Report

Specialization in Ocean Energy MODELLING OF WAVE ENERGY CONVERSION António F.O. Falcão Instituto Superior Técnico, Universidade de Lisboa 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 (low head 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 radiation 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) expi(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 iCX KX ei t 2mX Fr gScs X iCX KX Fe Frequency-domain analysis of heaving body Decompose radiation force coefficient: Fr (2 A i B) X (m A) i(B C) ( gS added mass X F cs K ) 2 e radiation 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: A added mass B radiationdampingcoefficient Fe excitationforceamplitudeper unit waveamplitude Aw Haskind relation: B 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: A added mass B radiationdampingcoefficient 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 iX 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: BC radiation damping = PTO damping 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 Haskind relation: 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 to model the radiation force. 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 eit 2 A() i B() X eit ( ) 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) ei sd s 0 Time-domain analysis of a single heaving body i A() A() B() gr (s) ei 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 Te4 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 Te4 5 exp 1052(Te ) 4 . H s significant waveheight Te energyperiod H s 4 m0 Te m1 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 expi(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 expi(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 )) iˆ 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 21 m1 2 f e,1 f r ,11 f r ,12 gScs,11 f PTO, dt m2 d 2 2 dt 2 f e,2 f r ,22 f r ,21 gScs,2 2 f PTO. Wave energy absorption by 2-body oscillating systems m1 m2 d 21 dt 2 d 2 2 dt 2 f e,1 f r ,11 f r ,12 gScs,11 f PTO, f e,2 f r ,22 f r ,21 gScs,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 ki 2 g D(kh) 2 3i 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 21 dt 2 2 0 Bij ( ) cos t d f e,1 t g r ,11 (t ) 1 ( ) d g S cs,11 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 0.6897b3 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 0.6 106 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 radius a 5 m 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 0.6 106 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 0.86 106 s/kg P 55.0 kW Regular waves Aw 0.667m, T 9 s Optimal latching R>1 R 16, G 7.7 106 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 0.7 106 s/kg P 10.3 kW/m2 Irregular waves Te = 9 s Optimal latching R>1 R 16, G 4.2 106 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