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H YDRODYNAMICS O F T HE IPS B UOY W AVE E NERGY C ONVERTER I NCLUDING T HE E FFECT O F N ON -U NIFORM A CCELERATION T UBE C ROSS S ECTION ANTÓNIO F.O. FALCÃO JOSÉ J. CÂNDIDO PAULO A.P. JUSTINO JOÃO C.C. HENRIQUES IPS Buoy Abhishek Mondal IIT Kharagpur nstituto Superior Técnico, Technical University of Lisbon, 1049-001 Lisbon, Portugal rio Nacional de Energia e Geologia, Estrada Paço do Lumiar, 1649-038 Lisbon, Portugal 1

IPS Buoy

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H Y D R O D Y N A M I C S O F T H E IPS B U O Y W AV E E N E R G Y C O N V E R T E R IN C LU D I N G TH E E F F E C T O F N O N -U N I F O R M A C C E L E R AT I O N T U B E

C R O S S SE C T I O N

A N T Ó N I O F. O . FA L C Ã OJ O S É J . C Â N D I D OPA U L O A . P. J U S T I N OJ O Ã O C . C . H E N R I Q U E S

IPS Buoy

Abhishek MondalIIT Kharagpur

IDMEC, Instituto Superior Técnico, Technical University of Lisbon, 1049-001 Lisbon, PortugalLaboratório Nacional de Energia e Geologia, Estrada Paço do Lumiar, 1649-038 Lisbon, Portugal

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What is IPS Buoy ?

A wave energy converter

Initiated by Swedish Company Inter Project Service (IPS)

Connected to fully submerged vertical acceleration tube oscillating in heave motion

Relative motion of piston and floater-tube system generates Power Take Off (PTO)

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Assumptions

The buoy-tube-piston system is mathematically modelled using the following assumptions :

Buoy-tube system has heave motion onlyThe tube is sufficiently below the water surface; thus the

excitation & radiation force become negligible.Negligible interaction between the wave fields at tube

endsFlow inside the tube is one dimensionalPiston has negligible length and mass

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Mathematical Modelling

V(t) : Piston VelocityA1 : Cross-section of inner tubeA2 : Cross-section of outer tube = α2 A1

A(ξ) : Cross-section at conical transition

Flow Velocity Pressure

where

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Hydrodynamic Analysis in Regular Waves

Force on piston: fp(t)= -Mxx - Myy+Ky+Cy

Power absorbed by PTO P = fp(t)y

Wave excitation force fe(t) = AwΓ(ω)eiωt

Force on the tube ft(t) = -mxx - myy

• x(t): floater-tube position• y(t): position of piston• K : spring stiffness• C : PTO damping coeff.• Aw: linear wave amplitude• ω: wave frequency • Mb: buoy mass (mb) +

added mass (μb)• Mt: tube mass (mt) + added mass (μt)• Γ(ω): excitation force coeff.• β : half-angle• l : added length

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Hydrodynamic Analysis in Regular Waves

Where Mx = ρA1(L+2l) My = ρA1(b1+α-2(b3+b4+2l)+2b2α-1) mx = ρA1[0.667b2(α2+α-2) + (α2 -1)(b3+b4+2l)] my = ρA1[2b2(1 - α-1) + (1 – α-2)

Equation of motion :

{x(t), y(t), fe(t)} = {X, Y, Fe}eiωt

(Mb+Mt)x + Bx + ρgSx = fe(t) + ft(t) + fp(t)

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Hydrodynamic Analysis in Regular Waves

Solving governing equation of motion of the system under the influence of linear sinusoidal wave field :

-ω2(Mb+Mt+mx+Mx)X + iωBX + ρgSX - ω2(my+My)Y = Fe --> (1)

-ω2MxX - ω2MyY + (K+iωC)Y = 0 --> (2)

Linear algebraic equations (1) & (2) is further solved to find X & Y and thus x(t) and y(t) are obtained

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Numerical Results in Regular Waves

For a cylindrical buoy of radius a submerged upto the depth a, following non-dimentional parameters are obtained :μb

* = μb/(ρπa3)B* = B/(ρπa3ω)T* = T(g/a)1/2

M1* = 1+ (Mt/mb)

M2* = ρA1(L+2l)/mb

C*(ω) = C/B(ω)X* = |X|/Aw

Y* = |Y/X|P* = P/Pmax For the case α = 1

_ _ _

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Numerical Results in Regular Waves

Time averaged wave power P = 0.5ω2C|Y|2

Pmax = (g3ρAw2)/4ω3

Maximum absorbed power attained for Xopt = |Fe|(2ωB)-1

For the case α = 1, T* = 10, P* = 1

___

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T* = 12 T* = 14

Numerical Results in Regular Waves

α = 1

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Comparison : α=1 & α=1.25

α =1(black dots)

α =1.25(white dots)

L*=L/ab1

*= b1/a =0.533

β =30o

T*=10P*=1

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Comparison : T*=10 & T*=12

T*=10(white)

T*=12(black)

b1*= (b1/a) =

0.2β =30o

P*=1

α = 4

_

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Numerical Analysis in Irregular Waves

Pierson-Moskowitz spectral distribution :

S(ω) = 526Hs2 Te

-4 ω-5 exp(-1054 Te-4 ω-4 )

[ Hs : Significant wave height ; Te: Energy period ]

Time averaged power in irregular wave :

Pirr(Hs , Te) = ∫ Preg(ω) S(ω) dω

Pirr,max = 149.5 Hs2 Te

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Non-dimensionalized parameters :

Te* = Te (g/a)1/2 ; Pirr

* = Pirr/Pirr,max ; D2* = D2/a

0

__

__

__

S(ω) = 526Hs2 Te

-4 ω-5 exp(-1054 Te-4 ω-

4 )

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Numerical Analysis in Irregular Waves

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Key Benefits of IPS Buoy

Renewable energy sourceProduces electricity for desalination

plants and remote areasCluster of buoys act as wave breakerEasily expandable by adding more

unitsEasy installation and maintenance

Low production cost/kWh50-100 MW annual power generationMeasures weather parameters and

forecast

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Global Distribution

Available Wave Energy

(kW/m)

IPS Buoy Installed Areas

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References

Falcão AF de O. Wave energy utilization: a review of the technologies. Renew

Sust Energy Rev 2010; 14:899-918. Masuda Y. Wave-activated generator. Int. colloq. exposition oceans,

Bordeaux, France; 1971. Noren SA. Apparatus for recovering the kinetic energy of sea waves.

US Patent No. 4,773,221; 1988 [original Swedish Patent No. 8104407; 1981]. Salter SH, Lin CP. Wide tank efficiency measurements on a model of

the sloped IPS buoy. In: Proc. 3rd European wave energy conf., Patras, Greece; 1998. p. 200-6. Evans DV. The oscillating water column wave-energy device. J Inst

Math Appl 1978;22:423-33. Munson BR, Young DF, Okiishi TH. Fundamentals of fluid mechanics.

2nd ed. New York: Wiley; 1994 Falnes J. Optimum control of oscillation of wave-energy converters.

Int J Offshore Polar Eng 2002;12:147-55.

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