PART II
INVISCID INCOMPRESSIBLE FLOW
In part II,we deal with the flow of a fluid which has
constant density– incompressible flow,This applies to the
flow of liquids,such as water flow,and to low-speed flow
of gases,The material covered here is applicable to low-
speed flight through the atmosphere—flight at a Mach
Chapter 3
Fundamentals of Inviscid,
incompressible Flow
Theoretical fluid dynamics,being a difficult subject,is for
convenience,commonly divided into two branches,one
treating frictionless or perfect fluids,the other treating of
viscous or imperfect fluids,The frictionless fluid has no
existence in nature,but is hypothesized by mathematicians in
order to facilitate the investigation of important laws and
principles that may be approximately true of viscous or natural
fluids,
From an aerodynamic point of view,at
air velocities between 0 to 360km/h the
air density remains essentially constant,
varying only a few percent,
Purpose of this chapter is to establish
some fundamental relations applicable to
inviscid,incompressible flows and to
discuss some simple but important flow
fields and applications,
3.2 Bernoulli’s Equation
c o n s tVp ?? 2
2
1 ?
Derivation of Bernoulli’s equation
Step 1,x component momentum equation
without viscous effect and body forces
x
p
Dt
Du
?
????
or
x
p
z
uw
y
uv
x
uu
t
u
?
???
?
??
?
??
?
??
?
? ????
x
p
z
uw
y
uv
x
uu
?
???
?
??
?
??
?
?
?
1
Multiply both sides with dx
dx
x
pdx
z
uwdx
y
uvdx
x
uu
?
???
?
??
?
??
?
? ???
Along a streamline in 3D space,there are
0
0
??
??
udyv d x
w d xudz
Substituting the differential equations of the
streamline into x component momentum equation
dx
x
pdz
z
uudy
y
uudx
x
uu
?
???
?
??
?
??
?
?
?
1
or
dx
x
p
dz
z
u
dy
y
u
dx
x
u
u
?
?
????
?
?
??
?
?
?
?
?
?
?
?
?
?
?
1
as
dz
z
udy
y
udx
x
udu
?
??
?
??
?
??
We have
dx
x
pudu
?
???
?
1
or
dx
x
pdu
?
???
?
1
2
1 2
In the same way,we can get
dy
y
pdv
?
???
?
1
2
1 2 dz
z
pdw
?
???
?
1
2
1 2
??
?
?
??
?
?
?
?
?
?
?
?
?
?
???? dz
z
p
dy
y
p
dx
x
p
wvud
?
1
)(
2
1 222
2222 Vwvu ???
as
and
dpdz
z
pdy
y
pdx
x
p ?
?
??
?
??
?
?
Then,we have
dpdV
?
1
2
1 2 ?? or V d Vdp ???
V d Vdp ???
The above equation is called Euler’s Equation,
precondition,inviscid,without body force,along a
streamline,
usage,setup the relation between dpa n ddV
Step 2,integration of Euler’s equation
For incompressible flows,
The integration from point 1 to point 2 along a
streamline is
c o n s t??
?? ??
2
1
2
1
V
V
p
p
V d Vdp ?
or
??
?
?
??
?
?
????
22
2
1
2
2
12
VV
pp ?
or
2
11
2
22 2
1
2
1 VpVp ?? ???
The above equation is called Bernoulli’s Equation,
body force,along a streamline,
usage,setup the relation between at point
1 on a streamline to at another point 2 on
the same streamline,
11 pa n dV
22 pa n dV
As point 1 and point 2 are arbitrarily chosen on a
streamline,so
c o n s tVp ?? 2
2
1 ?
Along a streamline
1,As there are no stipulation being made to whether
the flow is rotational or irrotational,it satisfies for
both irrotational flows and rotational flows as well,
2,The value of the constant will change from one
streamline to another,
3,For irrotational flows
c o n s tVp ?? 2
2
1 ? Through out the flow
Bernoulli’s equation is also a relation for mechanical
energy in an incompressible flow
c o n s tVp ?? 2
2
1 ?
where is the kinetic energy per unit volume 22V?
As Bernoulli’s equation can also be derived from the
energy equation,the energy equation is redundant
for the analysis of inviscid,incompressible flow,
The way for solving inviscid,incompressible flows,
1,Obtain the velocity field from the governing
equations,
2,Obtain the pressure field from Bernoulli’s equation
3.3 Incompressible flow in a duct,the
Venturi tube and low-speed wind tunnel
Assumption of quasi-one-dimensional flows
1,Flow-field properties are uniform across any cross
section,
2,All the flow-field properties are assumed to be
functions of x only,
Respect to continuity equation,and for steady case
0222111 ??? VAVA ??
or
222111 VAVA ?? ?
For incompressible flow
2211 VAVA ?
Applications of venturi tube
Measurements of flow speed in a duct,
2
212
2
1 )(
2 VppV ???
?
From Bernoulli’s equation
From continuity equation,we have
1
2
1
2 VA
AV ?
Then
2
1
2
2
1
12
2
1 )(
2
V
A
A
ppV ??
?
?
??
?
?
???
?
? ?1)(
)(2
2
21
12
1 ?
?
?
AA
pp
V
?
3.4 Pitot tube,measurement of airspeed
Definition of static pressure,the measurement of
the static pressure,
Definition of stagnation pressure,the measurement
of the stagnation pressure,
The usage of Pitot tube,
The pressure at point C is the stagnation pressure
0p
The pressure at point A is the static pressure
1p
The speed at point C is equal to zero
The speed at point A is equal to free stream velocity
1V
The speed and pressure and at point A and B are the
same,apply Bernoulli’s equation,we have
22
2
1
2
1
BBAA VpVp ?? ???
0
2
1
0
2
11 ??? pVp ?
or
?
)(2 10
1
pp
V
?
?
Combined instrument for the measurement of both
static and total(stagnation) pressure,
Definition of dynamic pressure,the relationship
between of the static pressure,total pressure and
dynamic pressure,
2
2
1 Vq ??
For incompressible flows
0
2
11 2
1 pVp ?? ?
static
pressure
dynamic
pressure
total
pressure
Keep in mind !!!!!
Any result derived from Bernoulli’s equation holds for
incompressible flow only
3.5 Pressure coefficient
Definition of the pressure coefficient
?
???
q
ppC
p
where
2
2
1
??? ? Vq ?
The pressure coefficient is a nondimensional value,
it can be used throughout aerodynamics,for both
compressible flow and incompressible flow,
For incompressible flow,the pressure coefficient
can be expressed in terms of velocity only
If and are defined as the freestream velocity
and pressure,And and are the velocity and
pressure at an arbitrary point in the flow,Then,with
Bernoulli’s equation,
?V ?
p
V p
22
2
1
2
1 VpVp ?? ???
??
or
)(
2
1 22 VVpp ???
?? ?
?
???
q
ppC
p
? )(
2
1 22 VVpp ???
?? ?
2
22
2
1
)(
2
1
?
?
?
?
?
?
?
?
V
VV
q
pp
C
p
?
?
or
2
1 ??
?
?
??
?
?
??
?
?
??
?
V
V
q
pp
C p
Valid for
incompressible
flow only
??
??
?
?
?
??
??
??
??
??
??
ppVV
Cqpp
CVV
CVV
CVV
CV
p
p
p
p
p
0
0
0
10
3.6 Condition on velocity for
incompressible flow
Contents in the center branch
1,Laplace’s equation
2,General philosophy and use in solving problems
3,Synthesis of complex flows from a superposition of
elementary flows
Basic condition on velocity in an incompressible
flow,
From the definition for the divergence of velocity
? ?
Dt
VD
V
V ?
?
1??? ?
If the density of the fluid is equal to a constant,then
? ? 0?
Dt
VD ?
That means,for incompressible flow,the velocity
must satisfies
0??? V?
If we consider the continuity equation
? ? 0????
?
? V
t
?
??
And apply it for the incompressible flows,then,the
continuity equation becomes
00 ???? V??
?
0??? V?
3.7 Governing equation for irrotational,
incompressible flow,Laplace’s Equation
The continuity equation for incompressible flow is
0??? V?
???V?
Therefore,for incompressible and irrotational flows
0)( 2 ?????? ?? Laplace’s equation
If the flow is irrotational,then,the velocity can be
expressed with velocity potential
Solutions of Laplace’s Equation are called harmonic
functions,
Solutions of Laplace’s Equation
in Cartesian coordinates,
02
2
2
2
2
2
2 ?
?
?
?
?
?
?
?
?
??
zyx
???
?
in cylindrical coordinates,
0
11
2
2
2
2
2
2 ?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
??
zrr
r
rr
?
?
??
?
For 2D incompressible flow,a stream function can
be defined,and the relation between velocity and
stream function is
x
v
y
u
?
???
?
?? ??,
The continuity equation,expressed in Cartesian
coordinates is,
0?
?
??
?
????
y
v
x
uV?
0
22
?
??
?
?
??
?
??
?
?
?
?
?
?
?
?
?
?
???
?
?
??
?
?
?
?
?
?
xyyxxyyx
????
The very definition and use of stream function is a
statement of the continuity equation of mass,and
therefore Equations bellow
x
v
y
u
?
???
?
?? ??,
Can be used in place of the continuity equation itself,
If the flow is irrotational,then
0?
?
??
?
?
y
u
x
v
? xvyu ?
???
?
?? ??,
0???
?
?
??
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
yyxx
??
?
02
2
2
2
?
?
?
?
?
?
yx
??
Laplace’s equation
Important conclusions
1,Any irrotational,incompressible flow has a velocity
potential and stream function(for 2D flow) that
both satisfy Laplace’s Equation
2,Conversely,any solution of Laplace’s Equation
represents the velocity potential or stream
function(2D) for an irrotational and incompresible
flow,
Superposition of the solutions of Laplace’s Equation
The Laplace’s Equation is a second-order linear
partial differential equation,The sum of any
particular solutions of a linear differential equation is
also a solution of the equation(Laplace’s Equation),
For example,if represent n separate
solutions of the Laplace’s equation,then the sum
n????,,,,321 ??
n????? ????? ??321
is also a solution of the Laplace’s equation
Conclusion respects to the superposition of the
solutions of Laplace’s equation
A complicated flow pattern for an irrotational,
incompressible flow can be synthesized by adding
together a number of elementary flows which are
also irrotational and incompressible,
How to make the difference between irrotational
and incompressible flows around different
aerodynamic shapes,--- Boundary conditions
3.7.1 Infinity boundary conditions
0,?
?
???
?
???
?
??
?
??
? xyvVyxu
????
At infinity in all directions
3.7.2 Wall boundary conditions
0?? nV ??
On the surface of a solid body,respecting to
or
0?
?
?
n
?
?
0?
?
?
s
?
or
c o n s tbyys u r f a c e ?? ???
On the surface of a solid body,respecting to ?
On the surface of a solid body,respecting to
streamline
s u r f a c e
b
u
v
dx
dy
?
?
?
?
?
??
3.8 Interim summary
General approach to solve the incompressible and
irrotational flows
1,Solve Laplace’s equation for or along with
the proper boundary conditions,These solutions
are usually in the form of a sum of elementary
solutions,
2,Obtain the flow velocity from or
3,Obtain the pressure from Bernoulli’s equation
? ?
???V?
xvyu ??????? ??,
22
2
1
2
1
?? ??? VpVp ??
3.9 Uniform flow,our first elementary flow
There are a series of elementary flows which can be
superimposed to synthesize more complex
incompressible flows,Right now,we are starting to
introduce them one by one,
Uniform flow,if the velocity is a constant at every
point in the flow field,then,such a flow can be
called as uniform flow,
It is easy to see,a uniform flow is a physical possible
incompressible flow and it is irrotational
0??? V? 0??? V?
Velocity vector is in x direction,and it is a constant
As this is a irrotational flow,then,the velocity
components can be expressed by
y
v
x
u
?
??
?
?? ??,
0,??
?
???
?
?
? vyVux
??
or
After integration being carried out
c o n s txV ?? ??
or
xV ???
Reason?
As this is a divergence free flow,then,the velocity
components can be expressed by
0,?
?
????
?
??
? xvVyu
??
0,???
?
???
?
?
? vxVuy
??
or
After integration being carried out
c o n s tyV ?? ??
yV ???
or
Characteristics of Uniform flow
0???? ?
C
sdV ?
?
1,Circulation around any closed curve in a uniform
flow is zero,
2,The velocity potential and stream function satisfies
Laplace’s equation,
3.10 Source flow,
our second elementary flow
Definition of source flow,
1,All the streamline are straight line emanating from
a center point O,
2,Velocity along the streamline varies inversely with
the distance from point O,
3,Source flow is a physical possible flow(that means
the divergence of velocity is zero at every point
except the origen.where it is infinite,Thus,the
origin is a singular point,
4,Source flow is irrotational at every point,
The origin,where the streamlines emanating from,
can be interpreted as a discrete source or sink,
The radial flow surrounding the origin can be
looked as the flow induced by the discrete source
or sink,
Superposition of distributions of singularities over
arbitrary bodies can be used to synthesize the
irrotational,incompressible flows around the bodies,
Definition of the strength of a discrete source or sink,
r
cV
r ?
0??V
Line source in three-dimensional space
Mass flow across the surface of the cylinder
rrr Vrldr lVlrdVm ??????
??
?? ???
2
0
2
0
2)(?
Volume flow across the surface of the cylinder
rr lV
mv ?
?
2??
??
The rate of volume flow per unit length along the
cylinder,
rrVl
v ?2??? ?
r
V r
?2
???
Velocity potential for a source
r
V
r r ?
?
2
???
?
?
01 ??
?
?
??
? V
r
)(ln
2
?
?
? fr ???
)( rgc o n s t ???
?
c o n s tr ??? ln
2 ?
?
?
rln
2 ?
? ??
Stream function for a source
r
V
r r ??
?
2
1 ???
?
?
0??
?
??
?
? V
r
)(
2
rf??? ?
?
?
)(?? gc o n s t ??
?
c o n s t??? ?
?
?
2
?
?
?
?
2
??
Description of stream function for a source
c o n s t?? ? c o n s t??
?
a straight line from the origin
Description of velocity potential for a source
c o n s t?? ? c o n s tr ?
?
a circle with its center at the origin
The velocity potential and stream function of a
source satisfy Laplace’s equation,
Conclusion,
source flow is a viable elementary flow for use in
building more complex flows
3.11 Combination of a uniform flow with a
source and sink
In a polar coordinate system,the stream functions of
a uniform flow and a source or sink can be written as,
?
?
?
2
???? s i nrV ?? and
Combination of two simple elementary flows,
Combination of a uniform flow and a source flow,
The stream function of the resulting flow is,
?
?
??
2
s i n ??? ? rV
as it satisfies Laplace’s equation,then,the stream
function above represents a viable irrotational,
incompressible flow,
Streamlines of the combined flow,
c o n s trV ???? ? ?
?
??
2
s i n
Velocity field of the combined flow,
r
V
r
V r
?
?
?
?
2
c o s1 ???
?
??
?
??? s i n???
?
??? V
r
V
Velocity field of a uniform flow,
?c o s?? VV r ?? s i n??? VV
Velocity field of a source flow,
r
V r
?2
?? 0??V
Conclusion,
The combined velocity filed of a uniform flow and
a source flow is simply direct sum of the two velocity
fields,Therefore,not only can we add the values of
velocity potential or stream function to obtain more
complex flows,we can add their derivatives,i.e.,the
velocities as well,
Analysis of the shape of the streamlines in the
combined flow field,
Equation of streamlines of the resulting flow,
c o n s trV ???? ? ?
?
??
2
s i n
Every streamline can be denoted as a solid wall
Among all the streamlines,which one we are
interested most?
A special point in the flow field --- stagnation point
At the stagnation point
0
2
c o s1 ????
?
??
? rVrV r ???
?
0s i n ???
?
???
? ?
?
? VrV
Solving for and, We find one stagnation point
located at
r ?
),2(),( ??? ??? Vr
The stagnation point is located directly upstream of
the source for a distance
)2( ?? V?
)2( ?? V??V increase ? decrease
? increase ? )2( ?? V?
increase
The value of stream function at the stagnation point is
c o n s t
V
V ?????
?
? ????? 2s in2
c o n s t???
2
?
Hence,the streamline that passing through the
stagnation point is described by,and it is
the curve ABC shown in the figure,
2???
Conclusion,(for inviscid flow)
1,Any streamline of the combined flow could be
replaced by a solid surface,
2,Curve ABC is a dividing streamline,
3,The entire region inside ABC could be replaced
with a solid body of the same shape,
4,The streamline ABC extends downstream to
infinity,and will not intersect with each other,that
means,the body is not closed,
Superposition of a uniform flow,a source flow,and
a sink flow,
The stream function of the resulting flow is,
21 22s in ??????
?????
? rV
or
)(
2
s in 21 ??
?
?? ???? ? rV
are functions of
21,?? ba n dr ?,
Setting,we can get two stagnation points at A
and B,And these two stagnation points are located
such that,
0?V
?
?
???
V
b
bOBOA
?
2
The equation of the streamline is
c o n s trV ????? ? )(
2
s i n 21 ??
?
??
At point A,
???? ??? 21
At point B,
021 ??? ???
So,the values of stream function at two stagnation
points are zero
The two stagnation points are on the same streamline,
and the equation the stagnation streamline is
0)(
2
s i n 21 ????? ??
?
?rV
It is the equation of an oval,
It is also the dividing streamline,
The region inside the oval can be replaced by a solid
body with the shape given by the equation above,
The region outside the oval can be interpreted as the
inviscid,potential,incompressible flow over the oval,
3.12 Doublet flow,our third elementary
flow
※ Source-sink pair degenerate a doublet
Stream function of a induced flow field by a source-
sink pair
?
?
??
?
? ???????
2
)(
2 21
Let distance,but remains constant,
then,the limit of the stream function denotes a
special flow called doublet
0?l ???l
?
?
?
?
?
? ???
??
?
?
?
?
d
c o n s tl 2
l im
,0
For an infinitesimal,the geometry of fig.3.19 yields ?d
badlrbla ???? ???,co s,s i n
Hence,
?
??
c o s
s i n
lr
l
b
?
??
?
?
?
?
?
? ???
??
?
?
?
?
d
c o n s tl 2
l im
,0
?
??
c o s
s i n
lr
l
b
????
?
?
?
?
?
?
?
???
?? ?
?
?
?
? c o s
s i n
2
lim
,0 lr
l
c o n s tl
or
?
?
?
?
?
?
?
??
?? ?
?
?
??
? c o s
s i n
2
lim
,0 lrc o n s tl
r
?
?
?? s i n
2
??
or Stream function of a doublet
In a similar way we can get
r
?
?
?? c o s
2
?
Velocity potential of a
doublet
Streamlines of a doublet flow
cc o n s t
r
???? ?
?
?? s i n
2
or
?
?
? s i n
2 c
r ??
Let,then we have an equation in polar
coordinate system
cd ?? 2??
?s i ndr ?
It is a circle with diameter d,on the vertical axis and
with the center located directly above the origin,
yddryxr ???? ?s i n222
?
222 )2()2( ddyx ???
Direction of a doublet flow,
the direction of a doublet is pointed from the sink to
the source
Detailed explanation for the limit as,
0?l
Doublet is a singularity that induces about it the
double-lobed circular flow,
3.13 Nonlifting flow over a circular cylinder
Some thing should be mentioned first
Uniform flow + a source
flow over a semi-infinite body
Uniform flow + a source-sink pair
flow over an oval-shaped body
Uniform flow + a doublet

Stream function combined with a uniform flow and
a doublet
r
rV ?
?
??? s i n
2
s i n ?? ?
or
??
?
?
??
?
?
??
?
? 221s in rVrV ?
?
??
Let
?? VR ?? 22
??
?
?
??
?
?
?? ?
2
2
1s in
r
R
rV ??
Stream function for a uniform flow + a doublet flow
Stream function for a flow over a circular cylinder of
Prove,
Step 1,Velocity field
??
?
?
c o s11)c o s(
11
2
2
2
2
?? ??
?
?
??
?
?
????
?
?
??
?
?
??
?
?
? V
r
R
r
R
rV
rr
V r
?
?
?
?
?
?
??
?
?
??
?
?
????
?
?
?? ?? )s in(1
2
)s in( 2
2
3
2
??
?
? Vr
R
r
R
rV
r
V
?
?
? s in1 2
2
???
?
?
??
?
?
???
?
?
?? V
r
R
r
V
Step 2,Stagnation points
set
0c o s1
2
2
???
?
?
??
?
?
? ? ?V
r
R 0s i n1
2
2
???
?
?
??
?
?
? ? ?V
r
R
There are two stagnation points,located at
)0,(),( Rr ?? ),(),( ?? Rr ?
and
Point A Point B
Step 3,The value of stram function at two stagnation
points
01s in
,02
2 Rr
r
R
rV
?
?
? ???
?
?
??
?
?
??
??
??
The streamline with passing through both
stagnation points,the equation of this streamline is
0??
01s in
2
2
???
?
?
??
?
?
?? ?
r
R
rV ??
The equation above is satisfied with for all Rr ? ?
The equation above is satisfied with for all ??,0? r
The streamline is a dividing streamline,
therefore,we can replace the flow inside the circle
by a solid body,and the external flow will not know
the difference,
The inviscid irrotational,incompressible flow over a
circular cylinder of radius R can be synthesized by a
uniform flow with velocity and a doublet of
strength,where R is related to and through
0??
?V
?
?V
?
?
?
V
R
?
?
2
The entire flow field is symmetrical about both the
horizontal and vertical axes through the center of
the circular cylinder
There is no net lift and net drag over the cylinder
d’Alembert paradox,drag due to the viscous effect
Velocity distribution over the circular cylinder
pressure distribution over the circular cylinder
Example 3.9
3.14 Vortex flow,
Our fourth elementary flow
The flows we have introduced with superposition
of the elementary flows,such as uniform flow,
source or sink flow,and doublet flow,can be used to
present several flows around semi-infinite body,oval,
and circular cylinder,
Now,one thing we have to keep in mind,
there are no net lift exerted on the bodies
mentioned above,
What will happen if a vortex flow joins into the
superposition
Description of a vortex flow
Streamlines,circles
0?rV
r
CV ?
?
Definition of vortex flow,
1,All the streamline are circles with its center located
at point O,
2,Velocity along the streamline varies inversely with
the distance from point O,
3,Vortex flow is a physical possible flow(that means
the divergence of velocity is zero at every
point).where it is infinite.,
4,Vortex flow is irrotational at every point except the
origin,Thus,the origin is a singular point
Evaluation of C,
r
C
r
c o n s tV ??
?
Circulation around a given circular streamline
)2( rVsdV
C
??? ?????? ?
?
or
r
V
?? 2
??? ?
?2
???C
For vortex flow,the circulation along any streamline
is the same,that’s to say,,And is
called as the strength of the vortex,
C?2??? ?
What is the value of vorticity at r=0
CSdV
S
?2)( ??????? ??
?? Stokes theorem
dSVSdVC
SS ????
??????
???
)(2 ?
The vorticity is perpendicular to the paper
Choose a circle with its center at the origin,and its
radius r approached to zero,Not matter how close
the circle approaches to the origin,the circulation
0?r
dSVdSV
S
??
???????
That is
dSVC ?????2
or
dS
CV ?2??? ?
??
dS
C?2
as 0?r
or
???? V?
as 0?r
Vortex flow is irrotational everywhere except at the
point r = 0,where the vorticity is infinite,
The origin,where the vortex is located,is a singular
point in the flow field,
Velocity potential and stream function of the vortex
flow
?
?
?
2
???
rln
2 ?
? ??
3.15 Lifting flow over a circular cylinder
??
?
?
??
?
?
?? ?
2
2
1 1s in r
R
rV ??
Nonlifting over a circular cylinder
c o n s tr ??? ln
22 ?
?
Vortex flow
with
Rc o n s t ln
2 ?
???
R
rln
22 ?
? ??
R
r
r
R
rV ln
2
1s in
2
2
21 ?????
?
???
?
?
??
?
?
???? ?
After superimpose the two flows
?? ofv a l u e sa l lf o rRr 0???
Therefore,r=R is part of a streamline of the flow,
Influence of the value of on the value of the
stream function along the circle r =R
?
is a special case for the flows around a
circular cylinder,
0??
Determined value of is another condition for
the flow around any cylinder being defined
uniquely,
?
the streamlines about the horizontal axis is no longer
symmetrical,There must be a normal force exerted
on the circular cylinder
the streamlines about the vertical axis is still
symmetrical,There would be no horizontal force
exerted on the circular cylinder
Velocity field
?c o s1 2
2
???
?
?
??
?
?
?? V
r
R
V r
r
V
r
R
V
?
??
2
s in1 2
2 ?
???
?
?
??
?
?
??? ?
Stagnation point
0c o s1 2
2
???
?
?
??
?
?
?? ? ?V
r
R
V r
0
2
s in1 2
2
?
?
???
?
?
??
?
?
??? ?
r
V
r
R
V
?
??
With r=R
??
?
?
??
?
? ?
??
? RV?
?
4
a r c s in
Note,
1
4
??
? RV?
1
4
??
? RV?
1
4
??
? RV?
1
4
??
? RV?
For case
14 ??
? RV?
0c o s1 2
2
???
?
?
??
?
?
?? ? ?V
r
R
V r
? 22 ??? or??
r
V
r
R
V
?
??
2
s in1 2
2 ?
???
?
?
??
?
?
??? ?
?
2
2
44
R
RVRV
r ???
?
?
??
?
? ?
?
?
?
?? ??
Velocity on the surface of the cylinder
R
VVV
?
??
2
s i n2 ????? ?
Pressure coefficient along the surface of the cylinder
?
?
?
?
?
?
?
?
??
?
?
??
?
? ?
?
?
????
??
2
2
2
s in2
s in41
RVRV
VC p
??
?
??
The lift per unit span of the circular cylinder
??? ?VL ?
Kutta-Joukowski theorem,
the lift per unit span is directly proportional to the
circulation,
Comparison between the theoretical result and
real case,
Creation of lift on a spinning cylinder
3.16 The Kutta-Joukowski theorem and
the generation of lift
The lift per unit span of the circular cylinder
??? ?VL ?
It also valid in general for cylinder bodies of arbitrary
cross section
If the airfoil is producing lift,the velocity field around
the airfoil will be such that the line integral of velocity
around A will be finite
? ??? A sdV ??
Based on Kutta-Joukowski theorem,the lift per unit
span on the airfoil is given as
??? ?VL ?
Kutta-Joukowski theorem is simply an alternate way
of expressing the consequence of the surface
pressure distribution,
It is not quite proper to say that circulation,causes”
lift,
In the theory of incompressible,potential flow,it is
generally much easier to determine the circulation
around the body rather than calculate the detailed
surface pressure distribution,
How to calculate the circulation for a given body in a
given incompressible,inviscid flow?
3.17 Nonlifting flows over arbitrary bodies,
the numerical source panel method
Construction of a source panel
1,Fig.(a),m = 5
2,Fig.(b),m=11,total strength is the same as case 1
3,Fig.(c),m=101,total strength is the same as case 1
4,Fig.(d),m=101,but the source strength is reduced
5,Fig.(e),
6,Fig.(f),Boundary conditions at inclined surface
???? Vm 2,?
Investigation of nonlifting flows with indirect method
1,Superposition of known elementary flows
2,Find the stagnation points
3,Find dividing streamline
4,Replace the inner region of the dividing stramline
with a solid body
Flows over semi-infinite body,oval,circular cylinder
Find a correct combination of elementary flows to
synthesize the flow over a given body,
------- direct method,
Purpose of this section,present a direct method for
nonlifting flows with source panel method,
------- numerical solution
Under which condition,this numerical method is
valid,
------- incompressible,potential flows
Concept of the source sheet,
infinite number of line sources distributed side by
side,the strength of each line source if infinitesimal
small,
Definition for the strength of source sheet,
source strength per unit length along s
Discontinuity of the normal velocity component
across the source sheet,
?
?
Velocity potential at point P contributed by a small
section of the source sheet,
rdsd ln
2 ?
?? ?
Velocity potential at point P contributed by the
whole source sheet,
rdsyx
b
a
ln
2
),( ??
?
??
A given body of arbitrary shape in free stream flow
A source sheet is used to cover the surface of the
body,Condition for the distribution of )(s?
The source sheet is approximated by a series of
source panels,
The source strength per unit length is a constant
over a given panel,but varies from one panel to the
next,
?
If there are n source panels distributed along the
body surface,then,there would be n unknowns,
that is the source strength per unit length
j?
The velocity potential induced at point P due to the
jth panel is
???
b
a
jpj
j
j dsrln2 ?
?
?
The velocity potential induced at point P due to all
the panels
? ??
??
???
n
j
b
a
jpj
j
n
j
j dsrP
11
ln
2
)(
?
?
??
where
22 )()(
jjpj yyxxr ????
Let point P is located at the control point on panel i
? ?
?
?
n
j
j
jij
j
ii dsryx
1
ln
2
),(
?
?
?
where
22 )()(
jijiij yyxxr ????
Component of normal to the ith panel is
?V
iin VnVV ?c o s,??? ???
??
Normal component of velocity induced at the
control point of the ith panel by the source panels
? ?),( ii
i
n yxnV ??
??
? ?? ?
?
?
?
?
??
n
ij
j
j
jij
i
ji
n dsr
n
V
)(
1
ln
22 ?
??
Boundary condition
0,??? nn VV
? ? 0c o sln
22
)(
1
??
?
?
? ?
?
?
? ? i
n
ij
j
j
jij
i
ji
Vdsr
n
?
?
??
Let
? ? j
j
ij
i
ji dsrnI ? ?
?? ln
,
0c o s
22
)(
1
,??? ?
?
?
? i
n
ij
j
ji
ji
VI ?
?
??
It is a linear algebraic equations with n unknowns
After the n unknowns been solved,we can
calculate the tangential velocity component on the
surface of the body
is VV ?s i n,?? ?
? ?? ?
?
?
?
?
?
?
?
n
j
j
jij
j
s dsr
ss
V
1
ln
2 ?
??
? ?? ?
?
??
?
?
????
n
j
j
jij
j
issi dsr
s
VVVV
1
,ln
2
s in
?
?
?
With Bernoulli’s equation
2
1 ??
?
?
??
?
?
??
?V
V
C ip