Simplified Numerical Model for the Study of Wave Energy Convertes (WECs)€¦ · Introduction...

Introduction Aproximation Identification Applications

Simplified Numerical Model for the Study of WaveEnergy Convertes (WECs)

J. A. Armesto, R. Guanche, A. Iturrioz and A. D. de Andres

Ocean Energy and Engineering Group“IH Cantabria”, Universidad de Cantabria

October 18th, 2013

Armesto, Guanche, Iturrioz and de Andres 18/X/2012 Simplified Numerical Model for WECs

1 Introduction

2 Aproximation

3 Identification

4 Applications

The movement of every body can be given by 6 Degrees ofFreedom (DOF), three displacements:

Surge, x

Sway, y

Heave, z

and three rotations:

Roll, φ

Pitch, θ

Yaw, ψ

Cummins equation in time domain:

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (1)

where:

x = [x , y , z , φ, θ, ψ]T is the array of DOFs.

M is the inertia matrix of the floating body and A∞ theinfinity added mass.

K is the transfern function.

C is the hydrostatic matrix.

fe are the external forces applied over the body,. . .

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (1)

where:

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (1)

where:

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (1)

where:

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (1)

where:

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (1)

where:

Ogilvie equation in frequency domain:

−ω2 (M + A(ω)) X(iω) + iωB(ω)X(iω) + CX(iω) = Fe(iω) (2)

Ogilvie Relations:

A(ω) = A∞ −1

∫ ∞0

K (t) sin(ωτ)dτ (3)

B(ω) =

∫ ∞0

K (t) cos(ωτ)dτ (4)

The transfern function can be written as:

K (t) =2

∫ ∞0

B(ω) cos(ωt)dω & A∞ = lımω→∞

A(ω) (5)

K (iω) =2

∫ ∞0

K (t)e−iωtdt = B(ω) + iω[A(ω)− A∞]. (6)

Ogilvie Relations:

A(ω) = A∞ −1

∫ ∞0

B(ω) =

∫ ∞0

K (t) =2

∫ ∞0

A(ω) (5)

K (iω) =2

∫ ∞0

Ogilvie Relations:

A(ω) = A∞ −1

∫ ∞0

B(ω) =

∫ ∞0

K (t) =2

∫ ∞0

A(ω) (5)

K (iω) =2

∫ ∞0

Data A and B is obtained by a Boundary Element Method (BEM)software WADAM/WAMIT for a given set of frequencies.

This program also provides M, C and wave forces for the given setof frequencies.

Then, we know:

A(ω) and B(ω) for ω in {ωi}Nn=1.

M and C .

Fw (ω) for ω in {ωi}Nn=1.

Then, we know:

M and C .

Then, we know:

M and C .

Convolution integrals in Cummins equation are computationallyexpensive, then we want to replace them for something moreefficient.

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (7)

We want to replace every convolution integral by a state-space:

α(t) = Ai ,jα(t) + Bi ,j xj(t)yi ,j(t) = Ci ,jα(t)

Such that:

yi ,j(t) ≈∫ t

0ki ,j(t − τ)xj(τ)dτ, i , j ∈ {1, 2, . . . , 6} (9)

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (7)

Such that:

yi ,j(t) ≈∫ t

0ki ,j(t − τ)xj(τ)dτ, i , j ∈ {1, 2, . . . , 6} (9)

(M + A∞) x(t) +

0K (t − τ)x(τ)dτ + Cx(t) = fe(t) (7)

Such that:

yi ,j(t) ≈∫ t

0ki ,j(t − τ)xj(τ)dτ, i , j ∈ {1, 2, . . . , 6} (9)

One Degree of Freedom

Cummins equation results:

(M + A∞) z(t) +

0K (t − τ)z(τ)dτ + Cz(t) = fe(t) (10)

Which can be written as:

(M + A∞) z(t) + yc(t) + Cz(t) = fe(t) (11)

Lets define k1(t) = z(t) y k2(t) = z(t)

k1(t) = k2(t)

k2(t) = z(t)(12)

(M + A∞) k2(t) + yc(t) + Cz(t) = fe(t) (13)

k2(t) = − yc(t)

M + A∞− C

M + A∞z(t) +

M + A∞(14)

(M + A∞) z(t) +

0K (t − τ)z(τ)dτ + Cz(t) = fe(t) (10)

(M + A∞) z(t) + yc(t) + Cz(t) = fe(t) (11)

k1(t) = k2(t)

k2(t) = z(t)(12)

(M + A∞) k2(t) + yc(t) + Cz(t) = fe(t) (13)

k2(t) = − yc(t)

M + A∞− C

M + A∞z(t) +

M + A∞(14)

(M + A∞) z(t) +

0K (t − τ)z(τ)dτ + Cz(t) = fe(t) (10)

(M + A∞) z(t) + yc(t) + Cz(t) = fe(t) (11)

k1(t) = k2(t)

k2(t) = z(t)(12)

(M + A∞) k2(t) + yc(t) + Cz(t) = fe(t) (13)

k2(t) = − yc(t)

M + A∞− C

M + A∞z(t) +

M + A∞(14)

(M + A∞) z(t) +

0K (t − τ)z(τ)dτ + Cz(t) = fe(t) (10)

(M + A∞) z(t) + yc(t) + Cz(t) = fe(t) (11)

k1(t) = k2(t)

k2(t) = z(t)(12)

(M + A∞) k2(t) + yc(t) + Cz(t) = fe(t) (13)

k2(t) = − yc(t)

M + A∞− C

M + A∞z(t) +

M + A∞(14)

(M + A∞) z(t) +

0K (t − τ)z(τ)dτ + Cz(t) = fe(t) (10)

(M + A∞) z(t) + yc(t) + Cz(t) = fe(t) (11)

k1(t) = k2(t)

k2(t) = z(t)(12)

(M + A∞) k2(t) + yc(t) + Cz(t) = fe(t) (13)

k2(t) = − yc(t)

M + A∞− C

M + A∞z(t) +

M + A∞(14)

Then, we have:

α(t) = Acα(t) + Bc x(t)

yc(t) = Ccα(t) ≈∫ t−∞ k(t − τ)x(τ)dτ

k2(t) = k1(t) (16)

k2(t) = − Cc

M + A∞α(t)− C

M + A∞x(t) +

M + A∞(17)

Then, we have:

k2(t) = k1(t) (16)

k2(t) = − Cc

M + A∞α(t)− C

M + A∞x(t) +

M + A∞(17)

Then, we have:

k2(t) = k1(t) (16)

k2(t) = − Cc

M + A∞α(t)− C

M + A∞x(t) +

M + A∞(17)

Definning χ = [α, k1, k2] = [α, x , x ] it is possible to rewriteprevious equations into an unique state-space:

χ(t) = Aχ(t) + Bfe(t)

y(t) = Cχ(t)(18)

where:

Ac... Bc

00 . . . 0 0 1

Cc/a C/a 0

0· · ·00−1/a

; C =[

0 . . . 1 0]

(19)Being a = M + A∞.We have written Cummins equation as an unique state-space!

More than one Degree of Freedom

The same procedure can be used for more than one degree offreedom.

χ(t) =[[αc1], [αc2], . . . [αcn], [ξ], [ξ]

]T(20)

[Ac1] 0 0 0 [Bc1]

0. . .

......

...... 0 [Acn] 0 [Bcn]

0... 0 0 [Id ]

[Cc1/M] . . . [Ccn/M] [G/M] 0

−[M−1

0 . . . 0 1 0]

With everything we known, how do we compute the matrix andvectors that approximate the convolution integrals?

Data: A(ω) and B(ω) for ω ∈ {ωi}Nn=1.

There are two options: Identification in time domain:

K (t) =2

∫ ∞0

B(ω) cos(ωt)dω (23)

or identificatio in frecuency domain:

K (iω) =2

∫ ∞0

K (t) =2

∫ ∞0

K (iω) =2

∫ ∞0

K (t) =2

∫ ∞0

K (iω) =2

∫ ∞0

K (t) =2

∫ ∞0

K (iω) =2

∫ ∞0

Using A(ω) and B(ω) it is possible to built the transfern functionfor every frequency ω ∈ {ωi}Nn=1:

K (iωj) = B(ωj) + iωj (A(ωj)− A∞) (25)

A Matlab Toolbox done by Perez and Fossen, compute valuesθ = [p0, p1, . . . , pn−1, q0, q1, . . . , qn−1] such that:

K (s, θ) =pn−1s

n−1 + pn−2sn−2 + . . .+ p1s

1 + p0sn + qn−1sn−1 + qn−2sn−2 + . . .+ q1s1 + q0

is the best possile approximation (least square sense) for K (iω),ω ∈ {ωi}Nn=1

This means that this toolbox provides element θ? that minimizesthe following operatos:

θ? = mınθ

(K (iωn)− K (iωn, θ)

)2)(27)

K (iωj) = B(ωj) + iωj (A(ωj)− A∞) (25)

K (s, θ) =pn−1s

n−1 + pn−2sn−2 + . . .+ p1s

1 + p0sn + qn−1sn−1 + qn−2sn−2 + . . .+ q1s1 + q0

θ? = mınθ

)2)(27)

K (iωj) = B(ωj) + iωj (A(ωj)− A∞) (25)

K (s, θ) =pn−1s

n−1 + pn−2sn−2 + . . .+ p1s

1 + p0sn + qn−1sn−1 + qn−2sn−2 + . . .+ q1s1 + q0

θ? = mınθ

)2)(27)

Properties the identification must :

lımω→0

K (iω) = 0⇒ It has ceros for s = 0 (28)

lımω→∞

K (iω) = 0⇒ It is strictly proper (29)

lımt→0+

K (t) 6= 0⇒ Relative order 1 (gr(Q) = gr(P)+1) (30)

lımt→∞

K (t) = 0⇒ BIBO Stable (31)

X 7→ y is passive⇒ Re(K (iω)) > 0 (32)

lımω→0

lımω→∞

lımt→0+

lımt→∞

lımω→0

lımω→∞

lımt→0+

lımt→∞

lımω→0

lımω→∞

lımt→0+

lımt→∞

lımω→0

lımω→∞

lımt→0+

lımt→∞

Once we know θ = [p0, p1, . . . , pn−1, q0, q1, . . . , qn−1]:

0 0 0 . . . −q01 0 0 . . . −q10 1 0 . . . −q2...

.... . .

...0 0 . . . 1 −qn−1

; Bc =

p0p1p2...

pn−1

; (33)

Cc = [0, 0, . . . , 0, 1]⇒ yc(t) = CXc(t) = xn(t) (34)

CatAir

(M + A∞) z(t)+

−∞K (t−τ)z(τ)dτ+Cz(t)+kl z+knl |z |z = fe(t)

(M + A∞) z(t)+

−∞K (t−τ)z(τ)dτ+Cz(t)+(kl + knl |z |) z = fe(t)

(M + A∞) k2(t) + yc(t) + Cz(t) + (kl + knl z) z = fe(t)⇒ (37)

k2(t) = − Cc

M + A∞Xc(t)− C

M + A∞z(t)− kl + knl |z |

M + A∞z +

M + A∞(38)

CatAir

(M + A∞) z(t)+

k2(t) = − Cc

M + A∞Xc(t)− C

M + A∞z +

M + A∞(38)

CatAir

(M + A∞) z(t)+

k2(t) = − Cc

M + A∞Xc(t)− C

M + A∞z +

M + A∞(38)

CatAir

We have defined X (t) such that Xn(t) = z(t), then:

Ac... Bc

00 . . . 0 0 1

Cc/a b/a (kl + knl |Xn|)/a

0· · ·00−1/a

0 . . . 0 1 0]

CatAir

mair = − d

dt(ρair (t)Vair (t)); Vair (t) = V0 − Sz(t) = S(L0 − z(t)).

(41)Assuming the proccess is isentropic, ρair (t)(P0 + ∆P(t))−1/γ = KThe air mass flux through the openning can be expressed asfollows:

mair (t) = cdAv sgn(∆P(t))

[2mair (t)|∆P(t)|

Vair (t)

]1/2(42)

CatAir

Lets define the relative pressureP∗(t) = P(t)/P0 = (∆P(t) + P0)/P0.The air pressure variation inside the chamber,

P∗(t) =

[−Avcdγkv

ρ0Ssgn(P∗(t)− 1)

(2P0ρ0P

∗(t)−1/γ |P∗(t)− 1|)1/2

+γz(t)]P∗(t) = ΓP∗(t) (43)

Cummins’ equation as solved using a state-space,

X (t) = AX (t) + Bu(t)y(t) = CX (t)

Cummins’ equation is modified to include pressure effects,

(m + A∞)z(t) + yc(t) + ρgSz(t) + kz(t)−∆P(t)S = fe(t) (45)

CatAir

Writting together equations (45) and (43), we can write a newstate-space.The first step is to redefine the space’s vector,

X (t) =[X (t) P∗(t)

]. (46)

Then matrix A and arrays B and C are redefined as:

A 0−S

0 . . . 0 Γ

;C =[C 0

]. (47)

CatAir

Figura: Laboratory and time-domain model free surface and chamberdynamic pressure time series. Regular waves (H = 0,08 m, T = 1,1 s),top opening=4.5 mm, cd = 0,8643

WEFIES

P = CPTO(z1 − z2)2.

WEFIES

Laboratory tests coming(hopefully) soon...

WEFIES

Thanks for yourattention!

Simplified Numerical Model for the Study of Wave Energy Convertes (WECs)€¦ · Introduction...

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Transcript of Simplified Numerical Model for the Study of Wave Energy Convertes (WECs)€¦ · Introduction...