Simplified Numerical Model for the Study of Wave Energy Convertes (WECs)

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1 Simplified Numerical Model for the Study of Wave Energy 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 18 th, 213

2 1 Introduction 2 Aproximation 3 Identification 4 Applications

3 The movement of every body can be given by 6 Degrees of Freedom (DOF), three displacements: Surge, x Sway, y Heave, z and three rotations: Roll, φ Pitch, θ Yaw, ψ

4 Cummins equation in time domain: t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (t) (1)

5 Cummins equation in time domain: t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (t) (1) where: x = [x, y, z, φ, θ, ψ] T is the array of DOFs.

6 Cummins equation in time domain: t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (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 the infinity added mass.

7 Cummins equation in time domain: t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (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 the infinity added mass. K is the transfern function.

8 Cummins equation in time domain: t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (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 the infinity added mass. K is the transfern function. C is the hydrostatic matrix.

9 Cummins equation in time domain: t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (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 the infinity added mass. K is the transfern function. C is the hydrostatic matrix. f e are the external forces applied over the body,...

10 Ogilvie equation in frequency domain: ω 2 (M + A(ω)) X(iω) + iωb(ω)x(iω) + CX(iω) = F e (iω) (2)

11 Ogilvie equation in frequency domain: ω 2 (M + A(ω)) X(iω) + iωb(ω)x(iω) + CX(iω) = F e (iω) (2) Ogilvie Relations: A(ω) = A 1 ω B(ω) = K(t) sin(ωτ)dτ (3) K(t) cos(ωτ)dτ (4)

12 Ogilvie equation in frequency domain: ω 2 (M + A(ω)) X(iω) + iωb(ω)x(iω) + CX(iω) = F e (iω) (2) Ogilvie Relations: A(ω) = A 1 ω B(ω) = The transfern function can be written as: K(t) = 2 π K(iω) = 2 π K(t) sin(ωτ)dτ (3) K(t) cos(ωτ)dτ (4) B(ω) cos(ωt)dω & A = ĺım A(ω) (5) ω K(t)e iωt dt = B(ω) + iω[a(ω) A ]. (6)

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

14 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 set of frequencies.

15 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 set of frequencies. Then, we know: A(ω) and B(ω) for ω in {ω i } N n=1. M and C. F w (ω) for ω in {ω i } N n=1.

16 Convolution integrals in Cummins equation are computationally expensive, then we want to replace them for something more efficient. t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (t) (7)

17 Convolution integrals in Cummins equation are computationally expensive, then we want to replace them for something more efficient. t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (t) (7) We want to replace every convolution integral by a state-space: α(t) = A i,j α(t) + B i,j ẋ j (t) y i,j (t) = C i,j α(t) (8)

18 Convolution integrals in Cummins equation are computationally expensive, then we want to replace them for something more efficient. t (M + A ) ẍ(t) + K(t τ)ẋ(τ)dτ + Cx(t) = f e (t) (7) We want to replace every convolution integral by a state-space: α(t) = A i,j α(t) + B i,j ẋ j (t) y i,j (t) = C i,j α(t) (8) Such that: y i,j (t) t k i,j (t τ)ẋ j (τ)dτ, i, j {1, 2,..., 6} (9)

19 One Degree of Freedom Cummins equation results: t (M + A ) z(t) + K(t τ)ż(τ)dτ + Cz(t) = f e (t) (1)

20 One Degree of Freedom Cummins equation results: t (M + A ) z(t) + K(t τ)ż(τ)dτ + Cz(t) = f e (t) (1) Which can be written as: (M + A ) z(t) + y c (t) + Cz(t) = f e (t) (11)

21 One Degree of Freedom Cummins equation results: t (M + A ) z(t) + K(t τ)ż(τ)dτ + Cz(t) = f e (t) (1) Which can be written as: (M + A ) z(t) + y c (t) + Cz(t) = f e (t) (11) Lets define k 1 (t) = z(t) y k 2 (t) = ż(t) k 1 (t) = k 2 (t) k 2 (t) = z(t) (12)

22 One Degree of Freedom Cummins equation results: t (M + A ) z(t) + K(t τ)ż(τ)dτ + Cz(t) = f e (t) (1) Which can be written as: (M + A ) z(t) + y c (t) + Cz(t) = f e (t) (11) Lets define k 1 (t) = z(t) y k 2 (t) = ż(t) k 1 (t) = k 2 (t) k 2 (t) = z(t) (12) (M + A ) k 2 (t) + y c (t) + Cz(t) = f e (t) (13)

23 One Degree of Freedom Cummins equation results: t (M + A ) z(t) + K(t τ)ż(τ)dτ + Cz(t) = f e (t) (1) Which can be written as: (M + A ) z(t) + y c (t) + Cz(t) = f e (t) (11) Lets define k 1 (t) = z(t) y k 2 (t) = ż(t) k 1 (t) = k 2 (t) k 2 (t) = z(t) (12) (M + A ) k 2 (t) + y c (t) + Cz(t) = f e (t) (13) k 2 (t) = y c(t) M + A C M + A z(t) + f e(t) M + A (14)

24 One Degree of Freedom Then, we have: α(t) = A c α(t) + B c ẋ(t) y c (t) = C c α(t) t k(t τ)x(τ)dτ (15)

25 One Degree of Freedom Then, we have: α(t) = A c α(t) + B c ẋ(t) y c (t) = C c α(t) t k(t τ)x(τ)dτ (15) k 2 (t) = k 1 (t) (16)

26 One Degree of Freedom Then, we have: α(t) = A c α(t) + B c ẋ(t) y c (t) = C c α(t) t k(t τ)x(τ)dτ (15) k 2 (t) = k 1 (t) (16) C c k 2 (t) = α(t) M + A C M + A x(t) + f e(t) M + A (17)

27 One Degree of Freedom Definning χ = [α, k 1, k 2 ] = [α, x, ẋ] it is possible to rewrite previous equations into an unique state-space: where: χ(t) = Âχ(t) + ˆBf e (t) y(t) = Ĉχ(t) (18) Â = A c. B c... 1 C c /a C/a ; ˆB = 1/a ; Ĉ = [... 1 ] Being a = M + A. We have written Cummins equation as an unique state-space! (19)

28 More than one Degree of Freedom The same procedure can be used for more than one degree of freedom. [ ξ]] T χ(t) = [α c1 ], [α c2 ],... [α cn ], [ξ], [ (2) Â = [A c1 ] [B c1 ] [A cn ] [B cn ]. [Id] [C c1 /M]... [C cn /M] [G/M] ; ˆB =. [ M 1] (21) Ĉ = [... 1 ] (22)

29 With everything we known, how do we compute the matrix and vectors that approximate the convolution integrals?

30 With everything we known, how do we compute the matrix and vectors that approximate the convolution integrals? Data: A(ω) and B(ω) for ω {ω i } N n=1.

31 With everything we known, how do we compute the matrix and vectors that approximate the convolution integrals? Data: A(ω) and B(ω) for ω {ω i } N n=1. There are two options: Identification in time domain: K(t) = 2 π B(ω) cos(ωt)dω (23)

32 With everything we known, how do we compute the matrix and vectors that approximate the convolution integrals? Data: A(ω) and B(ω) for ω {ω i } N n=1. There are two options: Identification in time domain: K(t) = 2 π or identificatio in frecuency domain: B(ω) cos(ωt)dω (23) K(iω) = 2 π K(t)e iωt dt = B(ω) + iω[a(ω) A ]. (24)

33 Using A(ω) and B(ω) it is possible to built the transfern function for every frequency ω {ω i } N n=1 : K(iω j ) = B(ω j ) + iω j (A(ω j ) A ) (25)

34 Using A(ω) and B(ω) it is possible to built the transfern function for every frequency ω {ω i } N n=1 : K(iω j ) = B(ω j ) + iω j (A(ω j ) A ) (25) A Matlab Toolbox done by Pérez and Fossen, compute values θ = [p, p 1,..., p n 1, q, q 1,..., q n 1 ] such that: ˆK(s, θ) = p n 1s n 1 + p n 2 s n p 1 s 1 + p s n + q n 1 s n 1 + q n 2 s n q 1 s 1 + q (26) is the best possile approximation (least square sense) for K(iω), ω {ω i } N n=1

35 Using A(ω) and B(ω) it is possible to built the transfern function for every frequency ω {ω i } N n=1 : K(iω j ) = B(ω j ) + iω j (A(ω j ) A ) (25) A Matlab Toolbox done by Pérez and Fossen, compute values θ = [p, p 1,..., p n 1, q, q 1,..., q n 1 ] such that: ˆK(s, θ) = p n 1s n 1 + p n 2 s n p 1 s 1 + p s n + q n 1 s n 1 + q n 2 s n q 1 s 1 + q (26) is the best possile approximation (least square sense) for K(iω), ω {ω i } N n=1 This means that this toolbox provides element θ that minimizes the following operatos: θ = mín θ ( n ) ( K(iω n ) ˆK(iω ) 2 n, θ) (27)

36 Properties the identification must : ĺım ˆK(iω) = It has ceros for s = (28) ω

37 Properties the identification must : ĺım ˆK(iω) = It has ceros for s = (28) ω ĺım ˆK(iω) = It is strictly proper (29) ω

38 Properties the identification must : ĺım ˆK(iω) = It has ceros for s = (28) ω ĺım ˆK(iω) = It is strictly proper (29) ω ĺım ˆK(t) Relative order 1 (gr(q) = gr(p)+1) (3) t +

39 Properties the identification must : ĺım ˆK(iω) = It has ceros for s = (28) ω ĺım ˆK(iω) = It is strictly proper (29) ω ĺım ˆK(t) Relative order 1 (gr(q) = gr(p)+1) (3) t + ĺım ˆK(t) = BIBO Stable (31) t

40 Properties the identification must : ĺım ˆK(iω) = It has ceros for s = (28) ω ĺım ˆK(iω) = It is strictly proper (29) ω ĺım ˆK(t) Relative order 1 (gr(q) = gr(p)+1) (3) t + ĺım ˆK(t) = BIBO Stable (31) t Ẋ y is passive Re( ˆK(iω)) > (32)

41 Once we know θ = [p, p 1,..., p n 1, q, q 1,..., q n 1 ]:... q 1... q 1 A c = 1... q ; B c = p p 1 p 2. ; (33)... 1 q n 1 p n 1 C c = [,,...,, 1] y c (t) = CX c (t) = x n (t) (34).

42 CatAir

43 CatAir

44 CatAir t (M + A ) z(t)+ K(t τ)z(τ)dτ +Cz(t)+k l ż+k nl ż ż = f e (t) (35)

45 CatAir t (M + A ) z(t)+ K(t τ)z(τ)dτ +Cz(t)+k l ż+k nl ż ż = f e (t) (35) t (M + A ) z(t)+ K(t τ)z(τ)dτ+cz(t)+(k l + k nl ż ) ż = f e (t) (36)

46 CatAir t (M + A ) z(t)+ K(t τ)z(τ)dτ +Cz(t)+k l ż+k nl ż ż = f e (t) (35) t (M + A ) z(t)+ K(t τ)z(τ)dτ+cz(t)+(k l + k nl ż ) ż = f e (t) (36) (M + A ) k 2 (t) + y c (t) + Cz(t) + (k l + k nl ż) ż = f e (t) (37) k 2 (t) = C c C X c (t) z(t) k l + k nl ż ż + f e(t) (38) M + A M + A M + A M + A

47 CatAir We have defined X (t) such that X n (t) = ż(t), then: A = A c. B c... 1 C c /a b/a (k l + k nl X n )/a ; B = 1/a (39) C = [... 1 ] (4)

48 CatAir

49 CatAir ṁ air = d dt (ρ air (t)v air (t)); V air (t) = V Sz(t) = S(L z(t)). (41) Assuming the proccess is isentropic, ρ air (t)(p + P(t)) 1/γ = K The air mass flux through the openning can be expressed as follows: [ ] 2mair (t) P(t) 1/2 ṁ air (t) = c d A v sgn( P(t)) (42) V air (t)

50 CatAir Lets define the relative pressure P (t) = P(t)/P = ( P(t) + P )/P. The air pressure variation inside the chamber, Ṗ (t) = [ A v c d γk ( ) v 1/2 sgn(p (t) 1) 2P ρ P (t) 1/γ P (t) 1 ρ S Cummins equation as solved using a state-space, +γż(t)] P (t) = ΓP (t) (43) Ẋ (t) = AX (t) + Bu(t) y(t) = CX (t) Cummins equation is modified to include pressure effects, (44) (m + A ) z(t) + y c (t) + ρgsz(t) + kż(t) P(t)S = f e (t) (45)

51 CatAir Writting together equations (45) and (43), we can write a new state-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 = A S... Γ ; B = B ; C = [ C ]. (47)

52 CatAir Figura: Laboratory and time-domain model free surface and chamber dynamic pressure time series. Regular waves (H =,8 m, T = 1,1 s), top opening=4.5 mm, c d =,8643

53 WEFIES P = C PTO (ż 1 ż 2 ) 2.

54 WEFIES Laboratory tests coming (hopefully) soon...

55 WEFIES Thanks for your attention!

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