2.20 - Marine Hydrodynamics, Spring 2005
Lecture 11
2.20 - Marine Hydrodynamics
Lecture 11
3.11 - Method of Images
• Potential for single source: φ =
m 2
ln x + y 2
2π
m
m
m
• Potential for source near a wall: φ =
2π
m
m
2
2
ln x2 + (y − b) + ln x2 + (y + b)
y
dφ
=0
dy
b
x
b
m
Added source for
symmetry
Note: Be sure to verify that the boundary conditions are satisfied by symmetry or
by calculus for φ (y) = φ (−y).
1
Γ
• Vortex near a wall (ground effect): φ = Ux +
2π
−1 y − b
−1 y + b
) − tan (
)
tan (
x
x
Γ
b
U
y
x
b
-Γ
Added vortex for
symmetry
Verify that
dφ
= 0 on the wall y = 0.
dy
• Circle of radius a near a wall: φ ∼
= Ux 1 +
a2
a2
+
x2 + (y − b)2 x2 + (y + b)2
y
U
b
y
x
b
∂φ
This solution satisfies the boundary condition on the wall ( ∂n
= 0), and the degree it
satisfies the boundary condition of no flux through the circle boundary increases as
the ratio b/a >> 1, i.e., the velocity due to the image dipole small on the real circle
for b >> a. For a 2D dipole, φ ∼ d1 , ∇φ ∼ d12 .
2
• More than one wall:
Example 1:
b
b
U
U
b'
b'
b'
Example 3:
Example 2:
b
b
b
b
-Γ
Γ
b'
b'
b'
b'
b'
b'
b'
b'
Γ
b
b
-Γ
b
b
3.12 Forces on a body undergoing steady translation
“D’Alembert’s paradox”
3.12.1 Fixed bodies & translating bodies - Galilean transformation.
y
o
y’
o’
x
z’
z
Fixed in space
Fixed in translating body
x = x` + Ut
3
x’
U
Reference system O’: v , φ , p
Reference system O: v , φ, p
U
X
O
S
O’
∇2 φ = 0
v · n̂ =
X’
S
U
∇2 φ = 0
∂φ
∂n
· n̂ = (U, 0, 0) · (nx , ny , nz )
=U
= U nx on Body
v · n̂ =
∂φ
∂n
=0
φ → 0 as |x| → ∞
φ → −U x as |x | → ∞
v → (−U, 0, 0) as |x | → ∞
v → 0 as |x| → ∞
Galilean transform:
v(x, y, z, t) = v (x = x − U t, y, z, t) + (U, 0, 0)
φ(x, y, z, t) = φ (x = x − U t, y, z, t) + U x ⇒
−U x + φ(x = x + U t, y, z, t) = φ (x , y, z, t)
Pressure (no gravity)
p∞ = − 12 ρv 2 + Co = Co = − 12 ρv 2 + Co = Co − 12 ρU 2
∴ Co = Co − 12 ρU 2
In O: unsteady flow
In O’: steady flow
∂φ 1
− 2 ρ v 2 +Co = Co
ps = −ρ
∂t
− 12 ρ v 2 +Co
ps = −ρ ∂φ
∂t
U2
∂φ
∂t
∂
∂x ∂
=(
+
) (φ + U x ) = −U 2
∂x
∂t
∂t
0
∴ ps = ρU
2
−U
− 12 ρU 2
0
0
+ Co = 12 ρU 2 + Co
ps − p∞ = 12 ρU 2 stagnation pressure
ps − p∞ = 12 ρU 2 stagnation pressure
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3.12.2 Forces
B
n̂
Total fluid force for ideal flow (i.e., no shear stresses): F = pˆ
ndS
B
For potential flow, substitute for p from Bernoulli:
⎛
⎞
⎜ ∂φ 1
⎟
⎟
⎜
ˆ
F = −ρ ⎜
+ |∇φ|2 + gy +c(t)⎟ ndS
2
⎝∂t ⎠
B
hydrodynamic
force
hydrostatic
force
For the hydrostatic case v ≡ φ ≡ 0 :
F s = (−ρgyn̂) dS
=
↑
Gauss outward
theorem normal
B
∇ (−ρgy) dυ = ρg∀ĵ
(−)
↑
Archimedes
principle
υB
We evaluate only the hydrodynamic force:
∂φ 1
2
+ |∇φ| n̂dS
F d = −ρ
∂t
2
For steady motion
∂φ
∂t
B
≡0 :
1
Fd = − ρ
2
B
5
v 2 ndS
ˆ
where ∀ =
dυ
υB
3.12.3 Example Hydrodynamic force on 2D cylinder in a steady uniform stream.
n̂
S
U
x
a
B
F d =
−ρ B
2
Fx =
F ·
î =
=
ρa
2
2π
0
|∇φ|2 nd
ˆ =
−ρa
2
2π
0
2π −ρ 2
n̂adθ
|∇φ|
r=a
2
0
dθ |∇φ|
2r=a n̂
·
î
− cos θ
2
|∇φ|
r=a
cos θdθ
Velocity potential for flow past a 2D cylinder:
a2
φ = U r cos θ 1 + 2
r
Velocity vector on the 2D cylinder surface:
∂φ
∂r
∇φ|r=a = ( vr |r=a , vθ |r=a ) =
0
r=a
Square of the velocity vector on the 2D cylinder surface:
|∇φ|2 r=a = 4U 2 sin2 θ
6
1 ∂φ
,
r ∂θ
r=a
−2U sin θ
Finally, the hydrodynamic force on the 2D cylinder is given by
ρa
Fx =
2
2π
0
1
dθ 4U 2 sin2 θ cos θ = ρU 2 (2a) 2
2 ps −p∞
diameter
or
projection
2π
2
dθ sin
θ cos
θ = 0
0
even
≡0
odd
π 3π
w.r.t ,
2 2
Therefore, Fx = 0 ⇒ no horizontal force ( symmetry fore-aft of the streamlines). Similarly,
1
Fy = ρU 2 (2a)2
2
2π
dθ sin2 θ sin θ = 0
0
In fact, in general we find that F ≡ 0, on any 2D or 3D body.
D’Alembert’s “paradox”:
No hydrodynamic force∗ acts on a body moving with steady translational (no circulation)
velocity in an infinite, inviscid, irrotational fluid.
∗
The moment as measured in a local frame is not necessarily zero.
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3.13 Lift due to Circulation
3.13.1 Example Hydrodynamic force on a vortex in a uniform stream.
φ = Ux +
Γ
Γ
θ = U r cos θ +
θ
2π
2π
Γ
U
Consider a control surface in the form of a circle of radius r centered at the point vortex.
Then according to Newton’s law:
d steady flow
LCV −→
dt
N ET
= 0 ⇔ F ≡ −FV = FCS + M
ΣF =
N ET
(FV + FCS ) + M
Where,
F
= Hydrodynamic force exerted on the vortex from the fluid.
FV = −F = Hydrodynamic force exerted on the fluid in the control volume from the vortex.
FCS
= Surface force (i.e., pressure) on the fluid control surface.
N ET
M
= Net linear momentum flux in the control volume through the control surface.
d L
dt CV
= Rate of change of the total linear momentum in the control volume.
y
θ
Fy
Γ
U
Fx
Control
volume
IN
The hydrodynamic force on the vortex is F = FCS + M
8
x
N ET .
a. Net linear momentum flux in the control volume through the control surfaces, M
Recall that the control surface has the form of a circle of radius r centered at the point
vortex.
a.1 The velocity components on the control surface are
u = U−
v =
Γ
sin θ
2πr
Γ
cos θ
2πr
The radial velocity on the control surface is therefore, given by
ur = U
∂x
= U cos θ = V · n̂
∂r
vθ =
Γ
2πr
U
θ
a.2 The net horizontal and vertical momentum fluxes through the control surface are
given by
2π
2π
Γ
sin θ U cos θ = 0
(MN ET )x = − ρ dθruvr = − ρ dθr U −
2πr
0
0
2π
2π
Γ
cos θ U cos θ
(M N ET )y = − ρ dθrvvr = − ρ dθr
2πr
0
=−
ρU Γ
2π
2π
0
cos2 θdθ = −
0
9
ρU Γ
2
b. Pressure force on the control surface, FCS .
b.1 From Bernoulli, the pressure on the control surface is
1
p = − ρ |v |2 + C
2
b.2 The velocity |v |2 on the control surface is given by
2 2
Γ
Γ
2
2
2
|v | =u + v = U −
sin θ +
cos θ
2πr
2πr
2
Γ
Γ
2
=U − U sin θ +
πr
2πr
b.3 Integrate the pressure along the control surface to obtain FCS
(FCS )x =
(FCS )y =
2π
dθrp(− cos θ) = 0
0
2π
0
ρ
dθrp(− sin θ) = − 2
ΓU
2π
− πr (−r)
0
dθ sin2 θ = − 12 ρU Γ
π
c. Finally, the force on the vortex F is given by
Fx = (FCS )x + (Mx )IN = 0
Fy = (FCS )y + (My )IN = −ρU Γ
i.e., the fluid exerts a downward force F = −ρU Γ on the vortex.
Kutta-Joukowski Law
2D : F = −ρU Γ
× Γ
3D : F = ρU
Generalized Kutta-Joukowski Law:
×
F = ρU
n
Γi
i=1
.
where F is the total force on a system of n vortices in a free stream with speed U
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