Lecture hydrodynamics

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

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

2

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

3

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.

4

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.

5

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)

6


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)


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.

7

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

8

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

9

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

10

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

11

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

12

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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