panel method
TRANSCRIPT
Contents
1 Introduction 11.1 Solids and Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Concept of Continuum . . . . . . . . . . . . . . . . . . . . . . . 11.3 Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3.1 Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3.2 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3.3 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 The Coefficient of Viscosity . . . . . . . . . . . . . . . . . . . . . . . 51.5 Variables of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 Closed and Open Systems . . . . . . . . . . . . . . . . . . . . . . . . 71.7 Forms of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.8 Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.9 Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.9.1 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.9.2 Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.9.3 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.10 The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . 121.11 Specific Heats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.12 The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . 141.13 Isentropic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.14 Sound Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.15 Leibnitz’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.16 The Fundamental Laws . . . . . . . . . . . . . . . . . . . . . . . . . . 191.17 The Transport Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 20
2 The Differential Approach to Flow Analysis 262.1 Overview of the Differential Approach . . . . . . . . . . . . . . . . . 262.2 The Eulerian Derivative . . . . . . . . . . . . . . . . . . . . . . . . . 272.3 Kinematics of a Fluid Particle . . . . . . . . . . . . . . . . . . . . . . 28
2.3.1 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3.2 Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.3.3 Angular Velocity and Vorticity . . . . . . . . . . . . . . . . . 322.3.4 Rate of Volumetric Strain . . . . . . . . . . . . . . . . . . . . 342.3.5 Shear Strain Rate . . . . . . . . . . . . . . . . . . . . . . . . . 352.3.6 The Strain Rate Generally . . . . . . . . . . . . . . . . . . . . 36
2.4 Differential Form of Continuity Equation . . . . . . . . . . . . . . . . 372.5 Elementary Cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
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2.6 Momentum Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 392.6.1 The Cauchy Equations . . . . . . . . . . . . . . . . . . . . . . 39
2.7 Navier-Stokes Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 422.8 Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.9 Mechanical Energy Equation . . . . . . . . . . . . . . . . . . . . . . . 472.10 The Flow Equations Overview . . . . . . . . . . . . . . . . . . . . . . 48
3 Fundamentals of Inviscid Flow 503.1 General Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.2 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.3 Newton Second Law of Motion . . . . . . . . . . . . . . . . . . . . . . 513.4 Conservation of Thermodynamic Energy . . . . . . . . . . . . . . . . 513.5 Equation of state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.6 Euler Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . 523.7 Vorticity and Circulation . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.7.1 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.7.2 Helmholtz Theorem . . . . . . . . . . . . . . . . . . . . . . . . 533.7.3 Kelvin Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 563.7.4 Vortex Line, Surface, Tube, and Filament . . . . . . . . . . . 593.7.5 Spatial Conservation of Vorticity . . . . . . . . . . . . . . . . 603.7.6 Vortex Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.7.7 Velocity Field due to Vorticity Distribution . . . . . . . . . . . 65
3.8 The Velocity Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 663.9 Irrotational Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.10 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.11 Joukowski Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.12 Joukowski Theorem for non-Lifting Surfaces . . . . . . . . . . . . . . 733.13 Small Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . 753.14 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . 783.15 Green’s Second Formulae . . . . . . . . . . . . . . . . . . . . . . . . . 793.16 The General Solution of Laplace Equation . . . . . . . . . . . . . . . 80
3.16.1 Internal Potential . . . . . . . . . . . . . . . . . . . . . . . . . 823.16.2 Contribution from S∞ . . . . . . . . . . . . . . . . . . . . . . 833.16.3 Contribution from SW . . . . . . . . . . . . . . . . . . . . . . 83
3.17 Biot-Savarat Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 843.18 2D General Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 883.19 Stream Line and Stream Function . . . . . . . . . . . . . . . . . . . . 89
3.19.1 Some Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 893.19.2 Streamlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 903.19.3 Stream Surfaces and Stream Tubes . . . . . . . . . . . . . . . 913.19.4 Stream Functions . . . . . . . . . . . . . . . . . . . . . . . . . 913.19.5 Stream Function for 2D space . . . . . . . . . . . . . . . . . . 933.19.6 Stagnation Points . . . . . . . . . . . . . . . . . . . . . . . . . 94
3.20 Elementary Solutions of Laplace Equation . . . . . . . . . . . . . . . 943.21 Uniqueness of the solution . . . . . . . . . . . . . . . . . . . . . . . . 963.22 Elementary Solutions of Small Disturbance Equation . . . . . . . . . 983.23 Physical Interpretation and Basic Solutions . . . . . . . . . . . . . . . 102
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3.23.1 Point Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.23.2 Point Doublet . . . . . . . . . . . . . . . . . . . . . . . . . . . 1033.23.3 Uniform Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
3.24 Two-Dimensional Version of Basic Solutions . . . . . . . . . . . . . . 1073.24.1 Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.24.2 Doublet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1083.24.3 Uniform Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 1093.24.4 Vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1093.24.5 Two-Dimensional Singularity Distributions . . . . . . . . . . . 110
3.25 Complex Variables Expressions . . . . . . . . . . . . . . . . . . . . . 1123.25.1 Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1123.25.2 Some Elementary Flows . . . . . . . . . . . . . . . . . . . . . 1143.25.3 Elementary Flows in Tabular Form . . . . . . . . . . . . . . . 116
3.26 Theorem of Blasius . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1163.26.1 Extension of the Theorem of Blasius . . . . . . . . . . . . . . 119
3.27 Lagally’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
4 Principle of Superposition 1224.1 Basic Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1224.2 Rankine’s Oval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1224.3 Source in a Uniform Stream . . . . . . . . . . . . . . . . . . . . . . . 1244.4 Sources in Uniform Flow . . . . . . . . . . . . . . . . . . . . . . . . . 1254.5 Doublet in Uniform Flow . . . . . . . . . . . . . . . . . . . . . . . . . 127
4.5.1 Cauchy Integral Formula . . . . . . . . . . . . . . . . . . . . . 1274.5.2 Flow about Cylinder . . . . . . . . . . . . . . . . . . . . . . . 129
4.6 Flow around Cylinder with Circulation . . . . . . . . . . . . . . . . . 1314.7 Kutta-Joukowski Theorem . . . . . . . . . . . . . . . . . . . . . . . . 1364.8 Aerodynamic Force on the Vortex Line . . . . . . . . . . . . . . . . . 1374.9 Strait Solid Wall in the Flow Field . . . . . . . . . . . . . . . . . . . 140
4.9.1 3D Source near the Infinite Plane Wall . . . . . . . . . . . . . 1404.9.2 2D Source near the Infinite Plane Wall . . . . . . . . . . . . . 1404.9.3 2D Doublet near the Impermeable Infinite Plane . . . . . . . . 1414.9.4 3D Doublet near the Impermeable Infinite Plane . . . . . . . . 142
4.10 The Circle Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 1434.10.1 A Few Examples . . . . . . . . . . . . . . . . . . . . . . . . . 145
4.11 Continuous Distributions of Singularities . . . . . . . . . . . . . . . . 1514.11.1 Integration of Source Distributions . . . . . . . . . . . . . . . 1524.11.2 Integration of Doublet Distribution . . . . . . . . . . . . . . . 1534.11.3 Integration of Vortex Distributions . . . . . . . . . . . . . . . 155
4.12 Extension of the Circle Theorem . . . . . . . . . . . . . . . . . . . . . 1554.12.1 Constant Vorticity Extension . . . . . . . . . . . . . . . . . . 1554.12.2 Singularities Distribution along Circle . . . . . . . . . . . . . . 156
5 Subsonic Singularity Panels 1605.1 Two-Dimensional Point Singularity Elements . . . . . . . . . . . . . . 160
5.1.1 Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1605.1.2 Doublet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
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5.1.3 Vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1635.2 Numerical Quadrature . . . . . . . . . . . . . . . . . . . . . . . . . . 165
5.2.1 Orthogonal Polynomials . . . . . . . . . . . . . . . . . . . . . 1655.2.2 Gaussian Quadrature . . . . . . . . . . . . . . . . . . . . . . . 169
5.3 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1735.4 Two-Dimensional Line Panels . . . . . . . . . . . . . . . . . . . . . . 176
5.4.1 Source panel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1765.4.2 Vortex Panels . . . . . . . . . . . . . . . . . . . . . . . . . . . 1925.4.3 Doublet Panels . . . . . . . . . . . . . . . . . . . . . . . . . . 1955.4.4 Some General Remarks . . . . . . . . . . . . . . . . . . . . . . 201
5.5 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2035.6 Three-Dimensional Point Singularity Elements . . . . . . . . . . . . . 214
5.6.1 Point Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2145.6.2 Point Doublet . . . . . . . . . . . . . . . . . . . . . . . . . . . 2145.6.3 Point Vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
5.7 Three-Dimensional line Elements . . . . . . . . . . . . . . . . . . . . 2155.7.1 3D Line Source Element . . . . . . . . . . . . . . . . . . . . . 2155.7.2 3D Line Vortex . . . . . . . . . . . . . . . . . . . . . . . . . . 216
5.8 Surface distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 2295.8.1 Rotation of Coordinates . . . . . . . . . . . . . . . . . . . . . 2295.8.2 Constant Source Distribution on Unswept Panel with Stream-
wise Taper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2305.8.3 Constant Source Distribution on Swept Panel with Span-wise
Taper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2345.8.4 Constant-Strength, Three-Dimensional Quadrilateral Source . 2365.8.5 Linearly Varying Source Distribution of Swept Panel with
Span-wise Taper . . . . . . . . . . . . . . . . . . . . . . . . . 2385.8.6 Constant Vortex Distribution on Swept Panel with Span-wise
Taper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2395.8.7 Linearly-Varying Vortex Distribution on Swept Panel with
Span-wise Taper . . . . . . . . . . . . . . . . . . . . . . . . . 2415.8.8 Quadrilateral Doublet . . . . . . . . . . . . . . . . . . . . . . 249
5.9 Three-Dimensional Higher-Order Elements . . . . . . . . . . . . . . . 252
6 Supersonic Singularity Panels 2556.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2556.2 Green’s Second Formulae Analog . . . . . . . . . . . . . . . . . . . . 2576.3 Two-Dimensional Point Singularities . . . . . . . . . . . . . . . . . . 2586.4 Parametric Differentiation of Integrals . . . . . . . . . . . . . . . . . 2596.5 Three-Dimensional Line elements . . . . . . . . . . . . . . . . . . . . 260
6.5.1 Small Disturbance Flow about Bodies of Revolution . . . . . . 2606.5.2 Slender Body in Axi-Symmetric Flow . . . . . . . . . . . . . . 2636.5.3 Boundary Condition . . . . . . . . . . . . . . . . . . . . . . . 2636.5.4 The Basic Solutions of the Equation (6.25) . . . . . . . . . . . 2646.5.5 Singularity Lines Passing Through Coordinate Origin . . . . . 269
6.6 Surface Distribution of the Singularities . . . . . . . . . . . . . . . . . 2706.6.1 Gothert’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 270
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6.6.2 Integration Procedure . . . . . . . . . . . . . . . . . . . . . . 2716.6.3 Constant Source Distribution on Un-swept Panel with Stream-
wise Taper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2726.6.4 Constant Source Distribution on Swept Panel with Span-wise
Taper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2746.6.5 Linearly Varying Source Distribution on Swept Panel with
Span-wise Taper . . . . . . . . . . . . . . . . . . . . . . . . . 2756.6.6 Constant Vortex Distribution on Swept Panel with Span-wise
Taper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2766.6.7 Linearly Varying Vortex Distribution on Swept Panel with
Span-wise Taper . . . . . . . . . . . . . . . . . . . . . . . . . 2766.6.8 Panel Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 2786.6.9 Coordinate Transformation Again . . . . . . . . . . . . . . . . 279
7 Applications 2817.1 Incompressible Flow about Airfoil . . . . . . . . . . . . . . . . . . . . 281
7.1.1 Flow about Circle with Circulation . . . . . . . . . . . . . . . 2817.1.2 Conformal Mapping . . . . . . . . . . . . . . . . . . . . . . . 2827.1.3 Transformation of a Circle into and Airfoil . . . . . . . . . . . 2827.1.4 The Joukowsky Transformation . . . . . . . . . . . . . . . . . 2847.1.5 The Karman–Trefftz Airfoils . . . . . . . . . . . . . . . . . . . 2897.1.6 The Point-Vortex Solution . . . . . . . . . . . . . . . . . . . . 2937.1.7 Solution using Constant-Strength Vortex Distribution . . . . . 3007.1.8 Linearly Varying Vortex Panel Application . . . . . . . . . . . 3097.1.9 Two Airfoil Problem . . . . . . . . . . . . . . . . . . . . . . . 3187.1.10 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . 326
7.2 Axi-Symmetric Supersonic Flow . . . . . . . . . . . . . . . . . . . . . 3307.2.1 Calculation of the Flow around Sharp-Nose Body of Revolution331
7.3 Vortex Lattice Method . . . . . . . . . . . . . . . . . . . . . . . . . . 338
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Chapter 1
Introduction
Aerodynamics as a part of fluid mechanics is the branch of engineering science thatis concerned with forces and moments that act on the vehicles in flight.
1.1 Solids and Fluids
Air being a fluid is a substance that deforms continuously under the action of shear-ing forces. Figure (1.1) illustrate basic difference between solids and fluids. Considerimaginary chunks of both a solid and a fluid (it can be easily imagined with steel orrubber but impossible with fluid, because fluid will spread away). Upper row demon-strate deformation of solid under constant shear force, while lower row demonstratedeformation of imaginary chunk of fluid. A short time after application of shearingforce the solid body deforms to equilibrium position maintaining the same shearingangle φ1 = φ2 at the moments to +∆t and to + 2∆t respectively.Contrary to solid body fluid deforms continuously under action of shear force Fs.Deformation angles φ1 and φ2 are not equal at two distinct moments t + ∆t andt+ 2∆t to be more specific φ2 > φ1.Application of force to a solid body assumes certain deformed shape and retaining ofthat shape as long as force is applied. Thus, for solids applied force can be connectedto deformation. Application of force to fluids continuously deforms fluid, so appliedshear force could be connected only to rate of fluid deformation. Fluid is a substancewhich can resist shear only when moving.We have just established basic difference between solids and fluids. A solid does noteasily change shape while a fluid change shape with relative ease. This concept offluids is valid both for liquids which easily change shape but not volume, and gaseswhich easily change both shape and volume.
1.2 The Concept of Continuum
All substances are composed of an extremely large number of discrete particles calledmolecules. Air is mechanical mixture of different types of molecules. Moleculesinteract with each other via collisions and inter molecular forces. Molecules of solidsare relatively close to each other (order of a molecular diameter), while the moleculesof air are relatively apart (order of magnitude greater then with solids). Air exerts
1
Fs
Fs
FsFs
Fs Fs
fluid
solidφ1
φ1
φ2
φ2
a)
b)
to to +∆t to + 2∆t
φ2 = φ1
φ2 > φ1
Figure 1.1: Response to a shear force a) solid and b) fluid.
relatively weak inter molecular forces. That is explanation why air easily changeshape and volume.In practice it is impossible to describe behavior of air in terms of molecule dynamics,because of extremely large number of molecules involved. Approximately 25 of airat normal conditions contains 1024 molecules. Specification of initial conditions(x, y, z – position) and (u, v, w – velocity) fore each molecule will require 1017 yearsif for specifying each data we spend one second.For most cases of practical interest, it is possible to ignore the molecular nature ofmatter and to assume that matter is continuous. This assumption is called contin-uum model and can be equally applied to solids and fluids. This model assumesthat molecular structure is so small relative to the dimensions involved in problemsof practical interest that we may ignore it.Continuum model requires description of fluids in terms of its properties, whichrepresent average characteristics of its molecular structure. Typical example is usageof density of the fluid instead of number of molecules, mass of molecule, and occupiedvolume. If matter were truly continuum, the properties would be continuous functionof time and space.
1.3 Fluid Properties
Fluid properties characterize the state of the fluid and microscopically representmolecular structure and motion.
1.3.1 Density
The density of a fluid at a point in space is the mass of the fluid, contained in theincremental volume surrounding the point, per incremental volume:
= lim∆V→0
∆m
∆V (1.1)
2
The incremental volume must be large compared to molecular dimensions yet verysmall relative to the dimensions of the vehicle whose flow field we seek to ana-lyze. Dimension of density is M/L3. Standard atmospheric density of the air is1.225Kg/m3.Several other properties can be derived directly from density. The specific volumeυ is defined by:
υ =1
while specific weight γ is defined as γ = g, where g is the local acceleration ofgravity (g = 9.80665m/s2).
Figure 1.2: Gas thermometer.
1.3.2 Temperature
We are all familiar with temperature (T ). However because of difficulty in quantita-tively defining the temperature, we define equality of temperature. Two bodies haveequality of the temperature, T2 = T1 when no change in any observable propertyoccurs when they are in thermal contact. Further, two bodies respectively equal intemperature to a third body must be equal in temperature to each other (T1 = T3and T2 = T3 means T1 = T2). Temperature of the body is measured indirectly andis expressed in K and T [C] = T [K] − 273.15. Standard atmospheric temperatureis 15C.The term temperature can, however, be described in more objective manner throughof number of easily measured physical parameters. Let us now introduce the con-cept of the thermostat. By term thermostat we mean a body whose heat capacity isinfinite compared to bodies of interest to which the thermostat will be brought tocontact. That means after bringing any body to contact with thermostat tempera-ture of the body will equalize to thermostat temperature, while the temperature ofthe thermostat will remain unchanged. Let us place in thermostat vessel contain-ing a gas with loaded piston. We will measure pressure inside the vessel, volume,and temperature of the thermostat which can be changed. Changing the load onthe piston for constant temperature we can plot the diagrams shown in the figure(1.2b).
3
Each line represent isotherm. Experimentally is shown that isotherms never inter-sects, so vessel can be used as thermometer. Temperature is thus unique functionof pressure and volume T = T (P, V )
Figure 1.3: Sketch for pressure definition.
1.3.3 Pressure
Because of the random motion due to their thermal energy, the individual moleculesof air would continuously strike a surface that is placed in the air. Pressure is themagnitude of the force per unit area of the surface. Collisions between moleculesand surface will occur even though the surface is at rest relative to the fluid. ByNewton’s second law, a force is exerted on the surface equal to the time rate of changeof the momentum of the rebounding molecules. Since the fluid at rest cannot resist totangential forces, the pressure on the surface must act in the direction perpendicularto that surface. Consider a small surface ∆A within a fluid, as shown in figure (1.3).A force ∆Fn acts normal to the surface. Tangential forces may also be present butare not relevant to the definition of pressure. the Pressure is defined by:
p = lim∆A→0
∆Fn
∆A
The limiting value ∆A→ 0 represents the lowest bound of continuum assumption.Furthermore the pressure acting at a point in a fluid at rest is the same in alldirections.Let us consider fluid particle at rest shown in figure (1.4). Let us suppose, also, thatpressure p acts on the fluid particle faces built by coordinate surfaces. We will provethat pressure p acting on the plane ABC is also equal to p. Since fluid particle isat rest, sum of all forces acting on this particle must be equal to zero. Forces actingon the coordinate surfaces, of the particle are:
POBC = 1/2p∆y∆zı , POCA = 1/2p∆z∆x , POAB = 1/2p∆x∆yk (1.2)
To calculate force acting on the ABC we must first determine the area of the triangleABC:
AABC = 1/2AC × AB =1
2
∣∣∣∣∣∣∣ı k
−∆x 0 ∆z−∆x ∆y 0
∣∣∣∣∣∣∣4
Figure 1.4: Pressure acting on fluid particle at rest.
from which flows:
AABC = −1/2∆y∆zı− 1/2∆x∆z − 1/2∆x∆y k
Sum of all forces should be equal to zero:
POBC + POCA + POAB + p AABC = 0
or:1/2(p− p)(∆y∆zı+∆x∆z +∆x∆y k) = 0 (1.3)
which can always be true only if p = p. We have proved that pressure must be equalin all directions if a fluid is at rest.Most engineering calculations end when pressure on a surfaces immersed in the fluidis found. Standard atmospheric pressure is equal to 101 325N/m2.
1.4 The Coefficient of Viscosity
A fundamental approach to viscosity shows that it is the property of a fluid whichrelates applied stress to the resulting strain rate. Here we consider a simple andwidely used example of a fluid sheared between two plates, as in figure (1.5). Thisgeometry is such that the shear stress τxy must be constant throughout the fluid(first index x means in x direction, while the second index y means in the planeto which y is normal). The motion is in the x direction only and varies with y,u = u(y) only. Thus there is only a single finite strain rate in this flow:
τxy = f(du/dy) (1.4)
Since, for a given motion V of the upper plate τxy is constant, it follows that inthese fluids du/dy, is constant, so that the resulting velocity profile is linear acrossthe plate, as sketched in figure (1.5). This is true regardless of the actual formof the functional relationship in equation (1.4). If the no-slip condition holds, thevelocity profile varies from zero at the lower wall to V at the upper wall. Repeated
5
experiments with various values of τxy will establish the functional relationship (1.4).For simple fluids such as water, oils, or gases, the relationship is linear or Newtonian:
τxy µdudy
= µV
h(1.5)
Everybody knows from experience that viscosity is associated with the ability ofa fluid to flow freely. Heavy oil takes a long time to flow out of a can. Lightoil flows out quickly. The idea of viscosity being proportional to time to flow hasbecome accepted practice in the petroleum industry. Thus the motorist purchasesoil with a viscosity labeled SAE 30. This means that 60 ml of this oil at a specifiedtemperature takes 30 s to run out of a 1.76cm hole in the bottom of a cup. Thisexperiment is convenient and reproducible for very viscous liquids such as oil, butthe time to flow is not viscosity, any more than the speed of sound is the time ittakes an echo to return from a mountainside. It is an intriguing fact that the flowof a viscous liquid out of the bottom of a cup is a difficult problem for which noanalytic solution exists at present.
Figure 1.5: Fluid sheared between two plates.
1.5 Variables of State
Let us define thermodynamic system as a quantity of matter separated from sur-roundings by an “enclosure”. The enclosure does not necessarily consist of a solidboundary, it is only necessarily that the enclosure forms a closed surface and thatits properties are defined everywhere.Let us consider system at equilibrium than all macroscopic quantities will be inde-pendent of time. Variables of the system which depend only on state of the systemare called variables of state. Before mentioned properties; pressure p, density ,volume υ and temperature T are variables of state. Experiments show that thesevariables are not independent. They are connected to each other through relationof the form:
p = p(, T ) (1.6)
which is called equation of state, or zeroth law of thermodynamics. In most casesair can be considered as ideal gas for which equation (1.6) has simple form:
p = RT (1.7)
6
where R is constant equal to the ratio Ru and molecular weight M.W. of the air.Ru being equal to:
Ru = 8.31434kJ
kmolKSince molar mass of air is 28.97kg/kmol then the gas constant for air is:
R =Ru
M.W.=
8.3143
28.97= 0.2870
kJ
kg K
1.6 Closed and Open Systems
In the previous section we have defined closed or thermodynamics system. The massor region outside the system is called surroundings. The real or imaginary surfacethat separates the system from its surroundings is called the boundary. These termsare illustrated in figure (1.6), the boundary of the system can be fixed or movable.The boundary of a system has zero thickness, and thus it can neither contain anymass nor occupy any volume in space.
Figure 1.6: Thermodynamic system (Control mass).
Systems may be considered to be closed or open, depending on whether a fixedmass or a fixed volume in space is chosen for study. A closed system (also knownas a control mass) consists of a fixed amount of mass, and no mass can cross itsboundary. If besides mass even energy is not allowed to cross the boundary, thatsystem is called an isolated system.An open system, or a control volume is a properly selected region in space. Bothmass and energy can cross the boundary of a control volume, which is called a controlsurface. A control volume is shown in the figure (1.7).The thermodynamic relations that are applicable to closed and open systems aredifferent. Therefore, it is extremely important that we recognize the type of systemwe have before we start analyzing it.
1.7 Forms of Energy
Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential,magnetic, electric, chemical, and nuclear, and their sum constitutes the total energy
7
Figure 1.7: A control volume, both mass and energy can cross the boundaries ofcontrol volume.
Et of the system. The total energy of the system per unit mass et is defined as:
et = ek + ep + ei =Et
m(kJ/kg) ,
where ek is kinetic, ep is potential, and ei is internal energy.Thermodynamics does not provide information about absolute value of the totalenergy of a system. It deals with the energy change within a system which iswhat matters in engineering problems. Thus value Et = 0 can be assigned to someconvenient reference point. The change of the total energy of a system is independentof the reference point selected.Generally total energy of the system is divided into two groups: macroscopic andmicroscopic forms of energy. System processes in which system takes part as awhole with respect to some outside reference frame contribute tomacroscopic energy,such as kinetic and potential energies. Microscopic forms of energy are relatedto molecular structure of the system and the degree of molecular activity. Themicroscopic forms of energy are independent of outside reference frame. The sumof all microscopic forms of energy is called internal energy. Symbol U is devoted tointernal energy in thermodynamics. But in aerodynamics symbols U and u are usedfor x-component of velocity, thus we will use symbols E and e for internal energy.The energy that the system possesses as a result of its motion relative to somereference frame is called kinetic energy :
K.E. =mV 2
2
where V is velocity of the system, and m is the mass of the system. Kinetic energyper unit mass is obtained from former equation after division by m:
k.e. =V 2
2
Potential energy does not play significant role in aerodynamic since density of theair is small under normal conditions. Anyway, energy of the system as a result ofits elevation in a gravitational field is called potential energy and is expressed as:
P.E. = −mg · zwhere g = 9.80665m/s2 and z is elevation. Potential energy of the system per unitmass is:
p.e. = −g · z
8
1.8 Internal Energy
To have better understanding of internal energy, let us consider a system at molec-ular level. In general internal energy is consisted of sensible energy, latent energy,chemical energy, and nuclear energy. The sensible energy is associated with thekinetic energies of the molecules, figure (1.8a)
Figure 1.8: Forms of internal energy.
The internal energy is also associated with the inter molecular forces between themolecules of a system. These are the forces that bind the molecules together. Theinternal energy associated with phase of a system is called latent energy, figure (1.8b).The internal energy associated with the atomic bonds in a molecule is called chemicalenergy. During chemical reactions some chemical bonds are destroyed while othersare formed. As a result, the internal energy is changed, figure (1.8c).Tremendous amount of internal energy is associated with the bonds within thenucleus of the atom. This internal energy is called nuclear energy, figure (1.8d). Sowe can write:
ei = e+ el + ech + en
where e is sensible energy (we can sense it by termometer), el is latent energy, ech ischemical energy, and en is nuclear energy. In most aerodynamics applications totalenergy of the air can be considered to be consisted of kinetic energy and sensibleenergy, only, since other forms of energy did not change:
et = e+ ek .
Forms of energies capable to cross system boundaries are called dynamic formsof energy. The only two forms of energy capable to cross closed system boundaryare heat transfer and work. Heat transfer is possible only if exists temperaturedifference between the system and surroundings. For control volume the mass isalso transferred and of course energy associated with that mass. If there are noenergy and mass transfer through boundary of the system it is said for that systemthat it is in equilibrium state. For equilibrium state all state variables are notfunction of time.Figure (1.9) shows relation between various forms of energy of the fluid.
9
Figure 1.9: Relation between various forms of energy.
1.9 Heat Transfer
People use fire for warmth and light about half million of years. It has been onlyin relatively recent time that people have understood that heat is energy and thattemperature is a measure of the amount of that energy. Being a from of energy heatcan be converted to work. We do not speak about amount of heat accumulated inbody since word heat is restricted to energy being transferred.Heat flows from one body to another as a result of a difference in temperature. Heatalways flow from the hotter body to the colder body. In absence of phase changes thetemperature of the colder body will always rise and the temperature of the hotterbody decrease.Since temperature gradient is essential for heat flow it is necessary to know temper-ature distribution whiting body to calculate that flow. Heat transfer is performedthrough three distinct mechanisms: conduction, convection, and radiation.
1.9.1 Conduction
The conduction law is formulated by Fourier based on experimental observationsmade by Biot. Fourier law states that the rate of heat flow by conduction in a givendirection is proportional to temperature gradient in that direction and area normalto that direction:
Q = −kSn∇Twhere S is the area of the surface, n is the unit normal to that area, and k isproportionality constant being material property. Minus sign is needed because heatflows form higher to lower temperature, and gradient (∇T ) is positive if temperatureincreases. Heat flow rate per unit area is called heat flux and is defined by:
q = −k∇T (1.8)
1.9.2 Convection
When fluid flow over a solid body, and the temperatures are different, heat transfertakes place between the fluid and the solid surface as a result of motion of the fluid.This mechanism of heat flow is called convection, since the motion of the fluid plays a
10
significant role in augmenting the hat transfer rate. Let us denote fluid temperaturewith Tf and wall temperature of the body with Tw than experimental relation ofheat flux from body to fluid is given by
q = h(Tw − Tf )n (1.9)
where h is heat transfer coefficient, and n is the unit normal to the body surface.Coefficient h depends on
• The type of flow (laminar, turbulent, transitional)
• Geometry of the body
• Physical properties of the fluid
• Temperature difference
• Position along the surface of the body
• Of the mechanism of convection (free or forced)
1.9.3 Radiation
All bodies emit energy due to their temperature. This energy is called thermalradiation. The radiation flux emitted by a real body at an absolute temperature Tis always less than the black body emissive power, and is given by:
q = ε(σT 4
)n (1.10)
where ε is emissivity of the body (always less than 1), terms in parenthesis representblack body emissive power, σ = 5.6697 10−8W/(m2K4) is Stefan-Boltzman constant.
Figure 1.10: Energy can cross the boundary of the closed system only in the formof heat or work.
11
1.10 The First Law of Thermodynamics
The principle based on experimental observations that energy can be neither creatednor destroyed is known as the first law of thermodynamics. As we already mentionedheat and work are dynamic forms of energy. There is no sense in saying that a bodystores some heat or work. Heat and work are only forms of energy transfer betweensystem and surroundings and vice versa, figure (1.10).In general case, when addition of a heat to a body results in a rise in the temperatureof the body, and external work is performed due to an increase in the volume of thebody, the heat ∆Q added to the body is expended to increase its internal energy∆E and to accomplish work W . This statement can be expressed by the formula:
∆Q = ∆Et +W (1.11)
or in differential form:
dQ = dEt + dW per unit mass: dq = det + dw (1.12)
The heat added to a system and work done by the system is considered to bepositive, figure (1.11). Signs can be chosen arbitrary, but those mentioned aboveare customary in engineering.
Figure 1.11: Heat added to a system, Q, and work done by the system, W , againstsurrounding pressure p.
1.11 Specific Heats
The specific heat is defined as the energy required to raise the temperature of a unitmass of a substance by one degree. This energy depend on the way how the process isexecuted. In thermodynamics we are interested in specific heat at constant pressureCP and specific heat at constant volume CV . Let us consider imaginary experimentshown in the figure (1.12a). Pressure is maintained constant by the movable piston.When heat is added to the system it is “spent” to the work against the pressure andto the raise of temperature:
q = w +∆e ,
12
in former equation our tothal energy is consisted of sensible energy only (et = e),since all other forms of internal energies are assumed unchanged.Let us denote such heat by ∆h, then by definition specific heat at constant pressuremust be equal to:
CP =
(∂h
∂T
)p
, ⇒ ∆h = CP∆T (1.13)
In the second case figure (1.12b) heat is spent only to temperature rising (i.e. inernal
Q
W
q = hq = e
Q
a) b)
Figure 1.12: Specific heat at constant pressure is the heat added to a system perunit mass which raise temperature by one degree when pressure is unchanged a),while specific heat at constant volume is the heat added to unit mass of a systemwhich raise temperature of the system by one degree when volume of the system ismaintained constant b).
energy q = e). Specific heat at constant volume is equal by definition to:
CV =
(∂e
∂T
)v
, ⇒ ∆e = CV∆T (1.14)
It is clear that CP > CV because h must be greater then e since heat h is spent alsoto work, and by definition temperature increment is equal to 1 in both cases. If wechoose that at T = 0 also e = h = 0 than equations (1.13) and (1.14) become
e = CV T (1.15)
h = CP T (1.16)
From equation equation (1.12) the first principle of thermodynamics for constantpressure system, per unit mass, follows
dh = dq = de+ pdVm
(p = const.) ⇒ dh = d
(e+
p
)(1.17)
since
dVm
=dVdm︸︷︷︸1/
·
d︷ ︸︸ ︷dm/Vm/V︸ ︷︷ ︸
=d
2
13
takeing into account sign convention for work, we can express equation (1.17) in theform:
h = e+p
(1.18)
where h is anoter state variable called enthalpy. For general execution of processeswe have, (from (1.12)):
dq = de+ pd
(1
)= d
(e+
p
)− dp
= dh− dp
(1.19)
since dw = p dV , and the following is taken into account:
d
(p
)= pd
(1
)+dp
Substituting definitions (1.15) and (1.16) into equation (1.18) for ideal gas thenfollows
CP T = CV T +p
⇒ (CP − CV )T = RT
which connects specific heats with gas constant. If we define ratio of specific heatsCP/CV = γ than former equation yields
CV =R
γ − 1CP =
γR
γ − 1(1.20)
Since e is also variable of state equation of state equation (1.7) can be expressedthrough this variable
p = RT = (CP − CV ) T = (γ − 1)CV T = (γ − 1) e (1.21)
from which follows:
e =1
γ − 1
p
(1.22)
Equation of state can be also expressed through other variables of state, for ehample:
h =γ
γ − 1
p
(1.23)
1.12 The Second Law of Thermodynamics
The second law of thermodynamics asserts that processes occur in a certain directionand that energy has quality as well as quantity. A process cannot take place unlessit satisfies both the first and second laws of thermodynamics.To ascertain the proper direction of a processes, let us define a new state variable,the entropy, as follows:
ds =δqrevT
where s is the entropy of the system, δqrev is an incremental amount of heat addedreversibly to the system, and T is the system temperature. Dispute the fact that wehave defined entropy s in terms of reversible addition of heat δqrev entropy is state
14
variable, and it can be used in conjunction with any type of processes, reversible orirreversible. For general processes the following relation is valid:
ds =δq
T+ dsirrev (1.24)
The dissipative phenomena always increase the entropy:
dsirrev ≥ 0 (1.25)
The equal sign denotes a reversible process, where by definition above dissipativephenomena is absent. Hence combination of (1.24) and (1.25) yields
ds ≥ δq
T(1.26)
If process is adiabatic δq = 0 equation (1.26) reduces to
ds ≥ 0 (1.27)
Equations (1.26) and (1.27) are forms of the second law of thermodynamics. Thislaw tell us in what direction a process will take place. A process will proceed in adirection such that entropy of the system and surroundings will not decrease.Let us, for a while, assume that process is reversible, than δq = T ds. First law ofthermodynamics (1.12) when only pressure performs work now reads
T ds = de+ p · d(1
)(1.28)
or in terms of enthalpy:
T ds = dh− dp
(1.29)
For thermally perfect gas we have dh = CP dT , substitution in equation (1.29) gives:
ds = CPdT
T− dp
T(1.30)
If from equation of state product T is replaced by p/R in the equation (1.30) thefollowing is obtained
ds = CPdT
T−R dp
p(1.31)
The equation (1.31) is valid for thermally perfect gas. It can be integrated if functionCP (T ) is known. In the case when CP is constant (calloricaly perfect gas) integrationbetween state 1 and 2 gives:
s2 − s1 = CP lnT2T1−R ln
p2p1
(1.32)
If instead of p2/p1 we substitute 2T2/1T1 the equation (1.32) gives
s2 − s1 = CV lnT2T1−R ln
21
(1.33)
15
If we eliminate temperature ratio T2/T1 from equation (1.32) by equation of statewe, after some manipulations, obtain
p2p1
=
(21
)γ
e(s2−s1)/(CV ) (1.34)
What is equation of state expressed in terms of pressure p, density , and entropy sas a variables of state.
1.13 Isentropic Relations
Process is concerned to be isentropic if it is adiabatic δq = 0 and reversible s2 = s1.From equation (1.32) then follows
0 = CP lnT2T1− (CP − CV ) ln
p2p1
(1.35)
Since:CP
R=
CP
CP − CV
=γ
γ − 1
CV
R=
CV
CP − CV
=1
γ − 1
from equation (1.35) the following relation between pressure and temperature isobtained:
p2p1
=(T2T1
)γ/(γ−1)(1.36)
From equation (1.33) follows equation (when ∆s = 0):
21
=(T2T1
)1/(γ−1)(1.37)
Summarizing Eqs. (1.36) and (1.37) following relation is obtained for isentropicrelations
p2p1
=
(21
)γ
=(T2T1
)γ/(γ−1)(1.38)
Equation (1.38) relates pressure, density and temperature, as well as equation ofstate. But there are significant difference between equation (1.38) and equation(1.7). Equation (1.7) is always valid as long as air can be considered to be idealgas, while equation (1.38) describes the process. At each point of the process (1.38)equation of state (1.7) is satisfied. If the process is known to be isentropic, then oneof the fundamental laws of fluid nature (to be discussed later) can be replaced bysimple relation (1.38) what greatly simplifies practical applications.
1.14 Sound Velocity
Let us consider, now, straight tube of fluid at rest. At some moment t = 0 in thetube of unit area, at location O, is introduced disturbance, figure (1.13). Fromthat moment on disturbance continues to propagate through tube. We consider theportion of the fluid which passed location A in unit time δt = 1, with density and
16
velocity a. After short time interval ∆t fluid will pass location B having density + ∆ and velocity a + ∆a. Since mass must be preserved the following relationholds
a︸︷︷︸A
·S = (+∆)(a+∆a)︸ ︷︷ ︸B
·S
Relation is accurate to the first order in time since there is no guarantee that thesame mass will pass point B in unit time. Neglecting higher order terms fromprevious equation and canceling finite term products we obtain
∆a = −a∆ (1.39)
Figure 1.13: Propagation of disturbance through air contained in long tiny tube.
We now apply Newton second law of motion to the same mass. Change in momentumof the mass must be caused by the forces applied to it. We neglect influences of tubewall, so only pressure force is to be considered
(+∆)(a+∆a)2︸ ︷︷ ︸B momentum
·S − a2︸︷︷︸A momentum
·S = −∆p · S
Again canceling finite order terms and neglecting higher order terms we obtain
a2∆+ 2a∆a = −∆p (1.40)
Substituting product ∆a from equation (1.39) yields
a2 =∆p
∆
or when the considered amount of mass is infinitesimally small
a2 =dp
d(1.41)
The relation between pressure and density is not unique, because p = p(, T ), sofraction dp/d can have various values. Newton has supposed that the disturbancesthrough gases are transmitted isothermally
p
= const or:
dp
d=p
17
But experiments has proved that another great scientist (Laplace) had better idea.Namely, he supposed that the disturbances are transmitted through gases adiabat-ically
p
γ= const from which follows:
dp
d= γ
p
Constant a is called velocity of sound and is defined as
a =
√γp
=√γRT (1.42)
or symbolically
a =
(∂p
∂
)S=const
(1.43)
1.15 Leibnitz’s Theorem
Integrals that involve a parameter often occur in fluid mechanics. In most casestime plays the role of a parameter and the integrals are of the form
I(t) =∫V(t)
B(t, x, y, z) dV (1.44)
Here B, stands for any scalar, vector, or tensor function of interest. Not only doestime change the integrand, but the region of integration V(t) may be moving. We letv be the velocity of the surface of V(t). In addition to translating, the surface maybe expanding or contracting. The velocity v is any prescribed function of positionon the surface. The theorem of Leibnitz allows us to find dI/dt in a convenientmanner. The theorem is
d
dt
∫V(t)
B(t, x, y, z) dV =∫V∂B∂tdV +
∮SnvB dS (1.45)
Equation (1.45) states that we may move the derivative with respect to time insidethe integral if we add a surface integral to compensate for the motion of the bound-ary. The surface integral tells how fast B is coming into V because of the surfacevelocity v, regions A and B in figure (1.15). If the boundary does not move, thenv = 0 and the theorem merely says it is permissible to interchange the order ofdifferentiation and integration.As a specific example take B as the constant scalar function one (B = 1). Theintegral on the left of (1.45) is the volume of the region. Since partial derivative ofconstant with respect to time is equal to zero, (1.45) becomes
d
dt
∫V(t)
dV =∮Sv · n dS (1.46)
The rate of change of the volume of a region is the integral of the normal componentof the surface velocity over the region. The sketch of the region V for two close timeinstances is shown in the figure (1.14).
18
Figure 1.14: Region V in two close time instances.
1.16 The Fundamental Laws
Experience has shown that all fluid motion must obey the following fundamentallaws of nature.
• The law of conservation of mass. Mass can be neither created nor destroyed;it can only be transported or stored.
• Newton’s three laws of motion.1. A mass remains in a state of equilibrium, that is, at rest or moving at
constant velocity, unless acted on by an unbalanced force. (First law)
2. The rate of change of momentum of a mass is proportional to the netforce acting on the mass. (Second law)
3. Any force action has an equal (in magnitude) and opposite (in direction)force reaction. (Third law)
• The first law of thermodynamics (law of conservation of energy). Energy,like mass, can be neither created nor destroyed. Energy can be transported,changed in form, or stored.
• The second law of thermodynamics. The second law deals with the availabilityof energy to perform useful work. The only possible natural processes arethose that either decrease or, in the ideal case, maintain, the availability ofthe energy of the universe. The science of thermodynamics defines a materialproperty called entropy, which quantifies the second law. The entropy of theuniverse must increase or, in the ideal case, remain constant in all naturalprocesses.
• The state postulate (law of property relations). The various properties of a fluidare related. If a certain minimum number (usually two) of a fluid’s propertiesare specified, the remainder of the properties can be determined from thesetwo.
The important thing to remember about these laws is that they apply to all flows.They do not depend on the nature of the fluid, the geometry of the boundaries, or
19
anything else. It is known, that they have always been true and will continue to betrue unless they are suspended by the Creator of the universe. Hence we can firmlybase analysis of all flows on these laws. In addition to these universal laws, severalless fundamental ”laws” apply in restricted circumstances. An example is Newton’slaw of viscosity, which states that:The shear stress in a fluid is proportional to she rate of deformation of the fluid.This “law” is true only for some fluids and does not apply at all to solids. Such“laws” are better termed constitutive relations. We must use constitutive relations tosolve most flow problems, but we must select them carefully to match the particularproblem.
1.17 The Transport Theorem
The transport theorem is a mathematical expression that connects the system andcontrol volume points of view for a global or finite control volume approach to flowanalysis. It provides a way to identify a finite system and to evaluate the rateof change of any property or characteristic of that system by examining the flowthrough a control volume.The fundamental laws deal with rates of change of certain properties of the systemto which they are applied; for example, Newton’s second law addresses the rate ofchange of system momentum, and the first law of thermodynamics addresses therate of change of system energy. In order to apply these laws, we have to figure outhow to identify a specific system in a moving, deforming fluid and how to calculatethe rate of change of this system’s properties. This is where a control volume comesin. Although fluid flows through a control volume continuously, a control volumecontains a specific mass of fluid at any instant. We designate our system as the fluidmass that instantaneously occupies the control volume. A different system occupies
Figure 1.15: Control volume and enclosed system in a flow field.
the control volume at each instant; however, we may apply the fundamental laws towhatever system occupies the control volume at that instant. The control volumebecomes a way to identify a specific system, even if only for a very short time. Thisidea is somewhat like taking motion pictures. We point the camera at a particularlocation (say, a busy street corner) and take pictures. We record frame-by-framethe pictures of people who occupy the location. When we run the film at regular
20
speed, we see a continuous flow of people; but each frame shows a particular personoccupying the location.Figure (1.15) shows a three-dimensional, three-directional flow field with a controlvolume superimposed. The mass contained in the control volume at the instant thatthe figure represents is chosen as a system. Figure (1.16) shows the passage of thatsystem out of the control volume over time.Figure (1.16a) shows the situation at time t, when the system exactly fills the controlvolume. Figure (1.16b) shows the situation at a later time, t+ δt, when a portion ofthe system has left the control volume: V1, is the portion of the control volume thathas been vacated by the system in time δt; V2 is the portion of the control volumethat contained part of the system at time t and still contains part of the system attime t + δt; and V3 lies outside the control volume and contains the portion of thesystem that has left the control volume in time δt.
Mass of fluid incontrol volume atinitial time t.
a)
Vsys = Vcv
Control volume andsustem after shortperiod dt.
b)
V1
V2
V3
Control volumeand system atsome later time
c)
Vcv
Vsys
Vsys = V2 + V3
Vcv = V1 + V2
Figure 1.16: A system flows through a control volume.
We can calculate the rate of change of an arbitrary property B of this system;B could be momentum, energy, or any other property. Property B may be nonuniformly distributed throughout the system, so we let b be the specific (per-unit-
mass) property. If B represents system momentum ( M = mV ), then b is velocity
(V ). If B represents system kinetic energy (B = mV 2/2), then b is V 2/2. The totalamount of B in the system is
Bsys =∫msys
b dm =∫Vsys
b dV (1.47)
where msys is the system mass and Vsys is the system volume.By definition, the rate of change of Bsys is
dBsysdt
= limδt→0
Bsys∣∣∣t+δt
− Bsys∣∣∣t
δt
Substituting definition from equation (1.47), we have
dBsysdt
= limδt→0
∫Vsys b dV|t+δt −
∫Vsys b dV|t
δt. (1.48)
21
At time t,Vsys
∣∣∣t= Vcv ,
and at time t+ δt,
Vsys∣∣∣t+δt
= V2 + V3 = Vcv − V1 + V3 ,so
dBsysdt
= limδt→0
∫Vcv b dV|t+δt −
∫V1b dV|t+δt +
∫V3b dV|t+δt −
∫Vcv b dV|t
δt.
We can rewrite this expression as
dBsysdt
= limδt→0
∫Vcv b dV|t+δt −
∫Vcv b dV|t
δt︸ ︷︷ ︸TermI
+ limδt→0
∫V3b dV|t+δt − 0
δt︸ ︷︷ ︸TermII
− limδt→0
∫V1b dV|t+δt − 0
δt︸ ︷︷ ︸TermIII
.
where zeros are the values of integrals in terms II and III at the moment t.Each of the three terms has its own meaning. Term I is similar to the right-handside of equation (1.48), with one important difference: In term I the integrationis over the control volume, whereas in equation (1.48), the integration is over thesystem volume. Using the definition of a derivative, we find that
Term I =d
dt
∫Vcv
b dV =d
dt(Bcv) ,
which represents the time rate of change of the amount of B contained in the controlvolume. Note that this amount of B could change because of changes in Vcv (amoving or deforming control volume) and changes of or b with time at pointswithin the control volume. We sometimes call term I the rate of accumulation.The integral in the numerator of term II represents the amount of B that flows outof the control volume between time t and time t + δt. Dividing by δt, we get theaverage rate of outflow, and passing to the limit, we get the instantaneous rate ofoutflow:
Term II = Bout .The integral in term III is the amount of B that flows into the control volumebetween time t and time t + δt. (Note that this B is not in the original system.)Dividing by δt, we get the average rate of inflow, and passing to the limit, we getthe instantaneous rate of inflow:
Term III = BinWe sometimes call terms II and III the flux terms. Collecting terms I, II and III,we get
dBsysdt
=d
dt
∫Vcv
b dV + Bout − Bin . (1.49)
In words:
22
The rate of change of any property of the system that occupies acontrol volume at any particular instant is equal to the instantaneousrate of accumulation of the property inside the control volume plusthe difference between the instantaneous rates of outflow and inflowof the property. This latter term is the net rate of outflow of theproperty across the control surface.
Sometimes we can use equation (1.49) directly, but we often must relate Bout andBin to the flow field. For Bout, where B is a property of the fluid mass and the fluidis flowing across the control surface, each unit of mass leaving the control volumecarries b units of B out with it. As b may not be uniform, we write
dBout = b(dm)outand
Bout =∫dBout =
∫b(dm)out . (1.50)
Similarly,
Bin =∫dBin =
∫b(dm)in . (1.51)
Now consider the rate of flow of mass out of the control volume. Figure (1.17) showsan enlargement of a small portion of the control surface. Mass flows out from thecontrol volume through the area dS. Figure (1.17a) shows the fluid velocity at dSat a particular time t; n is a unit vector drawn perpendicular to dS and pointing
Figure 1.17: Detail of outflow through small piece of the control surface (dS): a)velocity and unit out drawn normal vector at surface; b) fluid particles that were atdS at t are at dS ′ at t + δt; fluid in volume d(δV) has left control volume throughdS.
outward from the control volume. In general, the fluid velocity is not perpendicularto the control surface, so the angle θ between V and n is not zero. Figure (1.17b)shows the fluid that has left the control volume through dS in time δt. The fluidthat was at dS at time t has now moved to dS ′. All the fluid in the volume d(δV)between dS and dS ′ has flowed out of the control volume through dS in time δt.Thus we can write
d(δm)out = d(δV) . (1.52)
23
From figure (1.17b), we have
d(δV) = (vδt) cos θ dS . (1.53)
To obtain an expression for the rate of outflow through dS, we substitute equation(1.53) into equation (1.52), divide by δt, and take the limit as δt→ 0. The result is
dmout = V cos θ dS . (1.54)
To identify the outflow area, note that fluid flows outward across the control surfaceat any point where 0 ≤ θ ≤ 90. We can write in vector notation
dmout = V · n dS . (1.55)
Remember that the magnitude of vector n is unity.Next, consider the rate of mass flow into the control volume. Figure (1.18) illustratesa portion of the control surface through which mass enters the control volume.Figure (1.18a) shows the velocity vector V and the unit outward normal vector n atthe area dS at time t. Note that for inflow, the angle θ lies between 90 and 180.The angle θ′ is the complement of the angle θ. Figure (1.18b) shows the situationat time t+ δt. The fluid that was at dS at time t has now moved to dS ′, inside thecontrol volume. All the fluid in volume d(δV) flowed into the control volume acrossdS in time δt. We can write
d(δm)in = d(δV) . (1.56)
whered(δV) = (V δt) cos θ′ dS = −V δt cos θ dS .
Dividing by δt and taking the limit as δt→ 0 gives
Figure 1.18: Details of inflow through a small piece of control surface: a) velocityand unit out drawn normal vector at surface; b) fluid particles that were at dS attime t are at (dS ′) at time t + δt; fluid in volume d(δV) has entered the controlvolume.
dmin = −V cos θ dS . (1.57)
24
In vector notation,dmin = V · n dS . (1.58)
Substituting the expressions for mass flow rate [Eqs. (1.54), (1.55), (1.57) and (1.58)]into Eqs. (1.50) and (1.51) gives
Bout =∫Sout
bV cos θ dS =∫Sout
b(V · n
)dS . (1.59)
andBin =
∫Sin
bV cos θ dS =∫Sin
b(V · n
)dS . (1.60)
Using Eqs. (1.59) and (1.60), we write equation (1.49) in the following forms:
dBsysdt
=d
dt
∫Vcv
bdV +∫Sout
bV cos θ dS +∫Sin
bV cos θ dS , (1.61)
dBsysdt
=d
dt
∫Vcv
bdV +∫Sout
bVn dS −∫Sin
bVn dS , (1.62)
anddBsysdt
=d
dt
∫Vcv
b dV +∮Scv
b(V · n
)dS . (1.63)
In equation (1.62), Vn represents the velocity component normal (perpendicular) tothe control surface. We combine the two surface integrals in Eqs. (1.61) and (1.62)into a single integral over the entire surface, as in equation (1.63), because any point
on the surface has either inflow, outflow, or V · n = 0.Equation (1.61) to (1.63), in any of its forms, is the transport theorem. It allows usto relate the system and control volume points of view (alternatively, the Lagrangianand Eulerian methods of description) at a global level of detail.If the control volume is fixed in space we can move differentiation d/dt. inside theintegral and replace it with partial differentiation ∂/∂t. If control volume is movableor deformable than Leibntz’s rule must be applied, equation (1.46), to replace thederivative of the first term in equation (1.63). Thus,
dBsysdt
=∫Vcv
∂
∂t(b) dV −
∮Scv
b (v · n) dS +∮Scv
b(V · n
)dS .
or:dBsysdt
=∫Vcv
∂
∂t(b) dV +
∮Scv
b(V − v
)· n dS . (1.64)
where V − v is relative fluid velocity, and (V − v) · n is the local normal componentof relative velocity, Vrn.The system point of view is related to a Lagrangian description of flow. Its advantageis that all the fundamental laws may be expressed directly in terms of a specificcollection of mass. The control volume point of view is related to an Euleriandescription of flow. Its advantage is that control volumes are easier to keep trackof than systems. Unfortunately, Newton’s laws of motion and the state postulateapply only to samples of matter, not to volumes. Thus we need to adopt the systempoint of view to formulate the fundamental laws but use the control volume pointof view to apply them to flow problems! Fortunately, we have formally connect thetwo points of view by purely mathematical relationships.
25
Chapter 2
The Differential Approach to FlowAnalysis
In this chapter, we consider the differential approach to flow analysis. We concen-trate on developing the governing differential problems.
2.1 Overview of the Differential Approach
The differential approach to flow analysis begins with formulation of the law ofconservation of mass and Newton’s second law of motion as differential equations.Because these laws are valid at every point in a flow field, the resulting differentialequations are capable of describing every point in the field. Deriving the differentialequations is a relatively straightforward exercise, and they are quite general.Another necessary task is to prescribe boundary conditions and initial conditionsfor the equations. Boundary conditions are derived from the physical constrainsthat exist at the boundaries of the flow (for example, the velocity of a air particle incontact with a aircraft wing must have the same velocity as the wing (if the viscousmodel of the flow is considered). Initial conditions describe the flow field at theinitial instant of time. Boundary conditions and initial condition depend on thephysical model which describes the flow. Boundary conditions and initial conditionsare unique to a given problem. The main reason is variations in geometry: Flow overa streamlined airfoil is different from flow over a square rod and both are differentfrom flow through a circular pipe.The (general) differential equations, together with the (problem-dependent) bound-ary and initial conditions, are a complete mathematical model of a particular flow.If we can solve the differential equations and incorporate the correct boundary andinitial conditions, we are rewarded with information about the flow at every pointin the field for all time (or at least until the conditions change). We can calculateflow rates from the known velocities and forces from the known pressure and shearstress distributions.Detailed view in the flow field is paid by the fact that each different flow geometrymust be solved separately; only the differential equations are general. The analyticsolution to the general flow problems are hard to find or even does not exist. In-stead of solving problem analytically, we may tackle accurate approximations to the
26
differential equations with the methods of computational fluid dynamics. Advancesin digital computer hardware and software have made the differential approach afeasible alternative for many important aerodynamic flow problems.
2.2 The Eulerian Derivative
The Eulerian derivative is a mathematical expression that connects the system andcontrol volume points of view for a local or differential equation approach to flowanalysis. It relates the rate of change of any property of a fluid particle (system) tothe particle’s location in the flow field.
Figure 2.1: Property b at point (x, y, z) at time t and shortly after that (t+ dt).
Imagine a specific fluid particle located at a specific point in a flow field (see figure2.1). For generality, we assume a three-directional, three-dimensional, unsteady flow.We describe the flow in Cartesian coordinates. If we use an Eulerian description,the particle’s properties and velocity depend on its location and time. Then for anarbitrary property b (b may be density, pressure, velocity, etc.), we can write
bP = b(xP , yP , zP , t)
where the subscript P reminds us that we use the location of the particle.The fundamental laws usually require rates of change of certain properties of matter(e.g., momentum, mass, energy, entropy). We determine the rate of change of bP byusing the rule for calculating total derivatives:
dbPdt
=∂b
∂t+∂b
∂xP
dxPdt
+∂b
∂yP
dyPdt
+∂b
∂zP
dzPdt
.
However, xP , yP and zP are the position coordinates of the fluid particle, so theirtime derivatives are the particle’s velocity components,
dxPdt
= uP ,dyPdt
= vP ,dzPdt
= wP .
Substituting, the rate of change of b is
db
dt=∂u
∂t+ u
∂b
∂x+ v
∂v
∂y+ w
∂b
∂z(2.1)
where we have now dropped the subscript P .
27
The complicated form of equation (2.1), requiring four terms to express a single totaltime derivative, is a consequence of selecting the Eulerian method of description inwhich fluid properties depend on both location and time. This type of derivativeis variously called the Eulerian derivative (because it is necessary in an Euleriandescription), the material derivative (because it expresses the rate of change of aproperty of a particle of the material), the substantial derivative (substituting theword substance for material), or the total derivative.As the property b was arbitrary, we can drop it and write the derivative as a math-ematical operator. To remind us of its special nature, we often write the operatoras a D and rearrange it as follows:
D
Dt=∂
∂t+ u
∂
∂x+ v
∂
∂y+ w
∂
∂z. (2.2)
A shorthand notation, using vectors, is
D
Dt=∂
∂t+ (V · ∇) , (2.3)
where V is the fluid velocity vector and ∇ is the gradient operator.Some comments on the physical meaning of Eqs. (2.1–2.3) are in order. The termdb/dt (alternatively, D/Dt) is the total rate of change over time of the property bof the particle. That property may change because the particle moves to anotherlocation over time or because the value of the property at a specific location maychange over time. For example, the temperature of a specific particle of air in theatmosphere may change as winds carry it to a region of different temperature, orif the particle remains at a fixed location, the temperature may change from dayto night or season to season. The rate of change resulting from changes of locationis reflected in the terms containing the velocity. The rate of change resulting fromunsteady effects at a fixed location is reflected in the partial derivative with time.Remember the basic idea that, for any property of a fluid particle, we may write:
Total rate ofchange ofproperty of afluid particle
=
UNSTEADYTERMRate of changeat a fixedlocation ∂
∂t
+
CONVECTIVETERMVelocity times thederivatives withrespect to space(V · ∇)
The first term on the right-hand side of this word equation is called the local un-steady term. The right-most term is called the convective rate of change, becauseit reflects that the fluid is convected (carried) about in the field.
2.3 Kinematics of a Fluid Particle
Kinematics is the study of space-time relationships. In order to understand a compli-cated fluid motion, we must consider quantities such as fluid velocity, acceleration,rotation, and deformation. Fluid kinematics and dynamics interact strongly, be-cause force is related to acceleration and fluid shear stress is related to the rate ofshear deformation.
28
a)
x
y
V δt
TRANSLATION
V
x
y
ROTATION
1/2∇× Vb)
x
y
STRETCHING
εxy, εxz, εyz
x
y
∇ · V
SHEAR DEFORMATION
MOTION DEFORMATION
Figure 2.2: General particle motion and its components in x – y plane. a) total xyparticle solid body motion; b) components of particle deformation.
Figure 2.3: Fluid particle in three-dimensional, three-directional velocity field; par-ticle center is at particle point (x, y, z) with velocity (u, v, w) at time t.
29
Consider a small cubical fluid particle in a three-directional, three-dimensional, un-steady flow. As the particle moves from point to point, its motion is a combinationof translation, rotation, stretching, and shear deformation. Figure (2.2) illustratesall these components of motion. We can express them in terms of the velocity. field.Figure (2.3) shows the particle located at point (x, y, z) at time t. The fluid velocityat the center of the particle (in an Eulerian description) is
V = V (x, y, z, t) = u(x, y, z, t)ı+ v(x, y, z, t)+ w(x, y, z, t)k
The velocity is continuously variable so the velocities at the faces of the fluid particlemay be different from the velocity at the center. Figure (2.3) shows these velocities,as approximated by the first term of a Taylor series.
2.3.1 Acceleration
The acceleration of the particle is the total time rate of change of its velocity. It isconsisted of all possible transformations of elementary particle. We introduce thesubscript P to remind us that we are considering a specific fluid particle, and write
VP = VP (xP , yP , zP , t) .
for the velocity of the particle P . The acceleration for this particle is
aP =d
dt
(VP)=d
dt
(VP (xP , yP , zP , t)
).
We write the velocity and acceleration in x, y and z components:
aP,x =d
dt(uP (xP , yP , zP , t))
aP,y =d
dt(vP (xP , yP , zP , t))
aP,z =d
dt(wP (xP , yP , zP , t))
Considering aP,x, we have
aP,x =∂uP∂xP
(dxPdt
)+∂uP∂yP
(dyPdt
)+∂uP∂zP
(dzPdt
)+∂uP∂t
.
However,dxPdt
= uP ,dyPdt
= vP , anddzPdt
= wP .
so
aP,x =∂uP∂t
+ uP∂uP∂xP
+ vP∂uP∂yP
+ wP∂uP∂zP
At the instant in question, the fluid particle is at xP = x, yP = y, zP = z withuP = u, vP = v and wP = w, so
ax =∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
30
By a similar development,
ay =∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+ w
∂v
∂z
az =∂w
∂t+ u
∂w
∂x+ v
∂w
∂y+ w
∂w
∂z
Note that the derivatives on the right-hand side of the acceleration equations arethe Eulerian derivatives from equation (2.2), so
ax =Du
Dt, ay =
Dv
Dt, az =
Dw
Dt.
All three equations written together are:
ax =Du
Dt=∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z; (2.4)
ay =Dv
Dt=∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+ w
∂v
∂z; (2.5)
az =Dw
Dt=∂w
∂t+ u
∂w
∂x+ v
∂w
∂y+ w
∂w
∂z. (2.6)
Or we can write them in a single equation using vector notation:
a =DV
Dt=∂V
∂t+(V · ∇
)V . (2.7)
Figure 2.4: Comparison of rotation of rectangular particles of rigid material at a)time t and b) time t+ δt; and of fluid at c) time t and d) time t+ δt.
2.3.2 Translation
The first of transformations shown in the Figure (2.2) is translation. The translation
of the fluid particle is defined by the particle velocity, V . After small time intervalposition of the particle changes according to r = ro + V δt where ro is the initialposition of the particle, and r its position after δt.
31
2.3.3 Angular Velocity and Vorticity
Purely mathematical procedure yields the acceleration of a fluid particle. However,we cannot obtain the angular velocity quite so rigorously. To understand why,consider a particle of a rigid material (such as steel), as shown in figure (2.4a).Two perpendicular lines are marked on the particle. Because the particle is rigid,both lines rotate at the same angular velocity. If the angular velocity is not zero,a short time later the lines appear as in (2.4b). Next, consider a rectangular fluidparticle with two perpendicular lines marked on it (2.4c). The fluid is deformable,so a short time later the fluid and the lines appear as shown in part (2.4d). Becauseof deformation, the two lines may have rotated different amounts. To obtain aunique value for the angular velocity of the fluid particle, we must introduce adefinition: The instantaneous angular velocity (ω) of a fluid particle is the averageof the instantaneous angular velocities of two mutually perpendicular lines on thefluid particle.
vu
A A′
B
B′
u+ ∂u∂y
δy2
u− ∂u
∂yδy2
v − ∂v
∂xδx2
v + ∂v
∂xδx2
Figure 2.5: Rectangular fluid particle with two instantaneous perpendicular linesAA′ and BB′; velocities perpendicular to AA′ and BB′ are also shown.
Figure (2.5) shows a fluid particle with two lines AA′ and BB′. By definition,
ω ≡ ωAA′ + ωBB′
2, (2.8)
where
ωAA′ =vA′ − vAδx
=
(v + ∂v
∂xδx2
)−(v − ∂v
∂xδx2
)δx
=∂v
∂x(2.9)
and
ωBB′ = −uB′ − uBδy
= −(u+ ∂u
∂yδy2
)−(u− ∂u
∂yδy2
)δy
= −∂u∂y, (2.10)
so
ωz = 1/2
(∂v
∂x− ∂u
∂y
). (2.11)
32
We added the subscript z to ω to remind us that this is actually the angular velocityabout the z axis (perpendicular to the paper). The 1/2 is somewhat inconvenient,so instead of the angular velocity ωz, we often use the vorticity ζz, defined by
ζz = 2ωz . (2.12)
In three-directional, three-dimensional flow, the angular velocity and vorticity havethree components:
ζx = 2ωx =∂w
∂y− ∂v
∂z; (2.13)
ζy = 2ωy =∂u
∂z− ∂w
∂x; (2.14)
ζz = 2ωz =∂v
∂x− ∂u
∂y. (2.15)
in vector notation,ζ = 2ω = ∇× V . (2.16)
A flow in which angular velocity and vorticity are zero is called an irrotational flow.In irrotational three-dimensional flow, all three components of ω and ζ must equalzero.The simplest continuum motion is the solid body motion. We prove now that forsuch motion is also valid equation (2.16). Form the courses in Mechanics we nowthat velocity of the arbitrary point at the solid body is expressed as:
V = VP + ω × (r − rP ) (2.17)
where V is the velocity of the arbitrary point of the body defined by position vector r,velocity VP is translational velocity of the some point P at the body whose positionvector is rP , and ω is instantaneous angular velocity of the body (the same for allpoints of the body, i.e. does not depend on the spatial coordinates). Note that
application of any differential operator to constants VP , ω, and rP results to zero.Let us now apply ∇× on both sides of the equation (2.17) from the left-hand side:
∇× V = ∇× VP︸ ︷︷ ︸=0
+∇× (ω × r)−∇× (ω × rP )︸ ︷︷ ︸=0
in former equation differentiation of constant terms is marked as zero. Applying thevector differentiation rule
∇× (a×b) = a∇ ·b−b∇ · a+ (b∇)a− (a∇)b ,
to the remaining right-hand side term we obtain:
∇× V = ∇× (ω × r) = ω(∇ · r)︸ ︷︷ ︸3(ω
−r(∇ · ω)︸ ︷︷ ︸=0
+(r∇)ω︸ ︷︷ ︸=0
− (ω∇)r︸ ︷︷ ︸(ω
= 2ω .
what is exactly equation (2.16).
33
2.3.4 Rate of Volumetric Strain
A fluid particle can be “stretched” and “squeezed”, as illustrated in figure (2.6). Thisprocess can cause a change in the volume of the fluid particle. The rate of changeof volume, divided by the volume itself, is called the rate of volumetric strain orstretching:
1
δVd
dt(δV) ≡ 1
δVD
Dt(δV) .
The volume of this fluid particle is (refer also to figure 2.3)
δV = δx× δy × δz .In three-directional, three-dimensional flow, the particle stretches or shrinks in allthree directions, so
d(δV)dt
= δyδzd(δx)
dt+ δxδz
d(δy)
dt+ δxδy
d(δz)
dt.
We have δx as the distance between the left and right faces of the fluid particle;
δy
δx
t+ δtt
Figure 2.6: Fluid particle stretched in x direction and squeezed in y direction.
d(δx)/dt is the relative x velocity between the two faces:
d(δx)
dt= δ
(dx
dt
)= u|right face − u|left face
ord(δx)
dt=
(u+
∂u
∂x
δx
2
)−(u− ∂u
∂x
δx
2
).
Simplifying, we getd(δx)
dt=∂u
∂xδx .
By a similar argument,
d(δy)
dt=∂v
∂yδy and
d(δz)
dt=∂w
∂zδz .
34
so the rate of volumetric strain for a three-dimensional, three-directional flow is:
1
δVd(δV)dt
=∂u
∂x+∂v
∂y+∂w
∂z. (2.18)
In vector notation:1
δVd(δV)dt
= ∇ · V =1
δVD
Dt(δV) . (2.19)
Figure 2.7: Shear deformation of fluid particles, positive deformation angles φ1 andφ2 are shown.
2.3.5 Shear Strain Rate
In addition to rotation and volumetric strain, a fluid particle in a flow field alsoexperiences shear deformation. Next to acceleration, the rate of shear deformationis the most important kinematic property because the shear stresses in the fluid arerelated to this strain rate.A general definition of strain rate can be developed with the aid of a sketch. Figure(2.7) shows an initially rectangular fluid particle with two perpendicular lines on itand the deformed fluid particle at a later time. The average shear deformation, φ,is defined by
φ ≡ 1/2 (φ1 + φ2) ,
with φ1 and φ2 positive as shown. The rate of shear deformation, φ, is
φ =dφ
dt= 1/2
(dφ1dt
+dφ2dt
). (2.20)
Borrowing results from the discussion of angular velocity, we have
dφ1dt
= ωAA′ anddφ2dt
= −ωBB′
When we use Eqs. (2.9) and (2.10), equation (2.20) becomes
φ = 1/2
(∂u
∂y+∂v
∂x
)= εxy = εyx . (2.21)
35
This expression can be extended to a three-dimensional fluid particle, where thereare six component rates of shear strain, which are:
εxy = εyx = 1/2
(∂u
∂y+∂v
∂x
)
εxz = εzx = 1/2
(∂u
∂z+∂w
∂x
)(2.22)
εyz = εzy = 1/2
(∂v
∂z+∂w
∂y
)
We can place all deformation velocities (volumetric strain and shearing deformationvelocities) into one matrix:
S =
∂u∂x
εyx εzxεxy
∂v∂y
εzyεxz εyz
∂w∂z
=
ux 1/2(uy + vx) 1/2(uz + wx)1/2(uy + vx) vy 1/2(vz + wy)1/2(uz + wx) 1/2(vz + wy) wz
(2.23)
which is called the tensor of deformation velocities, the first index deontes the di-rection normal to the plane in which shearing acts, and the second index denotesdirection of shearing. It is customary to denote ux as εxx, vy as εyy, and wz as εzz
2.3.6 The Strain Rate Generally
The velocity deformation rate component is represented by the term ∂ui
∂xj, where ui is
the velocity component, and indices are one of i, j ∈ x, y, z. The symmetrical partof the deformation rate tensor is the strain rate tensor, and the anti-symmetricalpart is the rotation tensor. The strain rate tensor is denoted as εij and the rotationtensor as ωij. The deformation rate can be written in terms of εij and ωij as:
∂ui∂xj
= εij + ωij , i, j ∈ x, y, z . (2.24)
The components of the deformation velocities tensor can be, thus, always obtainedif the strain rate tensor and the rotation tensor are known.Cartesian expression for strain rate and rotational tensor are given below:
εij = εji =1
2
(∂ui∂xj
+∂uj∂xi
)(2.25)
ωij = −ωji = −ωk =1
2
(∂ui∂xj
− ∂uj∂xi
)i = j (2.26)
where indices i, j, and k denote components in coordinate directions. Summing upthe righthand sides of the former two equations we obtain ∂ui/∂xj what is assumedby expression (2.24).The general expression for particle acceleration is obtained when we apply Eulerderivative to the velocity vector V :
DV
Dt=∂V
∂t+ u
∂V
∂x+ v
∂V
∂y+ w
∂V
∂z
36
or in component form:
Du
Dt=
∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂zDv
Dt=
∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+ w
∂v
∂z(2.27)
Dw
Dt=
∂w
∂t+ u
∂w
∂x+ v
∂w
∂y+ w
∂w
∂z
If we, now, substitute expression (2.24) into former set of equations we obtain:
Du
Dt=
∂u
∂t+ uεxx + v (εyx − ωz) + w (εzx + ωy)
Dv
Dt=
∂v
∂t+ u (εxy + ωz) + vεyy + w (εzy − ωx) (2.28)
Dw
Dt=
∂w
∂t+ u (εxz − ωy) + (εyz + ωx) + wεzz
or in matrix form:
D
Dt
uvw
=∂
∂t
uvw
+
0 −ωz ωy
ωz 0 −ωx
−ωy ωx 0
uvw
+
εxx εxy εxzεyx εyy εyzεzx εzy εzz
uvw
The first two terms on the right-hand side represent the solid body acceleration,while the last term is the contribution of the deformation to the acceleration. Wecan shortly write the former equations as:
DV
Dt=∂V
∂t+ ω × V + V · S (2.29)
where V · S denotes product between vector and tensor which is performed in a suchway, that for each vector component of the left-hand side suitable row of the tensorS is chosen for dot product.
2.4 Differential Form of Continuity Equation
One of the fundamental laws of nature which states that mass can neither be creatednor destroyed will be used to derive the differential form of continuity equation.Mathematically this gives
D(δm)
Dt= 0 (2.30)
where δm = δV is the mass of a fluid particle, is the density and δV is thevolume of a particle. If former expression for δm is substituted into Equation (2.30)the following equivalent expression is obtained
D(δm)
Dt=D(δV)Dt
=D
DtδV +
D(δV)Dt
= 0
37
Let us now divide former equation by δV and substitute derivative D/Dt by theequation (2.1) in which b is replaced by
∂
∂t+ u
∂
∂x+ v
∂
∂y+ w
∂
∂z+
δVD(δV)Dt
= 0 (2.31)
The last term is the rate of volumetric strain. If we now substitute the last term onthe left-hand side by the expression given by (2.18) the following results:
∂
∂t+ u
∂
∂x+ v
∂
∂y+ w
∂
∂z+
(∂u
∂x+∂v
∂y+∂w
∂z
)= 0 (2.32)
orD
Dt+ (∇ · V ) = 0 (2.33)
what is the non-conservative form of the continuity equation. Former equation canbe easily transformed into conservative form:
∂
∂t+∂u
∂x+∂v
∂y+∂w
∂z= 0 (2.34)
or in the vector notation:∂
∂t+∇ ·
(V
)= 0 (2.35)
For steady flows derivative with respect to t is dropped and equation reduces tocompressible steady flow continuity equation:
∇ ·(V
)= 0 (2.36)
or∂u
∂x+∂v
∂y+∂w
∂z= 0 (2.37)
In the case that fluid can be considered incompressible equation reduces to:
∇ · V =∂u
∂x+∂v
∂y+∂w
∂z= 0 (2.38)
what means that the rate of volumetric strain in that case is equal to zero. Notethat continuity equation has the same form for steady and unsteady flow when fluidis incompressible. This fact can be used to adopt some steady state computationalprocedures to unsteady one.
2.5 Elementary Cube
Before going on further let us examine differential properties of fluid particle chosento be of a cubic shape, shown in the figure (2.8). Let the property ϕ be defined inthe center of the cube. On the faces of the cube which are ±δx/2, ±δy/2 or ±δz/2apart values of a property ϕ are little-bit different. We can estimate these valuesusing Taylor expansion truncated after first derivative term. Values of a property ϕ
38
are shown in the figure (2.8), for the faces of the cube visible to observer. For thefaces behind these values are determined similarly by the expressions:
ϕ− ∂ϕ
∂x
δx
2for −x direction
ϕ− ∂ϕ
∂y
δy
2for −y direction (2.39)
ϕ− ∂ϕ
∂z
δz
2for −z direction
Also notice that unit normal to cube faces are ı, −ı, , −, k, and −k for front,
Figure 2.8: Differential properties of the cubic particle.
back, right-hand, left-hand, top, and bottom face respectively. Practically always itis needed to find total “inflow” of the variable ϕ (which is usually specified per unitarea) through parallel faces, thus(
ϕ+∂ϕ
∂x
δx
2
)(ı δyδz) +
(ϕ− ∂ϕ
∂x
δx
2
)(−ı δyδz) = ı
∂ϕ
∂xδV(
ϕ+∂ϕ
∂y
δy
2
)( δxδz) +
(ϕ− ∂ϕ
∂y
δy
2
)(− δxδz) =
∂ϕ
∂yδV (2.40)(
ϕ+∂ϕ
∂z
δz
2
)(k δxδy
)+
(ϕ− ∂ϕ
∂z
δz
2
)(−k δxδy
)= k
∂ϕ
∂zδV
2.6 Momentum Equations
Next we will discuss some common forms of momentum equation. Type of forcestaken into account differentiate forms of equations.
2.6.1 The Cauchy Equations
The Cauchy equations are the differential equations of motion expressed in termsof stress. Consider a fluid particle located at point (x, y, z) in a three-dimensional,
39
three-directional flow field. The particle experiences forces from viscous stress aswell as gravity and pressure. Newton’s second law, for the x , y and z directions,respectively, gives
δmpax = δFx,pressure + δFx,gravity + δFx,stress
δmpay = δFy,pressure + δFy,gravity + δFy,stress (2.41)
δmpaz = δFz,pressure + δFz,gravity + δFz,stress
Mass of the particle is given by:
δmp = δxδyδz
Acceleration of the particle (its center of gravity) is obtained from Euler derivative:
ax =∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
ay =∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+ w
∂v
∂z
az =∂w
∂t+ u
∂w
∂x+ v
∂w
∂y+ w
∂w
∂z
Pressure force acts on particle faces (in direction opposite to surfaces unit normals)so we must to apply relation derived for general surface variable ϕ, equation (2.40),
substituting ϕ by −p(ı+ + k) and δV by δx δ yδz
δFx,pressure = −∂p∂xδx δy δz
δFy,pressure = −∂p∂yδx δy δz
δFz,pressure = −∂p∂zδx δy δz
Among various body forces we consider here only gravity force g δmp = g δV whoseprojections to coordinate axes are
δFx,gravity = gx δx δy δz ;
δFy,gravity = gy δx δy δz ;
δFz,gravity = gz δx δy δz .
In order to determine the net stress force acting on a fluid particle, we consideragain figure (2.8), and equation (2.40). In general two types of stresses may acton the fluid: tangential (stress) and normal stress. In a presence of volumetricstrain, besides pressure, appears also corresponding normal stress so normal stressis represented by two components: pressure p and additional normal stress σ. Weassume that at cubic center acts pressure p and three normal stresses σx, σy andσz. Shear stresses τxy, τxz, . . . , τzx and τzy are also assumed to act at cubic center.Figure (2.9) shows elementary cube with stresses on its faces.The first subscript on the shear stress identifies the coordinate direction perpendic-ular to the face of the particle on which the particular shear stress act. The second
40
Figure 2.9: Stresses acting on the particle surfaces.
subscript identifies the direction in which the particular shear stress acts. In three-dimensional flow, x-face have stresses σx, τxy, and τxz acting on it. So on each faceacts three stresses one normal and two in the plane of the face. Since the valuesof these stresses are little-bit different from values at cube center we must applyequation (2.40) to calculate net force. Application of equation (2.40) to each stresscomponent, and gathering components which acts in the same coordinate direction,yields
δFx,stress =
(∂σx∂x
+∂τyx∂y
+∂τzx∂z
)δx δy δz
δFy,stress =
(∂τxy∂x
+∂σy∂y
+∂τzy∂z
)δx δy δz
δFz,stress =
(∂τxz∂x
+∂τyz∂y
+∂σz∂z
)δx δy δz
Next, we substitute the particle’s mass and acceleration and all the forces into Eqs.(2.41), divide by δxδyδz and take the limit, shrinking the particle to the point(x, y, z). The results are
(∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
)= gx − ∂p
∂x+∂σx∂x
+∂τyx∂y
+∂τzx∂z
(∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+ w
∂v
∂z
)= gy − ∂p
∂y+∂τxy∂x
+∂σy∂y
+∂τzy∂z
(∂w
∂t+ u
∂w
∂x+ v
∂w
∂y+ w
∂w
∂z
)= gz − ∂p
∂z+∂τxz∂x
+∂τyz∂y
+∂σz∂z
(2.42)
Like deformation velocities stresses can also be put into matrix form. Let us intro-duce new symbols τxx, τyy, and τzz which resemble σx, σy, and σz respectively, and
41
build the matrix of the form
τij =
τxx τyx τzxτxy τyy τzyτxz τyz τzz
(2.43)
which is called stress tensor. Now we can write the system of equations (2.42) morecompact form
DV
Dt=
∂V
∂t+
(V · ∇
)V = g −∇p+∇ · τij (2.44)
where τij is the stress tensor, operator ∇ should be applied to each row of the stress
tensor for each coordinate direction. The term (V ·∇)V is not suitable for coordinatetransformation so it is often substituted using vector identity
(V · ∇
)V = ∇
(V 2
2
)− V ×
(∇× V
)(2.45)
Replacing this relation into (2.44) we obtain
DV
Dt=
∂V∂t
+∇(V 2
2
)− V ×
(∇× V
) = g −∇p+∇ · τij (2.46)
These are the Cauchy equations for three-dimensional, three-directional flow. Three-directional, three-dimensional viscous flow must satisfy the Cauchy equations andthe differential continuity equation. These four equations, however, contain thirteendependent variables (u, v, w, p, τxx, τyx, τzx, τyx, τyy, τzy, τxz, τyz and τzz). If weconsider the rotational equilibrium of a fluid, than
τxy = τyx , τxz = τzx , τyz = τzy .
Even so, we still have ten dependent variables and only four equations. To obtain aclosed set of equations, we must relate the fluid stresses to the velocity field.
2.7 Navier-Stokes Equation
In order to reduce number of equations we have to connect the stress tensor withthe tensor of deformation velocities:
τi,j = f(S) (2.47)
Navier-Stokes equations are obtained from Cauchy equations under following as-sumptions:
1. Stress tensor is symmetric. This is needed to be able to switch from onecoordinate system to another using six equations of equilibrium.
2. The viscous stress linearly depend on first derivatives of velocity. This impliesthat there are no viscous stress at rest.
42
3. The viscous stress vanish for a rigid body motion. This exclude vorticity fromconsideration.
4. Viscous stress vanish for homotetic deformation of a particle. Homotetic de-formation is such for which deformed body remains similar to the body beforedeformation.
5. For the velocity field u = u(y), w = v = 0 equation (1.5) from Section (1.4)must be satisfied.
All these requirements can be satisfied by the following relation between stressesand deformation velocities, as it is indicated by (2.47): τxx τyx τzx
τxy τyy τzyτxz τyz τzz
= a
uxuy+vx
2uz+wx
2uy+vx
2vy
vz+wy
2uz+wx
2vz+wy
2wz
+ λ
∇ · V 0 0
0 ∇ · V 0
0 0 ∇ · V
(2.48)
The first criterion is met since matrices on the right-hand side are symmetric. Secondcriterion is met since only first order derivative of the velocity components are used.Vorticity is not included on the right-hand side. Homotetic deformation is obtainedwhen ux = vy = wz and all other derivatives are equal zero. In that case viscousstress will be zero only if the following is satisfied:
a ux + 3λux = 0 , λ = −a/3From the last requirement follows that a = 2µ and thus λ = −2µ/3. So viscousstresses and deformation velocities are related by the equation:
τxx τyx τzxτxy τyy τzyτxz τyz τzz
= µ
2ux +
λµ∇ · V uy + vx uz + wx
vx + uy 2vy +λµ∇ · V vz + wy
uz + wx vz + wy 2wz +λµ∇ · V
(2.49)
or explicitly
τxx = 2µux + λ∇ · V τyx = µ (uy + vx) τzx = µ (uz + wx)
τxy = µ (uy + vx) τyy = 2µvy + λ∇ · V τzy = µ (vz + wy)
τxz = µ (uz + wx) τyz = µ (vz + wy) τzz = 2µwz + λ∇ · V(2.50)
In the equation (2.50) we have retained symbol λ instead of −2/3µ since assump-tion 4 is not always valid. That assumption is equivalent to demand that averagevalue of normal stresses equals zero (Stokes hypothesis). Experiments show that inregions with high density gradients (like shock waves) that value is of opposite signand greater then first coefficient. Out of shock waves Stokes hypothesis does not in-troduce significant errors. Cauchy equation, equation (2.42) together with relations(2.50) is called Navier-Stokes equation.Right-hand side of the Cauchy equation, equation (2.42) when expressions (2.50)are used is given in the form
∂τxx∂x
+∂τyx∂y
+∂τzx∂z
=∂
∂x
[2µux + λ
(∇ · V
)]+∂
∂y[µ(uy + vx)]
43
+∂
∂z[µ(uz + wx)]
∂τxy∂x
+∂τyy∂y
+∂τzy∂z
=∂
∂x[µ(vx + uy)] +
∂
∂y
[2µvy + λ
(∇ · V
)]+
∂
∂z[µ(vz + wy)]
∂τxz∂x
+∂τyz∂y
+∂τzz∂z
=∂
∂x[µ(wx + uz)] +
∂
∂y[µ(vz + wy)]
+∂
∂z
[2µwz + λ
(∇ · V
)]If we assume constant viscosity coefficients than we can rearrange Navier-Stokesequation (equation above)
∂τxx∂x
+∂τyx∂y
+∂τzx∂z
= µ∇2u+ (µ+ λ)∂
∂x
(∇ · V
)∂τxy∂x
+∂τyy∂y
+∂τzy∂z
= µ∇2v + (µ+ λ)∂
∂y
(∇ · V
)(2.51)
∂τxz∂x
+∂τyz∂y
+∂τzz∂z
= µ∇2w + (µ+ λ)∂
∂z
(∇ · V
)Equation (2.42) together with (2.51) gives
(∂u
∂t+ u
∂u
∂x+ v
∂u
∂y+ w
∂u
∂z
)= gx − ∂p
∂x+ µ∇2u+ (µ+ λ)
∂
∂x
(∇ · V
)
(∂v
∂t+ u
∂v
∂x+ v
∂v
∂y+ w
∂v
∂z
)= gy − ∂p
∂y+ µ∇2v + (µ+ λ)
∂
∂y
(∇ · V
)(2.52)
(∂w
∂t+ u
∂w
∂x+ v
∂w
∂y+ w
∂w
∂z
)= gz − ∂p
∂z+ µ∇2w + (µ+ λ)
∂
∂z
(∇ · V
)or in more compact form
DV
Dt=
∂V∂t
+(V · ∇
)V
= g −∇p+ µ∇2V + (µ+ λ)∇(∇ · V
)(2.53)
Or if viscosity λ is replaced by −2/3µ but both coefficients are assumed constant
DV
Dt=
∂V∂t
+(V · ∇
)V
= g −∇p+ µ∇2V +µ
3∇(∇ · V
)(2.54)
If the fluid can be assumed incompressible than last term on the right-hand side isequal to zero, thus:
DV
Dt=
∂V∂t
+(V · ∇
)V
= g −∇p+ µ∇2V (2.55)
Or when the term (∇ · V )V is replaced with equivalent expression
∂V∂t
+∇(V 2
2
)+ V ×
(∇× V
) = g −∇p+ µ∇2V (2.56)
44
2.8 Energy Equation
We derive energy equation applying the first law of thermodynamics to fluid particlewhich states
Q− W =d
dt
(e+ 1/2V
2 − g · r) δV
where Q is heat flow rate to the particle, W is work done by the particle, e is internalenergy per unit mass, (ech = en = el = 0), 1/2V 2 is kinetic energy per unit mass and−g · r is energy of position per unit mass. Since mass is preserved we can put termδV in front of the differential. We know the forces acting on the particle surface sowork done on the particle should be taken instead of minus sign with plus sign informer equation.To derive heat flow and work we apply consideration of Sec. (2.5) and equation(2.40). Heat flux q is little-bit different at cube surfaces then on its center and heatis increased if heat flux is directed opposite to direction of face unit normals of thefluid particle. Heat flux, thus, is taken with minus sign. Total heat flow to fluidparticle is thus:
Q = −(∂qx∂x
+∂qy∂y
+∂qz∂z
)δV = −∇q δV (2.57)
where (qx, qy, qz) are components of heat flux.Work rate done to the particle is expressed as product between forces and velocities(F · V ):
W =
((τxx − p) + ∂(τxx − p)
∂x
δx
2
)δyδz
(u+
∂u
∂x
δx
2
)−(
(τxx − p)− ∂(τxx − p)∂x
δx
2
)δyδz
(u− ∂u
∂x
δx
2
)+(
τyx +∂τyx∂y
δy
2
)δxδz
(u+
∂u
∂y
δy
2
)−(τyx − ∂τyx
∂y
δy
2
)δxδz
(u− ∂u
∂y
δy
2
)+(
τzx +∂τzx∂z
δz
2
)δxδy
(u+
∂u
∂z
δz
2
)−(τzx − ∂τzx
∂z
δz
2
)δxδy
(u− ∂u
∂z
δz
2
)+(
(τyy − p) + ∂(τyy − p)∂y
δy
2
)δxδz
(v +
∂v
∂y
δy
2
)−(
(τyy − p)− ∂(τyy − p)∂y
δy
2
)δxδz
(v − ∂v
∂y
δy
2
)+(
τxy +∂τxy∂x
δx
2
)δyδz
(v +
∂v
∂x
δx
2
)−(τxy − ∂τxy
∂x
δx
2
)δyδz
(v − ∂v
∂x
δx
2
)+(
τzy +∂τzy∂z
δz
2
)δxδy
(v +
∂v
∂z
δz
2
)−(τzy − ∂τzy
∂z
δz
2
)δxδy
(v − ∂v
∂z
δz
2
)+(
(τzz − p) + ∂(τzz − p)∂z
δz
2
)δxδy
(w +
∂w
∂z
δz
2
)−(
(τzz − p)− ∂(τzz − p)∂z
δz
2
)δxδy
(w − ∂w
∂z
δz
2
)+
45
(τxz +
∂τxz∂x
δx
2
)δyδz
(w +
∂w
∂x
δx
2
)−(τxz − ∂τxz
∂x
δx
2
)δyδz
(w − ∂w
∂x
δx
2
)+(
τyz +∂τyz∂y
δy
2
)δxδz
(w +
∂w
∂y
δy
2
)−(τyz − ∂τyz
∂y
δy
2
)δxδz
(w − ∂w
∂z
δy
2
)
Finite order term cancel each other in the previous equation and second order termscan be neglected compared to products between first order terms and finite values,so former expression reduces to
W
δV = u
(∂τxx∂x
− ∂p
∂x+∂τyx∂y
+∂τzx∂z
)+ (τxx − p)∂u
∂x+ τyx
∂u
∂y+ τzx
∂u
∂z+
v
(∂τyy∂y
− ∂p
∂y+∂τxy∂x
+∂τzy∂z
)+ (τyy − p)∂v
∂y+ τxy
∂v
∂x+ τzy
∂v
∂z+
w
(∂τzz∂z
− ∂p
∂z+∂τxz∂x
+∂τyz∂y
)+ (τzz − p)∂w
∂z+ τxz
∂w
∂x+ τyz
∂w
∂y
(2.58)
or after regrouping
W = δV[∂
∂x(τxxu− pu+ τxyv + τxzw)
+∂
∂y(τyyv − pv + τyxu+ τyzw) + ∂
∂z(τzzw − pw + τzxu+ τzyv)
](2.59)
If we denote with et total internal energy per unit mass, which in this case is consistedof sensible energy, kinetic energy, and potential energy:
et = e+ 1/2V2 − g · r (el = ech = en = 0)
and apply Euler derivative to it
DetDt
=∂et∂t
+ (V · ∇)et (2.60)
then we have completed derivation of all terms found in the first law of thermody-namics. Gathering (2.60), (2.59), and (2.57) into the first law of thermodynamics,the following is obtained
−∇q + V · F = DetDt
or:
∂et∂t
+ (V · ∇)et +∇q = ∂
∂x(τxxu− pu+ τxyv + τxzw)
+∂
∂y(τyyv − pv + τyxu+ τyzw) + ∂
∂z(τzzw − pw + τzxu+ τzyv)
(2.61)
what represents the first law of thermodynamics for the moving and deforming fluidparticle. We can move pressure from the right-hand side to the left-hand side noting
46
that on right hand side we have −∇(pV ) ≡ −(V · ∇)p− p∇ · V . After moving part
of the pressure term (V · ∇)p to the left-hand side and adding to both sides ∂p/∂twe obtain
DetDt
+Dp
Dt+∇q = ∂p
∂t− p∇ · V +
∂
∂x(τxxu+ τxyv + τxzw)
+∂
∂y(τyyv + τyxu+ τyzw) +
∂
∂z(τzzw + τzxu+ τzyv) (2.62)
We can now include Dp/Dt into single term et + p/ obtaining
D
Dt
(et +
p
)+p
D
Dt+∇q = ∂p
∂t− p∇ · V +
∂
∂x(τxxu+ τxyv + τxzw)
+∂
∂y(τyyv + τyxu+ τyzw) +
∂
∂z(τzzw + τzxu+ τzyv)
where we added the term (p/)(D/Dt) in order to maintain identity between this
equation and equation (2.62). Euler derivative D/Dt can be replaced by −∇Vfrom equation (2.33), so we obtain on both sides the same term −p∇ · V whichcancels out and we obtain
D
Dt
(et +
p
)+∇q = ∂p
∂t+∂
∂x(τxxu+ τxyv + τxzw)
+∂
∂y(τyyv + τyxu+ τyzw) +
∂
∂z(τzzw + τzxu+ τzyv) (2.63)
Note that total enthalpy h is defined as h = et+ p/. Thus, equation (2.63) is thefirs principle of thermodynamics in enthalpy form.
2.9 Mechanical Energy Equation
The equation which we are deriving now is not an independent law, but is derivedfrom the momentum equation. The dot product of the velocity V with momentumequation leads to the mechanical (kinetic) energy equation. Let us note that:
D
Dt
(1/2V · V
)= V · D
V
Dt
what is the left-hand side of the equation (2.44) multiplied by V , it must be, thus,
equal to the right-hand side of the equation (2.44) multiplied by V
D
Dt
(1/2V · V
)= V ·
∂V∂t
+ (V∇
)V
= V · [g −∇p+∇ · τij] (2.64)
or after expansion
D
Dt
(1/2V · V
)− V · g = u
(∂(τxx − p)
∂x+∂τyx∂y
+∂τzx∂z
)+
+v
(∂τxy∂x
+∂(τyy − p)
∂y+∂τzy∂z
)+ w
(∂τxz∂x
+∂τzy∂z
+∂(τzz − p)
∂z
)(2.65)
47
If we compare right-hand side of equation (2.65) to right-hand side of equation(2.58) than we observe that half of terms in the equation (2.65) are missing. Wecan interpret this situation in the following way: Terms given in the equation (2.65)contribute to acceleration of the fluid particle and increase in mechanic energy whilerest of the terms in equation (2.58) contribute to increase of thermal energy. Re-maining terms without pressure gradients are called the dissipation function and aredenoted by Φ
Φ = τxx∂u
∂x+ τyx
∂u
∂y+ τzx
∂u
∂z+
τyy∂v
∂y+ τxy
∂v
∂x+ τzy
∂v
∂z+ τzz
∂w
∂z+ τxz
∂w
∂x+ τyz
∂w
∂y
If we replace expression for viscous stresses from equation (2.50) into previous equa-tion, and after some regrouping we obtain
Φ = λ(∇ · V
)2+ 2µ
(u2x + v
2y + w
2z
)+
+µ (vx + uy)2 + µ (wy + vz)
2 + µ (uz + wx)2 (2.66)
All terms in equation (2.66) are positive so must be Φ, in accordance with the secondlaw of thermodynamics, since viscosity cannot add energy to the system. On theother hand, from Stoke’s hypothesis λ = −2/3µ and is negative. Since dissipationfunction Φ must be positive correct condition for µ and λ may follow from (2.66)when homotetic deformation is assumed:
µ ≥ 0 , 9λ+ 6µ ≥ 0
If we subtract mechanical energy equation (2.65) from energy equation, equation(2.63) we obtain
Dh
Dt+∇ · q = ∂p
∂t+∇pV + Φ
or
Dh
Dt+∇ · q = Dp
Dt+ Φ (2.67)
what is the energy equation in enthalpy form, where h = et + p/.
2.10 The Flow Equations Overview
We repeat here allready derived vector forms of the flow equations. The Continuityequation is given by (2.35):
∂
∂t+∇ ·
(V
)= 0 (2.68)
Momentum equation or Newton second law of motion is given by:
∂V
∂t+∇ ·
(V 2
2
)+ (∇× V )× V = g − ∇p
+µ
∇2V +
µ+ λ
∇(∇ · V
)(2.69)
48
First principle of thermodynamics is expressed as
D
Dt
(et +
p
)+∇q = ∂p
∂t+∂
∂x(τxxu+ τxyv + τxzw)
+∂
∂y(τyyv + τyxu+ τyzw) +
∂
∂z(τzzw + τzxu+ τzyv) (2.70)
or in enthalpy form
Dh
Dt+∇ · q = Dp
Dt+ Φ (2.71)
or in more compact form:
DetDt
+∇q = V · F (2.72)
where F is resultant of pressure force and surface stresses acting on the fluid particle.To conclude discussion about general physical law we repeat here also equation ofstate which can be choosen from one of the following expressions
p = RT
e =1
γ − 1
p
(2.73)
h =γ
γ − 1
p
49
Chapter 3
Fundamentals of Inviscid Flow
3.1 General Assumptions
Assume:
1. Fluid is continuum.
2. Fluid is inviscid and adiabatic.
3. Perfect gas or constant density fluid.
4. Shock waves are treated separately.
Consequences:
1. Can apply mathematical analysis.
2. No heat exchange, No boundary layer, no friction, no flow separation.
3. One of the following relationships hold:
p
= RT or: = const
4. Shock wave is the boundary of the problem.
3.2 Continuity Equation
Is expressed as:∂
∂t+∇
(V
)=D
Dt+ ∇V = 0 (3.1)
In subsequent expressions is density, p is static pressure, and T is temperature.Velocity V is defined as:
V = Uı+ V +Wk (3.2)
where ı,,k are the unit vector of the coordinate axes x, y, z.
50
3.3 Newton Second Law of Motion
Is expressed as:DV
Dt= F − ∇p
(3.3)
where F is force per unit mass (such as gravity). Often we write:
F = ∇Π (3.4)
If F = −gk, and g = 9.80665m/s2, then:
Π = −gz (3.5)
3.4 Conservation of Thermodynamic Energy
D
Dt
(e+
|V |22
)= −1
∇(pV
)+ F · V (3.6)
where e is internal energy per unit mass, |V | =∣∣∣V ∣∣∣ is magnitude of the vector V . If
we introduce enthalpy h = e+ p/ in former equation we get:
D
Dt
(h+
|V |22
)=∂p
∂t+ F · V (3.7)
The simplest form of this law is:Ds
Dt= 0
where s is entropy per unit mass.
3.5 Equation of state
For thermally perfect gases is given in the form:
p = RT and: CP , CV = const (3.8)
For calloricaly perfect gas:CP , CV = const (3.9)
For constant density fluid = const
Ratio of specific heats for air is given in the form:
γ =CP
CV
Equivalent expressions for equation of state:
e = e(p, ) p = p(e, s)
51
3.6 Euler Equations of Motion
For the following seven unknowns: u, v, w, p, , e, T we have corresponding numberof equations:
1. Vector equation of motion (3 scalar equations):
DV
Dt= F −∇p
2. Conservation of mass equation:
D
Dt+ ∇ · V = 0
3. Energy equation, if we from equation (3.6) subtract equation (3.3) multiplied
by V we obtain:
De
Dt= −p∇V
4. Equation of state
p = RT , p = p(e, ) , e = e(p, )
3.7 Vorticity and Circulation
3.7.1 Vorticity
Circulation is defined as:
Γ =∮c
V · dr =∮cUdx+ V dy +Wdz (3.10)
or:Γ =
∮c
V · dr = ∑i
∮-i
V · dr
If the square regions are of infinitesimal size (see figure (3.2)) we get:∮-i
V · dr ≈ U∆x+1
2
∂U
∂x∆x2︸ ︷︷ ︸
AB
+V∆y +∂V
∂x∆x∆y +
1
2
∂V
∂y∆y2︸ ︷︷ ︸
BC
−U∆x− ∂U
∂y∆x∆y − 1
2
∂U
∂x∆x2︸ ︷︷ ︸
CD
−V∆y − 1
2
∂V
∂y∆y1︸ ︷︷ ︸
DA
=
(∂V
∂x− ∂U
∂y
)∆x∆y
= (∇× V ) · k∆x∆y (3.11)
so: ∑i
∮-i
V · dr ≈ (∇× V ) · k∆x∆y︸ ︷︷ ︸(n∆S
52
or:Γc =
∮c
V · dr = ∑i
(∇× V
)· n∆S (3.12)
when number of areas (∆S) tends to infinity in a such way that each ∆S → 0 we
Figure 3.1: Circulation of the vector V around closed curve c.
can replace sum on the right-hand side by area integral:
Γc =∮c
V · dr =∫ ∫S
(∇× V )n dS (3.13)
The former equation is known as Stokes theorem. The quantity
ζ = ∇× V
is called vorticity of the flow field, physically it is equal to the circulation per unitof surface.
3.7.2 Helmholtz Theorem
Let us start from the Newton second law of motion for inviscid and incompressiblefluid
DV
Dt= F −∇
(p
)We can express acceleration of the fluid particle in the form
DV
Dt=∂V
∂t+ (V · ∇)V ≡ ∂V
∂t+∇
(V 2
2
)+ ζ × V
After replacing expression for acceleration into Newton’s second law of motion weget
∂V
∂t+∇
(V 2
2
)+ ζ × V = ∇Π−∇
(p
)(3.14)
53
u, v
u+ ∂u∂y∆y
v + ∂v∂y∆y
u+ ∂u∂x∆x
v + ∂v∂x∆x
u+ ∂u∂x∆x+
∂u∂y∆y
v + ∂v∂x∆x+
∂v∂y∆y
Figure 3.2: Circulation of the vector V around square i.
where it is assumed that body forces have potential F = ∇Π, what is true forgravitational force. If we group all the terms preceded by ∇ we get
∂V
∂t+ ζ × V = ∇
(Π− p
− V 2
2
)
Let us now apply operator ∇× to the former equation. Right-hand side is equal tozero since ∇×∇( ) is always zero
∂∇× V∂t
+∇×(ζ × V
)= 0 (3.15)
but ζ = ∇× V and
∇×(ζ × V
)= (∇ · ζ)V − (V · ∇)ζ + (ζ · ∇)V − (∇ · V )ζ (3.16)
The first and last terms are always equal to zero for incompressible inviscid flow. Soequation (3.15) is transformed to
∂ζ
∂t+ (V · ∇)ζ − (ζ∇)V = 0 (3.17)
or:Dζ
Dt= (ζ · ∇)V (3.18)
We now consider right-hand side of the former equation:
(ζx∂
∂x+ ζy
∂
∂y+ ζz
∂
∂z)(uı+ v+ wk) = (ζ · ∇u)ı+ (ζ · ∇v)+ (ζ · ∇w)k
54
what represents derivative of the velocity vector V in the direction of vorticity vecotrζ. So equation (3.18) states that the material rate of change of the vorticity is zerowhenever the vorticity is zero. Thus, we can conclude if vorticity of fluid element atany time is equal to zero it is equal to zero all times.Sommerfeld has proved the theorem for any vector A that:
D
Dt( A · n dS) =
∂ A∂t
+ V∇ · A−∇× (V × A)
· n dSLet us apply this theorem to vorticity vector ζ:
D
Dt(ζ · n dS) =
∂ζ∂t
+ V∇ · ζ −∇× (V × ζ) · n dS (3.19)
Since ∇ · ζ = ∇ · (∇× V ) ≡ 0 the former equation is reduced to:
D
Dt(ζ · n dS) =
∂ζ∂t−∇× (V × ζ)
· n dS (3.20)
Expanding the second term on the right-hand side (equation (3.16)), we have
∇× (V × ζ) = (∇ · ζ)V − (V∇)ζ + (ζ∇)V − (∇ · V )ζ= −(V · ∇)ζ + (ζ · ∇)V
Substituting previous expression into equation (3.20) we get
D
Dt(ζ · n dS) =
∂ζ∂t
+ (V · ∇)ζ − (ζ · ∇)V · n dS (3.21)
or:D
Dt(ζ · n dS) =
DζDt
− (ζ · ∇)V · n dS (3.22)
But right-hand side of the former equation must be zero if equation (3.18) is valid
D
Dt(ζ · n dS) = 0 (3.23)
Equation (3.23) is very important. If the body forces have potential the materialchange of the outflow of vorticity through any surface element moving with the fluidis permanently equal to zero. Alternatively we can said outflow vorticity through anysurface element moving with the fluid remains a constant for all times, so vorticityis convected with the fluid.Equation (3.23) can also be expressed in terms of the circulation around the surfaceelement. Denoting by dΓC the circulation around the boundary C of the elementn dS, we have
dΓc = ζ · n dS (3.24)
55
Figure 3.3: Circulation around the same fluid particles.
Therefore, equation (3.23) takes the form
D
Dt(dΓc) = 0 (3.25)
We can formulate similarly the rules already expressed for equation (3.23). Thematerial change of the circulation around any surface element moving with the fluidis permanently zero if the body forces have potential, or, the circulation around anysurface element moving with the fluid remains constant all the times. Theorems(3.18) and (3.23) are basic theorems to describe vortex motion. In two-dimensional
flow product (ζ · ∇)V is identically equal to zero therefore (3.18) reduces to
Dζ
Dt= 0 (3.26)
In two-dimensional flows of an ideal fluid under the action of irrotational body forces,the vorticity of a fluid element remains a constant for all times.
3.7.3 Kelvin Theorem
Let us consider closed curve C which enclose the same fluid particles all the times,and move with them, figure (3.3). Let at the moment t the value of the circulationaround curve C is Γ. During time interval ∆t fluid particles enclosed with the curveC move to a new position C1. So C1 is the position of the curve C at the momentt + ∆t, and circulation around that curve is Γ1. The rate of change in circulationalong C following its motion is then given by the material derivative
DΓ
Dt= lim
∆t→0
Γ1 − Γ
∆t(3.27)
If we substitute definition of circulation into former equation we have
DΓ
Dt≡ D
Dt
∮C
V · ds = lim∆t→0
∮C1
V1 · ds1 −∮C
V · ds (3.28)
56
Since we are following the same set of fluid particles, the limit on the right side ofthis equation is the same as the integral
∮C(D/Dt)(
V · ds) along the curve C at theinstant t. Therefore equation (3.28) becomes
DΓ
Dt=D
Dt
∮C
V · ds =∮C
D
Dt
(V · ds
)=∮C
DV
Dt· ds+
∮C
V · DDtds (3.29)
Where DV /Dt is the acceleration of an arbitrary point A on the curve C, as it isshown in the figure (3.3). Let us now define (D/Dt)ds. If ds and ds1 are smallvectors along curves C and C1 respectively then
D
Dtds = lim
∆t→0
ds1 − ds∆t
(3.30)
From the figure (3.3) it is clear that the vector−−→AB1 can be expressed in two different
ways −−→AB1 =
−−→AA1 + ds1 = ds+
−−→BB1
ords1 − ds = −−→
BB1 −−−→AA1 (3.31)
Since−−→AA1 = V∆t and
−−→BB1 = (V + dV )∆t after substitution in equation (3.31) we
haveD
Dtds = dV
It than follows that ∮C
V · DDtds =
∮C
V · dV =∮C
d
(V 2
2
)= 0 (3.32)
With former relation equation (3.29) becomes
DΓ
Dt=∮C
DV
Dt· ds =
∮C
a · ds (3.33)
where a is acceleration vector. Equation (3.33) states that at any instant the rateof change of circulation around any fluid curve is equal to the line integral of theacceleration around that curve taken at the instant considered. An immediate con-sequence of equation (3.33) is that if the acceleration is expressible as the gradient ofsome (single-valued) scalar function, that is, if the acceleration field is an irrotationalvector field, then the integral ∮
C
a · ds
is zero, and consequently DΓ/Dt is zero. Equation (3.33) is derived on strictlykinematical basis. We need dynamical considerations in order to evaluate formerintegral. If we restrict ourselves to an inviscid fluid and use Euler’s equation we get
DΓ
Dt=∮C
a · ds =∮C
F · ds−∮C
∇p· ds (3.34)
57
If we again assume that body force F has potential (F = ∇Π, it is irrotational) anduse identity dΠ = ∇Π · ds, so closed loop integral of uniquely defined quantity mustbe equal to zero
∮dΠ ≡ 0, we obtain
DΓ
Dt= −
∮C
∇p· ds (3.35)
If it is possible to assume that fluid is incompressible ( = const) then∮C
∇p· ds = 1
∮C
dp = 0 ⇒ DΓ
Dt= 0
If the fluid is assumed to be either compressible or incompressible and inhomoge-neous so that is a variable, the integral no longer vanishes automatically for anyarbitrary curve C. However if there is a single-valued relation between the densityand the pressure so that one can express
∇p
= ∇P ⇒∮C
∇p· ds =
∮C
∇P · ds =∮C
dP = 0 ⇒ DΓ
Dt= 0
Thus, when we assume that is constant or that there is a single-valued relationbetween and p, equation (3.35) becomes
DΓ
Dt= 0 (3.36)
This equation express the theorem of conservation of circulation. The rate of changeof circulation around any fluid curve is permanently zero if the body forces are irro-tational and if there is a single-valued, pressure-density relation, or, the circulationaround a fluid curve remains a constant for all times as the curve moves with thefluid.Consider now any infinitesimal surface element ndS that moves with the fluid. LetdΓc denote the circulation around the boundary c of such a surface element. Wehave from equation (3.24)
dΓc = ζ · n dSand from equation (3.36)
D
Dt( ζ · n dS ) = 0
which is exactly Hemholtz’s theorem, equation (3.23). So we concluded that theo-rems of Kelvin and Helmholtz are identical. The derivation in this subsection clearlyshows that the theorems are applicable not only to incompressible homogeneous flu-ids but also to compressible fluids in which there is a single-valued –p relation.Considering, again, at any instant a closed curve C surrounding a fluid, draw anarbitrary surface S which is bounded by such curve then
Γ =∮C
V · ds =∫∫S
ζ · n dS
58
Differentiating former equation, and taking into account (3.36) results
D
Dt
∫∫S
ζ · n dS = 0 (3.37)
Equation (3.37) states that outflow of voritcity through any surface bounded bycurve C remains constant for all times as the curve moves the fluid. It is clear fromconditions specified during derivation of theorems of Helmholtz and Kelvin thatthese theorems are valid only in the case when fluid has acceleration potential.
3.7.4 Vortex Line, Surface, Tube, and Filament
The field lines of voritcity filed are called vortex lines. Analytically they are describedby the differential equation
ζ × ds = 0 (3.38)
where ds is an element of a vortex line. Vortex line is represented as shown infigure (3.4). At any point in the flow field, the direction of the vorticity vector (orequivalently of the angular velocity vector) is given by the direction, at that point, ofthe vortex line passing through that point. In Cartesian coordinates the componentsof vorticity vector are
ζ = (ζx, ζy, ζz)
Equation (3.38) gives∣∣∣∣∣∣∣ı kζx ζy ζzdx dy dz
∣∣∣∣∣∣∣ = 0 , ⇒ dx
ζx=dy
ζy=dz
ζz
If, at any instant of time, we draw an arbitrary line in the flow field and draw thevortex lines passing through that line, a surface is formed. Such a surface is calleda vortex surface, see figure (3.5).
Figure 3.4: Vortex line.
If we consider a closed curve and draw all the vortex lines passing through it, atube is formed. Such a tube is called a vortex tube and is represented as shown infigure (3.6). A vortex tube of infinitesimal cross-sectional area is known as a vortexfilament.
59
Figure 3.5: Vortex surface.
Figure 3.6: Vortex tube.
3.7.5 Spatial Conservation of Vorticity
Since the vorticity is the curl of another vector field, we have
∇ · ζ = ∇(∇× V ) = 0 (3.39)
So vorticity is divergenceless field.Consider, at any instant, a region of space R enclosed by a closed surface S. Wehave ∫∫
S
ζ · n dS =∫∫R
∫∇ · ζ dR = 0 (3.40)
This equation states that the net outflow of vorticity through any closed surface iszero. This is true at every instant of time.
Strength of the Vortex Tube
Let us consider now, at any instant of time, a vortex tube drawn in the flow field.Denote by R the region of space enclosed between the wall of the tube and any twosurfaces S1 and S2 which cut the tube, see figure (3.7). Then according to (3.40)the outflow of vorticity through the surface S of the region R vanishes. Thereforewe have ∫∫
S1
ζ · n dS +∫∫S2
ζ · n dS +∫∫SW
ζ · n dS =∫∫S
ζ · n dS = 0 (3.41)
60
Here SW denotes the surface of the wall of the tube in the portion under considera-tion. On the wall of the tube, ζ lies in the surface SW . Hence the integral over SW
vanishes ∫∫SW
ζ · n dS = 0
Consequently we obtain ∫∫S1
ζ · n1 dS −∫∫S2
ζ · n2 dS = 0 (3.42)
where unit vector term n is substituted with proper unit vector from figure (3.7),which is directed outward with reference to the region R. Equation (3.42) is equiv-alent to ∫∫
S1
ζ · n1 dS =∫∫S2
ζ · n2 dS (3.43)
This equation states that the flow of vorticity through any cross-sectional surface S1of a vortex tube is equal to the flow of vorticity through any other cross-sectionalsurface S2 of the tube. This is true at any instant of time. If S denotes any cross-sectional surface of the vortex tube, equation (3.43) may be expressed as∫∫
S
ζ · n dS = const (3.44)
Figure 3.7: Illustrating the derivation of the spatial conservation of vorticity.
At any instant of time it is true that the flow of vorticity through any cross-sectionalsurface of a vortex tube is a constant all along the tube. Because vorticity andcirculation are related we can express (3.44) equivalently in terms of circulation.Let be any closed curve that encloses the vortex tube and lies on its wall, then
Γ- =∫ ∫
(ζ·(n dS
= const (3.45)
This states that at any instant of time circulation around any closed curve embracinga vortex tube is a constant all along the tube. For a vortex filament of variable cross-sectional area, this equation takes the form
Γc = ζ · n dS = const (3.46)
61
where n dS is any cross-sectional area of the filament and C is the boundary curveof n dS. If we take n in the direction of ζ former equation is reduced to
Γc = ζ dS = const
This shows that vorticity at any cross-sectional area of vortex filament is inverselyproportional to its cross-sectional area. Important consequence of the former is thatvortex tube and vortex filament or a vortex line, cannot begin or end abruptly ina fluid. It should either form a closed ring or end at infinity or at a solid or freesurface, figure (3.8). The circulation around any closed curve embracing a vortextube is the characteristic of the whole vortex tube and is called strength of vortextube.If we consider a vortex filament of variable cross-sectional area and shrink the areato zero in such a way that the vorticity goes to infinity as the area goes to zero,an the strength of the filament remains constant, we arrive at the conception of avortex filament with concentrated vorticity or vortex line.
Figure 3.8: Vortex tube.
3.7.6 Vortex Sheet
Vortex sheet is defined in a manner similar to that of a vortex filament with con-centrated vorticity. Consider the narrow region of vorticity enclosed between twoneighboring surfaces of vorticity S1 and S2 figure (3.9). Let P be a point on anintermediate vortex surface S. Choose at P an element n dS of surface S and anelemental cylinder with ndS as the cross section through P and with height equalto h (the distance at P between S1 and S2. If ζ(P ) denotes vorticity at P , we canwrite
ζ(P ) dV = ζ(P )h dS
where dV is the volume of the infinitesimal cylinder. We now wish to replace spatialdistribution of vorticity with with single surface distribution. We could introducethe new unknown γ which represents vorticity per unit surface which is defined as
ζ(P )h dS = γ dS (3.47)
In this way we shrink the narrow region of vorticity to a single surface on whichthe vorticity itself is infinite but γ defined by (3.47) is finite. Such a sheet is called
62
Figure 3.9: Definition of the vortex sheet.
vortex sheet of concentrated vorticity, or simply a vortex sheet, and γ the strength ofthe vortex sheet. The strength γ has the dimensions of vorticity per unit area.We now relate γ with the velocities on either side of the vortex sheet. According tothe definition of curl vector, we have
ζ dV = ∇× V dV =∫∫δS
n× V dSwhere V denotes fluid velocity. We choose dV as the infinitesimal cylinder describedbefore and write
γ dS = limh→0
∫∫δS
n× V dS= lim
h→0
∫∫dS1
n× V dS +∫∫
dS2
n× V dS +∫∫
wall
n× V dS
where dS1 and dS2 are respectively the two faces of the cylinder and ‘wall’ is thewall of the cylinder. The limit of the sum of the integrals over dS1 and dS2 isn× (V1 − V2) dS, and that of the integral over wall vanishes. Hence we obtain
γ dS = n× (V1 − V2) dSor
γ = n× (V1 − V2) (3.48)
We note that γ, n and (V1− V2) form a right-hand system of orthogonal vectors. It
follows that the vector (V1 − V2) is tangential to the vortex sheet. We see that
γ × n = V1 − V2 = ∆V (3.49)
sincen× (n×∆V︸ ︷︷ ︸
(γ
) = (n ·∆V︸ ︷︷ ︸=0
)n− (n · n︸ ︷︷ ︸=1
)∆V
the first term on the right-hand side is equal to zero since dot product of twoorthogonal vectors has that value
(V1 − V2) · n = 0 or V1 · n = V2 · n (3.50)
63
and|V1 − V2| = γ (3.51)
We therefore state that there is a velocity discontinuity across a vortex sheet; thediscontinuity occurs only in the tangential component of the velocity, whereas thenormal component remains continuous; the magnitude of the discontinuity is equalto the magnitude of the strength of the sheet. A vortex sheet is thus a surface oftangential discontinuity, and, conversely a tangential discontinuity is a vortex sheet.Vortex sheets and vortex filaments of concentrated vorticity are not physical possi-bilities. However, they form suitable analytical approximations when the vorticity isconfined to physically narrow regions. The concepts of vortex filaments and sheetsare widely applied in wing theory.We could assume that velocity of arbitrary point at the vortex sheet S is arithmeticmean of the velocities above and below vortex sheet
V =V1 + V2
2(3.52)
if we express V2 from this equation we have
V2 = 2V − V1 , ⇒ V2τ = 2Vτ − V1τWhere index τ denotes tangential components of mentioned velocities. Since acrossvortex sheet exists discontinuity of velocities only in tangential components to thesurface Σ, we could write equation (3.51) also as
V1τ − V2τ = γ
and finally
V1τ = Vτ +γ
2and V2τ = Vτ − γ
2(3.53)
If the vortex sheet does not move this equation is also valid in vector form
V1 = V +γ
2τ and V2 = V − γ
2τ (3.54)
where τ is unit vector tangential to the vortex sheet and oriented into direction ofaverage velocity.Consider at any time vortex sheet drawn in the fluid, figure (3.10). At that specificmoment must be
ζ · n = 0
for every element n dS of the vortex sheet. Choosing an element of the sheet anfollowing that element in its motion, we observe that as a consequence of equation(3.26)
ζ · n dS = 0
for all times although ζ and n dS may change. This means that ζ · n does notexist for all times for the surface element on the vortex sheet. From this follows asurface which is a vortex sheet at one instant remains a vortex sheet for all times.We could also conclude that fluid particle which is at one instant part of the vortex
64
sheet remains the part of it for all times. It also follows that fluid element whichis at some instance part of the vortex tube, vortex filament, or vortex line remainspart of it for all times. This also implies that the circulation around of vortex tube(strength of vortex tube) remains a constant for all times as the tube moves around,regardless of the changes experienced by the vortex tube.
Figure 3.10: Vortex sheet.
3.7.7 Velocity Field due to Vorticity Distribution
If the fluid could be considered incompressible than continuity equation for steadyand unsteady flow is reduced to
∇ · V = 0 (3.55)
while vorticity is defined as before
ζ = ∇× V (3.56)
Since ∇× A is another vector we can choose A such that
V = ∇× A (3.57)
Let us define another vector B such that
B = A+∇f
where f(x, y, z) is arbitrary differentiable scalar function chosen such that vector B
satisfy ∇ · B = 0∇ · B = ∇ · A+∇2f = 0 (3.58)
Equation (3.57) still holds since ∇× (∇f) ≡ 0
∇× B = ∇× ( A+∇f) = ∇× A = V , ⇒ V = ∇× B
65
If we now substitute instead velocity vector in equation (3.56) ∇× B we have
ζ = ∇× V = ∇× (∇× B) ≡ ∇(∇ · B)−∇2 B
The first term on the right-hand side is equal to zero since we have chosen vector Bto satisfy that condition which yields
∇2 B = ∇Bxı+∇By +∇Bzk = −ζ (3.59)
or∇2Bx = −ζx , ∇2By = −ζy , ∇2Bz = −ζz (3.60)
This system of equations allows determination of velocity field caused by vorticitydistribution form V = ∇× B.
Figure 3.11: Circulation over potential flow field.
3.8 The Velocity Potential
Vorticity of the flow field is defined as:
ζ = ∇× V =
∣∣∣∣∣∣∣ı k
∂/∂x ∂/∂y ∂/∂zU V W
∣∣∣∣∣∣∣or:
ζ =
(∂W
∂y− ∂V
∂z
)ı+
(∂U
∂z− ∂W
∂x
)+
(∂V
∂x− ∂U
∂y
)k (3.61)
If we neglect vorticity:ζ = 0 → ∇× V = 0
upper will be satisfied always only if V = ∇Φ:∇× (∇Φ) ≡ 0 (3.62)
Scalar function Φ is called velocity potential. We now find expression for the circu-lation of velocity if the velocity has potential, V = ∇Φ, figure (3.11)
Γ =
B∫A
V · dr =B∫
A
∇Φ · dr =B∫
A
dΦ = ΦB − ΦA (3.63)
66
Figure 3.12: Circulation for multiply connected bodies.
since ∇Φ · dr = dΦ.It is interesting to find the value of the circulation for the closed loop curves shownin the figure (3.12). From the Stokes theorem (3.13) it is clear that any closedloop integral for the loops wholly immersed in the fluid, which surround only fluidparticles, must be equal to zero, since ∇ × (∇Φ ) ≡ 0. But for the loops whichencompass some bodies in the flow field, figure (3.12) on the left, we have∑
Γi = Γ− Γ1 − Γ2 = 0 ⇒ Γ = Γ1 + Γ2
we needed to introduce the cut in the flow field by which bodies are connected withouter boundary, and by which newly created boundary encompass only fluid. Wecan express former equation generally as
Γ =∑i
Γi (3.64)
Consider now right-hand side of the figure (3.12) we have
Γ− Γ1 = 0 , ⇒ Γ = Γ1
Circulation around curve C is Γ, and around curve C1 is Γ1. Only ”one” point ismissing to close curve C and the same is true for the curve C2. Let us denote endingpoints of curve C by A and B, figure (3.12). Since only one point is missing wecould write
Γ =∮C
V · dr =B∫
A
dΦ = ΦB − ΦA
We conclude that at the cut there is jump in velocity potential values. Since Γ = Γ1the jump in the potential is constant along the cut. The way we introduce cut doesnot influence the value of the jump it is always equal to the circulation around thecurve C, ∆Φ = ΦB − ΦA = Γ.
3.9 Irrotational Flow
Condition of irrotationality adds one equation more to our system of differentialequations given i section (3.6). Since number of equations is not increased we can
67
forget one of equations given in section (3.6) and use irrotationality condition in-stead. According to Crocco’s theorem irrotational flow is isentropic (s = const), sowe can replace energy equation with:
p
po=
(
o
)γ
what is equivalent to s = const. So our system of equations now looks:
∂
∂t+∇
(V
)= 0
∂V
∂t+(V∇
)V = −∇p (3.65)
p
po=
(
o
)γ
Former system is consisted of one vector and two scalar equations what is equivalentto 5 scalar equations, so we can find 5 unknowns from such system of PDE-s. Theseunknowns are U, V,W, p, and .We still can use energy equations and equation of state, but that is not necessaryto solve the system. Energy equation in Bernoulli form is given by expression:
U2 + V 2 +W 2
2+
a2
γ − 1=
a2oγ − 1
(3.66)
If any of additional equations is used it simply replace one of equations of formersystem (3.65). For steady flow pressure term can be eliminated from equation ofmotion:
∇p =(∂p
∂
)s=const
∇ = a2∇ (3.67)
Multiplying momentum equation by ·V we obtain for the steady case:
U2∂U
∂x+ UV
∂U
∂y+ UW
∂U
∂z+ V 2∂V
∂y+ UV
∂V
∂x+
+VW∂V
∂z+W 2∂W
∂z+ UW
∂W
∂x+ VW
∂W
∂y= −U
∂p
∂x− V
∂p
∂y− W
∂p
∂z
= −a2
U∂
∂x− a2
V∂
∂y− a2
W∂
∂z
The expression on the right-hand side of the last row can be substituted from con-tinuity equation (∇V = −V∇) so:
U2∂U
∂x+ V 2∂V
∂y+W 2∂W
∂z+ UV
(∂U
∂y+∂V
∂x
)+
UW
(∂U
∂z+∂W
∂x
)+ VW
(∂V
∂z− ∂W
∂y
)= a2
(∂U
∂x+∂V
∂y+∂W
∂z
)(3.68)
68
or:(U2 − a2
) ∂U∂x
+(V 2 − a2
) ∂V∂y
+(W 2 − a2
) ∂W∂z
+
UV
(∂U
∂y+∂V
∂x
)+ UW
(∂U
∂z+∂W
∂x
)+ VW
(∂V
∂z+∂W
∂y
)= 0 (3.69)
If we introduce velocity potential Φ:
U =∂Φ
∂x= Φx, , V =
∂Φ
∂y= Φy , W =
∂Φ
∂z= Φz
we have:
(Φ2x − a2)Φxx + (Φ2
y − a2)Φyy + (Φ2z − a2)Φzz+
+2ΦxΦyΦxy + 2ΦxΦzΦxz + 2ΦyΦzΦyz = 0 (3.70)
or:
Φ2xΦxx + Φ2
yΦyy + Φ2zΦzz + 2ΦxΦyΦxy + 2ΦxΦzΦxz + 2ΦyΦzΦyz = a
2∆Φ (3.71)
where:∆Φ = Φxx + Φyy + Φzz
When a→∞ the equation (3.71) reduces to:
∆Φ = Φxx + Φyy + Φzz = 0 (3.72)
which is called Laplace equation.
3.10 Bernoulli Equation
We start again from Newton’s second law of motion which for fluids is expressed as
DV
Dt= F −∇p
or∂V
∂t+ ( V∇ )V︸ ︷︷ ︸
∇(V 2/2)+(ζ×(V
= F − ∇p
If we assume again that body force has potential, (F = ∇Π), what is true in thecase of gravitational force field, then
∂V
∂t+∇
(V 2
2
)+ ζ × V = ∇Π− ∇p
(3.73)
Assumption that between pressure and density exists unique relationship allow usto introduce again the quantity ∇P = ∇p/ so we have
∂V
∂t+ ζ × V +∇
(V 2
2− Π+ P
)= 0 (3.74)
69
For steady and irrotational flow field equation (3.74) reduce to
V 2
2− Π+ P = const (3.75)
The value of the potential is the same for whole flow field. Potential Π = −gz forthe axis z directed upward while the value of potential P depends on the type offluid we study
P =
p/ for = const (incompressible fluid).γp/(γ − 1) for p/γ = const (compressible fluid).
(3.76)
In the case when vectors ζ and V are mutually parallel and the flow is steady weagain obtain form equation (3.74) equation (3.75), but the constant of integrationvary from streamline to streamline. Note that vorticity lines and velocity linesoverlap in this case.There is one case more when equation (3.74) can be integrated. If velocity field has
potential V = ∇Φ then after substituting instead of V gradient of potential ∇Φ weget
∂(∇Φ)∂t
+∇(∇Φ · ∇Φ
2− Π+ P
)= 0
we can change order of operators ∂/∂t and ∇ in the first term, the second term in
the equation (3.74) vanishes since ζ = ∇× (∇Φ) ≡ 0 which yields
∇(∂Φ
∂t+∇Φ · ∇Φ
2− Π+ P
)= 0 (3.77)
After multiplying this equation by dr we obtain total differential which after inte-gration along some path gives equivalently
∂Φ
∂t+∇Φ · ∇Φ
2− Π+ P = C(t) (3.78)
constant C(t) is unique for the whole flow field, but generally is dependent on time.In the case of the air we can neglect contribution to the former equation from Π.Constant C(t), being constant for whole flow field, can be calculated from the valuesof flow parameters at infinity
∂Φ
∂t+∇Φ · ∇Φ
2+ P =
∂Φ∞∂t
+∇Φ∞ · ∇Φ∞
2+ P∞ (3.79)
We use this equation to calculate pressure from pressure function P which can beexpressed as
P =∂
∂t(Φ∞ − Φ) +
V 2∞ − V 2
2+ P∞
For incompressible unsteady flow we can express pressure in the following way
p
= −∂Φ
∂t− V 2
2(3.80)
For steady incompressible flow derivative with respect to time vanishes
p = const− ∞V 2
2(3.81)
70
Figure 3.13: Notation for Joukowski theorem.
Figure 3.14: Shading of vortices.
3.11 Joukowski Theorem
We will prove the Joukowski theorem ‘in small’ for arbitrary three-dimensional non-stationary motion. Let the infinitely thin lifting surface Σ be replaced by a vortexsheet with strength γ, continuously distributed over this surface. We will assumethat the medium is perfect and incompressible and that the surface Σ is sufficientlysmooth and not closed, so that the fluid encloses it on both sides. The flow aboutthe lifting surface will be assumed to be smooth, we will assume that a change withtime of the strength of the bound vortices can only be caused by the formation andshedding of free vortices. Shedding velocity of free vortices is equal to the velocityof the fluid particles.The pressure p at any point can be obtained from Bernoulli’s theorem, (3.81). Closeto the surface Σ but outside of the vortex trail the flow is potential
V = ∇ΦWe observe that velocities above and below the lifting surface Σ are different. Letus denote with V+ velocity above the lifting surface and with V− the velocity belowthe lifting surface. The velocity at arbitrary point P is assumed to be V .Pressure difference above and below the lifting surface is obtained from equation(3.80)
∆p
=∂(Φ+ − Φ−)
∂t+V 2+ − V 2
−2
(3.82)
71
where Φ+ and Φ− are the limits of the potential of perturbed velocities obtained byapproaching an arbitrary point P of the lifting surface from above and from below.We imagine closed curve C intersecting the surface Σ at the single point P , figure(3.13). Let the circulation of the disturbed velocity about the contour C be Γ then
Γ =∫C
V · dr =∫C
∂Φ
∂rdr = Φ− − Φ+
and so∂(Φ+ − Φ−)
∂t= −∂Γ
∂t(3.83)
To find the derivative ∂Γ/∂t, we consider the variation during the time ∆t in thecirculation Γ around the contour C for a fixed point P . A change in the circulationΓ by an amount ∆Γ can be produced only by the departure of free vortices, someof which, surrounding the contour C are carried away from the boundaries of thiscontour. These free vortices are of a completely non stationary nature and gaveaxes parallel to the axes of those bound vortices from which they were shed. Instationary motion there will be no free vortices of this type.Let V be the relative flow velocity at the point P which, in agreement with theassumption concerning the smoothness of the flow about the lifting surface, is tan-gential to Σ. The shading velocity of free vortices at each point of the lifting surfaceΣ will be also V . The total strength of the free vortices carried away during thetime ∆t from the limits of the contour C will be V∆tγn, where γn is the componentof the free vortex strength in the direction of the unit vector perpendicular to V ,figure (3.14). Hence axes of the free vortices γn are perpendicular to the velocity V .At the point P under consideration, we introduce the rectangular coordinate systemτ, P, ν in the plane tangent to the surface Σ. Let the ν be direction of the boundvortex at the point P , so there are no component of bound vortices in the directionof τ . Let us also denote with ”+” sign the components of the bound vortices,and with the ”−” sign components of the free vortices. The positive direction ofγn is shown in the figure (3.14). The total strength of the vortex sheet Σ will becalculated form the strength of the bound vortices γ+ and free vortices γ−, where
γ = γ+ + γ− γτ = γτ− (3.84)
If we by Vν , and Vτ denote components of velocity V along the axis of coordinatesystem than the unit vector along vector V is
ıV =ıνVνV
+ıτVτV
where ıν and ıτ are units vectors in the directions ν and τ . We can also express unitvector in the direction of γn in terms of unit vectors in the direction ν and τ
ın =ıνVτV−ıτ Vν
V
and so
γn− = γν−VτV− γτ−
VνV
72
andV γn− = γν−Vτ − γτ−Vν
So we can write the following for the change in the circulation ∆Γ during the time∆t
∆Γ = −(γν−Vτ − γτ−Vν)∆tand so
∂Γ
∂t= −(γν−Vτ − γτ−Vν)
On the other side equation (3.53) also holds, only we have to adjust it to currentneeds, so
Vν+ − Vν− = γτ Vτ+ − Vτ− = −γν(3.85)
Vν+ + Vν−2
= VνVτ+ + Vτ−
2= Vτ
Now remains only to express the term (V 2+ − V 2
−)/2 in the equation (3.82). If wesubstitute equations (3.84) into equations (3.85) and multiply these equations colum-nwise we get
V 2ν+− V 2
ν−
2= γτ−Vν
V 2τ+− V 2
τ−
2= −(γν+ + γν−)Vτ
After adding left-hand side and right-hand side of these equations yields
V 2+ − V 2
−2
= −γν−Vτ + γτ−Vν︸ ︷︷ ︸∂Γ/∂t
−γν+Vτ ≡∂Γ
∂t− γν+Vτ (3.86)
After replacing this equation, and equation (3.83) into equation (3.82) we finally getfor the pressure difference across the vortex sheet
∆p = p+ − p− = −γν+Vτ (3.87)
where positive direction of γν+ is shown in the figure (3.14). It is clear form (3.87)that the pressure difference generated on an element of the upper surface is deter-mined by the linear strength of the bound vortices only. Hence in particular it followsthat on the vortex sheets consisting of free vortices which are formed behind a bodyof finite span both in steady and in unsteady motion there are no pressure drops.Equation (3.87) also shows that any pressure distribution ∆p can be reproduced byvortex sheet.
3.12 Joukowski Theorem for non-Lifting Surfaces
As in previous section, we consider three-dimensional non-stationary motion of aninfinitely thin lifting surface Σ. Again we assume that the medium is perfect andincompressible fluid, and that the surface Σ is sufficiently smooth over both of itssides.
73
Figure 3.15: Joukowski theorem for the irrotational flow.
In contrast to section (3.11), however, we investigate here the case of irrotationalflow about the lifting surface. in this case no vortex trail will be formed behind thebody, and the theorem concerning the constancy of the circulation about a closedfluid contour holds because the circulation about any contour containing the liftingsurface will be zero. Hence the vortex layer replacing the lifting surface can bethought of as a vortex system formed by a combination of closed vortex lines of aconstant strength. This whole vortex system is fixed in the surface Σ, and it mustbe considered as a system of bound vortices. In flow without circulation about alifting surface, we may thus assume that no free vortices are formed either on thesurface Σ or behind it.We now determine pressure difference ∆p at an arbitrary point P of the lifting sur-face Σ, figure (3.15). We retain the same notation for the limiting relative velocities:V+ is the limit on the upper surface; V− is the limit on the lower surface; V is therelative velocity at the point P .With P as origin, we define arbitrary rectangular coordinate axis P -τ and P -ν inthe plane tangent to the surface Σ. The pressure difference ∆p between the lowerand upper surfaces is, in agreement with Bernoulli’s theorem given by (3.82). Weexpress the derivative in terms of the strengths of the vortices which, in the problemsstudied below, are determined by the solution of these problems. It is convenient toconsider the contour C shown in the figure (3.15) then
Φ− − Φ+ = −Γwhere, again, Γ is the circulation around the closed curve C which intersects thelifting surface at P . The circulation Γ can be obtained s the sum of the strengthsof vortices between the trailing edge and the point P . We can thus write
∂Γ
∂t=∂(Φ+ − Φ−)
∂t(3.88)
and, from (3.86), we have
V 2+ − V 2
−2
= γτVν − γνVτusing this relation, (3.88), and (3.82) yields
p− − p+ = ∆p =
(Vνγτ − Vτγν + ∂Γ
∂t
)(3.89)
74
Figure 3.16: Uniform flow and perturbed flow
This formula express Joukowsky theorem for irrotational flow. A comparison of(3.87) and (3.89) easily shows that the expression for the pressure difference in thefirst case, for rotational flow, is simpler than in the other case.This same theorem could also be obtained from the relations of the preceding sectionby nothing that
γν− = 0 , γν+ = γν
γτ− = γτ , Γ = −ΓliftingIt should be stressed that (3.87) and (3.89) can only be used in the determinationof the normal components of the aerodynamic loads, and that they cannot be usedto obtain the so-called suction force.In a perfect incompressible medium, the suction forces are generated by flow aroundsharp edges of the lifting surfaces, and are in the direction of the tangent to thesurface at the point of the sharp edge under consideration. Their existence is dueto the infinitely high velocities and rarefactions in the flow of a perfect fluid aroundsharp edges.
3.13 Small Perturbation Theory
Let us denote perturbed velocities with u, v, and w. Let us assume also that u/U <<1, v/U << 1, and w/U << 1. Substituting U + u, v, and w instead of U , V and Win equation (3.68) we get:
a2(∂u
∂x+∂v
∂y+∂w
∂z
)= (U + u)2
∂u
∂x+ v2
∂v
∂y+ w2∂w
∂z
+(U + u)v
(∂u
∂y+∂v
∂x
)+ vw
(∂v
∂z+∂w
∂y
)+ w(U + u)
(∂w
∂x+∂u
∂z
)(3.90)
The speed of sound can be expressed from (3.66):
(U + u)2 + v2 + w2
2+
a2
γ − 1=U2
2+
a2∞γ − 1
(3.91)
75
or:
a2 = a2∞ − γ − 1
2
(2uU + u2 + v2 + w2
)(3.92)
Replacing a2 in (3.90) by (18) and dividing by a2∞ after rearrangement we obtain:
(1−M2∞)∂u
∂x+∂v
∂y+∂w
∂z=
M2∞
[(γ + 1)
u
U+γ + 1
2
u2
U2+γ − 1
2
v2 + w2
U2
]∂u
∂x
+M2∞
[(γ − 1)
u
U+γ + 1
2
v2
U2+γ − 1
2
w2 + u2
U2
]∂v
∂y
+M2∞
[(γ − 1)
u
U+γ + 1
2
w2
U2+γ − 1
2
u2 + v2
U2
]∂w
∂z
+M2∞
[v
U
(1 +
u
U
)(∂u
∂y+∂v
∂x
)+w
U
(1 +
u
U
)(∂u
∂z+∂w
∂x
)+vw
U2
(∂w
∂y+∂v
∂z
)](3.93)
This is still exact equation, but written in perturbation component form!We can simplify this equation if we neglect value of second order of smallness:
(1−M2∞)∂u
∂x+∂v
∂y+∂w
∂z=
M2∞(γ + 1)
u
U
∂u
∂x+M2
∞(γ − 1)u
U
(∂v
∂z+∂w
∂y
)+
+M2∞v
U
(∂u
∂y+∂v
∂x
)+M2
∞w
U
(∂u
∂z+∂w
∂x
)(3.94)
Further reduction is possible if we neglect all terms with perturbed velocities, i.e.whole RHS:
(1−M2∞)∂u
∂x+∂v
∂y+∂w
∂z= 0 (3.95)
If we introduce perturbed potential φ:
u =∂φ
∂x, v =
∂φ
∂y, w =
∂φ
∂z,
then:
(1−M2∞)∂2φ
∂x2+∂2φ
∂y2+∂2φ
∂z2= 0 (3.96)
For M∞ = 0:∂2φ
∂x2+∂2φ
∂y2+∂2φ
∂z2= 0 (3.97)
For incompressible flow small disturbance equation (3.97) and full potential equation(3.72) have the same form!The pressure coefficient is defined as:
CP =p− p∞12∞U2
=2
γM2∞
p− p∞p∞
(3.98)
76
Figure 3.17: Effect of coordinate transformation from x to x′.
If we eliminate pressure with velocities:
CP =2
γM2∞
[1 +
γ − 1
2M2
∞
(1− (U + u)2 + v2 + w2
U2
)]γ/(γ−1)− 1
or after simplification:
CP = −(2u
U+ (1−M2
∞)u2
U2+v2 + w2
U2
)(3.99)
For 2D bodies it is sufficient to write:
CP = −2u
U(3.100)
but for axially symmetric bodies we must retain additional terms:
CP = −2u
U− v2 + w2
U2(3.101)
Equation (3.96) can be written in Laplace equation form. Let us define variable β:
β2 =
1−M2
∞ , M∞ < 1M2
∞ − 1 , M∞ > 1
Then equation (3.96) looks:
β2∂2φ
∂x2+∂2φ
∂y2+∂2φ
∂z2= 0 , M∞ < 1
(3.102)
−β2∂2φ
∂x2+∂2φ
∂y2+∂2φ
∂z2= 0 , M∞ > 1
Since β2 = const we can make coordinate transformation x = βx′ so dx = βdx′ and∂x2 = β2∂x′2. If we substitute this into equation (3.102) we obtain:
∂2φ
∂x′2+∂2φ
∂y2+∂2φ
∂z2= 0 , M∞ < 1
(3.103)
− ∂2φ
∂x′2+∂2φ
∂y2+∂2φ
∂z2= 0 , M∞ > 1
We will drop ′ further on for convenience. In the figure (3.17) is shown effect ofcoordinate transformation from x to x′.
77
The first of equations (3.103) is elliptic Laplace equation, but the second is of hy-perbolic nature called the second order wave equation. Both equations are linearthus they allow superposition of elementary solutions!It is interesting to note that equation (3.103) can be directly obtained from equation(3.102) when β = 1. Two values of Mach number give β = 1 for subsonic flowM∞ = 0, and for supersonic flow M∞ =
√2.
3.14 Boundary Conditions
Let us assume undisturbed velocity of the fluid to be W∞. Let the body of interesthave translational velocity V and angular velocity Ω about some point O. Figure(3.18) illustrate former notations. Let also Wν = W∞+ wν be the vector of absolute
Figure 3.18: Derivation of boundary conditions.
particle velocity, where wν denotes perturbed velocity. Let also W ∗ν be the vector of
velocity of the point ν at the body:
W ∗ν = V + Ω× rν (3.104)
then B.C. at the point ν can be specified as:(Wν − W ∗
ν
)· nν = 0 (3.105)
or: (W∞ + wν − V − Ω× rν
)· nν = 0
If W∞, V , and Ω are given in advance, the former system can be rearranged intoform:
wν · nν =(V + Ω× rν − W∞
)· nν , ν = 1, 2, . . . , n (3.106)
Equation (3.106) represents the “general” boundary condition case. If boundarycoordinate system is fixed to the body, and if body does not rotate than equation(3.106) is reduced to:
wν · nν = − W∞ · nν , ν = 1, 2, . . . , n (3.107)
78
where, again, wν is the vector of perturbed velocities at point ν:
wν = uνı+ vν+ wνk
If the perturbed velocity wν has potential φ than:
wν = ∇φν (3.108)
Boundary condition (3.107) can be expressed also:
∇φν · nν = − W∞ · nν (3.109)
or:∂φ
∂n
∣∣∣∣∣ν
= − W∞ · nν (3.110)
Figure 3.19: Terms appearing in Green theorem.
3.15 Green’s Second Formulae
From Gauss theorem which, for arbitrary vector function A, is expressed as:∮S
A · n dS =∫ ∫V
∫∇ AdV (3.111)
if vector A is assumed to be: A = Φ1∇Φ2, we obtain the following:∮S
Φ1∇Φ2 · n︸ ︷︷ ︸∂Φ2∂n
dS =∫ ∫V
∫∇ (φ1∇Φ2) dV
=∫ ∫V
∫ (∇Φ1∇Φ2 + Φ1∇2Φ2
)dV (3.112)
where dS is elementary surface, V is volume of the region, and n is unit normal to theelementary surface, as it is shown in the figure (3.19). If we now define A = Φ2∇Φ1
than equation (3.112) looks:∮S
Φ2∂Φ1
∂ndS =
∫ ∫V
∫ (∇Φ2∇Φ1 + Φ2∇2Φ1
)dV (3.113)
79
Equations (3.112) and (3.113) are valid for any scalar functions Φ1 and Φ2! If wenow subtract from equation (3.112) equation (3.113) we obtain:∮
S
(Φ1∇Φ2 − Φ2∇Φ1) · n dS =∫ ∫V
∫ (Φ1∇2Φ2 − Φ2∇2Φ1
)dV (3.114)
Equations (3.112) and (3.113) are called Green’s first formulae, while equation(3.114) is known as Green’s second formulae. In former formulas, when problems ofaerodynamics are concerned, the surface S, figure (3.20), is consisted of:
S = SB + S∞ + SW (3.115)
where SB is the surface of the solid body, S∞ is the boundary of a fluid at infinity,and SW is the vortex wake downwind of the solid body.
3.16 The General Solution of Laplace Equation
Let n points outside of the region of interest, and S∞, SB, SW , and Sε denote bound-ary of the region at far-field, body, wake, and about arbitrary point P respectively,see figure (3.20) Let r be the distance from an arbitrary point P (x, y, z) to some
Figure 3.20: Typical boundaries of an aerodynamics problem
point on boundary S and let the point P be outside volume V . Let us also assumethat:
Φ1 =1
rand: Φ2 = φ
where r denotes the distance between point P and arbitrary point on the boundary.Let also φ satisfy Laplace equation ∇2φ = 0. Since ∇2(1/r) ≡ 0 we obtain fromequation (3.114): ∮
S
(1
r∇φ− φ∇1
r
)· n dS = 0 (3.116)
If point P is inside region V than we must introduce additional boundary (smallsphere) Sε which is infinitely close to the point P , i.e. ε = r → 0, otherwise we
80
have in our expression term 1/r which is not defined when r = 0. So potential φis singular at point P . Since our point P is again outside volume V we can applyformulae (3.116): ∮
Sε+S
(1
r∇φ− φ∇1
r
)· n dS = 0 (3.117)
or: ∮Sε
(1
r∇φ− φ∇1
r
)· n dS +
∮S
(1
r∇φ− φ∇1
r
)· n dS = 0 (3.118)
To integrate about small sphere enclosed by Sε we need to introduce spherical co-ordinate system at point P . Since n points toward P (outside V) then n = −rε,and:
n · ∇φ = −∂φ∂r, ∇
(1
r
)= −rε
r2
We can assume that φ and ∂φ/∂n do not vary too much within infinitesimal sphere,and since dSε = 4πε2 we could replace the first integral in equation (3.118) with:
−(1
ε
∂φ(P )
∂r+Φ(P )
ε2
)4πε2 +
∮S
(1
r∇φ− φ∇1
r
)· n dS = 0
When ε→ 0 the former equation is reduced to:
φ(P ) =1
4π
∮S
(1
r∇φ− φ∇1
r
)· n dS (3.119)
Note that r is the distance from the point P inside volume V to the arbitrary pointon the boundary. Unit normal n is defined for the same point on the boundary towhich vector r is pointing. Expression (3.119) determine the value of the potential φat the point P inside the volume V in terms of the values of potential φ and gradient∂φ∂non the boundary S!
If the point P is exactly at the boundary S than we cannot construct whole sphere,but semi-sphere instead, as it is shown in figure (3.21). For this special case equation
Sε =4ε2π2
= 2ε2π
Figure 3.21: Point P on the boundary of the problem.
81
(3.119) looks:
φ(P ) =1
2π
∮S
(1
r∇φ− φ∇1
r
)· n dS (3.120)
We can imagine some specific boundaries such as those shown in figure (3.22). For
Sε =4ε2π4
= ε2π Sε =4ε2π2
= ε2π/2
Figure 3.22: Point P on specific type of boundaries.
the quarter of sphere (left illustration) formulae (3.119) looks:
φ(P ) =1
π
∮S
(1
r∇φ− φ∇1
r
)· n dS (3.121)
while for the illustration on the right in figure (3.22) follows:
φ(P ) =2
π
∮S
(1
r∇φ− φ∇1
r
)· n dS (3.122)
3.16.1 Internal Potential
Let us now consider volume bounded by the surface SB, figure (3.23) this volumeis outside the volume V . Let the potential of the volume enclosed by SB is denotedby φi. Point P inside V is outside of the volume enclosed by SB so must be:
1
4π
∮SB
(1
r∇φi − φi∇1
r
)· n dS = 0 (3.123)
If we add equation (3.123) to equation (3.119) we get:
φ(P ) =1
4π
∮SB
[1
r∇(φ− φi)− (φ− φi)∇1
r
]· n dS +
1
4π
∮SW+S∞
(1
r∆φ− φ∇1
r
)· n dS
(3.124)The − sign comes from different orientations of unit normal n!
82
Figure 3.23: Definition of the internal potential.
3.16.2 Contribution from S∞Exact value of the potential depends on selected coordinate system. If we choosestationary fluid at infinity and body which moves through it, then φ∞(P ) can beselected to be constant. Otherwise:
φ∞(P ) =1
4π
∮S∞
(1
r∆φ− φ∇1
r
)· n dS (3.125)
3.16.3 Contribution from SW
Generally normal component of velocity is continuous through wake (∇φ · n =∂φ/∂n), so: ∮
S
1
r∇φ · n dS ≡ 0
Then we finally obtain the expression for potential at arbitrary point P :
φ(P ) =1
4π
∮SB
[1
r∇(φ− φi)− (φ− φi)∇1
r
]· n dS −
1
4π
∮SW
φn · ∇1
rdS + Φ∞(P ) (3.126)
Let us denote:
−µ = φ− φi , −σ =∂φ
∂n− ∂φi
∂n(3.127)
then equation (3.126) is transformed to:
φ(P ) = − 1
4π
∮SB
[σ
r− µn · ∇1
r
]dS
+1
4π
∮SW
µn · ∇(1
r
)dS + Φ∞(P ) (3.128)
if we take into account that (∇ · n ≡ ∂/∂n) then:
φ(P ) = − 1
4π
∮SB
[σ
r− µ ∂
∂n
(1
r
)]dS +
1
4π
∮SW
µ∂
∂n
(1
r
)dS + Φ∞(P ) (3.129)
83
Equation (3.129) is general solution of Laplace equation, but it does not specifyunique distribution of values for σ and µ. Choice for the values of µ and σ is basedon physical reasoning.It is possible to require that ∂φi/∂n = ∂φ/∂n on SB what gives σ = 0, equation(3.127). In that case source term vanish and only doublet distribution remains.Also it is possible to specify φi = φ on SB, so that solution will be consisted only ofsource distribution along boundary.
3.17 Biot-Savarat Law
We already have derived, equation (3.60), that the following expressions hold
∇2 B = −ζ(r) ⇒∇2Bx = −ζx(r)∇2By = −ζy(r)∇2Bz = −ζz(r)
(3.130)
We assume that the field B vanishes sufficiently strongly at infinity. This vagueassumption will be defined later. It is necessary only to consider the solution of oneof the equations of the system (3.130). To construct the solution we use Green’ssecond theorem, equation (3.114)∮
S
(Φ1∇Φ2 − Φ2∇Φ1) · n dS =∫ ∫V
∫ (Φ1∇2Φ2 − Φ2∇2Φ1
)dV
Let us assume Φ1 ≡ φ = Bx(r), and
Φ2 =1
|r − r1|then
∇2φ = −ζx(r) , ∇φ = ∇Bx
(3.131)
∇2Φ2 = 0 , ∇Φ2 = − r − r1|r − r1|3
The last equation is valid at any point except at point r = r1 where potential isnot defined. Again we construct small sphere around arbitrary point P and applyGreen’s second formulae to the region between small sphere and outer boundary Σwhich is moved to infinity, figure (3.24). After applying Green’s second theorem andrelations given by (3.131) we have
∫∫R
∫ ζx(r)
|r − r1| dR = − limΣ→∞
∫∫Σ
(Bx
r − r1|r − r1|3 +
∇Bx
|r − r1|)· n dS
− lim→0
∫∫S
(Bx
r − r1|r − r1|3 +
∇Bx
|r − r1|)· n dS (3.132)
84
Figure 3.24: Computation of vector potential component.
The first integral over Σ is equal to zero since we choose B such that it dies suf-ficiently fast. The second integral on the right-hand side can be calculated usingmean value theorem
− lim→0
∫∫S
(Bx
r − r1|r − r1|3 +
∇Bx
|r − r1|)· n dS = −4πBx(r1)
where dS = 4π2. So complete right-hand side of the equation (3.132) is equal to4πBx(r1) or
4πBx(r1) =∫∫R
∫ ζx(r)
|r − r1| dR
We can now switch the role of variables r and r1, and write
Bx(r) =1
4π
∫∫R
∫ ζx(r1)
|r − r1| dR (3.133)
where r1 is a new variable of integration. Solutions for By and Bz are similar to
solution for Bx. From these we conclude that the solution for B(r, t) is given by
B(r, t) =1
4π
∫∫R
∫ ζ(r1, t)
|r − r1| dR (3.134)
It could be easily shown by choosing Σ to be sphere of large radius R, that B mustdecay faster than 1/R. Note also that integration is taken over vorticity distribution.Velocity field can be calculated from
V =1
4π∇×
∫∫R
∫ ζ(r1, t)
|r − r1| dR (3.135)
If we denote by δ B the contribution to B at r due to the vortex element ζ dR placedat r1 and similarly by δV the contribution to V at r, we have
δ B(r, t) =1
4π
ζ(r, t)
|r − r1| dR (3.136)
85
δV (r, t) =1
4π∇r ×
ζ(r, t)
|r − r1| dR (3.137)
where ∇r emphasize that the curl is to be taken with respect to the coordinates ofthe point r.
Figure 3.25: Illustration for Biot-Savart law.
Consider a vortex filament of strength Γ. Choose a volume elementR of this filamentas the cylinder formed by a cross-sectional surface n dS and an element of length dalong the filament, figure (3.25). Then the contribution to the vector potential Bat a field point r, from the vortex element r1 is given by
δ B(r) =1
4π
ζ(r1)
|r − r1| (n dS · d) (3.138)
Since we have
d =ζ
ζd
andζ · n dS = Γ
equation (3.138) may be rewritten as
δ B(r) =Γ
4π
d
|r − r1| (3.139)
The contribution to the velocity at the point r from the element of the filament isthen given by
δV (r) = ∇r × Γ
4π
d
|r − r1| (3.140)
In carrying out the curl operation we keep d and r1 fixed. Equation (3.140) reducesto
δV (r) =Γ
4π
d× (r − r1)|r − r1|3 (3.141)
86
where d = dx1ı + dy1 + dz1k. This expression is known as the Biot-Savart law.The velocity at r due to the whole vortex filament is obtained by integration of theexpression (3.141) over the length of the filament. We thus have
V (r) =Γ
4π
∫ d× (r − r1)|r − r1|3 (3.142)
Figure 3.26: Induced velocity due to vortex filament.
Since Γ is the strength of the filament, it is a constant and hence appears outside theintegral. We now assume that vortex line, figure (3.26), is defined parametrically
x1 = x1(ξ) ⇒ dx1 =dx1dξ
dξ
y1 = y1(ξ) ⇒ dy1 =dy1dξdξ
z1 = z1(ξ) ⇒ dz1 =dz1dξdξ
d =
(dx1dξı+
dy1dξ+
dz1dξk
)dξ
The distance between vortex line element d and arbitrary point M(x, y, z) is
= |r − r1| =√(x− x1)2 + (y − y1)2 + (z − z1)2
while
d× (r − r1) =
∣∣∣∣∣∣∣∣ı k
dx1
dξdξ dy1
dξdξ dz1
dξdξ
x− x1 y − y1 z − z1
∣∣∣∣∣∣∣∣87
After replacing former expressions into equation (3.142) we obtain the followingcomponents of induced velocity
u(r) =Γ
4π
ξ2∫ξ1
[dy1dξ
(z − z1)− dz1dξ
(y − y1)]dξ
3, (3.143)
v(r) =Γ
4π
ξ2∫ξ1
[dz1dξ
(x− x1)− dx1dξ
(z − z1)]dξ
3, (3.144)
w(r) =Γ
4π
ξ2∫ξ1
[dx1dξ
(y − y1)− dy1dξ
(x− x1)]dξ
3. (3.145)
where ξ1 and ξ2 are starting and ending values of the parameter ξ.
Figure 3.27: Solution at point P on 2D boundary.
3.18 2D General Solution
Let us start again from Green’s formula (3.114), and let us choose Φ1 = ln r andΦ2 = φ. Integration around point P is now about circle instead of sphere so equation(3.114) becomes:
−∮ε
(ln r
∂φ
∂r− φ
r
)dS +
∮S
(ln r∇φ− φ∇ ln r) · n dS = 0 (3.146)
The circumference of the small circle is 2πε so value of the potential at point P isexpressed through equation:
φ(P ) =1
2π
∮S
(ln r∇φ− φ∇ ln r) · n ds (3.147)
If the point P lies on the boundary then integration is around semicircle, figure(3.27), thus:
φ(P ) =1
1π
∮S
(ln r∇φ− φ∇ ln r) · n ds (3.148)
Two-dimensional version of equation (3.129) is:
φ(P ) =1
2π
∫SB
[σ ln r − µ ∂
∂n(ln r)
]dS − 1
2π
∫SW
µ∂
∂n(ln r) dS + Φ∞(P ) (3.149)
88
Figure 3.28: Definition of a streamline.
3.19 Stream Line and Stream Function
3.19.1 Some Definitions
Let us first give some definitions to clarify terminology.
Steady Flow
If at various points of the flow field all quantities (such as velocity, density, pressureetc.) associated with the fluid flow remain unchanged with time, the motion is saidto be steady ; otherwise it is called unsteady. Thus various flow parameters are onlyfunctions of position.
Path Line
The curve described in space by a moving fluid element is known as its trajectory orpath line. Such a line is obtained by giving the position of an element as a functionof time. The differential equation of path line is defined as:
dx
dt= u ,
dy
dt= v ,
dw
dt= w (3.150)
Streak Line
The Streak Line is the curve of all fluid particles which at some time have coincidedwith a particular fixed point of space. The streak line is observed when naturallybuoyant marker fluid is continuously injected into the flow at a fixed point of space.The marker fluid may be smoke if the main flow involves a gas such as air, or a dyeif the main flow involves liquid such as water. When the flow is steady path lines,streak lines and stream lines coincide.
89
Figure 3.29: Stream surface.
3.19.2 Streamlines
If we draw a curves such that each curve is tangent at each point to the velocitydirection at that point, we obtain streamlines. Let us consider a certain instant oftime, at any point r, let ds be an element of the streamline passing through thepoint an let V denote the velocity vector at that point at that instant figure (3.28).Then in view of the definition of a streamline, we state that the direction of ds isthe same as that of V , that is: ds is parallel to V ! Since the cross product of twoparallel vectors is equal to zero, we have:
ds× V = 0 (3.151)
what is differential equation of the streamline. Let the components of ds be ds =dxı+ dy+ dzk then:
ds× V =
∣∣∣∣∣∣∣ı kdx dy dzu v w
∣∣∣∣∣∣∣ =ı (w dy − v dz) + (u dz − w dx) + k (v dx− u dy) = 0
(3.152)
This vector equation is equivalent to three scalar equations:
w dy − v dz = 0
u dz − w dx = 0 (3.153)
v dx− u dy = 0
which can be put in the symmetric form:
dx
u=dy
v=dz
w(3.154)
Note that u, v, and w are functions of (x, y, z, t). Equations (3.153) and (3.154) areCartesian form of the differential equations for a streamline.At each instant of time we can construct a picture of the streamlines. If the motionis unsteady, the streamline picture will change from instant to instant. If the motion
90
Figure 3.30: Stream tube.
is steady, the picture remains the same for all times. In this case path lines, streaklines and streamlines are identical. A picture of the streamlines helps us to see,as it were, the flow field and therefore plays an important part in the analysis andunderstanding of fluid flow problems.
3.19.3 Stream Surfaces and Stream Tubes
If at any instant of time we draw an arbitrary line in the fluid region and draw thestreamlines passing through the line, a surface is formed. Such surface is called astream surface, figure (3.29). If we consider closed curve and draw all the streamlinespassing through it, a tube called a stream tube is formed, figure (3.30). In unsteadymotion, the shape of a stream surface drawn through an arbitrarily chosen curveand the shape of a stream tube drawn through a chosen closed curve will changewith time. In steady motion stream surfaces and stream tubes once drawn remainunchanged. No fluid can cross a stream surface or the walls of a stream tube, for thewalls and the stream surface are always parallel to the fluid velocity everywhere. Forsteady flow, a stream tube behaves like one with rigid boundaries inside which fluidflows. It follows that in steady flow, an impermeable solid-fluid boundary, acrosswhich there cannot be any flow of fluid, would be a stream surface.
3.19.4 Stream Functions
Let us again consider integration of the equation (3.151) for a streamline:
ds× V = 0
Although V = V (r, t), the integration of former equation involves only the spacevariables. As such, in what follows, we shall not exhibit explicitly the dependenceon time. Bear in mind, however, that the results obtained hold at every instantof time. From the system (3.154) we can construct two independent system ofequations whose general form is represented by:
a1(x, y, z) dx+ b1(x, y, z) dy + c1(x, y, z) dz = 0
91
Figure 3.31: Stream function and stream line.
(3.155)
a2(x, y, z) dx+ b2(x, y, z) dy + c2(x, y, z) dz = 0
The general solution of former system of equations is given in the form:
ψ(x, y, z) = c (3.156)
where c is constant. Equation (3.156) represents family of surfaces. We thus con-clude that the solution of the system (3.155) is presented in the form:
ψ1(x, y, z) = c1 , ψ2(x, y, z) = c2 (3.157)
where c1 and c2 are constants. Such conclusion could have been reached immediatelyfrom the observation that a line in space, such as a streamline, may be described asthe curve of intersection of two surfaces.The functions ψ1 and ψ2 should naturally be related to the velocity components u,v, and w. In the figure (3.31) is shown streamline which belongs simultaneously to
two stream surfaces. Velocity vector V (r) is tangent to streamline, and belongs toboth surfaces ψ1 = c1 and ψ2 = c2. Normal vectors to stream surfaces are obtainedfrom ∇ψ1 and ∇ψ2 so:
V · ∇ψ1 = 0 , V · ∇ψ2 = 0
This shows that vector V is normal to plane built by vectors ∇ψ1 and ∇ψ2, i.e. thatvector V is parallel to the cross product of two normal vectors:
µ(r)V = ∇ψ1 ×∇ψ2 (3.158)
where µ(r) is scalar function of position. Knowing that:
∇(∇ψ1 ×∇ψ2) = 0
and applying operator ∇ to the equation (3.158) we conclude that function µ(r)should satisfy the condition:
∇(µV ) = 0 (3.159)
92
Function µ have extremely simple form in the case of incompressible flow whenµ(r) ≡ 1. In the case of steady compressible flow we have∇(V ) = 0, so µ(r) = (r).Solving fluid flow problems by use of two stream functions when such use is possiblehas not received much attention so far. In certain problems where only two velocitycomponents appear, the third being zero, there would be only one single unknownstream function instead of two.If flow velocity have potential (V = ∇φ) it is clear from equation (3.158) thatpotential surface must be orthogonal to stream surfaces since gradients of streamsurfaces are orthogonal to gradient of potential. In two-dimensional case surfaces arereduced to lines, so potential lines and stream function lines are mutually orthogonal.
3.19.5 Stream Function for 2D space
If the flow pattern and various flow parameters are independent of one dimension(for example z) we said that the fluid motion is two-dimensional or planar. In thatcase:
∂
∂z( ) = 0
for any parameter instead of parenthesis, the motion in all planes normal to z-axisare the same. In Cartesian coordinate system we have:
ds = dxı+ dy+ dzkV = uı+ v
u = u(x, y) (3.160)
v = v(x, y)
Equation of the streamline is now in the form:
dx
u(x, y)=
dy
v(x, y)=dz
0
from which immediately follows that:
dz = 0 , z = const
Thus, one of the stream function is simply z. The other stream function is onlyunknown function. Let us denote it by ψ(x, y). From equation (3.158) follows:
µ(r)V = ∇ψ ×∇z =
∣∣∣∣∣∣∣∣ı k∂ψ∂x
∂ψ∂y
0
0 0 1
∣∣∣∣∣∣∣∣ (3.161)
From former equation we obtain:
µu =∂ψ
∂y, µv = −∂ψ
∂x(3.162)
note again that µ = 1 for incompressible flow, and µ = for compressible flow.From the definition of velocity potential:
u =∂φ
∂x, v =
∂φ
∂y
93
Figure 3.32: Examples of streamline patterns at a stagnation point in two-dimensional flow.
it follows:∂φ
∂x=∂ψ
∂y,
∂φ
∂y= −∂ψ
∂x(3.163)
in the case of incompressible flow, and:
∂φ
∂x=
1
∂ψ
∂y,
∂φ
∂y= −1
∂ψ
∂x(3.164)
for the case of compressible flow.
3.19.6 Stagnation Points
Points in the flow fields where the velocity becomes zero are known as stagnationpoints. In terms of the components of the velocity, this means that at a stagnationpoint all the velocity components are zero. If we express streamline equations inCartesian coordinates:
dy
dx=v(x, y, z)
u(x, y, z),
dz
dx=w(x, y, z)
u(x, y, z)(3.165)
At stagnation points these equations become
dy
dx=
0
0,
dz
dx=
0
0
Figure (3.32) shows streamline patterns for various types stagnation points.
3.20 Elementary Solutions of Laplace Equation
We will express Laplace equation in various coordinate systems. Laplace operatoris defined as:
∆ = ∇2 = ∇ · ∇ (3.166)
94
Cartesian form of this operator is given:
∆ϕ =∂2ϕ
∂x2+∂2ϕ
∂y2+∂2ϕ
∂z2= 0 (3.167)
In cylindrical coordinate system (r, θ, z), figure (3.33), this operator is expressed inthe form:
∆ϕ =1
r
∂
∂r
(r∂ϕ
∂r
)+
1
r2∂2ϕ
∂θ2+∂2ϕ
∂z2(3.168)
Figure 3.33: Cylindrical coordinate system.
In spherical coordinates r, θ, and ϕ it has the form:
∆φ =1
r2 sin θ
[∂
∂r
(r2 sin θ
∂φ
∂r
)+∂
∂θ
(sin θ
∂φ
∂θ
)+∂
∂ϕ
(1
sin θ
∂φ
∂ϕ
)]= 0 (3.169)
Solution of Laplace equation which exists everywhere is of the form:
φ = V∞ · r = U∞x+ V∞y +W∞z (3.170)
since:V = ∇φ ⇒ V = V∞ (3.171)
it appears that potential φ, (3.170) is the potential of uniform flow.Let us now find solution of equation (3.168) which depends only on distance r, soall derivatives with respect to z and θ are equal to zero, thus:
∆φ =1
r
d
dr
(rdφ
dr
)= 0 (3.172)
After integration we obtain:φ = σ ln (|r − ro|) (3.173)
where constant of integration is embedded in ro. In figure (3.34) it is shown thesense of the ∇φ.We can obtain from equation (3.168) solution which depends only on coordinate θ:
d2φ
dθ2= 0 ⇒ φ = Aθ +B
95
Figure 3.34: Cylindrical coordinate system.
or:
φ =Γ
2πarctan
y − yox− xo (3.174)
where xo and yo are constants of integration. Physically they represent the pointthrough singularity pass parallel with z axis. Velocity field of such potential isdefined as:
∇φ = Γ
2π∇ arctan
y − yox− xo =
Γ
2π
r − ro|r − ro|2 ×
k (3.175)
In 3D space we search solution again which depends only on distance r. Fromequation (3.169) follows:
d
dr
(r2 sin θ
dφ
dr
)= 0
which after integration gives:
φ = − σ
|r − ro| (3.176)
3.21 Uniqueness of the solution
Let us specify the problem:
∇2φ = 0∂φ
∂n= n · VB (3.177)
∇φ→ 0 , r →∞
This is the Neuman exterior problem. It is necessary to answer the question ofuniqueness of the solution for the problem (3.177).The answer is different for simply and multiply connected regions. Let us find firstthe answer for simply connected region, i.e. the region which is consisted only of theflow-field without any bodies. Assume that there are two independent solutions φ1
96
and φ2. Difference of these two potentials must also satisfy Laplace equation sincethat equation is linear:
∇2(φ1 − φ2) = ∇2φ1 −∇2φ2 ≡ 0
Let us apply now Green first formula to the function φ∇φ:∫ ∫V
∫∇φ∇φ dV =
∮S
φ∂φ
∂ndS (3.178)
where S = SB+S∞. But at the S∞ velocity is equal zero, as it is specified in (3.177).Since:
∂φ
∂n= ∇φ · n , when: r →∞ ,
∂φ
∂n→ 0 ⇒
∮S
φ∂φ
∂ndS = 0
If we now substitute φ with φ1 − φ2 into equation (3.178) we obtain:∫ ∫V
∫∇(φ1 − φ2)∇(φ1 − φ2) dV =
∫ ∫V
∫[∇(φ1 − φ2)]2 dV = 0
Function under integral is always positive function, so integral could be equal zeroonly if the function is equal to zero in each point of the space. That means thatdifference between solutions φ1 and φ2 is constant:
∇(φ1 − φ2) = 0 , ⇒ φ1 − φ2 = const
We can conclude that Neuman exterior problem is unique to within constant forsimply connected regions.Let us now consider doubly connected region exterior to body B. We could constructsimply connected region by introducing the cut which connects outer boundary
∞ with the boundary which surrounds the body (SB). Cut introduces two linesegments (C+ and C−) which form the new boundary and along which we don’tknow boundary conditions. We only know that velocities at corresponding pointsat the cut must be the same, so velocity normal to this cut must be the same also.Let us assume again that there exists two different solutions φ1 and φ2, and formdifference φD = φ1 − φ2. Let us again apply Green’s first formula for the problemsketched in the figure (3.35).
∫ ∫V
∫∇φD · ∇φD dV =
∮SB
φD∂φD
∂ndS +
∮S∞
φD∂φD
∂ndS
+∮C−
φD∂φD
∂ndS −
∮C+
φD∂φD
∂ndS (3.179)
the first two terms on the right-hand side are equal to zero, as before. The last twoterms Are not equal to zero. Let us denote with φD− the value of the potential atthe boundary C−, and with φD+ the value of the potential at the boundary C+,than: ∫ ∫
V
∫(∇φD)
2 dV =∫C−
φD−∂φD−
∂ndS −
∫C+
φD+
∂φD+
∂ndS (3.180)
97
Figure 3.35: Multiply connected region.
The normal derivative across cut is constant so we can write former equation:∫ ∫V
∫(∇φD)
2 dV =∫C
(φD− − φD+)∂φD
∂ndS (3.181)
If we restore the values of φD:
φD− − φD+ = φ−1 − φ+1︸ ︷︷ ︸Γ1
+φ+2 − φ−2︸ ︷︷ ︸−Γ2
so: ∫ ∫V
∫(∇φD)
2 dV = (Γ1 − Γ2)∫C
∂φD
∂ndS (3.182)
In general we could not require that integral on right-hand side is equal to zero. Soif we want unique solution circulation about body for both solutions φ1 and φ2 mustbe the same, in that case Γ1−Γ2 ≡ 0. The value of circulation Γ cannot be specifiedcompletely on mathematical grounds! We have to consider physical properties ofthe problem to correctly specify the value of the circulation.
3.22 Elementary Solutions of Small Disturbance
Equation
Small disturbance equation is given with equation (3.102):
(1−M2∞)∂2φ
∂x2+∂2φ
∂y2+∂2φ
∂z2= 0 , M∞ < 1 (3.183)
−(M2∞ − 1)
∂2φ
∂x2+∂2φ
∂y2+∂2φ
∂z2= 0 , M∞ > 1 (3.184)
98
Equation (3.183) is similar to Laplace equation so elementary solution for suchequation is:
φ =1√
(x− xo)2 + (1−M2∞) [(y − yo)2 + (z − zo)2](3.185)
Everything said so far for the uniqueness of Laplace equation is also valid for equation(3.183). Elementary solution of equation (3.184) is little bit different:
φ =1√
(x− xo)2 − (M2∞ − 1) [(y − yo)2 + (z − zo)2](3.186)
Uniqueness of the solution is much more difficult to prove since besides singularityat point (xo, yo, zo) elementary solution posses singularity at the surface:
x− xo = ±√M2∞ − 1
√(y − yo)2 + (z − zo)2 (3.187)
what represents the cone sketched in the figure (3.36)
Equation (3.187)
Figure 3.36: Dependence and influence regions.
To ensure uniqueness it is important to introduce the finite part of integral. Thiswill be illustrated by calculating mass flux from the source through cylinder shownin the figure (3.37). Let us introduce the following variables:
β2 = M2∞ − 1
r2 = (y − yo)2 + (z − zo)2 (3.188)
φ =1√
(x− xo)2 − β2r2
The volume flux through Sε is given by:
J(ε) =∫Sε
V · n dS =∫Sε
∇φ · n dS (3.189)
99
Figure 3.37: Calculating finite part of the integral.
where:
∇φ = ∇ 1√x2 − β2r2 , and: ∇φ · n = ∂φ
∂x(3.190)
so:∂
∂x
1√x2 − β2r2 = − x√
(x2 − β2r2)3at x = a gradient in x-direction is:
∇φ · n = − a√(a2 − β2r2)3
, while: dS = 2rπ dr
After substituting the former into equation (3.189) we obtain:
J(ε) = −rε∫0
a√(a2 − β2r2)3
2rπ dr (3.191)
We use definitions in figure (3.38) to calculate finite part of integral (3.191).It is clear from the figure (3.38) that:
ra = a tanµ =a
β
and:
rε = ra − ε
cosµ=a
β− εM∞
β
So our problem is to find integral:
J(ε) = −2πaa/β−εM∞/β∫
0
rdr
(a2 − β2r2)3/2
100
Figure 3.38: Definitions for calculating finite part of the integral.
Let us introduce substitution:
Z2 = a2 − β2r2 , ⇒ rdr = −Z dZβ2
which results in change of the limits:
r = 0 Z = a
and:
r =a
β− εM∞
β, Z =
√2aεM∞ − (M∞ε)
2
After performing integration we have:
J(ε) =2πa
β2
√2εM∞−ε2M2∞∫
a
dZ
Z2=
2π
β2− 2πa
β2√2aεM∞ − ε2M2∞
Note that first term on the right-hand side does not depend on the value a i.e. onthe radius of cylinder ra. So we have finite part which is constant and singularity(last term). If we modify expression for mass flux:
J∗ = J(ε) +2πa
β2√2aεM∞ − ε2M2∞
(3.192)
we have always finite expression for mass flux. Potential of supersonic source of unitintensity is defined thus:
φ =β2
2π
1√(x− xo)2 − β2r2
(3.193)
The constant β2/2π can be included in the arbitrary source strength! Uniquenessof the supersonic potential flow can be proved the same way as the uniqueness ofsubsonic potential solution when we take into account only finite part of mass fluxof the supersonic source.
101
3.23 Physical Interpretation and Basic Solutions
3.23.1 Point Source
Let the strength of the point source be σ than at some point away of the point thevelocity is calculated by:
V =σ
4πd2(3.194)
where d is distance between source and arbitrary point, and σ is intensity of thesource. Figure (3.39).
Figure 3.39: Point source.
Velocity is in the radial direction from the source outwards:
V =σ
4π |r − ro|2︸ ︷︷ ︸d2
· r − ro|r − ro|︸ ︷︷ ︸
unit vector
(3.195)
We can obtain potential of the velocity by formally integrating velocity since it isin radial direction. When the source is in the coordinate origin we have:
∂φ
∂r=
σ
4πr2⇒ φ = − σ
4πr(3.196)
It is clear that for r = const, |V | = const, and φ = const. Velocity drops with the
square of the distance r from origin. When r → ∞, |V | → 0. Mass flux throughany sphere with center in the source is constant.Let us review:
φ = − σ
4π|r − ro|V =
σ
4π
r − ro|r − ro|3
φ(x, y, z) = − σ
4π
1√(x− xo)2 + (y − yo)2 + (z − zo)2
(3.197)
u(x, y, z) =∂φ
∂x=
σ
4π
x− xo[(x− xo)2 + (y − yo)2 + (z − zo)2]3/2
102
v(x, y, z) =∂φ
∂y=
σ
4π
y − yo[(x− xo)2 + (y − yo)2 + (z − zo)2]3/2
w(x, y, z) =∂φ
∂z=
σ
4π
z − zo[(x− xo)2 + (y − yo)2 + (z − zo)2]3/2
Elementary source can be distributed along line, surface or volume. The potentialfor these three cases figure (3.40) can be expressed as:
φ(x, y, z) =1
4π
∫-
σ(xo, yo, zo) d√(x− xo)2 + (y − yo)2 + (z − zo)2
for line (3.198)
φ(x, y, z) =1
4π
∫∫S
σ(xo, yo, zo) dS√(x− xo)2 + (y − yo)2 + (z − zo)2
for surface(3.199)
φ(x, y, z) =1
4π
∫∫V
∫ σ(xo, yo, zo) dV√(x− xo)2 + (y − yo)2 + (z − zo)2
for vol. (3.200)
Velocities at point (x, y, z) can be obtained by differentiating expressions for velocity
Figure 3.40: Singular solution distributions.
potential given by equations (3.198) to (3.200):
(u, v, w) =
(∂φ
∂x,∂φ
∂y,∂φ
∂z
)(3.201)
3.23.2 Point Doublet
It is clear from equations (3.128) and (3.129) it is clear that exist the second basicsolution given the form:
µn · ∇(1
r
)= −∂φ
∂n
(1
r
)(3.202)
Since φ = σ/r is potential of the source, and ”−” comes from the fact that flow-fieldis on the opposite side of the unit vector n orientation. It can be said that:
φdoublet = − ∂
∂n(φsource) (3.203)
103
Figure 3.41: Point doublet.
This suggest that point doublet can be obtained from sources. Figure (3.41) illus-
trates the source at origin and sink at position . Potential at point P (x, y, z) dueto sink and source is:
φ = − σ
4π
(1
|r| −1
|r − |
)(3.204)
what again represents superposition of elementary solutions, or:
φ = − σ
4π
|r − | − |r||r| · |r − | (3.205)
Let us also choose σ such that σ→ µ when → 0. Then when → 0 :
|r| · |r − | → r2 , |r − | − |r| → cos θ
So potential becomes:
φ = − µ
4π
cos θ
r2(3.206)
The angle θ is the angle between position vector r and orientation of the doublet:
µ cos θ = µr
|r| (3.207)
and potential is finally expressed as:
φ = − µ · r4π r3
(3.208)
Vector µ points from sink to source direction. From equation (3.202) it is clear thatsurface normal n define the orientation of the doublet element:
φ(x, y, z) =µ
4πn · ∇
(1
|r − ro|)= − µ
4π
∂
∂n
(1
|r − ro|)
(3.209)
where ro denotes position of the doublet if it is not placed at coordinate origin.Taking ∂/∂n to be in x, y, or z direction yields:
φ(x, y, z) = − 1
4π
µx∂∂x
µy∂∂y
µz∂∂z
1√(x− xo)2 + (y − yo)2 + (z − zo)2
(3.210)
After performing differentiation we obtain:
104
If µ = µxı:
φ(x, y, z) =µx
4π
x− xo√[(x− xo)2 + (y − yo)2 + (z − zo)2]3
(3.211)
If µ = µy:
φ(x, y, z) =µy
4π
y − yo√[(x− xo)2 + (y − yo)2 + (z − zo)2]3
(3.212)
If µ = µzk:
φ(x, y, z) =µz
4π
z − zo√[(x− xo)2 + (y − yo)2 + (z − zo)2]3
(3.213)
We can obtain velocity components after differentiation of equations (3.211) to(3.213). If for short we use r2 = (x− xo)2 + (y − yo)2 + (z − zo)2 we obtain:
For x-direction
u1 = −µx
4π
r2 − 3(x− xo)2r5
v1 =3µx
4π
(x− xo)(y − yo)r5
(3.214)
w1 =3µx
4π
(x− xo)(z − zo)r5
For y-direction
u2 =3µy
4π
(x− xo)(y − yo)r5
v2 = −µy
4π
r2 − 3(y − yo)2r5
(3.215)
w2 =3µy
4π
(y − yo)(z − zo)r5
For z-direction
u3 =3µz
4π
(x− xo)(z − zo)r5
v3 =3µz
4π
(y − yo)(z − zo)r5
(3.216)
w3 = −µz
4π
r2 − 3(z − zo)2r5
If the doublet has general orientation, i.e. µ = µxı + µy + µzk the total velocities
component are obtained after summing corresponding velocities:
u = u1 + u2 + u3
v = v1 + v2 + v3 (3.217)
w = w1 + w2 + w3
105
Doublets also can be distributed along line, surface or volume. Potential for men-tioned distributions is expressed as:
φ(x, y, z) = − 1
4π
∫-
µx(x− xo) + µy(y − yo) + µz(z − zo)√[(x− xo)2 + (y − yo)2 + (z − zo)2]3
d
φ(x, y, z) = − 1
4π
∫∫S
µx(x− xo) + µy(y − yo) + µz(z − zo)√[(x− xo)2 + (y − yo)2 + (z − zo)2]3
dS (3.218)
φ(x, y, z) = − 1
4π
∫∫V
∫ µx(x− xo) + µy(y − yo) + µz(z − zo)√[(x− xo)2 + (y − yo)2 + (z − zo)2]3
dV
In the first integral µ is doublet strength per unit length, in the second strength perunit surface and in the last strength per unit volume!Using vector identity
A× ( B × C) = B( A · C)− ( A · B)C
we could write
µ×[∇×
(r
r3
)]= ∇
[µ ·
(r
r3
)]− (µ · ∇)
(r
r3
)
where A = µ, B = ∇, and C = r/r3, from this equation we could express
∇[µ ·
(r
r3
)]= µ×
[∇×
(r
r3
)]+ (µ · ∇)
(r
r3
)(3.219)
Since
∇×(r
r3
)=(∇ 1
r3
)× r + 1
r3∇× r ≡ 0
we finally get
∇[µ ·
(r
r3
)]= (µ · ∇)
(r
r3
)(3.220)
Velocity field is obtained from equation (3.208) as
V = ∇φ = − 1
4π∇[µ ·
(r
r3
)](3.221)
or after (3.220)
V = − 1
4π(µ · ∇)
(r
r3
)
= − 1
4π
(µx∂
∂x+ µy
∂
∂y+ µz
∂
∂z
)(x− xo)ı+ (y − yo)+ (z − zo)kr3
(3.222)
what is vector expression equivalent to expressions (3.214) to (3.216).
106
Figure 3.42: Uniform flow field.
3.23.3 Uniform Flow
Linear combination of coordinate directions:
φ = Ax+By + Cz
must be solution of Laplace’s partial differential equation since that equation is ofthe second order. Velocity components of such potential are:
U∞ =∂φ
∂x= A , V∞ =
∂φ
∂y= B , W∞ =
∂φ
∂z= C (3.223)
We can thus write potential also in the following form:
φ = U∞x+ V∞y +W∞z (3.224)
Since velocity components are constant at any point of the space flow field describedby potential (3.224) is uniform, figure (3.42)
3.24 Two-Dimensional Version of Basic Solutions
We develop now two-dimensional versions of basic solution. The difference betweentwo versions come from the fact that two-dimensional solutions have logarithmicsingularity while three-dimensional solutions have fractional singularity.
3.24.1 Source
In 2D version of unit source intensity of radial velocity is constant on the circleimagined around source. Let on some distance r the velocity is equal to:
Vr =∂φ
∂r=
σ
2πr(3.225)
where 2πr is the circumference of the circle. Than:
φ =σ
2πln r (3.226)
107
or if the point is not in the coordinate origin:
φ =σ
2πln√(x− xo)2 + (y − yo)2 (3.227)
Two-dimensional components of velocity (u, v) is obtained from former equation(3.227) by differentiating it with respect to x and y:
u =∂φ
∂x=σ
2π
x− xo(x− xo)2 + (y − yo)2 (3.228)
v =∂φ
∂y=σ
2π
y − yo(x− xo)2 + (y − yo)2 (3.229)
Stream function expression could be obtained similarly
ψ =σ
2πarctan
y − yox− xo (3.230)
3.24.2 Doublet
What is said for three-dimensional point doublet is also true for two-dimensionalpoint doublet. So potential of the point doublet is determined by differentiatingpotential for source (3.209) and (3.210):
φ(r) = − ∂
∂n
(σ
2πln r
)(3.231)
φ(x, y) = − 1
2π
(µx
∂∂x
µy∂∂y
)ln(√
(x− xo)2 + (y − yo)2)
(3.232)
where source strength σ is replaced by doublet strength µ. Taking n to point inx-direction vector µ is reduced to µx = µ and µy = 0, so:
φ(x, y) = − µ
2π
x− xo(x− xo)2 + (y − yo)2
or:
φ(r, θ) = − µ
2π
cos θ
r(3.233)
Two-dimensional components of velocity are also obtained by differentiating expres-sion for potential with respect to x and y:
u =∂φ
∂x=µ
2π
(x− xo)2 − (y − yo)2[(x− xo)2 + (y − yo)2]2
(3.234)
v =∂φ
∂y=µ
2π
2(x− xo)(y − yo)[(x− xo)2 + (y − yo)2]2
(3.235)
Starting from the expressions for velocity components we can integrate to obtainstream function
ψ =µ
2π
y − yo(x− xo)2 + (y − yo)2 or: ψ =
µ
2π
sin θ
r(3.236)
Similar to equations (3.233) and (3.236) can be obtained for µ = µ.
108
Figure 3.43: Vortex.
3.24.3 Uniform Flow
Uniform flow in two-dimensions is simply obtained from uniform flow in three-dimensions by simply dropping z-component:
φ = U∞x+ V∞y (3.237)
It is obvious that:u = U∞ , and: v = V∞
Stream function formulation is given by
ψ = U∞y − V∞x (3.238)
3.24.4 Vortex
Let us consider circular flow about some point in the flow-field, figure (3.43). Letalso velocity at some distance r be constant and tangent to a circle drown aboutvortex. Intensity of the velocity should satisfy:
V =Γ
2rπ(3.239)
so that circulation about any circle is constant and does not depend on the distancer.Velocity vector is tangential to the circle at point at which vector r is pointing. Wecan obtain velocity vector by the following expression:
V = − Γ
2rπ
k × rr
(3.240)
minus comes from the fact that the positive value of the circulation Γ is in oppositesense to mathematical positive sense. Vector k is unit vector normal to the plane ofthe vortex (z-direction):
V = − Γ
2π
∣∣∣∣∣∣∣ı k0 0 1
x− xo y − yo 0
∣∣∣∣∣∣∣(x− xo)2 + (y − yo)2 (3.241)
109
where (xo, yo) are coordinates of the vortex. Cartezian components of the velocityvector are given by:
u =∂φ
∂x=
Γ
2π
y − yo(x− xo)2 + (y − yo)2 (3.242)
v =∂φ
∂y= − Γ
2π
x− xo(x− xo)2 + (y − yo)2 (3.243)
Expression for potential φ can be obtained either integrating first equation (3.242)with respect to x, or the second equation with respect to y. Arbitrary constant ofintegration can be chosen such that both expressions are equivalent:
φ = − Γ
2πarctan
y − yox− xo (3.244)
Knowing that potential lines and stream lines are mutually orthogonal, and compar-ing the expression for stream function for the source, (3.227) and expression (3.244),we conclude that expression for vortex stream function must be similar to expressionfor source potential
ψ =Γ
2πln(√
(x− xo)2 + (y − yo)2). (3.245)
3.24.5 Two-Dimensional Singularity Distributions
We give now expressions for velocities, potential, and stream function for singularitydistributions over line and surface. Figure (3.44) illustrates singularity distributionsover two-dimensional line and over two-dimensional surface.
Figure 3.44: Singularity distributions over two-dimensional line and surface.
Source distributions
Let us give, first, expressions for source line distributions. With (dσ/d) d we denotesource strength of the infinitesimally small element of length d, as it is shown in thefigure (3.44). Expressions for velocities, and velocity potential, and stream function
110
are obtained by summing contributions from whole line
uP =∫L
dσ/d
2π· x− xo()(x− xo())2 + (y − yo())2 · d
vP =∫L
dσ/d
2π· y − yo()(x− xo())2 + (y − yo())2 · d
(3.246)
φP =∫L
dσ/d
2π· ln
√(x− xo())2 + (y − yo())2 · d
ψP =∫L
dσ/d
2π· arctan y − yo()
x− xo() · d
Influence of the source distributions in the x-y—plane can be considered similarly.Now dσ/dS represents intensity of the source per area. Velocities, velocity potential,and stream function are calculated as
uP =∫L
dσ/dS
2π· x− xo(x− xo))2 + (y − yo)2 · dS
vP =∫L
dσ/dS
2π· y − yo(x− xo)2 + (y − yo)2 · dS
(3.247)
φP =∫L
dσ/dS
2π· ln
√(x− xo)2 + (y − yo)2 · dS
ψP =∫L
dσ/dS
2π· arctan y − yo
x− xo · dS
where (xo, yo) are coordinates of elementary surface dS.
Doublet Distributions
We now proceed with giving expressions for calculating velocities, velocity potential,and stream function for line and surface distributions of two-dimensional doublets.Again we use expressions for point singularity for arbitrary doublet orientation µ =µxı+ µy
uP =1
2π
∫L
1
r4
[dµx
d(x− xo)2 − dµx
d(y − yo)2 + 2
dµy
d(x− xo)(y − yo)
]d
vP =1
2π
∫L
1
r4
[dµy
d(y − yo)2 − dµy
d(x− xo)2 + 2
dµx
d(x− xo)(y − yo)
]d
φP = − 1
2π
∫L
1
r2
[dµx
d(x− xo) + dµy
d(y − yo)
]d (3.248)
ψP =1
2π
∫L
1
r2
[dµx
d(y − yo)− dµy
d(x− xo)
]d
111
where r =√(x− xo)2 + (y − yo)2, and xo, yo, and zo are functions of line segment
length . We are not giving here expressions for surface distribution of doublet sinceexpressions are similar we need only to substitute dµx/d and dµy/d with dµx/dSand dµy/dS and to integrate over whole surface S. So we need to replace d with dS.Coordinates of the point P are (x, y), and they have to be considered as constant incalculations of integrals in equations (3.248). Doublet distributions dµx/d, dµy/d,dµx/dS, and dµy/dS are, in general, functions of position and cannot be extractedin front of the integrals.
Vortex Distribution
Vortex distribution per unit surface is called vorticity and is denoted with γ, we usethis symbol even for the distribution of vortex along arbitrary line L. Intensity ofthe vorticity is dependent on position on the line L or surface S, so it should beconsidered as variable in integrations. As in previous discussions we start from theexpressions for point vortex and obtain
uP =1
2π
∫L
γ() (y − yo)(x− xo)2 + (y − yo)2 d
vP = − 1
2π
∫L
γ() (x− xo)(x− xo)2 + (y − yo)2 d
φP = − 1
2π
∫L
γ() arctany − yox− xo d (3.249)
ψP =1
2π
∫L
γ() ln(√
(x− xo)2 + (y − yo)2)d
where xo and yo are functions of position, and (x, y) are constants and representcoordinates of the point P . Vorticity γ is also function of position along the curveL. Again we do not write here expression for surface distribution since only d haveto be substituted with dS. Now γ is really vorticity i.e. vortex strength per unitarea.
3.25 Complex Variables Expressions
3.25.1 Elements
It is customary to denote with z arbitrary complex variable z = x+ ıy, (ı =√−1),
and with W complex function of complex variable z: W =W(z). The special classof complex functions are called analytic functions. Function W(z) is called analyticinside some circle in the plane of complex variable z if it satisfy any of the followingconditions:
a) For function W(z) exists derivative W ′(z) at any point of the circle.
b) If W(z) = U(x, y) + ıV(x, y), where U(x, y) and V(x, y) are real functions of
112
real variables x and y and if these functions satisfy Laplace equation:
∂2φ
∂x2+∂2φ
∂y2= 0
and if among these functions exist the following relation:
∂U∂x
=∂V∂y
,∂U∂y
= −∂V∂x
(3.250)
c) The function W(z) is continuous inside circle and integral taken around anyclosed curve within the circle is equal to zero.
d) Inside any smaller concentric circle within the largest circle functionW(z) canbe approximated with arbitrary accuracy.
The theory of analytic functions can be based on any of the former statements. Ifwe compare equation (3.250) to the equation (3.163) we note the same expression,so we can construct analytic function using velocity potential (φ) instead of thereal function U , and stream function (ψ) instead of the real function V , for two-dimensional flow field:
F = φ+ ıψ (3.251)
Both equations (3.163) and (3.250) require:
∂φ
∂x=∂ψ
∂y, and:
∂φ
∂y= −∂ψ
∂x
If we differentiate the first equation with respect to x, and second with respect to yand add them side by side we obtain Laplace partial differential equation:
∆φ = ∇2φ =∂2φ
∂x2+∂2φ
∂y2= 0 .
Differentiation of the first equation with respect to y and the second equation withrespect to x and after subtraction of the second equation from the first we againobtain Laplace equation:
∆ψ = ∇2ψ =∂2ψ
∂x2+∂2ψ
∂y2= 0 .
As it is concluded in the subsection (3.19.4) equipotential lines ψ(x, y) = constand streamlines ψ(x, y) = const are mutually orthogonal set of curves. The similarproperty is exhibited by the velocity components, for we have:
∂u
∂x= −∂v
∂yincompressibility condition
∂u
∂y=∂v
∂xirrotationality condition
and also:∂2u
∂x2+∂2u
∂y2= 0 ,
∂2v
∂x2+∂2v
∂y2= 0 .
113
Curves u = const and v = const are also mutually orthogonal. Relation betweenreal and imaginary part of some complex variableW is exactly the same as relationbetween velocity potential lines and streamlines. So we can conclude that real andimaginary part of some analytic function represent velocity potential and streamfunction of some two-dimensional flow. It is customary to combine velocity potentialφ and stream function ψ into one analytic function and call it the complex potential.Let us denote it by F (z), than we could write:
F (z) = φ(x, y) + ıψ(x, y) (3.252)
The x, y-plane in this context becomes the plane of the complex variable z. Thederivative of the complex potential should not be dependent on the direction it iscalculated, thus
F ′(z) ≡ dF
dz=
∂φ
∂x+ ı∂ψ
∂x
=∂ψ
∂y− ı∂φ∂y
= u(x, y)− ıv(x, y) (3.253)
Comparing expressions for the derivative dF/dz we again arrive to the condition:
∂φ
∂x=∂ψ
∂y,
∂φ
∂y= −∂ψ
∂x
which must be satisfied in order to derivative be independent on the way derivativeis calculated. This condition is called Cauchy-Riemann condition. Right-hand sideof the equation (3.253) is the complex conjugate of u + ıv which is the velocityvector. It has, however, become customary to call the conjugate u− ıv the complexvelocity. It is usually denoted by W (z). We thus have
W (z) ≡ u(x, u)− v(x, y) = dF
dz= V (3.254)
3.25.2 Some Elementary Flows
We said that any analytic function can represent some two-dimensional flow. Wewill examine the flows represented by some simple functions. Let us consider theflow represented by the complex velocity of the form
W (z) = Azn
where A is complex constant. Velocity potential is obtained then:
F (z) =∫Azn dz =
A
n+1zn+1 n = −1
A ln z n = −1 (3.255)
Uniform Flow
Let complex constant A be A = U∞ − ıV∞, and let n = 0. In that case complexvelocity corresponds to homogeneous two-dimensional flow, while complex potentialF is expressed as:
F (z) = Az = (U∞ − ıV∞) z (3.256)
It is obvious that former equation represents uniform flow whose direction is deter-mined by constants U∞ and V∞.
114
Source and Vortex
Let us now assume n = −1 and that A is real and positive. Then we have
u− ıv = W (z) =A
z= Ar(cos θ − ı sin θ) = Are−ıθ (3.257)
where z = re−ıθ represents polar form of a complex number. Velocity potential isobtained by integrating former equation:
φ+ ıψ = F (z) = A ln z = A ln r + ıAθ (3.258)
we conclude that flow corresponds to the source at the origin. The strength of thesource is given by:
σ = 2πA
Consider next that A is purely imaginary, equal, say, to ıb with b > 0. We then have
u− ıv =W = ıb
z= ı
b
r(cos θ − ı sin θ)
After integrating of the former equation we obtain:
φ+ ıψ = F (z) = ib ln z = −bθ + ıb ln r (3.259)
We conclude that the flow is that due to a vortex at the origin. The circulation ofthe vortex is given by
Γ = 2πb , ⇒ b =Γ
2πWe thus observe that the complex potentials for the source and vortex flows are ofthe form
F (z) = A ln z
where A, in general, is a complex coefficient. If A is purely real we obtain a sourceflow, whereas if A is purely imaginary we have a vortex flow, thus A = (σ− ıΓ)/2π.In this sense the source and vortex belong to the same class of singular flows.
Doublet
If we now assume n = −2 and coefficient A to be real and positive we have
u− ıv = W =A
z2=A
r2e−2ıθ
After integrating the former equation we obtain
φ+ ıψ = F = −Az= −Acos θ
r+ ıA
sin θ
r(3.260)
We conclude that the flow is that due to a doublet situated at the origin. The axisof the doublet is in the direction of x and strength of the doublet is given by
µ = 2πA
Proceeding this manner we may obtain the corresponding flows for n < −2. Allthese flows are singular flows with the singularity at the origin.
115
Table 3.1: Elementary solutions in complex variable formulations.
Elementary solution Complex potential Velocity potential Stream function
Homogeneous flow Voe−ıα Vo(x cosα+ y sinα) Vo(y cosα− x sinα)
under α to x-axis
Source or sink ± σ2πln(z − zo) ± σ
2πln r ± σ
2πθ
at point zo
Doublet directed −µeıα
2π(z−zo)−µ2πr
cos(α− θ) µ2πr
sin(α− θ)under α to x-axis
Vortex of intensity Γ2πı
ln(z − zo) − Γ2πθ Γ
2πln r
Γ at point zo
Vortex and source σ−ıΓ2π
ln(z − zo) σ ln r−Γθ2π
σθ+Γ ln r2π
at point zo
3.25.3 Elementary Flows in Tabular Form
Let us shortly review all what is said about elementary flows expressed throughcomplex variables. Table (3.1) shows the most important elementary solutions oftwo-dimensional inviscid incompressible equations of motion. Both complex poten-tial and its real and imaginary part (which represents velocity potential and streamfunction) are given. We use in the next chapter these elementary solutions andcombine them to satisfy boundary conditions. In the expressions for r in the table(3.1) it is assumed that
r = |z − zo| =√(x− xo)2 + (y − yo)2 =
√(z − zo)(z − zo)
Also angle θ is calculated from
θ = arctany − yox− xo
It is said that there are two class of problems which can be solved using singularitymethod. The first class of problems determine the flow patterns about arbitrarydistributed singularities. The second class of problems require determination ofstrength of the singularities distributed in some portion of the flow field to satisfysome additional conditions.
3.26 Theorem of Blasius
Let F (z) be the complex potential of a two-dimensional inviscid flow past a bodyof any given shape. Let X and Y be the components in the x and y direction
116
respectively of the force per unit span on the body. Let M be the anti-clockwisemoment per unit span about the point z = 0. Then
X − ıY =ı
2
∮C
(dF
dz
)2
dz
and
M = −2
∮C
z
(dF
dz
)2
dz
where C is the curve representing the boundary of the body, and W = u − ı v =dF/dz is complex velocity. This result is known as the Theorem of Blasius, andmay be proved as follows, by considering a solid body in general flow, as depictedin figure (3.45). Consider the small element δs in the boundary of the solid. Thenδx = − sin β δs and δy = cos β δs, while
δX = −p δs cos β = −p δyδY = p δs sin β = p δx
ThenX − ıY = −
∮C
p(ı dx+ dy) = −ı∮C
p(dx− ı dy) (3.261)
Note that ∮C
dx =∮C
dy = 0
so that ∮C
const(dx− ı dy) = 0
Figure 3.45: Derivation of Blasius theorem.
Also from Bernoulli’s Theorem equation (3.81)
p = const− V 2
2
117
follows ∮C
p(dx− ıdy) =∮C
const(dx− ı dy)−
2
∮C
V 2(dx− ı dy) .
Equation (3.261) then becomes
X − ıY = ı
2
∮C
V 2(dx− ı dy) (3.262)
Observe now that dz = dx − ı dy = (dx + ı dy) − 2ı dy = dz − 2ı dy and thatV 2 = u2 + v2. Applying former to the terms in equation (3.64) we have
V 2(dx− ı dy) = (u2 + v2)(dx− ı dy)= (u2 + v2)dz − 2ı (u2 + v2)dy
= (u2 − v2)dz + 2v2dx− 2ı u2dy
= (u− ı v)2(dx+ ı dy) + 2v(v + ı u)dx− 2u(v + ı u)dy
But the contour C is a streamline of the flow, so that, on C, vdx = udy, and2vdx(v + ı u) = 2udy(v + ı u). Thus equation (3.262) becomes
X − ı Y = ı
2
∮C
(u− ı v)2(dx+ ı dy) = ı2
∮C
(dF
dz
)2
dz . (3.263)
Similarly, if M is measured anti-clockwise, the we can write δM = −δX · y+ δY ·x.Thus
M =∮C
p(xdx+ ydy) = −2
∮C
V 2(xdx+ ydy) , (3.264)
since, again
p = const−
2V 2
and ∮C
const(xdx+ ydy) = 0
we could expand, now, the term z(dF/dz)2
x
(dF
dz
)2
dz = (u− ı v)2(x+ ı y)(dx+ ı dy)
= (u2 − v2 − 2ı uv)(xdx− ydy + ı ydx+ ı xdy)
so that
z (dF
dz
)2
dz
= (u2 − v2)(xdx− ydy) + 2uv(xdy + ydx) (3.265)
Again, since C is a streamline, on C we have vdx = udy, so that the second termon the right-hand side of equation (3.265) is equal to
2uv(xdy + ydx) = 2v2xdx+ 2u2ydy
118
Equation (3.265) thus becomes
z (dF
dz
)2
dz
= (u2 + v2)(xdx+ ydy) = V 2(xdx+ ydy)
so that equation (3.264) now gives
M = −2∮C
z
(dF
dz
)2
dz
(3.266)
which, together with equation (3.263) constitutes the theorem of Blasius.
3.26.1 Extension of the Theorem of Blasius
We now extend the theorem of Blasius to unsteady flow. From equations (3.261)and (3.264) we have
dX − ı dY = −p dz , dM + ı dN = pz dz
Now let dz = ds eı α and therefore dz = dz e−2ı α. Therefore
X − ı Y = −ı∮C
p e−2ı αdz , M + ı N =∮C
pz e−2ı αdz (3.267)
For unsteady flow we can express P from equation (3.78)
P = C(t)− ∂Φ
∂t− ∇Φ · ∇Φ
2+ Π
For the incompressible flow this equation is reduced to
p = ∞
(C(t)− ∂Φ
∂t− ∇Φ · ∇Φ
2+ Π
)
For each instant of time C(t) is unique for the whole flow-field so circle integral ofit is equal to zero. It is customary to neglect influence of gravitational forces for airso we drop Π form the equation also, thus
p = ∞
(C(t)− ∂Φ
∂t− W ·W
2
)(3.268)
If we now substitute this value for pressure p into expression for Blasius force, andif we substitute Φ = (F + F )/2 we get
X − ı Y =
2
∮C
(dF
dz
)2
dz − ı2
∮C
(∂F
∂t+∂F
∂t
)dz (3.269)
The same way we obtain
M = −2∮C
z
(dF
dz
)2
dz
+
2∮C
z
(∂F
∂t+∂F
∂t
)dz
(3.270)
119
3.27 Lagally’s Theorem
Consider a uniform stream whose complex potential is (U−ıV )z and a source placedat point a of a strength σ. Complex potential of the combined flow-field is thus
F (z) = (U − ıV )z + σ
2πln(z − a)
If we introduce closed contour into the flow field C the complex potential is “dis-
Figure 3.46: Derivation of Lagally’s theorem.
turbed” by the addition of a function which must vanish at infinity, for the presenceof the finite body cannot affect the distant parts of the fluid. To make the case gen-eral, suppose there is a circulation of strength Γ about the body C, where positivevalue is considered to be clockwise . Then the complex potential at great distanceswill be of the form
F (z) = (U − ıV )z +m ln(z − a) + ıκ ln z + A
z+B
z2+ · · · (3.271)
where m = σ/2π and κ = Γ/2π. The last terms take into account circulation andeffects of introduced body C into flow field .The complex velocity is then given by
W =dF (z)
dz= (U − ıV ) + m
z − a +ıκ
z+A
z2+B
z3+ · · · (3.272)
where constants A, B, . . . are changed to avoid writing new constants besides arbi-trary constants.To find the force on the body, we have, form the theorem of Blasius,
X − ıY = ı
2
∮C
(dF (z)
dz
)2
dz .
Now let S be a circle of great radius which includes both the cylinder and the source.From the conclusion (3.64) we could enlarge contour S arbitrary and we may writefor any single valued function the the same expression. Thus∮
S
(dF
dz
)2
dz =∮C
(dF
dz
)2
dz +∮γ
(dF
dz
)2
dz
120
where γ is small contour drawn round the source. Thus
X − ıY =ı
2
∮S
(dF
dz
)2
dz − ı
2
∮γ
(dF
dz
)2
dz (3.273)
Now on the circle S, sice |z| is large, we can expand 1/(z− a) in powers of 1/z, andtherefore from (3.272),
dF
dz= (U − ıV ) + m
z
(1 +
a
z+a2
z2+ · · ·
)+ıκ
z+A
z2+B
z3+ · · ·
and hence (dF
dz
)2
= (U − ıV )2 + 2(U − ıV )(m+ ıκ)
z+A′
z2+B′
z3+ · · · ,
where A′, B′,. . . are certain constants. Thus by the residue theorem,
ı
2
∮S
(dF
dz
)2
dz = −2π(U − ıV )(m+ ıκ) . (3.274)
To calculate the second integral in (3.273), we see from (3.272) that
dF
dz= f(z) +
m
z − awhere
f(z) =d
dz[F −m ln(z − a)]
what is the complex velocity obtained by omitting the contribution from the sourceat a, now (
dF
dz
)2
= f 2(z) +m2
(z − a)2 +2mf(z)
z − a (3.275)
By the Taylor’s theorem
f(z) = f [(z − a) + a] = f(a) + (z − a)f ′(a) + · · ·Hence the residue of (dF/dz)2 at z = a is 4πımf(a). Thus
ı
2
∮γ
(dF
dz
)2
dz = −2πmf(a)
Now f(a) is the complex velocity induced at a by the remaining part of the complexpotential when we omit he source m. Thus calling this induced velocity um − ı vm,we have finally
X − ıY = −2πıκ(U − ıV )− 2πm(U − ıV + um − ı vm) (3.276)
this result is called Lagally’s theorem.Lagally’s theorem is reduced to simpler form when there is no free stream andcirculation. In such case we have
X − ı Y = −2πm(um − ı vm)
121
Chapter 4
Principle of Superposition
4.1 Basic Principle
Let Φ1, Φ2,. . . ,Φn are solutions of Laplace equation
∇2Φ1 = 0 , ∇2Φ2 = 0 , . . . ∇2Φn = 0
then linear combination of former solutions is also solution of Laplace’s partial dif-ferential equation
Φ =n∑
i=1
ciΦi , ∇2Φ = c1∇2Φ1︸ ︷︷ ︸=0
+c2∇2Φ2︸ ︷︷ ︸=0
+ · · ·+ cn∇2Φn︸ ︷︷ ︸=0
= 0
where c1, c2, . . . , cn are arbitrary constants, shorter this equation can be written inthe form
∇2Φ =n∑
i=1
ci∇2Φi = 0 (4.1)
Explained property is characteristics of linear equations to which Laplace equationbelongs. Constants ci could be selected to satisfy some additional requirement, suchas boundary conditions of the problem.
4.2 Rankine’s Oval
Consider the two-dimensional flow resulting from superimposing a uniform flow par-allel to the x-axis with velocity U∞, a source with a strength σ at x = −xo, and asink with a strength −σ at x = xo, figure (4.1). The velocity potential for this flowwill be
Φ(x, y) = U∞x+σ
2πln(√
(x+ xo)2 + y2)− σ
2πln(√
(x− xo)2 + y2)
(4.2)
where constants from equation (4.1) are c1 = U∞, c2 = σ, and c3 = −σ, while
Φ1 = x , Φ2 =1
2πln(√
(x+ xo)2 + y2), Φ3 =
1
2πln(√
(x− xo)2 + y2)
122
Figure 4.1: Superposition of uniform flow, source, and sink.
Since the stream function also satisfy Laplace equation, we can construct the streamfunction by superimposing expressions for stream function of each component
Ψ(x, y) = U∞y +σ
2πθ1 − σ
2πθ2 (4.3)
whereθ1 = arctan
y
x+ xoand θ2 = arctan
y
x− xoor, after substitution
Ψ(x, y) = U∞y +σ
2πarctan
y
x+ xo− σ
2πarctan
y
x− xo (4.4)
The velocity field due to this potential is obtained by differentiating either thevelocity potential or the stream function:
u =∂Φ
∂x= U∞ +
σ
2π
x+ xo(x+ xo)2 + y2
− σ
2π
x− xo(x− xo)2 + y2 (4.5)
v =∂Φ
∂y=
σ
2π
y
(x+ xo)2 + y2− σ
2π
y
(x− xo)2 + y2 (4.6)
Because of the symmetry about the x axis the stagnation points are located alongthe x axis, at points further out than the location of the source and sink, say atx = ±a. The distance a is then found by setting the u component of the velocity tozero
u(±a, 0) = U∞ +σ
2π
1
±a+ xo −σ
2π
1
±a− xo = U∞ − σ
π
xoa2 − x2o
= 0
and a is
a =
√σ xoπ U∞
+ x2o (4.7)
Consider the stagnation streamline (which pass through the stagnation points). Thevalue Ψ for the stagnation streamline can be found by observing the value of equation(4.3) on the left side stagnation point (where θ1 = θ2 = π, and z = 0). This resultsin Ψ = 0, which can be shown to be the same for the right-hand side stagnation
123
Figure 4.2: Streamlines for Rankine oval.
point as well (where θ1 = θ2 = 0). The equation for the stagnation streamline istherefore
Ψ(x, y) = U∞y +σ
2πarctan
y
x+ xo− σ
2πarctan
y
x− xo (4.8)
The streamlines of this flow, including the stagnating streamline, are sketched infigure (4.2). Note that the stagnating streamline includes a closed oval shape, calledRankine’s oval. Note that adding in the uniform flow source and sink of equalstrengths generate closed streamline. Since there is no cross-flow through streamline we can consider any streamline as the surface of some body. Closed streamlinesare of particular interest since they represent finite bodies.
4.3 Source in a Uniform Stream
Let us now apply principle of superposition on the solutions which represents ho-mogeneous flow parallel to the x axis, for which stream function looks
Ψ1 = U∞y
and source placed at coordinate origin
Ψ2 =σ
2πarctan
y
x
where σ is the strength of the source, (x, y) are coordinates of the arbitrary pointin the flow-field. Stream function for the former combination is obtained by simpleaddition
Ψ = Ψ1 +Ψ2 = U∞y +σ
2πarctan
y
x(4.9)
Similarly we could obtain potential of the combination
Φ(x, y) = U∞x+σ
2πln(√x2 + y2
)(4.10)
124
Constants ci are now c1 = U∞, and c2 = σ. Figure (4.3) shows streamlines forthis problem. It is obvious that generated body is of infinite size. To obtain closedcontour we must have additional sinks of equal strength distributed within uniformflow.We can calculate velocity components at arbitrary point (x, y) by definition
u =∂Φ
∂x= U∞ +
σ
2π
x
x2 + y2(4.11)
v =∂Φ
∂y=
σ
2π
y
x2 + y2(4.12)
Figure 4.3: Source in uniform stream.
4.4 Sources in Uniform Flow
We could apply any number sources (and sinks) in the uniform flow. We couldconstruct stream flow patterns similarly, as before. Let us find potential of n sources(some of them could be actually sinks) distributed at points (xi, yi), let also thestrengths to be σi. If we wish to obtain closed contour in the flow-field it must besatisfied
∑σi = 0. Thus
Φ(x, y) = U∞x+1
2π
n∑i=1
σi ln(√
(x− xi)2 + (y − yi)2)
(4.13)
and
Ψ(x, y) = U∞y +1
2π
n∑i=1
σi arctany − yix− xi (4.14)
125
Figure 4.4: Three sources in uniform stream.
The velocity components are obtained by differentiating expression (4.13) with re-spect to x and y
u =∂Φ
∂x= U∞ +
1
2π
n∑i=1
x− xi(x− xi)2 + (y − yi)2 (4.15)
v =∂Φ
∂y=
1
2π
n∑i=1
y − yi(x− xi)2 + (y − yi)2 (4.16)
We illustrate now the former by combining three sources in the uniform flow field.One of the sources is placed at the coordinate origin, while other two are placedon the y-axis symmetrically with respect to x-axis. Figure (4.4) shows stream lineswhen the strength of each source is equal to 5, and when U∞ = 1, and V∞ = 0.Picture shows clearly that the line pattern is similar to one obtained for the singlesource.Three dark, horizontal, lines from the sources to the left require explanation. Stream-lines are obtained requiring iso-lines plot from the software used to prepare pictures.But source has ‘double sheet’ property for stream function, i.e. stream function hasalong some line jump between 0 and some finite value, resulting in many isolinesbetween two close points where jump is occurred. We could, thus, neglect thesedark lines.Function arctan has jump when y → 0. Little bit above x-axis function arctan hasthe value π, and little bit below it has the value −π, as it is shown in the figure(4.5).Figure (4.6) shows two sources, one above x-axis, and one below symmetricallyplaced with respect to x-axis. The strength of the third source and uniform velocityare zero. We see that x-axis is the line of symmetry and also the streamline. Sinceany streamline can be treated as solid boundary we ‘discovered’ the way how to ob-tain the flow-field in the neighborhood of solid planes. Method by which singularitiesare distributed symmetrically about some line is called method of images.
126
Figure 4.5: Double sheet property of stream function for the source.
Final example with three sources is shown in the figure (4.7). Again sources areplaced along y-axis. The source below x-axis is placed at (0,−a) with strengthσ. The second source is placed at coordinate origin with strength also σ, and thirdsource is actually sink placed at (0, a) with strength −2σ. As it is clear total strengthof distributed sources is zero so we could expect closed contour if uniform flow comesin proper direction. The proper direction for such distribution of singularities isvertical upward, i.e. U∞ = 0 and V∞ = 0.
4.5 Doublet in Uniform Flow
4.5.1 Cauchy Integral Formula
What follows require application of Cauchy integral formula. We mention here themost important results. If the function f(z) is analytic everywhere within and on asimple closed contour C taken in the positive sense, and if zo is any point interiorto C, then
f(zo) =1
2πı
∮C
f(z)
z − zo dz (4.17)
This formula is called Cauchy integral formula. It tell us that if a unction f is to beanalytic within and on a simple closed contour C, then values of f interior to C arecompletely determined by the values of f on C. Proof of this theorem can be found
127
Figure 4.6: Two sources without uniform flow.
in any book about complex analysis. It can be proven that
f ′(zo) =1
2πı
∮C
f(z)
(z − zo)2 dz
we could prove this by formal differentiation of equation (4.17) with respect to zo.According to formula (4.17) we have
f(zo +∆zo)− f(zo)∆zo
=1
2πı
∮C
(1
z − zo −∆zo− 1
z − zo)f(z)
∆zodz
=1
2πı
∮C
f(z) dz
(z − zo −∆zo)(z − zo)
when ∆zo → 0 the left-hand side tends to f ′(zo) obtaining
f ′(zo) =1
2πı
∮C
f(z)
(z − zo)2 dz (4.18)
It can be generalized that
f (n)(zo) =n!
2πı
∮C
f(z)
(z − zo)n+1 dz (4.19)
where (n) means n-th derivative.
128
Figure 4.7: Sources and sink in uniform flow.
4.5.2 Flow about Cylinder
We could proceed the same way as before using equations (3.236) and (3.238) to con-struct such flow. But we will illustrate the same using complex variable formulation.In the sub-section (3.25.2) we have derived the following expressions
W (z) =µ
2π
1
z2, F (z) = − µ
2π
1
z(4.20)
for doublet oriented in positive x-direction and
W∞(z) = U∞ − ıV∞ , F (z) = (U∞ − ıV∞) z (4.21)
for uniform flow. Adding two expressions for complex potential (when uniform flowis parallel to the x-axis, and doublet oriented in −x direction) we get
F (z) = U∞z +µ
2π
1
z(4.22)
If the strength of the doublet is chosen to be µ = 2πR2U∞ then
F (z) = U∞
(z +
R2
z
), W (z) = U∞
(1− R2
z2
)(4.23)
Velocity potential and stream function are obtained from complex potential
Φ = (F (z)) , Ψ = (F (z)) (4.24)
129
Figure 4.8: Doublet in uniform stream.
So
Ψ = (F (z)) = [U∞
(z +
R2
zzz
)]= U∞
(y − R2
|z|2y)
when Ψ = 0 we get|z|2 = R2 , and y = 0
what is the equation of the zero-th streamline. Former expressions say that zero-th streamline is circle |z| = R and line y = 0 what is x-axis. Figure (4.8) showsstreamlines for the combination of uniform flow and doublet when U∞ = 2πR2.Again we have inner and outer flow. Flows are separated with zero-th streamline.The other equation y = 0 simply say that the flow-field is symmetrical about x axis.We now calculate force acting on the circle R, from Blaisus theorem, (3.263), wehave
X − ıY =ı
2
∮C
(dF
dz
)2
dz =ıU2
∞2
∮C
(1− R2
z2
)2
dz
=ıU2
∞2
∮C
(1− 2
R2
z2+R4
z4
)dz (4.25)
If we apply Cauchy integral formula to the former equation we have
X − ıY =ıU2
∞2
∮C
f1 dz −∮C
f2(z − 0)2
dz
︸ ︷︷ ︸=f ′
2(0)=0
+∮C
f3(z − 0)4
dz
︸ ︷︷ ︸=f ′′′
3 (0)=0
where f1(z) = 1, f2(z) = 2R2 = const, and f3(z) = R
4 = const. The first integral isnot Cauchy integral it is integral of analytic function. Any integration of analyticfunction along closed contour is equal to zero since value of the integral does not
130
Figure 4.9: Contour integral of analytic function.
depend on integration path. Let us imagine any two distinct points on some closedcontour, figure (4.9), then
∮C
f1(z) dz =
B∫A
f1(z) dz +
A∫B
f1(z) dz =
B∫A
f1(z) dz −B∫
A
f1(z) dz = 0 .
It is obvious that X − ıY ≡ 0. We, thus, concluded that there is no force acting onthe cylinder when only doublet singularity is used to form the cylinder.
4.6 Flow around Cylinder with Circulation
We now add circulation about the cylinder obtained in previous section. Our super-position has now three components: uniform flow, doublet oriented in −x direction,and vortex. We place both doublet and vortex at coordinate origin, thus
F (z) = (U∞ − ıV∞)z + µ
2π
1
z+0 + ıΓ
2πln z (4.26)
Let us again assume that the uniform flow is parallel to the x-axis, (V∞ = 0). Thancomplex velocity W is obtained by differentiating equation (4.26) with respect to z
W (z) =dF
dz= U∞ +
µ
2π
1
z2+ıΓ
2π
1
z(4.27)
where µ = 2πR2U∞. To calculate the force we need W 2(z)
W 2(z) = U2∞ +
(µ
2π
)2 1
z4−(Γ
2π
)2 1
z2+µU∞π
1
z2+ıU∞Γπ
1
z+ıΓµ
2π21
z3
or
W 2(z) = f1(z) +f2(z)
z+f3(z)
z2+f4(z)
z3+f5(z)
z4
where all fi are constants. For integration around circle |z| = R we need to consideronly the second term since all other are equal to zero based on the same argumentas in the previous section, thus
X − ıY =ı
2
∮|z|=R
ıU∞Γ/πz − 0
dz = −ıU∞Γ1
2πı
∮|z|=R
1
z − 0dz
︸ ︷︷ ︸=1
(4.28)
131
Figure 4.10: Flow about cylinder with circulation.
thusX = 0 , Y = U∞Γ (4.29)
As we see circulation generates vertical force component (lift), but horizontal forcecomponent is still equal to zero. Reality deny this conclusion since potential flowneglects influence of viscosity. Measurements confirm that pressure distributionaround cylinder is in a good agreement for the front part of cylinder but considerablydiffers on the rear side. We obtain streamlines taking imaginary part of complexpotential (4.26)
Ψ = [(U∞ − ıV∞)z + µ
2π
1
z+0 + ıΓ
2πln z
](4.30)
Streamlines are shown in the figure (4.10). If we add one singularity more to thisproblem, such as one additional vortex, circular shape is destroyed as it is shownin the figure (4.11). We need special treatment if we wish to preserve circularshape. Preservation of circular shape is important because we could apply conformalmapping which map circular shape to other shapes.
Example 4.1:Write program by which it is possible to generate data for plotting figures such as(4.11). Program should read undisturbed velocity components U and V , the number ofsingularities in the flow-field, and for each singularity:
1. Type of singularity: 1 – Source or sink, 2 – Vortex, and 3 – Doublet.
2. Strength of the singularity. Note that for source minus sign mean sink.
3. Coordinate where the singularity is placed.
The program should record the values of the stream function in the rectangular grid.
Solution 4.1:
132
Figure 4.11: Streamlines obtained when one more singularity is added in the flow-field shown in the figure (4.10).
module data_structuretype singl
integer tip ! type of the singularity =1 source or sink;!=2 vortex;!=3 doublet
real strength ! strength of the singularityreal xo,yo ! location of the singularity
end type!
end module data_structure! **************************************************************
program Singuluse data_structuretype(singl) s(20)integer, parameter :: Nx=100, Ny=100real psi(Nx,Ny)
! Read input dataCall ReadInput(U,V,s,N)
! Set ploting parametersxmax = 5.0ymax = 3.0dx = 2*xmax/(Nx-1)dy = 2*ymax/(Ny-1)
!! Record results into *.dat file! ready to be ploted by TECPLOT (C)
open(unit=1,file=’sing.dat’,form=’formatted’,status=’unknown’)write(1,101) Nx,Ny
101 format(’variables= "x", "y", "psi"’/’zone i=’,i4,’ j=’,I4,’ f=point’)
133
!do j = 1, Ny
y = (j-1)*dy - ymaxdo i = 1, Nx
x = (i-1)*dx - xmaxpsi(i,j) = CalculatePsi(U,V,s,N,x,y)write(1,102) x,y,psi(i,j)
end doend do
102 format(3F12.3)!
stop ’Singul - End of run!’end program Singul
! **************************************************************function CalculatePsi(U,V,s,N,x,y)use data_structuretype(singl) s(20)integer sitypereal, parameter:: pi=3.14159265Psi = U*y-V*x ! Uniform flowdo i = 1, N
xo = s(i)%xoyo = s(i)%yosi = s(i)%strengthsitype = s(i)%tipCheck_Singularity_Type: select case (sitype)case(1)
Psi = Psi + si/(2*pi)*atan2(y-yo,x-xo)case(2)
r = sqrt((x-xo)**2+(y-yo)**2)if(r<>0.0) then
Psi = Psi + si/(2*pi)*log(r)else
Psi = 0.0end if
case(3)rr = (x-xo)**2+(y-yo)**2if(rr <> 0.0) then
Psi = Psi - si/(2*pi)*(y-yo)/rrelse
Psi = 0.0end if
case defaultstop ’Undefined type of singularity’
end select Check_Singularity_Typeend doCalculatePsi = Psireturnend subroutine CalculatePsi
! **************************************************************subroutine ReadInput(U,V,s,N)use data_structuretype(singl) s(20)print 101
101 format(’ Uniform flow components (U,V) : ’,$)read *,U,V
134
print 102102 format(’ Number of point singularities in flow field : ’,$)
read *,Nprint 106do i = 1, N
print 103,iread *,s(i)%tipprint 104read *,s(i)%strengthprint 105read *,s(i)%xo,s(i)%yo
end do103 format(’ Type of the singularity no:’,I2,’ : ’,$)104 format(’ Strength of the singularity : ’,$)105 format(’ Location of the singularity (x,y) : ’,$)106 format(’ Avaliable types of singularities:’/ &
5x,’1 - source or sink’/ &5x,’2 - vortex’/ &5x,’3 - doublet’/ )returnend subroutine ReadInput
Example 4.2:In the figure (4.12) is shown cascade of source-vortex singularities (of the strength σ−ıΓ)distributed along y-axis. Distance between singularities is s. Calculate induced velocityat arbitrary point P (x, y).
Figure 4.12: Cascade of source-vortex singularities.
Solution 4.2:Complex potential of point source-vortex is given as
F (z) =σ − ıΓ2π
ln(z − zo)
while the complex velocity is obtained by differentiating this equation with respect to z
W = u− ıv = dF
dz=σ − ıΓ2π
1
z − zo
135
Summing contribution from complete cascade we get
u− ıv = σ − ıΓ2π
∞∑k=−∞
1
z − ısk
Mathematical tables supply us with formula
∞∑k=−∞
1
(k + a)n= (−1)n−1 π
(n− 1)!
d(n−1)
dan−1cot aπ
so we need to adjust our expression to this formula (n = 1)
u− ıv = ıσ − ıΓ2sπ
∞∑k=−∞
1
k + ızs
= ıσ − ıΓ2sπ
π cot(ız
sπ)
Since cot(ıx) = −ı coth(x) we finally have
u− ıv = −σ − ıΓ2s
coth(z
sπ)
If none of the singularities coincide with coordinate origin, but one of them is at ıh thenthe this equation looks
u− ıv = −σ − ıΓ2s
coth
(z − ıhs
π
)
4.7 Kutta-Joukowski Theorem
We consider the flow past 2D body whose contour is denoted by C, figure (4.13).Let the components of aerodynamic force acting on the body be X and Y in thedirection of x and y axes, respectively. According to the Blasius formula (3.263)this component of the forces can be calculated as
X − ı Y =ı
2
∮C
(dF
dz
)2
dz .
We could assume that complex velocity potential has the form
F (z) = (U∞ − ı V∞)z + σ + ıΓ
2πln z − µ
2π
1
z+
∞∑i=2
Ai1
zi
From which follows complex velocity W
W (z) =dF
dz= (U∞ − ı V∞) + σ + ıΓ
2π
1
z+µ
2π
1
z2+
∞∑i=2
Ai1
zi+1
thus
W 2(z) = (U∞ − ı V∞)2 + 2(U∞ − ı V∞)σ + ıΓ2π
1
z+
∞∑i=2
Bi
zi
136
Figure 4.13: Kutta-Joukowsky theorem.
It is clear from (4.19) that all terms 1/zi vanishes since Bi are constant except term1/z. So we concentrate our attention on
X − ıY =ı
2
∮C
2(U∞ − ı V∞)σ + ıΓ2π
1
z − 0dz
orX − ıY = −(U∞ − ı V∞)(σ + ıΓ)
since we are interested in closed finite body then must be σ = 0
X − ıY = −ı(U∞ − ı V∞)Γ (4.31)
Note also that ı = eı π/2, −ı = e−ı π/2 and
U∞ − ı V∞ = Q∞e−ıα , where: Q∞ =√U2∞ + V 2∞
ThusX − ıY = Q∞Γe−ı(α+π/2)
If we calculate complex conjugate from both sides of equation sign, we get
X + ı Y = X − ıY = Q∞Γeı (α+π/2) (4.32)
It is clear that the angle between velocity vector and force vector is π/2. Aero-dynamic force is always orthogonal to velocity vector in the case of potential flow.Force which is always orthogonal to velocity vector is called lift, thus
L = Q∞Γ (4.33)
this equation is known in aerodynamic community as Kutta-Joukowsky theorem.
4.8 Aerodynamic Force on the Vortex Line
Consider vortex line of the strength Γ and point P at we wish to determine inducedvelocities due to vortex line. We derive expression for induced velocity by consideringfirst influence of the vortex line segment between two arbitrary pointsA andB, figure
137
(4.14). Vortex line segment of infinitesimal length d induce velocity δV at point Paccording to Biot-Savart’s law, (3.141)
δV (r) =Γ
4π
dL× (r − rL)|r − rL|3 (4.34)
since = r − rL than δV must be orthogonal to the plane which contains vectors
Figure 4.14: Velocity induced by vortex line segment.
dL and . The intensity of the cross product is than |dL× | = −dL sin θ so
δV = − Γ
4π
sin θ dL
2(4.35)
where direction of δV is orthogonal to the plane ABP . From the figure (4.14) isalso clear that
h = sin θ , ⇒ =h
sin θL = r1 cos θA − cos θ = r1 cos θA − h cot θdL = − h
sin2 θdθ
after substitution of former expressions into equation (4.35), and after integratingfrom θA to θB we get
V = − Γ
4πh
θB∫θA
sin θ dθ = − Γ
4πh(cos θA − cos θB) (4.36)
If we are interested in induced velocity from whole vortex line than we must sub-stitute, in equation (4.36), instead θB, and θA values π and 0 respectively. Thus weget
V =Γ
2πh
138
what is exactly the same expression as one given by equation (3.239). We concludethat two-dimensional vortex and three-dimensional vortex line induce velocity ofsame magnitude. Other properties connected with two-dimensional vortex must bealso valid for vortex line, such as Kutta-Joukowsky theorem.In calculations it is necessary to to determine vector of induced velocity not onlymagnitude. We introduce vectors
ro =−→AB = rB − rA , r1 =
−→AP = r − rA , r2 =
−−→BP = r − rB
From the figure (4.14) and definitions of vector dot and cross products we have
h =|r1 × r2||ro|
cos θA =ro · r1|ro||r1|
cos θB =ro · r2|ro||r2|
it remains only to define unit vector of the plane which contains vectors r1 and r2
n =r1 × r2|r1 × r2|
We have now all elements to calculate induced velocity at any arbitrary point P
V = − Γ
4π
r1 × r2|r1 × r2|2
[ro ·
(r1|r1| −
r2|r2|
)](4.37)
Figure 4.15: Kutta-Joukowsky theorem for three-dimensional flow.
According to Joukowsky theorem for vortex surface pressure jump ∆p exists only ifthere is velocity component normal to bound vortex. Velocity component parallelto vorticity does not contribute to lift. We can, thus, generalize Kutta-Joukowskytheorem for three-dimensional flow-field, figure (4.15)
F = V∞ × Γwhere Γ = Γν, and is the vector in the direction of vortex tube, and equal to thelength of the vortex tube. Vector F is aerodynamic force per unit length.
139
4.9 Strait Solid Wall in the Flow Field
Consider a fluid containing a distribution of sources, sinks and doublets. If a surfaceS can be drawn in the fluid so as not to pass through any of the singularitiesdistributed on either side of it, then the two sets of singularities on opposite sidesof S are said to be image systems in the surface S.
Figure 4.16: Image system for the source.
4.9.1 3D Source near the Infinite Plane Wall
Figure (4.16a) shows a simple source of strength σ situated at A(a, 0, 0) at a distancea from the infinite rigid plane (x = 0). We show now that the appropriate image inthe rigid plane is an equal source of strength σ at A′(−a, 0, 0), what is the reflectionof point A in the plane.Consider figure (4.16b) in which we have two sources of strength σ at A(a, 0, 0) andA′(a, 0, 0) with no rigid boundary. Let P be any point on the plane x = 0. Thanfluid velocity at point P due to the two sources is
VP =σ
AP3
−→AP +
σ
A′P3
−−→A′P =
σ
AP3 (−→AP +
−−→A′P )
But −→AP +
−−→A′P = 2
−→OP
thenVP =
2σ
AP3
−→OP
This shows that at any point P at the plane x = 0 fluid flows vertically i.e. tangen-tially to the plane, and so there is no transport of fluid across this plane. Thus wemust have at all points in the plane ∂φ/∂n = 0.
4.9.2 2D Source near the Infinite Plane Wall
The illustration for two-dimensional version of source image is similar to that shownin figure (4.16). Only the expressions are those which correspond to two dimensions.
140
We again use complex variables. Let us consider two sources of equal strength σplaced at points z1 and −z1, then complex potential for those sources is
F (z) =σ
2πln(z − z1) + σ
2πln(z + z1)
Complex velocity W is obtained by differentiating this equation with respect to z
W =dF (z)
dz=σ
2π
(1
z − z1 +1
z + z1
)or
u− ı v = σ
2π
(− 1
z1 − z +1
z1 + z
)Note that for z = ı y first term in the parenthesis is conjugate complex value of thesecond term
u− ı v = σ
2π
(− 1
z1 − ı y +1
z1 + ı y
)so we have difference between complex conjugate variables what gives purely imag-inary value. Thus Flow is again tangent to the vertical axis x = 0, and it could beconsidered as a solid wall.
4.9.3 2D Doublet near the Impermeable Infinite Plane
Doublet singularity is obtained from the source—sink combination, so we could startfrom the definition of the doublet to determine image system. Let us now considerfigure (4.17). On the left is shown image system for the source—sink combination.On the right (4.17b) we show image system for the doublets which generate verticalstreamline x = 0 thus simulating solid plane wall.
Figure 4.17: Image system for the doublet.
From (3.260) we have for the doublet at coordinate origin
F = −Az
if we choose A = µe−ı α/2π we obtain the following expression for the complexpotential of the doublet at point z1
F = −µe−ı α
2π
1
z − z1141
The vertical plane is obtained from the superposition
F (z) =µeı α
2π
1
z + z1− µe−ı α
2π
1
z − z1or
F (z) =µ
2π
(eı α
z + z1− e−ı α
z − z1
)(4.38)
Let us check again velocity field induced by the doublet system
u− ı v = µ
2π
(− eı α
(z + z1)2+
e−ı α
(z − z1)2)
at the line z = i y
u− ı v = µ
2π
( −eı α[x1 + ı(y − y1)]2 +
e−ı α
[−x1 + ı(y − y1)]2)
denominators are conjugate complex numbers so we could write this equation as
u− ı v = µ
2Rπ
(−eı αeı θ
+e−ı α
e−ı θ
)=
µ
2Rπ
(−eı (α−θ) + e−ı (α−θ)
)where [x1+ ı(y−y1)]2 = Reı θ. Right-hand side is clearly imaginary number so thereis no component velocity in the x direction, thus line x = 0 could be considered assolid wall.
4.9.4 3D Doublet near the Impermeable Infinite Plane
Similarly to 2D doublet image system we could make the plane x = 0 to be solidwall. It is clear that vector µ, which represent doublet orientation and intensitymust be symmetric about plane x = 0. Figure (4.18) shows plane of symmetry(solid wall) and plane which contains doublet vector µ, which is also orthogonal tothe solid wall. Doublet is placed at point A(a, y, z), while image of the doublet isplaced A′(−a, y, z). Unit vector ı is orthogonal to solid wall so
n =ı× µ
vector n is orthogonal on the plane which contains both doublet orientation vectorµ and vector ı. That plane must be orthogonal to the solid wall (x = 0).Now we have to decompose vector µ into component parallel to intersection oftwo planes and component normal to intersection of these planes. Since vector nis orthogonal to the plane which contains doublet orientation vector µ, and ı isorthogonal to the plane x = 0 vector
τ =ı× n , τo =τ
|τ |must be parallel to the intersection of these two planes, so the component of thevector µ parallel to the intersection line is µ · τo. The remaining component µ− (µ ·
142
Figure 4.18: Image system for the three-dimensional doublet.
τo)τo is orthogonal to the intersection line. The normal component of the doubletorientation vector must be symmetric with respect to x = 0 line which is obtained bysimply reversing the sign of obtained orthogonal component (µ·τo)τo−µ. Finally thedoublet orientation vector at point A′(−a, x, z) is obtained after adding componentparallel to the intersection line (which is the same for both vectors).
µ′ = 2(µ · τo)τo − µ (4.39)
Intensity of the vector τ can be calculated from
τ =ı× (ı× µ) =ı(ı · µ)− (ı ·ı)µ = µxı− µ = −µy− µzk
where µ = µxı+ µy+ µzk, and we used vector identity A× ( B × C) = ( A · C) B −
( A · B)C to expand double cross product. Thus
τ = −µy− µzk , |τ | =
√µ2y + µ
2z , τo = −µy+ µz
k√µ2y + µ
2z
It is clear that if we replace this expressions into equation (4.39) that we obtain
µ′ = −µxı+ µy+ µzk
what we could write immediately, but described procedure is valid for any planeorientation.
4.10 The Circle Theorem
We discussed so far how to obtain solid wall in the flow field. It is of some importanceto obtain the images of singularities outside the circle in a such way that circularshape is maintained. Since we are talking about two-dimensional problem we useagain complex variables. Let complex potential of the flow with singularities outsidecircle be denoted by F (z) than
F (z) = φ(z) + ı ψ(z)
143
where both functions φ and ψ are real functions. It is important to differentiate thefollowing notations F (z), F (z), and F (z)
F (z) = φ(z)− ı ψ(z)F (z) = φ(z) + ı ψ(z)
F (z) = φ(z)− ı ψ(z) = F (z)
Now we can formulate the circle theorem. Let there be irrotational two-dimensionalflow of incompressible inviscid fluid in the z-plane. Let there be no rigid boundariesan let the complex potential of the flow be F (z), where the singularities of F (z) areall at a distance greater than a from the origin. If a circular cylinder typified by itscross-section the circle C, z = a, be introduced into the field of flow, the complexpotential becomes
f(z) = F (z) + F
(a2
z
)(4.40)
Proof. Since there are no rigid boundaries, the flow given by F (z) is determinate atevery point of the z-plane, except perhaps at the singularities of F (z) which ariseform the vortices, sources, doublets, etc.After the cylinder is inserted, C must become a streamline ψ = const, and withoutloss of generality we may assume that it is the streamline ψ = 0.Let us consider now position in the complex plane of the points z and a2/z. Ifz = reı θ is outside the circle |z| = a than point a2/z = a2e−ı θ/r is inside the circleC, as it is shown in the figure (4.19a).
Figure 4.19: Relationship between points z and a2/z.
Coordinate of the point S is a2eı θ/r, and point Q is symmetric to the point S withrespect to x-axis, i.e. points have conjugate complex coordinates. In the limitingcase, figure (4.19b), when r → a points P and Q fall on to the circle C. Coordinateof the point Q is z if the coordinate of the point P is z.Considering the expression (4.40) it is clear that on the circle C the function F (a2/z)is equal to F (z) thus complex function f(z) is completely real at the circle C,meaning that ψ = 0. Thus circle C is streamline.
144
4.10.1 A Few Examples
Flow about Circle
Let us first consider again flow about circle without circulation. Complex potentialof the uniform flow is expressed as
F (z) = (U∞ − ı V∞)zcomplex function F (z) is given as
F (z) = (U∞ + ı V∞)z , ⇒ F
(a2
z
)= (U∞ + ı V∞)
a2
z
and finally after adding expressions in previous two equations (superposition prin-ciple) we get
f(z) = (U∞ − ı V∞)z + (U∞ + ı V∞)a2
z(4.41)
Source and Vortex near the Circle
As it is given by equations (3.258) and (3.259) complex potential of the source andvortex at the zo are given as
F (z) =σ + ıΓ
2πln(z − zo)
We construct complex potential of the flow about circle again in two step process
F (z) =σ − ıΓ2π
ln(z − zo) ⇒ F
(a2
z
)=σ − ıΓ2π
ln
(a2
z− zo
)
Adding previous two functions (superposition principle) we get complex potentialf(z) of the source-vortex combination near the circle
f(z) = F (z) + F
(a2
z
)=σ + ıΓ
2πln(z − zo) + σ − ıΓ
2πln
(a2
z− zo
)(4.42)
Adding equations (4.41) and (4.42) (superposition principle) we get complex poten-tial of the uniform flow with source and vortex singularities near the circle
f(z) = (U∞ − ı V∞)z + (U∞ + ı V∞)a2
z+σ + ıΓ
2πln(z − zo) + σ − ıΓ
2πln
(a2
z− zo
)(4.43)
If we carefully look at the last term of this equation we can write it
σ − ıΓ2π
ln
(a2
z− zo
)=σ − ıΓ2π
ln (−zo) + σ − ıΓ2π
ln
(z − a2
zo
)− σ − ıΓ
2πln (z)
the first term on the right-hand side is the complex constant and can be discardsince velocity potential and stream function are undetermined up to some constant.The last two terms on the right-hand side represent singularities within the circularcylinder, last term being the singularity at the coordinate origin.
145
Doublet near the Circle
Complex potential of the doublet at the point zo is given by table (3.1)
F (z) =−µeıα2π
· 1
z − zowe, again, find F (a2/z) in two step process
F (z) =−µe−ıα
2π· 1
z − zo , ⇒ F
(a2
z
)=−µe−ıα
2π· 1
a2
z− zo
And, as before, adding former expressions (superposition principle) we get complexpotential of the flow due to doublet near the solid cylinder
f(z) =−µeıα2π
· 1
z − zo +−µe−ıα
2π· 1
a2
z− zo
(4.44)
Adding this expression with complex potential of the uniform flow about circle weget the complex potential of the flow about cylinder with doublet singularities inthe flow-field.
All Together
We again use superposition principle to obtain complex potential of the flow aboutcylinder in the flow-field. Let us find complex potential for the flow about circularcylinder placed at the coordinate origin if in the flow field exists n source-vortexsingularities of the intensity σi + ıΓi, placed at points zi outside circle, and mdoublets of the intensity µi, with orientation αi, and at the positions zi outsidecircle
f(z) = (U∞ − ı V∞)z + (U∞ + ı V∞)a2
z+
n∑i=1
σi + ıΓi
2πln(z − zi) +
n∑i=1
σi − ıΓi
2πln
(a2
z− zi
)+
m∑i=1
−µieıαi
2π· 1
z − zi +m∑i=1
−µie−ıαi
2π· 1
a2
z− zi(4.45)
If it is required to place cylinder instead at origin to some other place, say zc, thanwe have to substitute z in equation (4.45) with z − zc, and all zi with zi − zc.In the figure (4.20) is shown flow-field produced by the vortex in the vicinity of thecylinder. We can compare this flow-field with the one shown (4.11)In the figure (4.21) we show flow-field about cylinder with three singularities aroundit.
Example 4.3:Develop the program by which it is possible to calculate flow-field around circular cylinderin the presence of singularities in the flow-field.
Solution 4.3:Our task is to program equation (4.45). We need to read U∞, V∞, n, m, the com-plex constants for source vortex singularity (σi,Γi), and µi, αi constants for doubletsingularity. Program that follows is possible solution of our problem.
146
Figure 4.20: Vortex near the circular cylinder.
Program CircleTheoreminteger, parameter :: NSV=10, ND=10, Nx=100, Ny=100real, parameter :: xmin=-2.0, xmax=6.0, ymin=-2.0, ymax=3.0complex F,zi(NSV+ND),zc, zreal sigma(NSV),Gamma(NSV),Mu(ND),alpha(ND), Psi(Nx,Ny)
! Read necessary dataCall GetData(sigma,Gamma,Mu,alpha,zi,zc,Ub,Vb,a,NSV,ND,NoSv,NoD)
! Prepare for coordinate calculationsdx = (xmax-xmin)/(Nx-1)dy = (ymax-ymin)/(Ny-1)open(unit=1,file=’circ.dat’,form=’formatted’,status=’unknown’)write(1,101) Nx,Ny
101 format(’variables= "x", "y", "psi"’/’zone i=’,i4,’ j=’,I4,’ f=point’)!! Calcualte stream function
do i = 1, Nxx = xmin + dx*(i-1)do j = 1, Ny
y = ymin + dy*(j-1)z = cmplx(x,y)Call CalculatePotential(z,Ub,Vb,a,zc,zi,NoSV,NoD,sigma,Gamma,Mu,alpha,F)write(1,’(3F10.3)’) x,y,aimag(F)
end doend doclose(unit=1)stop ’CircleTheorem - E n d o f r u n !’end program CircleTheorem
! ***************************************************************************Subroutine GetData(sigma,Gamma,Mu,alpha,zi,zc,Ub,Vb,a,NSV,ND,NoSV,NoD)complex zi(NSV+ND),zcreal sigma(NSV),Gamma(NSV),Mu(ND),alpha(ND)
! Undesturbed velocitiesUb = 1.0
147
SOURCE
VORTICES
CYLINDER
Figure 4.21: Uniform flow, one source and two vortices in the vicinity of the circularcylinder.
Vb = 0.2! Radius of the cylinder
a = 1.0! Location of the cylinder
zc = (1.0,0.0)! Sources and Vortices
NoSV = 3sigma(1) = 0.0Gamma(1) = 8.0zi(1) = (2.0,2.0)sigma(2) = 0.0Gamma(2) = -5zi(2) = (-1.0,2.0)sigma(3) = 5.0Gamma(3) = 0.0zi(3) = (-2.0,-1.0)
! DoubletsNoD = 0returnend subroutine GetData
! ***************************************************************************Subroutine CalculatePotential(z,Ub,Vb,a,zc,zi,NoSV,NoD,sigma,Gamma,Mu,alpha,F)real ,parameter :: pi=3.14159265complex F,zi(*),zc,zreal sigma(NoSV),Gamma(NoSV),Mu(NoD),alpha(NoD)
! Contribution from undesturbed velocityif(z == (0.0,0.0)) then
F = (0.0,0.0)else
F = cmplx(Ub,-Vb)*(z-zc)+cmplx(Ub,Vb)*a**2/(z-zc)end if
! Contribution from sources and vortices
148
do i = 1, NoSVif(z <> zi(i)+zc) then
F = F + cmplx(sigma(i),Gamma(i))/(2*pi)*log(z-zc-zi(i))end ifif(z<>zc .and. (a**2/(z-zc)-conjg(zi(i))<>0.0)) then
F = F + cmplx(sigma(i),-Gamma(i))/(2*pi)*log(a**2/(z-zc)-conjg(zi(i)))end if
end do! Contribution from doublets
do i = 1, NoDii = i+NoSVif(z<>zc+zi(ii)) then
F = F + Mu(i)*exp(cmplx(0.0,alpha(i)))/(2*pi)/(z-zc-zi(ii))end ifif(z<>zc .and. a*a/(z-zc)<>conjg(zi(ii)) ) then
F = F + Mu(i)*exp(cmplx(0.0,-alpha(i)))/(2*pi)/(a*a/(z-zc)-conjg(zi(ii)))end if
end doreturnend subroutine CalculatePotential
Example 4.4:Find the force exerted on a circular cylinder by the source of intensity m = σ/2π placedat d at x-axis, outside the cylinder, figure (4.22).
Figure 4.22: Find force exerted by source at d on the cylinder at coordinate origin!
Solution 4.4:Theorem of Blasius from (3.263) is
X − ıY = ı
2
∮C
(df
dz
)2
dz
where df/dz is the complex velocityW on the cylinder surface. We again use Thomson’stheorem to find complex potential of the flow about circular cylinder. Complex potentialof the flow caused by the source place at d is given by
F (z) =σ
2πln(z − d) = m ln(z − d)
The next step is to find F (z) what is exactly the same as this equation. The secondstep to find F (a2/z)
F
(a2
z
)= m ln
a2 − zdz
149
Adding the last two expressions (superposition principle) we obtain required complexpotential f(z)
f(z) = m ln(z − d)−m ln z +m ln(a2 − zd)we could drop complex constant from the last term (m ln(1/− d)) so we finally get
f(z) = m ln(z − d)−m ln z +m ln(z − d′) , where d′ = a2/d
Complex velocity W is then
W =m
z − d −m
z+
m
z − d′thus (
df
dz
)2
= W 2 =m2
(z − d)2 +m2
(z − 0)2+
m2
(z − d′)2 −2m2
z(z − d)− 2m2
z(z − d′) +2m2
(z − d)(z − d′)In order to calculate force on the circular cylinder we wish to apply Cauchy integralformula (4.17). It is clear that all nominators are constants thus all terms of the form1/(z − c)2 will disappear. The remaining terms should be transformed such
− 2m2
z(z − d) =2m2
d(z − 0)− 2m2
d(z − d)− 2m2
z(z − d′) =2m2
d′(z − 0)− 2m2
d′(z − d′)2m2
(z − d)(z − d′) =2m2
(d− d′)(z − d) +2m2
(d′ − d)(z − d′)Point d is not inside the integration path so circular integral of the terms 1/(z − d) isequal to zero, we thus have
∮CW 2 dz =
∮C
(2m2
d(z − 0)+
2m2
d′(z − 0)− 2m2
d′(z − d′) +2m2
(d′ − d)(z − d′))dz
= 2πı(2m2)(1
d+
1
d′− 1
d′+
1
d′ − d)
From Blasius theorem we get
X − ı Y = 2πm2 d′
d(d− d′) = σ2
2π
a2
d(d2 − a2)Therefore the cylinder is attracted toward the source. An examination of the streamlineswill show that the pressure is greater on the side of the cylinder remote from the source.The same is true for the sink.
150
Figure 4.23: Continuous distribution of singularities over x-axis segment.
4.11 Continuous Distributions of Singularities
So far we apply principle of superposition to isolated point singularities. Now wego little bit further. Consider strait line segment overlapping with x-axis. Figure(4.23) shows continuous distribution of singularities along such segment. We areinterested to calculate velocities, potential and stream function values induced bysingularities distribution. With δ(x) we mean dσ(x)/dx = σ′(x) in the case ofsource distributions, dµ(x)/dx = µ′(x) in the case of doublet distribution, anddΓ(x)/dx = γ(x) in the case of vortex distribution. Vortex distribution is alsoknown as vorticity γ(x).Consider infinitesimal segment of the length d which belongs to segment x1x2, andarbitrary point P (x, y). Contribution of the infinitesimal segment to the total flowproperties at point P can be expressed as:
• If the singularity distribution δ(x) is source distribution:
d uP (x, y) =σ′(xo)2π
· x− xo(x− xo)2 + y2 · dxo
d vP (x, y) =σ′(xo)2π
· y
(x− xo)2 + y2 · dxo(4.46)
dφP (x, y) =σ′(xo)2π
· ln√(x− xo)2 + y2 · dxo
dψP (x, y) =σ′(xo)2π
· arctan y
x− xo · dxo
• If the singularity distribution δ(x) is doublet distribution:
duP (x, y) =1
2πr4
[µ′x(xo)(x− xo)2 − µ′x(xo)y2 + 2µ′y(xo)(x− xo)y
]dxo
dvP (x, y) =1
2πr4
[µ′y(xo)y
2 − µ′y(xo)(x− xo)2 + 2µ′x(xo)(x− xo)y]dxo
dφP (x, y) = − 1
2πr2
[µ′x(xo)(x− xo) + µ′y(xo)y
]dxo (4.47)
dψP (x, y) =1
2πr2
[µ′x(xo)y − µ′y(xo)(x− xo)
]dxo
• If the singularity distribution δ(x) is vorticity:
duP (x, y) =1
2π
γ(xo) y
(x− xo)2 + y2 dxo
151
dvP (x, y) = − 1
2π
γ(xo) (x− xo)(x− xo)2 + y2 dxo
dφP (x, y) = − 1
2πγ(xo) arctan
y
x− xo dxo (4.48)
dψP (x, y) =1
2πγ(xo) ln
(√(x− xo)2 + y2
)dxo
Equations (4.46) to (4.48) looks simpler than equations (3.246) to (3.249) since thereis no need to parametrically describe singular line segment, and because yo ≡ 0. Toobtain uP , vP , φP and ψP we need to integrate former expressions between x1 andx2
uP =
x2∫x1
duP , vP =
x2∫x1
dvP , φP =
x2∫x1
dφP , ψP =
x2∫x1
dψP . (4.49)
4.11.1 Integration of Source Distributions
Let us assume σ′(xo) = σ′ = const, and substitute
(x− xo)2 + y2 = t2 , ⇒ tdt = −(x− xo)dxothen
t1 =√y2 + (x− x1)2 and: t2 =
√y2 + (x− x2)2
so we can integrate equation (4.46) for velocity uP
uP = − σ′
2π
t2∫t1
dt
t
what gives
uP (x, y) =σ′
4πlny2 + (x− x1)2y2 + (x− x2)2 (4.50)
The second component of velocity (v) can be integrated after we introduce substi-tution
x− xoy
= t ⇒ dxo = −ydt
and
t1 =x− x1y
t2 =x− x2y
thus
vP (x, y) = − σ′
2π
t2∫t1
dt
1 + t2=σ′
2π
(arctan
x− x1y
− arctanx− x2y
)
=σ′
2π
(arctan
y
x− x2 − arctany
x− x1)
(4.51)
152
This integral has unique value at all points of the x-y–plane except at x-axis wheny = 0. If we approach to x-axis from the above we get
vP (x, 0+) =
σ′
2π
[π
2−(−π2
)]=σ′
2
Approaching to x-axis from below gives
vP (x, 0−) =
σ′
2π
(−π2− π
2
)= −σ
′
2
or shorter
vP (x, 0±) =
∂φ(x, 0±)∂x
= ±σ′
2(4.52)
4.11.2 Integration of Doublet Distribution
Note that expression for the velocity potential in equation (4.46) is identical in formto the expression for the velocity component v, (4.47), when µ′ = µ′y and µ′x = 0.Thus we conclude that potential must have jump across the doublet segment
φ(x, 0±) = ∓µ′(x)2
(4.53)
This leads to discontinuous tangential velocity component given by
u(x, 0±) =∂φ(x, 0±)∂x
= ∓1
2
dµ′(x)dx
(4.54)
Let us, now, calculate circulation about line shown in the figure (4.24). Since the
Figure 4.24: Sketch for determination of the circulation around doublet segment.
doublet distribution begins at x1, the circulation γ(x) around path surrounding thesegment x1x is
Γ(x) =
x∫x1
u(xo, 0+)dxo +
x1∫x
u(xo, 0−)dxo = −µ′(x) (4.55)
153
which is equivalent to the jump in the potential
Γ(x) = φ(x, 0+)− φ(x, 0−) = −µ′(x) = ∆φ(x) (4.56)
Let us calculate induced velocities at arbitrary point P (x, y). We introduce substi-tution of the form
t2 = y2 + (x− xo)2 , ⇒ tdt = −(x− xo)dxothus
uP (x, y) =µ′
π
x2∫x1
(x− xo)y dxo[y2 + (x− xo)2]2
= −µ′yπ
t2∫t1
dt
t3=µ′y2π
(1
t22− 1
t21
)
or after restoring to original variables
uP (x, y) =µ′
2π
(y
y2 + (x− x2)2 −y
y2 + (x− x1)2)
(4.57)
In order to calculate vP we need to split expression for it in the equation (4.47) intwo parts
vP =µ′
2π
x2∫x1
y2
[y2 + (x− xo)2]2 dxo −µ′
2π
x2∫x1
(x− xo)2[y2 + (x− xo)2]2 dxo
The both integrals can be reduced to the standard table form by the substitution
X = x− xo , ⇒ dX = −dxo , X1 = x− x1 , X2 = x− x2thus we have
vP =µ′
2π
x−x2∫x−x1
X2
(y2 +X2)2dX − µ′
2π
x−x2∫x−x1
y2
(y2 +X2)2dX
After integrating we get
µ′
2π
x−x2∫x−x1
X2
(y2 +X2)2dX =
µ′
2π
(1
2yarctan
X
y− X
2(X2 + y2)
)∣∣∣∣∣x−x2
x−x1
µ′
2π
x−x2∫x−x1
y2
(y2 +X2)2dX =
µ′
2π
(1
2yarctan
X
y+
X
2(X2 + y2)
)∣∣∣∣∣x−x2
x−x1
It is clear that after subtraction of this equations the first terms cancel each otherso we finally get
vP =µ′
2π
(x− x1
(x− x1)2 + y2 −x− x2
(x− x2)2 + y2)
(4.58)
154
4.11.3 Integration of Vortex Distributions
Comparing the equations (4.46) and (4.48) we conclude that the expression for uPfor vortex distribution is equal in form to the expression in (4.46) for vP , so we canwrite immediately from (4.51)
uP (x, y) =γ
2π
(arctan
x− x2y
− arctanx− x1y
)(4.59)
Following the same arguments we gave in section (4.11.1) we conclude that x-component of induced velocity must have jump across the vortex sheet
u(x, 0±) =∂φ(x, 0±)∂x
= ±γ(x)2
(4.60)
Figure 4.25: Sketch for determination of the circulation around vorticity segment.
We can integrate to obtain potential jump at the arbitrary place x, figure (4.25).Since potential jump is equal to the circulation
∆φ(x, 0) = φ(x, 0+)− φ(x, 0−) = Γ(x) =
x∫x1
γ(x)
2dx−
x∫x1
−γ(x)2dx
Comparing this equation and equation (4.60) with (4.55) and (4.54) we concludethat
Γ(x) = −µ′(x) , and: γ(x) = −dµ′(x)/dx . (4.61)
This indicates that a vortex distribution can be replaced by an equivalent doubletdistribution and vice verse.
4.12 Extension of the Circle Theorem
4.12.1 Constant Vorticity Extension
The circle theorem of (4.10) applies to irrotational motion. Now we extend thattheorem to the flow with constant vorticity.
155
Let the flow be described by the following stream function
ψo(z, z) = F (z) + F (z) +ωzz
4(4.62)
Let there be no rigid boundaries and let all the singularities of F (z) be at distancegreater then a from the origin. If a circular cylinder C, |z| = a is introduced intothe flow field than stream function of such flow looks
ψ(z, z) = F (z)− F(a2
z
)+ F (z)− F
(a2
z
)+ωzz
4(4.63)
Since on C, zz = a2 the stream function value on C is ψ(z, z) = ωa2/4 which isconstant so C is the streamline for the motion given by (4.63).Since all the singularities of F (z), and therefore also of F (z) are outside the circum-ference C, all the singularities of F (a2/z) and of F (a2/z) are inside C, so no newsingularities are introduced at infinity so motion given by (4.63) at infinity is thesame as that given by (4.62).The vorticity of the flow given by (4.63) is
4∂2ψ
∂z∂z= ω
Thus (4.63) satisfies all the conditions and is therefore the stream function of theperturbed motion.If instead of last term in (4.62) we write ωzz/4+Γ ln(zz)/4π we get for the perturbedflow
ψ(z, z) = F (z)− F(a2
z
)+ F (z)− F
(a2
z
)+ωzz
4+
Γ
4πln(zz)
This gives circulation Γ about C.
4.12.2 Singularities Distribution along Circle
We consider now velocities induced by source and vortex placed on a circle. In thefigure (4.26) shows singularities and some arbitrary point on the circle at which wecalculate induced velocity. Let r be the radius of the circle, d the distance betweensingularity and arbitrary point, and let θ be the angle between d and r as it is shownin the figure. Intensity of the velocity induced by the vortex is expressed as
v =Γ
2πd=
Γ
4πr cos θ
The tangential component to the circle is obtained
vτ = v cos θ =Γ
4πr cos θcos θ =
Γ
4πr= const
So vortex induces constant tangential component of velocity to the circle except atthe point where vortex is placed, since that point is singular.Intensity of velocity induced by source placed on the circle at arbitrary point on thecircle is
v =σ
2πd=
σ
4πr cos θ
156
Figure 4.26: Induced velocity from the singularity at the circle.
The normal component of the velocity to the circle is expressed as
vn = v cos θ =σ
4πr= const
So source induces constant normal component of velocity to the circle contour,except at the point where source is placed.We now express relations when source singularities and vorticity are distributedalong circle contour. Let γ(ϕ)r∆ϕ = ∆Γ where γ(ϕ) represents vorticity distribu-tion per unit angle. Also let σ(ϕ) represents distribution of source strength per unitangle. At any point ξ of the circle we can express the former components of velocityas
vτ (ξ) =γ(ξ)
2+
2π∫0
γ(ϕ)r dϕ
4πr
(4.64)
vn(ξ) =σ(ξ)
2+
2π∫0
σ(ϕ) dϕ
4π
From the mean value theorem we could write
γ
2=
2π∫0
γ(ϕ) dϕ
4π=γ
4π(2π − 0) and:
σ
2=
2π∫0
σ(ϕ) dϕ
4π(4.65)
From the definition of the mean value follows
2π∫0
σ(ϕ) dϕ =
2π∫0
(σ
2+σ(ϕ)
2
)dϕ ,
2π∫0
γ(ϕ) dϕ =
2π∫0
(γ
2+γ(ϕ)
2
)dϕ (4.66)
We now integrate equations (4.64), consider first of these equations
2π∫0
vτ (ξ) dξ =
2π∫0
γ(ξ)2
+
2π∫0
γ(ϕ) dϕ
4π
dξ157
=
2π∫0
(γ(ξ)
2+γ
2
)dξ
=
2π∫0
γ(ξ) dξ (4.67)
The same is true for source distribution
2π∫0
vn(ξ) dξ =
2π∫0
σ(ξ) dξ (4.68)
If we substitute results of equations (4.67) and (4.68) into equation (4.64) we get
vτ (ξ) =γ(ξ)
2+
1
4π
2π∫0
vτ (ϕ) dϕ
vn(ξ) =σ(ξ)
2+
1
4π
2π∫0
vn(ϕ) dϕ (4.69)
from which follows
γ(ξ) = 2vτ (ξ)− 1
2π
2π∫0
vτ (ϕ) dϕ , σ(ξ) = 2vn(ξ)− 1
2π
2π∫0
vn(ϕ) dϕ (4.70)
So if tangential or normal velocity is specified along circle contour we can calculatesingularity strength distribution from equation (4.70).We could prove that circulation around circle for the source singularity is equal tozero and that total flux across circle for the vortex singularity is also zero. Weonly mention that the outside singularity flow is potential so absence of circulationis natural consequence. The same is true for vortex singularity since we have notintroduced sources or sinks into flow field continuity equation must be satisfied andtotal outflow must be equal to zero.
Example 4.5:Find circumferential source distribution σ(θ) along circle to obtain potential non-liftingflow, figure (4.27).
Figure 4.27: Boundary condition for the flow about cylinder.
158
Solution 4.5:Required source distribution can be obtained from the equation (4.70). Uniform flowU∞ has the following component in the direction of unit normal n at the position θ:U∞ cos θ. Singularity distribution σ(θ) must induce the same velocity of opposite signto satisfy non-penetrating boundary condition, thus:
σ(θ) = −2U∞ cos θ − 1
2π
2π∫0
−U∞ cosϕdϕ
And finallyσ(θ) = −2U∞ cos θ
what is required source singularity distribution to satisfy required boundary condition.This value can be used to test panel codes.
159
Chapter 5
Subsonic Singularity Panels
Now we derive subsonic singularity elements for two-dimensional applications.
5.1 Two-Dimensional Point Singularity Elements
5.1.1 Source
We have already derived expressions for velocity potential, stream function, andvelocity components, for example (3.227) to (3.230), which we repeat here for con-venience, see also figure (5.9).
φ =σ
2πln√(x− xo)2 + (y − yo)2 = σ
2πln r
ψ =σ
2πarctan
y − yox− xo =
σ
2πθ
u =∂φ
∂x=σ
2π
x− xo(x− xo)2 + (y − yo)2 =
σ
2π
x− xor2
v =∂φ
∂y=σ
2π
y − yo(x− xo)2 + (y − yo)2 =
σ
2π
y − yor2
(5.1)
where r and θ are shown in the figure (5.1).
Figure 5.1: Point source.
In order to calculate φ, ψ, u, and v we need the following five variables (x, y), (xo, yo),and σ. It is easy to program expressions (5.1). One of possibilities is given in theexample which follows.
160
Example 5.1:Write the program by which it is possible for the source of strength σ, placed at (xo, yo)to calculate φ, ψ, u and v at the point (x, y).
Solution 5.1:We place all the details in one subroutine which is given below.
! File: PointSource2D.f90! 16-Dec-2000! Zlatko Petrovic!*****************************************************************!
Subroutine PointSource2D(x, y, sigma, xo, yo, phi, psi, u, v)! Calculates Velocity Potential, Stream Function! and velocity components u and v at arbitrary! point (x,y) due to point source placed at (xo,yo)! and strength sigma.!*********************************************
real x,y ! coordinates of the point in the flow-fieldreal xo,yo ! coordinates of the point where source is placedreal sigma ! strength of the sourcereal phi,psi ! Velocity potential and stream functionreal u,v ! x and y component of the induced velocityreal, parameter :: pi=3.14159265, TwoPi=2*pi
!rr = (x-xo)**2 + (y-yo)**2theta = atan2(y-yo,x-xo)if(theta < 0.0) theta = theta + TwoPi
!phi = sigma*log(sqrt(rr))/TwoPipsi = sigma*theta/TwoPiu = sigma/TwoPi * (x-xo)/rrv = sigma/TwoPi * (y-yo)/rr
!returnend subroutine PointSource2D
To test this subroutine we use program in which all input data are specified throughexpressions.
program test!! Specify input data!
x = 5.y = 3.xo = 5.yo = 1.sigma = 100.Call PointSource2D(x, y, sigma, xo, yo, phi, psi, u, v)
!! Show Result!
print 101, sigma, xo,yo, x,y
161
101 format(/’ sigma = ’,F8.3/’ xo = ’,F8.3/’ yo = ’,F8.3/’ x = ’,F8.3/ &’ y = ’,F8.3///)print 102, phi, psi, u, v
102 format(/’ phi = ’,F8.4/’ psi = ’,F8.4/’ u = ’,F8.4/’ v = ’,F8.4///)stop ’End of run !’end program test
5.1.2 Doublet
We give here expressions for velocity potential φ, stream function ψ, and inducedvelocities u and v, for general orientation of doublet strength µ = µxı+ µy, at thepoint (x, y) when the doublet is placed at the point (xo, yo).
φ(x, y) = −µx
2π
x− xo(x− xo)2 + (y − yo)2 −
µy
2π
y − yo(x− xo)2 + (y − yo)2
ψ(x, y) =µx
2π
y − yo(x− xo)2 + (y − yo)2 −
µy
2π
x− xo(x− xo)2 + (y − yo)2
u(x, y) =µx
2π
(x− xo)2 − (y − yo)2[(x− xo)2 + (y − yo)2]2 +
µy
π
(x− xo)(y − yo)[(x− xo)2 + (y − yo)2]2
v(x, y) =µx
π
(x− xo)(y − yo)[(x− xo)2 + (y − yo)2]2 −
µy
2π
(x− xo)2 − (y − yo)2[(x− xo)2 + (y − yo)2]2
(5.2)
Example 5.2:Write the subroutine by which it is possible to calculate at any point (x, y) velocitypotential φ, stream function ψ, and components of induced velocity u and v for thedoublet placed at point (xo, yo).
Solution 5.2:We have directly to apply the formulas (5.2), so required subroutine follows.
! File: PointDoublet2D.f90! 16-Dec-2000! Zlatko Petrovic!*****************************************************************!
subroutine PointDoublet2D(x,y, mux,muy, xo,yo, phi,psi, u,v)! Calcualtes Velocity Potential phi, Stream Function psi,! and velocity components u and v, at arbitrary! point (x,y) due to point doublet placed at (xo,yo)! and of the strength (mux,muy)!*********************************************
real x,y ! coordinates of the point in the flow-fieldreal xo,yo ! coordinates of the point where source is placedreal mux,muy ! strength of the doublet in x- and y-directionreal phi,psi ! Velocity potential and stream functionreal u,v ! x and y component of the induced velocityreal rr,rrrr,xx,yyreal, parameter :: Pi=3.14159265, TwoPi = 2*Pixx = x - xoyy = y - yo
162
rr = xx**2 + yy**2rrrr = rr*rrphi = -mux*xx/(TwoPi*rr) - muy*yy/(TwoPi*rr)psi = mux*yy/(TwoPi*rr) - muy*xx/(TwoPi*rr)u = mux*(xx**2 - yy**2)/(TwoPi*rrrr) + muy*xx*yy/(Pi*rrrr)v = mux*xx*yy/(Pi*rrrr) - muy*(xx**2 - yy**2)/(TwoPi*rrrr)returnend subroutine PointDoublet2D
We use the following program to test the former subroutine
program TestPointDoublet2Dreal mux, muy
! Specify initial data!
x = 4.0y = 3.0xo = 2.0yo = 0.0mux = 100.0muy = 250.0
! Calculate Flow-Field parameters!
Call PointDoublet2D(x,y, mux,muy, xo,yo, phi,psi, u,v)!! Show results of calculation!
print 10, x,y,xo,yo,mux,muy, phi,psi,u,v10 format(//’ (x,y) = ’,2F10.3/’ (xo,yo) = ’,2F10.3/’ mux,muy = ’,2F10.3/// &
’ phi = ’,F10.3/’ psi = ’,F10.3/’ u = ’,F10.3/’ v = ’,F10.3///)stop ’TestPointDoublet2D - E n d o f r u n !’end program TestPointDoublet2D
5.1.3 Vortex
We have already given the expressions for velocity potential φ, stream function ψ,and velocity components u and v at any arbitrary point (x, y) due to vortex placedat (xo, yo), (3.242) to (3.245). We repeat those equations for convenience.
φ = − Γ
2πarctan
y − yox− xo = − Γ
2πθ
ψ =Γ
2πln(√
(x− xo)2 + (y − yo)2)=
Γ
2πln r
u =Γ
2π
y − yo(x− xo)2 + (y − yo)2 =
Γ
2π
y − yor2
v = − Γ
2π
x− xo(x− xo)2 + (y − yo)2 = − Γ
2π
x− xor2
(5.3)
where r and θ have the same meaning as that defined in the figure (5.1).
Example 5.3:Write subroutine by which it is possible to calculate Velocity potential φ, stream function
163
ψ, and velocity components u and v at any arbitrary point (x, y) due to the vortex ofthe strength Γ placed at (xo, yo).
Solution 5.3:Our task is again simple, we have to program equations (5.3). Possible solution follow
! File: PointVortex2D.f90! 16-Dec-2000! Zlatko Petrovic!*****************************************************************!
subroutine PointVortex2D(x, y, Gamma, xo, yo, phi, psi, u, v)! Calculates Velocity Potential, Stream Function! and velocity components u and v at arbitrary! point (x,y) due to point VORTEX placed at (xo,yo)! and of strength Gamma.!*********************************************
real x,y ! coordinates of the point in the flow-fieldreal xo,yo ! coordinates of the point where source is placedreal Gamma ! strength of the vortexreal phi,psi ! Velocity potential and stream functionreal u,v ! x and y component of the induced velocityreal, parameter :: Pi=3.14159265, TwoPi=2*pireal xx,yy,rr,theta
!GammaOverTwoPi = Gamma/TwoPixx = x - xoyy = y - yorr = xx*xx + yy*yytheta = atan2(yy,xx)if(theta < 0.0) theta = theta + TwoPi
!phi = -GammaOverTwoPi * thetapsi = GammaOverTwoPi * log(sqrt(rr))u = GammaOverTwoPi * yy/rrv = -GammaOverTwoPi * xx/rrreturnend subroutine PointVortex2D
We test the former subroutine by the following program
program TestPointVortex2D!! Specify input data!
x = 3.0y = 5.0xo = 4.0yo = 2.0Gamma = 100.0
!Call PointVortex2D(x, y, Gamma, xo, yo, phi, psi, u, v)
!! Print calculated parameters!
print 1, x,y, Gamma, xo,yo, phi,psi, u,v
164
1 format(’ (x,y) = ’,2F10.3/’ (xo,yo) = ’,2F10.3/’ Gamma = ’,F10.3/// &’ phi = ’,F10.3/’ psi = ’,F10.3/’ u,v = ’,2F10.3///)
!stop ’TestPointVortex2D - E n d o f r u n !’end program TestPointVortex2D
5.2 Numerical Quadrature
Before we start with two-dimensional distributions we need to say a few words aboutnumerical quadrature. Here we explain the so called Gauss-Legendre Quadrature.Let us search approximate integral formula for the function f(x) in the form
b∫a
f(x) dx =n∑
i=0
wif(xi) (5.4)
where n+1 values wi are the weights to be given to the n+1 functional values f(xi).The xi are chosen so that the sum of the n + 1 appropriately weighted functionalvalues yields the integral exactly when f(x) is a polynomial of degree 2n+1 or less.We need some background material before we proceed further.
5.2.1 Orthogonal Polynomials
Two functions gn(x) and gm(x) selected from a family of related functions gk(x) aresaid to be orthogonal with respect to a weighted function w(x) on the interval [a, b] if
b∫a
w(x)gn(x)gm(x) dx = 0 , n = m,b∫
a
w(x)[gn(x)]2 dx = c(n) = 0 .
(5.5)
In general, c depends on n. If these relationships hold for all n, the family offunctions gk(x) constitutes a set of orthogonal functions. Some common familiesof orthogonal functions are the sets sin kx and cos kx. Orthogonality can beviewed as a generalization of the perpendicularity property of two vectors in ndimensional space where n becomes very large and the elements (coordinates) of thevectors can be represented as continuous functions of some independent variable.For our purpose, the definition (5.5) is adequate. The functions 1, x, x2, x3, . . . , xn
are not orthogonal. However, several families of well-known polynomials do possessa property of orthogonality. Four such sets are Legendre, Laguerre, Chebyshev, andHermite polynomials.
Legendre Polynomials
The Legendre polynomials Pn(x) are orthogonal on the interval [−1, 1] with respectot the weighting function w(x) = 1, that is,
165
+1∫−1Pn(x)Pm(x) dx = 0 , n = m,
+1∫−1
[Pn(x)]2 dx = c(n) = 0 .
(5.6)
The first few Legendre polynomials are:
P0(x) = 1 ,
P1(x) = x ,
P2(x) = (3x2 − 1)/2 ,
P3(x) = (5x3 − 3x)/2 ,
P4(x) = (35x4 − 30x2 + 3)/8 ,...
(5.7)
While the general recursion relation is
Pn(x) =2n− 1
nxPn−1(x)− n− 1
nPn−2(x) . (5.8)
Laguerre Polynomials
The Laguerre polynomials Ln(x) are orthogonal on the interval [0,∞] with respectto the weighting function e−x, that is,
∞∫0
e−xLn(x)Lm(x) dx = 0 , n = m,∞∫0
e−x[Ln]2 dx = c(n) = 0 .
(5.9)
The first few Laguerre polynomials are:
L0(x) = 1 ,
L1(x) = −x+ 1 ,
L2(x) = x2 − 4x+ 2 ,
L3(x) = −x3 + 9x2 − 18x+ 6 ,...
(5.10)
The general recursion relation is
Ln(x) = (2n− x− 1)Ln−1(x)− (n− 1)2Ln−2(x) . (5.11)
Chebyshev Polynomials
The Chebyshev polynomials Tn(x) are orthogonal on the interval [−1,+1] with re-spect ot the weighting function w(x) = 1/
√1− x2, that is,
166
+1∫−1
1√1− x2Tn(x)Tm(x) dx = 0 , n = m,
+1∫−1
1√1− x2 [Tn(x)]
2 dx = c(n) = 0 .
(5.12)
The first few polynomials are:
T0(x) = 1 ,
T1(x) = x ,
T2(x) = 2x2 − 1 ,
T3(x) = 4x3 − 3x ,
T4(x) = 8x4 − 8x2 + 1 ,...
(5.13)
The general recursion relation is
Tn(x) = 2xTn−1(x)− Tn−2(x) (5.14)
Hermite Polynomials
The Hermite polynomialsHn(x) are orthogonal on the interval [−∞,∞] with respectto the weighting function e−x2
, that is,
∞∫−∞
e−x2
Hn(x)Hm(x) dx = 0 , n = m,∞∫
−∞e−x2
[Hn]2 dx = c(n) = 0 .
(5.15)
The first few Hermite polynomials are:
H0(x) = 1 ,
H1(x) = 2x ,
H2(x) = 4x2 − 2 ,
H3(x) = 8x3 − 12x ,
H4(x) = 16x4 − 48x2 + 12 ,...
(5.16)
While the general recursion relation is
Hn(x) = 2xHn−1(x)− 2(n− 1)Hn−2(x) (5.17)
167
General Comments on Orthogonal Plynomials
The sequence of polynomials Pn(x), Ln(x), Tn(x), and Hn(x), respectivelysatisfying relationships (5.6), (5.9), (5.12), and (5.15) are unique. Each of the poly-nomials Pn(x), Ln(x), Tn(x), and Hn(x) is an nth-degree polynomial in x with realcoefficients and n distinct real roots interior to the appropriate interval of integra-tion; for example, all n roots of Pn(x) lie on the open interval (−1,+1).An arbitrary nth-degree polynomial pn(x) =
∑ni=0 αix
i may be represented by alinear function of any of the above families of orthogonal polynomials. Thus
pn(x) = β0Z0(x) + β1Z1(x) + · · ·+ βnZn(x) =n∑
i=0
βiZi(x) (5.18)
where Zi(x) is the ith-degree polynomial of one of the families of orthogonal poly-nomials.
Example 5.4:Expand the fourth-degree polynomial p4(x) = α0 + α1x+ α2x
2 + α3x3 + α4x
4 in termsof Legendre polynomials.
Solution 5.4:Let us substitute Pi(x) into (5.18), thus
p4(x) = β0 + β1x+ β23x2 − 1
2+ β3
5x2 − 3x
2+ β4
35x4 − 30x2 + 3
8
Collecting the coefficients of like powers of x,
p4(x) =
(β0 − β2
2+3β48
)+
(β1 − 3β3
2
)x+
(3β22− 15β4
4
)x2 +
5
2β3x
3 +35
8β4x
4
Thus
β4 =8
35α4
β3 =2
5α3
β2 =2
3
(α2 +
15
4β4
)=
2
3
(α2 +
6
7α4
)β1 = α1 +
3
2β3 = α1 +
3
5α3
β0 = α0 +1
2β2 − 3
8β4 = α0 +
1
3α2 +
1
5α4
It is now easy to show that polynomial p4(x) = x4+3x3− 2x2+2x− 1 is equivalent to
p4(x) = −22
15P0(x) +
19
5P1(x)− 16
21P2(x) +
6
5P3(x) +
8
35P4(x)
168
5.2.2 Gaussian Quadrature
We describe now procedure called Gauss-Legendre quadrature. We need to estimateintegral
b∫a
f(x) dx
by approximating function f(x) with an nth-degree interpolating polynomial pn(x),and integrate as follows:
b∫a
f(x) dx =
b∫a
pn(x) dx+
b∫a
Rn(x) dx (5.19)
Here, Rn(x) is the error term for the nth-degree interpolating polynomial. Since xiare not yet specified the Lagrangian form of the interpolating polynomial will beused, since Lagrangian polynomials permit arbitrary spacing between points
f(x) = pn(x) +Rn(x) =n∑
i=0
Li(x)f(xi) +f (n+1)(ξ)
(n+ 1)!
n∏i=0
(x− xi) (5.20)
where a < ξ < b and
Li(x) =n∏
i=0, j =i
(x− xjxi − xj
)
We could somewhat simplify development without removing any generality of theresult by changing interval of integration from [a, b] to [−1,+1]. Assume that allthe base points are in the interval of integration , that is, a ≤ x0, x, . . . , xn ≤ b. Letthe new variable, z, where −1 ≤ z ≤ +1, be defined by
z =2x− (a+ b)
b− a (5.21)
Define also a new function F (z) so that
F (z) = f(x) = f
((b− a)z + (a+ b)
2
)
then (5.20) becomes
F (z) =n∑
i=0
Li(z)F (zi) +F (n+1)(ξ)
(n+ 1)!
n∏i=0
(z − zi) (5.22)
where
Li(z) =n∏
i=0, j =i
(z − zjzi − zj
)and − 1 < ξ < +1 .
Here, zi is simply the base-point value xi transformed by (5.21). Now if f(x) isassumed to be a polynomial of degree 2n+1 then the term F (n+1)(ξ)/(n+1)! mustbe a polynomial of degree n, since
∑ni=0 Li(z)F (zi) is a polynomial of degree n at
most, and∏n
i=0(z − zi) is a polynomial of degree n+ 1. Let
f (n+1)(ξ)
(n+ 1)!= qn(z) ,
169
where qn(z) is a polynomial of degree n. Then
F (z) =n∑
i=0
Li(z)F (zi) +
[n∏
i=0
(z − zi)]qn(z) (5.23)
Now integration of both sides of this equation between integration limits -1 and +1gives
+1∫−1F (z) dz =
+1∫−1
n∑i=0
Li(z)F (zi) dz +
+1∫−1
[n∏
i=0
(z − zi)]qn(z) dz (5.24)
Since the F (zi) are fixed values, the summation operator can be taken outside theintegral sign. Dropping the right-most integral this equation becomes
+1∫−1F (z) dz =
n∑i=0
F (zi)
+1∫−1Li(z) dz =
n∑i=0
wiF (zi) , (5.25)
where
wi =
+1∫−1Li(z) dz =
+1∫−1
n∏i=0, j =i
(z − zjzi − zj
)dz (5.26)
Note that (5.25) is of the desired form, and that the second integral on the right-handside of (5.24),
ε =
+1∫−1
[n∏
i=0
(z − zi)]qn(z) dz (5.27)
is then the error term for the integration or quadrature formula of (5.25). The objectis to select zi in such a way that the error term (5.27) vanishes. The orthogonalityproperty of the Legendre polynomials, as already defined by (5.6) will be used toestablish such values zi.First we expand the two polynomials qn(z) and
∏ni=0(z−zi) in terms of the Legendre
polynomials is given as
n∏i=0
(z − zi) = b0P0(z) + b1P1(z) + · · ·+ bnPn(z) + bn+1Pn+1(z) =n+1∑i=0
biPi(z) (5.28)
and
qn(z) = c0P0(z) + c1P1(z) + · · ·+ cnPn(z) =n∑
i=0
ciPi(z) . (5.29)
The product qn(z)∏n
i=0(z − zi) is, from (5.28) and (5.29)
qn(z)n∏
i=0
(z − zi) =n∑
i=0
n∑j=0
bicjPi(z)Pj(z) + bn+1n∑
i=0
ciPi(z)Pn+1(z) (5.30)
The error term ε is from (5.27) given by
ε =
+1∫−1
n∑i=0
n∑j=0
bicjPi(z)Pj(z) + bn+1n∑
i=0
ciPi(z)Pn+1(z)
dz (5.31)
170
Because of the orthogonality properties of Legendre polynomials, all terms of thisintegral that are of the form
bicj
+1∫−1Pi(z)Pj(z) dz , i = j
will vanish. Thus the error term (5.27) for the quadrature formula (5.25) may bewritten
ε =
+1∫−1
[n∏
i=0
(z − zi)]qn(z) dz =
+1∫−1
n∑i=0
bici[Pi(z)]2 dz ++bn+1
+1∫−1
n∑i=0
ciPi(z)Pn+1(z) dz
=n∑
i=0
bici
+1∫−1
[Pi(z)]2 dz + bn+1
+1∫−1
n∑i=0
ciPi(z)Pn+1(z) dz .(5.32)
One way to make this expression vanish is to specify that the first n + 1 of the bi,i = 0, 1, 2, . . . , n are zero. The coefficient bn+1 of Pn+1(z) is still unspecified, butfrom (5.28) it must be given by
n∏i=0
(z − zi) = bn+1Pn+1(z) . (5.33)
The important feature of this formula is that it is already in factored form, thatis, it has n + 1 roots zi, i = 0, 1, . . . , n. Since bn+1Pn+1(z) is the same polynomial,the zi must be the roots of the bn+1Pn+1(z) as well, or equivalently Pn+1(z). Thusn + 1 base points to be used in the integration formula (5.25) are n + 1 roots ofthe appropriate (n+1)th-degree Legendre polynomial. The relative weight assignedto each function value F (zi) is given by (5.26). Table (5.1) gives the values of theappropriate base point coordinates and corresponding weight factors.Note also that from (5.21) dx = (b− a)dz/2, giving
b∫a
f(x) dx =b− a2
+1∫−1F (z) dz = (b− a)
n∑i=0
wiF (zi) (5.34)
Example 5.5:Use the two-point Gauss-Legendre quadrature formula to evaluate
+1∫−1F (z) dz =
+1∫−1
(z3 + z2 + z + 1) dz = 22
3
Solution 5.5:From table (5.1), the two-point formula is
+1∫−1F (z) dz =
1∑i=0
wiF (zi) = 1× F (0.57735 . . .) + 1× F (−0.57735 . . .)
171
Table 5.1: Roots and weight factors for Gauss-Legendre Quadrature.
zi
+1∫−1F (z) dz =
n∑i=0
wiF (zi) wi
Two-Point Formulan = 1
±0.57735 02691 89626 1.00000 00000 00000Three-Point Formula
n = 20.00000 00000 00000 0.88888 88888 88889±0.77459 66692 41483 0.55555 55555 55556
Four-Point Formulan = 3
±0.33998 10435 84856 0.65214 51548 62546±0.86113 63115 94053 0.34785 48451 37454
Five-Point Formulan = 4
0.00000 00000 00000 0.56888 88888 88889±0.53846 93101 05683 0.47862 86704 99366±0.90617 98459 38664 0.23692 68850 56189
Six-Point Formulan = 5
±0.23861 91860 83197 0.46791 39345 72691±0.66120 93864 66265 0.36076 15730 48139±0.93246 95142 03152 0.17132 44923 79170
Ten-Point Formulan = 9
±0.14887 43389 81631 0.29552 42247 14753±0.43339 53941 29247 0.26926 67193 09996±0.67940 95682 99024 0.21908 63625 15982±0.86506 33666 88985 0.14945 13491 50581±0.97390 65285 17172 0.06667 13443 08688
Fifteen-Point Formulan = 14
0.00000 00000 00000 0.20257 82419 25561±0.20119 40939 97435 0.19843 14853 27111±0.39415 13470 77563 0.18616 10001 15562±0.57097 21726 08539 0.16626 92058 16994±0.72441 77313 60170 0.13957 06779 26154±0.84820 65834 10427 0.10715 92204 67172±0.93727 33924 00706 0.07036 60474 88108±0.98799 25180 20485 0.03075 32419 96117
172
We have
F (−0.57735 . . .) = 0.566353297
F (0.57735 . . .) = 2.10313369 .
Since both weights are 1, we need to add these two values only, so that
+1∫−1
(z3 + z2 + z + 1) dz = 2.6666666
To the given number of figures, this result is exact (as would be expected), since thetwo-point formula (n = 1) is exact when F (z) is a polynomial of degree three (2n+ 1)or less.
5.3 Improper Integrals
We also need to talk about improper integral we met here and the way we solvethem. We use here development given in [1]. The term improper integral is usedto designate integrals having one or more infinite limits of integration or having afinit domain of integration with internal singularities. We are interested in improperintegrals of the second type.An integral of the type
b∫a
f(x) dx
(xo − x)αwhere f(x) is a bounded function in the interval and a ≤ xo ≤ b, converges only ifα < 1. This integral diverges if α ≥ 1, in which case the singularity is said to be ofthe “strong” type. For this kind of integrals Hadamard has established the conceptof “finite part” as an extension of Cauchy’s principal value of single-pole integrals.Mangler and Lighthill have given formulas for the computation of the finite partof some integrals that occur frequently in aerodynamics. For the case of poles ofinteger order, Mangler’s formula reads as follows:
FPb∫
a
f(x) dx
(xo − x)n+1 = limε→0
xo−ε∫a
f(x) dx
(xo − x)n+1 +b∫
xo+ε
f(x) dx
(xo − x)n+1
+ (−1)nn−1∑j=0
f (j)(xo)
j!
1− (−1)n−j
(n− j)εn−j
(5.35)
where FP means finite part of the integral, ε > 0, n is a non-negative integer,a < xo < b, and the superscript j of the function f denotes jth derivative of thisfunction with respect to x. Lighthill’s equivalent result for the case xo = 0 is givenby
FPb∫
a
f(x)
xn+1dx =
1
n!limε→0
−ε∫a
f (n)(x)
xdx+
b∫ε
f (n)(x)
xdx
−
n∑j=1
(j − 1)!
n!
[f (n−j)(b)
bj− f (n−j)(a)
aj
](5.36)
173
Mangler’s technique relies mainly on breaking up the integration interval into reg-ularly integrable regions and adding correction terms, Lighthill’s method is basedsolely on the technique of integration by parts. However, both formulas are simi-lar in nature, as each requires the evaluation of numerical limits. Brandao [1] hasdeveloped alternative approach which do not require evaluation of numerical limits.Let us consider the problem of evaluating the finite part of the following integral:
b∫a
f(x) dx
(xo − x)α+n+1
where a ≤ xo ≤ b, 0 < α < 1, and n ≥ −1. We will assume that the function f(x) isbounded in the interval a ≤ x ≤ b and has all necessary derivatives well defined andfinite at the singularity xo. With these hypotheses, we can use part of the Taylorseries expansion of f(x) about x = xo
f(x) =∞∑j=0
f (j)(xo)
j!(x− xo)j =
∞∑j=0
(−1)jj!
f (j)(xo)(xo − x)j
to write the following:
FPb∫
a
f(x) dx
(xo − x)α+n+1=
b∫a
f(x) + n+1∑j=0
(−1)j+1j!
f (j)(xo)(xo − x)j dx
(xo − x)α+n+1−
−n+1∑j=0
(−1)j+1j!
f (j)(xo)
FP b∫a
dx
(xo − x)α+n+1−j
(5.37)
For the case of integrals with poles of integer order (α = 0), we need one lessderivative and equation (5.37) reduces to
FPb∫
a
f(x) dx
(xo − x)n+1 =
b∫a
f(x) + n∑j=0
(−1)j+1j!
f (j)(xo)(xo − x)j dx
(xo − x)n+1 −
−n∑
j=0
(−1)j+1j!
f (j)(xo)
FP b∫a
dx
(xo − x)n+1−j
(5.38)
In these two equations, the finite part of a divergent integral has been transformedinto the sum of a regular integral and simpler finite-part integrals. We done thisby subtracting the terms in Taylor expansion from the function f(x) which are thesource of problems. But what is subtracted must be added to keep the integrationunchanged. These explicitely added terms are source of improperness, so we mustto apply special treatment. We will call these later integrals the finite-part integralsof elementary poles. The regular integral is designed to be evaluated numericallyas a piece. The terms added to and subtracted from the original problem are thoserequired to regularize the integrand at the singularity in the sense of L’Hopital.This approach has three advantages over previous methods. First, the regularizedintegral has a continuous integrand that can be handled by ordinary integrationschemes. Second, the finite part of the problem is reduced to analytical work, withimprovements in accuracy. Finally, no numerical limit is required. The answer isobtained directly, with considerable savings in computing time.
174
The Finite-Part Integral of Elementary Poles
Here we will derive formulas for the exact computation of the finite-part integral ofelementary poles. Let us start by considering the problem
F = FPb∫
a
dx
(xo − x)α+n+1
where a < xo < b. Following Hadamard, the answer is given by
F = limε→0
xo−ε∫a
dx
(xo − x)α+n+1+
b∫xo+ε
dx
(xo − x)α+n+1+G(ε)
(5.39)
where ε is a small positive real quantity and G(ε) is the function of ε canceling outexactly the terms resulting from the integrals that are singular in the limit.The first integral in equation (5.39) yields
xo−ε∫a
dx
(xo − x)α+n+1= − 1
α+ n
[1
(xo − a)α+n− 1
εn+α
]
In the interval (xo + ε) ≤ x ≤ b, the quantity (xo − x) is negative. So, we can write
(xo − x) = −(x− xo) = eıπ(x− xo)Using t as a dummy variable to denote (x − xo), the second integral in equation(5.39) can be worked out as follows:
b∫xo+ε
dx
(xo)α+n+1=
[eı(α+n+1)π
] b−xo∫ε
dt
tα+n+1= −cos(α+ n+ 1)π
α+ n
[1
(b− xo)α+n− 1
εα+n
]
where [·] denotes the real part of what comes between square brackets.Adding the previous results and choosing the function G(ε) to be equal to
G(ε) = −1 + cos(α+ n+ 1)π
(α+ n)εα+n
the limit in equation (5.39) becomes independent of ε and is given by
F = −(b− xo)α+n + (xo − a)α+n cos(α+ n+ 1)π
(α+ n) [(xo − a)(b− xo)]α+n (5.40)
For the particular case of poles of integer order, equation (5.40) reduces to
FPb∫
a
dx
(xo − x)n+1 = −(b− xo)n + (−1)n+1(xo − a)nn[(xo − a)(b− xo)]n (5.41)
This is not valid when n = 0. For this case, the result is given by
FPb∫
a
dx
(xo − x) = lnxo − ab− xo (5.42)
175
Now let us deal with the case where the singularity is placed either on the lower orupper limit of integration. in the former case, the answer comes from
FPb∫
a
dx
(a− x)α+n+1= lim
ε→0
b∫
a+ε
dx
(a− x)α+n+1+G(ε)
(5.43)
The integral here can be worked out as follows:
b∫a+ε
dx
(a− x)α+n+1=
[eı(α+n+1)π
] b−a∫ε
dt
tα+n+1
= −cos(α+ n+ 1)π
α+ n
[1
(b− a)α+n− 1
εα+n
]
where t is the dummy variable for the (x−a). Choosing the corresponding functionG(ε) as given by
G(ε) = −cos(α+ n+ 1)π
(α+ n)eα+n)
Equation (5.43) assumes the following explicit form:
FPb∫
a
dx
(a− x)α+n+1= − cos(α+ n+ 1)π
(α+ n)(b− a)α+n(5.44)
When α = 0, equation (5.44) reduces to
FPb∫
a
dx
(a− x)n+1 =(−1)n
n(b− a)n (5.45)
Again, this formula cannot be applied when n = 0. In this case, we have
FPb∫
a
dx
(a− x) = − ln(b− a) (5.46)
Similar development can be done with respect to singularities at the upper limit ofintegration.
5.4 Two-Dimensional Line Panels
5.4.1 Source panel
General Theory
We have already discussed constant strength source panel. Now we discuss problemmore general. Let again panel be placed along x-axis between points x1 and x2, asit is shown in the figure (5.2). Generally it could be assumed that source strengthper unit length is arbitrary function of position σ = f(x). In practice very restricted
176
number of functions f(x) are used. The most common choice is element of Taylorexpansion
σ(x) = Anxn dx (5.47)
The most interesting values for n are 0, 1, and 2. It is necessary to integratecontribution from whole line segment from x1 to x2 to velocity potential φ, streamfunction ψ, and velocity components u and v. Thus
φ =An
2π
x2∫x1
ξn ln√(x− ξ)2 + y2 dξ
ψ =An
2π
x2∫x1
ξn arctany
x− ξ dξ
u =An
2π
x2∫x1
ξn(x− ξ)(x− ξ)2 + y2 dξ
v =An
2π
x2∫x1
yξn
(x− ξ)2 + y2 dξ
(5.48)
where σ(ξ) = Anξn dξ. After integrating the first and second integral of the equa-
Figure 5.2: Line source panel along x-axis.
tions (5.48) by parts we have
φ =An
2π
ξn+1
n+ 1ln√(x− ξ)2 + y2
∣∣∣∣∣x2
x1
+An
4π(n+ 1)
x2∫x1
ξn+1(x− ξ)(x− ξ)2 + y2 dξ
ψ =Anξ
n+1
2π(n+ 1)arctan
y
x− ξ
∣∣∣∣∣x2
x1
− Any
2π(n+ 1)
x2∫x1
ξn+1
(x− ξ)2 + y2 dξ
u =An
2π
x2∫x1
ξn(x− ξ)(x− ξ)2 + y2 dξ =
Anx
2π
x2∫x1
ξn
(x− ξ)2 + y2 dξ −An
2π
x2∫x1
ξn+1
(x− ξ)2 + y2 dξ
v =Any
2π
x2∫x1
ξn
(x− ξ)2 + y2 dξ
(5.49)
177
After integration by parts all of the integrals are of the same type
Jm(x, y) =∫ ξm
ξ2 − 2xξ + x2 + y2dξ =
∫ ξm
aξ2 + bξ + cdξ
where a = 1, b = −2x, and c = x2+ y2. This integral can be integrated analytically,we use here expressions which can be found in any table of indefinite integrals.Before giving analytical expressions for required integrals let us replace characteristicintegrals in equations (5.49) by their symbolic names
φn =φn
An
=1
2π
ξn+1
n+ 1ln√(x− ξ)2 + y2
∣∣∣∣∣x2
x1
+1
4π(n+ 1)(xJn+1 − Jn+2)
∣∣∣∣∣x2
x1
ψn =ψn
An
=ξn+1
2π(n+ 1)arctan
y
x− ξ
∣∣∣∣∣x2
x1
− y
2π(n+ 1)Jn+1
∣∣∣∣∣x2
x1
un =unAn
=1
2π(xJn − Jn+1)
∣∣∣∣x2
x1
vn =vnAn
=1
2πyJn
∣∣∣∣x2
x1
(5.50)
Note also that4ac− b2 = 4x2 + 4y2 − 4x2 > 0
so ∫ dξ
aξ2 + bξ + c=
2√4ac− b2 arctan
2aξ + b√4ac− b2 (5.51)
We need expressions for two more integrals before we give recursive expression forgeneral case
∫ ξ dξ
aξ2 + bξ + c=
1
2aln(aξ2 + bξ + c
)− b
2a
∫ dξ
aξ2 + bξ + c∫ ξ2 dξ
aξ2 + bξ + c=
ξ
a− b
2a2ln(aξ2 + bξ + c
)+b2 − 2ac
2a2
∫ dξ
aξ2 + bξ + c
(5.52)
Now we are ready for recursive formula for general case (ξm)
∫ ξm dξ
aξ2 + bξ + c=
ξm−1
(m− 1)a− c
a
∫ ξm−2 dξaξ2 + bξ + c
− b
a
∫ ξm−1 dξaξ2 + bξ + c
(5.53)
In practice recursive programming is avoided, it is replaced by loops and use of stackstructure, but here we discuss about principles not actual design of programs so weuse recursive algorithm to calculate needed flow parameters.
Example 5.6:Write subroutine to calculate integral given by the expression (5.53)
Solution 5.6:Let us first define subroutines for integrals (5.53) when m = 0, 1, and m = 2. They aregiven below
178
! File: i0.f90! 18-Dec-2000! Zlatko Petrovic!*****************************************************************!function f0(a,b,c,x)
!! Finds integral:!! dx! int( --------------- )! a*x*x + b*x + c!
d = sqrt(4*a*c - b*b)f0 = 2/d * atan2(2*a*x + b,d)returnend function f0!******************************************************************!
function f1(a,b,c,x)!! Finds integral:!! x dx! int( --------------- )! a*x*x + b*x + c!!
f1 = ( log(a*x*x + b*x + c) - b*f0(a,b,c,x) )/(2*a)returnend function f1
!******************************************************************!
function f2(a,b,c,x)!! Finds integral:!! x*x dx! int( --------------- )! a*x*x + b*x + c!
f2 = x/a + ( (b*b-2*a*c)*f0(a,b,c,x) - b*log(a*x*x + b*x + c) )/(2*a*a)returnend function f2
This subroutines are tested by the following program.
program tst! Define input data!
x1 = 0.0x2 = 2.5
!x = 5.y = 1.
! a, b and c
179
a = 1.0b = -2*xc = x**2 + y**2
!! Calculate integrals!
print 101,x,y,x1,x2,0,f0(a,b,c,x2)-f0(a,b,c,x1)print 101,x,y,x1,x2,1,f1(a,b,c,x2)-f1(a,b,c,x1)print 101,x,y,x1,x2,2,f2(a,b,c,x2)-f2(a,b,c,x1)
101 format(’ x = ’,F10.3,5x,’ y = ’,F10.3/’ x1 = ’,F10.3,5x,’ x2 = ’,F10.3// &I3,’ int(x2-x1) = ’,F10.3///)stop ’tst - End of run !’end program tst
Example 5.7:Write subroutine which is capable to calculate integral (5.53)
Solution 5.7:Integral (5.53) is recursively defined so we use recursive feature of FORTRAN 90.
recursive function fr(a,b,c,x,m)!! Finds integral:!! x**m dx! int( --------------- )! a*x*x + b*x + c!
if(m<0) thenstop ’m < 0 !’
end ifselect case(m)
case(0)fr = f0(a,b,c,x)
case(1)fr = f1(a,b,c,x)
case(2)fr = f2(a,b,c,x)
case defaultfr = x**(m-1)/((m-1)*a) - ( b*fr(a,b,c,x,m-1) + c*fr(a,b,c,x,m-2) )/a
end selectreturnend function fr
Program to test former function is given below.
program testfr! Define arguments
x = 2.0y = 2.0
!x1 = 0.0x2 = 4.0
!
180
a = 1.0b = -2*xc = x*x + y*y
!m = 4print 101, x,y, x1,x2, m, fr(a,b,c,x2,m)-fr(a,b,c,x1,m)
101 format(’ x,y = ’,2F10.3/’ x1,x2 = ’,2F10.3/’ m = ’,I2/’ int = ’,F10.3//)stop ’testfr - End of run !’end program testfr
Figure 5.3: Linearly varying source strength panel along x-axis.
Example 5.8:Find expressions to calculate velocity potential φ, stream function ψ, and induced velocitycomponents u and v due to linearly varying strength of a source segment, placed on thex-axis, between points x1 and x2, figure (5.3).
Solution 5.8:We have to apply developed procedure for the case when strength intensity is varyingaccording to the formula
σ(ξ) =x2 − ξx2 − x1σ1 +
ξ − x1x2 − x1σ2
orσ(ξ) = (A′
0 + A′1ξ)σ1 + (A′′
0 + A′′1ξ)σ2
where
A′0 =
x2x2 − x1 A′
1 = − 1
x2 − x1A′′0 = − x1
x2 − x1 A′′2 =
1
x2 − x1So we need to consider function of the form A0 + A1ξ.
181
We consider, now, each parameter of the flow-field separately, first φ. From equation(5.48) we have
φ =1
2π
x2∫x1
(A0ξ
0 + A1ξ)ln√(x− ξ)2 + y2 dξ
=A0
2π
x2∫x1
ξ0 ln√(x− ξ)2 + y2 dξ + A1
2π
x2∫x1
ξ ln√(x− ξ)2 + y2 dξ (5.54)
After integration of this equation by parts or after applying equation (5.49) we get
φ =A0
2πξ ln
√(x− ξ)2 + y2
∣∣∣∣x2
x1
+A0
4π
x2∫x1
ξ(x− ξ)(x− ξ)2 + y2 dξ +
+A1
2π
ξ2
2ln√(x− ξ)2 + y2
∣∣∣∣∣x2
x1
+A1
4π
x2∫x1
ξ2(x− ξ)/2(x− ξ)2 + y2 dξ
or after rearrangement
φ =
(A0
2πξ +
A1
2π
ξ2
2
)ln√(x− ξ)2 + y2
∣∣∣∣∣x2
x1
+A0x
4π
x2∫x1
ξ dξ
(x− ξ)2 + y2 +
+(A1x
8π− A0
4π
) x2∫x1
ξ2 dξ
(x− ξ)2 + y2 −A1
8π
x2∫x1
ξ3 dξ
(x− ξ)2 + y2 (5.55)
Let us now give expression for the stream function ψ calculation. Again from equation(5.48) and (5.49) we get
ψ =A0ξ
2πarctan
y
x− ξ
∣∣∣∣∣x2
x1
− A0y
2π
x2∫x1
ξ dξ
(x− ξ)2 + y2 +
A1ξ2
4πarctan
y
x− ξ
∣∣∣∣∣x2
x1
− A1y
4π
x2∫x1
ξ2 dξ
(x− ξ)2 + y2 (5.56)
Expressions for velocity components follow directly from equation (5.48)
u =A0x
2π
x2∫x1
dξ
(x− ξ)2 + y2 +A1x− A0
2π
x2∫x1
ξ dξ
(x− ξ)2 + y2 −A1
2π
x2∫x1
ξ2 dξ
(x− ξ)2 + y2
v =A0y
2π
x2∫x1
dξ
(x− ξ)2 + y2 +A1y
2π
x2∫x1
ξ dξ
(x− ξ)2 + y2 (5.57)
We could obtain the same result more effectively by applying expressions given by equa-tion (5.50)
φ = (A′0φ0 + A
′1φ1)σ1 + (A′′
0φ0 + A′′1φ1)σ2
ψ = (A′0ψ0 + A
′1ψ1)σ1 + (A′′
0ψ0 + A′′1ψ1)σ2
u = (A′0u0 + A
′1u1)σ1 + (A′′
0u0 + A′′1u1)σ2 (5.58)
v = (A′0v0 + A
′1v1)σ1 + (A′′
0 v0 + A′′1 v1)σ2
182
where φi, ψi, ui, and vi are calculated according to expressions (5.50).
Considering previous example it is relatively easy to generalize expressions to higherorder polynomials.
Example 5.9:Write subroutine which is capable to calculate flow parameters φ, ψ, u, and v at anarbitrary point (x, y) from the source segment placed along x-axis between points x1and x2, if the source strength vary linearly from σ1 at x1 to σ2 at x2.
Solution 5.9:We have to program system (5.58) where
A′0 =
x2x2 − x1 A′
1 = − 1
x2 − x1A′′0 = − x1
x2 − x1 A′′2 =
1
x2 − x1and
a = 1, b = −2x, c = x2 + y2
Since we have not discussed procedure for calculating flow parameters when arbitrarypoint falls in the segment (x1, x2) we assume here that this point is outside that segment,i.e. when y = 0, x ∈ (x1, x2). The program which follow is far from an efficient way tosolve the problem but we believe that it is easy to understand it.
! File: SourceLin2D.f90! 02-Jan-2001! Zlatko Petrovic!*****************************************************************!
Subroutine SourceLin2D(x1,x2,sig1,sig2,x,y,phi,psi,u,v)! Subroutine calculates flow parameters phi,psi,! u, and v, at an arbitrary point (x,y), from! the source segment at x-axis between points x1! and x2. Source strength distribution varies! linearly from sig1 at x1 to sig2 at x2.
real x1,x2 ! The source segment is placed herereal sig1,sig2 ! The intensity of source distributionreal x,y ! Here we calculate flow parametersreal phi,psi,u,v ! These are output results
!! Here we assume that point (x,y) is outside the! segment (x1,x2) and y=0
A0P = x2/(x2-x1)A1P = -1/(x2-x1)A0S = -x1/(x2-x1)A1S = 1/(x2-x1)
!phi0b = phib(x,y,0,x2) - phib(x,y,0,x1)phi1b = phib(x,y,1,x2) - phib(x,y,1,x1)psi0b = psib(x,y,0,x2) - psib(x,y,0,x1)psi1b = psib(x,y,1,x2) - psib(x,y,1,x1)u0b = ubar(x,y,0,x2) - ubar(x,y,0,x1)u1b = ubar(x,y,1,x2) - ubar(x,y,1,x1)
183
v0b = vbar(x,y,0,x2) - vbar(x,y,0,x1)v1b = vbar(x,y,1,x2) - vbar(x,y,1,x1)phi = (A0P*phi0b + A1P*phi1b)*sig1 + (A0S*phi0b + A1S*phi1b)*sig2psi = (A0P*psi0b + A1P*psi1b)*sig1 + (A0S*psi0b + A1S*psi1b)*sig2u = (A0P*u0b + A1P*u1b)*sig1 + (A0S*u0b + A1S*u1b)*sig2v = (A0P*v0b + A1P*v1b)*sig1 + (A0S*v0b + A1S*v1b)*sig2returnend Subroutine SourceLin2D
! *****************************************************************function phib(x,y,n,z)real, parameter :: Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi
!np1 = n+1np2 = n+2a = 1.0b = -2*xc = x**2 + y**2
!phib = z**np1/(TwoPi*np1)*log(sqrt(a*z*z+b*z+c)) + &(x*fr(a,b,c,z,np1)-fr(a,b,c,z,np2))/(FourPi*np1)returnend function phib
! *****************************************************************function psib(x,y,n,z)real, parameter :: Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi
!np1 = n+1np2 = n+2a = 1.0b = -2*xc = x**2 + y**2
!theta = atan2(y,x-z)if(theta < 0) theta = theta + TwoPi
!psib = (z**np1*theta - y*fr(a,b,c,z,np1))/(TwoPi*np1)returnend function psib
! *****************************************************************function ubar(x,y,n,z)real, parameter :: Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi
!np1 = n+1a = 1.0b = -2*xc = x**2 + y**2
!ubar = (x*fr(a,b,c,z,n) - fr(a,b,c,z,np1))/TwoPireturnend function ubar
! *****************************************************************function vbar(x,y,n,z)real, parameter :: Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi
!a = 1.0b = -2*x
184
c = x**2 + y**2!
vbar = fr(a,b,c,z,n)/TwoPireturnend function vbar
Now it remains to say few words about the case when arbitrary point (x, y) lies atx-axis (y = 0), and x ∈ (x1, x2). In that case we have improper integral which is di-vergent. We must take finite part in order to obtain finite values of flow parameters.Our typical integral has now the form
Jm(x, 0) =∫ ξm
(x− ξ)2 dξ
We have integer power in denominator so equation (5.38) is appropriate for thiscase, and n = 1. Regularized integral of this equation is given below
JRm(x) =∫ ξm − xm +mxm−1(x− ξ)
(x− ξ)2 dξ (5.59)
It remains to add the finite-part integrals of elementary poles to complete the jobwhich are of the form
xmFPx2∫
x1
dξ
(x− ξ)2 −mxm−1
x2∫x1
dξ
x− ξSo finally we get
x2∫x1
ξm
(x− ξ)2 dξ =
x2∫x1
ξm + (m− 1)xm −mxm−1ξ(x− ξ)2 dξ
−xm x2 − x1(x− x1)(x2 − x) −mx
m−1 lnx− x1x2 − x (5.60)
The integral on the right-hand side is regularized, i.e. each of its components issingular but nominator of the fraction regularize this integral in L’Hopital sense.We, thus, must take this function without brakeing it into pieces what is possibleby using numerical quadrature.
Example 5.10:Regularize typical integral for the linear source distribution for m = 5 and show diagramsof regularized and non-regularized functions in the interval (0, 2) when x = 1.
Solution 5.10:Non-regularized function is defined as
f(ξ) =ξ5
(1− ξ)2while regularized function is obtained from (5.59)
fr(ξ) =ξ5 − 15 + 5 · 14(1− ξ)
(1− ξ)2 =ξ5 − 5ξ + 4
(1− ξ)2Diagrams of both regularized and non-regularized functions are shown in the figure (5.4)
185
0 0.5 1 1.5 2x
0
10
20
30
40
50
60
70
80
90
100
fr
0
10
20
30
40
50
60
70
80
90
100
f
Non-regularizedfunction
Regularizedfunction
Figure 5.4: Non-regularized and regularized graph of function f(ξ).
Example 5.11:Write program by which it is possible to calculate finite part of the characteristic integralas it is given by (5.60).
Solution 5.11:Input parameters for such program are x, m, x1, and x2. For regular part we have toapply Gauss-Legendre quadrature formula. In the case that m = 0 we have simple poleso no quadrature is required. The subroutine which calculates required integral follows
! File: RegularizedSource.f90! 04-Jan-2001! Zlatko Petrovic!*****************************************************************!
function RegularizedSource2D(x,m,x1,x2)real zi(6), wi(6)data zi /-0.2386192,0.2386192,-0.6612094,0.6612094,-0.9324695,0.9324695/data wi /2*0.4679139,2*0.3607616,2*0.1713245/
! Calculates integral int(x1,x2) z**m/(x-z)**2 dzF(z)=( z**m + (m-1)*x**m - m*x**(m-1)*z ) / (z-x)**2
!! Find the integral of regularized function using! Gauss-Legendre quadrature
if(m > 0) thenRegInt = 0.0do i = 1, 6
xi = ((x2-x1)*zi(i) +x1+x2)/2
186
RegInt = RegInt + wi(i)*F(xi)end doRegInt = (x2-x1)*RegInt/2 - x**m*(x2-x1)/((x-x1)*(x2-x)) &
-m*x**(m-1)*log((x-x1)/(x2-x))else if(m==0) then
RegInt = -x**m*(x2-x1)/((x-x1)*(x2-x))-m*x**(m-1)*log((x-x1)/(x2-x))else
stop ’RegularizedSource2D: m<0 !?’end ifRegularizedSource2D= RegIntreturnend function RegularizedSource2D
The former subroutine is tested by the following test program
! File: TestRegularizedSource.f90! 04-Jan-2001! Zlatko Petrovic!*****************************************************************!
program TestRegularizedSource2D! Define input values
x = 1.0x1 = 0.0x2 = 2.0print 101
101 format(’ Input m: ’,$)read *,mprint 102, x, m, x1, x2
102 format(’ x m x1 x2’/F8.3,I2,2F8.3/)print 103, RegularizedSource2D(x,m,x1,x2)
103 format(’ Finite part : ’,F10.3///)stop ’TestRegularizedSource -- End of run!’end program TestRegularizedSource2D
Now it is the time to calculate flow parameters for general case regardless whetheror not the point (x, y) lies in the interval (x1, x2).
Example 5.12:Correct the subroutine for calculating flow parameters from the source segment alongx-axis when the position of the point (x, y) is arbitrary.
Solution 5.12:We have to add the case when point (x, y) lies on the panel. Since in practical calcula-tions we need only velocities our subroutine will supply us only with u and v as answers.Also note that expression for v is not irregular since denominator and nominator tend tozero so we have to apply development from (4.52). Listing which follows is realizationof all so far mentioned
! File: SourceLin2Dfinal.f90! 04-Jan-2001! Zlatko Petrovic!*****************************************************************!
187
Subroutine SourceLin2D(x1,x2, sig1,sig2, x,y, u,v)real, parameter :: Pi=3.14159265, TwoPi=2*Pi
! Subroutine calculates flow parameters phi,psi,! u, and v, at an arbitrary point (x,y), from! the source segment at x-axis between points x1! and x2. Source strength distribution varies! linearly from sig1 at x1 to sig2 at x2.
real x1,x2 ! The source segment is placed herereal sig1,sig2 ! The intensity of source distributionreal x,y ! Here we calculate flow parametersreal u,v ! Induced velocities
!A0P = x2/(x2-x1)A1P = -1/(x2-x1)A0S = -x1/(x2-x1)A1S = 1/(x2-x1)
!! Check if point falls on the interval (x1,x2)
if(abs(y/(x2-x1)) < 1.e-6 .and. (x1-x)*(x2-x) < 0.0) then! Point is on the segment
u0b = (x*RegularizedSource2D(x,0,x1,x2)-RegularizedSource2D(x,1,x1,x2))/TwoPiu1b = (x*RegularizedSource2D(x,1,x1,x2)-RegularizedSource2D(x,2,x1,x2))/TwoPisig = sig1+(sig2-sig1)/(x2-x1)*(x-x1)u = (A0P*u0b + A1P*u1b)*sig1 + (A0S*u0b + A1S*u1b)*sig2v = sig/2return
elseu0b = ubar(x,y,0,x2) - ubar(x,y,0,x1)u1b = ubar(x,y,1,x2) - ubar(x,y,1,x1)v0b = vbar(x,y,0,x2) - vbar(x,y,0,x1)v1b = vbar(x,y,1,x2) - vbar(x,y,1,x1)u = (A0P*u0b + A1P*u1b)*sig1 + (A0S*u0b + A1S*u1b)*sig2v = (A0P*v0b + A1P*v1b)*sig1 + (A0S*v0b + A1S*v1b)*sig2
end ifreturnend Subroutine SourceLin2D
! *****************************************************************function ubar(x,y,n,z)real, parameter :: Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi
!np1 = n+1a = 1.0b = -2*xc = x**2 + y**2
!ubar = (x*fr(a,b,c,z,n) - fr(a,b,c,z,np1))/TwoPireturnend function ubar
! *****************************************************************function vbar(x,y,n,z)real, parameter :: Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi
!a = 1.0b = -2*xc = x**2 + y**2
!
188
vbar = fr(a,b,c,z,n)/TwoPireturnend function vbar
We use the following test program to test it.
program TestSourceLin2Dfinal! define input parameters
print 101101 format(’ (x,y) = ’,$)
read *,x,yx1 = 0.0x2 = 2.0print 102
102 format(’ sig1, sig2 : ’,$)read *, sig1, sig2Call SourceLin2D(x1,x2, sig1,sig2, x,y, u,v)print 103, x,y, x1,x2, sig1,sig2, u,v
103 format(’ (x,y) : ’,2F10.3/’ x1, x2 : ’,2F10.3/ &’ sig1, sig2 : ’,2F10.3///’ u,v : ’,2F10.3//)stop ’TestSourceLin2Dfinal -- End of run !’end program TestSourceLin2Dfinal
So far we have shown very general expressions for source distributions along strait-line segments. The question is do we really need so general expressions. For linearand constant strength distributions exist much simpler subroutines so we give themnow.
Constant-Strength Source Distribution
Consider a source distribution along the x-axis as shown in the figure (5.5). It isassumed that the source strength per length is constant such that σ(x) = σ = const.The influence of this distribution at a point P is an integral of the influences of thepoint elements along the segment x1 → x2.
φ =σ
2π
x2∫x1
ln√(x− ξ)2 + y2 dξ
ψ =σ
2π
x2∫x1
arctany
x− ξ dξ
u =σ
2π
x2∫x1
x− ξ(x− ξ)2 + y2 dξ
v =σ
2π
x2∫x1
y
(x− ξ)2 + y2 dξ
(5.61)
If we denote
θk = arctany
x− xk , k = 1, 2 (5.62)
rk =√(x− xk)2 + y2 , k = 1, 2 (5.63)
189
Figure 5.5: Constant-strength source panel along x-axis.
then integration of system (5.61) gives
φ =σ
4π
[(x− x1) ln r21 − (x− x2) ln r22 + 2y(θ2 − θ1)
]ψ =
σ
2π
[(x− x1)θ1 − (x− x2)θ2 + y
2lnr21r22
]
u =σ
4πlnr21r22
v =σ
2π(θ2 − θ1)
(5.64)
Of particular interest is the case when the point P is on the element (usually at thecenter). In this case y = ±0 and (θ1 = 0, θ2 = π), so flow parameters become
φ(x, 0±) =σ
4π
[(x− x1) ln(x− x1)2 − (x− x2) ln(x− x2)2
]ψ(x, 0±) = ∓ σ
2π(x− x2)
u(x, 0±) =σ
2πlnx− x1|x− x2|
v(x, 0±) = ±σ2
(5.65)
Coordinate y + 0 means that the point P lies infinitely close to the panel but fromthe upper side, and y = −0 that the point P is infinitely close to the panel from thelower side. As can be easily checked the velocity component u is zero at the panelcenter (x1 + x2)/2 and is infinite at the panel edges.
Linear-Strength Source Elements
The representation of a continuous singularity distribution by a series of constant-strength elements results in a discontinuity of the singularity strength at the paneledges. To overcome this problem, a linearly varying strength singularity elementcan be used. The requirement that the strength of the singularity remains the sameat the edge of two neighbor elements results in an additional equation. Therefore,with this type of element, for N collocation points 2N equations will be formed.Consider linear source distribution along the x-axis (x1 < x < x2) with sourcestrength of σ(x) = σ0 + σ1(x− x1), as shown in figure (5.6). Based on the principle
190
of superposition, this can be divided into a constant-strength element and a linearlyvarying strength element with the strength σ(x) = σ1x. Therefore, for the generalcase (as shown in the left-hand side of figure (5.6)) the results of this section mustbe added to the results of the constant-strength source element.
Figure 5.6: Decomposition of a linear strength element to constant-strength andlinearly varying strength element.
The influence of the simplified linear distribution source element, where σ(x) = σ1x,at point P is obtained by integrating the influences of the point elements betweenx1 → x2, figure (5.7).
φ =σ12π
x2∫x1
ξ ln√(x− ξ)2 + y2 dξ
ψ =σ12π
x2∫x1
ξ arctany
x− ξ dξ
u =σ12π
x2∫x1
ξ(x− ξ)(x− ξ)2 + y2 dξ
v =σ12π
x2∫x1
yξ
(x− ξ)2 + y2 dξ
(5.66)
Figure 5.7: Nomenclature for calculating influence of linearly varying strengthsource.
Without going into details of integration we give here final result
191
φ =σ14π
[x2 − x21 − y2
2ln r21 −
x2 − x22 − y22
ln r22 + 2xy(θ2 − θ1)− x(x2 − x1)]
ψ =σ12π
[xy
2lnr21r22
+y
2(x1 − x2) + x2 − x21 − y2
2θ1 − x2 − x22 − y2
2θ2
]
u =σ12π
[x
2lnr21r22
+ (x1 − x2) + y(θ2 − θ1)]
v =σ14π
[y ln
r22r21
+ 2x(θ2 − θ1)]
(5.67)
where r1, r2, θ1, and θ2 are defined by equations (5.62) and (5.63).When the point P lies on the element y = ±0 , x1 < x < x2, then equations (5.67)reduces to
φ =σ14π
[(x2 − x21) ln(x− x1)− (x2 − x22) ln |x− x2| − x(x2 − x1)
]ψ = ∓σ1
4(x2 − x22)
u =σ12π
[x ln
x− x1|x− x2| + (x1 − x2)
]
v = ±σ12x
(5.68)
5.4.2 Vortex Panels
Expressions needed to calculate flow parameters due to vortex panel, are very similarto those given for the source panels. We will not repeat general development fromprevious section, instead we give here expressions only for constant-strength paneland linearly varying strength panel, which are the most frequently used types ofpanels in applications.
Constant-Strength Vortex Panel
Once the influence terms of the constant-strength source element are obtained, owingto the similarity between the source and vortex distributions, the formulation for thispanel element becomes simple. The constant strength vortex distribution γ(x) =γ = const. is placed along the x-axis as shown in figure (5.8).The influence coefficients at a point P is an integral of the influences of the pointelements between x1 → x2.
192
φ = − γ
2π
x2∫x1
arctany
x− ξ dξ
ψ =γ
2π
x2∫x1
ln√(x− ξ)2 + y2 dξ
u =γ
2π
x2∫x1
y
(x− ξ)2 + y2 dξ
v = − γ
2π
x2∫x1
x− ξ(x− ξ)2 + y2 dξ
(5.69)
Note great similarity between this equations and equations (5.61), thus we can useexpressions for source panels directly. Let us aggain assume the same notation asthat given by (5.62) and (5.63), then
φ = − γ
2π
[(x− x1)θ1 − (x− x2)θ2 + y
2lnr21r22
]
ψ =γ
4π
[(x− x1) ln r21 − (x− x2) ln r22 + 2y(θ2 − θ1)
]u =
γ
2π(θ2 − θ1)
v =γ
4πlnr22r21
(5.70)
The influence of the element on itself at y = ±0 and (x1 < x < x2) can be found byapproaching from above the x-axis. In this case θ1 = 0, θ2 = π and
φ(x, 0±) = ∓ γ
2π[(x− x1)0− (x− x2)π] = ±γ
2(x− x2)
ψ(x, 0±) =γ
4π
[(x− x1) ln(x− x1)2 − (x− x2) ln(x− x2)2
]u(x, 0±) = ±γ
2
v(x, 0±) =γ
4πln(x− x2)2(x− x1)2
(5.71)
Figure 5.8: Constant-strength vortex panel along x-axis.
193
Linear Vortex Distribution
In this case the strength of vortex distribution varies linearly along the element
γ(x) = γ0 + γ1(x− x1)Again, for simplicity consider only the linear portion where γ(x) = γ1x and γ1 is aconstant. The influence of this vortex distribution at point P in the x–y plane isobtained by integrating the influences of the point elements between x1 → x2:
φ(x, y) = − γ12π
x2∫x1
ξ arctany
x− ξ dξ
ψ(x, y) =γ12π
x2∫x1
ξ ln√(x− ξ)2 + y2 dξ
u(x, y) =γ12π
x2∫x1
yξ
(x− ξ)2 + y2 dξ
v(x, y) = − γ12π
x2∫x1
ξ(x− ξ)(x− ξ)2 + y2 dξ
(5.72)
Since these expressions for flow parameters have corresponding expression in linearsource distributions we can simply transfer these results here, (5.66), to obtain
φ(x, y) = − γ12π
[xy
2lnr21r22
+ yx1 − x2
2+x2 − x21 − y2
2θ1 − x2 − x22 − y2
2θ2
]
ψ(x, y) =γ14π
[x2 − x21 − y2
2ln r21 −
x2 − x22 − y22
ln r22 + 2xy(θ2 − θ1)− x(x2 − x1)]
u(x, y) = − γ14π
[y ln
r21r22− 2x(θ2 − θ1)
]
v(x, y) = − γ12π
[x
2lnr21r22− (x2 − x1) + y(θ2 − θ1)
](5.73)
again rk and θk are defined by the equations (5.62) and (5.63).When the point P lies on the panel element (y = ±0; x1 < x < x2), then equation(5.73) reduces to
φ(x, 0±) = ∓γ14(x2 − x22)
ψ(x, 0±) =γ14π
[(x2 − x21) ln(x− x1)− (x2 − x2) ln |x− x2| − x(x2 − x1)
]u(x, 0±) = ±γ1
2x
v(x, 0±) = − γ12π
[x
2lnx− x1|x− x2| − (x2 − x1)
] (5.74)
as before + means above and − means below the panel, and ±0 means infinitelyclose to the panel, where signs means from which side is infinitely close.
194
5.4.3 Doublet Panels
Consider a doublet distribution along the x-axis consisting of elements pointing inthe y direction (µ = 0ı + µ), as shown in the figure (5.9). The influence of thepanel at the point P (x, y) is an integral of the influences of the point elements(5.2) distributed between x1 → x2. We use such orientation of doublets since Gausssecond formulae, section (3.15), states that orientation of doublets should be normalto the surface where they are distributed.
Figure 5.9: Doublet distribution along x-axis segment.
Constant-Strength Doublet Panels
If µ(x) = µ = const then from (5.2) follows
φ(x, y) = − µ
2π
x2∫x1
y
(x− ξ)2 + y2 dξ
ψ(x, y) = − µ
2π
x2∫x1
x− ξ(x− ξ)2 + y2 dξ
u(x, y) =µ
π
x2∫x1
y(x− ξ)[(x− ξ)2 + y2]2 dξ
v(x, y) = − µ
2π
x2∫x1
(x− ξ)2 − y2[(x− ξ)2 + y2]2 dξ
(5.75)
Note that the integral for the u component of the source distribution is similar tothe potential integral of the doublet distribution, and that the integral for the vcomponent of the source distribution is similar to the stream function integral ofthe doublet distribution. So we can use already derived expressions. Therefore, thepotential at point P (x, y) by using equation (5.61) for v is
φ(x, y) = − µ
2π(θ2 − θ1) (5.76)
where, again, angle θk is defined by equation (5.62). Comparing this expression tothe potential of a point vortex (5.3) indicates that this constant doublet distributionis equivalent to two point vortices with opposite sign at the panel edges such that
195
Figure 5.10: Equivalence between constant-strength doublet distribution and pairof the point vortex.
Γ = µ, see figure (5.10). Consequently velocity components are readily available byusing equation (5.3). So expressions for flow parameters follow
φ(x, y) = − µ
2π(θ2 − θ1)
ψ(x, y) = − µ
4πlnr21r22
u(x, y) =µ
2π
(y
r21− y
r22
)
v(x, y) = − µ
2π
(x− x1r21
− x− x2r22
)(5.77)
When the point P is on the panel (y = ±0, x1 < x < x2) then from (5.65) and (5.3)we have
φ(x, 0±) = ∓µ2
ψ(x, 0±) = − µ
2πlnx− x1|x− x2|
u(x, 0±) = ∓dµ(x)dx
= 0
v(x, 0±) = − µ
2π
(1
x− x1 −1
x− x2)
(5.78)
components of velocity v are singular at edges of the panel.
Figure 5.11: Linear varying strength doublet panel, Γ1 = −µ1x1, Γ2 = µ1x2.
196
Linear Doublet Distribution
Consider a doublet distribution along the x-axis with a strength µ(x) = µ0+µ1(x−x1), consisting of doublet elements oriented in the positive y-axis direction, µ =(0, µ), as shown in the figure (5.11). In this case, too, only the linear term µ(x) = µ1xis considered and the influence at the point P (x, y) is an integral of the influencesof the point doublet elements between x1 → x2.
φ(x, y) = −µ12π
x2∫x1
yξ
(x− ξ)2 + y2 dξ
ψ(x, y) = −µ12π
x2∫x1
(x− ξ)ξ(x− ξ)2 + y2 dξ
u(x, y) =µ1π
x2∫x1
yξ(x− ξ)[(x− ξ)2 + y2]2 dξ
v(x, y) = −µ1π
x2∫x1
[(x− ξ)2 − y2]ξ[(x− ξ)2 + y2]2 dξ
(5.79)
The integral for the velocity potential φ is similar to the v velocity component ofthe linear source, while the integral for the stream function ψ is similar to the ucomponent of the linear source, thus
φ(x, y) = −µ14π
[2x(θ2 − θ1) + y ln r
22
r21
]
ψ(x, y) = −µ12π
[x
2lnr21r22
+ (x1 − x2) + y(θ2 − θ1)] (5.80)
If we compare the expression for the velocity potential φ in this equation, andexpression for the velocity potential for the constant-strength vortex panel element,(5.70), we observe that these two potential can be easily related. Let us rearrangethe expression for velocity potential of the constant-strength vortex element
φV = − γ
2π
[(x− x1)θ1 − (x− x2)θ2 + y
2lnr21r22
]
=γ
4π
[2x(θ2 − θ1) + y ln r
21
r22
]︸ ︷︷ ︸Doublet φ when γ = −µ1
+γ
2π(x1θ1 − x2θ2)
Thus velocity potential of the linearly varying strength doublet panel, φD, is relatedto velocity potential of the constant-strength vortex panel, φV , through relationship
φD = φV +µ12π
(x1θ1 − x2θ2) (5.81)
the last two terms are velocity potentials of point vortices with strengths Γ1 =−µ1x1, and Γ2 = µ1x2. The velocity components therefore are readily available,either by differentiation of this velocity potential or by combining expressions for
197
the constant-strength vortex panel (5.69) and expressions for the induced velocitiesfor point vortices (5.3). Complete expressions for the velocity components are thus
u(x, y) = −µ12π
(θ2 − θ1) + µ1x22π
y
r22− µ1x1
2π
y
r21
v(x, y) = −µ14π
lnr22r21
+µ1x12π
x− x1r21
− µ1x2x− x2 r
22
(5.82)
The values of the potential, stream function, and the velocity components on thepanel itself (y = ±0, x1 < x < x2) are
φ(x, 0±) = ∓µ12x
ψ(x, 0±) = −µ12π
[x ln
x− x1|x− x2| + (x1 − x2)
]
u(x, 0±) = ∓µ12
v(x, 0±) = −µ14π
[ln(x− x2)2(x− x1)2 +
2x1x− x1 −
x2x− x2
](5.83)
Quadratic-Strength Doublet Distribution
Quadratic doublet distribution is similar to vortex linear distribution. However,in situations when the Dirichlet type boundary conditions is applied, it is moreconvenient to use the corresponding doublet distribution (instead of the linear vortexdistribution). We assume that doublet strength vary according to µ(x) = µ0+µ1(x−x1)
2 + µ2(x − x1)2 when x1 < x < x2. If we temporarily move coordinate origin tox1 and concentrate to the contribution of the quadratic term the flow parametersfrom that term are obtained from the following integrals:
φ(x, y) = −µ22π
x2∫x1
yξ2
(x− ξ)2 + y2 dξ
ψ(x, y) = −µ22π
x2∫x1
(x− ξ)ξ2(x− ξ)2 + y2 dξ
u(x, y) =µ2π
x2∫x1
y(x− ξ)ξ2[(x− ξ)2 + y2]2 dξ
v(x, y) = −µ22π
x2∫x1
[(x− ξ)2 + y2]ξ2]ξ2[(x− ξ)2 + y2]2 dξ
(5.84)
Let us integrate velocity potential integral. We introduce the new variable X = x−ξ(thus dX = −dξ), which transforms velocity potential integral of equation (5.84) tothe form
φ =µ22π
x−x2∫x−x1
y(x2 − 2xX +X2)
X2 + y2dX
198
=µ22π
yx2 x−x2∫x−x1
dX
X2 + y2− 2xy
x−x2∫x−x1
X
X2 + y2dX + y
x−x2∫x−x1
X2
X2 + y2dX
From the table of integrals, for example from [10], we have
x−x2∫x−x1
dX
X2 + y2=
1
y
(arctan
x− x2y
− arctanx− x1y
)x−x2∫
x−x1
X
X2 + y2dX =
1
2ln(x− x2)2 + y2(x− x1)2 + y2
x−x2∫x−x1
X2
X2 + y2dX = (x2 − x1) + y
(arctan
x− x1y
− arctanx− x2y
)
Figure 5.12: Quadratically varying strength doublet panel, Γ1 = µ2x21 and Γ2 =
−µ2x21.
It is clear from the figure (5.12) that
arctanx− xky
=π
2− θk , and r2k = (x− xk)2 + y2
thus
φ(x, y) =µ22π
[(x2 − y2)(θ1 − θ2)− xy ln r
22
r21+ y(x1 − x2)
]Expression for this potential is similar to expression for linearly varying vortex panel,equation (5.72). We could rearrange this equation such that
φD = φV +µ22π
(x21θ1 − x22θ2)where φV is expression for linearly varying vortex panel when µ2 = −γ1/2, andlast two terms are contributions from point vortices of intensity Γ1 = µ2x
21 and
Γ2 = −µ2x21 placed at the panel edges.Expressions for influence velocities could be obtained from expressions for linearlyvarying vortex panel and point vortices at the edges of the panel, thus
u =µ22π
[y ln
r21r22− 2x(θ2 − θ1) + yx22
r22− yx21r21
]
v =µ2π
[x
2lnr21r22
+ (x1 − x2) + y(θ2 − θ1) + x212· x− x1
r21− x22
2· x− x2
r22
]
199
It remains to calculate stream function ψ(x, y). Let us again introduce substitutionx− ξ = X ⇒ dξ = −dX, and when ξ = x1, X = x− x1; x = x2, X = x− x2.we can expand required integral
x2∫x1
(x− ξ)ξ2(x− ξ)2 + y2 dξ = −
x−x2∫x−x1
X(x−X)2
X2 + y2dX
into components:
−x−x2∫
x−x1
X(x−X)2
X2 + y2dX = −x2
x−x2∫x−x1
X dX
X2 + y2+ 2x
x−x2∫x−x1
X2 dX
X2 + y2−
x−x2∫x−x1
X3 dX
X2 + y2
The last integral on the right-hand side is addition to the list of already givenintegrals
x−x2∫x−x1
X3 dX
X2 + y2=x22 − x21
2− y2
2ln(x− x2)2 + y2(x− x1)2 + y2
what gives
ψ(x, y) = −µ22π
[2xy(θ2 − θ1) + y2 − x2
2lnr22r21
+ (x2 − x1)(2x− x1 + x2
2
)]
or all together:
φ(x, y) =µ22π
[(x2 − y2)(θ1 − θ2)− xy ln r
22
r21+ y(x1 − x2)
]
ψ(x, y) = −µ22π
[2xy(θ2 − θ1) + y2 − x2
2lnr22r21
+ (x2 − x1)(2x− x1 + x2
2
)]
u(x, y) =µ22π
[y ln
r21r22− 2x(θ2 − θ1) + yx22
r22− yx21r21
]
v(x, y) =µ2π
[x
2lnr21r22
+ (x1 − x2) + y(θ2 − θ1) + x212· x− x1
r21− x22
2· x− x2
r22
](5.85)
When the point P lies on the panel, y = ±0, x1 < x < x2 we have θ2− θ1 = π wheny = o+, and θ2 − θ1 = −π when y = 0−, thus
φ(x, 0±) = ∓µ2x2
2
ψ(x, 0±) = −µ22π
[2πxy + (x2 − x1)
(2x− x1 + x2
2
)+x2
2ln(x− x2)2(x− x2)2
]u(x, 0±) = ∓µ2xv(x, 0±) =
µ2π
[x
2ln(x− x1)2(x− x2)2 + x1 − x2 +
x212(x− x1) −
x222(x− x2)
](5.86)
Influence velocity v is singular at the edges of the panel.
200
Figure 5.13: Geometry approximated by the strait line segments.
5.4.4 Some General Remarks
So far we have developed our theory on the singularity distributions over panel of theform σ(x) = σ0 + σ1(x), or some other polynomial. Such representation is not themost convenient since we have additional unknowns. In the case of linear strengthdistribution we have two unknowns per each panel (σ0, and σ1). Let us clarify thisby the figure (5.13).In this figure we see three types of numbering. The first numbering belongs to panelnumbers, so each panel is identified by unique number. The second numbering isglobal panel-edge numbering where each edge point has unique number. This isnecessary because we associate unknown flow parameter values and coordinates ofthe panel with such numbering. And third numbering is local numbering where eachpanel has its edge number 1 and edge number 2. We have developed our expressionsin local coordinate system (when panel segment lies on the x-axis). Thus we musthave global to local and vice-verse coordinate transformations.Let us uniquely identify coefficients σ0 and σ1 to each panel by adding one additionalindex which indicate for which panel are these constants. For example σ0,i and σ1,iare constants of the i-th panel. Since there is no reason that intensity of singularityhas jumps in strength when passing from one panel to it neighboring, we shoulddemand that the values of singularity distribution at the common edge are the same.This makes additional as many equations as there are common edges (generally thatdoubles total number of equations). Singularity continuity condition can be writtenas
σ0,i−1 + σ1,i−1(x2,i−1 − x1,i−1) = σ0,iwhere xk,i means k-th edge of the i-th panel.We could avoid having additional unknowns if we uniquely associate unknown sin-gularity strengths to global panel edge numbering. Let us denote with σ(k) thevalue of the source singularity at the edge k. We could easily connect this unknownswith previous. At the local edge 1, x = x1,i of the panel i, the global value of thesingularity distribution at that edge is σ(i)
σ(i) = σ0,i + σ1,i(x1,i − x1,i) = σ0,iAt the local point 2 of the panel i we have global value σ(i+1) which is also equal to
σ(i+1) = σ0,i + σ1,i(x2,i − x1,i)
201
From this two equations we can express panel coefficients in terms of global nodevalues
σ0,i = σ(i), σ1,i =
σ(i+1) − σ(i)x2,i − x1,i (5.87)
So in all the developments we done so far we can quickly switch from panel coef-ficients to global node values by simply substituting σ0 ← σ(i) and σ1 ← (σ(i+1) −σ(i))/(x2,i−x1,i). With this transition from panel unknowns to global unknowns wereduce the size of our problem to half of its initial size in the case of linearly varyingsingularities.We now introduce procedure to combine constant strength and linearly varyingstrength panel elements. For constant strength, and for linearly varying sourcepanel element we have, (5.61) and (5.66)
φc =1
2π
x2∫x1
ln√(x− ξ)2 + y2 dξ (5.88)
φv =1
2π
x2∫x1
ξ ln√(x− ξ)2 + y2 dξ (5.89)
where φc is the velocity potential of the unit constant strength source element, andφv is the velocity potential of the linearly varying unit-strength source element. Butcomplete expression for the potential is of the form
φ =1
2π
x2∫x1
(σ0 + σ1(ξ − x1)
)ln√(x− ξ)2 + y2 dξ
orφ = (σ0 − σ1x1)φc + σ1φv (5.90)
the same rule applies to remaining flow parameters. We could now replace σ0 andσ1 by panel edge values σ(i) = σ0, and σ1 = (σ(i+1) − σ(i)/(x(i+1) − x(i)). Thus, forthe velocity u we have
u =
(σ(i) − σ(i+1) − σ(i)
x(i+1) − x(i))uc +
σ(i+1) − σ(i)x(i+1) − x(i) uv (5.91)
or
u =(1− uv − uc
x(i+1) − x(i))σ(i) +
uv − ucx(i+1) − x(i)σ
(i+1) = (1− α)σ(i) + ασ(i+1) (5.92)
where α = (uv − uc)(x(i+1) − x(i)). The similar expression should be applied toremaining flow parameters (φ, ψ, and v), and for vortex and doublet distributions.Even for higher oder singularity distributions, the number of unknowns could bereduced if we use instead of panel variables global node variables. We could in-crease the order of panel approximation either by increasing the order of singularitydistribution polynomial, or by refining geometry approximation, or both. In thefigure (5.57) actual geometry is approximated by strait line segments. If we wish toapproximate geometry with curved elements that generally highly complicate cal-culation of influence integral, and very often it is much easier and efficient to usenumerical quadrature instead of analytical.
202
Figure 5.14: Global and local coordinate system.
5.5 Coordinate Systems
We mentioned in previous section global and local node numbering. We could as-sociate to such numbering adequate coordinate system. Figure (5.14) shows pointP defined in two coordinate systems. Let xOy represents global coordinate system,and let x′O′y′ represents local coordinate system. Let us also define vectors R andRo in global coordinate system, and vector r in the local coordinate system. As itis clear from the figure we could write
R = Ro + r
orXı+ Y = Xoı+ Yo+ xı
′ + y ′ (5.93)
where ı and are unit vectors of the global coordinate system, and ı ′ and ′ areunit vectors of the local coordinate system. If we multiply the equation (5.93) by ıand respectively we obtain
X = Xo + xı′ ·ı+ y ′ ·ı
Y = Yo + xı′ · + y ′ ·
From the figure (5.14) it is clear that
ı ′ ·ı = cosϕ ′ ·ı = cos(ϕ+ π/2)ı ′ · = cos(π/2− ϕ) ′ · = cosϕ
or in matrix form X −Xo
Y − Yo
=
[ı ′ ·ı ′ ·ıı ′ · ′ ·
]xy
(5.94)
Multiplying equation (5.93) with ı ′ and ′ respectively we obtain x and yxy
=
[ı ′ ·ı ı ′ · ′ ·ı ′ ·
]X −Xo
Y − Yo
(5.95)
203
Figure 5.15: Find the coordinates of the point P (11, 8) in local coordinate systembonded to panel segment.
Example 5.13:Point P (11, 8) is given in global coordinate system xOy together with end-points of thepanel, as it is shown in the figure (5.15). Determine the coordinate of the point P inthe local coordinate system if the origin of such system is placed at the left edge of thepanel, and if the x′-axis pass through the second edge of the panel.
Solution 5.13:Let us first find the angle between x-axes of the global and local coordinate system.
tanϕ =Y2 − Y1X2 −X1
=3− 1
8− 2=
1
3
Using well known trigonometric identities
sinϕ =tanϕ√
1 + tan2 ϕ, cosϕ =
1√1 + tan2 ϕ
we have
sinϕ =1√10, cosϕ =
3√10
Now we could calculate local coordinates from (5.95)xPyP
=
[3/√10 1/
√10
−1/√10 3/√10
]XP −X1
YP − Y1
=
1√10
[3 1
−1 3
]11− 28− 1
After performing necessary calculations we getxPyP
=
34/
√10
12/√10
Example 5.14:Let the components of velocity (u, v)=(3, 2), at point P from previous example, are
204
given in local coordinate system. Find the velocity components (U, V ) in the globalcoordinate system!
Solution 5.14:The magnitude and orientation of velocity vector must be the same in both coordinatesystems, thus
Uı+ V = uı ′ + v ′
Multiplying this equation by ı and respectively we get the following transformationUV
=
[cosϕ − sinϕsinϕ cosϕ
]uv
(5.96)
what after substitution of numerical data givesUV
=
1√10
[3 −11 3
]32
=
7/√10
9/√10
Example 5.15:Find the velocity distribution and pressure coefficient CP at the cylinder surface placedin homogeneous flow parallel to x-axis. Let the undisturbed velocity be U∞ = 1m/sand let the radius of the cylinder be 1m, see figure (5.16). Assume that the fluid isincompressible and inviscid. Use constant-strength source panels to approximate cylindergeometry.
Figure 5.16: Flow about cylinder approximated by constant-strength source panels.
Solution 5.15:We have decided to use constant-strength source panels distributed as it is shown in thefigure (5.16). Let the number of panels be 2n than the angle ϕ = 2π/2n = π/n, andϕ/2 = π/(2n). We have to satisfy the following boundary condition
Vi · ni = 0, i = 1, 2, 3, . . . , 2n
which simply states that the normal component of the velocity to the i-th panel is equalto zero. With Vi we denote total velocity at control point i. In the middle of each panelis placed control point. Since the number of control points is equal to number of panelsand also equal to number of unknown source strengths, we have to satisfy this conditionat each control point. This will make us system of 2n equations for the 2n unknownsource panel strengths σi.
205
With (i) we denote edge points of the panels, thus panel i lies between points (i) and(i + 1). We assume that the first panel is at the rear side of cylinder and that thenumbering of panels and edge points is anti-clockwise. Let d denote distance betweencircle center and edge point of the panel. Since the radius of the cylinder is 1 we have
d cos(ϕ/2) = 1, d = 1/ cos(ϕ/2)
Angular position of edge point (i) is defined as
θ(i) = ϕ(i− 3
2
)Angular position of control point is also defined similarly
θi = ϕ(i− 1)
We could now calculate coordinates of the panel edge points, and coordinates of thecontrol points
x(i) = d cos θ(i), y(i) = d sin θ(i)
xi = d cos θi, yi = d sin θi
where (x(i), y(i)) denote coordinates of the i-th edge point, and (xi, yi) denote coordi-nates of the i-the control point. Knowing θi it is easy to define unit normal to the i-thpanel
ni = cos θiı+ sin θi
Let us now denote with wi,j influence velocity at the control point i from the sourcedistribution over the panel j. Since each panel contribute velocity to the control pointi we must add them all together
vi =2n∑j=1
wi,jσj, i = 1, 2, . . . , 2n
where vi is the influence velocity from all the panels. Total velocity Vi at control pointi is thus
Vi = U∞ı+2n∑j=1
wi,jσj, i = 1, 2, . . . , 2n
The boundary condition is now expressed as
Vi · ni =
U∞ı+2n∑j=1
wi,jσj
· (cos θiı+ sin θi) = 0
from which follows
2n∑j=1
(cos θiı+ sin θi) · wi,jσj = −U∞ cos θi, i = 1, 2, . . . , 2n
206
We can introduce a new coefficient Ci,j = (cos θiı + sin θi) · wi,j which representselement of aerodynamic influence matrix. Former system of equation can be writtenmore compacted
2n∑j=1
Ci,jσj = −U∞ cos θi, i = 1, 2, . . . , 2n
solving this system we obtain required source panel strengths (σj), than we can againdetermine total velocity at each control point, which must be tangential to the panelsince source strengths are determined to force normal component equal zero
Vi = U∞ı+2n∑j=1
wi,jσj, i = 1, 2, . . . , 2n
It remains only to calculate pressure coefficient. From Bernouli equation
1
2V 2
∞ + p∞ =1
2V 2
i + pi
we express pressure coefficient at control point i:
CPi=pi − p∞0, 5 V 2∞
= 1−(ViV∞
)2with this we conclude our solution algorithm it remains only to explain how to calculatevelocities wi,j.We have developed expressions to calculate induced velocities when the source paneloverlaps x-axis, (5.64), and when the point lies on the panel (5.65). We thus mustintroduce local coordinate system. Let the x-axis of such system starts from edge pointX(j), and is directed toward the pointX(j+1). then the angle betweenı andı ′ is obtainedfrom expression
cosı,ı ′ =ı ·ı ′ = X(j+1) −X(j)√(X(j+1) −X(j))2 + (Y(j+1) − Y(j))2
and
·ı ′ = Y(j+1) − Y(j)√(X(j+1) −X(j))2 + (Y(j+1) − Y(j))2
these values are obtained from the definition of dot product.Now we must uniquely define orientation of y′-axis. Let it be obtained from
′ = k×ı ′ = 1√(X(j+1) −X(j))2 + (Y(j+1) − Y(j))2
∣∣∣∣∣∣∣ı k0 0 1
X(j+1) −X(j) Y(j+1) − Y(j) 0
∣∣∣∣∣∣∣then
′ ·ı = − Y(j+1) − Y(j)√(X(j+1) −X(j))2 + (Y(j+1) − Y(j))2
207
and
′ · = X(j+1) −X(j)√(X(j+1) −X(j))2 + (Y(j+1) − Y(j))2
Coordinates of the control point (Xi, Yi) in the local coordinate system bounded to panelj are denoted by (xi, yi) and calculated from expression
xiyi
=
[ı ′ ·ı ı ′ · ′ ·ı ′ ·
]Xi −X(j)
Yi − Y(j)
Since our panel overlaps with x-axis of the local coordinate system, its end-point coor-
dinates are X1 = 0 and X2 =√(X(j+1) −X(j))2 + (Y(j+1) − Y(j))2. We obtain induced
velocities at control point (xi, yi) in local coordinate system when we apply (5.64)
ui,j =σj4π
lnr21r22
vi,j =σj2π
(θ2 − θ1)
where
θk = arctanyi
xi −Xk
, k = 1, 2
rk =√(xi −Xk)2 + y2i , k = 1, 2
We need one more transformation to obtain velocity components in global coordinatesystem
Ui,j
Vi,j
=
[ı ′ ·ı ′ ·ıı ′ · ′ ·
]ui,jvi,j
Finally, we have obtained influence of the panel j to the control point i. This procedurehave to be repeated for each control point and for each panel, thus 4n2 times. Theprogram which follows is the complete solution of this problem. Basic steps in calculationof the flow are separated in separate subroutine which names are self-explanatory.
! File: cyl.for! Z. Petrovic, 9-Feb-2001! Calulation of nonlifting flow about arbitrary bodies! ================================================================
program cylparameter (MM=50, MMM=MM*MM)real x(MM), y(MM) ! Panel edge coordinatesreal xc(MM), yc(MM) ! Control point coordinatesreal A(MMM), b(MM), Sig(MM) ! Aerodynamic influence mtrixreal nx(MM), ny(MM) ! Unit normals at control ptsreal uij(MMM), vij(MMM) ! Induced velocities due to
! unit strength source distr.equivalence (b(1), Sig(1))common /c1/ Acommon /c2/ uijcommon /c3/ vij
!Call Input_Data(N, x, y, alpha, VV)
208
Uinf = VV*cos(alpha)Vinf = VV*sin(alpha)
! Control point and unit normalCall Geometry(N, x, y, xc, yc, nx, ny)
! Calculate induced velocityCall Induced(N, x, y, xc, yc, uij, vij)
! Influence matrixCall Aero(N, uij, vij, nx, ny, Uinf, Vinf, N-1, A, b)
! Solve the equation systemCall SIMQ(A, Sig, N-1, Ierr)if(Ierr .ne. 0) then
stop ’*** Singular Influence Matrix *** !’end if
! Prepare results for plottingCall Output(xc, Sig, N-1)stop ’cyl - End of run !’end
! ****************************************************************subroutine Output(xc, Sig, NA)real xc(NA), Sig(NA)open(unit=1, file=’cyl.dat’,form=’formatted’,status=’unknown’)write(1,100) NAfi = 2*acos(-1.)/NA
100 format(’ variables = "x", "S", "Sa"’/’ zone i=’,I3)write(1,110)
& ((i-1)*fi*180/acos(-1.),Sig(i),-2*cos((i-1)*fi),i=1,NA)110 format(3f12.3)
close(unit=1)returnend
! ****************************************************************subroutine Aero(N, uij, vij, nx, ny, Uinf, Vinf, NA, A, b)real uij(N,N), vij(N,N), nx(N), ny(N), A(NA,NA), b(NA)do i = 1, NA ! For each control point
b(i) = -Uinf*nx(i) -Vinf*ny(i)do j = 1, NA ! For each panel
A(i,j) = uij(i,j)*nx(i) + vij(i,j)*ny(i)end do
end doreturnend
! ****************************************************************subroutine Induced(N, x, y, xc, yc, uij, vij)real x(N), y(N), xc(N), yc(N), uij(N,N), vij(N,N)do i = 1, N-1 ! For each control point
do j = 1, N-1 ! For each panelxf = x(j)yf = y(j)xs = x(j+1)ys = y(j+1)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.Call cs2d(x1, x2, xx, yy, uL, vL)
! Transform velocities to! global coordinate system
209
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,j) = u ! future referencevij(i,j) = v
end doend doreturnend
! ****************************************************************subroutine ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)
! Translates local velocity! components (uL,vL) to global
xx = xs - xfyy = ys - yfs = sqrt(xx**2 + yy**2)
! rotation cosinescost = xx/ssint = yy/s
! Global velocity componentsu = uL*cost - vL*sintv = uL*sint + vL*costreturnend
! ****************************************************************subroutine ToLocalCS(xf,yf, xs,ys, xc,yc, x1,x2, x,y)
! Transforms the panel and! contrlo point coordinates! from global to local CooSys
xx = xs - xfyy = ys - yfs = sqrt(xx**2 + yy**2)cost = xx/ssint = yy/s
! Panel coordinatesx1 = 0.0x2 = s
! Control point coordinatesx = cost*(xc-xf) + sint*(yc-yf)y = -sint*(xc-xf) + cost*(yc-yf)returnend
! ****************************************************************Subroutine Geometry(N, x, y, xc, yc, nx, ny)real x(N), y(N), xc(N), yc(N), nx(N), ny(N)do i = 1, N-1
xc(i) = (x(i) + x(i+1))/2yc(i) = (y(i) + y(i+1))/2tx = x(i+1) - x(i)ty = y(i+1) - y(i)t = sqrt(tx**2 + ty**2)nx(i) = -ty/tny(i) = tx/t
end doreturnend
! ****************************************************************
210
subroutine cs2d(x1, x2, x, y, u, v)parameter (Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi)
! Calculates induced velocities! u and v at the point (x,y)! due to constant unit-strength! source distribution along x-! axis between x1,x2.!
if(abs(y).lt.0.001 .and. (x-x1)*(x-x2).lt.0.0) then! x is between x1 and x2
u = log(abs((x-x1)/(x-x2)))/TwoPiv = -0.5
elseu = log(((x-x1)**2+y**2)/((x-x2)**2+y**2))/FourPiv = (atan2(y,x-x2)-atan2(y,x-x1))/TwoPi
end ifreturnend
! ****************************************************************subroutine Input_Data(N, x, y, alpha, VV)parameter (Pi=3.14159265)real x(*), y(*)
!NP = 48 ! Number of panels
!N = NP/2*2+1 ! Ensure odd number of pointsfih = Pi/(N-1) ! Notation example 5.15fi = 2*fihd = 1./cos(fih)
! Edge pointsdo i = 1, N
theta = fi*(i-1.5)x(i) = d*cos(theta)y(i) = d*sin(theta)
end do! Angle of attack and Vinf
VV = 1.0alpha = 0.0 ! Specify in degeesalpha = alpha*Pi/180. ! Convert to radiansreturnend
! ****************************************************************include ’simq.for’
In this program we used the following subroutine to solve system of equations
CC ..................................................................CC SUBROUTINE SIMQCC PURPOSEC OBTAIN SOLUTION OF A SET OF SIMULTANEOUS LINEAR EQUATIONS,C AX=BCC USAGE
211
C CALL SIMQ(A,B,N,KS)CC DESCRIPTION OF PARAMETERSC A - MATRIX OF COEFFICIENTS STORED COLUMNWISE. THESE AREC DESTROYED IN THE COMPUTATION. THE SIZE OF MATRIX A ISC N BY N.C B - VECTOR OF ORIGINAL CONSTANTS (LENGTH N). THESE AREC REPLACED BY FINAL SOLUTION VALUES, VECTOR X.C N - NUMBER OF EQUATIONS AND VARIABLES. N MUST BE .GT. ONE.C KS - OUTPUT DIGITC 0 FOR A NORMAL SOLUTIONC 1 FOR A SINGULAR SET OF EQUATIONSCC REMARKSC MATRIX A MUST BE GENERAL.C IF MATRIX IS SINGULAR , SOLUTION VALUES ARE MEANINGLESS.C AN ALTERNATIVE SOLUTION MAY BE OBTAINED BY USING MATRIXC INVERSION (MINV) AND MATRIX PRODUCT (GMPRD).CC SUBROUTINES AND FUNCTION SUBPROGRAMS REQUIREDC NONECC METHODC METHOD OF SOLUTION IS BY ELIMINATION USING LARGEST PIVOTALC DIVISOR. EACH STAGE OF ELIMINATION CONSISTS OF INTERCHANGINGC ROWS WHEN NECESSARY TO AVOID DIVISION BY ZERO OR SMALLC ELEMENTS.C THE FORWARD SOLUTION TO OBTAIN VARIABLE N IS DONE INC N STAGES. THE BACK SOLUTION FOR THE OTHER VARIABLES ISC CALCULATED BY SUCCESSIVE SUBSTITUTIONS. FINAL SOLUTIONC VALUES ARE DEVELOPED IN VECTOR B, WITH VARIABLE 1 IN B(1),C VARIABLE 2 IN B(2),........, VARIABLE N IN B(N).C IF NO PIVOT CAN BE FOUND EXCEEDING A TOLERANCE OF 0.0,C THE MATRIX IS CONSIDERED SINGULAR AND KS IS SET TO 1. THISC TOLERANCE CAN BE MODIFIED BY REPLACING THE FIRST STATEMENT.CC ..................................................................C
SUBROUTINE SIMQ(A,B,N,KS)DIMENSION A(1),B(1)
CC FORWARD SOLUTIONC
TOL=0.0KS=0JJ=-NDO 65 J=1,NJY=J+1JJ=JJ+N+1BIGA=0IT=JJ-JDO 30 I=J,N
CC SEARCH FOR MAXIMUM COEFFICIENT IN COLUMNC
IJ=IT+I
212
IF(ABS(BIGA)-ABS(A(IJ))) 20,30,3020 BIGA=A(IJ)
IMAX=I30 CONTINUE
CC TEST FOR PIVOT LESS THAN TOLERANCE (SINGULAR MATRIX)C
IF(ABS(BIGA)-TOL) 35,35,4035 KS=1
RETURNCC INTERCHANGE ROWS IF NECESSARYC
40 I1=J+N*(J-2)IT=IMAX-JDO 50 K=J,NI1=I1+NI2=I1+ITSAVE=A(I1)A(I1)=A(I2)A(I2)=SAVE
CC DIVIDE EQUATION BY LEADING COEFFICIENTC
50 A(I1)=A(I1)/BIGASAVE=B(IMAX)B(IMAX)=B(J)B(J)=SAVE/BIGA
CC ELIMINATE NEXT VARIABLEC
IF(J-N) 55,70,5555 IQS=N*(J-1)
DO 65 IX=JY,NIXJ=IQS+IXIT=J-IXDO 60 JX=JY,NIXJX=N*(JX-1)+IXJJX=IXJX+IT
60 A(IXJX)=A(IXJX)-(A(IXJ)*A(JJX))65 B(IX)=B(IX)-(B(J)*A(IXJ))
CC BACK SOLUTIONC
70 NY=N-1IT=N*NDO 80 J=1,NYIA=IT-JIB=N-JIC=NDO 80 K=1,JB(IB)=B(IB)-A(IA)*B(IC)IA=IA-N
80 IC=IC-1RETURNEND
213
5.6 Three-Dimensional Point Singularity Elements
We already have derived formula for this type of elements so we give here only finalexpressions for the sake of completeness. In three-dimensional space stream-functionis actually two of them. We will skip finding and writing such functions since so farthere were no practical use of them.
5.6.1 Point Source
Expressions for point source are given in (3.197)
φ = − σ
4π|r − ro|V =
σ
4π
r − ro|r − ro|3
(5.97)
or explicitly
φ(x, y, z) = − σ
4rπ
u(x, y, z) =∂φ
∂x=
σ
4π
x− xor3
v(x, y, z) =∂φ
∂y=
σ
4π
y − yor3
w(x, y, z) =∂φ
∂z=
σ
4π
z − zor3
(5.98)
where now r =√(x− xo)2 + (y − yo)2 + (z − zo)2, and (xo, yo, zo) are coordinates
of the point where source is placed. Source strength is denoted with σ if negativewe have sink.The velocity at any point P (x, y, z) induced by a point source located at x axis atthe distance ξ from the origin is given by
V (x, y, z) =σ
4π
1
(x− ξ)2 + y2 + z2 (5.99)
The velocity is directed along the line joining point source and the field point P .
5.6.2 Point Doublet
Expressions for point doublet are given by (3.208) and (3.221). Here we give vectorform notations
214
φ = − µ · r4π r3
V = − 1
4π∇[µ ·
(r
r3
)](5.100)
where r =√(x− xo)2 + (y − yo)2 + (z − zo)2, (xo, yo, zo) are coordinates of the point
where doublet is placed, µ is the intensity of the doublet oriented from the sink tothe source.
5.6.3 Point Vortex
There are no point vortex element in three-dimensional flow. Direct consequence ofKelvin and Helmholtz theorem is the spatial conservation of vorticity (3.7.5) whichstates that there could not be any free pieces of vorticity in the flow-field.
5.7 Three-Dimensional line Elements
We devote considerable amount of space to this type of elements since they arebuilding blocks for three-dimensional surface elements – key elements for real aero-dynamics applications.
5.7.1 3D Line Source Element
In practice this type of elements are the most frequently used either as strait lineelement or as a building block for generation of three-dimensional surface element.Figure (5.17) shows strait line source segment of the strength σ = const placed along
Figure 5.17: Calculation of induced velocities due to source-line along x-axis.
x-axis from the origin to the point L on the x-axis. The distance between point Pand origin is denoted by d1, and distance between point P and second edge of thesource segment (point L) is denoted by d2. Plane which pass through point P andx-axis is inclined to the horizontal xOy-plane by the angle θ. Running coordinate
215
ξ shows position along source segment of the source element. Angle between linewhich pass through this element and point P is denoted by ϕ. Distance betweenprojection of the point P to the plane yOz is denoted by r. Velocity vector V lies inthe plane build by x-axis and point P . Contribution to the total velocity from sourceelement at position ξ makes angle ϕ with x-axis. So induced velocity component inx direction is calculated by
u =
L∫0
V cosϕdξ =σ
4π
L∫0
(x− ξ) dξ[(x− ξ)2 + y2 + z2]3/2 =
σ
4π
(1
d2− 1
d1
)(5.101)
where cosϕ = (x− ξ)/d1, and
r =√y2 + z2 , d1 =
√x2 + y2 + z2 =
√x2 + r2, d2 =
√(x− L)2 + r2
The remaining two velocity components are calculated similarly
v =
L∫0
V sinϕ cos θ dξ =σy
4π
L∫0
dξ
[(x− ξ)2 + y2 + z2]3/2 =σy
4πr2
(x2d2− x1d1
)
w =
L∫0
V sinϕ sin θ dξ =σz
4π
L∫0
dξ
[(x− ξ)2 + y2 + z2]3/2 =σz
4πr2
(x2d2− x1d1
)(5.102)
where sinϕ = y/d1, sin θ = z/r, x1 = x, and x2 = x− L.
5.7.2 3D Line Vortex
We already have developed expressions for influences from elementary line vortexsegment, see figure (4.14), and equation (4.35), we repeat this expression for specialplacement of the vortex line segment shown in the figure (5.18)
V =Γ
4π
L∫0
sin ϕ dξ
(x− ξ)2 + y2 + z2 (5.103)
The velocity at the point P (x, y, z) induced by vortex line segment coincident withthe x-axis and having a length L is obtained by applying Biot-Savart’s law to eachinfinitesimal element of vortex line, and integrating contributions from all of them.The velocity vector is normal to the plane containing the x-axis and the pointP (x, y, z).It is clear from the figure (5.18) that r2 = y2 + z2 and
sin ϕ =r
d=
√√√√ y2 + z2
(x− ξ)2 + y2 + z2
Integration of (5.103) gives
V = − Γ
4π r
(x2d2− x1d1
)(5.104)
216
Figure 5.18: Calculation of induced velocities due to vortex-line along x-axis.
where again x1 = x, x2 = x− L, d2 =√(x− L)2 + r2, and d1 = sqrtx2 + r2.
Since there is no axial component of velocity induced by the vortex (u = 0), the vand w components are obtained by resolving V into its y and z components. Thus,
u = 0
v = −V sin θ =Γz
4πr2
(x2d2− x1d1
)w = V cos θ = − Γy
4πr2
(x2d2− x1d1
)(5.105)
where (x, y, z) are coordinates of the point P , Γ is the strength of the vortex line, andrest of the symbols have the same meaning as that in the equation (5.104). The threecomponents of velocity can be shown to satisfy Laplace’s equation. The elementaryline vortex solution is used as the basis of the more complex vortex distributionsfor surface three-dimensional panels. Care must be taken during these derivationsto ensure that all vortex line form closed rings and thus satisfy Helmholtz’s vortextheorem.But there are applications where vortex lines are used solely. For such applicationsit is better to use expressions which do not require coordinate transformations. Wealready gave this expression in vector form (4.37), and we repeat it for completeness
V = − Γ
4π
r1 × r2|r1 × r2|2
[ro ·
(r1|r1| −
r2|r2|
)]
where r1 and r2 are shown in the figure (4.14), and ro is the unit vector in thedirection of the vortex line.
Example 5.16:Write subroutine which will use former expression to calculate induced velocity at arbi-trary point P (x, y, z) due to vortex line segment between r1 and r2. The intensity ofthe vortex line is Γ.
Solution 5.16:Here is the possible solution of our problem
217
Subroutine LV3D(Gama, x,y,z, x1,y1,z1, x2,y2,z2, u,v,w)parameter (pi = 3.141592654, eps = 1.0e-10)real Gama ! Vortex filament strength per unit
! lengthreal x, y, z ! Control point coordinates for
! which influence velocities are calculatedreal x1, y1, z1 ! First point of the vortex filamentreal x2, y2, z2 ! Last point of the vortex filament
!real u, v, w ! Induced velocities at point P
! *************************************! Subroutine LV3D! calculates induced velocities (u,v,w)! at point (x,y,z) induced by vortex! filament of intensity Gama per unit! length for the segment, between points! (x1,y1,z1) and (x2,y2,z2)! *************************************!! Components R1 X R2!
R1R2X = (y-y1)*(z-z2) - (z-z1)*(y-y2)R1R2Y = -( (x-x1)*(z-z2) - (z-z1)*(x-x2) )R1R2Z = (x-x1)*(y-y2) - (y-y1)*(x-x2)
!! Calculate (R1 X R2)**2!
R1R2KV = R1R2X**2 + R1R2Y**2 + R1R2Z**2!! Calclate R1 i R2!
R1 = sqrt( (x-x1)**2 + (y-y1)**2 + (z-z1)**2 )R2 = sqrt( (x-x2)**2 + (y-y2)**2 + (z-z2)**2 )
!! Calculate R0*R1 i R0*R2!
R0R1 = (x2-x1)*(x-x1) + (y2-y1)*(y-y1) + (z2-z1)*(z-z1)R0R2 = (x2-x1)*(x-x2) + (y2-y1)*(y-y2) + (z2-z1)*(z-z2)
!! Calculate u, v, w!
if( R1.lt.eps .or. R2.lt.eps .or. R1R2KV.lt.eps) thenu = 0v = 0w = 0
elsegampi = Gama/(4*pi*R1R2KV)*(R0R1/R1 - R0R2/R2)u = gampi*R1R2Xv = gampi*R1R2Yw = gampi*R1R2Z
end if!
returnend
And here is the test program
218
Program TLV3D ! Z. Petrovic, 1-Feb-1993, LV3D.FOR! Program testira potprogram LV3D, koji! sracunava indukovane brzine (u,v,w) u! tacki (x,y,z) usled linijskog vrtloga! intenziteta Gama po jedinici duzine za! segment, koji se prostire od tacke! (x1,y1,z1) do tacke (x2,y2,z2)!
print 1001 ! Ucitavanje podatakaread *, Gamaprint 1002read *, x, y, zprint 1003read *, x1, y1, z1print 1004read *, x2, y2, z2
!Call LV3D(Gama, x,y,z, x1,y1,z1, x2,y2,z2, u,v,w)
!print 1005, Gama, x,y,z, x1,y1,z1, x2,y2,z2, u,v,w
!! I/O - formati!1001 format(2x,’Intenzitet vrtlozne linije Gama : ’,$)1002 format(2x,’Koordinata tacke P(x,y,z) : ’,$)1003 format(2x,’Koordinata tacke (x1,y1,z1) : ’,$)1004 format(2x,’Koordinata tacke (x2,y2,z2) : ’,$)1005 format(///5x,’Brzina indukovana linijskim vrtloznim segmentom’//
&5x,’Gama :’,F10.3/5x,’(x,y,z) :’,3F10.3/&5x,’(x1,y1,z1) :’,3F10.3/5x,’(x2,y2,z2) :’,3F10.3///&5x,’(u,v,w) :’,3F10.3//)end
!!
include ’lv3d.for’
Polygonal Line
Having developed expressions for strait line segment we can apply it to any shapeconsisted of strait line segments. Strait line vortex ring is interesting since it isequivalent to constant-strength doublet surface distribution. Figure (5.19) showsthe strait-line vortex ring defined by the quadrilateral corner coordinates. We applyhere, again, expression (4.37)
V1,2 = − Γ
4π
r1 × r2|r1 × r2|2
[r1,2 ·
(r1|r1| −
r2|r2|
)](5.106)
where ri represents positional vector of the point i, and ri,j = ri − rj. Velocityinduced by the strait line vortex ring is now simply
V = V1,2 + V2,3 + V3,4 + V4,1 (5.107)
where index i, j denotes contribution of corresponding strait line segment.
219
Figure 5.19: Quadrilateral vortex ring element.
We could generalize this result to any collection of strait line segments defined bycorner points. Let be defined n points by their coordinates (xi, yi, zi then inducedvelocity from the whole polygonal line is obtained by summing contribution fromeach segment:
V =n−1∑i=1
Vi,i+1 (5.108)
it is clear that for closed polygonal line must be satisfied (x1, y1, z1) ≡ (xn, yn, zn).
Example 5.17:Write subroutine by which it is possible to calculate induced velocity from polygonalstrait line segments.
Solution 5.17:Since we have developed subroutine for calculation of induced velocities from one straitline segment it is easy to ‘generalize’ this result. The subroutine which follows is possiblesolution to our problem.
Subroutine PolyVortex(Gama, x,y,z,N, xP,yP,zP, u,v,w)!
real Gama ! Intensity of the vortex line per! unit length
real x(N) ! x-coordinate of polygonal linereal y(N) ! y-coordinate of polygonal linereal z(N) ! z-coordinate of polygonal linereal xP, yP, zP ! Control point coordinates
!real u, v, w ! Induced velocities
!u = 0.0v = 0.0w = 0.0
! Call subroutine which calculates! contribution of one segment! required number of times
do i = 1, N-1Call LV3D( Gama,
& xP, yP, zP,& x(i), y(i), z(i),& x(i+1), y(i+1), z(i+1),& uu, vv, ww)
u = u + uu
220
v = v + vvw = w + ww
end do!
returnend
Horseshoe Vortex
The simplified case of strait line vortex ring is the horseshoe vortex. In this casethe vortex line is assumed to be placed in the xOy plane as shown in the figure(5.20). Two trailing vortex segments are placed parallel to the x-axis at y = yaand at y = yb, and the leading segment is placed parallel to the y-axis between thepoints (xa, ya) and (xa, yb). The induced velocity in the xOy plane will have onlya component in the negative z direction and can be computed by using equation(4.37) for a straight vortex segment:
w(x, y, 0) =−Γ4π
(cos β1 − cos β2) (5.109)
where the angles and their cosine are shown in the figure (5.20). For example,for the semi-infinite filament parallel to the x axis, and beginning at y = yb, thecorresponding angles are
cos β1 =x− xa√
(x− xa)2 + (y − yb)2, cos β2 = cos π = −1
Figure 5.20: Horseshoe vortex element.
For the vortex segment parallel to y-axis,
cos β1 =y − ya√
(x− xa)2 + (y − ya)2
cos β2 = − cos(π − β2) =y − yb√
(x− xa)2 + (y − yb)2
221
The down-wash due to the horseshoe vortex is now
w(x, y, 0) = − Γ
4π
1
x− xa
yb − y√(x− xa)2 + (y − yb)2
+yb − y√
(x− xa)2 + (y − yb)2
+
1
yb − y
1 + x− xa√(x− xa)2 + (y − yb)2
+
1
y − ya
1 + x− xa√(x− xa)2 + (y − ya)2
(5.110)
After some manipulations we get
w(x, y, 0) =−Γ
4π(y − ya)
1 +√(x− xa)2 + (y − ya)2
x− xa
+
Γ
4π(y − yb)
1 +√(x− xa)2 + (y − yb)2
x− xa
(5.111)
when x→ xa the limit of this equation gives
w(xa, y, 0) = − Γ
4π
(1
y − ya +1
y − yb
)(5.112)
where the finite-length segment does not induce down-wash on itself.The velocity potential of the horseshoe vortex may be obtained by reducing theresults of a constant strength doublet panel or by integrating of a point doubletelement. The potential of such point placed at (xo, yo, 0) and pointing in the zdirection, is
φ = − Γ
4π
z
r3
where r2 = (x− xo)2 + (y − yo)2 + z2. To obtain the potential due to the horseshoeelement at an arbitrary point P , this point doublet must be integrated over the areaenclosed by the horseshoe element:
φ = − Γ
4π
yb∫ya
dyo
∞∫xa
z dxo
[(x− xo)2 + (y − yo)2 + z2]3/2
This result is given by Moran [11], as
φ = − Γ
4π
yb∫ya
z(xo − x)dyo[(y − yo)2 + z2]
√(x− xo)2 + (y − yo)2 + z2
∣∣∣∣∣∣∞
xa
= − Γ
4π
yb∫ya
1 + x− xa√(x− xo)2 + (y − yo)2 + z2
z dyo(y − yo)2 + z2
= − Γ
4π
arctan yo − yz
+ arctan(yo − y)(x− xa)
z√(x− xa)2 + (y − yo)2 + z2
yb
ya
222
= − Γ
4π
arctan z
y − yb − arctanz
y − ya + arctan(yo − y)(x− xa)
z√(x− xa)2 + (y − yo)2 + z2
∣∣∣∣∣∣yb
ya
(5.113)
in practice we can use our polygonal line vortex subroutine to simulate horseshoevortex, instead of ∞ we could specify very large coordinate value so triangle willpractically give influence velocities similar as horseshoe if the third corner is movedto a great distance, figure (5.21).
Figure 5.21: Horseshoe vortex element approximated by triangle when one corneris far away.
Vortex Line of Arbitrary Shape
Rotational aerodynamical objects such as rotor blades or propellers require helicoidalvortex line shapes or even more general. We give here expressions for calculatinginduced velocities from such shapes. Biot-Savart law was derived in the section (3.17)and for arbitrary shaped vortex line specified parameterically equations (3.143) to(3.145) are repeated here for convenience
u(r) =Γ
4π
ξ2∫ξ1
[dy
dξ(z − z)− dz
dξ(y − y)
]dξ
3,
v(r) =Γ
4π
ξ2∫ξ1
[dz
dξ(x− x)− dx
dξ(z − z)
]dξ
3,
w(r) =Γ
4π
ξ2∫ξ1
[dx
dξ(y − y)− dy
dξ(x− x)
]dξ
3.
(5.114)
where ξ1 and ξ2 are starting and ending values of the parameter ξ, x(ξ), y(ξ) andz(ξ) are coordinates of vortex line element, and (x, y, z) are coordinates of pointwhere velocity vector is calculated. And is distance between point (x, y, z) andpoint (x, y, z)
=√(x− x)2 + (y − y)2 + (z − z)2
223
Strait-Line Segment Again
We illustrate procedure for calculating induced velocities by using equation (5.114)for strait line segment. Let strait line is parameterically be defined as:
x = xo + λxξ
y = yo + λyξ
z = zo + λzξ(5.115)
thendx
dξ= λx,
dy
dξ= λy,
dz
dξ= λz
After substitution of these expressions into equations (5.114) we get
u(r) =Γ
4π
ξ2∫ξ1
[λy(z − zo − λzξ)− λz(y − yo − λyξ)] dξ3,
v(r) =Γ
4π
ξ2∫ξ1
[λz(x− xo − λxξ)− λx(z − zo + λz)] dξ3,
w(r) =Γ
4π
ξ2∫ξ1
[λx(y − yo − λyξ)− λy(x− xo + λxξ)] dξ3.
(5.116)
where2 = (x− xo − λxξ)2 + (y − yo − λyξ)2 + (z − zo − λzξ)2
After squaring the terms in parenthesis we could write
2 = a+ bξ + cξ2
where
a = λ2x + λ2y + λ
2z
b = −2 [λx(x− xo) + λy(y − yo) + λz(z − zo)]c = (x− xo)2 + (y − yo)2 + (z − zo)2
substituting these expressions into (5.116) we get
u(x, y, z) =Γ
4π[λy(z − zo)− λz(y − yo)]
ξ2∫ξ1
dξ
3, (5.117)
v(x, y, z) =Γ
4π[λz(x− xo)− λx(z − zo)]
ξ2∫ξ1
dξ
3, (5.118)
w(x, y, z) =Γ
4π[λx(y − yo)− λy(x− xo)]
ξ2∫ξ1
dξ
3. (5.119)
224
So all three induced velocity components are dependent on the value of the sameintegral ∫ dξ
(aξ2 + bξ + c)3/2=
2(2aξ + b)
(4ac− b2)√aξ2 + bξ + cwhen we replace this integral into equations (5.117) to (5.119) we finally get
u =Γ
2π
λy(z − zo)− λz(y − yo)4ac− b2
2aξ2 + b√aξ22 + bξ2 + c
− 2aξ1 + b√aξ21 + bξ1 + c
v =
Γ
2π
λz(x− xo)− λx(z − zo)4ac− b2
2aξ2 + b√aξ22 + bξ2 + c
− 2aξ1 + b√aξ21 + bξ1 + c
w =
Γ
2π
λx(y − yo)− λy(x− xo)4ac− b2
2aξ2 + b√aξ22 + bξ2 + c
− 2aξ1 + b√aξ21 + bξ1 + c
(5.120)
Example 5.18:Write subroutine which calculates induced velocities at point P (x, y, z) from vortexline segment between points A(xA, yA, zA) and B(xB, yB, zB) using equation (5.120).Intensity of the vortex line is Γ.
Solution 5.18:We first need to specify equation of vortex line in parametric form, what is not reallycomplicated
x = xA + (xB − xA)ξ, y = yA + (yB − yA)ξ, z = zA + (zB − zA)ξ .so, when ξ = 0, (x, y, z) = (xA, yA, zA), and when ξ = 1, (x, y, z) = (xB, yB, zB).Directional gradients are simply
λx = xB − xA, λy = yB − yA, λz = zB − zA .One of the possible ways to make this subroutine is given below
! File: PointVortex2D.f90! 15-Jan-2001! Zlatko Petrovic!*****************************************************************!
subroutine VortexLine3D(Gamma, PA, PB, P, Vel)! Subroutine calculates induced velocities at point P! due to vortex line segment between points A and B! intensity of the vortex is Gamma.! Input arguments: Gamma, A, B, P! Output argument: Vel
real PA(3) ! (x,y,z) coordinate of the point Areal PB(3) ! (x,y,z) coordinate of the point Breal P(3) ! (x,yz,) coordinate of the point Preal Gamma ! Intensity of the vortex line segmentreal Vel(3) ! (u,v,w) components of the induced velocity
! ==============================================real lamx,lamy,lamz
225
real, parameter:: Pi=3.14159265! Define local function
F(xi) = (2*a*xi+b)/sqrt(a*xi**2+b*xi+c)! Make notation clear (This is not necessary)
xA = PA(1)yA = PA(2)zA = PA(3)
!xB = PB(1)yB = PB(2)zB = PB(3)
!x = P(1)y = P(2)z = P(3)
!lamx = xB - xAlamy = yB - yAlamz = zB - zA
!a = lamx**2 + lamy**2 + lamz**2b = -2*( lamx*(x-xA) + lamy*(y-yA) + lamz*(z-zA) )c = (x-xA)**2 + (y-yA)**2 + (z-zA)**2
!const = Gamma/(2*Pi) * (F(1.0) - F(0.0)) / (4*a*c-b**2)
!Vel(1) = const * ( lamy*(z-zA) - lamz*(y-yA) )Vel(2) = const * ( lamz*(x-xA) - lamx*(z-zA) )Vel(3) = const * ( lamx*(y-yA) - lamy*(x-xA) )
!returnend subroutine VortexLine3D
And here is also the test program
Program TestVortexLine3Dreal PA(3), PB(3), P(3), V(3)
! Specify input argumentsPA(1) = 0.0PA(2) = 0.0PA(3) = 0.0
!PB(1) = 5.0PB(2) = 4.0PB(3) = -3.0
!P(1) = 1.0 ; P(2) = 2.0; P(3) = 1.0
!Gamma = 1.0
!Call VortexLine3D(Gamma, PA, PB, P, V)print 101,PA,PB,P,Gamma,V
101 format(’ xA,yA,zA : ’,3F10.3/’ xB,yB,zB : ’,3F10.3/ &’ x, y, z : ’,3F10.3/’ Gamma : ’,F10.3///’ u, v, w : ’,3F10.3//)stop ’ TestVortexLine3D - End of run!’end program TestVortexLine3D
226
Vortex Ring
We now determine induced velocities due to vortex ring, figure (5.40). Let vortexring with radius R be placed at coordinate origin in the xOy-plane. Vector r pointsto an arbitrary point P (x, y, z), vector connects vortex element determined byangle ϕ with point P . Intensity of the vortex ring is Γ. Radial distance of the pointorthogonal projection of the P to the xOy plane is ro.We need to define vortex ring parameterically
x = R cos ϕ
y = R sin ϕ
z = 0 (5.121)
where role of parameter ξ is taken up by ϕ. Now
dx
dϕ= −R sin ϕ
dy
dϕ= R cos ϕ
dz
dϕ= 0
(5.122)
Figure 5.22: Vortex Ring notation.
We could now replace expressions (5.121) and (5.122) into general expression (5.114)to obtain
u(x, y, z) =Γ
4π
2π∫0
R cos ϕ(z − 0)− 0
3dϕ
v(x, y, z) =Γ
4π
2π∫0
0 +R sin ϕ(z − 0)
3dϕ
w(x, y, z) =Γ
4π
2π∫0
−R sin ϕ(y −R sin ϕ)−R cos ϕ(x−R cos ϕ)
3dϕ
(5.123)
where
2 = (x−R cos ϕ)2 + (y −R sin ϕ)2 + z2 = r2o +R2 + z2 − 2xR cos ϕ− 2yR sin ϕ
227
r2o = x2 + y2, cos θo =
x
ro, sin θo =
y
ro
then =
√r2o +R
2 + z2 − 2Rro cos(θo − ϕ) (5.124)
System of equations (5.123) now looks
u(x, y, z) =ΓRz
4π
2π∫0
cos ϕ
[r2o +R2 + z2 − 2Rro cos(θo − ϕ)]3/2 dϕ
v(x, y, z) =ΓRz
4π
2π∫0
sin ϕ
[r2o +R2 + z2 − 2Rro cos(θo − ϕ)]3/2 dϕ
w(x, y, z) = −ΓR
4π
2π∫0
R− ro cos(θo − ϕ)[r2o +R
2 + z2 − 2Rro cos(θo − ϕ)]3/2 dϕ
(5.125)
Note also that vr = u cos θo + v sin θo so we could write
vr(x, y, z) =ΓRz
4π
2π∫0
cos(θo − ϕ)[r2o +R
2 + z2 − 2Rro cos(θo − ϕ)]3/2 dϕ
w(x, y, z) = −ΓR
4π
2π∫0
R− ro cos(θo − ϕ)[r2o +R
2 + z2 − 2Rro cos(θo − ϕ)]3/2 dϕ
we could drop constant angle θo from this equations since there are no other functionsunder integral dependent on ϕ except cos, and since integration is performed aroundwhole ring, or we could choose some suitable value. Lets choose θo = π, thus
vr(x, y, z) = −ΓRz
4π
2π∫0
cos ϕ
[r2o +R2 + z2 + 2Rro cos ϕ]3/2
dϕ
w(x, y, z) = −ΓR
4π
2π∫0
R + ro cos ϕ
[r2o +R2 + z2 + 2Rro cos ϕ]3/2
dϕ (5.126)
substituting ϕ = 2φ we get
vr(x, y, z) = −ΓRz
2π
π∫0
cos 2φ
[(ro +R)2 + z2 − 4Rro sin2 φ]3/2
dφ
w(x, y, z) = −ΓR
2π
π∫0
R + ro cos 2φ
[(ro +R)2 + z2 − 4Rro sin2 φ]3/2
dφ
Another substitution of the form
k2 =4Rro
(ro +R)2 + z2
Puts the former expression into form
228
vr(x, y, z) = −ΓRz
2k3π
π∫0
cos 2φ
[1− k2 sin2 φ]3/2 dφ
w(x, y, z) = − ΓR
2k3π
π∫0
R + ro cos 2φ
[1− k2 sin2 φ]3/2 dφ(5.127)
5.8 Surface distributions
In the subsections (5.7.1) and (5.7.2) we have given expressions for three-dimensionalsource line and three-dimensional vortex line coincident with x-axis. Now we needplane coordinate transformation, similar to (5.5) in order to apply derived expres-sions for the general plane panel orientation.
5.8.1 Rotation of Coordinates
Figure (5.23) shows global coordinate system, and in plane rotated coordinate sys-tem. Let the angle between global y-axis and local x′-axis be Λ, and lets defineλ = tanΛ. According to (5.105) we have
Figure 5.23: Coordinate transformation.
x′
y′
=
[ı ′ ·ı ı ′ · ′ ·ı ′ ·
]xy
(5.128)
where
ı ′ ·ı = sin Λ, ı ′ · = cos Λ, ′ ·ı = − cos Λ, ′ · = sin Λ
Or equivalently
x′ = sin(Λ) x+ cos(Λ) y
y′ = − cos(Λ) x+ sin(Λ) y
It is easy to express former equations in terms of λ, so we have
229
x′ =λx+ y√1 + λ2
y′ =λy − x√1 + λ2
z′ = z
(5.129)
The distance d from the field point to the origin is unchanged in this transformation,but the perpendicular distance of the point from the line source or vortex is givenby
r′ =
√(x− λy)21 + λ2
+ z2 (5.130)
The velocity components calculated in the local coordinate system are transformedinto global by relation
u =λu′ − v′√1 + λ2
v =λv′ + u′√1 + λ2
w = w′(5.131)
5.8.2 Constant Source Distribution on Unswept Panel withStreamwise Taper
The velocity components induced at point P (x, y, z) by a constant source distribu-tion in the plane of the panel are derived by summing the influences of a series ofelementary line sources extending across the panel parallel to the leading edge. Thegeometry of the elementary line source located a distance ξ from the leading edgeand having a strength σ dξ is illustrated in the figure (5.24). In the derivation which
Figure 5.24: Constant Source Panel.
follows we assume that the panel lies in the xOy-plane. The distance of the point P
from the left end of the line source is d1 =√(x− ξ)2 + (y −m1ξ)2 + z2 and the dis-
tance from the right end of the line source is d2 =√(x− ξ)2 + (y − b−m2ξ)2 + z2.
230
The panel edge slopes m = dy/dx my be arbitrary. The velocity components areobtained by applying a 90 degree coordinate rotation to the line source velocityformulas given by equations (5.101) and (5.102). Integration across the panel chordgives total induced velocity components from whole panel
u = −v′ = − σ
4π
c∫0
(x− ξ) dξ(x− ξ)2 + z2
(y −m1ξ
d1− y − b−m2ξ
d2
)(5.132)
v = u′ =σ
4π
c∫0
(1
d1− 1
d2
)dξ (5.133)
w = w′ = −σz4π
c∫0
dξ
(x− ξ)2 + z2(y −m1ξ
d1− y − b−m2ξ
d2
)(5.134)
Each of these integrals we could split in two parts, for example for component u
u = − σ
4π
c∫0
y −m1ξ
d1
x− ξ(x− ξ)2 + z2 dξ +
σ
4π
c∫0
y − b−m2ξ
d2
x− ξ(x− ξ)2 + z2 dξ
the second integral is very similar to the first except that it has instead of constanty constant y− b. So we need to find integral only once, the value of second integralwill be obtained when y is replaced by y − b, and m1 by m2. It is also clear thatfirst integral traverse the side 1–2, and second traverse the side 3–4. The lower limitof the first integral correspond to “contribution” of the node 1, while upper limitcorrespond to contribution of the node 2. But we really do not need to evaluate thisupper limit.Induced velocities due to source panel do not depend on the coordinate systemchosen, so we can shift the origin of the coordinate system to the point 2. In thatcase contribution of point 2 will correspond to lower limit (i.e. 0). Expression forthe induced velocities will be the same except that instead old coordinate x we willhave the new which is shorter for the length of the chord (c). Thus, we could obtaincontribution of the node 2 from the expression obtained for node 1 if we replace xby x− c.
Figure 5.25: Semi-plane arithmetic.
Let us denote contributions from nodes by indices u1, u2, u3 and u4, so that index 1means contribution from node 1, index 2 means contribution from node 2 etc. Thentotal induced velocity is obtained by summing all contributions, as it is shown inthe figure (5.25)
231
u = u1 − u2 − u3 + u4v = v1 − v2 − v3 + v4w = w1 − w2 − w3 + w4
(5.135)
where
u1(x, y) =σ
4π
m1√1 +m2
1
sinh−1x+m1y√
(y −m1x)2 + (1 +m21)z
2− sinh−1
y√x2 + z2
v1(x, y) =
−14π√1 +m2
1
sinh−1x+m1y√
(y −m1x)2 + (1 +m21)z
2
w1(x, y) =1
4πarctan
z√x2 + y2 + z2
−x(y −m1x) +m1z2
(5.136)
then if we consider velocity components as functions u1(x, y,m), v1(x, y,m) andw1(x, y,m) we have
u2(x, y) = u1(x− c, y −m1c,m1)
v2(x, y) = v1(x− c, y −m1c,m1)
w2(x, y) = w1(x− c, y −m1c,m1)(5.137)
In the following relationships we assume instead of slope m1, slope m2
u3(x, y) = u1(x, y − b,m2),
v3(x, y) = v1(x, y − b,m2),
w3(x, y) = w1(x, y − b,m2),
u4(x, y) = u1(x− c, y − b−m2c,m2),
v4(x, y) = v1(x− c, y − b−m2c,m2),
w4(x, y) = w1(x− c, y − b−m2c,m2)
(5.138)
The velocity components given by equation (5.135) to (5.138) are expressed in termsof a local coordinate system which lies in the plane of the panel. Usually onemore coordinate transformation is necessary to connect velocities expressed in localcoordinate system to express them in the coordinate system of the vehicle. Let thepanel coordinate system be denoted by primes and is rotated through the angle δabout y-axis with respect to global coordinate system. The referenced coordinatesystem (global one) has its origin at the inboard corner of the panel leading edge,but the x-axis is parallel to the body reference axis, figure (5.26). Defining a = tan δ,we have the following coordinate transformations
x′
y′
z′
=
ı′ ·ı 0 ı ′ · k0 1 0
k′ ·ı 0 k
′ · k
xyz
(5.139)
232
Figure 5.26: Coordinate transformation by rotation about y-axis.
where
ı ′ ·ı = cos δ, ı ′ · k = sin δ, k′ ·ı = − sin δ, k
′ · k = cos δ.
Substituting instead sin δ and cos δ in terms of tan δ
sin δ =tan δ√
1 + tan2 δ=
a√1 + a2
, cos δ =1√
1 + tan2 δ=
1√1 + a2
.
into (5.139) we get
x′ =x+ az√1 + a2
y′ = y
z′ =z − ax√1 + a2
(5.140)
alsom′ =
m√1 + a2
(5.141)
Similarly, the velocity components become:
u =u′ − aw′√1 + a2
=σ
4π√1 + a2
(mG−H − aF )
v = v′ = −σG√1 + a2
4π
w =w′ + au′√1 + a2
=σ
4π√1 + a2
[F + a(mG−H)]
(5.142)
where
233
F = arctan(z − ax)√x2 + y2 + z2
−x(y −mx)− z(ay −mz)G =
1√1 + a2 +m2
sinh−1x+my + az√
(y −mx)2 + (ay −mz)2 + (z − ax)2
H = sinh−1y√
x2 + z2
(5.143)
5.8.3 Constant Source Distribution on Swept Panel withSpan-wise Taper
The velocity component induced at point P by constant source distribution in theplane of a swept panel are derived in a similar manner by summing the influences ofa series of elementary line sources extending across the panel parallel to the leadingedge. In this case, the line sources are swept back by the angle Λ. The geometry ofan elementary line source located a distance ξ from the leading edge , and havingstrength σdξ is shown in the figure (5.27).
Figure 5.27: Constant-source swept element panel.
Again we assume that panel lies in the xOy-plane. The distance of the point P
from the left end of the line source is d1 =√(x− ξ)2 + y2 + z2 and the distance
from the right end of the source line is d2 =√(x− ξ − λb)2 + (y − b)2 + z2, where
λ is the tangent of the leading edge sweep-back angle Λ, (λ = tanΛ). The velocitycomponents are obtained by rotating the coordinates of the line source velocityformulas through the angle Λ, and integrating across the panel chord as follows:
u =λu′ − v′√1 + λ2
=σ
4π√1 + λ2
c∫0
λ(1
d2− 1
d1
)+
+x− ξ − λy
r2
[λ(x− ξ) + y
d1− λ(x− ξ − λb) + y − b
d2
]dξ (5.144)
234
v =λv′ + u′√1 + λ2
=σ
4π√1 + λ2
c∫0
1
d2− 1
d1−
λ(x− ξ − λy)r2
[λ(x− ξ) + y
d1− λ(x− ξ − λb) + y − b
d2
]dξ (5.145)
w = w′
=σz√1 + λ2
4π
c∫0
[λ(x− ξ) + y
d1− λ(x− ξ − λb) + y − b
d2
]dξ
r2(5.146)
where r2 = (x− ξ − λy)2 + (1 + λ2)z2.In order to obtain the results in standard form the integrals are divided by
√1 + λ2
prior to their evaluation. As before, only those integrals associated with the in-board edge of the panel require evaluation, and then only at their lower limit. Theresulting velocity components correspond to the influence of the in-board corner ofthe panel leading edge. Denoting these results with subscript one,
u1 =−1
4π√1 + λ2
sinh−1λx+ y√
(x− λy)2 + (1 + λ2)z2
v1 =1
4π
λ√1 + λ2
sinh−1λx+ y√
(x− λy)2 + (1 + λ2)z2− sinh−1
x√y2 + z2
w1 =
1
4π
[arctan
z√x2 + y2 + z2
−xy + λ(y2 + z2) − arctanz
y
] (5.147)
The velocity components induced by the remaining three corners are obtained byapplying the above formulas with the origins shifted, and using the value of λ cor-responding to the leading or trailing edge. The influence of the complete panel isobtained by summing the influences of the four corner as indicated by equation sys-tem (5.135). Again if we assume expressions for velocity components as a functionof three arguments (x, y,Λ) then we have
u2(x, y) = u1(x− c, y,ΛT )
v2(x, y) = v1(x− c, y,ΛT )
w2(x, y) = w1(x− c, y,ΛT )
u3(x, y) = u1(x, y − b,Λ)v3(x, y) = v1(x, y − b,Λ)w3(x, y) = w1(x, y − b,Λ)u4(x, y) = u1(x− c− b tanΛT , y − b,ΛT )
v4(x, y) = v1(x− c− b tanΛT , y − b,ΛT )
w4(x, y) = w1(x− c− b tanΛT , y − b,ΛT )
(5.148)
235
5.8.4 Constant-Strength, Three-Dimensional QuadrilateralSource
The simplest and most basic general three-dimensional element will have quadri-lateral geometry and a constant-strength singularity distribution. The derivationis again performed in a local frame of reference, and for a global coordinate trans-formation is required. Consider a surface element with a constant-strength sourcedistribution σ per area bounded by four strait lines as described in figure (5.28). Theelement corner points are designated as (x1, y1, 0), . . . , (x4, y4, 0) and the potentialat an arbitrary point P (x, y, z), due to this element is
φ(x, y, z) = − σ
4π
∫S
dS√(x− ξ)2 + (y − η)2 + z2
(5.149)
and the velocity components can be obtained by differentiating the velocity poten-tial:
uvw
=
∂φ∂x∂φ∂y∂φ∂z
(5.150)
Execution of the integration within the area bounded by the four straight lines
Figure 5.28: Quadrilateral constant-strength source element.
requires considerable amount of time, so we use final expressions from [2]. Thuspotential for quadrilateral element is
φ(x, y, z) =−σ4π
[(x− x1)(y2 − y1)− (y − y1)(x2 − x1)
d12lnr1 + r2 + d12r1 + r2 − d12
+(x− x2)(y3 − y2)− (y − y2)(x3 − x2)
d23lnr2 + r3 + d23r2 + r3 − d23
+(x− x3)(y4 − y3)− (y − y3)(x4 − x3)
d34lnr3 + r4 + d34r3 + r4 − d34
+(x− x4)(y1 − y4)− (y − y4)(x1 − x4)
d41lnr4 + r3 + d41r4 + r3 − d41
]
236
+|z|[arctan
m12e1 − h1zr1
− arctanm12e2 − h2
zr2
+arctanm23e2 − h2
zr2− m23e3 − h3
zr3
+arctanm34e3 − h3
zr3− m34e4 − h4
zr4
+ arctanm41e2 − h4
zr4− m41e1 − h1
zr1
](5.151)
where
d12 =√(x2 − x1)2 + (y2 − y1)2 d23 =
√(x3 − x2)2 + (y3 − y2)2
d34 =√(x4 − x3)2 + (y4 − y3)2 d41 =
√(x4 − x1)2 + (y4 − y1)2 (5.152)
and
mi,j =yi − yjxi − xj , (i, j) = (1, 2), (2, 3), (3, 4), (4, 1) (5.153)
and
ri =√(x− xi) + (y − yi) + z2
ei = (x− xi)2 + z2hi = (x− xi)(y − yi)
i = 1, 2, 3, 4 (5.154)
The velocity components are given by
u =σ
4π
[y2 − y1d12
lnr1 + r2 − d12r1 + r2 + d12
+y3 − y2d23
lnr2 + r3 − d23r2 + r3 + d23
+y4 − y3d34
lnr3 + r4 − d34r3 + r4 + d34
+y4 − y1d41
lnr4 + r1 − d41r4 + r1 + d41
](5.155)
v =σ
4π
[x1 − x2d12
lnr1 + r2 − d12r1 + r2 + d12
+x2 − x3d23
lnr2 + r3 − d23r2 + r3 + d23
+x3 − x4d34
lnr3 + r4 − d34r3 + r4 + d34
+x1 − x4d41
lnr4 + r1 − d41r4 + r1 + d41
](5.156)
w =σ
4π
[arctan
m12e1 − h1zr1
− arctanm12e2 − h2
zr2
+arctanm23e2 − h2
zr2− arctan
m23e3 − h3zr3
+arctanm34e3 − h3
zr3− arctan
m34e4 − h4zr4
+ arctanm41e4 − h4
zr4− arctan
m41e1 − h1zr1
](5.157)
237
The u and v components of the velocity are defined everywhere, but at the edges ofthe quadrilateral they become infinite. In practice, usually the influence of the ele-ment on itself is sought, then near the centroid these velocity components approachzero. The jump in the normal velocity component as z → 0 inside the quadrilateralis
w(z = 0±) = ±σ2
(5.158)
When point P lies outside of the quadrilateral then
w(z = 0±) = 0 (5.159)
Far-Field Calculation
For improved computational efficiency, when the point of interest P lies far fromthe center of the element (ξ, η, 0) then the influence of the quadrilateral elementwith an area of A can be approximated by a point source. The term “far” iscontrolled by the programmer but usually if the distance between points is morethan 3–5 average panel diameters then the simplified approximation is used. Thepoint source influence for the velocity potential is
φ(x, y, z) =−σA
4π√(x− ξ)2 + (y − η)2 + z2
(5.160)
while the velocity components are
u(x, y, z) =σA(x− ξ)
4π[(x− ξ)2 + (y − η)2 + z2]3/2
v(x, y, z) =σA(y − η)
4π[(x− ξ)2 + (y − η)2 + z2]3/2
w(x, y, z) =σAz
4π[(x− ξ)2 + (y − η)2 + z2]3/2(5.161)
in previous expressions A is equal to the area of panel element, and σ is the sourcestrength per unit area.
5.8.5 Linearly Varying Source Distribution of Swept Panelwith Span-wise Taper
The velocity components induced by a source distribution having a linear variationin the chord-wise direction are derived in the same manner as described in thepreceding section for the constant source distribution. In this case, however, theexpressions under the integral signs in equations (5.144) to (5.146) are multipliedby ξ prior to integration. The velocity components induce d by the in-board cornerof the panel leading edge are given below:
u1 =−14π
x− λy√1 + λ2
sinh−1λx+ y√
(x− λy)2 + (1 + λ2)z2+ y sinh−1
x√y2 + z2
238
−z[arctan
z√x2 + y2 + z2
−xy + λ(y2 + z2) − arctanz
y
](5.162)
v1 =1
4π
(x− λy) λ√
1 + λ2sinh−1
λx+ y√(x− λy)2 + (1 + λ2)z2
− sinh−1x√
y2 + z2
+ x+
√x2 + y2 + z2 − λz
[arctan
z√x2 + y2 + z2
−xy + λ(y2 + z2) − arctanz
y
](5.163)
w1 =1
4π
(x− λy)
[arctan
z√x2 + y2 + z2
−xy + λ(y2 + z2) − arctanz
y
]
+ z
√1 + λ2 sinh−1 λx+ y√(x− λy)2 + (1 + λ2)z2
− λ sinh−1 x√y2 + z2
(5.164)
The velocity components induced by the remaining three corners are obtained byapplying the above formulas with the origin shifted, and using the appropriate valueof λ. The influence of the complete panel is obtained by summing the influences ofthe four corners as indicated by equations (5.148).
5.8.6 Constant Vortex Distribution on Swept Panel withSpan-wise Taper
The velocity components induced at a point P by a constant vortex distribution inthe plane of a swept panel are derived by summing the influences of elementary linevortices extending across the panel parallel to the leading edge, and concentratededge vortices extending back to infinity from the panel side edges. The geometryof an elementary line vortex located a distance ξ from the leading edge, and havingstrength Γ dξ, is illustrated on the figure (5.29).Let us consider first the influences of the bound vortex. The distance of the point
P (x, y, z) from the left end of the vortex is d1 =√(x− ξ)2 + y2 + z2, and the
distance from the right end of the vortex is d2 =√(x− ξ − λb)2 + (y − b)2 + z2,
where λ is the tangent of the leading edge sweep-back angle as before, (λ = tanΛ).The velocity components are obtained by rotating the coordinates of the line vortexvelocity formulas through the angle Λ, and integrating from the leading edge toinfinity as follows:
u =
∞∫0
λu′ − v′√1 + λ2
dξ =z
4π
∞∫0
K dξ
v = −λuw =
−14π
∞∫0
(x− ξ − λy)K dξ(5.165)
where
K =1
r2
[λ(x− ξ) + y
d1− λ(x− ξ − λb) + y − b
d2
]
239
Figure 5.29: Constant-vortex swept element panel.
andr2 = (x− ξ − λy)2 + (1 + λ2)z2
Only those integrals corresponding to the in-board edge of the panel require eval-uation, since the out-board edge can be obtained by a coordinate translation. Inthis case, however, both upper an lower limits of the integral must be evaluated toobtain the correct results. The resulting velocity components give the influences ofa semi-infinite region bounded by the leading edge and the x-axis, with origin at thein-board leading edge corner of the panel. They are identified by the subscript one.
u1 =1
4π
[arctan
z√x2 + y2 + z2
−xy + λ(y2 + z2) − arctanz
y
]v1 = −λu1w1 =
1
4π
√1 + λ2 sinh−1 λx+ y√(x− λy)2 + (1 + λ2)z2
− λ sinh−1 x√y2 + z2
− λ ln√y2 + z2
] (5.166)
It should be noted that the last term in the last equation of equations (5.166) isobtained by considering the influence of both in-board and out-board edges of thepanel simultaneously as the upper limit of the integral approaches infinity.The edge vortex contributes only to the v and w components of velocity. Thevelocity components are obtained by integrating equations (5.105) for a line vortex
240
of infinite length with respect to ξ, as follows:
∆v1 =z
4π
∫ ∞
0
1 + x− ξ√(x− ξ)2 + y2 + z2
=
z
4π
[x+
√x2 + y2 + z2
y2 + z2
](5.167)
∆w1 =−y4π
[x+
√x2 + y2 + z2
y2 + z2
](5.168)
Therefore, the u1 velocity component induced by the in-board leading edge cornerof the panel is given by the first of the equations (5.166), the remaining velocitycomponents are obtained by summing corresponding v component of the equation(5.166) and equation (5.167) and w component of equation (5.166) and equation(5.168). The velocity components induced by the remaining three corners are ob-tained by applying these equations with the origin shifted, and using the appropriatevalues for λ, (5.148). The influence of the complete panel is obtained by summingthe influences of the four corners as indicated by equations (5.135).
5.8.7 Linearly-Varying Vortex Distribution on Swept Panelwith Span-wise Taper
A vortex distribution is considered which has a linear variation in the chord-wisedirection, and lies within the triangular region bounded by the panel leading andtrailing edges extended to intersection, and the panel in-board edge. the velocitycomponents induced at a point P by this vortex distribution are derived in threesteps.
The First Step
In the first step, the velocities induced by a horse-shoe vortex of strength ξdξ havingits bound segment located along a radial line from the intersection of the leading andtrailing edges are evaluated and integrated across the panel chord. The geometryof the bound and trailing segments of the horse-shoe vortex are shown in the figure(5.30).The bound vortex is located a distance ξ from the panel origin. The point P is
located a distance d1 =√(x− ξ)2 + y2 + z2 from in-board end of the vortex, and
a distance d2 =√(x− ξ − λb2)2 + (y − b)2 + z2 from the out-board end. In this
derivation, the slope of the vortex is a linear function of ξ, λ = λ1 + aξ/c, wherea = λ2 − λ1, b = c/a, and λ1 = tanΛLE and λ2 = tanΛTE are the slopes ofthe leading- and trailing-edge of the panel respectively. The line vortex formulas arerotated through angle Λ, as before, to obtain expressions for the velocity componentsof the bound vortex prior to integration.Here are the velocity components in integral form
241
Figure 5.30: Linearly varying vorticity distribution.
u =z
4πc
c∫0
Kξ
r2dξ
v = −λuw −− 1
4πc
c∫0
(c−ξ−λy)Kξr2 dξ
(5.169)
where
λ = λ1 +a
cξ
K =λ(x− ξ) + y
d1− λ(x− ξ − λb) + y − b
d2r2 = (x− ξ − λy)2 + (1 + λ2)z2
These integrals are evaluated by making use of the following substitution in termsof the integration variable χ.
ξ = c(x− λ1y)(c− ay)− aλ1z2
(c− ay)2 + a2z2 − χ (5.170)
After a lengthy integration procedure, the velocity components induced by the in-board edge of the panel are obtained. In the following formulas, the parameter λ isredefined as the panel leading edge slope.
u =c
4πρ2
z
[ad
c+
(λ− 2as
ρ2
)G2
]
+z
ρ2
[(cλ− ax)as− (c− ay)e2
]G1
− 1
ρ2
[(cλ− ax)az2 + (c− ay)s
]F1
c
0
(5.171)
v = −(cλ− ax)(c− ay) uρ2− azt (5.172)
w = −(c− ay)t+ (cλ− ax)azuρ2
(5.173)
242
where
t =−14πρ2
d
[a(x+ ξ)
2c− (cλ− ax)(c− ay)
ρ2
]
+
[c(c− ay)(cλ− ax)s+ ae2z2
ρ4− y + ar2
2c
]G2
−e2
ρ4
[(c− ay)s+ (cλ− ax)az2
]G1
+zc
ρ4
[(c− ay)e2 − (cλ− ax)as
]F1
c
0
(5.174)
and
d =√(x− ξ)2 + r2
r2 = y2 + z2
ρ2 = (c− ay)2 + a2z2e2 = (cλ− ax)2 + ρ2s = (c− ay)(x− λy) + aλz2
F1 = arctanzd
(λ− aξ/c)r2 − y(x− ξ) (5.175)
G1 =1
esinh−1
(cλ− ax)(x− ξ) + y(c− ay)− az2c√[x− λy + (c− ay)ξ/c]2 + z2[1 + (λ− aξ/c]
(5.176)
G2 = sinh−1xξ
r(5.177)
Note that equations (5.175) to (5.177) differ from those given by (5.143) for thefunction F and G!The distribution of velocity corresponding to these functions can be determined byexamining the behaviour or th axial velocity u for z = 0. From equation (5.171),
u =−c(x− λy)4(c− ay)2
Along the panel leading edge, x = λy and therefore u = 0. Along the trailing edgex = c+ λ2y, therefore
u =−c
4(c− ay) (5.178)
Thus, the velocity distribution is seen to vary linearly chord-wise, and inversely asthe local chord span-wise.The contribution of the trailing vortex originating along the in-board edge of thepanel is considered next. This vortex contributes only v and w components ofvelocity, which are obtained by multiplying equations (5.105) for a line vortex ofinfinite length by ξ, and integrating. The results are as follows:
∆v =z
4πc
c∫0
1 + x− ξ√(x− ξ)2 + y2 + z2
ξ dξ
y2 + z2
243
=−z8πc
x− ξr2
[x− ξ +G2 +
√(x− ξ)2 + r2
]c
0
− z
4πr2
[x− c+
√(x− c)2 + r2
](5.179)
(5.180)
Similarly,
∆w =−y8πc
x− ξr2
[x− ξ +G2 +
√(x− ξ)2 + r2
]c
0
+y
4πr2
[x− c+
√(x− c)2 + r2
](5.181)
The first term in the braces gives the velocities induced by a pair of line vortices ofquadratic strength along the x-axis, and the last term gives the velocities inducedby a linearly varying vortex from the panel trailing edge. The combination gives thecontribution of a line vortex of quadratic strength to the trailing edge, followed bya constant vortex of strength c/2 extending downstream in the wake. A constantvortex of equal by opposite strength trails downstream from the out-board tip ofthe triangular panel.
The Second Step
In the second step, the velocities induced by a vortex distribution having a linearvariation in both chord-wise and span-wise directions is derived and subtractedfrom those given above to obtain the velocity components corresponding to a vortexdistribution having a linear variation chord-wise, but remaining constant span-wise.In this step, the bound vortex located along the radial line from the intersectionof the panel leading and trailing edges is given a linear variation in the span-wisedirection prior to performing the chord-wise integration. The linearly varying boundvortex is made up by superimposing a series of horseshoe vortices of strength ξ dξ dηwith in-board edge located at η, and out-board edge located at b. The geometry isillustrated in the figure (5.31)
Figure 5.31: Linearly varying vorticity in both span-wise and chord-wise direction.
244
The contribution of the bound segment of this elementary horseshoe vortex is ob-tained from the line vortex formulas, with origin shifted to the point (ξ + λη, η),and the coordinates rotated through the angle Λ. The point P is located a distance
d1 =√x− ξ − λη)2 + (y − η)2 + z2 from the in-board end of the bound vortex, and
d2 =√(x− ξλb)2 + z2 from the out-board end. The velocity components are given
first in integral form
u =z
4πc
c∫0
ξ dξ
r2
b∫0
K dη (5.182)
v = −λu (5.183)
w =−14πc
c∫0
ξ(x− ξ − λy)r2
dξ
b∫0
K dη (5.184)
where
λ = λ1 +a
cξ
a = λ2 − λ1K =
λ(x− ξ − λη) + y − ηd1
− λ(x− ξ − λb) + y − bd2
r2 = (x− ξ − λy)2 + (1 + λ2)z2
Only the first term in the K integral requires evaluation, as the second term cancelsin the superposition process. Integrating this with respect to η,
I =
b∫0
λ(x− ξ) + y − (1 + λ2)η
d1dη = d− d2 (5.185)
where d =√(x− ξ)2 + y2 + z2, and d2 is previously defined. The integral (5.182)
to (5.184) may now be written
u =z
4πc
c∫0
(d− d2)ξ dξr2
(5.186)
v = −λu (5.187)
w =−14πc
c∫0
d− d2)(x− ξ − λy)ξr2
dξ (5.188)
These integrals are evaluated, using the substitution given by equation (5.170). Thevelocity components induced by the in-board edge of the panel are given below,where λ is redefined as the panel leading edge slope.
u =−c4πρ2
z
[x− 2(cλ− ax) ( y(c− ay)− az2)
ρ2
]G2 − zd
− zρ2
[(cλ− ax)cs− e2
(y(c− ay)− az2
)]G1
245
+1
ρ2
[(cλ− ax)cz2 + s
(y(c− ay)− az2
)]F1
c
0
(5.189)
v = −(cλ− ax)(c− ay) uρ2− azt (5.190)
w = −(c− ay)t+ (cλ− ax)azuρ2
(5.191)
where
t =c
4πρ2
dc(cλ− ax)
(y(c− ay)− az2
)ρ2
− x+ ξ
2
+
[−1ρ4
[(cλ− ax)
(y(c− ay)− az2
)s+ ce2z2
]+r2
2c
]G2
+e2
ρ2
[(y(c− ay)− az2
)s+ c(cλ− ax)z2
]G1
+z
ρ2
[(cλ− ax)cs− e2
(y(c− ay)− az2
)]F1
c
0
(5.192)
and remaining functions and variables are defined following equation (5.174).The distribution of velocity corresponding to these new velocity functions is givenby the value of u for z = 0. From equation (5.189)
u =−cy(x− λy)4(c− ay)2
The axial velocity is zero along the leading edge, and along the trailing edge, wherex = c+ λ− 2y,
u =−cy
4(c− ay) (5.193)
If the new axial velocity function is multiplied by a/c and subtracted from theoriginal, the value of u along the trailing edge will be constant. This can be seenby multiplying equation (5.193) by a/c and subtracting from equation (5.178). Theresult is:
u =−c
4(c− ay) +ay
4(c− ay) = −1
4
Thus, the combined functions give the desired vortex distribution on the panel,which is zero along the leading edge, constant along the trailing edge, and varieslinearly in the chord-wise direction. The velocity components corresponding to thisvortex distribution are given below
u =−14πρ2
sF1 + z
[e2G1 − (cλ− ax)G2
]c
0(5.194)
v = −(cλ− ax)(c− ay) uρ2− azt (5.195)
w = −(c− ay)t+ (cλ− ax)azuρ2
(5.196)
246
where:
t =c
4πρ2
s
ρ2
[e2G1 − (cλ− ax)G2
]− ze2
ρ2F1 +
(y − ar2
c
)G2 +
cλ− axc
d
c
0
(5.197)and the remaining functions and variables are defined following the equation (5.174).It should be noted that the final velocity functions given by equations (5.194) to(5.197) are considerably simpler than either of the preceding sets of equations.The derivation of the velocity component formulas for this vortex distribution iscompleted by adding the contribution of the wake. Returning to the figure (5.31)on the page 244, it can be seen that the elementary horseshoe vortices generate atrailing vortex sheet of constant strength. This vortex sheet contributes only to thev and w components of velocity. The v component of velocity will be derived firstby integrating equation (5.105) for a line vortex of infinite length, as follows:
∆v =−az4πc2
c∫0
ξ dξ
b∫0
[1 +
x− ξ − ληd1
]dη
(y − η)2 + z2 (5.198)
where d1 =√(x− ξ − λη)2 + (y − η)2 + z2, and λ = λ1 + aξ/c.
The inner integral is evaluated fires, giving
∆v =−a4πc2
c∫0
arctan z√(x− ξ)2 + r2
−y(x− ξ) + λr2 − arctanz
y
ξ dξ=
−a8π
arctan z
√(x− ξ)2 + r2
−y(x− ξ) + λr2 − arctanz
y
− zt
c+(x− λy)(c− ay) + aλz2
(c− ay)2 + a2z2 u
c
0
(5.199)
where u and t are given by equations (5.194) and (5.197) respectively, r2 =√y2 + z2,
and λ is redefined as the leading edge slope λ1.The w component of velocity is derived in a similar manner, by integrating equation(5.105) for w velocity component, for a line vortex of infinite length. Here,
∆w =a
4πc2
c∫0
ξ dξ
b∫0
[1 +
x− ξ − ληd1
](y − η)dη
(y − η)2 + z2 (5.200)
where d1 is defined above, and λ = λ1 + aξ/c.The inner integral is evaluated first, giving
∆w =a
4πc2
c∫0
λ√1 + λ2
sinh−1y + λ(x− ξ)√
(x− ξ − λy)2 + (1 + λ2)z2− sinh−1
x− ξr
+ ln r
ξ dξ(5.201)
Only the last two integrals can be evaluated in closed form. Thus
∆w =a
4πc2(I1 − I2 + I3) (5.202)
247
where
I1 =
c∫0
λ√1 + λ2
sinh−1y + λ(x− ξ)√
(x− ξ − λy)2 + (1 + λ2)z2dξ (5.203)
I2 =1
4
[(3x+ ξ)
(d− (x− ξ)G2
)+ d2G2
]c0
(5.204)
I3 =c2
2ln r (5.205)
where λ = λ1 + aξ/c, and d, r, and G2 are defined following equation (5.174).Equation (5.203) must be integrated numerically.It should be noted that ∆v and ∆w as derived above have been multiplied by −a/cprior to integration in order to correctly account for the contribution of the wake.
The Third Step
The velocity components induced by a vortex distribution which has a linear varia-tion in the chord-wise direction, and remains constant in the span-wise direction gavenow been derived for a triangular region bounded by the panel leading and trailingedges, and the in-board side edge. In the third step of this analysis, these velocitycomponent formulas are combined to give the influence of a swept, tapered panelof arbitrary span. This is accomplished by super-imposing two of these triangularregions having common out=board intersections and equal values of the leading andtrailing edge slopes. The superposition process is illustrated in the figure (5.32).
Figure 5.32: Superposition principle on two triangular wings to obtain trapesoidalone.
The upper triangular panel has a concentrated vortex of strength c1 trailing from thein-board edge, and a vortex sheet of strength a behind the trailing edge. There is noconcentrated vortex shed from the out-board tip, since the circulation around the
248
railing vortex sheet is equal and opposite to that of the concentrated edge vortex.A similar vortex pattern is shed by the second triangular panel, except that theconcentrated vortex has a strength c2.The influence of a swept, tapered panel of finite span b can be obtained by super-imposing the two triangular panels as indicated. It should be noted that the con-centrated vortices trailing from the edges of this panel are of unequal strength if thepanel is tapered, the difference being made up by the vortex sheet in the wake. Thevortex distribution on the panel is zero along the leading edge, and varies linearlyin the chord-wise direction to a constant value along the trailing edge. The axialcomponent of velocity u is given by equation (5.194), the v component of velocityis given by the sum of equations (5.179), (5.195) and (5.199), and w component ofvelocity is given by the sum of equations (5.181), (5.196) and (5.202).If the influence of a triangular panel is required, special care must be taken in theevaluation of equations (5.194) and 5.197. In this case, the chord of the out-boardpanel subtracted in the superposition process is zero, and two terms in the equationsbecome indeterminate. The limiting values of these terms are given below.First, the function G1 becomes:
limc→0
[G1]c0 =
λ
2ln
(x− λ1y)2 + (1 + λ21)z2
(x− λ2y)2 + (1 + λ2)2z2(5.206)
where λ1 and λ2 are the slopes of the panel leading and trailing edges.Second, the last two terms in the expression for t become:
limc→0
[1
c
(r2G2 + xd
) ]c0=√x2 + r2 (5.207)
The remaining terms in the equations are unchanged.
Figure 5.33: Quadrilateral constant-strength doublet element.
5.8.8 Quadrilateral Doublet
Consider the quadrilateral element shown in the figure (5.33), with a constant dou-blet distribution. Using the doublet element which points in the z direction thevelocity potential can be obtained by integrating the point elements:
φ(x, y, z) =−µ4π
∫S
z dS
[(x− ξ)2 + (y − η)2 + z2]3/2 (5.208)
249
This integral for the potential is the same integral as the w velocity component ofthe quadrilateral source and consequently
φ =µ
4π
[arctan
m12e1 − h1zr1
− arctanm12e2 − h2
zr2
+arctanm23e2 − h2
zr2− arctan
m23e3 − h3zr3
+arctanm34e3 − h3
zr3− arctan
m34e4 − h4zr4
+ arctanm41e4 − h4
zr4− arctan
m41e1 − h1zr1
](5.209)
as z → 0φ = ∓µ
2(5.210)
The velocity components are obtained by differentiating the velocity potentialuvw
=
∂φ∂x∂φ∂y∂φ∂z
After differentiating (5.209) from, [2], we get
u =µ
4π
[z(y1 − y2)(r1 + r2)
r1r2r1r2 − [(x− x1)(x− x2) + (y − y1)(y − y2) + z2]+
z(y2 − y3)(r2 + r3)r2r3r2r3 − [(x− x2)(x− x3) + (y − y2)(y − y3) + z2]
+z(y3 − y4)(r3 + r4)
r3r4r3r4 − [(x− x3)(x− x4) + (y − y3)(y − y4) + z2]+
z(y4 − y1)(r4 + r1)r4r1r4r1 − [(x− x4)(x− x1) + (y − y4)(y − y1) + z2]
](5.211)
v =µ
4π
[z(x1 − x2)(r1 + r2)
r1r2r1r2 − [(x− x1)(x− x2) + (y − y1)(y − y2) + z2]+
z(x2 − x3)(r2 + r3)r2r3r2r3 − [(x− x2)(x− x3) + (y − y2)(y − y3) + z2]
+z(x3 − x4)(r3 + r4)
r3r4r3r4 − [(x− x3)(x− x4) + (y − y3)(y − y4) + z2]+
z(x4 − x1)(r4 + r1)r4r1r4r1 − [(x− x4)(x− x1) + (y − y4)(y − y1) + z2]
](5.212)
w =µ
4π
[[(x− x2)(y − y1)− (x− x1)(y − y2)](r1 + r2)
r1r2r1r2 − [(x− x1)(x− x2) + (y − y1)(y − y2) + z2]+
[(x− x3)(y − y2)− (x− x2)(y − y3)](r2 + r3)r2r3r2r3 − [(x− x2)(x− x3) + (y − y2)(y − y3) + z2]
+[(x− x4)(y − y3)− (x− x3)(y − y4)](r3 + r4)
r3r4r3r4 − [(x− x3)(x− x4) + (y − y3)(y − y4) + z2]+
[(x− x1)(y − y4)− (x− x4)(y − y1)](r4 + r1)r4r1r4r1 − [(x− x4)(x− x1) + (y − y4)(y − y1) + z2]
](5.213)
250
On the element, as z → 0, u = 0 and v = 0 while
w =µ
4π
[[(x− x2)(y − y1)− (x− x1)(y − y2)](r1 + r2)
r1r2r1r2 − [(x− x1)(x− x2) + (y − y1)(y − y2)]+
[(x− x3)(y − y2)− (x− x2)(y − y3)](r2 + r3)r2r3r2r3 − [(x− x2)(x− x3) + (y − y2)(y − y3)]
+[(x− x4)(y − y3)− (x− x3)(y − y4)](r3 + r4)
r3r4r3r4 − [(x− x3)(x− x4) + (y − y3)(y − y4)]+
[(x− x1)(y − y4)− (x− x4)(y − y1)](r4 + r1)r4r1r4r1 − [(x− x4)(x− x1) + (y − y4)(y − y1)]
](5.214)
Note that z = 0 must be used in the rk terms of equation (5.154) too.
Far Field
The far-field formulas for a quadrilateral doublet with area A can be obtained byusing the results for point doublet, equations (5.100). So we could apply theseresults when we replace point doublet strength with µA:
φ(x, y, z) =−µA4π
z
[(x− xo)2 + (y − yo)2 + z2]3/2
u =3µA
4π
(x− xo)z[(x− xo)2 + (y − yo)2 + z2]5/2
v =3µA
4π
(y − yo)z[(x− xo)2 + (y − yo)2 + z2]5/2
w = −µA4π
(x− xo)2 + (y − yo)2 − 2z2
[(x− xo)2 + (y − yo)2 + z2]5/2
(5.215)
where (xo, yo, 0) is some point within panel, usually centroid.
Constant Doublet Panel Equivalence to Vortex Ring
Consider the doublet panel with constant strength µ. Its potential (5.208) can bewritten as
φ(x, y, z) = − µ
4π
∫S
z dS
r3
where r = sqrt(x− xo)2 + (y − yo)2 + z2. The velocity is obtained by differentiatingthis equation
V = ∇φ = − µ
4π
∫S
∇(z
r3
)dS =
µ
4π
∫S
[ı∂
∂xo
z
r3+
∂
∂yo
z
r3+ k
(1
r3− 3z
r5
)]dS
where we have used
∂
∂x
1
r3= − ∂
∂xo
1
r3∂
∂y
1
r3= − ∂
∂yo
1
r3
251
Now, let C represent the curve bounding the panel in the figure (5.33) and consider avortex filament of circulation Γ along C. The velocity due to the filament is obtainedfrom the Bio-Savart law as
V =Γ
4π
∫C
d× rr3
and for d = (dxo, dyo) and r = (x− xo, y − yo, z) we get
V =∫C
[ız
r3dyo − z
r3dxo + k
((y − yo)dxo − (x− xo)dyo
)]
Stokes theorem for the vector A is∮C
A · d =∫S
n · ∇ × AdS
and with n = k this becomes∮C
A · d =∫S
(∂Ay
∂xo− ∂Ax
∂yo
)dS
Using Stokes’ theorem on the above velocity integral we get
V =Γ
4π
∫S
[ı∂
∂xo
z
r3+
∂
∂yo
z
r3− k
(∂
∂xo
x− xor3
+∂
∂yo
y − yor3
)]dS
Once the differentiation si performed, it is seen that the velocity of the filament isidentical to the velocity of the doublet panel if Γ = µ.The above derivation is a simplified version of the derivation by Hess, which relatesa general surface doublet distribution to a corresponding surface vortex distribution
V = − 1
4π
∫S
(n×∇µ)× r
r3dS +
1
4π
∫C
µd× rr3
whose order is one less that the order of the doublet distribution plus a vortex ringwhose strength is equal to the edge value of the doublet distribution.
5.9 Three-Dimensional Higher-Order Elements
The surface shape and singularity strength distribution over an arbitrarily shapedpanel can be approximated by a polynomial of a certain degree. The surface of suchan arbitrary panel as shown in figure (5.34) can be approximated by a “zero-order”flat platne
z = a0
by a first-order surfacez = z0 + b1x+ b2y
252
Figure 5.34: Higher-order three-dimensional surface element.
Figure 5.35: PANAIR singularity element.
by a second order surface
z = a0 + b1x+ b2y + c1x2 + c2xy + c3y
2
or any higher-order approximations. Evaluation of influence coefficients is possibleonly for first order surfaces as it is shown in the figure (5.35). This approach is usedin the code PANAIR.The source distribution for such element is approximated by the first order polyno-mial:
σ(x0, y0) = σ0 + σxx+ σyy
while doublet distribution is approximated by
µ(xo, yo) = µ0 + µxxo + µyyo ++µxxx2o + µxyxoyo + µyyy
2o
Velocity potential is obtained from
φs =1
4π
∫S
σ(ξ, η) dS√(x− ξ)2 + (y − η)2 + z2
φd = − 1
4π
∫S
µ(ξ, η) z dS
[(x− ξ)2 + (y − η)2 + z2]3/2
253
where index s denote velocity potential due the source distribution, and index dvelocity potential due to doublet distribution. At end velocities a given as linearfunction of node values
(u, v, w)s = fs(σ1, σ2, σ3, σ4, σ9) , (u, v, w)d = fd(σ1, σ2, σ3, σ4, σ5, σ6, σ7, σ8, σ9)
where, again, index s denote source, and index d denote doublet.
254
Chapter 6
Supersonic Singularity Panels
6.1 Introduction
We have already shown that small disturbance equation for compressible flow lookssimilar for both subsonic and supersonic flows. We repeat here for completenessequation (3.102)
β2∂2ϕ
∂x2+∂2ϕ
∂y2+∂2ϕ
∂z2= 0 , M∞ < 1 (6.1)
−β2∂2ϕ
∂x2+∂2ϕ
∂y2+∂2ϕ
∂z2= 0 , M∞ > 1 (6.2)
where
β2 =
1−M2
∞ , M∞ < 1M2
∞ − 1 , M∞ > 1
When the flow is incompressible, β = 1,M∞ = 0, equation (6.1) reduces to Laplace’sequation for which we have presented singular solutions, such as
ϕ = − σ
4π√(x− xo)2 + (y − yo)2 + (z − zo)2
(6.3)
which represents flow due to source of strength σ located at the point (xo, yo, zo).Expression (6.3) is regular everywhere except at the point (xo, yo, zo).For the subsonic case M∞ < 1 we could rearrange equation (6.1) so that it lookslike Laplace equation
∂2ϕ
∂x2+
∂2ϕ
∂(βy)2+
∂2ϕ
∂(βz)2= 0 (6.4)
Simple transformation y = βy and z = βz formally reduces this equation to Laplaceequation for which our singular solution (6.3) looks
ϕ = − σ
4π√(x− xo)2 + β2[(y − yo)2 + (z − zo)2]
(6.5)
We could rearrange this equation to obtain(x− xoσ/4πϕ
)2
+
(y − yoσ/4πβϕ
)2
+
(z − zoσ/4πβϕ
)2
= 1 (6.6)
255
The equipotential surfaces of this subsonic compressible source are ellipsoids of rev-olution. Equipotential lines obtained by intersection of this surfaces with xOy-planeare shown in the figure (6.1a).
a) b)
Figure 6.1: Equipotential lines in xOy-plane for a) subsonic source, and b) supersonicsource.
For the supersonic case, equation (6.2), we could write
∂2ϕ
∂x2+
∂2ϕ
∂(ıβy)2+
∂2ϕ
∂(ıβz)2= 0 (6.7)
where ı =√−1. By analogy with (6.4), a particular solution is
ϕ = − σ
4π√(x− xo)2 − β2[(y − yo)2 + (z − zo)2]
(6.8)
which may be rearranged in the form(x− xoσ/4πϕ
)2
−(y − yoσ/4πβϕ
)2
−(z − zoσ/4πβϕ
)2
= 1 (6.9)
It is evident that the equipotential surfaces of supersonic compressible source arehyperboloids of revolution in the upstream and downstream Mach cones. Intersec-tion of this surfaces with xOy plane is shown in the figure (6.1b). Only the part ofthe flow pattern laying in the downstream Mach cone is physically significant.It must be emphasized that equations (6.5) and (6.8) do not truly represent sourceflows, but acquire this title in a conceptual sense because of the formal analogy withthe solution of Laplace’s equation.All the singular solutions of the Laplace’s equation can be arranged to solutions ofsupersonic equation, by substitution instead of y − yo and (z − zo) the followingexpressions ıβ(y − yo) and ıβ(z − zo) respectively
256
6.2 Green’s Second Formulae Analog
We now extend Green’s second formula, (3.114), to be valid for the supersonic case.Let us introduce, first, the new operator :
ϕ = β2∂2ϕ
∂x2− ∂2ϕ
∂y2− ∂2ϕ
∂z2, = β2
∂2
∂x2ı− ∂2
∂y2− ∂2
∂z2k
Then the Second Green’s Theorem Analog for supersonic flow looks∫∫S
(ϕ1ϕ2 − ϕ2ϕ1) · ν dS =∫ ∫V
∫ (ϕ1
2ϕ2 − ϕ22ϕ1)dV (6.10)
where ν is so called unit co-normal of the surface S, connected to unit normal as
(νx, νy, νz) = (−nx, ny, nz)
where indices denote projection to corresponding axis, figure (6.2).
Figure 6.2: Definition of the co-normal of an arbitrary surface.
Again we can choose for ϕ2 = 1/√(x− xo)2 − β2(y − yo)2 − β2(z − zo)2 so that
2ϕ2 ≡ 0, and for ϕ1 = ϕ we choose velocity potential of the flow we consider,(6.2), thus 2ϕ ≡ 0 also. Our problem then, again, is reduced to surface integralequation: ∫∫
S
(ϕ1ϕ2 − ϕ2ϕ1) · ν dS = 0 (6.11)
The use of this equation depends on physical understanding of the nature of super-sonic flow problem. In accordance with these the disturbance can affect the pointsonly within the after Mach cone. That is, the cone which starts from the point ofdisturbance, and its axis extends in the direction of the flow-filed. Therefore thepoint in the flow-field can be influenced only by the disturbances emanating fromthe points within the fore-cone, as illustrated in the figure (6.3).
257
Figure 6.3: Part of the surface which influence an arbitrary point P .
6.3 Two-Dimensional Point Singularities
When the flow field is the same in all lines parallel to xOy-plane we can drop thederivatives with respect to z, thus equation (6.2) is reduced to
−β2∂2ϕ
∂x2+∂2ϕ
∂y2= 0 (6.12)
which can be formally written in the form
∂2ϕ
∂x2+
∂2ϕ
∂(ıβy)2= 0 (6.13)
From equation (5.1) we have
ϕ =σ
2πln√(x− xo)2 − β2(y − yo)2 = σ
2πln ρ (6.14)
where we have denoted ρ =√(x− xo)2 − β2(y − yo)2. Differentiating the former
equation with respect to x and y give us velocity components u and v respectively
u =σ
2π
∂
∂xln ρ =
σ
2π
x− xoρ2
= (6.15)
v =σ
2π
∂
∂yln ρ =
σ
2π
y − yoρ2
(6.16)
Two-dimensional singularity panels have not found place in aerodynamic solutions oftwo-dimensional supersonic flow problems since there exist more efficient and accu-rate methods, so we skip development of these elements directly to three-dimensionalline elements.
258
6.4 Parametric Differentiation of Integrals
Induced velocity components for the supersonic flow are obtained by differentiatingintegrals of the form
d
dx
g(x)∫a
F (x, y, z, ξ) dξ (6.17)
We develop this differentiation in consecutive steps. Let us start from definition
F (x) =d
dx
x∫a
F (ξ) dξ (6.18)
Now let us change the upper bound replacing x with g(x). In order to obtaindifferential of such integral we have to apply chain rule
d
dx
g(x)∫a
F (ξ) dξ =
d
dg(x)
g(x)∫a
F (ξ) dξ
dg(x)dx
= F (g(x)) · g′(x) (6.19)
We now have to differentiate integral with respect to parameter which does notappear in the limits of integration
∂
∂y
x∫a
F (y, ξ) dξ =
x∫a
Fy(y, ξ) dξ (6.20)
where indexed symbol Fy denotes partial differentiation with respect to y. We arenow ready to find the parametric differential of the form given by equation (6.17)
d
dx
g(x)∫a
F (x, y, z, ξ) dξ = F (x, y, z, g(x)) g′(x) +g(x)∫a
Fx(x, y, z, ξ) dξ (6.21)
where Fx is partial differential of the function F (x, y, z, ξ) with respect to x whenall remaining parameters are considered to be constant, and g′(x) = d g(x)/dx.
Problem 6.1:Potential of the supersonic axi-symmetric flow can be expressed in the form
φ =
x−βr∫0
f(ξ) dξ√(x− ξ)2 − β2r2
find the expression for the induced velocity component along the x-axis!
Solution 6.0:Velocity component along x-axis (u) is obtained by differentiating the expression for thepotential with respect to x
u =∂φ
∂x=∂
∂x
x−βr∫0
f(ξ) dξ√(x− ξ)2 − β2r2
259
Figure 6.4: Body of revolution coordinate system.
Application of equation (6.21) gives
u =f(ξ)√
(x− ξ)2 − β2r2· 1∣∣∣∣∣∣x−βr
−x−βr∫0
f(ξ)(x− ξ) dξ[(x− ξ)2 − β2r2]3/2
whered
dx(x− βr) = 1
Formally we have finished our job here but the first term is infinite, and also the secondterm since exponent of the expression in the bracket is greater than 1. If we partiallyintegrate the last term by choosing
t2 = (x− ξ)2 − β2r2 , tdt = −(x− ξ)dξ
we get
u =f(ξ)√
(x− ξ)2 − β2r2· 1∣∣∣∣∣∣x−βr
− f(ξ)√(x− ξ)2 − β2r2
· 1∣∣∣∣∣∣x−βr
0
+
x−βr∫0
f ′(ξ) dξ√(x− ξ)2 − β2r2
Usually f(0) = 0, thus
u =
x−βr∫0
f ′(ξ) dξ√(x− ξ)2 − β2r2
what is required expression for axial component of velocity at point (x, r) due to sourcedistribution f(x) along x-axis.
6.5 Three-Dimensional Line elements
6.5.1 Small Disturbance Flow about Bodies of Revolution
Cylindrical coordinate system is the most suitable coordinate system to study theflow problems about bodies of revolution. The x-axis is assumed coincident withthe axis of the body, and the r-axis normal to x-axis in the meridian plane in thedirection of radius of every circular cross-section of the body. The meridian planeis defined by the angle ϕ between the meridian plane which contains the direction
260
of the undisturbed velocity and the meridian plane considered, figure (6.4). Theangle of attack α of the body is the angle between the direction of the undisturbedvelocity and the x-axis of the body, in the meridian plane ϕ = 0.At every point the velocity V can be divided into three components: along the x-and y-axis, and normal to the meridian plane. The components of the velocity canbe defined in the following form:
u = u1 + u2 = u1 +∂φ2∂x
= V∞ cosα+∂φ2∂x
v = v1 + v2 = v1 +∂φ2∂r
= V∞ sinα cosϕ+∂φ2∂r
(6.22)
w = w1 + w2 = w1 +1
r
∂φ2∂ϕ
= −V∞ sinα sinϕ+1
r
∂φ2∂ϕ
where u1, v1, and w1 are components of the undisturbed velocity correspondingto the potential of the undisturbed stream φ1, and φ2 is the potential functionwhich defines the variation of the flow generated by the presence of the body. Thispotential function always exists, because the hypothesis of small disturbances isaccepted. Because with the hypothesis accepted α must be small, the differentialequation of motion reduces to:
(1−M2∞)∂2φ2∂x2
+∂2φ2∂r2
+1
r2∂2φ2∂ϕ2
+1
r
∂φ2∂r
= 0 (6.23)
terms v1 and w1 could also be considered small (of the same order as disturbances).The solution of equation (6.23) is found to consists of two parts
φ2 = φ′2 + φ
′′2 (6.24)
where φ′2 is the function of x and r only, and is obtained as the solution of thefollowing partial differential equation
(1−M2∞)∂2φ′2∂x2
+∂2φ′2∂r2
+1
r
∂φ′2∂r
= 0 (6.25)
which is the equation that defines the flow for the case of the body with axis in thedirection of the velocity. The second component, φ′′2 is given in by the expression
φ′′2 = F (x, r) cos ϕ (6.26)
where
F (x, r) =∂φ′2∂r
(6.27)
We could prove that this expression is the solution of the equation (6.23) if thepotential φ′2 is the solution of (6.25) and at the same time of the equation (6.23)since φ′2 does not depend on ϕ.Substituting the expression for φ′2 given by (6.26) into (6.23) we get
(1−M2∞)∂2F
∂x2+∂2F
∂r2− 1
r2F +
1
r
∂F
∂r= 0 (6.28)
261
Figure 6.5: Axi-symmetric flow about the body of revolution.
Differentiating equation (6.25) with respect to r we get
(1−M2∞)
∂2
∂x2
(∂φ′2∂r
)+∂
∂r
(∂φ′2∂r
)+1
r
∂
∂r
(∂φ′2∂r
)− 1
r2
(∂φ′2∂r
)= 0 (6.29)
Assuming∂φ′2∂r
= F
equation (6.28) is coincident with equation (6.29), therefore equation (6.26) is thesolution of equation (6.23).The velocity potential which describes the flow can therefore be expressed in theform
φ = φ1 + φ2
or
φ = φ′1 + φ′2 + φ
′′1 + φ
′′2
= xV∞ cos α+ φ′2 + yV∞ sin α cosϕ+∂φ′2∂r
cos ϕ (6.30)
Velocity potential φ′1+φ′2 describes the axi-symmetric flow about body of revolution.
To determine the effect of the angle of attack it is necessary to determine only thevariation of the phenomenon dependent on the presence of cross flow due to theangle of attack which is represented by the function φ′2+φ
′′1. The part φ1 is assumed
independent of the angle of attack, and u1 is assumed equal to V∞ (because α issmall). The determination of the flow around body of revolution can be consideredtherefore divided in two parts. The first part consists in determination of the valueof the potential function φ′2 which, with φ
′1, represents the flow under zero-th angle
of attack. The second part consists in determination of φ′′2 which, with φ′′1, gives the
variation of the flow parameters dependent on the angle of attack.This possibility to superpose the effects of the axial flow on the effects on cross flowis due to hypothesis of small disturbances. This greatly simplifies computationalprocedure. Procedure we are going to develop is applicable only for the small angleof attack.
262
6.5.2 Slender Body in Axi-Symmetric Flow
Figure (6.5) shows the flow-field as seen in any meridian plane, since the flow isaxi-symmetric. Let us distribute the sources along the axis of symmetry of the bodyof revolution. Potential at any point P (x, r) of the plane can be determined by
φ′2 =b∫
a
f(ξ) dξ√(x− ξ)2 − β2r2
(6.31)
It is clear from the figure (6.5) that only part of the source strength distributionf(ξ) can influence the point P (x, r), only the part within fore-cone which tracesare shown with lines passing under angle µ from the point P . Lower bound of theintegral is simply a = 0 since body of revolution is placed with the nose tip at theorigin. Upper bound is determined by the intersection of the lower characteristicsline with x-axis. From the Rectangular triangle shown in the figure we have
xo = x− βr , β = tanµ =√M2∞ − 1
thus, our upper bound b = x − βr depends on the position of the point P , i.e. ofthe coordinates (x, r). So we have
φ′2 =x−βr∫0
f(ξ) dξ√(x− ξ)2 − β2r2
(6.32)
6.5.3 Boundary Condition
So far we have approximated governing equation of motion. We have to specifyalso boundary conditions. We could also apply the approximations to the boundaryconditions of the same order as we used to approximate governing equations ofmotion. If the body is defined by r = R(x), the exact tangency condition for theoriginal problem defined by (6.25) is
R′(x) =v
U∞ + u, at: r = R(x)
where U∞ = V∞, u = ∂φ′2/∂x, and v = ∂φ′2/∂r. We now drop prime symbol andindex (′2) to facilitate writing of expressions but we still bear in mind that we aresolving only part of the problem, and we introduce the symbol φ = φ′2/U∞. Thusthis equation looks
R′(x) =φr
1 + φx
, or: φr = R′(1 + φx) (6.33)
if we neglect φx << 1 since (φ′2)x << U∞ we get approximate boundary conditionof our problem
φr(x) = R′(x) , at: r = R(x). (6.34)
Equation (6.25) looks
φrr +φr
r− β2φxx = 0 (6.35)
263
Figure 6.6: Superposition principle for the body of revolution.
After the velocity components are determined, the pressure coefficient is given by
CP =2
γM2∞
[1 +
γ − 1
2M2
∞(1− (1 + φx)
2 − φ2r)]γ/(γ−1)
− 1
(6.36)
6.5.4 The Basic Solutions of the Equation (6.25)
Any first-order solution may be regarded as resulting from continuous distributionof supersonic sources along the axis of the body. Therefore the first-order problemconsists simply in determining the source-distribution function f(x) which producesthe desired shape. However, substituting this expression into tangency conditionyields an integral equation which cannot be solved exactly.The Karman—Moore procedure for obtaining an approximate numerical solutioninvolves that the unknown continuous source function f(x) can be replaced by abroken line as indicated in the figure (6.6). Another (quite equivalent) viewpointis that the function f(x) is approximated by the sum of a number of linear sourcedistributions having various starting points along the body, as it is also shown in thefigure (6.6). The slope of each of these linear elements is determined in successionb imposing the tangency condition at corresponding points along the body.
Figure 6.7: Superposition principle for the body of revolution with quadratic ele-ments.
264
Figure 6.8: Square-root strength source distribution to simulate bodies with slopediscontinuities in the shape.
For some applications this broken line approximation to the source strength is toocrude. For higher order approximations we need higher order derivatives of thedistribution function. Since differentiation is roughening process, this means thatthe source distribution function must be one order higher than the order of requireddifferentiation. The continuity of the second derivatives is achieved by the quadraticfunction for source distribution, as it is shown in the figure (6.7). Only for the nosetip of the slender body we still need first order source distribution to ensure sharp,pointed body.Linear and quadratic elements are sufficient for the bodies of continuous shape. Butif the body has steps or discontinuities in the curvature we need additional elements.A corner is accounted for by adding a source distribution of square root strength,which produce discontinuity in streamline slope along its foremost Mach cone. Asindicated in the figure (6.8), this corner solution must be shifted up-stream so thatits effect first reaches the surface just at the corner. In the same way, a curvaturediscontinuity is accounted for by adding a source distribution of 3/2-power strength,which produces a discontinuity in streamline curvature along its foremost Mach cone.This curvature solution is required also at a corner, because an apparent curvaturediscontinuity remains after the corner solution is added.Step solution in streamline is achieved by adding inverse square root strength sourcedistribution, as it is shown in the figure (6.9). To summarize, the first-order solutionsis obtained by superposing the following five basic solutions:
1. Linear source solution – used at tip of pointed body.
2. Quadratic source solution – used thereafter for body having continuous cur-vature.
3. Corner solution – used to account for corner.
4. Curvature solution – used to account for curvature discontinuity.
5. Step solution – used to cancel steps in second order solutions at corners, if thesecond order solution is planned to be applied.
265
Figure 6.9: Source distribution to simulate bodies with steps in the shape.
Homogeneous Solutions
The required solutions are axially symmetric solutions of the wave equation, homo-geneous in the space variables. The order of homogeneity is integral (1 or 2) in thefirst two cases, and half-integral (1/2, 3/2, -3/2) in the other cases. Such solutions havebeen studied in detail by Hess. For present purposes φ(m), the solution homogeneousof orderm, can be obtained by taking the source distribution f(x) in equation (6.32)proportional to xm. It is convenient to choose the source strength as
f(x)(m) =c
m!xm (6.37)
where C is a normalization constant, so that solutions of various orders are relatedby
φ(m−p) =∂p
∂xpφ(m) (6.38)
For integral m, the solutions have simplest form if the normalization constant C istaken to be unity. The using various relations for the hyper-geometric functions thesolutions are found to be given by
φ(m)(x, r) = − xm
1 · 3 · · · (2m− 1)(1− t2)m+1/2 F
(m+ 1
2,m+ 2
2,m+
3
2, 1− t2
)(6.39)
here the conical variable
t =βr
x, β =
√M2∞ − 1 (6.40)
is the ratio of the tangent of the polar angle to the tangent of the Mach angle, ando varies from zero on the axis to unity at the Mach cone. For integral m, thehyper-geometric functions which occur in equation (6.39) can be expressed in termsof products of
√1− t2 and sech−1t with polynomials in t2. The first two required
basic solutions are obtained by setting m equal to 1 and 2 which gives:
Linear Source Solution m = 1
Here are the expressions for the linear source solution for velocity potential φ andits derivatives:
266
φ = −x(sech−1t−
√1− t2
)φx = −sech−1tφr = β
√1− t2t
φxx = −1
x
1√1− t2
φxr =β
x
1
t√1− t2
φrr = −β2
x
1
t2√1− t2
(6.41)
where
sech−1t = ln
1
t+
√1
t2− 1
Quadratic source solution m = 2
φ = −x2
2
[(1 +
t2
2
)sech−1t− 3
2
√1− t2
]φx = −x
(sech−1t−
√1− t2
)φr =
βx
2
(√1− t2t
− tsech−1t)
φxx = −sech−1tφxr = β
√1− t2t
φrr = −β2
2
(√1− t2t2
+ sech−1t
)
(6.42)
For half-integralm, it is convenient to choose the normalization constant C as√2/π,
so that the solutions have simple values at the Mach cone. The difference in nor-malization for integral and half-integral m is of no concern, because the connectionbetween them is never used. Transforming the hyper-geometric function into a moreuseful form for this case gives
φ(m)(x, r) = −xm√2(1− t)m+1/2
Γ(m+ 3/2)√1 + t
F(1/2,m+ 1,m+ 3/2,
1− t1 + t
)(6.43)
Corner Solution m = 1/2
The hyper-geometric function occurring here can be expressed in terms of products ofcomplete elliptic integrals and algebraic functions of t. The remaining three requiredbasic solutions are obtained by setting m equal to 1/2, 3/2, and -1/2. For convenience,asymptotic values valid just inside the Mach cone (where t = 1) are also given below
267
φ = −√x4√2
π
√1 + t(K − E) , t→ 1 φ→ 0
φx = − 1√x
√8
π
1√1 + t
K , t→ 1 φx → − 1√x
φr =β√x
√8
π
1√1 + t
(1 + t
tE −K
), t→ 1 , φr → β√
x
φxx =1
x3/2
√2
π
1
(1− t)√1 + t(K − E) , t→ 1 , φxx → 1
8
1
x3/2
φxr =β
x3/2
√2
π
1
(1− t)√1 + t(E
t−K
), t→ 1 , φxr → 3
8
β
x3/2
φrr = − β2
x3/2
√2
π
1
(1− t)√1 + t
(2− t2t2
E − 2− ttK
), t→ 1 , φrr → −7
8
β2
x3/2
(6.44)
Here K and E are the complete elliptic integrals of first and second kind with
modulus k =√(1− t)/(1 + t).
Curvature solution m = 3/2
φ = −x3/28√2
9π
√1 + t
[(3 + t)K − 4E
], t→ 1 , φ ∼ 0
φx = −√x 4√2
π
√1 + t(K − E) , t→ 1 , φx ∼ 0
φr = β√x4√2
3π
√1 + t
(E
t−K
), t→ 1 , φr ∼ 0
φxx = − 1√x
2√2
π
1√1 + t
K , t→ 1 , φxx ∼ − 1√x
φxr =β√x
2√2
π
1√1 + t
(1 + t
tE −K
), t→ 1 , φxr ∼ β√
x
φrr = − β2
√x
2√2
3π
1√1 + t
(21 + t
t2E − 2− t
tK), t→ 1 , φrr ∼ − β
2
√x
(6.45)
Again the K and E are the complete elliptic functions of the first and second kind
respectively with modulus k =√(1− t)/(1 + t).
268
Step Solution m = − 1/2
φ = − 1√x
2√2
π
1√1 + t
K , t→ 1 , φ ∼ − 1√x
φx =1
x3/2
√2
π
1
(1− t)√1 + t (K − E) , t→ 1 , φx ∼ 1
8
1
x3/2
φr =β
x3/2
√2
π
1
(1− t)√1 + t(E
t−K
), t→ 1 , φr ∼ 3
8
β
x3/2
(6.46)
All three second derivatives are necessary in order to carry out the second-ordersolution. Considerable manual labor can be avoided by calculating only one ofthem, say φxx. Then φxr and φrr can be obtained form the equation of motion andtangency condition. Thus the first-order equation of motion (6.35) gives immediatelyan expression for φrr:
φrr = β2φxx− φr
r(6.47)
Differentiating the tangency condition, equation (6.33) with respect to x gives anexpression for φxr on the surface of the body:
φxr = R′′(1 + φx) +R
′φxx , at: r = R(x) (6.48)
6.5.5 Singularity Lines Passing Through Coordinate Origin
Velocity potential of the source distribution along spatial line can be written as
φ(x, y, z) =∫ σ(t) dt√
(x− ξ)2 − β2(y − η)2 − β2(z − ζ)2(6.49)
where (ξ, η, ζ) are running coordinates along singularity line and are function ofparameter t. In the case that line pass through coordinate origin for the parametert we could choose coordinate ξ. Thus,
φ(x, y, z) =
ξo∫0
σξ dξ√(x− ξ)2 − β2(y − νξ)2 − β2(z − τξ)2
(6.50)
where ν = η/ξ and τ = ζ/ξ and ξo is the value of ξ for which the denominator of theintegrand becomes zero. We assumed that strength of the source varies linearly withthe coordinate ξ and that σ is the scale of this variation. Physically interpreted, therange of integration is from the origin to the last source point which can influencethe field point, as it is shown in the figure (6.10). Performing the integration yields
φ = σ
−√x2 − β2(y2 + z2)1− β2(ν2 + τ 2)
− x− β2(σy + τz)[1− β2(ν2 + τ 2)]3/2 ctnh
−1 x− β2(νy + τz)√[1− β2(ν2 + τ 2)][x2 − β2(y2 + z2)]
(6.51)
269
P (x, y, z)
y
z
xξη
ζ
Figure 6.10: Source distribution along line passing through coordinate origin.
where
ctnh−1(x) = ln
√x+ 1
x− 1
If two such source lines of opposite strength are brought together from the x-directionat the xy-plane while the product of source strength and the angle between themis kept constant, the potential of a line of doublets in the xy-plane at the an anglearctan ν from the x-axis is obtained. Thus differentiating with respect to τ andsetting τ = 0 gives
φ(x, y, z) = − µzβ2
(1− β2ν2)3/2(
δ
δ2 − 1− ctnh−1δ
)(6.52)
where
δ =x− β2νy√
(1− β2ν2[x2 − β2(y2 + z2)]and µ is the doublet strength. Differentiating the potential function with respect toz gives the velocity w:
w = − µβ2
(1− β2ν2)3/2(
δ
δ2 − 1− ctnh−1δ
)
+2µz2β2(x− β2νy)
√x2 − β2(y2 + z2)
(x− β2νy)2 − (1− β2ν2)[x2 − β2(y2 + z2)]2 (6.53)
Note that the line of doublets creates a conical flow field as the velocity is only afunction z/x and y/x.
6.6 Surface Distribution of the Singularities
6.6.1 Gothert’s Rule
The compressible velocity components induced by the source and vortex distribu-tions are obtained by applying Gothert’s rule to the incompressible velocity com-ponents derived in the previous chapter. Rule states that the velocity components
270
1 2
3d1
d2 d1
d2
d2
d1
Figure 6.11: For point 1 both distances d1 and d2 are imaginary. Only distance d2is imaginary for point 2. For point 3 both distances d1 and d2 are real.
u, v, and w at a point P (x, y, z) in a compressible flows are equal to the real partsof ui, βvi, and βwi, where ui, vi and wi are the incompressible velocity components
evaluated at a point P (x, βy, βz), and β =√1−M2∞. In subsonic flow, this rule
agrees exactly with that given by Gothert if each of the compressible velocity com-ponents are divided by the constant β2. In supersonic flow, the compressible velocitycomponents become complex functions, and care must be taken to extract the realparts of these functions in order to obtain correct results. However, this procedureis generally much simpler than formally evaluating the velocity components by inte-gration, and provides straightforward method for obtaining the supersonic velocityfields corresponding to any existing incompressible flow solution.A simple example of the extended rule is obtained by transforming the velocity com-ponents induced by a line source located along the x-axis. The velocity componentformulas given by equations (5.101) to (5.102) are unchanged by this transforma-
tion, except that d1 =√x2 + β2r2 and d2 =
√(x− L)2 + β2r2. Both these terms
are real in subsonic flow; but in supersonic flow, both are imaginary ahead of theMach cone from the origin, d1 is real but d2 is imaginary between the Mach conefrom the origin and rear Mach cone, and both are real behind the rear Mach cone,figure (6.5). Thus, the velocity components are zero ahead of the Mach cone fromthe origin, and the finite length of the source has no influence on the velocity fieldexcept within the rear Mach cone. Considerable advantage is take of this abilityto correctly define the regions of influence of each term in the velocity componentformulas in the following applications.The compressible velocity components for each of the five basic surface singularitydistributions used for subsonic approximation are presented in the following subsec-tions.
6.6.2 Integration Procedure
The velocity component integrals appearing in the text, below, may all be reducedto forms appearing in standard integral tables, and two non-standard integrals givenbelow:
J1 =∫ dξ
(ξ2 + e2)√aξ2 + 2bξ + c
271
J2 =∫ ξdξ
(ξ2 + e2)√aξ2 + 2bξ + c
The result of integration is summarized below:
J1 =γ
γ4 − b2e2(bF +
γ2
eG
)
J2 =γ
γ4 + b2e2
(γ2F − beG
)where γ is a real non-zero root of the equation
γ4 − (ae2 − c)γ2 − b2e2 = 0
and
F = arctanγ√aξ2 + 2bξ + c
γ2 − bξG = tanh−1
ξγ + eδ
e√aξ2 + 2bξ + c
= sinh−1ξγ + eδ√
(ae2 − γ2)ξ2 + (c− δ2)e2
where δ = be/γ.
6.6.3 Constant Source Distribution on Un-swept Panel withStream-wise Taper
The incompressible velocity components for this source distribution are given byequations (5.141) to (5.142). The corresponding compressible velocity componentsare:
u = − kσ
4π√1 + β2a2
[aF − β2mG−H
β2
](6.54)
v = −kσ√1 + β2a2
4πG (6.55)
w =kσ
4π√1 + β2a2
[F + a(β2mG−H)
](6.56)
where σ is the source distribution strength, and
F = arctan(z − ax)d
−x(y −mx)− β2z(ay −mz) (6.57)
G =1
esinh−1
x′
βr′(6.58)
H = β sinh−1βy√
x2 + β2z2(6.59)
272
z = ax
z
y
y = mx
y = (m−ae′)x1−e2
y = (m+ae′)x1−e2
x = my+az+|mz−ay|e′a2+m2
x = my+az+|mz−ay|e′a2+m2
Figure 6.12: Traces of the Mach waves and Mach cone.
and β2 = 1−M2∞, and
k =
1 for: M∞ ≤ 12 for: M∞ > 1
d =√x2 + β2r2
e =√1 + β2(a2 +m2)
d′ = de
r′ =√(y −mx)2 + (z − ax)2 + β2(ay −mz)2
x′ = x+ β2(my + az)
The k gives the correct scaling factor for the supersonic velocity components, sincecontribution to the up-stream cone is delivered to downstream.We have to determine real parts of the functions F , G, and H for the supersonicflow. The function F is zero everywhere ahead of the Mach cone from the origin,except for panels having supersonic side edges, when F = ±π within the “two-dimensional” region bounded by the Mach waves from the side edge and the Machcone from the origin. The boundaries of the two-dimensional region are given in thefollowing sketch, figure (6.12), which shows the traces of the Mach waves and Machcone from the origin on a plane perpendicular to the x-axis.The function G takes several different forms depending on the relative sweep-backon the side edges. Expressing the function in logarithmic form,
G =
1eln x′+d′
β′r′ for: e2 > 0dx′ for: e2 = 01e′ arccos
x′β′r′ for: e2 < 0, and: − β′r′ < x′ < β′r′
± πe′ for: e2 < 0, x′ ≤ −β′r′
and: x > my+az+|ay−mz|e′a2+m2
0 elsewhere
(6.60)
273
whereβ′ =
√M2∞ − 1 e′ =
√−1− β2(a2 +m2)
Finally, in supersonic flow, the function H becomes:
H =
β′ arctan d
β′y for: x > βr
0 elsewhere(6.61)
6.6.4 Constant Source Distribution on Swept Panel withSpan-wise Taper
Incompressible velocity components for this source distribution are given by systemof equations (5.147). The corresponding compressible velocity components are:
u = −kσG1
4π(6.62)
v =kσ
4π(λG1 −G2) (6.63)
w =kσ
4π(F1 − F2) (6.64)
where
F1 = arctanz d
−xy + λr2 (6.65)
F2 = arctanz
y(6.66)
G1 =1
esinh−1
λx+ β2y
β√(x− λy)2 + (β2 + λ2)z2
(6.67)
G2 = sinh−1x
βr(6.68)
and
k =
1 for: M∞ ≤ 10 for: M∞ > 1
d =√x2 + β2r2
e2 = β2 + λ2
r2 = y2 + z2
In supersonic flow, the real parts of the functions F1, F2, G1, and G2 must bedetermined. The function F2 is always real, and can be dropped from equation(6.64) without affecting the results since the contributions from the four corners ofthe panel always cancel. The function F1, is zero everywhere ahead of the Mach conefrom the origin, except for panel having supersonic leading edges, when F1 = ±πwithin the “two-dimensional” region bounded by the Mach waves form the leadingedge and the Mach cone from the origin. The boundaries of the two-dimensionalregion for this case are shown in the figure (6.13).
274
x = λy
y
z
y = λx/β′2
x = λy + e′|z|
Figure 6.13: Traces of the Mach waves and Mach cone for the supersonic leadingedge.
The function G1 takes several different forms depending on the relative sweep-backof the panel leading edge. Expressing the function in logarithmic form,
G1 =
1eln x′+d′
β′r′ for: e2 > 0dx′ for: e2 = 01e′ arccos
x′β′r′ for: e2 < a
and: − β′r′ < x′ < β′r′
± πe′ for: e2 < 0, x′ < −β′r′
and: x > λy + e′|z|0 elsewhere
(6.69)
where
β′ =√M2∞ − 1
x′ = λx+ β2y
e′ =√−β2 − λ2
r′ =√(x− λy)2 + e2z2
d′ = ed
The function G2 becomes:
G2 =
ln x+d
β′r for: x > βr
0 elsewhere(6.70)
6.6.5 Linearly Varying Source Distribution on Swept Panelwith Span-wise Taper
The incompressible velocity components for this source distribution are given by thesystem of equations (5.164), the corresponding compressible velocity components
275
are:
u = −kσ4π
[(x− λy)G1 + yG2 − z(F1 − F2)
](6.71)
v =kσ
4π
[(x− λy)(λG1 −G2) + x+ d− λz(F1 − F2)
](6.72)
w =kσ
4π
[(x− λy)(F1 − F2) + z(e2G1 − λG2)
](6.73)
where the functions F1, F2, G1, G2, d, e, k, and r are defined by equations (6.65)to (6.68). In supersonic flow, the behaviour of the functions F1 and F2 is describedfollowing equation (6.68) and the real parts of G1 and G2 are given by equations(6.69) and (6.70). In supersonic flow, the sum x+d appearing in the above equationsis replaced by d, and is real only within the Mach cone from the origin.
6.6.6 Constant Vortex Distribution on Swept Panel withSpan-wise Taper
The incompressible velocity components for the bound vortex distribution are givenby equations (5.166). The corresponding compressible velocity components are:
u =kΓ
4π(F1 − F2) (6.74)
v = −λu (6.75)
w =kΓ
4π(e2G1 − λG2) (6.76)
where the functions F1, F2, G1, G2, d, e, k, and r are defined by equations (6.65) to(6.68).The contribution of the edge vortices in compressible flow is given by
∆u = 0 (6.77)
∆v =kΓz
4π
x+ d
r2(6.78)
∆w = −kΓy4π
x+ d
r2(6.79)
In supersonic flow, the sum x + d appearing in the above equations is replaced byd, and is real only within the Mach cone from the origin.
6.6.7 Linearly Varying Vortex Distribution on Swept Panelwith Span-wise Taper
The incompressible velocity components for the bound vortex distribution are givenby equations (5.194) to (5.197) The corresponding compressible velocity componentsare:
u = − k
4πρ2
sF1 + z[e
2G1 − (cλ− ax)G2]c
0(6.80)
v = −(cλ− ax)(c− ay) uρ2− azt (6.81)
w = −(c− ay)t+ (cλ− ax)azuρ2
(6.82)
276
where
t =ck
4πρ2
s
ρ2[e2G1 − (cλ− ax)G2]− ze2
ρ2F1
+ β2(y − ar2
c
)G2 + (cλ− ax)d
c
0
(6.83)
and
k =
1 for: M∞ ≤ 12 for: M∞ > 1
a = λ2 − λ1d =
√(x− ξ)2 + β2r2
r2 = y2 + z2
ρ2 = (c− ay)2 + a2z2e2 = (cλ− ax)2 + β2ρ2s = (c− ay)(x− λy) + aλz2
and
F1 = arctanzd
−y(x− ξ) + (λ− aξ/c)r2 (6.84)
G1 =1
esinh−1
(cλ− ax)(x− ξ) + β2[y(c− ay)− az2]βc√[x− λy + (c− ay)ξ/c]2 + z2[β2 + (λ+ aξ/c)2]
(6.85)
G2 = sinh−1x− ξβr
(6.86)
In supersonic flow, the behaviour of function F1 is described following equation(6.68), and the real parts of G1 and G2 are given by equations (6.69) and (6.70),where
x′ = (cλ− ax)(x− ξ) + β2[y(c− ay)− az2]r′ = c
√[x− λy + (c− ay)ξ/c]2 + z2[β2 + (λ+ aξ/c)2]
d′ = ed
e′ =√−(cλ− ax)2 − β2ρ2
Finally, the contribution of the vortex sheet in the wake in compressible flow is givenby
∆u = 0 (6.87)
∆v = − a
8π
k(F1 − F2)− zt
c+(x− λy)(c− ay) + aλz2
(c− ay)2 + a2z2 u
c
0
(6.88)
where u and t are given by the equations (6.80) and (6.83) respectively. The contri-bution to the third component of velocity is given as
∆w =ka
4πc2(I1 − I2 + I3) (6.89)
where I1, I2, and I3 are given by the equations (5.203) to (5.205), with the Gotherttransformation applied.
277
Figure 6.14: Non-planar quadrilateral.
6.6.8 Panel Geometry
In all our derivations we assumed that panel is flat i.e. it lies in the xOy-plane. Realconfigurations are far from being flat, so we need procedure to ensure flattens of thethree-dimensional surface panels we use in our calculations.Let the four node points which define quadrilateral lies arbitrary in the space. Weshow here procedure how to make them planar. The figure (6.14) we show such
panel. We also introduce vectors which connects diagonal points T1 =−→13 and
T2 =−→24. The components of these vectors are
T1x = x3 − x1 T1y = y3 − y1 T1z = z3 − z1T2x = x4 − x2 T2y = y4 − y2 T2z = z4 − z2
We could now generate vector N orthogonal to both vectors T1 and T2
N = T2 × T1 =
∣∣∣∣∣∣∣ı kT2x T2y T2zT1x T1y T1z
∣∣∣∣∣∣∣The unit normal vector n of the element can be now defined as
nx =Nx
N, ny =
Ny
N, nz =
Nz
N
where (Nx, Ny, Nz) are components of the vector N and N =√N2
x +N2y +N
2z . The
plane of the element is completely defined if we besides unit normal specify also onepoint in the plane. Let this point be the centroid of the four corner nodes:
x =x1 + x2 + x3 + x4
4
y =y1 + y2 + y3 + y4
4
z =z1 + z2 + z3 + z4
4
We need now to project along unit vector n the four corner nodes to the newlydefined plane. The resulting points are the corner points of the required quadrilateralelement. Let us write the equations of the plane with unit vector n passing throughpoint (x, y, z)
nx(x− x) + ny(y − y) + nz(z − z) = 0
278
and equation of the line passing through the corner node i in the direction of theunit vector n
x− xinx
=y − yiny
=z − zinz
= t
Let us express point coordinates (x, y, z) laying on this line as a function of parametert
x = nxt+ xi , x = nyt+ yi , x = nzt+ zi
Substituting this expression into equation of the plane we get
nx(nxt+ xi − x) + ny(nyt+ yi − y) + nz(nzt+ zi − z) = 0
from which follows
ti =nx(x− xi) + ny(y − yi) + nz(z − zi)
n2x + n2y + n
2z
where we added the index to parameter t to denote that it corresponds to cornernode i. Changing the coordinates of corners (xi, yi, zi), i = 1, 2, 3, 4 we calculate allfour values of parameter t. The new i-th node coordinates are
Xi = nxti + xi , Yi = nyti + yi , Zi = nzti + zi
where (Xi, Yi, Zi) are the new planar coordinates of the quadrilateral.All expressions given so far are given in global coordinate system, we need to buildlocal coordinate system. Let us choose point 2 for coordinate origin, and vector
−→21
for the direction of the local x-axis. Local unit vector ı ′ is defined thus
′ =−→21
|−→21| =(X1 −X2)ı+ (Y1 − Y2)+ (Z1 − Z2)k√(X1 −X2)2 + (Y1 − Y2)2 + (Z1 − Z2)2
If we orient local z-axis in the direction of the unit vector n, then the third unitvector ′ is obtained as
′ = n×ıWe have expressed all three unit vectors, so we can apply coordinate transformationsbetween global and local coordinate system and vice verse to calculate inducedvelocity components in the global coordinate system.
6.6.9 Coordinate Transformation Again
We now extend the coordinate transformation procedure developed in the section(5.5) for two-dimensional flows. Again primed symbols denote the values of unitvectors in the local coordinate system.The new local coordinates of the panel corners (in accordance to the node numberingdefined for vortex panels) are given below, first corner number 1:
x1y1z1
=
000
279
coordinates of the point 2 are
x2y2z2
=
ı ′ ·ı ı ′ · ı ′ · k ′ ·ı ′ · ′ · kk
′ ·ı k ′ · k′ · k
X1 −X2
Y1 − Y2Z1 − Z2
coordinates of the point 3 are
x3y3z3
=
ı ′ ·ı ı ′ · ı ′ · k ′ ·ı ′ · ′ · kk
′ ·ı k ′ · k′ · k
X3 −X2
Y3 − Y2Z3 − Z2
and finally coordinates of the point 4 are
x4y4z4
=
ı ′ ·ı ı ′ · ı ′ · k ′ ·ı ′ · ′ · kk
′ ·ı k ′ · k′ · k
X4 −X2
Y4 − Y2Z4 − Z2
We could perform similarly transformation of coordinates for source panels. Wechoose again local point 2 for local coordinate origin, and direction of the vector
−→23
for the direction of local y-axis. So unit vector along y-axis ( ′) is defined as
′ =(X3 −X2)ı+ (Y3 − Y2)+ (Z3 − Z2)k√(X3 −X2)2 + (y3 − y2)2 + (Z3 − Z2)2
Again we choose vector n for the direction of the z-axis, remaining unit vector ı ′ isobtained as
ı ′ = ′ × nAgain point two is origin (that is point 1 in discussions about source panels) so(x1, y1, z1) = (0, 0, 0). Local point 2 is obtained from transformation
x2y2z2
=
ı ′ ·ı ı ′ · ı ′ · k ′ ·ı ′ · ′ · kk
′ ·ı k ′ · k′ · k
X1 −X2
Y1 − Y2Z1 − Z2
Coordinates of the points (x3, y3, z3) and (x4, y4, z4) are obtained the same way asthe points for vortex panel. The only difference is the definition of the coordinateaxes unit vectors. We could reduce computational time by specifying z coordinateto 0, since surface panels are assumed to be planar. Thus, we reduce the problemto calculate only two coordinates.
280
Chapter 7
Applications
In this chapter we illustrate some of the basic concepts developed so far on a fewsimple examples.
7.1 Incompressible Flow about Airfoil
We generate, first, the analytical case which will be used to compare results ofcalculation using panel method with analytically exact solution. We plan to ap-ply complex mapping of the flow about circle to the airfoil shape. Thus, we needexpression for the velocities on the circle placed out the coordinate origin.
7.1.1 Flow about Circle with Circulation
Equation (4.45) give us general expression for the flow about the circle with singular-ities around it. We have to adjust that equation for our needs keeping only vortexat the center of the circle and moving the circle at the position zc. So, complexvelocity potential for our case looks
F (z) = (U∞ − ı V∞)(z − zc) + (U∞ + ı V∞)a2
z − zc +Γ
2πıln(z − zc) (7.1)
Differentiating this equation with respect to z give us complex velocity W
W = u− ı v = dF
dz= U∞ − ı V∞ − (U∞ + ı V∞)
a2
(z − zc)2 +Γ
2πı
1
z − zc (7.2)
It is possible to define any point on the circle (z − zc = ae−ıβ) as stagnating point(W = 0). Thus, Circulation Γ which ensures that this point is stagnating point isobtained from (7.2) when W = 0
Γ = −4πa(U∞ sin β + V∞ cos β)
= −4πa√U2∞ + V 2∞ sin(α+ β) (7.3)
where
α = arctanV∞U∞
281
is the angle of attack. From the Kutta-Joukowsky theorem we have
1/2(U2∞ + V 2
∞)CL · 1 = √U2∞ + V 2∞Γ
or:CL = −8πa
sin(α+ β) (7.4)
Gradient of the lift slope is obtained from this equation after differentiating it withrespect to α.
dCL
dα= −8πa
cos(α+ β) (7.5)
where = (ζ)max −(ζ)min|, and (·) means real part of the complex number.
7.1.2 Conformal Mapping
Flow about airfoil-like shapes can be obtained by conformal mapping of the flowabout circular cylinder. Let the circle be defined in the z-plane, and airfoil-likeshape in the ζ-plane. Also, let us assume one to one mapping of the form ζ = ζ(z).The value of the velocity potential of the flow must be the same in the correspondingpoints. This property will be used to find velocity distribution around airfoil-likeshape from the velocity distribution around circular cylinder. Complex velocity is,thus
W =dF
dζ=dF
dz
dz
dζ(7.6)
If the mapping function ζ = ζ(z) is given than former equation should be arrangedin the more suitable form
W =dF
dζ=dF
dz
1
dζ/dz(7.7)
Derivative dF/dz is defined by the equation (7.2), while equation (7.3) defines circu-lation Γ for the flow about circle. Circulation is chosen such that when z = zT , anddζ/dz = 0 at the same tame is dF/dz = 0 so we have the ratio 0/0 in the equation(7.7) for z = zT . To obtain the velocityW in such case we need to apply L’Hopital’srule:
W =d2F/dz2
d2ζ/dz2
It remains to define 1/(dζ/dz).
7.1.3 Transformation of a Circle into and Airfoil
We now consider the problem of generating airfoil-like profiles by choosing a circleand transformation function. The transformation from a circle to an airfoil may beexpressed as
ζ(z) = z +∞∑n=1
Cn
zn(7.8)
Such a function meets the requirements at infinity: that the point at infinity in theζ-plane goes into the point at infinity in the z-plane, and that the complex velocityat infinity be mapped into itself.
282
Derivative of the transformation (7.8) is
dζ
dz= 1− C1
z2− 2
C2
z3− 3
C3
z4− · · · (7.9)
This becomes infinite at the point z = 0, that is, at the origin of the coordinates. Itmay become zero at a number of points. Let us say that there are k such points andlet us denote them by z1, z2,. . . ,zk. These are then the solutions of the equation(
1− z1z
)(1− z2
z
)· · ·
(1− zk
z
)=dζ
dz= 0
In the expansion of the left-hand side of this equation, the coefficient of the 1/zterm is
k∑i=1
zi
This must equal the coefficient of 1/z in the expansion given in (7.9). This lattercoefficient is, however, zero. Therefore we should have
k∑i=1
zi = 0 (7.10)
This means that the centroid of the critical points of the transformation is the originof coordinates in the z-plane.We now have sufficient information to choose the circle and construct the transfor-mation. The procedure is as follows:
1. Choose k points in the z-plane to be the zeros of dζ/dz. We denote thesezeroes, as done above, by the indices z1, z2, . . . , zk.
2. The transformation is then given by
dζ
dz=(1− z1
z
)(1− z2
z
)· · ·
(1− zk
z
)(7.11)
The integral of this equation may be expressed as
ζ(z) = z +C1
z+C2
z2+ · · ·+ Ck
zk(7.12)
where the complex coefficients C1, C2, . . . , Ck are determined by z1, z2, . . . , zk.
3. Choose the centroid of the zeros of dζ/dz as the origin O of the coordinates.
4. Choose one of these zero points to be the trailing-edge point, and denote it byzT .
5. Draw the circle passing through zT which enclose all other zeros of dζ/dz.Denote the center of the circle by zc. The radius of the circle is then given by
a = |zT − zc|6. Choose the line O to zT as the real axis x. The imaginary y-axis is automati-
cally fixed. Denoting by −β the argument of zT − zc we haveae−ıβ = zT − zc
So we have the parameter to fix the value of the circulation.
283
7.1.4 The Joukowsky Transformation
We now apply preceding ideas to obtain the simplest transformation. For this pur-pose we look for a function ζ(z) whose derivatives has only two zeros. Denotingthem by z1 and z2 we should have, according to (7.10),
z1 + z2 = 0
orz2 = −z1
Now, one of these points should be the trailing-edge point. Let us set
z1 = zT (7.13)
Then it follows thatz2 = −zT (7.14)
Equation (7.11) now takes the form
dζ
dz=
(1− zT
z
)(1 +
zTz
)= 1− z2T
z2(7.15)
Integrating with respect to z we get the transformation
ζ(z) = z +z2Tz
(7.16)
Choosing the real axis along the vector zT we set
zT = C (7.17)
where C is real positive number. The transformation (7.16) then takes the form
ζ(z) = z +C2
z(7.18)
This simple transformation is known as Joukowsky transformation.The points z = −C and z = C map into the points ζ = −2C and ζ = 2C re-spectively. Consider the circle described by z = Ceıθ, substituting into (7.16) weget
ζ = Ceıθ + Ce−ıθ = 2C cos θ
Thus, the points on the circumference of this circle go into points on the real axisalong the strip −2C ≤ ζ ≤ 2C. Figure (7.1) sows the circles and the result ofmapping. Let us now choose circle of the larger radius then C in z-plane, then inthe ζ-plane we have
ζ =
(r +
C2
r
)cos θ + ı
(r − C2
r
)sin θ
284
circle circle
Figure 7.1: Mapping of the circles to the elliptic shapes.
If we denote ξ to be the real part of ξ = (ζ), and η to be imaginary part of η = (ζ)than we can eliminate θ by squaring and adding these terms to get
ξ2(r + C2
r
)2 + η2(r − C2
r
)2 = 1
what represents ellipse in the ζ-plane.We could now imagine that circle which simultaneously touch from outer side circle|z| = C, and from inside circle |z| = ro, where ro > C, must have common pointwith these circles also in the ζ-plane, and that rest of points must lay between theimages of these circles. This is clarified in the following subsection.
Joukowsky Airfoils
We now apply Joukowsky transformation to a circle. We choose the circle that passthrough point zT = C, and encloses the point z = −zT = −C. If, as before, zcdenotes the center of the circle and a is its radius, we have
zc = C − ae−ıβ = meıδ (7.19)
The quantities m and β may be taken as parameters defining the circle. Generallym and β are chosen small.To calculate velocity distribution around airfoil contour we need derivative (7.15)and expression for velocity around circle (7.2). As we said, exception to this is thetrailing edge point for which we need to apply L’Hopital’s rule
W =d2Fdz2
d2ζdz2
=2(U∞ + ıV∞) a2
(z−zc)3− Γ
2πı1
(z−zc)2
2z2T
z3
(7.20)
285
So, we have all components of equation (7.6) ready. Pressure coefficient can beobtained from incompressible form of Bernouli equation
CP = 1− |W |2U2∞ + V 2∞
(7.21)
Example 7.1:Write program to generate Joukowsky airfoil coordinates and to calculate pressure co-efficient data. Suppose that circle pass through point C = (1, 0) and that this pointshould be mapped to the trailing-edge. Take the radius of the circle to be a = 1.08,and β = 5. Let U∞ = 1.0 and V∞ = 0.
Solution 7.1:First we have to determine circulation so that velocity at the trailing edge is finite,equation (7.3). Than we have to distribute NP points on the circle to be mapped tothe airfoil
ϕi =2π(i− 1)
NP − 1− β , i = 1, 2, . . . , NP
andzi = zc + ae
ıϕi
The first and the last point of the circle fall on the point to be mapped on the trailing-edge. Then, apply mapping of the circle to the airfoil-like shape
ζi = zi +C2
zi, i = 1, 2, . . . , NP
and calculate modulus of the mapping
dζ
dz= 1− C2
z2i, i = 1, 2, . . . , NP
where C = 1 for our example. Now, we have to calculate velocity Wi around the circleby using equation (7.2) for each of the points zi. Velocity at the airfoil-like shape isobtained form Wi when we divide it by modulus of transformation. Pressure coefficientis obtained from (7.21). Everything said so far is realized in the short program givenbelow.
! File : jouk.for! Zlako Petrovic, 5-Feb-2001, Baghdad! Program calculates pressure distribution around Joukowski airfoil! =============================================================
program JOUKparameter (Pi=3.14159265, TwoPi=2*Pi)complex zc, z, zeta, zT, W, dzetadz, CircW, AirfWreal x(200), y(200), xi(200), eta(200), CP(200)
! Specify parametersCall InputData(Uinf, Vinf, a, zT, zc, beta, NP)dfi = TwoPi/(NP-1)C = real(zT)
! Calculate circulation for
286
! stagnating point at zTalpha = atan2(Vinf,Uinf)UV = sqrt(Uinf**2+Vinf**2)Gamma = -4*Pi*a*UV*sin(alpha+beta)ximax = 0.0ximin = 0.0
!do i = 1, NP
! Points on the circlefi = dfi*(i-1) - betaz = zc + a*exp(cmplx(0.0,fi))
! Calculate airfoil coordinates! and modulus of transformation
Call Joukowsky(C, z, zeta, dzetadz)! Velocity around circle
W = CircW(Uinf, Vinf, a, zc, Gamma, z)! Velocity around airfoil
W = AirfW(W, Uinf, Vinf, a, z, zc, Gamma, C, dzetadz)! Pressure coefficient
CP(i) = 1 - abs(W)**2/(Uinf**2+Vinf**2)! Remember coordinates
x(i) = real(z)y(i) = aimag(z)xi(i) = real(zeta)eta(i) = aimag(zeta)
! Needed for chord lengthximax = max(ximax,xi(i))ximin = min(ximin,xi(i))
end do! Chord length
b = ximax-ximin! Calculate CL, dCL/d Alpha
CL = 8*Pi*a/b*sin(alpha+beta)dCLda = 8*Pi*a*cos(alpha+beta)/bprint 101, alpha*180./Pi,CL,dCLda
! Write data to a fileCall Output(x, y, xi, eta, CP, NP)stop ’JOUK - End of run!’
101 format(’ alpha = ’,F5.1/’ CL = ’,F5.2/’ dCL/da = ’,F5.2//)end
! *************************************************************complex function AirfW(W,Uinf,Vinf, a, z, zc, Gamma, C, dzetadz)parameter (Pi=3.14159265, TwoPi=2*Pi)complex z, zc, W, dzetadz, WWif(abs(dzetadz) .lt. 0.001) then
! L’Hopital’s ruleWW = 2*cmplx(Uinf,Vinf)*a**2/(z-zc)**3WW = WW + cmplx(0.0,Gamma/TwoPi)/(z-zc)**2WW = WW/(2*C**2/z**3)
elseWW = W/dzetadz
end ifAirfW = WWreturnend
! *************************************************************
287
complex function CircW(Uinf, Vinf, a, zc, Gamma, z)parameter (Pi=3.14159265, TwoPi=2*Pi)complex z, zcCircW = cmplx(Uinf,-Vinf) - cmplx(Uinf,Vinf)*a**2/(z-zc)**2
& - cmplx(0.0,Gamma/TwoPi)/(z-zc)returnend
! *************************************************************subroutine Joukowsky(C, z, zeta, dzetadz)complex z, zeta, dzetadz
! Points on the airfoilzeta = z + C**2/z
! Modulus of transformationdzetadz = 1. - (C/z)**2returnend
! *************************************************************subroutine Output(x, y, xi, eta, CP, NP)real x(NP), y(NP), xi(NP), eta(NP), CP(NP)open(unit=1,file=’JOUK.DAT’,form=’formatted’,status=’unknown’)write(1,101) NP
101 format( ’ variables = "xi", "eta", "Cp", "x", "y" ’/& ’ zone i =’,I3)write(1,111) (xi(i),eta(i),CP(i),x(i),y(i),i=1,NP)
111 format(5F12.3)close(unit=1)returnend
! *************************************************************subroutine InputData(Uinf, Vinf, a, zT, zc, beta, NP)complex zT, zcparameter (Pi=3.14159265, toRad=Pi/180.)zT = (1.0,0) ! Pozition of the trailing-edgebeta = 5. * toRad ! Small angle in radiansa = 1.08*zT ! Radius of the circlezc=zT-a*exp(cmplx(0.0,-beta)) ! Center of the circleV = 1.0 ! unidisturbed Velocity
! Number of points around airfoilprint 101
101 format(’ Number of points around airfoil : ’,$)read *, NP
! Angle of attackprint 111
111 format(’ Angle of attack : ’,$)read *, alphaalpha = alpha*toRad
! Find undisturbed componentsUinf = V*cos(alpha)Vinf = V*sin(alpha)returnend
! *************************************************************
Pressure distribution for the data defined in the program for α = 4 are given in thefigure (7.2).
288
-2 -1 0 1 2xi
-2
-1.5
-1
-0.5
0
0.5
1
Cp
Joukowsky Airfoil
α=4
Figure 7.2: Pressure coefficient distribution around Joukowsky airfoil.
7.1.5 The Karman–Trefftz Airfoils
The Joukowsky family and the Mises family of airfoils all have cusped trailing edges.Naturally it is not possible to construct such edges in practice. Hence one seeks atransformation that will generate from a circle an airfoil profile whose trailing-edgeangle τ is not zero. A simple transformation of this kind is set up as follows.We had seen that if z = zo is a critical point (specially zero of the first derivative) ofthe transformation ζ = ζ(z), then we may express ζ(z) in a series expansion aboutthe point zo and write
ζ(z)− ζ(zo) = (z − zo)nf(z) (7.22)
where n is the order of the first nonzero derivative of ζ(z) at zo, and f(z) and itsfirst derivative does not become zero or infinite at zo. At zo angles are not preservedbut are multiplied by the factor n.In application to the airfoil problem, the trailing edge point z = zT is a zero ofdζ/dz. Hence the transformation may be expressed in the form
ζ(z)− ζ(zT ) = (z − zT )nf(z) (7.23)
Now, if we choose only the zeros for dζ/dz, as done in the case of the Joukowskytransformation, the other zero is at the point
z = −zTbecause the origin of coordinates is to be at the centroid of the zeros. Then thetransformation may also be expressed as
ζ(z)− ζ(−zT ) = (z + zT )nf(z) (7.24)
Dividing corresponding sides of equations (7.23) and (7.24) we obtain
ζ(z)− ζ(zT )ζ(z) + ζ(−zT ) =
(z − zT )n(z + zT )n
289
Setting ζT = ζ(zT ) and −ζT = ζ(−zT ), we gaveζ(z)− ζTζ(z) + ζT
=(z − zT )n(z + zT )n
(7.25)
This equation can be also expressed in the form
ζ =ζTnzT
z +n2 − 1
3
z2Tz+ · · · (7.26)
We require that the transformation should be such that for z →∞, dζ/dz = 1. Tosatisfy this requirement, from (7.26), we should set
ζT = nzT (7.27)
Then the transformation takes the form
ζ = z +n2 − 1
3
z2Tz+ · · ·
or the formζ − nzTζ + nzT
=(z − zT )n(z + zT )n
(7.28)
Finally, if we set zT = C, where C is a real positive number, we may write thisequation as
ζ − nCζ + nC
=(z − C)n(z + C)n
(7.29)
This is the simplest transformation that will produce a sharp trailing edge with afinite angle. The trailing-edge angle is given by
τ = π(2− n)If the trailing edge is required to be cusp, we set τ = 0, whence equation (7.29)reduces to
ζ − 2C
ζ + 2C=
(z − C)2(z + C)2
(7.30)
what is exactly Joukowsky transformation we already studied. So equation (7.29)is the simplest extension of Joukowsky transformation which generates airfoil-likeshapes with finite trailing-edge angles. The airfoils generated by it are known as theKarman–Trefftz family of airfoils.Performing logarithm on both sides of equation (7.29) and differentiating we canexpress modulus of transformation in the form
dζ
dz=ζ2 − n2C2
z2 − C2(7.31)
We also express ζ in terms of z from equation (7.29)
ζ = nC1 +
(z−Cz+C
)n1−
(z−Cz+C
)n (7.32)
290
Example 7.2:Write program to generate Karman–Trefftz airfoil coordinates and to calculate pressurecoefficient data. Suppose that circle pass through point C = (1, 0) and that this pointshould be mapped to the trailing-edge. Take the radius of the circle to be a = 1.08,and β = 5. Let V∞ = 1.0 is magnitude of undisturbed velocity and α = 4 deg be theangle of attack. Specify parameter n = 1.97, so τ = π · 0.03.Solution 7.2:The solution procedure closely follows one given for Joukowsky airfoils. We changedlittle bit the the input data. Instead of components of undisturbed velocity we readmagnitude of it and angle of attack, then u∞ = V∞ cos α and v∞ = V∞ sin α.The only essential difference is the subroutine which calculates airfoil-like coordinates andmodulus of transformation. In this program that is done in the subroutine KarmanTrefftz,by employing equations (7.31) and (7.32). The following listing is possible solution ofour problem.
! File : Treftz.for! Zlako Petrovic, 6-Feb-2001, Baghdad! Program calculates pressure distribution around Karman-Trefftz airfoil! =================================================================
program TREFFTZparameter (Pi=3.14159265, TwoPi=2*Pi)complex zc, z, zeta, zT, W, dzetadz, AirfW, CircWreal x(200), y(200), xi(200), eta(200), CP(200), n
! Specify parametersCall InputData(V, alpha, a, n, zT, zc, beta, NP)dfi = TwoPi/(NP-1)C = real(zT)
! Calculate circulation for! stagnating point at zT
Gamma = -4*Pi*a*V*sin(alpha+beta)Uinf = V*cos(alpha)Vinf = V*sin(alpha)ximax = 0.0ximin = 0.0
!do i = 1, NP
! Points on the circlefi = dfi*(i-1) - betaz = zc + a*exp(cmplx(0.0,fi))
! Points on the airfoilCall KarmanTrefftz(n, C, z, zeta, dzetadz)
! Velocity around circleW = CircW(Uinf, Vinf, a, zc, Gamma, z)
! Velocity around airfoilW = AirfW(W, Uinf, Vinf, a, z, zc, Gamma, C, dzetadz)
! Pressure coefficientCP(i) = 1 - abs(W)**2/(Uinf**2+Vinf**2)
! Remember coordinatesx(i) = real(z)y(i) = aimag(z)xi(i) = real(zeta)
291
eta(i) = aimag(zeta)! Needed for chord length
ximax = max(ximax,xi(i))ximin = min(ximin,xi(i))
end do! Chord length
b = ximax-ximin! Calculate CL, dCL/d Alpha
CL = 8*Pi*a/b*sin(alpha+beta)dCLda = 8*Pi*a*cos(alpha+beta)/bprint 101, alpha*180./Pi,CL,dCLda
! Write data to a fileCall Output(x, y, xi, eta, CP, NP)stop ’Treftz - End of run!’
101 format(’ alpha = ’,F5.1/’ CL = ’,F5.2/’ dCL/da = ’,F5.2//)end
! *************************************************************complex function AirfW(W,Uinf,Vinf, a, z, zc, Gamma, C, dzetadz)parameter (Pi=3.14159265, TwoPi=2*Pi)complex z, zc, W, dzetadz, WWif(abs(dzetadz) .lt. 0.001) then
! L’Hopital’s ruleWW = 2*cmplx(Uinf,Vinf)*a**2/(z-zc)**3WW = WW + cmplx(0.0,Gamma/TwoPi)/(z-zc)**2WW = WW/(2*C**2/z**3)
elseWW = W/dzetadz
end ifAirfW = WWreturnend
! *************************************************************complex function CircW(Uinf, Vinf, a, zc, Gamma, z)parameter (Pi=3.14159265, TwoPi=2*Pi)complex z, zcCircW = cmplx(Uinf,-Vinf) - cmplx(Uinf,Vinf)*a**2/(z-zc)**2
& - cmplx(0.0,Gamma/TwoPi)/(z-zc)returnend
! *************************************************************subroutine Output(x, y, xi, eta, CP, NP)real x(NP), y(NP), xi(NP), eta(NP), CP(NP)open(unit=1,file=’TREF.DAT’,form=’formatted’,status=’unknown’)write(1,101) NP
101 format( ’ variables = "xi", "eta", "Cp", "x", "y" ’/& ’ zone i =’,I3)write(1,111) (xi(i),eta(i),CP(i),x(i),y(i),i=1,NP)
111 format(5F12.3)close(unit=1)returnend
! *************************************************************subroutine InputData(V, alpha, a, n, zT, zc, beta, NP)complex zT, zcreal nparameter (Pi=3.14159265, toRad=Pi/180.)
292
zT = (1.0,0) ! Pozition of the trailing-edgebeta = 5. * toRad ! Small angle in radiansa = 1.08*zT ! Radius of the circlezc=zT-a*exp(cmplx(0.0,-beta)) ! Center of the circleV = 1.0 ! Uniform velocitytau = 5. ! Trailing-edge anglen = 2. - tau/180. ! Mapping parameter -- f(tau)
! Number of points around airfoilprint 100
100 format(’ Number of points around airfoil : ’,$)read *, NPprint 110
110 format(’ Angle of attack : ’,$)read *, alphaalpha = alpha*toRad ! Angle of attackreturnend
! *************************************************************subroutine KarmanTrefftz(n, C, z, zeta, dzetadz)complex z, zeta, dzetadz, zzreal n, Czz = ((z-C)/(z+C))**n
! Airfoil coordinatezeta = n*C*(1. + zz)/(1. - zz)
! Modulus of transformationdzetadz = (zeta**2 - (n*C)**2)/(z**2-C**2)returnend
The shape of the Karman-Trefftz airfoil for the input data defined in the program isshown in the figure (7.3).
Figure 7.3: Karman–Trefftz airfoil.
7.1.6 The Point-Vortex Solution
Let us now summarize the procedure of calculation basic airfoil parameters whenthe point vortices are used to simulate flow about airfoil, figure (7.4). Note againthat we don’t need to solve Laplace’s equation
∂2φ
∂x2+∂2φ
∂y2= 0
293
since it is already solved, because we are combining singular solutions, and becauseLaplace’s equation is linear one. We only have to worry not to place singularities inthe flow field. Placing singularities inside airfoil contour or on it doesn’t brake thisrequirement.
Figure 7.4: Airfoil contour approximated by point vortices.
The first step is to read input parameters which define airfoil shape: number of pointswhich define that shape, and coordinates of the points (xi, yi), i = 1, 2, . . . , NP .It will be assumed that the points are defined in anti-clockwise direction. We mustissue the warning that the first and the last point of the airfoil must not coincidesince they will then generate the same influence coefficients what will make oursystem of equations singular, since we will have two rows of the system of equationswith identical coefficients. So we must have small gap at the trailing edge. We alsoneed to specify undisturbed velocity V∞ and angle of attack, (α). Thus
u∞ = V∞ cos α , v∞ = V∞ sin α . (7.33)
In the step two we have to define control point coordinates and components of theunit normal to the contour of the airfoil at the control points. We assume that theairfoil shape is approximated by the strait-line segments drawn between (xi, yi) and(xi+1, yi+1), obviously there are one segment less than number of points. Also thereare one control point less then the number of coordinates (xi, yi)
xci =xi + xi+1
2, yci =
yi + yi+12
, i = 1, 2, . . . , NP − 1. (7.34)
To calculate unit normal we need first the unit vector in the direction of the panelsegment
ti =√(xi+1 − xi)2 + (yi+1 − yi)2
and
txi=xi+1 − xi
ti, tyi
=yi+1 − yi
ti, i = 1, 2, . . . , NP − 1
The unit normal is then obtained from the expression
ni = k × t =∣∣∣∣∣∣∣ı k0 0 1txi
tyi0
∣∣∣∣∣∣∣ = −tyiı+ txi
= nxiı+ nyi
(7.35)
for i = 1, 2, . . . , NP − 1.The third step is consisted of determination two matrices ui,j, and vi,j which repre-sents induced velocity components at the control point i due to the unit point vortex
294
at the panel-segment edge j. Obviously there are NP − 1×NP such values, sincethere are one control point less the number of the edges of the strait-line segments. Itis advisable to place induced velocity calculation in one subroutine with parameters(xci, yci, xj, yj, ui,j, vi,j), where (xci, yci)are coordinates of the control point (xj, yj)are the coordinate of the point vortex placed at the j-th panel edge, and ui,j, vi,jare induced velocity components. The induced velocity components due to the unitvortex are calculated as follows
r2i,j = (xci − xj)2 + (yci − yj)2
ui,j =yci − yjr2i,j
(7.36)
vi,j = −xci − xjr2i,j
, i = 1, 2, . . . , NP − 1 , j = 1, 2, . . . , NP.
The fourth step is consisted of generation Aerodynamic influence matrix and right-hand side of the system of equations. Since any combination of the point vorticeswill satisfy Laplace’s equation the system of equations is build from the boundaryconditions of the problem
Vi · ni = 0 (7.37)
where
Vi = V∞ +NP∑j=1
(ui,jı+ vi,j)Γj , i = 1, 2, . . . , NP − 1 (7.38)
where Γj are unknown values of the point vortex strength at point (xj, yj) andui,j and vi,j are induced velocities (7.36) at the control point (xci, yci), due to unitstrength vortex placed at (xj, yj). If we substitute this equation into (7.37) andsubstitute (7.35) instead of ni we get
NP∑j=1
(ui,jnxi+ vi,jnyi
)Γj = −u∞nxi− v∞nyi
, i = 1, 2, . . . , NP − 1 (7.39)
what represents the system of NP − 1 equations for NP unknowns. Let us denote
ai,j = ui,jnxi+ vi,jnyi
, and: bi = −u∞nxi− v∞nyi
then we have generated so far everything, but last row of the system of equationsfor the unknown strengths of the point vortices
a1,1 a1,2 · · · a1,NP
a2,1 a2,2 · · · a2,NP...
......
...aNP−1,1 aNP−1,2 · · · aNP−1,NP
0 0 · · · 0
Γ1Γ2...
ΓNP−1ΓNP
=
−u∞nx1 − v∞ny1
−u∞nx2 − v∞ny2
...−u∞nxNP−1
− v∞nyNP−1
0
The last row of the equation system is filled from the Kutta-Joukowsky condition.Numerical equivalent of the Kutta-Joukowsky condition is given as
Γ1 + ΓNP = 0 (7.40)
295
which states that the vortex strength at the trailing edge should be zero. Thus, ourcomplete system of equations looks
a1,1 a1,2 · · · a1,NP−1 a1,NP
a2,1 a2,2 · · · a2,NP−1 a2,NP...
......
......
aNP−1,1 aNP−1,2 · · · aNP−1,NP−1 aNP−1,NP
1 0 · · · 0 1
Γ1Γ2...
ΓNP−1ΓNP
=
b1b2...
bNP−10
(7.41)
Next step (fifth) is to solve the system of equations (7.41). We use library routinesfor this purpose so we will not devote space for this step.The sixth step is to calculate tangential velocities at the control points, for thenow known point vortex strengths. Since we have found these strengths from thecondition of zero-th normal component of the velocities at the control points, weneed to find total velocities which are at the same time tangential components ofvelocity at the control points. Thus
Vi = V∞ +NP∑j=1
(ui,jı+ vi,j)Γj , i = 1, 2, . . . , NP − 1
or
Ui = u∞ +N∑j=1
ui,jΓj ,
Vi = v∞ +N∑j=1
vi,jΓj , i = 1, 2, . . . , NP − 1.
(7.42)
The pressure coefficient is determined from
CPi= 1− U2
i + V2i
V 2∞, i = 1, 2, . . . , NP − 1. (7.43)
The last step (seventh) is to calculate remaining aerodynamic parameters by inte-grating pressure distribution. We show alternative approach for the lift coefficientdetermination. Total circulation about airfoil is obtained after summing all vortexstrengths
Γ =NP∑j=1
Γj (7.44)
From Kutta-Joukowsky theorem we have
1/2V2∞SCL = V∞Γ
or
CL =2Γ
V∞(7.45)
where characteristic area S is replaced by ·1, and is the chord length of the airfoil.It remains only to plot the obtained results.
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Example 7.3:Write the program which will be capable to calculate aerodynamic parameters for thearbitrary airfoil in the incompressible inviscid flow.
Solution 7.3:The program which follows fulfill all what is said about solution procedure in this sub-section.
! File: PMV.FOR! Z. Petrovic, 7-Feb-2001, Baghdad! Point vortex simulation of the flow about airfoil! ****************************************************************
program PMVparameter (MP=60,MM=MP*MP)real x(MP), y(MP) ! Panel edge coordinatesreal xc(MP), yc(MP) ! Control point coordinatesreal nx(MP), ny(MP) ! Unit normal componentsreal CP(MP) ! Pressure coefficientreal uij(MM), vij(MM) ! Induced velocity componentsreal A(MM), b(MP) ! Influence matrix and RHSreal Gamma(MP) ! Vortex strengths
!equivalence (Gamma(1), b(1))
! This is necessary becausecommon /c1/ uij ! of the bad compilercommon /c2/ vijcommon /c3/ a
!Call Spec_Input(NP, x, y, alpha, VV)
! Undisturbed componentsUinf = VV*cos(alpha)Vinf = VV*sin(alpha)
! Control points and nx, nyCall Define_Geometry(NP, x, y, xc, yc, nx, ny, bb)
! Induced velocity componentsCall Induced_Vel(NP, x, y, xc, yc, uij, vij)
! Aerodynamic matrix and RHSCall Aero(NP, uij, vij, nx, ny, Uinf, Vinf, A, b)
! Solve system of equationsCall SIMQ(A,Gamma,NP,KS)
! Is solution O.K.if(KS .ne. 0) then
stop ’ *** Singular matrix *** !’end if
! Calculate CPCall CP_Calculation(NP, uij,vij, x,y, bb,Gamma, Uinf,Vinf, CP,CL)
! Write data for plottingCall Output(NP, xc, CP, Gamma)print 101, alpha*180/acos(-1.0), CL
101 format(’ alpha = ’,F10.3/’ CL = ’,F10.3//)stop ’PMV - End of run !’end
! ****************************************************************subroutine Spec_Input(NP, x, y, alpha, V)parameter (Pi=3.14159265, ToRad=Pi/180.)
297
real x(*), y(*)character FileName*20
!alpha = 4.*ToRad ! Angle of attackV = 1.0 ! Undisturbed Velocity
!print 101
101 format(’ File with airfoil coordinates : ’,$)read ’(A)’, FileNameopen(unit=1,file=FileName,form=’formatted’,status=’old’)read(1,111) NNNP = NN - 2 ! Trailing edge must NOT be
111 format(/9x,I3) ! closed soread(1,’(1x)’) ! Skip first pointread(1,121) (x(i),y(i),i=1,NP)
121 format(2F12.3)close(unit=1)returnend
! ****************************************************************subroutine Define_Geometry(NP, x, y, xc, yc, nx, ny, b)real x(NP), y(NP), xc(NP), yc(NP), nx(NP), ny(NP)xmax = x(1)xmin = x(NP)do i = 1, NP-1
! Control point coordinatesxc(i) = (x(i) + x(i+1))/2yc(i) = (y(i) + y(i+1))/2xmin = min(xmin, x(i))
! Unit normal componentstx = x(i+1) - x(i)ty = y(i+1) - y(i)t = sqrt(tx**2 + ty**2)nx(i) = -ty / tny(i) = tx / t
end do! Chord length
b = xmax-xminreturnend
! ****************************************************************subroutine Induced_Vel(NP, x, y, xc, yc, uij, vij)real x(NP), y(NP), xc(NP), yc(NP), uij(NP,NP), vij(NP,NP)
! For each control pointdo i = 1, NP-1
! From each vortexdo j = 1, NP
Call Vortex(xc(i), yc(i), x(j), y(j), u, v)uij(i,j) = uvij(i,j) = v
end doend doreturnend
! ****************************************************************subroutine Vortex(xc, yc, x, y, u, v)
298
parameter (Pi=3.14159265, TwoPi=2*Pi)xx = xc - xyy = yc - yrr = xx*xx + yy*yyu = yy /(TwoPi * rr)v = -xx /(TwoPi * rr)returnend
! ****************************************************************subroutine Aero(NP, uij, vij, nx, ny, Uinf, Vinf, A, b)real A(NP,NP), b(NP), uij(NP,NP), vij(NP,NP), nx(NP), ny(NP)do i = 1, NP-1
b(i) = -Uinf*nx(i) - Vinf*ny(i)do j = 1, NP
A(i,j) = uij(i,j)*nx(i) + vij(i,j)*ny(i)end do
end do! Kutta-Joukowsky condition
A(NP,1) = 1.0A(NP,NP) = 1.0b(NP) = 0.0do j = 2, NP-1
A(NP,j) = 0.0end doreturnend
! ****************************************************************subroutine CP_Calculation(NP,uij,vij,x,y,b,Gamma,Uinf,Vinf,CP,CL)real uij(NP,NP), vij(NP,NP), Gamma(NP), CP(NP), x(NP), y(NP)VV = Uinf**2 + Vinf**2
! Kutta-Condition if the! airfoil is closed
Gamma(1) = 0.0Gamma(NP) = 0.0Gam = 0.0do i = 1, NP-1
u = Uinfv = Vinfdo j = 1, NP
u = u + uij(i,j)*Gamma(j)v = v + vij(i,j)*Gamma(j)
end dot = sqrt((x(i+1)-x(i))**2 + (y(i+1)-y(i))**2)gh = (Gamma(i)+Gamma(i+1))/(4*t)ug = gh*(x(i+1)-x(i))/tvg = gh*(y(i+1)-y(i))/tu = u - ugv = v - vgCP(i) = 1. - (u**2 + v**2)/VV
! Total circulationGam = Gam + Gamma(i)
end doCL = 2*Gam/(b*sqrt(VV))returnend
! ****************************************************************
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subroutine Output(NP, xc, CP, Gamma)real xc(NP), CP(NP), Gamma(NP)open(unit=1,file=’PMV.dat’,form=’formatted’,status=’unknown’)write(1,101) NP-1
! Tecplot header101 format(’ variables = "x", "Cp", "G"’/’ zone i= ’,I3)
write(1,111) (xc(i), CP(i), Gamma(i), i = 1, NP)close(unit=1)
111 format(3F12.3)returnend
! ****************************************************************include ’simq.for’
7.1.7 Solution using Constant-Strength Vortex Distribution
We must issue immediate warning that this distribution leads to ill conditionedmatrix, so it is not of practical use, but we develop here procedure as a logicalextension of the point vortex solution procedure. This development will be a niceintroduction to linearly-varying vorticity distribution solution. Comparing this pro-cedure to the previous one we find a lot of common points. There are two seriousweakness of this method why it is unusable for applications:
1. The impossibility to specify Kutta-Joukowsky condition. Since constant vor-ticity distribution is unknown of the panel we have the same number of un-knowns as we have the number of control points. There are no room foradditional condition to fix the value of circulation at the trailing-edge.
2. The property of the constant vorticity distribution which does not inducevelocity in normal direction when the control point is in the middle of thepanel. The consequence is that the generated aerodynamic influence matrixhave all zeros along main diagonal. This property disqualify the matrix for thesolution of the large system of equations. Some usable results can be obtainedif we move control point from the middle of panel to some side, but in thisway we can obtain only the solution for symmetric airfoils.
Let us review the necessary steps for this method.Step one is exactly the same as one we gave in previous subsection. We need to readthe number of points (NP ) by which the contour is defined, the coordinates of thelifting contour, intensity of the undisturbed velocity, and angle of attack. Never-theless, this step is greatly revised in the following program to allow determinationof the pressure distributions around airfoils of the NACA series, besides ones wegenerated analytically. After competition of this step we will have:
1. N – the number of points by which airfoil is defined.
2. (Xi, Yi) , i = 1, 2, . . . , N – coordinates of the airfoil given from the uppertrailing edge in the anti-clockwise direction. Capital letters denote coordinatesin the global coordinate system.
300
3. α – Angle of attack.
4. V∞ – Velocity of the uniform flow field.
In step two we calculate control point coordinates and components of the unit normalat the control point in the exact way we done in the previous section. Actually weuse the same subroutine.Step three is greatly revised. The reason is that we have expressions for inducedvelocities when the panel coincides with x-axis. Since in the practice this is rarelythe case we need coordinate transformation to place the panel and control point inthe local coordinate system. Than we need to calculate induced velocities parallelto local coordinates, and finally to determine the induced velocity components inthe global coordinate system. Expressions for induced velocities, in local coordinatesystem, are given by equations (5.70) for arbitrary position of the control point(x, y), and with equation (5.71) for the case when control point is at the x-axis suchthat x ∈ (x1, x2), we repeat these expressions for reference
u =1
2π(θ2 − θ1)
v =1
4πlnr22r21
(7.46)
where γ is the strength of vorticity distribution, and
θ1 = arctany
x− x1 , θ2 = arctany
x− x2and
r21 = (x− x1)2 + y2 , r22 = (x− x2)2 + y2 .where (x, y), x1, and x2 are the coordinates in the local coordinate system. For thecase when y = ±0, and at the same time x1 < x < x2 the former expressions reducesto
u(x,±0) = ±1
2
v(x,±0) =1
2πln∣∣∣∣x− x2x− x1
∣∣∣∣ (7.47)
In the figure (7.5) we show the first sub-step – the transformation of coordinatesbetween global and local coordinate system. The x-axis of the local coordinatesystem is directed form the point (Xi, Yi) to the point (Xi+1, Yi+1). With capitalsymbols we denote coordinates defined in global coordinate system. Let us expressthe unit vector of this axis in the terms of global coordinate system unit vectors.Let
t = (Xi+1 −Xi)ı+ (Yi+1 − Yi)be the vector along x-axis and t is its length
t =√(Xi+1 −Xi)2 + (Xi+1 −Xi)2 , tx = Xi+1 −Xi , ty = Yi+1 − Yi
301
Figure 7.5: Sketch for the transformation between coordinate systems.
thenı ′ = τxı+ τy (7.48)
is the unit vector along x-axis, where τ = t/t, and τx = tx/t and τy = ty/t arethe components of unit vector along x-axis. Components of the unit vector ′ aredetermined by
′ = k ×ı ′ =∣∣∣∣∣∣∣ı k0 0 1τx τy 0
∣∣∣∣∣∣∣ = −τyı+ τx (7.49)
Let us return to our coordinate transformation. With R = Xı + Y we denote theposition of the control point in the global coordinate system, with r = xı ′+ y ′ theposition of the same point in the local coordinate system, and with Ro = Xiı+ Yithe position of the coordinate origin. Note that we place coordinate origin of thelocal coordinate system to coincide with the first point of the panel, so x1 = 0 thex2 coordinate is simply the distance between points (Xi, Yi) and (Yi+1, Yi+1)
x2 =√(Xi+1 −Xi)2 + (Yi+1 − Yi)2 ≡ t
Since the control point is defined in two ways it must be
R = Ro + r
or in component form
(X −Xi)ı+ (Y − Yi) = xı ′ + y ′ (7.50)
We obtain x coordinate in the local coordinate system when we multiply this equa-tion by ı ′, and y coordinate when we multiply it by ′. Or in matrix form
xy
=
[ı ·ı ′ ·ı ′ı · ′ · ′
]X −Xi
Y − Yi
or when we substitute expressions (7.48) and 7.49 in this equation to getxy
=
[τx τy−τy τx
]X −Xi
Y − Yi
(7.51)
We use x1, x2 and (x, y) coordinates to calculate local velocity components u andv. Transformation of these components in global components is done in reverse way
302
to equation (7.51). We use here property of transformation matrix that its inversematrix is equal to its transpose matrix. Thus,
UV
=
[τx −τyτy τx
]uv
(7.52)
where U and V are induced velocity components in the global coordinate system,and u and v are induced velocities calculated in the local coordinate system.The remaining steps follow exactly the procedure described for the calculation ofthe flow-filed by using point vortices. The only difference is how we could calculatecirculation around airfoil. By definition circulation is given by
Γ =∮C
V · dr
what could be approximated as
Γ = −NP−1∑
1
Vi ·∆t
where ∆t = (Xi+1−Xi)ı+(Yi+1−Yi)ı is the vector along panel side, Vi is the velocityat the control point, minus sign comes from the way we define our contour (anti-clock wise direction) since that numbering is in opposite direction the circulation isdefined. Once circulation is defined we can calculate the lift coefficient as
CL =2Γ
V∞
where, again, is the chord length, V∞ is the free-stream velocity, and Γ is circulationaround airfoil.Note again that the results obtained using this approach are of poor quality. Programwhich follows realize what is said in this subsection.
! File: pmvv.for! Z. Petrovic, 10-Feb-2001, Baghdad! Calulation of theflow about lifting bodies - constant strength vorticity! ================================================================
program PMVVparameter (MM=99, MMM=MM*MM)real x(MM), y(MM) ! Panel edge coordinatesreal xc(MM), yc(MM) ! Control point coordinatesreal A(MMM), b(MM), G(MM) ! Aerodynamic influence mtrixreal nx(MM), ny(MM) ! Unit normals at control ptsreal uij(MMM), vij(MMM) ! Induced velocities due to
! unit strength source distr.equivalence (b(1), G(1))common /c1/ Acommon /c2/ uijcommon /c3/ vij
!Call Input_Data(N, x, y, alpha, VV)Uinf = VV*cos(alpha)Vinf = VV*sin(alpha)
303
! Control point and unit normalCall Geometry(N, x, y, xc, yc, nx, ny)
! Calculate induced velocityCall Induced(N, x, y, xc, yc, uij, vij)
! Influence matrixCall Aero(N, uij, vij, nx, ny, Uinf, Vinf, N-1, A, b)
! Solve the equation systemCall SIMQ(A, G, N-1, Ierr)Call Test(N-1, xc, G)if(Ierr .ne. 0) then
stop ’*** Singular Influence Matrix *** !’end if
! Calculate pressure coeff.Call CP_Calc(N, G, uij, vij, Uinf, Vinf, CP)
! Prepare results for plottingCall Output(xc, CP, N-1)stop ’PMVV - End of run !’end
! ****************************************************************subroutine Test(N, x, y)real x(N), y(N)open(unit=1,file=’test.dat’,form=’formatted’,status=’unknown’)write(1,101) N
101 format(’variables = "x", "y"’/’zone i =’,I3)write(1,105) (x(i),y(i),i=1,N)close(unit=1)
105 format(2F12.3)returnend
! ****************************************************************subroutine CP_Calc(N, G, uij, vij, Uinf, Vinf, CP)real G(N), uij(N,N), vij(N,N), CP(N)VV = Uinf**2 + Vinf**2
! For each control pointdo i = 1, N-1
u = Uinfv = Vinf
! From each paneldo j = 1, N-1
u = u + uij(i,j)*G(j)v = v + vij(i,j)*G(j)
end doCP(i) = 1.0 - (u**2 + v**2)/VV
end doreturnend
! ****************************************************************subroutine Output(x, CP, NA)real x(NA), CP(NA)open(unit=1, file=’PMVV.dat’,form=’formatted’,status=’unknown’)write(1,100) NA
100 format(’ variables = "x", "G"’/’ zone i=’,I3)write(1,110) (x(i),CP(i), i=1,NA)
110 format(2f12.3)close(unit=1)return
304
end! ****************************************************************
subroutine Aero(N, uij, vij, nx, ny, Uinf, Vinf, NA, A, b)real uij(N,N), vij(N,N), nx(N), ny(N), A(NA,NA), b(NA)do i = 1, NA ! For each control point
b(i) = -Uinf*nx(i) -Vinf*ny(i)do j = 1, NA ! For each panel
A(i,j) = uij(i,j)*nx(i) + vij(i,j)*ny(i)end do
end doreturnend
! ****************************************************************subroutine Induced(N, x, y, xc, yc, uij, vij)real x(N), y(N), xc(N), yc(N), uij(N,N), vij(N,N)do i = 1, N-1 ! For each control point
do j = 1, N-1 ! For each panelxf = x(j)yf = y(j)xs = x(j+1)ys = y(j+1)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, uL, vL)! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,j) = u ! future referencevij(i,j) = v
end doend doreturnend
! ****************************************************************subroutine ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)
! Translates local velocity! components (uL,vL) to global
xx = xs - xfyy = ys - yfs = sqrt(xx**2 + yy**2)
! rotation cosinescost = xx/ssint = yy/s
! Global velocity componentsu = uL*cost - vL*sintv = uL*sint + vL*costreturnend
! ****************************************************************subroutine ToLocalCS(xf,yf, xs,ys, xc,yc, x1,x2, x,y)
! Transforms the panel and! contrlo point coordinates! from global to local CooSys
305
xx = xs - xfyy = ys - yfs = sqrt(xx**2 + yy**2)cost = xx/ssint = yy/s
! Panel coordinatesx1 = 0.0x2 = s
! Control point coordinatesx = cost*(xc-xf) + sint*(yc-yf)y = -sint*(xc-xf) + cost*(yc-yf)returnend
! ****************************************************************Subroutine Geometry(N, x, y, xc, yc, nx, ny)real x(N), y(N), xc(N), yc(N), nx(N), ny(N)o = 0.8oo = 1-odo i = 1, N-1
if(i .gt. N/2) o = ooxc(i) = o*x(i) + x(i+1)*(1.-o)yc(i) = o*y(i) + y(i+1)*(1.-o)tx = x(i+1) - x(i)ty = y(i+1) - y(i)t = sqrt(tx**2 + ty**2)nx(i) = -ty/tny(i) = tx/t
end doreturnend
! ****************************************************************subroutine cv2d(x1, x2, x, y, u, v)parameter (Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi)
! Calculates induced velocities! u and v at the point (x,y)! due to constant unit-strength! VORTEX distribution along x-! axis between x1,x2.!
if(abs(y).lt.0.001 .and. (x-x1)*(x-x2).lt.0.0) then! x is between x1 and x2
u = -0.5v = log(abs((x-x2)/(x-x1)))/TwoPi
elseu = (atan2(y,x-x2)-atan2(y,x-x1))/TwoPiv = log(((x-x2)**2+y**2)/((x-x1)**2+y**2))/FourPi
end ifreturnend
! ****************************************************************subroutine Input_Data(N, x, y, alpha, VV)parameter (Pi=3.14159265, ToRad = Pi/180.)character FileName*20real x(*), y(*)
!! Angle of attack and Vinf
306
VV = 1.0alpha = 4.0alpha = alpha*ToRad ! Convert to radians
!print 100
100 format( 10x,’1- NACA airfoil’ /& 10x,’2- Joukowsky or Trefftz airfoil ’///& 10x,’Your choice: ’,$)read *, Ichoiceif(Ichoice .eq. 1) then
! NACA 4 or 5 digit airfoilsprint 131
131 format(’ Number of the points around airfoil : ’,$)read *, NN = N/2*2+1 ! Make it odd numberprint 141
141 format(’ Chord length : ’,$)read *, bprint 151
151 format(’ NACA airfoil : ’,$)read *, NACAdt = Pi/(N/2) ! Angle incrementdo i = 1, N/2
xx = ( 1. + cos(dt*(i-1)) )/2Call NACA45(NACA, xx, xg, yg, YGp, xd, yd, YDp)x(i) = xg*by(i) = yg*bx(N-i+1) = xd*by(N-i+1) = yd*b
end doxx = 0.0Call NACA45(NACA, xx, xg, yg, YGp, xd, yd, YDp)x(N/2+1) = xg*by(N/2+1) = yg*b
else if(Ichoice .eq. 2) then! Trefftz or Joukowsky airfoil! Analytical pressure distri-! bution exists.
print 101101 format(’ File with airfoil coordinates : ’,$)
read ’(A)’, FileNameopen(unit=1, file=FileName, form=’formatted’, status=’old’)read(1,111) N
111 format(/9x,I3)read(1,121) (x(i), y(i), i = 1, N)
121 format(2F12.3)close(unit=1)
elsestop ’*** Wrong choice ***’
end ifreturnend
! ****************************************************************include ’simq.for’include ’naca45.for’
307
This program needs subroutine for the generation of NACA airfoil coordinates. Wegive this subroutine without explanation.
SUBROUTINE NACA45(NACA, X, XG, YG, YGp, XD, YD, YDp)real mt(5), kt(5), k1, m
! Calculates coordinates and slope for the! 4 and 5 digit NACA airfoils
integer NACA ! Designation of NACA airfoil 0012=12real X ! We need data for this coordinatereal XG, YG, YGp ! Data for upper surface of the airfoilreal XD, YD, YDp ! Data for lower surface of the airfoil
!data mt /0.0580, 0.1260, 0.2025, 0.2900, 0.3910/data kt /361.4, 51.64, 15.957, 6.643, 3.23/
im = NACA/1000m = 0.01*im ! Maximum ordinate of the camberline (4)ip = NACA/100 - 10*imp = 0.1*ip ! Position of the maximum camberline (4)if(p.eq.0.0) p = -1.ideb = NACA - 1000*im - 100*ipt = 0.01*ideb ! Relative thickness of the airfoil
yt = t*(1.4845*sqrt(X)-0.63*X-1.758*X**2+1.4215*X**3-0.5075*X**4)!
if(x.gt.1.e-37) thenytp = t*(.74225/sqrt(X) -0.63-3.516*X +4.264*X**2 -2.03*X**3)
elseytp = 1.0E37
end if!
if(im .gt. 9) then ! Five digit airfoilm = mt(im-20)k1 = kt(im-20)if(X .gt. m) then
yc = k1*m**3*(1-X)/6 ! Camberline ordinateycp = -k1*m**3/6 ! Slope of the camberlineycs = 0 ! yc"teta = atan(ycp) ! Angle wich corresponds to
else ! slopeyc = k1*(X**3-3*m*X**2+m**2*(3-m)*X)/6ycp = k1*(3*X**2-6*m*X+m**2*(3-m))/6ycs = k1*(X - m)teta = atan(ycp)
end ifelse ! Four digit airfoil
if(X .le. p) thenyc = m*(2*p*X - X**2) / p**2ycp = 2*m*(p - X) / p**2ycs = -2*m/p**2teta = atan(ycp)
elseyc = m*((1-2*p) + 2*p*X - X**2)/(1-p)**2ycp = 2*m*(p - X)/(1-p)**2ycs = -2*m/(1-p)**2teta = atan(ycp)
end if
308
Figure 7.6: Decomposition of trapesoidal distribution to constant vortex distributionand linearly varying vortex distribution.
end if! Upper surface data
XG = X - yt*sin(teta)YG = yc + yt*cos(teta)YGp = ycp + ytp*cos(teta) - yt*sin(teta)*ycs/(1 + ycp**2)
! Lower surface dataXD = X + yt*sin(teta)YD = yc - yt*cos(teta)YDp = ycp - ytp*cos(teta) + yt*sin(teta)*ycs/(1 + ycp**2)
!returnend
7.1.8 Linearly Varying Vortex Panel Application
Constant strength panel we extend in this subsection to linearly varying vortexpanel. We keep the same overall structure of the solution method and concentrateonly on differences. All the steps mentioned for the constant panel vorticity distri-bution are the same except the step in which we calculate induced velocities. Thatstep require additional extension.We now have panel edge variables instead of panel variables, so there are one un-known more than we have control points. Thus, we need Kutta-Joukowsky conditionto close the equation system. Compared to the constant strength vorticity distri-bution procedure we need the extension of the step where we generate aerodynamicinfluence matrix, but that step is equal to the step described by the point vortexapproximation.Let us now consider what is really new for this panels. Figure (7.6) shows, al-ready explained decomposition of the trapesoidal vorticity distribution to constantvorticity distribution and triangular vorticity distribution.We have considered constant vorticity influence given by equations (7.46) and (7.47).Now we need to extend this to the linearly varying distribution. In figure (7.7) weshow that value at some panel edge i generally belongs to two panels, so to obtaininfluence of this value to the some control point we need to consider both panels.The first panel have at the edge i − 1 value of vorticity 0, and at edge i vorticityis equal to 1 what corresponds to x1 and x2 respectively. For the panel betweenedge i and i+ 1 at the edge x1 we have value of vorticity 1 and at edge x2 we have
309
Figure 7.7: Value of the vorticity at node i is common to two neighboring panels.
zero. This is known the figure (7.8), from which it is clear that we need to performsimilar calculation twice (for each) panel in order to obtained contribution of theunit vorticity at arbitrary edge i. Exception to this are first and last edge. Bothedges belongs only to one panel.
Figure 7.8: Influence of the edge i must be calculated by considering both panelswhich share this edge.
First we consider the case shown in the figure (7.9). Vorticity distribution for thiscase is given by
γ(ξ) =ξ
x2 − x1 −x1
x2 − x1when ξ = x2 vorticity γ(x2) = 1, and when ξ = x1 then γ(x1) = 0. Substitutingthis expression for vorticity strength in the equation for influence velocities we get
u =1
x2 − x1
y
2π
x2∫x1
ξ dξ
(x− ξ)2 + y2− x1
x2 − x1
y
2π
x2∫x1
dξ
(x− ξ)2 + y2 (7.53)
v =1
x2 − x1
−12π
x2∫x1
ξ(x− ξ)dξ(x− ξ)2 + y2
− x1x2 − x1
−12π
x2∫x1
(x− ξ) dξ(x− ξ)2 + y2
(7.54)or
u =1
x2 − x1 u- −x1
x2 − x1 uc
v =1
x2 − x1 v- −x1
x2 − x1 vc(7.55)
where u- and v- represents contribution from the triangular distribution given byequations (5.73) for arbitrary position of the control point, we repeat expressionsfor velocities here for reference
310
Figure 7.9: Left-hand panel influence of the common edge.
u- = − 1
4π
[y ln
r21r22− 2x(θ2 − θ1)
]
v- = − 1
4π
[x
2lnr21r22− (x2 − x1) + y(θ2 − θ1)
] (7.56)
for the case when the control point falls on the x-axis (y = ±0) and x1 < x < x2 wehave
u-(x,±0) = ±γ12x
v-(x,±0) = − γ12π
[x ln
x− x1|x− x2| − (x2 − x1)
](7.57)
expressions for uc and vc are already given for constant vorticity distribution by(7.46) and (7.47).Now it is time to take into account contribution from the second panel. Figure (7.10)shows the vorticity distribution used for the calculation of the contribution of thevorticity of unit strength at the common node. This distribution can be expressedas
γ(ξ) =x2
x2 − x1 −ξ
x2 − x1Substitution of this expression into the integrals for influence velocities we get
u =−1
x2 − x1u- +x2
x2 − x1uc
v =−1
x2 − x1 v- +x2
x2 − x1vc(7.58)
where u-, uc, v-, and vc are defined as before.The remaining steps to complete solution of our airfoil problem are the same alreadydescribed with point vortex, and constant vorticity distribution examples. Programwhich follows is possible realization what is said in this subsection.
! File: pmlv.for! Z. Petrovic, 11-Feb-2001, Baghdad
311
Figure 7.10: Right-hand panel influence of the common edge.
! Calulation of theflow about lifting bodies - linear strength vorticity! ================================================================
program PMLVparameter (MM=99, MMM=MM*MM)real x(MM), y(MM) ! Panel edge coordinatesreal xc(MM), yc(MM) ! Control point coordinatesreal A(MMM), b(MM), G(MM) ! Aerodynamic influence mtrixreal nx(MM), ny(MM) ! Unit normals at control ptsreal uij(MMM), vij(MMM) ! Induced velocities due to
! unit strength source distr.equivalence (b(1), G(1))common /c1/ Acommon /c2/ uijcommon /c3/ vij
!Call Input_Data(N, x, y, alpha, VV)Uinf = VV*cos(alpha)Vinf = VV*sin(alpha)
! Control point and unit normalCall Geometry(N, x, y, xc, yc, nx, ny)
! Calculate induced velocityCall Induced(N, x, y, xc, yc, uij, vij)
! Call WriteTable(vij,N)! Call Test(N,x,y)
! Influence matrixCall Aero(N, uij, vij, nx, ny, Uinf, Vinf, A, b)
! Solve the equation systemCall SIMQ(A, G, N, Ierr)if(Ierr .ne. 0) then
stop ’*** Singular Influence Matrix *** !’end if
! Calculate pressure coeff.Call CP_Calc(N, x, y, G, uij, vij, Uinf, Vinf, CP, CL)
! Prepare results for plottingCall Output(xc, CP, N-1)print *,’CL=’,CLstop ’PMLV - End of run !’end
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! ****************************************************************subroutine LifCalc(N, x, y, G, V, CL)real x(N), y(N), G(N)xmin = x(1)xmax = x(1)GG = 0.0do i = 2, N
t = sqrt( (x(i)-x(i-1))**2 + (y(i)-y(i-1))**2 )GG = GG + (G(i)+G(i-1))*t/2xmin = min(xmin,x(i))xmax = max(xmax,x(i))
end dob = xmax-xminCL = 2*GG/(b*V)returnend
! ****************************************************************subroutine Induced(N, x, y, xc, yc, uij, vij)real x(N), y(N), xc(N), yc(N), uij(N,N), vij(N,N)do i = 1, N-1 ! For each control point
! First edge calculationxf = x(1)yf = y(1)xs = x(2)ys = y(2)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, ucL, vcL)! linear vorticity distribut.
Call lv2d(x1, x2, xx, yy, ulL, vlL)! Calculate local vel. comp.
uL = -ulL/(x2-x1) + x2*ucL/(x2-x1)vL = -vlL/(x2-x1) + x2*vcL/(x2-x1)
! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,1) = u ! future referencevij(i,1) = vdo j = 2, N-1 ! For each edge
! LEFT-HAND PANEL CONTRIBUT.xf = x(j-1)yf = y(j-1)xs = x(j)ys = y(j)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, ucL, vcL)! linear vorticity distribut.
Call lv2d(x1, x2, xx, yy, ulL, vlL)! Calculate local vel. comp.
uL = ulL/(x2-x1) - x1*ucL/(x2-x1)
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vL = vlL/(x2-x1) - x1*vcL/(x2-x1)! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,j) = u ! future referencevij(i,j) = v
! RIGHT-HAND PANEL CONTRIB.xf = x(j)yf = y(j)xs = x(j+1)ys = y(j+1)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, ucL, vcL)! linear vorticity distribut.
Call lv2d(x1, x2, xx, yy, ulL, vlL)! Calculate local vel. comp.
uL = -ulL/(x2-x1) + x2*ucL/(x2-x1)vL = -vlL/(x2-x1) + x2*vcL/(x2-x1)
! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,j) = u + uij(i,j) ! future referencevij(i,j) = v + vij(i,j)
end do! Last Point Calclation
xf = x(N-1) !========================yf = y(N-1)xs = x(N)ys = y(N)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, ucL, vcL)! linear vorticity distribut.
Call lv2d(x1, x2, xx, yy, ulL, vlL)! Calculate local vel. comp.
uL = ulL/(x2-x1) - x1*ucL/(x2-x1)vL = vlL/(x2-x1) - x1*vcL/(x2-x1)
! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,N) = u ! future referencevij(i,N) = v
end doreturnend
! ****************************************************************subroutine cv2d(x1, x2, x, y, u, v)parameter (Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi)
314
! Calculates induced velocities! u and v at the point (x,y)! due to constant unit-strength! VORTEX distribution along x-! axis between x1,x2.!
if(abs(y).lt.0.001 .and. (x-x1)*(x-x2).lt.0.0) then! x is between x1 and x2
u = -0.5v = log(abs((x-x2)/(x-x1)))/TwoPi
elseu = (atan2(y,x-x2)-atan2(y,x-x1))/TwoPiv = log(((x-x2)**2+y**2)/((x-x1)**2+y**2))/FourPi
end ifreturnend
! ****************************************************************subroutine lv2d(x1, x2, x, y, u, v)parameter (Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi)r1=(x-x1)**2 + y**2r2=(x-x2)**2 + y**2t1=atan2(y,x-x1)t2=atan2(y,x-x2)if(abs(y) .lt. 0.0001 .and. (x-x1)*(x-x2).lt.0.0) then
u = -x/2v = -(x*log(r1/r2)-2*(x2-x1))/FourPi
elseu = -(y*log(r1/r2)-2*x*(t2-t1))/FourPiv = -(x*log(r1/r2)/2-(x2-x1)+y*(t2-t1))/TwoPi
end ifreturnend
! ****************************************************************subroutine CP_Calc(N, x, y, G, uij, vij, Uinf, Vinf, CP, CL)real G(N), uij(N,N), vij(N,N), CP(N), x(N), y(N)VV = Uinf**2 + Vinf**2xmin = x(N)xmax = x(N)Gam = 0.0
! For each control pointdo i = 1, N-1
u = Uinfv = Vinfxmax = max(xmax,x(i))xmin = min(xmin,x(i))tx = x(i+1)-x(i)ty = y(i+1)-y(i)
! From each edgedo j = 1, N
u = u + uij(i,j)*G(j)v = v + vij(i,j)*G(j)
end doCP(i) = 1.0 - (u**2 + v**2)/VVGam = Gam - tx*u - ty*v
end dob = xmax-xmin
315
CL= 2*Gam/(b*sqrt(VV))returnend
! ****************************************************************subroutine Output(x, CP, NA)real x(NA), CP(NA)open(unit=1, file=’PMLV.dat’,form=’formatted’,status=’unknown’)write(1,100) NA
100 format(’ variables = "x", "G"’/’ zone i=’,I3)write(1,110) (x(i),CP(i), i=1,NA)
110 format(2f12.3)close(unit=1)returnend
! ****************************************************************subroutine Aero(N, uij, vij, nx, ny, Uinf, Vinf, A, b)real uij(N,N), vij(N,N), nx(N), ny(N), A(N,N), b(N)do i = 1, N-1 ! For each control point
b(i) = -Uinf*nx(i) -Vinf*ny(i)! b(i) = -Uinf*nx(i)/ny(i) -Vinf
do j = 1, N ! For each panel edge! A(i,j) = uij(i,j)*nx(i)/ny(i)+ vij(i,j)
A(i,j) = uij(i,j)*nx(i) + vij(i,j)*ny(i)end do
end do! Kutta-Joukowsky condition
A(N,1) = 1.0A(N,N) = 1.0b(N) = 0.0do j = 2, N-1
A(N,j) = 0.0end doreturnend
! ****************************************************************subroutine ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)
! Translates local velocity! components (uL,vL) to global
xx = xs - xfyy = ys - yfs = sqrt(xx**2 + yy**2)
! rotation cosinescost = xx/ssint = yy/s
! Global velocity componentsu = uL*cost - vL*sintv = uL*sint + vL*costreturnend
! ****************************************************************subroutine ToLocalCS(xf,yf, xs,ys, xc,yc, x1,x2, x,y)
! Transforms the panel and! contrlo point coordinates! from global to local CooSys
xx = xs - xfyy = ys - yf
316
s = sqrt(xx**2 + yy**2)cost = xx/ssint = yy/s
! Panel coordinatesx1 = 0.0x2 = s
! Control point coordinatesx = cost*(xc-xf) + sint*(yc-yf)y = -sint*(xc-xf) + cost*(yc-yf)returnend
! ****************************************************************Subroutine Geometry(N, x, y, xc, yc, nx, ny)real x(N), y(N), xc(N), yc(N), nx(N), ny(N)o = 0.5oo = 1-odo i = 1, N-1
if(i .gt. N/2) o = ooxc(i) = o*x(i) + x(i+1)*(1.-o)yc(i) = o*y(i) + y(i+1)*(1.-o)tx = x(i+1) - x(i)ty = y(i+1) - y(i)t = sqrt(tx**2 + ty**2)nx(i) = -ty/tny(i) = tx/t
end doreturnend
! ****************************************************************subroutine Input_Data(N, x, y, alpha, VV)parameter (Pi=3.14159265, ToRad = Pi/180.)character FileName*20real x(*), y(*)
!! Angle of attack and Vinf
VV = 1.0alpha = 4.0alpha = alpha*ToRad ! Convert to radians
!print 100
100 format( 10x,’1- NACA airfoil’ /& 10x,’2- Joukowsky or Trefftz airfoil ’///& 10x,’Your choice: ’,$)read *, Ichoiceif(Ichoice .eq. 1) then
! NACA 4 or 5 digit airfoilsprint 131
131 format(’ Number of the points around airfoil : ’,$)read *, NN = N/2*2+1 ! Make it odd numberprint 141
141 format(’ Chord length : ’,$)read *, bprint 151
151 format(’ NACA airfoil : ’,$)read *, NACA
317
N2 = N/2dt = Pi/N2 ! Angle incrementdo i = 1, N2
! xx = ( 1. + cos(dt*(i-1)) )/2gama = float(i-1)/N2Ci = FjaNabiranja(.5, 1.08, gama)xx = 1.0 - Ci
! xx = 1. - 1./(N/2)*(i-1)Call NACA45(NACA, xx, xg, yg, YGp, xd, yd, YDp)x(i) = xg*by(i) = yg*bx(N-i+1) = xd*by(N-i+1) = yd*b
end doxx = 0.0Call NACA45(NACA, xx, xg, yg, YGp, xd, yd, YDp)x(N/2+1) = xg*by(N/2+1) = yg*b
else if(Ichoice .eq. 2) then! Trefftz or Joukowsky airfoil! Analytical pressure distri-! bution exists.
print 101101 format(’ File with airfoil coordinates : ’,$)
read ’(A)’, FileNameopen(unit=1, file=FileName, form=’formatted’, status=’old’)read(1,111) N
111 format(/9x,I3)read(1,121) (x(i), y(i), i = 1, N)
121 format(2F12.3)close(unit=1)
elsestop ’*** Wrong choice ***’
end ifreturnend
! ****************************************************************function FjaNabiranja(alfa, beta, gama)bb = ((beta+1)/(beta-1))**((gama-alfa)/(1-alfa))FjaNabiranja = ((2*alfa+beta)*bb+2*alfa-beta)/((2*alfa+1)*(1+bb))returnend
! ****************************************************************include ’simq.for’include ’naca45.for’
7.1.9 Two Airfoil Problem
We continue to continually modify our examples, to take a new things one by one.Now our problem is aerodynamic interference between two airfoils. Better saiddemonstration of aerodynamic interference. We could approach to this problem invarious way. The most general one is the substructure approach, in which we keeptracking various components of the our problem. Let us assume that we wish tocalculate flow about a few aerodynamic contours at the same time, and let the first
318
Figure 7.11: Two airfoils as single contour.
contour be defined by N1 points, second by N2 points and so on. Each contour hasexactly Ni unknowns where index i denotes contour numbering, and Ni − 1 controlpoints. Missing equation is obtained from the Kutta-Joukowsky condition which isexpressed as
γ1i+ γNi
= 0, i = 1, 2, . . . , K
Here we introduced local numbering of the unknowns for each of K lifting contours.In order to solve our aerodynamic problem we need to assemble our problem intosingle system of equations, or single aerodynamic matrix and single right-hand side
[A1,1] [A1,2] · · · [A1,K ]
[A2,1]. . . · · · [A1,K ]
......
. . ....
[AK,1] [AK,2] · · · [AK,K ]
γ1γ2...
γK
=
b1b2...
bK
(7.59)
where [Ai,j] is the sub-matrix of the aerodynamic influence coefficients for the controlpoints at structure i from the unknown vorticity strengths at the structure j. Thelast row of these matrices is either empty or contains Kutta-Joukowsky condition.Only sub-matrices [Ai,i] have first and last column of the last row filled with 1, allother [Ai,j] when i = j contain in the last row only zeros. Our problem is mainlybookkeeping one. The question is where to put in this matrix coefficient which isobtained for the control point k at the i-th structure, from the unknown n at thej-th structure. We have to determine row (r) and column (c) in the matrix wherewe have to place our coefficient
r = k +∑m=1
i− 1Nm , c = n+j−1∑m=1
Nm
But in our two airfoils problem we take different approach. We connect both contoursin our mind, as it is shown in the figure (7.11). So we have added non-existing portionof the contour in the flow field, we must take care about it.Let the number of points defining the first contour is N1, and N2 is the numberof points by which the second contour is defined. Then the panel between points
319
(XN1 , YN1) and (XN1+1, YN1+1) should not exist. The total number of unknowns isN = N1 + N2. We keep this fictional panel in order to calculate induced velocitiestable in one loop. But, we must calculate induced velocities form edge-points 1 andN1 + 1 the same way because there are no panels preceding this edge points.The same is true for the contributions of the vorticity strengths at points N1 andN = N1 +N2. In this case there are no panels following this edge point.Row N1 and N of the aerodynamic matrix is left for Kutt-Joukowsky condition.The N1 row is filled with zeros except at column 1 and N1 which contain 1. Similaris true for row N which is also filled with zeros except the columns N1 + 1 and Nwhich contain 1. The right-hand side of these columns is zero.We give below program which solves our two airfoil problem. For simplicity we takethe same contour for both airfoils, and define them with the same number of points.
! File: pmlv2.for! Z. Petrovic, 11-Feb-2001, Baghdad! Calulation of theflow about lifting bodies - linear strength vorticity! ================================================================
program PMLV2parameter (MM=150, MMM=MM*MM)real x(MM), y(MM) ! Panel edge coordinatesreal xc(MM), yc(MM) ! Control point coordinatesreal A(MMM), b(MM), G(MM) ! Aerodynamic influence mtrixreal nx(MM), ny(MM) ! Unit normals at control ptsreal uij(MMM), vij(MMM) ! Induced velocities due to
! unit strength source distr.equivalence (b(1), G(1))common /c1/ Acommon /c2/ uijcommon /c3/ vij
!Call Input_Data(N, x, y, alpha, VV)Uinf = VV*cos(alpha)Vinf = VV*sin(alpha)
! Control point and unit normalCall Geometry(N, x, y, xc, yc, nx, ny)
! Calculate induced velocityCall Induced(N, x, y, xc, yc, uij, vij)
! Call WriteTable(vij,N)! Call Test(N,x,y)
! Influence matrixCall Aero(N, uij, vij, nx, ny, Uinf, Vinf, A, b)
! Solve the equation systemCall SIMQ(A, G, N, Ierr)if(Ierr .ne. 0) then
stop ’*** Singular Influence Matrix *** !’end if
! Calculate pressure coeff.Call CP_Calc(N, G, uij, vij, Uinf, Vinf, CP)
! Prepare results for plottingCall Output(xc, CP, N/2-1)stop ’PMLV2 - End of run !’end
! ****************************************************************subroutine Induced(N, x, y, xc, yc, uij, vij)
320
real x(N), y(N), xc(N), yc(N), uij(N,N), vij(N,N)do i = 1, N-1 ! For each control point
do j = 1, N ! For each edge! LEFT-HAND PANEL CONTRIBUT.! First edge calculation
if(j.eq.1 .or. j.eq. N/2+1) thenxf = x(j)yf = y(j)xs = x(j+1)ys = y(j+1)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, ucL, vcL)! linear vorticity distribut.
Call lv2d(x1, x2, xx, yy, ulL, vlL)! Calculate local vel. comp.
uL = -ulL/(x2-x1) + x2*ucL/(x2-x1)vL = -vlL/(x2-x1) + x2*vcL/(x2-x1)
! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,1) = u ! future referencevij(i,1) = v
else if (j.eq.N .or. j.eq.N/2) then! Last Point Calclation
xf = x(j-1) !========================yf = y(j-1)xs = x(j)ys = y(j)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, ucL, vcL)! linear vorticity distribut.
Call lv2d(x1, x2, xx, yy, ulL, vlL)! Calculate local vel. comp.
uL = ulL/(x2-x1) - x1*ucL/(x2-x1)vL = vlL/(x2-x1) - x1*vcL/(x2-x1)
! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,N) = u ! future referencevij(i,N) = v
elsexf = x(j-1)yf = y(j-1)xs = x(j)ys = y(j)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.
321
! constant vorticity distrib.Call cv2d(x1, x2, xx, yy, ucL, vcL)
! linear vorticity distribut.Call lv2d(x1, x2, xx, yy, ulL, vlL)
! Calculate local vel. comp.uL = ulL/(x2-x1) - x1*ucL/(x2-x1)vL = vlL/(x2-x1) - x1*vcL/(x2-x1)
! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,j) = u ! future referencevij(i,j) = v
! RIGHT-HAND PANEL CONTRIB.xf = x(j)yf = y(j)xs = x(j+1)ys = y(j+1)
! Coordinate transformationCall ToLocalCS(xf,yf, xs,ys, xc(i),yc(i), x1,x2, xx,yy)
! Induced vel. in local C.S.! constant vorticity distrib.
Call cv2d(x1, x2, xx, yy, ucL, vcL)! linear vorticity distribut.
Call lv2d(x1, x2, xx, yy, ulL, vlL)! Calculate local vel. comp.
uL = -ulL/(x2-x1) + x2*ucL/(x2-x1)vL = -vlL/(x2-x1) + x2*vcL/(x2-x1)
! Transform velocities to! global coordinate system
Call ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)! Remember velocities for
uij(i,j) = u + uij(i,j) ! future referencevij(i,j) = v + vij(i,j)
end ifend do
end doreturnend
! ****************************************************************subroutine cv2d(x1, x2, x, y, u, v)parameter (Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi)
! Calculates induced velocities! u and v at the point (x,y)! due to constant unit-strength! VORTEX distribution along x-! axis between x1,x2.!
if(abs(y).lt.0.001 .and. (x-x1)*(x-x2).lt.0.0) then! x is between x1 and x2
u = -0.5v = log(abs((x-x2)/(x-x1)))/TwoPi
elseu = (atan2(y,x-x2)-atan2(y,x-x1))/TwoPiv = log(((x-x2)**2+y**2)/((x-x1)**2+y**2))/FourPi
end if
322
returnend
! ****************************************************************subroutine lv2d(x1, x2, x, y, u, v)parameter (Pi=3.14159265, TwoPi=2*Pi, FourPi=4*Pi)r1=(x-x1)**2 + y**2r2=(x-x2)**2 + y**2t1=atan2(y,x-x1)t2=atan2(y,x-x2)if(abs(y) .lt. 0.0001 .and. (x-x1)*(x-x2).lt.0.0) then
u = -x/2v = -(x*log(r1/r2) - 2*(x2-x1))/FourPi
elseu = -(y*log(r1/r2) - 2*x*(t2-t1))/FourPiv = -(x*log(r1/r2)/2 - (x2-x1) + y*(t2-t1))/TwoPi
end ifreturnend
! ****************************************************************subroutine CP_Calc(N, G, uij, vij, Uinf, Vinf, CP)real G(N), uij(N,N), vij(N,N), CP(N)VV = Uinf**2 + Vinf**2
! For each control pointdo i = 1, N/2-1 ! of the first airfoil
u = Uinfv = Vinf
! From each edgedo j = 1, N
u = u + uij(i,j)*G(j)v = v + vij(i,j)*G(j)
end doCP(i) = 1.0 - (u**2 + v**2)/VV
end doreturnend
! ****************************************************************subroutine Output(x, CP, NA)real x(NA), CP(NA)open(unit=1, file=’PMLV2.dat’,form=’formatted’,status=’unknown’)write(1,100) NA
100 format(’ variables = "x", "G"’/’ zone i=’,I3)write(1,110) (x(i),CP(i), i=1,NA)
110 format(2f12.3)close(unit=1)returnend
! ****************************************************************subroutine Aero(N, uij, vij, nx, ny, Uinf, Vinf, A, b)real uij(N,N), vij(N,N), nx(N), ny(N), A(N,N), b(N)do i = 1, N-1 ! For each control point
b(i) = -Uinf*nx(i) -Vinf*ny(i)do j = 1, N ! For each panel edge
A(i,j) = uij(i,j)*nx(i) + vij(i,j)*ny(i)end do
end do! Kutta-Joukowsky condition
323
N2 = N/2do j = 1, N
A(N2,j) = 0.0A(N,j) = 0.0
end doA(N2,1) = 1.0A(N2,N2) = 1.0b(N2) = 0.0A(N,N2+1) = 1.0A(N,N) = 1.0b(N) = 0.0returnend
! ****************************************************************subroutine ToGlobalCS(xf,yf, xs,ys, uL,vL, u,v)
! Translates local velocity! components (uL,vL) to global
xx = xs - xfyy = ys - yfs = sqrt(xx**2 + yy**2)
! rotation cosinescost = xx/ssint = yy/s
! Global velocity componentsu = uL*cost - vL*sintv = uL*sint + vL*costreturnend
! ****************************************************************subroutine ToLocalCS(xf,yf, xs,ys, xc,yc, x1,x2, x,y)
! Transforms the panel and! contrlo point coordinates! from global to local CooSys
xx = xs - xfyy = ys - yfs = sqrt(xx**2 + yy**2)cost = xx/ssint = yy/s
! Panel coordinatesx1 = 0.0x2 = s
! Control point coordinatesx = cost*(xc-xf) + sint*(yc-yf)y = -sint*(xc-xf) + cost*(yc-yf)returnend
! ****************************************************************Subroutine Geometry(N, x, y, xc, yc, nx, ny)real x(N), y(N), xc(N), yc(N), nx(N), ny(N)o = 0.5oo = 1-odo i = 1, N-1
if(i .gt. N/2) o = ooxc(i) = o*x(i) + x(i+1)*(1.-o)yc(i) = o*y(i) + y(i+1)*(1.-o)tx = x(i+1) - x(i)
324
ty = y(i+1) - y(i)t = sqrt(tx**2 + ty**2)nx(i) = -ty/tny(i) = tx/t
end doreturnend
! ****************************************************************subroutine Input_Data(N, x, y, alpha, VV)parameter (Pi=3.14159265, ToRad = Pi/180.)character FileName*20real x(*), y(*)
!! Angle of attack and Vinf
VV = 1.0alpha = 4.0print 160, alpha, VV
160 format(’ Two airfoil interference’//& ’ alpha = ’,F5.1,5x,’V=’,F5.1//)alpha = alpha*ToRad ! Convert to radians
!print 170
170 format(’ (x,y) shift of the seccond airfoil : ’,$)read *, xs, ysprint 100
100 format( 10x,’1- NACA airfoil’ /& 10x,’2- Joukowsky or Trefftz airfoil ’///& 10x,’Your choice: ’,$)read *, Ichoiceif(Ichoice .eq. 1) then
! NACA 4 or 5 digit airfoilsprint 131
131 format(’ Number of the points around airfoil : ’,$)read *, NN = N/2*2+1 ! Make it odd numberprint 141
141 format(’ Chord length : ’,$)read *, bprint 151
151 format(’ NACA airfoil : ’,$)read *, NACAN2 = N/2do i = 1, N2
gama = float(i-1)/N2Ci = FjaNabiranja(.5, 1.08, gama)xx = 1.0 - CiCall NACA45(NACA, xx, xg, yg, YGp, xd, yd, YDp)x(i) = xg*by(i) = yg*bx(N-i+1) = xd*by(N-i+1) = yd*b
end doxx = 0.0Call NACA45(NACA, xx, xg, yg, YGp, xd, yd, YDp)x(N/2+1) = xg*by(N/2+1) = yg*b
325
else if(Ichoice .eq. 2) then! Trefftz or Joukowsky airfoil! Analytical pressure distri-! bution exists.
print 101101 format(’ File with airfoil coordinates : ’,$)
read ’(A)’, FileNameopen(unit=1, file=FileName, form=’formatted’, status=’old’)read(1,111) N
111 format(/9x,I3)read(1,121) (x(i), y(i), i = 1, N)
121 format(2F12.3)close(unit=1)
elsestop ’*** Wrong choice ***’
end if! Generate the second airfoil
do i = 1, Nx(i+N) = x(i) + xsy(i+N) = y(i) + ys
end doN = 2*Nreturnend
! ****************************************************************function FjaNabiranja(alfa, beta, gama)bb = ((beta+1)/(beta-1))**((gama-alfa)/(1-alfa))FjaNabiranja = ((2*alfa+beta)*bb+2*alfa-beta)/((2*alfa+1)*(1+bb))returnend
! ****************************************************************include ’simq.for’include ’naca45.for’
Note the great similarity with the program for isolated airfoil, with linearly varyingvorticity application.
7.1.10 Numerical Integration
We illustrate in this subsection procedure for numerical integration. It is possibleto calculate induced velocities in the global coordinate system directly. But thatis not the biggest advantage of numerical integration over exact integration, if any.For some applications numerical integration is only possible since exact integrationdoes not exists.Let us consider procedure to calculate induced velocities at arbitrary point (X,Y )by numerical integration for the panel given between points (Xi, Yi) and (Xi+1, Yi+1).Assume that vorticity vary from value γi at point i to γi+1 at point i + 1, figure(7.12). Coordinates of arbitrary point along panel are defined as
X(ξ) =(Xi+1 −Xi)ξ + (Xi +Xi+1)
2
Y (ξ) =(Yi+1 − Yi)ξ + (Yi + Yi+1)
2(7.60)
326
Figure 7.12: Panel definition for numerical integration.
γ(ξ) =(γi+1 − γi)ξ + (γi + γi+1)
2
Induced velocities at arbitrary point (X,Y ) are given as
u =
√(Xi+1 −Xi)2 + (Yi+1 − Yi)2
4π
+1∫−1
γ(ξ)Y dξ
[X −X(ξ)]2 + [Y − Y (ξ)]2 (7.61)
v = −√(Xi+1 −Xi)2 + (Yi+1 − Yi)2
4π
+1∫−1
γ(ξ)[X −X(ξ)] dξ
[X −X(ξ)]2 + [Y − Y (ξ)]2 (7.62)
where γ(ξ), X(ξ), and Y (ξ) are defined by (7.60). In order to compare resultsobtained from numerical integration with those obtained with exact integration wedeveloped small program given below. For numerical integration we use Gauss-Legendre quadrature procedure.
Example 7.4:Illustrate application of numerical integration technique to strait linearly varying vorticitydistribution panels.
Solution 7.4:We use described technique to calculate induced velocities with three similar programsof different order of accuracy. So we could test the influence of order of integrationformula used on accuracy of the computed induced velocities. Program which followsillustrates the numerical integration technique.
! File: NU.FOR! Z. Petrovic, 13-Feb-2001! This program compares numerical integration results with! exact expressions for induced velocities! ****************************************************************
program NUreal X(2), Y(2), uij(2,2), vij(2,2), xc(2), yc(2)print 100
100 format(’ First edge coordinates of the panel (X,Y) : ’,$)
327
read *, X(1), Y(1)print 110
110 format(’ Second edge coordinates of the panel (x,Y) : ’,$)read *,X(2), Y(2)print 120
120 format(’ Vorticity intensities at panel edges G1 and G2 : ’,$)read *, G1, G2print 130
130 format(’ Control point coordinates (X,Y) : ’,$)read *, XC(1), YC(1)
! Exact integrationCall Induced(2,X,Y,XC,YC,uij,vij)u = uij(1,1)*G1 + uij(1,2)*G2v = vij(1,1)*G1 + vij(1,2)*G2
! Exact rezultsprint 140, u, v
140 format(’ Exact results’/’ u = ’,F10.3,5x,’v = ’F10.3/)! Numerical integration
Call InducedN(X(1),Y(1),X(2),Y(2),XC(1),YC(1),G1,G2,u,v)print 150, u,v
150 format(’ Approximate results’/’ u = ’,F10.3,5x,’v = ’F10.3/)Call InducedNN(X(1),Y(1),X(2),Y(2),XC(1),YC(1),G1,G2,u,v)print 150, u,vCall InducedNNN(X(1),Y(1),X(2),Y(2),XC(1),YC(1),G1,G2,u,v)print 150, u,vstopend
! ****************************************************************subroutine InducedNNN(x1, y1, x2, y2, X, Y, G1, G2, u, v)parameter (Pi=3.14159265, FourPi=4*Pi, NPT=6)real W(NPt), Z(NPt)data Z/0.2386192,-0.2386192,0.6612094,-0.6612094,0.9324695,
& -0.9324695/data W/0.4679139,0.4679139,0.3607616,0.3607616,0.1713245,
& 0.1713245/! Internal function
f(a1,a2,xi)=((a2-a1)*xi+(a1+a2))/2s = sqrt((x2-x1)**2+(y2-y1)**2)/FourPisu = 0.0sv = 0.0do i = 1, 6
xi = Z(i)g = f(G1,G2,xi)*(Y-f(y1,y2,xi))/
& ((X-f(x1,x2,xi))**2+(Y-f(y1,y2,xi))**2)su = su + g*W(i)g = f(G1,G2,xi)*(X-f(x1,x2,xi))/
& ((X-f(x1,x2,xi))**2+(Y-f(y1,y2,xi))**2)sv = sv + g*W(i)
end dou = su*sv = -sv*sreturnend
! ****************************************************************subroutine InducedNN(x1, y1, x2, y2, X, Y, G1, G2, u, v)parameter (Pi=3.14159265, FourPi=4*Pi, NPt=3)
328
real W(NPt), Z(NPt)data Z/0.0, 0.7745967, -0.7745967/data W/0.88888889, 0.55555556, 0.55555556/
! Internal functionf(a1,a2,xi)=((a2-a1)*xi+(a1+a2))/2s = sqrt((x2-x1)**2+(y2-y1)**2)/FourPisu = 0.0sv = 0.0do i = 1, 3
xi = Z(i)g = f(G1,G2,xi)*(Y-f(y1,y2,xi))/
& ((X-f(x1,x2,xi))**2+(Y-f(y1,y2,xi))**2)su = su + g*W(i)g = f(G1,G2,xi)*(X-f(x1,x2,xi))/
& ((X-f(x1,x2,xi))**2+(Y-f(y1,y2,xi))**2)sv = sv + g*W(i)
end dou = su*sv = -sv*sreturnend
! ****************************************************************subroutine InducedN(x1, y1, x2, y2, X, Y, G1, G2, u, v)parameter (Pi=3.14159265, FourPi=4*Pi, NPt=4)real W(NPt), Z(NPt)data Z/-0.8611363, -0.3399810, 0.3399810, 0.8611363/data W/0.3478548, 0.6521452, 0.6521452, 0.3478548/
! Internal functionf(a1,a2,xi)=((a2-a1)*xi+(a1+a2))/2s = sqrt((x2-x1)**2+(y2-y1)**2)/FourPisu = 0.0sv = 0.0do i = 1, 4
xi = Z(i)g = f(G1,G2,xi)*(Y-f(y1,y2,xi))/
& ((X-f(x1,x2,xi))**2+(Y-f(y1,y2,xi))**2)su = su + g*W(i)g = f(G1,G2,xi)*(X-f(x1,x2,xi))/
& ((X-f(x1,x2,xi))**2+(Y-f(y1,y2,xi))**2)sv = sv + g*W(i)
end dou = su*sv = -sv*sreturnend
! ****************************************************************include ’induced.for’ ! Subroutines used for analytic calcula-
! tion of induced velocities.
329
7.2 Axi-Symmetric Supersonic Flow
We now illustrate solution procedure for the potential defined by equation (6.32)together with boundary condition (6.33). We repeat these equations for reference
φ =
x−βr∫0
f(ξ) dξ√(x− ξ)2 − β2r2
(7.63)
where β2 = M2∞ − 1, and φ is velocity potential. If the body of the axi-symmetric
shape is defined as r = ρ(x), then velocity potential should satisfy the followingboundary condition
ρ′(x) =v
U∞ + u, or: v = ρ′(U∞ + u) (7.64)
where u = φx, and v = φr. After the velocity components are determined wecalculate pressure coefficient as
CP =2
γM2∞
[1 +
γ − 1
2M2
∞(1− (1 + u/U∞)2 − (v/U∞)2
)]γ/(γ−1)− 1
(7.65)
Substituting for the variable ξ the expression
ξ = x− βr cosh zequation (7.63) becomes
φ =
0∫cosh−1 x/βr
f(x− βr cosh z) dz (7.66)
We assumed as before that our body starts at point x = 0. Any point P can beinfluenced by source distribution along x-axis contained within fore cone with apexin P , what is clearly given by boundaries of integration. Sources which influencearbitrary point P can be distributed between coordinate origin and the point B atthe x-axis determined from relation
xB = xP − βrPIf we assume f(x− βr cosh z) be proportional to its argument we have
φ = Ko
0∫cosh−1 x/βr
(x− βr cosh z) dz (7.67)
wherecosh−1 z = ln
(z +
√z2 − 1
)after integration yields the following
φ = Ko
(−x cosh−1 x
βr+√x2 − β2r2
)(7.68)
330
where Ko is the constant of integration having the dimensions of velocity. Thedisturbance velocities are obtained by differentiation
u = φx = Ko
0∫cosh−1 x/βr
dz (7.69)
v = φr = Ko
0∫cosh−1 x/βr
−β cosh z dz (7.70)
The velocity components are
u = U∞ −Ko cosh−1 xβr
(7.71)
v = Koβ
√x2
β2r2− 1 (7.72)
Equations (7.71) and (7.72) shows that the components of the velocity are constantfor constant ration x/r, and therefore are constant along every strait line that passesthrough the point x = 0, r = 0. Equations (7.71) and (7.72) can be used to determinethe pressure distribution around a circular cone with axis parallel to the flow. If theε is the semi-vertex angle of the cone, than the value of x/r at the cone is given by
r
x= tan ε
The ratio between velocity components v/u must also be equal to tan ε or
tan ε =Koβ
√x2
β2r2 − 1
U∞ −Ko cosh−1 cot ε
β
(7.73)
and
Ko =U∞ tan ε√
cot2 ε− β2 + tan ε cosh−1(cot ε/β)(7.74)
in order to obtain the real values for Ko in this equation, the value of cot ε must belarger than β, or the value of ε must be smaller than the Mach angle µ, (cotµ = β).
7.2.1 Calculation of the Flow around Sharp-Nose Body ofRevolution
Figure (7.13) shows sharp-nose slender body of revolution with the axis parallel tothe free-stream velocity. For any point P on the body we can determine the point Bat the x-axis which determines which part of the singularity distribution influencesthe flow at point P . So point Pi(xi, ρi) is influenced by the source distributionbetween coordinate origin and point Bi on the x-axis, where
xBi= xi − βρi (7.75)
331
Figure 7.13: Determination of points Bi along x-axis.
where symbol ρi denotes that we are talking about the point at the surface of thebody, xBi
denotes the point on x-axis where Mach line drawn from the point Pi
intersects x-axis, and xi is the abscisa of the point Pi.Let us distribute our source singularities to start from points Bi. It is clear thatsource distribution starting from the point Bi cannot influence point on the surfacePi, but it influence points behind that point, i.e. Pi+1, Pi+2, . . . . If we need tocalculate influence velocities due to singularity distribution along x-axis we have toconsider all points Bi from which singularity starts between origin and point Bi−1.
Example 7.5:Develop program to calculate distribution of the pressure coefficient for the Ogive-cylinder body. For the approximation at the nose use linear strength singularity (m = 1),for other points use basic solution corresponding to m = 2, section (6.5.4).
Solution 7.5:We solve our problem in few steps. The first step is as usually definition of geometricdata. We denote by λ = /d, figure (7.14). From the notation in the figure it is clearthat
R =(1
4+ λ2
)d
and
rc =(1
4− λ2
)d , = λd
Equation of the ogive part is given as
ρ(x) =√R2 − x2 −
(1
4− λ2
)d
The slope at arbitrary point on the body is
ρ′(x) =
− x√R2−x2 − < x < 0
0 0 < x < L
The semi-vertex angle is determined from
tan ε =λ√
(0.25 + λ2)2 − λ2
332
Figure 7.14: Definition of geometric data.
To define our problem we need λ, d, L, number of points at which we wish to satisfyboundary conditions on ogive part of the body, No, and number of points at which wewish to satisfy boundary conditions at cylindrical part, Nc. Since we are only illustratingprocedure we will distribute points evenly. We also need free-stream Mach number to
calculate β =√M2∞ − 1.
We define the points on the body contour at which we satisfy boundary conditions inthe step two. First we define coordinates of the points Pi. Coordinate increment ∆xalong x-axis and x coordinates of the points Pi, for the ogive part we determine from
∆x =
No
, xPi= −+∆x · (i− 1) , i = 1, 2, . . . , No (7.76)
The radius of the ogive body is given by
R =(1
4+ λ2
)d
And ordinates for selected points
ρi =√R2 − x2Pi
− rc , rc = (1/4− λ2) dSlope at the body contour at points Pi is obtained by differentiating this equation withrespect to x
ρ′(xPi) = − xPi√
R2 − x2Pi
Now we define points xBiat x-axis which correspond to each of the points Pi
xBi= xPi
− βρiFor the cylindrical part again we first distribute points along x-axis
∆x =L
Nc− 1, xPi
= ∆x · (i− 1−No) , i = No + 1, No + 2, . . . , No +Nc
Ordinates are the same for all points ρi = d/2, ad the slope at these points is ρ′i = 0.Points Bi for the cylindrical part are determined the same way as for ogive part
xBi= xPi
− βd2, i = No + 1, No + 2, . . . , No +Nc
333
In the third step we decide upon the type singularities used to start from the points xBi.
To simplify the procedure we use for all points the same type of singularity except for thefirst point, from which two types of singularity will start. Linear distribution is used toapproximate semi vertex angle of the ogive body, while the quadratic distribution (whichalso starts from point 1) will be determined so that the boundary condition is satisfiedat point P1. Semi-vertex angle is determined from
tan ε =λ√
(1/4+ λ2)2 − λ2
The intensity for the linear distribution is determined from equation (7.74)
Ko =U∞ tan ε√
cot2 ε− β2 + tan ε cosh−1(cot ε/β)
Linear singularity starting from the nose of the body influence all other points on thebody. We calculate contribution of this distribution to any point with the expressions(7.71) and (7.72)
ui = uPi= −Ko cosh
−1 xiβρi
vi = vPi= Koβ
√√√√ x2iβ2ρ2i
− 1
where cosh−1 z = ln(z +
√z2 − 1
). One index symbol is used to differentiate contribu-
tion of this linear distribution from the contributions of the quadratic distributions. Tocalculate contribution of the distribution starting from the point Bj(xj, 0) to the controlpoint Pi(xi, ρi) we define
ti,j =βρi
xi − xj , again: β2 =M2∞ − 1
so induced velocities are calculated from (6.42) for unit strength coefficient
ui,j = −(xi − xj)(sech−1ti,j −
√1− t2i,j
)vi,j =
β(xi − xj)2
√1− t2i,jti,j
− ti,jsech−1ti,j
At the first glance it may seem that we need to solve system of equations in order todetermine unknown strengths Ki but that is not necessary. At the first control point(P2) we determine the strength of the first quadratic distribution starting from the pointB1. At point B2 starts the second source distribution its strength is determined fromthe conditions at point (P3), and so on. Let us determine velocity at arbitrary point Pi.
u(Pi) = U∞ + ui +i−1∑j=1
Kjui,j
v(Pi) = vi +i−1∑j=1
Kjvi,j (7.77)
334
In this expression only Ki−1 coefficient is unknown because all others are determined ina such way that flow tangency condition is satisfied in previous points. We determinethis coefficient from flow tangency condition at point Pi:
v(Pi)
u(Pi)= ρ′i =
vi +∑i−2
j=1Kjvi,j +Ki−1vi,i−1U∞ + ui +
∑i−2j=1Kjui,j +Ki−1ui,i−1
We introduce now the new symbols
U = 1 +uiU∞
+i−2∑j=1
Kjui,jU∞
and
V = vi +i−2∑j=1
Kjvi,jU∞
thus
ρ′i =V +Ki−1
vi,i−1
U∞U +Ki−1
ui,i−1
U∞
From this equation we have:
Ki−1 =ρ′iU − V
vi,i−1
U∞ − ρ′i ui,i−1
U∞
After the coefficient Ki is known we are ready to calculate influence velocities at pointPi
u(Pi)
U∞= U +Ki−1
ui,i−1U∞
,v(Pi)
U∞= V +Ki−1
vi,i−1U∞
than we calculate the pressure coefficient either form the expression (7.65) or fromsimplified expression
CPi = −2uPi
U∞From preceding explanations it is clear that velocity U∞ is not necessary to be specifiedto determine CP (Pi) distribution. So we can assume that U∞ = 1 in this example.Program which follow is possible realization of explained algorithm.
! File: OGIVE.FOR! Z. Petrovic, 17-Feb-2001, Baghdad! Supersonic flow about combination tangent ogive cylinder! ****************************************************************
program Ogiveparameter (NN=100)real xP(NN), ro(NN), rop(NN), xB(NN), CP(NN)real lam, Lc
! Read data about ogive-cylCall InputData(beta, lam, Lc, d, No, Nc)
! Generate control point coord.Call Geometry(beta, lam, Lc, d, No, Nc, N, xP, ro, rop, xB, tge)
! Pressure coefficient calculationCall Aero(beta,tge, N,xP,ro,rop,xB, CP)
! Prepare for plotting
335
Call Output(xP, CP, N)stop ’Ogive -- End of run !’end
! ****************************************************************subroutine Output(xP, CP, N)real xP(N), CP(N)open(unit=1,file=’ogive.dat’,form=’formatted’,status=’unknown’)write(1,100) N
100 format(’variables = "X", "Cp" ’/’zone i=’,I4)write(1,110) (xP(i), CP(i), i = 1, N)close(unit=1)
110 format(2F12.3)returnend
! ****************************************************************subroutine Aero(beta,tge, N,xP,ro,rop,xB, CP)real xP(N), ro(N), rop(N), xB(N), CP(N), Ko, K(100), Ma2
! Strength of linear distributioncge = 1./tgeMa2 = beta**2+1. ! Mach squaredKo = tge/( sqrt(cge**2 - beta**2) + tge*acosh(cge/beta) )
! Velocities at the vertexuo = 1. - Ko*acosh(cge/beta)vo = Ko*sqrt(cge**2 - beta**2)
! Pressure distribution at vertexCP(1) = CpCalc(Ma2, uo, vo)do i = 2, N
! Contribution from conet = beta*ro(i)/(xP(i)-xB(1))u = 1.-Ko*acosh(1./t)v = Ko*beta*sqrt(1./t**2-1.)
! Contribution from quadraticdo j = 1, i-2
t = beta*ro(i)/(xP(i) - xB(j))u = u - K(j)*(xP(i)-xB(j))*(asech(t)-sqrt(1.-t**2))v = v + K(j)*beta*(xP(i)-xB(j))*
& (sqrt(1./t**2-1.) - t*asech(t))/2end dot = beta*ro(i)/(xP(i)-xB(i-1))uij = -(xP(i)-xB(i-1))*(asech(t) - sqrt(1.-t**2))vij = beta*(xP(i)-xB(i-1))*(sqrt(1./t**2-1.) - t*asech(t))/2
! Strength of singularity at xB(i-1)K(i-1)=(rop(i)*u - v)/(vij - rop(i)*uij)
! Velocity at xP(i)u = u + K(i-1)*uijv = v + K(i-1)*vijCP(i) = CpCalc(Ma2, u, v)
end doreturnend
! ****************************************************************function CpCalc(Ma2, u,v)parameter(g=1.4, f=(g-1.)/2, h=g/(g-1.))real Ma2CpCalc = -2*(u-1.) ! Small disturbance expression
! Complete expression
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c = 2/(g*Ma2)! CpCalc = c*( ( 1. + f*Ma2*(1. - u**2 - v**2) )**h - 1.)
returnend
! ****************************************************************function acosh(x)acosh = log( x + sqrt(x**2-1.) )returnend
! ****************************************************************function asech(x)asech = log( 1./x + sqrt(1./x**2 - 1.) )returnend
! ****************************************************************subroutine InputData(beta, lam, Lc, d, No, Nc)real lam, Lc, Machlam = 3.5 ! Slenderness of the ogive partLc = 2.0 ! Length of the cyllindrical partd = 1.0 ! Calibar of the bodyNo = 10 ! Number of points defining ogiveNc = 10 ! Number of points defining cyl.Mach = 2.0 ! Free-stream Mach numberbeta = sqrt(Mach**2 - 1)returnend
! ****************************************************************subroutine Geometry(beta, lam,Lc,d,No,Nc,N, xP, ro, rop, xB, tge)real lam, Lc, xP(*), xB(*), ro(*), rop(*)
! Data for the ogive part ******el = lam*d ! Ogive lengthDx = el/NoR = (0.25 + lam**2)*d ! Ogive shape radiusrc = (0.25 - lam**2)*ddo i = 1, No
xP(i) = -el + Dx*(i-1) ! Control point abscisa! Distance of the C.P. from x-axis
ro(i) = sqrt(R**2 - xP(i)**2) + rc! The slope of the contour at C.P.
rop(i)= -xP(i)/sqrt(R**2 - xP(i)**2)! Here starts the new singularity
xB(i) = xP(i) - beta*ro(i)end do
! The data for cylindrical partDx = Lc/(Nc-1)N = No + Ncdo i = No+1, N
xP(i) = Dx*(i - 1 - No)ro(i) = d/2xB(i) = xP(i)-beta*ro(i)rop(i) = 0.0
end do! Tangent of the semi-vertex angle
tge = lam/sqrt( (.25 + lam**2)**2 - lam**2 )returnend
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7.3 Vortex Lattice Method
We illustrate now method which is simple enough to be presented as example com-pletely, and at same time by which is possible to study three-dimensional configura-tions. The key building block for the application of this method is subroutine whichcalculates induced velocity at arbitrary point from the polygonal, and in generalcase three-dimensional, vortex line. It is expected that the polygonal line is definedby its corner points.
Subroutine PolyVortex(Gama, x,y,z,N, xP,yP,zP, u,v,w)!
real Gama ! Intensity of the vortex line per! unit length
real x(N) ! x-coordinate of polygonal linereal y(N) ! y-coordinate of polygonal linereal z(N) ! z-coordinate of polygonal linereal xP, yP, zP ! Control point coordinates
!real u, v, w ! Induced velocities
!u = 0.0v = 0.0w = 0.0
! Call subroutine which calculates! contribution of one segment! required number of times
do i = 1, N-1Call LV3D( Gama,
& xP, yP, zP,& x(i), y(i), z(i),& x(i+1), y(i+1), z(i+1),& uu, vv, ww)
u = u + uuv = v + vvw = w + ww
end do!
returnend
This subroutine repeatedly calls subroutine which calculates induced velocity fromone line segment, and which is given below
Subroutine LV3D(Gama, x,y,z, x1,y1,z1, x2,y2,z2, u,v,w)parameter (pi = 3.141592654, eps = 1.0e-10)real Gama ! Vortex filament strength per unit
! lengthreal x, y, z ! Control point coordinates for
! which influence velocities are calculatedreal x1, y1, z1 ! First point of the vortex filamentreal x2, y2, z2 ! Last point of the vortex filament
!real u, v, w ! Induced velocities at point P
! *************************************! Subroutine LV3D! calculates induced velocities (u,v,w)
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! at point (x,y,z) induced by vortex! filament of intensity Gama per unit! length for the segment, between points! (x1,y1,z1) and (x2,y2,z2)! *************************************!! Components R1 X R2!
R1R2X = (y-y1)*(z-z2) - (z-z1)*(y-y2)R1R2Y = -( (x-x1)*(z-z2) - (z-z1)*(x-x2) )R1R2Z = (x-x1)*(y-y2) - (y-y1)*(x-x2)
!! Calculate (R1 X R2)**2!
R1R2KV = R1R2X**2 + R1R2Y**2 + R1R2Z**2!! Calclate R1 i R2!
R1 = sqrt( (x-x1)**2 + (y-y1)**2 + (z-z1)**2 )R2 = sqrt( (x-x2)**2 + (y-y2)**2 + (z-z2)**2 )
!! Calculate R0*R1 i R0*R2!
R0R1 = (x2-x1)*(x-x1) + (y2-y1)*(y-y1) + (z2-z1)*(z-z1)R0R2 = (x2-x1)*(x-x2) + (y2-y1)*(y-y2) + (z2-z1)*(z-z2)
!! Calculate u, v, w!
if( R1.lt.eps .or. R2.lt.eps .or. R1R2KV.lt.eps) thenu = 0v = 0w = 0
elsegampi = Gama/(4*pi*R1R2KV)*(R0R1/R1 - R0R2/R2)u = gampi*R1R2Xv = gampi*R1R2Yw = gampi*R1R2Z
end if!
returnend
We develop program which accepts only flat-plate trapezoidal panels. Inclinationsof such panels are realized through definition of unit normal. Figure (7.15) showspossible combinations of trapezoidal shapes.Each trapezoidal shape has unique unit normal and associated number of rows NRi
and columns NCi by which trapezoidal element is divided into elementary panels.To calculate forces on various configurations we need
1. To read parameters of the problem, such as number of trapezoidal elements;For each element NRi and NCi and coordinates of the corner points (x1, y1)i,(x2, y2)i, (x3, y3)i, and (x4, y4)i. For each element unit normal components nxi
,nyi
and nzi. Free-stream velocity components U∞, V∞, and W∞. Flag which
differentiate symmetric from asymmetric problem.
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Figure 7.15: Combination of trapezoidal elements.
2. In this step we generate all required geometry data such as: coordinates ofthe control points, and unit normal components for each control point. Boundvortex coordinates.
3. We generate tables of induced vortices for each control point and from eachvortex filament.
4. Next, we build equation system by requiring satisfaction of boundary condi-tions.
5. Finally we calculate forces at each bound vortex.
Everything said so far is implemented in the program which follow.
! File: VLMN.FOR! Z. Petrovic, 18-Feb-2001! Vortex lattice method for planars tructures! ****************************************************************
program VLMparameter (ND=125, NDND=ND*ND)real xB(2,ND), yB(2,ND), xCP(ND), yCP(ND), nx(ND), ny(ND), nz(ND)real uij(NDND), vij(NDND), wij(NDND)real A(NDND), G(ND), Lift(3), Mom(3), CG(3), VV(3)structure /welement/
real x(4), y(4)integer NR, NCreal nx,ny,nz
end structurerecord /welement/ trapez(20)logical symm
! Compiler problemscommon /c01/ uijcommon /c02/ vijcommon /c03/ wijcommon /c04/ G
! Read data about the geometry! of the problem
Call Input(N, trapez, VV, symm, alpha, ro, CG)Ui = VV(1)Vi = VV(2)
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Wi = VV(3)! Calculate geometry of the prob.
Call Geometry(N, trapez, xB, yB, xCP, yCP, nx, ny, nz, NCP)! Induced velocity calculation
Call Induced(NCP, xCP, yCP, xB, yB, uij, vij, wij, symm, Ui, Vi)! Build system of equations
Call Aero(uij, vij, wij, nx, ny, nz, A, G, NCP, Ui, Vi, Wi)! Solve system of equations
Call SIMQ(A, G, NCP, Ierr)if(Ierr .ne. 0) then
stop ’Singular matrix !’end ifCall Forces(NCP, xB, yB, G, VV, ro, CG, Lift, Mom, symm)print *,’ Lift : ’,Liftprint *,’ Moment : ’,MomCall Output(xCP, yCP, G, NCP)stop ’VLMN --- End of run !’end
! ****************************************************************subroutine Forces(N, xB, yB, G, V, ro, CG, Lift, Mom, symm)real xB(2,N), yB(2,N), V(3), CG(3), Lift(3), Mom(3), G(N), F(3)real t(3), M(3), FF(3)logical symmt(3) = 0.0Call ScalVec(0.0, Lift, Lift, 3)Call ScalVec(0.0, Mom, Mom, 3)do i = 1, N
! Bound vortex vectort(1) = xB(1,i) - xB(2,i)t(2) = yB(1,i) - yB(2,i)Call VecPr3(t, V, F)
! Lift of the bound vortexCall ScalVec(ro*G(i), F, F, 3)
! Total liftCall VecAdd(F, Lift, Lift, 3)if(symm) then
t(2) = -t(2)Call VecPr3(t, V, FF)Call ScalVec(ro*G(i), FF, FF, 3)Call VecAdd(FF, Lift, Lift, 3)
end if! Distance between CG and! bound vortex force
t(1) = (xB(1,i) + xB(2,i))/2 - CG(1)t(2) = (yB(1,i) + yB(2,i))/2 - CG(2)t(3) = -CG(3)
! Bound vortex contribution to! moment about CG
Call VecPr3(t, F, M)! Total Moment
Call VecAdd(M, Mom, Mom, 3)if(symm) then
t(1) = (xB(1,i) + xB(2,i))/2 - CG(1)t(2) = -(yB(1,i) + yB(2,i))/2 - CG(2)t(3) = -CG(3)Call VecPr3(t, FF, M)
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Call VecAdd(M, Mom, Mom, 3)end if
end doreturnend
! ****************************************************************subroutine Aero(uij, vij, wij, nx, ny, nz, A, G, N, Ui, Vi, Wi)real uij(N,N), vij(N,N), wij(N,N), nx(N), ny(N), nz(N)real G(N), A(N,N)do i = 1, N
G(i) = -(Ui*nx(i) + Vi*ny(i) + Wi*nz(i))do j = 1, N
A(i,j) = uij(i,j)*nx(i) + vij(i,j)*ny(i) + wij(i,j)*nz(i)end do
end doreturnend
! ****************************************************************subroutine Output(x,y,G, N)real G(N), x(N), y(N)open(unit=1,file=’VLMN.DAT’,form=’formatted’,status=’unknown’)write(1,100) N
100 format(’variables = "x", "y", "G" ’/’zone i=’,I4,’ f=point’)write(1,110) (x(i), y(i), G(i), i = 1, N)
110 format(3G15.6)returnend
! ****************************************************************subroutine Induced(NCP,xCP,yCP,xB,yB,uij,vij,wij,symm,Ui,Vi)real xCP(NCP), yCP(NCP), xB(2,NCP), yB(2,NCP), uij(NCP,NCP)real vij(NCP,NCP), wij(NCP,NCP)real x(10), y(10), z(10)logical symmDt = 1.e15do i = 1, NCP
xp = xCP(i)yp = yCP(i)do j = 1, NCP
x(1) = xB(1,j) + Ui*Dty(1) = yB(1,j) + Vi*Dtx(2) = xB(1,j)y(2) = yB(1,j)x(3) = xB(2,j)y(3) = yB(2,j)x(4) = xB(2,j) + Ui*Dty(4) = yB(2,j) + Vi*DtN = 4if(symm) then
x(5) = x(4)y(5) = -y(4)x(6) = x(3)y(6) = -y(3)x(7) = x(2)y(7) = -y(2)x(8) = x(1)y(8) = -y(1)
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N = 8end ifdo k = 1, N
z(k) = 0.0end doCall PolyVortex(1., x,y,z,N, xP,yP,0.0, u,v,w)uij(i,j) = uvij(i,j) = vwij(i,j) = w
end doend doreturnend
! ****************************************************************subroutine Geometry(N,trapez,xB,yB,xCP,yCP,nx,ny,nz,NCP)real xB(2,*), yB(2,*), xCP(*), yCP(*), nx(*), ny(*), nz(*)real x(4), y(4)structure /welement/
real x(4), y(4)integer NR, NCreal nx,ny,nz
end structurerecord /welement/ trapez(20)L = 0do i = 1, N
NR = trapez(i).NRNC = trapez(i).NCdo j = 1, 4
x(j) = trapez(i).x(j)y(j) = trapez(i).y(j)
end dodo ir = 1, NR
do ic = 1, NCL = L + 1Call PanelData(ir,ic,NR,NC,x,y,xB(1,L),yB(1,L),
& xCP(L),yCP(L))nx(L) = trapez(i).nxny(L) = trapez(i).nynz(L) = trapez(i).nz
end doend do
end doNCP = Lreturnend
! ****************************************************************subroutine PanelData(ir,ic,NR,NC,x,y,xB,yB,xCP,yCP)real x(4), y(4), xB(2), yB(2)real xp(4), yp(4)
! Find the row dataxp1 = x(1)+(x(2)-x(1))/NR*(ir-1)! 1 4yp1 = y(1)+(y(2)-y(1))/NR*(ir-1)! +-----+-----+-----+-----+xp2 = x(1)+(x(2)-x(1))/NR*ir ! ! ! ! ! !yp2 = y(1)+(y(2)-y(1))/NR*ir ! ! ! ! ! !xp4 = x(4)+(x(3)-x(4))/NR*(ir-1)! ! ! ! ! !yp4 = y(4)+(y(3)-y(4))/NR*(ir-1)! ! ! ! ! !
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xp3 = x(4)+(x(3)-x(4))/NR*ir ! +-----+-----+-----+-----+yp3 = y(4)+(y(3)-y(4))/NR*ir ! 2 3
! Find the column dataxp(1) = xp1+(xp4-xp1)/NC*(ic-1)yp(1) = yp1+(yp4-yp1)/NC*(ic-1)xp(4) = xp1+(xp4-xp1)/NC*icyp(4) = yp1+(yp4-yp1)/NC*icxp(2) = xp2+(xp3-xp2)/NC*(ic-1)yp(2) = yp2+(yp3-yp2)/NC*(ic-1)xp(3) = xp2+(xp3-xp2)/NC*icyp(3) = yp2+(yp3-yp2)/NC*ic
! Bound Vortex coordinatesxB(1) = xp(1) + (xp(2)-xp(1))*.25yB(1) = yp(1) + (yp(2)-yp(1))*.25xB(2) = xp(4) + (xp(3)-xp(4))*.25yB(2) = yp(4) + (yp(3)-yp(4))*.25
! Control point coordinatesxCP = (xp(1) + (xp(2)-xp(1))*.75 + xp(4) + (xp(3)-xp(4))*.75)/2yCP = (yp(1) + (yp(2)-yp(1))*.75 + yp(4) + (yp(3)-yp(4))*.75)/2returnend
! ****************************************************************subroutine Input(N, trapez, V, symm, alpha, ro, CG)parameter (Pi=3.14159265, ToRadians=Pi/180.)real V(3), CG(3), nx, ny, nzlogical symm
!structure /welement/
real x(4), y(4)integer NR, NCreal nx,ny,nz
end structurerecord /welement/ trapez(*)
!open(unit=1,file=’vlmn.inp’,form=’formatted’,status=’old’)read(1,*) symm ! .true. for xz-symetric bodies
! .false. for unsimetric caseread(1,*) N, ! Number of trapez shapes
& alpha, ! Angle of attack& Vinf, ! Free-stream velocity& ro, ! Density of the air& CG ! X,Y,Z of the center of grafity
!alpha = alpha*ToRadiansUi = Vinf*cos(alpha)Vi = 0.0Wi = Vinf*sin(alpha)
!V(1) = UiV(2) = ViV(3) = Wido i = 1, N
do k = 1, 4read(1,*) x,y ! Read coordinates of trapeztrapez(i).x(k) = x ! corners 1-----4trapez(i).y(k) = y ! ! !
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end do ! 2-----3read(1,*) NR, NC ! Divide it to NR rows and NC colstrapez(i).NR = NRtrapez(i).NC = NCread(1,*) nx, ny, nz ! Unit normal components for trapeztrapez(i).nx = nxtrapez(i).ny = nytrapez(i).nz = nz
end do!
returnend
! ****************************************************************include ’polivrt.for’include ’lv3d.for’include ’simq.for’include ’vecopn.for’
The program requires input file given below.
.false.24.5594531.1.2250 0 00 00.2 00.7 0.50.5 0.51 40 0 10.5 -0.50.7 -0.50.2 0.00.0 0.01 40 0 1
read(1,*) N, ! Number of trapez shapes& alapha, ! Angle of attack& Vinf, ! Free-stream velocity& ro, ! Density of the air& CG ! X,Y,Z of the center of grafity
do i = 1, Ndo k = 1, 4
read(1,*) x,y ! corners of the trapezend doread(1,*) NR, NC ! Number of rows, Number of columnsread(1,*) nx, ny, nz ! Unit normal components
end do
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Bibliography
[1] Brandao, M. P., Improper Integrals in Theoretical Aerodynamics: The ProbelmRevisited, AIAA Journal, Vol. 25, No. 9, September 1987.
[2] Katz, J., Plotkin, A., Low-Speed Aerodynamics – From Wing Theory to PanelMethods, McGraw-Hill, Inc., New York, 1991.
[3] Van Dyke, M., First- and Second-Order Theory of Supersonic Flow Past Bodiesof Revolution, Journal of the Aeronautical Sciences, Vol. 18, No. 3, March 1951.
[4] Milne-Thompson, L. M., Theoretical Hydrodynamics, MacMillan Co. Ltd, Lon-don, 1955.
[5] Milne-Thompson, L. M., Theoretical Aerodynamics, Dover Publications, Inc.,New Yourk, 1958.
[6] Karamacheti, K., Principles of Ideal-Fluid Aerodynamics,
[7] Belotserkowski, S. M., The Theory of Thin Wings in Subsonic Flow, PlenumPress, New York, 1967.
[8] Belotserkowski, S. M., Modelirovanie Plosko-Paralelnogo Otrivnogo ObtekaniaTel, Nauka, Moscow, 1988.
[9] Krasnov, N. F., Aerodynamics 2 – Methods of Aerodynamic Calculations, MirPublishers Moscow, 1985.
[10] Murray, R. S., Mathematical Handbook – of Formulas and Tables, Schaum’sOutline Series in Mathematics, McGraw Hill Book Company, New York, 1968.
[11] Moran, J., An Introduction to Theoretical and Computational Aerodynamics,Willey, New York, 1984.
[12] Glauert, H., The Elements of Aerofoil and Airscrew Theory, 2d edition, Cam-bridge University Press, 1959.
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