AER307 Aerodynamics - Skuleexams.skule.ca/exams/AER307H1_20149_661418446443aer307...Aerodynamics Ronald Hanson [email protected] UTIAS AER307: Aerodynamics c Philippe Lavoie

University of Toronto Institute for Aerospace Studies

AER307

Aerodynamics

Ronald [email protected]

UTIAS

AER307: Aerodynamics c©Philippe Lavoie 2012 0-1

Introductions

• Book: John D. Anderson Jr., Fundamentals of Aerodynamics. Current edition is 5th, but

any previous editions will do.

• My office hours: Upon request. Preferably one hour before or after classes. ‘Virtual’ office

times (e.g. via email discussion) can also be organized if convenient.

• Email policy: Should be used to setup a meeting time or short questions. Replies typically

within 1-2 open days.

• Course on Blackboard: I will not print hand-outs. They will be posted on Blackboard for

download. I would encourage you not to be wasteful in you printing.

• Tutorial: Tutorials will be agglomerated into 2-hour sessions spread-out strategically during

the year. The TA is Hamed Sadeghi ([email protected] – above email policy

applies).

• Grading:

– Total mark based on 5 assignments (2% each),

– 2 term tests (15% each), and

– A final (60%).

Term tests are closed book, no aids, non-programmable calculator.

Assignments will be posted on Blackboard.

Due back one week after they have been given out.


• Honesty policy: You can discuss ideas with colleagues but you must solve the assignment

on your own. The solutions you submit must reflect your understanding of the material.

Late policy is 10% of grade per day.

• Course Outline. Table of content of my notes included below. Numbers in brackets refer to

corresponding sections in the textbook. Section 1, 2, 3 and 9 are independent of the

textbook and the notes for these sections will be posted on Blackboard (after the lectures).


Contents

1 Fundamentals 13

1.1 What is Aerodynamics? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.2 Aerodynamic Forces: How do Planes Fly? (1.4) . . . . . . . . . . . . . . . . 14

1.3 The Airfoil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.4 Frame of Reference and Steady Flow Assumption . . . . . . . . . . . . . . . 15

1.5 Aerodynamic Forces and Moments (1.5) . . . . . . . . . . . . . . . . . . . . 16

1.5.1 Dimensionless Coefficients . . . . . . . . . . . . . . . . . . . . . . . 18

1.6 Pressure Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.7 Centre of Pressure (1.6) . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.8 Types of Flow (1.10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.8.1 Inviscid vs. Viscous . . . . . . . . . . . . . . . . . . . . . . . . . . 23


1.8.2 Viscous Flows: Boundary Layers and Wakes . . . . . . . . . . . . . . 24

1.8.3 Laminar, Transitional and Turbulent Flows . . . . . . . . . . . . . . . 25

1.8.4 Attached vs. Separated Flows . . . . . . . . . . . . . . . . . . . . . 26

1.8.5 Compressible vs incompressible . . . . . . . . . . . . . . . . . . . . . 29

2 The Basic Equations of Fluid Dynamics 31

2.1 General Form of a Conservation Law . . . . . . . . . . . . . . . . . . . . . 31

2.1.1 Scalar Conservation Law . . . . . . . . . . . . . . . . . . . . . . . . 31

2.1.2 Vector Conservation Law . . . . . . . . . . . . . . . . . . . . . . . 34

2.2 The Equation of Mass Conservation . . . . . . . . . . . . . . . . . . . . . . 38

2.3 The Equation of Momentum Conservation . . . . . . . . . . . . . . . . . . . 40

2.4 The Equation of Energy Conservation . . . . . . . . . . . . . . . . . . . . . 45

2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47


3 The Dynamic Levels of Approximation 48

3.1 The Navier-Stokes Equations . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.2 Closure of the System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.4 A Brief Review of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . 52

3.4.1 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . 53

3.4.2 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . 54

3.5 The Speed of Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.5.1 Mach Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3.6 Classification of Simplifications . . . . . . . . . . . . . . . . . . . . . . . . 59

3.7 Incompressible Fluid Model . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.8 The Reynolds-average N-S Equations . . . . . . . . . . . . . . . . . . . . . 63

3.9 The Boundary-Layer Approximation . . . . . . . . . . . . . . . . . . . . . . 66


3.10 The Euler Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

3.11 Potential Flow Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3.11.1 Pathlines & Streamlines of a Flow . . . . . . . . . . . . . . . . . . . 75

3.11.2 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3.11.3 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3.11.4 Stream Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

3.11.5 Velocity Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

3.11.6 The Potential Flow Model . . . . . . . . . . . . . . . . . . . . . . . 81

3.12 Incompressible Potential Flow . . . . . . . . . . . . . . . . . . . . . . . . . 83

4 Fundamentals of Inviscid, Incompressible Flows 86

4.1 Bernoulli’s Equation (3.2) . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

4.2 Incompressible Potential Flow (3.7) . . . . . . . . . . . . . . . . . . . . . . 90

4.3 Elementary Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92


4.3.1 Uniform flow (3.9) . . . . . . . . . . . . . . . . . . . . . . . . . . 92

4.3.2 Source/Sink Flow (3.10) . . . . . . . . . . . . . . . . . . . . . . . . 93

4.3.3 Uniform Flow + source + sink (3.11) . . . . . . . . . . . . . . . . . 95

4.3.4 Doublet Flow (3.12) . . . . . . . . . . . . . . . . . . . . . . . . . . 96

4.3.5 Non-lifting Flow over a Cylinder (3.13) . . . . . . . . . . . . . . . . . 98

4.3.6 Vortex flow (3.14) . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

4.3.7 Lifting Flow over a Cylinder (3.15, 3.16) . . . . . . . . . . . . . . . . 101

5 Incompressible Flow Over Airfoils 105

5.1 Airfoil Characteristics (4.3) . . . . . . . . . . . . . . . . . . . . . . . . . . 105

5.2 Source and Vortex Sheets (4.4) . . . . . . . . . . . . . . . . . . . . . . . . 106

5.3 Kutta Condition (4.5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

5.4 Kelvin’s Circulation Theorem (4.6) . . . . . . . . . . . . . . . . . . . . . . 109

5.5 Classical Thin Airfoil Theory . . . . . . . . . . . . . . . . . . . . . . . . . 110


5.5.1 Symmetric Airfoil (4.7) . . . . . . . . . . . . . . . . . . . . . . . . 112

5.5.2 Cambered Airfoil (4.8) . . . . . . . . . . . . . . . . . . . . . . . . . 115

5.6 Panel method (3.17, 4.10) . . . . . . . . . . . . . . . . . . . . . . . . . . 120

6 Incompressible Flow over Finite Wing 121

6.1 Downwash and Induced Drag (5.1) . . . . . . . . . . . . . . . . . . . . . . 121

6.2 The Vortex Filament, Biot-Savart Law, and Helmholtz’s Theorem (5.2) . . . . . 123

6.3 Prandtl’s Classic Lifting Line Theory (5.3) . . . . . . . . . . . . . . . . . . . 124

6.3.1 Elliptical Lift Distribution . . . . . . . . . . . . . . . . . . . . . . . 128

6.3.2 General Lift Distribution . . . . . . . . . . . . . . . . . . . . . . . . 131

6.4 Reduction of Lift Slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

7 Subsonic Compressible Flow over Airfoils 137

7.1 Energy Equation in a Steady, Inviscid Flow . . . . . . . . . . . . . . . . . . 137


7.2 The Full Velocity Potential Equation (11.2) . . . . . . . . . . . . . . . . . . 140

7.3 The Linearized Velocity Potential Equation (11.3) . . . . . . . . . . . . . . . 143

7.3.1 Coefficient of Pressure in Compressible Flows . . . . . . . . . . . . . . 146

7.4 The Prandtl-Glauert Compressibility Correction (11.4) . . . . . . . . . . . . . 148

7.5 Critical Mach Number (11.6) . . . . . . . . . . . . . . . . . . . . . . . . . 149

7.6 Drag Divergence Mach Number (11.7) . . . . . . . . . . . . . . . . . . . . 151

7.6.1 Supercritical Airfoils (11.9) . . . . . . . . . . . . . . . . . . . . . . 152

8 Normal Shock Waves, Oblique Shock Waves and Expansion Waves 153

8.1 Equation of 1D Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

8.2 Stationary Normal Shock Wave Properties . . . . . . . . . . . . . . . . . . . 160

8.3 Oblique Shock Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

8.4 Prandtl-Mayer Expansion Waves . . . . . . . . . . . . . . . . . . . . . . . 174


9 Linearised Supersonic Flow 184

9.1 Some Preceding Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

9.1.1 The Speed of Sound in 1D Flow (8.3) . . . . . . . . . . . . . . . . . 184

9.1.2 Propagation of Disturbances in Subsonic and Supersonic Flows (9.1) . . 188

9.1.3 The Wave Equation (12.2) . . . . . . . . . . . . . . . . . . . . . . 189

9.2 Linearised BCs and Pressure Coefficient (12.2) . . . . . . . . . . . . . . . . . 191

9.3 Application to Supersonic Airfoils (12.3) . . . . . . . . . . . . . . . . . . . . 192

10 Introduction to Viscous Flow 194

10.1 Qualitative Characteristic of Viscous Flow . . . . . . . . . . . . . . . . . . . 194

10.2 Laminar versus Turbulent Boundary Layer . . . . . . . . . . . . . . . . . . . 197

10.3 Transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

10.4 Boundary layer properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

10.5 Incompressible Flow Over a Flat Plate: The Blasius Solution . . . . . . . . . . 201


10.6 Momentum Integral Boundary Layer Equation for a Flat Plate . . . . . . . . . 205

10.6.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

10.6.2 Turbulent boundary layer approximation . . . . . . . . . . . . . . . . 210

10.7 Computing laminar & turbulent boundary layer . . . . . . . . . . . . . . . . 215

10.7.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216


1 Fundamentals

Why are you taking this course? What do you expect to learn?

Today, we’ll get a brief overview of the basics.

1.1 What is Aerodynamics?

Concerned with the flow of air... More broadly, refers to the flow of fluid. (Almost synonymous

with fluid dynamics.)

What is this course about? Focus on external flow. And airplanes in particular.

What do we need to know about the aerodynamics of an ’object’ inorder to design it?

1. How much lift

2. drag force

3. moments

I.e. What are the forces acting on the body due to the fluid flow around it?

This course can be summarized a single concept:• No spontaneous generation of stuff (i.e. mass, momentum or energy)!


1.2 Aerodynamic Forces: How do Planes Fly? (1.4)

A fluid moving relative to a body exerts forces on this body. Consider a fluid particle moving

around a cylinder. What physically is happening such that this particle moves along this path?

Hint: think Newton...

Given this particular path, what can we say about the pressure on the surface of the cylinder?

The same can be done for a shape like the cross-section of

a bird wing:

But why should the fluid stay attached to the surface? One word: viscosity.


1.3 The Airfoil

Let us consider a generic airfoil.

1.4 Frame of Reference and Steady Flow Assumption

1. For fixed point in space,∂

∂t6= 0

2. For coordinate system attached to the airfoil (or frisbee)

∂

∂t= 0

• Steady flow

• large simplifications

⇒ Similar to a wind tunnel’s frame of reference.


1.5 Aerodynamic Forces and Moments (1.5)

R: Resultant force.

M : Resultant moment.

L: Lift, component of R ⊥ to U∞.

D: Drag, component of R ‖ to U∞.

N : Normal force, component of R ⊥ to c.

A: Axial force, component of R ‖ to c.

From geometry, L = N cosα− A sinα

D = N sinα+ A cosα.

A fluid exerts forces (and moments) on a body via:1. Pressure p (normal to surface)2. Shear stress τ (tangential to surface)

θ +ve is rotation clockwise

(i.e. increasing α).


Consider an element of length ds and assume unit span so that dA = ds.

dN ′u = −pu cosθ dsu − τu sinθ dsu

dA′u = −pu sinθ dsu + τu cosθ dsu

upper surface

dN ′l = pl cosθ dsl − τl sinθ dsl

dA′l = pl sinθ dsl + τl cosθ dsl

lower surface

The prime denotes 2D (i.e. per unit span quantity).

Total normal and axial forces per unit span are given by

N′

=

∫ TE

LE

(−pu cosθ − τu sinθ) dsu +

∫ TE

LE

(pl cosθ − τl sinθ) dsl

A′

=

∫ TE

LE

(−pu sinθ + τu cosθ) dsu +

∫ TE

LE

(pl sinθ + τl cosθ) dsl

Now, lets consider the pitching moment about the leading edge. By definition (not a right-hand rule!),

M′LE =

∫ TE

LE

[(pu cosθ + τu sinθ) x− (pu sinθ − τu cosθ) y] dsu

+

∫ TE

LE

[(−pl cosθ + τl sinθ) x+ (pl sinθ + τl cosθ) y] dsl


Note: moment force changes depending on reference point and angle of attack. Often forces

are given relative to the c/4 point. This will become clear later in this course.

Note: L/D ∝ 102 for airfoil

∝ 101 for aircraft

1.5.1 Dimensionless CoefficientsNormalize forces using free-stream velocity, U∞, and density, ρ∞.

⇒ Free-stream dynamic pressure, q∞ ≡ 1/2ρ∞U2∞.

Let us use the planform area, S, and cord length, c, as reference area and length, respectively.

(3D) CL =L

q∞S, CD =

D

q∞S, CM =

M

q∞Sc

(2D) cL =L′

q∞c, cD =

D′

q∞c, cM =

M ′

q∞c2

Pressure coefficient: Cp =p− p∞q∞

.

Skin friction coefficient: cf =τ

q∞.


Typical behaviour of CL, CD, and

CM with α.

1.6 Pressure Distributions

• Boundary layer is very thin for high Re. Thus, p1 = pw (to be derived rigorously later in

the course).

• p1 can be obtained from inviscid calculations (also covered in later chapters).


Pressure distributions.

Distribution for a symmetric airfoil at α = 0 (same on both side).

Adverse pressure gradient:

• ∂p/∂x > 0

• prone to separation

• almost always turbulent

Favourable pressure gradient:

• ∂p/∂x < 0

• attached BL.

Distribution for a cambered airfoil.


1.7 Centre of Pressure (1.6)

Where to place N ′ and A′ such that they have the same effect as the distributed loads?

⇒ Where they produce the same moment about the leading edge (i.e. where Mcp = 0).

By definition M ′LE = −xcpN ′

(No contribution due to A′ since it

is colinear to the moment arm.)

xcp: centre of pressure.

Recall L′ = N ′ cosα− A′ sinα

Small angle α gives cosα ' 1

sinα ' α ' 0

∴ L′ ' N ′

Hence, xcp = −M ′

LE

N ′' −

M ′LE

L′

xcp is dependent on lift - for L′ → 0, xcp → −∞!! Not very convenient...


∴ Convenient to specify at c/4

Small α

M′LE ' −

c

4L′+M

′c/4

• For symmetric airfoil in inviscid, incompressible flow, CMc/4 = 0 or xcp = xc/4 at all α.

(to be demonstrated in §5).

• For cambered airfoil, xcp a bit further downstream than xc/4.


1.8 Types of Flow (1.10)1.8.1 Inviscid vs. Viscous

• High Re, streamlined, low α (< 10).

– Viscous effects confined to attached boundary layer.

– Inviscid assumption good for L, M , but not D.

• High α.

– Inviscid solution not good.

• Bluff body

– Inviscid solution useless!!!


1.8.2 Viscous Flows: Boundary Layers and Wakes

Most of the flows in aerodynamics can be assumed inviscid. That is not to say viscosity is not

important! Let us consider the different types of viscous flows.

The importance of viscosity is related to the Reynolds number:

Re =ρU`

µ=U`

ν∝

inertial forces

viscous forces=(ρU

2)(µU

`,

)−1

where U and ` are the velocity and length scales of the flow. For an airfoil, U ≡ U∞ and

` ≡ c, where c is the cord of the airfoil.• For a glider, Re ∼ 106.

• For large commercial aircraft, Re ∼ 108.

For high Re, viscocity is relatively unimportant to the flow globally.

Within boundary layers and wakes, gradients are high, so viscous effects are important,

regardless of Re.

⇒ Viscous forces resist gradients.

→ leads to the generation of shear stress at the wall (skin friction drag).


1.8.3 Laminar, Transitional and Turbulent Flows

Laminar flow: – Steady.

– Smooth streamlines.

Turbulent flow: – Unsteady (though it can be statistically steady).

– Irregular/chaotic-like behavior.

– Higher ∂u/∂y than laminar boundary layers.

– Thus higher shear stress (∼ 7 times more).

– Thicker boundary layer (at same Reynolds number).

Transition. Complex process, which depends mainly on:• Perturbation environment (surface roughness, free stream turbulence, acoustic waves).

• Reynolds number (less stable at higher Re).

• Pressure gradient.

– ∂p/∂x < 0→ favorable pressure gradient.

– ∂p/∂x > 0→ adverse pressure gradient.


1.8.4 Attached vs. Separated Flows

• The flow is said to be attached if the streamlines close to the surface follow the surface.

• The flow is separated if some streamlines change directions.

e.g. streamlined body:

or bluff body:


For attached flow.

∂u

∂y

∣∣∣∣y=0

> 0

Verge of separation.

∂u

∂y

∣∣∣∣y=0

= 0

Boundary layer separation

(flow reversal).

∂u

∂y

∣∣∣∣y=0

< 0


Separation can lead to

• oscillatory flow due to vortex shedding,

• increase drag, and

• loss of lift on airfoils (stall).

Particular examples: Leading vs trailing edge separation.

Round LE:

TE separation

Gradual stall

Sharp LE:

LE separation

Sudden stall

Turbulent boundary layers have increased drag because of increased momentum transfer, which

also leads to increase resistance to separation. ∴, turbulence can also be good!


1.8.5 Compressible vs incompressible

The difference between compressible and incompressible flow is ’obvious’ - either ρ is constant

or it’s not! But what governs whether ρ is constant in open aerodynamics?

It all depends on how quickly the fluid is moving relative to the speed at which information

about what’s coming can be transmitted. This is quantified via the Mach number, defined as

M =U

a,

where U is the characteristic speed and a is the speed of sound (speed at which information

moves upstream). The speed of sound in an ideal gas (like air) is given by

a =√γRT ,

where γ is the specific heat ratio, R is the perfect gas constant and T is the ambient

temperature (absolute).

Mach number regimes

(a) M < 0.3 Incompressible

subsonic

e.g.: • General aviation aircraft.

• Sail planes.

• Marine applications.


(b) M < 1

everywhere

Compressibility

effects

subsonic

e.g.: deHavilland commuter aircraft

0.4 ≤M∞ ≤ 0.6

(c) M ' 1 transonic e.g.: Commercial aircraft.

• M∞ < 1

• M∞ > 1

(d) 1 < M∞ < 5 supersonic e.g.: High speed civil aircraft (Concord)


2 The Basic Equations of Fluid Dynamics2.1 General Form of a Conservation Law2.1.1 Scalar Conservation Law

We consider the scalar quantity per unit volume, U , that exists in

a volume. A volume element is given by dΩ and a surface element

is given by dS. The unit outward normal for the surface element

is n. The fluxes passing through Ω are contained in the vector ~F .

Surface and volume sources are given by ~QS and QV respectively.

Note that n is a row vector, while ~F and ~Q are column vectors.

The variation per unit time of the scalar quantity U within Ω is given by

d

dt

∫Ω

UdΩ.

This variation is given by contributions from incoming fluxes and source terms. The net

contribution from the incoming fluxes through S is

−∮S

n · ~FdS.

Contributions from the volume and surface source terms are given by∫Ω

QV dΩ +

∮S

~QS · ndS.


The net balance is given by

d

dt

∫Ω

UdΩ = −∮S

n · ~FdS +

∫Ω

QV dΩ +

∮S

~QS · ndS.

We assume ~QS and ~F are continuous in space. Given Gauss’ Theorem, which states that the

flux of a vector field through a closed surface is equal to the volume integral of the divergence

of the region inside the surface (also known as the divergence theorem), viz.∫Ω

~∇ · ~FdΩ =

∮S

n · ~FdS,

we getd

dt

∫Ω

UdΩ = −∫

Ω

~∇ · ~FdΩ +

∫Ω

QV dΩ +

∫Ω

~∇ · ~QSdΩ.

This is referred to as the volume integral form of the scalar conservation law. With Ω

assumed to be fixed in space, we get∫Ω

∂U

∂tdΩ = −

∫Ω

~∇ · ~FdΩ +

∫Ω

QV dΩ +

∫Ω

~∇ · ~QSdΩ.

Since Ω is arbitrary, we can obtain a differential form given by

∂U

∂t+ ~∇ · ~F = QV + ~∇ · ~QS.


The flux vector ~F is composed of convective fluxes and diffusive fluxes as follows

~F = ~FC + ~FD.

→ Convective fluxes are simply given by

~FC = ~vU,

where ~v is the velocity vector. It is attributed to changes in U due to the motion of the fluid.

→ In contrast, diffusive fluxes are defined as the contribution present in fluids at rest (e.g.

due to molecular motion). We use the generalized gradient law of Fick (i.e. the flux is from

region of high to low concentration, and proportional to the concentration gradient) to describe

the diffusive flux. With Uρ describing the quantity per unit mass, we have

~FD = −Xρ~∇(U

ρ

),

where X is the constant of proportionality. Substituting this into the scalar diffential form, we

get the generalized scalar law for U . The differential form becomes

∂U

∂t+ ~∇ · (~vU) = ~∇ ·

[Xρ~∇

(U

ρ

)]+QV + ~∇ · ~QS.


In two-dimensional Cartesian coordinates, we have

~v = u ı+ v

~FC = uU ı+ vU

~FD = −Xρ[∂

∂x

(U

ρ

)ı+

∂

∂y

(U

ρ

)

]~∇ · ~FC =

∂uU

∂x+∂vU

∂y

~∇ · ~FD = −∂

∂x

[Xρ

∂

∂x

(U

ρ

)]+

∂

∂y

[Xρ

∂

∂y

(U

ρ

)]

2.1.2 Vector Conservation Law

Typically, we wish to express several equations together in an efficient manner. For example, the

momentum conservation law is conveniently expressed as a vector conservation law. The

conserved vector quantity can be expressed in a coordinate system by its components. For

example, in three-dimensional Cartesian coordinates, the conserved vector quantity ~U is given by

~U = U1 ı+ U2 + U3 k.

We will use three-dimensional Cartesian coordinates in this section to express vector and tensor

quantities. We will use subscripts 1, 2, and 3 to express components of vector and tensor

quantities. Using this notation we are able to easily describe the vector conservation law by its


components. Using the scalar conservation law from the previous section, the conservation of

component U1 is given by

d

dt

∫Ω

U1dΩ = −∮S

n ·(~fC1 + ~fD1

)dS +

∫Ω

QV 1dΩ +

∮S

n · ~qS1dS,

where ~fC1 and ~fD1 are the convective and diffusive fluxes of U1. Specifically,

~fC1 = ~vU1

= uU1 ı+ vU1 + wU1 k

~fD1 = −Xρ[∂

∂x

(U1

ρ

)ı+

∂

∂y

(U1

ρ

)+

∂

∂z

(U1

ρ

)k

].

The surface source vector for U1 is given by ~qS1 and QV 1 is the volume source component of

U1.

In a similar manner, we can write the conservation of U2 and U3. Collectively this is a system

of equations that looks like

d

dt

∫Ω

U1dΩ = −∮S

n ·(~fC1 + ~fD1

)dS +

∫Ω

QV 1dΩ +

∮S

n · ~qS1dS

d

dt

∫Ω

U2dΩ = −∮S

n ·(~fC2 + ~fD2

)dS +

∫Ω

QV 2dΩ +

∮S

n · ~qS2dS

d

dt

∫Ω

U3dΩ = −∮S

n ·(~fC3 + ~fD3

)dS +

∫Ω

QV 3dΩ +

∮S

n · ~qS3dS.


We can use matrices to compactly describe the system as

d

dt

∫Ω

U1dΩ

U2dΩ

U3dΩ

= −∮S

n ·(~fC1 + ~fD1

)dS

n ·(~fC2 + ~fD2

)dS

n ·(~fC3 + ~fD3

)dS

+

∫Ω

QV 1dΩ

QV 2dΩ

QV 3dΩ

+

∮S

n · ~qS1dS

n · ~qS2dS

n · ~qS3dS

.Furthermore, we can describe the conserved vector quantity ~U by

~U =

U1

U2

U3

and the volume source term by

~QV =

QV 1

QV 2

QV 3

.Hence the system becomes

d

dt

∫Ω

~UdΩ = −∮S

n ·

~fC1 + ~fD1

~fC2 + ~fD2

~fC3 + ~fD3

T

dS +

∫Ω

~QV dΩ +

∮S

n ·

~qS1

~qS2

~qS3

T

dS,

where the rows of the two matrices [] are the flux and surface source vectors.


We now introduce the flux tensor F which contains the convective and diffusive flux

components. Furthermore, we introduce the surface source tensor QS. Using these tensors we

can finally express the vector integral form of the vector conservation law as

d

dt

∫Ω

~UdΩ = −∮S

n · FdS +

∫Ω

~QV dΩ +

∮S

n ·QSdS.

Invoking Gauss’ Theorem, we arrive at the differential form

∂~U

∂t+ ~∇ ·

(F −QS

)= ~QV .

We continue with some more discussion about the flux tensor F . It is comprised of the

convective and diffusive flux tensors as

F = FC + FD.

The convective flux tensor is given by

FC = ~v ⊗ ~U.

Using index notation, viUj defines a matrix FCij as

FCij = viUj =

v1U1 v1U2 v1U3

v2U1 v2U2 v2U3

v3U1 v3U2 v3U3

.


The columns of FCij are the vectors ~fC1, ~fC2, and ~fC3 respectively. The diffusive flux tensor is

also an extension of the diffusive flux vector. The tensor is given by the matrix

FDij = −Xρ∂(Ujρ

)∂xi

.

Again, the columns of FDij are the vectors ~fD1, ~fD2, and ~fD3 respectively. For example,

~fD1 = −Xρ~∇(U1

ρ

)= −Xρ

∂

(U1ρ

)∂x1

∂

(U1ρ

)∂x2

∂

(U1ρ

)∂x3

.

2.2 The Equation of Mass ConservationWe use the scalar conservation law to derive the equation of mass conservation. The conserved

scalar quantity is simply the density of the fluid (i.e. unit mass per unit volume1)

U = ρ.

Hence, the convective and diffusive fluxes are

~FC = ~vU

= ρ~v1In fluids, quantities are expressed in unit volume (compared to the definition given in mechanics). This is typically assumed to be

known implicitly and will not be stated explicitly from this point on.


~FD = −Xρ~∇(U

ρ

)= −Xρ~∇(1) = 0

The diffusive flux is zero since the ratio of U/ρ is a constant. Furthermore, there are no surface

or volume sources of mass. Hence,

~QS = ~0

QV = 0.

The integral form of mass conservation is therefore given by

d

dt

∫Ω

ρdΩ +

∮n · (ρ~v) dS = 0.

The differential form is given by (again Gauss’ Theorem is used here)

∂ρ

∂t+ ~∇ · (ρ~v) = 0.

We introduce the familiar total derivative operator (i.e. substantial derivative) to find an

alternative form of the differential form of the mass continuity equation. Recalling,

D

Dt≡∂

∂t+ ~v · ~∇,

we have ∂ρ

∂t+ ~v · ~∇ρ+ ρ~∇·~v = 0, or


Dρ

Dt+ ρ~∇·~v = 0.

This states that rate of change of density of a fluid particle that is being tracked, varies

proportionally with the negative of the dilatation of that fluid particle.

One particular case of this equation is what is referred to as the incompressible continuityequtation. For incompressible flow, ρ is constant both in space and time, consequently

~∇ · ~v = 0.

For an incompressible fluid, the divergence of the velocity field is zero.

2.3 The Equation of Momentum Conservation

The momentum conservation equation requires the vector conservation law. Here the conserved

quantity is the momentum given by~U = ρ~v.

→ The convective flux tensor is given by

FC = ~v ⊗ ρ~v

and the diffusive flux tensor is zero since there is no momentum diffusion for a fluid at rest (as

per our definition of diffusive fluxes).


→ Sources of variation of momentum are forces, ~fe. For example, if gravity is the only external

force that the fluid experiences we can write

~QV = ρ~fe = ρ~fg.

→ Surface sources of momentum are caused by internal forces in the fluid including pressureand viscous forces. They are dependent on the nature of the fluid, specifically the relation

between internal deformations and internal stresses. They only appear on the surface because

internally they cancel out. Therefore,

QS ≡ σ = −pI + τ,

where σ is the stress tensor, −pI is the normal stress tensor associated with pressure, and τ

is the tensor describing viscous forces. The unit tensor I means that the pressure stress acts

normal to the surface of the fluid particle.

The viscous stress tensor has some special properties. The most important arises due to

rotational equilibrium. In three-dimensional Cartesian coordinates, the stress tensor τij is2

τij =

τ11 τ21 τ31

τ12 τ22 τ32

τ13 τ23 τ33

=

τxx τyx τzx

τxy τyy τzy

τxz τyz τzz

=

~τx

~τy

~τz

.2This tensor notation is somewhat different than what we defined on page 37. As we will see, τij is symmetric, thus τij = τji. The

notation used here is related to the physical orientation and location of the force as defined on page 42.


We write two forms here.The first index indicates what surface the

stress acts on (i.e. the direction of the normal of the surface) and

the second index indicates in what direction the stress acts. For

example, τxy, would be a viscous stress on the side with normal

colinear to ı and in the direction colinear to . When the indices are

the same, the stress acts normal to the surface and is referred to as

the normal viscous stress. When the indices are different, then the

stress is called a shear (viscous) stress.

Generally, shear stresses are much more important than normal viscous stresses. In aerodynamics

we are concerned primarily with air which is a Newtonian fluid. It can be shown that3

τij = µ

[(∂vj

∂xi+∂vi

∂xj

)−

2

3

(~∇ · ~v

)δij

],

where µ is the dynamic viscosity and δij is the Kronecker delta function, which simply states

δij =

1 if i = j

0 if i 6= j.

Clearly, τij is a symmetric tensor.

3The normal stress may seem a bit odd, but they express the viscous forces that appear due to the gradients ∂vi/∂xi, which arenormally very small (unless through a shock wave). The first term is due to the standard Fick Law, while the second term means thatthe normal stress also have a component proportional to the dilatation of the fluid element. The constant of proportionality is roughlyequal to -2/3µ based on empirical evidence, which is a good enough assumptions for all our purposes (but if you are curious google ‘bulkviscosity coefficient’).


Let us consider for example τij in two-dimensional Cartesian coordinates. Here, x1 = x,

x2 = y, v1 = u, and v2 = v. Hence,

τxx = µ

[4

3

∂u

∂x−

2

3

∂v

∂y

]τxy = µ

[∂v

∂x+∂u

∂y

]= τyx = µ

[∂u

∂y+∂v

∂x

]τyy = µ

[4

3

∂v

∂y−

2

3

∂u

∂x

].

The momentum conservation equation in integral form is given by

d

dt

∫Ω

ρ~vdΩ +

∮S

n · (ρ~v ⊗ ~v) dS =

∫Ω

ρ~fedΩ +

∮S

n · σdS.

The surface source term can be divided into∮S

n · σdS = −∮S

n·(pI)dS +

∮S

n · τdS.

To arrive at the differential form, we apply Gauss’ Theorem. Hence,

∂

∂t(ρ~v) + ~∇ ·

(ρ~v ⊗ ~v + pI − τ

)= ρ~fe.


Using the total derivative and conservation of mass (which is a nontrivial exercise) we can also

write

ρD~v

Dt= −~∇p+ ~∇ · τ + ρ~fe.

This equation is Newton’s Second law for a fluid particle. The left-hand side is the mass per

unit volume times acceleration, and the right hand side is the net force that acts on the fluid

particle.

For a Newtonian fluid with a constant viscosity, µ, we find

ρD~v

Dt= −~∇p+ µ

[∇2~v +

1

3~∇(~∇ · ~v

)]+ ρ~fe

which are referred to as the Navier-Stokes equations. Taking µ out of the derivatives results

in negligible error. The operator ∇2 is the Laplacian operator. For example, in

three-dimensional Cartesian coordinates

∇2 ≡∂2

∂x2+

∂2

∂y2+∂2

∂z2.

Furthermore, we recall that for incompressible flow ~∇ · ~v = 0. For constant µ and

incompressible flow, the equations reduce to

ρD~v

Dt= −~∇p+ µ∇2

~v + ρ~fe.


2.4 The Equation of Energy Conservation

The scalar conservation law is used to derive the energy equation. The conserved quantity per

unit volume, U , is energy and is given by

U = ρE,

where E is the energy per unit mass. (Multiplying E by ρ gives us the necessary energy per

unit volume dimensions.) This energy can be divided into

E = e+~v · ~v

2,

where e is the internal energy per unit mass and ~v · ~v/2 is the kinetic energy per unit mass.

→ The convective flux is given by~FC = ρE~v,

and the diffusive flux is given by~FD = −k~∇T,

which is Fourier’s law of heat conduction. The constant k is called the thermalconductivity.

→ Volume sources for the variation of total energy per unit volume are the work done by forces

acting on the system in addition to heat transmitted to the system. Hence,

QV = Wf + qH

= ρ~fe · ~v + qH,

where Wf is the work done by the body forces and qH is the heat addition other than heat

conduction.


→ Surface sources come from work that is done by the internal forces (i.e. pressure and viscous

stresses). Therefore,~QS = σ·~v

= −pI ·~v + τ ·~v.

Substituting the above equations into the volume integral form of the general scalar

conservation law we find

d

dt

∫Ω

ρEdΩ+

∮S

n·(ρE~v) dS =

∮S

n·(k~∇T

)dS+

∫Ω

(Wf + qH) dΩ+

∮S

n·(σ · ~v

)dS.

Using Gauss’ Theorem, we acquire the differential form, viz.

∂

∂t(ρE) + ~∇ · (ρE~v) = ~∇ ·

(k~∇T

)+Wf + qH + ~∇ ·

(σ · ~v

).

We define the total enthalpy per unit mass as

H = E +p

ρ

to find an alternative form of the energy equation where the stress tensor is simplified to the

viscous stress tensor. The energy equation becomes

∂

∂t(ρE) + ~∇ ·

(ρH~v − k~∇T − τ · ~v

)= Wf + qH.


2.5 SummaryIn this chapter we introduced the scalar and vector conservation laws. Index and tensor notation

proved to be valuable in describing the latter. The scalar conservation law was used to derive

the mass and energy conseravtion laws for fluids. The vector conservation law was used to

derive the momentum equation. We also introduced various quantities along the way such as

viscosity and thermal conductivity.

In three-dimensional Cartesian coordinates, the system consists of 5 equations. However, in its

current form with k and µ, there are 15 unknowns. The next chapter is entirely focused on

closing the system and exploring various approximations to it.


3 The Dynamic Levels of Approximation

3.1 The Navier-Stokes Equations

∂

∂t

ρ

ρ~v

ρE

+ ~∇ ·

ρ~v

ρ~v ⊗ ~v + pI − τρ~vH − τ · ~v − k~∇T

=

0

ρ~fe

Wf + qH

.e.g. 3D Cartesian coordinates gives:

k~∇T = k

[∂T

∂xı+

∂T

∂y+

∂T

∂zk

]

τ · ~v =

~τx · ~v~τy · ~v~τz · ~v

(vector),

where, for instance, one row of τ is given by

~τx =

τxx

τyx

τzx

and ~v =

u

v

w

.AER307: Aerodynamics c©Philippe Lavoie 2012 3-48

Therefore,

~τx · ~v = τxxu+ τyxv + τzxw

~τy · ~v = τxyu+ τyyv + τzyw

~τz · ~v = τxzu+ τyzv + τzzw.

Also,

pI =

p 0 0

0 p 0

0 0 p

, τ =

τxx τyx τzx

τxy τyy τzy

τxz τyz τzz

, ~v ⊗ ~v =

u2 uv uw

vu v2 vw

uw vw w2

.System can be rewritten to look like:

∂U

∂t+∂f

∂x+∂g

∂y+∂h

∂z= Q,

where

U =

ρ

ρ~v

ρE

=

ρ

ρu

ρv

ρw

ρE

→ “conservative state variables”


f =

ρu

ρu2 + p− τxxρuv − τxyρuw − τxzρuH − (τxxu+ τyxv + τzxw)− k∂T

∂x

g =

ρv

ρuv − τyxρv2 + p− τyyρwv − τyzρvH − (τxyu+ τyyv + τzyw)− k∂T

∂y

h =

ρw

ρuw − τzxρvw − τzyρw2 + p− τzzρwH − (τxzu+ τyzv + τzzw)− k∂T

∂z


3.2 Closure of the System

We have 5 equations, 15 unknowns (underdetermined!): ρ, ~v, E, p, τ,H, k, T.

(Remember, ~v counts for three unknowns and τ only for 6 since it is symmetric.)

Assumptions

1. Newtonian Fluid (down to 9 unknowns)

τ = τ (~v, µ) ← lose 6 unknowns (τ)

← gain 1 unknown (µ)

H = E +pρ ← lose 1 unknown.

2. Thermodynamics laws (down to 7 unknowns)

e = e (ρ, T )

p = p (ρ, T )

]← lose 2 unknowns.

3. Dependence of µ and k on fluid properties (down to 5 unknowns)

Sutherland’s law for viscosity

µ =1.45× 10−6T 3/2

T + 110(S.I. unites).

Prandtl numberPr =

µCp

k,

where Cp is the heat capacity for constant pressure, and Pr = 1.45 for air.

Above two equations are approximate and don’t work at high T . Use tables for these cases.


3.3 Boundary Conditions1. No-slip condition: ~v = 0 at a solid surface. More rigorously, ~v = ~vw, where ~vw is the

velocity of the surface.

2. Wall tangency: ~v is along a solid surface (inviscid boundary condition).

3. Isothermal: T = Tw at solid surface, where Tw is the temperature of the surface.

4. Adiabatic: ∂T/∂n = 0 at solid surface. I.e. no heat transfer through surface.

3.4 A Brief Review of Thermodynamics

E = e+ ~v · ~v2 : energy per unit mass

→ e : internal energy per unit mass (i.e. specific energy)

h = e+pρ : enthalpy

= e+ pυ

→ υ =1

ρ: specific volume

H = E + pρ : total enthalpy per unit mass

Ideal gas law (Thermally Perfect)

p = ρRT ⇒ pυ = RT

Specific heat at constant volume: Cv = ∂e∂T

∣∣∣v


Specific heat at constant pressure: Cp = ∂h∂T

∣∣∣p

For a thermally perfect gas

h = h(T ) ⇒ dh = Cp dT

e = e(T ) ⇒ de = Cv dT

For a calorically perfect gasCv

Cp

are constant

γ =CpCv

; Cp − Cv = R

Cp =γ

γ − 1R ; Cv = 1γ − 1R

γ = 1.4 for air

3.4.1 First Law of Thermodynamics

δq + δW = de

δq is the heat added to system,

δW is work done on system, and

de is the change in energy (dependent on initial and final states).


Some special cases of interest:

1. Adiabatic (no heat transfer)⇒ δq = 0.

2. Reversible (hypothetical)⇒ no dissipative/diffusive mechanism (i.e. no viscosity, thermal

conduction or mass diffusion).

3. Isentropic⇒ cases 1 and 2.

For reversible processes,δW = −p dv,

where dv is an incremental ∆ in volume due to displacement of body.

∴ δq = p dv + de

3.4.2 Second Law of Thermodynamics

Change in entropy→ ds =δqrevT

.

ds =δq

T+ dsirrev

• 2nd Law⇒ dsirrev ≥ 0 (= 0 for reversible).

• ⇒ ds ≥ δqT

(=δqT

for reversible).

• Adiabatic⇒ ds ≥ 0, ∵ δq = 0.

• Isentropic⇐ Adiabatic + Reversible.


Assume reversible for now (i.e. dsirrev = 0), ∴

ds =δq

T⇒ Tds = δq

But δq = pdv + de (from 1st law)

∴ Tds = de+ pdv (3.1)

Recall h = e+ pv ⇒ dh = de+ pdv + vdp,

∴ Tds = dh− vdp (3.2)

Thermally perfect gas ⇒ de = CvdT

dh = CpdT

pv = RT ⇒ vT

= Rp

(3.1)⇒ ds = CvdTT

+pdvT

(3.2)⇒ ds = CpdTT− vdp

T= Cp

dTT − R

dpp

s2 − s1 =

∫Cp

dT

T−∫R

dp

p


Calorically perfect gas⇒Cp = const.

Cv = const.

∴ R = const.

∴ s2 − s1 = Cp ln T2T1− R ln

p2p1

Isentropic⇒ s2 = s1 (not true across shocks or in boundary layers or wakes)

∴ Cp lnT2

T1

= R lnp2

p1

∴p2

p1

=

(T2

T1

)CpR

=

(T2

T1

) γγ−1

⇒p2

p1

=

(T2

T1

)3.5

for air

Ideal/thermally perfect gas: p = ρRT . Therefore, we get the following isentropic relations,

ρ2

ρ1

=p2T1

p1T2

=

(T2

T1

) γγ−1(T2

T1

)−1

=

(T2

T1

) 1γ−1

p2

p1

=

(ρ2

ρ1

)γp2 =

(p1

ργ1

)ργ2 = Cρ

γ2,

where C is a constant.


3.5 The Speed of Sound

The speed of sound a is given by (this relationship will be derived in §9)

a2

=∂p

∂ρ

∣∣∣∣s

.

For a perfect gas, p = Cργ, such that

dp

dρ= Cγρ

(γ−1)= γ

p

ρ= γRT

∴ a =√γRT

The speed of sound is a function of T only!!!

3.5.1 Mach Number

The Mach number refers to the speed of the fluid with respect to the speed of sound. The local

Mach number is given by

M =|~v|a


Compressibility.

Consider the ratio→ kinetic energy/unit massinternal energy/unit mass

=v

2/2

CvT=

v2/2

RT/(γ − 1)=

v2/2

a2/γ(γ − 1)

=γ(γ − 1)

2v

2

a2

=γ(γ − 1)

2 M2

M is a measure of compressibility.

• M . 0.3 incompressible.

• M > 0.3 compressible.


Mach Regions

M Classification

M ∼ 0, M < 0.3 incompressible

M < 1 subsonic

M . 1, M & 1 transonic

M > 1 supersonic

M 1 hypersonic

3.6 Classification of Simplifications

Get back to thinking about how to simplify our equations.

Simplifications that can be made:

1. 3D→ 2D w = 0, ∂/∂z = 0

2. Steady ∂/∂t = 0

3. Incompressible ρ = const.

4. Inviscid µ = 0

5. no heat conduction k = 0


Classification diagram


Validity of Simplifications

(a) Inc. N.-S. Eq.: • Viscous flow, low M.

• Energy eqn. decoupled (only solve if you need T)

• Numerically a little cheaper than N-S.

(b) B.L. Eqn.: • Low α, high Re.

• Not valid for complete flow field (must solve p field

from different equation).

• Can be coupled to inviscid solution to provide viscous effects.

• Numerically expensive.

(c) Euler: • Compressible, inviscid.

• Cruise at low α

• Attached flow

when viscous effects are small.

• Not good for drag prediction.

unless wave drag→ shocks (only when no separation).

• Numerically cheap.

(d) Full potential: • Similar to Euler but for weak shocks.

• Cheaper than Euler.


(e) Linearized F.P.: • Small perturbations.

• Thin, streamlined bodies at low α.

(f) Inc. potential flow: • Same as Euler + F.P. but M∞ < 0.3.

• Numerically cheapest.

3.7 Incompressible Fluid ModelM < 0.3 (small Mach number, no shocks). ∴ ρ = const.

MassDρ

Dt+ ρ~∇ · ~v = 0

∴ ~∇ · ~v = 0

Momentum ρD~v

Dt= −~∇p+ µ∇2

~v + ρ~fe

D~vDt

= −~∇pρ + ν∇2~v + ~fe

Energy If you want T , then solve this decoupled

⇒ T comes via p.

ν ≡ µρ

µ: dynamic viscosity

ν: kinematic viscosity


3.8 The Reynolds-average N-S Equations

In high Re, high shear regions of the flow, turbulent velocity fluctuations appear (e.g. in

boundary layer).

Recall

B.L. profiles

For turbulent:

•∂u

∂yhigher on average at the wall than

laminar.

• B.L. thicker.

Why? answer at end of section

Reynolds decompositionA = A+ A′,

where A is a time averaged quantity, and

A′ is the fluctuating component.

A (~x) = 1T

∫ T

0

A (~x, t+ τ) dτ

A′ ≡ 0

Example: Apply approach to the incompressible x-momentum equation.4

∂

∂tρu+

∂

∂x

(ρu

2+ p− τxx

)+

∂

∂y(ρuv − τyx) +

∂

∂z(ρuw − τzx) = ρfex.

4Note that the time dependence of the momentum equation is kept in the following analysis for generality since the mean can vary intime if the variation is much slower than the period T over which the average is taken. This is justified if the turbulent quantity A′ varieson a time scale much smaller than T .


Apply ρ = const.

⇒ ρ∂u

∂t+ ρ

∂u2

∂x+ ρ

∂uv

∂y+ ρ

∂uw

∂z= −

∂p

∂x+∂τxx

∂x+∂τyx

∂y+∂τzx

∂z+ ρfex.

Apply5u = u+ u

′, v = v + v

′, w = w + w

′, p = p+ p

′

ρ∂

∂t

[u+ u

′]+ ρ

∂

∂x

[u

2+ 2uu

′+ u

′2]

+ ρ∂

∂y

[u v + uv

′+ u

′v + u

′v′]

+ρ∂

∂z

[u w + uw

′+ u

′w + u

′w′]

= −∂

∂x

[p+ p

′]+∂τxx

∂x+∂τyx

∂y+∂τzx

∂z+ ρfex.

Take average: u′ = 0 u′w = 0 2uu′ = 0 uw′ = 0

uv′ = 0 u′v = 0 p′ = 0

Non-zero terms:

u2 = u2, u v = u v, u w = u w, p = p.

New terms⇒ u′2 6= 0, u′v′ 6= 0, u′w′ 6= 0. Such that,5Note that τij terms are linear with respect to velocity. For simplicity, we do not expend the terms here since only the mean

component remains.


ρ∂u

∂t+ ρ

∂

∂x

[u

2+ u′2

]+ ρ

∂

∂y

[u v + u′v′

]+ ρ

∂

∂z

[u w + u′w′

]= −

∂p

∂x+∂τxx

∂x+∂τyx

∂y+∂τ zx

∂z+ ρfex.

Reynolds stresses

τRxx ≡ −ρu′2

τRyx ≡ −ρu′v′

τRzx ≡ −ρu′w′

Model these terms

ρ∂u

∂t+ ρ

∂

∂xu

2+ ρ

∂

∂y(u v) + ρ

∂

∂z(u w) = −

∂p

∂x

+∂

∂x

(τxx + τ

Rxx

)+

∂

∂y

(τyx + τ

Ryx

)+∂

∂z

(τ zx + τ

Rzx

)+ ρfex

* Derivation for v- and w-momentum equation is left as an exercise.

⇒ Answer to above question: new terms, the turbulent shear stress terms, produce an increase

in momentum transfer, leading to the thicker boundary layer with a ‘fuller’ velocity profile.


3.9 The Boundary-Layer Approximation

We perform here an order of magnitude analysis. To make analysis easier, assume steady, 2D,

no body forces (~fe = 0). Equation remains valid for viscous and compressible flows.

Continuity∂ (ρu)

∂x+∂ (ρv)

∂y= 0

x-momentum

ρu∂u

∂x+ ρv

∂u

∂y= −

∂p

∂x+

∂

∂x

[µ

(4

3

∂u

∂x−

2

3

∂v

∂y

)]+

∂

∂y

[µ

(∂v

∂x+∂u

∂y

)].

Lets make these equations non-dimensional, by defining

ρ′=

ρ

ρ∞, u

′=

u

U∞, v

′=

v

U∞, p

′=

p

p∞

y′=y

c, x

′=x

c, µ

′=

µ

µ∞Continuity

∂(ρ∞ρ

′U∞u′)

∂ (cx′)+∂(ρ∞ρ

′U∞v′)

∂ (cy′)=

[ρ∞U∞

c

]∂(ρ′u′)

∂x′+

[ρ∞U∞

c

]∂(ρ′v′)

∂y′= 0

∴∂(ρ′u′)

∂x′+∂(ρ′v′)

∂y′= 0


x-momentumTerms of equation become,

ρu∂u

∂x= ρ

′ρ∞u

′U∞

∂(u′U∞

)∂ (x′c)

=ρ∞U

2∞

cρ′u′∂u

′

∂x′

ρv∂u

∂y=ρ∞U

2∞

cρ′v′∂u

′

∂y′

−∂p

∂x= −

p∞

c

∂p′

∂x′

∂

∂x

[µ

(4

3

∂u

∂x−

2

3

∂v

∂y

)]=µ∞U∞

c2

∂

∂x′

[µ′(

4

3

∂u′

∂x′−

2

3

∂v′

∂y′

)]∂

∂y

[µ

(∂v

∂x+∂u

∂y

)]=µ∞U∞

c2

∂

∂y′

[µ′(∂v′

∂x′+∂u′

∂y′

)]Introduce into momentum equation and multiply equation by c

ρ∞U2∞

to get

ρ′u′∂u

′

∂x′+ρ′v′∂u

′

∂y′= −

p∞

ρ∞U2∞

∂p′

∂x′

+µ∞

ρ∞U∞c

∂

∂x′

[µ′(

4

3

∂u′

∂x′−

2

3

∂v′

∂y′

)]+

∂

∂y′

[µ′(∂v′

∂x′+∂u′

∂y′

)].


But

Re =ρ∞U∞c

µ∞,

andp∞

ρ∞U2∞

=γp∞

γρ∞U2∞

=γp∞

ρ∞

1

γU2∞

= γRT∞1

γU2∞

=a2∞

γU2∞

=1

γM2∞

p∞

ρ∞U2∞

=1

γM2∞

,

therefore,

ρ′u′∂u

′

∂x′+ ρ

′v′∂u

′

∂y′= −

1

γM2∞

∂p′

∂x′

+1

Re

∂

∂x′

[µ′(

4

3

∂u′

∂x′−

2

3

∂v′

∂y′

)]+

∂

∂y′

[µ′(∂v′

∂x′+∂u′

∂y′

)].

Consider an airfoil with a growing boundary layer on the surface

1. Streamlined (i.e. no separation)

2. High Re.

3. Boundary layer thickness, δ(x), such

that δ/c 1 (verify this next page).


Under these conditions, an order of magnitude analysis suggests that

ρ′=

ρ

ρ∞= O(1) , u

′=

u

U∞= O(1) , v

′=

v

U∞= O(?) p

′=

p

p∞= O(1)

x′=x

c∈ [0, 1] = O(1) , y

′=y

c∈[0,δ

c

]= O

(δ

c

), µ

′=

µ

µ∞∼ 1± 0.1

Applying order of magnitudes to continuity equation gives,

∂(ρ′u′)

∂x′+∂(ρ′v′)

∂y′= 0

y′= O

(δ

c

)⇒

O(1)O(1)

O(1)+O(1)O(??)

O(δ/c)= 0

∴ v′ = O

(δ

c

)⇐ small !!

Doing the same to the x-momentum equation yields,

ρ′u′∂u

′

∂x′= O(1)O(1)

O(1)

O(1)= O(1)

ρ′v′∂u

′

∂y′= O(1)O

(δ

c

)O(1)

O(δ/c)= O(1)

1

γM2∞

∂p′

∂x′=

O(1)

O(1)O(1)

O(1)

O(1)= O(1)


∂

∂x′

[µ′(

4

3

∂u′

∂x′−

2

3

∂v′

∂y′

)]=

1

O(1)

[O(1)

(O(1)

O(1)−O(δ/c)

O(δ/c)

)]= O(1)

∂

∂y′

[µ′(∂v′

∂x′+∂u′

∂y′

)]=

1

O(δ/c)

[O(1)

(O(δ/c)

O(1)+

O(1)

O(δ/c)

)]= O

(1

(δ/c)2

).

Net effect is that x-momentum equation will look like,

O(1) +O(1) = O(1) +1

Re

[O(1) +O

(1

(δ/c)2

)]

The dominating term in this equation is the one corresponding to∂

∂y′

(µ′∂u

′

∂y′

)(i.e. this is

the only viscous term to survive in the momentum equation).

∴1

Re

[O

(1

(δ/c)2

)]= O(1) ⇒

δ

c= O

(Re−1/2

)“large Re implies small δ/c”


y-momentum

ρ′u′∂v

′

∂x′+ ρ

′v′∂v

′

∂y′= −

1

γM2∞

∂p′

∂y′

+1

Re

∂

∂x′

[µ′(∂v′

∂x′+∂u′

∂y′

)]+

∂

∂y′

[µ′(

4

3

∂v′

∂y′−

2

3

∂u′

∂x′

)].

ρ′u′∂v

′

∂x′= O(1)O(1)

O(δ/c)

O(1)= O

(δ

c

)ρ′v′∂v

′

∂y′= O(1)O

(δ

c

)O(δ/c)

O(δ/c)= O

(δ

c

)∂

∂x′

[µ′(∂v′

∂x′+∂u′

∂y′

)]=

1

O(1)

[O(1)

(O(δ/c)

O(1)+

O(1)

O(δ/c)

)]= O

(1

(δ/c)

)∂

∂y′

[µ′(

4

3

∂v′

∂y′−

2

3

∂u′

∂x′

)]=

1

O(δ/c)

[O(1)

(O(δ/c)

O(δ/c)−O(1)

O(1)

)]= O

(1

(δ/c)

)

Net effect on y-momentum equation looks like

O

(δ

c

)+O

(δ

c

)= O(1)O(??) +O

(δ

c

)2

O

(1

(δ/c)

)


∴∂p′

∂y′= O

(δ

c

)or smaller

∂p′

∂y′∼

∆p′

∆y′= O

(δ

c

)∴ ∆p

′= O

(δ

c

)2

⇐ VERY small

Pressure does not change across a boundary layer. ∴ p = pe(x)

Continuity∂ (ρu)

∂x+∂ (ρv)

∂y= 0

x-momentum

ρu∂u

∂x+ ρv

∂u

∂y= −

dpe

dx+

∂

∂y

(µ∂u

∂y

)y-momentum→ Not useful ∵ p not a function of y.


Boundary conditions

u = ue(x) at y = δ

u = v = 0 at y = 0

Numerical solution is found via time marching (parabolic solution).

3.10 The Euler EquationsAssumptions used to arrive at Euler equations:• Inviscid: µ = 0.

• No heat conduction: k = 0.

Equations apply outside boundary layers and wakes for streamlined bodies at high Re. Since

the boundary layer is very thin, that applies for most of the flow. As we’ll see later, the Euler

and boundary layer equations can be solved separately in an iterative fashion.

Mass, momentum and energy (see page 3 of notes and set all viscous term to zero)

∂U

∂t+∂f

∂x+∂g

∂y+∂h

∂z= Q,

where

U =

ρ

ρ~v

ρE

, Q =

0

ρ~fe

wf + qh


f =

ρu

ρu2 + p

ρuv

ρuw

ρuH

, g =

ρv

ρuv

ρv2 + p

ρvw

ρvH

h =

ρw

ρuw

ρvw

ρw2 + p

ρwH

Boundary conditionsFlow tangency at the wall (i.e. normal velocity component to the surface is zero). This

contrast the no-slip condition when viscosity is considered.

For fully attached flow, solution provide:

• lift,

• pitching moment, and

• pressure, but

• not drag! (except wave drag)

Can use the Euler equation to solve for pe(x) and then solve boundary layer equation to get

viscous drag.


3.11 Potential Flow EquationsHere, we want to simplify the Euler equations. First, we need some tools.

3.11.1 Pathlines & Streamlines of a FlowPathlines follow particles in the flow. Streamlines are based on the velocity field. If the flow is

steady, they are the same.

d~s× ~v = 0 along a streamline.

d~s = dx ı+ dy + dz k

Example in 2D:

d~s = dx ı+ dy

~v = uı+ v

d~s× ~v =

∣∣∣∣∣∣∣∣ı k

dx dy 0

u v 0

∣∣∣∣∣∣∣∣ = (vdx− udy) k = 0

∴ vdx = udy

dy

dx=

v

uI.e. the slope of the streamline is the ratio of the velocity component.


3.11.2 Angular Velocity

~ω =1

2~∇× ~v. Recall ~∇ =

∂

∂xı+

∂

∂y+

∂

∂zk

3.11.3 Vorticity

~ξ = 2~ω = ~∇× ~v

→ If ~ξ 6= ~0 at every point in the flow⇒ the flow is said to be rotational.

→ If ~ξ = ~0 at every point in the flow⇒ the flow is said to be irrotational.

(∴ only translational movement of fluid elements.)

2D condition for irrotationality

~ξ =

(∂v

∂x−∂u

∂y

)k = ~0

∴∂v

∂x=∂u

∂y


3.11.4 Stream Function

• Assume 2D, steady flow.

• Consider u = u(x, y) and v = v(x, y) are known.

• Starting from the streamline function, viz.dy

dx=v

u

• Integrate streamline function to get ψ(x, y) = C, where C is a constant of integration.

The streamline function is defined as

ψ (x, y) = C

The mass flow between the 2 streamlines is constant,

since no flow through streamlines. Let us define the

stream function such that ∆ψ(

= ψ2 − ψ1

)is this

mass flow rate.

∆ψ = ρ~v · ~∆n = ρu∆y − ρv∆x

As ∆n→ 0,

dψ = ρudx− ρvdy.


But dψ =∂ψ

∂xdx+

∂ψ

∂ydy (chain rule).

By inspection,

∂ψ

∂x= −ρv

∂ψ

∂y= ρu

For incompressible flow, ψ ≡ψ

ρ

∴ u =∂ψ

∂y, v = −

∂ψ

∂x

For cylindrical/polar coordinates,

vr =1

r

∂ψ

∂θ, vθ = −

∂ψ

∂r

3.11.5 Velocity Potential

For irrotational flow,~ξ = ~∇× ~v = 0

⇒ Consider a scalar φ. Scalar identity gives

~∇×(~∇φ)

= 0


(i.e. the curl of the scalar gradient is zero). By comparison, we get

~v = ~∇φ ⇐ defining the velocity potential.

⇒ For an irrotational velocity field, one can define the velocity vector field with a scalar field!

In Cartesian coordinates,

~v = uı+ v+ wk =∂φ

∂xı+

∂φ

∂y+

∂ψ

∂zk

or u =∂φ

∂x, v =

∂φ

∂y, w =

∂φ

∂z


Streamlines are prependicular to potential lines

(or equipotential lines), for 2D incompressible

and irrotational flow.


3.11.6 The Potential Flow Model

• Assume inviscid, irrotational and isentropic.

• Same as Euler + no shocks (except weak ones).

• Good to calculate lift, moment, pressure, but not drag (not even wave).

Irrotational⇒ ~∇× ~v = 0⇒ ~v = ~∇φ← i.e. potential function

Isentropic⇒∂φ

∂t+H = const. = H, where H = h+

|~v|2

2(total specific enthalpy).

Mass∂ρ

∂t+ ~∇·

(ρ~∇φ

)= 0

Isentropicρ

ρA=

(T

TA

) 1γ−1

=

(CpT

CpTA

) 1γ−1

=

(h

hA

) 1γ−1

where A is a reference point.

∴ρ

ρA=

H −

|~v|2

2−∂φ

∂thA

1

γ−1

∴ ρ = f (φ)


Noting that |~v|2 =∣∣∣~∇φ∣∣∣2, gives

∴ ρ = ρA

H −

∣∣∣~∇φ∣∣∣22−∂φ

∂thA

1γ−1

Steady potential flow∂ρ

∂t=∂φ

∂t= 0⇒ ~∇·

(ρ~∇φ

)= 0

Using stagnation point as reference

H = h +|~v|2

2= h

∴ρ

ρ=

1−

∣∣∣~∇φ∣∣∣22H

1

γ−1

Boundary conditions

• Flow tangency (inviscid) vn =∂φ

∂n= 0, where n is the direction normal to the wall.


3.12 Incompressible Potential Flow

Incompressibility is the final assumption! (For M . 0.3)

Mass∂ρ

∂t+ ~∇·

(ρ~∇φ

)= 0

Incompressible ⇒∂ρ

∂t= 0 ⇒ ~∇ρ·~∇φ = 0

∴ ∇2φ = 0 ⇐ Laplace’s equation

e.g., 2D Cartesian

∂2φ

∂x2+∂2φ

∂y2= 0

Linear eqn.

Can be solved using elliptic solution technique.

Stream function

• Conservation of mass (2D steady)

∂ (ρu)

∂x+∂ (ρv)

∂y=

∂

∂x

(∂ψ

∂y

)−

∂

∂y

(∂ψ

∂x

)= 0

∴ continuity is satisfied automatically by stream function.


• From irrotational flow,∂u

∂y−∂v

∂x= 0

∴ 0 =∂

∂y

(∂ψ

∂y

)−

∂

∂x

(−∂ψ

∂x

)=∂2ψ

∂y2+∂2ψ

∂x2

∴ ∇2ψ = 0

Laplace’s equation is linear. Therefore, if φ1, φ2, φ3, ..., φn are solutions, then φ =∑

i φi is

also a solution.

Boundary conditions for inviscid flows Equation is the same for different geometries. The

exact flow field changes due to variations in boundary conditions.

1. Infinity BC

∴

u =∂φ

∂x=∂ψ

∂y= U∞

v =∂φ

∂y=∂ψ

∂x= 0


2. Wall boundary condition

~v·~n =(~∇φ)·~n = 0 (“Wall tangency”)

∴∂φ

∂n= 0 ,

∂ψ

∂s= 0,

where n and s are the direction normal and tangential to the surface.


4 Fundamentals of Inviscid, Incompressible Flows4.1 Bernoulli’s Equation (3.2)

Steady, inviscid, incompressible

p+1

2ρV

2= const.,

where V = |~v|.

The equation applies for:

1. along streamlines, and

2. everywhere in irrotational flow.

Lets prove (1).

Momentum equation: ρD~v

Dt= −~∇p+

=0︷︸︸︷~∇·τ +

=0︷︸︸︷ρ~fe

x-momentum: ρu∂u

∂x+ ρv

∂u

∂y+ ρw

∂u

∂z= −

∂p

∂x

y-momentum: ρu∂v

∂x+ ρv

∂v

∂y+ ρw

∂v

∂z= −

∂p

∂y

z-momentum: ρu∂w

∂x+ ρv

∂w

∂y+ ρw

∂w

∂z= −

∂p

∂z


u∂u

∂x+ v

∂u

∂y+ w

∂u

∂z= −

1

ρ

∂p

∂x

u∂v

∂x+ v

∂v

∂y+ w

∂v

∂z= −

1

ρ

∂p

∂y

u∂w

∂x+ v

∂w

∂y+ w

∂w

∂z= −

1

ρ

∂p

∂z

Along a streamline, d~s× ~v = 0,

∴ v dx− u dy = 0 v dx = u dy

u dz − w dx = 0 or u dz = w dx

w dy − v dz = 0 w dy = v dz

∴ u∂u

∂xdx+ u

∂u

∂ydy + u

∂u

∂zdz = −

1

ρ

∂p

∂xdx

but du =∂u

∂xdx+

∂u

∂ydy +

∂u

∂zdz

→ udu = −1

ρ

∂p

∂xdx.

Thus,1

2du

2= −

1

ρ

∂p

∂xdx. (4.1)


Similarly1

2dv

2= −

1

ρ

∂p

∂ydy (4.2)

1

2dw

2= −

1

ρ

∂p

∂zdz. (4.3)

Sum (4.1), (4.2) and (4.3) to get

1

2

(du

2+ dv

2+ dw

2)

= −1

ρ

(∂p

∂xdx+

∂p

∂ydy +

∂p

∂zdz

)

⇒1

2dV

2= −

1

ρdp

or

∫V dV =

∫−

1

ρdp

V 2

2+p

ρ= const.


Recall, Cp =p− p∞

1/2ρ∞U2∞

. Incompressible flow, Bernoulli⇒ p∞ +1

2ρ∞U

2∞ = p+

1

2ρV

2, so

p− p∞ =1

2ρ∞(U

2∞ − V

2)

∴ Cp =1/2ρ∞

(U2∞ − V

2)

1/2ρ∞U2∞

Cp = 1−(V

U∞

)2

← Only a function

of velocity!!

Note: At a stagnation point, the local velocity is zero. Hence, Cp = 1− (0/U∞)2 = 1, for

incompressible flow at a stagnation point (maximum pressure).


4.2 Incompressible Potential Flow (3.7)

Recall the velocity potential function, φ, viz.

~v = ~∇φ since it is for irrotational flows.

Conservation of mass⇒ ~∇·~v = ~∇·~∇φ = ∇2φ = 0.

Laplacian operator in 3D Cartesian + cylindrical polar coordinates.

∇2φ =

∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2

~∇φ =

(∂φ

∂r,

1

r

∂φ

∂θ,∂φ

∂z

)

∇2φ = ~∇·~∇φ =

1

r

∂

∂r

(r∂φ

∂r

)+

1

r2

∂2φ

∂θ2+∂2φ

∂z2

For 2D, steady, irrotational, inviscid, incompressible flow

u =∂ψ

∂y, v = −

∂ψ

∂x.

Continuity gives,~∇·~v = 0⇒ ∇2ψ = 0 (satisfied by definition).


Laplace’s equation is linear ∴ Sum of any particular solutions is also a solution!!

• Infinity boundary conditions:

– assume for away U∞ is aligned with x-axis.

u =∂φ

∂x=∂ψ

∂y= 0, v =

∂φ

∂y= −

∂ψ

∂x= 0

• Wall boundary conditions:

– wall tangency.

– normal velocity component = 0

~v·n = ~∇·n = 0

⇒∂φ

∂n= 0

∣∣∣∣∣∣∣∣∣∣∣∣ ∂ψ

∂s= 0

Strategy

1. Solve ∇2φ = 0 or ∇2ψ = 0.

2. Get ~v from ~v = ∇φ or u =∂ψ

∂y, v = −

∂ψ

∂x.

3. Get p from Bernoulli’s equation.

Note: (2D polar) ~v = (vr, vθ)

~∇·~v =1

r

∂

∂r(rvr) +

1

r

∂vθ

∂θ

~∇×~v = k

[1

r

∂

∂r(rvθ)−

1

r

∂vr

∂θ

]


4.3 Elementary Solution

4.3.1 Uniform flow (3.9)

u = U∞ incompressible: ~∇·~v = 0√

v = 0 irrotational: ~∇×~v = 0√

u =∂φ

∂x= U∞

v =∂φ

∂y= 0

⇒ φ = φ (x) ,dφ

dx= U∞

∴ φ = U∞x

u =∂ψ

∂y= U∞

v = −∂ψ

∂x= 0

⇒ ψ = ψ (y) ,dψ

dy= U∞

∴ ψ = U∞y

Note: you can verify that Γ = −∮c~v·d~s = 0!


4.3.2 Source/Sink Flow (3.10)

Direction of streamline

• towards centre (sink)

• away from centre (source)

vθ = 0 (no tangential velocity)

φ = φ(r)

∇2φ =

1

r

d

dr

(r

dφ

dr

)= 0, r 6= 0

rdφ

dr= c⇒

dφ

dr=c

r⇒ φ = c ln r

vr =∂φ

∂r=c

r

vθ =1

r

∂φ

∂θ= 0

← Since streamlines diverge, fluid particles

must slow down as r increases.

c⇒[

m2/s]

~∇·~v = 0 except at origin.~∇×~v = 0 everywhere.


Mass flow rate (per unit depth)

m =

∫ 2π

0

dm =

∫ 2π

0

ρ

(c

r

)rdθ︸︷︷︸

ρvA⇒ m =

∫ 2π

0ρcdθ

= 2πρc

Volume flow rate per unit mass: Λ ≡ m/ρ = 2πc.

∴ c =Λ

2π

Λ > 0⇒ source.

Λ < 0⇒ sink.

Hence,

φ =Λ

2πln r

vr =Λ

2πrvθ = 0

Note: vr =1

r

∂ψ

∂θvθ = −

∂ψ

∂rbut vθ = 0

ψ = ψ (θ)

vr =c

r⇒

c

r=

1

r

∂ψ

∂θ

∴ ψ = cθ ⇒ ψ =Λθ

2π


4.3.3 Uniform Flow + source + sink (3.11)(simulates flow around a solid body)

ψ = U∞y, ψ =Λθ

2π(for uniform and sour/sink, respectively)

Total stream functionψ = U∞r sin θ +

Λ

2πθ1 −

Λ

2πθ2

θ1, θ2 are found from geometry (r, θ, b)

vr =1

r

∂ψ

∂θ, vθ = −

∂ψ

∂r


Stagnation streamline

U∞r sin θ +Λ

2π(θ1 − θ2) = 0

(Oval) Rankine oval

4.3.4 Doublet Flow (3.12)(Source-sink pair)

ψ =Λ

2π(θ1 − θ2) =

Λ

2π∆θ

Let L shrink, while ΛL = const. = κ (doublet strength)Limiting case

ψ = limL→0κ=const.

[Λ

2πθ −

Λ

2π(θ + dθ)

]


(−

Λ

2πdθ

)


sin θ =a

L⇒ a = L sin θ

b ' r − L cos θ

dθ 'a

b=

L sin θ

r − L cos θ


[−

Λ

2π

L sin θ

r − L cos θ

]

Recall that ΛL = κ (constant)


[−κ

2π

sin θ

r − L cos θ

]⇒ ψ = −

κ

2π

sin θ

r

Streamlines are circles

ψ = const. = c

r = −κ

2πcsin θ

r = d sin θ is a circle of diameter d,

centered at ±(0, d/2).


4.3.5 Non-lifting Flow over a Cylinder (3.13)(Uniform + doublet)

ψ = U∞r sin θ −κ

2π

sin θ

r

ψ = U∞r sin θ

(1−

κ

2πU∞r2

)

ψ = U∞r sin θ

(1−

R2

r2

),

where the circle radius is R2 ≡ κ/2πU∞, since at r = R, ψ = 0.

vr =1

r

∂ψ

∂θ⇒ vr =

(1−

R2

r2

)U∞ cos θ

vθ = −∂ψ

∂r⇒ vθ = −

(1 +

R2

r2

)U∞ sin θ

• vr = 0, and vθ = 0 at stagnation points (i.e. (R, 0) and (R, π).

• Stagnation streamline is given by ψ = 0.

• On cylinder surfacce, vr = 0, and vθ = −2U∞ sin θ.

Cp = 1−v2θ

U2∞

= 1− 4 sin2θ

Symmetric about x-axis.

⇒ no lift, no drag (d’Alembert’s paradox)

→ Drag due to viscosity.


4.3.6 Vortex flow (3.14)(our fourth and last elementary flow) vr = 0

vθ =const.

r=C

r

φ = φ (θ)

⇒ ∇2φ =

1

r2

d2φ

dθ2= 0⇒ φ = Cθ, ψ = C ′ ln r

vr =∂φ

∂r= 0

vθ =1

r

∂φ

∂θ=C

r


Can verify ~∇·~v = 0 everywhere.~∇×~v = 0 everywhere except at origin.

To evaluate C, take circulation along a streamline of radius r, viz.

Γ = −∮~v·d~s = −vθ2πr

∴ Γ = −2πC, or C = −Γ

2πindependent of r

But, if closed surface does not include center:

Γ = −∮c∗~v·d~s

= 0

∴ all circulation is generated at origin.

∵ vθ = −∂ψ

∂r= −

C ′

r

⇒ C′= −C

φ = −Γ

2πθ ψ =

Γ

2πln r

vr = 0

vθ = −Γ

2πr


4.3.7 Lifting Flow over a Cylinder (3.15, 3.16)(doublet + uniform︸︷︷︸

non-lifting

+vortex)

ψ1 = U∞r sin θ

(1−

R2

r2

), ψ2 =

Γ

2πln r + const.

Set constant to −Γ

2πlnR ⇒ ψ2 =

Γ

2πln

(r

R

)

ψ = U∞r sin θ

(1−

R2

r2

)+

Γ

2πln

(r

R

)

(one of 3 cases 3rd Ed., p.235)

⇒ Magnus effect 3rd Ed., p.241.

vr =

(1−

R2

r2

)U∞ cos θ

vθ = −(

1 +R2

r2

)U∞ sin θ −

Γ

2πr︸︷︷︸new term


On cylinder surface, r = R,

⇒ vr = 0, vθ = −2U∞ sin θ −Γ

2πR

Cp = 1−V 2

U2∞

= 1−(−2 sin θ −

Γ

2πRU∞

)2

∴ Cp = 1−[

4 sin2θ +

2Γ sin θ

πRU∞+

(Γ

2πRU∞

)2]

CL =1

c

∫ c

0

Cp,ldx−1

c

∫ c

0

Cp,udx,

where x = R (1 + cos θ), dx = −R sin θ dθ, and c = 2R.

⇒ CL =1

2R

∫ 2π

π

Cp,l (−R sin θ) dθ −1

2R

∫ 0

π

Cp,u (−R sin θ) dθ,

which simplifies to CL = −1

2

∫ 2π

0

Cp sin θdθ.

CL = −1

2

∫ 2π

0

1−

[4 sin

2θ +

2Γ sin θ

πRU∞+

(Γ

2πRU∞

)2]

sin θdθ∫ 2π

0

sin θ dθ = 0,

∫ 2π

0

sin3θ dθ = 0,

∫ 2π

0

sin2θ dθ = π

⇒ CL =Γ

RU∞


CL =L′

1/2ρ∞U2∞c

, where L′ is lift per unit span, and the cord length c = 2R.

∴ L′ = ρ∞U∞Γ Kutta-Joukowski Theorem.

• General 2D shape.

• Steady, incompressible, inviscid, irrotational

Cp symmetric fore and aft (formal derivation in Anderson – 3rd Ed., p.237).

∴ CD = 0 D’Alembert’s Paradox

Stagnation points (optional)

vr =

(1−

R2

r2

)U∞ cos θ = 0 ← satisfied by R = r or θ = ±π/2.

vθ = −(

1 +R2

r2

)U∞ sin θ −

Γ

2πr= 0

Therefore, stagnation points are given by

θ = arcsin

(−

Γ

4πU∞R

)if r = R ⇐ undefined for Γ > 4πU∞R, then

use θ = ±π/2 ⇒ r =Γ

4πU∞±

√(Γ

4πU∞

)2

− R2


Γ < 4πU∞R Γ = 4πU∞R Γ > 4πU∞R

⇒ Magnus effect→ CL/CD not very big!

⇒ Viscosity is required in real life to get lift!!


5 Incompressible Flow Over AirfoilsThis chapter⇒ airfoils (i.e. wing sections)

Next chapter⇒ finit wing.

5.1 Airfoil Characteristics (4.3)

• CL varies linearly with α for low-moderate α.

• αL=0 for symmetric airfoil is α = 0

• Inviscid results over-predict CL and cannot

predict stall.

• CD requires viscous analysis (not in this

chapter).

• CM < 0 and roughly constant.


5.2 Source and Vortex Sheets (4.4)

Point vortex (2D)

Vθ = −Γ

2πr

Vortex filament (3D)

→ Straight vortex filament

equivalent to point vortex.

← Vortex sheet (3D)

Edge view of vortex sheet→with camber.

(Panel methods.)

• γ = γ(s) is the strength of vortex sheet per unit length.

• Element ds, of strength γds, induces a velocity dV normal to r given by

dV = −γds

2πr⇐ Thin airfoil theory

• Integration would require vectors, so use potential function φ instead, viz.

dφ = −γds

2πθ


Induced velocity potential due to vortex sheet:

φ(x, z) = −1

2π

∫ b

a

θγds ⇐ Panel Method

Circulation around vortex sheet:Γ =

∫ b

a

γds

The circulation around dash box is

Γ = (u1 − u2) ds+(v1 − v2) dn = γds

In the limit as dn→ 0

γds = (u1 − u2) ds

γ = u1 − u2

I.e. local jump in tangential velocity across the vortex sheet is equal to the local sheet strength!

5.3 Kutta Condition (4.5)

• Sharp TE is important for lift generation.

• Infinite number of solutions satisfy Laplace’s equations.

• Above solution is valid for inviscid flow, yet physically unrealistic.


• Choose the solution for which the flow leaves smoothly, and does not go around TE. This

can be stated several ways:

1. Cusped TEContinuity of pressure

⇒ Pu = PlIrrotational flow⇒ P+1/2ρV 2 = const. everywhere!

⇒ Vu = Vl

2. Finite TE angle (as special case of 1)

Only smooth when

⇒ Vu = Vl = 0

∴ TE is a stagnation point.

3. Various solutions have different values for Γ.

In region ds, Γ = γds; γ = Vu − VlI.e. γ(TE) = γ(a) = Vu − VlConditions (1) and (2) both satisfied by γ(TE) = 0 .

Note: This implies, 0 ≤ Cp < 1 at the trailing edge (condition 1), or Cp = 1 for condition 2.

The Kutta condition is essentially a model for the effect of viscosity, which is the reason why

the flow leaves the trailing edge in a smooth fashion (if the boundary layer does not separate -

an obvious assumption given that we are assuming inviscid flow).


5.4 Kelvin’s Circulation Theorem (4.6)How is circulation conserved? Consider an arbitrary inviscid, incompressible flow.

It can be shown from conservation of mass that

−∫C1

~v · d~s = −∫C2

~v · d~s, or

Γ1 = Γ2.

DΓ

Dt= 0

“Time rate of change of circulation

around a closed curve consisting of

the same fluid elements is zero”

Material derivative gives rate of change following fluid particles.• Inviscid, irrotational flow

Start flow

V = 0

At t > 0

V = U∞ Γ3 + Γ4 = Γ2

DΓ

Dt= 0⇒ Γ1 = Γ2 ⇒ Γ2 = 0

∴ Γ3 = −Γ4

Circulation around airfoil is equal and opposite to starting vortex.


5.5 Classical Thin Airfoil TheoryIntroduction Required items: 1. Laplace’s equation

2. Kutta condition

3. Flow tangency (BC), and

4. Far-field conditions (at∞).

Thin airfoil: • Simulated by a vortex sheet along the camber line.

• Force camber line to be a streamline of the flow.

∗ Place vortex sheet on chord line instead:

γ = γ(x) instead of γ = γ(s)

γ(c) = 0 Kutta condition

Let w′ = w′(s) be the velocity induced normal to the camber line. However, the camber is a

streamline.∴ U∞,n + w′(s) = 0 ,

where U∞,n is the component of U∞ normal to the camber line.


From geometry,

U∞,n = U∞ sin

[α+ tan

−1

(−

dz

dx

)]For thin airfoil and small α,

⇒ U∞,n = U∞

(α−

dz

dx

)

Let w(x) be the velocity normal to the chord line. For thin airfoils, camber and chord lines are

approximately equal⇒ w′(s) ' w(x).

Can obtain w(x) from γ(x)

(Recall that dV = −

γds

2πr

)

dw = −γ(ξ)dξ

2π (x− ξ)Velocity induced by elemental

vortex dξ at x.

Total induced velocity is:

w(x) = −∫ c

0

γ(ξ)dξ

2π (x− ξ)


But w(x) + U∞,n = 0,

∴1

2π

∫ c

0

γ(ξ)dξ

x− ξ= U∞

(α−

dz

dx

)Fundamental equation of thin airfoil theory.

5.5.1 Symmetric Airfoil (4.7)∴ no camber (average of upper and lower surface is the line z = 0)

⇒dz

dx= 0

∴1

2π

∫ c

0

γ(ξ)dξ

x− ξ= U∞α

(inviscid, incompressible

flow over a flat plate)

Let us apply a change of variable such that ξ, x =c

2(1− cos θ) and dξ, dx =

c

2sin θ dθ.

A fixed x location corresponds to a fixed θ, thus let

x =c

2(1− cos θ)

0 ≤ ξ ≤ c⇒ 0 ≤ θ ≤ π

∴1

2π

∫ π

0

γ(θ) sin θdθ

cos θ − cos θ= U∞α


Solution is not trivial, but we eventually get

⇒ γ(θ) = 2αU∞

(1 + cos θ

sin θ

)

Kutta condition... γ(π) = limθ→π

2αU∞

(1 + cos θ

sin θ

)= lim

θ→π2αU∞

(− sin θ

cos θ

)= 0 (l’Hospital’s rule)

Calculate total circulation:

Γ =

∫ c

0

γ(ξ)dξ =c

2

∫ π

0

γ(θ) sin θdθ

⇒ Γ = αcU∞

∫ π

0

(1 + cos θ) dθ

⇒ Γ = παcU∞

Kutta-Joukowski theorem gives the lift per unit span:

L′= ρ∞U∞Γ = παcρ∞U

2∞

CL ≡L′

q∞c, q∞ =

1

2ρ∞U

2∞


∴ CL =παcρ∞U

2∞

1/2 ρ∞U2∞c

= 2πα

CL(α) = 2παdCL

dα= 2π, Lift slope

For thin airfoils in general (don’t know

this yet, but we will soon see!!)Moment about LE:

Element vortex with strength γ(ξ)dξ located at ξ from LE gives

dΓ = γ(ξ)dξ

⇒ dL′= ρ∞U∞dΓ = ρ∞U∞γ(ξ)dξ

∴ dM′LE = −ξdL

′= −ρ∞U∞ξγ(ξ)dξ

→M′LE = −

∫ c

0

ρ∞U∞ξγ(ξ)dξ = −ρ∞U∞∫ c

0

ξγ(ξ)dξ

⇒M′LE = −q∞c2πα

2

CM,LE =M ′

LE

q∞c2= −

πα

2

But CL = 2πα⇒ πα =CL

2

∴ CM,LE = −CL

4Recall CM,c4

= CM,LE +CL

4⇒∴ CM,c4

= 0 .


Centre of Pressure: point about which moments are zero.

Aerodynamic Centre: point where moments are independent of α.

Here, c/4 (quarter-chord) is both the centre of pressure and the aerodynamic centre!

5.5.2 Cambered Airfoil (4.8)

1

2π

∫ c

0

γ(ξ)dξ

x− ξ= U∞

(α−

dz

dx

).

Now,dz

dx6= 0.

Apply the same transformation as for the symmetric case and using Fourier sine series.

γ(θ) = 2U∞

[A0

(1 + cos θ

sin θ

)+

∞∑n=1

An sin(nθ)

].

Makes the camber line a streamline of theflow. (Derivation outside scope of course)

A0 = α−1

π

∫ π

0

dz

dxdθ, and

An =2

π

∫ π

0

dz

dxcos(nθ) dθ,

where dz/dx is a function of θ. Note, γ(π) = 0 satisfies the Kutta condition.

Total circulation:

Γ =

∫ c

0

γ(ξ)dξ =c

2

∫ π

0

γ(θ) sin θ dθ


⇒ Γ = cU∞

[A0

∫ π

0

(1 + cos θ) dθ +∞∑n=1

An

∫ π

0

sin(nθ) sin θdθ

].

Since ∫ π

0

(1 + cos θ) dθ = π, and∫ π

0

sin(nθ) sin θdθ =

π/2 for n = 1

0 for n 6= 1,

then

Γ = cU∞

(πA0 +

π

2A1

).

Lift per unit span:

L′= ρ∞U∞Γ = ρ∞U

2∞c

(πA0 +

π

2A1

)⇒ CL =

L′

12ρ∞U

2∞c

= π (2Ao + A1)

CL = 2π

[α+

1

π

∫ π

0

dz

dx(cos θ − 1) dθ

].

Note that CL 6= 0 at α = 0!


But,dCL

dα= 2π ← still! ∴ CL =

dCL

dα(α− αL=0)

αL=0 = −1

π

∫ π

0

dz

dx(cos θ − 1) dθ.

Moment: (problem 4.9) see appendix to this chapter (to be posted online).

CM,LE = −π

2

(A0 + A1 −

A2

2

).

Recall, CL = π (2A0 + A1),

∴ CM,LE = −[CL

4+π

4(A1 − A2)

]

CM,c/4 = CM,LE +CL

4=π

4(A2 − A1) .

c/4 is not the centre of pressure, but it is the aerodynamic centre since CM,c/4 is independent

of α. Centre of pressure:

xcp = −M ′

LE

L′= −

CM,LE · cCL

∴ xcp =c

4

[1 +

π

CL(A1 − A2)

]


Example

z = 4zm

[x

c−(x

c

)2]

dz

dx=

4zm

c

(1− 2

x

c

)

Use change of variable

x =c

2(1− cos θ)

[Don’t use θ now, since derivation is complete(got rid of ξ also).]

∴dz

dx=

4zm

ccos θ

A0 = α−1

π

∫ π

0

dz

dxdθ = α

A1 =2

π

∫ π

0

dz

dxcos(nθ)dθ =

4zm

cfor n = 1

0 for n ≥ 2

∴ CL = π (2A0 + A1) = 2π (α− αL=0)

αL=0 = −A1

2= −

2zm

c, CL = 2π

(α+

2zm

c

)


CL = 2π

(α+

2zm

c

)

Moment force:

CM,LE = −π

2

(A0 + A1 −

A2

2

)= −

π

2

(α+

4zm

c

)

CM,c/4 =π

4(A2 − A1) =

π

4

(0−

4zm

c

)= −

πzm

c

Centre of pressure:

xcp = −CM,LE · cCL

=c

4

[1 +

π

CL(A1 − A2)

]xcp =

c

4+

zm

2α+ 4zm/c


5.6 Panel method (3.17, 4.10)

• Numerical technique for solving potential flows (arbitrary shape, thickness and α).

• No computational grid.

• Next step up is CFD, and N-S or Euler.

• Can apply compressibility correction, but no shocks, (say M . 0.6).

• Simulate a vortex sheet over an airfoil by straight panels, with midpoints being control

points.

• Enforce flow tangency at each point and solve the resulting linear system for the local vortex

strength.

• Can then get aerodynamic forces.

Show an example using XFOIL.


6 Incompressible Flow over Finite WingN-S ⇒ 2D Laplace ⇒ 3D (Inviscid, incompressible)

(Chapter 6) (Chapter 7)

6.1 Downwash and Induced Drag (5.1)

Flow around wing tip causes

wing-tip vortex.

Wing-tip vortices induce a small downward velocity on the wing called downwash, W .


U∞ and W combine to make a local relative wind.

αeff(= α− αi) is what is actually seen by the airfoil section. the component of local lift

vector L in the direction of U∞ is the induced drag Di, and lift is reduced.

(Inviscid, incompressible⇒ D’Alembert’s paradox: in 2D⇒ D = 0, but not for 3D!!)

CD = CD,profile + CD,induced ,

where CD,profile is the profile drag due to skin friction and pressure (essentially 2D) and

CD,induced is the induced drag due to the generation of lift (3D).


6.2 The Vortex Filament, Biot-Savart Law, and Helmholtz’s Theorem(5.2)

• Consider a curved vortex filament.

• Circulation around any closed path containing the filament is constant and equal to Γ.

• Strength of vortex filament is thus Γ.

d~V =Γ

4π

d~l×~r|~r|3

Biot-Savart Law(similar to electromagnetic theory for induced

magnetic field due to a current.)

For a straight filament,

∫ ∞−∞

gives 2D results as before

(i.e. vθ = −

Γ

2πr

).

Basic vortex behaviour described by Helmholtz’s Theorems:

1. Vortex strength remains constant along its length.

2. Vortex filament cannot end in a fluid.

(a) Extend to boundaries.

(b) Closed loop.


6.3 Prandtl’s Classic Lifting Line Theory (5.3)

But this implies an infinite induced

velocity at the wing tip. In order for

W to be finite, need a distribution of

Γ along span.


With an infinite number of horseshoe vortices, we have Γ = Γ(y) (a continuous distribution).

Strength is dΓ =

(dΓ

dy

)dy

dw = −dΓ

4π (y − y)= −

(dΓ/dy) dy

4π (y − y)

IfdΓ

dy

∣∣∣∣y< 0, then dw > 0

for our configuration

w(y) = −1

4π

∫ b/2

−b/2

(dΓ/dy) dy

y − yTotal downwash

But remember that Γ distribution is the unknown we are after.


αi(y) = tan−1

(−w(y)

U∞

)' −

w(y)

U∞

Induced angle of attack. (negative sign

consistent with our previous picture)

∴ αi(y) =1

4πU∞

∫ b/2

−b/2

(dΓ/dy) dy

y − y

From thin airfoil theory,

CL = 2π[αeff(y)− αL=0

]. (6.1)

• With aerodynamic twist, αL=0 varies with the span (i.e. y), but known characteristic of airfoil at y.

• Cord for a given section can also vary with span, thus c = c(y).

Kutta-Joukowsky⇒ L′ = 12ρ∞U

2∞c(y)CL = ρ∞U∞Γ(y)

∴ CL =2Γ(y)

U∞c(y). (6.2)

Solve (6.1) for αeff and substitute into (6.2) to give

αeff =Γ(y)

πU∞c(y)+ αL=0(y).


But αeff = α− αi, therefore

α(y) =Γ(y)

πU∞c(y)+ αL=0(y)︸︷︷︸

αeff

+1

4πU∞

∫ b/2

−b/2

(dΓ/dy) dy

y − y︸︷︷︸αi

Fundamental equation of Prandtl’s lifting line theory(Implicitly-defined solution for Γ(y))

Once you know Γ(y) (drop the subscript for simplicity),

1. L′(y) = ρ∞U∞Γ(y)

2. Total lift: L =

∫ b/2

−b/2L′(y)dy = ρ∞U∞

∫ b/2

−b/2Γ(y)dy

CL =L

q∞S=

2

U∞S

∫ b/2

−b/2Γ(y)dy

where S is the planform area of the wing.

3. D′i = L′i sinαi ' L′iαi, so that

Di =

∫ b/2

−b/2L′i(y)αi(y)dy = ρ∞U∞

∫ b/2

−b/2Γ(y)αi(y)dy

CDi =Di

q∞S=

2

U∞S

∫ b/2

−b/2Γ(y)αi(y)dy

Need to get Γ(y) — This is the key to finite wing theory!!


6.3.1 Elliptical Lift Distribution(A special case to study)

Γ at origin.

Γ(y) = Γ

√1−

(2y

b

)2

L′(y) = ρ∞U∞Γ

√1−

(2y

b

)2

Downwash, w(y):dΓ

dy= −

4Γ

b2

y

(1− 4y2/b2)1/2

∴ w(y) =Γ

πb2

∫ b/2

−b/2

y

(1− 4y2/b2)1/2

(y − y)dy

Change of variables⇒ y =b

2cos θ, dy = −

b

2sin θ dθ

∴ w(y) = −Γ

2b(Constant over the span!Both interesting and important!)

Induced angle of attack, αi:αi = −

w

U∞=

Γ

2bU∞Note: for b→∞⇒ αi = 0 (2D result).


L = ρ∞U∞

∫ b/2

−b/2Γ(y)dy

L = ρ∞U∞Γ

∫ b/2

−b/2

(1−

4y2

b2

)1/2

dy = ρ∞U∞Γb

4π

∴ Γ =4L

ρ∞U∞bπ=

2U∞SCL

bπ

(recall L =

1

2ρ∞U

2∞SCL

)

αi =Γ

2bU∞=

2U∞SCL

(2bU∞) (bπ)=SCL

πb2

Aspect ration: AR ≡b2

S∴ αi =

CL

πAR

Induced drag:

CDi =2

U∞S

∫ b/2

−b/2Γ(y)αi(y)dy, where αi is constant.

CDi =2αi

U∞S

∫ b/2

−b/2Γ(y)dy︸︷︷︸

Γbπ

4

⇒ CDi =παiΓb

2U∞S


CDi =

(πb

2U∞S

)(CL

πAR

)︸︷︷︸

αi

(2U∞SCL

bπ

)︸︷︷︸

Γ

∴ CDi =C2L

πAR

Drag due to lift.Varies with the square of lift, but

reduced with higher AR.

Most airplanes AR = 6− 8, but gliders can be between 10-226.

Point of interest:

• Assume no geometric twist (i.e. α constant along span).

• Assume no aerodynamic twist (i.e. αL=0 constant along span).

• Thin airfoil theory⇒ CL = a(αeff − αL=0

).

(a = 2π)⇒ CL is also constant along span.

L′(y) = q∞c(y)CL ⇒ c(y) =

L′

q∞CL∴ chord varies elliptically along the span.

6For gliders, increasing AR is limited by structural integrity.


6.3.2 General Lift Distribution

Recall, α(y) =Γ(y)

πU∞c(y)+ αL=0(y)︸︷︷︸αeff

+1

4πU∞

∫ b/2

−b/2

(dΓ/dy) dy

y − y︸︷︷︸αi

.

Transform y → θ

−b

2≤ y ≤

b

2→ 0 ≤ θ ≤ π

y = −b

2cos θ

dy =b

2sin θdθ

Represent Γ(θ) as a finite Fourier sine series (cosines cancel due to BC).

Γ(θ) = 2bU∞

N∑n=1

An sinnθ

Note Γ(0) = Γ(π) = 0 at wing tips.

(For elliptical distribution, Γ(θ) = Γ sin θ ⇒ A1 =Γ

2bU∞& An = 0, n > 1)

dΓ

dy=

dΓ

dθ

dθ

dy= 2bU∞

N∑1

nAn cosnθdθ

dy


Substitute Γ and dΓ/dy into Prandtl’s equation to give

α(θ) =2b

πc(θ)

N∑1

An sinnθ + αL=0(θ) +1

π

∫ π

0

∑N1 nAn cosnθ

cos θ − cos θdθ︸︷︷︸

standard integral linein thin airfoil theory.

α(θ) =2b

πc(θ)

N∑1

An sinnθ + αL=0(θ) +

N∑1

nAn

sinnθ

sin θ

By choosing locations for θ, we obtain N algebraic equations in terms of N unknowns,

A1, A2, ..., AN. Once we solve for the An’s,

CL =L

q∞S=ρ∞U∞

∫ b/2−b/2 Γ(y)dy

1/2ρ∞U2∞S

=2

U∞S

∫ π

0

Γ(θ)b

2sin θdθ

CL =2b2

S

N∑1

An

∫ π

0

sinnθ sin θdθ︸︷︷︸n = 1 survives only.

∴ CL =2b2

S

(A1

π

2

)=A1πb

2

S= A1πAR

CL = A1πAR(Although CL depends only on A1, we need to

solve for all A’s in order to get A1.)


CDi =2

U∞S

∫ b/2

−b/2Γ(y)αi(y)dy =

2b2

S

∫ π

0

(N∑1

An sinnθ

)αi(θ) sin θ dθ.

But, αi(y) =1

4πU∞

∫ b/2

−b/2

(dΓ/dy)

y − ydy =

1

π

N∑1

nAn

∫ π

0

cosnθ

cos θ − cos θdθ.

Since

∫ π

0

cosnθ

cos θ − cos θdθ = π

sinnθ

sin θ, then αi(θ) =

N∑1

nAn

sinnθ

sin θ

∴ CDi =2b2

S

∫ π

0

(N∑1

An sinnθ

)(N∑1

nAn sinnθ

)dθ

Note:

∫ π

0

sinmθ sin kθ =

0 ,m 6= k

π/2 ,m = k

CDi =2b2

S

(N∑1

nA2n

)π

2= πAR

(N∑1

nA2n

)

∴ CDi = πAR A21 [1 + δ] , δ =

N∑2

n

(An

A1

)2

Therefore, δ ≥ 0 and CDi is minimal with δ = 0 (i.e. for elliptical distribution).


Define a span efficiency factor, e =1

1 + δ≤ 1. e = 1 for elliptical lift distribution.

CDi =C2L

πAR(1 + δ) =

C2L

πAR e

Compromise for elliptical wing is the tapered wing.

6.4 Reduction of Lift Slope dCL

d (α− αi)= a

⇒ CL = a (α− αi) + const.

For elliptical wing,

∴ CL = a

(α−

CL

πAR

)+const.

dCL

dα= a −

a

πAR

dCL

dαdCL

dα= a =

a

1 +a

πAR

⇒ a =a

1 +

(a

πAR

)(1 + τ)

,

where τ is a function of Ai.


Finite Wing ExampleAssume untwisted, uncambered, rectangular wing with AR = 6.

α(θ) =2b

πc(θ)

N∑1

An sinnθ + αL=0(θ) +

N∑1

nAn

sinnθ

sin θ

1. Untwisted⇒ α(θ) = const..

2. Uncambered⇒ αL=0(θ) = 0.

3. Rectangular⇒ c(θ) = const. = c.

∴ α =

N∑1

(2AR

πAn sinnθ +

nAn sinnθ

sin θ

)For a symmetrically-loaded wing, An = 0 for even n. Truncate series at N = 8.

α = A1 sin θ

(1

sin θ+

2AR

π

)+ A3 sin 3θ

(3

sin θ+

2AR

π

)+A5 sin 5θ

(5

sin θ+

2AR

π

)+ A7 sin 7θ

(7

sin θ+

2AR

π

)4 unknowns⇒ A1, A3, A5, A7. Choose 4 θ values; 0 < θ ≤ π/2, due to symmetry.

θ =

π

8,π

4,3π

8,π

2


Solve 4×4 systemA1

A3

A5

A7

=

α

α

α

α

⇒

A1 = 0.2402α

A3 = 0.0289α

A5 = 0.0057α

A7 = 0.0010α

⇒ CL = A1πAR = 4.53α⇒ CL = 4.53α

CDi =C2L

πAR(1 + δ) , δ =

N∑2

n

(An

A1

)2

δ =3A2

3 + 5A25 + 7A2

7

A21

= 0.046

CDi = 1.14α2

Note:dCL

dα= 4.53 < 2π = a (∼28% drop in lift slop!)


AER307 Aerodynamics - Skuleexams.skule.ca/exams/AER307H1_20149_661418446443aer307...Aerodynamics Ronald Hanson [email protected] UTIAS AER307: Aerodynamics c Philippe Lavoie

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