Chapter 6
Basic flows
6.1
Uniform Flow at An Angle α
Given velocity field is:
~ = (V∞ cos α, V∞ sin α)
V
Check if conservation of mass is satisfied first to test if it is a physically possible flow?
?
~ =0
∇·V
∂u ∂v ?
+
=0
∂x ∂y
~ =0
Since u and v are both constants, ∇ · V
Therefore ψ exists. From conservation of mass,
∂ψ
∂ψ
and v = −
∂y
∂x
∂ψ
u = V∞ cos α =
∂y
ψ = V∞ cos α y + f (x)
∂ψ
= −v = 0 + f 0 (x)
∂x
f 0 (x) = −V∞ sin α or f (x) = −V∞ sin α x + g(y)
u=
ψ = −V∞ sin α x + V∞ cos α y
ψ = const. = −V∞ sin α x + V∞ cos α y
ψ
= − sin α x + cos α y
V∞
ψ
= − tan α x + y
V∞ cos α
Equation of streamlines:
y = tanα x +
ψ
V∞ cos α
Check if the given flow is a potential flow?
?
~ =
∇×V
0
∂u ?
∂v
=0
−
∂x ∂y
44
CHAPTER 6. BASIC FLOWS
6.1. UNIFORM FLOW AT AN ANGLE α
~ =0
Since V∞ and α are constant throughtout the flow, ∇ × V
~
Therefore φ exists and V = ∇φ.
∂φ
∂φ
V~ = ∇φ =
ı̂ +
̂
∂x
∂y
∂φ
= V∞ cos α = u
∂x
φ = V∞ cos α x + f (y)
∂φ
= 0 + f 0 (y) = v
∂y
f 0 (y) = V∞ sin α or f (y) = V∞ sin α y + f (x)
φ = V∞ cos α x + V∞ sin α y (uniform flow at an angle α)
φ = const. = V∞ cos α x + V∞ sin α y
φ
x
=
+y
V∞ sin α
tan α
Equation of Equipotential lines:
y=−
φ
1
x+
tan α
V∞ sin α
φ constant lines are orthogonal to ψ constant lines.
6.1.1
Γ: Contour Integral over a Close CurveC
Y
3
V
8
h
es
et
4
2
s
8
V
8
V
1
X
I
~ · d~l
Γ=− V
 2

Z
Z3
Z4
Z1
~ · d~l + V
~ · d~l + V
~ · d~l
= −  V~ · d~l + V
1
2
3
4

 2
Z1
Z4
Z3
Z
= −  (V∞ ês ) · (ds ês ) + (V∞ ês ) · (dh êt ) + (V∞ ês ) · (−ds ês ) + (V∞ ês ) · (−dh êt )
1
3
2
= − [(V∞ s) + 0 + (−V∞ s) + 0] ≡ 0
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4
CHAPTER 6. BASIC FLOWS
6.2
6.2. 2-D SOURCE (LINE SOURCE)
2-D Source (Line Source)
Definition: A source is a point from which fluid issues along radial lines. Streamlines are straight lines
emanating from a central point. Velocity varies inversely with distance from the origin.
From the definition of the source the velocity vector can be written as:
~ = vr êr
V
where vr ∝ 1r or vr = rc , and vθ = 0 where C is a constant.
Check if the assumed flow is physically possible.
c
V~ = êr + 0êθ
r
?
~ =
∇·V
0
~ = 1 ∂(rvr ) + ∂(vθ ) = 1 ∂(c) + ∂(0) ≡ 0
∇·V
r
∂r
∂θ
r ∂r
∂θ
Flow is physically possible and ψ exists.
6.2.1
Evaluation of c
From mass conservation for a steady flow we know
Z
dṁ = 0
From continuity the mass of fluid per unit time crossing any circle centered at the source is a constant and
equal to the mass of fluid issuing per unit time from the source. Consider a cylinder centered on the source.
There is mass flowing out only from the sides of the cylinder.
~ r = hθ hz dθ dzêr = r dθ dzêr
dA
ṁ =
ZL Z2π
0
~ · dA
~=
ρV
ZL Z2π
ρ(Vr êr ) · (r dθ dz)êr
0
0
0
It is a 2-D flow and hence the integral can be reduced to:
ṁ = L
Z2π
ρVr r dθ
0
Vr is not a function of θ. Vr is only a function of r.
Z2π c
ṁ = L ρ
r dθ = ρ L c 2π
r
0
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CHAPTER 6. BASIC FLOWS
6.2. 2-D SOURCE (LINE SOURCE)
Y
X
L
Z
Cylinder centered at a source
Volume flow per second is:
ṁ
= c 2π L
ρ
Define K as the source strength. It is physically the rate of volume flow from the source per unit depth into
the page (2-D).
K
K = 2πc or c =
2π
then the velocity becomes:
K
vr =
2πr
~ = 0 is satisfied, the flow is physically possible and from the definition of ψ in polar coordinates,
Since ∇ · V
ψ can be found.
∂ψ
∂ψ
vr =
and vθ = −
r∂θ
∂r
K
∂ψ
=
r∂θ
2πr
∂ψ
K
=
∂θ
2π
K
ψ=
θ + f (r)
2π
∂ψ
= −vθ = 0 + f 0 (r) = 0
∂r
f (r) = const.
K
ψ=
θ
2π
Since the source strength, K is a constant, ψ constant lines are radial lines.
ψ=
K
θ = const.
2π
?
~ =
Is ∇ × V
0.
~ =
∇×V
1
r
êr
∂
∂r
vr
rêθ
∂
∂θ
rvθ
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êz
∂
∂z
vz
=
47
1 ∂(rvθ ) ∂(vr )
êz ≡ 0
−
r
∂r
∂θ
CHAPTER 6. BASIC FLOWS
6.3. COMBINATION OF POTENTIAL FLOWS
therefore φ exists.
K
1 ∂φ
∂φ
=
and vθ =
=0
∂r
2πr
r ∂θ
K
φ=
ln r + g(θ)
2π
∂φ
= 0 + g 0 (θ) = 0
∂θ
g(θ) = const.
K
... φ =
ln r
2π
vr =
6.3
Combination of Potential Flows
Uniform flow and source/sink satisfy Laplace equation and therefore superposition is possible.
6.3.1
Combination of a Uniform Flow to The Right (α = 0) and A Source at
The Origin
Y
Y
+
Quantity
~
V
φ
ψ
=
X
Uniform flow
V∞ ı̂
V∞ x
V∞ y
Source/Sink
K
± 2πr
êr
K
± 2πr
ln r
K
θ
± 2πr
X
Combination
K
V∞ ı̂ ± 2πr
êr
K
V∞ x ± 2πr
ln r
K
θ
V∞ y ± 2πr
Stagnation Point:
~ ≡0
At the stagnation point, V
K
=0
2πr
vθ = −V∞ sin θ = 0
vr = V∞ cos θ +
Solve for θ and r at the stagnation point to get (rstag , θstag ). Proceed to find ψstag to get the shape of the
body. From vθ = 0:
sin θ = 0
θ = 0 or ± π
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CHAPTER 6. BASIC FLOWS
6.3. COMBINATION OF POTENTIAL FLOWS
Case 1: θs = 0. Sove for rs from vr =0
vr = V∞ cos θ +
K
=0
2πr
if θ = θs = 0
cos θs = 1
K
=0
vr = V ∞ +
2πrs
K
or rs = −
2πV∞
Impossible solution as rs < 0; (... K and V∞ are positive)
Case 2: θs = ±π.
vr = −V∞ +
rs =
K
=0
2πrs
K
2πV∞
θs = +π for the upper half of the body
θs = −π for the lower half of the body
Coordinates of the stagnation point:
(rs , θs ) =
K
, ±π
2πV∞
Body Shape (Stagnation Streamline):
A general expression for the streamfunction for the combined flow is:
ψ = V∞ r sin θ +
K
θ
2π
Find ψs (= the body shape) by substituting (rs , θs ).
ψs = V∞ r sin(±π) +
K
K
(±π) = ± = const.
2π
2
In Cartesian coordinate, the general expression for the body streamfunction becomes:
y
K
K
±
= V∞ y +
tan−1
2
2π
x y K
K
= ± − V∞ y
tan−1
2π
x
2
y
2πV∞ y
−1
= ±π −
tan
x
K
2πV∞ y
y
= tan ±π −
x
K
y
x=
tan ±π − 2πVK∞ y
To find maximum y value, consider ψ = K/2 (upper half of the body).
y
K
K
= V∞ y +
tan−1
2
2π
x
K
ymax = y@x=∞ =
2V∞
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CHAPTER 6. BASIC FLOWS
6.3. COMBINATION OF POTENTIAL FLOWS
Y
P(x,y)
θ1
6.3.2
θ2
θ
X
(b,0)
(-b,0)
Combined Flow of a Source at (−b, 0) and a Sink at (b, 0)
K
K
θ1 −
θ2
2π
2π
where θ1 and θ2 are measured from the center of the source and sink respectively.
y
y
, θ2 = tan−1
θ1 = tan−1
x+b
x−b
y
y
θ2 − θ1 = tan−1
− tan−1
x−b
x+b
2by
θ2 − θ1 = tan−1
x2 + y 2 − b 2
ψ=
K
(θ1 − θ2 )
2π 2by
−1
θ1 − θ2 = − tan
x2 + y 2 − b 2
K
2by
−1
ψsource+sink = −
tan
2π
x2 + y 2 − b 2
2by
2πψ
−1
= − tan
K
x2 + y 2 − b 2
2by
2πψ
tan
=− 2
K
x + y 2 − b2
2πψ
2
2
x + y + 2by cot
= b2
K
2
2πψ
2πψ
= b2 1 + cot2
(x − 0)2 + y + b cot
K
K
2
2πψ
2πψ
(x − 0)2 + y + b cot
= b2 csc2
K
K
Equation of a circle with center at 0, ±b cot 2πψ
and radius of b csc 2πψ
. When y = 0, x = ±b. All
K
K
streamlines go through ±b.
ψ = ψ 1 + ψ2 =
6.3.3
Uniform Flow to The Right+Source (−b, 0)+ Sink (b, 0) (Rankine oval)
• Source of strength K placed at (−b, 0)
• Sink of strength K placed at (b, 0)
• Uniform flow to the right (α = 0)
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CHAPTER 6. BASIC FLOWS
6.3. COMBINATION OF POTENTIAL FLOWS
ψ = V∞ r sin θ +
K
K
θ1 −
θ2
2π
2π
Problem:
Analyze Rankine oval.
6.3.4
2-D Doublet
Definition: A doublet is obtained when a source and sink of equal strength approach each other so that the
product of their strength and the distance apart remains a constant.
B = K(2b) = constant
K
2by
−1
ψsource+sink = −
tan
2π
x2 + y 2 − b 2
2by
−1
x2 +y 2 −b2
2bK tan
=−
4π
b


2by
−1
tan
x2 +y 2 −b2
µ

lim ψsource+sink = ψdoublet = −
lim 
2bK→µ
b→0
4π
b
Using L’Hospital’s rule:
ψdoublet

2by
d
db x2 +y2 −b2



“
”2
(x2 +y 2 −b2 )(2y)−(2by)(−2b)
2by


µ
µ
µ
2y
(x2 +y 2 −b2 )2
 1+ x2 +y2 −b2 


=−
lim 
lim 
=−
=−
2
db

4π b→0 
4π b→0
4π x2 + y 2
2by
db
1 + x2 +y
2 −b2
ψdoublet = −
µ sin θ
2π r
Streamlines
µy
=0
2πψ
2 2
µ
µ
(x − 0)2 + y +
=
4πψ
4πψ
x2 + y 2 +
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CHAPTER 6. BASIC FLOWS
6.3. COMBINATION OF POTENTIAL FLOWS
µ µ
from the x-axis with a radius of 4πψ
Streamlines are circles centerd on the y-axis a distance − 4πψ
. All
circles pass through the origin.
Problem:
Show that φ =
6.3.5
µ cos θ
2π r
for a 2-D doublet.
Uniform Flow to The Right + A 2-D Doublet
Quantity
~
V
φ
ψ
Vr =
Uniform flow
V∞ ı̂
V∞ x
V∞ y
2-D doublet
Combination
µ cos θ
2π r
µ sin θ
− 2π
r
µ cos θ
V∞ x + 2π
r
µ sin θ
V∞ y − 2π r
1 ∂ψ
µ cos θ
1
V∞ r cos θ −
=
r ∂θ
r
2π r


"
2 #

1
R
µ


= V∞ cos θ 1 −
 = V∞ cos θ 1 −
2

2πV r 
r
| {z∞}
1/R2
Where R2 =
µ
2πV∞ .
"
2 #
h
µ i
R
∂ψ
= − V∞ +
sin θ = −V∞ sin θ 1 +
Vθ = −
∂r
2πr2
r
~ = 0)
Stagnation Points (V
Set Vθ = 0.
"
0 = −V∞ sin θ 1 +
R
r
2 #
sin θ = 0 or θs = (0 or π)
Now set Vr = 0.
"
0 = V∞ sin θ 1 −
R
r
2 #
For θ = 0 or π, cos θ 6= 0
"
2 #
µ
... 1 − R
≡ 0 or r2 = R2 =
r
2πV∞
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CHAPTER 6. BASIC FLOWS
6.4. 2-D VORTEX FLOW (POTENTIAL VORTEX)
The stagnation points are located at
(rs , θs ) ≡ (R, 0) and (R, π)
For Cylindrical System

R2


φ = V∞ r cos θ 1 + 2


r




2

R



ψ = V∞ r sin θ 1 − 2

r
(r ≥ R)

R2


Vr = V∞ cos θ 1 − 2


r




2
R 


Vθ = −V∞ sin θ 1 + 2 
r
µ
2πV∞
Substitute (rs , θs ) = (R, 0) or (R, π) in the expression for ψ.
Where R2 =
ψs = 0
at r = R (surface of the cylinder)
R2
Vr = V∞ cos θ 1 − 2 = 0 (no flow out of the cylinder)
r
Vθ = −2V∞ sin θ
Cp |r=R =
p − p∞
=1−
1
2 ρ ∞ V∞
V
V∞
2
=1−
Vr2 + Vθ2
=1−
V ∞2
Cp (2-D cylinder) = 1 − 4 sin2 θ
6.4
Vθ
V∞
2
= 1 − 4 sin2 θ
2-D Vortex Flow (Potential Vortex)
A 2-D point vortex is a mathematical concept that induces a velocity field given by
Vr = 0, Vθ =
const.
C
=
r
r
1. Check if the flow satisfies conservation of mass (Is it a physically possible flow?)
?
~ =0
∇·V
1 ∂(Vr r) ∂Vθ
~
∇·V =
= 0 → ψ exist.
+
r
∂r
∂θ
∂ψ
= 0 → ψ = g(r)
Vr =
r∂θ
∂ψ
C
Vθ = −
=
→ ψ = −C ln r + f (θ)
∂r
r
∂ψ
= f 0 (θ) = 0
∂θ
f (θ) = const.
... ψ = −C ln r + const.
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CHAPTER 6. BASIC FLOWS
6.4. 2-D VORTEX FLOW (POTENTIAL VORTEX)
When r → 0, Vθ = ∞ and ψ → ∞. To eliminate the infinite velocity it is arbitrary assumed that
ψ = 0 at r = R
... ψ = −C ln R + const. = 0
2. Check if the flow is irrotational
const. = C ln R
r
ψ = −C ln
for (r ≥ R)
R
?
~ =
∇×V
0
1
~ =
∇×V
h1 h2 h3
h1 ê1
∂
∂q1
h1 V1
h2 ê2
∂
∂q2
h2 V2
êr
h3 ê3 1
∂
= ∂
∂q3 r ∂r
Vr
h3 V3 rêθ
∂
∂θ
rVθ
Vorticity or (∇ × V~ ) in the r − θ plane
1 ∂rVθ
∂Vr
∂0
∂C
−
−
=
= 0 → φ exist.
r
∂r
∂θ
∂r
∂θ
Problem:
Show that φ = −Cθ.
Evaluate the Constant C
Evaluate the circulation Γ around the point vortex.
C
C1
C2
1. Around closed curve C1 that does not include the point vortex
I
ZZ
~=0
~ ) · dA
ΓC1 = − V~ · d~l =
(∇ × V
| {z }
C1
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54
0
êz ∂ ∂z Vz CHAPTER 6. BASIC FLOWS
6.4. 2-D VORTEX FLOW (POTENTIAL VORTEX)
2. Around C2 that includes the point vortex.


I
ΓC2 = −  (Vr êr + Vθ êθ ) · (dr êr + r dθêθ )
C2

= −

I
(Vr êr + Vθ êθ ) · (dr êr + r dθêθ ) + 
C

= −
= −
C2−C
(Vr êr + Vθ êθ ) · (dr êr + r dθêθ ) + 0
I
Vr dr +
C
ΓC2
I

I
C


I
C


Vθ r dθ = − 0 +
Z2π
Γ
= −2πC or −
=C
2π
0
C
r

(Vr êr + Vθ êθ ) · (dr êr + r dθêθ )

r dθ = −2πC
This implies that the circulation evaluated for a curve enclosing the 2-D vortex is a constant and not
equal to zero.
Γ
and ψ = −C ln Rr .
For a potential vortex, Vθ = − 2πr
Γ
r
... ψ =
ln
for r ≥ R
2π R
ψ = const , then ln Rr =
2πψ r
Γ , R
= e2πψ/Γ , r = Re2πψ/Γ
Streamlines are concentric circles with center at the 2D point vortex.
Vr =
∂φ
∂r
, Vθ =
1 ∂φ
r ∂θ
Γ
= − 2πr
Γ
Γ
θ + C or φ = − 2π
θ (straight lines form the origin).
φ = − 2π
A line vortex can be described as a string of rotating particles. A chain of fluid particles are spinning on
their common axis and carrying around with them a swirl of particles which flow around in circles.
A cross-section of such a string of particles and its associated flow shows a spinning point ’outside’ of which
is streamline flow in concentric circles.
Vortices are common in nature, the difference between a real vortex as opposed to a theoretical line vortex
is that the former has a core of fluid which is rotating as a ’solid’, although the associated ’swirl’ outside is
the same as the flow ’outside’ the point vortex.
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CHAPTER 6. BASIC FLOWS
6.4.1
6.4. 2-D VORTEX FLOW (POTENTIAL VORTEX)
Uniform Flow to The Right (α = 0) + A 2-D Doublet + A 2-D Point
Vortex
• As we all know, uniform flow to the right + 2-D Doublet = non-lifting over a cylinder
• Uniform flow to the right + 2-D Doublet + 2-D Point Vortex = Lifting flow over a cylinder
The parameters for lifting flow over a cylinder are as follow (spinning cylinder):
Quantity
ψ
φ
Vr
Vθ
Non-lifting flow over a cylinder
2
V∞ r sin θ(1 − Rr2 )
2
V∞ r cos θ(1 + Rr2 )
2
V∞ cos θ(1 − Rr2 )
2
−V∞ sin θ(1 + Rr2 )
Vortex of Strength Γ
Γ
r
2π ln R
Γ
− 2π
θ
0
Γ
− 2πr
Combination
2
Γ
r
V∞ r sin θ(1 − R
r 22 ) + 2π ln R
Γ
V∞ r cos θ(1 + Rr2 ) − 2π
θ
R2
V∞ cos θ(1 − r2 )
2
Γ
−V∞ sin θ(1 + Rr2 ) − 2πr
• Flow satisfies continuity at every point r ≥ R.
~ = 0.
... ∇ · V
• Flow satisfies irrotationality at every point r ≥ R.
... ∇ × V~ = 0.
Determine the stagnation points for the combined flow
~ = 0,Vr = 0 = Vθ . If we set Vr = 0, we get rs = R or θs = ± π ,
At the stagnation points, V
2
Case(1): r = Rs = R
Vθ = −V∞ sin θs (1 + 1) −
sinθs = −
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Γ
=0
2πR
Γ
≤0
4πRV∞
56
CHAPTER 6. BASIC FLOWS
6.4. 2-D VORTEX FLOW (POTENTIAL VORTEX)
Γ
Γ
Because Γ > 0, 4πRV∞ > 0, 4πV
> 0. When 4πV
< R, θs has one value in the third quadrant and one in
∞
∞
the fourth quadrant that will satisfy the above relation.
The coordinates of the stagnation point are:
ys = Rsinθs = −
When
Γ
4πV∞
p
xs = ± R2 − ys2 = ±
r
Γ
4πV∞
R2 − (
Γ 2
)
4πV∞
= R, there is only one solution. However, the method fails when
Γ=0
0<
Γ
4πV∞
<R
Γ
4πV∞
> R.
Γ
4πV∞
=R
Case(2): θ = ± π2
Case(2a): θ =
π
2,
r = rs
R2
π
Vr = V∞ cos( )(1 − 2 ) = 0
2
r
π
R2
Γ
Vθ = −V∞ sin( )(1 + 2 ) −
=0
2
r
2πr
Γ
rs + R 2 = 0
2πV∞
r
Γ 2
Γ
rs = −
± (
) − R2
4πV∞
4πV∞
rs 2 +
Γ
When 4πV
> R, rs results in negative number for all cases. Because both roots are negative, the solution
∞
is impossible.
Case(2b) : θ = − π2 , r = rs
π
R2
Vr = V∞ cos(− )(1 − 2 ) = 0
2
r
R2
Γ
π
=0
Vθ = −V∞ sin(− )(1 + 2 ) −
2
r
2πr
Γ
rs + R 2 = 0
2πV∞
r
Γ 2
Γ
rs =
± (
) − R2
4πV∞
4πV∞
rs 2 −
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57
CHAPTER 6. BASIC FLOWS
When
Γ
4πV∞
6.4. 2-D VORTEX FLOW (POTENTIAL VORTEX)
However, θs = − π2 and
Γ
4πV∞
Γ
4πV∞
q
Γ
)2 − R2 < R. So we can’t use this solution.
( 4πV
∞
q
Γ
Γ
> R is an acceptable solution when rs = 4πV
±
)2 − R 2 > R
( 4πV
∞
∞
> R, we get rs =
−
Force on a Cylinder with Circulation in a Uniform Steady Flow
Force on an elemental distance on the surface of the cylinder:
dF~ = −pb Rdθeˆr
dF~ = −pb Rdθ(cos θî + sin θĵ)
F~ =
Z2π
−pb Rdθ(cos θî + sin θĵ)
0
The drag per unit span is
D0 = F~ · ĵ =
Z2π
−pb cosθRdθ
0
The lift per unit span is
L = F~ · î =
0
Z2π
−pb sinθRdθ
0
2
As we know, in incompressible flow the total pressure po = p + ρV2 , which is a constant throughout the flow.
pb = p o −
ρ(Vr2 +Vθ2 )
2
. Besides, there is no flow normal to the surface, Vr = 0.
ρ
pb = po − Vθ2
2
ρ
Γ 2
pb = po − (−2V∞ sin θ −
)
2
2πR
pb = po − 2ρV∞ 2 (sin θ)2 − ρV∞ sinθ
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58
ρΓ2
Γ
− 2 2
Rπ 8π R
CHAPTER 6. BASIC FLOWS
6.4. 2-D VORTEX FLOW (POTENTIAL VORTEX)
... D0 = R
Z2π
−(po − 2ρV∞ 2 (sin θ)2 − ρV∞ sinθ
ρΓ2
Γ
− 2 2 )cosθdθ = 0
Rπ 8π R
0
which means that d’Alembert’s paradox still prevails.
0
L =R
Z2π
−(po − 2ρV∞ 2 (sin θ)2 − ρV∞ sinθ
Γ
ρΓ2
− 2 2 )sinθdθ = ρV∞ Γ
Rπ 8π R
0
which is the Kutta-Joukowski theorem.
In inviscid, incompressible flow, the resultant force per unit span acting on a 2-D body of any cross section
is equal to ρV∞ Γ and acts perpendicular to V∞ .
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