Solution of exercises in the book -Sealoads on ship and offshore structures- by O. Faltinsen

Solution of exercises in the book

'Sea1oads on ship and offshore structure&'

by

Prof. O.Faltinsen

This work has been carried out by Jan Kvãlsvold

Department of Marine Hydrodynamics Norwegian Institute of Technology

Solution exercise 2.1

a) Want to show that the following expression satisfies Laplace equation and a bottom boundary condition.

A cosh(k(zh)) cosky coswt

Laplace equation (2-dimensions)

z2

= - k2Acosh(k(z+h)) cosky coso.)t

y2

T

= k2Acosh(k(z+h)) cosky coso)t az.2

Hence;

aT

₌

ay2 az2

Bottom boundary condition

acp

az lz=h

-kAsinh(k(zh)) cosky cosot (7)

(8)

b) Boundary condition at the walls of the tank

Approximate formulation as h/b-0. Then the trigonometric

function can be

= - kAcosh(k(zh)) siy cosot

satisfy the condition for y = -b or y = b, the following must hold In order to sin(kb) = 0 (11)

;n=1,2,

(12) Hence n = 1,2 (13) b

c) Free. surface condition: Iz=O g---- = 0

(14.)

Inserting values for and we obtain

0 (15)

- o)2Acosh(k(zh)) cosky coscit + gkAsinh(k(z+h)) cosky cosO)t =

Assuming A, cosky as well as cosot to be non-zero, we get:

= kg tanhkh (16)

Use that T =

and insert values of k, we get: ci)

2it/

.

tanh( nh

_;n=1,2,

(17)

d) Fluid motion at the free surface as function of time. Free surface elevation

- _IZrO= Acoshkh cosky sinwt

gat

g

(20)

We can see that the free surface oscillates harmonically with an amplitudeç given as

= . Acoshkh cosky (21)

g

We observe that the amplitude varies in space, and is maximum at the endsof the tank. tanh fl7th = 2b (18) (19) b

Further;

= 2r/ b b b (gh)"2

Velocity potential:

( -t \112

= Ae .±. cos(wt-kr)

r

'1

Laplace equation is not satisfied everywhere. This is for r = 0 where the velocity potentiaI is not defined.

Propagation direction of waves. Have to investigate the argument of the

trigonometric function, in time and space. Assume

ci)t

kr = constant = 0

(2)

ot = kr

(3)

Hence; increased time corresponds to increased radius. Then the waves are

propagating outward to larger radius r.

Variation of wave amplitude in space. (z=0)

c

gt

1 a zO

Wave amplitude is given as:

=A('!

\J12

g

r1

Then, decays as

H

₁

Solution exercise 2.2

Ct) =

-.

_{AI_}

sin(ot-kr) g r1 (1) (4) (5)

0)2 _{= kg =>} Then C

=JI_=!=1.561mIs

(3) g 2 0) 2 2it Propagation time TA 100

_=64s

₍₄₎ 1.561

b) Transportation of a cork. Due to equation 2.21 in the text book, the Stoke's drift velocity is

0)=g

k 0)

Solution exercise 2.3

a) Assume infinite depth. A wave front propagates with the group velocity which is equal to the half of the phase velocity on deep waters. Then

10)

(1)

g

2k

Free surface condition:

(2)

Assume z0 = 0, then the drift time TB is given as

TB 100 100 = 506,2 s

ok

.Q).°)

g

c) Maximum fluid velocity in the tank. Fluid velocity amplitude as function of depth:

ae (7)

(10)

__ =

- _?:_ = 1.51 s

Vff 2 Vff

V= -im/s=Vff=2,123m/S

= 2,94 S

The phase shift is unchanged. This subject is called the frequency of encounter effect and will be investigated later in the book.

U It is left to the reader to obtain numerical answers, but note the followingfacts: - Energy as well as the wave front moves with the group velocity.

= 0.7854 rn/s (8)

The wave crests move to the wave front with the phase velocity. The time between two crests is then

7'. = - = = T = 2 s (9)

ci) Ct)

Phase of wave 1.5 m close to the wave

maker. Investigate the argument of

trigonometric function which is wt - kx. The k termcontrols the ph.ase variation in space.

2 (c 2

kx

= -.-

-x = ±.

-I

x = 1Y_ = 1.509[-]

g

gT)

g

This is approximately .IE. which corresponds to a phase shift of X14

Time between two passing wave crests. Have to consider the relative velocity between the wave crest and observer.

V=

1m/sVff=9i 1.Om/s=4.123m/s

(11)

- The wave crests moves with the phase velocity which is higher than the group

velocity. Hence; the wave crests propagate to the wave front were they break. The group velocity for deep water is half the phase velocity. For shallow waters the group and phase velocities are equal.

Solution exercise 2.4

a) Definition of T. Note that

m = ufS(odo (a-0.5)b

H3 T1

T1 T1

=27c_=21r

=T

m1 =2T1 1

a 0.5)

Hence; a + 0.5 = 2.0 a 1.5 Further;

H3

= 16m0 = 16 (a-0.5) b - - H,3 T1 = 32ir

- b H

(4)

--

(5) 32ir

b) Relationship between T2 and TT

3 ....(a2_0.52) 2ir 2 T = 0.961. - T1

b-H2

-T1

H3 -T1

(1) (2) (6) 4.' _{1.5 -0.5} T2 =

2t

\m2

_Ø53) 2ir T1 (a-0.5)b - .

H3 T1

= 2ir

a) Standard deviation of horizontal velocity as well as acceleration. =

Solution exercise 2.5

(3) K k1

s(,ek)

(5)

(\

--

S()dci. (1) I \2 = S(o)dw (2) J Note that A b) Assume -

as Noo ,

LW -40

For z = 0 the standard deviation is given as

= _{w4S(o) d} (4)

The behavior of the integrand near w 0 is -.1. Hence; the integral does not

exist for z = 0. Elsewhere the integral exists since the integrand behaves like

o2z

1

_{e1 which is finite near ci=0.}

(1)

In order to get the standard deviation of fluid acceleration to exist in the whole fluid domain, the wave spectrum has to be proportional to O where n

c) Consistent with equation 2.38 in the text book we assume a seastate build up by N frequency components and K wave directions. Then we have:

N, K, Aw-O,

Aeko

Then we obtain standard deviation of velocity as well as

crested sea given as:

I '2

=

s(,e)

d d9

S(co,8) dw d

Assume that wave spectru.m can be written as

S,e) = S() . fte

This means that the wave frequency and the directions are independent of each other.

acceleration in short

= =l S(o1,Bk)A w A ek

(6)

I

Increasing number of frequency components as vell as number of wave direction we have:

Assuming constant mass density. Incident wave is expressed : = çsin(cüt-kx).

Velocity potential for wave propagating in negative x-direction on deep water kz

- ...--e cos(cz)t--kx) 0)

a) Vertical excitation force on a small body compared to the wave length.

F3 = _ffpn3ds + A31a1 + A32a2 + A33a3 (2)

where

ap1

p=-p--T

a1= , a2= , a3=

xat

azt

Excitation force on a strip dx:

dF

= -pn3Bdx + A a3

Where a2 = 0. Further, A31 0 since the barge is symmetric about the yz- plane. Total excitation force is then expressed as

F

= -

esin(ot

kx)(-1)B dx x -

JA Ck

e Tsin(wt kx)dx x

= gpB e

-kD fsin(0)t _kx)dx - A

çgk e

fsiI(0)t + kx) dx x

Calculating the integral:

Solution exercise 3.1

(6) (1) (3) (4) (5)

Tsin(w t + kx) dx 1J2 Ltcos(o t + kx) = - COSIO) t + kL I - cos t cosk = - Sifl(i) t

sini.

k 2

= {pgç B e -kD -

gA' kca

4}

1in

SiflU)

(8) b) We have: 0.8 pBD B33 = 0 M= pLED C33 = pgLB

Equation of motibn (no damping):

(M + A33)i3 + C33i3 _{= 3es} SiflCi) t

- cos

2)

- sin t I _L oi

t

-2 si - cOSW t COS -L 2 (7)

The formula is valid for long waves compared to a characteristicbody dimension.

Itis also assumedthatk

whichisnotsatiSfiedWhen 0)

0.When

*0,

then a3 0, and the hydrodynamic pressure is approaimatedby:

(9) p = pçg sin(wt)

Then the total excitation force can be written as

Assume solution : = sin(cot) F:33 ₍₁₂₎ + A3.3) 4 C.33 = [1000 9.81 10 30 e15 - 9.81 0.8 1000 - 30 15 k 10 e75kJ ! sin(100 k) k k 27r 2it =

0.0209 (m')

X 300 Hence = 1.255 10 [N]

Heave motion in center of gravity is found as 1.255 .108 sino t _{= 4.9 sinu tErn]}

-3.32 iO + 5.89 - iO

Heave motion is in phase with the incident wave, and heave amplitude is

approximately half the wave amplitude. c) Pitch excitation moment:

F5 = zF1 - xF3 -xF3

L/2 (16)

F5

-

f x sin(ot

kx)

dx B e - gA

k6 e

kDI2}

-L.P2

Jxsino)t +kx)dx

= 5pg B (-x2 5) dx L/2

Bi5 [- x3]

=

--1.T Bpg5

-L/23

32)

/ 2 sin"kL 2 _I

!

..

COS..

{

cos cos cot

kD

0.8 k De

where damping is neglected. No coupling between heave and pitch since the barge is symmetric about both xz and yz- plane.

Assume : T1 = liSa cos(cot)

T15

_{_2(I}

cos(cot) (19)

+ A55) C55

Excitation moment:

= - 1.327 1010 Nm

Restoring moment of a strip dx: pg B dx (T3 - XT5) xdx

where the other terms are zero

cos cot (20) 2 _I (17) (18) = F cos(cot) pgç2B k k d) Equation of motion: (155 -- A55) 115 + _C551i5

dç

-

x=O dx = k cos(wt-kx)I0 = 10 155 2(L

32

cos(cot) = 0.21 cos(cot)

BD

(22)

Pitch motion is in face with the wave steepness amidship (pitch and wave

steepness are positive in opposite directions). Maximum pitch is approximately2/3 of maximum wave steepness.

e) Vertical acceleration in bow (x = L/2) is to be calculated. Heave motion i sincot - liSa cOscL)t

Acceleration 2

(i

SiflQ)t

_-

T5 coscüt) =

asin(c't?)

-A5fl55 L2

th

C- 5) dx -L12 =

-T

0.8 p

B D

(21)

32)

=A55={4J0.8

-BD

L5i

Constant mass density ofthe barge is assumed. Hence

C55 = 196,2 1O Nmlrad A55 = 240 iO kgm2

153 300 iO kgm2

()2 _{= 0.2054 s2}

Ti5 -0.15 cos (wt) [i-ad]

Wave steepness amidship (i.e x=0)

c = çsin(wt+kx)

22t

=;' acosy = U)2T(30

asiny =

= a

2J 2

= 0,34

f) Relative motion (between wave and barge) in bov:

fl3R = flsinO)t .: Tcoswt - TlaSlfl(0)t + ₍₂₃₎

9,86 sincot + 6,72 coso)t

where the integral

5o

Amplitude: = /9.862 + 6.722 11.93 m

One can see that a draft of D = 15 m do not cause the bow to exit water. I addition freeboard has to be 12 meter or more to avoid green water on deck.

g)

1. Contribution fromexcitation force

Il = fxsin(cot kx)dx pgçBe -

gA° k0

L/2

x sin(ot

kx) dx

= x coskx

sinot dx

L2

x six cost

dx

_sint7x

coskx dx - cosWt

x SiX

dx

L/2x -sinO)t +

= -

sincot t2k 2 -coscotI-__coski_

t2k

2 cos kxj £12 1

-coscot [ - .coskx 4- __sin)x]

Ok

k L12 =

_Jxsin(ct

kx) dx 0 1

cosk -

L

-

1 k2 2 k2

--

sin k2 2 and

Ar

= 0.8. pBD

Substitute (25) into (24) and we get the correct expression of I.

2. Contribution from restoring force 12 - xfl ) dx = BpgJ( - 5x2) dx L/2

= Bpg [x3 -

.±x3T5]

o2

3 =Bpg

(4J3

3. Contribution from added mass L/2 13 = 5 xA (113 - x115)dx =

A)[LZ113

- _X15J

o2

3

= 0.8 .p

B D

i13

(4J

151

4. Contribution from. mass inertia

14 = 5x pBL) ( - x15) dx 0

L./2 i

= -pBD ._x13

±x3115]

where

/

Po Pa

a

IP,

P,1_

Want to verify the expression of the continuity equation for the air in the air cushion. The mass in the air cushion is assumed constant.

= constant (6)

Solution, exercise 3.2 a) What to show that

p p[1--pJ [y (1+p/p0)]] (1)

For an adiabatic process the following must hold

p = a p( _{= p =} I

\1Jr

(2)

Taylor expansion of p about Po + _Pa: I [P - (p0 _Pafl

-'(I [i

--(p

_mr']

-

(p0

p)1''

a1 (3) LP = p- (p0 Use that p.-p(p) = p,2 = ;1 Then: 1 = p [1 - p1 [y (1 + PiP0 )]] (4) (5) a n .1-n p(p) = £0 £a

a

Hence

pa

= 0

dt dt

adt

Evaluate the different terms:

dc dt = Pa[1 + uJ [-y(l + pjp0)J]) = Pa -IL p.3(t)- xr(t) - ra(t)]dx 1 _d!1 + PcZP) dt (7) (8) b

dr(td

_p

Jb

dr1(t)d (9) dt aL dt dii.3(t) dV dt + dt

where fla(t) is the surface elevation inside the air cushion and the integral of the tei-rn proportional to 1](t) is equal to zero due symmetry (x=0 amidship). L is the length of the vesse]Jcushion. Hence

Want to find an expression of the heave natural period. Then two equations are set up, namely mass continuity in the air cushion as well as vertical equilibrium of the SES. Neglect the forcing (excitation force) since we want to solve the

homogenous equation system when predicting the eigenperiod. This means

d = 0 Ab dV Q dj..i

=0

(10) dt

1-p/p

dt Newtons 2nd law =

F

_{= =} (11) dt2 Hence

-p4j.i=O

(12)

dV _{= 0. Then:} dt

M

A5 P

= A5h5 (13)

g

Substitute into equations (10) and (12). Then:

A_+

A5h5 b dt y (1

PIP) dt

A5p,, d.,

- PP0Ab - 0

g

dt-Assume harmonic solutions

113 - 113A e

= e"'

Insert equation (16) into the equation (14)/(15) and require a non-trivial solution. This means that the coefficient determinant has to be zero. Hence;

h 1OTA J.1 - c JA 5 iou = 0 (17) g

y(l -p/p)

Finally:

C) T . = = 2rr '1. cy gy (1 pip,,)

b) Neglect buoyancy of the hulls. Then the weight of the vessel must be balanced by a pressure difference between aircushion pressure and atmospheric pressure.

p,BL = 200000 9.81 [NI

(19) 200000 . 9.81 _{= 4905 [N/rn2]} PC) - 10-40 (18) (20)

p0 = pg(D-d) ' D .2 = 0.488 Em]

pg Natural heave period:

hb

T3 = 2

(1+p/p0)

c) Model scale, X = LJL . Then by assuming Pa'PO'

= ( 1 +p/p') ( \1i2 h, 2ir (

2t

yg(1rp/p)) / = h,

i-p/p)

_/ 'JJ2 ,' \Ji2 h,

_1PjPt)

= 2t

I 1.4.9.81(1-101300/4905) , h, 1-'-pjpt / Pt) = = L,JL 2.0 (21) \112 22 0.515 Es] (23)

Usually one does not scale the atmospheric pressure. Hence.; the relationbetween heave natural period in model- and full scale is eqial to the model scale X. This gives very small natural periods in model scale (T < 0.05 s).

Transferring results from model to full scale, it is very common to use Froude scaling which means that the Froude number F is kept constant. (F1 = U/VTh). Hence ( _{_} M \1i2 .

PiP

\L2 PJPo _, I M \112 T3 'r' rrM rt3 T3

TM L MIUM _U

- sJL t/L = _J5

L/U UM L

Froude scaling gives wrong full scale resonance period. This may cause error when predicting accelerations in particular sea states.

Heave resonance occurs when period of encounter is equal to heave natural period. From equation 3.52 in the text book

T, = T3 T0

1 _cosI3

T0g

(T0-T3)gT (2.5 - 0.515) 9.812.5 _{= 15.04 [mIs] = 29.2 [knop] (28)}

- 27rT 2ir 0.515

e) Changing the plenum height hh, one may change heave natural period. The plenum height is changed when the cushion pressure is changed.

U Want to verify the given expression of the volume pumping. We have:

r(t)

çsin(u.' kx)

dV.

- = _f

_dx =

w.J cos(w,t kx)dx dt dt

LLdt

L L A, L2

= -

Jcos(ot

kx)dx Hence dV.1 x

.7tL

- A6.. Ci) cos(ci.t)

Solution exercise 3.3

Start with the velocity potential and the velocities u and w in x and zdirection; respectively: =

_(ot-kx)

u =

= w0e

sin(ct -kx) ax w = . = wocae cos(w,t-kx) az

Recall that the total velocity potential t' is expressed by = Ux +

where U is the forward speed. 2D lift of a foil segment is given as:

FL = .._p

C1U

2

CL is the lift coefficient (function of angle of attack a) and I is the cordlength of

the foil.

a) We want to derive an expression ofthe dynamic lift force on the foils. Before we do that we have to say something about the foilmotion as well as the flow field around the foil. The surge, heave and pitch motionzof the foil are given as follows:

flijbiI = + _zi11.

113fnil = 113 - Xifl.

fls;;ri = 115 (8)

z-WR

is small arctan WR WE? WE?

-:;- Ut? U

Dynamic component of the lift FLE) can then be written as:

The first and second terms are the static and the dynamic lift force; respectively. Further, assume that:

(16) a

a

- _fl arctan I wR (13) UR Then: (14)

CLa.)

C(a). -

I

(aa,)

-Substitute (14) into equation (5 and find an expression of the lift force:

C(a)

I U2 _- I a-a0) I U2 (15)

=

±.

= U + sin(wt-kxL) (9)

Jx

= w0çe cos(ot-kx) (10)

Relative velocity in x-direction between fluid and foil is expressed as:

UR = U.- flI1QLl = U wocae sin(w,t - kx1) In z-direction:

WR WTfl.JQi1 = o0c,e coS(Q),,tkX)

-(12)

In equation (5), only CL(a) is a function of time. We carry out a Taylor expansion of CL(a) about a=c. a0 is the angle ofattack of the incident flow in still water.a can be decomposed into components as follows:

FLD CL acL acx .U2 U - a0) I

U (w

UT1c)

F1 = P

acL

_u

_{[co0çe cos(wt-kx1)}

.2 aa

.I.U2

(17)

+ xii5

Ur] (18)

b) Assume the foil to be small compared to the wave length. Then the force per unit length due to added mass (radiation and diffraction part) is written as:

F3A = w (19)

where a3R is the relative acceleration of the foil. The foil is approximated by a flat plate. Assume cix, p oc, then:

=ir[.2

(20)

Further the vertical acceleration a:R is determined as:

a.3 ₌ _..(wR) = -C oo(ek

sin(ut-kx) -

-'- Xt (21)

F: = pit

[_cc0çe

sincot-kx) - 113 + xi]

(22)

Note that the first term in equation above is the diffraction force. The Froude -Kriloff component is negligible since, the volume of the foil is close to zero. The two latter terms are forces due to forced oscillation (radiation problem)

C) Generally the hydrodynamic force (radiation part) is expressed as:

F

dq

dn

Jdt2

'dt

where AkJ is the component of hydrodynamic force proportional to acceleration.

Equation (23) is a 2-dimensional force in z-direction where parts of it are

proportional to the acceleration. A33 for both foils is then given as:

A = 2p7t

_)

S

(24)

Where s is the span of the foil. Damping is a force proportional to velocity. From equation (18) one may find the damping coefficient B::3 as the part of the force proportional to the heave velocity:

___ I I

U S

2 = p

acL

2

a

3cz

B3. = Bc.3 = A3 = 0

due to the symmetric location of the foils. Further Bc5 and A are determined by multiplying equations (18) and (22) by -x, (which will give a moment). Then

Aç = Bcc ₌ I pit s x = A. 2

I,

2 acL 3cx (.L-

.1.0

= B:5. (23) (27) (28)

Restoring force is proportional to a position. From equation (18) one may

determine C,, as:

(29)

2

acLJ aa

=

pr (i.j

s wc0çe . cos[k...J sin(ot)

aC

I I

U s

unekz cosk cos(wt)

2

The corresponding excitation moment in pitch can be evaluated from:

= (-x1

F

f plc Sin(cQt-kx)] I I . . . cc.J((Qt-kx)J (J'c (

r\

1. I - -- . T .__I

Pit

-1 &CL - P

_2cz

I

IUs

cos(cot-kx.)] U S

sin k_ sin(u,t.

2 I (30) (31)

Excitation force is the one proportional to particle velocity as well as acceleration. Hence

F:3 ₌ sin(wt-kx)J

p.

aCL

d) Want to determine relative damping due to the foils. C = p g 30 m2 = 294300 N/rn2 C 60 m2 C. T3 = 27rJ

MA33

= 4.3 [s] T,75 = 2it N

-=, 5.3 :1 C5r;

Critical damping in heave and pitch; respectively:

2(M A) . 2t

Cj,ç = 2(M A33)

T,1.3

2(Mr

+ A) 2t

Ck = 2(Mr,ç 4 Aç

Relative damping is defined as: = 5.2 Is]

Ck.

where B. is the damping coefficient in heave or pitch. Then

(35)

(38)

Note that these damping coefficients are very high. For a conventional vessel the relative damping is 1-3 %. For high speed vessels with transom stern the relative damping may be as high as 10 %, which is due to lift effects on the hull. Anyway:

a foilcatamaran where 80-90 % of the mass supported by lift from the foils, the relative damping is much higher. In general. the hydrodynamic force on those foilcatamarans are dominated by the forces on the foils. This is the reason why the hulls are not taken into account when calculating the hydrodynami coefficientsA;, and B,1.

Explaination of the sy.boIs: M 144 A22 A42,A24 A44 F2 F4 C44

112 and r are sway and roll motion in center of gravity; respectively. Hence equation of motion has to be calculated in center and gravity. Therefore one has to transform the z-coordinate to a new z-coordinate, z' where

(1)

where Bt is the distance between center of buoyancy and gravity. The velocity potential due to the incident waves is expressed as (deep

water):

ecos(ot-ky)

(2)

Solution exercise 3.4

- Total mass

- Roll inertia moment

- Added mass in sway due to sway motion

- Coupled added mass sway/roll

- Added inertia moment in roll - Sway excitation force

- Roll excitation moment - Restoring coefficient in roll

The 2-dimensional added mass on the strip is expressed as:

Where o is the wave frequency,

is the wave amplitude. The wave

is

propagating in positive y- direction

a) Investigate forces and moment due to forced motion in sway and roll (radiation problem). Then:

(3) F2= - A22112 - A24fl4

(4) F4= - A44i4 - A421'i2

Horizontal acceleration on a strip, dz is given as

(5)

2D rD2

A99 _- p

The force F2, is equal to mass times acceleration integrated over the body.

fAz'

[112 ( d - 12 + (-2 + BG)Ji14]dz .

(a

+

t) tl4dz

2 + BG)zJId 14 (6) (7)

where A is the cross section area of the body. Comparing equation (7) with (3) We do the following findings:

A22 = oAd (8)

A24

= -4

(9)

The added mass (moments) A42 and A are found when calculating the moment

about center of gravity due to the force F2.

(y-acceleration) dz = fdA2h12 -

fAz' (

+ ( + BG) 114 dz (10) = pAdBGt2 -

A_

dBG)

114 F2 = - JA&12d2 +

fAf

L2

= -

pAdl2 +

pAL2

+ (.

- pAdt2 pABUdfl4

Hence

and V are the waterplane area and the volume displacement; respectively. Excitation force F9 (diffraction problem) is determined from (see pages 58-62 in the text book):

F2 = fpA + A)adz

(15) where (16) a t3y

Then:

F2 = 2pAco2coswtf

edz = 2pgA(1 - e)cosot

(17)

(11) A42 =

-pAdU

(12)

A44 = pA_ +

dBG')

The restoring moment F4 is equal to the hydrostatic moment. Then

(13) F4. = -C44ri4

where

Excitation moment F4 in roll is calculated as the moment about centernf gravity due to the force F9.

F4 =

_f2pA

z ( BG)la. dz

2

- pAdTh12

['41

PA] 14

Assume harmonic response:

= TCOSO)t 1(4 - rLcoscot pgVGM1(4 = -2pgAçcoswt(C -De ) (24) = -2pA(cx2coswt

f

z +

edz

(18) =

-2pgAcosot (C

De)

where

c=+H

-?:. 2 k (19)

D=-Bt+!

(20) 2 k

b) Knowing all coefficients, equations of motions are obtained as

(M+pAd)fl.2- pAdHGfi4 = pgA( 1 -e kd)cost

The sine terms in equation (22) and (23) are

skipped due to zero damping.

Substitute into equations of motions and obtain

2(M + pAd) o2pAdBthi4 =

2pgA(1-e)

The natural frequency is calculated with no excitation (i.e right-hand side vanishes). Non-trivial solution do exist if and only if the determinant of the coefficient matrix is equal to zero. Hence

= w4(M - pAd)(144 -A44) - w2(M - pAd)pgVGM - w4(pAdBU)2 = 0 (25)

Ci) = (M

+ pAd)pgVM

'

(M pAd)(144 + A44) - (pAdHU)2

gVGM

(I

A4) -

!MBG

i) The following data are given:

k = = 0.3142 rn' = kg 3.08 s_I = 0.0432

M

= pAd = 31416 kg (p = 1000 kg/rn3) pActBt = 31416 kgm = 0.15L2 = 923630 kgm2 A44 =

M_

BG2J = 293215 kgm2

GM

= - = 1.025

m

V

2pgAç(1-e)

= 14744 N C

= (.

BU) - = 2.818

m

2 k

D =(-Bt.)=7.183m

2 k 2pgAç(C+De1) = 48175

Nm

Then the natural period can be calculated as:

T = 12.3 S

Substitute known data into equation (24)

0.5i = 0.076

(28)

- 35.5 = - 0498

and find after some algebra

112 = - 0.07

(29)

= 0.012 (30)

= O.Ol2cosot [rad] (31)

112 = - 0.O7coswt [mJ (32)

Wave profile due to incident waves:

= -

1Li0

= ç6sin(cüt-ky) = 0.25sin(ot-ky)

gat

Wave slope (positive upwards);

= -0.25k cos(wt) = -0.079 cos(ot) (34)

dy

Compared with equation (31) we can see that roll is in phase with wave slope. (Note that positive 114 is opposite directed compared to c) The roll amplitude is

dy less than the wave slope.

Sway motion is 900 degrees out of phase (cosine - sine) with the wave elevation,. with an amplitude less than the wave amplitude.

Horizontal motion of arbitrary point on buoy:

flhor =

-Set flhor = 0 and find z = -11.83 m, which is located off the buoy.Hence; no point

on the buoy has motions equal to zero.

iv) Horizontal acceleration at top of buoy. (z=4.O m)

ihor = 12 -

c

+ BU]

(36)

= 2(oo7 + (4.0 + + .)0.012) cosot = 0.59 coswi [mis2]

2 ₎ d 2 .4- n4 = -I 0.07 (z - )0.012 coswt

(3)

/

Solution exercise 3.5

Roll equation of motion:

+ Bi14 + C44i14 F4 sinwt (1)

a) All moments are to be taken about the origin of the reference frame used. It is most common to calculate motionabout center of gravity. Hence all moments have to be taken with respect to center of gravity.

A

14 - mass inertia due to roll

- added mass inertia due to roll - potential damping in roll B

- restoring coefficient

C44

- roll excitation moment F4

b) Want to develop a relation between damping and wave radiation. Assuming harmonic motion; only the damping term can be connected to energy. This energy is taken care of by wave radiation. The work done during one oscillation period of the force due to damping can be calculated as:

On pages 45 - 49 in the text book energyrelations in regular waves are discussed.

Due to equation (3.25) in the text

book the energy of the radiated wave (both

sides) is

E2=pgAi_2.2

(5)

2o2

d114d114 (2) dt di AssumeS (3) 114 = 114 SiflCJ)t

Then we may rewrite (2) as:

Substitute (8) into (7):

(7)

d) Want to verify the given expression of equivalent linearized damping coefficient. Find K such that the work done during one oscillation period of the terms

K14 (10)

and

14 1141

becomes equal to each other.

(12) = / pg2 \1J2 4.4 I

F-

4a

-

Pg2D.

2 44 B44 _{= p} A3 g2 (6) 14

where A3 is the amplitude of the radiated waves. c) Roll amplitude at resonance. (co=co)

-

F41C44 F4

114_j -

___

2

Due to equation 3.45 in the text book there is a relation between damping and excitation at resonance.

Assume:

114 = sinwt

Then we may rewrite (12) as:

"'

-

.,

K i or cosO)tdt = K iL

f

144I 14

= 4 B,

=4B0) .---

₃₀₎

Hence; by setting equation (14) equal to equation (15) one finally get:

K=B2

(16)

e) Locally the flow in the vicinity of thebilge keel can be approximated by theflow around a flat plate of breadth 2b where b is the dimension of the bilge keel. Added mass in a direction normal to the bilgekeel:

= ..!: pit (17)

2

The added moment of inertia is then expressed through:

A=cr2.L=.Ptb2[.J

L (18)

where B is the breadth of the vessel. Further; assume b = 0.02 B. Then:

A = 0.00005pic

B4 L

(19) (13) , 1

or -

-

2ic ₍₁₄₎ cos3wt dt

2w

Q)3 (15)

The effect on the damping can be implemented in the viscous way. Normal force

(viscous) on the bilge keel:

f) Standard deviation of roll motion may be estimated through

the following procedure

= -

O.5pC0 b U U (20)

where u is the local velocity near the keel given as

B ₍₂₁₎

U = fl

The following equality must hold:

-Be, 141141 =

F

B (22)

Hence;

B = 0.5p CDb B (23)

2

C0 is not yet known but can be estimated through the relation:

(24) Cb = 8.0 KC

where

KC Um = j1(4 it 25 (25)

(29)

( '2

0 S(o)do)

= S(o)dw

where r4/C

and fl4ia denote the

transferfunctions of roll motion and velocity; respectively.

Horizontal motions at a point with arbitrary z-coordinate:

(28)

lhor = 1]2 - 2T14

By requiring Tlhr = 0 one may

find the z-position. Note

that this point not

necessarily has to be located on the ship.

Lateral acceleration due to sway, roll and yaw motion, may be evaluated by taking the time derivative of equation 3.9 in the text book and extracting the y-component. Then

Motion in a moonpool.

Bernoulli's equation (linear theory):

p + pgz + p- = 0

= + pg + = 0

On the free surface i =

.t

Then by rewriting equation (2):

d!i= -!.E.-g

dt2 p z

Want to verify the given expression. Integrate term by term

fddZ_(flh)diih.d11

-h _{dt2 dt2 cit2}

consistent with linear theory.

-

1dz

p J- az

by use of equation (1). Further:

-f g dz

= -

g(rh)

Solution exercise 3.6

=

-!

fp(z=) -

p(z=-h)] = . p(z=-h)

p p

-

-g

h

dt2 h h t

C) The natural period is characterized by the homogenious solutions of(7). Assume

-

(8)

11 = 11 sino)t

Then: by substituting into equation (7) we obtain + .:. = o

h

\lg

d) Want to find natural period of a U-tube following the same procedure as in C). Then

d2q 1 a

_-g

ct2 p a

Integrate this equation from 1" = -. to 1" = rl. this corresponds to an integration in z from z z' to z = fl where z' corresponds to 1' = -.. . 1" is an integration

variable aiong the tube.

dzrl I _{d7m (12)} dl' = IT1+--1

-!

f .. dz = -

[p(z=) - p(z=zrn

-pz' _{- ---L=} (13)

-f gdz = - g(-z')

(14) (7)

5 dt2

2) dt2

2 dt2

(15)

2 dt2 t

In order to calculate the natural period one has to solve the homogenioUs equation. Assume solution

(16)

= srncot

Substitute into equation (15) we fmd:

(17) 2

+g=O

T,, - 27t.J (18) e) Explaination of symbols SF.(0) - excitation spectrum

m - is the mass (= 1 here)

K - equivalent linearized damping coefficient

Soludon exercise 3.7

When the frequency of encounter approaches zero the wave crest is to be considered as static with respect to the ship. Then the fluid velocities around the hull becomes very small.

The rudder can be considered as a lifting surface with low aspect ratio and where the lift (side force on the rudder) can be approximated by:

FL = PCL .s V2 (1)

when CL is the lift coefficient, I and s are the cord length and span; respectively. V is

the incident velocity on the rudder. Due to the low velocity V the force on the rudder becomes very small. Hence; the effect of the rudder vanishes.

The incident velocity is dependent on the position of the waves reiave to the ship. The x - component of the fluid velocity follows the wave elevation. Hence; if the rudder s located under a wave crest or a wave trough there may be a small incident velocity on the rudder. If there is a node of the wave elevation located above the rudder, the incident velocity is zero.

Anyway; if the rudder is located in the jet flow behind the propeller there will be an incident velocity to the rudder.

A wave spectrum represents energy. Hence; over a small band do)., in the frequency of encounter spectrum it has to be the same amount of energy as over the band dw0 in the wave spectrum. Then

So

do)j S(w0) do)0 I

Assuming head sea the following relation exists betweenfrequency of encou.nter and wave frequency, w.

S;(coe) =

I,

=r

S(o) dco0

g

Determination of heave transfer function. We have

Se(co) dWe

Solution exercise 3.8

(1) (6)

O)U

(2) = coo + g Differentiating 2w0 dw0 U 2w0U (3) dWe = dco0 + g g

Substitute into equation (1) and find: S ( w) dw0 [ 2co0 U (4) = S(w0) dw0 g S(w0)

S(w

(5) 2co0U

We have

Se(We) dcoI = S(w0) Idco0l (7)

were . is the unknown. This is an integral equation to determine and S(w0) is to be considered as known.

C) In following sea one has to take care since cob, is not a one-to-one function of w-,. The figure below shows the relation between w, and a.

(8)

O)U

1

When e

_{< - three different wave components correspond to the same}

g 4

frequency of encou.nter. o is small and represents long waves. w is less than _, but as the phase velocity is A_. it means that the phase velocity is larger than

CL).,

g

For (1)3 the following relation holds:

g which means: d0)0 1 dCOe 4U0)

1+

g

Substituting into equation (9) we finally get:

Want to calculate dw0 For w and 0)9 the following relations hold:

(10) dCOe ci;, = -Then

_g

_1±1-

4Uo

(11) -g

where the plus sign corresponds to Further:

dco0 1 (12) dw

_4Uo

S(c1)) = S(w1 dw° dw dw, dw ₍₉₎ dw - S(o).3)

S(

S(0)1) S(oi,) S(0):3)

4Uo,

4Uw.

g

(15)

When > there is a one-to-one relation between w and we,. Then

g S(w3)

S(w'

(16)

4Uo

Ii +

g

This means that Se(a) _{is singular as}

(17)

g

Consider a source distribution over a plane circular element radius R (see Fig. 4.16 in the text book ). Assume the fluid has infinite extent and the source density is constant equal to one. Show that the vertical velocity V at a point P on the z-axis induced by the source distribution can be written as:

= - 0.5 I

(R2 - z2)"2 zj

Solution exercise 4.1

(1)

The velocity potential at a point (x,y,z) due to a point source with strength Q, in

the point (x1,y,z1) is given as

(xy,.zx11y1,z1) =

-

4R

(2)

R = _{I(x._x1)2 (y-y1)2 (z-z1)2}

The velocity potential in a point (x.y,z) due to a source distribution with density q(x1,y1,z1), over the surface S is given as

(xy,z) =

us

4itR

(4)

The vertical velocity is given as : (q(x1,y1,z)=q=constantl)

(3)

Vz

(5)

V2

=1.. 11

dS (6)

47t JJS((x - x1)2 + (y - y1)2

-The source distribution is located in the xy-plane = = 0. Further; the field

=

L Jf.

dS

L

CR

i

z rdrd8 4it _5(x -- y _z2)3'2 4ir .h Cr2 +

z22

v_ = Z 1R

rdr

c-' L ₂ '32 - (r

z-)

= q Z((R2

zV

-2 jzj

Discuss the velocity behavior close to the source element both for negative and positive z-values.

Investigate the limit of V when z=O and z=0

lim1.

=

lim1.

1im..

= lim2 -lim121.V1 = urn121 q _{Z(r2 + z2)} -z z 2LjR2 + z2)uJ2

ii

What is the behavior like far away from the source element? Investigate the behavior of V1 as R/z - 0 Then

1( z 2

V1

- 2(R2 _z2)"2

ii

_1(

1 z z

21((R/z)2 + 1)"

i;-i

ii

Rh << 1 when z are increasing. Hence; we do a Taylor expansion with respect to

R/z z

z_1

2IjR2 z2)"2

1i))

2 z

=1

IzI 2

(i(

1 z z 1.

2((R/z)

+ 1)1/2 j

-T

(7) (8) (9) (10) (12) J (13)

lim2V

= urn. R2

=

sgn(z)_

We can see that as z -

the velocity potential due to a source distribution

becomes more and, more equal to the velocity potential due to a point source located in the origin. Induced velocity due to a point source located in the origin.

q7tR2

(x111)

= -

_{4(x -x1)}

+ y1) = = - sgn(z)__ 4V? R2 = vz =

sgn(z)__

I ₁₍ E

- - 1 -

_(R/z)2 O((RIz)4)

-2 (14) (16)

z - z)2

₍₁₅₎

Solution exercise 4.2

a) Irregular frequency is not a physical phenomena. It appears when there exists an inner solution _{which satisfies Laplace equation inside the body as well as} a certain body boundary condition.

The following boundary value problem is to be solved

Assume solution

= A sinh(k(z + D)) sinky sinO)t

Boundary condition at y = B

4(y=B) = Asinh(k(z _{D)) sinhB coscot = 0}

kB =jir

_j

1,2... Further;

= kAcosh(k(z + D)) siithy - cosot

Then the free surface condition _{(combined) can be written as}

-&

+ = 0

dz

- w2 AsinithD siay

cos ot kgA coshD sin)ey cosot = 0

B B

where X is the wave length.

Distribution of elements. Assume we want to calculate first order linear wave induced motion of a barge. It is thenimportant that the element density increases in areas where the flow changes rapidly in space. That may be around sharp corners and so on When studying second order motion it is important to have a high element density near thefree surface since the second order velocitypotential

decays in depth as e2. In

addition there is not the same need of high element

density around sharp corners as long as they are deeply submerged.

w=kg

coshJD (7)

sinhkD

Insert value of k. Hence:

(8)

= j--

cothLE

B j=1,2...

g

b) The irregular frequencies are veryhigh, and it can be shown that

irr > 1.0

(9)

Ng

Hence; we are sure that there are no irregular frequencies when

d) Symmetry planes of a barge

Surge motion: yz is a asymmetry plane xz is a symmetry plane Sway motion: xz is a asymmetry plane

yz is a symmetry plane

a) Note the misprint in equation 4.47 in the text book. The field coordinate should be (y1, z1). Want to verify the given expression. Start with Green's second _identity

LI

_{L an}

L-!1ds=O

_an]

LI

[

_{L an}

Solution exercise 4.3

Choose w = logr which satisfies Laplace equation. Integration over S1

- i

_--1

ds

JJ

[

__1ogr -

1ogr±1

anj _L &n anj

=

Jf

[-

r -

1ogri1ds

_anj

=

Jj!ds

-

Jf1ogrids

(y11)

!

2r

_-

logr 2r

r

= 2it ,(y1,z1) (when r 0)

(1)

Hence; (1) can be written as: - 2it ô(y,z) = = = -b) 1ogr _-

_{1ogr.i_ ds}

an an logr _-

_1ogril

_ds 2tJJS L an

anj

Green's second identity

(5)

- = 0

where S = S +S,, +S - Assume the following

*

_{Move S so far away that}

the contribution from Se,, vanishes. Note

that

the integrand decays faster

than S increases.

(6)

0

onS

'V = logr 1ogi

-y)2 ( z)2

The contribution to the integral over S1 is derived in question a). Then

f(!

- 'V ds = 2ir (y1,z1) (9)

s1 an

an)

Note that the selection of , satisfies

..!

= 0 on S. 1og expresses the mirror an

image of logr.

Finally:

- 2ir p(y1,z1)

=

Note : First term on right side represents a dipole where is the strength. The second term is the source where is the source density.

an

c) Combined free surface condition

_2

_{+ = 0}

when co -* 0 we have

az (12)

Equation (12) expresses no vertical velocity. Hence; the free surface

_{can be}

replaced by a rigid wall and the velocity potential can be expressed as a

distribution of sinks/sources and dipoles over the body. Hence;

21 Zl

;-+

y:+z; The term (logr - logi) It -JJS9

logyz'

. 11

.tds

The source strei.gth is given as:

1

JJs8 TL

2 logyZ

(15)

has a leading term WhiCh decays as Then the term dies out very rapidly

y

and we are left with terms proportiOflal to Hence; by using (13) and(14):

(yz1) ₌

--N'

ds

(16)

(17) logJy2+z

+ log yIzt

2 2

+ y-2

-y1

-_yI+zl -Yi

- -2

-yl+ZI (14) -2 O(y1.1) = Jj3 -ds (13) where N' is given in b)

Taylor expansion of N' about y=z=O. Note that y1,z1 is the field point (assumed constant when taking the derivative) and y,z is the coordinate on the body.

-k

=+

yfg24üug

2U U2 U2

Use that

O)Uif

g

Solution exercise 4.4

a) Velocity potential e = Aek

Free surface boundary condition

cT_O

aX) ax ax1 az Substitute for T where it is assumed c Solve for k:

g -

2cU± /(2wU - g)2 + 4U2 co 2U2

and rewrite Assume k1., g - 2t ±

1 -

4t]

2U2 = Ae2*

Then by substituting into the free surface condition:

-O) 20e Uk - U2k2 + gk = 0

k3 [1 + 2-c ± 4t

2U2

b) The discribed moving coordinate system satisfies the free surface boundary condition as given in a). Hence; the results from a) may be.used.

Group velocity: Ct2 1 g 1 g 2U2 g 2 k1.2

2 ' g (1 - 2t

±

(1 -

4t)112)

=U.

2

c;<u

c;>u

Further; c3.4_g = 2 1 - 2-t ± (1 - 4t)''2 1 (1 + 2-c ± 11 +

4t

)I/2 (14)

From figure: Wave system AA

Dl

3 D2 284 D3 170

There are good agreement between figure and calculations. It is left to the reader to check the phase velocities

x 1.0 500 = 2.2 46 Wave system, i k 1 1,134,4 5,5 2 0,0352 178,5 3 1.949 3.2 4 0,02052 306

c<u

(15) (16)

This means that wave systems 2 and 4 are upstreams while

1 and 3 are

downstream the body. If

t

= wave system 1 and 2 propagate with a group

velocity Cg = U. Wave system 3 and 4 are downstream. and upstream; respectively.

c)

Numerical values corresponding to

figure 4.10

is U = 2.5

rn/s and

t = 0.1274

d Asymptotic expressions as t-40

k. = __

1.2 2U2 L k = A_

(1 - 2t)

k9=O

k34

g_[1

+2t±

2U2L k3 = .i_ [1. + 2t U2 k4 = 0

e) Wave system 2 and 4 have X - = Cg

o and

vanish. Wave system 3 and 1 approach both wavenumber k = which corresponds to Cg = U. This is the case for steady motion.

1-

1 (17)

4t1

2 (18) (19) - (20) 1. + 4t 2 (21) (22)

T1=2lrM

=27tMh1

\j Cii _J

Restoring coefficient of heave motions is dominated by the stiffliess of the legs. One may therefore consider the contribution from the water plane as negligible. Let r be the deviation in heave from the equilibrium position. Then the force is given as:

Solution exercise 5.1

a) Want to calculate surge natural period. The restoring coefficient C11

is calculated by the following procedure. Let P() be the pretension of the legs and let1 be the deviation in surge from the equilibrium position. The restoring force is then given as (assume r small):

F1 =

-Psin(1/1)

-P0/1

= -C11,

Cfl =

Natural period in surge is then given by:

EA _{= - (4)}

I

C. = EAII (5)

Natural period in heave:

T.1=2]tl

Ft.

I M-'-A. Zv1A.

(6)

b Asymptotic formulation of drift force for small wave lengths is given on page 144, example 2 in the text book

= pgc2r _cos (7)

a

Neglecting hydrodynarnic interaction between the columns, the total drift-force is given as : (4 colunms with cos 8 = 1.0)

= 4pgr = 2.28

iO [N/m2j 3

(8)

This formulation overestimates the drift-force in figure 5.24. most probably due to interaction effects between the columns.

The bottom at w = 0.9 is due to canceling effects between the different columns. For very long waves the drift-force approaches zero since, the body do not reflect or generate waves.

The wave drift force is proportional to .Whether or not the viscous component

is to be taken is to consideration, is then a question of both wave amplitude and wave frequency.

Assume that the slow-drift motion is governed by the following equatiofl of motion. (Equation 5.51 in the text book)

(A'!

A)

B

± ±

- CILx = F" (9)

Non-linear damping is selected since its origin is due to

drag. The standard

deviation of the velocity (assume small damping) is then given as (equation 5.54 in the text book)

\L3

=

S.i)

(10)

8C11B0

Where in our case is the natural frequency of surge motion. Table 5.1 gives

i,. = 0.0587 [radls].

S-(u)

(C0 A

PtrTh2 S(fl (11) 2ir

Wind spectrum S(f) is taken from equation 2.40 in the text book.

S(f) L f/U10

Uo 4 . L U1,,

(2 + (L . f/U10)2)516 (2 + (L f/U10)2)

S(f,) =

4 .

. L U10 4 . 0.003 . 1800 40 = 452 [m2/s] (13) (2 (L f/U)2)516 (2 (1800 0.0093/40)2)

Values of L,U10,C

are consistent with the ones at page

31 in the text book. Assume that all 4 columns contribute to the frontal area. Density of air and water

is given in table 6.1 on page 175 in the text book. A crude estimate

of the excitation spectrum is then obtained as: (U

S) = (Cr) A

P2irTh2

S(f)

27r = [0.7 (4 17 25 1.0 . 10 87) 1.2928 . 40] 452 = 6.226 10 kg2 m2 Restoring coefficient C11: C11 = = 19000 [ton]! 600[m] = 3.11 10 [N/rn] Use a drag coefficient C[, = 0.7. Then B11 is taken as:

= C0!pA

= 0.7 0.5 1025(4 17 32 2 (87-2 17)

ii)

= 1.199 10 Ns2 ,n

Another approach to calculate the damping is to estimate the mean wavedrift damping in the seastate. This procedure is described in exercise 5.2 b) in the text book.

(12)

1

Solution exercise 5.2

a) Want to verifSr the given expression of mean wave force in irregular sea. Both

spectrum and transfer

functions are given.

The transfer function

is to be

interpreted as follows : It expresses the response made non-dimensional by wave amplitude squared. But what wave amplitude is to be used in irregular sea, since there exist an infinitive number of wave frequencies and wave amplitude is a function of wave frequency. Therefore one introduces the wave spectrum which multiplied by a small frequency range d is proportional to the wave amplitude squared in the actual frequency raage. Hence; the mean force is obtained as a weighted sum (integral) of the transfer function over all frequencies. The wave spectrum then plays the role as a weight factor. Due to equation 5.28 in the text book the mean wave force is given as:

-

e..

T(o)

₁

But what about the integrationlimits ? Wave spectrum is defined in the frequency

range:

_!E<<

(3)

T1 T1

while transfer function has limits (co2 = kg = .

o

Assuming both ...!L and 2V less than Then the upper limit is:

T1 L T1 (4)

FS2IS(o)).

Divide both do and I F(cü) obtain dw (2) side with pgLHT1 = 2 S(w) S F1 pgLH,3T1 HJT1

pgL1

T1

Fi.irther; the lower limit, 0) is maximum of _!L and

.2i.

Then

T1 L

- o.03(0)max

-32ir

pgLHT1

O.00O6pgHTj((0x - O)miz)

b) Excitation spectrum S(ji,) is calculated by means of equation 5.45 in thetext book

I

S,(p) = 8S(w) S(o

Divide both sides with (HT1pgL)2 and obtain:

"2 I

(8)

Si)

=

S(co) S(0) + i.i,) 2 / dco

(HT1pgL)2

HT1

HT1

pgL

The difficulty is to determine the integration limits S(0)) and S(co

j) are

defined in the following frequency. ranges; respectively:

it 3it

< Ci) <

-T1 T1 it 3ic

_1n<(o<__ _1rt

dco (6) (7) (.9) ( 2 3ir (.5)

In addition, the transfer function is defined in the frequency range:

b) Slow drift damping in irregular sea is to be determined. Use same formulation as in equation (2) but replace force transfer function with wavedrift damping transfer function. Note that wavedrift damping is defined in same frequency range as the force. Then the calculations will be similar.

BWL) 11 =

2f

S(co)

B(co)

dco

HT1p(Lg)

HJT1 p(Lg)/2 = 2 1 . 0.7(COTflaX - 0tnin

32t

Hence B1WD _{= 0.014 H T1p(Lg)"2(co}

-

(15) Then 2 Then

the lower limit co1

will be the mimum

Of

!_ and

-(10) (11) (12) (13)

____

T1 L 2 (0.0006)2 - _< Ci) <

Further;

-L 2

the upper limit co2 is given as:

3ic = -5;- - J1, we have:

Si)

_{= 8} 1 _0.032 = 2 (HT1pgL)2 (32t)2

Finally:

S.(j.i) = 2 (0.0006pgLHT1)2 (Co2

-Solution exercise 5.3

a) A crude approximation of added resistance of a ship in ship motion range:

where and are the relative motion in bow and stern; respectively. In order to describe what parameters added resistance is dependent on, one needs to discuss ship motion as heave and pitch. Important hail parameters are CB (block coefficient),

BID and L where B,D,L are breadth, draft and length; respectively

In order to construct proper dimensionless quantities one needs to incorporate as many relevant parameters as possible. Hence; in order tO construct dimensionless added resistance, RAW one should use B/D,L,C3 as parameters. The expression RAW! (pg B) do not include the ship length L while RAv /(pg B2/L) does. It is

therefore expected that the latter expression is least sensitive to main hull

characteristics. Anyway; either of the expressions include the draft, 0.

Realistic values of 0) , , L and tJ. Due to the joint frequency table T0 = 8 s is

realistic This corresponds to a wave length A. = 100 m. Added resistance is large

when X/L = 1 which means L = 100 m. Further; one has to be aware of the

frequency of encounter effect. T0 = 8 s 0)) 0.79 s which together with a

forward speed U = 7,5 rn/s represents a frequency of encounter

=1.26s

-I (2)

g

Hence; w is selected to be 1.26 s'. The wave spectrum is given as: ( \-5

011 coT

S(w) = H,3 T1

2t

2t

The wave spectrum is proportional to and T. Realistic values of H113 may vary from 1 to 4 m which represents a factor of change in F1s equal to 16; T1 may vary in the range 6 to 12 s which represents a factor of 16 in calculation of F.

0.44

( \-4

coT1 2ir

d) Start out with equation 5.23 in the text book.

k

20) JL B U__ Adx V(x) dx

I

Neglect forward speed U and express the damping by equation 3.26 in the textbook, one may rewrite:

R

ff

A32

dx AW

2 0)2

where V is the relative velocity amplitude and T3j is to be interpreted as the local relative motion amplitude, _3R

Then v

w Rewrite and find per unit

length

R°

_AW pg A2 2

This is the same form as Maruo's formula which is given as:

-p. = 42 2 R

where A is the amplitude of the reflected wave. Equation (6) does not take into account the effect of sway and roll. This is due to the neglection of the asymmetric part of the diffraction potential which excites sway and roll motions.

(4)

U = 1 - 0.95

[rex-For

_{' <X}

_{and fzI <.}

The velocity behind a cylinder (twine) of diameter D in uniform current is given as (equation 6.49 in the text book):

/ '-1J2 (

y2

.._ =

1.0 - 0.95 exp

- 0.0888

CD . D CDD1

The part of the argument of the exponential function equal to y is generally

written as y

_-y0

Then equation (2) can be interpreted as the velocity in a point (x,y) due to a cylinder located in x=O and y =y01. The defect velocity u1 is defined as

Solution exercise 6.1

a) Want to show that the velocity ditribution in the wake of a netting panel can be written as (there is a misprint in the text book):

+ EN exp

I (iX-z)2 O.O888CDDx (1) (2)

Assuming the cylinders independent of each other the total velocity defect can be obtained by -superposition of the velocity defect of each cylinder.. it is worth to point out that superposition is only valid for the defect velocity and not for the velocity u.

a1=U-u=0.95U

where 11= - \-X L/2

.exp(_n2

4 (3) (4) CDD /0.0222C0Dx

In our case we have (2N+1) parallel cylinders in y-and z-direction; respectively. Th cylinders are located in x=O, y=y01,

and x0, zz. The cylinders are equally

space with a istance

X from each other. Then we have that y01=jX and z01=iX.

Total velocity defect due to twines parallel with the z- axis is then obtained as:

= 0.95 1.0 - 0.95 - 0.95 -x CDD exp / exp (JX-y)2 O.0888 C0 D .

Same kind of expression can be outlined for twines parallel with the y- axis. Subsequently, the total velocity defect is obtained by superposition.

The total velocity reduction factor u/U, is then expressed as

(

(jX-y)2

exp ( (iX-z)2 0.0888 CD . D x

b) The following fortran code calculates the velocity reduction factor.

PROGRAM WAKE C IMPLICIT NONE C INTEGER N,I C

REAL D, CD, LAMBDA, K, Y, Z, SUM1, SUN2 , DENUN, VELDEF

C D = 0.002 CD 1.0 LAMBDA = 0.02 N 25 (5) UI = + 0.95 0.95 U U ( x _exp ,exp ( -5" / 5" "S (IX -y)2 (6) CDD1 I \-1J2 x

0.0888.CD*Dx

_I 'S (iX-z)2 CDL? 0.0888 CD .

D x

₁ 5" 0.0888 .00

. D x

(7)

C

VELDEF = 1.0 - O.95*(X/(CD*D)) WRITE(*,*) 'FAC',FAC

GOTO 10

END

Results from calculations:

Consider the wake in the figure.

The velocity will be constant behind the netting panels for those x-values where the wake is fully developed. A velocity gradient will appear behind the edge ofthe panel as indicated in the figure. The velocity does also have a very small variation with the distance x from the netting panel.

x y z 2.0 0.0 0.0 0.8996146

2.0

0.0

0.01

0.8996459

2.0

0.01

0.0

0.8996459

2.0

0.01

0.01

0.8996773 2.0 0.005 0.005 0.8996459 WRITE (*, *) _{.t. /Y,zI} READ (*, *) _{x, Y, Z} SUM1 = 0.0 STJM2 = 0.0 DO I=-N,N DENTJN = 0.0888*CD*D*X

St.Thfl = SUM1 + EXP(_(I*LAMBDA - y)**2/DENTJM)

ST.Th12 = StJN2 + EXP(_(I*LAMBDA - Z)**2/DENUM)

END DO

c) The velocity in the wake of a netting panel will decay as a geometric series

(8) u.

where u1 is the velocity behind netting panel i.

d) The drag force on the netting panel can be calculated from

(9)

FD = 0.5pCUA

where CD is the drag coefficient for the whole netting panel. Want to justify the given drag coefficient. Assume

the twine diameter d =

0.002 [m] as well as distance between each twine X = 0.02. Then the netting panel consists of 310.02

=50twinesinyandzdireCtiOfleach1thfr0nt'

= d 1 [ml = 0.002, and

drag coefficient C=1.0. Total drag is obt.ined by summatiofl

=

!pE:00

C A1'U =

!pCA U

Hence;

CD=

Total drag force on 4 netting panels in tandem:

= 0.5pC1,AU ₌ (13)

F'

= 0.5 - 1025 0.204 - 1.0 - 1.02 (1 + 0.902+ 0.90k + 0.906) = 313.4 [NI A 1

-

1.0 100 0.002 _0.2 (10) (12)

a) Want to calculate the natural periods, expressed as (equation 6.72 in the text book)

Solution exercise 6.2

Tr=

Im A22 4L I2 726 kg/rn = 0.068 s/rn (6) n + v'1250 kN + 11250 kN

Calculate values for different combinations of length, L and mode n.

n 1 2 3 4 L=500m 34s 17s

us

8.5s L=l000m 68s 34s 23s 17s L=1500m 102s 51s 34s 26s L=2000m 136s 68s 45s 34s

T==

(1) F'

i+ /:Ei;;-.

WL=(T-T5)

(2)

Assume that the riser is approximately neutral in water, which means, that

W = 0. Hence;

T = T8 = 1250 kN

(3)

Further we have the following expressions of mass and added mass; respectively

m = p = 1025

kg/m095

rn)2 = 726 kg/rn (4)

= 1025 tkg/m3] 1.0 it(0.95 m) = 726 kg/rn (5)

The values in the table are all in good agreement with the values in figure 6.28 in 'e text book.

b) The oscillation mode is given through:

= sin(zX-zB) eU&( n=1,2,..

z = ____ + A22) (T3 Wx)

= -'- A22) TB

Want tO examine the expression

zx - zB =

.2.J(m

+ A22) (TB Wx) - + A22)

Taylor expansion of /1 Wx/T8 for small values of WXITB.

w

Consequently: 2w, I zx = (m A22) =

2/(m

A22) TB I Wx

1. __

-2T8 x 2T8 (13) zx = + A2) TB

(i -WT8-1)

WxJT (12)

Natural frequency: 2L / vm - A29 Hence; Tt fl X -

-L

Eigenmodes are expressed as

Pn(x)=sin(

2

X)

Maximum oscillation amplitude is given in equation 6.54 in the text book.

(misprint in the text book)

= P(XIL)

fLqJ2d

)dx (14) A. (2ic . St)2 2(m + (27r)2 St2 (19) pD

where St is the Strouhal number and is the relative structure damping. 2

(18)

(AID )Tfl8X

0.32 y

(17)

/

1L. 27tnx

_sin

_ax=

L 2 CL Jo Cr) CDI) fl X _{dx =} / 1 L L 1 8 1.51 (20) (21) . sin

i1x

(26)

L )

sin-

1dx 2.92 (27)

L]

(AID) 0.32 1.155 (23) /0.06 (4t3 St2 . r .

Neglecting the structual damping, on may get a crude estimate of the maximUm oscillation amplitude as =

(AID = 1.51 (24)

C) An estimate of the increase of the local, drag coefficient appears through the relation.: (see text book p. 212)

r'

1 -

2(AJD)''

(25)

C°'

DI

Note that (A/D)' varies along the riser. The

total integrated effect is obtained

as

= P(x/L)

L12 \L'2 = 1.0 = 1.155 (22) 3L/8

__ = .

4' 2(AJD), r)I)

Inserting n=1:

Solution exercise 6.3

Equations of motions: + Cyy T Cyw T = 0 C?112 + m dail6

+ Cii6 = 0

Assumptions jn equations of motions:

- No hydrodynatnic damping.

- 112 and fl are small. This is consistent with linear theory.

- Effect of wind and waves are negligible.

- The mooring system is to be considered as infinite stiff. - The motion of the bouy is considered as negligible. Force due to current is given in equation 6.24 in the text book.

F1c 0.075 _p

su cos

jcos13

(log10R - 2)2 2

where

R ULIcos13I

V

S is wetted surface, iJ is current velocity and (3 is the angle between the current direction and the x-axis.

In our case

In addition we have truster forces, T. Equilibrium of the ship then requires:

T - F1c - F,) = 0

c) Assume:

12 =

116 = e

Substitute into equations

of motions and require a

non-trivial solution. The determinant of the coefficient matrix must be zero.

Then: C i +

(m& C)

= 0 Further:

(ma2

C)(ma

+

C) - C

. C 0 +

(mC

m)2

(C - C = 0 a4c34 - a9a2 + a0 = 0

where a4, a2 and a are defined in the exercise text. Substituted for C , ,

C,

and C, we obtain:

a4 = m a9 = m,4a F0 .. +

.)+

m (10) (15) F0 _(1.6) I

2 2 F0 a - F0 a F0 F1a

->2

I,

F0 3M

= -

-

a-1

d) Want to verify the given expression: a0 > 2(a0 a4)

m4aFo+1J-l+m

.

wi

I ( F0 3M 3F

--

+

a-l314f

Rewrite and get:

[m.a2

m ]

__2.

- 2

"1J'2

[m m]U2

-myy / 3M 3M

m (aF0-___)>O

_yy

e) The new variable is t = L!. The roots of equation (21) are denoted t1 and t.,; respectively. Rewrite (21) in a simplified version:

iF1,

-s1 / 3w1

Fa

aF,) 3F I I 3F

- a

F0 I \112 (17) (20) (21) 12 12 I ₀ 12

- 4m m

aM / -

4(m.a2 m)

[aFo -

.:j

< 0 (31)

At2 - Bt

(t

- t)(t

where:

C>

0 - t2) > 0 2 (22) (23) (24) (25) (26) (27) (28)

B±v/B24AC

1.2 tI <t2 A = m B = -2 C =

maF0

2A a2

-m m

aM

-If the discriminant

B2 - 4AC is negative the two roots

become complex

conjugate. Then (23) can be rewritten as:

(t_C1_iC2)(t_C1iC2)(t-C1)2C2>0

(29)

where

t12 = ± iC2 (30)

and i is the complex unit. (29) is always positive and the system

is stable.

t2 I m, _M I a2 m,,.,,

In order to satisfy (33) we must have

I

I

C1J2 2(rna2 +

m)

a2 m,

I<F

₀₍

m m7,

\av

/2 ( \L2 I4L m = I._Cr (35) (36) = F, a

--aw = F,.,. iVI m,41 t3F,. a2 m,

Another way to ensure the stability is to have

F0

> t2

I

where t2 is the largest root of the polynomial. Use the values of A, B and C, we may find that:

B2-4AC=4m

(34).

where is the discriininant expressed by 6.80 in the text book. Then we have:

2

- m,

2 L'2 m I + m _w1.

U Neglecting thruster forces. Then we can write:

F0=F

Further; F can be written as:

F = F0 0.075 J_ps U

(log10 R - 2)2 2

Then; due to equation 6.79 in the text book:

1.r C S

c=

I =lm

cr 0.075 .

!

ui2 2 f m

D a2

m yy / 112 I '4i2 m M

_-a

m ,

and i is given equation 6.80 in the text book.

Typical values of C is 0.032. This is shown in reference Liapis (1985). Then we have:

(41)

In that way a full loaded tanker in current is usually unstable. (39) also shows the scale effect. A model may be stable in current while the corresponding ship may be unstable.

L (log10 R - 2)2 _ALS

g) Influence on the stability of hull geometry and propeller:

Hull geometry: Center of gravity should be located far away from the fairlead. Hull geometry influences as well as and thus the stability.

Propeller: Increased propeller thrust (negative) means increased F0 and better stability.

a) To get an idea when the total force has its maximum, let us again consider incident deep sea regular sinusoidal waves. We can write

=

!.ecos(wt

- kx)

U1

= (çesin(U)t - kx)

a1 = ____ = (±2çecos(wt + kx)

atax

Force on a strip dz is then gven as (x=0)

dF 3dZCM cJ)2çecos o)t ₊ *CD D dzco2 (e sincot sinot

where k is the wave number, 0) the circular frequency of oscillation, t the time variable and the wave amplitude of the undisturbed waves. The absolute sign has no influence on the maximum force, only on the direction of the force. In that way one may ignore the absolute sign.

Differentiate with respect to time and set the expression equal to zero. The

absolute sign will not cause trouble. Then:

-_(dF) =

-

'dzC w3esinwt

dt 4

p_CD Ddz w e 2sincotcoswt . Co ₍₃₎

Solution exercise 7.1

pDdz CDO3

esinCot

There are two stationary points (Ut =0

CM D

_{_____ + COSO)t}

T z;

=0

ItC%. D COS(L)t = - _____

,_T P p- kz

'-'1) ..ae

Find the corresponding local maximum forces as:

dF. =

'13dz Co2ç ekz

7C1 D

p7rD2 _dzC1

_{()2 Ce_}

4 -r 2. C0 Ddzo)2 e 2 = 2. C0 Ddz (02 e2

Investigate condition when

dFr&.r: >

1-'2

itC., D

C1, P

dzC,2

_{e >} 2.00dz w2

1-

D (7) (8) = pCr,Ddz& e

In the last equality one has used that coswt =

- wben consideting

dF. In addition 1--cos2wt has the maximum value equal to 2. Hence:

(9)

4C0

In the inequality above çe is the radius of the circular path of an undisturbed

/ '2

1-

TCC.%f D