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?
