morison's equation

7/28/2019 Morison's Equation

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2/25

8-2

Fig. 8.2 Area projection on a vertical plane

21

2D

F Au (8.2)

Where =mass density of fluid

A= area of object projected on a plane held normal to flow direction

u = flow velocity

Introducing the constant of proportionality, CD, and assuming a steady, uniform flow in a

viscous fluid, we have

21

2D DF C Au= (8.3)

whereDC is coefficient of drag. Its value depends on body shape, roughness, flow

viscosity and several other parameters.

Fig. 8.3 Particle velocities


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8-3

Because the direction of wave induced water particle velocity reverses after every half

cycle, we write,

1

2D DF C Au u= (8.4)

The force of inertia is proportional to mass times the fluid acceleration:

.

IF v u

uVFI &..

where V = volume of fluid displaces by the object.

.

u = acceleration of fluid

Hence,

.

I mF C v u= (8.5)

Wherem

C =Coefficient of Inertia. It depends on shape of the body, its surface

roughness and other parameters.

Most of the structural members are circular in cross section. Hence,

1'

2D DF C DL u u=

2 .

'4I m

dF C L u

=

Because u and.

u vary along L and further considering unit pile length i.e. L=1. Hence,

2 .1

2 4T D m

dF C Du u C u

= + (8.6)

where,T

F = in-line (horizontal) force per meter length at member axis at given time

at given location.

1

2DC Du u = in-line (horizontal) water particle velocity at the same time at the

same location.


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8-4

2 .

4m

dC u

is in-line (horizontal) water particle acceleration at the same time at

the same location.

Note that ( )cosu f = and ( ).

sinu f = . Hence u and

.

u are out of phase by0

90 and

are not maximum at the same time.

Basically CD

and Cm

are functions of size and shape of the object. If that is fixed

then they depend on Keulegan-Carpenter number, Reynolds number as well as

roughness factor.

Keulegan-Carpenter number: KC

It is basically a ratio of maximum drag to maximum inertia. We have,

( ) 2maxmax

1

2F C Du

D D=

where, 2maxu

( )2 2 2 22 2

coshcos

sinh

H k d z

T kd

+=

( ) maxmax

2 .

4

dF C u

I m

=

where( )2

max 2

. 2 coshsin

sinh

H k d zu

T kd

+=

At z=0,

( )( )

max

max

1 cosh

sinh

F C H kdD D

C D kd F mI

= = max2

1C u TD

C Dm

(8.7)

The ratio of maximum drag to maximum inertia can thus be taken as proportional to

maxu T

D= Where

maxu =Maximum velocity in the wave cycle

T= wave period

D= Diameter

The above ratio also stands for (Total horizontal motion of the particle / Diameter).


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8-5

If KC < 5 then inertia is dominant,

If KC >15 then drag is dominant and regular eddies are shed at downstream section.

Fig. 8.4 Eddy shedding

at frequency ofSv

feD

= where S = Strouhal No. 0.2.

Alternate eddy shedding gives rise to alternate lift forces due to pressure gradient across

the wake.


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8-6

Fig. 8.5 Variation of CD and CM against KC

Reynolds Number, Re:

It is the ratio of the inertia force to the viscous force, i. e., R e:

maxu D

v= (8.8)

10 100

1

0

CD

Kc

Cm

0

Re X 10

1


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8-7

Fig. 8.6 Variation of CD and CM against Re

Roughness Factor:

Fig. 8.7 Encrustation around cylindrical members

Structural members are in course of time covered by sea weeds, barnacles, shell fish etc.Due to this, effective diameter changes, effective mass increases, flow pattern, eddy

structure changes . Finally the wave force also changes. Lab studies have shown that

mC does not change much.

DC changes appreciably and can become 2 to 3 times more

than the initial value.

2

1

0

CD

Re X 1055


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8-8

Fig. 8.8 Effect of roughness on CD and CM

Scatter inD

C ,m

C values: Many laboratory and field studies have been made to

assess the effects of all unaccounted factors like eddy shedding , past flow history, initial

turbulence , wave irregularity directionality, local conditions , data reduction techniques.

But experiments are inconclusive.

Experiments to evaluateD

C ,m

C are performed in the following way.

1

2

Cm

0Re X 10

5

Rough

Smooth

20

CD

Re X 10

5

Rough

Smooth


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8-9

Fig. 8.9 Flow chart to obtain CD and CM through lab measurement

Almost all experiments suffer from widely scattered values. Major reasons of the scatter

are: (1) use of either steady/ oscillatory / wavy flow, (2) difficulty in achieving high Re

( 710 ), (3) wave theories over predict velocity, (4) definition of Re is arbitrary, (5) waves

are irregular, henceD

C ,m

C are large, (6) use ofu

t

(not

du

dt) overestimate forces, (7) no

accounting for directionality, current, 3-D flow.

Recommendations:

1) For Indian conditions DC =0.7 ; Cm=2 are generally used.

2) DnV :D

C =0.7-1.2 ; Cm=2

3) A.P.I. : DC =0.6-1.0 ; Cm=1.5-2

4) Shore Protection Manual: DC -Refer Fig. ; Cm=1.5 if Re>5 x510

=2 if Re


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8-10

4

= , at a location 10m below SWL . The water depth is 60m.

TakeD

C =1; Cm=2; Use linear theory.

=1030 3kg

m

Solution :

max max max

1

2D D

F C D u u=

( )coshcos

sinh

H k d zu

T kd

+=

( )2

(5) cosh 5080

.12(10) sinh (60)80

=

=0.717 m/s

max

1(1)1030(1) 0.717 0.717

2DF =

=264.76 N/m.

( ) maxmax

2 .

4

dF C u

I m

=

( )2

max 2

. 2 coshsin

sinh

H k d zu

T kd

+=

( )2

2

22 (5) cosh 50

80 .12

(10 )sinh (60)80

=

=0.452

m

s


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8-11

maxIF =

2(1 )2(1030) (0.45)

4

=728 N/m.

( )coshcos

sinh

H k d zu

T kd

+=

=(5)(25.3869)

.cos(10)(55.1544) 4

=0.5067 m/s

( )2

max 2

. 2 coshsin

sinh

H k d zu

T kd

+=

=2

2

2 (5)(25.3869).sin

(10 )(55.1544) 4

=0.3182

m

s

F2 .1

2 4D m

dC Du u C u

= +

( ) ( )( ) ( )( )

( )2

2 111 1030 1 0.5067 2 1030 0.318

2 4

= +

=646.72 N/m


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8-12

8.2 Total Wave Force on the Entire Member Length

Fig. 8.10 Variation of drag and inertia over a vertical

Consider a vertical located at x=0 as shown above.

Consider Linear Theory

2

cos2

H t

T

=

as x=0

( )cosh 2cos

sinh

H k d z tu

T kd T

+ =

( )2.2

2 cosh 2sin

sinh

H k d zu tu

t T kd T

+ =

( )tt

hd

zdk

T

HDCuDuCF

TTDDD

22

2

2

2

22

coscossinh

cosh

2

1

2

1 +==

When t=0, ( )maxDD

FF

SWL

0

Z=-d

Drag

X

Z

Inertia

Z =-z


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8-13

( )( )t

hd

zdk

T

HdCu

dCF

TmmI

2

2

22.2

sinsinh

cosh2

44

+==

When t=0, ( )maxII

FF

But

( )maxI

F When ( )tT2sin =1 or when t

T2 =

2

Or when t=4

T

At this time tT2cos =

2cos

=0 Hence DF =0

Note: When ( )maxDF occurs IF =0

When ( )maxI

F occurs DF =0

( )maxD

F occurs after time4

Twhen ( )

maxIF occurs.

If is small, =0 , if is not small , =T

tH 2cos

2

+=

d

I

d

DT dzFdzFF

( ) ( )

+++=

d

I

d

D dzzdFdzzdFM

Hence total horizontal force on entire member length at any time t : TITDT FFF +=

( )2 2 2

2 2

cosh1 2 2cos cos

2 sinhTD D D

d d

H k d z t tF F dz C D dz

T kd T T

+ = =

( )( )

2 2

2

2 2

cos cos1cosh

2 sinhD

d

H t tC D k d z dz

T kd

= +


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8-14

( ) ( ) ( )2 2

2 2

cos cos sinh 21

2 sinh 2 4D

d

H t t k d z k d zC D

T kd k k

+ + = +

[Using 2sinh2

cosh2 4

x xx = + ]

( )( ) ( ){ }

2 2

2 2

cos cos1 12 sinh 2

2 sinh 4D

d

H t tC D k d z k d z

T kd k

= + + +

( ) ( ){ }( )

2 2

2

2 sinh 21 1cos cos

2 4 4 sinhD

d

k d z k d zHC D t t

k kd

+ + +=

( )( ) ( ){ }

( )2

2

2 sinh 2cos cos

32 sinh

DTD

z

k d z k d zC DF H t t

k kd

=

+ + +=

(8.9)

( )( )

22

2

2 coshsin

4 sinhTI I m

d d

H k d zdF F dz C t dz

T kd

+= =

( ) ( )2 2 sin sinh2

4 4 sinhm

d

t k d zd HC

kd k

+ =

Hence ( ) ( ) ( )

22 sinh

1 sin4 2 sinh

TI m

z

k d zdF C H t

k kd

=

+ =

(8.10)

Similarly,T DT IT M M M= +

( )( ) ( ) ( ) ( ){ } ( )

22

2 22 sinh 2 cosh 2 2 1 cos cos

64 sinh

DDT

HC DM k d z k d z k d z k d z t t

k kd

= + + + + + +

( ) ( ) ( ) ( )( )

22

2

sinh cosh 1sin

2 4 sinh sinh

mIT

H k d z k d z k d zC dM t

k kd kd

+ + + + =

(8.11)


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8-15

EXAMLPE:

Obtain variation of total horizontal force and moment at the sea bed with time for a

circular vertical pile of diameter 1.22 m extending into a water depth of 22.9 m. The

wave height is 10.67m and the wavelength is 114.3m. TakeD

C =1 andm

C =2 .

=10.063

KN

m.

What are the maximum force and moment values?

Use Linear Theory.

Consider two cases (a) Integration up to SWL.

(b) Integration up to Free surface.

SOLUTION:

K=2 2

114.3L

= =0.05497cycles/m

{ } ( ){ }1122tanh 9.81(0.05497) tanh 0.05497 22.9gk kd = = =0.6773 rad/s

( )( ) ( ){ }

( )2

2

2 sinh 2cos cos

32 sinh

DTD

z

k d z k d zC DF H t t

k kd

=

+ + +=

= ( )( ) ( ){ }

( )2

2

2 sinh 2cos cos

32 sinh

Dk d k d C D

H t tk kd

+

( )( )( ){ }

( )( )

2

2

2(0.05497)(22.9) sinh2(0.05497) 22.91(10.06)1.220.6773 10.67 cos cos

32(9.81)(0.05497) sinh (0.05497) 22.9t t

+=

=123.022 ( )cos cost t KN

( )

( )

( )

22 sinh1

sin4 2 sinhTI mz

k d zd

F C H t k kd

=

+ =

( )( )

( )2

2 sinh1sin

4 2 sinhm

k ddC H t

k kd

=


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8-16

( )( )( )

( )( )

221.3310.06 1

2 10.67 0.6773 sin9.81 4 2 0.05497

t

=

=-106.74 ( )sin t KN

Hence

TF =123.022 ( )cos cost t -106.74 ( )sin t

Vary t=0, T

2 t

T

=0,

2.T

T

t=0, 2

=0, 6.284=0, 1,2, ,7

tTD

F (KN)TI

F (KN)T

F (KN)

0 123.02 0 123.02

1 35.01 -89.82 -53.91

2 -21.31 -97.06 -118.37

3 -120.57 -15.06 -135.64

4 -52.56 80.78 28.22

5 9.9 102.86 112.26

6 113.42 29.83 143.24

7 69.92 -70.13 -0.21


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8-17

-200

-150

-100

-50

0

50

100

150

200

0 2 4 6 8

wt

Ft

Series1

We can express: cos cos sinF C K = + . For maximum conditions this equation

can be worked out using as: 10 sin2

F K

c

= =

.

8.3 Wave Forces Using Stokes (V) Theory

Water particle kinematics are calculated at every m length of the vertical

structural member (at its center along the immersed length of the member axis) using the

Stokes Fifth Order theory. Corresponding forces are worked out using the Morrisons

equation at every such segment and then they are added up to cover the full member

length. For a typical case of wave attack shown below, the results are further indicated in

the following figure:


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8-18

Fig. 8.11 Calculation of total wave force

DC =1 and

mC =2

8.4 Calculation Of Wave Forces Using Deans Theory

For circular vertical piles, based on Deans theory and Morisons equation, it is possible

to express approximately the total maximum force within the wave cycle as:

2

m m DF gC H D= (8.12)

Where,m

= Coefficient to be read from curves plotted for various values of

m

D

C DW

C H= (8.13)

Similarly the total maximum moment at the base is:

2m m DM gC H Dd = (8.14)

Wherem

=Coefficient to be read from curves

SWL

35

375

75

4

Force


19/25

8-19

LIFT FORCES:

For high drag ( 15c

K > ) there is regular and alternate eddy shedding on the

downstream side on both sides of cylinder at a frequency.

eddyshedding

svf

D=(8.15)

where s = Strouhal No. 0.2, = kinematic viscosity of sea water, D = diameter.

This gives rise to lift force given by:

1

2L LF C Du u= (8.16)

WhereLC is Lift coefficient = ( )cf K

DC If cK >20

If cK


20/25

8-20

n n n nF C V V Ka= + (8.17)

Where2

D

DC C =

2

4m

dK C =

Wheren

V andn

a are normal components of total velocity (V) and acceleration (a)

Fig. 8.13 Normal to axis force

nV (or na ) lies along the line of intersection of the two planes

( )'nV cx V xc=

( )'na cx a xc=

If c is unit vector along axis andx

c , yc , zc are its direction cosines and if,

n nx ny nzV V i V j V k = + +

cV

0

Plane containing

cylinder axis and totalvelocity vector V

Plane normal to cylinder

axis at 0

x

yz


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8-21

n nx ny nza a i a j a k = + +

x y zc c i c j c k = + +

then evaluating the products,

2

2

2

1

1

1

nx x x y x z x

ny x y y y z y

nz x z y z z z

V c c c c c V

V c c c c c V

V c c c c c V

=

and { }1

2 22 2 2 2 2 2

n nx ny nz x y z x x y y z zV V V V V V V c c c c c c = + + = + + + +

Thus we get,

x nx nx

y n ny ny

z nz nz

F V a

F c V V K aF V a

= +

(8.18)

Where

2

2

2

1

1

1

nx x x y x z x

ny x y y y z y

nz x z y z z z

a c c c c c a

a c c c c c a

a c c c c c a

=

Note: If wave theory is used then 0y y

V a= = .

Fig. 8.14 Special case of a pipeline

Further for a horizontal member, 0x zc c= = and 1yc =

z

y

x


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8-22

Note on calculation of direction cosines,

Fig. 8.15 Direction cosines

c c i c j c k x y z

= + + ; where,

sin cos

sin cos

cos

x

y

z

c

c

c

=

=

=

A note on flexible cylinders

The previous discussion was based on the assumption that the cylinder on which the

force was exerted was rigidly held at its bottom. On the contrary if it is free to move

appreciably with waves, not only the exact volume of water displaced by the cylinder

contributes to the inertia force but also some volume surrounding it behaves as one withthe cylinder and contributes to the force due to inertia. This volume is some fraction of

the displaced volume V. The resulting inertia force is thus:

x

y

z

c

Vn0

1c =

coszc =

sin coscx

=

sin siny

c =


23/25

8-23

Fig. 8.16 Added mass effect

.

( )I F aF C C V u= + (8.19)

WhereF

C = Froude-Crylov coefficient and

aC = Coefficient of added mass =1 (theoretically)

Hence Total IF = Froude-Crylov Force + Added mass force

The Froude-Crylov Force is the force required to accelerate the fluid particles within the

volume of cylinder in its absence, whereas, the added mass force is the force due to

acceleration of water surrounding the cylinder and oscillating with it.


24/25

8-24

8.6 Wave Slam:

Fig. 8.17 Wave attack on a Jacket

When wave surface rises, it slams underneath horizontal members near the SWL and then

passes by them. The resulting slamming force (nearly vertical) due to sudden buoyancy

application is given as follows

21

2z s zF C Du= (8.20)

Wheres

C (theoretically for circular cylinder)

The American Petroleum Institute (API) suggests that it should be taken into

consideration to calculate total individual member loads and not to get the global

horizontal base shear and overturning moments. Impulsive nature of this force however

can excite natural frequency of the members creating resonant condition and large

dynamic stresses.

8.7 Limitations of the Morrisons Equation:

1) Physics of wave phenomenon is not well represented in it.


25/25

2) The drag force formula and the inertia force formula involve opposite

assumptions. The former assume that the flow is steady while the latter implies

that the flow is unsteady

3) Real sea effects like transverse forces, energy spreading (directionality) are

unaccounted for.

4) There is a high amount of scattering in values ofD

C andm

C .

5) Inaccuracies in the wave theory based values of water particle kinematics get

reflected in the resulting force estimates.

morison's equation

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