2.20 - Marine Hydrodynamics, Spring 2005
Lecture 10
2.20 - Marine Hydrodynamics
Lecture 10
3.7 Governing Equations and Boundary Conditions for P-Flow
3.7.1 Governing Equations for P-Flow
(a) Continuity �2 φ = 0
�
�
1
2
(b) Bernoulli for P-Flow (steady or unsteady) p = −ρ φt + |�φ| + gy + C(t)
2
3.7.2 Boundary Conditions for P-Flow
Types of Boundary Conditions:
(c) Kinematic Boundary Conditions - specify the flow velocity �v at boundaries.
∂φ
= Un
∂n
(d) Dynamic Boundary Conditions - specify force F� or pressure p at flow boundary.
�
�
1
2
p = −ρ φt + (�φ) + gy + C (t) (prescribed)
2
1
The boundary conditions in more detail:
• Kinematic Boundary Condition on an impermeable boundary (no flux condition)
�
�v ·n̂ = ����
U
·n̂ =
����
fluid velocity
�v =�φ
boundary velocity
Un
����
= Given
nornal boundary velocity
�φ · n̂ = Un ⇒
(n1
∂
∂
∂
+ n2
+ n3
)φ = Un ⇒
∂x1
∂x2
∂x3
∂φ
=
∂n
Un
v
U
v
n = (n1 , n 2 , n 3 )
v
(
)
• Dynamic Boundary Condition: In general, pressure is prescribed
�
�
1
2
p = −ρ φt + (�φ) + gy + C (t) = Given
2
2
3.7.3 Summary: Boundary Value Problem for P-Flow
The aforementioned governing equations with the boundary conditions formulate the
Boundary Value Problem (BVP) for P-Flow.
The general BVP for P-Flow is sketched in the following figure.
KBC : (Lecture 19)

Free surface DBC : − ρ (φ + 1 (∇φ ) 2 + gy ) + C (t ) = GIVEN
t
23

21
non − linear

∇ 2φ = 0
1
2
p = − ρ (φt + (∇φ ) + gy ) + C (t )
2
Solid boundary KBC :
∂φ
= U n = GIVEN
∂n
It must be pointed out that this BVP is satisfied instantaneously.
3
3.8 Linear Superposition for Potential Flow
In the absence of dynamic boundary conditions, the potential flow boundary value
problem is linear.
• Potential function φ.
∇ 2 φ = 0 in V
∂φ
= U n =f on B
∂n
• Stream function ψ.
∇2 ψ = 0 in V
ψ=g on B
Linear
Superposition: if φ1 , φ2 , . . . are harmonic functions, i.e., �2 φi = 0, then φ =
�
αi φi , where αi are constants, are also harmonic, and is the solution for the boundary
value problem provided the kinematic boundary conditions are satisfied, i.e.,
∂φ
∂
=
(α1 φ1 + α2 φ2 + . . .) = Un on B.
∂n
∂n
The key is to combine known solution of the Laplace equation in such a way as to satisfy
the kinematic boundary conditions (KBC).
The same is true for the stream function ψ. The K.B.C specify the value of ψ on the
boundaries.
4
3.8.1 Example
���
�
Let φi x denote a unit-source flow with source at xi , i.e.,
�� � �
���
�� � �
1
φi x ≡ φsource x, xi =
ln �x − xi �
(in 2D)
2π
� �� � ��−1
= − 4π �x − xi �
(in 3D),
then find mi such that
φ=
�
�
mi φi (x) satisfies KBC on B
i
Caution: φ must be regular for x ∈ V , so it is required that �x ∈
/ V.
v
•x 2
v
•x 1
∇ 2 φ = 0 in V
v
v
•x 3 •x 4
∂Φ
=f
∂n
Figure 1: Note: �xj , j = 1, . . . , 4 are not in the fluid domain V .
5
3.9 - Laplace equation in different coordinate systems (cf Hildebrand §6.18)
3.9.1 Cartesian (x,y,z)
�
�v =
ˆ
j
î
ˆ
k
�
�
∂φ ∂φ ∂φ
,
,
∂x ∂y ∂z
u, v, w
= �φ =
�2 φ =
∂ 2φ ∂ 2φ ∂ 2φ
+
+ 2
∂x2 ∂y 2
∂z
z
�
êz
P ( x, y , z )
y
O
x
êx
6
ê y
3.9.2 Cylindrical (r,θ,z)
r 2 = x2 + y 2 ,
θ = tan−1 (y/x)
� ê ê ê � � ∂φ 1 ∂φ ∂φ �
r
z
θ
v = vr , vθ , vz =
,
,
∂r r ∂θ ∂z
�
∂ 2 φ 1 ∂φ 1 ∂ 2 φ ∂ 2 φ
+
+ 2 2 + 2 ⇔
2
∂z
�∂r �� r ∂r� r ∂θ
1 ∂φ
r ∂φ
r ∂r ( ∂r )
�
�
1 ∂φ
∂φ
1 ∂2φ ∂ 2φ
2
�φ =
r
+ 2 2 + 2
r ∂r
∂r
r ∂θ
∂z
�2 φ =
z
êz
P (r , θ , z )
O
y
θ
r
x
êx
7
ê y
3.9.3 Spherical (r,θ,ϕ)
r 2 = x2 + y 2 + z 2 ,
θ = cos−1 (z/r) ⇔ z = r (cos θ)
ϕ = tan−1 (y/x)
�
�
v = �φ =
êr êθ êϕ
vr , vθ , vϕ
�
�
=
∂φ 1 ∂φ
1
∂φ
,
,
∂r r ∂θ r(sin θ) ∂ϕ
�
�
�
∂ 2φ
∂ 2 φ 2 ∂φ
1
∂
1
∂φ
�φ =
+
+
⇔
sin
θ
+
2
2
∂θ
r2 sin2 θ ∂ϕ2
�∂r �� r ∂r� r sin θ ∂θ
1 ∂
(r2 ∂φ
∂r )
r 2 ∂r
�
�
�
�
1 ∂
1
∂
∂φ
1
∂ 2φ
2
2 ∂φ
�φ = 2
r
+ 2
sin θ
+ 2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂ϕ2
2
z
êz
P (r , θ , φ )
θ
r
O
y
φ
x
êx
8
ê y
3.10 Simple Potential flows
1. Uniform Stream �2 (ax + by + cz + d) = 0
1D:
2D:
φ = U x + constant ψ = U y + constant;
�
φ = U x + V y + constant ψ = U y − V x + constant;
�
v = (U, 0, 0)
v = (U, V, 0)
�
v = (U, V, W )
3D: φ = U x + V y + W z + constant
2. Source (sink) flow
2D, Polar coordinates
1 ∂
� =
r ∂r
2
�
�
�
∂
1 ∂2
r
+ 2 2 , with r = x2 + y 2
∂r
r ∂θ
An axisymmetric
solution: φ = a ln r + b. Verify that it satisfies �2 φ = 0, except at
�
r = x2 + y 2 = 0. Therefor, r = 0 must be excluded from the flow.
Define 2D source of strength m at r= 0:
φ=
�φ =
m
ln r
2π
∂φ
m
m
êr =
êr ⇐⇒ vr =
, vθ = 0
∂r
2πr
2πr
y
source
(strength m)
x
9
Net outward volume flux is
�
�
v · n̂ds =
��
C
�
� · vds =
��
S
�
�
v · n̂ds =
Sε
�2π
vr rε dθ =
����
0
Cε
�
� · vds
m
2πrε
m
����
source
strength
y
n̂
C
Sε
x
ε
S
If m < 0 ⇒ sink. Source m at (x0 , y0 ):
�
m
φ =
ln (x − x0 )2 + (y − y0 )2
2π
m
m
φ =
ln r (Potential function) ←→ ψ =
θ (Stream function)
2π
2π
m
θ
2π
Ψ=
y
θ
Vr =
ψ=0
1
10
m
2π
x
3D: Spherical coordinates
1 ∂
� = 2
r ∂r
2
�
∂
r
∂r
2
�
�
+
�
�
∂ ∂
,
, · · · , where r = x2 + y 2 + z 2
∂θ ∂ϕ
A spherically symmetric solution: φ =
a
+ b. Verify �2 φ = 0 except at r = 0.
r
Define a 3D source of strength m at r = 0. Then
φ=−
m
∂φ
m
⇐⇒ vr =
=
, vθ = 0, vϕ = 0
4πr
∂r
4πr2
Net outward volume flux is
��
m
� vr dS = 4πrε2 ·
= m (m < 0 for a sink )
4πrε2
11
3. 2D point vortex
1 ∂
� =
r ∂r
2
�
�
∂
1 ∂2
r
+ 2 2
∂r
r ∂θ
Another particular solution: φ = aθ + b. Verify that �2 φ = 0 except at r = 0.
Define the potential for a point vortex of circulation Γ at r = 0. Then
Γ
Γ
∂φ
1 ∂φ
θ ⇐⇒ vr =
= 0, vθ =
=
and,
2π
∂r
r ∂θ
2πr
1 ∂
(rvθ ) = 0 except at r = 0
ωz =
r ∂r
φ=
Stream function:
ψ=−
Γ
ln r
2π
Circulation:
�
�
�
�
v · dx =
C1
�
�
�
v · dx +
C2
�2π
�
�
v · dx =
C1 −C2
�
� R R ��
ωz dS=0
S
12
0
Γ
rdθ =
2πr
Γ
����
vortex
strength
4. Dipole (doublet flow)
A dipole is a superposition of a sink and a source with the same strength.
2D dipole:
� �
�
�
m
2
2
2
2
φ=
ln (x − a) + y − ln (x + a) + y
2π
�
�
�
µ
∂
2
2
lim φ =
ln (x − ξ) + y ��
a→0
2π
∂ξ
ξ=0
����
µ = 2ma
constant
µ
x
µ x
=−
=
−
2π x2 + y 2
2π r2
2D dipole (doublet) of moment µ at the origin oriented in the +x direction.
∂
NOTE: dipole = µ ∂ξ
(unit source)
13
ξ
unit
source
α
x
φ=
−µ x cos α + y sin α
−µ cos θ cos α + sin θ sin α
=
2
2
2π
x +y
2π
r
3D dipole:
⎛
1
⎠ where µ = 2ma fixed
−�
a→0
2
2
2
2
2
2
(x − a) + y + z
(x + a) + y + z
�
�
�
µ ∂
1
µ
x
µ x
�
�
=−
=
−
=
−
�
4π ∂ξ
4π (x2 + y 2 + z 2 )3/2
4π r3
(x − ξ)2 + y 2 + z 2 �
φ = lim −
m⎝
�
4π
⎞
1
ξ=0
3D dipole (doublet) of moment µ at the origin oriented in the +x direction.
14
5. Stream and source: Rankine half-body
It is the superposition of a uniform stream of constant speed U and a source of
strength m.
U
m
2D: φ = U x +
m � 2
ln x + y 2
2π
U
m
U
x
D
v
stagnation point v = 0
Dividing
Streamline
∂φ
m
x
=U+
2
∂x
2π x + y 2
m
u|y=0 = U +
, v |y=0 = 0 ⇒
2πx
m
V� = (u, v) = 0 at x = xs = −
, y=0
2πU
u=
For large x, u → U , and U D = m by continuity ⇒ D =
15
m
.
U
3D: φ = U x −
4π
�
m
x2 + y 2 + z 2
stagnation point
div. streamlines
∂φ
m
x
=U+
2
2
∂x
4π (x + y + z 2 )3/2
m x
u|y=z=0 = U +
, v |y=z=0 = 0, w|y=z=0 = 0 ⇒
4π |x|3
�
m
�
V = (u, v, w) = 0 at x = xs = −
, y=z=0
4πU
u=
For large x, u → U and U A = m by continuity ⇒ A =
16
m
.
U
6. Stream + source/sink pair: Rankine closed bodies
y
S
U
+m
-m
S
x
a
dividing streamline
(see this with PFLOW)
To have a closed body, a necessary condition is to have
�
min body = 0
2D Rankine ovoid:
m
φ = U x+
2π
�
�
�
� �
�
m
(x + a)2 + y 2
2
2
2
2
ln (x + a) + y − ln (x − a) + y = U x+ ln
4π
(x − a)2 + y 2
3D Rankine ovoid:
⎡
φ = Ux −
m⎣
�
4π
⎤
1
2
−�
(x + a) +
y2
17
+
z2
1
2
(x − a) +
⎦
y2
+
z2
For Rankine Ovoid,
�
�
∂φ
m
x+a
x−a
u=
=U+
−
∂x
4π �(x + a)2 + y 2 + z 2 �3/2 �(x − a)2 + y 2 + z 2 �3/2
�
�
m
1
1
u|y=z=0 =U +
−
4π (x + a)2 (x − a)2
m (−4ax)
=U +
4π (x2 − a2 )2
�2 � m �
�
u|y=z=0 =0 at x2 − a2 =
4ax
4πU
At x = 0,
u=U+
m
2a
where R = y 2 + z 2
4π (a2 + R2 )3/2
Determine radius of body R0 :
�R0
2π
uRdR = m
0
18
7. Stream + Dipole: circles and spheres
r
U
µx
2D: φ = U x +
2πr2
µ
θ
�
=
↑
x=r cos θ
µ �
cos θ U r +
2πr
The radial velocity is then
ur =
�
∂φ
µ �
= cos θ U −
.
∂r
2πr2
Setting the radial velocity vr = 0 on r = a we obtain a =
for a stationary circle of radius a. Therefore, for
µ = 2πU a2
the potential
�
µ �
φ = cos θ U r +
2πr
is the solution to ideal flow past a circle of radius a.
• Flow past a circle (U, a).
19
�
µ
.
2πU
This is the K.B.C.
�
φ = U cos θ r +
Vθ =
a2
r
�
�
�
2
= −U sin θ 1 + ar2
�
= 0 at θ = 0, π
− stagnation points
= −2U sin θ
π 3π
= �2U at θ = 2 , 2 − maximum tangential velocity
1 ∂φ
r ∂θ
Vθ |r=a
2U
θ
2U
Illustration of the points where the flow reaches maximum speed around the circle.
3D: φ = U x +
�
µ cos θ
µ �
=
U
r
cos
θ
1
+
4π r2
4πr3
y
µ
r
θ
U
x
z
The radial velocity is then
vr =
�
∂φ
µ �
= cos θ U −
∂r
2πr3
20
Setting the radial velocity vr = 0 on r = a we obtain a =
for a stationary sphere of radius a. Therefore, choosing
�
3
µ
2πU
µ = 2πU a3
the potential
�
µ �
φ = cos θ U r +
2πr
is the solution to ideal flow past a sphere of radius a.
• Flow past a sphere (U, a).
�
�
a3
φ = Ur cos θ 1 + 3
2r
�
�
1 ∂φ
a3
vθ =
= −U sin θ 1 + 3
r ∂θ
2r
�
3U
= 0 at θ = 0, π
vθ |r=a = −
sin θ
at θ = π2
= − 3U
2
2
3/
2
U
θ
3/
2
U
21
x
. This is the K.B.C.
8. 2D corner flow Velocity potential φ = rα cos αθ; Stream function ψ = rα sin αθ
� 2
�
∂
1 ∂
1 ∂2
(a) �2 φ = ∂r
+
+
φ=0
2
r ∂r
r2 ∂θ2
(b)
∂φ
= αrα−1 cos αθ
∂r
1 ∂φ
uθ =
= −αrα−1 sin αθ
r ∂θ
∴ uθ = 0 { or ψ = 0} on αθ = nπ, n = 0, ±1, ±2, . . .
ur =
i.e., on θ = θ0 = 0, απ , 2π
, . . . (θ0 ≤ 2π)
α
i. Interior corner flow – stagnation point origin: α > 1. For example,
α = 1, θ0 = 0, π, 2π,
u = 1, v = 0
y
x
ψ=0
22
π
3π
α = 2, θ0 = 0, , π, ,2π u = 2x, v = −2 y
2
2
o
(90 corner)
ψ=0
ψ=0
θ=2π/3, ψ = 0
α = 3 2 , θ0 = 0,
(120o corner)
2π 4π
, ,2π
3 3
120o
120o
120
θ=4π/3, ψ = 0
23
o
θ=0, ψ = 0
θ=2π, ψ = 0
ii. Exterior corner flow, |v| → ∞ at origin:
α<1
π
θ0 = 0, only
α
Since we need θ0 ≤ 2π, we therefore require
π
α
≤ 2π, i.e., α ≥ 1/2 only.
1/2 ≤ α < 1
π
θ0 = 0,
α
For example,
α = 1/2, θ0 = 0, 2π (1/2 infinite plate, flow around a tip)
θ=0, ψ = 0
θ=2π, ψ = 0
α = 2/3, θ0 = 0, 3π
(90o exterior corner)
2
θ=0, ψ = 0
θ=3π/2, ψ = 0
24
Appendix A1: Summary of Simple Potential Flows
Cartesian Coordinate System
Flow
Streamlines
Uniform flow
m
2π
2D Source/Sink (m) at (xo , yo )
Stream function
ψ(x, y)
U∞ x + V∞ y + W∞ z
U∞ y − V∞ x
ln((x − xo )2 + (y − yo )2 )
1
m q
− 4π
3D Source/Sink (m) at (xo , yo , zo )
(x−xo )2 +(y−yo )2 +(z−zo )2
Γ
2π
Vortex (Γ) at (xo , yo )
α
µ
− 4π
25
y−yo
arctan( x−x
)
o
m
2π
y−yo
arctan( x−x
)
o
NA
Γ ln((x − x )2 + (y − y )2 )
− 2π
o
o
(x−xo ) cos α+(y−yo ) sin α
(x−xo )2 +(y−yo )2
µ (y−yo ) cos α+(x−xo ) sin α
2π
(x−xo )2 +(y−yo )2
(x−xo )
((x−xo )2 +(y−yo )2 +(z−zo )2 )3/2
NA
µ
− 2π
2D Dipole (µ) at (xo , yo ) at an angle α
3D Dipole (+x) (µ) at (xo , y0 , zo )
Potential
φ(x, y, z)
Appendix A2: Summary of Simple Potential Flows
Cylindrical Coordinate System
Flow
Streamlines
Uniform flow
Potential
φ(r, θ, z)
Stream function
ψ(r, θ)
U∞ r cos θ + V∞ r sin θ + W∞ z
U∞ r sin θ − V∞ r cos θ
2D Source/Sink (m) at (xo , yo )
m
2π
3D Source/Sink (m) at (xo , yo , zo )
m
− 4πr
NA
Γ θ
2π
Γ ln r
− 2π
µ cos θ cos α+sin θ sin α
− 2π
r
µ sin θ cos α+cos θ sin α
2π
r
µ cos θ
− 4π
2
NA
Vortex (Γ) at (xo , yo )
2D Dipole (µ) at (xo , yo ) at an angle α
α
3D Dipole (+x) (µ) at (xo , yo , zo )
ln r
r
26
m θ
2π
Appendix A3: Combination of Simple Potential Flows
(2D)
φ = U∞ x +
m
2π
ln r
Rankine Half Body
(3D)
φ = U∞ x −
m
4π
√
Stream + Source + Sink
(2D)
φ = U∞ x +
m
2π
Rankine Closed Body
(3D)
φ = U∞ x +
m √
1
4π ( (x+a)2 +y 2 +z 2
Stream + Dipole
(2D)
φ = U∞ x +
µx
2πr 2
if µ = 2πa2 U∞
φ = U∞ cos θ(r +
a2
r )
Circle (Sphere) R = a
(3D)
φ = U∞ x +
µ cos θ
4πr 2
if µ = 2πa3 U∞
φ = U∞ cos θ(r +
a3
2r 2 )
2D Corner Flow
(2D)
φ = Crα cos(αθ)
ψ = Crα sin(αθ)
θ0 = 0, nπ
α
Stream + Source
m
xs = − 2πU
∞
D=
m
U∞
A=
m
U∞
=
�
1
x2 +y 2 +z 2
xs = −
�
m
4πU∞
ln((x + a)2 + y 2 ) − ln((x − a)2 + y 2 )
�
=
−√
1
)
(x−a)2 +y 2 +z 2
=
27
Appendix B: Far Field Behavior of Simple Potential Flows
Far field behavior
r >> 1
φ
(2D)
∼ ln r
�v = �φ
∼
1
r
Source
(3D)
∼
1
r
∼
1
r2
(2D)
∼
1
r
∼
1
r2
(3D)
∼
1
r2
∼
1
r3
(2D)
∼1
∼
1
r
Dipole
Vortex
28