Disibuyuk, N. B.

Korobkin, A. A.

English

Korobkin, A Alexander

Dişibüyük, Nazile Buğurcan

Dokuz Eylul University Research Information System

Wave forces on a vertical cylinder with non circular cross section

University of East Anglia digital repository

The 30th International Workshop on Water Waves and Floating BodiesWave Forces On A Vertical Cylinder With Non-Circular CrossSectionN.B. Dis¸ibu¨yu¨k1, A.A. Korobkin2,1 Department of Mathematics, Dokuz Eylul University, Izmir, Turkeye-mail: bugurcan.ruzgar@deu.edu.tr2 School of Mathematics, University of East Anglia, Norwich, UK,e-mail: a.korobkin@uea.ac.uk1 IntroductionWe consider the linear problem of water waves scattering by a vertical cylinder with non-circular crosssection extending from the sea bottom to the free surface in water of finite depth h. We assume aplane wavetrain incident from x ∼ −∞ and propagating at an angle α to the positive x−directiontoward a vertical cylinder whose cross section is described by the equation r = R+ εf(θ) with ε 1.The function f(θ) describes the deviation of the shape of the cylinder from the circular one, f(θ) = 0corresponds to the circular cylinder with radius R. The problem of wave scattering by a nearly circularcylinder was formulated in [1]. The top view of the problem is shown in Figure 1. The problem ofwave diffraction by a vertical cylinder has been solved by a number of researchers for many differentshapes. The challenge of the present study is to solve the complex body geometries with less effort.The abstract presents the results which have been obtained for one simple geometry: a cylinder withelliptic cross section.Figure 1: Top view of the problem configuration.2 Mathematical Formulation of The ProblemThe linear boundary problem is formulated with respect to the velocity potential Φ(r, θ, z, t)Φ(r, θ, z, t) = <{gAωcosh[k(z + h)]cosh(kh)φ(r, θ)e−iwt},where φ satisfies the Helmholtz equation (∇2 + k2)φ = 0 in the flow region, A is the incident waveamplitude, k = 2piλ is the wave number, λ is the incident wave length, ω is the wave frequency relatedto the wave number k by the dispersion relation ω2 = gk tanh(kh), where g is the gravitationalacceleration. The coordinate system (r, θ, z) is used with the origin at the free surface and the z-axisdirected upwards. The axis of the corresponding circular cylinder with ε = 0 coincides with the z-axis.The boundary condition on the cylinder r = R+ εf(θ) is∂φ∂n= 0 on r = R+ εf(θ), −h < z < 0, (1)where ~n is the unit normal vector on the cylinder. This boundary condition can be written as∂φ∂r(R+ εf(θ), θ)− εf′(θ)[R+ εf(θ)]2 ∂φ∂θ (R+ εf(θ), θ) = 0. (2)We approximate the boundary condition (2) up to O(ε5) using the Taylor expansion at r = R andsubstitute the fifth order asymptotic expansion of the potential φφ(r, θ) = φ0(r, θ) + εφ1(r, θ) + ε2φ2(r, θ) + ε3φ3(r, θ) + ε4φ4(r, θ) +O(ε5), (3)into the boundary condition (2) with the resultφ0,r + ε[φ1,r + f(θ)φ0,rr − f′(θ)R2φ0,θ]+ ε2[φ2,r + f(θ)φ1,rr − f′(θ)R2φ1,θ +f2(θ)2φ0,rrr +2f(θ)f ′(θ)R3φ0,θ − f(θ)f′(θ)R2φ0,rθ]+ ε3[φ3,r + f(θ)φ2,rr − f′(θ)R2φ2,θ +f2(θ)2φ1,rrr +2f(θ)f ′(θ)a3φ1,θ − f(θ)f′(θ)R2φ1,rθ+2f2(θ)f ′(θ)R3φ0,rθ +f3(θ)6φ0,rrrr − 3f2(θ)f ′(θ)R4φ0,θ − f2(θ)f ′(θ)2R2φ0,rrθ]+ ε4[φ4,r + f(θ)φ3,rr − f′(θ)R2φ3,θ +f2(θ)2φ2,rrr − f(θ)f′(θ)a2φ2,rθ +2f(θ)f ′(θ)R3φ2,θ− 3f2(θ)f ′(θ)R4φ1,θ − f2(θ)f ′(θ)2a2φ1,rrθ +2f2(θ)f ′(θ)a3φ1,rθ +f3(θ)6φ1,rrrr− 3f3(θ)f ′(θ)R4φ0,rθ +f3(θ)f ′(θ)a3φ0,rrθ − f3(θ)f ′(θ)6a2φ0,rrrθ +f4(θ)24φ0,rrrrr +4f3(θ)f ′(θ)R5φ0,θ]= 0,(4)where the functions are computed at r = R. Since the right hand side of this equation is zero, thecoeffcients of εi, i = 0, 1, 2, 3, 4, on the left hand side of (4) are equal to zero. Using (4) we derive fiveboundary conditions for five unknown potentials φi(r, θ), i = 0, 1, 2, 3, 4, first two of them are:φ0,r(R, θ) = 0, (5)φ1,r(R, θ) =1R2f ′(θ)φ0,θ(R, θ)− f(θ)φ0,rr(R, θ). (6)It is clear that φ0(r, θ) is the velocity potential of the diffraction problem for the circular cylinderr = R with the solution (see [1])φ0(r, θ) =∞∑m=0mim[Jm(kr)− J′m(kR)H(1)′m (kR)H(1)m (kr)]cos[m(θ − α)],which satisfies equation (5), where m is the Neumann symbol which is given by 0 = 1, m = 2,m ≥ 1. The series converges exponentially as m→∞.The most general representations of φi(r, θ), i = 1, 2, 3, 4, which satisfy the radiation condition atinfinity areφi(r, θ) =∞∑m=0[Ci,m cos[m(θ − α)] +Di,m sin[m(θ − α)]]H(1)m (kr),where the evanescent modes are not included (see [1]) and Ci,m and Di,m, i = 0, 1, 2, 3, 4, are unknowncoeffcients. The coefficients can be determined using the boundary conditions (5),(6) and the other 3conditions and hence we can find the velocity potentials with the accuracy O(ε5).We also assume that f(θ) can be written as a Fourier seriesf(θ) =f c02+∞∑m=1[f cm cos(mθ) + fsm sin(mθ)],where the coeffcients f ci and fsi , i = 0, 1, 2, . . . depend on a particular shape of the vertical cylinderin waves. If the function f(θ) is independent of ε then the right hand side of the conditions (5), (6)and the other 3 conditions depends only on θ. So we can write these conditions as φi,r(a, θ) = Gi(θ),i = 0, 1, 2, 3, 4, where Gi(θ) are represented by their Fourier series. After writing Gi(θ) as Fourierseries, we can find the unknown coefficients Ci,m and Di,m, i = 0, 1, 2, 3, 4.If the function f(θ, ε) depends on θ and ε, then we can use the asymptotic expansion of f(θ, ε) asε→ 0:f(θ, ε) = f0(θ) + εf1(θ) + ε2f2(θ) + ε3f3(θ) + ε4f4(θ) +O(ε5),or higher order, and substituting this into (4) and then applying the same procedure as in the previouscase we can find the unknown coefficients. As an example of this case, we have solved a problem forthe cylinder with elliptic cross section of small eccentricity in the next section and calculated thehydrodynamic forces acting on this cylinder. The x and y components of the hydrodynamic force dueto the fluid motion are given byFx = <{−iρgA tanh(kh)k[∫ 2pi0φ(a+ εf(θ), θ)[εf ′(θ) sin θ + [a+ εf(θ)] cos θ] dθ]e−iwt}, (7)Fy = <{−iρgA tanh(kh)k[∫ 2pi0φ(a+ εf(θ), θ)[−εf ′(θ) cos θ + [a+ εf(θ)] sin θ] dθ]e−iwt}. (8)Dividing Fx and Fy by ρgApia2 tanh(kh), we arrive at the non-dimensionalized force components F˜xand F˜y.3 Example: Elliptic cylinderThe ellipse’s equation in the polar coordinates with the origin at the focus readsr =a(1− e2)1− e cos θ , (9)where e =√1− b2a2, 0 < e < 1, is the eccentricity of the ellipse, a is semi-major axis, b is semi-minoraxis. Assuming e 1 and setting e = ε we can write (9) in the form of Fourier series and then usingTaylor expansion about ε = 0 we obtainr = R+ εf(θ) = a√1− ε2 + 2a√1− ε2∞∑n=1[ε1 +√1− ε2]ncos(nθ)= a+ εa cos θ − ε2a sin2 θ − ε3a cos θ sin2 θ + ε4a cos2 θ sin2 θ +O(ε5).Hence,R = a,f(θ) = a cos θ − εa sin2 θ − ε2a cos θ sin2 θ + ε3a cos2 θ sin2 θ +O(ε4) (10)in (2). Rewriting conditions (5),(6) and the other 3 conditions for the function (10) we get:φ0,r(a, θ) =0,φ1,r(a, θ) =− 1asin θφ0,θ(a, θ)− a cos θφ0,rr(a, θ),φ2,r(a, θ) =− a cos θφ1,rr(a, θ) + a sin2 θφ0,rr(a, θ)− a2 cos2 θ2φ0,rrr(a, θ)− 1asin θφ1,θ(a, θ).In the same manner, we can write the conditions for φ3,r(a, θ) and φ4,r(a, θ).Substituting (10) into the equations (7), (8) and dividing them by ρgApia2 tanh(kh), we get non-dimensionalized force components F˜x, F˜y for elliptic cylinder.4 Results and ConclusionsWe have applied the asymptotic analysis described above to scattering problems of one-dimensionalwaves by a vertical cylinder with non-circular cross section. For an elliptic cylinder, we comparedour results with the results by Williams [2] who used the expansion of the exact expressions for theforces which are given by the Mathieu functions for small values of the elliptic eccentricity parameter.The present asymptotic approach provides a good approximation for the forces exerted on the ellipticcylinder with eccentricity ε = 0.5 to the incident wave for values of α = 0◦ and α = 90◦ (see Figure 2,Figure 3). We found that, if the incident wave makes zero angle with the positive x-axis and ka→ 0,thenF˜x =[2− 32ε2 − 18ε4 +O(ε6)]cosωt,and if the incident wave makes the angle of pi2 with the positive x-axis and ka→ 0, thenF˜y =[2− 12ε2 − 18ε4 +O(ε6)]cosωt,which coincide with the asymptotic formulae from [2].In conclusion, for different shapes of a vertical cylinder, r = R+ εf(θ), we can find the wave forcesacting on the cylinder by using the formulae (7) and (8).Figure 2: The x-component of the non-dimensionlized force on elliptic cylinder for ε =0.5. (Solid curve is from [2], dotted curve is bythe present method).Figure 3: The y-component of the non-dimensionlized force on elliptic cylinder for ε =0.5. (Solid curve is from [2], dotted curve is bythe present method).Acknowledgement: This research was performed while N. Bug˘urcan Dis¸ibu¨yu¨k was visiting theSchool of Mathematics, University of East Anglia as a visiting research scientist during the periodfrom October 2014 to February 2015. Her visit was supported by YO¨K (http://www.yok.gov.tr), TheCouncil of Higher Education. This support is greatly acknowledged.References1. Mei C.C., Stiassnie M., K. Dick, Yue P. (2005) Theory And Applications of Ocean SurfaceWaves, Part1: Linear Aspects. World Scientific. Vol. 23, pp. 363-375.2. Williams A. N. (1985) Wave Forces On An Elliptic Cylinder. Journal of Waterway, Port, Coastaland Ocean Engineering. Vol. 111, pp. 433-449.3. Gradshteyn, I.S. & Ryzhik I.M. (1996) Tables of Integrals, Series, and Products. AcademicPress.

Wave Forces On A Vertical Cylinder With Non-Circular Cross Section

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Wave Forces On A Vertical Cylinder With Non-Circular Cross Section

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