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2.20 Marine Hydrodynamics, Fall 2014
Lecture 13
Copyright c 2014 MIT - Department of Mechanical Engineering, All
rights reserved.
2.20 - Marine HydrodynamicsLecture 13
3.16 Some Properties of Added-Mass Coefficients
1. mij = [function of geometry only]
F, M = [linear function of mij] [function of instantaneousnot of
motion history
U, U ,]
2. Relationship to fluid momentum.
F(t)
where we define to denote the velocity potential that
corresponds to unit velocityU = 1. In this case the velocity
potential for an arbitrary velocity U is = U.
The linear momentum L in the fluid is given by
L =
V
vdV =
V
dV =
Greenstheorem
B
+
0 at
ndS
Lx (t = T ) =
B
UnxdS = U
B
nxdS
The force exerted on the fluid from the body is F (t) = (mAU) =
mAU .
1
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T0
dt [F (t)] =T
0
mAUdt = mA U ]T0
mAU
Newtons Law= Lx (t = T )Lx (t = 0) = U
B
nxdS
Therefore, mA = total fluid momentum for a body moving at U = 1
(regardless ofhow we get there from rest) = fluid momentum per unit
velocity of body.
K.B.C. n
= n = (U, 0, 0) n = Unx, n = Unx Un
= Unx
n= nx
mA =
B
nxdS
mA = B
ndS
For general 6 DOF:
mjijforce/moment
idirection of motion
=
B
ipotential due to body
moving with Ui=1
njdS =
B
ijn
dS = j fluid momentum due toi body motion
3. Symmetry of added mass matrix mij = mji.
mji =
B
i
(jn
)dS =
B
i (j n)dS =
DivergenceTheorem
V
(ij) dV
=
V
i j + i2j =0
dVTherefore,
mji =
V
i jdV = mij
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4. Relationship to the kinetic energy of the fluid. For a
general 6 DoF body motionUi = (U1, U2, . . . , U6),
= Uiinotation
; i = potential for Ui = 1
K.E. =1
2
V
dV = 12V
Uii UjjdV
=1
2UiUj
V
i jdV = 12mijUiUj
K.E. depends only on mij and instantaneous Ui.
5. Symmetry simplifies mij. From 36 symmetry
21 ?. Choose such coordinate systemthat some mij = 0 by
symmetry.
Example 1 Port-starboard symmetry.
mij =
m11 m12 0 0 0 m16
m22 0 0 0 m26m33 m34 m35 0
m44 m45 0m55 0
m66
FxFyFzMxMyMz
U1 U2 U3 1 2 3
12 independent coefficients
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Example 2 Rotational or axi-symmetry with respect to x1
axis.
mij =
m11 0 0 0 0 0
m22 0 0 0 m35m22 0 m35 0
0 0 0m55 0
m55
where m22 = m33,m55 = m66and m26 = m35, so 4 different
coefficients
Exercise How about 3 planes of symmetry (e.g. a cuboid); a cube;
a sphere?? Workout the details.
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3.17 Slender Body Approximation
Definitions
(a) Slender Body = a body whose characteristic length in the
longitudinal direction isconsiderably larger than the bodys
characteristic length in the other two directions.
(b) mij = the 3D added mass coefficient in the ith direction due
to a unit acceleration in
the jth direction. The subscripts i, j run from 1 to 6.
(c) Mkl = the 2D added mass coefficient in the kth direction due
to a unit acceleration in
the lth direction. The subscripts k, l take the values 2,3 and
4.
1x
2x
6x
5x
4x
3x
O
x
L
3x
2x
4x
Goal To estimate the added mass coefficients mij for a 3D
slender body.
Idea Estimate mij of a slender 3D body using the 2D sectional
added mass coefficients(strip-wise Mkl). In particular, for simple
shapes like long cylinders, we will use known 2Dcoefficients to
find unknown 3D coefficients.
mij3D
=
[Mkl(x)2D
contributions]
Discussion If the 1-axis is the longitudinal axis of the slender
body, then the 3D addedmass coefficients mij are calculated by
summing the added mass coefficients of all the thinslices which are
perpendicular to the 1-axis, Mkl. This means that forces in
1-directioncannot be obtained by slender body theory.
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Procedure In order to calculate the 3D added mass coefficients
mij we need to:
1. Determine the 2D acceleration of each crossection for a unit
acceleration in the ith
direction,
2. Multiply the 2D acceleration by the appropriate 2D added mass
coefficient to get theforce on that section in the jth direction,
and
3. Integrate these forces over the length of the body.
Examples
Sway force due to sway accelerationAssume a unit sway
acceleration u3 = 1 and all other uj, uj = 0, with j = 1, 2, 4, 5,
6.It then follows from the expressions for the generalized forces
and moments (Lecture12, JNN 4.13) that the sway force on the body
is given by
f3 = m33u3 = m33 m33 = f3 = L
F3(x)dx
A unit 3 acceleration in 3D results to a unit acceleration in
the 3 direction of each2D slice (U3 = u3 = 1). The hydrodynamic
force on each slice is then given by
F3(x) = M33(x)U3 = M33(x)
Putting everything together, we obtain
m33 = L
M33(x)dx =L
M33(x)dx
Sway force due to yaw accelerationAssume a unit yaw acceleration
u5 = 1 and all other uj, uj = 0, with j = 1, 2, 3, 4, 6.It then
follows from the expressions for the generalized forces and moments
that thesway force on the body is given by
f3 = m35u5 = m35 m35 = f3 = L
F3(x)dx
For each 2D slice, a distance x from the origin, a unit 5
acceleration in 3D, resultsto a unit acceleration in the -3
direction times the moment arm x (U3 = xu5 = x).The hydrodynamic
force on each slice is then given by
F3(x) = M33(x)U3 = xM33(x)
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Putting everything together, we obtain
m35 = L
xM33(x)dx
Yaw moment due to yaw accelerationAssume a unit yaw acceleration
u5 = 1 and all other uj, uj = 0, with j = 1, 2, 3, 4, 6.It then
follows from the expressions for the generalized forces and moments
that theyaw force on the body is given by
f5 = m55u5 = m55 m55 = f5 = L
F5(x)dx
For each 2D slice, a distance x from the origin, a unit 5
acceleration in 3D, resultsto a unit acceleration in the -3
direction times the moment arm x (U3 = xu5 = x).The hydrodynamic
force on each slice is then given by
F3(x) = M33(x)U3 = xM33(x)
However, each force F3(x) produces a negative moment at the
origin about the 5 axis
F5(x) = xF3(x)
Putting everything together, we obtain
m55 =
L
x2M33(x)dx
In the same manner we can estimate the remaining added mass
coefficients mij - notingthat added mass coefficients related to
the 1-axis cannot be obtained by slenderbody theory.
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In summary, the 3D added mass coefficients are shown in the
following table. The emptyboxes may be filled in by symmetry.
m22 =L
M22dx m23 =L
M23dx m24 =L
M24dx m25 =L
xM23dx m26 =L
xM22dx
m33 =L
M33dx m34 =L
M34dx m35 =L
xM33dx m36 =L
xM32dx
m44 =L
M44dx m45 =L
xM34dx m46 =L
xM24dx
m55 =L
x2M33dx m56 =L
x2M32dx
m66 =L
x2M22dx
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3.18 Buoyancy Effects Due to Accelerating Flow
Example Force on a stationary sphere in a fluid that is
accelerated against it.
(r, , t) = U (t)
r + a32r2dipole for sphere
cos
t
r=a
= U3a
2cos
|r=a =(0,3
2U sin , 0
)1
2||2
r=a
=9
8U2 sin2
Then,
Fx = ()(2r2
) 0
d sin ( cos )[U3a
2cos +
9
8U2 sin2
]
= U3a3
0
d sin cos2
=2/3
+U29
4a2
0
d cos sin3
=0
Fx = U (2a3
) 43a3
=
+ 23a3
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Part of Fx is due to the pressure gradient which must be present
to cause the fluid toaccelerate:
x-momentum, noting U = U (t) :U
t+ U
U
x0
+ vU
y0
+wU
z 0
= 1
p
x(ignore gravity)
dp
dx= U for uniform (1D) accelerated flow
Force on the body due to the pressure field
F =
B
pndS = VB
pdV ; Fx = VB
p
xdV = U
Buoyancy force due to pressure gradient = U
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Analogue: Buoyancy force due to hydrostatic pressure gradient.
Gravitational accelera-tion g U= fluid acceleration.
ps = gyps = gj
F s = gj Archimedes principleSummary: Total force on a fixed
sphere in an accelerated flow
Fx = U
(
Buoyancy
+ m(1)added mass
12
)= U 3
2 = 3Um(1)
In general, for any body in an accelerated flow:
Fx = Fbuoyany + Um(1),
where m(1) is the added mass in still water (from now on, m)
Fx = Um for body acceleration with U
Added mass coefficient
cm =m
in the presence of accelerated flow Cm = 1 + cm
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Appendix A: More examples on symmetry of added mass tensor
Symmetry with respect to Y (= X-Z plane symmetry) 12 non-zero,
independent coefficients
Symmetry with respect to X and Y (= Y-Z and X-Z plane symmetry)
7 non-zero, independent coefficients
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Axisymmetric with respect to X-axis 4 non-zero, independent
coefficients
Axisymmetric with respect to X axis and X (=Y-Z plane symmetry)
3 non-zero, independent coefficients
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