This document discusses potential flow theory and its applications. It defines irrotational flow, introduces the velocity potential and stream function, and describes several elementary plane flows including uniform flow, source/sink flow, vortex flow, and doublet flow. It also discusses how more complex flows can be modeled through superposition of these elementary flows, providing examples of combining sources, sinks, and uniform flow and combining doublets, vortices, and uniform flow.
2. Irrotational Flow
Fluid elements moving in the flow field do
not undergo any rotation
0,0 =×∇=ω V
0=
∂
∂
−
∂
υ∂
=
∂
∂
−
∂
∂
=
∂
υ∂
−
∂
∂
y
u
xx
w
z
u
zy
w
0
111
=
θ∂
∂
−
∂
∂
=
∂
∂
−
∂
∂
=
∂
∂
−
θ∂
∂ θθ rzrr V
rr
rV
rr
V
z
V
z
VV
r
Rectangular coordinate
Cylindrical coordinate
3. Bernoulli Equation
If the flow is irrotational,
( )VVkgp
ˆ1
∇⋅=−∇
ρ
− ( ) ( ) ( )VVVVVV
×∇×−⋅∇=∇⋅
2
1
= 0
( ) ( )2
2
1
2
1ˆ1
VVVkgp ∇=⋅∇=−∇
ρ
−
[ ] rd
⋅
dzkdyjdxird ˆˆˆ ++=
z
k
y
j
x
i
∂
∂
+
∂
∂
+
∂
∂
=∇ ˆˆˆ
( )2
2
1
Vdgdz
dp
=−
ρ
− 0
2
2
=
++
ρ
gz
Vp
d
constant
2
2
=++
ρ
gz
Vp valid between any two points
in an irrotational flow
0=×∇ V
4. Example: Flow field with tangential motion
- Forced Vortex
( )rfVVr == θand0
Forced Vortex (rigid body rotation)
( ) rrfV ω==θ
( )
θ∂
∂
−
∂
∂
=×∇=ω θ r
zz
V
rr
rV
r
V
11
2
1
2
1
( ) ω=ω=
∂
ω∂
= r
rr
r
r
2
2
11
2
1 2
a
b ra
rb
∫ ρω=∫
∂
∂
=− ar
br
ar
brba rdrdr
r
p
pp 2
( ) 0
2
22
2
>−
ρω
= ba rr
5. Example: Flow field with tangential motion
- Free Vortex
( )rfVVr == θand0
Free Vortex (irrotational vortex)
( )
r
r
r
C
rfV a ω
===θ
2
( )
θ∂
∂
−
∂
∂
=×∇=ω θ r
zz
V
rr
rV
r
V
11
2
1
2
1
0
1
2
1
=
∂
∂
=
r
C
r
a
b ra
rb
( )22
2
abba VVpp −
ρ
=−
−
ρ
=
22
2 11
2 ab rr
C
thenIf 2ω= arC ( ) 0
2
222
2
>−
ρω
=− babba rrrpp
6. Velocity Potential
If the flow is irrotational, a potential function, φ,
can be formulated to represent the velocity field.
From the vector identity, 0=φ∇×∇
The velocity field of an irrotational
flow can be defined by a potential
function so that
φ−∇≡V
z
w
yx
u
∂
φ∂
−=
∂
φ∂
−=υ
∂
φ∂
−=
z
V
r
V
r
V zr
∂
φ∂
−=
θ∂
φ∂
−=
∂
φ∂
−= θ
1
7. Stream Function and Velocity Potential
For a two-dimensional, incompressible, irrotational
flow, the velocity field can be expressed in terms of
both ψ and φ.
xy
u
∂
ψ∂
−=υ
∂
ψ∂
=
yx
u
∂
φ∂
−=υ
∂
φ∂
−=
According to the irrotationality condition,
0=
∂
∂
−
∂
υ∂
y
u
x
02
2
2
2
=
∂
ψ∂
+
∂
ψ∂
yx
According to the continuity equation,
0=
∂
υ∂
+
∂
∂
yx
u
02
2
2
2
=
∂
φ∂
+
∂
φ∂
yx
} Laplace’s
equation
Solution of Laplace’s equation represents a possible 2-
D, incompressible, irrotational flow field.
8. Slope of Velocity Potential Line
Along a line of constant φ, dφ=0 and
0=
∂
φ∂
+
∂
φ∂
=φ dy
y
dx
x
d
The slope of a line of constant φ is given by υ
−=
∂φ∂
∂φ∂
−=
φ
u
y
x
dx
dy
uuy
x
dx
dy υ
=
υ−
−=
∂ψ∂
∂ψ∂
−=
ψ
1−=
υ
×
υ
−=
×
ψφ u
u
dx
dy
dx
dy
The slope of a line of constant ψ is given by
Lines of constant ψ and constant φ are orthogonal.
9. Example: Determine the Velocity Potential
Line of a Flow
The flow streamline function is .2axy=ψ
∂
ψ∂
∂
∂
−
∂
ψ∂
−
∂
∂
=
∂
∂
−
∂
υ∂
=ω
yyxxy
u
x
z2 ( ) ( ) 022 =
∂
∂
−−
∂
∂
= ax
y
ay
x
ay
x
ax
y
u 22 −=
∂
ψ∂
−=υ=
∂
ψ∂
=
The flow is irrotational.
In term of velocity
potential, the velocity
components are
ay
y
ax
x
u 22 −=
∂
φ∂
−=υ=
∂
φ∂
−=
( ) ( ) 2
22
22 caxxgax
dx
dg
uxgayay
y
+−=⇒=−=⇒+=φ⇒−=
∂
φ∂
−=υ
( ) ( ) 1
22
22 cayyfay
dy
df
yfaxax
x
u +=⇒−=−=υ⇒+−=φ⇒=
∂
φ∂
−=
caxay +−=φ 22
10. Elementary Plane Flows
Uniform Flow
3c=ψ
2c=ψ
1c=ψ
0=ψ
1c−=ψ
2c−=ψ
3c−=ψ
=φ 3k 2k 1k 0 1k− 2k− 3k−
U
x
y
x
y
Uu =
0=υ
Uy=ψ
U
y
u =
∂
ψ∂
=
0=
∂
ψ∂
−=υ
x
Ux−=φ
U
x
u =
∂
φ∂
−=
0=
∂
φ∂
−=υ
y
Γ=0 around any closed curve
11. Inclined uniform flow
=φ 3k
2k
1k 0
1k−
2k− 3k−
x
yα= cosVu
α=υ sinV
( ) ( )xVyV α−α=ψ sincos
α=
∂
ψ∂
= cosV
y
u
α=
∂
ψ∂
−=υ sinV
x
( ) ( )xVyV α−α−=φ cossin
α=
∂
φ∂
−= cosV
x
u
α=
∂
φ∂
−=υ sinV
y
V
x
y
α
0
3c
2c
1c
1c−
2c−
3c−
=ψ
Γ=0 around any closed curve
16. Superposition of Elementary Plane Flows
∀ φ and ψ satisfy Laplace’s equation.
• Laplace’s equation is linear and homogeneous.
• Solutions of Laplace’s equation may be added
together (superposed) to develop more complex
flow patterns.
• If ψ1 and ψ2 satisfy Laplace’s equation, so does ψ3
= ψ1 + ψ2 .
• Any streamline contour can be imagined to
represent a solid surface (there is no flow across a
streamline).
17. Source + Uniform Flow
θ+θ
π
=+θ
π
=ψ+ψ=ψ sin
22
flowuniformsource Ur
q
Uy
q
θ−
π
−=−
π
−=φ+φ=φ cosln
2
ln
2
flowuniformsource Urr
q
Uxr
q
x
y
U
Stagnation point
Flow past a half body
Solid surface formed
by two streamlines
18. Source + Sink Flow
( )2121sinksource
222
θ−θ
π
=θ
π
−θ
π
=ψ+ψ=ψ
qqq
1
2
21sinksource ln
2
ln
2
ln
2 r
rq
r
q
r
q
π
=
π
+
π
−=φ+φ=φ
Source and sink with equal
strength, origin separated 2a
apart
19. Source + Sink+Uniform Flow
( ) θ+θ−θ
π
=+θ
π
−θ
π
=ψ+ψ=ψ sin
222
2121flowuniformsource Ur
q
Uy
qq
θ−
π
=−
π
+
π
−=φ+φ=φ cosln
2
ln
2
ln
2 1
2
21flowuniformsource Ur
r
rq
Uxr
q
r
q
x
y
U
Stagnation point
(Flow past a Rankine body)
Solid surface formed
by two streamlines
Stagnation point
20. Doublet+Uniform Flow
Λ
−θ=θ+
θΛ
−=+
θΛ
−=ψ+ψ=ψ 2flowuniformdoublet 1sinsin
sinsin
r
U
UrUr
r
Uy
r
Λ
+θ−=θ−
θΛ
−=−
θΛ
−=φ+φ=φ 2flowuniformdoublet 1coscos
coscos
r
U
UrUr
r
Ux
r
U
Stagnation point
Flow past a cylinder
Solid surface formed
by two streamlines
Stagnation point
a
U
a
Λ
=
24. Example: Flow past a cylinder
Λ
−θ=ψ+ψ=ψ 2flowuniformdoublet 1sin
r
U
Ur
Λ
+θ−=φ+φ=φ 2flowuniformdoublet 1cos
r
U
Ur
U
Stagnation point (a,π)
Stagnation point (a,0)
a U
a
Λ
=
−θ=
+θ
∂
∂
=
∂
φ∂
−= 2
2
2
2
1cos1cos
r
a
U
r
a
Ur
rr
Vr
+θ−=
+θ
θ∂
∂
−=
θ∂
φ∂
−=θ 2
2
2
2
1sin1cos
11
r
a
U
r
a
Ur
rr
V
25. Example: Flow past a cylinder-
pressure distribution on the surface
U
a
U
a
Λ
=
−θ= 2
2
1cos
r
a
UVr
+θ−=θ 2
2
1sin
r
a
UV
∞p
gz
V
pgz
U
p ++=++∞
22
22
] ] θ=+= =θ=
22222
sin4UVVV arrar
= 0
( ) ( ) ( )θ−ρ=θ−ρ=−ρ=− ∞
2222222
sin41
2
1
sin4
2
1
2
1
UUUVUpp
θ−=
ρ
− ∞ 2
2
sin41
2
1
U
pp
26. Pressure distribution on the surface
of the cylinder
2
2
1
U
pp
ρ
− ∞
θ
θ−=
ρ
− ∞ 2
2
sin41
2
1
U
pp
27. Pressure drag force acting on the cylinder
U
∞p
ApdFd
=
a
−
ρ
+= ∞ 2
22
1
2 U
VU
pp
] ( )∫
π
∞ θθ
θ−ρ+−=
2
0
22
pressure cossin41
2
1
dabUpF Drag
]
ππ
π
∞
θρ+
θρ−θ−=
2
0
32
2
0
22
0
sin
3
4
2
1
sin
2
1
sin abUabUabp
0=
θ
] ( ) ∫∫
π
θθ−=θ−=
2
0
pressure coscos dpabpdAF
ADrag
29. Example: Flow past a cylinder with circulation
r
K
r
a
Ur ln
2
1sin 2
2
flowuniformvortexdoublet
π
+
−θ=ψ+ψ+ψ=ψ
θ
π
+
+θ−=φ+φ+φ=φ
2
1cos 2
2
flowuniformvortexdoublet
K
r
a
Ur
a
U
a
Λ
= UaUK Λπ=π< 44
−θ−=
∂
φ∂
−= 2
2
1cos
r
a
U
r
Vr
r
K
r
a
U
r
V
π
−
+θ−=
θ∂
φ∂
−=θ
2
1sin
1
2
2
Stagnation points
]
a
K
UV ar
π
−θ−==θ
2
sin2
4
sinandat0 1-
π
=θ==θ
Ua
K
arV
30. Flow past a cylinder with circulation
- Surface pressure distribution
a
U
a
Λ
= UaUK Λπ=π< 44
]
a
K
UV ar
π
−θ−==θ
2
sin2
] ( )]
2
222
2
sin2
π
−θ−=+= =θ=
a
K
UVVV arrar
] 01cos 2
2
=
−θ−==
a
a
UV arr
−
ρ
+= ∞ 2
22
1
2 U
VU
pp
2
2
2
2
2
sin
2
sin4
π
+θ
π
+θ=
Ua
K
Ua
K
U
V
π
−θ
π
−θ−
ρ
+= ∞
2
2
2
2
sin
2
sin41
2 Ua
K
Ua
KU
pp
31. Pressure distribution on the surface
of the cylinder
2
2
1
U
pp
ρ
− ∞
θ
aUK π=
aUK π= 2
aUK π= 3
π
−θ
π
−θ−
ρ
+= ∞
2
2
2
2
sin
2
sin41
2 Ua
K
Ua
KU
pp
32. Pressure drag force acting on the cylinder
∞p
−
ρ
+= ∞ 2
22
1
2 U
VU
pp
] ∫
π
∞ θθ
π
−θ
π
−θ−ρ+−=
2
0
2
22
pressure cos
2
sin
2
sin41
2
1
dab
aU
K
aU
K
UpF Drag
ApdFd
=
θ
] ( ) ∫∫
π
θθ−=θ−=
2
0
pressure coscos dpabpdAF
ADrag
a
( )∫
π
∞ θθ
θ−ρ+−=
2
0
22
cossin41
2
1
dabUp ∫
π
θθ
π
+θ
π
ρ+
2
0
2
2
cos
2
sin
2
2
1
dab
Ua
K
Ua
K
U
0=
0sin
22
sin2
2
0
22
=
θ
π
+
θ
π
=
π
ab
Ua
K
Ua
Kab
33. Lift force acting on the cylinder
∞p
−
ρ
+= ∞ 2
22
1
2 U
VU
pp
] ∫
π
∞ θθ
π
−θ
π
−θ−ρ+−=
2
0
2
22
pressure sin
2
sin
2
sin41
2
1
dab
aU
K
aU
K
UpF Lift
ApdFd
=
θ
] ( ) ∫∫
π
θθ−=θ−=
2
0
pressure sinsin dpabpdAF
ALift
a
( )∫
π
∞ θθ
θ−ρ+−=
2
0
22
sinsin41
2
1
dabUp ∫
π
θθ
π
+θ
π
ρ+
2
0
2
2
sin
2
sin
2
2
1
dab
Ua
K
Ua
K
U
0=
ρ
=
π
π
ρ
=
θ
π
+
θ
−
θ
π
=
π
b
U
KU
b
U
KU
ab
Ua
K
Ua
Kab 2
22
2
2
cos
24
sin
2
2 22
2
0
22
34. Circulation around on the cylinder
∞p
( ) ∫∫∫
π
θθθ
π
θ
π
θθθ ⋅θ+⋅θ=θ⋅+=Γ
2
0
2
0
2
0
ˆˆˆˆˆˆˆ eerdVeerdVerdeVeV rrrrr
∫ ⋅≡Γ sdV
a
∫
π
θ
π
−θ−=
2
0 2
sin2 rd
Ur
K
U
0=
r
K
UV
π
−θ−=θ
2
sin2
−θ−= 2
2
1cos
r
a
UVr
] K
K
Ur −=
θ
π
−θ=
π
π
2
0
2
0
2
cos2
ρ=
U
K
U
b
FLift 2
2
1 2
Γρ−=ρ= UUK
