 0
r 2 r r 2 r 2 2 z2
dr 2 r dr 2 r 2
(4.5.1)
The operator acting on 1 is of the Bessel type. The solution that is finite within the jet
is 1 = AIn kr , where In kr is a modified Bessel function given by the series
ikr s + 2m
−1 m
In kr = i−n Jn ikr = i−s
2
m = 0 m!m + n + 1
(4.5.2)
n
2m
1
kz
kr
=
2
2
m = 0 m!m + n + 1
The boundary conditions to be satisfied on the free surface are
+U
=
t
z
r
on r = a
and
p − p0
= − −U
=
t
z
1
1
=
−
+
R1 R2
a
The kinematic condition gives
−i d1
=
eikz + n
kU +
dr r = a
+ t
=
The pressure condition gives
−iAkU +
Is ka e
ikz + n + t
or
kU +
i kA
=
kU +
2
=
a3
1
2
2
− 2 + 2 − 2
on r = a
a
z
−ikA dIs
eikz + n
kU +
d = ka
+ t
(4.5.3)
2
1 dIs
n + a2 k2 − 1 eikz + n
a2 d = ka
kaIs′ ka 2
n + a2 k2 − 1
Is ka
+ t
(4.5.4)
There is thus a possibility that the right-hand side of equation (4.5.4) can be zero
in the range 0 < ka < 1. This is a case where the wavelength is greater than the
circumference of the jet. The right-hand side takes its largest negative value when
130
Surface and Interfacial Waves
k2 a2 = 04858, or when the wavelength equals = 2/k = 4508 × 2a. Thus, the jet
takes on a series of enlargements and contractions with continually increasing amplitude,
until it breaks up into a series of separated drops.
4.6 Local Surface Disturbance on a Large Body
of Fluid—Kelvins Ship Wave
The waves caused by creating a local disturbance to the surface of a large body of fluid
is of interest in studying the wave pattern associated with ships. Considering the fluid to
be very deep and linearizing the boundary conditions as before, the basic equations are
2 = 0
p
for − ≤ z ≤ 0
− g on z = 0
t
=
on z = 0
t
z
(4.6.1)
=−
Working in cylindrical polar coordinates and assuming independence of the angle ,
separation of variables gives
kz
r z t = At e J0 kr
−1 j
J0 kr =
2
j = 0 j!
kr
2
2j
(4.6.2)
where J0 kr is the Bessel function of order zero. Inserting equation (4.6.2) into the
boundary conditions gives
1
1 dA
=−
=
= kAJ0 kr
(4.6.3)
J kr and
=
g t z = 0 g dt 0
t
z z = 0
Having in mind a problem where there is an initial displacement given to the free surface
at r = 0, choose as a solution the forms
r z t = fk
g kz
e sin gktJ0 kr
k
r t = fk cos gktJ0 kr
(4.6.4)
To find the nature of this solution, first consider a very basic initial disturbance
where the solution is concentrated near the origin. This is a disturbance much like our
Green’s functions of Chapter 2, or like concentrated loads in elementary beam theory,
and is akin to a Dirac delta function. Use the property of integral transforms that
fk =
0
fr J0 kr 2rdr
and
fr =
and choose the initial disturbance to be
⎧
1
⎨lim
a→0
a2
fr =
⎩
0
1
2
0
fa J0 ar ada
for 0 ≤ r ≤ a
for a < r
(4.6.5)
4.6
131
Local Surface Disturbance on a Large Body of Fluid—Kelvins Ship Wave
Then lim
a→0 0
fr 2rdr = 1 and
r z t =
=
r t =
=
0
0
0
0
g kz
e sin gktJ0 kr kdk
fa J0 ka ada
k
0
g kz
e sin gktJ0 kr kdk
k
cos gktJ0 kr kdk
0
(4.6.6)
fa J0 ka ada
cos gktJ0 kr kdk
(4.6.7)
It is not possible to evaluate these integrals in closed form. However, if in the
expression for the displacement the cosine is expanded in a Taylor series, it is found that
3
5
1
12 gt2 12 · 32 gt2
12 · 32 · 52 gt2
r t =
−
+
−···
(4.6.8)
2r 2 2! r
6!
r
10!
r
This tells us that for any particular phase of the displacement, the quantity gt2 /r is
constant. Thus, the waves travel radially out from where they originated, each phase
traveling with a constant acceleration.
The more important problem would be to determine what the wave pattern would
be as the source moves, emitting wave pulses along the way. That would involve
superposition of the preceding solution over a time integral. The problem was first solved
by Kelvin (1891) using the method of stationary phase that he had in fact originated.
The paper, however, leaves out many details and does not tell how he performed the
analysis. Lamb (1932, §256) provides a figure, but, again, the presentation is sketchy and
difficult to follow. Later papers by Peters (1949) and Ursell (1960) clarified the details
omitted by the previous authors. In any case, the mathematics at this point involves
quite complicated use of asymptotic methods and becomes very involved. Details are
left to the discussions in Lamb, Ursell, and Peters.
Using the notation of Peters, the results can be summarized as follows:
• The waves are contained within an angle of 2 · sin−1 13 = 38 56′ .
• Two wave patterns exist within this triangle: a set of transverse waves and a set
of divergent waves. The patterns of constant phase are shown in Figure 4.6.1.
They are given by the following:
g
1
Transverse waves 2 rh1 = 2n +
n > 0
(4.6.9)
c
4
g
3
Divergent waves 2 rh2 = 2n +
n > 0
(4.6.10)
c
4
Here, r and are the usual cylindrical coordinates centered at the present location
of the disturbance, and c is the speed at which it travels. The various functions
are given by
hi = sin coth pi cosh2 pi
√
√
with sinh p1 = 41 cot − cot 2 − 8 sinh p2 = 41 cot + cot 2 − 8 .
(4.6.11)
132
Surface and Interfacial Waves
30
20
10
Y
0
–10
–20
–30
0
10
20
30
40
50
60
70
X
Figure 4.6.1 Kelvin ship wave—waves of constant phase
The pi ’s are real if cot2 ≥ 8—that is, −1947 ≤ ≤ 1947 . Outside of this
region the disturbance is greatly diminished.
The preceding pattern is a reasonable depiction of the bow wave on a slender ship or
the pattern due to a tree branch or fish line in the water. Understanding the pattern also
has a practical use. Resistance to the motion of ships is about equally divided between
viscous shear and wave drag. (The waves have a considerable amount of potential
energy, all provided by the ship.) Simple models of a ship use a single source-sink
pair—the source modeling the ship’s bow—to generate the pattern given above. In the
1950s Inui (1957) and others discovered that if somehow another “source” could be
placed in front of the ship’s bow, it could create waves out of phase with the bow wave
and thus cancel them. Experiments conducted in the towing tank at The University of
Michigan by Inui and others confirmed this theory, leading to the now-famous bulbous
bow. This is an underwater rounded protuberance extended from the bow. This idea
was embraced by the world’s tanker fleet, which added bulbous bows and midsections
to existing tankers, enabling greater capacity using the original power supplies. Similar
wave-cancellation ideas are used in noise-cancelling headphones.
4.7 Shallow-Depth Free Surface Waves—Cnoidal
and Solitary Waves
Many years ago, Russell (1838, 1894) pointed out that for wavelengths that were large
with respect to the depth, waves are observed for which shape changes only occur very
gradually as the wave proceeds along a shallow channel. The wave height of these waves
is not necessarily small. To investigate this situation, consider a two-dimensional case
where the wave pattern varies along the channel but not with the direction transverse to
the channel walls. To make the wave pattern steady, the undisturbed flow in the channel
is taken to move at a speed c, this speed being whatever is necessary to make the wave
pattern stationary.
4.7
133
Shallow-Depth Free Surface Waves—Cnoidal and Solitary Waves
First, consider a periodic wave pattern in an infinitely long channel. The x-axis is
taken as being the direction of wave propagation, and the coordinate origin is at the
bottom of the channel. The governing equations and boundary conditions are
2 2
+ 2 = 0
x2
y
= 0 on y = 0
y
=
x x
y
and
1
2
(4.7.1)
x
2
1
+
2
y
2
+ g = C
on y =
Here, C is a constant that represents the total energy at a point, to be determined later.
Next, following Lamb (1932, §252), assume that a solution can be found for the
velocity potential in the form of a Taylor series in y. Then to satisfy the Laplace
equation, let
x y = G −
y2 ′′ y4 ′′′′
G + G ∓ · · ·
2!
4!
where G is a function of x and primes denote x derivatives.
Substituting equation (4.7.2) into the boundary conditions gives
2
4
3
G′ − G′′′ + G′′′′′ ∓ · · · ′ = −G′′ + G′′′′ ∓ · · ·
2!
4!
3!
2
2
4
2
3
G′ − G′′′ + G′′′′′ ∓ · · · + −G′′ + G′′′′ ∓ · · · = 2C − 2g
2!
4!
3!
(4.7.2)
(4.7.3)
At this point, since long waves are being considered, assume that the more G is
differentiated, the smaller its effect becomes. Then, under this assumption, equation
(4.7.3) becomes after rearrangement
′ G′ + G′′ − ′
2 ′′′
3
G − ′ G′′′′ ∓ · · · = 0
2!
3!
2
(4.7.4)
2
− 2C + 2g + G′ + −G′′ − 2 G′ G′′′ ± · · · = 0
The first of these can be integrated, giving
G′ −
3 ′′′
G ± · · · = D
3!
or GF ′ =
D 2 ′′′
+ G ∓ · · ·
3!
(4.7.5)
where D is a constant of integration. Inserting this into the pressure condition and
simplifying gives
−2C + 2g
1
2 ′′
′2
+
+
−
= 0
D2
2 3
32
(4.7.6)
Equation (4.7.6) is a form of the Korteweg-deVries equation (1895). It can be
integrated by first multiplying by ′ . The result is
−2C + g2 1 ′2
− +
= H
D2
3
where H is still another constant of integration.
(4.7.7)
134
Surface and Interfacial Waves
The constants of integration can be evaluated by noting that at both the crests and
the troughs of the waves the slope is zero. Thus,
3 − htrough hcrest − − L
hcrest htrough L
′2 =
2C
where L =
− hcrest + htrough
g
(4.7.8)
and L ≤ htrough ≤ ≤ hcrest
The magnitude restriction on L comes about because the right-hand side must always
be positive to match the left-hand side.
Applying the conditions at the crest and trough of the waves, the constants D and
H are found to be given by
D2 = gLhtrough hcrest
H =−
htrough + hcrest 1
−
htrough hcrest
L
(4.7.9)
The solution can be advanced further by letting = htrough + hcrest − htrough cos2 .
Then the equation for the displacement becomes
d
dx
2
hcrest − htrough
3hcrest − L
2
=
sin
1−
hcrest htrough L
hcrest − L
This can be integrated to give x as a function of and also the wavelength. The result is
x=
0
dp
1 − k2 sin2 p
where = wavelength = 2
and k2 =
hcrest − htrough
hcrest − L
(4.7.10)
= Fk
/2
dp
= 2Fk/2
1 − k2 sin2 p
hcrest htrough L
and =
3hcrest − L
0
Here, x is measured from the wave crest, and Fk is the incomplete elliptic function
of the first kind. (When = /2, it becomes the complete elliptic integral of the first
kind.) The function that is essentially the inverse of F , giving ! as a function of x, is
= htrough + hcrest − htrough
x
cn
2
(4.7.11)
where cnx/ = cos is called the cnoidal function, and the wave described by
equation (4.7.11) is the cnoidal wave.
The remaining task is to determine the constant C. From equation (4.7.1) it is seen
that C can be expressed in the form C = c2 /2 + gh, where h is the average depth and c is
the speed necessary to hold the wave steady. The amount of fluid in one wavelength is
/2
h =
0
dx =
0
dx
dp = 2
dp
/2
0
hcrest − hcrest − htrough sin2 p
dp
1 − k2 sin2 p
4.7
135
Shallow-Depth Free Surface Waves—Cnoidal and Solitary Waves
or
h = 2 LFk/2 + hcrest − L Ek/2 /
(4.7.12)
/2
1 − k2 sin2 pdp is the complete elliptic integral of the secwhere Ek/2 = 0
ond kind.
The procedure for completing the solution for a specific case is as follows. First,
choose three of the variables—say, htrough hcrest , and L—since their relative magnitudes
are known. From equation (4.7.10) and (4.7.12) k can be found, as well as x as a function
2
of the
local wave elevation. Since C = c /2 + gh, it follows from equation (4.7.8) that
c = gL + htrough + hcrest − h . Thus, all of the parameters of the problem have been
determined. Plots of the elliptic functions are given in Figures 4.7.1 and 4.7.2, where
= sin−1 k is expressed in degrees. Values of these functions are shown in tables in the
Appendix.
A more elaborate presentation of this material is given in Wehausen and Laitone
(1960, §31). Their procedure is to make x dimensionless using the wavelength x/ and
y dimensionless by a representative depth h y/h and then writing equation (4.7.2) as
x y = Gx/ +
−1 n "2n y 2n d2n Gx/
2n !
h
dx/ 2n
n=1
"=
h
This form clarifies what is meant by the statement that the more G is differentiated, the
smaller the terms become, and also allows better assessment of the order of accuracy
of the results.
As approaches unity, equation (4.7.10) can be integrated in a simpler form. This
value of # gives a wave whose wavelength approaches infinity. It is called the solitary
wave, with L → htrough = h . The result is
1.6
5
30
40
50
70
E(k, φ)
1.2
90
k
0.8
0.4
0.0
0
20
60
40
φ
Figure 4.7.1 Elliptic function Ek
80
136
Surface and Interfacial Waves
5
90 k
4
85
80
F(k, φ)
3
70
60
50
5
2
1
0
0
20
40
60
80
φ
Figure 4.7.2 Elliptic function Fk
′2 =
3 − h 2 hcrest −
, which can be integrated to give
hcrest h2
3hcrest − h
hcrest − h
with $ =
= h +
2
h2 hcrest
cosh 05$x
(4.7.13)
The speed necessary to keep this wave stationary can be found in this case by realizing
that far upstream the constant C in equation (4.6.13)
is C = c2 /2 + gh . Putting this in
the expression for L in equation (4.6.18) gives c = gh . Conversely, this would be
the speed at which the wave travels in otherwise still water.
Wehausen and Laitone (1960, §31) show that there is a theoretical limit to the
height of finite amplitude waves, given by
hcrest − htrough
82
(4.7.14)
= 2
h
9 + 2
max
At this amplitude theory shows that there is a reversal in the vertical velocity near the
crest. For the solitary wave this has a value of 8/11 = 07273. Experiments with these
waves show that as the wave amplitude approaches this limit, the wave starts to break.
4.8 Ray Theory of Gravity Waves for Nonuniform Depths
When the depth of the flow is variable, particularly in shallow regions where depth is
comparable to wavelength, observations of waves from the shoreline show waves both
steepening and changing direction. To understand this phenomena, the approach must
be altered a bit from the previous.
4.8
137
Ray Theory of Gravity Waves for Nonuniform Depths
Since a three-dimensional problem is now being considered, choose z as being the
direction perpendicular to the free surface. The waves will be considered small so that
the linearized theory can be used. The governing equations are then
2 = 0
in 0 ≤ z ≤ −hx y
+ g = 0
t
=
t
z
(4.8.1)
on z = 0
(4.8.2)
on z = 0
h h
= vz = −
+
z
x x y y
(4.8.3)
on z = −hx y
(4.8.4)
Since equation (4.8.4) involves products of function of x and y, methods such as
separation of variables cannot be used for general functions h(x,y).
To simplify the equations, certain assumptions must be made concerning the relative
values of the various quantities. The easiest way to do this is to deal with dimensionless
quantities. L will be taken to be an appropriate length, possibly a wavelength, measure
in the x-y plane, and changes in the z direction will be taken to occur at a faster rate
than in the x-y directions. Thus, "L will be the appropriate scale in the z direction. Our
dimensionless coordinates will then be X = x/L√Y = y/L Z = z/"L, and H = "h/L.
The appropriate time dimensionalization is T = t "g/L.
As a result of this,
1
=
x L X
1
=
y L Y
1
=
z "L Z
and
=
t
"g
L T
Thus, our equations become
1 2
= 0 in 0 ≤ Z ≤ −Hx y
"2 Z2
2
2
+ j 22 =
+ 2
where 2 = i
2
X
Y
X
Y
22 +
(4.8.5)
Also,
1
"g
"g
+ g = 0
=
on Z = 0
L T
L T
"L Z
1
H H
on Z = −H
= −"
+
" Z
X X Y Y
(4.8.6)
(4.8.7)
If these are rearranged and is solved for, the result is
"2 22 +
=−
2
=0
Z2
"
gL T
in 0 ≤ Z ≤ −Hx y
1
=
on Z = 0
T
"3 gL Z
(4.8.8)
(4.8.9)
138
Surface and Interfacial Waves
The displacement can be eliminated from consideration by differentiating the first
result in equation (4.8.9) and using it in the second. Thus,
H H
= −"2
+
on Z = −H
(4.8.10)
Z
X X Y Y
2
+
= 0 on Z = 0
(4.8.11)
T 2 Z
To proceed with a solution, recall that in the constant-depth case there were solutions
like eikx−%t to represent traveling waves. Here, in correspondence with this, write
XYZT = fX Y Z eisXYT /" . Substituting this into our equations yields
2
f
2
f
= 0 in 0 ≤ Z ≤ −Hx y
−
s
"2 22 f + i"22 s · 2 f + f 22 s +
2
Z2
2
s
2 s
f
−
f + i" 2 f = 0 on Z = 0
Z
T
T
"2
f
+ i"f 2 s · 2 H + "2 2 f · 2 H = 0 on Z = −H
Z
As yet neither L nor " have been defined. In the constant depth case, it was seen
that the quantity kx − t corresponds to our sX Y T /", where k is the wave number
and the circular frequency. In analogy with this, it is customary
to choose the scaling
√
so that s/T = 1, which assumes the circular frequency g/"L is independent of time,
and to write k = 2 s as a wave number that depends on X and Y . This equation for k
is called the eikonal equation in geometric optics. (Eikonal is the Greek word for “ray.”
It sounds much more impressive than the short word ray.) The surfaces s = constant
are wave fronts, or lines of constant phase, and 2 s is a vector normal to the wave front
and represents a ray.
If everything has been scaled properly, the terms not multiplied by " should be
handled first. From them it is found as the lowest approximation to f that
f0 = A0 x y cosh kZ + H
with k tanh kh = 1
(4.8.12)
Checking back with the constant depth case, the results are the same, but now the wave
number k is a function of the horizontal coordinates.
The amplitude solution can be continued by writing
f = f0 +
n=1
"n cosh kZ + H An XYZ
and substituting into the equations and collecting terms on powers of &. The process is
tedious but straightforward. The approach is discussed in Meyer (1979).
Of main interest, however, is how a wave front moves. To carry out a solution
numerically, one would start with a given wave front (i.e., a given value of s). At any
given depth, H is known, so k can be found. The eikonal equation plus the known
directionality of 2 s at any point (perpendicular to the wave front) is then integrated
to find the ray direction and establish a new wave front. The process repeats until a
boundary is encountered.
Note that it is possible that rays can cross one another, just as light rays cross when
focusing a beam of light. The intersection of two rays is called a caustic. Where they
intersect, our theory predicts infinite amplitude of the wave. In fact, our theory breaks
Problems—Chapter 4
down there, just as it does at the envelopes of the rays. The same happens in geometric
optics theory. In both cases a different scaling is needed.
To summarize, in dimensional terms the various quantities in terms of dimensional
units are given by
k
1
,
dimensional wave number → kD = 2 s =
"
"L
g
1 s
dimensional circular frequency → D =
=
,
" T
"L
√
"gL
,
dimensional wave speed → cD = D =
kD
k
2"L
2
.
=
dimensional wave length → D =
kD
k
The monograph by Mader (1988) presents more details as well as numerical programs for calculating the progress of such waves.
Problems—Chapter 4
4.1 For a monochromatic (a single wavelength) traveling wave on a deep layer
of fluid, the potential and the elevation are given by xt = −aceky cos kx − ct ,
xt = a sin kx − ct . The kinetic and potential energies contained in one wavelength
are
0
1
2
1
2
KE =
2 dx − ct
+
dx − ct dy and PE =
g
2 − 0
x
y
2
0
Show that the two energies are equal. This is called the principle of equi-partition of
energy.
4.2 Find the wavelengths to be expected for standing waves on the free surface of
a rectangular tank a by b, with depth d.
4.3 Find the wavelengths to be expected for standing waves on the free surface of
a cylindrical tank of radius a and depth d.
4.4 Two fluids of different densities are separated by an interface at y = 0. The
fluids can each be considered to have infinite extent. Find the speed at which interfacial
waves travel, including the effects of surface tension. The linearized surface tension
2
.
condition can be taken as p =
x2
4.5 Find the possible interfacial waves in a canal, where the bottom layer of depth
a has a higher density than the top layer of thickness b. The top layer is open to
the atmosphere. Neglect surface tension. (Ships moving in fjords where the top layer
is brackish and the lower layer is salty can notice increased ship drag due to waves
generated at the interface of the two fluids.)
139
Chapter 5
Exact Solutions of the Navier-Stokes
Equations
5.1 Solutions to the Steady-State NavierStokes Equations When Convective
Acceleration Is Absent 140
5.1.1 Two-Dimensional Flow
Between Parallel Plates 141
5.1.2 Poiseuille Flow in a
Rectangular Conduit 142
5.1.3 Poiseuille Flow in a Round
Conduit or Annulus 144
5.1.4 Poiseuille Flow in Conduits
of Arbitrarily Shaped CrossSection 145
5.1.5 Couette Flow Between
Concentric Circular
Cylinders 147
5.2 Unsteady Flows When Convective
Acceleration Is Absent 147
5.2.1 Impulsive Motion of a Plate—
Stokes’s First Problem 147
5.2.2 Oscillation of a Plate—
Stokes’s Second Problem 149
5.3 Other Unsteady Flows When
Convective Acceleration Is Absent 152
5.3.1 Impulsive Plane Poiseuille
and Couette Flows 152
5.3.2 Impulsive Circular Couette
Flow 153
5.4 Steady Flows When Convective
Acceleration Is Present 154
5.4.1 Plane Stagnation Line
Flow 155
5.4.2 Three-Dimensional Axisymmetric Stagnation Point Flow 158
5.4.3 Flow into Convergent or
Divergent Channels 158
5.4.4 Flow in a Spiral Channel 162
5.4.5 Flow Due to a Round Laminar
Jet 163
5.4.6 Flow Due to a Rotating
Disk 165
Problems—Chapter 5 168
5.1 Solutions to the Steady-State Navier-Stokes Equations
When Convective Acceleration Is Absent
Because of the mathematical nonlinearities of the convective acceleration terms in the
Navier-Stokes equations when viscosity is included, and also because the order of the
Navier-Stokes equations is higher than the order of the Euler equations, finding solutions
is generally difficult, and the methods and techniques used in the study of inviscid
flows are generally not applicable. In this and the following chapters, a number of cases
140
5.1
Solutions to the Steady-State Navier-Stokes Equations When Convective Acceleration Is Absent
where exact and approximate solutions of the Navier-Stokes equations can be found are
discussed.
In particular, for flows where the velocity gradients are perpendicular to the velocity, the convective acceleration terms vanish. The results are then independent of the
Reynolds number. Several such cases will be considered.
5.1.1
Two-Dimensional Flow Between Parallel Plates
Consider flow between parallel plates as shown in Figure 5.1.1. Write the velocity in
the form v = uy 0 0, which automatically satisfies the continuity equation. With
gravity acting in the plane of the figure, the Navier-Stokes equations then become
d2 u p
=
− g cos
dy2
x
p
0=
− g sin
y
p
0=
y
(5.1.1)
where is the angle between the x-axis and the direction of gravity. Since the viscous
terms are functions of at most y, and since from the second and third equation p must
u (y )
b
0.8b
0.6b
y
0.4b
0.2b
–0.3
–0.2
–0.1
Figure 5.1.1 Plane Poiseuille-Couette flow
0.0
0.1
0.2
0.3
0.4
141
142
Exact Solutions of the Navier-Stokes Equations
be linear in y and independent of z p/ x must be a constant. Thus, integration of
equation (5.1.1) gives
1 p
− g cos y2 + c1 y + c2
u=
2 x
where c1 and c2 are constants of integration.
If there is an upper plate at y = b moving with velocity Ub , and a lower plate at y = 0
moving with velocity U0 , then applying these conditions to our expression for u yields
c2 p
− g cos
U0 =
2 x
and
1
Ub =
2
p
− g cos
x
b2 + c1 b + c2
Solving for c1 and c2 gives
y
1 p
− g cos y2 − by + U0 + Ub − U0
uy =
2 x
b
(5.1.2)
with a shear stress given by
xy y =
du 1
=
dy
2
a pressure by
p
− g cos
x
2y − b +
Ub − U0
b
(5.1.3)
p
x + gy sin + p0
(5.1.4)
x
and a volumetric discharge per unit width of
b 1 p
b
y
2
u dy = a
− g cos y − by + U0 + Ub − U0 dy
Q=a
2 x
b
0
0
3
U − U0
−ab p
− g cos + ab b
(5.1.5)
=
12 x
2
px y =
The terms in the velocity that are associated with the pressure gradient and body forces
are called the Poiseuille flow terms. The terms due to the motion of the boundaries are
called the Couette flow terms.
5.1.2
Poiseuille Flow in a Rectangular Conduit
For flows in a rectangular conduit, the analysis is similar to that in the previous section
with of course one more space dimension, since the velocity depends on both y and z.
Letting v = uy z 0 0, the Navier-Stokes equations for this flow are
2
u 2 u
p
+
0 = − + g cos +
x
y2 z2
p
+ g sin
y
p
0=−
y
0=−
(5.1.6)
5.1
Solutions to the Steady-State Navier-Stokes Equations When Convective Acceleration Is Absent
z /b
0.6
0.4
0.2
y /b
0.0
–0.2
–0.4
–0.6
0.0
0.2
0.4
0.6
0.8
1.0
Figure 5.1.2 Rectangular conduit—stream surfaces
with u = 0 on the (stationary) walls at 0 ≤ y ≤ b −05a ≤ z ≤ 05a as shown in
Figure 5.1.2.
Separation of variables suggests taking
1 p
− g cos y2 − by + YyZz
us =
2 x
The terms quadratic in y are needed to make the equation homogeneous so that the
method of separation of variables works. Inserting this into equation (5.1.6) gives
0=
d2 Z
d2 Y
Z
+
Y
, which after dividing by YZ becomes
dy2
dz2
1 d2 Y
1 d2 Z
=
−
= a constant—say, − k2
Y dy2
Z dz2
Solving these equations leads to Y being made up of sines and cosines of ky and Z
being hyperbolic sines and hyperbolic cosines of kx. Applying the no-slip boundary
conditions and symmetry in z eliminates the cosine and sinh functions. To meet the
remaining condition, it is necessary that k = nb , where n is any positive integer. After
summing these in a Fourier series, the result is
1 p
n z
n y
− g cos
cosh
y2 − by + An sin
(5.1.7)
ux =
2 x
b
b
n=1
with shear stress components
n
1 p
n y
n z
ux
=
−
g
cos
cos
cosh
2y
−
b
+
An
=
xy
y
2 x
b
b
b
n=1
n
1 p
n y
n z
ux
An
=
−
g
cos
sin
sinh
(5.1.8)
=
xz
z
2 x
b
b
b
n=1
143
144
Exact Solutions of the Navier-Stokes Equations
Use of the orthogonality conditions for the sine gives
b2
n y
4 3 3 1 − −1n
dy
yb − y2 sin
b
= n
An =
n a b 2n y
n a
sin
dy
cosh
cosh
0
b
b
b
b
0
The volumetric discharge is given by
b 05b
1 p
ux dz =
dy
− g cos
Q=
2 x
−05a
0
−ab3
n a
2b2
×
+ An 2 2 1 − −1n sinh
6
n
2b
n=1
5.1.3
(5.1.9)
(5.1.10)
Poiseuille Flow in a Round Conduit or Annulus
For Poiseuille flow in a circular annulus as shown in Figure 5.1.3, adapting the previous
forms to cylindrical polar coordinates suggests letting v = 0 0 wr. Again, this form
for the velocity automatically satisfies the incompressible continuity equation. Then
the Navier-Stokes equations in cylindrical polar coordinates (see Appendix) in the z
direction (along the tube) reduces to
2
d w 1 dw
p
− g cos =
+
z
dr 2 r dr
Since by use of the product rule of calculus
d2 w 1 dw 1 d
+
=
dr 2 r dr
r dr
dw
r
dr
this can be integrated twice to obtain
1 p
− g cos r 2 + c1 ln r + c2
w=
4 z
b
a
Figure 5.1.3 Rotating Couette flow—inner and outer cylinders
(5.1.11)
5.1
Solutions to the Steady-State Navier-Stokes Equations When Convective Acceleration Is Absent
If w = Wa at r = a and w = Wb at r = b, proceeding as in the plane Poiseuille flow
case, we have
1 p
ln r/b
w=
− g cos
r 2 − b2 − b2 − a2
4 z
ln a/b
(5.1.12)
ln r/b
+ Wb + Wb − Wa
ln a/b
with a shear stress
rz =
dw
1
=
dr
4
p
− g cos
z
2r −
b2 − a2
W − Wa
+ b
r ln a/b
r ln a/b
(5.1.13)
If there is no inner cylinder a = 0, all logarithmic terms in equation (5.1.12) must
be ruled out, since they would make both the velocity and the shear stress infinite at
r = 0. In that case the solution reduces to
1 p 2
w=
r − b2 + Wb
(5.1.14)
4 z
with
zr
=
dw r
=
dr
2
p
− g cos
z
(5.1.15)
If, instead, the outer cylinder is removed, no solution exists that is bounded at large
r. Thus, if a long cylinder is moved longitudinally through a fluid, one-dimensional
motion, depending only on r is impossible.
5.1.4
Poiseuille Flow in Conduits of Arbitrarily Shaped Cross-Section
Except for conduits of a few simple geometries, such as the circle, ellipse, rectangle, and
equilateral triangle, exact solutions of equation (5.1.6) are impossible to find. However,
the discharge can still be estimated very accurately. From the definition of Q we have
Q=
udS
(5.1.16)
S
Rewrite equation (5.1.6) as
2
u 2 u
p
− g cos =
+
x
y2 z2
− g cos , where K is a positive constant. Dividing
For convenience, let K = −/ p
x
this equation by the negative of the pressure gradient and gravity term, we have
2
u 2 u
1 = −K
+
= −K 2 u
(5.1.17)
y2 z2
Multiply equation (5.1.17) by u, and integrate over the conduit area, giving
u2 dS
· uu − u2 dS = K
u 2 udS = −K
udS = − K
Q=
S
S
S
S
(5.1.18)
where we have used the product rule of calculus, then the divergence theorem, and
finally the no-slip boundary condition.
145
146
Exact Solutions of the Navier-Stokes Equations
To get the first estimate on Q, choose an approximation to the velocity that
satisfies the continuity equation and the no-slip boundary conditions but not necessarily the momentum equation (5.1.17). Then, by some elementary algebra and
rearrangements,
1
1
v2 − u2 dS = K
v − u2 + 2v · u − 2u2 dS
K
2
2
S
S
1
= K
v − u2 + 2u · v − udS
2
S
1
= K
v − u2 + 2v − u · u − 2v − u 2 udS
2
S
1
v − udS
v − udS ≥
v − u2 dS +
= K
2
S
S
S
Rewriting this and using equation (5.1.17), we have
1
1
v2 − u2 dS − Q ≥
vdS − Q
K
2
2
S
S
or after rearrangement,
Q≥2
S
vdS − K
S
v2 dS
(5.1.19)
the equal sign holding only if v = u everywhere.
A different bound on Q can be found by choosing an approximation w for the velocity that satisfies equation (5.1.6) but not necessarily the no-slip boundary conditions.
Proceeding as before,
1
1
w2 − u2 dS = K
w − u2 + 2u · w − udS
K
2
2
S
S
1
= K
w − u2 + 2u · w − u − 2u 2 w − udS
2
S
1
w − u2 dS ≥ 0
= K
2
S
by virtue of both u and w satisfying equation (5.1.6) and u also satisfying the no-slip
conditions. Expanding the term on the left of the inequality, and using equations (5.1.16)
and (5.1.18), the result is
1
1
w2 dS − Q ≥ 0
K
2
2
S
or finally,
Q≤K
S
w2 dS
(5.1.20)
Again, the equal sign holds only if w equals u everywhere. Thus, from equations (5.1.19)
and (5.1.20) Q is bounded on both sides according to
K
2v − Kv2 dS
(5.1.21)
w2 dS ≤ Q ≤
S
S
By refining the approximating functions v and w, we can come as close to u, or at least
to Q, as we desire.
5.2
5.1.5
147
Unsteady Flows When Convective Acceleration Is Absent
Couette Flow Between Concentric Circular Cylinders
For flow between concentric circular cylinders, again use cylindrical polar coordinates
and let v = 0 vr 0. This satisfies continuity automatically. Then the Navier-Stokes
equations reduce to
2
v
v 1 v
+
−
0=
r 2 r r r 2
(5.1.22)
p
v2
=
r
r
The first equation is of Euler type, and its solutions are therefore of the form r n .
Substituting this form into equation (5.1.22a), we find that n − 1n + 1 = 0. Thus, the
solution is
v = Ar + B/r
(5.1.23)
The term multiplied by the constant A represents a rotational rigid body rotation. The
term multiplied by B is an irrotational vortex, which we saw in Chapter 2. If v = bb
at r = b and v = aa at r = a, applying these boundary conditions to equation (5.1.23)
gives for A and B
b2 b − a2 a
b2 − a2
(5.1.24)
a2 b2 a − b
B=
b2 − a2
Notice that if a2 a = b2 b , the flow is irrotational, so viscous flows can in rare cases
be irrotational. Generally, however, they are rotational because, except for very special
boundary conditions such as we have here, vorticity is generated at boundaries by the
no-slip condition and then is diffused throughout the flow by viscosity.
If the outer cylinder is absent, A = 0, which again is an irrotational flow. If the
inner cylinder is absent, B = 0, and the motion is a rigid body rotation.
A=
5.2 Unsteady Flows When Convective Acceleration Is Absent
Next consider two important cases of unsteady flow that were first solved by Stokes.
Both of the problems have a velocity of the form v = uy t 0 0 and an infinite plate
located at y = 0. For these flows, since the flow is due solely to the motion of the plate
and no other forcing is imposed, the pressure gradient is absent, and the x momentum
equation becomes
5.2.1
2 u
u
= 2
t
y
(5.2.1)
Impulsive Motion of a Plate—Stokes’s First Problem
Stokes considered the case where at an initial time a plate is suddenly caused to move
in the x direction with a velocity U . Considering the various parameters in the problem,
we see that u must depend only on y, t, U, , and . Thus, we have u = uy t U .
In dimensionless form this can be written as
Uy
u
=f y
(5.2.2)
U
t
148
Exact Solutions of the Navier-Stokes Equations
Since the momentum equation (5.2.1) in this case is linear, and since U appears only in
the boundary condition, expect that U will appear only as a multiplying factor, so the
dimensionless form (5.2.2) can be reduced to
u
=f y
(5.2.3)
U
t
2
U
′ u
To verify this, note that u
= − Uy
f y = U t f ′ , and yu2 = t f ′′ , where a prime
t
2t
t
√
denotes differentiation with respect to the dimensionlessvariable = y /t. Inserting
′
U
f = t f ′′ , or
the preceding into equation (5.2.1), the result is − Uy
2t
t
f ′′ + 05 f ′ = 0
(5.2.4)
The boundary conditions on f are as follows: Because of the no-slip condition,
u = U on the flat plate, which requires that f0 = 1. At very large values of y u
must go to zero, since the velocity is transported away from the plate only by viscous
diffusion, and it will take a very large amount of time for the effect to be noticed away
from the plate. Therefore, f = 0.
Equation (5.2.4) can be integrated once to give f ′ = A exp−2 /4, where A is a
2
constant of integration. Integrating this once again gives f = A 0 e− /4 d + B, where
B is a second constant of integration. Since f 0 = 1 B = 1. Since f = 0,
2
−1
0=A
e− /4 d + 1giving A = −2 /4
e
d
0
0
√
√
; thus, A = −1/
The integral appearing in the denominator of A has the value
and finally
u = U 1 − Err = U ∗ Errc
(5.2.5)
√
−2 /4
where Err = 1/
e
d. The integral appearing in equation (5.2.5) is a form
0
of the well-known error function, and 1 minus the error function is the complementary
error function. Typical values for the solution are seen in Table 5.2.1 and in Figure 5.2.1.
The combination 1 − u/U corresponds to the case where the plate is at rest and the
outer fluid is impulsively started in motion.
TABLE 5.2.1 The Error Function
00
01
02
03
04
05
06
07
08
09
10
11
12
u/U
1 − u/U
u/U
1 − u/U
u/U
10000
09436
08875
0832
07773
07237
06714
06206
05716
05245
04795
04367
03961
00000
00564
01125
01680
02227
02763
03286
03794
04284
04755
05205
05633
06039
13
14
15
16
17
18
19
20
21
22
23
24
25
03580
03222
02888
02579
02293
02031
01791
01573
01376
01198
01039
009
0077
06420
06778
07112
07421
07707
07969
08209
08427
08624
08802
08961
09103
09229
26
28
30
32
34
36
38
40
42
44
46
48
50
00660
00477
00339
00236
00162
00109
00072
00047
00030
00018
00011
00007
00004
1 − u/U
09340
09523
09661
09764
09838
09891
09928
09953
09970
09982
09989
09993
09996
5.2
149
Unsteady Flows When Convective Acceleration Is Absent
1.0
Error Function
0.8
0.6
0.4
Complementary
Error Function
0.2
0.0
0
1
3
2
4
5
Zeta
Figure 5.2.1 Error and complementary error functions
The variable that appears in this solution is called a similarity variable, and f is
called a similarity solution. By similarity we mean that at two different times, t1 and t2 ,
we have geometric similarity in our flow (one velocity profile is merely an enlargement
of the other) at locations y1 and y2 , providing that
y1
= y2
t1
t2
or equivalently, 1 = 2 .
(5.2.6)
Similarity solutions play an important role in fluid mechanics. It reduces the NavierStokes partial differential equations to ordinary differential equations that are much
easier to solve, with a solution that is much easier to understand physically. The existence
of a similarity solution depends on simplifying conditions such as the domain having
infinite extent, which allows the number of dimensionless parameters to be small. Thus,
the similarity solutions are very special solutions and exist only in a relatively few flow
cases. We will see more of similarity solutions in later sections.
This Stokes problem is a reasonably good approximation of the flow on a flat plate if
we consider Ut to be a “distance” from the leading edge of the plate. To see how “deep”
the viscous effects penetrate, note that for = 3 the fluid velocity is reduced to less
than 4% of the plate velocity. Consequently, we can consider this to be approximately
the outer edge of the effects of viscosity. This thickness is termed the boundary layer
thickness. In terms of distance from the leading edge of the plate, we have
(5.2.7)
boundary layer thickness = 3 t/ = 3 x/U = 3x/ Ux/,
We will consider the problem of flow on a flat plate again in Chapter 6 and find it to
be in general agreement with those results.
5.2.2
Oscillation of a Plate—Stokes’s Second Problem
A second problem considered by Stokes is one where an infinite plate oscillates back
and forth with a circular frequency , the motion having taken place sufficiently long
that the starting conditions can be ignored. Then, u 0 t = U cos t, with U being the
maximum speed of the plate.
150
Exact Solutions of the Navier-Stokes Equations
If we were to try to solve this problem using separation of variables, the “traditional”
method of solving partial differential equations, we would find phase differences in time
between the boundary velocity and the fluid velocities, and we would need both the
sine and cosine of t. The result would be that we would end up having to solve two
coupled second order equations.
An easier approach is to let u0 t = Ueit , where i is
√
the imaginary number −1, and by DeMoivre’s theorem eit = cost + i sint.
Since the real part of eit is cos t, our solution is found by taking only the real part
of u at the very end of our analysis.
Proceeding in this manner, let
u = fyeit
(5.2.8)
and substitute this into equation (5.7.8). The result is
if = f ′′
(5.2.9)
f = Ae−1+ iy/a + Be1+ iy/a
(5.2.10)
with the solution
where a = 12/ is a characteristic length for the problem. The parameter a tells
us how far the viscous effects penetrate from the boundary into the fluid.
Because of the condition at y = 0, we have A + B = U . If we consider the only solid
boundary to be at y = 0, then u must vanish as y becomes large. Since the term that B
multiplies grows exponentially as y increases, then the requirement is met if B = 0 is to
be everywhere finite. Then it follows that A = U and uy t = Ue−1+iy/a + it . Taking
the real part of this gives
uy t = Ue−y/a cos t − y/a
(5.2.11)
If there were a solid boundary at y = b instead, then ub t = 0 and 0 = Ae−1 + ib/a +
. Solving for A and B yields
Be
1 + ib/a
1 + ib − y
sinh
e1+ ib−y/a − e−1+ ib−y/a
a
fy = U
=U
1 + ib
e1+ ib/a − e−1+ ib/a
sinh
a
(5.2.12)
Notice that if a is small, the limit of equation (5.2.12) is f ≈ Ue−1+iy/a , which says
that the fluid motion is confined to a thin region near the wall with thickness of order a.
In fact, equation (5.2.12) shows us that when b ≥ 2a, and since e−2 ≈ 01, for practical
purposes the upper plate can be thought of as being at infinity for such values of a.
The results of Stokes’s second problem are important to the study of pulsatile flows.
There are few analytical tools available to deal with these flows, and Stokes’s solution
at least indicates some of the primary effects that can occur.
Example 5.2.1 Work done in an oscillating flow
An acoustic pressure p0 sin x/c sin t is set up in a gas that has a speed of sound
c. This in turn induces a velocity field U0 cos x/c cos t, where U0 = p0 /c. No
useful work is produced by this, since the pressure and velocity are 90 degrees out of
phase. Show that by introducing a flat plate into the flow and parallel to the velocity
field, useful work is produced.
5.2
Unsteady Flows When Convective Acceleration Is Absent
Solution. Assume the following: The plate is long enough that edge effects can be
ignored; the viscous terms in the Navier-Stokes equations that involve second derivatives
with respect to x can be neglected compared to the terms involving second derivatives with respect to y. The justification of this is that a boundary layer exists. The
equation that will govern the flow and use the preceding assumption on the viscous
terms is
u
p
2 u
= − + 2
t
x
x
This is the counterpart of equation (5.2.1) that, however, also includes the pressure
gradient term. As suggested by Stokes’s second solution and the far field, take
ux y t = U0 1 + fy cosx/ceit px t = −ip0 sinx/ceit
the −i in front of the pressure taking care of the time phase difference. Again, real
values will be taken after solving for f .
Substituting these forms into the governing equation, we have, after cancelling
out the common x and t factors, that iU0 1 + f = −p0 /c + U0 f ′′ . By virtue
of the relation between U0 and p0 , the constant terms cancel. Thus, we are left with
if = f ′′ the same as equation (5.2.9). Again, take the solution that decays as y
increases. Thus, f = Ae−1+iy/a , where a is defined as in Stokes’s second problem.
Applying the boundary condition gives A = −1. Thus,
ux y t = U0 1 − e−1+ iy/a cosx/ceit
which satisfies all boundary conditions. Taking the real part, we have for the final form
of the velocity
ux y t = U0 1 − e−y/a cosy/acost − e−y/a siny/asintcosx/c
The useful local work LW performed over a period T = 2 / is
1 T
LW =
pu dt.
T 0
Putting our expressions for pressure and velocity, we have
1 T
p sinx/c sintU0 1 − e−y/a cosy/a cost
LW =
T 0 0
− e−y/a siny/a sint cosx/cdt
The time terms that involve the product of the sine and cosine will average to zero over
the interval, while the average of the sine squared terms is T/2. Thus, we have
LW = −05e−y/a siny/a cosx/c.
Notice that the term independent of y, which represents the work done in the far field,
dropped out of LW. The total work TW per unit width and length is
TW =
e−y/a siny/a dy = −025p0 U0 cosx/c
LW dy = −05p0 U0 cosx/c
0
0
Thus useful work is done due to the phase shift introduced in the velocity field by
the no-slip condition and the presence of the plate. The results given here are a much
simplified version of a device known as the “acoustic refrigerator.”
151
152
Exact Solutions of the Navier-Stokes Equations
5.3 Other Unsteady Flows When Convective Acceleration
Is Absent
5.3.1
Impulsive Plane Poiseuille and Couette Flows
Several of the flows in the first section of this chapter can also be solved for if they
are unsteady flows. For the plane Poiseuille flow, suppose that the fluid is initially at
rest, and then suddenly the pressure gradients and plate motions are applied with the
walls held still. Then v = uy t 0 0, and a u/ t term must be included in the
acceleration. It is convenient now to let u = us + ut , where us is given by equation
(5.1.2) and ut is the transient solution. Then, by the linearity of the equations, and
getting = /,
ut
2 u
= 2t
t
y
(5.3.1)
and ut = −us at t = 0, with ut = 0 at y = 0 and y = b.
Separation of variables along with application of the boundary conditions yields
ut =
2
An e−n
n=1
2 t/b2
sin
n y
b
(5.3.2)
Applying the initial condition and using the orthogonality of the sine function gives
n y
dy
us sin
b
An =
b
2 n y
sin
dy
0
b
1 p
For stationary walls ux =
− g cos
y2 − by , giving
2 x
b 3
2 p
1 − −1n
An = −
− g cos
b x
n
−
b
0
(5.3.3)
A plot of the developing flow showing velocity profiles as time passes is shown in
Figure 5.3.1.
The solution given by equation (5.3.2) is for a step function increase in the driving
force p/ x − g sin . From this solution the velocity for a driving force that is an
arbitrary function of time can easily be found. Let uu = usu y + utu y t, where the
subscript u stands for our solution (5.3.2) with p/ x − g sin set to unity. For an
arbitrary forcing function p/ x − g sin = ft, Duhamel’s superposition theorem1
can be used to give the velocity
uy t = uu t0 +
t u y
u
ft − d
t0
(5.3.4)
Thus, only a single time integration is needed to find the velocity for the arbitrary
forcing function.
1
Duhamel’s theorem is similar to the convolution theorem encountered in the use of Fourier integrals.
5.3
153
Other Unsteady Flows When Convective Acceleration Is Absent
1.0
0.8
0.6
y /b
0.4
0.2
0.0
u ( y, t )/Umax
Figure 5.3.1 Developing plane Couette flow—two dimensions
5.3.2
Impulsive Circular Couette Flow
For impulsively started transient Couette flow in a circular annulus a ≤ r ≤ b, take
v = 0 vs r + vt r t 0 with vs given by equation (5.2.1). Using the unsteady NavierStokes equations in cylindrical polar coordinates together with separation of variables,
a procedure similar to the preceding yields
2
vt = e−k t Ak J1 kr + Bk Y1 kr
k
where J1 and Y1 are Bessel functions of the first and second kind. Application of the
boundary conditions yields
Ak J1 ka + Bk Y1 ka = 0
Ak J1 kb + Bk Y1 kb = 0
Solving these for A and B, find that
Bk = −Ak J1 ka/Y1 ka
(5.3.5)
J1 kbY1 ka − J1 kaY1 kb = 0
(5.3.6)
and
Equation (5.3.6) determines the characteristic values of k that are to be used in the
summation.
154
Exact Solutions of the Navier-Stokes Equations
For the case where no inner wall is present, 0 ≤ r ≤ b, and as a approaches zero in
equation (5.3.5), Bk also approaches zero and equation (5.3.6) becomes J1 kb = 0. Ak
is determined as before, using vt r 0 = −vs r.
For flow outside a cylinder and extending to infinity b ≤ r ≤ , the problem is
not as simple. Since both Bessel functions J1 and Y1 go to zero as r goes to infinity,
the condition at infinity is automatically satisfied. Thus, only the condition at the wall
need be imposed, which gives equation (5.3.5).
In this case, rather than a Fourier series solution as in equation (5.3.4), there is a
continuous spectrum of k rather than a discrete one, and the summation over the finite
spectrum is replaced by an integral over a continuous spectrum of ks. The result is
2
vt = e−k t
Ak Y1 kaJ1 kr − Y1 krJ1 kadk
(5.3.7)
0
where we have implicitly assumed that k is real—that is, that the solution decays
exponentially in time with no oscillations. Ak is again determined by the initial
condition, which this time requires solution of an integral equation. Some zeros of the
Bessel functions have been listed in tables. Others can be found by numerical integration.
5.4 Steady Flows When Convective Acceleration Is Present
There are a few special flows that allow similarity solutions of the full Navier-Stokes
equations even when the convective acceleration terms are present. Generally these
problems involve specialized geometry and are of an idealized nature. These similarity
solutions, even though they are for idealized geometries, do, however, provide much
of our basic understanding of laminar viscous flows. They also serve both as a starting
point, as well as a validation, of approximate and numerical solutions.
Our analysis of these flows starts with a stream function in all cases. To have a
similarity solution, the stream function will generally be of the form = x y U .
Putting this into dimensionless form, it becomes
n
xU yU
= x F
(5.4.1)
where n = 0 for two-dimensional problems (Lagrange’s stream function) and n = 1 for
three-dimensional axisymmetric problems (Stokes stream function). Here, U = Ux is
a characteristic velocity of the flow and is generally given by the tangential slip velocity
at the boundary found from an inviscid solution of the problem. In the flows for which
similarity solutions hold, F is generally taken to be of the form
xU
F=
f
(5.4.2)
where, reminiscent of Stokes’s first problem, the dimensionless variable is defined by
y xU
yU
xU
=
=
(5.4.3)
x
This form will reduce the nonlinear partial differential Navier-Stokes equations to a
nonlinear ordinary differential equation, which then usually must be solved numerically.
We can show the application of this with several examples.
5.4
155
Steady Flows When Convective Acceleration Is Present
6
5
4
3 y
2
1
–6
–4
–2
0
x
2
4
6
0
Figure 5.4.1 Streamlines for plane stagnation point flow
5.4.1
Plane Stagnation Line Flow
The flow impinging normally against a plane located at y = 0, with axes set so that
there is a stagnation point at (0,0) as in Figure 5.4.1, was first solved by Hiemenz in
1911. The simplest irrotational flow that corresponds to this case is = axy, with a
slip velocity U = ax on the boundary. The constant a has dimensions of reciprocal
time.
The irrotational flow = axy by itself is not particularly interesting. However,
it is a good model for the stream function in the vicinity of a stagnation point. For
instance, it can be shown to be the local value of the stream function for any flow past
an obstacle—for example, a uniform flow past a circular cylinder.
Taking U = ax as the representative velocity from the preceding, we have from
our general form (5.4.1)
ax2
a
xU
f =
f or = x
f
(5.4.4)
=
√
√
where = y a/. We see then that vx = /y = axf ′ vy = −/x = − a/f ,
where primes denote differentiation with respect to .
The no-slip boundary conditions are that vx = vy = 0 at y = 0, requiring that f0 =
f ′ 0 = 0. We will also require that vx approach U far from the plate, so that f ′ must
approach 1 as approaches infinity.
The following quantities are needed for use in the Navier-Stokes equations:
vx
= af ′
x
vy
= 0
x
vx
= xa a/f ′′
y
vy
= −af ′
y
2 vx
= 0
x2
2 vy
= 0
x2
2 vx
xa2 ′′′
=
f
2
y
2 vy
= −a a/f ′′
2
y
156
Exact Solutions of the Navier-Stokes Equations
Notice the following order of magnitude relations for later reference:
⎛ 2 ⎞
⎜ x2 ⎟
vx
⎟
O
= x2 a/ = xU/ and O ⎜
(5.4.5)
⎝ 2 ⎠ = 0
vy
y2
√
Here, xU/ is the local Reynolds number for the flow. These orderings will be used
in Chapter 6 in developing the boundary layer approximation.
Substituting the above into the two-dimensional Navier-Stokes equations, we have
p
axf ′ af ′ + − a/f xa a/f ′′ = − + xa2 f ′′′
x
and
p
axf ′ · 0 + − a/f −af ′ = − + −a a/f ′′
y
Since we started with a stream function, continuity is automatically satisfied.
Solving for the pressure gradients from the preceding equations, we find that
1 p
= xa2 −f ′ 2 + ff ′′ + f ′′′
x
(5.4.6)
and
√
1 p
1 p
= a/
= −a aff ′ + f ′′
y
(5.4.7)
Again, note for later reference that
⎛
⎞
p
⎜ x ⎟
⎟
O⎜
⎝ p ⎠ = Ux/
The second of these equations can be integrated exactly, giving p = −a05f 2 +
f + function of x. For now, let the function of x be written as hx. Substituting the
pressure into equation (5.4.6), the result is
′
dh
= xa2 −f ′ 2 + ff ′′ + f ′′′
dx
Comparing this with equation (5.4.6), we see that the only possibility of having the
variables x and separate is for
h = −05x2 a2 A
giving
f ′′′ + ff ′′ − f ′ 2 = −A = integration constant
(5.4.8)
For the potential flow corresponding to = axy, Bernoulli’s theorem gives the
pressure as
p = −05vx2 + vy2 = −05a2 x2 + y2
(5.4.9)
5.4
157
Steady Flows When Convective Acceleration Is Present
TABLE 5.4.1 Two-dimensional stagnation point flow against a plate (Hiemenz)
00
01
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19
f
f′
f ′′
f
f′
00000
00060
00510
00881
01336
01867
02466
03124
03835
04592
05389
06220
07081
07967
08873
09798
10737
11689
12650
00000
01183
03252
04145
04946
05663
06299
06859
07351
07779
08149
08467
08738
08968
09162
09323
09458
09568
09659
12326
11328
09386
08463
07583
06752
05973
05251
04587
03980
03431
02938
02498
02110
01770
01474
01218
01000
00814
20
22
26
28
30
32
34
36
38
40
42
44
46
48
50
55
60
65
70
13620
15578
19538
21530
23526
25523
27522
29521
31521
33521
35521
37521
39521
41521
43521
48521
53521
58520
63520
09732
09839
09946
09970
09984
09992
09996
09998
09999
10000
10000
10000
10000
10000
10000
10000
10000
10000
10000
f ′′
00658
00420
00156
00090
00051
00028
00014
00007
00004
00002
00001
00000
00000
00000
00000
00000
00000
00000
00000
Since far from the wall the condition f ′ approaches zero implies that f approaches
plus a constant, Bernoulli’s equation and our viscous result for the pressure are seen to
agree far from the wall, providing A is taken to be unity. Then we are left with
vx
′′
2
(5.4.10)
y=0 = xa a/f 0 = 12326U / xU/
xy wall =
y
with
f ′′′ + ff ′′ − f ′ 2 = −1
(5.4.11)
The numerical solution of equation (5.4.8) subject to the boundary conditions was
originally obtained by Hiemenz in 1911 and is given in Table 5.4.1 and Figure 5.4.2.
5
4
3
f
2
f′
1
f″
0
1
3
2
eta
Figure 5.4.2 Hiemenz’s solution for plane stagnation point flow
4
5
158
Exact Solutions of the Navier-Stokes Equations
√
The table shows that for y greater than 24 /a, the difference between the
irrotational solution and the Navier-Stokes solution is less than 1 percent. Thus, when
we required our viscous solution to approach the inviscid solution as approached
infinity, this limit was, for practical purposes, achieved at = 24. It follows that a
constant thickness boundary layer exists in this case, with the boundary layer thickness
in the form
= 24x/ Ux/
(5.4.12)
Inside this boundary layer, viscosity affects the flow in an important manner. Outside
of this region, viscous effects have a negligible effect on our flow.
5.4.2
Three-Dimensional Axisymmetric Stagnation Point Flow
In 1936 Homann provided a three-dimensional analog of the previous two-dimensional
stagnation point case. This time use
vr = −
1
r z
vz =
1
r r
The axisymmetric potential flow = −ar 2 z will be used as our starting point, since it
has a downward velocity toward the plate of vz = −2′ ar and a radial velocity of vr = az.
Thus, we have U = ar for the slip velocity along the plate and by Bernoulli’s equation
a pressure of
4r 2 + z2
2
Again, a is a constant with dimensions of reciprocal time. Letting
√
= −r 2 af = z a/
p = −a2
we can carry out the calculations in the same manner as in the two-dimensional case.
The final result of these manipulations is the equation
f ′′′ + 2ff ′′ − f ′ 2 = −1
(5.4.13)
which differs from equation (5.4.8) only by the number 2 multiplying the term containing
the second derivative. The numerical results satisfying the boundary conditions f0 =
f ′ 0 = 0 f ′ approaching 1 as gets large, differ little from the preceding case, as can
be seen from Table 5.4.2 and Figure 5.4.3.
5.4.3
Flow into Convergent or Divergent Channels
One interesting flow that shows a form of flow separation is that between two nonparallel
plates an angle apart, as shown in Figure 5.4.4. This flow was first studied by Jeffrey
(1915) and elaborated further by Hamel (1917). A source or a sink is placed where the
two planes meet. To use our standard form (5.4.2) for the stream function, we work in
cylindrical polar form and recognize that the inviscid flow in this case is like a source or
sink, with streamlines being straight radial lines passing through the source/sink. From
continuity considerations our radial velocity component then is proportional to 1/r, and
the stream function for the channel must therefore be of the form
= F Q/
(5.4.14)
5.4
159
Steady Flows When Convective Acceleration Is Present
TABLE 5.4.2 Three-dimensional stagnation point flow against a plate (Homann)
00
01
02
03
04
05
06
07
08
09
10
11
12
13
f
f′
f ′′
f
f′
00000
00064
00249
00545
00943
01432
02003
02647
03354
04116
04925
05774
06655
07564
00000
01262
02424
03487
04451
05317
06088
06767
07359
07868
08300
08663
08962
09205
13121
12121
11123
10129
09147
08183
07247
06347
05494
04696
03961
03295
02702
02182
14
15
16
17
18
19
20
22
24
26
28
30
32
08495
09443
10405
11377
12357
13343
14334
16323
18319
20318
22318
24319
26321
09401
09555
09675
09766
09835
09886
09923
09968
09989
09999
09999
09999
09999
5
4
3
f
2
f′
1
f″
0
1
3
2
eta
Figure 5.4.3 Homann’s solution for axisymmetric stagnation point flow
α
Source/sink
Figure 5.4.4 Flow in a converging–diverging channel
4
5
f ′′
01736
01359
01046
00793
00591
00433
00221
00154
00072
00032
00014
00007
00004
160
Exact Solutions of the Navier-Stokes Equations
the radial coordinate r dropping out of the list, since it cannot be combined with the other
variables to form a dimensionless parameter. Thus, only a radial velocity component is
present.
This velocity is of the form vr = r
f , with f = dF
, which automatically
d
satisfies continuity. Using primes to denote differentiation with respect to , find then that
vr
= − 2 f
r
r
vr
= f ′
r
r
2 vr
2
= 3 f
r 2
r
2 vr
= f ′′
r 2
r
The Navier-Stokes equations in cylindrical polar form in this case are
2
1 p
p
vr 1 vr
1 2 vr vr
2 vr
v
+
+
−
+
0
=
−
vr r = − +
r
r
r 2
r r
r2 2
r2
r
r2
giving for the pressure gradients
2
1 p
= 2 3 f 2 + f ′′
r
r
1 p
22
= 2 3 f ′
r
r
Integration of these gives
2
p 2 2
= 2 f + hr = − 2 f 2 + f ′′ + H
r
2r
(5.4.15)
Comparing terms on the left and right sides of this equation, we can see that hr
has to be of the form 2 A/r 2 , where A is a constant, and H has to be a constant that
can arbitrarily be set to zero. The 1/r 2 dependency of p is consistent with the pressure
field for an inviscid flow far from a source/sink. Thus,
p=
2
2
2f
f 2 + f ′′
+
A
=
−
r 2
2r 2
(5.4.16)
and
f ′′ + f 2 + 4f + 2A = 0
(5.4.17)
05f ′ 2 + 1/3f 3 + 2f 2 + 2Af + B = 0
(5.4.18)
Equation (5.4.17) can be integrated once if we first multiply it by f ′ . The result is
where B is a second constant of integration.
Before going further with the solution, consider the boundary conditions. From the
no-slip conditions, f must vanish at both walls of the wedge. Thus, from equations
′
′
′′
(5.4.16) and (5.4.18), A = −fwall
nor
2 . Unfortunately neither fwall
and B = −05fwall
′′
fwall are known.
We could try to express our constants in terms of the discharge per unit length
into the paper, Q = rvr d = fd , the integral being over the angle of the wedge.
Q would normally be a given quantity but we would first have to know f to be able to
carry out the integration.
Alternately, we could interpret A in terms of the pressure along a wall, so A =
r 2 pwall /2 , where pwall varies like 1/r 2 . This perhaps makes the most sense, as it is
the pressure gradient that drives the flow.
Traditional procedure has been to note that equation (5.4.30) is a first-order equation
and is one where the variables can be separated. This gives
df
=
+ C
(5.4.19)
2
−2B − 4Af − 4f 2 − f 3
3
5.4
161
Steady Flows When Convective Acceleration Is Present
where C is still another constant of integration. The three constants A B, and C are to
be determined such that the no-slip and discharge conditions
f± = 0 Q =
rvr d
(5.4.20)
−
where Q would normally be a given quantity.
This solution can be rewritten in the form of elliptic integrals, but the interpretation
of the result is still complicated, since it gives as a function of f rather than vice
versa, and either numerical integration or tables are necessary to interpret the elliptic
integrals. Further, at a point of maximum f the integrand become singular and the sign
of the square root changes.
Two separate approaches have been used to carry out computations of equation
(5.4.19). In 1940 Rosenhead made the substitution
A = −ab + ac + bc/6
B = −abc/3
where a + b + c = −6
This enabled him to factor the cubic terms under the square root in equation (5.4.3).
The following conclusions have been reached for various ranges of these constants:
• If the constant a is real and positive, the constants b and c are complex conjugates
of each other. In this case, the flow will have a positive velocity (sink flow)
throughout.
• If the constants a b, and c are real, with a > b > c, and a > 0 c < b < 0, there
can be one or more regions of reversed flow.
• Rosenhead also found that for a given wedge angle and value of Q/, there is an
infinite number of solutions. Also, for a given number of inflow-outflow regions
and wedge angle there is a critical value of Q/ above which the solution
doesn’t exist.
• Millsaps and Pohlhausen (1953), using a slightly different approach, made further
calculations and found that for pure outflow problems as Q/ is increased, the
flow concentrates more and more in the center of the wedge.
• If Q < 0 (sink flow), f is symmetric about a center line and always negative.
There is no flow reversal (Figure 5.4.5a).
• If Q > 0 (source flow), too large a value of Q/ for a given wedge angle can
result in unsymmetric flow with reversed flow near one or more walls. The nature
of the flow has to do with the question of whether the roots of the cubic term
under the square root sign in equation (5.4.30) are real or imaginary equation
(Figure 5.4.5b).
α
Sink
Figure 5.4.5a Velocity profile for a converging channel
162
Exact Solutions of the Navier-Stokes Equations
α
Source
Figure 5.4.5b Velocity profile for a diverging channel
• Most of the simpler two-dimensional flows have three-dimensional counterparts.
Interestingly, this one does not! (Try a radial velocity inversely proportional to
R2 in spherical coordinates to convince yourself.) This suggests that flow into an
ideal conical funnel must be much more complicated than the simple radial flow
one might expect.
This seemingly simple problem with an exact solution (if you are willing to accept
that elliptic integrals are “well-known functions”) illustrates the complexities that can
arise from solutions of the Navier-Stokes equations.
An alternate approach to the solution of this equation for those who have access to
personal computers involves straightforward numerical integration of equation (5.4.18)
after suitable choice of the constants A and B. Select values for A and B, and then
integrate equation (5.4.18) numerically, using, say, a Runge-Kutta scheme, continuing
the integration until f ′ becomes zero. You may consider the corresponding angle to
be an acceptable value of the wedge angle, or you may continue if you wish to allow
reversed flow. Numerical integration of f according to equation (5.4.31) will give you
the discharge for your choice of parameters. Since there are no square roots in this
procedure, it is not necessary to test for signs or patch solutions together.
If you want your constants to give a pure source flow, the integration should proceed
until f ′ becomes zero, which would be at the location of the center line of the wedge.
The flow past the center line is a mirror image of this flow. After you have done many
computations for values of the constants, you should have a better understanding of
these flows.
5.4.4
Flow in a Spiral Channel
Hamel also considered flow into a channel, which is in the shape of a logarithmic spiral
whose shape is given in cylindrical polar coordinates by A + B ln r = constant. Like
the approach in the previous section, he let
2
P(), where = A + B ln r
(5.4.21)
= f and p = p0 +
r
Letting primes denote differentiation with respect to , the radial and theta velocity
components are
vr =
1 A ′
=
f
r
r
v =−
B
B
= − f ′ = − vr
r
r
A
5.4
163
Steady Flows When Convective Acceleration Is Present
Substitution of these into the Navier-Stokes equations leads to
2
−2P + BP ′ = A2 + B2 f ′ + Af ′′′
AP ′ = A2 + B2 2f ′′ − Bf ′′′
(5.4.22)
From these the pressure is found to be
P=
A2 + B 2
2
−A f ′ + 2Bf ′′ − A2 + B2 f ′′′
2A
(5.4.23)
and the governing equation for the stream function is
A2 + B2 f iv − 4Bf ′′′ + 4 + 2Af ′ f ′′ = 0
This can be integrated once to give
A2 + B2 f ′′′ − 4Bf ′′ + 4 + Af ′ f ′ = c1
(5.4.24)
which is second order in f ′ . The boundary conditions are that f ′ vanish on the walls—
say, = 0 and 1 . The remainder of the solution procedure is much the same as in the
previous section.
5.4.5
Flow Due to a Round Laminar Jet
The solution for the round laminar axisymmetric jet was first presented by Landau
(1944) and later rediscovered by Squire (1951). Working in spherical coordinates they
considered a (vanishing) thin tube injecting fluid at the origin with a stream function
=
Rf where = cos
(5.4.25)
(The use of cos rather than itself simplifies the following equations considerably.)
Then the velocity components are given by
vR =
1
df
=−
R2 sin
R d
f
1
=−
v = −
R sin R
R sin
(5.4.26)
For axisymmetric flow the Navier-Stokes equations are
v vR v 2
v
p
2v
2 v 2v cot
vR R +
−
+ 2 vR − 2R − 2
−
=−
R
R
R
R
R
R
R2
v
v v v vR v
2 vR
1 p
2
+
+
+ v + 2
−
=−
vR
R
R
R
R
R R2 sin2
(5.4.27)
1 R2 vR
1 v sin
+
= 0
R2 R
R sin
1
1
2
2
R
+ 2
sin
where = 2
R R
R
R sin
164
Exact Solutions of the Navier-Stokes Equations
These equations can be considerably simplified by recognizing for this problem that
from the form of the stream function, the following holds:
vR
v
= − R
R
R
v
v
=−
R
R
Using these and some tedious but straightforward calculus, the Navier-Stokes equations
become
vR2 + v2 v sin vR
p
2 vR
−
−
+ 2
=−
R
R
R R 2
(5.4.28)
v
p vR
−v
=− −
R
The second of these can be integrated to obtain
v2 vR v c1
+ 2
p − p0 = − +
2
R
R
(5.4.29)
Here, p0 is the constant pressure at infinity. Since R was held constant during the
integration, the “constant of integration” c1 is in fact a function of R. The radial
momentum equation dictates that it be proportional to R−2 .
Using the form for the pressure, inserting it into the radial momentum equation,
and using the stream function, the result is
′
2
f ′ + ff ′′ = 2f ′ + 1 − 2 f ′′ − 2c1
Primes here denote differentiation with respect to . Integrating this once gives
ff ′ = 2f + 1 − 2 f ′′ − 2c1 − c2
and once more gives
1 2
f = 2f + 1 − 2 f ′ − c1 2 − c2 − c3
2
(5.4.30)
with boundary conditions f0 = f = 0.
Squire solved this for the special case c1 = c2 = c3 = 0, obtaining
f=
21 − 2
,
a+1−
(5.4.31)
where a is a parameter representing the strength of the jet.
The momentum flux M in the x direction that passes through the surface of a sphere
centered at the origin is
M=
vR vR cos − v sin 2 R2 sin d
0
(5.4.32)
a+2
2 32a + 1
− 16a + 1 + 8aa + 2 ln
=2
3aa + 2
a
5.4
165
Steady Flows When Convective Acceleration Is Present
and the force component for the same sphere is
"
!
v 1 vR
v
Fz = −
+
sin 2 R2 sin d
−p + 2 R cos − R
R
R R
R
0
(5.4.33)
a+2
2 2
2
24a + 1 − 12a + 24a + 4 ln
=
a
The ratio
M + Fz
=2
2 /
32a + 1
a+2
+ 8a + 1 − 4a + 12 ln
3aa + 2
a
(5.4.34)
gives a measure of the momentum strength of the jet. For values of a = 1, 0.1, and 0.01,
this ratio takes on the values 34.76, 314.0, and 3,282. Whether there are other important
combinations of the three constants of integration is not resolved by this analysis.
5.4.6
Flow Due to a Rotating Disk
One of the very few similarity solutions that involve all three velocity components was
found by von Kármán in 1921 for flow caused by a large rotating disk in a stagnant
fluid. We alter our previous “general” form for the stream function by replacing U
by r, where is the angular speed of the disk. Following our form for the general
similarity solution and remembering that for axisymmetric flows the stream function
has dimensions of length cubed over time, one length dimension higher than in the
two-dimensional case, we have
rU/f = −r 2 /f
(5.4.35)
= −r
where
= zU// rU/ = z /
(5.4.36)
v = rg
(5.4.37)
(We have also altered our general solution in one other trivial detail: We have put a
minus sign in because we expect to have a downward vertical velocity component
directed toward the disk, and it is convenient to have f positive.) The appropriate form
for the swirl velocity component to accompany the stream function was found by von
Kármán to be
Differentiation of the velocity components gives
vr = rf ′ v = rg vz = −2 /f
vr
r
2 vr
r 2
vr
z
2 vr
z2
v
vz
= g
= 0
r
r
2 v
2 vz
= 0
=
0
= 0
r 2
r 2
v
vz
= r /f ′′
= r /g ′
= −2f ′
z
z
r2 ′′′
2 v
r2 ′′
2 vz
=
f
=
g
= −2 /f ′′
2
2
z
z
= f ′
166
Exact Solutions of the Navier-Stokes Equations
Substituting the preceding results into the Navier-Stokes equations in cylindrical polar
form gives
p
= r2 −f ′ 2 + 2ff ′′ + g 2 + f ′′′
r
0 = −2f ′ g + 2fg ′ + g ′′
1 p
p
=
= −2 / 2ff ′ + f ′′
z
(5.4.38)
f ′′′ + 2ff ′′ − f ′ 2 + g 2 = 0
(5.4.39)
The no-slip boundary conditions at the wall give vz 0 = vr 0 = 0, or equivalently
f0 = f ′ 0 = 0 g0 = 1. As becomes large (far away from the wall), the viscous
solution must approach the inviscid flow in having vr and v approach zero (equivalently,
f ′ and g approach zero). Notice that the relative scalings of the various velocities and
the velocity and pressure gradients are as previously noticed. Since the flow is entirely
driven by the disk, the flow away from the disk is very weak. The disk is in fact acting
as a rather inefficient centrifugal pump.
Elimination of the pressure in the preceding forms results in
g′′ + 2fg′ − 2gf ′ = 0
(5.4.40)
and
p = −2f 2 + f ′
(5.4.41)
The numerical solution of this system of equations (5.4.39) and (5.4.40) subject
to the boundary conditions is shown in Figure 5.4.6. and Table 5.4.3 The unknown
derivatives at the wall are found to have the values g ′ 0 = −06159 and f ′′ 0 = 0510.
1.0
g
0.5
f′
f
0.0
g′
f″
–0.5
–1.0
0
1
2
3
η
Figure 5.4.6 Von Kármán’s solution for a rotating disk
4
5
6
5.4
167
Steady Flows When Convective Acceleration Is Present
TABLE 5.4.3 Rotating disk boundary layer (von Kármán)
f
f′
f ′′
g
00
01
02
03
04
00000
00024
00090
00189
00314
00000
00462
00836
01133
01364
05102
04163
03338
02620
01999
10000
09386
08780
08190
07621
06159
06112
05987
05803
05577
05
06
07
08
09
10
00459
00620
00790
00967
01147
01327
01536
01660
01742
01789
01807
01802
07076
06557
06067
05605
05171
04766
05321
05047
04763
04476
04191
03911
11
12
13
14
15
16
01506
01682
01853
02019
02178
02331
01777
01737
01686
01625
01559
01487
01467
01015
00635
00317
00056
−00157
04389
04038
03712
03411
03132
02875
03641
03381
03133
02898
02677
02470
17
18
19
20
22
24
26
02476
02613
02743
02866
03089
03284
03454
01413
01338
01263
01188
01044
00910
00788
02638
02419
02218
02033
01708
01433
01202
02276
02095
01927
01771
01494
01258
01057
28
30
30
32
34
36
03600
03726
03726
03834
03925
04003
00678
00581
00581
00496
00422
00358
01008
00845
00845
00708
00594
00498
00888
00745
00745
00625
00525
00440
38
40
42
44
46
48
04069
04125
04172
04212
04246
04274
00303
00257
00217
00183
00154
00129
00417
00349
00293
00245
00205
00172
00369
00309
00259
00217
00182
00152
50
04298
00109
00144
00128
−00327
−00461
−00564
−00640
−00693
−00728
−00747
−00754
−00751
−00739
−00698
−00643
−00580
−00517
−00455
−00455
−00397
−00343
−00296
−00253
−00216
−00184
−00156
−00132
−00112
−00095
−g ′
√
Since f = 04422, the flow far above the plate is vz = −08845 . The boundary
layer thickness, again constant since is independent of r, is approximately
(5.4.42)
= 5 /
168
Exact Solutions of the Navier-Stokes Equations
Even though our result is for an infinite disk, we can find the moment over a finite
radius R of one side of the disk by
R
R
r 3 dr = −05 R4 /g ′ 0
r z z = 0 2 r dr = − 2 /g ′ 0
M =−
0
0
Since − g 0 = 1935, we have for a moment coefficient
′
CM =
M
1
2 R5
2
=
1935
R2 /
(5.4.43)
In practice, the flow is found to be laminar for values of the Reynolds number
R2 / less than 105 , and turbulent for larger Reynolds numbers.
Problems—Chapter 5
5.1 Two fluids of different densities and viscosities are flowing down a plane with
slope . The lower layer of fluid occupies the region 0 ≤ y ≤ a and has density and
viscosity 1 and 1 , while the upper fluid occupies the region a ≤ y ≤ b and has density
and viscosity 2 and 2 . The lower fluid has the higher density. The surface at y = b is
open to the atmosphere. Find the pressure and velocity distributions.
5.2 A layer of viscous fluid flows down a vertical plate under the action of gravity.
The density of the fluid varies linearly from 1 at the wall to 2 < 1 at y = b, the free
surface. Find the velocity distribution and the velocity at the free surface.
5.3 Find the flow in a rotating pipe of radius a. A pressure gradient is present.
Take the flow to be axially symmetric and steady.
5.4 Verify that the velocity in a conduit with an elliptic cross-section with semimajor
and semiminor axes a and b is given by
2
z2
1 p a2 b2
y
+
−
1
where
K
=
vx = K
a2 b 2
2 x a2 + b2
5.5 For flow in the conduit with an elliptic cross-section (problem 5.4), find the ratio
b/a that gives the maximum flow rate for a given pressure gradient and cross-sectional
area.
5.6 Verify that the laminar velocity in a pipe with the cross-section of an equilateral
triangular of side b is given by
√
√
√
√
√
3 p
3b
3b
z z + 3y −
z − 3y −
vx = −
6b x
2
2
5.7 Using the solution for flow between concentric rotating cylinders, find the
velocity distribution caused by a circular cylinder that is rotating in an infinite fluid.
Show that this is the same as that due to a line vortex along the z-axis. What is the
strength of this vortex?
5.8 Find the flow of a layer of viscous fluid on an oscillating plate U0 cos t at
y = 0. The fluid is of constant thickness a; the surface at y = a is stress-free.
5.9 A large disk is rotated sinusoidally at a frequency in a semi-infinite fluid.
Find the steady-state fluid velocity, assuming vr z t = 0 rfz t 0.
Problems—Chapter 5
5.10 A concentrated line vortex of the type discussed in Chapter 2 is suddenly
introduced into a viscous fluid. The action of viscosity is to diffuse the vorticity. The
problem to be solved for is thus the following:
#
2
1
0 for r > 0
r
+
2 rdr =
=
0
=
t
r 2
r r
for r = 0
0
a. Show the form of a similarity solution = r t that is a possible
solution for this problem.
b. Solve for the vorticity.
5.11 Show that the form r t = At f, where = r
, when inserted into the
t
2
1
=
+
leads to an ordinary differential equation.
vorticity equation
t
r 2
r r
169
Chapter 6
The Boundary Layer Approximation
Introduction to Boundary Layers 170
The Boundary Layer Equations 171
Boundary Layer Thickness 174
Falkner-Skan Solutions for Flow Past
a Wedge 175
6.4.1 Boundary Layer on a
Flat Plate 176
6.4.2 Stagnation Point Boundary
Layer Flow 178
6.4.3 General Case 178
6.5 The Integral Form of the Boundary
Layer Equations 179
6.6 Axisymmetric Laminar Jet 182
6.1
6.2
6.3
6.4
6.7 Flow Separation 183
6.8 Transformations for Nonsimilar
Boundary Layer Solutions 184
6.8.1 Falkner Transformation 185
6.8.2 von Mises Transformation 186
6.8.3 Combined Mises-Falkner
Transformation 187
6.8.4 Crocco’s Transformation 187
6.8.5 Mangler’s Transformation for
Bodies of Revolution 188
6.9 Boundary Layers in Rotating
Flows 188
Problems—Chapter 6 191
6.1 Introduction to Boundary Layers
As we saw in Chapter 5, solutions to the full Navier-Stokes equations are few in number
and difficult to obtain. In the exact solutions of the Navier-Stokes equations, it was
repeatedly seen that when a local Reynolds number was large, viscous effects are felt
mainly in the immediate vicinity of a solid boundary. In 1904 Prandtl introduced an
approximate form of the Navier-Stokes equations that holds in the thin boundary layer
near the wall, where viscous effects are comparable to the inertia effects. Here, the
relationship between the boundary layer equations and the full Navier-Stokes equations
is investigated by demonstrating how the boundary layer equations can be derived as a
limiting form of the Navier-Stokes equations.
To understand how inviscid flow theory and boundary layer theory fit with the
Navier-Stokes equations, consider the following example, first presented by Friedrichs
in 1942. His model equation is
170
d2 f df
+
= a
dx2 dx
f0 = 0
f1 = 1
(6.1.1)
6.2
171
The Boundary Layer Equations
The second-order derivative can be thought of as the viscous terms with a small viscosity
, the first derivative as a “momentum,” and the constant a as a pressure force. When the
highest-order derivative is neglected (“inviscid” approximation), the first-order equation
that remains has the solution f = ax + c c being a constant of integration. Clearly, only
one of the boundary conditions can be satisfied. Our version of relaxing the “no-slip”
condition would be to impose f1 = 1, giving c = 1 − a. Thus, our “inviscid” solution is
f1 x = 1 + ax − 1
(6.1.2)
To take care of the unsatisfied boundary condition, recall that in all of the similarity
solutions the coordinate was “stretched” by dividing by the small viscosity, giving a
thin layer at the wall. For this example, let = x/, so that our equation becomes
1 d2 f 1 df
+
= a
d2 d
The temptation now is to disregard the right-hand side and find
f = b + ce−
To satisfy the boundary condition at x = 0, select c = −b, so
f2 = b 1 − e−
(6.1.3)
To “match” the two solutions, require next that there is some region where the two
solutions overlap. In that region, write both solutions in one of the two variables (x and
), and then take a limit, the limit being either x → 0 or → , depending on which
variable was chosen.
Frequently, it is simpler to choose as the variable of choice. In that case our first
solution is f1 x = 1 + a − 1. To the lowest-order, b = 1 − a and
1 + ax − 1 away from the wall
−
f2 = 1 − a 1 − e ×
(6.1.4)
1 − a 1 − e− near the wall
Putting this in the context of fluid mechanics, to the lowest-order approximation
as the Reynolds number becomes large = 1/Re → 0, a good approximation to the
solution of the Navier-Stokes equations is to solve the Euler equations for the slip
velocity on the boundary and use Prandtl’s boundary layer approximation for the flow
near the wall.
Other forms of boundary layers exist as well. As we will see in Chapter 8, slow
flows at low Reynolds numbers also require similar handling. Boundary layers can also
exist as shear layers in the interior of a region—for example, the Gulf Stream where
it departs the U.S. coast and crosses the Atlantic Ocean. Other areas of physics exhibit
similar phenomena, such as the behavior of the free sides of a thin elastic plate when
it is bent and current flowing in a solid wire at high frequencies. For more detailed
explanations of the matching process, see Lagerstrom and Cole (1955), Goldstein (1960),
and Van Dyke (1964).
6.2 The Boundary Layer Equations
In all of our exact solutions of the Navier-Stokes equations it was seen that the pressure
gradient along a wall was of greater magnitude than that of the pressure gradient
172
The Boundary Layer Approximation
perpendicular to the wall, and that the viscous terms involving second derivatives
along the wall were smaller than those involving derivatives taken perpendicular to
the wall. The continuity equation was always satisfied in full, whereas the momentum
equation in the direction perpendicular to the wall introduced only very low orders of
magnitude terms.
To help in deriving the boundary layer equations, scale the various terms in the
Navier-Stokes equations so that these orders of magnitude hold true, particularly when
the Reynolds number is large. As a bookkeeping scheme it is convenient to build these
notions as to the orders of magnitude of the various terms into our dimensionalization so that these orderings appear automatically. To do this, introduce dimensionless
coordinates
√
(6.2.1)
x = xD /L y = yD Re/L t = tD /T
with the Reynolds and Strouhal numbers given by
Re = UD L/
and
St = L/UD T
(6.2.2)
Here, xD yD are the dimensional lengths along and perpendicular to a wall, UD is a
representative constant body or stream velocity, T is a time scale for unsteady effects,
and L is a body length along the wall. The existence of constant L and UD , and hence
of a constant Re, is vital to what follows, as will be discussed later.
Since changes along the body occur on a length scale L, whereas those perpendicular
to the body
√ occur over a distance of the order of the (thin) thickness of the boundary
introduced in equation (6.2.1) “stretches”
layer L/ Re, the biased dimensionalization
√
the thin yD by multiplying it by Re to make the partial derivatives reflect their true
orders of magnitude.
Introducing a dimensionless stream function (as done in the similarity solutions of
the full Navier-Stokes equations) by
√
Re
(6.2.3)
= D/
gives dimensionless velocities
1
u
= D
y
UD y
UD
√
√ v
Re
=−
= Re D
vy = −
x
UD x
UD
vx =
=
(6.2.4)
(6.2.5)
These dimensionless velocities bring out the physical fact that, at the outer edge of the
boundary
layer uD will be of magnitude UD , and vD will be of the (small) magnitude
√
UD / Re. Our stretching of the coordinates, then, along with the nondimensionalization
of , has stretched the velocity components appropriately as well.
Making pressure dimensionless by
p = pD / UD2
(6.2.6)
the dimensionless Navier-Stokes equations for incompressible flow with constant density
and viscosity in two-dimensional Cartesian coordinates become
St
2
v
v
p
v
1 2 vx
vx
+ vx x + vy x = − + 2x +
t
x
y
x
y
Re x2
(6.2.7)
6.2
173
The Boundary Layer Equations
and
St
2
vy
vy
vy
vy
1 2 vy
p
+ vx
+ vy
= −Re + 2 +
x
y
Re x2
t
x
y
(6.2.8)
For large values of the Reynolds number, the limit of equations (6.2.7) and (6.2.8) is
St
2
vx
v
v
p
v
+ vx x + vy x = − + 2x
t
x
y
x
y
(6.2.9)
and
p
y
0=−
(6.2.10)
Equation (6.2.10) states that the pressure gradient across the thin boundary layer is
negligible, and thus the pressure gradient at the outer edge of the boundary layer, as
obtained from the Euler equations, can be used. Since the inviscid velocity component
perpendicular to a wall vanishes, this means that the Euler equations at the wall are
p
U
U
+U
(6.2.11)
= St
t
x
x
where U is the velocity along the wall (made dimensionless by UD ) as predicted from
inviscid theory. Thus, in equation (6.2.9) the pressure gradient is known, and, since v is
related to u through continuity, there is only one equation in one unknown to be solved.
The appropriate boundary conditions are the no-slip conditions at the wall,
−
vx = vy = 0
at
y = 0
u=v=0
at
y = 0
(6.2.12)
and a joining to the inviscid velocity—that is,
vx → U
(6.2.13)
at the outer edge of the boundary layer.
Having developed the boundary layer equations in dimensionless form, usually it
is more convenient when doing problems to start with the dimensional form. This is, in
two dimensions with v = u v 0,
2
u
u
u
U
U
u
+u +v =
+U
+ 2
t
x
y
t
x
y
p
= 0
y
(6.2.14)
(6.2.15)
and
u
v
+
= 0
x
y
(6.2.16)
with u = v = 0 on the wall and u approaches U at the outer edge of the boundary layer.
Mathematically, the outer edge of the boundary layer can be defined as being at
an infinite value of dimensionless y, since in the stretching of the y coordinate the
Reynolds number has been made large. In practice, a value for dimensionless y of five
or so can often be regarded as infinity for practical purposes, since, as has been seen in
the previous similarity solutions, u will not vary much for values of y beyond that point.
Our boundary layer equations have been derived for a flat boundary. For curved
boundaries, providing x is regarded as being locally tangent to the boundary and y as
174
The Boundary Layer Approximation
being locally normal to the boundary, the equations still hold, since curvature effects
are of higher order unless the curvature is extreme.
In approximating the Navier-Stokes equations by the boundary layer equations,
the mathematical character of the equations have changed. The original Navier-Stokes
equations are of elliptic nature, while the boundary layer equations are parabolic. (The
boundary layer equations are sometimes referred to as the parabolized Navier-Stokes
equations.) Equations of parabolic type have solutions that can be “marched” in the
x direction, and the numerical methods appropriate for parabolic equations are quite
different from those used for elliptic equations. The implication now is that upstream
conditions completely determine downstream behavior. When this is not true, for example, where a flow starts to separate, the boundary layer assumption breaks down.
6.3 Boundary Layer Thickness
In Chapter 5 where exact solutions were being considered, boundary layer thickness
was established by first finding the solution and then considering for what value of
the horizontal velocity component approached some percentage of the outer velocity.
For both theoretical and experimental purposes, it is preferable to have a less subjective
definition of “thickness.” Two such definitions have been found to be useful.
The displacement thickness of the boundary layer is defined by
1
U − udy = displacement thickness
(6.3.1)
D =
U 0
where is a value “large enough” so that choosing a slightly larger value would not
significantly affect the value of D . Looking at Figure 6.4.1 (on page 177), we can
see that the integral represents the area U minus the actual discharge in the layer of
thickness . As the value of is increased, there comes a point where no significant area
is added to the integral, so whether 90%, 99%, or evaluation by eye of experimental
data is used, the value of the displacement thickness remains virtually unchanged.
Another interpretation of the displacement
thickness is that because the volumetric
flow in the boundary layer is actually 0 udy, the same flow would be achieved if the
wall was displaced upward into the flow in an amount D , since the definition can be
written in the form
− D U =
udy
0
Similarly, a momentum thickness definition has been found to be useful. It is
given by
1
uU − udy = momentum thickness
(6.3.2)
M =
U2 0
Again, its value is relatively insensitive to increases in as long as a large enough
is chosen. Its physical interpretation is a bit more difficult to explain than the previous
definition, as it involves the momentum difference between the actual flow and an
inviscid one with the same discharge, as follows from
U
u2 dy
udy − U 2 M =
0
0
As we will see in Section 6.5, both these quantities arise naturally when the boundary
layers are integrated over the boundary layer thickness.
6.4
175
Falkner-Skan Solutions for Flow Past a Wedge
6.4 Falkner-Skan Solutions for Flow Past a Wedge
To illustrate the behavior of the boundary layer equations, next consider a solution of
similarity type first credited to Falkner and Skan (1930). This is a solution where U
is proportional to xm . The irrotational flow that corresponds to this is the flow past a
wedge of angle radians with a complex potential and velocity given by
Ar m + 1 cos m + 1 + i sin m + 1
Azm + 1
=
(6.4.1)
m+1
m+1
dw
= u − iv = Ar m cos m + i sin m
(6.4.2)
dz
Evaluating equation (6.4.1) on the surface of the wedge = 0 gives U = Axm . From
the Euler equations,
w = +i =
p
dU
= U
= mA2 x2m − 1
x
dx
For the similarity form of the solution use
Ux/ f with = y U/x
=
−
as in the full Navier-Stokes equations. Then with
m−1 d
=
x
2 x d
from differentiation of
y
=
U d
x d
find that
u = Uf ′
1
v=−
2
u U
= mf ′ + 05m − 1f ′′
x
x
U
x
m + 1f + m − 1f ′
u
U/ xf ′′
=U
y
2
u
U 2 ′′′
=
f
y2
x
Substituting all of these into the boundary layer equations, the result is
2
m f ′ − 05m + 1ff ′′ = m + f ′′′
The terms on the left-hand side result from the convected acceleration, the constant on
the right side results from the pressure gradient, and the third derivative is the viscous
stress gradient. Rearranging to put all terms on the left-hand side, the result is
f ′′′ + 05m + 1ff ′′ + m1 − f ′ 2 = 0
(6.4.3)
Equation (6.4.3) is to be solved together with the no-slip boundary conditions
f0 = f ′ 0 = 0
(6.4.4)
f ′ approaches 1 as becomes large
(6.4.5)
and the outer condition
The shear stress at the wall is
wall =
u
U/x f ′′ 0
= U
y y=0
Consider next several special cases of equation (6.4.6).
(6.4.6)
176
The Boundary Layer Approximation
6.4.1
Boundary Layer on a Flat Plate
The first solution of equation (6.4.3) was carried out by Blasius (1908) for a semi-infinite
flat plate, where m = 0. Then equation (6.4.3) reduces to
f ′′′ + 05ff ′′ = 0
(6.4.7)
Equation (6.4.7) has the minimum number of terms that equation (6.4.3) can have for a
boundary layer flow, but still no exact solution is possible. A numerical solution must
be resorted to, and since boundary conditions must be applied at both the wall and the
outer edge of the boundary layer, generally our choice is either to use a shooting method
(first guess f ′′ 0, then integrate out from the wall to see if f ′ approaches unity, then
iterate on this procedure) or use a two-point method that has the boundary conditions
at both ends included. Blasius found that f ′′ 0 = 0332.
For this special problem only, numerical integration using a shooting method can
be simplified. First, make the change of variables = c and f = cF so that equation
(6.4.7) is replaced by
F ′′′ + 05FF ′′ = 0
(6.4.8)
where this time a prime is a derivative with respect to .
Set F ′′ 0 arbitrarily to unity,√and integrate to find that F ′ approaches some value
b at large . Then, letting c = 1/ b will make f ′ approach unity as becomes large,
completing the desired solution. The results of such a numerical integration are shown
in Table 6.4.1 and Figure 6.4.1.
Several interesting features of boundary layer flows can be seen from the results of
this problem.
1. Since the outer edge of the boundary layer is at some constant value of , the
exact value √
depending on how the boundary layer thickness has been defined,
then ∼ x/ Ux/ .
2. As becomes large, f approaches the asymptotic value − 17208. Thus, the
displacement thickness is
1
=
/ Ux
U − udy = 17208x
D
U 0
3. The shear stress is proportional to x−1/2 , so the shear stress decreases going
downstream from the plate leading edge. This decrease is due to the thickening
of the boundary layer, with a resulting decrease in velocity gradient.
4. The drag force per unit width over a length L of the plate is
L
wall dx = 2 Uf ′′ 0 UL/
F=
0
5. From the asymptotic value of f , the stream function is seen to approach
= Uy − 17208 Ux / + · · ·
Thus, the displacement thickness effect does not result in an irrotational flow
away from the wall.
6.4
177
Falkner-Skan Solutions for Flow Past a Wedge
TABLE 6.4.1 Flat plate boundary layer (Blasius)
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
19
f
f′
f ′′
f
f′
f ′′
00000
00017
00066
00149
00266
00415
00597
00813
01061
01342
01656
02002
02379
02789
03230
03701
04203
04735
05295
05884
00000
00332
00664
00996
01328
01659
01989
02319
02647
02974
03298
03619
03938
04252
04563
04868
05168
05461
05748
06027
03320
03320
03320
03318
03315
03309
03301
03289
03274
03254
03230
03201
03166
03125
03079
03026
02967
02901
02829
02751
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
55
60
65
70
06500
07812
09223
10725
12310
13968
15691
17469
19295
21160
23057
09670
26924
28882
30853
32833
37806
42796
47793
52792
06298
06813
07290
07725
08115
08460
08761
09018
09233
09411
09555
24980
09759
09827
09878
09915
09969
09990
09997
09999
02668
02484
02281
02065
01840
01614
01391
01179
00981
00801
00642
00505
00390
00295
00219
00159
00066
00024
00008
00002
y
δ
u
U
Figure 6.4.1 Blasius’s solution for boundary layer flow on a semi-infinite plate
6. The shear stress is infinite for all values of y at x = 0. We could be content with
this if it were infinite only at the leading edge, but it is perhaps surprising that it
should be singular along a line perpendicular to the plate. Also, a solution does
not exist for negative x, for in that case becomes imaginary.
The criticisms raised in the last two points may seen to be minor ones, but they are
needless errors that have in fact been introduced by our choice of a coordinate system.
In performing any mathematical solution, it is always best to use a natural coordinate
system (Kaplun, 1954)—that is, one where setting one of the coordinates to a constant
value yields the desired boundary. For our choice of a coordinate system, the boundary
is at y = 0 for x > 0.
178
The Boundary Layer Approximation
Having to qualify the value of x by saying that it must be positive means that this
is not a natural coordinate system for this problem. A coordinate system that is a natural
coordinate system for the semi-infinite flat plate is a parabolic system, given by
2 = x + x2 + y2 2 = −x + x2 + y2
(6.4.9)
with the inverse relations
2 − 2
y =
(6.4.10)
2
Use of these coordinates eliminates both problems raised in points e and f , leaves
equation (6.4.7) unchanged so the problem does not have to be re-solved, and the
parabolic behaves as the originally
√ introduced,√since far away from the leading
edge along the plate it is seen that ≃ 2x ≃ y/ 2x.
You may have noticed another point that involves general principles in what has
been done. In deriving the boundary layer approximation in the first place, it was stated
that there was a definite boundary length scale L upon which our Reynolds number was
based. In the Falkner-Skan problems, however, there is no such L, and our results come
out with a Reynolds number based on the local distance x. The distinction is perhaps
subtle, but it is important as far as an understanding of the implications is concerned.
The question was not answered until the 1960s, and the lack of attention to the problem
resulted in some confusing and erroneous publications prior to that time.
When there is a fixed-length scale L in a problem, what is being done in the
boundary layer approximation is expanding the solution to the Navier-Stokes equations
in terms of negative powers (and perhaps logarithms) of the Reynolds number. This is
then an expansion in the constant parameter Re. Near a boundary, the first term in the
expansion is the boundary layer equation.
Away from boundaries, the first term in the expansion is the Euler equation.
Presumably, it is possible to go to higher-order terms in the expansion (albeit with
increasing difficulty and work) and still have the solutions match at the overlap of the
Euler and boundary layer regions.
When there is no geometric length scale L, the expansion is in terms of a coordinate such as x, but it may not be able to find solutions in the various regions that
mathematically match together properly, although they might numerically join together
satisfactorily. The question of how to obtain higher-order approximations is thus left
unanswered, and we may have to be satisfied with just the lowest-order approximation.
(We usually settle for this in any case, but it would be nice to at least know that
higher-order approximations could be carried out if we wanted them.) A more thorough
discussion of this matter can be found in Van Dyke (1964).
x=
6.4.2
Stagnation Point Boundary Layer Flow
When = , the inviscid flow becomes flow perpendicular to a flat plate, and our
equation reduces to that found in the full solution of the Navier-Stokes equations. In
this case, then, the boundary layer solution is an exact solution of the Navier-Stokes
equations for the velocity (but not the pressure).
6.4.3
General Case
Values for f ′′ 0 are given in Table 6.4.2 for a number of wedge angles. The last
18 sets of values in were computed by Hartree in 1937. He found that if > 0, the
6.5
179
The Integral Form of the Boundary Layer Equations
TABLE 6.4.2 f ′′ 0 as a function of and m
radians
m
f ′′ 0
degrees
−056549
−047124
−031416
−015708
−007854
−324
−27
−18
−9
−45
−008257
−006977
−004762
−002439
−001235
−00657
−009002
−00973
−007543
−0052
−062202
−059692
−05655
−050262
−043982
−031417
0
031415
062831
094248
125664
157078
188495
251312
314159
376991
502655
628318
−3564
−342
−324
−288
−252
−18
0
18
36
54
72
90
108
144
180
216
288
360
−009008
−008676
−008257
−007407
−006542
−004762
0
005263
011111
017647
025
033333
042857
066667
1
15
4
0
005811
00871
012928
016338
022013
033206
042585
051131
059363
067586
075689
084177
10224
12326
149669
240491
solution of equation (6.4.7) was unique and that for > 20, the velocity components
are imaginary. He carried the solution out for = 24 to facilitate interpolation and for
values of < 0 for which f ′ approached one exponentially from below.
Stewartson (1954) computed the first five values in the table. These flows show
reversed flow regions near the wall and also demonstrate that for a given value of m,
the solution of equation (6.4.3) is not necessarily unique.
The fact that solutions of the Falkner-Skan equation may not be unique is interesting
in itself. It also is a caution that simply formulating a similarity solution in terms of a
nonlinear ordinary differential equation along with appropriate boundary conditions is
not the end of the story. At least one solution must be found and then investigated over
at least some useful parameter range so that there is reasonable cause to believe that
there are no other solutions for those parameters.
6.5 The Integral Form of the Boundary Layer Equations
Next, an integral form for the boundary layer approximation will be derived by integrating the boundary layer equation. First, rearrange equation (6.2.14) in the form
U
u
u
U − u
+U
−u −v
=−
t
x
x
y
2
u
y2
(6.5.1)
180
The Boundary Layer Approximation
From integration of the continuity relation equation (6.2.16), find that
u
dy
v =−
x
0
(6.5.2)
Integrating equation (6.5.1) with respect to y and using the continuity equation and the
product rule of calculus, the result is
wall
U
u
u
U − u
=
+U
−u −v
dy
t
x
x
y
0
U D
U
u
uv
v
=
U
+
−u −
+u
dy
t
x
x
y
y
0
U
u
u
v
U D
U
=
+
− 2u + u
+
dy − Uv
t
x
x
x
y
0
Using equation (6.5.2) to bring the last term back under the integral sign gives
wall
U D
U
u2
u
=
+
U
−
+U
dy
t
x
x
x
0
U
u2
Uu
U
U D
+
U
−
+
−u
dy
=
t
x
x
x
x
0
(6.5.3)
U D
U
=
+
U − u + u U − u dy
t
x
x
0
=
U
D
t
+U
U
x
D
+
U 2 M
x
To use equation (6.5.3) to estimate the nature of the boundary layer a form for uy
must be chosen. For illustration of the method, choose first a linear approximation,
⎧
⎨U y 0 ≤ y ≤
uy =
⎩U
y≥
Then, D = 05 ,
(6.5.3) gives
D
U
Multiplying this by
U
= 0167 , and wall =
=
t
U
+
0167U 2
x
x
05U + 05U
gives
=
t
025U 2 + 05U
U/ . Inserting these into equation
2
U
+
00835U 2 2
x
x
Provided U is known, this is a linear equation
in 2 . For
steady flow past
the case of the
x/00835 U =
a flat √
plate it reduces to U / = d/dx 00835U 2 2 , giving =
/ xU .
346x
We can see that this has the correct Reynolds number dependency. Using this to
determine the shear stress gives
U
2
= 0289 U
wall =
√
Ux
346
x/ U
6.5
181
The Integral Form of the Boundary Layer Equations
Again, the Reynolds number behavior is correct, and the 0.289 is only 13% apart from
the Blasius value of 0.332.
The linear approximation used here is, of course, unnecessarily crude. Other profiles
that could have been used follow, where = y/ :
U 2 − 2 0 ≤ ≤ 1
Parabolic u =
U
≥ 1
U sin
0 ≤ ≤ 1
2
(6.5.4)
Sinusoidal u =
U
≥ 1
U 2 − 23 + 4 + 2 − 32 + 33 − 4 0 ≤ ≤ 1
Pohlhausen u =
U
≥ 1
The results for the flat plate are given in Table 6.5.1.
The Pohlhausen profile1 (1921) was selected to agree with all solutions that were
known by 1921. Pohlhausen used the following boundary conditions in the selection
process:
2
At = 0
dU
u
=− U
At = 1
y2
dx
u = 0
u
= 0
y
u = U
2
u
= 0
y2
2
dU
The parameter =
must lie in the range −12 ≤ ≤ 12, since at = −12 the
dx
shear stress becomes zero at the wall, implying flow separation, and at = 12 becomes
greater than U within the boundary layer.
If the Kármán-Pohlhausen approximation is inserted into our equations, the result is
D
3
−
10 120
=
M
=
63
2
37
−
−
5
15 144
and wall = 2
U
1+
(6.5.5)
12
TABLE 6.5.1 Comparison of various boundary layer profiles with the Blasius results
Velocity
Profile
Linear
Parabolic
Sinusoidal
KármánPohlhausen
Blasius
M
D
1/2 = 05
1/3 = 0333
1/3 = 0167
2/15 = 0133
1 − 2/ = 0363
2/ − 1/2 = 0137
0.344
0.1328
0.3
37/315 = 0117
wall
√
U U/x
√
x/U
√
30 = 548
2/4 − = 48
1260/37 = 5836
√
1/ 12 = 0288
2/15 = 0365
√
2 − /2/2 = 0328
37/315 = 0343
5.0
0.332
√
12 = 346
Error in
wall
13%
9.9%
1.5%
3.3%
0%
1
Pohlhausen was acting on a suggestion by von Kármán (1921), so the method is usually referred to as
the Kármán-Pohlhausen method.
182
The Boundary Layer Approximation
Putting these into the integrated Navier-Stokes equations results (after a reasonable
amount of algebraic manipulation) in
U d 2M
= HK where
dx
FK =
GK =
D
M
M
U
63 3 − /12
10 37/5 − /15 − 2 /144
2
u
2
37
=
−
−
1+
y wall 63
12
5
15 144
=
HK = 2GK − 2K 2 + FK
K=
2
M
2
37
−
−
5
15 144
1
dU
=
dx
3969
(6.5.6)
2
While this looks formidable, plotting H versus K shows that over a good range the
relationship is practically linear and given by HK = 047 − 6K. Thus,
d U 6 2M
U d 2M
≈ 047 − 6K or
≈ 047
dx
U5
dx
After integration of this it becomes
2
M
=
2
M 0+
047 x 5
U dx
U6 0
(6.5.7)
As we will see in the next section, this result is useful in predicting separation.
Because of the present day availability of ready access to powerful personal computers, the integral formulation of the boundary layer equations is no longer as important
as it once was. However, the integrated form of the boundary layer is still convenient
for providing an approximate solution of the boundary layer equations with a minimum
amount of labor with reasonable accuracy. It is also useful for providing starting values
for the wall shear when more accurate numerical methods are used.
6.6 Axisymmetric Laminar Jet
Using the boundary layer equations, it is also possible to solve for a laminar jet, similar
to Squire’s solution in Section 5.4.5. Working in cylindrical polar coordinates, let
=
4 z
f with =
1r
8z
3 M
2
Here, M is the momentum of the jet given by
w2 2rdr
M=
0
1/2
(6.6.1)
The boundary layer momentum equation in the z direction then becomes, after a bit of
rearranging,
d
d
4f df
d
=
d
d
1 df d2 f
−
d d2
(6.6.2)
6.7
183
Flow Separation
with the boundary conditions
df
= 0
d 0
f0 = 0
df
=
Integrating equation (6.6.2) gives 4f d
integrated to give
df
d
2f 2 = 2f −
df
= 0
d
(6.6.3)
2
2
d f
df
df
− d
2 . Since 4f d = 2 d , this can be
df
+ C1
d
From the boundary conditions the constant of integration C1 = 0. Then
This can be integrated to give f =
finally
C2 2
.
1+C2 2
f =
and
1
u=
2
3M
z2
1/2
1 − 2
1 + 2 2
df
−f 2 +f
Using equation (6.6.1) find C2 = 1, so
2
1 + 2
w=
= 2 d
.
3M
1
8 z 1 + 2 2
(6.6.4)
(6.6.5),
(6.6.6)
6.7 Flow Separation
When the pressure gradient along a body surface is negative (favorable pressure gradient), the pressure acts to locally accelerate the flow along the body and to overcome
the viscous forces that act to decelerate. However, when the pressure gradient becomes
positive (adverse pressure gradient), both pressure and viscous stresses act to decelerate
the flow, and
the boundary layer thickness rapidly increases. The flow soon reverses
u
<
0
, and flow separation is said to occur.
y wall
The point at which the flow reversal starts is called the separation point and is
usually a short distance downstream from the point where the pressure gradient changes
sign. Boundary layer theory is capable of predicting where the separation point is located
but is not able to predict the flow past this point.
Thwaites (1949) correlated a number of solutions of the integrated boundary
layer equations and found that separation could be closely predicted by the result
009
2
M = − dU/dx . Inserting this into equation (6.5.7) gives
009
047 x 5
−
= 6
U dx
(6.7.1)
dU/dx
U 0
where the integration has been started from a stagnation point.
A second criteria suggested by Stratford (1954) suggests that separation will occur
at the point where
x − x0 2 cp
dcp
dx
2
= 00104
2
where cp = 1 − U 2 /Umax
(6.7.2)
Both criteria require knowledge of U at the boundary. The Stratford criterion is somewhat easier to use in that it doesn’t involve an integration.
Even though the Thwaites and Stratford criteria are for two-dimensional flows, they
are frequently used in three-dimensions in situations where the flow is predominately
two-dimensional in behavior.
184
The Boundary Layer Approximation
Example 6.7.1 Calculation of the separation point on a circular cylinder in a uniform
stream.
Schmidt and Wenner (1941) found experimentally that a good approximation to the
laminar surface flow on a circular cylinder of radius a was given by
U = U0 2s − 0451s3 − 000578s5 where
s = x/a and x is the distance along the circumference of the cylinder.
Find the predicted separation point using the Stratford and Thwaite criteria.
Solution using Stratford’s criterion. Taking the derivative of U , find that U has a
maximum where 2 − 1353s2 − 00289s4 = 0. By the quadratic formula, find that root to
be smax = 119765, giving Umax = 16063U0 . This is 68.62 degrees from the stagnation
point. Then
−2U dU
s − smax 2 1 − U/Umax 2
2
Umax
dx
2
− 00104 = 0
By successively substituting values of s greater than smax find the value s = 1447, which
corresponds to a separation point at 82.97 degrees from the stagnation point.
Solution using Thwaite’s criterion. Multiplying U to the fifth power and then
carrying out the integration in equation (6.7.1) is tedious, to say the least. Instead
the integration is best carried out numerically using something like the Newton-Cotes
formula. A program for doing this is given in Chapter 10. Separation is predicted at
s = 1576, corresponding to an angle of 90.30 degrees from the stagnation point.
These criteria both assume that the separation point does not move in time. In
reality, the region beyond the separation points usually develop vortices and a wake
that contains a vortex street as modeled by von Kármán (Section 3.8). If the flow is
started gradually from rest, the vortices may first align side by side. Soon, however,
they switch to the staggered position, as found in the von Kármán vortex street. As a
vortex grows and is shed, the separation point can move back and forth. As yet, no
simple method is available for predicting this analytically. For the most part, description
of wakes appears to be best done by numerical methods.
Even wakes with minimal large-scale vortices have proven difficult for analytical
methods. In 1930 Goldstein attempted an expansion solution of the wake behind a flat
plate. Some progress was made, but the formidable computations required had to be
done by hand, and progress was small. In two papers in 1968 and 1969, Stewartson
elaborated further on the structure of the flow near the trailing edge of the plate and
found that while Goldstein’s viscous solution held in a very thin layer in the rear of the
body, there was also a thin inviscid layer containing vorticity outside the Goldstein layer
before inviscid irrotational flow was achieved. Complexity thus reigns in wake flow!
6.8 Transformations for Nonsimilar Boundary Layer Solutions
The solutions of the boundary layer equations presented so far are limited to very
special external flows—namely, those outer-velocity profiles given in the Falkner-Skan
problem. Other outer pressure gradients require different methodology and numerical
techniques for their solution. Several transformations have been introduced over the
6.8
185
Transformations for Nonsimilar Boundary Layer Solutions
years in attempts to condition the boundary layers to simplify solution methods. These
transformations offer a possible starting point for the utilization of numerical methods.
6.8.1
Falkner Transformation
The transformations used in the solution for flow past a wedge can also be used for
nonsimilar solutions. Start with the steady form of equation (6.2.14)—namely,
u
u
u
+v
x
y
= U
2
u
y2
U
+
x
(6.8.1)
It is convenient to first suppress as much as is easily possible the growth of the boundary
layer thickness in the x direction. Introduce variables suggested by our Falkner-Skan
solution according to
x dU
Ux
Ux
= x = y
=
fx P =
(6.8.2)
U dx
Then, since
1 1 dU
=
+
− +
x 2
x U dx
and
y
=
U
x
it follows that
u=
y
=
U
x
Ux
f
=U
f
(6.8.3)
and
v=−
= − U x/
x
1
f
f
1 + P f +
+
−1 + P
2x
2x
(6.8.4)
Substituting these into equation (6.8.1), we find, after some arranging, that
U
2
2
f dU f
f
U
f
+U
+
−1 + P 2
dx
2x
−U2
2
f 1
dU
U 2 3f
f
f
1
+
−1
=
+U
+
P
f
+
+
P
2
3
2x
2x
x
dx
Multiplying the above equation by x/U 2 and simplifying, the result is
3
f 1 + P 2f
f 2
f 2f
f 2f
+
f
+
P
1
−
−
=
3
2
2
2
(6.8.5)
The left-hand side of the equation is essentially the equation seen in the similarity
solution, except now P can depend on x. Since a stream function is the starting point,
continuity is automatically satisfied.
It is seen that equation (6.8.5) is a third-order partial differential equation. In
numerical solutions, it is somewhat easier to deal with first-order partial differential
equations. To get to that point here, make the change of variables
f
F=
V=
U
186
The Boundary Layer Approximation
where F is then the dimensionless u velocity component, and V is a scaled dimensionless
version of the v velocity component. Then equation (6.8.3) becomes
1+P
f P − 1 f
f + +
2
2
V =−
Differentiating this with respect to gives
F P − 1 F
V
= − PF +
+
2
The momentum equation (6.8.4) can now be written as
2
F
F
F
V
+
=
−P
+
F
PF
+
P
−
1F
+
2
2
6.8.2
(6.8.6)
(6.8.7)
von Mises Transformation
Von Mises (1927) suggested the transformation = U 2 −u2 and used as the dependent
variable and x and as the independent variables. Then
(6.8.8)
d = −vdx + udy
d = 2 U
dU
u
u
u
−u −v
dx − 2 d
dx
y
x
y
(6.8.9)
It is seen from this that
dU
u
u
=2 U
−u −v
x
dx
y
x
= −2
2
= −2
u
y2
u
d
y
(6.8.10)
(6.8.11)
where the term on the far right of equation (6.8.10) comes from the momentum equation.
Taking further differentials gives
d
2
v 2u
u
u
+
d
−
2
x y u y2
y2
2
= −2
From this it follows that
2
2
=−
2 2u
u y2
(6.8.12)
Combining equations (6.8.10) and (6.8.12) gives the final form of the transformed
equation as
U2 −
2
2
=
x
(6.8.13)
This is to be solved subject to the conditions = U 2 at = 0 = 0 at = .
2
Once equation (6.8.13) is integrated,
u can be found from u = U − . Since
dy = d + udx/u y is found from y = 0 d /u, the integration being
performed
on a
line of constant x. The v velocity component is found from v = 0 yv dy = − ux dy =
− u1 ux d on a line of constant x. Since there is a singularity in the integrand at the
wall, care must be used in carrying out the integration.
6.8
187
Transformations for Nonsimilar Boundary Layer Solutions
6.8.3
Combined Mises-Falkner Transformation
The previous two transformations can be combined into one step by the following:
m
√
1 x
=
u = U w
Uds = f
(6.8.14)
U
U /
At this point m is a constant and f is a function, both to be chosen. For convenience,
note that
U
f m
mu
=
= 0
=
−
=
x U
y
x
f
y
x
=
U
1 f mv
+
−
U
f
y
=
mu
Using these in the boundary layer equation yields
√
2 dU
df w
w
=
1 − w −
+ m2 2−2/m f 2/m w
U d
f d
2
w 1 − 1/m w
(6.8.15)
+
2
There are two choices for m and f that have proven to be useful in the past.
1. m = 1/2 f = −1/4 . In this case equation (6.8.15) reduces to
√
2
w
w 1 w
w
w 2 dU
=
1 − w −
+ 2
−
U d
4 4
2
(6.8.16)
2. m = 1 f = 1. In this case equation (6.8.15) reduces to
√ 2w
2 dU
w
=
1 − w + w 2
U d
(6.8.17)
Integration in both cases proceeds as in the previous two cases.
6.8.4
Crocco’s Transformation
Crocco’s transformation (1939) is unusual in that it uses shear stress rather than a
velocity function as a dependent variable. Letting = uy , since
du =
dy =
d =
u
u
u
dx + dy = dx + dy then
x
y
x
du −
u
dx Also
x
dx + dy =
x
y
u
=
−
x
x x y
u
−
du so that
dx +
x x y
y
=
u
y
Then
d
u
1
1
du −
dx so that
dx +
x y
y y
x
2
2
2
2
1
2
= 2
−
u +v
Similarly 2 =
2
2
u
y
y
y
x
y
u
=
188
The Boundary Layer Approximation
With further algebra, the preceding can finally be reduced to
2
+
u2
u
x
1
dU
+U
dx u
1
= 0
(6.8.18)
The accompanying boundary conditions are / u = 0 at u = 0 and = 0 at u = U .
In using any of the previous transformations it is necessary to first decide whether
the dependent variable is the one best suited to the method. The Falkner transformation
results in an equation for x, the von Mises an equation for essentially u y,
the combined method an equation for u, and Crocco’s method for an equation for
xu. The equations are of about the same order of complexity, and all require a guess
of either the dependent variable or its derivative on the wall. The choice of method to
be used then depends on the end result sought and the suitability of the method to the
boundary conditions.
6.8.5
Mangler’s Transformation for Bodies of Revolution
Mangler’s transformation differs in intent from the four previous transformations in that
its purpose is to change an axisymmetric flow problem into a two-dimensional flow
problem rather than to condition the equations.
Consider the flow on a body generated by revolving the curve y = rx about the
x-axis. Providing the body radius r is much greater than the boundary layer thickness,
the boundary layer equations are
2
u
dU
ru
rv
u
u
+v = U
+
+
= 0
x
y
dx
y2
x
y
x
dr
Using the transformation x′ = 0 r 2 dx y′ = ry, then u′ = u v′ = 1r v + yur dx
, so that
′
dr
+ rv′ , then
u = u′ v = − yur dx
u
u′
u′
u′
+ v′ ′
′
x
y
= U
dU
+
dx′
v′
u′
+
= 0
x
y
2 ′
u
y′2
(6.8.19)
Thus, any solution procedure suited to two-dimensional boundary layer problems can
readily be transferred to flow past axisymmetric bodies.
6.9 Boundary Layers in Rotating Flows
The flows previously studied are caused by either pressure gradients, body forces, or
moving boundaries. In these cases, the resulting flow is in the direction of the gradient,
or body force, or boundary motion. In the atmosphere, however, as is apparent from
weather maps that show isobars, the flow tends to be along the isobars, due to Earth’s
rotation.
Studying flow on a sphere is difficult, so to simplify matters, adopt the viewpoint
of the meteorologists and oceanographers and use the beta plane approach. That is,
look at a relatively small segment of Earth—say, a square region 1,000 kilometers on a
side—and consider that portion of Earth to be flat, rotating at the rate 0 sin , where
is the latitude of the center of the region and 0 is Earth’s rotation speed of 2
radians per day. Introducing this into equation (1.17.4), the principal term of interest
6.9
189
Boundary Layers in Rotating Flows
is the Coriolis acceleration term 20 sin × v, where the angular velocity vector is
perpendicular to Earth’s surface, and the velocity vector is in plane of the tangent to
the Earth’s surface. The neglect of the centripetal acceleration is justified, since it adds
essentially a small body force, and the remaining terms are considered small compared
to the Coriolis terms.
If the vertical velocity component is neglected, the simplified Navier-Stokes equations, first presented by Ekman (1905), are then
−2 v0 sin = −
p
+ 2 u
x
2 u0 sin = −
p
+ 2 v
y
(6.9.1)
When viscosity is neglected, it is seen that the flow is along the lines of constant
pressure.
Ekman considered the problem where the wind shear at the surface of the ocean
was in the x direction, no pressure gradients were present, and the velocity components
only depended on the vertical coordinate. Then equations (6.9.1) reduce to
−2 v0 sin =
d2 u
dz2
2 u0 sin =
d2 v
dz2
(6.9.2)
Multiplying the second equation by i and adding it to the first gives
2i u + iv0 sin =
d2 u + iv
dz2
With z negative downward, this has the simple solution with
√
u + iv = U + iVe1+iz 0 sin / or
√
u = ez 0 sin / U cos z 0 sin / − V sin z 0 sin /
√
v = ez 0 sin / V cos z 0 sin / + U sin z 0 sin /
(6.9.3)
In particular, if the wind shear is such that the surface water is in the x direction, then
V = 0 and
√
u = Uez 0 sin / cos z 0 sin /
(6.9.4)
√
v = Uez 0 sin / sin z 0 sin /
It is seen from equations (6.9.4) that the path of the tip of the velocity vector traces
a spiral as we move down from the free surface. In the northern hemisphere it starts
pointing eastward, say, then turns north, then west, then southward, and repeating the
process and decreasing in magnitude as we proceed downward. Computing the discharge
by integrating the first of equations (6.9.3), find that
− √
−
U1 − i
Qx + iQy =
e 0 sin / 1+iz dz =
u + ivdz = U
(6.9.5)
0
0
2 0 sin /
Thus, equal quantities of fluid are transported perpendicular to the wind shear stress as
in the direction of the wind stress.
Boundary layers associated with rotating flows are referred to as Ekman
layers. From equations (6.9.4) we see that the thickness of the layer is of the
/ 0 sin .
order
190
The Boundary Layer Approximation
Another example of rotating boundary layers that does not necessarily need Earthmagnitude dimensions is for a fluid in weak rotation above a stationary plate. A stirred
cup of tea with a few loose tea leaves would be an example of such a flow. In this case
there is a radial pressure gradient p/ r = r2 outside of the boundary layer. Taking
radial and circumferential velocity components in the form u = rfz v = rhz,
the equations are
d2 f
dz2
d2 h
2 rf2 = r 2
dz
−2 rh2 − r2 = r
(6.9.6)
Again, multiplying the second of equations (6.9.6) by i and adding it to the first, the
result is
2if + ih − =
d2 f + ih
dz2
With z = 0 at the bottom of the cup, the solution is
d2 f + ih
− 2 if + ih = −
dz2
therefore,
√
f + ih = −i 1 − e−1+iz /
√
= −i 1 − e−z / cos z / − i sin z /
Splitting into real and imaginary parts gives
√
h = −1 + e−z / cos z /
√
f = e−z / sin z /
(6.9.7)
Above the boundary layer the almost neutrally buoyant tea leaves move in concentric
circles, the centripetal force balanced by the pressure gradient. As they move into the
boundary layer, the centripetal force diminishes, while the pressure gradient remains
unchanged, pushing the tea leaves to the center of the cup.
There is a popular belief that when draining a bathtub the water swirls one way in
the northern hemisphere and the opposite way in the southern hemisphere. Since Earth’s
rotation is small, its effects are mostly felt when lengths of hundreds of kilometers are
involved. For something like a bathtub, the length scale is so small that the Coriolis
acceleration is negligible. The behavior instead depends strongly on what swirl was
introduced when the tub was filled and on the geometry of the tub. Experiments with
well-designed circular containers with carefully centered outlets show that it is necessary
to wait many hours—perhaps even an entire day—for initial effects to die out. After
that, the swirl is hardly noticeable until the depth is of the order of the Ekman layer.
When that happens, the direction of swirl reverses several times as the layer becomes
thinner and thinner.
191
Problems—Chapter 6
Problems—Chapter 6
6.1 The similarity solution for flow into a sink between two inclined plates can be
obtained from the Falkner-Skan equations by putting
=
dU
= 0
dx
=
2 dU
= 1
dx
giving the equation
2
f ′′′ + 1 − f ′ = 0
f0 = f ′ 0 = 0
x = y/ =
x/
f ′ → 1 as →
Show that this equation may be integrated to give
f ′ = 3 tanh2 √ + 1146 − 2
2
where the prime denotes differentiation with respect to r.
6.2 The equation for source flow between inclined plates f ′′ + f 2 + 4f + 4a = 0
can be linearized if in some sense f can be considered to be small. In that case, the
equation becomes f ′′ + 4f + 4a ≈ 0. Solve this equation subject to f vanishing at 1
and at 2 .
6.3 Find the similarity solution for a two-dimensional submerged
jet
emitting at the
1/3
origin. Show that the form x y = = f = y M/ 2 x2 , reduces the
Navier-Stokes equations to a third-order ordinary differential equation. This equation
can be integrated twice, reducing it to w first-order differential equation.
6.4 The following form has been suggested for flow in the wake of a flat plate:
1/3
x y = 2 x2 / 2
f = y / x1/3
√ 2 1/3
with U = 3C x
. Here, represents the constant vorticity at the edge of the
boundary layer, and C is also a constant. Show that this is a suitable similarity solution
of the boundary layer equations. Do not attempt to solve the equation.
6.5 The boundary layer equations place severe restrictions on the possibility of
having a similarity solution. To make a general search for similarity solutions, start with
y
the form x y = ex + gxf = hx
, and substitute it into the boundary
layer equations. Find the restrictions that must be placed on ex gx, and hx to
obtain a differential equation solely in terms of f and .
6.6 Show that a solution of the boundary layer equation (in fact, the full NavierStokes equation) can be found for an infinite flat plate in a uniform stream (speed U )
with constant suction at the wall. The suction induces a downward velocity of strength
V0 . Such suctions have been used as a method for controlling boundary layer separation.
Use a velocity form of the type v y = u y V.
6.7 Fluid enters a channel of width 2b at a uniform speed U0 . Because of the
no-slip condition, the wall shear changes gradually from the uniform flow to the standard
Poiseuille parabolic flow. The flow will consist of two wall boundary layers plus an
inner core where the speed changes with x but is uniform across its domain. Since the
flow can be expected to be symmetric with respect to the centerline, choose y as the
distance measured from the bottom wall.
192
The Boundary Layer Approximation
U2 − 2 0 ≤ ≤ 1
U 1 ≤ ≤ hx/ = y/
with U being the speed outside of the boundary layer, find U and the distance needed
to establish the flow.
Using the profile vx =
6.8 The potential flow against an infinite plate is described by the velocity potential
= Ax2 − y2 , the plate being at y = 0. Using a parabolic velocity profile, solve for
the boundary layer thickness and wall shear.
UA + B cos C 0 ≤ ≤ 1
6.9 The form u =
U > 1 = y
is proposed as an approximation for the velocity profile in a boundary layer. Find
appropriate choices for A B, and C, and calculate the displacement and momentum
thicknesses.
6.10 A thin viscous layer of liquid flows down a vertical wall
√ under the force of
gravity. The flow at the outer edge of the
layer has the speed U = 2gx. Assuming the
U2 − 2 0 ≤ ≤ 1
layer in the liquid to be of the form vx =
U 1 ≤ ≤ hx/ = y/
find hx and x.
6.11 Viscosity does play a small role in the damping of free surface gravity waves.
A simple model for computing this is to say dtd KE + PE = −, where is the
dissipation function discussed in Chapter 1. For the two-dimensional √
traveling wave
given by x z t = cA sin kx − ctekz x t = A cos kx − cti c = g/k, compute
the potential and kinetic energies and the dissipation function to find the rate of decay
of the wave. Assume A depends exponentially on time.
6.12 The profile U = U0 1 − Lx has been used to study separated flows. Calculate
the location of the separation point using both the Thwaites and Stratford criteria.
Chapter 7
Thermal Effects
7.1 Thermal Boundary Layers 193
7.2 Forced Convection on a Horizontal
Flat Plate 195
7.2.1 Falkner-Skan Wedge Thermal
Boundary Layer 195
7.2.2 Isothermal Flat Plate 195
7.2.3 Flat Plate with Constant
Heat Flux 196
7.3 The Integral Method for Thermal
Convection 197
7.3.1 Flat Plate with a Constant
Temperature Region 198
7.3.2 Flat Plate with a Constant
Heat Flux 199
7.4 Heat Transfer Near the Stagnation
Point of an Isothermal Cylinder 200
7.5 Natural Convection on an
Isothermal Vertical Plate 201
7.6 Natural Convection on a Vertical
Plate with Uniform Heat Flux 202
7.7 Thermal Boundary Layer on
Inclined Flat Plates 203
7.8 Integral Method for Natural
Convection on an Isothermal
Vertical Plate 203
7.9 Temperature Distribution in an
Axisymmetric Jet 204
Problems—Chapter 7 205
7.1 Thermal Boundary Layers
If temperature gradients exist in a flow, the momentum boundary layer for large Reynolds
numbers discussed in Chapter 6 can be accompanied by a thermal boundary layer. Recall
equation (1.14.7), where it was found from the first law of thermodynamics that
dr
Du
= − · q +
+
(1.14.7)
Dt
dt
Taking u = cp T and q = −kT for the internal energy and heat flux, and considering
only cases where there is no internal heat generation and dissipation and compressibility
effects are secondary, this becomes
cp
DT
= k 2 T
Dt
(7.1.1)
Here, cp is the heat capacity at constant pressure J/kg · K, and k is the thermal
conductivity W/m · K.
193
194
Thermal Effects
Choose dimensionless parameters in the manner used for the flat plate in steady
flow. Using the form
√
x = xD /L y = yD Re/L
√
vx = uD /U0 vy = vD Re/L
T = TD − Twall /T − Twall
Pr = cp /k
then equation (7.1.1) becomes
vx
x
+ vy
y
=
1 2
Pr y2
(7.1.2)
The resemblance to the form of equation (7.1.2) and the momentum equations of
Chapter 6 suggest that similarity solutions are possible.
An important dimensionless parameter that comes up often in thermal flows is the
Prandtl number, defined as Pr = cp /k. It represents the ratio of viscous diffusivity
to thermal diffusivity. The Prandtl number is strictly a function of the properties of
the fluid and not on flow properties and plays a strong role in determining the ratio of
the thermal boundary layer to the momentum boundary layer. Representative Prandtl
numbers are shown for various fluids in Tables 7.1.1, 7.1.2, and 7.1.3.
TABLE 7.1.1 Prandtl numbers for liquid metals
Metal
Bismuth
Lead
Lithium
Temperature (K)
589
1033
644
977
477
1255
Pr
0014
00083
0024
0016
0065
0027
Metal
Temperature (K)
Mercury
273
589
422
977
366
400
Potassium
Sodium
Pr
0029
00084
00066
00030
00011
00163
TABLE 7.1.2 Prandtl numbers for gases at 300 K
Gas
Air
Ammonia NH3
Carbon dioxide CO2
Pr
0707
0887
0766
Gas
Pr
Carbon monoxide (CO)
Helium (He)
Hydrogen H2
073
068
0701
Gas
Nitrogen N2
Oxygen O2
Pr
0716
0711
TABLE 7.1.3 Prandtl numbers for liquids at 300 K
Liquid
Engine oil
Ethylene glycol C2 H4 OH2
Pr
6400
151
Liquid
Glycerin C3 H5 OH3
Water H2 O
Pr
6,780
5.83
7.2
195
Forced Convection on a Horizontal Flat Plate
7.2 Forced Convection on a Horizontal Flat Plate
7.2.1
Falkner-Skan Wedge Thermal Boundary Layer
Consider again the Falkner-Skan solution for flow past a wedge. Take the wedge surface
to be at a temperature Twall , and the ambient temperature to be T . If buoyancy effects are
negligible, the momentum equation is unaffected by the presence of thermal gradients.
wall
, equation (7.1.2) becomes
Then, with as before and letting = TT −T
−T
wall
m+1
d
d2
+
Pr f
= 0
d2
2
d
(7.2.1)
This is to be solved subject to the isothermal plate condition = 0 at = 0 → 1
as → .
The momentum boundary layer is unchanged from what was discussed in Chapter 6,
and the solution found there for the stream function is valid here. The energy equation
again requires a numerical solution, with Runge-Kutta-type methods among the easiest
to use. They require guessing 0 and then integrating after choosing values for m and
the Prandtl number.
7.2.2
Isothermal Flat Plate
A formal solution of equation (7.2.1) was found by Pohlhausen (1921) for the case
of the flat plate m = 0 by choosing
as a function of and the Prandtl number.
Then, from equation (6.4.7), f = −2f ′′′ /f ′′ . Substitution of this into equation (7.2.1),
integrating and applying the thermal boundary conditions gives
F
F
Pr
f ′′ d
(7.2.2)
For the special case Pr = 1, this gives = f ′ = u/U .
The rate of heat transfer per unit area from the plate is given by
U d
T
′′
qwall = − k = kTwall − T
y wall
x d
(7.2.3)
Pr =
where F =
0
=0
Detailed information of the heat transfer rate thus requires the integration of equation
(7.2.2) for a number of Prandtl numbers. Results show that to a good approximation
⎧
√
⎪
0564 Pr as Pr → 0
⎨
d
= 0332 Pr 1/3 for 06 < Pr < 15
d =0 ⎪
⎩
0339 Pr 1/3 as Pr →
The thickness of the thermal boundary layer can be defined analogous to the
displacement and momentum thicknesses as
T =
1
U Twall − T
0
h
uT − T dy
(7.2.4)
196
Thermal Effects
Since the Prandtl number represents the rate of momentum diffusivity to thermal
diffusivity,1 for low Prandtl numbers (e.g., liquid metals) momentum transfer in the
boundary layer is more efficient than heat transfer. The converse is true if the Prandtl
number is greater than unity (e.g., water). Therefore, in general expect that
T > if
Pr < 1
T = if
Pr = 1
T < if
Pr > 1
The previous solution was extended by Pohlhausen (1921) to include the effects
of dissipation. The dissipation function in the case of the boundary layer reduces to
u/ y2 , so equation (7.2.1) becomes
d
m+1
d2
+
Pr f
= A Pr
2
d
2
d
u
y
2
(7.2.5)
2
with A = c T 2U −T . Using our knowledge of the solution of the homogeneous equation,
p wall
the particular integral can easily be found. The solution now becomes
F
Pr
2−Pr
1 − KF1 + KF1 where F1 =
f ′′
f ′′
=
d d
F
0
(7.2.6)
The function F is the same as that found in the case where dissipation was neglected.
Equation (7.2.6) satisfies the conditions 0 = 0 = 1. The heat flux also goes
to zero at the outer edge of the boundary layer.
Note that if K is large compared with unity, the temperature inside the boundary
layer can be higher than either the wall or exterior temperatures.
7.2.3
Flat Plate with Constant Heat Flux
The case of constant heat flux emanating from the plate is also of some interest. For
the isothermal plate the correct choice for temperature dependency was a function of .
To meet the constant heat flux condition, the proper choice of boundary condition is
vx
′′
(7.2.7)
T = T − qwall
U
q ′′ d
d
This makes Ty = − wall
= 1. Insertion
,
so
the
boundary
condition
at
the
wall
is
k d
d
wall
of equation (7.2.7) into the energy boundary layer equation gives
d2
Pr
df
d
+
−
f
= 0
d2
2
d
d
along with the boundary conditions dd = 1 → 0 as → .
wall
1
Thermal diffusivity is defined as = k/cp .
(7.2.8)
7.3
197
The Integral Method for Thermal Convection
7.3 The Integral Method for Thermal Convection
The integral method introduced in Chapter 6 can be easily extended to thermal problems.
In using it the temperature dependence of the density is often taken in the simplified form
≃ 1 − T − T
where = −
1
d
dT
(7.3.1)
p
Density variations are usually neglected in the multiplier of the acceleration. Then, the
integrated momentum equation including buoyancy force becomes
d h
dU h
vx U − vx dy +
U − vx dy
dx 0
dx 0
(7.3.2)
h
vx
T − T dy +
= g cos
y
0
wall
Here, is the angle the x-axis makes with gravity. The constant part of the gravity
force has been included with the pressure gradient as the driving force for U outside
the boundary layer—that is,
U
p
dU
= − + g cos
dx
x
Carrying out a procedure in the manner used in Chapter 6 to obtain the integrated
momentum equation, find for the integrated first law equation
T
d h
cp vx T − Tdy = k
(7.3.3)
dx 0
y wall
In using the integral methods for thermal problems, the methodology employed in
the previous chapters work well as long as they are adopted to the circumstances at hand.
For example, if there is an external flow, the velocity profiles from the previous chapter
are satisfactory, but if the external flow is stagnant, the velocity profiles used must
vanish at the edge of the momentum boundary layer. An example of such a profile is
y 2
y
vx x y = U
1−
(7.3.4)
2
x
ux = 0 uy x = 0.
This satisfies the conditions ux 0 = 0 yu2 = − 4u
2
The quantity U can be a function of x and is usually related to buoyancy terms.
For problems where wall temperatures are prescribed, forms such as
y 2
Tx y = T + Twall − T 1 −
(7.3.5)
are suitable since this ensures that Tx 0 = Twall , whereas if heat flux is specified,
the form
q ′′
y 2
Tx y = T + wall 1 −
(7.3.6)
2k
′′
. As we saw in the similarity solutions, the
is a proper choice, since k Ty x 0 = −qwall
choice of appropriate velocity and temperature profiles first requires that the approximating functions satisfy the most important boundary conditions.
198
Thermal Effects
7.3.1
Flat Plate with a Constant Temperature Region
Suppose that a region of the flat plate of length x0 extending from the leading edge is
at temperature T . The remainder of the plate is at temperature Twall . For the velocity
and temperature profiles in the thermal boundary layer use
vx x y = U
3 y 1 y 2
−
2 2
3 y
1
−
Tx y = Twall + T − Twall
2 T 2
(7.3.7)
y
T
2
(7.3.8)
Then
1
d T
3
1
1
d
3
− 3
T
T − T vx dy = cp UT − Twall
1 − T + T3
dx 0
dx
2
2
2
2
0
3
3 2
T
3 k
d
−
T − Twall
T
=k =
= cp UT − Twall
dx
20 280
y wall 2 T
cp
dT
where = T /.
Since at least in the front portion of the plate the thermal boundary is thinner than
the momentum layer, the squared term in can be neglected, leaving
10k
d 2
=
dx
Ucp
√
Since = 464 x/U , with some manipulation this can be rewritten as
4 d3
0929
3 + x
=
3 dx
Pr
The solution of this first-order differential equation in 3 is then
3 = Cx−3/4 +
0929
Pr
C is a constant determined from 0 = 0 at x = x0 , so C = − 0929
x03/4 .
Pr
Hence, the boundary layer thickness ratio is given by
= 0976 Pr −1/3 1 −
The local heat flux is given as
x0 3/4
x
1/3
(7.3.9)
T
T
Ux
x
k
−
T
wall
′′
qwall
= − k = 0331
Pr 1/3
1−
y wall
x
x0
−3/4
−1/3
(7.3.10)
General wall temperature distributions can be handled using Duhamel’s superposition method. Let x y be the solution for
vx
k 2
+ vy
=
x
y
cp y2
7.3
199
The Integral Method for Thermal Convection
with boundary conditions
at y = 0
Then
as y →
x y = 0 0 < x <
1 x >
→ 0
at x =
Tx y = T + Twall 0 − T 0 x y +
7.3.2
= 0
x
x y
0
dTwall
d
d
(7.3.11)
Flat Plate with a Constant Heat Flux
The case of a plate with constant heat flux, solved using similarity methods in the
previous section, can also be solved by the integral method. With the integral method,
however, it is possible to have the heat flux start at a distance of x0 from the leading
edge of the plate. Using
3 y 1 y 2
vx x y = U
(7.3.12)
−
2 2
3 y
1
Tx y = Twall + T − Twall
−
2 T 2
y
T
Proceeding as before, find that
d T
d
1
1
′′
T − Tvx dy = cp Uqwall
T 2 − 2
cp
dx 0
10
dx
14
2
=k
giving
1
d
2T − 2
dx
14
=
10k
cpU
(7.3.13)
T
′′
= −qwall
y wall
Again, since this is in the thermal boundary layer and, at least in the front portion of
it, the thermal layer is thinner than the momentum layer, the cubic term in can be
neglected, leaving
d 3
10k
=
dx
Ucp
Integration of this gives 2 3 = C + c10kU , where C is a constant determined from 0 = 0
p
at x = x0 , so C = − c10kU x0 . Using our previous knowledge of , the boundary layer
p
thickness ratio is given by
x 1/3
= 0774 Pr −1/3 1 − 0
(7.3.14)
x
The local temperature is given as
Twall = 2394
′′
Ux
qwall
Pr −1/3
k
−1/2
1−
x0 1/3
x
(7.3.15)
Superposition as used in arriving at equation (7.3.10) can be used for heat flux that
varies along the plate.
200
Thermal Effects
7.4 Heat Transfer Near the Stagnation Point
of an Isothermal Cylinder
The irrotational flow past a circular cylinder predicts a velocity U = 2U sin along the
surface of a cylinder of radius a. Here, is the angle measured from the stagnation point.
As long as we don’t venture too far from the stagnation point, a good approximation
is sin ≃ x/a. The Kármán-Pohlhausen velocity profile discussed in Chapter 6 was of
the form vx = U F + G , with F = 2 − 23 + 4 and G = 16 1 − 3 ,
37
2
and M = 63
5 − 15
− 144
.
giving D = 10 3 − 12
2
U dM
Recalling from equation (6.5.6) that dx = H, since U is zero at the stagnation point and d2M /dx must be finite, H must be zero at the stagnation point. This will
be true if
3 + 14742 − 16704 + 9072 = 0
(7.4.1)
The three roots of this equation are 7.052, 17.75, and −70. Only the first of these
is in the range −12 < < 12, which is required for the flow to be nonseparated. From
the definition of G in equation (6.5.6) it follows that
2 =
=
dU/dx 2U
(7.4.2)
Hence,
2stagnation =
3526a
U
(7.4.3)
In the thermal boundary layer, using the cubic temperature profile, we again find that
=
3 y
1
−
2 T 2
y
T
3
(7.4.4)
From the integrated energy equation find
3k
d
U M + N =
dx
2
3
1
1
where = T / M = − 3 + 4
5
70
80
1 1
1 2
9 3
1 4
N=
− +
−+
6 10
8
140
80
cp
Again, neglecting higher powers √
of and√using U = 2U x/a along with 2 =
d
a /2U , this simplifies to dx
x2 12 + = 90
. Using the stagnaPr
3 3
10048
d3
tion value of 7.052 for reduces this to x dx + 2 = Pr . The solution of this that
keeps finite at x = 0 is
=
0875
Pr 1/3
(7.4.5)
7.5
201
Natural Convection on an Isothermal Vertical Plate
Then
′′
qwall
3 k
T
k
= 0645 Twall − T Pr 1/3 Re1/2
=
= −k
y wall 2
a
(7.4.6)
where Re = 2U a/ .
7.5 Natural Convection on an Isothermal Vertical Plate
In natural convection (also referred to as free convection) flow is due solely to buoyancy
forces. The boundary layer momentum equation for these flows is
vx
vx
v
+ vy x
x
y
=−
p
+ g +
x
2
vx
y2
(7.5.1)
The pressure gradient here is solely hydrostatic and balances the average buoyancy
force. The coordinate x is directed upward in the vertical direction.
Using the density form of equation (7.3.1), this becomes
0 vx
v
vx
+ vy x
x
y
2
= 0 gT − T +
vx
y2
(7.5.2)
Pohlhausen (1921a) found a similarity solution for the flow due to a uniformly
heated flat plate. Since there is no velocity outside of the boundary layer, there is no
reference velocity to use in the solution, so the previous general form, dependent as it
is on a Reynolds number, does not hold. By choosing stream and temperature functions
in the form of a power of x times a function of , a variable that is of the form y times
another power of x, he found that the momentum and energy equations along the plate
could be put in similarity form if he chose
x = 4 Cx3/4 f
T = T + Twall − T Pr
where = Cyx−1/4
1/4
wall −T
. Choosing the temperature as being
The constant C is given by C = gT
2
4 /
solely a function of the similarity variable is dictated by the requirement that the
temperature be constant on the boundary.
Insertion of these into the momentum and energy equations gives
d2 f
df
d3 f
+
3f
−2
3
2
d
d
d
2
+
= 0
(7.5.3)
d2
d
+ 3 Pr f
= 0
d2
d
(7.5.4)
subject to the boundary conditions
f0 = 0
df
0 = 0
d
0 = 1
df
= 0
d
= 0
(7.5.5)
Pohlhausen (1921a and b), Schmidt and Beckmann (1930), and Ostrach (1953)
all contributed to the solution of this problem. Their principle results are shown in
Table 7.5.1.
202
Thermal Effects
TABLE 7.5.1 Wall values for natural convection on an isothermal vertical wall
Pr
d
d
wall
d2 f
d2
wall
0.01
0.72
0.733
1.0
2.0
10.0
100
1000
0080592
050463
050789
056714
0716483
1168
21914
397
09862
06760
06741
06421
05713
04192
02517
01450
The rate of heat transfer from the wall per unit area is given by
√
k 1/4 d
T
′′
= − 2Twall − T Grx
qwall x = −k
y wall
x
d 0
(7.5.6)
where Grx is the local Grashof number given by
g Twall − T x 3
(7.5.7)
2
The Grashof number is a dimensionless quantity that represents the ratio of buoyancy
forces to viscous forces in natural convection.
Grx =
7.6 Natural Convection on a Vertical Plate with Uniform
Heat Flux
Sparrow and Gregg (1956) were able to find the flow on a vertical plate in a manner
similar to that used in the previous section. This time the issue in determining the
similarity variables is that the temperature gradient normal to the wall be constant.
Again, choosing stream and thermal functions in the form of a power of x times a
function of , a variable that is of the form y times another power of x, they found
the momentum and energy equations along the plate could be put in similarity form if
they chose
x = 4 Cx3/4 f/
T = T + Twall − T Pr
(7.6.1)
1/5
′′
/5k 2
and C2 =
where = Cyx−1/4 . The constants are given by C1 = 2 gqwall
4
′′
3
3 1/5
5 gqwall / k . This time the choice of the various powers is dictated by the
requirement that the temperature gradient be constant on the boundary,
Insertion of this stream function and temperature distribution into the momentum
and energy equations gives
d2 f
df
d3 f
+
4f
−3
d3
d2
d
2
d
d2
+ Pr 4f
−
d2
d
df
d
−
= 0
(7.6.2)
= 0
(7.6.3)
subject to the boundary conditions
f0 = 0
df
0 = 0
d
d
0 = 1
d
The principle results are given in Table 7.6.1.
df
= 0
d
= 0
(7.6.4)
7.8
203
Integral Method for Natural Convection on an Isothermal Vertical Plate
TABLE 7.6.1 Wall values for natural convection on a constant heat
flux vertical wall
Pr
0.1
wall
d2 f
d2 wall
−27507
16434
1.0
−13574
072196
10.0
100.0
−076746
−046566
030639
012620
The temperature at the wall is given by
Twall = T −
′′
xqwall
Grx−1/5 0
k
(7.6.5)
where Grx is the local Grashof number given by
Grx =
′′
x4
gqwall
5k /2
(7.6.6)
7.7 Thermal Boundary Layer on Inclined Flat Plates
The results of the previous two sections can also be used for nonvertical plates by
replacing the gravitational acceleration g by g cos , where is the angle that the
plate makes with the vertical. As the departure angle becomes too large, gravity also
tends to induce motion normal to the plate, which acts to thicken the momentum
boundary layer and causes separation. Details on some of the investigations can be
found in the works of Rich (1953), Vliet (1969), Fujii and Imura (1972), and Pern and
Gebhart (1972).
7.8 Integral Method for Natural Convection on an Isothermal
Vertical Plate
The integral method used in previous sections can also be adapted to free convection
problems. Not having an outer flow makes the definitions of displacement and momentum thicknesses useless, so we return to the integrated momentum and energy equation
and take the limit as U goes to zero. The result is
d
vx
2
0 gT − T dy −
v dy =
(7.8.1)
dx 0 0 x
y wall
0
d
T
0 cp
vx T − T dy = −k
(7.8.2)
dx
y wall
0
For the velocity and temperature profile choose
y 2
y
vx = u 1 −
T = T + Twall − T 1 −
(7.8.3)
y 2
(7.8.4)
204
Thermal Effects
These satisfy the constant temperature and no-slip conditions at the wall. Both are zero
and have zero derivatives at the edge of the boundary layer, meaning that the shear
stress and heat flux are both zero there.
Substituting these forms into equation (7.8.1) and (7.8.2) yields
1 d 2
u =
105 dx
1 d
u =
30 dx
u
1
gTwall − T −
3
2k
0 cp
(7.8.5)
1 d
2k
u =
30 dx
0 cp
(7.8.6)
From previous results it seems reasonable to assume that u and both behave as
powers of x. Choose = Cxm , and substitute this into equation (7.8.7) to find
ux =
60kx−m+1
−m + 10 cp C
or
(7.8.7)
60 x−2m+1
ux =
1 − m Pr C 2
Substituting this into equation (7.8.5), and requiring that the power of x in each term is
the same, gives m = 1/4. With this at hand, the constant C can be found to be
C = 2401/4 Pr −1/2
= 3936Pr
−1/2
1/4
20
+ Pr
21
20
+ Pr
21
2 gTwall − T
−1/4
2
1/4
2 gTwall − T
−1/4
2
(7.8.8)
Then
= 3936Pr
ux = 5164
−1/2
1/4
20
+ Pr
21
20
+ Pr
21
2 gTwall − T
2
−1/2
2 gTwall − T
2
−1/4
x1/4
(7.8.9)
1/2
x1/2
(7.8.10)
7.9 Temperature Distribution in an Axisymmetric Jet
The solution for a submerged jet given in Section 5.4.5 can have a temperature distribution added to it. Let
vx TR2 sin d
(7.9.1)
Q = 2cp
0
be the total heat flux in the jet, and also let the temperature be given in the form
TR =
1
G
R
= cos
(7.9.2)
205
Problems–Chapter 7
The energy boundary layer equation then becomes
cp
dG
df
G+f
d
d
=k
dG
d
1 − 2
d
d
(7.9.3)
Here, the stream function f is as given in Section 5.4.5. The boundary conditions to be
applied to G are
dG
=0
d
at
=0
and
(7.9.4)
=
These conditions ensure that there is no heat transfer across the axis of the jet.
Equation (7.9.3) can be integrated once to give
dG
= Pr f G
1 − 2
d
and again to give
a
a+1−
G = A
(7.9.5)
Pr
(7.9.6)
where equation (5.4.31) has been used for f . The constant of integration in equation
(7.9.5) was chosen as zero to satisfy the condition on = 0.
The second constant of integration, A, is determined from equation (7.9.1) to be
1
Gd. Since f and G are both known and are relatively simple,
A = Q/2cp −1 df
d
the integration can be carried out, giving, finally,
A=
Q
4cp
a+2
a
1−
2 Pr +1
a+2
2 Pr +1
a
a
1−
−
2 Pr −1
a+2
2 Pr −1
(7.9.7)
In the special case where a ≪ 1, this simplifies to
A ≃ Q
2 Pr +1
8cp
(7.9.8)
Problems—Chapter 7
7.1 A point source of heat such as a candle produces a rising thermal plume. Using
cylindrical polar coordinates and starting with the basic boundary layer equations in
the form
v
vr vr
+ + z = 0
r
r
z
v
v
vr z + v z z =
r
z
cp vr
T
T
+ vz
r
z
2
vz 1 v z
+
+ − g
r2
r r
2
=k
T 1 T
+
r2 r r
206
Thermal Effects
show that a similarity solution can be found in the form
4
f
=
G
d
T − T =
− g = g
dT
z
=
2 G
4 3
1/4
r
z1/4
Here, G is a constant related to the strength of the heat source, defined by
G=
2rvz − gdr
0
Give the similarity equations and the boundary conditions. Do not solve the equations.
(Closed form solutions have been found for Prandtl numbers of 1 and 2.)
7.2 A line source of heat produces a rising thermal plume in the form of a sheet.
Using Cartesian coordinates and starting with the basic boundary layer equations in
the form
2
vy
v
v
vx
vx
+
= 0 vx x + vy x =
+ − g
x
y
x
y
y2
cp vx
T
T
+ vy
x
y
2
=k
T
y2
show that a similarity solution can be found in the form
=
Gx3
3
2
− g = g
1/5
f
=
1
5
1
d
T − T =
dT
125
2 G
4 3
1/5
3 G4
x3 2
y
x2/5
1/5
Here,
G is a constant related to the strength of the heat source, defined by G =
v − gdy. Give the similarity equations and the boundary conditions. Do not
− x
solve the equations. (Closed form solutions have been found for Prandtl numbers of 5/9
and 2.)
7.3 For a plane
Poiseuille flow between parallel planes, the velocity distribution is
2
vx = U0 1 − ay 2 +Ulower + Uupper − Ulower ay . Find the temperature distribution including
the effects of dissipation. The upper wall temperature is Twall , and the lower wall heat
.
flux is q = k dT
dy
7.4 A horizontal semi-infinite plate is submerged in a uniform stream of a very low
Prandtl number fluid (a liquid metal). The fluid is at a temperature of T with constant
properties, and the wall is maintained at a constant temperature Twall . Using the integral
method with a temperature distribution in the form T = A + B sinCy, find A B, and
C; the thermal boundary layer thickness; and the heat transfer at the wall. (Hint: The
very low Prandtl number means that the thermal boundary layer is much thicker than
the momentum boundary layer.)
7.5 Use the integral method and the following velocity and temperature profiles to
find the boundary layer on a vertical flat plate with constant heat flux q.
y
y 2
y 2
q
1−
T = T +
vx = u 1 −
2k
Hint: As in the temperature-specified natural convection case, expect that u and will
be of the form of powers of x.
Chapter 8
Low Reynolds Number Flows
8.1 Stokes Approximation
8.2 Slow Steady Flow Past a Solid
Sphere
8.3 Slow Steady Flow Past a Liquid
Sphere
8.4 Flow Due to a Sphere Undergoing
Simple Harmonic Translation
207
209
210
212
8.5 General Translational Motion of
a Sphere
8.6 Oseen’s Approximation for Slow
Viscous Flow
8.7 Resolution of the Stokes/Whitehead
Paradoxes
Problems—Chapter 8
214
214
216
217
Flows that occur at small values of the Reynolds number are important for studying the
swimming of microorganisms, motion of small particles, determining the viscosity of
suspensions, motion of glaciers, micro- and nanotechnology, and many other applications. The mathematical difficulties involved are somewhat subtler than those found in
large Reynolds number flows but are fundamentally much the same.
8.1 Stokes Approximation
For very small values of the Reynolds number Stokes (1851) proposed that the convective acceleration terms could be neglected and the Navier-Stokes equations replaced by
v
= −p + 2 v
t
· v = 0
(8.1.1)
Taking the divergence of the first of equation (8.1.1) gives
2 p = 0
(8.1.2)
and taking the curl gives
= 2
t
with = × v
(8.1.3)
207
208
Low Reynolds Number Flows
If the time dependence of the velocities is of the form et , then it follows from
equation (8.1.1) that
p = 2 v − v
(8.1.4)
Since the pressure satisfies Laplace’s equation, a particular solution of equation (8.1.4)
is then
1
vparticular = − p
(8.1.5)
This is useful in solving for pressure once the velocity is known.
Stokes originally solved the problem of flow about a sphere in terms of the stream
function. Lamb (1932) presented a general solution for the pressure in terms of a Taylor
series. He also presented a general solution for the velocity in the form
v R = + × × A + R × where
2
2
= 0
− 2 A = 0 and
t
2
−
= 0
t
(8.1.6)
Since attention will be restricted here to simple geometries, these three approaches
are too complicated for our purposes. Instead, Stokes original solution supplies three
fundamental solutions of equation (8.1.1): a doublet, a stokeslet, and a rotlet.1
1. Doublet. This is the same doublet as that found previously for irrotational inviscid
flow—namely,
r =
A·R
R2
R = xi + yj + zk
(8.1.7)
A×R
Its velocity is given by v = =
. Here, A is a vector that can either be
R3
constant or a function of time.
2a. Stokeslet for steady flows. The velocity field
v=
B B·R
+ 3 R
R
R
(8.1.8)
where B is a constant vector, satisfies the time-independent case of (8.1.1). It
corresponds to the second term in (8.1.4). It contributes a force on a body, but
not a moment.
2b. Stokeslet for unsteady flows. The velocity vector
k2 t
−B 2 f + B · f where
(8.1.9)
1 − eikR
1
f = 2 ik +
k
R
√
and where k = 1 + i /2 and B are constants, satisfies equation (8.1.1)
for unsteady flows. The form for f has been chosen so that in the limit as k
approaches zero, the steady-state form equation (8.1.8) is recovered.
v = e−i
1
The terms Stokeslet and rotlet were coined in the later half of the twentieth century.
8.2
209
Slow Steady Flow Past a Solid Sphere
3a. Rotlet for steady flows. The velocity vector
C·R
v = R×
R3
(8.1.10)
where C is a constant vector, has been termed a rotlet. There is no pressure field
associated with it, nor does it contribute to the force acting on a body. It can,
however, exert a turning moment on a body.
3b. Rotlet for unsteady flows. The velocity vector
C·R
2
v = R×
h where h = eikR−k t/ 1 − ikR
(8.1.11)
R3
satisfies equation (8.1.1) for unsteady flows. Notice that in the limit as k
approaches zero, this velocity becomes equal to that given in equation (8.1.10).
8.2 Slow Steady Flow Past a Solid Sphere
For slow flow past a sphere of radius a, Stokes superimposed a uniform stream, doublet,
and stokeslet, all oriented in the z direction. Thus,
1 R
v = U k + B
+ C × × kR
R2 z
(8.2.1)
1
3z
z
1
= U + 3 B − C k + − 5 B − 3 C R
R
R
R
R
On R = a the velocity must vanish. Thus,
1
1
B − C = −U
a3
a
−
3
1
B − 3 C = 0
a5
a
yielding
B=−
a3
U
4
C=
3a
U
4
(8.2.2)
Thus,
3
3a
3a z 3az
a3
− 2 eR
k+U
v = U 1− 3 −
4R
4R
4R4
4R
3a
a3
a3
3a
= U 1−
+ 3 eR cos + U −1 +
+ 3 e sin
2R 2R
4R 4R
From this and equation (8.1.1) the pressure can be found as
p = p0 −
3a
U cos
2R2
(8.2.3)
and the stress components as
9a
3a3
−
cos
2R2
R4
1 vR v v
3a3
+
−
U sin
=−
R =
R
R
R
2R4
RR = −p + 2
vR
= U
R
(8.2.4)
210
Low Reynolds Number Flows
On the surface of the sphere, these stresses become RR a =
3
− 3a2a U sin The force on the sphere is thus given by
Fz =
0
3
U
2a
cos R a =
RR cos − R sin 2R sin a ad
= 3aU
0
cos2 + sin2 sin d
(8.2.5)
= 6Ua
The preceding solution has had many uses. Perhaps the most noteworthy was
determination of the charge on an electron by Robert A. Millikan, who was awarded
the 1923 Nobel Prize in physics for his work. First he sprayed a few tiny drops of oil
between two parallel plates. He then measured the diameter of the drops and, using the
fact that Fdrag = Weight − Buoyant Force, determined the diameter of the drops. He
then applied an electric charge on the plates in a direction opposite to gravity and again
measured velocity. The electric charge induces a force qE/D, where q is the charge on
the electron, E is the strength of the electric field, and D is drop diameter. Since the
number of electrons attached to a given drop is unknown, a number of measurements
had to be taken. The experiment does not give an exact answer, since because of small
droplet size Brownian motion causes an appreciable percentage of error. Nevertheless,
his results gave a far more accurate estimate than was previously obtainable.
Stokes’s result for the force on a sphere is also used in viscometry. By dropping
a small sphere of known diameter in a fluid and measuring the terminal velocity, the
viscosity can be deduced.
Einstein (1906, 1911) also used Stokes’s equations to provide an estimate of the
viscosity of dilute suspensions. He found that the equivalent viscosity was given by
equivalent = fluid 1 + 25
where
4
volume of spheres
= R3 c =
3
volume of suspension
Here, c is the concentration of the suspended material. (The preceding formula is a corrected version of Einstein’s results, as found in the text by Landau and Lifshitz (1959)).
When there are clusters of spheres falling in a fluid—an unusual behavior where
some particles pass others and the passed particles speed up and pass the original
passers—has been observed and explained in part using Stokes’s solution for the drag
force (Hocking, 1964).
8.3 Slow Steady Flow Past a Liquid Sphere
A result similar to the preceding for liquid spheres is also useful, as was demonstrated by
Millikan’s experiments. Expect that equation (8.2.1) can be used outside of the sphere,
but because it is infinite at the origin, a different nonsingular solution must be found
inside the sphere. Also, the doublet and stokeslet are not applicable inside the sphere
for the same reason. The rotlet can be used if C is chosen as
1
C= k
6
2
x2 + y2 + 15z2 x2 + y2 −
13 4
z
3
(8.3.1)
8.3
211
Slow Steady Flow Past a Liquid Sphere
This satisfies equation (8.1.1) and has the velocity field zer − 2r 2 k expressed in cylindrical polar coordinates. Then, using this rotlet plus a uniform stream of strength D in
the z direction, the result is
⎧
2B C
C
B
⎪
⎨ U−
−
cos eR + −U − 3 +
sin e R ≥ a
R3 R
R
R
v=
(8.3.2)
⎪
⎩
2
2
R ≤ a
D − R E cos eR + −D + 2R E sin e
Requiring that both of the v ’s be the same at R = a gives U + aB3 − Ca = D − 2a2 E.
Requiring that both of the vR ’s be zero at R = a (so that the sphere does not change
in size) gives
U−
2B C
− = D − a2 E = 0
a3
a
Thus
(8.3.3)
3U
C
3U
C
+ E = − 2 + 3
2
4a
2a
4a
The constant C is determined by requiring that R be continuous on the surface of
v
v
the sphere. Using R = R1 vR + R − R gives
B = a3 U − Ca2
D=−
C=−
inner
Ua3
4 inner + outer
(8.3.4)
Then
⎧
2outer + 3inner a
inner
a3
⎪
⎪
U cos 1 −
⎪
+
e
⎪
⎪
outer + inner 2R outer + inner 2R3 R
⎪
⎪
⎪
⎪
⎪
⎪
⎨
inner
a3
2
+ 3inner a
+
e
+U sin −1 + outer
v=
outer + inner 4R outer + inner 4R3
⎪
⎪
⎪
⎪
⎪
⎪
2
⎪
⎪
R
2R2
outer
⎪
⎪
U
cos
−
1
e
+
sin
1
−
e
⎪
R
⎩2
+
a2
a2
outer
for R ≥ a
for R ≤ a
inner
(8.3.5)
The portion of the pressure field is found by substituting equation (8.3.5) into
the steady form of equation (8.1.2). A body force must be included to provide the
mechanism for the flow. Here, gravity has been chosen as the most common body force.
The result is
⎧
outer a 2inner + 3outer
⎪
⎪
cos
−U
−
Rg
for R ≥ a
outer
⎪
⎨
2 R3 inner + outer
p=
(8.3.6)
⎪
5outer inner R
⎪
⎪
− inner Rg
for R ≤ a
⎩cos U
inner + outer a2
Having the velocity field, the stresses can be found as
⎧
⎪
⎪−3U outer inner
⎪
⎪
⎨
2 outer + inner
1 R
R =
− +
=
⎪
R
R R
⎪
outer inner
⎪
⎪
⎩−3U
2 outer + inner
a3
sin
R4
R ≥ a
R
sin
a2
R ≤ a
212
Low Reynolds Number Flows
RR = −p + 2
= cos
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
R
R
3
1 2outer + 3inner a
inner
a
3outer U
−
2
2
outer + inner
R
outer + inner R4
+outer Rg R ≥ a
R
inner
+ inner Rg R ≤ a
−3outer U
outer + inner a2
Therefore, on R = a,
⎧
2outer + 3inner
3U
⎪
⎪
− inner outer − outer ag R = a+
⎪
outer
⎨
2a
outer + inner
outer + inner
RR = cos
⎪
⎪
3U outer inner
⎪
⎩−
− inner ag
R = a−
a outer + inner
Requiring the two forms of the normal stress RR be the same at R = a gives
3U
2outer + 3inner
− inner outer − outer ag
outer
2a
outer + inner
outer + inner
=−
3U outer inner
− inner ag
a outer + inner
yielding for the liquid sphere’s terminal velocity
outer + inner
2 outer − inner a2 g
U=
3
outer
2outer + 3inner
(8.3.7)
Limiting cases of this which are of interest are
U→
2ga2 outer − inner
as inner → (solid sphere)
9outer
U→
ga2 outer − inner
as inner → 0 (vapor sphere).
3outer
(8.3.8)
In the preceding analysis, surface tension has been neglected because the combination of a perfectly spherical liquid drop and surface tension is incompatible with
gravity. Generally speaking, providing the Reynolds numbers are small, surface tension
will play a minor role. However, for very small gas bubbles rising in a liquid, it has
been found that their velocity is closer to that predicted for a solid sphere rather than the
second of equation (8.3.8). In this case, it is believed that the presence of surface-active
impurities in the liquid can form a mesh of large molecules on the interface, causing it
to behave like a rigid surface (Levich, 1962).
8.4 Flow Due to a Sphere Undergoing Simple
Harmonic Translation
If a solid sphere undergoes simple harmonic motion along the z-axis in a fluid that is
otherwise at rest, the flow can be described by a stokeslet-doublet combination. A stream
function approach can be used and the development in Lamb (1932, page 643) followed.
8.4
213
Flow Due to a Sphere Undergoing Simple Harmonic Translation
The calculations are a bit lengthy, so to simplify things, complex numbers are used for
the stream function, velocity components, and stresses, with the understanding that the
real portions of the expressions are intended.
Inserting the general form = sin2 f r U with U = U0 eit into equation (8.1.1)
gives solutions
A B
+ 1 + kR e−k R−a
R R
√
√
Here, k = i/ = 1 + i with = /2. In obtaining solutions of equation
(8.1.1) those portions that die out as R becomes large have been selected.
Applying the no-slip boundary conditions on the surface of the sphere, the constants
A and B are found to be
3a
3a
1 2 2
A = 2 1 + ka + k a B = − 2
2k
3
2k
fR =
giving
3 sin2 a
=
2k2 R
1 2 2
1 + ka + k a − 1 + kR e−k R−a U
3
This yields velocity components
1
3 cos a 3
vR = 2 2
U 1 + ka + k2 a2 − 1 + kR e−k R−a
ka
R
3
3 sin a 3
1
v = 2 2
U 1 + ka + k2 a2 − 1 + kR + k2 aR e−k R−a
2k a R
3
(8.4.1)
(8.4.2)
Having found the stream function and velocity, the pressure can be found from
equation (8.1.5). Since the pressure satisfies the Laplace equation, it must be associated
with the irrotational portion of the flow: the doublet. Thus, the pressure satisfies
p = −
vdoublet
t
The pressure is thus given by
p = iUA cos /R2
(8.4.3)
The force needed to produce this motion is found as before by integrating the shear
stresses over the surface of the sphere as in equation (8.2.5). The result is
Fz = −
RR cos − R sin a 2a2 sin d
0
4
1
9
1
1
iU + 3a3
= a3
+
1+
U
3
2 4a
a
a
(8.4.4)
2
2
2
dU
1 9
4
+
+ 3a3
= a3
1+
U
3
2 4 a2 dt
a2
a2
214
Low Reynolds Number Flows
The one-half in the acceleration term represents the added mass found in potential flow
theory, while the accompanying term represents a viscous correction to the added mass.
We can see that as the frequency becomes small, the result reduces to that found for the
stationary sphere.
8.5 General Translational Motion of a Sphere
The results of the previous section can be generalized to arbitrary translational motion
by use of the Fourier transform. If the velocity of the sphere is taken to be U t , then
Fourier transform theory tells us that there is a Fourier representation of the velocity in
terms of the pair of functions
1
U t eit dt
(8.5.1)
F e−it d and F =
Ut =
2 −
−
Landau and Lifshitz (1959, page 96) showed that the drag for each harmonic component
for the previous problem is given by
2
6
2
3 it
a e
F +
+3 1+i
Ḟ
(8.5.2)
a2
3
where Ḟ is the Fourier transform of dU
. The integration of this to produce the force
dt
requires some careful handling of the path of integration (a not uncommon problem
with Fourier integrals), the net result being
3 t dU d
1 dU
(8.5.3)
+3 2U +
F t = 2a3
√
3 dt
a
a − d
t−
8.6 Oseens Approximation for Slow Viscous Flow
If the solution of Section 8.2 for slow flow past a sphere were to be investigated further,
it would be found that at large distances from the sphere the neglected convective
acceleration terms were actually more important than the viscous terms retained in our
analysis. The ratio of the inertia to viscous forces is in fact of the order of Ur/.
Thus, any efforts made to try to use the Stokes approximation as a starting point for
solutions of the Navier-Stokes equations for somewhat larger values of the Reynolds
number would rapidly run into insurmountable difficulties. This has been termed the
Whitehead paradox, after the mathematician Alfred North Whitehead.
Even more embarrassing, if an attempt were made to solve the Stokes equations
for the slow flow past a circular cylinder, it would be found that the solution for the
stream function is of the form
D
3
sin
(8.6.1)
= Ar + Br ln r + Cr +
r
The constants A and B have to be rejected because they give velocities that grow
faster with r than does a uniform stream. C represents a uniform stream, so what is
left is one constant, D, to satisfy two boundary conditions! This is known as Stokes
paradox. (A dictionary definition of paradox is “an argument that apparently derives
8.6
215
Oseens Approximation for Slow Viscous Flow
self-contradictory conclusions by valid deduction from acceptable premises.” As we
will see in the next section, in this case the “acceptable premises” were not valid.)
Oseen (1910) suggested an alternative approach, approximating the Navier-Stokes
equations by
v
v
+U
= −p + 2 v
(8.6.2)
t
x
U being the free stream velocity in the x direction. Oseen obtained a solution for the
circular cylinder in terms of
vr =
1
+
− cos
r 2k r
1
1
+
+ sin
v =
r 2kr
(8.6.3)
where
r
1 a n
r
= Ua
cos n
cos + A0 ln +
A
a
a n=1 n n r
= Ue
kr cos
(8.6.4)
Bn Kn kr cos n
n=0
Here, k = U/ and the Kn ’s are the modified Bessel functions. The An and Bn are
determined from the no-slip boundary conditions.
Oseen also used his approximation to find the slow flow past a sphere. The solution
is more complicated than the Stokes equations and involves much tedious computation.
His result for the drag force is
3
(8.6.5)
F = 6aU 1 + Re where Re = Ua/
8
Stokes’s formula is seen to be the limit as Re goes to zero.
Goldstein (1929) carried out a series expansion in the Reynolds number for the
sphere using Oseen’s equations. His results for the drag coefficient gave
24
19
71
30179
3
Re2 +
Re3 −
Re4
1 + Re −
CD =
Re
16
1280
20480
34406400
(8.6.6)
122519
5
Re ± · · ·
+
550502400
Here the Reynolds number is based on the diameter. As we will see in the next section,
this result is not correct, since an expansion solely in powers of Re is inadequate.
The drag force per unit length of a circular cylinder as determined by the Oseen
equation was found by Lamb (1911) to be
Fper
unit length
=
4U
Ua
1
− − log
2
4
where is Euler’s constant (also called Mascheroni’s constant)
n
1
= lim
− log n = 0577215664901533
n→
m=1 m
(8.6.7)
(8.6.8)
216
Low Reynolds Number Flows
Another version of the drag force on a circular cylinder was given by Tomotika
and Aoi (1950) as
8
1
1
5
CD =
− + S Re2 + O Re4 ln2 Re
(8.6.9)
1−
ReS
32 16S 2
8
where Re = UD
and S = 17 − + ln Re
. Their result was achieved by using matched
asymptotic expansions.
For all of the formidable theory and numerous calculations that went into these
results, comparison with experiments (Van Dyke, 1965, page 164) seems best for the
result given in (8.6.9) up to a Reynolds number based on diameter of about 1.4.
8.7 Resolution of the Stokes/Whitehead Paradoxes
While Lamb (1932) points out many of the questions raised by the Stokes/Oseen equations, it wasn’t until the mid-1950s that the questions were finally resolved. Proudman
and Pearson (1957) and Kaplun (1957) independently pointed out that the Stokes equation hold only in the vicinity of the boundary and the Oseen equations hold only away
from the boundary. Using matched asymptotic expansions, they found that an intermediate region could be found where both Stokes and Oseen were valid. By matching
terms in that region, they were able to develop a solution. The result was an equation
for the drag coefficient of the sphere given by Proudman and Pearson as
9
3
6
CD =
(8.7.1)
1 + Re + Re2 ln Re + O Re2
Re
8
40
The logarithmic term is forced by the matching process and the logarithms of the
radius that appear in both the Stokes and Oseen solutions. Any expansions involving
the Navier-Stokes equations should be aware that logarithms of the Reynolds numbers
must be anticipated. This is presaged in equation (8.6.7).
Thus, both the large Reynolds number and the small Reynolds number solutions
have a common nature. For the large Reynolds number case, Prandtl’s boundary layer
equations must be used near the boundary, and the Euler equations must be used away
from the boundary. For the small Reynolds number case, Stokes equations must be used
near the boundary, and Oseen’s equations must be used away from the boundary. This
is true for the lowest-order expansions in each category. As more than one term in the
expansion is considered, the necessary equations become modified and, as should be
anticipated, more complex.
One final remark must be made about the order of the various terms in these
expansions, particularly terms involving ln Re. As pointed out by Van Dyke (1964,
Sections 1.3, 10.5), when the next unknown term in an expansion is of order Ren , then
even though mathematically it is smaller than Ren ln Re, since it numerically can be of
equal magnitude, it should be included. That is, even though
Ren
→ 0 for Re = 001— say
Re→0 Ren ln Re
Re2
10−4
1
= −4
=−
= −0217
2
4605
Re ln Re 10 ln 001
lim
which means that for practical purposes Re2 ln Re and Re2 can contribute equally to a
calculation.
217
Problems—Chapter 8
Problems—Chapter 8
8.1 Show that the flow field given by v = ReR × ez p = 0 is a solution of the
Stokes equations. Use it along with a steady rotlet to find the velocity and pressure for a
fluid contained between two concentric spheres of radii a and b, where the outer sphere
is rotating with angular velocity outer about the x-axis and the inner sphere is rotating
with angular velocity inner about the same axis and in the same direction.
8.2 Millikan’s experiment for making the first accurate measurement of the charge
on an electron involved spraying a few tiny droplets of oil between two electrically
charged plates. He first determined the diameter of the small drops by measuring their
terminal velocity without an electric field VwithoutE and using Stokes’s formula. He
then applied a voltage E (the electric potential in volts N − m/C ) across the plates,
resulting in an upward force Eq/d on the drop and a terminal velocity VwithE . Here, d
is the plate spacing in meters, and q is the electron charge in Coulombs (C). Making
repeated measurements, he found that the values of q he determined were always integer
multiples of a specific number q1 , the charge due to a single electron. Determine first a
formula for the droplet diameter when no electric field is present. Then determine the
terminal velocity when the electric field is present. Solve for q in terms of the weight
of the droplet minus the buoyancy force, the terminal velocities with and without an
electric field, and the quantity E/d.
8.3 In a repeat of Millikan’s experiment (Problem 8.2), the voltage was 115 volts,
the plate spacing 4.53 mm, the oil density was 920 kg/m3 , and the air viscosity was
182 × 10−4 poise. Two horizontal lines were drawn 1.73 mm apart, and the time of
traverse of these two lines was measured with a stopwatch. By changing the direction
of the electric filed, it was possible to use the same droplet for each measurement,
although each set of measurements did have different numbers of electrons attached to
the droplet. Times of transit were as follows:
72.2
41.1
23.5
16.5
12.5
8.06
seconds
seconds
seconds
seconds
seconds
seconds
averaged
averaged
averaged
averaged
averaged
averaged
over
over
over
over
over
over
10 measurements, no electric field, direction down
8 measurements, electric field present, direction up
2 measurements, electric field present, direction up
5 measurements, electric field present, direction up
3 measurements, electric field present, direction up
7 measurements, electric field present, direction up
Find the terminal velocities for the six sets of data, and estimate the charge due to a
single electron. Be sure to observe the proper sign on the velocities.
Note: Millikan also found that Brownian motion due to the small size of these
droplets gives errors in measuring velocity, resulting in charge estimates that are about
27% too high.
Chapter 9
Flow Stability
9.1
9.2
Linear Stability Theory of Fluid
Flows 218
Thermal Instability in a Viscous
Fluid—Rayleigh-Bénard
Convection 219
9.3
Stability of Flow Between Rotating
Circular Cylinders—Couette-Taylor
Instability 226
9.4 Stability of Plane Flows 228
Problems—Chapter 9 231
9.1 Linear Stability Theory of Fluid Flows
Chapters 3 and 4 examined several examples that involved the stability of inviscid
flows, such as the round jet, the vortex street, and several cases of interfacial waves. In
those cases an infinitesimal disturbance was added to a primary flow, and then it was
determined whether there was a possibility of the disturbance growing, decaying, or
remaining unchanged. The driving mechanisms involved in those examples were gravity
and surface tension interacting with the momentum of the flow.
In this chapter the same type of analysis is used and applied to laminar viscous
flows as well. The presence of viscosity in some cases will dampen the destabilizing
influences and in others act to enhance them and thus destabilize the primary flow. Flows
frequently are laminar for sufficiently small values of a dimensionless parameter, such
as the Reynolds number, and become turbulent once the parameter becomes sufficiently
large. While this transition value of the parameter can be found experimentally, as
Reynolds did for pipe flow, in many cases a flow has so many describing parameters
that a complete experimental study becomes too costly. An analytic approach then can
be desirable, with the added advantage of possibly shedding some light on the physical
mechanisms involved in the transition.
The classical approach in studying such transitions, introduced in the nineteenth
century in the study of inviscid flows, is to take the solution for the laminar flow and
add to it a very small disturbance. There is usually some experimental knowledge of
the form of the instability, and the disturbance is tailored to reflect this. This combined
flow is then put into the Navier-Stokes equations, and the original laminar flow terms
are subtracted. By themselves, the primary flow satisfies the Navier-Stokes equations.
218
9.2
219
Thermal Instability in a Viscous Fluid—Rayleigh-Bénard Convection
However, some interaction terms between the primary and secondary flows remain
behind by this process. The remaining equations are then linearized in the disturbance
quantities. This set of partial differential equations is then solved to determine whether
the disturbance will grow or decay in time. For viscous flows, the mathematics for
finding the solution of the stability equations is generally much more complicated than
it is for inviscid flows, due to the higher order of the resulting differential equations
and also possible additional physical mechanisms for supporting the instability.
The use of infinitesimal disturbances results in linear equations, which by themselves have proven difficult to solve. In some of the cases where nonlinear formulations
have been made and solutions found, the efforts naturally increase the difficulty of
finding a solution. Landau and Lifshitz (1964) present some of the procedures and
assumptions involved with these nonlinear equations.
Because a linear approach is being used, if our solution predicts an unstable solution
of the primary flow, this by itself cannot predict the nature of the secondary flow. In
some cases a direct transition to turbulent flow exists, and in others, secondary laminar
flows evolve, more complicated in nature than the original primary flow but still not the
fully complex character associated with turbulent flow. Some flows also can evolve to
have tertiary flows that are still laminar. Such is the complexity of fluid flow behavior!
9.2 Thermal Instability in a Viscous Fluid—
Rayleigh-Bénard Convection
One of the least mathematically complicated examples of the analysis of viscous flow
instability is due to Lord Rayleigh (1916a). Several investigators in the nineteenth
century had noted that tesselated structures were frequently seen in the drying of
horizontal thin layers of fluids. Bénard (1900, 1901) performed quantitative experiments
on this phenomenon, most of the details later explained by Rayleigh’s analysis. This
instability can be seen in the drying of paints with metallic particles and also in the
stratified atmosphere, where it results in interesting cloud patterns.
While there are many versions of the boundary conditions that can be studied,
only the case of a fluid confined between two horizontal plates, each with horizontal
extent large compared to their vertical spacing, will be considered here. (A thorough
description of the extensive literature available on this problem is given in Koschmeider
(1993)). The plates are each heated to a constant temperature, with the lower plate being
the hottest, as shown in Figure 9.2.1. Thus, there is a basic gravitational instability,
with lighter fluid on the bottom. Initially, the fluid between the plates is at rest, with a
z
Tcold
d /2
d /2
Thot
Figure 9.2.1 Physical layout for studying Bénard stability
220
Flow Stability
vertical hydrostatic pressure gradient. The density will be taken as varying linearly with
temperature according to
Thot + Tcold
(9.2.1)
2
Here, reference density mean and reference temperature Tmean have been taken as the
density and temperature midway between the plates. Beta is given by
= mean 1 − T − Tmean
=−
with Tmean =
1 d
1 d dz
1
=−
=−
dT
dz dT
T
(9.2.2)
The plate spacing will be chosen as b, and the origin will be taken halfway between the
plates.
Since there is no mean flow for this problem, the Navier-Stokes and energy equations for the primary undisturbed flow are given by
z
z
T0 = Tmean −
T 0 = mean 1 − T
with T = Thot − Tcold
(9.2.3)
b
b
Here, primary flow components p0 z 0 , and T0 z are denoted by the subscript 0,
and k is a unit vector in the direction of gravity z. From these and the boundary
conditions, and taking the origin halfway between the plates, find for the quiescent
primary flow that
0 = − p0 − gk0
(9.2.4a)
d T0
dy2
(9.2.4b)
2
0=k
To investigate the stability of this flow, introduce a small disturbance and see
whether it grows or decays. For the disturbed flow, velocities, pressure, density and
temperatures will be denoted by primes. The combined primary and disturbance flows
must satisfy the continuity, Navier-Stokes, and energy equations.
It shall be assumed that even though the basic density does vary with z, the main
influence of the density variation on the flow stability is going to be felt in the body
force and not in the acceleration term. Thus, the 0 multiplying the acceleration term
is taken at its mean value and is considered constant, and the flow will be regarded as
being incompressible. This is called the Boussinesq approximation.
After linearizing on the small disturbance terms and then subtracting out the primary
flow terms, the terms’ first order in the primed quantities are
mean · v′ = 0
(9.2.5)
′
v
= − p′ + mean gT ′ k +
t
′
T
′ T
mean cp
− vz
= k 2T ′
t
b
mean
2 ′
v
(9.2.6)
(9.2.7)
Together with the conditions that v′ and T ′ vanish at each plate, the mathematical
statement of the problem has been completed. Now these equations must be made
manageable.
The first step is to eliminate the pressure. Taking the divergence of equation (9.2.6)
and using the continuity condition equation (9.2.5) gives
0=−
2 ′
p + mean g
T′
z
(9.2.8)
9.2
221
Thermal Instability in a Viscous Fluid—Rayleigh-Bénard Convection
Similarly, operating on the z component of equation (9.2.6) with the Laplace operator
gives
2 ′
2 ′
2 ′
2
vz = −
p + mean g 2 T ′ +
vz
mean
(9.2.9)
t
z
Using equation (9.2.8) to eliminate the pressure in equation (9.2.9) gives
2 ′
T′
2 ′
2
vz = −
mean g
+ mean g 2 T ′ +
vz
mean
t
z
z
which simplifies to
mean
where
t
−
2
2
1
2
=
−
2 ′
vz
2
z2
2 ′
1T
= mean g
2
=
x2
2
+
y
(9.2.10)
2
Next, rewrite equation (9.2.7) in the form
T ′
mean cp − k 2 T ′ = mean cp
v
t
b z
(9.2.11)
Equations (9.2.10) and (9.2.11), then, are two equations in the two unknowns vz′ and T ′ .
Since the plates are considered to be large in extent, no boundary conditions will
be imposed at the edges of the plates. The boundary conditions imposed are simply
b
on z = ±
2
At this point the velocity components and the temperature can be taken as being of
the form
vx′ = vy′ = vz′ = T ′ = 0
v′ x y z t = Uz Vz Wzfx yest
T ′ x y z t = zfx yest
(9.2.12)
(9.2.13)
where
2
2
f
f
a2
+
=
−
f
x2
y2
b2
(9.2.14)
This form allows separation of variables and is sufficiently general to describe the
observed flows.
If you have studied separation of variables in a mathematics course, you many
be surprised that this separable form had to be assumed rather than have it follow
from the analysis. The reason is that classical presentations of the method of separation
of variables all deal with second-order partial differential equations, while here our
system is actually of sixth order. Generally, systems of order higher than the second
do not separate except under very special boundary conditions. The governing equation
introduced for f allows for separating the variables and is compatible with a number of
boundary conditions. Further comments on the nature of f will be made a bit later.
Putting these forms into our system of equations, using equation (9.2.9) and cancelling f where it appears in every term of the equation, equations (9.2.5), (9.2.10), and
(9.2.11) become
U
f
dW
f
+V
+f
=0
x
y
dz
(9.2.15)
222
Flow Stability
mean s −
a2
d2
−
dz2 b2
mean cp s − k
a2
a2
d2
−
g
W
=
−
dz2 b2
b2 mean
d2
a2
− 2
2
dz
b
T
W
b
=
(9.2.16)
(9.2.17)
= 0 on
Note from equation (9.2.15) that U = V = 0 on the boundaries implies that dW
dz
the boundaries also.
It is possible that the time constant s could be complex. The real portion would be
associated with a growth or decay rate, and the imaginary part with an oscillatory behavior.
To show that the oscillatory behavior cannot occur, perform the following operations:
1. The momentum equation (9.2.16) is multiplied by the complex conjugate of W ,
denoted as W ∗ , and integrated over the gap between the plates. Using integration by
parts and the boundary conditions then gives
a2
smean
J
+
J
=
−
g
mean
b2 1 b4 2
b2
b/2
W ∗ dz
(9.2.18)
−b/2
where
J1 = b2
J2 = b 4
b/2
−b/2
b/2
−b/2
dW
dz
d2 W
dz2
2
+
a2
W 2 dz
b2
2
+2
a2 dW
b2 dz
2
+
a4
W 2 dz
b4
are both positive definite.
2. The energy equation (9.2.17) is multiplied by the complex conjugate of , denoted
as ∗ , and integrated over the gap between the plates. Using integration by parts and
the boundary conditions gives
mean bcp T 2
T b/2
k
W∗ dz
smean cp I0 + 2 I1 = −mean cp
k
b
b
−b/2
2
1 b/2
k
where I0 =
2 dz
mean bcp T
C 2 −b/2
(9.2.19)
2 b/2
2
bk
a2
d
I1 =
+ 2 2 dz
mean bcp T
dz
b
−b/2
are both positive definite. The rather awkward-appearing combination of parameters
appearing before the Is are to make the dimensions coincide in the final result.
Notice that, except for multiplying constants, the right-hand sides of equations
(9.2.18) and (9.2.19) are the complex conjugates of one another. From this it follows that
∗
smean cp b2 1
s mean cp b2
2
J + J2 = a Ra
I0 + I1
(9.2.20)
k
Pr 1
k
b3 g2
c T
mean p
is called the Rayleigh number.
where Pr is the Prandtl number and Ra =
k
Taking the imaginary portion of both sides of equation (9.2.20) gives
si J1 = −si PrRaI0
giving si J1 + PrRaI0 = 0
(9.2.21)
9.2
223
Thermal Instability in a Viscous Fluid—Rayleigh-Bénard Convection
It follows from this that if Ra is positive (Ra is positive if T is positive, so heavier
fluid is on the bottom), s must be real.
If the real portion of both sides of equation (9.2.20) is taken, the result is
sr mean cp b2
k
1
J1 − a2 RaI0 = a2 RaI1 − J2
Pr
(9.2.22)
This states that the real part of s can be either positive or negative if T is positive.
The results of the previous paragraph are often referred to as the principle of
exchange of stabilities. While the appropriateness of this title is unclear, the usefulness
of the principle is without doubt.
Since equations (9.2.16) and (9.2.17) have constant coefficients, their solution is in
the form of exponentials and they can be found in a standard manner. That is, letting
W = Aecz/b = Becz/b , equations (9.2.16) and (9.2.17) become
mean sb2 − c2 − a2 c2 − a2 A = −a2 b2 mean gB
mean cp b2 s − k c2 − a2 B = mean cp b TA
Eliminating A and B between these two equations, find that c must satisfy
2
2
mean cp b2
mean
2
2
2
2
s
c −a −
s + a2 Ra = 0
c −a −
c −a
k
(9.2.23)
where Ra is the Rayleigh number as just defined.
Equation (9.2.23) is a cubic equation for c2 , whose solution is somewhat messy.
The calculations can be simplified by saying that only the neutrally stable case is desired
to be solved; that is, only the results where the real part of s vanishes will be studied.
In that case the roots become
cn2
=a
2
Ra
1−
a
1/3
2n
2n
cos
+ i sin
3
3
n=0 1 2
(9.2.24)
The solution can now be expressed in terms of these roots. Since they occur in
complex conjugate pairs, write
W=
=
2
n=0
An cosh
cn z
c z
+ Bn sinh n
b
b
2
cn z
cn z
−
2
2 2
A
cosh
−
a
sinh
+
B
c
n
n
ga2 b2 n=0 n
b
b
Require that U V W , and vanish on z = ±b/2, or equivalently, from continuity
require that W, dW/dz, and vanish at these boundaries. The result is
224
Flow Stability
0=
0=
0=
0=
0=
0=
2
An cosh
c
cn
+ Bn sinh n
2
2
An cosh
cn
c
− Bn sinh n
2
2
n=0
2
n=0
2
c
c
cn An sinh n + Bn cosh n
2
2
n=0
2
n=0
cn
(9.2.25)
c
c
−An sinh n + Bn cosh n
2
2
2
c
c
cn2 − a2 An cosh n + Bn sinh n
2
2
n=0
2
c
c
cn2 − a2 An cosh n − Bn sinh n
2
2
n=0
These are six homogeneous equations in the six sets of unknown constants An Bn .
There is only a trivial solution unless the determinant of the coefficients of the An and
Bn vanishes. In this case the 6 by 6 determinant can be written as the product of two 3
by 3 determinants. The result of the calculation is that either
c
c
c
sinh 1
sinh 2
sinh 0
2c
2c
2c
c0 cosh 0
c1 cosh 1
c2 cosh 2
= 0
(9.2.26)
2 c1 2
2 c2
2 c0 2
2
2 2
2 2
2 2
c1 − a sinh
c2 − a sinh
c0 − a sinh
2
2
2
or
c
c
c
cosh 1
cosh 2
cosh 0
2c
2c
2c
c0 sinh 0
c1 sinh 1
c2 sinh 2
= 0
(9.2.27)
2 c
2 c
2 c
2
2
2
2
0
1
2
c0 − a2 cosh
c12 − a2 cosh
c22 − a2 cosh
2
2
2
The first 3 by 3 determinant came from the hyperbolic cosine terms in W and is referred
to as the even disturbance because of the hyberbolic cosine terms in W . The second 3
by 3 determinant came from the hyperbolic sine terms in W and is referred to as the
odd disturbance.
Expansion and simplification of the determinants in equations (9.2.26) and (9.2.27)
give
2
c
c
c
(9.2.28)
c0 − a2 c0 tanh 0 + c12 − a2 c1 tanh 1 + c12 − a2 c2 tanh 2 = 0
2
2
2
for even disturbances, and
√
√
c0 1 + i 3
c1 −1 + i 3
c
c1 coth +
c2 coth 2 = 0
c0 coth +
2
2
2
2
2
(9.2.29)
for odd disturbances. Further simplification is possible by noting that c0 is pure imaginary
and c2 is the complex conjugate of c1 . Therefore, both equations (9.2.28) and (9.2.29)
are real.
9.2
225
Thermal Instability in a Viscous Fluid—Rayleigh-Bénard Convection
Even disturbances
20,000
Rayleigh number
15,000
unstable
10,000
5,000
stable
0
0
1
2
3
4
5
Wave number
6
7
8
Figure 9.2.2 Bénard stability curve for even disturbances
Odd disturbances
60,000
Rayleigh number
50,000
unstable
40,000
30,000
20,000
10,000
stable
1
2
3
4
5
Wave number
6
7
8
Figure 9.2.3 Bénard stability curve for odd disturbances
Plots of the roots of these equations are shown in Figure 9.2.2 for even disturbances
and Figure 9.2.3 for odd disturbances. Even disturbances occur at the lowest value of the
Rayleigh number—approximately 1,707.76—occurring at a wave number a of 3.117.
For values of the Rayleigh number below this, according to our analysis the disturbances
are all stable. For values of the Rayleigh number above this critical value, there will be
some range of wave number a for which the disturbance will grow. Thus the flow will
be unstable. For odd disturbances, the corresponding values of the critical Rayleigh and
wave numbers are 17,610.39 and 5.365.
Other boundary conditions that have been used in the solution include the case where
the top surface is free Racritical = 1 1007 acritical = 2682 and when both surfaces are
free Racritical = 6575 acritical = 2221.
As yet, the shape factor f has not been found. The actual form of f will be determined by the plate geometry. There are, in fact, many geometric possibilities for the cell
shape. For instance, f proportional to either sin ax
or sin ayb will give long cylindrical
b
226
Flow Stability
sin qyb , where
rollers observed frequently in the atmosphere; f proportional√ to sin px
b
ay
3ax
p2 +q 2 = a2 , will give rectangular cells; f proportional to 2 cos 2b
cos 2b +cos ayb gives
hexagonal cells. Bénard observed hexagonal cells in his experiments, but this shape was
likely due to surface tension on the free surface at the top of his apparatus. The cell
shape is influenced by the boundary shape, and circular rollers can be observed if round
plates are used (Koschmeider, 1993).
Research has shown that the secondary flow just described leads to at least one
further laminar flow if the temperature distance between the plates is increased beyond
the critical limit.
The preceding is an exact solution for the neutrally stable curves for the linear stability problem. While the computations are tedious, the mathematical questions
encountered are relatively uncomplicated. This generally is not true when the primary
flow is nonquiescent, as shall be seen.
9.3 Stability of Flow Between Rotating Circular
Cylinders—Couette-Taylor Instability
Couette (1890) studied the possibility of using flow between rotating cylinders to
determine the viscosity of liquids. His general formula for the flow in cylindrical polar
coordinates was
B
v = Ar +
(9.3.1)
r
where
r 2 r 2 −
r 2 − r 2
B=− 1 2 2 1 2 2
(9.3.2)
A = 1 21 22 2
r 1 − r2
r 1 − r2
The subscript 1 refers to the inner cylinder and 2 to the outer cylinder. Couette built an
apparatus with a rotating outer cylinder and an inner cylinder supported on a fine torsion
wire to measure torque. His measurements were not particularly accurate, but he did
notice that as the angular speed was increased, the graph of torque versus angular speed
departed from the straight line expected for laminar flow. His results prompted Rayleigh
(1916b) to examine the flow and find that for the flow to be stable, it is necessary that
2
2
router
inner
outer > rinner
(9.3.3)
Since Couette’s apparatus did not meet this criterion, departure of the flow from the
state predicted by equation (9.3.1) was a possibility.
Taylor (1923) constructed an apparatus similar to that used by Couette but that
allowed for visualization of the flow between the cylinders. He also provided a theoretical
stability analysis and thus was able to confirm his findings. The analysis proceeded as
follows.
In the manner used in Rayleigh’s convection problem, a small disturbance is added
to the flow of equation (9.3.1) and introduced onto the Navier-Stokes equations. The
disturbance, based on Taylor’s observations, was taken to be axisymmetric and of
the form
ur = ur cos kz est
u = vr cos kz est
uz = wr sin kz e
st
(9.3.4)
9.3
227
Stability of Flow Between Rotating Circular Cylinders—Couette-Taylor Instability
Insertion of these into the Navier-Stokes equations gives
2
2V
dp
d u 1 du
u
2
su −
v =−
+
+
−k u− 2
r
dr
dr 2 r dr
r
2
v
d v 1 dv
dV V
2
+
+
−
k
v
−
u =
sv +
dr
r
dr 2 r dr
r2
2
d w 1 dw
2
+
−
k
w
sw = kp +
dr 2 r dr
(9.3.5)
(9.3.6)
(9.3.7)
along with the continuity condition
du u
+ − kw = 0
dr r
(9.3.8)
These are to be solved subject to the boundary conditions
u = v = w = 0 at r = r1
and at r = r2
(9.3.9)
Solution of the system of equations (9.3.5) through (9.3.9) is complicated by the
order of the system (6th), the cylindrical polar coordinates that introduce nonconstant
coefficients and suggest the need for Bessel functions, and also the number of parameters
needed to describe the problem. (Remember that in 1923 the most sophisticated computer
available was an adding machine.) Taylor restricted the problem to the case of small
gap spacing and therefore was able to obtain an approximate solution in terms of
trigonometric and hyperbolic functions. His and later results are shown in Figure 9.3.1.
The dashed line in the lower right-hand corner is the Rayleigh stability criterion. The top
1,800
1,600
1,400
2
1,000
ν
R inner Ω inner
1,200
800
600
400
200
0
–1,200 –1,000 –800
–600
–400
–200
2
R outer Ω outer
ν
Figure 9.3.1 Stability curves for Taylor/Couette stability
0
200
400
600
228
Flow Stability
curve is for r2 /r1 = 11. In successive curves this ratio increases by 0.05. The curves
are from empirical formulae in Coles (1967). Taylor’s small-gap theory showed that as
the gap size goes to zero, for the case where the outer cylinder is stationary,
T = Taylor number =
4 router − rinner 2
≈ 1708
2
2
−1
2 router
/rinner
(9.3.10)
The instability that Taylor observed in Couette flow consisted of a series of stacked
rings, with the flow occurring along helical paths in each ring. The flow is complicated
but still laminar. Subsequent experimenters have found that higher speeds result in the
forming of wavy, ropey cells before turbulent flow is achieved and that the pattern
observed can depend on the history of how the final pattern is obtained. See Koschmeider
(1993) for a summary.
Note that, as in the previous case of convection flow, the cause of the flow is
inertial effects. In this case it is the centrifugal and Coriolis acceleration, as contrasted
with gravity and buoyancy in the Rayleigh problem.
As an interesting sidelight, Taylor, a descendant of George Boole, who discovered
Boolean algebra, on which computer logic is based, was given a lifetime stipend by the
British government upon completing his studies with the freedom to do whatever he
wished in science. He chose to do his work at Cambridge University, where Rayleigh
was located at the time Taylor worked on this problem. Taylor has said that Rayleigh
discouraged him from pursuing this work, as it was “unlikely to be fruitful.” Fortunately,
Taylor did not heed this advice, as it was one of his early successes. Taylor’s collected
works compile six volumes.
While Couette pursued this work initially for viscometric purposes, these types of
flows occur in many other situations. In high-speed situations in instruments such as
gyroscopes it is necessary to maintain constant speed, which may require temperature
control to ensure constant viscosity. This requires an understanding of the heat transfer
occurring in the flow, which differs greatly if the cellular rings form. Other situations
where this type of inertial instability forms are in flows between stationary curved
parallel plates and curved pipes, where similar cellular patterns occur. The curved plate
geometry was investigated by Dean (1928).
9.4 Stability of Plane Flows
The stability of plane flows has a long and interesting history. The equations were first
formulated by Orr (1906–1907), and exact solutions were presented by him for the
special case where the velocity varied linearly between two plates. The results, however,
were indefinite, since even though the solutions were in terms of well-known (but
complicated) functions, the calculations that had to be done were still formidable. Later
efforts by such famous physicists as Sommerfeld (1908) and Heisenberg (1924) found
asymptotic solutions for the stability of two-dimensional flow between parallel plates.
Their results were incorrect because of the difficult nature of the solution, involving
the proper way to traverse turning points. The first successful asymptotic solution was
by C. C. Lin (1945). His theoretical results were confirmed by a series of elaborate
experiments performed at the National Bureau of Standards by Schubauer and Skramstad
(1943). Since that first successful analysis, with the advent of digital computers and
guided by Lin’s results, numerical solutions have produced more accurate numerical
results without the need for elaborate asymptotic analysis.
9.4
229
Stability of Plane Flows
The physical nature of the instability of parallel flows is quite different than for the
Rayleigh-Bénard instability just studied. Here, the disturbance is of the form of a traveling wave rather than a standing cellular pattern. For plane Poiseuille flow with plates
at y = 0 and y = b, the primary flow is v = Uy 0 0, with Uy = Um y/b − y2 /b2
and a pressure P = −vxUm /b2 . Take a disturbance of the form u′ x y t v′ x y t 0,
where
u′ x y t = uyeikx−ct
v′ x y t = vyeikx−ct
(9.4.1)
and
p′ x y t = pyeikx−ct
The preceding is a two-dimensional disturbance that is sufficient to study the stability of
this flow, since it was shown by Squire (1933) that two-dimensional disturbances always
became unstable at lower Reynolds numbers than do three-dimensional disturbances.
Here, c = cr + ici , where cr is the speed at which the wave travels in the x direction, and
kci is the growth rate of the disturbance. If kci < 0, the disturbance decreases with time
and the wave is stable. If kci > 0, the wave grows with time and there is instability. If
kci = 0, the wave neither grows nor decays with time, and there is neutral stability.
From substituting this form for the velocities into the incompressible continuity
equation in the form
U + u′
v′
+
=0
x
y
find that
iku′ = −
v′
y
(9.4.2)
The applicable Navier-Stokes equations for this flow are
U + u′
U + u′
U + u′
+ U + u′
+ v′
t
x
y
2
′
2
′
U + u
U + u′
P + p
+
+
=−
x
x2
y2
2 ′
′
2 ′
v′
v′
v
v
P + p′
v
+ U + u′
+ v′
+
+
=−
t
x
y
y
x2
y2
Substituting the forms of equation (9.4.2), subtract out the terms from the primary flow
equilibrium
0=−
P
+
x
2
U
y2
Linearizing any terms of order greater than the first in u, v, and p, we are left with
2
du
dU
2
−k u
(9.4.3)
= −ikp +
−ikcu + Uiku + v
dy
dy2
and
dp
+
−ikcv + Uikv = −
dy
d2 v
2
−k v
dy2
(9.4.4)
230
Flow Stability
Solving for the pressure p from equation (9.4.3) gives
2
du
dU
p
2
− k u − ikU − cu −
ik =
dy2
dy
(9.4.5)
Differentiating this expression for p with respect to y and using the result together with
equation (9.4.4) to eliminate p and u, after some rearrangement the result is
2
2
2
d
d
d2 U
d
2
2
2
−
k
−
k
v
=
ikU
−
c
−
k
v − ik 2 v
(9.4.6)
2
2
2
dy
dy
dy
dy
This is to be solved subject to the boundary conditions u = v = 0 at y = 0 and y = b, or
equivalently,
v=
dv
= 0 at y = 0 and at y = b
dy
(9.4.7)
If equation (9.4.6) is made dimensionless by letting U ′ = U/Um c′ = c/Um k′ = kb
= y/b, and Re = Um b/ , the result is
2
2
2
d
d2 U ′
d
′
′
′
′2
′2
v
−
v
(9.4.8)
v
=
ik
Re
U
−
c
−
k
−
k
d2
d2
d2
with the boundary conditions
v=
dv
= 0 at = 0
d
and at = 1
(9.4.9)
This dimensionless form of the stability equation for parallel flows is referred to as
the Orr-Sommerfeld equation. It is also used as an approximation for nearly parallel
flows, such as the flow in a boundary layer. (Notice that in the preceding dimensionalization Um was any convenient velocity—that is, either the mean or maximum of U or
any other value could in fact be used.)
The differential equation (9.4.8) and the boundary conditions of equation (9.4.9)
are all homogeneous, so there is only the trivial solution v = 0, except for special
combinations of the parameters R k′ , and c′ . These combinations are what must be
determined to answer the flow stability question.
At first glance, the solution of equation (9.4.8) may not appear to be all that difficult.
The equation is fourth order, compared to the sixth-order equations governing RayleighBénard and Couette stability. There are, however, two important differences in the two
problems that far outweigh the difference in order. One is the fact that the coefficients
in the equation are no longer constant U ′ = U ′ y. The second, less obvious one is
that the real part of the coefficient U ′ − c′ can, and in fact does, change sign in the
region 0 ≤ ≤ 1 for the conditions of interest. If you think of the difference between the
solutions of the two more familiar equations y′′ + y = 0 (simple harmonic oscillations)
and y′′ − y = 0 (exponential behavior), it can be imagined that both behaviors are
represented in various parts of the y region. The disturbance in fact tends to be fairly
rapidly oscillating between the walls and where U ′ = cr′ (this is, of course, the place
where momentum transfer between the primary flow and the disturbance is easiest), and
more gradually varying in the central core.
Viscosity in these flows plays a dual role. On the one hand, it has its traditional
dissipative role, whereby it dissipates the energy of the disturbance. The second role of
viscosity is to set up phase differences between the pressure and the various velocity
231
Problems—Chapter 9
1.4
1.2
1.0
0.8
k2
0.6
0.4
0.2
0.0
0
10
20
30
40
Re1/3
50
60
70
80
Figure 9.4.1 Stability curve for plane Poiseuille flow
components that enhance the energy transfer from the primary flow. It is this role that
is responsible for allowing the growth of the instability. Experimentally it is known that
the instability that occurs is a transition from laminar flow to turbulent flow.
The technical details of the asymptotic solutions are more complicated than can
be dealt with here. See Lin’s works listed in the References for more details. Use
of matched asymptotic expansions to compute higher-order terms in the asymptotic
expansion are given in Graebel (1966). A plot of the neutral stability curve k′2 versus
Re1/3 for plane Poiseuille flow is given in Figure 9.4.1. Inside the curve the flow is
unstable, while outside it is stable. The minimum value of Re is approximately 5,772
at a dimensionless wave number of 1.0206.
A similar curve for the Blasius boundary layer profile is given in Figure 9.4.2.
The figure is based on data in Shen (1954), which was found to agree well with the
classic experiments of Schubauer and Skramstad (1947) at the U.S. Bureau of Standards.
Their experiments used a wand that oscillated at a variable rate as a disturbance
source. R1 is a Reynolds number based on boundary layer thickness as introduced by
Lin (1946). The minimum Reynolds number for stability was found by Shen to be
about 416.
Problems—Chapter 9
9.1 If viscousterms on equation
(9.4.5) are neglected, the Orr-Sommerfeld equation
d2 v
d2 U
2
becomes U − c dy2 − k v − dy2 v = 0, with fa = fb = 0. (This is sometimes
called the Rayleigh equation.) The real part of the complex number c is the wave speed.
The imaginary part relates to the rate of growth or decay of the disturbance. Because
of the difficulty in solving the full Orr-Sommerfeld equation, much early work was
done on the inviscid version. The problem is complicated that U − c likely has a zero
somewhere in the interval. To investigate this, first divide the inviscid equation by
U − c. Next, multiply it by v∗ , the complex conjugate of v, and then integrate over the
232
Flow Stability
400
(ωμ ⁄ρ U 2) × 106
300
200
100
2,000
1,000
Re1
Figure 9.4.2 Stability curve for Blasius flow
range—say, a to b. Form positive-definite integrals, and then, after splitting into real
and imaginary parts, make some conclusions concerning c.
9.2 An alternate
approach to solving the preceding Rayleigh equation
d2 v
d2 U
2
U − c dy
−
k
v
v = 0 was introduced by Howard (1961). He first made
−
2
dy2
the change of
variables f = c − Uv. Then the Rayleigh equation becomes
d
2 df
c
−
U
− k2 c − U2 f = 0. To investigate this, multiply it by f ∗ , the complex
dy
dy
conjugate of f , and then integrate over the range—say, a to b. Form positive-definite
integrals, and then, after splitting into real and imaginary parts, make some conclusions
concerning c.
9.3 Show that the equations governing stability of flow between rotating cylinders
(Couette-Taylor instability, equations 9.3.5, 9.3.6, 9.3.7) reduce to
D2 − 2 − sRv = 2RA′ u
D2 − 2 − sRD2 − 2 u = 22 Rv
where
r
k= 1
r2 − r1
=
R=
r2 − r1 2
s
1 r2 − r1 2
=
r − r1
r2 − r1
A
B
r −r
=
1+ 2 1
+ 2
1 r 1 1
r1
D=
−2
d
d
and A and B are as in equation (9.3.2). To aid in your analysis, let = r2 − r1 /r1 be
a very small parameter.
Chapter 10
Turbulent Flows
10.1
10.2
10.3
10.4
10.5
10.6
The Why and How of Turbulence 233
Statistical Approach—One-Point
Averaging 234
Zero-Equation Turbulent Models 240
One-Equation Turbulent Models 242
Two-Equation Turbulent Models 242
Stress-Equation Models 243
10.7
Equations of Motion in Fourier
Space 244
10.8 Quantum Theory Models 246
10.9 Large Eddy Models 248
10.10 Phenomenological Observations 249
10.11 Conclusions 250
tur-bu-lence (circa 1598): the quality or state of being turbulent: as a: wild commotion b: irregular atmospheric motion esp. when characterized by up-and-down
currents c: departure in a fluid from a smooth flow.
tur-bu-lent (1538) 1: causing unrest, violence, or disturbance 2 a: characterized
by agitation or tumult; TEMPESTUOUS b: exhibiting physical turbulence
turbulent flow (circa 1922): a fluid flow in which the velocity at a given point
varies erratically in magnitude and direction; compare LAMINAR FLOW
10.1 The Why and How of Turbulence
These dictionary definitions lay out the problems with the study of turbulent flow:
wild, commotion, irregular, unrest, violence, disturbance, agitation, tumult, tempestuous,
erratic. To that list could be added adjectives such as chaotic, random, confused,
unorganized—well, you see the point. Clearly, while the study of laminar flow is
difficult, the study of turbulent flow requires a completely different approach.
Experience tells us that turbulence is often, but not necessarily, associated with high
Reynolds numbers and is all too common in the atmosphere, oceans, rivers, lakes—even
the plumbing in our buildings and our bodies. Sometimes it serves good purposes. If
you want to mix together two liquids, cool off your coffee by stirring, or stir a can of
paint, the introduction of turbulence comes in handy. On the other hand, if you want
233
234
Turbulent Flows
to increase lift on airplane wings or decrease drag on autos, ships, and planes, keeping
things laminar as long as possible may be better.
Thus, it is necessary to understand the rudiments of turbulence, both from theory
and experiments, and to find ways to predict what turbulence does to fluid flow. This
will not be as clear-cut as in the case of laminar flow, but some information can be
obtained.
Several approaches have been used to develop equations suited for the study of
turbulent flow. Since the Navier-Stokes equations are valid for both laminar and turbulent
flows, they usually serve as a general starting point. The complexity and rich detail
of what transpires in turbulent flow, however, limits our abilities to fully tackle the
Navier-Stokes equations directly. This is because the major effect of turbulence is that of
mixing. As the Reynolds number increases, the range of the length scales of the various
eddies in the flow also increases. Such a range of detail in the flow cannot be dealt
with entirely by analytic means, and it poses a challenge to the largest supercomputers
available. Computer simulations of the full Navier-Stokes equations are usually limited
to flows where periodicity of the flow can be assumed and the boundaries are of
particularly simple shape, usually rectangular. Numerical grids used must be small
enough to resolve the smallest significant eddy scale1 present in the flow, and the
simulation must be carried out for a sufficiently long time that initial conditions have
died out and significant features of the flow have evolved. The necessary memory and
speed requirements on computers has in fact been one of the principle driving factors
in the evolution of supercomputers and distributed computing architecture.
Covering the complete spectrum of eddy length scales is necessary, since different
physical mechanisms are important at various places in the spectrum. Large-scale eddies
contain much of the kinetic energy and little of the vorticity. The eddies tend to be
anisotropic (i.e., the turbulence behaves differently in different spatial directions), and
viscous effects are not important. These large eddies gradually break down into smaller
eddies that have little kinetic energy but much vorticity. The small eddies tend to be
isotropic, and viscous effects are important for them because it is the mechanism for
converting the kinetic energy of the turbulence finally into heat.
10.2 Statistical Approach—One-Point Averaging
The vast amount of detail in turbulent flows requires that many features be looked at in
a statistical sense. That is, for any turbulent quantity (e.g., velocity or pressure), split
the flow into a time mean quantity and a turbulent fluctuation quantity. For example,
letting unmarked quantities denote the total value, superposed bars denote an average,
and primes denote the turbulent portion, for velocity, temperature, and pressure, write
v = v + v′
p = p + p′
where the average of a quantity (pressure for example) is given by
1
pdt
p=
0
(10.2.1)
(10.2.2)
(10.2.3)
1
An eddy is defined as an entity that contains vorticity. See Lighthill, in Rosenhead (1963), Introduction.
Boundary Layer Theory, Chapter II.
10.2
235
Statistical Approach—One-Point Averaging
This is also called the first moment of the pressure. The averaging time is a time scale
sufficiently long compared to the fluctuation times present in the flow. Notice that for
any fluctuating quantity, it follows by definition that
p′ = 0
(10.2.4)
The root mean square value (designated by rms, and also called the second moment)
gives some idea as to the size of a fluctuation quantity. It is defined in the case of
pressure, for example, by
1 T ′ ′
′
′
p p dt
(10.2.5)
rms p = p p =
T 0
This term is never zero unless pressure fluctuations are completely absent from the
flow. This second-order moment is a correlation of p′ with itself and is thus called an
autocorrelation.
Reynolds first looked at the Navier-Stokes equations in this manner and introduced
what are now called the Reynolds averaged Navier-Stokes equations. To see how the
equations are found, start first with the continuity equation for a constant density flow.2
Splitting the velocity as in equation (10.2.1) results in
· v = · v + · v′ = 0
(10.2.6)
Averaging equation (10.2.6) and using the fact that the average of a fluctuation is zero,
this reduces to
· v = 0
(10.2.7)
Substituting this into equation (10.2.6) gives also
· v′ = 0
(10.2.8)
Consider next the component of the Navier-Stokes equations in the xi direction.
It is convenient to use index notation for much of what follows, so we will alternate
between using v = uvw and v = v1 v2 v3 in what follows. It is also convenient to
write the Navier-Stokes equation in the index notation form (with the help of continuity
equation (10.2.6)):
vi vj
vj
vi
p
vi
+
+
+
=−
(10.2.9)
t
xj
xi
xj
xj
xi
The summation convention on repeated subscripts is being used here—that is,
vi vj
vi vj
vi v1
vi v2
vi v3
3
= j =1
=
+
+
i = 1 2 or 3
xj
xj
x1
x2
x3
2
In this presentation, to avoid a large degree of additional complication temperature and fluid properties
such as density and viscosity will be assumed to be constant. Variations in these quantities can be included
in a similar fashion with substantial additional effort.
236
Turbulent Flows
Decomposing the velocities into mean and turbulent parts, inserting these into
equation (10.2.9) and then averaging, results in
vi ′ vj ′
vi vj
vi
+
+
t
xj
xj
or upon rearranging,
vi vj
vi
p
+
+
=−
t
xj
xi
xj
=−
p
+
xi
2
vi
xj2
vi
′
′
− vi vj
xj
(10.2.10)
(10.2.11)
The terms in equation (10.2.11) that involve averages of the products of the turbulent
fluctuations
− v1′2 −v2′2 −v3′2 −v1′ v2′ −v1′ v3′ −v2′ v3′ or
(10.2.12)
− u′2 −v′2 −w′2 −u′ v′ −u′ w′ −v′ w′
are called the Reynolds stresses. They are the terms through which the turbulence
fluctuations interact with and change the mean flow. At a solid boundary they must
vanish, but usually they have large gradients near a boundary and reach maximum
values very close to the wall. They can be measured by means of hot wire anemometers
or laser doppler anemometry, with suitable processing of the signals. They represent
what are called second-order correlations of the velocity at a given point.
Notice that in the averaging process, we have gone from a fully determinate system,
one where the number of unknowns is equal to the number of equations, to one that is
indeterminate, in that we now have more unknowns than equations. To attempt to correct
this, we could first multiply the Navier-Stokes equations by some quantity such as a
velocity component, then average the equations so obtained. In the process we would
introduce averages of three velocity fluctuation components (third-order correlations)
and find that the system is even more indeterminate than before. This game is one
that can’t be won, as this statistical averaging process will continue to generate new
unknowns faster than it generates new equations. As the order of a correlation increases,
the difficulties of both accurate measurement and physical understanding decrease,
and seldom is it necessary or useful to go beyond fourth-order correlations—if indeed
that far!
It should be noticed that although the statistical approach has many advantages
and is one of the few well-developed mathematical procedures available for analyzing
turbulent flows, it also has disadvantages. Averaging hides rather than reveals the
underlying physical processes that take place in such flows. Mollo-Christiansen (1971)
has given the example of a blind person using a single sensor on a single road as an
attempt to discover what kinds of vehicles used that road. “Happening to use a road
traveled only by limousines and motorcycles, he concluded that the average vehicle is
a compact car with 2.4 wheels.”3
At first it may seem strange that although the Navier-Stokes equations are closed
mathematically, the statistical equations obtained from them are not. Batchelor (1971)
has given two complementary answers to this. He notes that the Navier-Stokes equations
3
The remark was made to point out the inadequacy of using a single sensor.
10.2
237
Statistical Approach—One-Point Averaging
are open in time, since any one flow realization must be integrated over the whole
of future time before exact values can be found for the various correlations. This is
contrasted with the equations for the correlations, which can be made independent of
time. Second, additional correlations give less and less additional information so that
only a finite number of correlations is required to determine the properties of a turbulent
field to sufficient accuracy.
The problem of finding a way to get the same number of unknowns as equations is
called the closure problem of turbulence. In the elementary fluids course you may have
taken, you might have seen algebraic attempts at closure such as mixing length and eddy
viscosity. There, the Reynolds stresses were approximated in terms of mean quantities
and their gradients. While these closure models are useful for providing a correlation of
experimental data in simple flows such as pipe flow, they often fail when they are applied
to situations outside of the range of the original experiments. They have nevertheless
been developed to a high degree of complexity (see, for instance, Section 5.2.3 of
Bradshaw (1978)) and can give good results for boundary layer–type flows.
Before going to a discussion of closure formulas, three more correlation
equations will be developed. These are the transport equations for the important
correlations—namely, Reynolds stresses, kinetic energy, and dissipation. Deriving the
equations is an involved process, and it is necessary to use index notation because of the
nature of the terms involved in the needed multiplications. Also, it eases the physical
interpretation of the various terms.
Derivation of the first correlation equation starts with equation (10.2.9) with the
velocity written in decomposed form. This is next subtracted from equation (10.2.11),
giving
′
v′
Dv′
vi
+ vk i ≡ i
t
xk
Dt
(10.2.13)
vi′
p′
′
′ ′
′ ′
+
− vi vk + vi vk + vi vk
=−
xi
xk
xk
Multiplying equation (10.2.13) by vj′ and adding to this the same quantity but with the
i and j interchanged,4 we get
′
vj
Dvi′ vj′
vj
vi′
vi′ vj′
′ ′
′
′ ′ vi
+ vi vk
+
+p
−2
= − vj vk
Dt
xk
xk
xi
xj
xk xk
(10.2.14)
′ ′
v
v
j
i
−
+ vi′ vj′ vk′
p′ vj′ ik + vi′ jk −
xk
xk
Time averaging equation (10.2.14) then gives
Jijk
DRij
= Pij + Tij − Dij −
Dt
xk
I
II
III
IV
(10.2.15)
V
4
For example, for an equation ai = bi , multiplication gives vj ai = vj bi , so after the interchange and
addition, we have vj ai + vi aj = vj bi + vi bj .
238
Turbulent Flows
where
vi′ vj′
= dissipation of Reynolds stress,
xk xk
Rij
1
′
′
′
′
p vj ik + p vi jk −
Jijk =
+ vi′ vj′ vk′ = flux of Reynolds stress,
xk
v
v
Pij = − Rik k + Rjk k = production of Reynolds stress,
xj
xi
Dij = 2
Rij = vi′ vj′ = −Reynolds stress/
′
vj′
vi
1 ′
+
= pressure strain rate.
Tij = p
xj
xi
The roman numerals under the various terms indicate the source of that term and
the role it plays in affecting the turbulence. The physical interpretation of the terms are
as follows:
• Transport of the correlation—in this case, the Reynolds stresses. This term
represents convection of the correlation with the mean velocity.
• Generation or production. The presence of mean velocity and temperature gradients allows energy to be transferred between the mean and fluctuating field.
The energy can in principal flow either way—that is, from the mean flow into
the turbulence or vice versa. The first is generally more prevailing. These terms
arise from the fluctuating portion of the convective term.
• Redistribution (also called pressure-strain). In an incompressible flow the interactions of the pressure fluctuations can neither create nor destroy energy. Rather,
they cause it to be redistributed among the various components of the correlation.
• Destruction. This term comes from the viscous stresses and results in a decrease
of the correlation expressed in transport of the correlation.
• Diffusion. These terms arise from the fluctuating convection, pressure, and viscous
terms. They are characterized by being expressible as the gradient of a tensor.
For the second correlation equation, the disturbance velocity vector is dotted with
the Navier-Stokes equations and then time averaged. The result can be obtained from
the previous result by contracting on i and j, giving
J∗
Dk
= P− D− k
Dt
xk
I
II IV
(10.2.16a)
V
where
1
vi′ vi′
D = Dii =
= homogeneous dissipation of Reynolds stress,
2
xk xk
1
1
k
1
Jk∗ = Jiik = p′ vk′ + vi′ vi′ vk′ −
= flux of Reynolds stress,
2
2
xk
′
vi
1 ′ ′ 1 ′ ′ ′
vk′
′
Jk = p vk + vi vi vk − 2 vi
+
2
xk
xi
10.2
239
Statistical Approach—One-Point Averaging
1
1
1
k = q 2 = Rii = vi′ vi′ = turbulent kinetic energy,
2
2
2
vi
1
P = Pii = −Rij
= rate of kinetic energy production,
2
xj
′
′
vj′
vj′
vi
vi
1
+
+
= total rate of energy dissipation.
=
2 xj
xi
xj
xi
This equation is also found in the literature written in the alternate form
J
Dk
= P− − k
Dt
xk
I
(10.2.16b)
II IV V
The Roman numerals under the various terms are interpreted as for equation (10.2.16).
The third equation is found
by first differentiating the Navier-Stokes equations
v′
and then multiplying it by xi , adding the interchanged equation, and averaging the
j
result. This gives
H
DD
= −W − k
Dt
xk
(10.2.17)
where
W =2
+2
Hj =
vi′ vj′ vk′
v ′ v ′ vj
v′ vj′ vi
v ′ 2 vm
+ vj′ i
+ i i
+ i
xj xk xi
xk xj xk
xj xk xk
xk xk xj
2
vj′
2 ′
vi
2 ′
vi
xj xj xk xk
(10.2.18)
vj′ p′
vi′ vj′
D
+
−
xk xk
xk xk
xk
The various correlations introduced so far are called one-point correlations. This
means that every velocity and/or pressure term in the correlation is measured at the same
point. At times two-point correlations are also used in turbulence studies, when one is
trying to ascertain how what happens at a downstream point is affected by what happens
at an upstream point.5 Similar correlation transport equations can be developed for such
correlations, but, of course, even more algebra is involved. Two-point correlations have
been used in turbulence calculations but to a much more limited extent than one-point
correlations.
At least five different approaches have been used to provide closure formulae for
our equations:
1. Zero-equation models. This approach uses equation (10.2.11) as the only
partial differential equation. The Reynolds stresses are modeled by algebraic
5
One example of such a correlation could be the degree to which the proverbial butterfly flapping his
wings in Beijing affects the climate in your location.
240
Turbulent Flows
2.
3.
4.
5.
expressions, based largely on eddy viscosity and mixing length concepts,
and equation (10.2.11) is used to determine mean flow. This is widely used
in practical applications, and it also crops up in the one- and two-equation
models.
One-equation models. Another partial differential equation involving a turbulence velocity scale is used along with equation (10.2.11), often equation
(10.2.16). The length scale often used is based on q or .
Two-equation models. Two partial differential equations involving two turbulence velocity scales are used along with equation (10.2.11), usually equations
(10.2.16) and (10.2.17). Length scales often are based on q and .
Stress-equation models. Partial differential equations are used for all Reynolds
stress components and often for a length scale as well.
Large eddy simulations. Computations are made for both large-scale and smallscale eddies.
In the following sections these methods will be discussed in more detail. A number
of reviews are available in the literature and should be consulted for more information.
Among them are Bradshaw (1972), Mellor and Herring (1973), Reynolds (1974, 1976),
Reynolds in Bradshaw (1978), and Bradshaw, Cebeci, and Whitelaw (1981).
Much of the early work in theoretical turbulence models was done for special
models: homogeneous turbulence, where the turbulence is the same in every location
(e.g., Batchelor 1967); isotropic turbulence, where at a given point the turbulence is
both homogeneous and independent of direction (e.g., Taylor, 1935–1938); and turbulent
shear flows (e.g., Townsend 1956).6 While the concepts seem simple in hindsight, the
results together with Taylor’s experiments demonstrated the power of the statistical
approach (Kármán 1937b, 1938; Komogoroff 1941; Batchelor 1947).
10.3 Zero-Equation Turbulent Models
The eddy viscosity
and mixing length ℓ formulas
du du
−u′ v′ = ℓ2
and − u′ v′ =
dy dy
T
T
du
dy
(10.3.1)
form the basis for most of these models. Many variations on them exist. For instance,
at large Reynolds numbers Re > 5000, it is common to neglect the viscous sublayer
and split the boundary layer region into two parts. An example of this is
04y
y 0 ≤ y ≤ yc
ℓ=
0075 yc ≤ y ≤
⎧
du
⎪
⎨ℓ2
y0 ≤ y ≤ y c
dy
(10.3.2)
T =
⎪
x
⎩
00168 uc − u dy yc ≤ y ≤
0
6
It should be noted that although Reynolds introduced the concept of what are now called Reynolds
stresses, the concept of statistical analysis of turbulent flows languished until Taylor in 1935 revived the idea
and brought it to the attention of researchers.
10.3
241
Zero-Equation Turbulent Models
Here, y0 = 40 /uc is the thickness of the viscous sublayer measured from the
wall, and yc is another distance found from making the mixing length continuous. The
parameter is a boundary layer thickness, defined, for example, where the mean velocity
is at 0.5% of the outer flow.
A number of empirical corrections of these formulae have become feasible over
the years as computing memory and speed have increased. One due to Cebeci and
Smith (1967, 1974) gives improved results for low Reynolds numbers, transition, mass
transfer, pressure gradients, and boundary curvature. The eddy viscosity is given by
⎧
du
⎪
⎨ℓ2
y ≤ ycrossover
T i
dy
(10.3.3)
=
T =
x
T o ⎪
⎩
00168 uc − u dy FKlebanoff y > ycrossover
0
The quantity ycrossover is where the outer eddy viscosity is greater than the inner viscosity.
The other quantities are
+
+
ℓ = 04y 1 − e−y /A
y dp −1/2
(10.3.4)
A+ = 26 1 +
wall dx
y 6 −1
FKlebanoff = 1 + 55
FKlebanoff being a factor introduced by Klebanoff to account
for intermittancy in the flow.
The quantity y+ is given by y+ = u∗ y/ , where u∗ = wall /. The expression for the
mixing length is the van Driest fit to velocity measurements near the wall. This model
was extended to three dimensions and also accounts for compressibility effects. It was
developed by Cebeci (1974), Cebeci and Abbott (1975), and Cebeci et al. (1975).
Because of its robustness, a model by Baldwin and Lomax (1978) is commonly
used in quick design iterations. It does not capture all details of the turbulence but still
gives useful design information. Its details are given by
⎧
⎧
⎨ T i
⎨ℓ2
y ≤ ycrossover
(10.3.5)
=
=
T
⎩
⎩
T 0
KCcp Fwake FKlebanoff y > ycrossover
The y+ is as given in equation (10.3.4). Other quantities are
uj
ui
1
+
+
= ij ij ℓ = 1 − e−y /A
+
ij =
2
xj
xi
Fwake = min ymax Fmax Cwk ymax udif 2 /Fmax
+
+
ymax and Fmax are the maximum of the function Fy = y 1 − e−y /A ,
−1
yCKlebanoff 6
FKlebanoff = 1 + 55
ymax
udif = max
ui ui − min
ui ui
242
Turbulent Flows
The various constants are given by
A+
Ccp
CKlebanoff
Cwk
26
1.6
0.3
0.25
0.4
0.0168
Granville (1987) has suggested improvements in these constants. Further comments
have been made by Wilcox (1998).
10.4 One-Equation Turbulent Models
An early model due to Prandtl (1945) suggested that
√
kℓ
T =
(10.4.1)
with k being the turbulent kinetic energy and the mixing length coming from an algebraic
formula. The turbulent kinetic energy is usually determined from equation (10.2.16b)
after suitable modeling of the terms on the right-hand side. A model used at Stanford
University in the STAN-5 code let
T
= c2 qℓ
1
Rij = q 2
3
ij
−
D = c3 q 2 /ℓ
Jj∗ = − c4
T
T
+
uj
ui
+
xj
xi
(10.4.2)
q2
xj
where the length scale is modeled by means described in the zero-equation model.
Typically the zero-equation model is used for y+ ≤ 2A+ , and the preceding is used for
greater values of y+ . STAN-5 uses c2 = 038 c3 = 0055 c4 = 059, all determined from
experimental results. A similar model was developed at Imperial College (Wolfshtein,
1969). Further forms for one-equation models can be found in the review by Mellor and
Herring (1973).
10.5 Two-Equation Turbulent Models
The previous models required specification of a turbulent-length scale for their realization. To avoid this, equation (10.2.17) is frequently used to compute the dissipation. For
the case of isotropic turbulence,
1
DD
D = − q 2 and
= −W
(10.5.1)
2
Dt
For large values of the Reynolds number, after making the assumption that W is of the
form W = c7 D2 /q 2 , the solution of equation (10.5.1) is
q 2 = q02 1 + t/a−n
nq02 /2D0
D = D0 1 + t/a−n−1
(10.5.2)
n = 2/c7 − 2. Early experiments suggest that c7 = 4 and n is in
where a =
the range 1.1 to 1.3. Later arguments by Reynolds (1976) suggest that this model works
10.6
243
Stress-Equation Models
best for short wavelength disturbances. If the entire spectrum is included, Reynolds
suggests that an improvement on equation (10.5.2) is
2/3 −2/3m + 5
3m + 5/2m + 3
3
aD
1
+
D = C q2
(10.5.3)
q2 =
m+1 2
A
where m = 2 seems the most reasonable choice. This corresponds to c7 = 11/3 and
n = 6/5 in the previous result.
A number of other models have been introduced, each with pluses and minuses.
A heuristic model by Saffman and Wilcox (1974) introduces a pseudovorticity and
uses the transport equation
D
ui ui
2
2
+ T
− +
(10.5.4)
=
Dt
xj xj
xj
xj
in place of equation (10.2.17). For equation (10.2.15) they use
uj
uj 1/2
ui
ui
1
P = ∗
+
+
D = ∗ q 2 /2
2
xj
xi
xj
xi
(10.5.5)
with recommended values
= 01638 ∗ = 03 = 015 ∗ = 009 c4 = 05 and c4∗ = 05
This model has been tested against a number of examples and has given reasonable
results—in some cases better than the more theoretical models proposed by others. For
updates, see Wilcox (1998).
Since the two-equation models frequently use equations (10.2.15) and (10.2.17),
which involve the kinetic energy and dissipation, they are frequently referred to in the
literature as k − models. The Saffman-Wilcox model is referred to as a k − model.
10.6 Stress-Equation Models
The one- and two-equation models still require modeling of the various terms in the
equations, and the results are often at best mixed. In view of these difficulties, attention
has also gone back to the original equation for the Reynolds stresses, equation (10.2.15).
For a given mean velocity field, the production term does not need to be modeled, but
the remaining ones do.
The type of modeling needed depends on the flow being studied. For instance, at
high Reynolds numbers the small-scale dissipative structures are practically isotropic,
and it has become customary to use
Dij =
2
3
ij D
(10.6.1)
The pressure strain rate term for incompressible flow has the property that
Tii = 0
(10.6.2)
and for flows with no mean strain rate it is responsible for the return to isotropic
conditions. It is known, however, that for deforming flows the situation is much more
complex, and a number of diverse methods have been proposed to model it. Examples
244
Turbulent Flows
are found in Tennekes and Lumley (1972); Rotta (1951); Hanjalic and Launder (1972);
Launder, Reece, and Rodi (1975); Norris and Reynolds (1975); Kwak, Reynolds, and
Ferziger (1975); and many others. A similar situation exists for the J term.
To establish the validity of a model, we compare the results of well-established
experiments. For incompressible flows, one of the gold standards is the rearward-facing
step, where reattachment occurs behind the eddy formed at the base of the step. This
was selected as a predictive case at a conferences at Stanford University in 1980 and
1981 (Kline et al., 1981). The algebraic stress model of Baldwin and Lomax (1978)
appeared to produce the most satisfactory results at that time.
Several classes of flows have provided the data used for determining the empirical
constants. These provide an indication of the flow situations that might be computed
from these closure models with the highest degree of confidence. It is, of course,
hoped that an even broader class of flows is covered by the models. Confident realization of that hope is only possible by calculation and subsequent comparison with
experiment
1. Decay of grid turbulence: Batchelor and Townsend (1948), Lin and Lin (1973).
2. Return to isotropy of distorted turbulence: Rotta (1962), Tucker and Reynolds
(1968), Uberoi (1957).
3. Free shear flows (jets, wakes, mixing layers): Bradbury (1965), Bradshaw et al.
(1964), Champagne et al. (1977), Chevray and Kovaszney (1969), Tailland and
Mathield (1967), Webster (1964), Wygnanski and Fiedler (1970).
4. Wall turbulence: Arya and Plate (1969), Hanjalic and Launder (1972), Johnson
(1959), Klebanoff (1955).
Much research is still being done today to find improvements on these models.
Results so far indicate the following:
Zero-equation models: These generally are simple, efficient, and numerically stable.
The concept of a length scale, however, is difficult to define. Also, history effects are
not accounted for.
One-equation models: These are naturally more difficult to code than the zeroequation models. Again, a length scale is not well defined. Often it is only marginally
better than zero-equation models.
Two-equation models: The length scale here is easily defined, but the equations
that result are numerically stiff (“stiff” implies extreme sensitivity to coefficients in
equations—see Chapter 11 and the Appendix). They do, however, provide better results
than the other two cases.
10.7 Equations of Motion in Fourier Space
Knowing the distribution of length scales in turbulence is important because the various
length scales behave in different ways. For example, if you want to reduce the level of
turbulence in a wind tunnel, the air is passed through arrays of fine screens or closely
spaced bars. After passing through these filters, the air turbulence is of fine length scale,
whose energy is dissipated rapidly by viscosity. Fourier analysis is used to determine
the distribution of length scales. The formal definition of the Fourier transform in three
dimensions is
10.7
245
Equations of Motion in Fourier Space
1
e−ixkx + yky + zkz fx y zdxdydz
V
23
fx y z =
eixkx + yky + zkz Fkx ky kz dkx dky dkz
Fkx ky kz =
(10.7.1)
V
where F is the Fourier transform of f , and f is “retrieved” or inverted by the second
of equation (10.7.1).
Differential equations for transformed quantities can be found by taking the Fourier
transform of equations such as (10.2.15), (10.2.16), and (10.2.17). Fourier transforms
are used with two-point correlations such as
Rij r = vi xtvj x + rt
(10.7.2)
the integration performed on r while x is held fixed, or with two-time correlations
such as
Rij = vi xtvj xt +
(10.7.3)
The purpose of the two-point correlations is to investigate the effects of either
different spatial scales or different time scales in the turbulence and to find the region of
coherence of the correlations. When dealing with the equations for two-point correlations, it is customary to make three basic assumptions: (1) mean velocity and temperature
gradients are constant, (2) the turbulence is homogeneous, and (3), the presence of walls
can be ignored in the equations. Under these assumptions Fourier transforms are useful.
The advantage of using Fourier transforms is that there are no space derivatives
involved in the equations. Certainly, a disadvantage is that we are now “transformed”
to a space where we have even less physical understanding of the new quantities.
Signals from hot wire or hot film anemometers can be transformed by signal
processors (dedicated computers called spectrum analyzers) capable of time averaging,
multiplication, time delay, and fast Fourier transforming (FFT). FFT is an approximation
to the integration process, which is well suited for use in a computer.
One of the triumphs of isotropic turbulence studies has been the explanation of the
energy content as a function of wavelength. When the log (base 10) of the energy spectrum is plotted versus log10 k/kKolmogoroff , where kKolmogoroff = / 3 1/4 , the plot has a linear region with downward slope k −3/5 in the range −4 < log10 k/kKolmogoroff < −1. This
is known as the inertial subrange. Below this range, from −1 < log10 k/kKolmogoroff < 0,
is the dissipation range. The combination of the two is called the universal equilibrium range.
It is easiest to describe what is happening by denoting wave number k by its
reciprocal, the wave length , so the more familiar dimension of length is being dealt
with. Then log10 k/kKolmogoroff = − log10 /Kolmogoroff . An eddy introduced into the flow
tends to cascade down this energy-length curve in the following manner. This process
is called, naturally, the energy cascade.
For large wavelengths—say, for argument, wavelengths of the order greater than
104 Kolmogoroff —the disturbance is highly direction-dependent and nonisotropic. This is
the range where energy production takes place. Eddies of a given size are stretched by
large eddies, and mixing and stretching takes place. Eddy size tends to be reduced in
the process, and bigger eddies are converted into smaller eddies. In the process, they
also tend more toward the isotropic turbulence state.
When an eddy gets into the inertial range, dissipation is still minor, but straining
of eddies continues (the Tij in our earlier discussion), and the eddy size continues to
246
Turbulent Flows
diminish until it reaches the Kolmogoroff length scale, at which point viscous dissipation
takes over and eventually destroys the eddy. This process seems to have the universality
of being applicable at all length scales, whether consideration is of a hurricane or a
“tempest in a teapot.”
10.8 Quantum Theory Models
Quantum theory approaches have interested a number of theoretical researchers and have
resulted in several models. It has a particular appeal in that it strives to provide a model
without empirical constants and thus endeavors to be universal and absolute. Most work,
however, is mainly intent on investigating general consequences of a given model rather
than on predicting the characteristics of a particular turbulent flow. Few if any actual
flow calculations appear to have been made using these models, and the theory does
not appear to be at a point where such is feasible except in the simplest possible case.
In the following, the basic equations are presented along with brief statements
about the approaches used in their derivation so readers may provide their own details.
No one reference exists that deals with all the equations simultaneously, but partial
versions of the equations can be found in the standard references dealing with turbulence
(e.g., Bradshaw (1978), Craya (1958), Favre et al. (1976), Hinze (1975), Leslie (1973),
and Lesieur (1978)).
The quantum theory starts with the Fourier-transformed one-point averaged version
of the Navier-Stokes equations, with the assumption that the turbulence is isotropic. In
that case they reduce to
+ k2 vj k = Mjmn k vm pvn r + fj k
(10.8.1)
t
where
vi pvn r =
r
vi kvun r
p
k − p − rd3 pd3 r and
1
Mjmn k = − i kn jm k + km jn k
2
jm k = jm − kj km /k2
(10.8.2)
The f function is a hypothetical “stirring force,” which models the interaction of the
mean flow with the Reynolds stress, and is theDirac delta function, infinite when
the argument is zero, zero everywhere else. The term arises from the convective
acceleration and pressure terms and illustrates the triad interaction that is characteristic
of this model. In other words, two wave numbers, p and r, combine algebraically to
form a third wave number, k = p + r.
The first attempts at closure of this model (Millionshchtikov, 1941; Proudman and
Reid, 1954) termed it the quasi-normal approximation (QN). An equation is formed
from the ensemble average of the triple correlation vj −kvm pvn r, which involves
fourth-order correlations. The fourth-order correlation is assumed to be expressible as
products of the second-order correlations, which would be exactly true if the correlations
were Gaussian in nature. The equations representing time derivatives of the second- and
third-order correlations thus represent a closed set. While the assumptions made in
developing the theory appear reasonable, calculations show that the ensemble average
of vi kvi −k becomes negative, which is physically impossible.
10.8
Quantum Theory Models
To correct this defect in the theory, the QN model was modified (Orszag, 1970) to
the eddy-damped quasi-normal approximation (EDQN). This theory adds linear combinations of the third-order correlations to the QN forms for the fourth-order correlation.
This necessarily introduces vector parameters (eddy viscosities) that multiply the thirdorder correlations. To assure that the second-order correlation remains positive, the
additional assumption is made that the relaxation time of the triple correlations is much
weaker than the evolution time of the energy spectrum. The model has now become the
EDQNM, or EDQN-“Markovianized,” model (Leith, 1971). A further model is needed
for specification of the eddy viscosity vectors (Kraichnan, 1971; Sulen et al., 1975).
Another quantum theory model that has received attention and that is at a high
level of development is the direct interaction (DI) model (Kraichnan, 1958, 1959;
Leslie, 1973). A velocity field in the form vi + vi , where vi is an infinitesimally
small velocity perturbation, is introduced into equation (10.8.1). This equation is then
linearized and a Green’s function is introduced. The resulting equation is called the
response equation, and the Green’s function is called the response function. The DI
model is then determined by proceeding in the following manner:
1. The nonlinear terms in equation (10.8.1) are taken to be multiplied by a small
parameter.
2. The velocity vi and the response function are expanded as power series in the
small parameter.
3. The lowest-order term of the velocity expansion is assumed to be Gaussian in
nature.
4. The expansion is stopped at the lowest-order terms that do not vanish when
averaged statistically. At that stage, the small parameter is set to unity!
5. In the equations that result from this process, the lowest-order terms in the expansion for the velocity and the response function are replaced by their exact values.
The justification for this process is necessarily indirect because of the complexity of
the original equations. Kraichnan (1961) introduced a model related to, but much simpler
than, the response equation. This model can be solved exactly. He shows that DI works
reasonably well for this model. This equation is then modified by a process that does not
depend on the nonlinear term being small into another equation for which DI gives the
exact statistical average. The same process can be used to generate a similar set of model
equations from the Navier-Stokes equations. The DI approximations to the Navier-Stokes
equations are then the exact statistical average of the second set of model equations.
This process is unorthodox and does not provide any assurance that the DI approximation lies close to the exact solution. In fact, Kraichnan put forth the statement that
existing mathematical techniques are incapable of providing such assurance. However,
to quote Leslie (1973, page 60), one of the proponents of the method, “DI, and other
methods which are broadly like it, do give answers where the earlier methods (e.g., QN)
gave nothing. These answers are certainly interesting and may well be relevant, but at
the moment it cannot be proven that the whole structure rests on a secure basis.” Perhaps
more confidence can be gained from the fact that DI has been shown to give reasonable
agreement with some computer simulations and that the number of investigators of these
ideas, and interest in them, has grown.
The DI approximation as first presented was Eulerian in concept (it is sometimes
labeled EDI to emphasize this) and suffered from the flaw that it was not invariant to
a Galilean transformation. As a result, one of the response integrals diverges at one of
247
248
Turbulent Flows
its limits of integration. The correction of this (Kraichnan, 1964, 1965, 1966; Edwards,
1968) is most easily carried out in a Lagrangian reference frame and has been termed
the Lagrangian history direct interaction (LHDI) model.
A direct Lagrangian approach is difficult, and closure approximations become
virtually impossible in such a frame. To simplify these matters, Kraichnan (1966) has
introduced a further model, the abridged Lagrangian interaction (ALI) model, which
introduces an approach somewhere between the Eulerian and Lagrangian models. To
illustrate this, consider the notation fxt0 t, which reads as “the property f at time t
of a fluid particle that was at x at time t0 .” Thus, if z is the current position of a fluid
particle and v its velocity,
vxt0 t =
zxt0 t
t
(10.8.3)
As
corner. Then z =
a simple example, consider an irrotational vortex and flow in a square
0
= tan−1 xx1
x12 + x22 cos + sin + 0 for the vortex, with = t−t
2
x1 +x22
2
−kt−t0
−kt−t0
as the circulation, and z = x1 e
x2 e
0 for the square corner.
Eulerian mechanics integrates along a line where t = t0 , while Lagrangian mechanics
integrates along the line t0 = constant. ALI is an approach that integrates the equations
first along the line t = t0 and then along a line of constant t. This results in a model
that is approximately Lagrangian in character but yet is simple enough to allow a DI
closure approximation.
ALI is extremely complicated and difficult to work with. To get around this,
Kraichnan (1970, 1971) modified EDI in a purely Eulerian framework to arrive at an
almost-Markovlan model that is invariant to a Galilean transformation. He called it the
test field model (TFM). TFM is much more efficient in computing time than ALI, in
that TFM implies an exponential time dependency, whereas ALI must compute the time
dependency. It is thought also that TFM may represent the diffusion of a scalar passive
better than does ALI (Leslie, 1973, Chapter 12). In a series of computations (Herring
and Kraichnan, 1972) where the predictions of these various models were compared
with computer solutions (Orszag and Patterson, 1972) and experiments (Grant, Stewart,
and Moillet, 1962), the following conclusions were reached:
1. At low turbulent Reynolds numbers, all quantum theory models are usable. The
accuracy depends on which quantity is being compared, and no one model stands
out as being obviously better. The skewness factor, which is particularly sensitive
to differences in the models, turns out to be best predicted by a version of TFM.
2. At high turbulent Reynolds numbers, the models that are not Galilean-invariant
fail completely, and only TFM and ALI are usable. These seem to agree quite
well with each other. Because the TFM model is much simpler than ALI, the
comparison favors TFM.
10.9 Large Eddy Models
In 1973 a group at Stanford University begin a long-term project to study the large-scale
end of the turbulence spectrum (Kwak, Reynolds, and Ferziger 1975) by simulation on
a supercomputer. The original calculations were done on a grid with 163 points, but
this was later increased to 323 points and more. Reasonable agreement was made with
previous experimental results, as long as strong filtering was done on the numerical
results. Experiments such as this are useful for testing turbulent flow models and also to
10.10
249
Phenomenological Observations
illustrate local behavior that might not be simply visualized. The memory requirements,
however, make this a research tool rather than a practical tool for applications.
10.10 Phenomenological Observations
Since O. Reynolds’s original dye experiments on the stability of pipe flow, it has been
known that for a given point in a flow at a given (sufficiently large) Reynolds number,
the flow would alternate between the laminar and turbulent states. Investigators defined
an intermittency factor as the percent of time at which a flow is turbulent at a fixed
point. Experimentally, this is done when measuring a flow by circuitry that produces an
intermittency signal I(t), where
0 if the flow is nonturbulent,
It =
(10.10.1)
1 if the flow is turbulent.
Then the intermittency is found as
= It = probability of occurrence of turbulence at that point.
(10.10.2)
Measurements made by Townsend (1951) in the wake behind a thin wire7 boundary
layer show that near the center line of the wake is approximately unity. Starting
around 18%, or the averaged wake half-diameter, the intermittency appears to drop in an
exponential manner, so at 60%, or the averaged wake half-diameter, the intermittency
has reduced to less than 0.1. Generally, it is necessary to correct measurements to
account for the local degree of intermittency.
In many ways turbulent flow can be thought of as a tangle of vortex filaments,
stretching and moving as they interact. (Think of a container filled with squirming
earthworms!) Vorticity must start at a wall and then be convected to the interior flow.
It has been observed that vortex filaments are created at the wall and oriented perpendicular to the flow. These filaments are generated in bursts in the sublayer, the bursting
phenomenon appearing to be cyclical. It has been suggested that the bursting process is
part of a cycle in which interaction between the wall region and the outer region of the
boundary layer is important. New bursts are created as a result of a disruption of the
primary flow by the previous burst (Willmarth, 1979). Once created, the filament can
then develop kinks and distort into a hairpin vortex, a U with long arms. The base of
the U then lifts from the wall and is carried off into the interior flow.
The interactions between the filaments are governed by the Biot-Savart law, first
developed to explain the magnetic field around an electric current-carrying wire. If
is the circulation for a given vortex filament of length s, then at a point a distance R
from the filament a speed v is induced, where
v=
sin
s
4 R2
(10.10.3)
The angle is the angle between the vortex filament and the line connecting the filament
and the point. The proof of the law comes from the use of the velocity potential of the
vortex and is purely kinematic in nature.
7
The wire diameter was approximately 1/16 inch, and the Reynolds number based on wire diameter was
1,360.
250
Turbulent Flows
Vortex filaments can be both desirable and undesirable in a flow. Recently, high
altitude research by NCAR (National Center for Atmospheric Research) indicates that
clear air turbulence, the bane of air travelers, is due to rapidly spinning horizontal vortex
tubes, presumably on the scale of modern aircraft. On the other hand, Boeing airplanes
in the 7X7 family have long used vortex generators on the upper wing to create artificial
turbulence that acts to stabilize the boundary layer, reducing drag. Winglets are used
to change the vortex pattern in a plane’s wake, reducing induced drag. And airplane
engine designers are turning to jet engines with chevron rims on the exit nozzles. The
serrated, or scalloped, edges on these rims introduce a sheet of vorticity that allows
the outer and inner flows to interact more rapidly than if a straight rim were used. The
result is a reduction in perceived engine noise with little loss in engine efficiency.
10.11 Conclusions
In the previous material a number of different models were used. The zero-, one-, and
two-equation models are used in many commercial CFD packages, although often the
details of the parameters are not made known to the user. To be optimistic, there may
come a day where one turbulence model fits all—every flow situation, every value of the
Reynolds number, every geometry, every Mach number—but that day is still far away.
In a lecture given before a turbulence class some years ago, Dave Wilcox of
DCW Industries, an active turbulence modeler, stated that the following is required in
turbulence modeling:
Tools of the trade for a turbulence modeler:
1.
2.
3.
4.
5.
Ingenuity
Computer literacy
Knowledge of numerical methods
Knowledge of perturbation methods
Lots of self esteem
Fundamental premises of a turbulence modeler:
• Try to model the physics, not the partial differential equation.
• Avoid adjusting coefficients from one class of flow to the next.
• Strive for elegance and simplicity.
• Know what other workers in the field are doing, but keep at arm’s length.
Good advice!
In summary, turbulence is a difficult subject. No one really understands turbulence!
Chapter 11
Computational Methods—Ordinary
Differential Equations
11.1 Introduction 251
11.2 Numerical Calculus 262
11.3 Numerical Integration of Ordinary
Differential Equations 267
11.4 The Finite Element Method 272
11.5 Linear Stability Problems—Invariant
Imbedding and Riccati Methods 274
11.6 Errors, Accuracy, and Stiff
Systems 279
Problems—Chapter 11 281
11.1 Introduction
Many of the problems that were solved in the previous chapters required exacting
calculations that were originally done either by hand computation or at best with the aid
of an adding machine. Numerical methods were invented as far back as Newton, and
techniques suited to the computer, such as the use of Green’s functions, were known
in the nineteenth century, but it wasn’t until the mid-twentieth century that mainframe
computers were available to researchers. Even then, the need to use storage devices such
as punched cards required significant time and effort for all but the smallest calculation.
Today, the personal computer has made the power of the most powerful computer
of mid-nineteenth century available to all. Many languages have been invented to make
calculating the problems previously covered relatively easy. The first of these was
FORTRAN (Formula Translator), developed at IBM in the 1940s. This was followed by
similar languages such as BASIC, QUICKBASIC, PASCAL, VISUAL BASIC, C ++,
and many others. BASIC originally was included as a chip in the original IBM PC,
and QUICKBASIC was provided as part of the disk operating system (DOS) until
version 4.5 of DOS. These languages required a fair degree of programming skills on
the part of the user. Today, MATHCAD, MATLAB, and symbolic manipulators such as
MAPLE and MATHEMATICA have greatly increased the availability of mathematical
computations.1
1
Newer languages are not necessarily more productive than the newer versions of FORTRAN, such as
versions FORTRAN 90 and 95. See www.nr.com/CiP97.pdf for comments on this matter.
251
252
Computational Methods—Ordinary Differential Equations
A few of the concepts and techniques used in computational fluid mechanics are
presented in this Chapter and Chapter 12, along with sample programs that illustrate
the methodology. The programs are written in FORTRAN, the most persistent of these
languages, and can be easily transformed to most of the BASIC class of language, as
well as others. In all of these programs, lines preceded by either a capital C or an
exclamation mark are comments to the reader and are not used in computations. (The
C is used for FORTRAN 90, and the ! for FORTRAN 95 programs.) The ampersand
(&) is used to denote line continuation.
There are a number of versions of the FORTRAN compiler available from several
sources. Programs presented in this book use the extensions .FOR (FORTRAN 90 fixed
source form); .F90 (FORTRAN 90 free source form); or .F95 (FORTRAN 95). Most
present-day compilers are capable of handling these files.
As the first example, Program 11.1.1 is a program for determination of the separation
point on a circular cylinder using both Thwaites’s and Stratford’s criteria. The velocity
PROGRAM
THWAITES
C
Thwaite - Stratford separation prediction of separation on a circular cylinder
C
Schmitt-Wenner velocity profile
C.............................................................
DIMENSION SX(10)
A=-0.451
B=-.00578
SMAX=1.1976
UMAX=1.606299
U=0.
UP=2.
DS=.001
G=0.
G1=0.
FINT=0.
FOLD=0.
GOLD=0.
DO I=1,1600
S=I*DS
S2=S*S
U=S*(2.+S2*(A+B*S2))
UP=2.+S2*(3.*A+5.*B*S2)
F0=-2.*(U**6)/UP
G2=U**5
FINT=FINT+0.5*(G1+G2)*DS
F=F0-FINT
DEG=S*180./3.14159
G1=((S-SMAX)**2)*(1.-(U/UMAX)**2)*((-2.*U*UP/(UMAX*UMAX))**2)
&
-.0104
IF (S.GT.SMAX) G=G1
WRITE(6,100) S,U,UP,F,G,DEG
G1=G2
IF ((F*FOLD).LT.0.) THEN
SX(1)=S
SX(2)=F
SX(3)=FOLD
SX(4)=DEG
WRITE(*,*) "THWAITES-----------------------------------"
END IF
IF ((G*GOLD).LT.0.) THEN
SX(5)=S
SX(6)=G
SX(7)=GOLD
SX(8)=DEG
WRITE(*,*) "STRATFORD----------------------------------"
END IF
FOLD=F
GOLD=G
END DO
Program 11.1.1—Separation criteria for a circular cylinder—program by author
11.1
253
Introduction
100
102
WRITE(*,*)
WRITE(*,*) "
S
F
FOLD
WRITE(*,*) "Thwaites Criterion"
WRITE(*,102) SX(1),SX(2),SX(3),SX(4)
WRITE(*,*) "Stratford Criterion"
WRITE(*,102) SX(5),SX(6),SX(7),SX(8)
FORMAT(6F10.4)
FORMAT(4F10.4)
END
DEGREES"
Program 11.1.1—(Continued)
profile is a fifth-order polynomial used by Schmitt and Wenner (1941) and derived from
their experiments. The quantity S is the arc distance along the surface of the cylinder,
measured from the stagnation point. The criteria are put in the form FS = 0, and FS
is calculated at progressively increasing values of S until F changes sign, indicating
that the zero point has been passed. Values of S and F are printed to the screen, and
zero-crossing is pointed out for each criteria.
A more complicated programming example is Program 11.1.2, the calculation of
the various regions for the behavior of the Mathieu functions. When encountering such
sophisticated problems, it is worthwhile (and the better part of valor) to first determine if
prepared programs are available. Good starting points for this are the IMSL Math/Library
(1987) and Press, Flannery, Teukolsky, and Vetterling (1986). In this case, however,
PROGRAM Mathieu
============================================================
Purpose: This program computes a sequence of characteristic
values of Mathieu functions using subroutine CVA1
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
y'' + (m^2 - 2q cos 2t) y = 0
Input :
&
10
11
m
--- Order of Mathieu functions
q --- Parameter of Mathieu functions
KD --- Case code
KD=1 for cem(x,q) ( m = 0,2,4,...)
KD=2 for cem(x,q) ( m = 1,3,5,...)
KD=3 for sem(x,q) ( m = 1,3,5,...)
KD=4 for sem(x,q) ( m = 2,4,6,...)
It is adapted from programs given in Computation of Special Functions
by S. Zhang and J. Jin, Wiley, 1996.
============================================================
IMPLICIT DOUBLE PRECISION (A-H,O-Z)
DIMENSION CV1(200),CV2(200),CVE(30,30),CVS(30,30),A(30,30),
IC(30),IB(65)
OPEN(7,FILE='A:MATHIEU.DAT',STATUS='UNKNOWN')
DO 19 MMAX=1,12
DO 11 K=1,28
Q=K
CALL CVA1(1,MMAX,Q,CV1)
CALL CVA1(2,MMAX,Q,CV2)
DO 10 J=1,MMAX/2+1
CVE(2*J-1,K)=CV1(J)
CVE(2*J,K)=CV2(J)
END DO
END DO
Program 11.1.2—Mathieu function stability regions adapted from program given in Computation of
Special Instruction by S. Zhang and J. Jin, Wiley, 1996. (The first adaptation was by Ben Barrowes
(barrowes@alum.mit.edu) and can be found on the internet.)
254
Computational Methods—Ordinary Differential Equations
15
16
17
19
22
C
DO 16 K=1,28
Q=K
CALL CVA1(3,MMAX,Q,CV1)
CALL CVA1(4,MMAX,Q,CV2)
DO 15 J=1,MMAX/2+1
CVS(2*J,K)=CV1(J)
CVS(2*J+1,K)=CV2(J)
END DO
END DO
J=MMAX
WRITE(*,*)'------------------------------------------'
WRITE(*,*) J
WRITE(*,*) " ALPHA Q
a(J)
b(J)"
WRITE(*,*)'------------------------------------------'
DO 17 K=1,28
Q=K
A(2*J-1,K)=CVE(J+1,K)
A(2*J,K)=CVS(J+1,K)
WRITE(*,50)Q,CVE(J+1,K),CVS(J+1,K)
END DO
END DO
DO 22 L=1,28
IC(L)=L
IB(2*L-1)=L*L
IB(2*L)=L*L
END DO
J=0
IXB=0
WRITE ROWS
WRITE(*,46)J,IXB,IB(1),IB(2),IB(3),IB(4),IB(5),IB(6),IB(7),
&
IB(8),IB(9),IB(10),IB(11),IB(12),IB(13),IB(14),IB(15)
WRITE(7,46)J,IXB,IB(1),IB(2),IB(3),IB(4),IB(5),IB(6),IB(7),
&
IB(8),IB(9),IB(10),IB(11),IB(12),IB(13),IB(14),IB(15)
WRITE(*,46)J,IXB,IC(1),IC(2),IC(3),IC(4),IC(5),IC(6),IC(7),
&
IC(8),IC(9),IC(10),IC(11),IC(12),IC(13),IC(14),IC(15)
WRITE(7,46)J,IXB,IC(1),IC(2),IC(3),IC(4),IC(5),IC(6),IC(7),
&
IC(8),IC(9),IC(10),IC(11),IC(12),IC(13),IC(14),IC(15)
DO 24 J=1,14
WRITE(*,45)J,IB(J),A(J,1),A(J,2),A(J,3),A(J,4),A(J,5),A(J,6),
&
A(J,7),A(J,8),A(J,9),A(J,10),A(J,11),A(J,12),A(J,13),A(J,14),
&
A(J,15),A(J,16),A(J,17),A(J,18),A(J,19),A(J,20),A(J,21),
&
A(J,22),A(J,23),A(J,24),A(J,25),A(J,26),A(J,27),A(J,28)
WRITE(7,45)J,IB(J),A(J,1),A(J,2),A(J,3),A(J,4),A(J,5),A(J,6),
&
A(J,7),A(J,8),A(J,9),A(J,10),A(J,11),A(J,12),A(J,13),A(J,14),
&
A(J,15),A(J,16),A(J,17),A(J,18),A(J,19),A(J,20),A(J,21),
&
A(J,22),A(J,23),A(J,24),A(J,25),A(J,26),A(J,27),A(J,28)
END DO
WRITE COLUMNS
DO 25 J=1,28
.
24
C
WRITE(*,30)J,IB(J),A(1,J),A(2,J),A(3,J),A(4,J),A(5,J),A(6,J),
A(7,J),A(8,J),A(9,J),A(10,J),A(11,J),A(12,J)
WRITE(7,30)J,IB(J),A(1,J),A(2,J),A(3,J),A(4,J),A(5,J),A(6,J),
&
A(7,J),A(8,J),A(9,J),A(10,J),A(11,J),A(12,J)
END DO
CLOSE(7,STATUS='KEEP')
FORMAT(2(I10,","),11(F10.5,","),F10.5)
FORMAT(2(I10,","),27(F10.5,","),F10.5)
FORMAT(16(I10,","),I10)
FORMAT((F6.1,","),F10.4,",",F10.4)
END
&
25
30
45
46
50
Program 11.1.2—(Continued)
11.1
Introduction
C===================================================================
C
SUBROUTINE CVA1(KD,M,Q,CV)
C
C
============================================================
C
Purpose: Compute a sequence of characteristic values of
C
Mathieu functions
C
Input : M --- Maximum order of Mathieu functions
C
q --- Parameter of Mathieu functions
C
KD --- Case code
C
KD=1 for cem(x,q) ( m = 0,2,4,úúú )
C
KD=2 for cem(x,q) ( m = 1,3,5,úúú )
C
KD=3 for sem(x,q) ( m = 1,3,5,úúú )
C
KD=4 for sem(x,q) ( m = 2,4,6,úúú )
C
Output: CV(I) --- Characteristic values; I = 1,2,3,...
C
For KD=1,CV(1), CV(2), CV(3),..., correspond to
C
the characteristic values of cem for m = 0,2,4,...
C
For KD=2,CV(1), CV(2), CV(3),..., correspond to
C
the characteristic values of cem for m = 1,3,5,...
C
For KD=3,CV(1), CV(2), CV(3),..., correspond to
C
the characteristic values of sem for m = 1,3,5,...
C
For KD=4,CV(1), CV(2), CV(3),..., correspond to
C
the characteristic values of sem for m = 0,2,4,...
C
============================================================
10
15
20
25
30
IMPLICIT DOUBLE PRECISION (A-H,O-Z)
DIMENSION G(200),H(200),D(500),E(500),F(500),CV(200)
EPS=1.0D-14
ICM=INT(M/2)+1
IF (KD.EQ.4) ICM=M/2
IF (Q.EQ.0.0D0) THEN
IF (KD.EQ.1) THEN
DO 10 IC=1,ICM
CV(IC)=4.0D0*(IC-1.0D0)**2
ELSE IF (KD.NE.4) THEN
DO 15 IC=1,ICM
CV(IC)=(2.0D0*IC-1.0D0)**2
ELSE
DO 20 IC=1,ICM
CV(IC)=4.0D0*IC*IC
ENDIF
ELSE
NM=INT(10+1.5*M+0.5*Q)
E(1)=0.0D0
F(1)=0.0D0
IF (KD.EQ.1) THEN
D(1)=0.0D0
DO 25 I=2,NM
D(I)=4.0D0*(I-1.0D0)**2
E(I)=Q
F(I)=Q*Q
E(2)=DSQRT(2.0D0)*Q
F(2)=2.0D0*Q*Q
ELSE IF (KD.NE.4) THEN
D(1)=1.0D0+(-1)**KD*Q
DO 30 I=2,NM
D(I)=(2.0D0*I-1.0D0)**2
E(I)=Q
F(I)=Q*Q
Program 11.1.2—(Continued)
255
256
Computational Methods—Ordinary Differential Equations
35
40
45
50
55
60
65
70
75
ELSE
D(1)=4.0D0
DO 35 I=2,NM
D(I)=4.0D0*I*I
E(I)=Q
F(I)=Q*Q
ENDIF
XA=D(NM)+DABS(E(NM))
XB=D(NM)-DABS(E(NM))
NM1=NM-1
DO 40 I=1,NM1
T=DABS(E(I))+DABS(E(I+1))
T1=D(I)+T
IF (XA.LT.T1) XA=T1
T1=D(I)-T
IF (T1.LT.XB) XB=T1
CONTINUE
DO 45 I=1,ICM
G(I)=XA
H(I)=XB
DO 75 K=1,ICM
DO 50 K1=K,ICM
IF (G(K1).LT.G(K)) THEN
G(K)=G(K1)
GO TO 55
ENDIF
CONTINUE
IF (K.NE.1.AND.H(K).LT.H(K-1)) H(K)=H(K-1)
X1=(G(K)+H(K))/2.0D0
CV(K)=X1
IF (DABS((G(K)-H(K))/X1).LT.EPS) GO TO 70
J=0
S=1.0D0
DO 65 I=1,NM
IF (S.EQ.0.0D0) S=S+1.0D-30
T=F(I)/S
S=D(I)-T-X1
IF (S.LT.0.0) J=J+1
CONTINUE
IF (J.LT.K) THEN
H(K)=X1
ELSE
G(K)=X1
IF (J.GE.ICM) THEN
G(ICM)=X1
ELSE
IF (H(J+1).LT.X1) H(J+1)=X1
IF (X1.LT.G(J)) G(J)=X1
ENDIF
ENDIF
GO TO 60
CV(K)=X1
CONTINUE
ENDIF
RETURN
END
Program 11.1.2—(Continued)
a search of the World Wide Web provided the needed information in the form of a
downloadable subroutine.
The next program (11.1.3) uses FORTRAN’s capability to deal with complex
numbers, avoiding tedious computations to arrive at the real and imaginary portions of
11.1
Introduction
PROGRAM Joukowsky
C=========================================================
C
GENERATES A JOUKOWSKY AIRFOIL
C=========================================================
C
w = zU* + Ua^2/z + i*GAMMA/2*pi ln z/a
C
C
B = GAMMA/4*pi*a*|U| = sin(alpha + sin-1 YCP/a)
C
C
dw/dz = U* - U*a^2/z^2 + iGAMMA/2*pi*z
C
C
ZP = Z'
ZPP = Z''
C
C
Z' =Zc' + Z
Z'' = Z' + b^2/Z'
C=========================================================
COMPLEX AI,DWDZ,U,USTAR,W,Z,ZCP,ZP,ZPP,ZSP,ZPSP,ZPPSP,ZTE,ZPTE,ZPPTE
PI=3.1415927
A=1.
NPOINT=180
NP=360/NPOINT
AI=CMPLX(0,1.)
DTHETA=2.*PI/NPOINT
OPEN(1, FILE="A:JOUKOWSKY.DAT", STATUS='REPLACE')
WRITE(*,*)
WRITE(*,*) "In this program a and Uinfinity are taken as unity."
WRITE(*,*)
1
WRITE(*,*) "Input Zc':"
WRITE(*,*)
WRITE(*,*) " Thickness increases with Xc'."
WRITE(*,*) " Camber increases with Yc'."
WRITE(*,*) " Xc' = Yc' = 0 stops the program.'"
WRITE(*,*)
WRITE(*,*) "
Xc' first. It must lie in -1< Xc' <0. "
READ(*,*) XCP
WRITE(*,*) " Now input Yc'. It must lie in -1< Yc' <1. "
READ (*,*) YCP
WRITE(*,*)
WRITE(*,*) " Enter the angle of attack <degrees>. "
READ(*,*) ALPHA
B=XCP+SQRT(1.-YCP*YCP)
B2=B**2
BETA=ALPHA+ASIN(YCP/A)
GAMMA=4.0*PI*A*SIN(BETA)
R=XCP*XCP+YCP*YCP
SING=(B+XCP)**2+YCP**2-A**2
IF (R.EQ.0.) THEN
STOP
ELSE IF (R.GT.1.0) THEN
WRITE(*,*) " The point Zc' must lie within the unit circle.
&Try again."
GOTO 1
ELSE IF (SING.GE.0) THEN
WRITE(*,*) " The point Z'=-b must lie within the unit circle.
&
You will have a singularity in the flow field. Try again."
GOTO 1
ELSE IF (B.LE.0.) THEN
WRITE(*,*) "The parameter b <= 0, try again."
GOTO 1
END IF
Program 11.1.3—Joukowsky Airfoil (program by the author)
257
258
Computational Methods—Ordinary Differential Equations
5
10
20
100
101
102
103
ALP=ALPHA*PI/180.
U=CMPLX(COS(ALP),SIN(ALP))
USTAR=CMPLX(COS(ALP),-SIN(ALP))
ZCP=CMPLX(XCP,YCP)
WRITE(1,5)
DO I=1, NPOINT+1
THETA=(I-1)*DTHETA
Z=A*CMPLX(COS(THETA),SIN(THETA))
ZP=Z+ZCP
ZPP=ZP+B2/ZP
CALL SPEED(B2,GAMMA,PI,Z,ZP,U,USTAR,UM)
CP=1.-0.5*UM*UM
XPP=REAL(ZPP)
YPP=AIMAG(ZPP)
J=(I-1)*NP
WRITE(*,*) J,XPP,YPP,UM,CP
WRITE(1,100) J,XPP,YPP,UM,CP
END DO
WRITE(*,*)
WRITE(*,*) "Note: The speed calculations are not accurate in "
WRITE(*,*) " the vicinity of the trailing edge. Use"
WRITE(*,*) " l'Hospital's rule here."
WRITE(*,*)
WRITE(*,*) "Press <ENTER> to continue."
READ(*,*)
ZTE=CMPLX(A*COS(ALPHA-BETA),A*SIN(ALPHA-BETA))
ZPTE=ZTE+ZCP
ZPPTE=ZPTE+B2/ZPTE
XTE=REAL(ZPPTE)
YTE=AIMAG(ZPPTE)
ZSP=CMPLX(-A*COS(ALPHA+BETA),-A*SIN(ALPHA+BETA))
ZPSP=ZSP+ZCP
ZPPSP=ZPSP+B2/ZPSP
XSP=REAL(ZPPSP)
YSP=AIMAG(ZPPSP)
WRITE(*,*)
WRITE(*,*) " Trailing edge:
",XTE,YTE
WRITE(*,*) " Stagnation point: ",XSP,YSP
WRITE(*,*) " Xc' , Yc':
",XCP,YCP
WRITE(*,*) " b, beta :
",B,BETA
WRITE(*,*) " Gamma:
",GAMMA
WRITE(1,10) XCP,YCP
WRITE(1,20) B,BETA
WRITE(1,101) XTE,YTE
WRITE(1,102) XSP,YSP
WRITE(1,103) GAMMA
WRITE(*,*)
CLOSE(1)
FORMAT("Angle",',',"x''",',',"y''",',',"Speed",',',"Cp")
FORMAT(" Xc' = ",F10.7,',',", Yc' = ",F10.7)
FORMAT(" b = ",F10.7,',',", beta = ",F10.7," radians")
FORMAT (I4,2(',',F10.7),',',F15.7,','F15.7)
FORMAT(" Trailing edge
:
",2F10.7)
FORMAT(" Stagnation point:
",2F10.7)
FORMAT(" Gamma:
",F15.7)
END PROGRAM
Program 11.1.3—(Continued)
11.1
Introduction
C===================================================
C
SUBROUTINE SPEED
C===================================================
SUBROUTINE SPEED(B2,GAMMA,PI,Z,ZP,U,USTAR,UM)
COMPLEX AI,DWDZ,Z,ZP,U,USTAR
AI=CMPLX(0,1.)
DWDZ=USTAR-U/(Z**2)+AI*GAMMA/(2.*PI*Z)
DWDZ=DWDZ/(1.-B2/(ZP*ZP))
UM=ABS(DWDZ)
RETURN
END SUBROUTINE
Program 11.1.3—(Continued)
complicated complex expressions. The user inputs the three parameters involved in the
Joukowsky transformations described in Chapter 3, as well as the angle of attack of the
airfoil. The program then generates the shape of the airfoil, as well as the velocity and
pressure distributions on its surface.
Program 11.1.4 is an example of the generation of a potential flow about a halfbody consisting of a source in a uniform stream. After specifying source strength and
PROGRAM
RANKINE
! RANKINE.FOR
!-----------------------------------------------------------------------!
Generates a Rankine half-body in two dimensions
!
Potential consists of a source in a uniform stream
!
This program is easily generated to the axisymmetric
!
half-body with a change in the potential
!
from a 2-dimensional source to a three-dimensional source.
!
More general shapes can be generated by the addition of more sources
!
and/or sinks
!
Remember that to generate a closed body the sum of the source
!
and sink strengths must be zero.
!-----------------------------------------------------------------------!
Data generated here was used in generating data for Figure 2.2.9
!
Values used:
!
A=m/2*PI*U
XSP = Stag point @ -A
Half-width = WH = pi*A
!
The data in File 1 is easily inserted in any spreadsheet.
!
This was made possible by including the commas in FORMAT statement 200.
!-----------------------------------------------------------------------!
Configure maximum data size for storage.
DIMENSION S(8),X0(26),Y
0(26),XXS(26),YYS(26),XX(26),YY(11,71)
!
Open file for data storage
OPEN(1, FILE='A:RANKINE.DAT', STATUS='UNKNOWN')
A=.125
PI=3.14159
!
Generates PSI = 0 streamline on x axis, infinity to source
DO I=1,25
X0(I)=-6.+(12./24.)*(I-1)
Y0(I)=0.
END DO
!
Generates PSI = 0 streamline off of x axis, starts at stagnation point
CALL BODYSHAPE(A,PI,XXS,YYS)
!
Generates other streamlines
XSTART=-6.
YSTART=0.
Program 11.1.4—Potential flow past a half-body (program by the author)
259
260
Computational Methods—Ordinary Differential Equations
YS=0.
DX=0.5
DO M=1,25
! X LOOP
XX(M)=XSTART+(M-1)*DX
X=XX(M)
DO N=1,8
! PSI LOOP, PSI incremented by 0.2
PSI=0.2*N
CALL SEARCH(PSI,A,X,Y,PSIS,ATD,IS,PI)
YY(N,M)=Y
FER=PSIS-PSI
WRITE(*,100) IS,PSI,FER,X,Y,ATD
END DO
WRITE(*,101)
END DO
WRITE(*,*) "
IS
PSI
PSIS
X
Y"
DO I=1,25
SM=-YYS(I)
DO J=1,8
S(J)=-YY(J,I)
END DO
WRITE(1,200) X0(I),Y0(I),XXS(I),YYS(I),SM,XX(I),YY(1,I),
&
YY(2,I),YY(3,I),YY(4,I),YY(5,I),YY(6,I),YY(7,I),YY(8,I),
&
S(1),S(2),S(3),S(4),S(5),S(6),S(7),S(8)
END DO
100
FORMAT(I6,F10.4,F10.6,3F10.4)
101
FORMAT("============================================")
200
FORMAT(21(F8.4,','),F8.4)
CLOSE (1)
END PROGRAM
!=====================================================================
SUBROUTINE BODYSHAPE(A,PI,XXS,YYS)
!--------------------------------------------------------------------!
Generates body shape for a 2-D Rankine half-body
!
Generates PSI = 0 streamline off of x axis, starts at stagnation point
!
Fixes Y, searches for X
!--------------------------------------------------------------------DIMENSION XXS(26),YYS(26)
WH=A*PI
XSP=-A
DY=WH/25
XXS(1)=XSP
YYS(1)=0.
II=1
XOLD=XSP
Y=DY
X=XOLD-.01
FOLD=Y+A*(ATAN(-X/Y)-.5*PI)
WRITE(*,*) A,WH
!
Y loop covers a range of Y's, staring with 0.04*halfwidth
!
and continuing to 0.96* half-width.
DO I=1,24
Y=I*DY
X=XOLD
IG=0
WRITE(*,*) "============================="
WRITE(*,*) X,Y,FOLD
!
X loop covers a range of X. Since Y is fixed, a starting point is used (XOLD)
!
and then incremented a small amount. The search procedure is much as used
!
in the subroutine SEARCH, except the roles of X and Y are reversed.
!
Comments made in the SEARCH subroutine are true here also regarding ATAN.
Program 11.1.4—(Continued)
11.1
Introduction
DO J=1,1500
DX=.0001
IF(X.GT.0.234) DX=.0009
IF(I.EQ.24) DX=0.0015
X=DX*(J-1)+XOLD
AT=ATAN(-X/Y)
F=Y+A*(AT-.5*PI)
!
Test for convergence.
FTEST=FOLD*F
ATD=AT*180./PI
IF((FTEST.LE.0.).AND.(IG.EQ.0)) THEN
WRITE(*,200)I,J,X,Y,F,ATD
II=I+1
XXS(II)=X
YYS(II)=Y
XS=X
YS=Y
IG=1
ENDIF
FOLD=F
END DO
XOLD=XS-.01
FOLD=YS+A*(ATAN(-(XS-.1)/YS)-.5*PI)
END DO
XXS(25)=6.
YYS(25)=WH
WRITE(*,*) " X
Y
FOLD"
WRITE(*,*) " I
J
X
Y
F
ATD"
100
FORMAT(F10.5,','F10.5)
200
FORMAT(2I5,3F10.5,F12.5)
RETURN
END SUBROUTINE
!---------------------------------------------------------------------SUBROUTINE SEARCH(PSI,A,X,YS,PSIS,ATD,IS,PI)
!
Generates streamlines
!
Fixed X, searches for Y
!
The search procedure used is not very elegant, but it works (a lot to be
!
said for that).
!
A desired value of PSI is inputted, along with a value of x.
!
A value of y is guessed at, then a value for PSI based on that is computed.
!
The first guess for y is always taken above the desired streamline so it
!
gives a value of PSI which is too large.
!
The y value is decremented a certain amount dy and the computation is
!
repeated, until the error changes sign.
!
Then x is incremented and the process repeated.
!
Two considerations arise in the computation:
!
1. At and near the stagnation point the slope of the streamline is
!
steep, thus small dy (0.0001) must be used. Later on larger values
!
of dy (0.001) are used.
!
2. Arctangents are multivalued, and FORTRAN assumes that you want
!
values between - pi/2 and + pi/2. Thus in using the ATAN function
!
care has to be taken.
!---------------------------------------------------------------------WH=A*PI
XSP=-A
DY=0.0001
IF(X.GE.-.6) DY=.001
Program 11.1.4—(Continued)
261
262
Computational Methods—Ordinary Differential Equations
!
!
!
FOLD=Y+A*(ATAN(-X/(Y-DY))-.5*PI)
IG=0
Y=PSI-.015
This is the Y loop, where Y is continually incremented.
DO I=1,1500
Y=Y+DY
PSIOLD=PSI-.05
AT=ATAN(-X/Y)
F=Y+A*(AT-.5*PI)
FTEST=F-PSI
Test to see if the desired streamline has been crossed.
If it has, linear interpolation is used to improve result.
IF((FTEST.GE.0.).AND.(IG.EQ.0)) THEN
IS=I
FF=(F-PSIOLD)/(PSI-PSIOLD)
PSIS=F
YS=Y
IG=1
ATD=AT*180./PI
ENDIF
PSIOLD=F
END DO
RETURN
END SUBROUTINE
Program 11.1.4—(Continued)
uniform stream velocity, the body shape and surrounding streamlines are found. The
computation is complicated by the fact that on a streamline it is not possible to solve
directly for y as a function of x. An iteration procedure must be used. The appropriate
source strength on the panel is computed so as to make the velocity tangent to the body.
11.2 Numerical Calculus
Many schemes are available for computing derivatives and integrals of functions,
depending on the degree of accuracy needed. Many methods are based on the Taylor
series expansion of a function—that is, for yx = fx, on
yx =
1 n
f x0 x − x0 n
n!
n=0
(11.2.1)
Here f n x0 stands for the nth order derivative of f evaluated at x0 . If y is known at
neighboring points and knowledge of its derivative is required, then to lowest order in
the point spacing, write yx + x = yx + f ′ xx + Ox2 , giving
yx + x − yx
+ Ox
x
If greater accuracy is needed, start with
f ′ x =
1
yx + x = yx + f 1 xx + f 2 xx2 + Ox3
2
1
yx − x = yx − f 1 xx + f 2 xx2 + Ox3
2
(11.2.2)
(11.2.3a)
(11.2.3b)
11.2
263
Numerical Calculus
Subtracting one of the two equations from the other gives
f 1 x =
yx + x − yx − x
+ Ox2
2x
(11.2.4)
Further accuracy can be found by including more terms and more points in the Taylor
expansion. Some examples follow:
1
dfx
≈
fx + x − fx − x + O x2
dx
zx
1
dfx
(11.2.5)
≈
Forward differences
fx + x − fx + Ox
dx
x
dfx
1
Backward differences
≈
fx − fx − x + Ox
dx
x
Central differences
Newton’s method is frequently used for finding the roots of an equation of the form
fx = 0. (Other methods can be found in Press, Flannery, Teukolsky, and Vetterling
(1986), and in Carnahan, Luther, and Wilkes (1969).) Newton suggested making an
initial guess and then repeatedly improving it according to the formula
xn+1 = xn −
fxn
f ′ xn
(11.2.6)
the prime denoting the derivative of f . This is merely a rearrangement of equation
(11.2.2). Providing that the first guess is reasonably close to the root sought, the iteration
proceeds quickly to the desired root.
Newton’s rule can be used for the computation of the stability curves for
Couette-Rayleigh
instability. First, to simplify the calculations, introduce
=
− c02 − a2 /a2 . Then
2 n
2 n
cn2 = a2 1 − cos
+ i sin
n = 0 1 2
3
3
Since we expect that > 1,
√
− 1,
c0 = ai
c1 = a 1 +
c2 = a 1 +
with
= tan−1
√
1
3
−i
2
2
√
1
3
+i
2
2
√
3
2+
= a 1+ +
2
e−i
/2
= a 1+ +
2
ei
/2
To start the computation, use the known minimum critical values for even disturbances to compute the Rayleigh number, and then move first to slightly smaller values
of the wave number, understanding that the corresponding Rayleigh number is always
slightly larger than the previous value. Use Newton’s method repeatedly to guess at the
Raleigh number for this new wave number. (The derivative is computed using equation
(11.2.2).) Repeat the procedure for wave numbers larger than critical, and finally the
264
Computational Methods—Ordinary Differential Equations
process is repeated for odd disturbances. Successive iteration improves the accuracy by
narrowing the interval at which the derivative is computed.
In Program 11.2.1 a subroutine is used to calculate the stability criteria to simplify
understanding of the program.
C.......................BENARD.FOR........................
C..............Rayleigh-Benard Stability..................
C............R = Rayleigh NUMBER, a = Wave Number.........
C.........................................................
DIMENSION AE(200),AO(200),RE(200),RO(200)
OPEN(1,FILE='A:BENARD.DAT', STATUS='UNKNOWN')
C.........................................................
C..............Even Disturbances..........................
C.........................................................
WRITE(6,*) "Even Disturbances"
AE(28)=3.117
RE(28)=1707.762
CALL CALC(AE(28),RE(28),FEVEN,FODD)
WRITE(6,*) " a
Ra
FEVEN"
WRITE(6,*) AE(28),RE(28),FEVEN
WRITE(6,*) "
J
a
RSAVE
FEVEN"
FEMIN=FEVEN
RMIN=RE(28)
DR=0.1
C..............Even - Less Than Minimun...................
FEVEN=FEVEN-0.1
ROLD=1700.-DR
DO I=1,27
J=0
a=3.1-(I-1)*0.1
DO WHILE (FEVEN<0.)
J=J+1
R=ROLD+DR
CALL CALC(a,R,FEVEN,FODD)
rat=-FEOLD/(-FEOLD+FEVEN)
RSAVE=R+rat*DR
IF(rat<0.) WRITE(6,*)FEVEN,FEOLD,rat
FEOLD=FEVEN
ROLD=R
END DO
WRITE(6,100) J,a,RSAVE,FEVEN
FEVEN=FEVEN-1.
AE(28-I)=a
RE(28-I)=RSAVE
END DO
C..............Even - Greater Than Minimun...................
FEVEN=FEVEN-0.1
ROLD=1700.-DR
DO I=1,59
J=0
a=3.2+(I-1)*0.1
DO WHILE (FEVEN<0.)
J=J+1
R=ROLD+DR
CALL CALC(a,R,FEVEN,FODD)
rat=-FEOLD/(-FEOLD+FEVEN)
RSAVE=R+rat*DR
FEOLD=FEVEN
ROLD=R
END DO
Program 11.2.1—Computation of the neutral stability curves for Rayleigh-Bénard stability (program by the
author)
11.2
265
Numerical Calculus
WRITE(6,101) J,a,RSAVE,FEVEN
FEVEN=FEVEN-1.
AE(28+I)=a
RE(28+I)=RSAVE
END DO
WRITE(1,104)
WRITE(1,105)
DO I=1,87
WRITE(1,106) I,AE(I),RE(I)
END DO
C........................................................
C..............Odd Disturbances..........................
C........................................................
WRITE(6,*) "Odd Disturbances"
AO(38)=5.365
RO(38)=17610.39
CALL CALC(AO(38),RO(38),FEVEN,FODD)
WRITE(6,*) " a
Ra
FODD"
WRITE(6,*) AO(38),RO(38),FODD
WRITE(6,*) "
J
a
RSAVE
FODD"
FOMIN=FODD
C..............Odd - Less Than Minimun...................
FODD=FOMIN-0.1
ROLD=17600.-DR
DO I=1,37
J=0
a=5.3-(I-1)*0.1
DO WHILE (FODD<0.)
J=J+1
R=ROLD+DR
CALL CALC(a,R,FEVEN,FODD)
rat=-FOOLD/(-FOOLD+FODD)
RSAVE=R+rat*DR
FOOLD=FODD
ROLD=R
END DO
WRITE(6,102) J,a,RSAVE,FODD
FODD=FODD-1.
AO(38-I)=a
RO(38-I)=RSAVE
END DO
C..............Odd - Greater Than Minimun...................
FODD=FOMIN-0.1
ROLD=17600.-DR
DO I=1,38
J=0
a=5.3+(I-1)*0.1
DO WHILE (FODD<0.)
J=J+1
R=ROLD+DR
CALL CALC(a,R,FEVEN,FODD)
rat=-FOOLD/(-FOOLD+FODD)
RSAVE=R+rat*DR
FOOLD=FODD
ROLD=R
END DO
Program 11.2.1—(Continued)
266
Computational Methods—Ordinary Differential Equations
WRITE(6,103) J,a,RSAVE,FODD
FODD=FODD-1.
AO(38+I)=a
RO(38+I)=RSAVE
END DO
WRITE(1,107)
WRITE(1,105)
DO I=1,75
WRITE(1,106) I,AO(I),RO(I)
END DO
CLOSE(1, STATUS='KEEP')
100 FORMAT(" Even 1 ",I5,F8.4,F12.4,F10.6)
101 FORMAT(" Even 2 ",I5,F8.4,F12.4,F10.6)
102 FORMAT("
Odd 1 ",I5,F8.4,F12.4,F10.6)
103 FORMAT("
Odd 2 ",I5,F8.4,F12.4,F10.6)
104 FORMAT("
EVEN DISTURBANCE")
105 FORMAT("
WAVE NUMBER
RAYLEIGH NUMBER")
106 FORMAT(I3,F12.4,F10.6)
107 FORMAT("
ODD DISTURBANCE")
END
C................Subroutine.................................
SUBROUTINE CALC(a,R,FEVEN,FODD)
TAU=((R/a)**(1./3.))/a
S3=SQRT(3.)
Q0=a*SQRT(TAU-1.)
Q1=a*SQRT(.5*SQRT(1.+TAU*(1.+TAU))+.5*(1.+.5*TAU))
Q2=a*SQRT(.5*SQRT(1.+TAU*(1.+TAU))-.5*(1.+.5*TAU))
P1=Q1+Q2*S3
P2=Q1*S3-Q2
FEVEN=Q0*TAN(.5*Q0)+(P1*SINH(Q1)+P2*SIN(Q2))/(COSH(Q1)+COS(Q2))
FODD=-Q0/TAN(.5*Q0)+(P1*SINH(Q1)-P2*SIN(Q2))/(COSH(Q1)-COS(Q2))
RETURN
END SUBROUTINE
Program 11.2.1—(Continued)
Numerical integration is performed in much the same manner as differentiation,
again using the Taylor series expansion and integrating it term by term, stopping when
the needed order of accuracy is obtained. The lowest order of accuracy is termed the
trapezoidal rule, in the form
x+x
1
(11.2.7)
fxdx ≈ fx + x + fx x
2
x
This approximates the true area of the integral by the area of a trapezoid with corners x 0 x fx x + x fx + x x + x 0. Integration over a finite interval
amounts to adding up the contributions of the small intervals.
An example of the use of the trapezoidal rule is given in Program 11.2.2, where the
integration of the Elliptic Integrals of the first and second kind are carried out. Tables
available traditionally use degrees instead of k degrees = sin−1 k. The program uses
step sizes of 0.01 degrees and prints results at every degree. Agreement with published
tables (given to three significant figures) is very good.
Another often used method for integration is Simpson’s rule. It essentially fits
a polynomial to the integrand and deals with points at x x, and 2x in one step.
Letting f0 = fx f1 = fx + x and f2 = fx + 2x, the integral of f from x to
x + 2x is
x+2x
1
(11.2.8)
fxdx ≈ x f0 + 4f1 + f2 + Ox5
I=
3
x
11.3
267
Numerical Integration of Ordinary Differential Equations
C
PROGRAM EllfUN
C
Calculates incomlete and complete Elliptic Integrals of the first & second kind.
C=======================================================================================
DOUBLE PRECISION, DIMENSION (18,100) :: AE,AF
DOUBLE PRECISION AET,AFT,AKDI,DI,G,GI,GOLD,GIOLD,PHI,PI
PI=3.1415927
DI=0.01*PI/180.
OPEN(1, FILE='a:EllFun.dat', STATUS='REPLACE')
WRITE(1,90)
C
VARIABLE INTEGRATION
DO I=1,18
IDEGR=5*I
AK=SIN(IDEGR*PI/180.)
GOLD=1.
GIOLD=1.
AE(I,1)=0.
AF(I,1)=0.
AET=0.
AFT=0.
ICOUNT=0
JPRINT=0
C
INTEGRATION ITERATION
DO J=1,9000
ICOUNT=ICOUNT+1
PHI=J*DI
G=SQRT(1.-AK*AK*SIN(PHI)*SIN(PHI))
GI=1./G
AET=AET+0.5*(G+GOLD)*DI
AFT=AFT+0.5*(GI+GIOLD)*DI
GOLD=G
GIOLD=GI
IF(ICOUNT.EQ.100) THEN
JPRINT=JPRINT+1
JDEGR=JPRINT
AF(I,JPRINT)=AFT
AE(I,JPRINT)=AET
WRITE(*,*) AK,IDEGR,JDEGR,AF(I,JPRINT),AE(I,JPRINT)
WRITE(1,100) AK,IDEGR,JDEGR,AF(I,JPRINT),AE(I,JPRINT)
ICOUNT=0
ENDIF
END DO
END DO
CLOSE(1)
90
FORMAT(" K
ARCSIN(K)
PHI F(K,PHI) E(K,PHI)")
100
FORMAT(F10.7,’,’,I10,’,’,F10.7,’,’,F10.7)
END
Program 11.2.2—Elliptic Integrals of the first and second kind (program by the author)
The formula
I=
x+3x
x
3
fxdx ≈ x f0 + 3f1 + 3f2 + f3 + Ox5
8
(11.2.9)
referred to as Simpson’s second rule, is also useful.
11.3 Numerical Integration of Ordinary Differential Equations
Frequently a first-order ordinary differential equation (or set of equations if y and f are
vectors) of the type
dy
= fx y
dx
(11.3.1)
cannot be integrated in closed form, and numerical methods must be resorted to if the
solution is to be found. One of the simplest methods of doing this is Euler’s method,
where the solution is found by repeating the process
268
Computational Methods—Ordinary Differential Equations
yx + x = yx + f yx x x
(11.3.2)
This technique is easy to implement in a computer code and is readily extended to the
case where f is an array. It is not, however, particularly accurate, the error being of the
order of x squared.
An example of more accurate methods is the family of Runge-Kutta methods. They
require evaluation at intermediate points in the interval to achieve their accuracy. The
method is used for systems of first-order differential equations, which can be linear or
nonlinear. Two versions of Runge-Kutta methods are listed here.
Second-order Runge-Kutta (the error is of order x cubed)
yx + x = yx + 1 − bk1 + bk2 x with
k1 = f yx x
(11.3.3)
k2 = f yx + pk1 x + p
p = x/2b
The constant b can be chosen arbitrarily in the range 0 to 1. Common choices are 1/2
and 1. When b is chosen as 1 (modified Euler-Cauchy method), then
yx + x = yx + k2 x with
k1 = f yx x
(11.3.4)
k2 = f yx + k1 x/2 x + x/2
When b is chosen as 1/2 (Heun’s method), equation (11.3.3) becomes
yx + x = yx + 05k1 + k2 x
with
(11.3.5)
k1 = f yx x
k2 = f yx + k1 x x + x
Both choices of b (1/2 and 1) give results with the same order of accuracy.
Fourth-order Runge-Kutta (the error is of order x to the fifth power.)
The general form for a fourth-order accurate Runge-Kutta scheme is
yx + x = yx +
4
wi k i
(11.3.6)
j=1
There are many different combinations of the w and k that all give the same accuracy.
All are easy to program. Here are three sets of commonly used formulas:
1. (Credited to Kutta.)
1
yx + x = yx + k1 + 2k2 + 2k3 + k4 with
6
k1 = f yx x x
k2 = f yx + k1 x/2 x + x/2 x
k3 = f yx + k2 x/2 x + x/2 x
k4 = f yx + k3 x x + x x
(11.3.7a)
11.3
269
Numerical Integration of Ordinary Differential Equations
2. (Credited also to Kutta.)
1
yx + x = yx + k1 + 3k2 + 3k3 + k4 with
8
k1 = f yx x x
k2 = f yx + k1 x/3 x + x/3 x
(11.3.7b)
k3 = f yx − k1 x/3 + k2 x x + 2x/3 x
k4 = f yx + k1 x − k2 x + k3 x x + x x
3. (Credited to Gill. Minimizes storage space needed.)
1
1
1
k + 2 1 − √ k2 + 2 1 + √ k3 + k4 with
yx + x = yx +
6 1
2
2
k1 = f yx x x
(11.3.8)
k2 = f yx + k1 x/2 x + x/2 x
√
√
k3 = f yx + −1 + 2k1 x/2 + 1 − 2k2 x/2 x + x/2 x
√
√
k4 = f yx − k2 x/ 2 + 1 + 1/ 2k3 x x + x x
Again, all three of these choices give results with the same order of accuracy.
Not many of the ordinary differential equations encountered so far have been
couched as a set of first order equations, but they all can be easily converted to such.
For instance, an equation such as
f ′′′ + af ′′ + bf ′ + cf = 0
(11.3.9)
can be converted by letting
y1 = f
y2 = f ′
y3 = f ′′
Then
⎧ ⎫ ⎧
⎫
y2
y
⎬
d ⎨ 1⎬ ⎨
y3
y2 =
dx ⎩y ⎭ ⎩−ay − by − cy⎭
2
1
3
(11.3.10)
Note that it is not necessary that the a b c be constants, or the right-hand side
of equation (11.3.1), be linear in y. It is, however, necessary that all y be known at the
starting point of the integration.
The similarity solutions encountered in investigating the boundary layer are all good
candidates for a Runge-Kutta scheme. Since some of the conditions are known at the
wall, it is necessary to guess them and then see if they do converge to the proper values
away from the boundary. Also, it is necessary to make a choice for “infinity.” As we
saw in the previous solutions, good choices can be 5, 7, or even 10. After reasonably
satisfactory values for the unknown boundary conditions are found, moving the infinite
boundary further out can be used to improve the solution.
Program 11.3.1 was used to compute the values for the Falkner-Skan class of
boundary layer solutions. Suggestions are made for the unknown condition (f ′′ 0 in
this case) based on published values, and interpolations from these can be easily made
for other values of m. Screen prints of f f ′ , and f ′′ are made. A good test of whether
270
Computational Methods—Ordinary Differential Equations
PROGRAM FALKNER
!=============================
!
Infinity is set at 7
!=============================
DIMENSION F0(3),Fk1(3)
H=.001
WRITE(*,*)
WRITE(*,*) " Solves the Falkner-Skan equation f''' + 0.5(m + 1)ff'' + m[1 - (f')^2] = 0"
WRITE(*,*) "
using a 4th order Runge-Kutta scheme."
WRITE(*,*)
WRITE(*,*) "Suggested value pairs"
WRITE(*,*) "Beta:
0 54
18 72
36 90
108
144
180
216 "
WRITE(*,*) "m:
0
0.0526 0.1111 0.1765 0.25
0.3333 0.4287 0.6667 1
1.5 "
WRITE(*,*) "f''(0): 0.332 0.4259 0.5113 0.5936 0.6759 0.7569 0.8418 1.022 1.232 1.497"
WRITE(*,*)
WRITE(*,*) "
Enter m.
"
READ (*,*) AM
WRITE(*,*)
DO 3 JJ=1,50
WRITE(*,*) " Enter f''(0). "
WRITE(*,*) "
A ZERO stops the program."
READ(*,*) FG
IF (FG.EQ.0.) GOTO 4
F0(3)=FG
F0(1)=0.
F0(2)=0.
X=0.
A=0.-.5
WRITE(*,*) "
eta
f
f'
f''"
DO 1 J=1,10000
IF (X.GE.A) THEN
WRITE(*,100) X,F0(1),F0(2),F0(3)
A=A+.1
ENDIF
CALL RK4(F0,H,AM)
X=X+H
IF (X.GT.7.05) GOTO 2
1
END DO
2
WRITE(*,*)" m
f''(0)
f'(large)"
WRITE(*,*) AM,FG,F0(2)
3
END DO
100 FORMAT(4F12.7)
4 END PROGRAM
!==================================================
!
Runge_Kutta solver - Press, Flannery, et al
!==================================================
Subroutine RK4(F0,H,AM)
DIMENSION F0(3),F1(3),F2(3),F3(3),Fk1(3),Fk2(3),Fk3(3),Fk4(3)
CALL DERIVS(F0,Fk1,AM)
DO 11 I=1,3
F1(I)=F0(I)+0.5*H*Fk1(I)
11
END DO
CALL DERIVS(F1,Fk2,AM)
DO 12 I=1,3
F2(I)=F0(I)+0.5*H*Fk2(I)
12
END DO
CALL DERIVS(F2,Fk3,AM)
DO 13 I=1,3
F3(I)=F0(I)+H*Fk3(I)
13
END DO
CALL DERIVS(F3,Fk4,AM)
DO 14 I=1,3
F0(I)=F0(I)+H*(Fk1(I)+2.*Fk2(I)+2.*Fk3(I)+Fk4(I))/6.
14
END DO
RETURN
END SUBROUTINE
Program 11.3.1—Falkner-Skan boundary layers (main program by the author, subroutine from Press,
Flannery, et al. (1986))
11.3
Numerical Integration of Ordinary Differential Equations
!===================================================
!
Defines equation
!===================================================
SUBROUTINE DERIVS(FF,DFF,AM)
DIMENSION FF(3),DFF(3)
DFF(1)=FF(2)
DFF(2)=FF(3)
DFF(3)=-0.5*(AM+1.)*FF(1)*FF(3)+AM*(FF(2)*FF(2)-1.)
RETURN
END SUBROUTINE
Program 11.3.1—(Continued)
the choice of f ′′ 0 was suitable is how f ′ behaves as the independent variable becomes
large. The program regards = 7 to be infinity.
As long as there is a suitable first guess for the wall shear term, the procedure is
simple to use and can be made to converge rather quickly by using Newton’s method
or, even simpler, just repeating the calculation over a range of points. In problems such
as von Kármán’s rotating disk or the thermal boundary layers where there are more
than one unknown at the origin, the search is over a two-dimensional set of parameters
and becomes more difficult. A starting point would be to use the Kármán-Pohlhausen
method to calculate a first guess. Generally, if one starts close to the desired pair of
values, improving the guess, no matter how crude the method, is doable.
Another possible approach would be a version of the steepest descent method,
whereby one sets one of the parameters and then considers the results of a series of
computations performed by varying the second parameter. Finding which value of the
second parameter gives the “best result,” the first parameter could then be incremented
and then the process repeated. Defining “best result” might not, however, be obvious,
and the convergence and programming just might be challenging.
A method that has been proposed in IMSL (1987, pages 660–671, programs
BVPFD/DBVPFD) is to multiply the nonlinear terms by a parameter p and then starting
with p = 0, and slowly increase p until it reaches unity—in which case the full original
equation is reached. This method is designed for two-point boundary value problems so
initial guessing of derivatives at the wall is avoided.
A method similar to this would be to insert time derivatives into each of the firstorder equations, thus making an initial value problem. The hope would be, of course,
that a steady state would eventually be reached.
Still another method for integrating a system of first-order ordinary differential equations is based on the method of cubic splines. Cubic splines became useful and popular with the advent of computer graphics. The idea is to take a function defined over an
interval and break the interval into a number of subintervals. Within each subinterval the
function is defined by a cubic polynomial. The polynomials are joined together in such
a manner that the function and its first and second derivatives are continuous throughout
the larger interval. To accomplish this, within each subinterval the function is defined by
x − x3
x − xi 3
yxi 1 ′′
yx = y′′ xi i+1
+ y′′ xi+1
+
− y xi i xi+1 − x
6i
6i
i
6
yxi+1 1 ′′
− y xi+1 i x − xi
+
(11.3.11)
i
6
where the subinterval extends from xi to xi+1 and is of length i = xi+1 − xi .
271
272
Computational Methods—Ordinary Differential Equations
To use this in integrating the system of equations
dy
= fx y in a ≤ x ≤ b
dx
first divide the interval into n subintervals and let
n
1 for xi ≤ x ≤ xi+1
yx = i gi x where i =
0 otherwise
i=1
(11.3.12)
(11.3.13)
The gi are the cubic splines as given in equation (11.3.11). Now insert equation (11.3.13)
into equation (11.3.12), multiply each side of the resulting equation by j and integrate
over the interval, where
1 for xj−1 ≤ x ≤ xj+1
j =
0 otherwise
The result is
yj+1 − yj−1 =
xj+1
xj−1
f x
n
i gi x dx
(11.3.14)
i=1
The integrals on the right-hand side can be carried out by any of the previously discussed
integration schemes.
There are a number of other methods for numerically integrating a system of firstorder ordinary differential equations, such as predictor-corrector schemes like AdamsBashforth-Moulton. Discussions of these can be found in Press, Flannery, Teukolsky,
and Vetterling (1986) or Carnahan, Luther, and Wilkes (1969).
To improve accuracy and speed, some integration approaches use variable step
size. For instance, one technique used for the Runge-Kutta fourth-order scheme is to
first perform the integration over the interval x = x2 − x1 , and then the integration is
repeated by returning to x1 and repeating the integration by doing it in two steps and
using a step size of one-half
1/nThe integration proceeds now from x2 but this time
y −yx.
2
1
xprevious , where y2 − y1 is the computed change in
using a step size xnew = DMC
the solution, DMC is the desired maximum change in the solution, and n is the order
of the error of the method—5 in this case. The choice of DMC is at the discretion of
the user.
11.4 The Finite Element Method
A method that is capable of extension to higher dimensional space is the finite element
method. To illustrate this method, consider the equation
dy
d
px
− qxy = fx where y0 = y1 = 0
(11.4.1)
dx
dx
This equation is referred to as the Sturm-Liouville equation, and many of the special
functions of mathematics are generated from it. It has the following important property.
Multiplication of it by yx (if the coefficients are complex, use the complex conjugate
of y) and subsequent integration of the result over the interval yields
1
1
dy 2
dy 1
2
p
+ qy dx =
yf dx
py −
dx 0
dx
0
0
11.4
273
The Finite Element Method
By virtue of the boundary conditions, the first term vanishes. Providing that p and q are
both positive and nonzero, it also tells us that the integral on the right-hand side must
be negative. Rearranging, find that
1
dy 2
2
+ qy + yf dx = 0
(11.4.2)
p
dx
0
Considering now the function Fy defined by
1
dy 2
2
Fy =
+ qy + 2yf dx
p
dx
0
(11.4.3)
it can be demonstrated that the exact solution of equation (11.4.1) is one that minimizes
equation (11.4.3). To do this, let z be an approximation to the exact solution y such that
zx = yx + x and z0 = z1 = 0. Introduce this into equation (11.4.3), and find
after integration by parts and use of the boundary conditions that
1 dy
1
dz 2
2
−p + qy + f dx
+ qz + 2zf dx + 2
Fz =
p
dx
dx
0
0
1
d 2
+ q2 dx
p
+
dx
0
The second integral is zero by virtue of equation (11.4.2), and the third is equal or
greater than zero. Thus,
Fz ≥ Fy
For the finite element method (FEM), break the interval into n regions (elements)
and let
yx =
where
n
yj Nj x
j=1
⎧
⎪
⎨x − xj −1 /xj − xj−1 for xj−1 ≤ x ≤ xj
Nj x = x − xj +1 /xj − xj+1 for xj ≤ x ≤ xj+1
⎪
⎩
0
elsewhere.
(11.4.4)
These chosen functions are triangles of a maximum height of one and are defined as
nonzero over just two intervals. Inserting equation (11.4.4) into equation (11.4.3) and
differentiating with respect to the yj to minimize F , the result is a set of n by n linear
algebraic equations in the yj in the form Ay = b, where
1 dN dN
1
j
aij =
p i
fNi dx
(11.4.5)
+ qNi Nj dx bi = −
dx dx
0
0
If n is not too large, the set of linear equations can be solved by elementary methods
such as Gaussian elimination, whereby line by line, the terms to the left of the diagonal
are eliminated, thereby giving a set of equations solvable by back substitution. However,
when this is applied to more than one dimension, or when n is very large, methods
dedicated to the solution of a large set of algebraic equations must be used.
274
Computational Methods—Ordinary Differential Equations
In this example, the approximating functions have been chosen to be linear. This
may be adequate in many cases. If not, higher-order polynomials or other functions may
be used that cover the interval in a similar fashion.
11.5 Linear Stability Problems—Invariant Imbedding and
Riccati Methods
In Chapter 9 several flow stability problems were considered—namely, RayleighBénard, Couette and Poiseuille viscous flows, and several interfacial-wave-type problems for inviscid flows. With the exception of the Rayleigh-Bénard convection flows,
all of the viscous flow problems involved rather serious and sophisticated computations, even though the equations were linear and therefore traditional series expansion
methods and superposition could be used. When numerical methods are considered,
however (sometimes surprisingly), there is no advantage to having linear equations, and
exchanging them for nonlinear equations may have some advantage.
To introduce the concept of invariant imbedding, consider a very simple equation
that often appears when discussing mechanical problems (the Euler column theory is
one example):
d2 y
+ 2 y = 0
dx2
y0 = y1 = 0
(11.5.1)
This is a classic eigenvalue (characteristic value) problem with eigenvalues , where
the solution is zero unless takes on the values = n n = 1 2 3
. In that case
the solution is yx = A sinx, where A is indeterminate.
If equation (11.5.1) is changed to
d2 y
+ 2 y = 1
dx2
y0 = y1 = 0
the solution becomes determinate, with the solution
1
sin x
yx = 2 1 − cos x + −1 + cos
sin
(11.5.2)
(11.5.3)
This is a well-behaved solution as long as does not equal one of the eigenvalues
associated with the homogeneous problem (11.5.)!
Similarly, if equation (11.5.1) had been changed to
d2 y
+ 2 y = 0
dx2
y0 = 0
y1 = 1
(11.5.4)
the solution would be
yx =
sin x
sin
(11.5.5)
Thus, either change from the homogeneous problem, whether it be in the differential
equation or the boundary conditions, results in a solution that is well defined except
when is one of the eigenvalues.
Of the two techniques of the section title, the Riccati method is the easiestto describe.
dy
dy
d2y
dR
dR
= Rxy. Then dx
+ R2 y.
Start with equation (11.5.1) and let dx
2 = dx y + R dx =
dx
11.5
275
Linear Stability Problems—Invariant Imbedding and Riccati Methods
The
derivative of y is known from equation (11.5.1). Thus,
dR second
2
+
R
y, or, upon dividing by y, the result is
dx
dR
= −2 − R2
dx
d2 y
dx2
= −2 y =
(11.5.6)
For later reference, note that in this simple case, equation (11.5.6) can be integrated to
give
Rx =
cos x
sin x
(11.5.7)
Equation (11.5.6) falls in the general form of what is termed a Riccati equation.
(The general form of a Riccati equation is R′ = A + BR + RC + RDR, where R can be a
vector and A, B, C, D square matrices.) It is nonlinear and ideally suited to a numerical
method of integration such as Runge-Kutta.
To start the integration, the value of R at x = 0 is needed. From the boundary
conditions, since y is to be set to zero and its derivative must be nonzero, it is seen
that R0 is singular. This can be gotten around in at least two ways. First, looking at
−R2 , so
equation (11.5.6), notice that when R and its derivative are large, then dR
dx
with a little imagination it can be concluded that R behaves like 1/x at the origin.
Alternately, we can work with the reciprocal (more generally, the inverse) of R,
= − R12 dR
= − R12 −2 − R2 = 1 + 2 S 2 . If R is infinite at
letting S = 1/R, so that dS
dx
dx
the origin, S will be zero there, so the integration can proceed.
To do so, however, a value for must be set first. Choose an arbitrary value that
seems in a reasonable range, and then carry out the integration until R becomes infinity
(defined here as being outside the bounds of numbers your computer is capable of
handling). Then this value where R becomes infinite is the length at which the chosen
value of is the correct eigenvalue. Thus, the problem has been turned around, finding
the value of the spacing associated with an eigenvalue rather than the usual reverse
of this.
One thing to note is that while S is finite throughout most of the region of integration,
it probably will be infinite at an intermediate point. (From equation (11.5.7), we can
see that this is true at x = /2.) Thus, programming requires that we start with the S
equation, then shift to the R equation, then (to improve accuracy) shift back to the S
equation as the singularity in R is approached. To accomplish this, fortunately, does not
require particularly skilled programming abilities.
The invariant imbedding technique follows very closely the Riccati approach. It
recognizes that the value of the eigenvalue ( in this case) depends on the length and
along the lines of equation (11.5.4) changes equation (11.5.1) to
2 yx z
+ 2 yx z = 0
x2
y0 z = 0
yz z = 1
(11.5.8)
R is chosen as Rz = yz z/z, differentiation of R with respect to z and using
equation (11.5.6) to eliminate the derivative, leads to an equation practically identical
to equation (11.5.6).
In the problem considered here, there is not much flexibility in the choice of R.
In higher-order differential systems, depending on the boundary conditions, it may be
possible to avoid having to deal with singularities and the need to invert R.
276
Computational Methods—Ordinary Differential Equations
These methods have been successfully used in computing the neutral stability
curves for Couette-Taylor stability (Curl and Graebel, 1972; Wilks and Sloan, 1976),
for Poiseuille flow between parallel plates (Curl and Graebel, 1972; Davey, 1977;
Sloan, 1977) and for the Blasius flat plate solution (Wilks and Bramley, 1977). They are
at least as accurate as previous methods (Orszag, 1971) and (perhaps even better) easier
to use.
To illustrate the method for more complex flows, consider the Orr-Sommerfeld
equation for plane Poiseuille
flow. Taking the origin at y = 0, the primary flow is
given by U = Umax 1 − y2 , where Umax is the centerline velocity and y has been made
dimensionless by the half-spacing of the plates. First, recall equation (9.4.8) in the
expanded form
2
2
d4 v
dv
d2 U
2d v
4
2
−
2k
+
k
v
=
ikR
U
−
c
−
k
v
−
v
dy4
dy2
dy2
dy2
(9.4.8)
To simplify the notation, the primes have been dropped and replaced by y.
Only disturbances of u even in y will be considered, as they are known to be less
stable. By the continuity relation, disturbances even in u are odd in v—thus,
d3 v
dv
= 3 = 0 at y = 0
dy dy
and u = v =
dv
= 0 at y = 1
dy
Following Davey (1977), let
Z1 = v Z3 = v′′ − k2 v
Z2 = v ′
(11.5.9)
Z4 = v′′′ − k2 v′
and
Z2
R 1 R2
Z1
=
Z4
R3 R4
Z3
(11.5.10)
From the boundary conditions at y = 0 Z2 and Z4 are both zero at y = 0. Since Z1 and
Z3 are not necessarily zero there, all of the Rs must be set to zero at y = 0. Since Z1
and Z2 must both be zero at y = 1 R2 = 0 at y = 1.
Next, differentiate the first of equation (11.5.10),
Z2 = R1 Z1 + R2 Z3
(11.5.11)
Z2′ = R′1 Z1 + R1 Z1′ + R′2 Z3 + R2 Z3′
(11.5.12)
obtaining
where here primes denote derivatives with respect to y. From equation (11.5.9) note
that
Z1′ = v′ = Z2
Z2′ = v′′ = Z3 − k2 Z1
Z3′ = Z4
(11.5.13)
11.5
277
Linear Stability Problems—Invariant Imbedding and Riccati Methods
so equation (11.5.12) can be rewritten as
Z2′ = Z3 + k2 Z1 = R′1 Z1 + R1 Z2 + R′2 Z3 + R2 Z4
(11.5.14)
Z2 and Z4 can be eliminated from equation (11.5.14) by using equation (11.5.10), which,
upon collection of terms, gives
(11.5.15)
Z1 R′1 + R21 + R2 R3 − k2 + Z3 R′2 + R1 R2 + R2 R4 − 1 = 0
From equation (9.4.8), the fourth derivative of v is given by
d2 U
v′′′′ = 2k2 v′′ − k4 v + ik Re U − c v′′ − k2 v − 2 v or equivalently
dy
(11.5.16)
2
dU
′′′′
2
4
v = 2k Z3 + k Z1 + ik Re U − c Z3 − 2 Z1
dy
Repeating the previous procedure with the second equation of (11.5.2) and using equation
(11.5.16) gives
Z1 R′3 + R1 R3 + R3 R4 − 2ik Re
+ Z3 R′4 + R2 R3 + R24 − k2 − ik Re 1 − y2 − c = 0
(11.5.17)
Equations (11.5.15) and (11.5.17) are two homogeneous equations in the two
unknowns Z1 and Z3 which must hold over a range of y. The conclusion is then that
the coefficients of Z1 and Z3 must vanish, leading to the system of four differential
equations
R′1 = −R21 − R2 R3 + k2
R′2 = −R1 R2 − R2 R4 + 1
R′3 = −R1 R3 − R3 R4 + 2ik Re
(11.5.18)
R′4 = −R2 R3 − R24 + k2 + ik Re 1 − y2 − c
These are the Riccati equations associated with the problem.
Notice that the second and third equations are the same except for the magnitude
of the constant. Thus,
R3 = 2ik Re R2
(11.5.19)
The procedure requires that values of the wave number, Reynolds number, and the wave
speed be set a priori. Integration then precedes until R2 = 0, or at least nearly zero. If
the initial guess is close to a true eigenvalue, then this will occur near to 1. Using the
dimensions of the various quantities, once the location of the eigenvalue is known, it is
possible to correct the starting parameters to make the zero occur at 1. Since integration
is in the complex plane, it is entirely possible that a zero of R2 will not be found for a
particular choice of the three input parameters. It is best to start near known results and
make small steps away from them. A discussion of dealing with more general problems
using the Riccati method is given in the Appendix.
Program 11.5.1 illustrates one realization of a program for this problem. Table 11.5.1
suggests several combinations of input data near the minimum Reynolds number.
278
Computational Methods—Ordinary Differential Equations
PROGRAM Orr
COMPLEX :: R(3)
DIMENSION AMR(3)
DY=.001
Y=0.
DO I=1,3
R(I)=0.
END DO
R2R=0.
R2I=0.
R2ROLD=.00001
WRITE(*,*)
WRITE(*,*) " Solves the Orr-Sommerfeld equation using a Riccati method and also"
WRITE(*,*) "
using a 4th order Runge-Kutta scheme."
WRITE(*,*)
WRITE(*,*) "
Enter the real part of the wave speed c.
"
READ (*,*) C
WRITE(*,*) "
Enter the wave number k.
"
READ (*,*) AK
WRITE(*,*) "
Enter the Reynolds number Re.
"
READ (*,*) Re
WRITE(*,*)
WRITE(*,*) "
y
R(1)
R(2)
R(3)
R(2)real
R(2)imag"
WRITE(*,100) Y,AMR(1),AMR(2),AMR(3),R2R,R2I
DO I=1,1050
Y=Y+DY
CALL RK4(R,AK,C,DY,Re,Y)
DO J=1,3
AMR(J)=ABS(R(J))
END DO
R2R=REAL(R(2))
R2I=AIMAG(R(2))
WRITE(*,100) Y,AMR(1),AMR(2),AMR(3),R2R,R2I
IF((R2R*R2ROLD)<0.) THEN
RN=RE*(Y**3)
CN=(C-1.+Y**2)/Y**3
YN=Y+R2R*DY/(R2ROLD-R2R)
R2RS=R2R
R2IS=R2I
END IF
R2ROLD=R2R
END DO
WRITE(*,*)
WRITE(*,*) "
Input c, k, Re: ",C,AK,Re
WRITE(*,*) " Computed y, c, k, Re: ",YN,CN,AK,RN
WRITE(*,*) " R(2) real, imaginary: ",R2RS,R2IS
WRITE(*,*)
100 FORMAT(6(F12.7,','))
END PROGRAM
!==================================================
!
Runge_Kutta solver - Press, Flannery, et al
!==================================================
SUBROUTINE RK4(R,AK,C,DY,RE,Y)
COMPLEX :: R(3),RT(3),DR(3),DRM(3),DRT(3)
DYH=0.5*DY
YH=Y+DYH
YPH=Y+DY
CALL DERIVS(R,DR,AK,C,RE,Y)
DO I=1,3
RT(I)=R(I)+DYH*DR(I)
END DO
! 11
CALL DERIVS(RT,DRT,AK,C,RE,YH)
DO I=1,3
RT(I)=R(I)+DYH*DRT(I)
END DO
! 12
Program 11.5.1—Neutral stability curves for the Orr-Sommerfeld equation (program by the author, subroutine
from Press, Flannery, et al. (1986))
11.6
279
Errors, Accuracy, and Stiff Systems
CALL DERIVS(RT,DRM,AK,C,RE,YH)
DO I=1,3
RT(I)=R(I)+DY*DRM(I)
DRM(I)=DRT(I)+DRM(I)
END DO
! 13
CALL DERIVS(RT,DRT,AK,C,RE,YPH)
DO I=1,3
R(I)=R(I)+DY*(DR(I)+DRT(I)+2.*DRM(I))/6.
END DO
! 14
RETURN
END SUBROUTINE
!===================================================
!
Defines equations
!===================================================
SUBROUTINE DERIVS(R,DR,AK,C,RE,Y)
COMPLEX :: AI,R(3),DR(3)
AI=CMPLX(0.,1.)
!
R(3)=2*I*K*Re*R(2)
DR(1)=-R(1)*R(1)-R(2)*(2.*AI*AK*RE*R(2))+AK**2
DR(2)=-R(1)*R(2)-R(2)*R(3)+1.
DR(3)=-R(2)*(2.*AI*AK*RE*R(2))-R(3)*R(3)+AK**2+AI*AK*RE*(1.-Y**2-C)
RETURN
END SUBROUTINE
Program 11.5.1—(Continued)
TABLE 11.5.1 Values of Re, k, and c near the minimum Reynolds number
Re
57739761
5773728
57732251
57729077
57726406
57724238
57722617
57721499
k
c
Re
k
10163
10168
10173
10178
10183
10188
10193
10198
02634788
02635388
02635988
02636588
02637188
02637788
02638388
02638988
57720947
57719966
57720869
57720879
57720908
57724077
57720591
10203
10205564
10205617
10205617
10208
10209
102056
c
02639588
02639906
02639905
02639905
02640188
02640288
02639905
11.6 Errors, Accuracy, and Stiff Systems
There are many issues that affect computations made with computers. Computers can
deal only with a finite number of digits when making computations, usually some
multiple of 8, depending on the computer and the program. This means that truncation
errors naturally occur during an operation, as well as round-off errors. These errors
can sometimes be minimized by avoiding the operation of subtraction and also by
minimizing the numbers of operation. If the error introduced by one round-off is ,
after N operations
you might be lucky and find that the accumulated error is of the
√
order of N , providing that the errors accumulated randomly. (On the other hand, if
the errors do not accumulate randomly, the error could be of the order N.) Subtraction
between√two nearly equal numbers can be disastrous. That is why use of the formula
√
2
2
y = −b± 2ab −4ac should be avoided if 4ac ≈ b2 . A better choice is y1 = − b+sgnb2a b −4ac ,
y2 = − b+sgnb2c√b2 −4ac , which avoids the subtraction.
The numerical method used in approximating a given operation introduces further
errors. In the case of finite difference formulations of integral and differential calculus
280
Computational Methods—Ordinary Differential Equations
operations, as we have seen, these errors are determined by the order of approximation
made in deriving the finite difference form. It is often suggested that after performing
a procedure at one step size, the procedure should be repeated using half the original
step size. This, of course, may not be practical in large calculations, and reducing the
step size too far may increase the errors discussed in the previous paragraph.
Sometimes calculations can become unstable. An example of this would be the
solution of an ordinary differential equation where a parameter in the equation means
that there are a family of solutions generated by solving the differential equation for
various values of the parameter. If at one point in the solution space the members of the
family lie close together, round-off errors can cause the procedure to leave the original
solution and drift to a neighboring solution.
A simple model of this is to compute integer
√
powers of the Golden Mean = 5−1
.
While
powers can be computed by successive
2
multiplications, it is easy to verify that n+1 = n−1 − n . Thus, knowing 0 = 1 and
1 = 061803398, successive powers can be computed by simple subtraction.
√
Unfortunately, while higher powers of rapidly decrease, the function − 5+1
2
satisfies the same recursion relation, and this function in absolute value is greater than
unity. Unavoidable computational errors indicate that a little round-off error means
that use of the recurrence relationship will soon give completely wrong errors, usually
around the power 16 (Press et al., 1986).
Another situation where instability occurs is in dealing with stiff systems. These
systems are those where there is more than one length scale to the problem, and at least
two of the scales differ greatly in magnitude. An example is the ordinary differential
d2 y
10x
equation dx
+Be−10x , with length
2 −100y = 0. This has the general solution yx = Ae
scales differing by a factor of one hundred. So, even if the boundary conditions are such
that A is zero—for example, if y0 = 1 y = 0—round-off errors would result in a
small nonzero value for A, and the solution would grow rapidly.
d2 y
2
The general form of this example equation is dx
2 − y = 0. Here, the values of
are the eigenvalues of the problem, and the equation is stiff because in the example the
two values of differ substantially. A similar example occurs in the solution of algebraic
equations. For a set of algebraic equations Ay = b, where A is an n by n matrix and y and
b are column vectors, the eigenvalues of the matrix A are the solution of the determinant
A − I = 0, I being the identity matrix. Again, if the values of the eigenvalue differ
too greatly, the usual solution methods can run into trouble. Since finite difference
methods frequently reduce the computation to a set of algebraic equations, it should not
be surprising that stiffness can occur in both algebra and calculus.
Should the Runge-Kutta scheme not work for a particular set of equations, another
method that has been successful in some cases2 is the Bulirsch-Stoer method, introduced in 1966. The method is based on three different concepts, and so programming
it is reasonably complicated. Programs can be found in Press et al. (1986) or on the
Web. The concepts are Richardson’s deferred approach to the limit (an extrapolation
procedure), choice of a proper fitting functions (the original choice was rational polynomials, but ordinary polynomials have been used), and the use of a method whose
error function is even in the step size. Further details can be found in the preceding
sources.
2
One may hope—but never expect—to find one method that works for all problems. See
www.nr.com/CiP97.pdf for some comments on this.
281
Problems—Chapter 11
Problems—Chapter 11
If available, use one of the programing languages mentioned in the book. If no such
languages are available to you, the problems can be done (if less conveniently) through
the use of spreadsheets such as Excel or Quattro Pro.
11.1 Use the trapezoidal rule to integrate the function cos x over the range 0 ≤
x ≤ 1. Perform the integration three times, using successively the intervals 0.2, 0.1,
and 0.5. Compare your computed answer with the exact result at each step of the
integration.
11.2 A technique known either as Richardson’s deferred approach to the limit or
as Richardson’s extrapolation enables improvement of the accuracy of the integration
carried out in the previous problem. For half-stepping, as was done there, if I1 is the
result for a step size of h and I2 , the result for a step size h/2, then the improved result
is given by Iimproved = 4I23−I1 . Use this method to improve your results of problem 11.1.
11.3 Compare the accuracy of the finite difference approximation to the derivative
of the sine of x for the following three formulas:
dy
dx
dy
dx
dy
Centered difference
dx
dy
Backward difference
dx
dy
dx
Forward difference
≃
≃
≃
≃
≃
yx + h − yx
h
−yx + 2h + 4yx + h − 3yx
2h
yx + h − yx − h
2h
yx − yx − h
h
3yx − 4yx − h + yx − 2h
2h
Do the computation at x = /4 with h = 01 .
dy
= y + x y0 = 0. Carry out the
11.4 Use Euler’s method with x = 01 to solve dx
integration to x = 10, and compare your result with the exact solution.
11.5 Repeat the above problem, this time using a fourth-order Runge-Kutta scheme
for integration.
11.6 The Runge-Kutta scheme of fifth-order accuracy is given by
5
27
125
1
k + k + k +
k h where
yx + h = yx +
24 1 48 4 56 5 336 6
k1 = fx y
1
1
k2 = f x + h y + k1
2
2
1
1
1
k3 = f x + h y + k1 + k2
2
4
4
k4 = fx + h y − k2 + 2k3
282
Computational Methods—Ordinary Differential Equations
2
7
10
1
k5 = f x + h y + k1 + k2 + k3
3
27
27
27
1
28
1
546
54
378
k6 = f x + h y +
k1 − k2 +
k3 +
k4 −
k5
5
625
5
625
625
625
Modify the procedure used in the previous problem and solve the same equation set.
2
d y
2
11.7 Use the finite element method to solve the equation dx
2 − y = fx y0 =
y2 = 0. Use the triangular elemental functions of equation (11.4.4) to find that yj+1 −
2
2 2
2yj + yj−1 − h 6 yj+1 + 4yj + yj−1 = h6 fj+1 + 4fj + fj−1 j = 1 2 · · · N − 1. The
parameter h is the constant spacing, with h = 2/N for this problem. Solve with =
1 fx = x N = 10.
11.8 Find exact solutions to the following equations, and consider how the small
parameter > 0 might affect a numerical solution.
a
b
c
dy
d2 y
+ 2 − y = 0
2
dx
dx
dy
d2 y
+ 2 + y = 0
2
dx
dx
dy
d2 y
2 + 2 + y = 0
dx
dx
y0 = 1
y = 0
y0 = 1
y = 0
y0 = 1
y = 0
4
d y
dy
4
11.9 Find the eigenvalues of the system dx
4 + y = 0 y0 = dx 0 = y1 =
dy
1 = 0 using the invariant imbedding method. This equation is associated with the
dx
vibration of a clamped-clamped beam.
Hint: Let Z1 = y Z2 = y′ Z3 = y′′ Z4 = y′′′ , and let Z1 = R1 Z3 + R2 Z4 Z2 =
R3 Z3 + R4 Z4 . The motivation for this choice is that y and y′ are known at the origin.
Chapter 12
Multidimensional Computational
Methods
12.1
12.2
12.3
12.4
Introduction 283
Relaxation Methods 284
Surface Singularities 288
One-Step Methods 297
12.4.1 Forward Time, Centered
Space—Explicit 297
12.4.2 Dufort-Frankel Method—
Explicit 298
12.4.3 Crank-Nicholson Method—
Implicit 298
12.4.4 Boundary Layer Equations—
Crank-Nicholson 299
12.4.5 Boundary Layer Equation—
Hybrid Method 303
12.4.6 Richardson Extrapolation 303
12.4.7 Further Choices for Dealing
with Nonlinearities 304
12.4.8 Upwind Differencing for
Convective Acceleration
Terms 304
12.5 Multistep, or Alternating Direction,
Methods 305
12.5.1 Alternating Direction Explicit
(ADE) Method 305
12.5.2 Alternating Direction Implicit
(ADI) Method 305
12.6 Method of Characteristics 306
12.7 Leapfrog Method—Explicit 309
12.8 Lax-Wendroff Method—Explicit 310
12.9 MacCormack’s Methods 311
12.9.1 MacCormack’s Explicit
Method 312
12.9.2 MacCormack’s Implicit
Method 312
12.10 Discrete Vortex Methods (DVM) 313
12.11 Cloud in Cell Method (CIC) 314
Problems—Chapter 12 315
12.1 Introduction
When dealing with differential equations involving more than one dimension, it is
important to know the classification of the equation, as the method of numerical computation must be tied to the behavior of the equation’s solution. The three archetypical
equations are the following:
Elliptic:
2V = 0
Parabolic:
2V =
V
t
(12.1.1)
283
284
Multidimensional Computational Methods
Hyperbolic: 2 V =
1 2 V
c2 t2
The wave equation is the most familiar form of hyperbolic equation. A propagation
velocity c is associated with the time derivative. Thus, a disturbance of an existing
condition at any point takes a finite time before its effect is noted at distant points.
Whereas solutions of the other two classes tend to be smooth, hyperbolic equations can
have abrupt discontinuities (shocks) in the solutions.
Parabolic equations are associated with diffusion processes such as heat, mass, and
concentration diffusion. There is no wave speed associated with this, so mathematically
a point an infinite distance from a place of change of conditions knows of the change
instantly. The Prandtl boundary layer equations, wherein the x second derivative term
is neglected, is frequently referred to as the parabolized Navier-Stokes equations, since
the highest order of the stream-wise derivative is one.
The elliptic type of equation, of which Laplace’s equation is the prime example,
has each location communicating at all times with all other locations in the domain,
as indicated by the mean value theorem that states that the value of the function at
the center of a circle or sphere is the average of the values on the surface. Thus, a
change at one point in the domain instantly affects every other point. Formally, the
steady-state Navier-Stokes equations belong in this class because of the order of the
viscous terms. At large Reynolds numbers, however, the magnitude of the convective
acceleration essentially overcomes this, and the behavior becomes either more like the
parabolic or hyperbolic classes.
Even for inviscid flows with free surfaces, the fact that surface and interfacial waves
can propagate at finite speeds indicates behavior more of a hyperbolic than elliptic
nature. This is brought about by the boundary conditions. So while classifications are
useful, one should keep in mind that there are other influences on the nature of the
solution.
Elliptic Partial Differential Equations
12.2 Relaxation Methods
Since the Laplace equation is perhaps the oldest and most used of the equations of
engineering physics, there are quite naturally the greatest number of methods for numerical calculation—many existing long before the advent of the computer. In the case
of rectangular boundaries, a rectangular grid can be superimposed on the boundary
containing rectangles of size x by y. Using either the numerical differentiation
procedures from Chapter 11 or the mean value theorem, Laplace’s equation can be
approximated by
0 = 2V ≈
Vj+1k − 2Vjk + Vj−1k Vjk+1 − 2Vjk + Vjk−1
+
x2
y2
(12.2.1)
to second order accuracy in the grid spacing, or
Vj+2k − 16Vj+1k + 30Vjk − 16Vj−1k + Vj−2k
x2
Vjk+2 − 16Vjk+1 + 30Vjk − 16Vjk−1 + Vjk−2
+
y2
0 = 2 V ≈
(12.2.2)
12.2
285
Relaxation Methods
k+1
k
1
1
4
1
k–1
1
j–1
j+1
j
Figure 12.2.1a Computational molecule for the relaxation method
k+2
1
k+1
–16
k
1
–16
60
–16
1
j+1
j+2
–16
k–1
k–2
1
j–2
j–1
j
Figure 12.2.1b Computational molecule for the SOR method
to fourth-order accuracy. The computational molecules for the two cases are shown in
Figures 12.2.1a and b. When x = y, these equations reduce to
Vj+1k + Vj−1k + Vjk+1 + Vjk−1
(12.2.1a)
4
−Vj+2k + 16Vj+1k + 16Vj−1k − Vj−2k + −Vjk+2 + 16Vjk+1 + 16Vjk−1 − Vjk−2
Vjk ≈
60
(12.2.2a)
Vjk ≈
If the values of the function V are known on the boundary (Dirichlet problem),
such as when posing a flow situation in terms of a stream function, the boundary conditions are easily handled. If the normal derivatives of the function V are known on
the boundary (Neumann problem), such as when solving for the velocity potential,
the boundary conditions require additional equations to accommodate the derivatives.
In analyzing a given flow using the velocity potential, the body shape and the normal derivatives are known on the surface, so this is a Neumann problem and the
286
Multidimensional Computational Methods
pressure coefficient can be found. In the case of a design problem, the body shape is
unknown except for the fact that it is a stream surface, and dealing with it as a Dirichlet problem has advantages. Often the pressure coefficient is also known in a design
situation.
The resulting algebraic equations can be handled in a number of ways.
1. Use traditional methods such as Gaussian elimination. Unfortunately this will
require N + 1 ! multiplications (N being the number of nodes), and possible round-off
errors can be introduced in the solution process.
2. Use the relaxation method. For the case with Dirichlet conditions and a square
grid, for second-order accuracy use equation (12.2.1a) in the form
i+1
Vjk
≈
i
i
i
i
+ Vjk−1
+ Vjk+1
+ Vj−1k
Vj+1k
4
(12.2.1b)
The superscript denotes the number of the sweep through all of the nodes. Start off
by assigning arbitrary values to all nodes. This is sweep number 0. Use your best
judgement in assigning these first values, but it is not necessary to be perfectly accurate. Next, go from point to point, changing the value of the point you are at to
the average of its neighbors (sweep number 2). After you have swept through every
point, repeat the process again and again until the change is negligible. In practice, rather than using old values of V in computing the right-hand side of equation
(12.2.1b), the most recently computed values are used, thus speeding up the process
slightly.
While such a boring procedure is perfectly designed for a computer, it is sobering
to reflect that in days gone by this was done with an adding machine (or maybe not),
pencil, paper, eraser, and a human!
3. Use the successive over-relaxation method (SOR). In this case equation (12.2.1a)
is replaced by
i+1
i
Vjk
+
≈ Vjk
4
i
i
i
i
i
+ Vjk+1
+ Vjk−1
− 4Vjk
Vj+1k + Vj−1k
(12.2.1c)
Again, the i superscript denotes the number of the integration, and
is a parameter
between 1 and 2 used to speed up the calculations. The “best” value of
to use for a
particular problem is determined by making a few trial runs for various values of ,
which can be time-consuming. Choosing a value somewhere around 1.7 or so does a
good job.
4. Use the successive line over-relaxation method (SLOR). This is the same
as the SOR method, but rather than going around from point to point, a row
of points (line) is solved using previous values and a method such as Gauss
elimination.
5. If information is sought only in a particular region, random walk techniques can
be useful. If you want to find the value at a point j k , for example, start at that point
and randomly choose one of the numbers 1 to 4 (1 to 6 for three-dimensional problems).
Do this until a boundary point is reached, where the value of V is Bi , the i standing for
the ith iteration. Repeat this process N times. Then Vjk ≈ N1 Ni=1 Ni Bi , where Ni is the
T
12.2
Relaxation Methods
number of steps needed to get to the boundary point with value Bi and NT = Ni=1 Ni
is the total number of steps. The accuracy increases as NT−4 , but clearly many, many
steps must be taken.
6. Write the N algebraic equations in N unknowns, and use a traditional algebraic
solver. The algebraic equations are sparse, which helps, but the fact that the matrix
of the coefficients is not narrow-banded means that special methods tailored to such a
problem must be used. These and other procedures can be found in much more detail
in Smith (1978), for example.
In the preceding, attention has been paid only to the case where boundaries are
rectangular, a fairly restricted case. For irregular boundaries, one could rephrase equation
(12.2.1) in a mesh of unequal sides, but then a good part of the computational problem
is to determine which boundary point you are near and which variation of equation
(12.2.1) is needed. Also, near corners, where changes in the solution can be rapid,
accuracy can be lost unless the grid mesh is shrunk. Two (at least) methods have been
introduced to overcome this problem.
The ideas of conformal mapping introduced in Chapter 3 are ideally suited to
generate grids to fit boundaries of any shape. Thomson, Warsi, and Mastin (1985)
present techniques useful in both two and three dimensions for computer generation
of grids for all three classes of partial differential equations. Basically, they use the
conformality of analytic functions to map the flow space into a rectangle. Control
functions can be used to adapt the grid spacing so that spacing is small and cell count
denser where the gradient of the function can be expected to be large. After the space is
transformed to the rectangle, the equations of interest are also transformed to the new
coordinate system and then solved.
The finite element method (FEM, also sometimes FEA for finite element analysis),
introduced for one-dimensional problems in Chapter 11, can also be adapted to two- and
three-dimensional problems. Programs usually come as a package, including grid generation and solvers. Grid generation is to some degree usually automatic, with provisions
for intervention by the user where refinements in the grid are needed. Elements used
can be rectangular, triangular, semi-infinite, and a variety of others. The polynomials
used on the sides of the elements vary in complexity, depending on the accuracy and
order of the derivatives needed.
FEM was originally developed for solution of problems in the linear theory of elasticity, where the equations are strongly elliptic. In elasticity theory energy is conserved,
and only the “laminar” state exists. Thus, one would expect that, unless special provisions are made for fluid flow problems, there would be a Reynolds number limitation
on computational accuracy. Upwind differencing, described later, has made it possible
to extend this limitation somewhat, and great claims have been made for the commercial programs. Many even claim to handle turbulent flows. However, since many of
the companies are secretive as to how Reynolds number limitations and turbulence are
treated, it is difficult to assess their claims.
FEM programs can be used for irrotational flows with cavities. A simple approach
is to first estimate the shape of the cavity, then correct the shape to make it tangent
to the computed velocity. In the process, all nodes on the cavity are moved by the
process. It is to be repeated until some error norm such as all cavity nodes ynew − yold 2
is less than some value. Other methods for cavity flows have been suggested (e.g.,
Brennen, 1969).
287
288
Multidimensional Computational Methods
12.3 Surface Singularities
To illustrate the use of surface singularities in two-dimensional flows, the basic starting
point is Cauchy’s integral formula, which states that for any analytic function (i.e., one
that satisfies the Cauchy-Riemann conditions)
1 f
d
(12.3.1)
fz =
2 i
−z
where the integration is about a closed path traversed in the positive sense. That is, as
the path of integration is traversed, the direction taken is such that the interior is always
to the left. If z is within the closed path, the integral is zero.
is an analytic function, we can write
Since the complex velocity dw
dz
u − iv =
1 u − iv
d
2 i
−z
Letting d = ei ds, where ds is real and
equation (12.3.2) becomes
(12.3.2)
represents the slope of the integration path,
u − iv d = u − iv ei ds = qtangent − iqnormal ds
where qtangent and qnormal are the tangent and normal velocity components with respect
to the path. Thus, equation (12.3.2) becomes
q
q
1 qtangent − iqnormal
tangent ds
normal ds
u − iv =
ds =
+i
(12.3.3)
2 i
−z
2
z−
2
z−
Recalling that
u − iv
source
=
m 1
2 z−
and
u − iv
vortex
=
i 1
2 z−
it is seen that the first integral represents a source distribution, while the second integral
can be interpreted as a vortex distribution. Notice as shown in Chapter 2, that a vortex
distribution and a doublet distribution are equivalent.
In Chapter 2 the panel method was discussed for finding the flow about a submerged
body. We next show how for two-dimensional inviscid flows this can be implemented
numerically for lifting bodies.
First, consider a closed two-dimensional body made up of a series of N flat panels.
An example is shown in Figure 12.3.1, where 12 panels are inscribed within the body.
Each panel has a source and a vortex on it. The strength of each source and vortex is
constant on a panel but can differ from panel to panel. Two such panels are shown in
Figure 12.3.2 to illustrate the geometry. The velocity potential then contains the uniform
stream plus the contributions from each of the N panels—that is,
N
N
y − yj′
mj
j
ln x − xj′ 2 + y − yj′ 2 dsj +
tan−1
ds
x − xj′ j
sj
sj
j=1 2
j=1 2
(12.3.4)
The control points where the boundary will be taken are at the center of each panel and
designated by xi yi . (Note: Panels can be constructed so either their endpoints lie on
the surface of the body or the control points lie on the body. Some evidence suggests
that the latter is more accurate.)
x y = Ux +
12.3
289
Surface Singularities
P3
P2
P4
P1
P5
P12
P6
P11
P7
P10
P8
P9
Figure 12.3.1 Panel method—numbering of panels
ni
i th panel
βi
θi
(X i + 1, Y i + 1)
(X i, Y i)
(x i, y i)
r ij
nj
βj
(X j + 1, Y j + 1)
(x∼j, y∼j)
Sj
(x j, y j)
θj
(X j, Y j)
j th panel
Figure 12.3.2 Panel method—definitions
Applying the boundary conditions, find that
n
Let
xi yi
= U cos i +
N
mj x − xj′ cos i + y − yj′ sin i
mi
dsj
+
2
x − xj′ 2 + y − yj′ 2
sj
j=1 2
j=i
N
j x − xj′ sin i − y − yj′ cos i
dsj
+
x − xj′ 2 + y − yj′ 2
sj
j=1 2
Iij =
x − x′ cos + y − y′ sin
i
i
j
j
sj
x − xj′ 2 + y − yj′
2
dsj
(12.3.5)
(12.3.6)
290
Multidimensional Computational Methods
for the sources and
Iij′ =
x − x′ sin − y − y′ cos
i
i
j
j
x − xj′ 2 + y − yj′
sj
2
dsj
(12.3.7)
for the vortices. Then
N
mj
j ′
mi i
+ +
Iij +
I = −U cos i
2
2 j=1 2
2 ij
(12.3.8)
j=i
The first thing to notice is that by applying the boundary conditions, there are N
equations in 2N unknowns. The Kutta condition also has not yet been applied, which
is necessary for a lifting body. Notice also that in using either source or vortex panels,
the tangency condition is satisfied only at the control points, and the velocity is infinite
at every panel edge.
There are a number of approaches that can be used to model a lifting surface and
balance the number of equations and unknowns in the process.
1. Use a unique source strength and the same vortex strength on every panel,
giving N + 1 unknowns. Impose the tangency condition at N control points, and
also impose one Kutta condition at the trailing edge. This a fully determinate
system.
2. Use a unique source and a parabolic vorticity distribution on the top and bottom
panels. Let top and bottom maximum vortex strengths have the same magnitude
so that there are N + 1 unknowns. Tangency conditions at N control points and
one Kutta condition make this a fully determinate system.
3. Use a unique source and vortex strength on each panel, giving 2N unknowns. Use
two control points on each panel and one Kutta condition. This an indeterminate
system, requiring a least squares procedure or something similar to resolve the
inconsistency.
4. Use a unique source and vortex on each panel, giving 2N unknowns. Use two
control points on each panel and one Kutta condition. This an indeterminate
system, requiring a least squares procedure or something similar.
5. Use a unique source and the same vorticity on each panel, giving N + 1
unknowns. Use two control points on each panel and one Kutta condition.
This an indeterminate system, requiring a least squares procedure or a similar
technique.
6. Use a unique vortex on each panel, giving N unknowns. Satisfy tangency at
one control point on each panel and one Kutta condition. This an indeterminate
system, requiring a least squares procedure or something similar.
7. Use a unique vortex on each panel, giving N unknowns. Use two control points
on each panel and one Kutta condition. This an indeterminate system, requiring
a least squares procedure or something similar.
8. Use curved panels, perhaps parabolic or cubic in shape, along with singularity
distributions that vary on each panel. This would be particularly advantageous
12.3
291
Surface Singularities
near the rounded nose of an airfoil, where otherwise the number of panels
must be increased to fit the geometry. The number of variations on this is
unlimited.
Indeterminate systems can be handled as follows:
If the given system is Nj=1 Aij xj = bj i = 1 2 N N + 1 N + M, let
2
E =
Then, to minimize E 2 , let
N
+M
N
j=1
i=1
2
Aij xj − bi
(12.3.9)
N
N
+M
E 2
Aij xj − bi = 0
=2
Aik
xk
j=1
i=1
Then
N
j=1
N
+M
i=1
Aij Aik xj =
N
+M
i=1
(12.3.10)
bi Aik k = 1 2 N
(12.3.11)
This system is determinate.
There are several approaches possible for satisfying the Kutta condition. For
example, if a body shape with a sharp trailing edge is to be modeled, the Kutta
condition could be imposed at the trailing edge. Consider the two panels surrounding the trailing edge, as shown in Figure 12.3.3. Locally the complex potential will
look like
(12.3.12)
w = Azn−1
where n = 2 2 − and is the wedge angle. Denoting the tangent velocities on the top
panel (panel #1, length S1 ) and the bottom panel (panel #N, length SN ), the Kutta
condition requires that the velocities be the same on these two panels. The velocities at
the control points are
n−1
n−1
1
1
vt1 = nA
S1
SN
vtN = −nA
2
2
(12.3.13)
1−n
1−n
1
1
S
S
+ vtN
= 0
therefore vt1
2 1
2 N
y
π – κ /2
κ
Figure 12.3.3 Kutta condition—first method
x
292
Multidimensional Computational Methods
y
CP1
α
CP3
x
CP2
β
Figure 12.3.4 Kutta condition—second method
An alternate way of meeting the Kutta condition is shown in Figure 12.3.4. Here
an extra panel of length S3 has been added at the trailing edge. Take the lengths of the
three panels be given as s1 s2 s3 . The result is then
⎧
⎪
⎪
⎨Az
cos
= Ar
+ i sin
0 ≤ ≤
wz =
⎪
/−1
/−1
⎪
+ i sin
− ≤ ≤ 0
cos
= Br
⎩Bz
/
/
(12.3.14)
giving
⎧
⎪
⎨ Az
dw
=
⎪ Bz
dz ⎩
/−1
/−2
0 ≤ ≤
(12.3.15)
− ≤ ≤ 0
Then at control point 3 there is a discontinuity in of magnitude
= A
1
2s3
/
−B
1
2s3
/
(12.3.16)
The velocities at control points 1 and 2 are
/ −1
1
s3
= V1
= A
2
cp1
/−1
1
dw
s
= V2
= A
dz cp2
2 3
dw
dz
and
Thus A =
V1
1
s
2 1
1−
/
B=
s
= V1 1
2
V2
s3
s1
1
s
2 2
/
1−
(12.3.17)
/
, so
s
+ V2 2
2
s3
s2
/
(12.3.18)
12.3
293
Surface Singularities
It is next necessary to carry out the integrations and put our equations in final form.
With the help of Figure 12.3.2, letting x̃j = Xj + sj cos j ỹj = Yj + sj sin j , then
Aj = Xj − xj cos
2
j
+ Yj − yj sin j
Bj = Xj − xj + Yj − yj 2
Cj = −Xj − xj sin
j
(12.3.19)
+ Yj − yj cos j
and
Iij =
1
sin i −
2
Sj Sj + 2Aj
ln 1 +
− cos i −
Bj
j
j
tan−1
Aj
Sj + Aj
− tan−1
Cj
Cj
(12.3.20)
Similarly
Iij′
1
= − cos i −
2
j
Sj Sj + 2Aj
ln 1 +
− sin i −
Bj
j
tan
−1
Aj
Sj + Aj
−1
− tan
Cj
Cj
(12.3.21)
With these at hand, a choice can be made as to the number of sources and vortices to
retain.
Note that from the point of view of the calculations, just having a determinate system
may not be sufficient for obtaining good results. Some of the combinations of sources
and vortices result in stiff systems. This means that when a system of equations such
as (in matrix form) Ax = B is being solved, such a system has a set of eigenfunctions
and eigenvalues given by Ayj = j yj . If the eigenvalues are all of the same order of
magnitude, the problem is not stiff, and the equation can be solved by standard methods.
If, however, there is a large disparity in the magnitude of the eigenvalues, then the
problem is stiff and is said to be ill-conditioned—that is, it is very sensitive to small
changes in the magnitudes of terms in either the matrix A or the known vector B.
Generally, it is not practical to first compute all of the eigenvalues to see whether
the system is stiff. It is better to solve the problem and then use back substitution or
some other check to see whether the solution is accurate. If it turns out that the method
you selected results in a stiff system, there are specialized methods that are designed
for dealing with stiff systems. It may, however, be much easier to select a different
combination of sources and vortices.
Notice that having a wide range in magnitude of the eigenvalues is reminiscent of
the boundary layer, where there are large gradients within the boundary layer and much
smaller ones outside it. It should come as no surprise that stiffness questions arise there
as well.
Here are some tips for maintaining accuracy. Rubbert and Saaris (1972) recommend
using double sheets of vortices inside the body. The following should be observed:
1. It is not necessary to place singularities on the boundary; they may be placed
within the boundary. If a single vortex is inserted in a thin body, the sources
above and below the vortex, the sources tend to be of opposite signs and act
like a doublet, with the strength inversely proportional to the distance between
them. This gives a strong gradient in source strength, which is to be avoided.
2. A series of vortices that are inserted with a strength approximating the load
distribution generally gives good results. However, if the angle of attack is to be
294
Multidimensional Computational Methods
varied, and the airfoil is highly cambered, two such rows may be needed, one to
take care of each problem individually.
A program using the preceding procedure with source panels only to develop the
body shape is given as Program 12.3.1. Once the panel strengths are known, there is
enough information to find the velocity and pressure at any point in the exterior flow.
PROGRAM PANELBOOK
REAL*8 LAMBDA
DIMENSION ACS(40),ASN(40),BETA(40),RHS(40),THETA(40),XCP(40),
&
XLEN(40),XP(40),YCP(40),YP(40),VEL(40)
DIMENSION AA(40,40),AA1(40,40),LAMBDA(40),SUM1(40,40),SUM2(40,40)
PI=3.14159265
WRITE(*,*)"This is a panel program designed to calculate the inviscid flow about"
WRITE(*,*)" an arbitrary body. The body shape is determined by the user. "
WRITE(*,*)"The program first asks for the number of panels to be used."
WRITE(*,*)"Next it asks for the X,Y coordinates of the end points of the panels."
WRITE(*,*)" These are called the nodes of the panel. For the program to run"
WRITE(*,*)" properly the nodal points should be numbered in a clockwise manner"
WRITE(*,*)" as one goes around the body."
WRITE(*,*)"'Finally the free stream velocity is requested."
WRITE(*,*)
WRITE(*,*)"The output will consist ofthe velocities calculated at the control points."
WRITE(*,*)"The control points were chosen so as to be located at the center of the panels."
WRITE(*,*)
WRITE(*,*)"Enter the number of panels you wish to use. Maximum allowable is 40. "
READ(*,*) N
WRITE(*,*)"The coordinates of the end points of the panels must now be entered."
WRITE(*,*)"his program is two dimensional, so only x and y coordinates should be entered."
WRITE(*,*)
WRITE(*,*)"You will be told which node you are entering. Panel number"
WRITE(*,*)"one has nodes one and two, panel two has nodes two and three,"
WRITE(*,*)"and the last panel has nodes n and one."
WRITE(*,*)
WRITE(*,*)"Results are in file A:PANELBOOK.DAT"
WRITE(*,*)
!============================================================
!
ENTER NODES
!============================================================
DO I=1,N
WRITE(*,*) "Enter node number ",I
READ(*,*) XP(I),YP(I)
END DO
OPEN(1, FILE='A:PANELBOOK.DAT', STATUS='UNKNOWN')
WRITE(*,*) "Your entered points are listed below:"
DO I=1,N
WRITE(*,100)I,XP(I),I,YP(I)
WRITE(1,100)I,XP(I),I,YP(I)
END DO
WRITE(*,*) " Enter the free steam velocity. "
READ(*,*) U
!============================================================
!
CALCULATE CONTROL POINTS
!============================================================
L=N-1
WRITE(1,*) "LOCATION OF CONTROL POINTS:"
DO I=1,L
XCP(I)=(XP(I)+XP(I+1))/2.
YCP(I)=(YP(I)+YP(I+1))/2.
WRITE(1,100)I,XCP(I),I,YCP(I)
END DO
XCP(N)=(XP(N)+XP(1))/2.
YCP(N)=(YP(N)+YP(1))/2.
WRITE(1,110)N,XCP(N),I,YCP(N)
Program 12.3.1—Panel method for flow past a nonlifting body (program by the author)
12.3
Surface Singularities
!============================================================
!
CALCULATE ANGLES
!============================================================
DO I=1,L
IF(XP(I+1).EQ. XP(I).AND. YP(I+1).LT.YP(I)) THEN
THETA(I)=-PI/2.
ELSE IF(XP(I+1).EQ.XP(I).AND.YP(I+1).GT.YP(I)) THEN
THETA(I)=PI/2.
ELSE
THETA(I)=ATAN((YP(I+1)-YP(I))/(XP(I+1)-XP(I)))
END IF
IF (XP(I+1).LT.XP(I)) THETA(I)=THETA(I)-PI
END DO
IF(XP(1).EQ.XP(N).AND.YP(1).LT.YP(N)) THEN
THETA(N)=-PI/2.
ELSE IF(XP(1).EQ.XP(N).AND.YP(1).GT.YP(N)) THEN
THETA(N)=PI/2.
ELSE
THETA(N)=ATAN((YP(1)-YP(N))/ (XP(1)-XP(N)))
END IF
IF( XP(1) .LT. XP(N)) THETA(N)=THETA(N)-PI
!============================================================
!
CALCULATE THE LENGTH AND ANGLES FOR ALL PANELS
!============================================================
DO I=1,L
XLEN(I)=SQRT( (XP(I+1)-XP(I))**2 + ( YP(I+1)-YP(I))**2 )
ASN(I)= SIN( THETA(I) )
ACS(I)=COS( THETA(I))
END DO
XLEN(N)=SQRT( (XP(1)-XP(N))**2 + (YP(1)-YP(N))**2 )
ASN(N)=SIN( THETA(N) )
ACS(N)=COS( THETA(N) )
!=======================================================================
!
CALCULATE I(I,J)=INTEGRAL ( D/DN LN(R) DR(J) ) OF THE Ith PANEL
!=======================================================================
DO I=1,N
DO J=1,N
IF(J.EQ.I) THEN
SUM1(I,J)=PI
SUM2(I,J)=0.
ELSE
A=-(XCP(I)-XP(J))*ACS(J)-( YCP(I)-YP(J))*ASN(J)
B=( XCP(I)-XP(J))**2 + ( YCP(I)-YP(J) )**2
C=SIN( THETA(I)-THETA(J) )
D=COS( THETA(I)-THETA(J) )
E=( XCP(I)-XP(J))*ASN(J) - ( YCP(I)-YP(J) )*ACS(J)
F=LOG( 1.0 +XLEN(J)*( XLEN(J)+2.*A)/ B )
G=ATAN( (XLEN(J)+A)/E )-ATAN(A/E)
SUM1(I,J)=C*F/2.-D*G
SUM2(I,J)=-D*F/2.-C*G
END IF
END DO
END DO
!============================================================
!
ASSIGN THE MEMBERS OF THE MATRIX EQUATION TO SOLVE FOR
!
THE INDIVIDUAL PANEL STRENGTH'S.
!============================================================
DO I=1,N
DO J=1,N
AA(I,J)=SUM1(I,J)
END DO
END DO
DO I=1,N
DO J=1,N
AA1(I,J)=AA(I,J)/AA(1,1)
END DO
END DO
DO I=1,N
BETA(I)=THETA(I)+PI/2.
RHS(I)=-U*COS( BETA(I) )/AA(1,1)
END DO
Program 12.3.1—(Continued)
295
296
Multidimensional Computational Methods
!============================================================
!
SOLVE THE MATRIX SYSTEM FOR THE PANEL STRENGTHS
!
USING A GAUSS-SIEDEL ROUTINE
!============================================================
DO I=1,N
LAMBDA(I)=0.
END DO
IMAX=20
DO K=1,IMAX
ICOUNT=1
DO I=1,N
XSTAR=LAMBDA(I)
LAMBDA(I)=RHS(I)
DO J=1,N
IF (J.GT.I) LAMBDA(I)=LAMBDA(I)-AA1(I,J)*LAMBDA(J)
END DO
!============================================================
!
TEST FOR CONVERGENCE
!============================================================
IF( DABS(XSTAR-LAMBDA(I)).GT.0.00001) THEN ICOUNT=0
END DO
IF (ICOUNT.EQ.1) GOTO 10
END DO
!============================================================
!
CALCULATE PANEL STRENGTHS
!============================================================
10 DO I=1,N
VI=U*ACS(I)
DO J=1,N
VI=VI+LAMBDA(J)*SUM2(I,J)
END DO
VEL(I)=VI
END DO
WRITE(1,120)
DO I=1,N
WRITE(1,130) I,LAMBDA(I)
END DO
WRITE(1,140)
DO I=1,N
WRITE(6,150) I,VEL(I)
END DO
100 FORMAT(1X,'XP(',I2,')=',F10.5,1X,'YP(',I2,')=',F10.5,/)
110 FORMAT(1X,'XCP(',I2,')=',1X,F10.5,2X,'YCP(',I2,')=',F10.5)
120 FORMAT("CALCULATED STRENGTHS OF THE PANELS:")
130 FORMAT(1X,'PANEL',1X,I2,'=',1X,F10.5)
140 FORMAT(//1X,'THE SOLUTION IS GIVEN BELOW'//)
150 FORMAT(1X,'THE VELOCITY AT CONTROL POINT',1X,I3,1X,'IS',F10.5)
CLOSE(1)
END PROGRAM
Program 12.3.1—(Continued)
The use of surface singularity distributions can be continued into three dimensions
with, however, the loss of the use of complex functions. Green’s second identity in
three dimensions is
f
h
h −f
h 2 f − f 2 h dV
(12.3.22)
dS =
n
n
c
V
Let h = , where 2 = 0, and f = 1/R, where R = x − x0 i + y − y0 j + z − z0 k.
Inserting this into equation (12.3.22) gives
1
1
1
+
−
dS
(12.3.23)
=
4
R n
n R
S
12.4
297
One-Step Methods
The 1/R term represents a surface distribution of sources of strength /n, and the
second term represents a surface doublet distribution of strength . Thus, in principal,
the two- and three-dimensional methodologies are the same.
Singularity distribution techniques have been widely used by naval architects and
aircraft designers. The approach was pioneered by Hess and Smith (1962, 1967) and
widely adapted and used by others. In those industries the primary emphasis is the
external flow past bodies. The methodology chosen can either suppress the “flow”
within the body or allow for that flow. For these applications the internal flow is ignored
as being without physical interest. It is also possible to use these methods for wave
problems where there are free surfaces with or without solid boundaries (Schwartz
(1981), Forbes (1989), Zhou and Graebel (1990)).
A combination of the surface integral method and the FEM is the boundary integral
method (BIM). It uses the idea of surface singularities to solve for the interior flow and
uses paneling as in FEM to solve for the magnitude of the singularities.
Parabolic Partial Differential Equations
12.4 One-Step Methods
One-step methods take the results from a previous computation in the space dimension(s)
and advance it in the time dimension.
12.4.1
Forward Time, Centered Space—Explicit
In one space dimension, the diffusion equation becomes
2 T
T
= 2
t
x
(12.4.1)
Approximating the time derivative by
T n+1 − Tin
T
≈ i
t
t
(12.4.2)
n
n
Ti+1
− 2Tin + Ti−1
2 T
≈
x2
x2
(12.4.3)
and the space derivative by
equation (12.4.1) can be approximated by
n
T n − 2Tin + Ti−1
Tin+1 − Tin
≈ i+1
t
x2
(12.4.4)
Here, the superscript refers to the time dimension, and the subscript refers to the space
dimension (Figure 12.4.1). Solving for T at the most recent time gives
Tin+1 ≈ Tin +
t n
n
T − 2Tin + Ti−1
x2 i+1
(12.4.5)
This is an explicit (i.e., it’s not necessary to solve a system of equations, just solve one
equation at a time) equation for T in terms of its value at three points at the previous
time. It is a stable method as long as t/x2 < 1/2. This stability criterion is called the
298
Multidimensional Computational Methods
i+1
1
i
–2
i–1
1
n
1
n+1
Figure 12.4.1 Computational molecule for the forward time centered space explicit method
Courant condition. It guarantees that the solution does not “blow up,” but it does not
tell us anything about the accuracy of the method. Since equations (12.4.2) and (12.4.3)
come about from Taylor series expansions, it can be shown that accuracy is of first
order in t and second order in x. If more terms in the Taylor expansions are taken,
providing we choose t/x2 < 1/6, the accuracy improves to second order in t and
fourth order in x.
12.4.2
Dufort-Frankel Method—Explicit
Changing the derivatives somewhat in the previous method leads to another explicit
method. Instead of equation (12.4.2), use
T
T n+1 − Tin−1
≈ i
t
2t
and instead of equation (12.4.3), use
n
n
− Tin+1 + Tin−1 + Ti−1
Ti+1
2 T
≈
x2
x2
Then, with = t/x2 , equation (12.4.5) is replaced by
n
1
1
n
n+1
+ ≈ Ti+1 + Ti−1 +
− Tin−1
Ti
2
2
(12.4.6)
(12.4.7)
(12.4.8)
The computational molecule is shown in Figure 12.4.2. This method is explicit and
unconditionally stable (i.e., stable for all values of ). It does, however, require another
method to start up and generate the first line, and must be small, or the method is not
even first-order accurate.
12.4.3
Crank-Nicholson Method—Implicit
If we stay with equation (12.4.2) for the time derivative but replace equation (12.4.3) by
n+1
n+1
n
n
Ti+1
− 2Tin + Ti−1
Ti+1
− 2Tin+1 + Ti−1
2 T
≈
+
x2
2x2
2x2
instead of equation (12.4.5), we have
n+1
+2
−Ti+1
1 + n+1
1− n
n+1
n
n
Ti − Ti−1
T + Ti−1
≈ Ti+1
+2
i
(12.4.9)
(12.4.10)
12.4
299
One-Step Methods
n+1
n
n–1
i–1
i
i+1
Figure 12.4.2 Computational molecule for the Dufort-Frankel method
where = t/x2 . This is an implicit method, meaning a system of equations must be
solved simultaneously, but since the system is tridiagonal, the solution method is easy.
Depending on boundary conditions, it is usually unconditionally stable (i.e., stable for
all values of ), but it can be unstable if the boundary conditions include derivatives of
the independent variable. Also, if there are two or more space dimensions, it becomes
block tridiagonal, increasing the difficulty of solution.
12.4.4
Boundary Layer Equations—Crank-Nicholson
We saw earlier that similarity solutions of the boundary layer equations could be
obtained, reducing a partial differential equation to an ordinary equation. This, however,
depended on having a minimum number of dimensions and also special outer-velocity
profiles. With the present-day easy access to personal computers, a numerical program
can easily be written to solve the two-dimensional steady equations with nonsimilar
solutions with little difficulty.
An easy procedure for the two-dimensional boundary layer equations uses the
Crank-Nicolson method. Start with the steady form of the boundary layer equations—
namely,
u
u U
U 2 u
u
+v =
+U
+
x
y
t
x
y2
(12.4.11)
The second-order derivative in y and the first in x tell us that this is a parabolic partial
differential equation. It is convenient to first suppress as much as is easily possible the
growth of the boundary layer thickness in the x direction. Introduce variables suggested
by our Falkner-Skan solution according to
Ux
Ux
=
f
(12.4.12)
= x/L = y
Then, since
1
1 1 dU
=
+
and
− +
x L 2
x U dx
=
y
U
x
300
Multidimensional Computational Methods
we have
U Ux
f
f =U
(12.4.13)
x
1
1 Ux f Ux
1 1 dU f
x dU
U
v=−
=−
+
+
f+
− +
x
2
U dx
x
L
2
x U dx
=
u=
y
Substituting these into equation (12.4.1), find, after some arranging, that
f dU f U 2 f
U
1 1 dU 2 f
U
+
+
− +
dx L
2
x U dx 2
−U2
dU
1 f
1 1 dU f
U 2 3 f
2 f 1 1 dU 1
+
+
+
+U
f
+
−
=
2
3
2 U dx x
L 2
x U dx
x
dx
For convenience in writing introduce the parameter = x/U dU/dx. Multiply the
preceding equation by x/U 2 and simplify. The result is
f
2 f
f
f
2 f + 1
3 f
f +
(12.4.14)
+
− 2
= 3 +
2
It is seen that equation (12.4.14) is a third-order partial differential equation. In
numerical solutions, it is somewhat easier to deal with first-order partial differential
equations. To get to that point here, make the further change of variables
x
f
V =v
F=
U
where F is then the dimensionless x velocity component and V is a scaled dimensionless
version of the y velocity component. Then, from the second of equation (12.4.13)
find that
f − 1 f
+1
f + +
(12.4.15)
V =−
2
2
Differentiating this with respect to gives
F − 1 F
V
= − F +
+
2
Our momentum equation (12.4.4) can now be written as
2 F
F
F
V + − 1 F = 2 +
+
F F +
2
(12.4.16)
(12.4.17)
To put these equations in numerical form, set up a grid with spacing by ,
with grid points at
= n − 1
n = 1 2 N
= m − 1
m = 1 2 M
12.4
301
One-Step Methods
Using centered differences and keeping in mind that we want to finish up with a set of
linear algebraic equations, the various derivatives can be replaced by
V
1
=
− Vm+1n−1 + Vmn − Vmn−1
V
2 m+1n
F =
m+1
Fm+1n + Fm+1n−1 + Fmn + Fmn−1
4
F
= m+05 Fm+1n − Fm+1n−1 + Fmn − Fmn−1
2
−1
− 1 F
= n−05 m+05
Fm+1n − Fm+1n−1 + Fmn − Fmn−1
2
4
1 − F 2 = m+1 1 − Fm+1n Fmn
(12.4.18)
F
= m+05 Fmn Fm+1n − Fmn
1
F
n m − 1
− 1
F
=
Fmn Fm+1n − Fm+1n−1 + Fmn − Fmn−1
Vmn +
V+
2
4
2
F
1
2 F
=
Fm+1n+1 − 2Fm+1n + Fm+1n−1 + Fmn+1 − 2Fmn + Fmn−1
2
2
2
Putting these into equation (6.6.5) find that
Vm+1n = Vm+1n−1 − Vmn + Vmn−1 + 2Pm Fm+1n + Qm Fm+1n−1
+ Rm Fmn + Sm Fmn−1
with
Pm = −
m+1 m+05 n−05
−
−
−1
4
2
4 m+05
Qm = −
m+1 m+05 n−05
−
+
−1
4
2
4 m+05
Rm = − m+1 + m+05 − n−05 m+05 − 1
4
2
4
Sm = −
(12.4.20)
m+1 m+05 n−05
+
+
−1
4
2
4 m+05
For the momentum equation (12.4.14), similar substitutions give
− m+1 1 − Fm+1n Fmn + m+05 Fmn Fm+1n − Fmn
− 1
1
Fmn Fm+1n − Fm+1n−1 + Fmn − Fmn−1
Vmn + n m
+
4
2
1
−
− 2Fm+1n + Fm+1n−1 + Fmn+1 −2Fmn + Fmn−1
F
22 m+1n+1
= 0
(12.4.19)
(12.4.21)
302
Multidimensional Computational Methods
These two equations can be rearranged in a form suitable for solution as
follows:
Amn Fm+1n−1 + Bmn Fm+1n + Cmn Fm+1n+1 = Dmn
Vm+1n+1 = Vm+1n + Emn
(12.4.22)
(12.4.23)
with 1 < n < N , and the parameters defined by
Vmn + 05n m − 1 Fmn
1
−
4
22
m+05 Fmn
1
+
Bmn = m+1 Fmn +
2
Amn = −
Cmn =
Vmn + 05n m − 1 Fmn
1
−
4
22
(12.4.24)
2
m+05 Fmn
Vmn + 05n m − 1 Fmn Fmn+1 − Fmn−1
−
4
Fmn+1 − 2Fmn + Fmn−1
+
22
Dmn = m+1 +
Emn = − Vmn + Vmn−1 + 2Pm Fm+1n + Qm Fm+1n−1 + Rm Fmn + Sm Fmn−1
This numerical integration procedure requires knowledge of F and V at the initial
line m = 1. These starting conditions usually would come from a Falkner-Skan or other
of the similarity solutions previously discussed. We also must know as input data at
the various values of .
For a given value of m, the Amn Bmn , and Cmn are all known. Equation (12.4.22)
is then a tridiagonal set of algebraic equations, which is an algebraic system that is
particularly amenable to solution by a number of methods. From the boundary conditions we know that Fm1 = 0 Vm1 = 0, and FmN = 1, so equation (12.4.13) represents N − 2 equations in terms of the N − 2 unknowns Fmn . Solve for the Fmn
first, and then go to equation (12.4.24) and solve for the Vmn step by step. Then
increment m by one, and repeat the procedure. As the solution proceeds downstream,
it is a good idea to check the wall velocity gradient as the integration goes forward in . When it becomes zero, flow reversal will occur downstream from this
point, and the boundary layer approximation is no longer valid. This would normally
determine M.
Numerical solution of these boundary layer equations would usually proceed
by taking a relatively coarse grid spacing (small M and N ), solving the algebraic system, and then repeating the solution after halving the grid spacing (doubling M and N ). This procedure would repeat until the changes are acceptably
small.
The numerical procedure just outlined is not the only numerical scheme that could
have been used. It has, however, the virtue of leading to a tridiagonal matrix, which
is about the simplest algebraic system to solve. It also is a method well within the
capabilities of a personal computer and provides results that appear to be at least as
good as those found by other numerical methods.
12.4
303
One-Step Methods
12.4.5
Boundary Layer Equation—Hybrid Method
The previous methods can be combined in various ways into a hybrid method for the
boundary layer equations. Write
un+1j − unj + un+1j−1 − unj−1 vn+1j−v − vn+1j−1
+
= 0
2x
y
un+1j − unj
run+1j + 1 − r unj
x
un+1j+1 − un+1j−1
unj+1 − unj−1
+ rvn+1j
+ 1 − r vnj
2y
2y
Un+1 − Un
= rUn+1 + 1 − r Un
x
+
r un+1j+1 − un+1j − un+1j − un+1j−1
2
y
+ 1 − r unj+1 − unj − un+1j − un+1j−1
(12.4.25)
(12.4.26)
Notice the following:
1. r = 0: The method is explicit, and the error is first order in x and second in y.
For stability,
2x
≤ 1
unj y2
2
vnj
x
unj
≤ 2
2. r = 1/2: This becomes the Crank-Nicholson method. The error is second order
in both x and y. There is no stability constraint.
3. r = 1: The method now is fully implicit. The error now is first order in x and
second order in y. There is no stability constraint.
12.4.6
Richardson Extrapolation
From the preceding it is seen that there is always a trade-off in stability, accuracy, and
ease of solution. For the case of nonlinear differential equations, such as the boundary
layer equations, the nonlinearity adds additional complexity. Sometimes, however, it is
possible to make some gains in accuracy without necessarily sacrificing computational
time and complexity.
Suppose that we have a nonlinear first order system of equations
dy
= fx y ya = y0
(12.4.27)
dx
Let uj be a finite difference solution at point xj ; for example, let uj+1 = uj + hfxj uj .
0 does
Further, let ej be the error at point
xj ej = uj − yxj . If e0 = h 0 , where
f
x
=
not depend on h, then ej = h xj + Oh2 where xj solves d
yx −
dx
y
1 d2 y
2 dx2
2
a = 0 . Then uj h =
j + hxj 1+ Oh . If the2 problem is solved again
1yx
with half the step size, then uj 2 h = yxj + 2 hxj + Oh . Subtracting these two
results to eliminate the first-order term in h gives
1
uj ≡ 2uj
h − uj h = yxj + Oh2
(12.4.28)
2
304
Multidimensional Computational Methods
Thus, by doing the numerical solution twice with different step sizes, the solution
represented by equation (12.4.28) has an extra order of accuracy.
12.4.7
Further Choices for Dealing with Nonlinearities
What makes solution techniques in computational fluid mechanics unique is how they
deal with the convective acceleration terms u u
+ v u
. There are a number of different
x
y
ways to do this.
1. Lagging computation. Evaluate u, v at their old position. Then
u
un+1j − unj
unj+1 − unj
u
u
+ v ≈ unj
+ vnj
x
y
x
y
(12.4.29)
This is first order in the “marching coordinate” x and will cause trouble when x
derivatives become large.
2. Simple iterative update. Use the lagging computation, and then repeat the computation using equation (12.4.29) but with the newly computed un+1j . The process
can be repeated several times.
3. Newton linearization. The convergence of a method is improved if the continuity
and momentum equations are solved in a coupled manner.
4. Extrapolation of coefficients. When using un+1j compute it using
u
x y x where
x n j
unj − un−1j
u
xn yj =
x
x
un+1j = unj +
12.4.8
(12.4.30)
Upwind Differencing for Convective Acceleration Terms
Perhaps the most popular method for improving calculation of convective terms is that
of upwind differencing. Traditional methods for calculating Lf = f
use
+ u f
x
t
Lf ≈
fn+1j − fnj
fnj+1 − fnj
+u
t
x
(12.4.31)
In upwind differencing, provided u is positive, instead use
Lf ≈
fn+1j − fnj
fnj − fnj−1
+u
t
x
(12.4.32)
Comparing these two equations shows that the only change has been in the x differencing, which is now in the upwind direction rather than the current (downwind) direction.
The effect of this change is to introduce an artificial viscosity into the problem so that
if we were trying to solve Lf = 0, using upwind differencing we would actually be
solving
f
uf
2 f
=
+ artificial
+ higher-order terms and differences.
t
x
x2
The artificial viscosity is given by artificial = 21 u x − ut . Providing this artificial
viscosity is positive, the use of upwind differencing is necessary for stability of the
computations.
12.5
305
Multistep, or Alternating Direction, Methods
12.5 Multistep, or Alternating Direction, Methods
12.5.1
Alternating Direction Explicit (ADE) Method
Consider the differential equation
f
2 f
= 2 =
t
x
x
f
x
(12.5.1)
Since in the previous section we saw that it was advantageous at times to take derivatives
in different ways, one suggestion for approximating equation (12.5.1) is
fn+1j − fnj
fnj+1 − fnj − fn+1j − fn+1j−1
=
(12.5.2)
t
x2
Solving for updated values gives
fn+1j =
1−
f +
f
+
f
1 + nj 1 + nj+1 1 + n+1j−1
(12.5.3)
If we are clever enough to sweep through our equations in the direction of increasing
j, the last term on the right-hand side is known, and we have a simple explicit method.
The good news is that, for equation (12.5.1), this method is unconditionally stable,
with errors of second order in both t and x. The bad news is that, if convection terms
are added to equation (12.5.1), the method may be unconditionally unstable.
12.5.2
Alternating Direction Implicit (ADI) Method
As our starting point we will be more ambitious than in the previous section, dealing
with one time dimension and two space directions. For our model equation let
2
f
f
f
f 2 f
= −u − v +
+
(12.5.4)
t
x
y
x2 y2
To do our time differencing proceed in two steps:
2
fn+1/2 − fn
fn+1/2
fn+1/2 2 fn
fn
Step1
= −u
−v
+
+ 2
t/2
x
y
x2
y
2
fn+1/2
fn+1/2 2 fn+1
f
fn+1 − fn
= −u
− v n+1 +
+
Step2
t/2
x
y
x2
y2
(12.5.5)
(12.5.6)
Notice that although each of the two steps are implicit, the resulting equations are
tridiagonal and thus easy to solve. The error is second order in each of the three
incremental deltas.
Formally, this method has unconditional stability. However, it can become unstable
if too big a time step is used. This is especially true if the boundary conditions are
allowed to lag in time. See Roache (1976, page 92).
In equations (12.5.5) and (12.5.6) details on the x and y differencing have been
omitted. One might use
2 fnij
fni+1j − 2fnij + fni−1j
≈
and
x2
x2
fnij+1 − 2fnij + fnij−1
2 fnij
≈
y2
y2
Further, to keep second-order accuracy, use un+1/2 vn in equation (12.5.5) and un+1/2 vn+1
in equation (12.5.6).
306
Multidimensional Computational Methods
To use this method to solve the full Navier-Stokes equations would require an
implicit coupled solution of an equation like (12.5.4) in the stream function and another
one in vorticity, both solutions to be carried out at the two times n + 1/2 and n + 1.
This would be a formidable task! These two methods, wherein the differencing of space
derivatives is split into two time steps, are the first of many splitting methods which
have been introduced.
Hyperbolic Partial Differential Equations
12.6 Method of Characteristics
Early in this chapter, we briefly examined the three basic classes of partial differential
equations and saw examples. We will now examine them more generally and derive a
method suitable for solving hyperbolic differential equations.
The forms of the three equations given in equation (12.1.1) can in two dimensions
be generalized by the form
A
2 f
2 f
2 f
+
2B
+
C
= E
x2
xy
y2
(12.6.1)
Here A, B, C, and E are real and may depend on f and its first derivatives. Coordinates
x and y are not necessarily space coordinates; one could also be a time dimension. Next,
ask the question, When and where are infinitesimal variations of the first derivatives of
f allowed, or alternatively, Where are first derivatives of f continuous? This question
can be phrased in terms of the solution of three equations in three unknowns—that is,
A
2 f
2 f
2 f
+ 2B
+ C 2 = E
2
x
xy
y
2
2 f
f
f
=d
dx 2 + dy
x
xy
x
2 f
f
2 f
+ dy 2 = d
dx
xy
y
y
(12.6.2)
where the second and third lines represent the differentials of the first derivatives.
Using Cramer’s rule, this system is easily solved, giving
N
2 f
= 1
2
x
D
2 f
N
= 2
xy
D
2 f
N
= 3
2
y
D
(12.6.3)
with
A 2B C
D = dx dy 0
0 dx dy
A
E 2B C
f
dy 0
d
x
N1 =
f
dx dy
d
y
E C
f
0
dx d
x
N2 =
f
0 d
dy
y
A 2B
E
f
dx dy d
x
N3 =
f
0 dx d
y
(12.6.4)
12.6
307
Method of Characteristics
By the usual restriction of Cramer’s rule, this is an acceptable solution except where
the denominator D vanishes. Expanding this determinant, we find it vanishes where
Ady2 + Cdx2 − 2Bdxdy = 0; that is, along the lines of slope
√
dy
B ± B2 − AC
=
(12.6.5)
dx
A
This pair of lines are called the characteristic lines, or just characteristics, of
equation (12.6.1). Notice the following:
1. If B2 − AC < 0, both of the characteristics are imaginary and play no decisive
role in the solution. This is the general definition of elliptic partial differential
equations.
2. If B2 − AC = 0, there is only one real characteristic. This is the general definition
of parabolic partial differential equations.
3. If B2 − AC > 0, both of the characteristics are real. This is the general definition
of hyperbolic partial differential equations. We will concentrate further attention
on this latter case.
If a region covered by characteristics is to have a solution, it is necessary that the
numerators in equation (12.6.3) also vanish, so l’Hospital’s rule can be used. Expanding
all three of the determinants, we find with the help of equation (12.6.5) that
2
dy
dy
N1 =
(12.6.6)
N3 and N2 = − N3
dx
dx
so all three numerators are zero on the characteristics provided N3 = 0, which
occurs when
f
f
A dy
E
d
d
(12.6.7)
=−
+ dy
y
C dx
x
C
With the aid of equations (12.6.5) and (12.6.7), a solution can now be constructed.
Suppose in Figure 12.6.1 that the derivatives of the function f is known on curve
1-2. Using equation (12.6.5), construct a series of characteristics on a closely spaced
number of points on this curve. Label the characteristics + if the plus sign in front of
the square root was used, and − otherwise. Notice that in the roughly triangular region
1-2-3 the + and − characteristics intersect, whereas outside this region there are no
intersections.
Next, magnify the small region a-b-c in Figure 12.6.1 as in Figure 12.6.2. Point c
is at the position xc yc , determined by solving
√
√
B − B2 − AC
B + B2 − AC
xc − xa yc − yb =
xc − xb
yc − ya =
A
A
a
b
(12.6.8)
Similarly, from equation (12.6.7) write
√
B + B2 − AC
fc fa
E
fc fa
y − ya
−
=−
−
+
y
y
C
x
x
C a c
a
(12.6.9)
√
B − B2 − AC
fc fb
E
fc fb
y − yb
=−
−
+
−
y
C
x
x
C b c
y
b
308
Multidimensional Computational Methods
_
_
_
_
_
_
_
_
_
_
_
+
3
+
+
+
y
+
+
+
1
+
+
c
a
+
b
2
x
Figure 12.6.1 Method of characteristics—region of solubility
(xc , yc )
(xa , ya )
(xb , yb )
Figure 12.6.2 Method of characteristics—detail of two intersecting characteristic lines
thereby solving for the derivatives of f at c. This process can be continued repeatedly
throughout the region 1-2-3, obtaining all derivatives of the function within this region.
Outside this region, no solution is possible due to insufficient data.
Note that if there is a rigid boundary (wall) or a line of symmetry, then a point
c can land on a wall and there is no − characteristic to meet it. The wall boundary
condition replaces this, however, so a solution is still possible.
In inviscid compressible isentropic flow, an equation similar to (12.6.1) governs
the flow with f the velocity potential, and
A = a2 − u2 B = −uv
C = a2 − v 2
a2 = a20 −
k−1 2
u + v2
2
(12.6.10)
u and v being the x and y velocity components (derivatives of “f ”) and a the speed of
sound. Where the flow is supersonic, the equations become hyperbolic.
The method of characteristics predates computers by almost a century. If one is
analyzing or designing flow in a nozzle, it is a problem that can easily be solved
12.7
309
Leapfrog Method—Explicit
with a pencil, eraser, and adding machine. With modern computers, however, programs
using this methodology become somewhat complicated, since even if the original data
is given on a straight line, the subsequent line will more than likely be curved. Often
interpolation is used to keep a straight grid in the region of solution. Characteristics can
be lines of discontinuity of derivatives of f , but not of f itself.
Even first-order differential equations can have characteristics. If the same procedure
is pursued on the equation
u
u
+ u = fu x t
t
x
(12.6.11)
for example, the result is
u fdx − udu
=
t
dx − udt
u du − fdt
=
x dx − udt
(12.6.12)
The equation of the characteristic is dx = u dt, and along the characteristic udu = f dx.
As an example of the use of the method of characteristics, the equations for analysis
of a water hammer are
V V
f
H
+V
+
+
V V = 0
g
x
x
t
2D
AV + A = 0 with
(12.6.13)
x
t
p
p
=
K = bulk modulus.
gH = + gz
t
K t
The method of characteristics has been used for this problem with much success.
12.7 Leapfrog Method—Explicit
For the one-dimensional wave equation
2 f
2 f
= a2 2
2
t
x
(12.7.1)
a method that gives interesting results is the leapfrog method. If centered differences
are used for the two second derivatives, equation (12.7.1) becomes
fn+1j − 2fnj + fn−1j
fnj+1 − 2fnj + fnj−1
= a2
t
x
Solving for f at the latest time step gives
fn+1j = 2 1 − A2 fnj − fn−1j + A2 fnj+1 + fnj−1
(12.7.2)
(12.7.3)
where A = at/x is the Courant number. Notice that when A = 1, the value at the
same position but preceding time drops out of the calculation.
The exact solution to equation (12.7.1) is known to be
fx t = Fx − at + Gx + at
(12.7.4)
where F and G depend on initial conditions according to
fx 0 = Fx + Gx
f
x 0 = a −F ′ 0 + G′ 0
t
(12.7.5)
310
Multidimensional Computational Methods
Taking x1 t1 as a starting point and letting xi = x1 +i−1 x tn = t1 +n−1 t, then
x − at = + ix − nat
= x1 − x − at1 − t and
x + at = + ix + nat
= x1 − x + at1 − t
(12.7.6)
Inserting these into equation (12.7.3) with A = 1 find that
−fn−1j + fnj+1 + fnj−1 = F + i − j − 1 x + G + i + j − 1 x
+ F + i − j + 1 x + G + i + j + 1 x
− F + i − j − 1 x − G + i + j − 1 x
(12.7.7)
= F + i − j − 1 x + G + i + j + 1 x
= fn+1j
Thus, for a Courant number of unity, the method gives the exact solution.
The reason for this happy result can be found by comparing the calculation procedure
with the characteristics of equation (12.7.1), which are given by dx
= ±a. The slopes of
dt
=
±a/A,
so
when the Courant
the lines connecting xn+1j with xnj−1 and xnj+1 are x
t
number is unity, the leapfrog method corresponds to the method of characteristics. If
we were to choose A < 1, we are within the triangle of the characteristic lines, and our
computation is using valid information and the result is stable. If, however, we were to
choose A > 1, we would be using information that is known not to be pertinent to the
calculation, and the calculation becomes unstable.
One drawback of the leapfrog method is that to start the calculations, there is a
need to know the solution at t0 . To handle this, place a fictitious row at t0 and let
f1j = F x1 − at1 + G x1 + at1 and
f0j = f2j − 2a −F ′ x1 − at1 + G′ x1 + at1 t
(12.7.8)
Then
f2j =f1j−1 + f1j+1 − f0j = f1j−1 + f1j+1 − f2j
+ 2a −F ′ x1 − at1 + G′ x1 + at1 t or
f2j =
(12.7.9)
1
f1j−1 + f1j+1 + a −F ′ x1 − at1 + G′ x1 + at1 t
2
This is the starting formula for the method.
12.8 Lax-Wendroff Method—Explicit
Many of the equations of fluid mechanics are of the type
U F
+
= 0
t
x
for instance, the one-dimensional compressible flow equation where
⎞
⎛
u
U=
F = ⎝ u2
a2 ⎠
u
+
2
k−1
(12.8.1)
(12.8.2)
12.9
311
MacCormacks Methods
The first of these equations is the continuity equation, the second the momentum
equation. The form of equation (12.8.1) is often referred to as a conservative form. The
Navier-Stokes equations and the inviscid compressible flow equations are examples that
fit this general form.
For the Lax-Wendroff method, let
F
F U
F
=
=A
t
U t
x
(12.8.3)
Differentiating equation (12.8.1) with respect to time yields
2 U
=−
t2
t
F
x
=−
x
F
t
=−
F
A
x
x
(12.8.4)
From equation (12.8.2) find that
⎛ F F ⎞
⎛
1
1
u
⎜ U1 U2 ⎟
⎟ = ⎝ 2a da
A=⎜
⎝ F F ⎠
2
2
k − 1 d
U1 U2
u
⎞
⎠
(12.8.5)
Then
2 U
t2 nj
nj
F
F
t2
≈Unj + t −
A
+
x nj
2 x
x nj
2
t
Fnj−1/2
Fnj+1/2
t
≈Unj −
− Anj−1/2
Fnj+1 − Fnj−1 +
Anj+1/2
2x
2x x
x
x
(12.8.6)
Un+1j ≈Unj + t
U
t
+
t2
2
t
− Fnj−1
F
2x nj+1
t2
+
Anj+1 + Anj Fnj+1 − Fnj − Anj + Anj−1 Fnj − Fnj−1
4x2
≈Unj −
3
t
≤ 1 with an error of order 61 t2 tU3 +
This method is explicit and stable providing 0 ≤ x
3
1
x2 xU3 . It is usually easy to program, although nonlinear problems require some
6
adjustment to the method. More details can be found in Lax (1954), Smith (1978), and
Ferziger (1981).
12.9 MacCormacks Methods
Over the years, R. W. MacCormack has investigated a series of approaches to the
conservative form (equation (12.8.1)), which have proven to be useful for computations.
Some of these methods are outlined here.
312
Multidimensional Computational Methods
12.9.1
MacCormack’s Explicit Method
This method uses a predictor-corrector approach, where an estimate of one of the
variables is first made and then updated in a later calculation. For equation (12.8.1)
MacCormack used
Predictors: U n+1j = Unj − Fnj+1 − Fnj t/x F n+1j = F U n+1j
(12.9.1)
1
Corrector: Un+1j =
Unj + U n+1j − F n+1j − F n+1j−1 t/x
2
the superposed bars denoting the predicted quantities.
Examples of the application of these include the following:
1. The one-dimensional wave equation
For
f
t
+ a f
= 0, application of MacCormack’s explicit method (1969) gives
x
Predictor: f n+1j = fnj − a fnj+1 − fnj t/x
Corrector: fn+1j =
1
fnj + f n+1j − a f n+1j − f n+1j−1
2
(12.9.2)
2. Burger’s equation
2
+ F
= xf2 F = af + 21 bf 2 , application of MacCormack’s explicit method gives
x
Predictors: f n+1j = fnj − Fnj+1 − Fnj t/x + r fnj+1 − 2fnj + fnj−1
For
f
t
1 2
F n+1j = af n+1j + bf n+1j
2
1
Corrector: fn+1j =
f + f n+1j − F n+1j − F n+1j−1 t/x
2 nj
+r f n+1j+1 − 2f n+1j + f n+1j−1
(12.9.3)
where r = t/x 2 .
12.9.2
MacCormack’s Implicit Method
In 1981 MacCormack suggested that for Burger’s equation the predictor-corrector procedure could be varied according to
t
Predictor 1 + t/x f n+1j = fnj explicit +
f n+1j+1 where
x
f n+1j = f n+1j − fnj
fnj explicit = −a fnj+1 − fnj t/x + r fnj+1 − 2fnj + fnj−1
t
fn+1i−1 where
Corrector 1 + t/x fn+1j = f nj explicit +
x
1
fn+1j =
fnj + f n+1 j +fn+1j
(12.9.4)
2
f nj explicit = −a f n+1j − f n+1j−1 t/x + r f n+1j+1 − 2f n+1j + f n+1j−1
12.10
Discrete Vortex Methods (DVM)
Here, is a constant such that ≥ max a + 2/x − x/t 0. This method is unconditionally stable and second-order accurate provided t/x 2 is bounded as t and
x approach zero.
12.10 Discrete Vortex Methods (DVM)
The methods previously discussed all start with an equation and then use finite differences or a similar procedure to transfer the mathematics from calculus to algebra. They
deal with flows that are well defined both physically and mathematically throughout a
region that is generally fixed. In flows that are separated, however, such as in the wake
of a bluff body, a wing, or a propellor, the portion of the flow that is of greatest interest
is in the vortices that are shed and find their way downstream. The Kármán vortex street
is an example of this.
Discrete vortex methods try to model, or simulate, these effects much in the manner
of the Kármán vortex street. Vorticity is generated at a boundary, often in the boundary
layer. At the separation point the vorticity tends to leave the boundary and move into
a flow region that is more or less inviscid. There the vorticity will move according to
D
≈ 0. The position of this bit of vorticity will change according to dr
= v, where
Dt
dt
the velocity is a combination of the inviscid flow plus the induced velocity from all
previous bits of shed vorticity.
To make this description more specific, consider a two-dimensional flow in the
wake of a cylinder. The boundary layer flow is solved and the separation point
determined by some separation criteria such as Stratford’s. To get the vortex out
of the boundary layer, a random walk procedure can be used. Chorin (1978) suggested that, to avoid the singularities in the vortexes such as discussed in the chapters on inviscid flows, “vortex blobs” could be introduced, with stream functions of
the form
⎧
r
⎪
⎪
⎨ 2 ! for r < !
(12.10.1)
=
⎪
⎪
⎩ ln r for r ≥ !
2
This gives constant velocity inside the vortex. The choice of the parameter ! is left up to
the user. As the blobs are released from the boundary layer, they are moved according
to the induced velocities.
There are some matters that remain, such as dealing with combining of vortices
if or when they collide, what to do when they leave the computational region, when
they reenter the boundary layer, and so forth. These can be dealt with in many ways
that fit the specific situation at hand and that deal with practical matters, such as the
capacity of the computer and time available. For instance, instead of the description
of the vortex blob just given, the formulation v = 2 z−z
F , where F is a vorticity
0
99r/! n
modification function such as F = 1+99r/! n n being an integer greater than or equal
to 2. This makes the self-induced velocity zero at the center of the blob and still
has the velocity behave as a line vortex away from the flow. More possibilities concerning the implementation of this method can be found in Alexandrou (1986) and
Hong (1988).
313
314
Multidimensional Computational Methods
12.11 Cloud in Cell Method (CIC)
This method is a variation on the discrete vortex method and was first introduced by
Christiansen (1973). Instead of tracking individual vortices as in DVM, a grid is placed
on the region, and as a vortex moves within a sector of the grid, its vorticity is distributed
to the corners of the sector. (See Figure 12.11.1.) For a sector x by y, if the vortex
is at a location !x !y with respect to the corner of the sector, the vorticity " might
be distributed to the corners according to
!y
!x
"ij = 1 −
1−
" = A1 "
x
y
!x !y
"ij+1 = 1 −
" = A2 "
x y
(12.11.1)
!x
!y
"i+1j =
1−
" = A3 "
x
y
"i+1j+1 =
!x !y
" = A4 "
x y
After this distribution, the stream function can be solved from
2 = "
(12.11.2)
The induced velocities and positions for finding how fast the vortex travels are then
found from
d xn
un
u
u
u
u
=
= A1
+ A2
+ A3
+ A4
(12.11.3)
vn
v ij
v i+1j
v ij+1
v i+1j+1
dt yn
For 1,000 vortices the CIC method appears to be about 20 times faster than the DVM
method. More references to this method and its implementation can be found in Roberts
and Christiansen (1972), Milinazzo and Saffman (1977), Chorin (1978), Alexandrou
(1986), and Hong (1988).
(i + 1, j + 1)
(i, j + 1)
ω
∆y
δy
(i + 1, j )
(i, j )
δx
∆x
Figure 12.11.1 Computational molecule for the cloud-in-cell method
315
Problems—Chapter 12
Problems—Chapter 12
In the following problems use either the program languages specified in the text or a
spreadsheet.
12.1 Use simple relaxation (equation (12.2.1a)) to find the values on the center line
of the elbow. On the inner boundary of the el (filled dots) the value is 0, on the outer
boundary it is 1, and the end points 1 and 10 average the two boundary values. Do at
least 10 iterations.
5
6
7
8
9
10
11
12
17
18
19
20
21
22
4
16
3
15
2
14
1
13
P12.1 Flow in an elbow
12.2 Repeat problem 12.1, this time using successive overrelaxation (SOR). Take
the relaxation parameter to be 1.7.
12.3 Use the leapfrog method to solve the one-dimensional wave equation
1 2 f
subject to the conditions
c2 t2
2 f
x2
=
f
x 0 = 0 0 ≤ x ≤ 1 f0 t = f1 t = 0
t
To start the solution, introduce a row of fictitious grid points at t = −t, where
fx 0 = sin x
f
x 0
t
Let x = 01 and t = x/c The wave speed can be taken as unity. Compute for at
least 20 time steps.
fx −t = fx 0 − 2t
12.4 An explicit method for the diffusion equation f /t = 2 f /x2 is given by
fi j + 1 = fi j + fi − 1 j − 2fi j + fi + 1 j , where = t/x2 ≤ 05.
Solve this for the conditions
$
4x
0 ≤ x ≤ 04
f0 t = 0 f1 t = 1 fx 0 =
−x + 2 04 ≤ x ≤ 1
Take = 025 and x = 01.
316
Multidimensional Computational Methods
12.5 An implicit formula for solving the unsteady flow in a channel, given
2
by the parabolic equation u
= − p
+ yu2 , is ui j − ui j − 1 = −t p
+
t
x
x
ui − 1 j − 2ui j + ui + 1 j , where is defined by = t/x2 . This scheme
is apparently stable for all positive values of . Solve for flow starting from rest with
$
p
0
t ≤ 0
u0 = u1 = 0
=
x
−1 t > 0
Take = 05 x = 01 t = 01, and do 200 time steps. Compare with the exact
solution (parabolic profile).
12.6 Repeat the previous problem, this time using the Crank-Nicholson method, with
the more accurate finite difference equation in the form
t p 1
+ ui − 1 j − 2ui j + ui + 1 j
x 2
1
+ ui − 1 j − 1 − 2ui j − 1 + ui + 1 j − 1
2
ui j − ui j − 1 = −
12.7 Repeat problem 12.5, this time using the DuFort-Frankel method for solving the
+ui+1j
problem. This method is explicit and uses ui−1j −uij−1x−uij+1
for the second
2
uij+1 −uij−1
for the time derivative.
derivative in x and
2t
a. Put these approximations into the Navier-Stokes equation and rearrange to
obtain a form suited to solving for the velocity at the grid points.
b. Draw the computational molecule for this method. Indicate round dots where
space derivatives are taken and squares where time derivatives are taken.
12.8 For the diffusion equation in two space derivatives and one time derivative
the alternating direction implicit method (ADI) is unconditionally stable. Starting with
2
2
f
= xf2 + yf2 , the method uses the form
t
f ∗ i j − fi j k
f ∗ i + 1 j − 2f ∗ i j + f ∗ i − 1 j
=
t/2
x2
+
fi j + 1 k − 2fi j k + fi j − 1 k
y2
and follows it with
fi j k + 1 − f ∗ i j
f ∗ i + 1 j − 2f ∗ i j + f ∗ i − 1 j
=
t/2
x2
+
fi j + 1 k + 1 − 2fi j k + 1 + fi j − 1 k + 1
y2
The first solves for the intermediate values f ∗ , the second completing the solution for
f at the next time step. Rearrange the equations to put them into a form suitable for
programming.
12.9 One way of dealing with the nonlinearities of the Navier-Stokes equations is
to treat steady-state flows as transient flows starting from a quiescent state. For natural
317
Problems—Chapter 12
convection on a vertical semi-infinite plate, start with the boundary layer equations in
the nondimensional form
u
u
2 u
u
+u +v = T + 2
t
x
y
y
u v
+
= 0
x y
T
T
2 T
T
+u
+v
= Pr 2
t
x
y
y
and write them in an explicit finite difference form. Propose a scheme for solution.
2
2
12.10 For the linear wave equation xu2 = a12 t2u use the method of characteristics to
solve for a disturbance moving from left to right and striking a wall at x = 0. The x
interval is in the range −020 ≤ x ≤ 0. The disturbance is initially zero except in the
interval −009 ≤ x ≤ −005 and a = 343 meters/sec. Continue the calculation until the
disturbance moves out of the domain.
Appendix
A.1
A.2
A.3
A.4
Vector Differential Calculus 318
Vector Integral Calculus 320
Fourier Series and Integrals 323
Solution of Ordinary Differential
Equations 325
A.4.1 Method of Frobenius 325
A.4.2 Mathieu Equations 326
A.4.3 Finding Eigenvalues—The
Riccati Method 327
A.5 Index Notation 329
A.6 Tensors in Cartesian Coordinates 333
A.7 Tensors in Orthogonal Curvilinear
Coordinates 337
A.7.1 Cylindrical Polar
Coordinates 339
A.7.2 Spherical Polar
Coordinates 340
A.8 Tensors in General Coordinates 341
I am very well acquainted, too, with matters mathematical,
I understand equations, both the simple and quadratical:
About binomial theorem I’m teeming with a lot of news,
With many cheerful facts about the square of the hypotenuse.
William S. Gilbert
A.1 Vector Differential Calculus
Derivatives of a vector function of the space coordinates can be performed in any
combination of the coordinate directions. A useful differentiation operator is the del
operator, denoted by the symbol , or del, defined in Cartesian coordinates as
= del = i
+j +k
x
y
z
(A.1.1)
The operator often appears in partial differential equations in its “squared” form
2 = · = Laplacian = harmonic operator =
318
2
2
2
+
+
x2 y2 z2
(A.1.2)
A.1
319
Vector Differential Calculus
To illustrate some of the uses of the del operator, consider a unit vector a with
direction cosines ax ay az . Then, the operation a · acting on a vector F gives the
derivative of F in the direction of a, or
a · F = ax
F
F
F
+ ay
+ az
x
y
z
The operator v · is encountered often in fluid mechanics, particularly where v is the
velocity vector. The operator itself is not a vector, since v · = · v, and is referred to
as a pseudo-vector.
Frequent uses of the operator include its operations on a scalar and also in vector
multiplications. For instance, if is a scalar function of the coordinates, then
= grad = i
+j
+k
x
y
z
(A.1.3)
is called the gradient of . Its magnitude tells us relative information about the spacing
of lines of constant . Where the magnitude of grad is large, the constant lines are
closer together than where it is small. The direction of grad is locally normal to the
surface = constant.
The scalar quantity
· F = div F =
Fx Fy Fz
+
+
x
y
z
(A.1.4)
is called the divergence of F. It tells us the quantity of F that passes through a surface.
The vector quantity
Fy Fx
Fz Fy
Fx Fz
× F = curl F = i
−
−
−
+j
+k
(A.1.5)
y
z
z
x
x
y
is called the curl of F and gives information on how much F is twisting or rotating. It
also tells the direction of this rotation.
Some useful formulas that involve the del operator follow. They can easily be
verified by expanding left- and right-hand sides and comparing the results.
f F = f · F + F · f
× f F = f × F + f × F
· A × B = B · × A − A · × B
× A × B = B · A − A · B + A · B − B · A
A · B = A · B + B · A + A × × B + B × × A
× f = 0 for any scalar f
· × F = 0 for any vector F
× × F = · F − 2 F
B · A =
1
× A × B − A × × B − B × × A + A · B
2
−A · B + B · A
A theorem due to Helmholtz states that any vector F can be expressed in the form
F = grad + curl A
where div A = 0
(A.1.6)
320
Appendix
Here, is called the scalar potential of F, and A is the vector potential of F. Since
· F = 2
(A.1.7)
× F = − 2 A
(A.1.8)
and
the scalar potential represents the irrotational part of F (the “curl-less” portion of F),
and the vector potential represents the rotational portion of F.
The Helmholtz decomposition (equation (A.1.6)) results in the two Poisson equations (A.1.7) and (A.1.8) (a Poisson equation is a Laplace equation with a nonhomogeneous “right-hand side”). These can in principle be solved, giving
· Fx′ gx x′ dV ′
x =
V′
(A.1.9)
× Fx′ gx x′ dV ′
Ax = −
V′
where
⎞
1
ln x − x′ in two dimensions
⎜
⎟
gxx′ = Green’s function = ⎝ 2 −1
⎠
in
three
dimensions
4 x − x′
⎛
(A.1.10)
Notice that the Green’s function is our potential for a source. The solution equation
(A.1.9), however, in general does not satisfy the constraint · A = 0. To take care of
this, replace equation (A.1.6) by
F = grad + curl A′ + grad a
(A.1.11)
Since curl grad a = 0 for any a, F has not been affected in any manner. Thus, A′ is
given by equation (A.1.9). Since A′ + grad a now replaces A, the requirement div A = 0
is replaced by divA′ + grad a = 0. Thus, if a is defined by
2 a = −div A′
(A.1.12)
giving
a=−
V′
· A′ x′ gx x′ dV ′
(A.1.13)
the constraint has been satisfied. (Note: In two dimensions the indicated volume integrals
become surface integrals.)
A.2 Vector Integral Calculus
There are several interesting and useful theorems regarding the operator and integration that are useful in fluid mechanics, both in deriving the basic equations and in putting
them in a form suitable for numerical calculation. They will be listed without proof.
These theorems are all closely related, and the names Gauss, Green, and Stokes are
intimately connected with them. In their use, they are closely related to the concept of
integration by parts of elementary calculus and can be thought of as a multidimensional
extension of that concept.
A.2
321
Vector Integral Calculus
Gauss’s theorem states the following:
In three dimensions, with S a closed surface,
S
n·
0
dS
=
4
R − R0 3
if R0 is outside of S
if R0 is inside of S
R − R0
(A.2.1a)
In two dimensions, with C a closed curve,
0
n·
ds =
2
2
C
r − r0
r − r0
if r0 is outside of S
if r0 is inside of S
(A.2.1b)
In both dimensions the integrand will be recognized as the radial velocity component
of a source located at R0 in three dimensions and r0 in two dimensions.
Stokes’s theorem is useful for changing line integrals to surface integrals and vice
versa. In its simplest form it is
C
t · F ds =
S
n · × F dS =
S
n × · FdS
(A.2.2)
where C is a closed curve bounding the surface S, n is a unit normal to the surface S,
and t is a unit tangent to the curve C. Note that C and S do not have to lie in a plane
and that C could bound an infinite number of different surfaces S.
Variations of Stokes’s theorem include the following:
C
tf ds =
S
n × f dS
n × × F dS
n · F − n × F dS =
S
S
C
n × · F dS
n · × F dS =
t · F ds =
S
S
C
−f + nn · f + f · n dS
t × nf ds =
t × F ds =
C
S
with
1
1
+
·n = −
R1 R2
R1 and R2 being the principal radii of curvature of the surface S.
The divergence theorem, also called Green’s theorem, is used for transforming
surface integrals to volume integrals and is stated as
S
n · F dS =
V
· F dV
(A.2.3)
where S is a surface enclosing the volume V , and n is again the unit outward normal.
The theorem states that the net outflow of F through S is made up of the sum of the
outflows from all of the regions inside of S.
322
Appendix
Variations of this theorem include the following:
× F dV
n × F dS =
V
S
n · F dS =
2 F dV
V
S
n · f dS =
2 f dV
S
V
f dV
nf dS =
V
S
On a surface,
n × × F dS = − t × F ds
C
n × · F dS =
n · × F dS =
S
S
· F dS = t × n · F ds
C
S
n × f dS = f ds
S
C
t · F ds
C
S
From these theorems several results follow that are useful in inviscid flow theory.
Green’s first identity
f
S
h
f · h + f 2 hdV
dS =
n
V
(A.2.4)
Here f and h are any two scalar functions of the coordinates, and S is the surface
enclosing V . This identity follows from Gauss’s theorem with F = f h.
Green’s second identity
f
h
h 2 f − f 2 hdV
dS =
h −f
n
n
V
C
(A.2.5)
Here f and h are any two scalar functions of the coordinates. This identity follows
from Green’s first identity, written first as in equation (A.2.3) and again with f and h
interchanged. The result follows by taking the difference of the two.
Green’s third identity
P = −
+
gP, Q 2 (Q) dV
V
S
(A.2.6)
gP Qn · Q − Qn · gP Q dS
where in equation (A.2.5)
⎞
1
ln P − Q for two dimensional problems
⎟
⎜
gP Q = Green’s function = ⎝ 2 −1
⎠
for three dimensional problems
4 P − Q
⎛
(A.2.7)
Here P is the position vector of a point in the interior of V and Q is the position vector
of a dummy point of integration, on the surface for the surface integral, and in the
A.3
323
Fourier Series and Integrals
interior for the volume integral. The function g is called the Green’s function for the
Laplace operator and is seen to be the velocity potential for a source.
A.3 Fourier Series and Integrals
Many times it is useful to represent a function in terms of an infinite series of simple
functions. A classic example of this is the Fourier series, where if ft is defined over
an interval 0 ≤ t ≤ T , the Fourier series representation of ft is
1
nt
nt
ft = a0 +
an cos + bn sin
.
2
T
T
n=1
(A.3.1)
This representation makes ft periodic with period T ; that is, ft + nT = ft for any
integer n.
The coefficients an and bn are determined by the property of the trigonometric
functions that, for integers m and n,
T
nt
mt
cos
dt = 0
T
T
0
T
T
nt
mt
mt
T
nt
cos cos
dt =
dt = 0 if n = m if n = m
sin sin
T
T
T
T
2
0
0
sin
(A.3.2)
The trigonometric functions are said to be orthogonal to one another, as the operation
of taking a product of two of them and then integrating is analogous to taking a dot
product of two vectors.
To use the orthogonality property, multiply both sides of equation (A.3.1) by either
the sine or cosine of m t/T , and then integrate over the period T . Interchanging
integration and summation, the result is
2 T
nt
ft cos dt
T 0
T
T
nt
2
ft sin dt for n = 0 1 2
bn =
T 0
T
an =
(A.3.3)
If the function ft is discontinuous at either a point interior to, or at either
end of, the interval, the series will converge to the average value at that point. Near
the discontinuity, the series sum will show oscillations and overshoots, the Gibbs
phenomenon.
The Fourier series representation of a function as given by equations (A.3.1) and
(A.3.3) is sometimes also called the finite Fourier transform. By DeMoivre’s theorem,
the sines and cosines can be replaced by exponentials so that the Fourier series (equation
(A.3.1)) becomes
1
int/T
−int/T
ft =
a +
a − ibn e
+
an + ibn e
2 0 n=1 n
n=1
cn eint/T
=
n = −
324
Appendix
where
⎛
05an − ibn for n > 0
⎞
⎟
⎜
⎟
⎜
cn = ⎜ 05a0 for n = 0
⎟.
⎠
⎝
05an + ibn for n < 0
There are a number of means of reducing the work needed to determine the coefficients
cn . These are termed fast Fourier Transform (FFT) methods.
If the interval T becomes infinite, the Fourier series becomes a Fourier integral, with
ft =
Fp e−2ipt dp
−
with F given by
Fp =
(A.3.4)
fp e2ipt dp
−
The equations given in (A.3.4) are analogs of equations (A.3.1) and (A.3.2). The Fourier
transform becomes the Laplace transform if ft = 0 for t < 0, and if 2 p is replaced
by the notation is, i being the root of −1.
The Fourier series expansion is a special case of expansions in terms of orthogonal
functions. These expansions are called spectral representations of the functions, since
the frequencies involved (2 n/T in the case of the Fourier series) make up a spectrum,
with the cn ’s being the amplitudes of the harmonics at these frequencies. For the general
case of expansion in terms of orthogonal functions, the expansion is
fx =
cn n x
n=−
a ≤ x ≤ b
(A.3.5)
where the n can be shown to be orthogonal easiest if they are solutions of the
equation
d pxd
+
dx
2
rx − qx = 0
(A.3.6)
with p and q both positive throughout the interval a ≤ x ≤ b. This is referred to as
the Sturm-Liouville equation. The boundary conditions which accompany equation
(A.3.6) are
d
a +
= 0 + > 0
dx a
(A.3.7)
d
b +
= 0 + > 0
dx b
Providing pd/dx = 0 at x = a and x = b, it can be shown that
b
rm n dx = 0
(A.3.8)
a
if m is different from n. The ’s are said to be orthogonal to one another with respect
to the weight function r. Special cases of orthogonal functions, along with the equations
they satisfy, are given in Table A.1.
A.4
325
Solution of Ordinary Differential Equations
TABLE A.1 Some well-studied orthogonal functions
Equation
Function name
Simple form
Sturm-Liouville form
Trigonometric
y′′ + n2 y = 0
y′′ + n2 y = 0
√
2
1 − x 2 y ′ ′ + √ n
Chebeyshev
Legendre
Laguerre
Hermite
1 − x2 y′′ − xy′ + n2 y = 0
2
y − 2xy + ny = 0
2 ′′
′
2
2
Bessel
x y + xy + x − n y = 0
Mathieu
y + a − b cos 2xy = 0
′
1 − x y − 2xy + ny = 0
−1 ≤ x ≤ 1
′′
′
xe−x y′ ′ + nxex y = 0
′
2
2
e−x y′ + ne−x y = 0
2
xy′ ′ + x − nx y = 0
xy′′ + 1 − xy′ + ny = 0
′′
−1 ≤ x ≤ 1
1−x2
′
y + a − b cos 2xy = 0
′′
0≤x≤T
y=0
2
1 − x y − 2xy + ny = 0
′′
Interval
′′
0≤x≤
− ≤ x ≤
0≤x≤T
0≤x≤
A.4 Solution of Ordinary Differential Equations
A.4.1
Method of Frobenius
Solutions for many of the equations listed above can be found in the form of power
s+n
series. The procedure is to assume a solution in the form y =
; substitute
n = 0 an x
this into the differential equation, and collect on terms. The lowest power of x is used
to determine s, and higher powers are used to give recursion relations to determine the
coefficients an . This is best demonstrated by considering an example.
Consider Bessel’s equation x2 y′′ + xy′ + x2 − n2 y = 0. Substituting the series form
for y into the equation gives
aj
j =0
xs
s2 + 2js + j 2 − n2 xj + s + xj + s+ 2 = 0 or
a0 s2 − n2 + a1 s + 12 − n2 x +
j =2
2
x aj s + 2js + j 2 − n2 + aj−2 = 0
j
Since the s power has been inserted to make a0 the starting term, it cannot be zero and
hence s = ±n. The recursion relationship is aj = − s + j12 − n2 aj−2 j ≥ 2, as long as n is
not an integer, in which case it would eventually involve division by zero. For the case
where n is not an integer, the solution is
−1j x n + 2j
n + 2j j!n + j + 1
j =0 2
y = Jn x and J−n x where Jn x =
(A.4.1)
Thus, two distinct solutions have been found with one effort.
Should n be an integer, we still have one solution Jn x but are missing the second.
The loss of this second solution can be easiest seen for the case n = 0, since J0 x =
J−0 x. For this case, let the second solution
be of the form J0 x ln
x + y2 x. Substituting
this into Bessel’s equation gives ln x x2 J0′′ + xJ0′ + x2 − n2 J0 + 2xJ0′ + x2 y2′′ + xy2′ +
x2 − n2 y2 = 0 The J0 terms multiplied by ln x vanish, and we are left with
326
Appendix
x2 y2′′ + xy2′ + x2 − n2 y2 = −2xJ0′ = −2
2j−1j x2j
4j−1j + 1 x2j
=
2j
22j j!2
j = 0 2 j!j + 1
j =1
We can now use a Taylor series to find the particular integral of this equation.
A.4.2
Mathieu Equations
2
d y
2
The solution of Mathieu’s equation dx
2 + p 1 − 2 cos 2x y = 0 for general values of
p and is not periodic and is therefore not of much interest. Numerical computation
is possibly the best option for solution. However, on each of the boundaries separating
the stable and unstable regions, there is one and only one periodic solution. Referring
to the Strutt diagram presented in Chapter 3, the solution on these boundaries are given
by the following:
p an even integer, period
:
ce2p x p =
A2n p cos 2nx
a2n
n=0
2
p2 A0 − p2 A2 = 0
p − 4 A2 − p2 2A0 + A4 = 0
2
p − 4n2 A2n − p2 A2n−2 + A2n+2 = 0 n = 2 3 4
se2p+2 x p =
n=0
B2n+2 p sin2n + 2x
2
p − 4 B2 − p2 B4 = 0
2
p − 4n2 B2n − p2 B2n−2 + B2n+2 = 0
(A.4.2)
b2n
n = 2 3 4
p an odd integer, period 2 :
ce2p+1 x p =
n=0
A2n+1 p cos2n + 1x
2
p − 4 A2 − p2 2A0 + A4 = 0
p2 − 2n + 12 A2n+1 − p2 A2n−1 + A2n+3 = 0
se2p+1 x p =
n=0
B2n+1 p sin2n + 1x
2
p − 1 B1 − p2 B3 − B1 = 0
p2 − 2n + 12 B2n+1 − p2 B2n−1 + B2n+3 = 0
a2n+1
n = 1 2 3
(A.4.3)
b2n+1
n = 1 2 3
There are several notations and conventions that are used in the literature and differences
in the value for the starting coefficient in each series. The notation chosen here is to
emphasize that in the Strutt diagram the curves denoted by a2n and b2n intersect for
= 0, as do the curves denoted by a2n+1 and b2n+1 .
To determine the boundary that separates stable from unstable regions, we are
interested in the shape of the dividing curve rather than the solution as a function of
the independent variable x. To determine this boundary curve, realize that if we use the
recursion relations to set up the infinite set of equations needed to find the An ’s and
A.4
327
Solution of Ordinary Differential Equations
Bn ’s, the equations can be solved unless the determinant of the coefficients vanishes.
Setting this determinant to zero is in fact what is needed to determine this curve!
This is illustrated for the function ce2p . From the recursion relations,
p2 A0 − p2 A2 = 0
−2p2 A0 + p2 − 4 A2 − p2 A4 = 0
−p2 A2 + p2 − 16 A4 − p2 A6 = 0
−p2 A4 + p2 − 36 A6 − p2 A8 = 0
Setting the determinant
2
p
−2p2
0
0
· · ·
of the coefficients for this function to zero thus gives
−p2 0
0
0
0 · · ·
p2 − 4 −p2 0
0
0 · · ·
−p2 p2 − 16 −p2 0
0 · · · = 0
0
−p2 p2 − 36 −p2 0 · · ·
···
···
···
···
· · · · · ·
Similar operations can clearly be carried out on the other three functions.
Notice that the terms on the diagonal of the determinant are growing at a rapid rate!
This could have been avoided by dividing each equation by a suitable constant (perhaps
the terms which appear on the diagonal) before forming the determinant.
A.4.3
Finding Eigenvalues—The Riccati Method
When using techniques such as separation of variables and when dealing with equations
such as Sturm-Liouville systems, determinations of the eigenvalues and eigenfunctions
are a key part of the solution. Indeed, in many engineering situations the determination
of frequencies involved in flow instabilities, natural frequencies of vibrating structures,
acoustics, and many other areas, the determination of frequency is the sole object.
As discussed in Chapter 11, the Riccati method is a powerful tool that is available
for determining the characteristic frequency1 of a system of differential equations. The
technique exhibited by examples in Chapter 11 will be explained more generally here.
For further discussion, see Scott (1973a and b) and Davey (1977).
Start with the n by n system of linear ordinary differential equations
y′ = Fy
(A.4.4)
where y and y′ are column vectors of order n, and F is an n by n matrix. Suppose that
m less than n of the boundary conditions are known at x = 0, and p = n − m are known
at x = L. Being an eigenvalue problem, these are necessarily homogeneous boundary
conditions.
Since the system will have n independent solutions for y, represent the family of
the n solutions y, which are involved in the boundary conditions at x = 0 by u (of
size m) and the remaining of the y by v (of size p). In general form, these boundary
conditions are expressed as
Gu = 0 at x = 0
Hu = Jv at x = L
(A.4.5)
1
The terms characteristic frequency, natural frequency and eigenfrequency are interchangeable. Eigen is
the German word for “characteristic.”
328
Appendix
G is a matrix of order m by n H of order p by m, and J of order p by p.
The method starts by rewriting the original system in the form
u′ = Au + Bv,
v′ = Cu + Dv
(A.4.6)
where A, B, C, D are matrices of order m by m m by p p by m and p by p, respectively.
Also, there will be a matrix R of order m by p such that
u = Rv
(A.4.7)
It follows from this that there exists the inverse relationship
v = Su
where S = R−1
(A.4.8)
is the matrix inverse of R. Differentiation of equation (A.4.6) and use of equation
(A.4.7) gives
u′ = Au + Bv = AR + B v
= Rv′
= R′ v + Rv′ = R′ v + RCu + Dv
= R′ + RCR + RD v
Since the equation is true for any v, it follows that
R′ = B + AR − RD − RCR
(A.4.9)
This must satisfy the boundary condition R0 = 0. At the second boundary,
HR − J v = 0 at x = L. This is satisfied providing
det HR − J = 0
(A.4.10)
Generally, it is easiest to set all parameters in F and integrate equation (A.4.9) until
equation (A.4.10) is satisfied. This gives the length at which that set of parameters gives
the correct eigenvalues. Frequently scaling can be performed so that the situation can
be changed to the more usual one of starting with a given L.
In the course of integrating equation (A.4.9), it is not unusual to find that R becomes
singular before the condition of equation (A.4.10) is met. Although it is possible to go
to a complex path of integration, it has often been found easier to switch to the inverse
problem as R becomes large. Repeating the process used in developing equation (A.4.9),
but this time starting with equation (A.4.8) instead of equation (A.4.7), the result is
S′ = C + BS − SA − SBS
(A.4.11)
Thus, at some point the switch is made from the R to S equations, and then after the
singularity in R (zero in S) is passed, the switch back to the R equations is made.
The expense in programming and computation time is small, requiring some logical
decisions and the inversion of the matrix. Runge-Kutta methods tend to perform well
on these problems. There have been reports in the literature that finding eigenvalues
higher than the first has produced difficulties and that backward integration to determine
eigenfunctions has proven to be unstable.
A.5
329
Index Notation
A.5 Index Notation
Considerable simplicity in writing physical equations is obtained by use of vector
notation. Vector notation has the ability to express briefly very powerful ideas, the
brevity of the notation in many cases being of great help in understanding terms in
the equation. Nevertheless, there are times when vector notation is not useful, either
when the quantities of interest are of a more general nature or when you want to deal
with an actual problem, when it is always necessary to deal with the components of the
vector rather than the vector itself. Index notation has been invented to accommodate
both of these cases. It is an adaptation of matrix notation and for physical problems
where the indexed quantities are tensors is the same as the notation of Cartesian tensors.
Students familiar with matrix algebra and/or computer programming languages such as
FORTRAN will already be familiar with many of the concepts, as it is used in most
programming languages.
To motivate the notation, consider as an example the vector equation F = ma. In
component form, this becomes, in x y z Cartesian coordinates,
Fx = max
Fy = may
Fz = maz
(A.5.1)
Now the three equations in component form just written are similar except for the
subscript. The fundamental information is displayed in the first equation, with the
remaining equations just giving further detail. In this example, the equations have few
terms, and there is no objection to the extra work involved in writing three equations
when the first one gives the sense of the rest. Of course, one could write
(A.5.2)
Fx = max
and add “and similarly for the y and z direction,” but it can become confusing if this
phrase must be written after each equation. Also, if the equation is complicated, perhaps
this brief description will not convey all the necessary information, and it will be
necessary to give further verbal description. And, certainly, computer programs would
prefer to deal with numbers rather than alphabets.
In looking at the component equations again, we can see that the need to write the
three equations was due to the original labeling of the axes through the use of different
letters. Instead of using alphabetical names, it is more convenient to adopt number
names for the coordinate axes, such as x1 x2 x3 . Then, Fx = Fx1 Fy = Fx2 Fz = Fx3 , or,
more briefly, since the “x” is now common to all three axes, Fx = F1 Fy = F2 Fz = F3 .
Then the component equations become
F1 = ma1
F2 = ma2
F3 = ma3 ,
(A.5.3)
or, taking full advantage of the new notation,
Fi = mai
i = 1 2 or 3.
(A.5.4)
Thus, one equation stands for three equations, as in vector language. Since the
phrase “i = 1 2, or 3” will usually be there, it is the custom to omit it and add a
comment only when this phrase is not true.
To tie this concept more closely to vector analysis, let us adopt the notation g1 g2 g3 ,
for the base vectors usually denoted by i, j, k. Then
3
F=
g i Fi
i=1
(A.5.5)
330
Appendix
Now, in (A.5.4), i appeared exactly once in every term of the equation. It is a free
index—that is, it can take on three different values at the user’s choice. In equation
(A.5.5), i appears as a repeated, or dummy, index and takes on the values denoted by
the summation sign. Now, in many applications where index notation is used, whenever
an index is a dummy index, it appears twice and only twice in that term and always with
the summation sign. To make full use of this, henceforth the summation convention
(attributed to A. Einstein) will be adopted. That is, whenever an index appears twice in
a term, summation on that index is implied. Thus, equation (A.5.1) could be written as
(A.5.6)
F = gi Fi
A repeated (dummy) index will always imply summation unless otherwise explicitly
stated.
Exercise A.5.1 Index notation
Write out the following in full component form:
A = B i Ci
Ai = Bi Cj Dj
Ai = Bi Cjj
Ai = Bij Cj
Ai = Bijj + Cik Dk
A2 = A i A i
Solution.
A = Bi Ci ⇒ A = B1 C1 + B2 C2 + B3 C3
Ai = Bi Cj Dj ⇒ A1 = B1 C1 D1 + C2 D2 + C3 D3
A2 = B2 C1 D1 + C2 D2 + C3 D3
A3 = B3 C1 D1 + C2 D2 + C3 D3
Ai = Bi Cjj ⇒ A1 = B1 C11 + C22 + C33
A2 = B2 C11 + C22 + C33
A3 = B3 C11 + C22 + C33
Ai = Bij Cj ⇒ A1 = B11 C1 + B12 C2 + B13 C3
A2 = B21 C1 + B22 C2 + B23 C3
A3 = B31 C1 + B32 C2 + B33 C3
Ai = Bijj + Cik Dk ⇒ A1 = B111 + B122 + B133 + C11 D1 + C12 D2 + C13 D3
A2 = B211 + B222 + B233 + C21 D1 + C22 D2 + C23 D3
A3 = B311 + B322 + B333 + C31 D1 + C32 D2 + C33 D3
2
2
A = A = Ai Ai ⇒ A2 = A1 A1 + A2 A2 + A3 A3
A.5
331
Index Notation
Since many of the quantities used in mechanics are vectors, it is helpful to see
how the new notation can be used to represent familiar quantities and operations. By
inspection,
grad = = gi
xi
div A = · A =
Ai
xi
(A.5.7)
and
A · B = A i Bi
In working with the cross product and curl, we can see that the application of
index notation is not as obvious. In fact, it is necessary to introduce a new symbol, the
e symbol or alternating tensor, to accomplish this operation. The components of the
alternating tensor are defined by
⎞
⎛
+1 if i j k is an even permutation of 1,2,3,
⎟
⎜
eijk = ⎝ −1 if i j k is an odd permutation of 1,2,3, ⎠
(A.5.8)
0 otherwise.
Thus,
e123 = e231 = e312 = 1
e132 = e213 = e321 = −1
e111 = e112 = e113 = e121 = · · · = 0.
Comparison of components then indicates
A × B = g i eij k Aj Bk
(A.5.9)
In particular,
× B = curl B = g i eij k
Bk
xj
(A.5.10)
Another symbol that is of use is the Kronecker delta, with components ij ,
defined by
1 if i = j
ij =
.
(A.5.11)
0 otherwise
Thus, 11 = 22 = 33 = 1 12 = 21 = 23 = 32 = 13 = 31 = 0. In matrix terminology,
the Kronecker delta forms the components of the identity matrix.
Notice that since we are dealing with base vectors that are orthogonal to one another
and are each of unit length, then
gi · gj = ij
Contraction is the process of multiplying a quantity by the Kronecker delta and
then summing on one of the indices. For instance,
Bij jk = Bik
Bij ji = Bii .
(A.5.12)
All of the operations of vector and matrix analysis (and many more as well) can now
be written directly in index notation.
332
Appendix
Notice that for the Kronecker delta, the value of the component is independent of
the order in which the indices are written—that is, ij = ji . Such a quantity is said to be
symmetric in those indices. Similarly, the components of the alternating tensor change
sign if two indices are interchanged. That is, eijk = −ejik . Such quantities are said to be
skew-symmetric in those indices. Note that any quantity with two or more indices can
be written as the sum as a symmetric and skew-symmetric form by the decomposition
Aij = 05 Aij + Aji + Aij − Aji .
Exercise A.5.2 Verify the following expressions by expanding them:
ii = 3
ij Aj = Ai
eijk e1jk = 2i1
Solution.
ii = 11 + 22 + 33 = 1 + 1 + 1 = 3
ij Aj = i1 A1 + i2 A2 + i3 A3 Thus, since
1j Aj = A1
2j Aj = A2
3j Aj = A3
the expression can be summarized as ij Aj = Ai .
eijk e1jk = ei12 e112 + ei13 e113 + ei23 e123 + ei21 e121 + ei31 e131 + ei32 e132
= 0 + 0 + ei23 e123 + 0 + 0 + ei32 e132
= 2ei23 e123 = 2i1
As a last bit of notation, since the typing of a partial differential is awkward on a
typewriter or even on a word processor, a comma is frequently used to denote partial
differentiation, the subscript after the comma denoting that coordinate is involved in the
differentiation. That is,
A—
= A—k
xk
(A.5.13)
As a final example of the use of index notation, the divergence and curl theorems
will be shown in index notation. With V as an arbitrary volume, and S the surface of
V , the divergence theorem becomes
A
i
Ai ni dS
(A.5.14)
dV =
V xi
S
Similarly, take surfaces drawn in space with the boundary curve on the periphery
of S ′ and n as the unit normal to S ′ . Further, take t as the unit tangent to C, with
integration around C performed in the direction of t and n chosen so that on C n × t
points inward toward S. Then, if the derivatives exist in the regions chosen, and if Ai
are single-valued, the curl theorem is
A
ni eijk k dS ′ =
Ai ti ds
(A.5.15)
xj
S′
C
A.6
333
Tensors in Cartesian Coordinates
A.6 Tensors in Cartesian Coordinates
In fluid mechanics, as indeed in most branches of physics, we are dealing with fundamental quantities in space that have properties such as magnitude and direction.
Familiar examples are scalars (e.g., density that has magnitude but no direction), vectors
(e.g., velocity that has magnitude and direction) and more general quantities (e.g., stress
that has magnitude and two directions, one for the force and the second for orientation
of the area). The generic term that contains all of these categories is a tensor. Thus, a
scalar is a tensor of order zero, a vector is a tensor of order one, and stress is a tensor
of order two. The order of a tensor refers to the number of directions that we associate
with the physical quantity. While any order is possible, tensors of order greater than
two are rare and usually arise as the derivatives of lower-order tensors.
Tensors are written in the following fashion:
=
v = vi gi
= ij gi gj
zero order tensor
(A.6.1)
(first order tensor)
(A.6.2)
(second order tensor)
(A.6.3)
For the second-order tensor, we have two base vectors, one for each direction, and
they appear naturally as a product in the right-hand side. (Notice that the appearance
of two dummy indices means that there are two summations present, one on i and one
on j.) Since we have not put a dot or a cross between the two base vectors, the product is
neither a dot product nor a cross product but rather an indefinite product. (It’s not really
indefinite because we know what the two directions mean. The terminology here is just
to identify this new type of product as having a name different from the other two.)
In dealing with tensors of any order, it is convenient to deal with just the components. A test to determine whether a set of quantities are in fact the components of a
tensor is to observe how these quantities transform as we rotate axes. (Note: A fundamental point that is all too easily forgotten in learning tensor analysis is that the tensor
itself does not change as we rotate the coordinate axes, only the components of the
tensor change, since they, and not the tensor itself, are axis-related.) To see this, at a
point in space introduce two Cartesian coordinate systems, x and y, one being obtained
from the other by a rigid rotation of axes. Then yi = aij xj , and
xj
y
= i = aij = aji = gi y · gj x
yi
yi
(A.6.4)
where the aij are the direction cosines of one set of axes with respect to the other, and
the letters in parentheses following the unit base vectors tell the reference frame they
refer to. By virtue of the orthogonality of the axes,
gi y · gj y = gi x · gj x = ij
(A.6.5)
By the Pythagorean theorem, and the fact that the angle between the axes xi and yk are
the same,
aik ajk = aki akj = ij
(A.6.6)
Since any tensor T must be independent of the choice of axes, then Ty = Tx, where
Ty means the tensor quantity T referred to the y axes and similarly for T(x). By
equation (A.6.4) the base vectors transform according to
g i y = aij gj x
(A.6.7)
334
Appendix
and so for a tensor of any order
= Tij··· xaim gm xajn gn x
Ty = Tij··· ygi ygj y
But
Tx = Tij··· xgi xgj x
Thus, to have Tx = Ty—that is, to have the tensor independent of the coordinate
system—it is necessary then that the components transform according to
Tmn— x = aim ajn —Tij —
or, multiplying both sides by direction cosines, contracting, and using equation (A.6.5)
several times,
Tij — y = aim ajn —Tmn— x
(A.6.8)
When there are no subscripts (that is, the tensor is a scalar), then the direction cosines
disappear from equation (A.6.8).
To emphasize this transformation law for the orders used the most, we write the
following transformation laws:
Zero order tensor Ty:
Tx = Ty
First order tensor components Tj y:
Ti x = aim Tm y
Second order tensor components Tij y:
Tij x = aim ajn Tmn y
Third order tensor components Tijk y:
Tijk x = aim ajn ako Tmno y
(A.6.9)
(A.6.10)
(A.6.11)
(A.6.12)
For a second-order tensor, if the third axis is not rotated, then
ai3 = a3i = i3
Then the equation
Tij x = aim ajn Tmn y
i = 1 2
j = 1 2
is the equation of a circle, called Mohr’s circle. It affords an easy graphical interpretation
of the transformation law in two dimensions.
Tensor components can have symmetry properties. If, for instance, for a secondorder tensor the components obey the rule
Tij = Tji
(A.6.13)
it is said that the second order tensor is symmetric. The Kronecker delta is an example
of a symmetric tensor. If, on the other hand, the components obey the rule
Tij = −Tji
(A.6.14)
it is said that the second order tensor is antisymmetric, or skew symmetric. Any secondorder tensor can be written as the sum of a symmetric part and a skew symmetric
part—that is,
Tij = 05Tij + Tji + 05Tij − Tji
(A.6.15)
The first two terms are clearly symmetric, and the last two are clearly skew symmetric.
Components of tensors of order higher than two can be symmetric or skew symmetric
in any pair of their indices.
A.6
335
Tensors in Cartesian Coordinates
Example A.6.1 Unit vectors
Show that gi x = aji gj y, and thus verify equations (A.6.4) and (A.6.6).
Solution. This result is simply the familiar decomposition of a vector into its components. In the y-axes, the three unit vectors ei x make angles with the axes y1 y2 , and
y3 , whose direction cosines are a1i a2i , and a3i . Applying the decomposition law, the
result follows.
Example A.6.2 Direction cosines
Show by a suitable dot product that the aij are direction cosines of the vectors gi y
referred to the gi x vectors.
Solution. Taking the dot product gi x · gj y and using the results of Example
A.6.1, we have gi x · gj y = aki gk y · gj y = aki kj = aji by the orthogonality of the
y-axes. QED.
Example A.6.3 Tensor properties of the Kronecker delta and alternating tensors
Show that the alternating tensor and the Kronecker delta are tensors in a Cartesian
coordinate system.
Solution. Assume that the Kronecker delta is a second-order tensor. From the transformation law for its components equation (A.6.11),
ij x = aim ajn mn y
From the summation property of the Kronecker delta,
ij x = aim ajn mn y = aim ajm = ij x
Thus, the transformation law is valid.
Assume that the alternating tensor is a third-order tensor. From the transformation
law for its components equation (A.6.12),
eijk x = aim ajn akp emnp y
By expansion of both sides, it is seen that
aim ajn akp emnp y = eijk y
times the determinant of a.
By the orthogonality of the axes, the determinant of the direction cosines is one. Thus,
eijk x = eijk y
and the transformation law is valid.
Besides the inherent invariant property of a tensor, tensor components also possess
invariant properties. For instance, Ai Ai is the same in any coordinate system, as can be
easily seen in equation (A.6.6) and in fact is the square of the magnitude of the vector
A. This is the only independent invariant that can be formed for a first-order tensor. For
a second-order tensor, there are at most three independent invariants:
Tii = IT
(A.6.16)
336
Appendix
05IT2 − Tij Tji = IIT
determinate of Tij = IIIT
(A.6.17)
(A.6.18)
That these are invariants can be shown through use of the transformation laws.
Example A.6.4 Invariants
Verify that equation (A.6.9) is invariant under rotation of axes.
Solution. To do this, we use the transformation of components relation (equation
(A.6.11))—that is,
Tij x = aim ajn Tmn y Then
Tii x = T11 x + T22 x + T33 x = a1m a1n Tmn y + a2m a2n Tmn y + a3m a3n Tmn y
= a1m a1n + a2m a2n + a3m a3n a1p a1q + a2p a2q + a3p a3q Tpq y
= aip aiq Tpq y
Tij x = aim ajn Tmn y Then
Tii x = T11 x + T22 x + T33 x = a1m a1n Tmn y + a2m a2m Tmn y + a3m a3n Tmn y
= a1m a1n + a2m a2n + a3m a3n Tmn y = aim ain Tmn y
Tij x = aim ajn Tmn y Then
Tii x = T11 x + T22 x + T33 x = a1m a1n Tmn y + a2m a2n Tmn y + a3m a3n Tmn y
= a1m a1n + a2m a2n + a3m a3n Tmn y = aim ain Tmn y
But from equation (A.6.6),
aim ain = mn Tii x = mn Tmn y = Tmm y Thus Tii x = Tii y = IT
From the preceding result, to prove 05IT2 − Tij Tji = IIT is invariant, we need only
show that
Tij xTji x = Tij yTji y
Using equation (A.6.11), we have
Tij xTji x = aim ajn Tmn yajp aiq Tpq y
= aim aiq ajn ajp aiq Tmn yTpq y after rearranging
= mq np Tmn yTpq y using equation A66
= Tqn yTnq y.
This is the desired result.
For the third result, we note from writing out
T11 T12
determinate of Tij x = T21 T22
T31 T32
the result in detail that
T13 e e T T T
ijk pqr ip jq kr
T23 =
6
T33
Using the transformation law (equation (A.6.11)),
A.7
337
Tensors in Orthogonal Curvilinear Coordinates
eijk epqr Tip xTjq xTkr x
= eijk epqr ais apt Tst yaju aqv Tuv yakw arz Twz y
= eijk epqr ais apt aju aqv akw arz Tst yTuv yTwz y.
As can be seen by writing out both sides,
eijk ais aju akw = esuw
epqr apt aq var z = etvz
determinant of a and
determinant of a
But the determinant of a is one, since the coordinates are orthogonal. Then
eijk epqr Tip xTjq xTkr x = esuw etvz Tst yTuv yTwz y
= esuw esuw times the determinant of Ty
= 6 times the determinant of Ty
that is the desired result.
That there are at most three invariants of a second-order tensor follows from the
Cayley-Hamilton theorem. This theorem states that the products of components of a
tensor are related by
Tim Tmn Tnj − IT Tim Tmj + IIT Tij − IIIT Tij = 0.
(A.6.19)
Proof of the Cayley-Hamilton theorem is as follows. By contraction (letting i equal j
and summing),
Tim Tmn Tni = IT IIT = IT IT2 − 2IIT − IIT IT + 3IIIT = IT3 − 3IT IIT + 3IIIT
(A.6.20)
Premultiplication of equation (A.6.10) by Tjk or higher powers of T with the implied
contraction shows the desired result. When there is only one subscript, equation (A.6.8)
is the familiar form for the decomposition of a vector. When there are two subscripts,
equation (A.6.8) is the three-dimensional generalization of Mohr’s circle.
A.7 Tensors in Orthogonal Curvilinear Coordinates
The mathematics of tensors in orthogonal curvilinear coordinates is only a bit more
complicated than that in Cartesian coordinates, although the application of the general
results to a particular coordinate system can be tedious. In the following, we use
orthogonal coordinate systems with coordinate axes y1 y2 y3 and with orthogonal base
vectors
(A.7.1)
gi = ds/dyi
where ds is an elemental distance. Define
√
√
h1 = g1 · g1 h2 = g2 · g2
h3 =
√
g3 · g3
(A.7.2)
so that the distance between two infinitesimally distant points is
ds2 = h1 dy1 2 + h2 dy2 2 + h3 dy3 2
(A.7.3)
338
Appendix
It is usually easier to use equation (A.5.4) to find the hs than equation (A.5.3) for a
particular coordinate system.
The gradient and Laplacian of a vector are given by
1 1 1
(A.7.4)
grad = =
h1 x1 h2 x2 h3 x3
1
h2 h3
h3 h1
h1 h2
2 =
+
+
(A.7.5)
h1 h2 h3 x1
h1 x1
x2
h2 x2
x3
h3 x3
The divergence of a vector is given by
h2 h3 V1 h3 h1 V2 h1 h2 V3
1
+
+
div V = · V =
h1 h2 h3
x1
x2
x3
(A.7.6)
The curl of a vector is given by
h3 V3 h2 V2
1
h1 V1 h3 V3
1
−
−
i+
j
curl V = × V =
h2 h3
x2
x3
h3 h1
x3
x1
(A.7.7)
h2 V2 h1 V1
1
−
k
+
h1 h2
x1
x2
Equations (A.7.6) and (A.7.7) are useful in expressing the continuity equation and
the vorticity vector in orthogonal curvilinear coordinates, and equation (A.7.5) for the
continuity equation for an irrotational flow. For the rate of deformation the expressions
for the tensor components are
1 v1 v2 h1 v3 h1
+
+
d11 =
h1 x1 h2 h2 h3 h3
1 v2 v3 h2 v1 h2
d22 =
+
+
h2 x2 h3 h3 h1 h1
1 v3 v1 h3 v2 h3
d33 =
+
+
h3 x3 h1 h1 h2 h2
(A.7.8)
h3
v3
h2
v2
2d12 = 2d21 =
+
h2 x2 h3
h3 x3 h2
h
v1
h
v3
2d23 = 2d32 = 1
+ 3
h3 x3 h1
h1 x1 h3
v2
h1
v1
h2
+
.
2d31 = 2d13 =
h1 x1 h2
h2 x2 h1
The stress components are given by
11 = −p + ′ · v + 2d11
33 = −p + ′ · v + 2d33
12 = 21 = 2d12
22 = −p + ′ · v + 2d22
23 = 32 = 2d23
31 = 31 = 2d31
or in index notation ij = −p + · vij + 2dij
′
(A.7.9)
A.7
339
Tensors in Orthogonal Curvilinear Coordinates
The acceleration is arrived at by first writing the convective terms in a correct
vector form. As can be verified by writing out the components in Cartesian coordinates
and comparing, that form is
v · v = −v × × v + 05 v2 = −v × curl v + 05grad v2 .
(A.7.10)
The first term on the right is found in an arbitrary orthogonal system by applying
equation (A.7.7), the second by use of equation (A.7.4).
The continuity equation is given by
1
h2 h3 v1 h3 h1 v2 h1 h2 v3
+
+
= 0
(A.7.11)
+
t h1 h2 h3
y1
y2
y3
Since the Navier-Stokes equations involve taking the second derivative of a vector,
necessary for the viscous terms, the procedure is much longer and more tedious than for
the continuity equation or the Laplace equation. The acceleration terms can be found
from equation (A.7.10), and the pressure gradient from equation (A.7.4). The remaining
terms involve the operation 2 v. Writing this out is a long process and would require
several pages of equations. Instead, note that
2 v = · v − × × v
(A.7.12)
and equations (A.5.7) and (A.5.8) tell us how to compute the divergence and curl. The
Navier-Stokes equations then are of the form
v
1 2
−v×+
= −p + g + · v − ×
(A.7.13)
v
t
2
To illustrate the previous, consider the next two familiar coordinate systems.
A.7.1
Cylindrical Polar Coordinates
The element of length in this coordinate system is
ds2 = dr 2 + rd2 + dz2
with
y1 y2 y3 = r z.
The polar coordinates are expressed in terms of Cartesian coordinates by r = x12 + x22
= tan−1 x2 /x1 z = x3
Comparing equation (A.7.10) with equation (A.7.3), we see that h1 = h3 = 1 h2 =
y1 = r. Then
2 =
1
r r
1 2 2
r
+ 2 2 + 2
r
r
z
1 rVr 1 V Vz
+
+
r r
r
z
1 Vz V
Vr Vz
1 rV Vr
×V =
−
−
−
er +
e +
ez
r
z
z
r
r
r
·V =
(A.7.14)
(A.8.15)
(A.8.16)
340
Appendix
1 v u
w
+ dzz =
r r
z
v 1 u
u w
2dr = 2dr = r
+
2drz = 2dzr =
+
r r
r
z r
1 w v
+
2dz = 2dz =
r z
y
r 2 = x2 + y2 = tan−1 z = z
x
drr =
u
r
d =
(A.8.17)
The continuity equation is then
1 ru 1 v w
+
+
+
= 0
t r r
r
r
(A.7.18)
The Navier-Stokes equations for constant density and viscosity are
u v u
u v2
u
2 v
p
u
2
+u +
+ w − 2 = − + gr + u − 2 − 2
t
r r
z r
r
r
r
v v v
v uv
v
2 u
1 p
v
+u +
+w + 2 = −
+ g + 2 v − 2 + 2
t
r r
z r
r
r
r
(A.7.19)
w
w v w
w
p
+u
+
+w
= − + g + 2 w
t
r r
z
z
where the Laplacian 2 is as given in equation (A.8.11) and u v w = vr v vz .
A.7.2
Spherical Polar Coordinates
The element of length in this coordinate system is
ds2 = dR2 + Rd2 + R sin d2 with y1 y2 y3 = R and so
h1 = 1 h2 = y1 = R h3 = y1 sin y2 = R sin .
In terms of Cartesian coordinates, R2 = x12 + x22 + x32 = cos−1
x
x3
= tan−1 2 . Then
R
x
1
1
e +
e +
e
R R R R sin
2
1
1
1
R2
+ 2
sin
+
2 = 2
R R
R
R sin
R2 sin2 2
=
(A.7.20)
(A.7.21)
1 v sin
1 w
1 R2 u
+
+
(A.7.22)
2
R R
R sin
R sin
1
w sin v
1
u wR sin
×v =
−
−
eR +
e
R sin
R sin
R
1 Rv u
−
e
(A.7.23)
+
R
R
·v =
dRR =
u
R
d =
1 v u
+
R R
d =
1 w u v cot
+ +
R sin R
R
A.8
341
Tensors in General Coordinates
w
1 v
+
sin
R sin
1 u
w
2dR = 2dR =
+R
R sin
R R
v 1 u
2dR = 2dR = R
+
R R
R
2d = 2d =
The continuity equation is given by
1 w
1 R2 u
1 v sin
+
+
+
= 0
t R2
R
R sin
R sin
(A.7.24)
(A.7.25)
The Navier-Stokes equations for constant density and viscosity are
u v u
w u v2 + w2
u
+u
+
+
−
t
R R R sin
R2
p
2u
2 v 2v cot
2
w
=−
+ gR + 2 u − 2 − 2
−
−
R
R
R
R2
R2 sin
v
v v
w v uv − w2 cot
v
+u
+
+
+
t
R R R sin
R2
1 p
v
2 u
2 cot w
2
=−
+ g + v − 2
+
−
(A.7.26)
R R
R sin R2 R2 sin
w v w
w w uw + vw cot
w
+u
+
+
+
t
R R R sin
R2
2
1 p
w
2 cot v
u
=−
+ g + 2 w − 2 2 + 2
+ 2
R sin
R sin R sin R sin
where the Laplacian 2 is as given in equation (A.8.19) and u v w = vR v v .
A.8 Tensors in General Coordinates
In Cartesian coordinates, dealing with tensors is particularly simple because the base
vectors are constants and thus for a vector (as an example)
V = Vj gj
(A.8.1)
In general coordinates, the base vectors vary from point to point, and when differentiating
a tensor, the base vectors have to be differentiated as well. This leads to greater
complications than might at first be expected, since more than one set of base vectors
can be defined and their direction cosines will generally differ. One starts to deal
with what are called contravariant and covariant base vectors and tensor components,
depending on the type of base vector that is used. Further, the various tensor components
and base vectors will generally differ in their physical dimensions. (For example, in
cylindrical polar coordinates two of the covariant components of the velocity vector will
have dimensions of length/time, while the third will have dimensions of just reciprocal
time.) In our work, we are interested in dealing with the physical components—that
342
Appendix
is, components that all have the same physical dimensions. Unfortunately, physical
components do not obey the simple transformation laws of tensors.
The starting point for a study of general tensors is geometry. As in studying rates of
deformation, consider two adjacent points separated by a vector distance dr. (Boldface
will be used to denote both vectors and tensors of higher order.) If we have a coordinate
system x1 x2 x3 , we can write
dr =
r i
dx
xi
(A.8.2)
This implies that the vectors r/xi are locally tangent to the x coordinate system. So
a good choice for base vectors tangent to our coordinate system is
gi =
r
xi
(A.8.3)
Since the magnitude ds of the distance between the two neighboring points is
given by
ds2 = dr · dr = gi dxi · gj dxj = gi · gj dxi dxj = gij dxi dxj
(A.8.4)
the gij given by gij = gi · gj is suitable as the covariant components of the symmetric
metric tensor. Depending on the coordinate system, the dimensions of the various gij
components can vary. In spherical polar coordinates, for instance, one has dimensions
of length, while two are dimensional.
Returning to equation (A.8.4), we could have written
ds2 =
s s i j
dx dx
xi xj
(A.8.5)
Comparison with equation (A.8.4) implies that an alternate definition of the covariant
metric tensor components is
gij =
s s
xi xj
(A.8.6)
If we introduce a second coordinate system y at the same point we are considering,
notice that using the product rule of calculus, we can write
dr = gi xdxi =
r xi j
dy = gj ydyj
xi yj
Thus,
gj y =
xi
g x
yj i
(A.8.7)
This is the transformation law for covariant components of a vector. In general, the
transformation law for covariant components of a tensor of order n is given by
A··· y =
xi xj xk
· · · Aijk··· x
y y y
(A.8.8)
where there are n subscripts on the A and n partial derivatives. The partial derivatives
act like direction cosines.
We have assumed here that a metric measuring the distance between two neighboring points exists in our space. Such a space is called a Riemannian space. Also,
A.8
343
Tensors in General Coordinates
our discussion is limited to three-dimensional spaces. Nonmetric spaces exist, but their
applicability to fluid mechanics is doubtful. Generalization to higher dimensions is
trivial.
Contravariant base vector components can also be defined using cross-products and
the alternating tensor. Letting
gi × gj = eijk gk
(A.8.9)
this definition serves the purpose of having the new base vectors orthogonal to the old,
but they will not necessarily be tangent to the coordinate system. Equation (A.8.10) can
be solved to explicitly give the contravariant base vectors as
√
√
√
g1 = g2 × g3 / g g2 = g3 × g1 / g g3 = g1 × g2 / g
(A.8.10)
√
where g = g1 · g2 × g3 g = detgij
In a manner similar to that just used, the components of the symmetric conjugate metric
tensor are defined by
g ij = gi · gj
(A.8.11)
The general transformation law for contravariant components is given by
y y y
· · · Aij k··· x
(A.8.12)
xi xj xj
What is the relationship between the contravariant and covariant tensor components?
With g denoting the determinant of gij as previously, let ij denote the cofactor of gij .
From equation (A.8.10) we have g ij = ij /g, so
A··· y =
gij g jk = gij jk /g = ki
(A.8.13)
where the are the components of the Kronecker delta. Thus, the contravariant gs play
the role of the inverse of the covariant gs. It can be shown in a similar fashion that
gi = gij gj
and gi = g ij gj
(A.8.14)
Vi = gij V j
and V i = g ij Vj
(A.8.15)
and V ij = g ik g jl Vkl
(A.8.16)
and Vki = g jl Vkl
(A.8.17)
The metric tensors can also be used to raise and lower indices. Since it can be
shown from the transformation laws that V = Vi gi = V i gi , it follows that
Similarly for a second-order tensor,
Vij = gik gjl V kl
and so on for higher-order tensors. Mixed components could also be introduced by
operations such as
Vik = gil V lk
Notice that in all of the preceding, the superscripts are the same on both sides of
the equals sign, as are the subscripts. This must always be true for the equation to be
valid. For example, the invariants used in the Cayley-Hamilton theorem for second-order
symmetric tensors in general coordinates are
1
I = Aii II =
Aij Aji − I 2
2
1
j
2Aij Ak Aki − 3Aij Aji I + I 3 = det Aij
III =
6
344
Appendix
They can be used to find the inverse of A (the inverse is defined such that A−1 A = ) by
−1 k
A i = Aij Ajk − Aki I − ki II /III
Components of tensors obey all of the familiar associative and distributive laws.
With this somewhat tedious explanation out of the way, we can proceed to the
i
+
calculus of tensors. Starting with the vector A = Ai gi , it follows that dA = gi A
xj
j
i gi
A xj dx .
It must be that the derivatives of the base vectors can be written in terms of the
base vectors, so let
gi
=
g
ij
xj
We then have
i
Ai
A
i
+
A
g dxj = gi
dxj
dA = gi j + Ai
ij
j
x
xj
We see that the quantities in parentheses are a valid components of a tensor and therefore
define the covariant derivative of contravariant components by
Ai
i
(A.8.18)
Ai j = j + A
j
x
By a similar calculation, find that define the covariant derivative of covariant components by
Ai
(A.8.19)
Ai j = j − A
i
j
x
Contravariant derivatives are defined by Ai j = Ai k g kj .
The symbols introduced to relate the derivatives of the base vectors to the base
vectors themselves are called Christoffel symbols and are given by
k
k
Christoffel symbol of second kind
=
= i j g k
ij
ji
1 gik gjk gij
+ i − k
Christoffel symbol of first kind i j k = j i k =
2 xj
x
x
(A.8.20)
Christoffel symbols are not components of any tensor, but when combined with the
partial derivatives as in equations (A.8.18) and (A.8.19), they produce tensor components. They are responsible for the additional terms we have seen when working in
cylindrical and spherical coordinates. Since even with symmetry, there are possibly
21 different components to be calculated, when necessary, the wise person resorts to
symbolic processors for their calculation!
With this in hand, representing the Navier-Stokes equations in general coordinates
is elementary: Just replace partial derivatives by covariant or contravariant derivatives.
The continuity equation then becomes
i
+ v i = 0
t
(A.8.21)
A.8
345
Tensors in General Coordinates
and the momentum equation is
vi
+ vj vi j = ji j + g i
t
i
j = −p + ′ vk k ij + vi j + vj i
(A.8.22)
It is easy to write equations in general coordinates: Start with the equations in
Cartesian coordinates, replace partial spatial derivatives with covariant derivatives, and
pay attention to being consistent in subscripts and superscripts. And it is relatively easy
to operate on them and manipulate them, but reducing them to any given coordinate
system than the Cartesian one (the Christoffel symbols vanish in Cartesian coordinates)
requires a good deal of patience.
One convenient device for computation of the Christoffel symbols is to start with
equation (A.8.4) and to write it as
2
dxi
ds
(A.8.23)
= gij vi vj where vi =
T=
dt
dt
Next, compute
fi =
T
1 d T
−
2 dt vi
xi
(A.8.24)
d 2 xj
+ j k i vj vk
dt2
(A.8.25)
It can be shown that
fi = gij
Comparison of the coefficients of vj vk in equations (A.8.24) and (A.8.25) yield the
Christoffel symbols more conveniently than by computation directly from the definition.
An account of how to determine physical components of tensors can be found in
Truesdell (1953). Briefly, physical base vectors are defined by
gi = gi / gii = gij gj / gii
(A.8.26)
where the summation convention is suspended on indices enclosed in parentheses. For
a vector, then, since
V = V i gi = V i gii gi = Vj g ij gii gi
(A.8.27)
it follows that the physical components of V are given by
Vi = V i gii = Vj g ij gii
(A.8.28)
Physical components of higher-order tensors can be computed in a similar manner.
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355
Index
Acceleration
advective 5
Cartesian coordinates 30
convective 5
cylindrical coordinates 131
orthogonal curvilinear coordinates 336
spherical coordinates 343
temporal 5
Acoustic refrigerator 151
Added mass
cylinder 113
sphere 79
tensor 79
three-dimensional flows 112
two-dimensional flows 112
Adverse pressure gradient 183
Aeolian tones 107
Alternating direction methods 305
Analytic functions 113
Anti-symmetric see Skew-symmetric
Apparent mass 60
Auto-correlation 235
natural coordinate system 177
parabolic coordinate system 178
rotating flows 188
similarity solutions
axisymmetric jet 204
flat plate–Blasius 176
general form 213
stagnation point–Holmann 178
wedge–Falkner-Skan 175
theory 183
thermal effects see Thermal boundary layers
thickness 149
transformations
combined Falkner-Mises 187
Crocco 187
Falkner 185
Mangler 188
von Mises 186
Boussinesq approximation 220
Bulbous bow 132
Bulirsch-Stoer method 280
Burger’s equation 311
Butler sphere theorem 84
Bénard cells 226
Bernoulli equation 66
Bessel functions 227
Bingham fluid 40
Biot-Savart law 249
Blasius flow (plate) 232
Blasius theorem 113
Blobs, vortex 312
Boundary conditions 28
Boundary layer
Crank-Nicholson method 298
displacement thickness 174
equations 173
hybrid method 303
integral form 179
Kármán-Pohlhausen profile 200
momentum thickness 203
Cartesian coordinates 3
Cant 103
Cauchy integral theorem 113
Cauchy-Riemann equations 89
Cavity flow 110
Cayley-Hamilton theorem 40, 336
Characteristics 306
Characteristics, method of 306
Christoffel symbols 343
Circle theorem 115
Circular frequency 121
Circulation
conservation of 85
definition of 47
Closure problem 237
Cloud-in-cell method 313
Cnoidal function 134
356
357
Index
Cnoidal waves 134
Complex conjugate 104
Complex variables 89
Complex velocity potential 89
Conformal mapping 108–109
Conservation
energy 37
mass 1, 2
momentum 2, 13
Conservative form 18, 310
Constitutive equation 24
Continuity equation 5
Continuum 4
Control volume 5
Control surface 5
Convection see Thermal boundary layer
Convolution theorem 152
Coordinate system
cylindrical 9
parabolic 178
moving 41
spherical 12
Correlation 236, 237
Courant condition 298
Couette flow
circular cylinders 147
parallel plates 141
parallel plates–impulsive 147
Couette-Taylor instability 226
Crank-Nicholson method 298
Crocco’s transformation 187
Cubic spline 271
Curl
Cartesian coordinates 332
cylindrical coordinates 131
orthogonal curvilinear coordinates 336
spherical coordinates 343
Cylindrical coordinates 12
Deformation rates see Rate of deformation
Dilatational strain rate 22
Dimensional analysis 39
Dirichlet problem 285
Discharge 8, 9
Discrete vortex method 312
Dispersive waves 121
Displacement thickness 174, 176
Dissipation range 245
Dissipation function 37
Divergence 207, 220, 320
cylindrical coordinates 131
orthogonal curvilinear coordinates 336
spherical coordinates 343
Divergence theorem 2, 320
Doublet–dipole
three dimensional–point 81
two dimensional–line 59, 112
Downwash 102
Drag coefficient 39
Dufort-Frankel method 298
Duhamel’s superposition theorem 152, 198
Eddy scale 234
Eddy viscosity 237, 240, 241
Edge waves, 149, 160
Eikonal equation 138
Ekman boundary layer 189
Elliptic differential equations 174, 284
Elliptic integral, elliptic functions 134, 266
Energy
cascade 245
equation 36
specific 37
spectrum 245
Equation of continuity 5
Euler equations 173
Euler’s method 267, 281
Eulerian description 247
Exchange of stabilities 223
Falkner-Skan solution 175, 195
Finite element method 272
First law of thermodynamics 37
Flow separation see Separation
Flow stability see Stability
Forward time–centered space method 297
Fourier
fast transform 245
finite transform 246
integral 121, 214
law of heat conduction 26
transform 184
Fourier-space turbulent equations 244
Free convection see Thermal boundary layer
Free streamline 108
Free surface 28, 132–6
Frobenius, method of 324
Froude number 39
Gauss theorems 321
Golden mean 280
Gradient
Cartesian coordinates 344
cylindrical coordinates 131
orthogonal curvilinear coordinates 336
spherical coordinates 343
Grashof number 202
Green’s functions 51, 130
Green’s theorem 251
Ground effect 63
Group velocity 123
Hairpin vortex 249
Heat flux 37
HeleShaw flows 57
Helmholtz vector decomposition 319
358
Index
Helmholtz vorticity theorems 34
Hodograph plane 108
Homogeneous turbulence 240
Hyperbolic differential equations 284, 306
Ideal (perfect) gases 24, 26, 27
Images, method of 59
Impulsive motion of a plate–Stokes first 147
Incompressible flow 6
Indefinite product 13, 332
Index notation (Indicial notation) 3, 328
free indices 329
repeated (dummy) indices 329
summation convention 329
Induced drag 102
Induced velocity 104
Inertial subrange 245
Intermittency 249
Invariant imbedding method 274
Invariants 334
Inverse point 64
Inviscid flows 46
Inviscid irrotational flow examples
three-dimensional
circular cylinder/uniform stream 68
Rankine half-body 65
Rankine oval 67
slender bodies 69
source/wall 59
sphere–force on 77
two-dimensional
circular arc/uniform
stream/circulation 109
circular cylinder/uniform stream 91
circular cylinder/uniform
stream/circulation 92
elliptic cylinder/uniform
stream/circulation 93
flat plate/uniform stream/circulation 110
Joukowski airfoil/uniform
stream/circulation 95
Kármán vortex street 103
Rankine half-body 65
Rankine oval 67
slender bodies 69
source/wall 59
vortex/wall 61
vortex pair/tea cup 64
Irrotational flows 31, 47
Irrotationality, persistence of 47
Isotropic turbulence 240
Isotropy, material 244
Inviscid flows 46
Jones-McWilliams airfoil 98
Joukowsky airfoil 96, 97
Joukowski transformation 98
Kármán-Pohlhausen profile 181
Kármán-Trefftz airfoil 98
Kármán vortex street 103
Kelvin ship wave 132
Kolmogoroff wave length 246
Korteweg-DeVries equation 133
Kutta condition 77, 290
Lagally theorem 79, 113
Lagrange stream function 11
Lagrangian description 7
Laplace equation 48
Laplace operator, Laplacian
Cartesian coordinates 344
cylindrical coordinates 131
orthogonal curvilinear coordinates 336
spherical coordinates 343
Laplace transform 323
Lax-Wendroff method 310
Laurent series 114
Leapfrog method 309
Lift
coefficient 39
forces 69
Lifting line theory 101
Low Reynolds number flow
Oseen’s approximation 214–16
cylinder 214, 215, 216
sphere 214, 215
Stokes’ approximation 207, 209
doublet 208
rotlet–steady 208, 209
rotlet–unsteady 209
sphere–liquid 210–12
sphere–solid–general translation
209–10
sphere–solid–simple harmonic motion
212–14
sphere–solid 210
Stokeslet–steady 208
Stokeslet–unsteady 208
MacCormack’s methods 310, 311
Magnus effect 97
Mangler’s transformation of boundary layer
equations 188
Mass density 5
Material derivative 4–5
Material description 4
Material isotropy 26, 31
Material surface 28
Mathieu equation, functions 325–6
Maximum kinetic energy theorem–Kelvin 83
Maximum modulus theorem 81
Maximum-minimum potential theorem 81
Maximum-minimum speed theorem 82
Minimum kinetic energy theorem–Kelvin 82–3
359
Index
Mean value theorem 284
Method of images 59, 68, 116
Moment coefficient 39
Momentum equation 13–14
Momentum thickness 174
Moving coordinate system 41–3
Multiple valued functions 91
NACA airfoil 99–101
Natural coordinate system 177–8
Natural convection see Free convection
Navier Stokes equations
Cartesian coordinates 3, 332–6
cylindrical coordinates 12, 131
orthogonal curvilinear coordinates 336–40
spherical coordinates 12, 162, 163, 343
Navier-Stokes equations–exact solutions
convective acceleration absent–steady 140–7
annulus 144–5
arbitrary cross-section conduit 145–6
circular conduit 145
circular cylinders–Couette 147
parallel plates–Couette 141–2
rectangular conduit 142–4
convective acceleration absent–unsteady
concentric cylinders–impulsive
start 168
flat plate–impulsive start 149
flat plate–oscillating 149
parallel plates–impulsive start 141–2
similarity solutions
axially symmetric stagnation point 168
flow–Homann 158
convergent/divergent channel–Hamel
158–62
rotating disk–von Kármán 271
round jet–Squire 163
spiral channel–Hamel 162
stagnation line flow–Hiemenz 155–8
Neumann problem 285
Newton’s method 263
Newtonian fluid 25
Normal stress 16
No-slip condition 29
One point correlation 239
Orr-Sommerfeld equation 230, 276
Orthogonal functions 323
Oscillating motion of a plate–Stokes
second 150
Oseen flow see Low Reynolds number flow
Panel method 71–7
Parabolic coordinate system
Parabolic differential equations 297
Parabolized Navier-Stokes equations 174
Path lines 7–13
Periphractic region 82
Persistence of circulation theorem–
Kelvin 84
Persistence of irrotationality 47
Pitot tube 62
Plane flow instability 228–31
Plate flow
infinite
impulsive motion–Stokes first 147–9
oscillating–Stokes second 150
semi-infinite
boundary layer–Blasius 176
isothermal vertical plate 201–202
constant heat flux vertical plate
202–203
inclined plate 191
Poiseuille flow
arbitrary cross-section conduit 145–6
circular conduit 144
elliptic conduit 145, 161
equilateral triangle conduit 145
impulsive 147
rectangular conduit 142–4
round tube 144
Potential flows 48
Prandtl number 194
Pressure coefficient 39
Principle of exchange of stabilities 223
Principle of material objectivity 31
Principle of material frame
indifference 31
Pseudo-vector 318
Quaternion 3
Random walk method 218
Rankine bodies
half-body 65–7
oval 67–8
Rate of deformation
Cartesian coordinates 8
cylindrical coordinates 338
orthogonal curvilinear coordinates 337
spherical coordinates 162, 163, 343
Ray theory 136–9
Rayleigh-Bénard instability 229
Rayleigh number 222
Rays 138
Reiner-Rivlin fluid 40
Relaxation method 284–6
Reynolds number 39
Reynoldsaveraged Navier Stokes
equations 235
Reynolds stress 236
Rheogoniometry 24
Riccati method 274–8
Richardson extrapolation 303–304
360
Index
Richardson number 39
Richardson’s deferred approach to the
limit 281
Root mean square average 235
Rotational flows 31
Runge-Kutta integration 268
Scalar potential 319
Scale of turbulence 240
Schwarz-Christoffel transformation 108
Second order correlation 236
Separation
adverse pressure gradient 183
favorable pressure gradient 183
Stratford’s criterion 184
Thwaite’s criterion 184
Shear stress 16
Shallow water theory see Waves
Ship wave–Kelvin 130–2
Similarity solutions–boundary layer
axisymmetric jet 204
flat plate–Blasius 176–8
stagnation point 178
wedge–Falkner-Skan 175–9
Similarity solution of flows–exact see
Navier-Stokes equation solutions
Simpson’s rule–integration 266
Single valued functions 91
Singularity distribution 69–77
Singularity distribution–surface 296
Skew-symmetric 30
Slender body approximation 69, 71, 72
two-dimensional 69
three-dimensional 69
Solitary wave 135
Sources and sinks
three dimensional–point 112–13
two dimensional–line–monopole, 52, 58
Spatial description 4
Specific energy 37
Specific heat 27
Spectral representation 323
Spectrum analyzer 245
Sphere theorems 84
Spherical coordinates 12
Stability
boundary layer 301
Couette-Taylor cells 276
plane Poiseuille flow 276
Rayleigh-Bénard thermal cells 274
round jet 128–30
Standing waves 123, 139
Stationary waves 132
Stiff systems 77
Stokes flow see Low Reynolds number flow
Stokes paradox 214
Stokes paradox–resolution 216
Stokes stream function 12
Stokes theorem 33
Stokesian fluid 40
Streamlines 7
Streamlines–free 111
Stream functions
Lagrange 7–10
Stokes for two dimensional flows 7
Three dimensions 11–12
Stream surface 7
Stream tube 7
Stress
Cartesian coordinates 3
cylindrical coordinates 343
orthogonal curvilinear coordinates 336–40
spherical coordinates 343
vector 14
Strouhal number 39
Strutt diagram 128
Sturm-Liouville system 326
Substantial derivative 5
Successive line over-relaxation method 286
Successive over-relaxation method 286
Superposition 59–61
Surface tension 28
Surface waves of small amplitude see
Waves
Taylor cells 226
Taylor series 113
Tensors 2
alternating 330
base vectors 340
Cartesian 328
Cayley-Hamilton theorem 40, 336
contraction 330
contravariant components 342, 343
covariant components 340–3
general definition of 2, 332
indefinite product 332
invariants 334
Kronecker delta 330
Mohr’s circle 333
physical components 344
summation convention 329, 344
symmetric 20, 23, 333, 342
skew-symmetric 30, 331
Thermal boundary layers–forced
convection
axially symmetric thermal jet 182–3
integral solutions
circular cylinder stagnation point 92
flat plate–constant heat flux region 199
flat plate–constant temperature region
198–9
similarity solutions
flat plate–constant heat flux 196, 199
flat plate–isothermal 195
wedge 195
theory 192–4
361
Index
Thermal boundary layers–natural convection
integral solution
isothermal vertical plate 201–202
similarity solutions
constant heat flux vertical plate
202–203
inclined plate 203
isothermal vertical plate 201–202
Thwaite’s flap 97
Toe 103
Trapezoidal rule 266
Traveling waves 121
Turbulence
autocorrelation 235
correlation 236
destruction 238
diffusion 238
eddy scale 234
eddy viscosity 237, 240, 241, 247
energy cascade 245
energy spectrum 245
generation 238
homogeneous 240
intermittency 249
isotropic 240
Kolmogoroff scale 246
redistribution 238
Reynolds stresses 238
transport 238
two-point correlations 239
Turbulence models
abridged Lagrangian interaction 248
eddy-damped quasi-normal
approximation 247
Lagrangian history direct interaction 248
large eddy simulation 240
one equation model 140
quantum theory model 246–8
quasi-normal approximation 246, 247
stress equation model 240, 243–4
test field model 248
two equation model 240
zero equation model 239, 244
Uniform stream 51, 58
Uniqueness theorem 84
Universal equilibrium range 245
Upwind differencing 304
Vectors
curl 337
differential calculus 317–19
divergence 337
divergence theorem 2
Gauss theorem 320
gradient 318
Green’s function 51, 319, 329
Green’s identities 321
Green’s theorem 320
Helmholtz decomposition 319
integral calculus 319–22
irrotational 319
rotational 319
scalar potential 319
Stokes theorem 320
vector potential 319
Velocity, induced 61
Velocity potential 48
Virtual mass 79
Viscoelastic fluids 26
Viscometric flows 40
Viscometry 24
Viscoplastic fluid 40
Viscosity
bulk–volume 25
second 25
viscosity 25
Vortex
Biot-Savart law 249
blobs 312
bound 101
definition of 31
hairpin 249
horse-shoe 101
lattice 102
line 32, 55
street 103
stretching and turning of 34
vortex sheet 32
vortex tube 32
Vortices
starting 101
tip 101
Vorticity
Cartesian coordinates 332
cylindrical coordinates 340
equation 34
orthogonal curvilinear coordinates 336
spherical coordinates 339
tensor 30
vector 30
Vorticity vector 30
Wakes 184
Wave
frequency 121
front 138
length 121
number 121
Waves
dispersive 121
equi-partition of energy 139
linear theory 118
container 122–3
container–accelerating 127
interfacial 118
362
Index
Waves (continued)
long channel 133
ray theory 136
round jet 128
ship wave 130
surface tension 126
shallow water theory
cnoidal 134
solitary 135
standing 123
traveling 121
Weber number 39
Weiss sphere theorem 84
Whitehead paradox 214
Whitehead paradox–resolution 214
Winglets 103
Work-energy equation 36