PART I

FUNDAMENTAL PRINCIPLES

（基本原理）

In part I,we cover some of the basic principles

that apply to aerodynamics in general,These are

the pillars on which all of aerodynamics is based

Chapter 2

Aerodynamics,Some Fundamental

Principles and Equations

There is so great a difference between a fluid and a collection

of solid particles that the laws of pressure and of equilibrium

of fluids are very different from the laws of the pressure and

equilibrium of solids,

Jean Le Rond d’Alembert,1768

2.1 Introduction and Road Map

Preparation of tools for the analysis of

aerodynamics

Every aerodynamic tool we developed in

this and subsequent chapters is

important for the analysis and

understanding of practical problems

Orientation offered by the road map

2.2 Review of Vector relations

2.2.1 to 2.2.10 Skipped over

2.2.11 Relations between line,surface,and

volume integrals

The line integral of A over C is related to the surface integral

of A(curl of A) over S by Stokes’ theorem,

? ? SAsA dd

SC

????? ???

Where aera S is bounded by the closed curve C,

The surface integral of A over S is related to the volume

integral of A(divergence of A) over V by divergence’ theorem,

? ?dVd

VS

????? ???? ASA

Where volume V is bounded by the closed surface S,

If p represents a scalar field,a vector relationship analogous

to divergence theorem is given by gradient theorem,

dVppd

VS

????? ??S

2.3 Models of the fluid,control

volumes and fluid particles

Importance to create physical feeling from

physical observation,

How to make reasonable judgments on difficult

problems,

In this chapter,basic equations of aerodynamics

will be derived,

Philosophical procedure involved with the

development of these equations

1,Invoke three fundamental physical principles which are

deeply entrenched in our macroscopic observations of

nature,namely,

a,Mass is conserved,that’s to say,mass can be neither

created nor destroyed,

b,Newton’s second law,force=mass? acceleration

c,Energy is conserved; it can only change from one form to

another

2,Determine a suitable model of the fluid,

3,Apply the fundamental physical principles listed in item 1

to the model of the fluid determined in item2 in order to

obtain mathematical equations which properly describe

the physics of the flow,

Emphasis of this section,

1,What is a suitable model of the fluid?

2,How do we visualize this squishy substance in

order to apply the three fundamental principles?

3,Three different models mostly used to deal with

aerodynamics,

finite control volume （ 有限控制体）

infinitesimal fluid element （ 无限小流体微团）

molecular （ 自由分子）

2.3.1 Finite control volume approach

Definition of finite control volume,

a closed volume sculptured within a finite region of

the flow,The volume is called control volume V,

and the curved surface which envelops this region

is defined as control surface S,

Fixed control volume and moving control volume,

Focus of our investigation for fluid flow,

2.3.2 Infinitesimal fluid element approach

Definition of infinitesimal fluid element,

an infinitesimally small fluid element in the flow,

with a differential volume,

It contains huge large amount of molecules

Fixed and moving infinitesimal fluid element,

Focus of our investigation for fluid flow,

The fluid element may be fixed in space with fluid moving

through it,or it may be moving along a streamline with velocity

V equal to the flow velocity at each point as well,

2.3.3 Molecule approach

Definition of molecule approach,

The fluid properties are defined with the use of

suitable statistical averaging in the microscope

wherein the fundamental laws of nature are

applied directly to atoms and molecules,

In summary,although many variations on the theme

can be found in different texts for the derivation of

the general equations of the fluid flow,the flow

model can be usually be categorized under one of the

approach described above,

2.3.4 Physical meaning of the

divergence of velocity

Definition of,

is physically the time rate of change of

the volume of a moving fluid element of fixed

mass per unit volume of that element,

V???

V???

Analysis of the above definition,

Step 1,Select a suitable model to give a frame

under which the flow field is being described,

a moving control volume is selected,

Step 2,Select a suitable model to give a frame

under which the flow field is being described,

a moving control volume is selected,

Step 3,How about the characteristics for this

moving control volume?

volume,control surface and density will be

changing as it moves to different region of the

flow,

Step 4,Chang in volume due to the movement of

an infinitesimal element of the surface dS over,

? ?? ? ? ? SdtVdSntVV ???? ???????

t?

The total change in volume of the whole control

volume over the time increment is obviously

given as bellow t?

? ??? ??

S

SdtV

??

Step 5,If the integral above is divided by,the

result is physically the time rate change of the

control volume

t?

? ? ???? ????

?

?

SS

SdVSdtV

tDt

DV ????1

Step 6,Applying Gauss theorem,we have

??? ???

V

dVV

Dt

DV ?

Step 7,As the moving control volume approaches

to a infinitesimal volume,,Then the above

equation can be rewritten as

V?

? ?

??? ???

V

dVV

Dt

VD

?

? ?

Assume that is small enough such that is the

same through out, Then,the integral can be

approximated as,we have

V? V???

V?

? ? VV ????

? ? VV

Dt

VD ?? ???? ? ?

Dt

VD

V

V ?

?

1??? ?or

Definition of,

is physically the time rate of change of

the volume of a moving fluid element of fixed

mass per unit volume of that element,

V???

V???

Another description of and,

?? ?

S

SdV ?? V

???

Assume is a control surface corresponding to control

volume,which is selected in the space at time,

At time the fluid particles enclosed by at time will

have moved to the region enclosed by the surface,

The volume of the group of particles with fixed identity

enclosed by at time is the sum of the volume in region

A and B,And at time,this volume will be the sum of the

volume in region B and C,

As time interval approaches to zero,coincides with,

If is considered as a fixed control volume,then,the

region in A can be imagined as the volume enter into the

control surface,C leave out,

V

S

t

1t S t

1S

S

t

1t

1S

S

S

Based on the argument above,the integral of can

be expressed as volume flux through fixed control surface,

Further,can be expressed as the rate at which fluid

volume is leaving a point per unit volume,

?? ?

S

SdV ??

V???

The average value of the velocity component on the right-

hand x face is

)2)(( xxuu ????

The rate of volume flow out of the right-hand x face is

? ? zyxxuu ?????? )2)((

That into the left-hand x face is

? ? zyxxuu ?????? )2)((

The net outflow from the x faces is

zyxxu ????? )( per unit time

The net outflow from all the faces in x,y,z directions per

unit time is

? ? zyxzwyvxu ??????????? )()()(

The flux of volume from a point is

? ?

zyx

zyxzwyvxu

V

flu xin flo wflu xo u tflo w

V ???

?????????????

?

)()()(lim

0

)()()(lim 0 zwyvxuV f l u xi n f l o wf l u xo u t f l o wV ???????????

2.4 Continuity equation

In this section,we will apply fundamental

physical principles to the fluid model,More

attention should be given for the way we

are progressing in the derivation of basic

flow equations,

Derivation of continuity equation

Step 1,Selection of fluid model,A fixed finite

control volume is employed as the frame for the

analysis of the flow,Herein,the control surface and

control volume is fixed in space,

Step 2,Introduction of the concept of mass flow,

Let a given area A is arbitrarily oriented in a flow,

the figure given bellow is an edge view,If A is small

enough,then the velocity V over the area is uniform

across A,The volume across the area A in time

interval dt can be given as

AdtVV o l u m e n )(?

The mass inside the shaded volume is

AdtVM a s s n )(??

The mass flow through is defined as the mass

crossing A per unit second,and denoted as m?

dt

AdtVm n )(???

or

AVm n???

The equation above states that mass flow through A

is given by the product

Area X density X component of flow velocity normal

to the area

mass flux is defined as the mass flow per unit area

nVA

mf l u xM a s s ??? ?

Step 3,Physical principle Mass can be neither

created nor destroyed,

Step 4,Description of the flow field,control volume

and control surface,

),,,(),,,,( tzyxVVtzyx ?? ?? ??

:Sd?

Directional elementary surface area on the control surface

:dV Elementary volume inside the finite control volume

Step 5,Apply the mass conservation law to this

control volume,

Net mass flow out of control

volume through surface S

Time rate decrease of mass

inside control volume V ?

or

CB ?

Step 6,Mathematical expression of B

The elemental mass flow across the area is

SdVSdV n ??? ?? ??

The physical meaning of positive and negative of

SdV ?? ??

Sd?

The net mass flow out of the whole control surface S

?? ??

S

SdVB

??

?

Step 7,Mathematical expression of C

The mass contained inside the elemental volume V is

dV?

The mass inside the entire control volume is

???

V

dV?

The time rate of increase of the mass inside V is

????

?

V

dV

t

?

The time rate of decrease of the mass inside V is

CdV

t V

?

?

?

? ??? ?

Step 8,Final result of the derivation

Let B=C,then we get

????? ?

?

???

VS

dV

t

SdV ??

??

or

0???

?

?

?????

SV

SdVdV

t

??

??

Derivation with moving control volume

Mass at time

)()( tMtM BA ?

Mass at time

t

1t

)()( 11 tMtM CB ?

Based on mass conservation law

? ? ? ? 0)()()()( 11 ???? tMtMtMtM BACB

?

? ? ? ? 0)()()()( 11 ???? tMtMtMtM ACBB

Consider the limits as tt ?

1

????????? ???

C

C

B

B

A

A dVMdVMdVM ???,,

? ?

????

?

?

?

?

?

V

BB

tt

dV

ttt

tMtM

?

1

1 )()(lim

1

? ?

?? ???

?

?

S

AC

tt

SdV

tt

tMtM ??

?

1

1 )()(lim

1

Then we get the mathematical description of the mass

conservation law with the use of moving control volume

0???

?

?

?????

SV

SdVdV

t

??

??

Why the final results derived with different fluid model are

the same

Step 9,Notes for the Continuity Equation above

The continuity equation above is in integral form,it gives the

physical behaviour over a finite region of space without

detailed concerns for every distinct point,

This feature gives us numerous opportunities to apply the

integral form of continuity equation for practical fluid

dynamic or aerodynamic problems,

If we want to get the detailed performance at a given point,

then,what shall we deal with the integral form above to get

a proper mathematic description for mass conservation law?

Step 10,Continuity Equation in Differential form

0???

?

?

?????

SV

SdVdV

t

??

??

?

0???

?

?

?????

SV

SdVdV

t

??

?

?

Control volume is fixed in space

? ? 0????

?

?

?????? dVVdVt

VV

?

?

?

? ? ?dVVSdV

VS

????? ???? ??? ??

The integral limit is not

the same

The integral limit is the

same

or

? ? 0??

?

?

??

? ???

?

?

???

V

dVV

t

?

?

?

A possible case for the integral over the control volume

If the finite control volume is arbitrarily chosen in the space,

the only way to make the equation being satisfied is that,

the integrand of the equation must be zero at all points

within the control volume,That is,

? ? 0????

?

? V

t

?

??

That is the continuity equation in a partial differential form,

It concerns the flow field variables at a point in the flow with

respect to the mass conservation law

It is important to keep in mind that the continuity equations

in integral form and differential form are equally valid

statements of the physical principles of conservation of

mass.they are mathematical representations,but always

remember that they speak words,

Step 11,Limitations of the equations derived

Continuum flow or molecular flow

As the nature of the fluid is assumed as Continuum

flow in the derivation so

It satisfies only for Continuum flow

Steady flow or unsteady flow

It satisfies both steady and unsteady flows

viscous flow or inviscid flow

It satisfies both viscous and inviscid flows

Compressible flow or incompressiblw flow

It satisfies both Compressible and

incompressiblw flows

Difference between steady and unsteady flow

Unsteady flow,

The flow-field variables are a function of both spatial

location and time,that is

),,,(),,,,( tzyxVVtzyx ?? ?? ??

Steady flow,

The flow-field variables are a function of spatial location

only,that is

),,(),,,( zyxVVzyx ?? ?? ??

For steady flow,0??? t

0???

?

?

?????

SV

SdVdV

t

??

?? 0????

S

SdV

??

?

?

? ? 0????

?

? V

t

?

??

? ? 0??? V???

For steady incompressible flow,

? ? 0??? V?? 0??? V??

2.5 Momentum equation

Newton’s second law

amF ?? ?

where

:F?

:m

:a?

Force exerted on a body of mass m

Mass of the body

Acceleration

Consider a finite moving control volume,the mass

inside this control volume should be constant as the

control volume moving through the flow field,So

that,Newton’s second law can be rewritten as

dt

VmdF )(

??

?

Derivation of momentum equation

Step 1,Selection of fluid model,A fixed finite

control volume is employed as the frame for the

analysis of the flow,

Step 2,Physical principle

Force = time rate change of momentum

Step 3,Expression of the left side of the equation of

Newton’s second law,i.e.,the force exerted on the

fluid as it flows through the control volume,

Two sources for this force,

1,Body forces,over every part of V

2,Surface forces,over every elemental surface of S

Body force on a elemental volume

dVf??

Body force over the control volume

???

V

dVf

?

?

Surface forces over the control surface can be divided into two

parts,one is due to the pressure distribution,and the other is

due to the viscous distribution,

Pressure force acting on the elemental surface

Spd ??

Note,indication of the negative sign

Complete pressure force over the entire control surface

???

S

Spd

?

The surface force due to the viscous effect is simply expressed

by

viscousF

?

Total force acting on the fluid inside the control

volume as it is sweeping through the fixed control

volume is given as the sum of all the forces we have

analyzed

v i s c o u s

SV

FSpddVfF

????

??? ????? ?

Step 4,Expression of the right side of the equation of

Newton’s second law,i.e.,the time rate change of

momentum of the fluid as it sweeps through the

fixed control volume,

Moving

control

volume

Let be the momentum of the fluid within region A,

B,and C,for instance,

CBA MMM,,

????????? ???

C

C

B

B

A

A dVVMdVVMdVVM

???

???,,

At time,the momentum inside is

)()( tMtM BA ?

t S

At time,the momentum inside is

1t 1S

)()( 11 tMtM CB ?

The momentum change during the time interval

tt ?1

)()()()( 11 tMtMtMtM BACB ???

or

? ? ? ?)()()()( 11 tMtMtMtM ACBB ???

As the time interval approaches to zero,the region B will

coincide with S in the space,and the two limits

? ?

?????? ?

??

?

??

?

?

?

VV

BB

tt

dV

t

VdVV

ttt

tMtM )()()(lim

1

1

1

??

??

? ? ?? ??

?

?

?

S

AC

tt

VSdV

tt

tMtM ??? )()()(lim

1

1

1

?

?? ?

S

VSdV

???

)( ?

?

Net momentum flow out of control

volume across surface S

?????

V

dVV

t

?

?

?

Time rate change of momentum due

to unsteady fluctuations of flow

properties inside V

The explanations above helps us to make a better

understanding of the arguments given in the text

book bellow

Net momentum flow out of control

volume across surface S

Time rate of change of momentum due

to unsteady fluctuations of flow

properties inside control volume V

?

?

G?

H?

Step 5,Mathematical description of

G?

mass flow across the elemental area dS is

SdV ?? ??

momentum flow across the elemental area dS is

? ?VSdV ??? ??

The net flow of momentum out of the control volume

through S is

? ??? ??

S

VSdV

????

?G

Step 6,Mathematical description of H?

The momentum in the elemental volume dV is dVV??

The momentum contained at any instant inside the control

volume V is

???

V

dVV

?

?

Its time rate change due to unsteady flow fluctuation is

????

?

?

V

dVV

t

??

?H

Be aware of the difference between

????

?

V

dVV

t

?

? ???

V

dVV

dt

d ?

?

and

Step 7,Final result of the derivation

Combine the expressions of the forces acting on the fluid and

the time rate change due to term and,respectively,

according to Newton’s second low

H?G?

? ? ?????

?

?

?????

VS

dVV

t

VSdV

dt

Vmd ??????

?

??HG

)(

F

dt

Vmd ?

?

?)(

? ?

v i s c o u s

SV

SV

FSpddVf

VSdVdVV

t

???

????

??

???

?

?

?????

?????

?

??

It’s the momentum equation in integral form

It’s a vector equation

Advantages for momentum equation in integral form

Step 8,Momentum Equation in Differential form

? ?

v i s c o u s

SV

SV

FSpddVf

VSdVdVV

t

???

????

??

???

?

?

?????

?????

?

??

Try to rearrange the every integrals to share the same limit

????? ????

VS

pdVSpd

?

gradient theorem

?????? ?

?

?

?

?

VV

dV

t

V

dVV

t

)(

?

? ?

? control volume is fixed in space

? ?

v i s c o u s

VV

SV

FpdVdVf

VSdVdV

t

V

??

???

?

???

???

?

?

??????

?????

?

?

? )(

Then we get

Split this vector equation as three scalar equations with

kwjviuV ???? ???

Momentum equation in x direction is

? ?

v i s c o u sx

VV

x

SV

FdV

x

p

dVf

uSdVdV

t

u

)(

)(

?

?

?

?

???

?

?

??????

?????

?

?

? ??

? ? ? ? ? ?dVVuSdVuuSdV

VSS

?????

??? ??????? ??????

divergence theorem

? ? 0)()( ??

?

?

??

? ??

?

?

????

?

????

V

v i s c o u sxx dVfx

p

Vu

t

u

F??

? ?

As the control volume is arbitrary chosen,then the integrand

should be equal to zero at any point,that is

? ? v i s c o u sxxf

x

pVu

t

u )()( F??

?

??????

?

? ??? ?

? ? v i s c o u sxxf

x

pVu

t

u )()( F??

?

??????

?

? ??? ?

? ? v i s c o u szzf

z

pVw

t

w )()( F??

?

??????

?

? ??? ?

? ? v i s c o u syyf

y

pVv

t

v )()( F??

?

??????

?

? ??? ?

x direction

y direction

z direction

These equations can applied for unsteady,3D flow of any

fluid,compressible or incompressible,viscous or inviscid,

? ? v i s c o u s

SVVS

FSpddVfdVV

t

VSdV

???????

???

?

??? ?????????? ???

? ? ???? ???

SS

SpdVSdV

????

?

Steady and inviscid flow without body forces

? ?

x

pVu

?

????? ??

? ?

y

pVv

?

????? ??

? ?

z

pVw

?

????? ??

Euler’s Equations and Navier-Stokes equations

Whether the viscous effects are being considered or not

Eulers Equations,inviscid flow

Navier-Stokes equations,viscous flow

Deep understanding of different terms in

continuity and momentum equations

0???

?

?

?????

SV

SdVdV

t

??

??

? ?

v i s c o u s

SV

SV

FSpddVf

VSdVdVV

t

???

????

??

???

?

?

?????

?????

?

??

????

?

V

dV

t

?

????

?

V

dVV

t

?

?

Time rate change of mass inside

control volume

Time rate change of momentum

inside control volume

?? ?

S

SdV

??

?

? ??? ?

S

VSdV

???

?

Net flow of mass out of the control

volume through control surface S

?? ?

S

SdV

??

Net flow of volume out of the control

volume through control surface S

Net flow of momentum out of the

control volume through control

surface S

???

V

dVf

?

?

???

S

Spd

?

Body force through out the control

volume V

Surface force over the control surface

S

What we can foresee the applications for

aerodynamic problems with basic flow

equations on hand?

0??? V?

If the steady incompressible inviscid flows are

concerned

? ?

x

pVu

?

?????

?

1?

? ?

y

pVv

?

?????

?

1?

? ?

z

pVw

?

?????

?

1?

Partial differential equation for velocity

Partial differential equation for velocity

and pressure

2.6 An application of the momentum

equation,drag of a 2D body

How to design a 2D wind tunnel test?

How to measure the lift and drag exerted on

the airfoil by the fluid?

A selected control volume around an airfoil

Descriptions of the control volume

1,The upper and lower streamlines far above and below the

body (ab and hi),

2,Lines perpendicular to the flow velocity far ahead and

behind the body(ai and bh)

3,A cut that surrounds and wraps the surface of the

body(cdefg)

1,Pressure at ab and hi,

2,Pressure at ai and bh,,velocity,

3,The pressure force over the surface abhi

?? pp

?? pp c o n s ta n tu ?1 )(22 yuu ?

???

a b h i

Spd

?

4,The surface force on def by the presence of the body,this

force includes the skin friction drag,and denoted as per

unit span,

5,The surface forces on cd and fg cancel each other,

6,The total surface force on the entire control volume is

R??

R ???? ??

a b h i

Spdf o r c es u r f a c e

?

7,The body force is negligible

Apply to momentum equation,we have

? ? R ??????

?

?

???????

a b h iSV

SpdVSdVdVV

t

?????

??

for steady flow

? ? ???? ?????

a b h iS

SpdVSdV

????

?R

Note,it’s a vector equation,

If we only concern the x component of the equation,

with represents the x component of, D? R?

? ? ? ????? ?????

a b h i

x

S

SpduSdVD

???

?

As boundaries of the control volume abhi are chosen

far away from the body,the pressure perturbation

due to the presence of the body can be neglected,

that means,the pressure there equal to the

freestream pressure,If the pressure distribution

over abhi is constant,then

? ? 0???

a b h i

xSpd

?

So that

? ??? ????

S

uSdVD

??

?

As ab,hi,def are streamlines,then

? ? ? ? ? ? 0?????? ??????

d e fhiab

uSdVuSdVuSdV

??????

???

As cd,fg are are adjacent to each other,then

? ? ? ????? ????

fgcd

uSdVuSdV

????

??

The only contribution to momentum flow through

the control surface come from the boundaries ai and

bh,For dS=dy(1),the momentum flow through the

control surface is

? ? ???? ????

b

h

a

i

S

dyudyuuSdV 222211 ???

??

Note,

1,The sign in front of each integrals on the right

hand side of the equation

2,The integral limits for each integrals on the right

hand side of the equation

Consider the integral form of the continuity equation

for steady flow,

02211 ??? ??

b

h

a

i

dyudyu ??

or

?? ?

b

h

a

i

dyudyu 2211 ??

As is a constant

1u

?? ?

b

h

a

i

dyuudyu 122211 ??

? ? ???? ????

b

h

a

i

S

dyudyuuSdV 222211 ???

??

?

? ? ???? ????

b

h

b

h

S

dyudyuuuSdV 222122 ???

??

?

? ? ??? ????

b

h

S

dyuuuuSdV )( 2122??

??

The final result gives the drag per unit span

? ???

b

h

dyuuuD )( 2122?

? ? ??? ??????

b

h

S

dyuuuuSdVD )( 2122??

??

?

The drag per unit span can be expressed in terms of

the known freestream velocity and flow-field

properties,across a vertical station

downstream of the body,

1u

22 uand?

Physical meaning behind the equation

? ???

b

h

dyuuuD )( 2122?

?

b

h

dyu 22? )( 21 uu ?

Mass flow out of the

control volume

Velocity decrement

? ?

b

h

dyuuu )( 2122?

Momentum decrement per second

For incompressible flow,that is,the density is constant

? ???

b

h

dyuuuD )( 212?

2.6.1 Comments

With the application of momentum principle to a

large,fixed control volume,an accurate result for

overall quantity such as drag on a body can be

predicted with knowing the detailed flow properties

along the control surface,That to say,it is

unnecessary to know the the details along the

surface of the body,

2.7 Energy equation

0??? V?

? ?

x

pVu

?

?????

?

1?

? ?

y

pVv

?

?????

?

1?

? ?

z

pVw

?

?????

?

1?

? Continuity equation

? Momentum equation

Unknowns,Vp ?,

c o n s ta n t??

For steady incompressible invicid flows

For compressible flows

? is an additional variable,and therefore we need an additional fundamental equation to complete the

system,This fundamental equation is the energy

equation,which we are going to develop,

Two additional flow-field variables will appear to the

energy equation,that is internal energy and

temperature,

e

T

Energy equation is only necessary for compressible

flows,

Physical principle(first law of thermodynamics)

Energy can be neither created nor destroyed; it can

only change in form

Definitions of system and internal energy per

unit mass e

Definition of surroundings

Heat transferred from the

surroundings to the system

Work done on the

surroundings by the system

q?

w?

Change of internal energy in system due

to the heat transferred and the work done

de

As energy is conserved,so

dewq ?? ??

Apply the first law to the fluid flowing

trough the fixed control volume,and let

B1 = rate of heat added to fluid inside control volume from

surroundings,

B2 = rate of work done on fluid inside control volume,

B3 = rate of change of energy of fluid as it flows through

control volume,

As first law should be satisfied,then

B1+B2 = B3

Actually speaking,the equation above is a power

equation,

Rate of volumetric heating

???

V

dVq ??

If the flow is viscous

B1 =

v i s c o u s

V

QdVq ?? ???? ?

Rate of volumetric heating = VF ?? ?

The force includes three parts F?

Pressure force,body force and skin friction force

Rate of work done on fluid inside V

due to pressure force on S ?? ??

S

VSpd

??

)(

Rate of work done on fluid inside V

due to body force ??? ?

V

VdVf

??

)( ?

B2 =

v i s c o u s

VS

WVdVfVSpd ?

????

????? ????? )()( ?

Since the fluid inside the control volume is not

stationary,it is moving at the local velocity with a

consequent kinetic energy per unit mass,so,

the total energy per unit mass is

V?

2/2V

2/2Ve ?

Net rate of flow of total energy

across control surface S ?? ???

S

V

eSdV )

2

()(

2??

?

Time rate change of total energy

inside V due to transient variations

of flow-field properties

??? ??

?

V

dV

V

e

t

)

2

(

2

?

B3 =

????? ??????

?

SV

V

eSdVdV

V

e

t

)

2

()()

2

(

22 ??

??

B1+B2 = B3

?????

????????

????

?

?

?

??????

SV

v i s c o u s

VS

v i s c o u s

V

SdV

V

edV

V

e

t

WVdVfVSpdQdVq

??

?

????

??

)

2

()

2

(

)()(

22

??

??

Energy equation in integral form

Notes in the text book

''

22

)()(

)

2

()

2

(

v i s c o u sv i s c o u s

WQVfVpq

V

V

e

V

e

t

??

???

?

?

????????

?

?

?

?

?

?

?????

?

?

?

?

?

?

?

?

??

??

Energy equation in partial differential form

If the flow is steady,inviscid,adiabatic,without body

force

???? ??????

SS

VSpdSdV

V

e

????

)()

2

(

2

?

)()

2

(

2

VpV

V

e

??

?????

?

?

?

?

?

??? ?

After apply three fundamental physical principles,we

have derived three basic equations for fluid flow,And

there are three variables,such as eVp,,,??

For calorically perfect gases

Tce v?

Then,one more property is added,but with perfect

gas equation

RTp ??

Continuity,momentum and energy equation with two

additional equations are five independent equations,

and there five unknowns, So that we have

got a closed system for the flow problems,

TeVp,,,,??

2.8 Interim summary

2.9 Substantial derivatives

Focus our eye on a infinitesimal fluid

element moving through a flow field,The

velocity field can be given as kwjviuV ???? ???

),,,(

),,,(

),,,(

tzyxww

tzyxvv

tzyxuu

?

?

?

The density can be given as

),,,( tzyx?? ?

),,,( 111111 tzyx?? ?

),,,( 222222 tzyx?? ?

With the use of Taylor series expansion about point 1

)()()()( 12

1

12

1

12

1

12

1

12 tttzzzyyyxxx ???

??

?

?

?

????

?

??

?

?

?

???

???

?

???

?

?

????

?

??

?

?

?

??? ??????

Dividing by )( 12 tt ?

112

12

112

12

112

12

112

12 ?

?

??

?

?

?

??

?

??

?

??

?

?

?

??

?

?

???

?

???

?

?

??

?

??

?

??

?

?

?

??

?

?

ttt

zz

ztt

yy

ytt

xx

xtt

??????

Dt

D

tttt

??? ?

?

?

?

12

12

12

lim

u

tt

xx

tt

?

?

?

?

12

12

12

lim v

tt

yy

tt

?

?

?

?

12

12

12

lim wtt

zz

tt

?

?

?

?

12

12

12

lim

tz

w

y

v

x

u

Dt

D

?

??

?

??

?

??

?

?? ?????

tz

w

y

v

x

u

Dt

D

?

??

?

??

?

??

?

??

in cartesian coordinates,

z

k

y

j

x

i

?

??

?

??

?

??? ???

then

? ????

?

?? V

tDt

D ?

t?

?

Dt

D

??V?

Substantial derivative

Local derivative

convective derivative

2.10 Fundamental equations in term

of substantial derivative

In this section,the continuity,momentum

and energy equations will be given in terms

of substantial derivative

? ? ??? ???????? VVV ???

The continuity equation in differential form is

? ? 0????

?

? V

t

?

??

0???????

?

? ??? VV

t

??

or

Dt

DV

t

??? ????

?

? ?

Since

So

0???? V

Dt

D ???

This is the continuity equation in terms of

substantial derivative

The x component of the momentum

equation in differential form is

? ? v i s c o u sxxf

x

pVu

t

u )()( F??

?

??????

?

? ??? ?

t

u

t

u

t

u

?

??

?

??

?

? ??? )(

? ? ? ? ? ? uVVuVu ???????? ??? ???

? ? ? ?

v i s c o u sxx

f

x

p

uVVu

t

u

t

u

)( F??

?

?

?

???????

?

?

?

?

?

?

??

?

?

??

or

? ? ? ?

v i s c o u sxx

f

x

p

uVV

t

u

t

u

)( F??

?

?

?

????

?

?

?

?

?

?

???

?

?

?

?

?

?

??

?

?

??

? ? 0??

?

?

??

? ???

?

? V

t

?

??

Continuity Equation

? ? v i s c o u sxxf

x

puV

t

u )( F??

?

??????

?

? ??? ?

hence

?

v i s c o u sxxfx

puV

t

u )( F??

?

???

?

?

?

?

?

? ???

?

? ?? ?

?

v i s c o u sxxfx

p

Dt

Du )( F??

?

??? ??

In the same way we can get

v i s c o u syyfy

p

Dt

Dv )( F??

?

??? ??

v i s c o u szzfz

p

Dt

Dw )( F??

?

??? ??

these are the momentum equations in terms of

substantial derivative in x,y,z directions respectively

Energy equation in terms of substantial derivative

''

2

)()()2( v i s c o u sv i s c o u s WQVfVpqDt VeD ??

???

? ????????? ???

Detailed descriptions for the comparison

between the basic flow equations in different

forms,refer to the text book

2.11 Pathlines and streamlines of a flow

Skipped over

2.12 Angular velocity,vorticity and strain

In this section,more attention will be paid to

examine the orientation of the fluid element

and its shape as it moves through a

streamline in the flow field,An important

quantity,vorticity,will be introduced,

Motion of a fluid

element along a

streamline

Try to set up the relationships between

with and 21

,?? ??

xvyu ????,t?

Distance in y direction that A moves

during time increment tv?

Distance in y direction that C moves

during time increment tdxxvv ??????? ???

Net displacement in y direction of

C relative to A

tdx

x

v

tvtdx

x

v

v

??

?

?

?

?

?

?

?

?

????

?

?

?

?

?

?

?

?

? ?? ? t

x

v

dx

tdxxv ?

?

???????

2t a n ?

Since is a small angle

2?? 22t a n ?? ???

t

x

v ?

?

???

2?

Similarly

t

y

u ?

?

????

1?

y

u

tdt

d

t ?

???

?

??

??

1

0

1 lim ??

x

v

tdt

d

t ?

??

?

??

??

2

0

2 lim ??

Definition,angular velocity of the fluid

element is the average angular velocity of

lines AB and AC,they are perpendicular to

each other at the time t

??

?

?

??

?

?

?

?

?

?

?

??

?

?

?

?

? ??

y

u

x

v

dt

d

dt

d

z 2

1

2

1 21 ??

?

k

y

u

x

vj

x

w

z

ui

z

v

y

w ????

???

?

???

?

?

??

?

??

?

?

??

?

?

?

??

?

??

???

?

???

?

?

??

?

??

2

1

2

1

2

1?

?? ?? 2? vorticity

k

y

u

x

v

j

x

w

z

u

i

z

v

y

w ????

??

?

?

??

?

?

?

?

?

?

?

??

?

?

?

?

?

?

?

?

?

?

???

?

?

??

?

?

?

?

?

?

?

??

In a velocity field,the curl of the velocity is

equal to the vorticity

1,If at every point in a flow field,the

flow is called rotational,This implies that the

fluid elements have a finite angular velocity

2,If at every point in a flow,the flow is

called irrotational,This implies that the fluid

element have no angular velocity; their

motion through space is a pure translation

0??? V?

0??? V?

Definition of strain,the strain of the fluid

element in xy plane is the change of in k,

where the positive strain corresponds to a

decreasing k,and k is the angle between

sides AB and AC,they are perpendicular to

each other at the time t

Strain =

12 ??? ??????

The time rate of strain in xy plane is

y

u

x

v

dt

d

dt

d

dt

d

xy ?

??

?

?????? 12 ????

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

z

w

y

w

x

w

z

v

y

v

x

v

z

u

y

u

x

u

In the matrix above which is composed of velocity

derivatives,the diagonal terms represent the

dilatation(扩张 ) of a fluid element,The off diagonal

terms are associated with rotation and strain of fluid

element,

Relations between viscous effect and rotation of a

fluid element,

Irrotational and rotational flows in practical

aerodynamic problems

2.13 Circulation

Important tool for we to obtain solutions for some

very practical and exciting aerodynamic problems,

Circulation can be used to calculate lift exerted on

an airfoil with unit span,

Definition of circulation

? ???? C sdV ??

Note:the negative sign in front of the line integral

Stokes’ theorem

? ? SdVsdV

SC

????

???????? ???

Referring to the vector analysis,what is the physical

meaning that the equation bellow speak?

? ?

dS

dnV ?????? ??

2.14 Stream function

In a steady 2D steady flow,the differential form of

streamlines can be expressed as

u

v

dx

dy ?

If are known functions of,then,after the

equation above being integrated,we can get the

algebraic equation of the streamline

vu,yx,

cyxf ?),(

For each streamline,is a constant,Its value varies

with different streamlines,Replacing the symbol

with,then we have

c

f

?

cyx ?),(?

The function is called stream function,

Different value of the,i.e,,represents

different streamlines in the flow field,

),( yx?

c

321,,ccc

Two streamlines respecting with different values of c

Physical meaning of the stream function

Arbitrariness of the integrand constant c

Difference in stream function between two individual

streamlines

?????? 1212 cc???

Mass flow between the two

streamlines ab and cd,

(per unit depth

perpendicular to the page)

How to remove the arbitrariness of the constant of

integration?

What will be the mass flow through an arbitrary

curve connecting two points on a streamline?

For a steady flow,the mass flow inside a given

streamtube is constant,

For a steady flow,the continuity equation should be

satisfied,then,the mass flow through a closed curve

C is zero,That means the mass flow through L1 is

the same to that of L2,

For a steady flow and physical possible flows,the

integration of mass flow between two points is

independent of the path of the integration,

Important properties of the stream function

If is small,is a constant value across n? V n?

)1(nV ??? ??

or

V

n

?? ?

?

?

as 0?? n

nn

V

n ?

??

?

??

??

???

0

l i m

The equation above states that,if we know the

distribution of,then we can obtain the product

by differentiating in the direction normal to

? )( V?

?

V

Note,is a scalar variable,its direction is defined

by the streamlines,

V

Relationship between and differential of vu ??,

v d xudyd ??? ??

As,then the chain rule of calculus states ),( yx?? ?

dy

y

dx

x

d

?

??

?

?? ???

Comparing the two equations above,we have

x

v

y

u

?

???

?

?? ????,

If is known for a given flow field,then at any

point in the flow the products and can be

obtained by differentiating in the directions

normal to and,respectively,

),( yx?

u? v?

?

u v

In polar coordinates

r

V

rr

V r

?

???

?

?? ??

?

??

?,

1

),( yx? is equal to mass flow,for incompressible flow,

that is, Then,the velocity along a

streamline can be written as

con stan t??

? ?

n

V

?

?? ??

if we define,then ??? ?

n

V

?

?? ?

The features for the function of, For incompressible

flow,is constant along a streamline,

Represents the equation of a streamline,

?

? co ns ta nt??

??

??

Mass flow between two streamlines

Volume flow between two streamlines

Two primary reasons for the concept of the stream

function to be a powerful tool in aerodynamics,

1,(or ) gives the

equation of a srteamline,

2,The flow velocity can be obtained by differentiating

co ns ta nt?? co ns ta nt??

)( ?? or

How to give the determination of the stream function?

2.15 Velocity potential

For an irrotational flow (no matter it is

compressible or incompressible)

0???? V???

Then,there must exists a scalar function,for which ?

???V?

where,is called velocity potential ?

another explanation for the existence of the vlocity

potential for a irrotational flow

0???

C

ldV

?? ?

0???? ??

A

B

B

A

ldVldV

????

?? ??? 21 LL ldVldV ????

The equation above shows that the line integral of the

velocity is independent with the path of the integration,

Or to say,the result of the integral is a function of the

start and end positions only,If the start point(A) is

fixed somewhere in the space,we have

?? ????? BABA w d zv d yudxldVzyx ??),,(?

dz

z

dy

y

dx

x

d

?

??

?

??

?

?? ????

x d zv d yu d xd ????

?

z

w

y

v

x

u

?

??

?

??

?

?? ???

The features for the velocity potential analogous to the

stream function ----- derivatives of yield the flow-

velocities,

Distinction between and

?

? ?

1,The flow-field velocity are obtained by

differentiating in the same direction as the

velocities,whereas is differentiated normal to

the velocity direction

2,The velocity potential is defined for irrotational flow

only,In contrast,the stream function can be used

in either rotational or irrotational,

3,The velocity potential applies to three dimensional

flows,whereas the stream function for 2D only,

?

?

Why the analysis of irrotational flow is simpler than

that of rotational flows?

Irrotational flows is usually called potential flows

How to give the determination of the velocity potential?

FUNDAMENTAL PRINCIPLES

（基本原理）

In part I,we cover some of the basic principles

that apply to aerodynamics in general,These are

the pillars on which all of aerodynamics is based

Chapter 2

Aerodynamics,Some Fundamental

Principles and Equations

There is so great a difference between a fluid and a collection

of solid particles that the laws of pressure and of equilibrium

of fluids are very different from the laws of the pressure and

equilibrium of solids,

Jean Le Rond d’Alembert,1768

2.1 Introduction and Road Map

Preparation of tools for the analysis of

aerodynamics

Every aerodynamic tool we developed in

this and subsequent chapters is

important for the analysis and

understanding of practical problems

Orientation offered by the road map

2.2 Review of Vector relations

2.2.1 to 2.2.10 Skipped over

2.2.11 Relations between line,surface,and

volume integrals

The line integral of A over C is related to the surface integral

of A(curl of A) over S by Stokes’ theorem,

? ? SAsA dd

SC

????? ???

Where aera S is bounded by the closed curve C,

The surface integral of A over S is related to the volume

integral of A(divergence of A) over V by divergence’ theorem,

? ?dVd

VS

????? ???? ASA

Where volume V is bounded by the closed surface S,

If p represents a scalar field,a vector relationship analogous

to divergence theorem is given by gradient theorem,

dVppd

VS

????? ??S

2.3 Models of the fluid,control

volumes and fluid particles

Importance to create physical feeling from

physical observation,

How to make reasonable judgments on difficult

problems,

In this chapter,basic equations of aerodynamics

will be derived,

Philosophical procedure involved with the

development of these equations

1,Invoke three fundamental physical principles which are

deeply entrenched in our macroscopic observations of

nature,namely,

a,Mass is conserved,that’s to say,mass can be neither

created nor destroyed,

b,Newton’s second law,force=mass? acceleration

c,Energy is conserved; it can only change from one form to

another

2,Determine a suitable model of the fluid,

3,Apply the fundamental physical principles listed in item 1

to the model of the fluid determined in item2 in order to

obtain mathematical equations which properly describe

the physics of the flow,

Emphasis of this section,

1,What is a suitable model of the fluid?

2,How do we visualize this squishy substance in

order to apply the three fundamental principles?

3,Three different models mostly used to deal with

aerodynamics,

finite control volume （ 有限控制体）

infinitesimal fluid element （ 无限小流体微团）

molecular （ 自由分子）

2.3.1 Finite control volume approach

Definition of finite control volume,

a closed volume sculptured within a finite region of

the flow,The volume is called control volume V,

and the curved surface which envelops this region

is defined as control surface S,

Fixed control volume and moving control volume,

Focus of our investigation for fluid flow,

2.3.2 Infinitesimal fluid element approach

Definition of infinitesimal fluid element,

an infinitesimally small fluid element in the flow,

with a differential volume,

It contains huge large amount of molecules

Fixed and moving infinitesimal fluid element,

Focus of our investigation for fluid flow,

The fluid element may be fixed in space with fluid moving

through it,or it may be moving along a streamline with velocity

V equal to the flow velocity at each point as well,

2.3.3 Molecule approach

Definition of molecule approach,

The fluid properties are defined with the use of

suitable statistical averaging in the microscope

wherein the fundamental laws of nature are

applied directly to atoms and molecules,

In summary,although many variations on the theme

can be found in different texts for the derivation of

the general equations of the fluid flow,the flow

model can be usually be categorized under one of the

approach described above,

2.3.4 Physical meaning of the

divergence of velocity

Definition of,

is physically the time rate of change of

the volume of a moving fluid element of fixed

mass per unit volume of that element,

V???

V???

Analysis of the above definition,

Step 1,Select a suitable model to give a frame

under which the flow field is being described,

a moving control volume is selected,

Step 2,Select a suitable model to give a frame

under which the flow field is being described,

a moving control volume is selected,

Step 3,How about the characteristics for this

moving control volume?

volume,control surface and density will be

changing as it moves to different region of the

flow,

Step 4,Chang in volume due to the movement of

an infinitesimal element of the surface dS over,

? ?? ? ? ? SdtVdSntVV ???? ???????

t?

The total change in volume of the whole control

volume over the time increment is obviously

given as bellow t?

? ??? ??

S

SdtV

??

Step 5,If the integral above is divided by,the

result is physically the time rate change of the

control volume

t?

? ? ???? ????

?

?

SS

SdVSdtV

tDt

DV ????1

Step 6,Applying Gauss theorem,we have

??? ???

V

dVV

Dt

DV ?

Step 7,As the moving control volume approaches

to a infinitesimal volume,,Then the above

equation can be rewritten as

V?

? ?

??? ???

V

dVV

Dt

VD

?

? ?

Assume that is small enough such that is the

same through out, Then,the integral can be

approximated as,we have

V? V???

V?

? ? VV ????

? ? VV

Dt

VD ?? ???? ? ?

Dt

VD

V

V ?

?

1??? ?or

Definition of,

is physically the time rate of change of

the volume of a moving fluid element of fixed

mass per unit volume of that element,

V???

V???

Another description of and,

?? ?

S

SdV ?? V

???

Assume is a control surface corresponding to control

volume,which is selected in the space at time,

At time the fluid particles enclosed by at time will

have moved to the region enclosed by the surface,

The volume of the group of particles with fixed identity

enclosed by at time is the sum of the volume in region

A and B,And at time,this volume will be the sum of the

volume in region B and C,

As time interval approaches to zero,coincides with,

If is considered as a fixed control volume,then,the

region in A can be imagined as the volume enter into the

control surface,C leave out,

V

S

t

1t S t

1S

S

t

1t

1S

S

S

Based on the argument above,the integral of can

be expressed as volume flux through fixed control surface,

Further,can be expressed as the rate at which fluid

volume is leaving a point per unit volume,

?? ?

S

SdV ??

V???

The average value of the velocity component on the right-

hand x face is

)2)(( xxuu ????

The rate of volume flow out of the right-hand x face is

? ? zyxxuu ?????? )2)((

That into the left-hand x face is

? ? zyxxuu ?????? )2)((

The net outflow from the x faces is

zyxxu ????? )( per unit time

The net outflow from all the faces in x,y,z directions per

unit time is

? ? zyxzwyvxu ??????????? )()()(

The flux of volume from a point is

? ?

zyx

zyxzwyvxu

V

flu xin flo wflu xo u tflo w

V ???

?????????????

?

)()()(lim

0

)()()(lim 0 zwyvxuV f l u xi n f l o wf l u xo u t f l o wV ???????????

2.4 Continuity equation

In this section,we will apply fundamental

physical principles to the fluid model,More

attention should be given for the way we

are progressing in the derivation of basic

flow equations,

Derivation of continuity equation

Step 1,Selection of fluid model,A fixed finite

control volume is employed as the frame for the

analysis of the flow,Herein,the control surface and

control volume is fixed in space,

Step 2,Introduction of the concept of mass flow,

Let a given area A is arbitrarily oriented in a flow,

the figure given bellow is an edge view,If A is small

enough,then the velocity V over the area is uniform

across A,The volume across the area A in time

interval dt can be given as

AdtVV o l u m e n )(?

The mass inside the shaded volume is

AdtVM a s s n )(??

The mass flow through is defined as the mass

crossing A per unit second,and denoted as m?

dt

AdtVm n )(???

or

AVm n???

The equation above states that mass flow through A

is given by the product

Area X density X component of flow velocity normal

to the area

mass flux is defined as the mass flow per unit area

nVA

mf l u xM a s s ??? ?

Step 3,Physical principle Mass can be neither

created nor destroyed,

Step 4,Description of the flow field,control volume

and control surface,

),,,(),,,,( tzyxVVtzyx ?? ?? ??

:Sd?

Directional elementary surface area on the control surface

:dV Elementary volume inside the finite control volume

Step 5,Apply the mass conservation law to this

control volume,

Net mass flow out of control

volume through surface S

Time rate decrease of mass

inside control volume V ?

or

CB ?

Step 6,Mathematical expression of B

The elemental mass flow across the area is

SdVSdV n ??? ?? ??

The physical meaning of positive and negative of

SdV ?? ??

Sd?

The net mass flow out of the whole control surface S

?? ??

S

SdVB

??

?

Step 7,Mathematical expression of C

The mass contained inside the elemental volume V is

dV?

The mass inside the entire control volume is

???

V

dV?

The time rate of increase of the mass inside V is

????

?

V

dV

t

?

The time rate of decrease of the mass inside V is

CdV

t V

?

?

?

? ??? ?

Step 8,Final result of the derivation

Let B=C,then we get

????? ?

?

???

VS

dV

t

SdV ??

??

or

0???

?

?

?????

SV

SdVdV

t

??

??

Derivation with moving control volume

Mass at time

)()( tMtM BA ?

Mass at time

t

1t

)()( 11 tMtM CB ?

Based on mass conservation law

? ? ? ? 0)()()()( 11 ???? tMtMtMtM BACB

?

? ? ? ? 0)()()()( 11 ???? tMtMtMtM ACBB

Consider the limits as tt ?

1

????????? ???

C

C

B

B

A

A dVMdVMdVM ???,,

? ?

????

?

?

?

?

?

V

BB

tt

dV

ttt

tMtM

?

1

1 )()(lim

1

? ?

?? ???

?

?

S

AC

tt

SdV

tt

tMtM ??

?

1

1 )()(lim

1

Then we get the mathematical description of the mass

conservation law with the use of moving control volume

0???

?

?

?????

SV

SdVdV

t

??

??

Why the final results derived with different fluid model are

the same

Step 9,Notes for the Continuity Equation above

The continuity equation above is in integral form,it gives the

physical behaviour over a finite region of space without

detailed concerns for every distinct point,

This feature gives us numerous opportunities to apply the

integral form of continuity equation for practical fluid

dynamic or aerodynamic problems,

If we want to get the detailed performance at a given point,

then,what shall we deal with the integral form above to get

a proper mathematic description for mass conservation law?

Step 10,Continuity Equation in Differential form

0???

?

?

?????

SV

SdVdV

t

??

??

?

0???

?

?

?????

SV

SdVdV

t

??

?

?

Control volume is fixed in space

? ? 0????

?

?

?????? dVVdVt

VV

?

?

?

? ? ?dVVSdV

VS

????? ???? ??? ??

The integral limit is not

the same

The integral limit is the

same

or

? ? 0??

?

?

??

? ???

?

?

???

V

dVV

t

?

?

?

A possible case for the integral over the control volume

If the finite control volume is arbitrarily chosen in the space,

the only way to make the equation being satisfied is that,

the integrand of the equation must be zero at all points

within the control volume,That is,

? ? 0????

?

? V

t

?

??

That is the continuity equation in a partial differential form,

It concerns the flow field variables at a point in the flow with

respect to the mass conservation law

It is important to keep in mind that the continuity equations

in integral form and differential form are equally valid

statements of the physical principles of conservation of

mass.they are mathematical representations,but always

remember that they speak words,

Step 11,Limitations of the equations derived

Continuum flow or molecular flow

As the nature of the fluid is assumed as Continuum

flow in the derivation so

It satisfies only for Continuum flow

Steady flow or unsteady flow

It satisfies both steady and unsteady flows

viscous flow or inviscid flow

It satisfies both viscous and inviscid flows

Compressible flow or incompressiblw flow

It satisfies both Compressible and

incompressiblw flows

Difference between steady and unsteady flow

Unsteady flow,

The flow-field variables are a function of both spatial

location and time,that is

),,,(),,,,( tzyxVVtzyx ?? ?? ??

Steady flow,

The flow-field variables are a function of spatial location

only,that is

),,(),,,( zyxVVzyx ?? ?? ??

For steady flow,0??? t

0???

?

?

?????

SV

SdVdV

t

??

?? 0????

S

SdV

??

?

?

? ? 0????

?

? V

t

?

??

? ? 0??? V???

For steady incompressible flow,

? ? 0??? V?? 0??? V??

2.5 Momentum equation

Newton’s second law

amF ?? ?

where

:F?

:m

:a?

Force exerted on a body of mass m

Mass of the body

Acceleration

Consider a finite moving control volume,the mass

inside this control volume should be constant as the

control volume moving through the flow field,So

that,Newton’s second law can be rewritten as

dt

VmdF )(

??

?

Derivation of momentum equation

Step 1,Selection of fluid model,A fixed finite

control volume is employed as the frame for the

analysis of the flow,

Step 2,Physical principle

Force = time rate change of momentum

Step 3,Expression of the left side of the equation of

Newton’s second law,i.e.,the force exerted on the

fluid as it flows through the control volume,

Two sources for this force,

1,Body forces,over every part of V

2,Surface forces,over every elemental surface of S

Body force on a elemental volume

dVf??

Body force over the control volume

???

V

dVf

?

?

Surface forces over the control surface can be divided into two

parts,one is due to the pressure distribution,and the other is

due to the viscous distribution,

Pressure force acting on the elemental surface

Spd ??

Note,indication of the negative sign

Complete pressure force over the entire control surface

???

S

Spd

?

The surface force due to the viscous effect is simply expressed

by

viscousF

?

Total force acting on the fluid inside the control

volume as it is sweeping through the fixed control

volume is given as the sum of all the forces we have

analyzed

v i s c o u s

SV

FSpddVfF

????

??? ????? ?

Step 4,Expression of the right side of the equation of

Newton’s second law,i.e.,the time rate change of

momentum of the fluid as it sweeps through the

fixed control volume,

Moving

control

volume

Let be the momentum of the fluid within region A,

B,and C,for instance,

CBA MMM,,

????????? ???

C

C

B

B

A

A dVVMdVVMdVVM

???

???,,

At time,the momentum inside is

)()( tMtM BA ?

t S

At time,the momentum inside is

1t 1S

)()( 11 tMtM CB ?

The momentum change during the time interval

tt ?1

)()()()( 11 tMtMtMtM BACB ???

or

? ? ? ?)()()()( 11 tMtMtMtM ACBB ???

As the time interval approaches to zero,the region B will

coincide with S in the space,and the two limits

? ?

?????? ?

??

?

??

?

?

?

VV

BB

tt

dV

t

VdVV

ttt

tMtM )()()(lim

1

1

1

??

??

? ? ?? ??

?

?

?

S

AC

tt

VSdV

tt

tMtM ??? )()()(lim

1

1

1

?

?? ?

S

VSdV

???

)( ?

?

Net momentum flow out of control

volume across surface S

?????

V

dVV

t

?

?

?

Time rate change of momentum due

to unsteady fluctuations of flow

properties inside V

The explanations above helps us to make a better

understanding of the arguments given in the text

book bellow

Net momentum flow out of control

volume across surface S

Time rate of change of momentum due

to unsteady fluctuations of flow

properties inside control volume V

?

?

G?

H?

Step 5,Mathematical description of

G?

mass flow across the elemental area dS is

SdV ?? ??

momentum flow across the elemental area dS is

? ?VSdV ??? ??

The net flow of momentum out of the control volume

through S is

? ??? ??

S

VSdV

????

?G

Step 6,Mathematical description of H?

The momentum in the elemental volume dV is dVV??

The momentum contained at any instant inside the control

volume V is

???

V

dVV

?

?

Its time rate change due to unsteady flow fluctuation is

????

?

?

V

dVV

t

??

?H

Be aware of the difference between

????

?

V

dVV

t

?

? ???

V

dVV

dt

d ?

?

and

Step 7,Final result of the derivation

Combine the expressions of the forces acting on the fluid and

the time rate change due to term and,respectively,

according to Newton’s second low

H?G?

? ? ?????

?

?

?????

VS

dVV

t

VSdV

dt

Vmd ??????

?

??HG

)(

F

dt

Vmd ?

?

?)(

? ?

v i s c o u s

SV

SV

FSpddVf

VSdVdVV

t

???

????

??

???

?

?

?????

?????

?

??

It’s the momentum equation in integral form

It’s a vector equation

Advantages for momentum equation in integral form

Step 8,Momentum Equation in Differential form

? ?

v i s c o u s

SV

SV

FSpddVf

VSdVdVV

t

???

????

??

???

?

?

?????

?????

?

??

Try to rearrange the every integrals to share the same limit

????? ????

VS

pdVSpd

?

gradient theorem

?????? ?

?

?

?

?

VV

dV

t

V

dVV

t

)(

?

? ?

? control volume is fixed in space

? ?

v i s c o u s

VV

SV

FpdVdVf

VSdVdV

t

V

??

???

?

???

???

?

?

??????

?????

?

?

? )(

Then we get

Split this vector equation as three scalar equations with

kwjviuV ???? ???

Momentum equation in x direction is

? ?

v i s c o u sx

VV

x

SV

FdV

x

p

dVf

uSdVdV

t

u

)(

)(

?

?

?

?

???

?

?

??????

?????

?

?

? ??

? ? ? ? ? ?dVVuSdVuuSdV

VSS

?????

??? ??????? ??????

divergence theorem

? ? 0)()( ??

?

?

??

? ??

?

?

????

?

????

V

v i s c o u sxx dVfx

p

Vu

t

u

F??

? ?

As the control volume is arbitrary chosen,then the integrand

should be equal to zero at any point,that is

? ? v i s c o u sxxf

x

pVu

t

u )()( F??

?

??????

?

? ??? ?

? ? v i s c o u sxxf

x

pVu

t

u )()( F??

?

??????

?

? ??? ?

? ? v i s c o u szzf

z

pVw

t

w )()( F??

?

??????

?

? ??? ?

? ? v i s c o u syyf

y

pVv

t

v )()( F??

?

??????

?

? ??? ?

x direction

y direction

z direction

These equations can applied for unsteady,3D flow of any

fluid,compressible or incompressible,viscous or inviscid,

? ? v i s c o u s

SVVS

FSpddVfdVV

t

VSdV

???????

???

?

??? ?????????? ???

? ? ???? ???

SS

SpdVSdV

????

?

Steady and inviscid flow without body forces

? ?

x

pVu

?

????? ??

? ?

y

pVv

?

????? ??

? ?

z

pVw

?

????? ??

Euler’s Equations and Navier-Stokes equations

Whether the viscous effects are being considered or not

Eulers Equations,inviscid flow

Navier-Stokes equations,viscous flow

Deep understanding of different terms in

continuity and momentum equations

0???

?

?

?????

SV

SdVdV

t

??

??

? ?

v i s c o u s

SV

SV

FSpddVf

VSdVdVV

t

???

????

??

???

?

?

?????

?????

?

??

????

?

V

dV

t

?

????

?

V

dVV

t

?

?

Time rate change of mass inside

control volume

Time rate change of momentum

inside control volume

?? ?

S

SdV

??

?

? ??? ?

S

VSdV

???

?

Net flow of mass out of the control

volume through control surface S

?? ?

S

SdV

??

Net flow of volume out of the control

volume through control surface S

Net flow of momentum out of the

control volume through control

surface S

???

V

dVf

?

?

???

S

Spd

?

Body force through out the control

volume V

Surface force over the control surface

S

What we can foresee the applications for

aerodynamic problems with basic flow

equations on hand?

0??? V?

If the steady incompressible inviscid flows are

concerned

? ?

x

pVu

?

?????

?

1?

? ?

y

pVv

?

?????

?

1?

? ?

z

pVw

?

?????

?

1?

Partial differential equation for velocity

Partial differential equation for velocity

and pressure

2.6 An application of the momentum

equation,drag of a 2D body

How to design a 2D wind tunnel test?

How to measure the lift and drag exerted on

the airfoil by the fluid?

A selected control volume around an airfoil

Descriptions of the control volume

1,The upper and lower streamlines far above and below the

body (ab and hi),

2,Lines perpendicular to the flow velocity far ahead and

behind the body(ai and bh)

3,A cut that surrounds and wraps the surface of the

body(cdefg)

1,Pressure at ab and hi,

2,Pressure at ai and bh,,velocity,

3,The pressure force over the surface abhi

?? pp

?? pp c o n s ta n tu ?1 )(22 yuu ?

???

a b h i

Spd

?

4,The surface force on def by the presence of the body,this

force includes the skin friction drag,and denoted as per

unit span,

5,The surface forces on cd and fg cancel each other,

6,The total surface force on the entire control volume is

R??

R ???? ??

a b h i

Spdf o r c es u r f a c e

?

7,The body force is negligible

Apply to momentum equation,we have

? ? R ??????

?

?

???????

a b h iSV

SpdVSdVdVV

t

?????

??

for steady flow

? ? ???? ?????

a b h iS

SpdVSdV

????

?R

Note,it’s a vector equation,

If we only concern the x component of the equation,

with represents the x component of, D? R?

? ? ? ????? ?????

a b h i

x

S

SpduSdVD

???

?

As boundaries of the control volume abhi are chosen

far away from the body,the pressure perturbation

due to the presence of the body can be neglected,

that means,the pressure there equal to the

freestream pressure,If the pressure distribution

over abhi is constant,then

? ? 0???

a b h i

xSpd

?

So that

? ??? ????

S

uSdVD

??

?

As ab,hi,def are streamlines,then

? ? ? ? ? ? 0?????? ??????

d e fhiab

uSdVuSdVuSdV

??????

???

As cd,fg are are adjacent to each other,then

? ? ? ????? ????

fgcd

uSdVuSdV

????

??

The only contribution to momentum flow through

the control surface come from the boundaries ai and

bh,For dS=dy(1),the momentum flow through the

control surface is

? ? ???? ????

b

h

a

i

S

dyudyuuSdV 222211 ???

??

Note,

1,The sign in front of each integrals on the right

hand side of the equation

2,The integral limits for each integrals on the right

hand side of the equation

Consider the integral form of the continuity equation

for steady flow,

02211 ??? ??

b

h

a

i

dyudyu ??

or

?? ?

b

h

a

i

dyudyu 2211 ??

As is a constant

1u

?? ?

b

h

a

i

dyuudyu 122211 ??

? ? ???? ????

b

h

a

i

S

dyudyuuSdV 222211 ???

??

?

? ? ???? ????

b

h

b

h

S

dyudyuuuSdV 222122 ???

??

?

? ? ??? ????

b

h

S

dyuuuuSdV )( 2122??

??

The final result gives the drag per unit span

? ???

b

h

dyuuuD )( 2122?

? ? ??? ??????

b

h

S

dyuuuuSdVD )( 2122??

??

?

The drag per unit span can be expressed in terms of

the known freestream velocity and flow-field

properties,across a vertical station

downstream of the body,

1u

22 uand?

Physical meaning behind the equation

? ???

b

h

dyuuuD )( 2122?

?

b

h

dyu 22? )( 21 uu ?

Mass flow out of the

control volume

Velocity decrement

? ?

b

h

dyuuu )( 2122?

Momentum decrement per second

For incompressible flow,that is,the density is constant

? ???

b

h

dyuuuD )( 212?

2.6.1 Comments

With the application of momentum principle to a

large,fixed control volume,an accurate result for

overall quantity such as drag on a body can be

predicted with knowing the detailed flow properties

along the control surface,That to say,it is

unnecessary to know the the details along the

surface of the body,

2.7 Energy equation

0??? V?

? ?

x

pVu

?

?????

?

1?

? ?

y

pVv

?

?????

?

1?

? ?

z

pVw

?

?????

?

1?

? Continuity equation

? Momentum equation

Unknowns,Vp ?,

c o n s ta n t??

For steady incompressible invicid flows

For compressible flows

? is an additional variable,and therefore we need an additional fundamental equation to complete the

system,This fundamental equation is the energy

equation,which we are going to develop,

Two additional flow-field variables will appear to the

energy equation,that is internal energy and

temperature,

e

T

Energy equation is only necessary for compressible

flows,

Physical principle(first law of thermodynamics)

Energy can be neither created nor destroyed; it can

only change in form

Definitions of system and internal energy per

unit mass e

Definition of surroundings

Heat transferred from the

surroundings to the system

Work done on the

surroundings by the system

q?

w?

Change of internal energy in system due

to the heat transferred and the work done

de

As energy is conserved,so

dewq ?? ??

Apply the first law to the fluid flowing

trough the fixed control volume,and let

B1 = rate of heat added to fluid inside control volume from

surroundings,

B2 = rate of work done on fluid inside control volume,

B3 = rate of change of energy of fluid as it flows through

control volume,

As first law should be satisfied,then

B1+B2 = B3

Actually speaking,the equation above is a power

equation,

Rate of volumetric heating

???

V

dVq ??

If the flow is viscous

B1 =

v i s c o u s

V

QdVq ?? ???? ?

Rate of volumetric heating = VF ?? ?

The force includes three parts F?

Pressure force,body force and skin friction force

Rate of work done on fluid inside V

due to pressure force on S ?? ??

S

VSpd

??

)(

Rate of work done on fluid inside V

due to body force ??? ?

V

VdVf

??

)( ?

B2 =

v i s c o u s

VS

WVdVfVSpd ?

????

????? ????? )()( ?

Since the fluid inside the control volume is not

stationary,it is moving at the local velocity with a

consequent kinetic energy per unit mass,so,

the total energy per unit mass is

V?

2/2V

2/2Ve ?

Net rate of flow of total energy

across control surface S ?? ???

S

V

eSdV )

2

()(

2??

?

Time rate change of total energy

inside V due to transient variations

of flow-field properties

??? ??

?

V

dV

V

e

t

)

2

(

2

?

B3 =

????? ??????

?

SV

V

eSdVdV

V

e

t

)

2

()()

2

(

22 ??

??

B1+B2 = B3

?????

????????

????

?

?

?

??????

SV

v i s c o u s

VS

v i s c o u s

V

SdV

V

edV

V

e

t

WVdVfVSpdQdVq

??

?

????

??

)

2

()

2

(

)()(

22

??

??

Energy equation in integral form

Notes in the text book

''

22

)()(

)

2

()

2

(

v i s c o u sv i s c o u s

WQVfVpq

V

V

e

V

e

t

??

???

?

?

????????

?

?

?

?

?

?

?????

?

?

?

?

?

?

?

?

??

??

Energy equation in partial differential form

If the flow is steady,inviscid,adiabatic,without body

force

???? ??????

SS

VSpdSdV

V

e

????

)()

2

(

2

?

)()

2

(

2

VpV

V

e

??

?????

?

?

?

?

?

??? ?

After apply three fundamental physical principles,we

have derived three basic equations for fluid flow,And

there are three variables,such as eVp,,,??

For calorically perfect gases

Tce v?

Then,one more property is added,but with perfect

gas equation

RTp ??

Continuity,momentum and energy equation with two

additional equations are five independent equations,

and there five unknowns, So that we have

got a closed system for the flow problems,

TeVp,,,,??

2.8 Interim summary

2.9 Substantial derivatives

Focus our eye on a infinitesimal fluid

element moving through a flow field,The

velocity field can be given as kwjviuV ???? ???

),,,(

),,,(

),,,(

tzyxww

tzyxvv

tzyxuu

?

?

?

The density can be given as

),,,( tzyx?? ?

),,,( 111111 tzyx?? ?

),,,( 222222 tzyx?? ?

With the use of Taylor series expansion about point 1

)()()()( 12

1

12

1

12

1

12

1

12 tttzzzyyyxxx ???

??

?

?

?

????

?

??

?

?

?

???

???

?

???

?

?

????

?

??

?

?

?

??? ??????

Dividing by )( 12 tt ?

112

12

112

12

112

12

112

12 ?

?

??

?

?

?

??

?

??

?

??

?

?

?

??

?

?

???

?

???

?

?

??

?

??

?

??

?

?

?

??

?

?

ttt

zz

ztt

yy

ytt

xx

xtt

??????

Dt

D

tttt

??? ?

?

?

?

12

12

12

lim

u

tt

xx

tt

?

?

?

?

12

12

12

lim v

tt

yy

tt

?

?

?

?

12

12

12

lim wtt

zz

tt

?

?

?

?

12

12

12

lim

tz

w

y

v

x

u

Dt

D

?

??

?

??

?

??

?

?? ?????

tz

w

y

v

x

u

Dt

D

?

??

?

??

?

??

?

??

in cartesian coordinates,

z

k

y

j

x

i

?

??

?

??

?

??? ???

then

? ????

?

?? V

tDt

D ?

t?

?

Dt

D

??V?

Substantial derivative

Local derivative

convective derivative

2.10 Fundamental equations in term

of substantial derivative

In this section,the continuity,momentum

and energy equations will be given in terms

of substantial derivative

? ? ??? ???????? VVV ???

The continuity equation in differential form is

? ? 0????

?

? V

t

?

??

0???????

?

? ??? VV

t

??

or

Dt

DV

t

??? ????

?

? ?

Since

So

0???? V

Dt

D ???

This is the continuity equation in terms of

substantial derivative

The x component of the momentum

equation in differential form is

? ? v i s c o u sxxf

x

pVu

t

u )()( F??

?

??????

?

? ??? ?

t

u

t

u

t

u

?

??

?

??

?

? ??? )(

? ? ? ? ? ? uVVuVu ???????? ??? ???

? ? ? ?

v i s c o u sxx

f

x

p

uVVu

t

u

t

u

)( F??

?

?

?

???????

?

?

?

?

?

?

??

?

?

??

or

? ? ? ?

v i s c o u sxx

f

x

p

uVV

t

u

t

u

)( F??

?

?

?

????

?

?

?

?

?

?

???

?

?

?

?

?

?

??

?

?

??

? ? 0??

?

?

??

? ???

?

? V

t

?

??

Continuity Equation

? ? v i s c o u sxxf

x

puV

t

u )( F??

?

??????

?

? ??? ?

hence

?

v i s c o u sxxfx

puV

t

u )( F??

?

???

?

?

?

?

?

? ???

?

? ?? ?

?

v i s c o u sxxfx

p

Dt

Du )( F??

?

??? ??

In the same way we can get

v i s c o u syyfy

p

Dt

Dv )( F??

?

??? ??

v i s c o u szzfz

p

Dt

Dw )( F??

?

??? ??

these are the momentum equations in terms of

substantial derivative in x,y,z directions respectively

Energy equation in terms of substantial derivative

''

2

)()()2( v i s c o u sv i s c o u s WQVfVpqDt VeD ??

???

? ????????? ???

Detailed descriptions for the comparison

between the basic flow equations in different

forms,refer to the text book

2.11 Pathlines and streamlines of a flow

Skipped over

2.12 Angular velocity,vorticity and strain

In this section,more attention will be paid to

examine the orientation of the fluid element

and its shape as it moves through a

streamline in the flow field,An important

quantity,vorticity,will be introduced,

Motion of a fluid

element along a

streamline

Try to set up the relationships between

with and 21

,?? ??

xvyu ????,t?

Distance in y direction that A moves

during time increment tv?

Distance in y direction that C moves

during time increment tdxxvv ??????? ???

Net displacement in y direction of

C relative to A

tdx

x

v

tvtdx

x

v

v

??

?

?

?

?

?

?

?

?

????

?

?

?

?

?

?

?

?

? ?? ? t

x

v

dx

tdxxv ?

?

???????

2t a n ?

Since is a small angle

2?? 22t a n ?? ???

t

x

v ?

?

???

2?

Similarly

t

y

u ?

?

????

1?

y

u

tdt

d

t ?

???

?

??

??

1

0

1 lim ??

x

v

tdt

d

t ?

??

?

??

??

2

0

2 lim ??

Definition,angular velocity of the fluid

element is the average angular velocity of

lines AB and AC,they are perpendicular to

each other at the time t

??

?

?

??

?

?

?

?

?

?

?

??

?

?

?

?

? ??

y

u

x

v

dt

d

dt

d

z 2

1

2

1 21 ??

?

k

y

u

x

vj

x

w

z

ui

z

v

y

w ????

???

?

???

?

?

??

?

??

?

?

??

?

?

?

??

?

??

???

?

???

?

?

??

?

??

2

1

2

1

2

1?

?? ?? 2? vorticity

k

y

u

x

v

j

x

w

z

u

i

z

v

y

w ????

??

?

?

??

?

?

?

?

?

?

?

??

?

?

?

?

?

?

?

?

?

?

???

?

?

??

?

?

?

?

?

?

?

??

In a velocity field,the curl of the velocity is

equal to the vorticity

1,If at every point in a flow field,the

flow is called rotational,This implies that the

fluid elements have a finite angular velocity

2,If at every point in a flow,the flow is

called irrotational,This implies that the fluid

element have no angular velocity; their

motion through space is a pure translation

0??? V?

0??? V?

Definition of strain,the strain of the fluid

element in xy plane is the change of in k,

where the positive strain corresponds to a

decreasing k,and k is the angle between

sides AB and AC,they are perpendicular to

each other at the time t

Strain =

12 ??? ??????

The time rate of strain in xy plane is

y

u

x

v

dt

d

dt

d

dt

d

xy ?

??

?

?????? 12 ????

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

z

w

y

w

x

w

z

v

y

v

x

v

z

u

y

u

x

u

In the matrix above which is composed of velocity

derivatives,the diagonal terms represent the

dilatation(扩张 ) of a fluid element,The off diagonal

terms are associated with rotation and strain of fluid

element,

Relations between viscous effect and rotation of a

fluid element,

Irrotational and rotational flows in practical

aerodynamic problems

2.13 Circulation

Important tool for we to obtain solutions for some

very practical and exciting aerodynamic problems,

Circulation can be used to calculate lift exerted on

an airfoil with unit span,

Definition of circulation

? ???? C sdV ??

Note:the negative sign in front of the line integral

Stokes’ theorem

? ? SdVsdV

SC

????

???????? ???

Referring to the vector analysis,what is the physical

meaning that the equation bellow speak?

? ?

dS

dnV ?????? ??

2.14 Stream function

In a steady 2D steady flow,the differential form of

streamlines can be expressed as

u

v

dx

dy ?

If are known functions of,then,after the

equation above being integrated,we can get the

algebraic equation of the streamline

vu,yx,

cyxf ?),(

For each streamline,is a constant,Its value varies

with different streamlines,Replacing the symbol

with,then we have

c

f

?

cyx ?),(?

The function is called stream function,

Different value of the,i.e,,represents

different streamlines in the flow field,

),( yx?

c

321,,ccc

Two streamlines respecting with different values of c

Physical meaning of the stream function

Arbitrariness of the integrand constant c

Difference in stream function between two individual

streamlines

?????? 1212 cc???

Mass flow between the two

streamlines ab and cd,

(per unit depth

perpendicular to the page)

How to remove the arbitrariness of the constant of

integration?

What will be the mass flow through an arbitrary

curve connecting two points on a streamline?

For a steady flow,the mass flow inside a given

streamtube is constant,

For a steady flow,the continuity equation should be

satisfied,then,the mass flow through a closed curve

C is zero,That means the mass flow through L1 is

the same to that of L2,

For a steady flow and physical possible flows,the

integration of mass flow between two points is

independent of the path of the integration,

Important properties of the stream function

If is small,is a constant value across n? V n?

)1(nV ??? ??

or

V

n

?? ?

?

?

as 0?? n

nn

V

n ?

??

?

??

??

???

0

l i m

The equation above states that,if we know the

distribution of,then we can obtain the product

by differentiating in the direction normal to

? )( V?

?

V

Note,is a scalar variable,its direction is defined

by the streamlines,

V

Relationship between and differential of vu ??,

v d xudyd ??? ??

As,then the chain rule of calculus states ),( yx?? ?

dy

y

dx

x

d

?

??

?

?? ???

Comparing the two equations above,we have

x

v

y

u

?

???

?

?? ????,

If is known for a given flow field,then at any

point in the flow the products and can be

obtained by differentiating in the directions

normal to and,respectively,

),( yx?

u? v?

?

u v

In polar coordinates

r

V

rr

V r

?

???

?

?? ??

?

??

?,

1

),( yx? is equal to mass flow,for incompressible flow,

that is, Then,the velocity along a

streamline can be written as

con stan t??

? ?

n

V

?

?? ??

if we define,then ??? ?

n

V

?

?? ?

The features for the function of, For incompressible

flow,is constant along a streamline,

Represents the equation of a streamline,

?

? co ns ta nt??

??

??

Mass flow between two streamlines

Volume flow between two streamlines

Two primary reasons for the concept of the stream

function to be a powerful tool in aerodynamics,

1,(or ) gives the

equation of a srteamline,

2,The flow velocity can be obtained by differentiating

co ns ta nt?? co ns ta nt??

)( ?? or

How to give the determination of the stream function?

2.15 Velocity potential

For an irrotational flow (no matter it is

compressible or incompressible)

0???? V???

Then,there must exists a scalar function,for which ?

???V?

where,is called velocity potential ?

another explanation for the existence of the vlocity

potential for a irrotational flow

0???

C

ldV

?? ?

0???? ??

A

B

B

A

ldVldV

????

?? ??? 21 LL ldVldV ????

The equation above shows that the line integral of the

velocity is independent with the path of the integration,

Or to say,the result of the integral is a function of the

start and end positions only,If the start point(A) is

fixed somewhere in the space,we have

?? ????? BABA w d zv d yudxldVzyx ??),,(?

dz

z

dy

y

dx

x

d

?

??

?

??

?

?? ????

x d zv d yu d xd ????

?

z

w

y

v

x

u

?

??

?

??

?

?? ???

The features for the velocity potential analogous to the

stream function ----- derivatives of yield the flow-

velocities,

Distinction between and

?

? ?

1,The flow-field velocity are obtained by

differentiating in the same direction as the

velocities,whereas is differentiated normal to

the velocity direction

2,The velocity potential is defined for irrotational flow

only,In contrast,the stream function can be used

in either rotational or irrotational,

3,The velocity potential applies to three dimensional

flows,whereas the stream function for 2D only,

?

?

Why the analysis of irrotational flow is simpler than

that of rotational flows?

Irrotational flows is usually called potential flows

How to give the determination of the velocity potential?