taller energías oceánicas - actividad 4

Taller Virtual: Energías Oceánicas. Aprovechamiento energético del oleaje

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RED DE EXPERTOS EN ENERGÍA

TALLER

“ENERGÍAS OCEÁNICAS: APROVECHAMIENTO ENERGÉTICO DEL OLEAJE”

MODERADOR: MARCOS LAFOZ PASTOR

COLABORACIÓN: LUIS GARCÍA-TABARÉS RODRÍGUEZ Y MARCOS BLANCO AGUADO

PARTE 4: THE HEAVING POINT ABSORBER FOR

WAVE ENERGY EXTRACTION

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Objetivos El objetivo de esta actividad del taller de energías oceánicas es estudiar en detalle un tipo concreto de captador de energía de las olas. El denominado ABSORBEDOR PUNTUAL o POINT ABSORBER.

Se desarrollarán las ecuaciones físicas que rigen su comportamiento, así como un modelo sencillo para entender el funcionamiento del dispositivo.

Dado que esta parte del taller es quizá la más compleja, se va a apoyar en un ejercicio práctico que, aunque es optativo, es totalmente recomendable para entender bien los conceptos que se explican.

El ejercicio es guiado, igual que en la actividad 2, y en él se seguirán los pasos para calcular la dinámica de un absorbedor puntual según varias estrategias de control.

Siga la presentación para entender los conceptos teóricos y después realice el ejercicio siguiendo los pasos, utilizando el archivo de Excel adjunto con la plantilla preparada para ello.

4. The heaving point absorber for wave energy extraction


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One-Body Point Absorber

Two-Body Point Absorber

It is based on the action-reaction between two bodies.

If one body is fixed to the sea-bed, it is a One-Body Point Absorber, while if both bodies move, it is a Two-Body Point Absorber.

A Heaving Point Absorber extracts energy from the wave movement. It oscillates in a vertical direction and its size is much smaller than the wave length.

4. The point absorber. Operation principles and modelling


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The main parts for a general Two Moving Body Heaving Point Absorber, are:

FLOAT(Body 1)

SPAR +INERTIA TANK(Body 2)

POWER TAKE-OFF (PTO)

OSCILLATING WATERSURFACE

The FLOAT is the main moving body. It is usually ballasted with water to achieve the required nominal draft. It is mechanically connected to the moving part of the Power Take-Off.

The PTO is the component where the mechanical energy is converted into electricity from the relative movement between Body 1 & Body 2. There are different types, like direct electrical generators, hydraulic or pneumatic actuators, etc.

The Spar+ the Inertia Tank constitute the reaction body. It is also usually ballasted to increase the mass and stability. Some times the Inertia Tank is substituted with a plate to increase the added water mass. The Spar usually also hosts some of the ancillaries, the control system, the power electronics, etc..



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PRACTICAL CASES OF HEAVING POINT ABSORBERS (HPA)

ONE-BODY HPA: The Lysekil

Project TWO-BODY HPA: OPT Power Buoy

TWO-BODY HPA: WEDGE W-200



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A general Two-Body Point Absorber has a conceptually simple mechanical model as described in the figure. (The meaning of the different forces will be explained in later slides )

FLOATBODY 1

SPAR + INERTIA TANKBODY 2

POWER TAKE-OFF

OSCILLATING WATER SURFACE

BODY 1 WAVE EXCITATION FORCE & BOUYANCY

BODY 1 RADIATION FORCE: Damping

BODY 2 WAVE EXCITATION FORCE & BOUYANCY

BODY 2 RADIATION FORCE: Damping

PTO FORCE: To be converted into electrical power

BODY 1 TOTAL MASS

BODY 2 TOTAL MASS

GROUND

GROUND



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Small Radius FloatLarge Radius Float

Displaced Volume

Following forces act on an HPA when oscillating in a wavy sea

WAVE EXCITATION FORCE: It corresponds to the incremental displaced volume of water caused by the wave passing through the body. Calculation is cumbersome even for simple geometries, requiring the use of CFD software (AQWA, WAMIT, ...). Since the wavelength depends on the frequency ω, if the frequency or the float radius are increased, the net wave displaced volume becomes relatively smaller. A fair approximation for the wave excitation force as a function of ω, R and the wave amplitude A is given as:

BUOYANCY: It is simply the weight of the incremental displaced volume of water by the floating body movement from its equilibrium position. For a cylindrical Float with radius R, and vertical velocity Vf, it can be expressed as:

For sinusoidal excitations at a frequency ω, the Buoyancy Force can be expressed in a complex form as:

A

4. The point absorber. Forces acting on a heaving point absorber (HPA)


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INERTIA FORCE: Inertia force is associated to the mass of the moving body and can be espressed as the product of the mass times the vertical acceleration. The issue is that the mass to be considered is not only the float mass but also the added water mass. A simple expression for this term is:

RADIATION FORCE: Radiation force is a dissipative term related to the energy that the oscillating Float is releasing to generate waves. It is associated to open seas were all the power is transmitted to the water and no reflected power is recovered back by the float (equivalence with an electrical transmission line).

Radiation Damping Coefficient

FLOAT

C

R

Again, calculation of the radiation term is cumbersome and requires CFD tools. A simple expression can be used, based on a transmission line model (TLModel):



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PTO FORCE: Is the forced exerted by the Power Take Off between the two moving bodies. In the most advanced cases this force is the sum of three terms, one proportional to the vertical displacement, a second one proportional to the vertical velocity and the third term proportional to the vertical acceleration in the following manner: FPTO

F Active

F Re

activ

e

vf

In a generalized form, the PTO force can be expressed in a complex or in a vector from (equivalent). The real part is the so called active force in phase with the velocity and the only one able to produce energy. The second one is out of phase with the float velocity, it is called reactive force , it is unable to produce any usable work but can be very convenient for modifying the dynamic properties of the Float: in essence reactive force modifies inertia and/or buoyancy forces. It is interesting to note that for a given PTO, its vector force can be located inside a circle with a radius equal to the maximum force that the PTO is able to provide. Also important to note that this force can be located in any of the 4 quadrants (motor up/motor down/generator up/generator down).

Electric PTO Based on a Switched Reluctance Machine



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For the sake of simplicity we shall consider a One Moving Body Point Absorber. Motion equation will impose that the sum of all forces acting on the Float must be null.

MECHANICAL MODEL

FPTOFW

Inertia (Fi)

Mf +Mw

FRFB

ELECTRICAL EQUIVALENT CIRCUIT:

4. The point absorber. Simplified motion equations of a HPA


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An ideal Heaving Point Absorber means:* No viscous damping *Lossless PTO* Simplified Force Expressions *Regular waves at frequency ω R=4

mC=2 m

CASE STUDY

PTO

CONCLUSIONS* A fairly constant amount of power can be extracted at any wave period.* There is always a frequency (period) for which the required force in minimum (natural Float resonance).* Achieving the maximum power at any wave period, requires force levels completely inadmissible.* In practice, HPA are equipped with PTOs able to harvest maximum available power around natural Float resonance. Energy capture reduces drastically far from that frequencies.

4. The point absorber. Power generatrion in an ideal one-body HPA


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Direct-Drive Electric PTO - IntroductionDifferent types of PTO can be used for wave energy extraction: Hydraulic, Mechanical (rack & pinion) or electrical (linear generator directly connected to the moving body)There are several options of direct-drive electrical PTOs based on different types of electrical machines such as permanent magnets, synchronous or switched reluctance machines.Electric Linear PTOs have been selected by companies such as AWS, WEDGE GLOBAL; SEABASED or TRIDENT.

AWS Electric PTOFuente: https://www.researchgate.net/figure/27342929_fig1_Figure-2-Photograph-of-the-linear-permanent-magnet-generator-in-the-AWS

SEABASED Electric PTOFuente: http://www.mdpi.com/2075-1702/2/1/73/htm

Uppsala University Wave Energy Converter Electric PTOFuente: http://www.mdpi.com/2075-1702/2/4/219/htm

Now, we shall consider the presence of a real PTO . Unless an ideal one, four limitations should be taken into account:

* PTO Losses * Force Limit * Stroke Limit * Speed Limit



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Now, we shall consider the presence of a real PTO . Unless an ideal one, four limitations should be taken into account:

* PTO Losses * Force Limit * Stroke Limit * Speed LimitDirect-Drive Electric PTO – Loss Model1. For a PTO based on a Electric Linear Generator, electrical transients can be

considered order of magnitude faster than mechanical ones. I consequence, the force command could be considered immediately imposed and a loss model could be used instead of a detailed PTO model.

2. Like in most of electrical machines, Force can be proportional to the current flowing in the machine. IPTO=KFI*FPTO.

3. Since PTO speeds are low, the required forces to produce significant power are very high, leading to high currents as well. In these circumstances, Joule losses (P loss_cu) in the windings dominate over the rest of the PTO losses . Ploss_cu=Rcu*IPTO.

4. Low speeds also means low electrical and mechanical frequencies, thus mechanical losses (Ploss_mec) can be neglected due to the low PTO speed as well as magnetic losses (Ploss_mag) due to the low frequency oscillations.

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2

PTOFIfPTOPTOfPTO

IVFP

FKVFIVF

PTOfPTOele

cucu

R

loss_cumec

P

loss_mecloss_magloss_cumecPTO RRP-PP-P-P-PP

model loss PTO domain Time

culoss

Point AbsorberPW=FW·Vf Pmec=FPTO·Vf Electrical PTO PPTO

Ploss=Ploss_cu+Ploss_mag+Ploss_mec



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222 1

FIcuPTO

PTO

PTOPTOFIculosslossPTOPTOPTO KR

RRUUKRP ;P]*iUReal[P

model loss PTO circuit Equivalent

PW PmecPloss

Electrical PTO losses are modelled

proportional to the square of FPTO

Electrical PTO losses are modelled as a electrical resistance in parallel with UPTO

PPTO

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2

PTOFIfPTOPTOfPTO

IVFP

FKVFIVF

PTOfPTOele

cucu

R

loss_cumec

P

loss_mecloss_magloss_cumecPTO RRP-PP-P-P-PP

model loss PTO domain Time

culoss

4. The point absorber. Power generatrion in an ideal one-body HPANow, we shall consider the presence of a real PTO . Unless an ideal one, four limitations should be taken into account:



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RZReal4

URUcos

ZU-

ZUUP

:powermaximum the extract to able modulus Force PTO the is ZcosR

R2U=U

RUcos

ZU-

ZUUP

power extracted the maximizes which argument Force PTO the is -=

RUcos

ZU-)cos(

ZUU]*iUReal[P

R

U

Z

U-0U=2i-1i= i

PTO THE AT DEVELOPEDPOWER (SPEED) CURRENT PTO

PTO

2W

PTO

2PTOX

2PTOXPTOXw

PTOXX

PTO

PTOwPTOX

PTO

2PTO

2PTOPTOw

PTOX

PTO

2PTO

2PTOPTOw

PTOPTOPTOPTO

PTOPTOwPTO

2

0

Z

4. The point absorber. Power generatrion in an ideal one-body HPANow, we shall consider the presence of a real PTO . Unless an ideal one, four limitations should be taken into account:



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Electrical PTO losses are taken into account with a loss model

Which implies a new definition of the Maximum Power Extracted

PTO

PTOPTOPTOPTO R

U]*iUReal[P2


Electrical PTO force limit (the PTO force is limited at its rated value, FLIM)are taken into account limiting the modulus of FPTO

-=;

RZReal4

UP

PTO

2W

PTOXX

2Z

-=;;

RZReal4

UminP

PTO

2W

PTOXX

LIMFZ2

PTO active velocity and stroke limitations lead to complex control strategies (out of the scope of the present course). For more information review the following paper.


http://www.sciencedirect.com/science/article/pii/S0029801813001200


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R=3,5mC=1,75 m

CASE STUDY

PTO

CONCLUSIONS* Loses could be modelled proportional to the square of the PTO force. This PTO loss model could be suitable for different electric machine types such as switched reluctance machine, or permanent magnet machine.* Force is proportional to the current flowing in the machine, so the use of the force leads to Joule power losses in the machine windings.* The use or reactive power always improve the mechanical power extracted results, but the PTO losses should be take into account to avoid use excessive PTO force values which penalise the extracted electric power.* Ignore PTO loses could leads to worse results in terms of electric power extracted.

3 5 7 90

100000

200000

300000

400000

500000

600000P_mec_1(W)P_mec_2(W)P_mec_3(W)

Period (s)

Pow

er (W

)

3 5 7 90

100000

200000

300000

400000

500000

600000P_ele_1 (W)P_ele_2 (W)P_ele_3 (W)

Period (s)

Pow

er (W

)

Extracted POWER


http://www.mdpi.com/1996-1073/8/10/11203/

http://www.sciencedirect.com/science/article/pii/S0196890413006729

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Ejercicio Actividad 4: Energía obtenida por un Absorbedor Puntual (HPA)

Siguiendo las instrucciones que se indican a continuación para obtener el modelo de un absorbedor puntual de un cuerpo y evaluar su capacidad energética.

• Se usarán las ecuaciones dinámicas del cuerpo flotante, el modelo de pérdidas del PTO y las estrategias de control de extracción de energía de cara a obtener velocidad y desplazamiento del captador y evaluar la potencia extraída (tanto mecánica como eléctrica), todo ello bajo la suposición de oleaje regular.

INSTRUCCIONES

Una vez terminado el ejercicio envíe el Excel cumplimentado al profesor

• Se calcularán en primer lugar los coeficientes hidrodinámicos para distintas frecuencias

• Después se evaluarán los términos de potencia y fuerza siguiendo tres estrategias de control distintas, para poder analizar los resultados y sacar conclusiones de ellos.


Preparación del modelo de absorbedor puntual en formato Excel. Paso 1/4- Partir de la plantilla Excel ejercicio2_HPA_TEMPLATE_v03.xlsx para el ejercicio y

de las ecuaciones presentadas en la presentación.1. Calcular en la hoja “INPUT DATA & RESULTS” los parámetros hidrodinámicos

necesarios para el modelo a partir de los datos de entrada que se incluyen en la misma hoja.

Los parámetros básicos ya aparecen debidamente cumplimentados:

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Radius R 3,5 m PTO Force Limit F_lim 160000 N

Draft c 1,75 m PTO Rated Force F_nom 160000 N

Wave Amplitude A 1 m PTO Rated Velocity v_nom 1 m/s

Sea Water Density ρ 1025 kg/m^3 PTO Efficiency η_nom 0,75 p.u.

Seguir para ello las ecuaciones que se indican a la derecha



Preparación del modelo de absorbedor puntual en formato Excel. Paso 2/42. A partir de las expresiones que también se incluyen en la hoja “INPUT DATA &

RESULTS” ,completar la hoja donde se calculan los parámetros hidrodinámicos que determinan la dinámica del flotador “Hidrodinamic Coeficients”. Calcular los parámetros para las frecuencias que aparecen en la tabla Excel.

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3. En cada una de las hojas que corresponden a cada estrategia de control, “Control Strategy 1”, “Control Strategy 2” y “Control Strategy 3” respectivamente, introducir las ecuaciones que permiten calcular las variables de la dinámica del flotador para todas las frecuencias.

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Extracted Mechanical Power P_mec_i

Extracted Electrical Power P_ele_i

Floater heave Velocity displacement V_i calculando su parte real, real(V), imaginaria, imag(V), y modulo, abs(V_i)

Floater heave Excursion s_i calculando su modulo, abs(s_1)


Preparación del modelo de absorbedor puntual en formato Excel. Paso 3/4

Se va a evaluar la dinámica del absorbedor puntual considerando 3 estrategias de control diferentes:

Control Strategy 1 Maximization of the mechanical power (P_mec_1)

Control Strategy 2 Maximization of the mechanical power (P_mec_1) with the force limited at F_lim

Control Strategy 3 Maximization of the electrical power (P_mec_1) with the force limited at F_lim

Las ecuaciones a introducir en cada columna son iguales independientemente del tipo de estrategia de control que se desee implementar, la cual se introducirá en las columnas F_i y α_i, siendo “i” el número que denota las distintas estrategias de control.


Usar las expresiones que aparecen en la parte superior en cada hoja de estrategia de control



- Una vez completadas las columnas con los valores de Fuerza y ángulo que aparecen en la parte superior, realizar también el cálculo en las celdas correspondientes de las potencias mecánica, eléctrica y resto de variables, de acuerdo a las expresiones que aparecen en la hoja inicial “INPUT DATA & RESULTS”

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4. Por último, analizar los resultados que quedan representados en la hoja inicial “INPUT DATA & RESULTS”.La potencia eléctrica es siempre menor que la mecánica debido al rendimiento, y la limitación de la fuerza supone una reducción en la potencia extraida, ¿qué más conclusiones puede sacar de este ejercicio?


taller energías oceánicas - actividad 4

Engineering