specialization in ocean energy

Specialization in Ocean Energy

MODELLING OF WAVE ENERGY CONVERSION

António F.O. FalcãoInstituto Superior Técnico,

Universidade de Lisboa2014

http://www.google.pt/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=x8rfOAuKQFVSLM&tbnid=Dp3KaAL-0TIWOM:&ved=0CAUQjRw&url=http://cfcul.fc.ul.pt/&ei=9Q0CU6vAGci80QXZzYEQ&bvm=bv.61535280,d.bGQ&psig=AFQjCNEoSWCNq6AG3VPfOne_sETxesQXRA&ust=1392729972868138

http://www.google.pt/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&docid=x8rfOAuKQFVSLM&tbnid=Dp3KaAL-0TIWOM:&ved=0CAUQjRw&url=http://cfcul.fc.ul.pt/&ei=9Q0CU6vAGci80QXZzYEQ&bvm=bv.61535280,d.bGQ&psig=AFQjCNEoSWCNq6AG3VPfOne_sETxesQXRA&ust=1392729972868138

https://www.google.pt/imgres?imgurl=http://cfcul.fc.ul.pt/Templates/img/logos/ULISBOA_VERTICAL_RGB.jpg&imgrefurl=http://cfcul.fc.ul.pt/&docid=x8rfOAuKQFVSLM&tbnid=Dp3KaAL-0TIWOM:&w=1202&h=1692&ei=9Q0CU6HeGeKy0QWUlYDACg&ved=0CAIQxiAwAA&iact=c

PART 3MODELLING OF OSCILLATING

BODY WAVE ENERGY CONVERTERS

Oscillating Water Column(with air turbine)

Oscillating body(hydraulic motor, hy-draulic turbine, linear electric generator)

Overtopping(low headwater turbine)

Floating

Submerged

Heaving: Aquabuoy, IPS Buoy, Wavebob, PowerBuoy, FO3

Pitching: Pelamis, PS Frog, Searev

Heaving: AWS

Bottom-hinged: Oyster, Waveroller

Fixed structure

Shoreline (with concentration): TAPCHAN

In breakwater (without concentration): SSG

Floating structure (with concentration): Wave Dragon

Fixed structure

Floating: Mighty Whale, BBDB

Isolated: Pico, LIMPET, Oceanlinx

In breakwater: Sakata, Mutriku

Wave Energy Converter Types

RESONANCE

x

y

z

heave 3

yaw 6

The six modes of oscillation of a rigid body

Characteristic scales

awA

)1(2 :bodies Large Oaka

Most ships“Large” WECs

1

)1(

aA

Oka

w Inviscid linearized diffraction theory applicable

Wave field of a single heaving body

• m = body mass• mg = body weight• In the absence of waves mg = buoyancy force

and• We ignore mooring forces (may be considered

later)• In the dynamic equations, we consider only

disturbances to equilibrium conditions; body weight does not appear

0


Wave field II: Diffracted wave field due to the presence of the fixed body• satisfies bottom condition and free-surface condition

d

Wave field I: Incident wave field• satisfies bottom condition and free-surface condition

i

Wave fields I + II: • satisfies also condition on fixed body wetted surface

di

Snnid on

tp di

e

)(

S eze Spnf dbody on force Excitation

due to wave fields I and IIdue to wave fields I and II


Wave field III: Radiated wave field of moving body• satisfies bottom condition, free-surface condition and

condition on wetted surface of heaving body

Sntn z

r dundisturbeon dd

due to wave field IIIdue to wave field III

S ezr Spnf dbody on forceRadiation

tp r

r

If, in the absence of incident waves, the body is fixed at , the buoyancy force does not balance the body weight.The difference is a hydrostatic restoring force .

For small displacement , it is

Hydrostatic restoring force

0

stf

csst Sgf csSarea

Dynamic equation for heaving body motion

PTO2

2

dd ffff

tm stre

massacceleration

excitationradiation

hydrostatic/restoring

PTO

Frequency-domain analysis of wave energy absorption by a single heaving body

LINEAR SYSTEM

input outputtXx i

0etYy i

0eamplitudescomplex

0

0

YX

output andinput between difference phasea is therepositive, realnot is If 00 XY

Our WEC is a linear system if the PTO is linear

Linear PTO: linear spring and/or linear damper

,dd

PTO Kt

Cf

damping coef. spring stiffness

Frequency-domain analysis of heaving body

)i(exp)( xktzii Incident wave

The system is linear:

trr

tee

t FtfFtfXt iii e)(,e)(,e)(

Complex amplitudes

tcsre

t KXCXXgSFFXm ii2 eie

PTO2

2

dd ffff

tm stre

Kt

Cf dd

PTO

ecsr FKXCXXgSFmX i2


Decompose radiation force coefficient: XBAFr )i( 2

ecs FXKgSCBAm )()(i)(2

added mass radiation damping coef.

ExerciseShow that the radiation damping coefficient B cannot be negative.


The hydrodynamic coefficients are related to each other:

mass addedA tcoefficien dampingradiation B

amplitude unit waveper amplitude force excitationw

e

AF

body

d)(

)(42

2 khDgkBHaskind relation:

khkh

khkhD tanh2sinh

21)(

d)(

4 1)( water deepIn 2

3

3

gBkhD

yyyBAA d)(2)()(

0 22

yyAyAB d)()(2)(

0 22

2

Kramers-Kronig relations:


Calculation of hydrodynamic coefficients:mass addedA tcoefficien dampingradiation B

amplitude unit waveper amplitude force excitationw

e

AF

• They are functions of frequency

• Analytical methods for simple geometries: sphere, horizontal cylinder, plane vertical and horizontal walls, etc.

• Commercial codes based on Boundary-Element-Method BEM for arbitrary geometries, several degrees of freedom and several bodies: WAMIT, ANSYS/Aqua, Aquaplus, …

Absorbed power and power outputInstantaneous power absorbed from the waves = vertical force component on wetted surface times vertical velocity of body

t

gStftftP csre dd)()()(abs

Instantaneous power available to PTO = force of body on PTO times vertical velocity of body

tK

tCtP

dd

dd)(PTO

PPP PTOabs :average In time 22PTO 2

1 XCPP

22

abs 2281

BFUBF

BPP e

e

itybody veloc of amplitudecomplex is i XU

Conditions for maximum absorbed power2

2abs 228

1B

FUBFB

PP ee

Given body, fixed wave frequency and amplitude

fixed

eFB

KBU , scoeficient PTOon depends amplitudeVelocity

BFXUP e2

imax velocity in phase with excitation force

ecs FXKgSCBAm )()(i)(2

BFX e2

i

BKgSCBAm cs 2i)()(i)(2

Conditions for maximum absorbed power

BKgSCBAm cs 2i)()(i)(2

Separate into real and imaginary parts:

AmKgScs

CB radiation damping = PTO damping

resonance condition

Analogy

mK

nsoscillatio free offrequency

22

abs 2281

BFUBF

BPP e

e 2max 8

1eF

BP

Capture or absorption width

Avoid using efficiency of the wave energy absorption process, especially in the case of “small” devices.

wavePPL Capture or absorption width

Incident waves

capture width

L

May be larger than the physical dimension of the body

Axisymmetric heaving body

d)(

)(42

2 khDgkB

Haskind relation:

khkh

khkhD tanh2sinh

21)(

d)(

4 1)( water Deep 2

3

3

gBkhD

offunction anot is :body icAxisymmetr

)(2 2

2

khDgkB

3

23

2 gB

(deep water)


)(2 2

2

khDgkB

2max 8

1eF

BP

kkhDAgP w

4)(22

max

Maximum capture width foran axisymmetric heaving buoy

)(41 22

wave khDAgP w

21

wave

maxmax

kPPL

Maximum capture width foran axisymmetric surging buoy

kL 2

max

Axisymmetric body with linear PTO

2

Max. capture width


Axisymmetric surging body

Incident waves

Incident waves

Exercise 3.1Hemispherical buoy

in deep water

ka )(* kaA )(* kaB 0 0.8310 0

0.05 0.8764 0.1036 0.1 0.8627 0.1816 0.2 0.7938 0.2793 0.3 0.7157 0.3254 0.4 0.6452 0.3410 0.5 0.5861 0.3391 0.6 0.5381 0.3271 0.7 0.4999 0.3098 0.8 0.4698 0.2899 0.9 0.4464 0.2691 1.0 0.4284 0.2484 1.2 0.4047 0.2096 1.4 0.3924 0.1756 1.6 0.3871 0.1469 1.8 0.3864 0.1229 2.0 0.3884 0.1031 2.5 0.3988 0.0674 3.0 0.4111 0.0452 4.0 0.4322 0.0219 5.0 0.4471 0.0116 6.0 0.4574 0.0066 7.0 0.4647 0.0040 8.0 0.4700 0.0026 9.0 0.4740 0.0017

10.0 0.4771 0.0012 0.5 0

332

332

)()(*

)()(*

aBkaB

aAkaA

222 )*2(* Tgaka

ga *

agTT *

2125*ga

CC

Dimensionless quantities

2125*ga

CC

agTT *

No spring K = 0

• Reproduce the curves plotted in the figures by doing your own programming.

• Compute the buoy radius a and the PTO damping coefficient C that yield maximum power from regular waves of period T = 9 s. Compute the time-averaged power for wave amplitude .

• Assume now that the PTO also has a spring of stiffness K that may be positive or negative. Compute the optimal values for the damping coefficient C and the spring stiffness K for a buoy of radius a = 5 m in regular waves of period T = 9 s. Explain the physical meaning of a negative stiffness spring (in conjunction with reactive control).

m1wA

Exercise 3.2. Heaving floater rigidly attached to a deeply submerged body

WaveBob, Ireland

Time-domain analysis of a single heaving body

• If the power take-off system is not linear

then the frequency-domain analysis cannot be employed.

• This is the real situation in most cases.

• In particular, even in sinusoidal incident waves, the body velocity is not a sinusoidal function of time.

• In such cases, we have to use the time-domain analysis to model the radiation force.

,dd

PTO Kt

Cf


• When a body is forced to move in otherwise calm water, its motion produces a wave system (radiated waves) that propagate far away.

• Even if the body ceases to move after some time, the wave motion persists for a long time and produces an oscillating force on the body which depends on the history of the body motion.

• This is a memory effect.

This dependence can be expressed in the following form:


)()(d)()()( tAtgtft

rr

How to obtain the memory function ?)(rg

ttrr XBAFtf i2i e)(i)(e)(

.e)(,ei)( i2i XtX

Take

de)(ie)(i)()( ii2

tr

t tgXXBAAWe obtain

tsChanging the integration variable from to , we have

ssgBAAi sr de)()()()( i

0

see later why


ssgBAAi sr de)()()()( i

0

Since the functions A, B and are real, we may writerg

sssgB r dcos)()(0

)(rg Note that, since if finite, the integrals vanish as , which agrees with .0)(,0)( AB

Invert Fourier transform

dcos)(2)(0

sBsgr

ssgB sr de)(

21)( i

Assume to be an even function)(sgr

sssgAA r dsin)()()(0


PTO)(d)()()()()( ftgStgtftAm cst

re

dcos)(2)(0

sBsgr

This has to be integrated in the time domain from initial conditions . and for

)()(d)()()( tAtgtft

rr


re


Note: since the “memory” decays rapidly, the infinite integral can be replaced by a finite integral. In most cases, three wave periods (about 30 s) is enough.

Adopted time steps are typicall between 0.01 s and 0.1 sThe convolution integral must be computed at every time step

Integration procedure:

• Set initial values (usually zero)

• Compute the rhs at time

• Compute from the equation

• Set

•

• Compute etc.

. and for

0t

)( 0t

ttt 01

tttttttt )()()(,)()()( 001001

)( 1t

Wave energy conversion in irregular waves

• Real ocean waves are not purely sinusoidal: they are irregular and largely random.

• Here, we consider only frequency spectra.

• The distribution of the energy of these wavelets when plotted against the frequency and direction is the wave spectrum.

• In linear wave theory, they can be modelled as the the superposition of an infinite number of sinusoidal wavelets with different frequencies and directions.


A frequency spectrum is a function )( fS f s)m (units 2

ffS f d)( is is the energy content within a frequency band of width equal to df


A frequency spectrum is a function )( fS f s)m (units 2

is is the energy content within a frequency band of width equal to df

ffSg f d)(


The characteristics of the frequency spectra of sea waves have been fairly well established through analyses of a large number of wave records taken in various seas and oceans of the world.

Goda proposed the following formula for fully developed wind waves, based on an earlier formula proposed by Pierson and Moskowitz

.)(675.0exp1688.0)( 4542 fTfTHfS eesf

.)(1052exp6.262)( 4542 ees TTHS

periodenergy height t wavesignifican

e

s

TH

d)(d)( SffS f


.)(1052exp6.262)( 4542 ees TTHS

periodenergy height t wavesignifican

e

s

TH

04 mH s 0

1mmTe

ffSfmnm nnn d)(:order ofmoment spectral is

0

Wave energy absorption from irregular waves

In computations, it is convenient to replace the continuum spectrum by a superposition of a finite number of sinusoidal waves whose total energy matches the spectral distribution.

Simulation of excitation force in irregular waves

)ˆ(iexp)ˆ()( ,1

iii

N

iie tAtf

)ˆcos()ˆ()( ,

1iii

N

iie tAtf

Divide the frequency range of interest into N small intervals of width and set

1 ii

iii 1

iiiiwii SAAS ,,2

,, 2or 21

)ˆ(, ii SS

)(ˆ 121

iii

or

)2,0( interval in phases random are i

Wave energy absorption from irregular waves

Simulation of excitation force in irregular waves

)ˆ(iexp)ˆ()( ,1

iii

N

iie tAtf

)ˆcos()ˆ()( ,

1iii

N

iie tAtf

Oscillating body with linear PTO and linear damping coefficient C . Averaged power over a long time:

)())ˆ((ˆi))ˆ((ˆ

)ˆ()ˆ(

2,

KgSCBAm

AX

csiiii

iii

,.if0

if21

d)ˆsin()ˆsin(1lim0

ji

jitttt

tjjii

tNote that:

2

1

22

)(21

dd

i

N

iiPTO XC

tPP

In singe-body WECs, the body reacts against the bottom. In deep water (say 40 m or more), this may raise difficulties due to the distance between the floating body and the sea bottom, and also possibly to tidal oscillations.

Wave energy absorption by 2-body oscillating systems

• Two-body systems may then be used instead.• The energy is converted from the relative motion between two bodies

oscillating differently.• Two-body heaving WECs: Wavebob, PowerBuoy, AquaBuoy


• The coupling between bodies 1 and 2 is due firstly to the PTO forces and secondly to the forces associated to the diffracted and radiated wave fields.

• The excitation force on one of the bodies is affected by the presence of the other body.

• In the absence of incident waves, the radiated wave field induced by the motion of one of the bodies produces a radiation force on the moving body and also a force on the other body.

,dd

PTO11,12,11,1,21

2

1 fSgffft

m csrre

.d

dPTO22,21,22,2,2

22

2 fSgffft

m csrre


,dd

PTO11,12,11,1,21

2

1 fSgffft

m csrre

.d

dPTO22,21,22,2,2

22

2 fSgffft

m csrre

Linear system. Frequency domain analysis

).(d

)(d21

21PTO

Kt

Cf

)2,1,(e)(,e)(,e)( i,,

i,,

i jiFtfFtfXt tijrijr

tieie

tii

)2,1,()i( 2, jiXBAF jijijijr Decompose radiation force:

21122112 , that proved becan It BBAA

negative becannot and :Note 2211 BB

Wave energy absorption by 2-body oscillating systems.Linear system. Frequency domain analysis

,)(i

)()(i)(

1,212122

11111112

e

cs

FXKCBA

XKSgCBAm

.)(i

)()(i)(

2,112122

22222222

e

cs

FXKCBA

XKSgCBAm

))(()( :power ousInstantane 21212

21PTO KCP

. :power averaged-Time 221

221

PTO XXCPP

d)(

)(42

2 iii khDgkB

)(2 2 khDgkB i

ii

water).(deep2 3

3

gB i

ii

water)(deep d)(4

23

3

iii gB

Relationships between coefficients: radiation damping force and excitation force

Axisymmetric systems:

Wave energy absorption by 2-body oscillating systems.Non-linear system. Time domain analysis

Excitation forces:

)()(d)()()( ,, tAtgtf jijjt

ijrijr

dcos)(2)()(0,, tBsgsg ijjirijr

,)()(d)()(

d)()(dd))((

PTO212212,

11,111,1,21

2

111

ftAtg

Sgtgft

Am

tr

cst

re

.)()(d)()(

d)()(d

d))((

PTO112112,

22,222,2,22

2

222

ftAtg

Sgtgft

Am

tr

cst

re

Exercise 3.3. Heaving two-body axisymmetric wave energy converter

Bodies 1 and 2 are axisymmetric and coaxial.

The draught d of body 2 is large:

oft independen is 0,0,0

22

2212122

ABBAFe

3031.3 :1body ofpart submerged of volume

60 angle-semi cone4.0

a

abca

322 6897.0 bA

The PTO consists of a linear damper, and no spring.


311*

11311*

11 ,a

BBa

AA

agagTT 12*


Discuss the advantages and limitations of a wave energy converter based on this concept

2125

2

*

tcoefficien damping PTO and ratio of valuesoptimal thefind *given For

draught offunction as mass Compute domainfrequency in the equations governing the Write

gaCC

adTdm

Oscillating systems with several degrees of freedom

The theory can be generalized to single bodies with several degrees of freedom or groups of bodies.

For the general theory, see the book by J. Falnes

Time-domain analysis of a heaving buoy with hydraulic PTO

Hydraulic circuit:• Conventional equipment• Accommodates large forces• Allows energy storage in gas

accumulators (power smoothing effect)

• Relatively good efficiency of hydraulic motor

• Easy to control (reactive and latching)

• Adopted in several oscillating-body WECS

• PTO is in general highly non-linear (time-domain analysis)



re

raccumulato LPin pressureraccumulato HPin pressure

2

1

PTO

pp

Sf

pc

damping" Coulomb" control) reactivein (except if movecannot piston andBody 21

ppp

HPin volumegas of change of rate)()(:ibleincompress is oil and rigid are r wallsaccumulato andduct If

motor) (hydraulic oil of rate flow volume)(

(piston) oil of rate flow volumedd)(

tqtq

tqt

Stq

m

m

c


process) isentropictely (approximapressure gas torelated is volumegas rs,accumulatoIn

LPin volumegas of change of rate

HPin volumegas of change of rate)()(

tqtq m

parameter control )()()(

:)()( difference pressure andmotor in )( rate flow oilbetween iprelationsh define :control PTO

212

21

GGtptpStq

tptptq

cm

m

nitrogen) (air, 4.1 constant,1

1 v

p

ccp


(gas) m17.5,kg30kg,150

s/kg,106.0,m01767.0m,53

021

62

Vmm

GSa c

s11,m5.1 es TH

s11 m,3 state Seam5 radius Sphere

es THa

kW4.178 kN 647 force External

dampedOptimally

PkW1.83 kN 200 force External

dampedUnder

P kW0.97 kN 1000 force External

dampedOver

P

s11 m,3 state Seam5 radius Sphere

es THa

(m) x

(kW) P

(m) x (m) x

(kW) P (kW) P(kW) P

(m) x

HP gasaccumulator

LP gasaccumulator

Cylinder

Valve

B

A

Buoy

Motor

HP gasaccumulator

LP gasaccumulator

Cylinder

Valve

B

A

Buoy

Motor


Underdamping and overdamping

s11,m5.1 es TH


(gas) m17.5,kg30kg,150

s/kg,106.0,m01767.0m,53

021

62

Vmm

GSa c

2output sHP

2absorbed sHP

PTO power


4800 5000 5200 5400 5600 5800 600010 ts0

10

20

30

40

P tH s2Wkm2

m 1sH

m 4sH2

output sHP

The smoothing effect decreases for more energetic sea states


Phase control by latching

Kjell Budall (1933-89)

Johannes Falnes

Pioneers in control theory of wave energy converters.

J. Falnes, K. Budal, Wave-power conversion by power absorbers. Norwegian Maritime Research, 6, 2-11, 1978.

They introduced the concept of phase-control by latching:

How to achieve phase-control by latching in a floating body with a hydraulic power-take-off mechanism?

Introduce a delay in the release of the latched body.

How?

Increase the resisting force the hydrodynamic forces have to overcome to restart the body motion.

Time-domain analysis of a heaving buoy with hydraulic PTOPhase control by latching

Phase-control by latching: release the body when )1()( exceedsbody on force ichydrodynam 21 RppRSc


Phase control by latching

Two control variables

G control of flow rate of oil through hydraulic motor

R release of latched body

No latchingR = 1

Regular waves

kW0.55s/kg1086.0,1 6

PGR

s9,m667.0 TAw

Optimal latchingR > 1

Regular waves

kW1.206s/kg107.7,16 6

PGR

s9,m667.0 TAw

NO LATCHING OPTIMAL LATCHING

Irregular waves, Te = 9 s

Irregular wavesTe = 9 s

No latchingR = 1

2

6

kW/m3.10

s/kg107.0,1

P

GR

Irregular wavesTe = 9 s

2

6

kW/m5.28

s/kg102.4,16

P

GR

Optimal latchingR > 1

NO LATCHING OPTIMAL LATCHING

Latching control

• May involve very large forces

• May be less effective in two-body WECs

END OFPART 3

MODELLING OF OSCILLATING BODY WAVE ENERGY

CONVERTERS

specialization in ocean energy

Documents

single heaving body

diffracted wave field

wave field iiidue

body massmg

body weightin

wetted surface of heaving

incident wave fieldsatisfies

fixed body wetted surface