### Time-domain analysis of a single heaving body

```Specialization in Ocean Energy
MODELLING OF WAVE
ENERGY CONVERSION
António F.O. Falcão
Instituto Superior Técnico,
2014
PART 3
MODELLING OF OSCILLATING
BODY WAVE ENERGY
CONVERTERS
Wave Energy Converter Types
Oscillating Water
Column
(with air turbine)
Fixed
structure
Oscillating body
Overtopping
water turbine)
In breakwater: Sakata, Mutriku
Floating: Mighty Whale, BBDB
Floating
(hydraulic motor, hydraulic turbine, linear
electric generator)
Isolated: Pico, LIMPET, Oceanlinx
Heaving: Aquabuoy, IPS Buoy, Wavebob,
PowerBuoy, FO3
Pitching: Pelamis, PS Frog, Searev
Heaving: AWS
Submerged
Fixed
structure
Bottom-hinged: Oyster, Waveroller
Shoreline (with concentration): TAPCHAN
In breakwater (without concentration): SSG
Floating structure (with concentration): Wave Dragon
The six modes of oscillation of a rigid body
z
yaw 6
heave 3
y
x
Characteristic scales
Aw
a

Large bodies : ka  2
ka  O(1)
Aw
 1
a
a

 O(1)
Most ships
“Large” WECs
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   0
• We ignore mooring forces (may be considered
later)
• In the dynamic equations, we consider only
disturbances to equilibrium conditions;
body weight does not appear
Wave field of a single heaving body
Wave field I: Incident wave field i
• satisfies bottom condition and free-surface condition
Wave field II: Diffracted wave field d due to
the presence of the fixed body
• satisfies bottom condition and free-surface condition
Wave fields I + II: i  d
• satisfies also condition on fixed body wetted surface
d
i

on S
n
n
Excitationforceon body f e    nz pe dS
S
due to wave fields I and II
 (i  d )
pe   
t
due to wave fields I and II
Wave field of a single heaving body
Wave field III: Radiated wave field of moving
body
• satisfies bottom condition, free-surface condition and
condition on wetted surface of heaving body
r d 

nz on undisturbe d S
n
dt
Radiationforceon body f r   nz pe dS
S
r
pr   
t
due to wave field III
due to wave field III
Hydrostatic restoring force
If, in the absence of incident waves, the body is
fixed at   0 , the buoyancy force does not balance
the body weight.
The difference is a hydrostatic restoring force f st .
For small displacement  , it is
f st   g Scs
area S cs
Dynamic equation for heaving body motion
d
2
m
dt
2
 f e  f r  f st  f PTO
excitation
mass
acceleration
hydrostatic/
restoring
PTO
Frequency-domain analysis of wave energy
absorption by a single heaving body
input
x  X 0 ei t
LINEAR output
SYSTEM y  Y0 ei t
X0
complexamplitudes
Y0 
If Y0 X 0 is not real positive,thereis a
phasedifferencebetween input and output
Our WEC is a linear system if the PTO is linear
Linear PTO: linear spring and/or linear damper
d
f PTO  C
 K ,
dt
damping coef.
spring stiffness
Frequency-domain analysis of heaving body
Incident wave
i  i ( z) expi(t  k x)
The system is linear:
 (t )  X ei t ,
f e (t )  Fe ei t ,
f r (t )  Fr ei t
Complex amplitudes
m
d 2
dt
2
 f e  f r  f st  f PTO
f PTO
d
 C
 K
dt
  2m Xei t  Fe  Fr   gScs X  iCX  KX ei t
  2mX  Fr   gScs X  iCX  KX  Fe
Frequency-domain analysis of heaving body
Decompose radiation force coefficient: Fr  (2 A  i B) X
  (m  A)  i(B  C)  ( gS
X  F
cs  K )
2
e
damping coef.
Exercise
Show that the radiation damping coefficient B cannot be negative.
Frequency-domain analysis of heaving body
The hydrodynamic coefficients are related to each other:
Fe

 excitationforceamplitudeper unit waveamplitude
Aw
k
4  g 2 D(kh)

2



(

)
d


2kh 

D(kh)  1 
tanhkh

sinh
2
kh


In deep water D(kh)  1 
body
3 
2


B

(

)
d
3 
4  g
Kramers-Kronig relations:
A( )  A() 
2

 0
 B( y )
 y
2
dy
2
B( ) 
2 2


0
A( y)  A()
 y
2
2
dy
Frequency-domain analysis of heaving body
Calculation of hydrodynamic coefficients:
Fe

 excitationforceamplitudeper unit waveamplitude
Aw
• 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 output
Instantaneous power absorbed from the waves = vertical force
component on wetted surface times vertical velocity of body
d
Pabs (t )   fe (t )  f r (t )   gS cs 
dt
Instantaneous power available to PTO = force of body on PTO
times vertical velocity of body
 d
 d
PPTO (t )  C
 K 
 dt
 dt
In timeaverage: Pabs  PPTO  P
1
P  PPTO   2C X
2
1
Fe
2 B
P  Pabs 
Fe  U 
8B
2
2B
2
2
U  iX is complexamplitudeof body velocity
Conditions for maximum absorbed power
1
F
2 B
P  Pabs 
Fe  U  e
8B
2
2B
Given body, fixed wave
frequency and amplitude
2
B
 fixed
 Fe 
VelocityamplitudeU dependson PT Ocoeficients B, K
Pm ax  U  i X 
Fe
2B
velocity in phase with excitation force
  (m  A)  i(B  C)  ( gS
X  F
cs  K )
2
i X 
e
Fe
2B
  2 (m  A)  i ( B  C )  (  gScs  K )  i2 B
Conditions for maximum absorbed power
  2 (m  A)  i ( B  C )  (  gScs  K )  i2 B
Separate into real and imaginary parts:
BC

 gScs  K
m A
resonance condition
Analogy
frequencyof free oscillations  
1
F
2 B
P  Pabs 
Fe  U  e
8B
2
2B
2
Pm ax 
K
m
1
2
Fe
8B
Capture or absorption width
Avoid using efficiency of the wave energy absorption process,
especially in the case of “small” devices.
Capture or absorption width
Incident
waves
L
P
Pwave
L
capture width
May be larger than the physical dimension of the body
Axisymmetric heaving body

k
2


B

(

)
d

2


4  g D(kh)
2kh 

D(kh)  1 
tanhkh

sinh
2
kh


3 
2


Deep water D(kh)  1  B 

(

)
d
3 
4  g
Axisymmetr
ic body :  is not a functionof 
B
 k 2
2 g 2 D(kh )
 3 2
B
2 g 3
(deep water)
Axisymmetric heaving body
Pm ax 
B
1
2
Fe
8B
 k 2
2 g 2 D(kh )
 g 2 Aw2 D(kh)
Pmax 
4 k
Pwave 
1
 g 2 Aw2 D(kh )
4
Maximum capture width for
an axisymmetric heaving buoy
Maximum capture width for
an axisymmetric surging buoy
Lm ax 
Pmax 1 
Lmax 
 
Pwave k 2
2 

k 
Axisymmetric body with linear PTO
Incident
waves

2
Axisymmetric
heaving body
Max. capture
width
Incident
waves


Axisymmetric
surging body
ka
0
0.05
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.4
1.6
1.8
2.0
2.5
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0

A * (ka)
0.8310
0.8764
0.8627
0.7938
0.7157
0.6452
0.5861
0.5381
0.4999
0.4698
0.4464
0.4284
0.4047
0.3924
0.3871
0.3864
0.3884
0.3988
0.4111
0.4322
0.4471
0.4574
0.4647
0.4700
0.4740
0.4771
0.5
B * (ka)
0
0.1036
0.1816
0.2793
0.3254
0.3410
0.3391
0.3271
0.3098
0.2899
0.2691
0.2484
0.2096
0.1756
0.1469
0.1229
0.1031
0.0674
0.0452
0.0219
0.0116
0.0066
0.0040
0.0026
0.0017
0.0012
0
Exercise 3.1
Hemispherical buoy
in deep water
Dimensionless quantities
A * (ka) 
B * (ka) 
A( )
2  a3 
3
B( )
2  a3 
3
ka  2a g   *2  (2 T *)2
*   a g
T*  T g a
C* 
C
a 5 2  g1 2
C* 
C
a 5 2  g1 2
T*  T g a
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 Aw  1m .
• 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).
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
d
f PTO  C
 K ,
dt
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
Time-domain analysis of a single heaving body
• 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.
Time-domain analysis of a single heaving body
This dependence can be expressed in the following form:
fr (t )  
t

gr (t   ) ( ) d  A()(t )
see later why
How to obtain the memory function gr ( ) ?
Take


fr (t )  Fr eit  2 A()  i B() X eit
( )  i X ei ,
We obtain
(t )   2 X ei .
 A()  A()  i B()X e
i t
2
t
 i X 

gr (t  ) ei d
Changing the integration variable from  to s  t   , we have

i A()  A()  B()   gr (s) ei sd s
0
Time-domain analysis of a single heaving body

i A()  A()  B()   gr (s) ei sd s
0
Since the functions A, B and gr are real, we may write

 A()  A()   gr (s) sin  s d s
0

B()   gr (s) cos s d s
0
Note that, since gr ( ) if finite, the integrals vanish as    , which
agrees with B()  0, A()  0.
1 
B ( )   g r ( s ) ei s d s
2 
Assume gr (s) to be an even function
Invert Fourier transform
g r ( s) 
2

 0
B ( ) cos  s d
Time-domain analysis of a single heaving body
fr (t )  
t

gr (t   ) ( ) d  A()(t )
t


m  A() (t )  fe (t )   gr (t  ) ( ) d   gScs (t )  fPTO

g r ( s) 
2

 0
B( ) cos  s d
This has to be integrated in the time domain from initial conditions
for and.
Time-domain analysis of a single heaving body
t


m  A() (t )  fe (t )   gr (t  ) ( ) d   gScs (t )  fPTO

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.
Integration procedure:
• Set initial values (usually zero) for and.
• Compute the rhs at time t 0
• Compute (t0 ) from the equation
• Set t1  t0  t
• (t1)  (t0 )  (t0 )t,  (t1)   (t0 )  (t0 )t
• Compute (t1) etc.
Adopted time steps are typicall between 0.01 s and 0.1 s
The convolution integral must be computed at every time step
Wave energy conversion in irregular waves
• Real ocean waves are not purely sinusoidal: they are irregular
and largely random.
• In linear wave theory, they can be modelled as the the
superposition of an infinite number of sinusoidal wavelets with
different frequencies and directions.
• The distribution of the energy of these wavelets when plotted
against the frequency and direction is the wave spectrum.
• Here, we consider only frequency spectra.
Wave energy conversion in irregular waves
A frequency spectrum is a function S f ( f ) (units m2s)
S f ( f ) df
equal to df
is is the energy content within a frequency band of width
Wave energy conversion in irregular waves
A frequency spectrum is a function S f ( f ) (units m2s)
 g S f ( f ) df is is the energy content within a frequency band of
width equal to df
Wave energy conversion in irregular waves
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


S f ( f ) 0.1688H s2 Te 4 f 5 exp  0.675(Te f )  4 .


S ( ) 262.6 H s2 Te4 5 exp  1052(Te  ) 4 .
H s  significant waveheight
Te  energyperiod
S f ( f )df  S ()d
Wave energy conversion in irregular waves


S ( ) 262.6 H s2 Te4 5 exp  1052(Te  ) 4 .
H s  significant waveheight
Te  energyperiod
H s  4 m0
Te 
m1
m0

mn is spectralmomentof order n : mn   f n S ( f ) df
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.
Divide the frequency range of interest into N small intervals i    i 1
of width i  i 1  i and set
S,i  S (ˆi )
1 2
S ,i  i  Aw,i or A ,i  2S ,i  i
2
ˆi  12 (i  i 1)
Simulation of excitation force in irregular waves
N
fe (t )   (ˆ i ) A ,i expi(ˆ i t   i ) or
i 1
N
f e (t )   (ˆ i ) A ,i cos(ˆ i t   i )
i 1
i are randomphasesin interval (0, 2 )
Wave energy absorption from irregular waves
Simulation of excitation force in irregular waves
N
fe (t )   (ˆ i ) A ,i expi(ˆ i t   i )
i 1
N
f e (t )   (ˆ i ) A ,i cos(ˆ i t   i )
i 1
Oscillating body with linear PTO and linear damping
coefficient C . Averaged power over a long time:
2
P  PPTO
X (ˆ i ) 
1 N 2
 2
 d 
    C  i X (i )
2 i 1
 dt 
(ˆ i ) A ,i
 ˆ i2 (m  A(ˆ i ))  iˆ i ( B(ˆ i )  C )  (  gScs  K )
Note that:
1

1 t

if i  j
ˆ
ˆ
lim
sin(

t


)
sin(

t


)
d
t

,

i
i
j
j
2

0
t  t

0 if i  j.
Wave energy absorption by 2-body oscillating systems
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.
• 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
Wave energy absorption by 2-body oscillating systems
• 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.
d 21
m1 2  f e,1  f r ,11  f r ,12  gScs,11  f PTO,
dt
m2
d 2 2
dt
2
 f e,2  f r ,22  f r ,21  gScs,2 2  f PTO.
Wave energy absorption by 2-body oscillating systems
m1
m2
d 21
dt
2
d 2 2
dt
2
 f e,1  f r ,11  f r ,12  gScs,11  f PTO,
 f e,2  f r ,22  f r ,21  gScs,2 2  f PTO.
Linear system. Frequency domain analysis
f PTO  C
i (t )  X i ei t ,
d(1   2 )
 K (1   2 ).
dt
f e,i (t )  Fe,i ei t ,
f r ,ij (t )  Fr ,ij ei t
Decompose radiation force: Fr ,ij  ( 2 Aij  i Bij ) X j
Note: B11 and B22 cannotbe negative
It can be provedthat A12  A21, B12  B21
(i, j  1, 2)
(i, j  1, 2)
Wave energy absorption by 2-body oscillating systems.
Linear system. Frequency domain analysis
  (m  A )  i(B  C)  ( g S  K )X
   A  i ( B  C )  K X  F ,
  (m  A )  i(B  C)  ( g S  K )X
   A  i ( B  C )  K X  F .
2
1
11
11
cs1
1
2
12
12
2
e,1
2
2
22
22
cs2
2
2
12
12
1
e, 2
Instantane
ous power : PPTO  C (1  2 )2  K (1  2 )(1  2 )
2
Time - averaged power : P  PPTO  12 C 2 X1  X 2 .
Relationships between coefficients: radiation damping force and excitation force
Bii 
k
4  g
2

i ( )



D(kh)
Axisymmetric systems:
2
3 
2


Bii 

(

)
d (deep water)
3  i
4  g
d
Bii 
 ki
2 g D(kh)
2
 3i
Bii 
2 g 3
(deep water).
Wave energy absorption by 2-body oscillating systems.
Non-linear system. Time domain analysis
Excitation forces:
t
f r ,ij (t )  

gr ,ij (t   ) j ( ) d  Aij ()j (t )
g r ,ij ( s)  g r , ji ( s ) 
(m1  A11 ())

t

d 21
dt
2
2

 0
Bij ( ) cos  t d
 f e,1  
t

g r ,11 (t   ) 1 ( ) d  g S cs,11
g r ,12 (t   ) 2 ( ) d  A12 ()2 (t )  f PTO ,
t
d 2 2
(m2  A22 ()) 2  f e, 2   g r , 22 (t   ) 2 ( ) d  g Scs, 2 2

dt

t

g r ,12 (t   ) 1 ( ) d  A12 ()1 (t )  f PTO.
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:
Fe2  0,
A12  0, B12  B22  0
A22 is independent of 
a  c b  0.4a conesemi - angle 60
volumeof submerged part of body 1 : 3.031a3
A22  0.6897b3
The PTO consists of a linear
damper, and no spring.
Exercise 3.3. Heaving two-body axisymmetric wave energy converter
*
A11

A11
 a
,
3
*
B11

B11
 a3
T *  T g a  2 1 g a
Exercise 3.3. Heaving two-body axisymmetric wave energy converter
 Writethegoverningequationsin thefrequencydomain
 Computemass m2 as functionof draught d
 For given T * find theoptimalvaluesof ratio d a and PT O
dampingcoefficient
C* 
C
a 5 2  g1 2
 Discuss the advantages and limitations of a
wave energy converter based on this concept
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 oscillatingbody WECS
• PTO is in general highly nonlinear (time-domain analysis)
Time-domain analysis of a heaving buoy with hydraulic PTO
m  A()(t )  fe (t )   gr (t  ) ( ) d   gScs (t )  fPTO
t
p 
f PTO
Sc
p1  pressurein HP accumulator
p2  pressurein LP accumulator
Body and pistoncannotmoveif p  p1  p2
(exceptin reactivecontrol) " Coulombdamping"
d
 volumeflow rateof oil (piston)
dt
qm (t )  volumeflow rateof oil (hydraulicmotor)
q(t )  Sc
If duct and accumulator walls are rigid and oil is incompressible :
q(t )  qm (t )  rateof changeof gas volumein HP
Time-domain analysis of a heaving buoy with hydraulic PTO
q (t )  qm (t )  rateof changeof gas volumein HP
  rateof changeof gas volumein LP
In accumulators, gas volumeis relatedto gas pressure
(approximatelyisentropicprocess)
cp
p1
 constant, 
 1.4 (air, nitrogen)

c
1
v
P T Ocontrol: define relationship between oil flow rateqm (t )
in motorand pressuredifferencep1 (t )  p2 (t ) :
qm (t )  Sc2  p1 (t )  p2 (t )G  G  controlparameter
Time-domain analysis of a heaving buoy with hydraulic PTO
a  5 m, Sc  0.01767m 2 , G  0.6 106 s/kg,
m1  150 kg, m2  30 kg, V0  5.17 m3 (gas)
H s  1.5 m, Te  11s
Time-domain analysis of a heaving buoy with hydraulic PTO
Underdamping and
overdamping
Sphere
Sea state H s  3 m, Te  11s
Sea state H s  3 m, Te  11s
Sphere radius a  5 m
x (m)
P (kW)
x (m)
P (kW)
x x(m)
(m)
PP(kW)
(kW)
Under damped
External force 200 kN
Optimally damped
External force 647 kN
Over damped
External force 1000 kN
P  83.1 kW
P  178.4 kW
P  97.0 kW
Time-domain analysis of a heaving buoy with hydraulic PTO
a  5 m, Sc  0.01767m 2 , G  0.6 106 s/kg,
m1  150 kg, m2  30 kg, V0  5.17 m3 (gas)
H s  1.5 m, Te  11s
PTO power
Pabsorbed H s2
Poutput H s2
Time-domain analysis of a heaving buoy with hydraulic PTO
Poutput H s2
Hs  4 m
Hs 1m
The smoothing effect decreases for more energetic sea states
Time-domain analysis of a heaving buoy with hydraulic PTO
Phase control by latching
Pioneers in control theory of wave
energy converters.
They introduced the concept of phasecontrol by latching:
Johannes
Falnes
Kjell Budall
(1933-89)
J. Falnes, K. Budal, Wave-power conversion by power absorbers.
Norwegian Maritime Research, 6, 2-11, 1978.
Time-domain analysis of a heaving buoy with hydraulic PTO
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.
Phase-control by latching: release the body when
hydrodynamic forceon body exceeds RSc ( p1  p2 ) ( R  1)
Time-domain analysis of a heaving buoy with hydraulic PTO
Phase control by latching
G
Two control
variables
control of flow rate of oil
through hydraulic motor
R
release of latched body
Regular waves
Aw  0.667m, T  9 s
No latching
R=1
R  1, G  0.86  106 s/kg
P  55.0 kW
Regular waves
Aw  0.667m, T  9 s
Optimal latching
R>1
R  16, G  7.7  106 s/kg
P  206.1kW
NO LATCHING
OPTIMAL LATCHING
Irregular waves, Te = 9 s
Irregular waves
Te = 9 s
No latching
R=1
R  1, G  0.7  106 s/kg
P  10.3 kW/m2
Irregular waves
Te = 9 s
Optimal latching
R>1
R  16, G  4.2  106 s/kg
P  28.5 kW/m2
NO LATCHING
OPTIMAL LATCHING
Latching control
• May involve very large forces
• May be less effective in two-body WECs
END OF
PART 3
MODELLING OF OSCILLATING
BODY WAVE ENERGY
CONVERTERS
```