Seakeeping of Ships By Professor Masashi KASHIWAGI Problem Set: No. Date: 27 January 2010 Due: 28 February 2010 By using the computer program (iem2d.f) for the boundary-element (Green function) method introduced in the lecture note (see the lowest bottom of this assignment for your download), let us understand various hydrodynamic relations proven theoretically and the frequency dependence in hydrodynamic forces, particularly in the added-mass and damping coefficients. [ 1 ] As an assignment No., work on the following items one by one. As a 2-D body to be computed, let us consider a so-called Lewis form, which can be represented by the following conformal mapping: z = x + i y = M ζ + a 1 ζ + a ζ, ζ = ξ + i η (1) Here M, a 1, and a are unknown three parameters (in particular M is the expansion-contraction coefficient or scale factor), which may be determined by specifying three physical quantities; they are usually taken as half breadth B/2 (= b), draft d, and sectional area S. As summarized in page 2 of the lecture note, once the half-breadth-to-draft ratio H 0 = b/d and the sectional area ratio σ = S/(Bd) are given, the nondimensional coordinates (x = x/b, y = y/b) can be calculated. Summarize the calculation procedure for the coordinates of a Lewis form (if possible, explain mathematical transformation of each equation in the procedure). Then draw the geometry (contour) of a Lewis form by specifying values of H 0 and σ as you like. (Note that σ is usually less than 1.0 by definition and H 0 is also around 1.0 for conventional section forms.) The coordinates are actually computed in Subroutine OFFSET in the computer program. So you may just plot computed results using a software available in your computer. [ 2 ] For the Lewis form that you selected in the above, compute the sway added mass (A 11 ) and damping coefficient (B 11 ) for various values of the frequency ω. In the computer program, the nondimensional wavenumber Kb = ω 2 b/g (denoted as AKB in the computer code) is supposed to be input, and the added-mass and damping coefficients will be given in nondimensional forms of A 11 /ρb 2 and B 11 /ρb 2 ω. So, you may plot these nondimensional values against Kb in a graph, using a software available in your computer. Note that large Kb implies short wavelength, and thus for larger values of AKB, you need to increase the number of division (NB) over the section contour to keep numerical accuracy. The accuracy may be confirmed by checking the degree of satisfaction in the energy-conservation principle, Haskind- Newman relation, and so on. You should realize that there must be a practical upper limit for the value of Kb to be specified in numerical computations, because of a limitation in the numerical resolution (accuracy) with practical number of panels (NB) on the body surface. 1
[ ] In the limit of Kb 0 (ω 0), the corresponding physical situation is such that the free surface is rigid and no waves are generated and thus the flow around a body is symmetric about y = 0, which is equivalent to a flow around the so-called double-body model (the body shape for y > 0 is reflected about y = 0 to form the image body with the same geometrical form as if the plane y = 0 is a mirror) which moves in the horizontal direction in an unbounded fluid. In this special case, the sway added mass can be analytically obtained even for a Lewis form, and the result takes the following form: A 11 ρb 2 = 2σ H 0 (1 a 1 ) 2 + a 2 1 a 2 1 a2 (2) Therefore, check whether computed results approach the above result as ω 0. results you computed may be something wrong.) [ 4 ] (Otherwise the By following the procedure written below, derive (prove) the analytical expression shown above for the sway added mass. First, we note that the complex velocity potential for the uniform flow f(z) = U z in the z-plane corresponds to f(ζ) UM ζ at infinity in the ζ-plane. Then let us consider the uniform flow around a circular cylinder with unit radius (r = 1) in the ζ-plane, in which the strength (velocity) of this uniform flow is assumed to be M (i.e. U = 1) and the flow direction is from positive to negative of the ξ-axis. In this case, the complex velocity potential is written as f(ζ) = () In order to consider the corresponding flow in the z-plane in terms of the conformal mapping and to extract only the component of disturbance due to the presence of a body, the only thing to do is simply to superimpose the uniform flow in the z-plane directed to the positive x-axis with unit strength (because the flow at infinity must be at rest). Therefore in terms of Eq. (1), the complex velocity potential in the z-plane representing the disturbance flow may be obtained as F (z) f(ζ) + z = (4) Substituting ζ = i e iθ with r = 1 into the above, we can obatin the disturbance velocity potential on the body surface, ϕ(x, y), with unit velocity (U = 1) in the form ϕ(x, y) = Re [ F (z) ] = M (1 a 1 ) sin θ + a sin θ With this velocity potential, the sway added mass in the z-plane can be computed by A 11 = ρ ϕ(x, y) n x dl (6) S H (This equation is essentially the same as that explained in the lecture note, but if you cannot understand, you should take this as the definition.) Although we are considering computations in the z-plane, actual integrals may be analytically performed in terms of θ by using the following relation n x dl = x y dl = dθ = (7) n θ (5) 2
Performing required integrals and noting that the actual Lewis form to be considered is just a half of the double-body model, we may obtain the following result: A 11 = (8) = 1 2 ρπm 2 (1 a 1 ) 2 + a 2 (9) (Show the details of your analytical transformation to be written as Eq. (8).) M 2 can be recast in terms of H 0, σ, a 1, a and b as follows: 1 2 πm 2 = (10) from which you may confirm that Eq. (2) is true for the sway added mass in the limit of ω = 0. [ 5 ] In the same way as in [ 2 ], plot the heave added mass (A 22 ) and damping coefficient (B 22 ) against the nondimensional wavenumber Kb. In this case of heave, the asymptotic behavior of A 22 as ω 0 must be different from that in A 11. A reason for this is as follows. At the low-frequency limit, as already mentioned, the plane y = 0 becomes rigid and the flow (the velocity potential) must be symmetric with respect to y = 0. Therefore, symmetry in the heave motion requires the phase of the image to be opposite to that of the real body. In this case, the motion of the body plus its image corresponds to an oscillatory dilation of the body; which is obviously different from the motion of the rigid double body as in sway. If we consider a flow around the rigid double body in heave, the flow must be antisymmetric about y = 0 (odd function in y) and thus the condition on y = 0 for the velocity potential becomes just ϕ = 0 (not the rigid wall any more), which corresponds to another limit of ω in the free-surface boundary condition. In order to understand these situations, consider asymptotic forms of the linearized free-surface boundary condition for both cases of ω 0 and ω. ϕ + K ϕ = 0 on y = 0 (11) y Note: The computer program (iem2d.f) can be downloaded from http://www.naoe.eng.osaka-u.ac.jp/kashi/iem2d.f
Seakeeping of Ships Comments on Assignment-No. First, let me apologize a mistake in the assignment (a sort of confusion) in taking the direction of θ in the conformal mapping. (The positive direction can be defined as you like, but that definition must be consistent throughout the work.) According to the explanation in pages 41 42 of the lecture note, ζ is defined as ζ = i e iθ and then we obtain x = M (1 + a 1 ) sin θ a sin θ y = M (1 a 1) cos θ + a cos θ Therefore in the derivation of the velocity potential on the body surface, Eqs. (4) and (5) in the assignment, ζ = i e iθ should be substituted (ζ = e iθ written in the assignment is wrong). In this case, the integration range in evaluating A 11 must be from π/2 to π/2 (or twice the integral from 0 to π/2). Next, some of you seemed to be unable to confirm good coincidence of the sway added mass for Kb 0 with the corresponding analytical value: A 11 ρb 2 = 2σ H 0 (1 a 1 ) 2 + a 1 a 2 1 a2 After checking the value with a pocket calculator, I found that a mistake in your report occurred in calculating the analytical value above. Please check again if you could not see good coincidence (computed result must be very accurate). Last, let me write some comments regarding asymptotic values of the heave added mass (A 22 ) and damping coefficient (B 22 ) for Kb 0. You may understand that the wave-exciting force in heave as Kb 0 (i.e. in a very long incident wave) becomes the same as the restoring force, and thus it follows from Haskind-Newman s relation that (12) E 2 = ρgζ ab = ρgζ ah + 2 (B = 2b) (1) Therefore H + 2 = 2b as Kb 0 (14) On the other hand, from the energy conservation principle the damping coefficient can be computed by B 22 = ρω H + 2 2 = ρω(2b) 2 B22 = 4.0 as Kb 0 (15) ρb 2 ω You should confirm again that the nondimensional value of B 22 for Kb 0 approaches 4.0 in the numerical computation, too. It should be noted that Eqs. (1) to (15) hold irrespective of the body shape (only the breadth is concerned). The analytical analysis for the heave added mass as Kb 0 is somewhat complicated. cylinder, it was shown by Ursell (also shown in my private note, known as Kashiwagi s note) that A 22 ρb = 4 ( ln Kb ) 2 π 2 + 2 ln 2 + γ as Kb 0 where γ denotes Euler s constant equal to 0.5772156. For a circular This characteristic of becoming logarithmically infinite as Kb 0 is the same irrespective of the body shape, and I suppose you have realized this characteristic through your results computed. Regarding analytical expressions for the added masses in sway and roll (including cross terms) for Kb = 0, although details are not described here, it is known that the following results hold: A 11 = 1 2 ρπm 2 (1 a 1 ) 2 + a 2 4
l 1 (0) d A 1 = A 1 = A 11 l 1 (0) ( 4 = 16 a 1(1 a 1) + a 5 + 4 ) ( 5 a1 a2 1 + a 2 5 a1 12 ) 7 π (1 a1 ) 2 + a 2 (1 a1 + a ) ( κ(0) d ) 2 = 2 π 2 A = A 11 κ 2 (0) a 2 1(1 + a ) 2 + 8 9 a 1a (1 + a ) + 16 (1 a1) 2 + a 2 (1 a1 + a ) 2 9 a2 Therefore H + 1 1 a 1 b iπ(kb) as Kb 0 (1 + a 1 + a ) 2 B 11 = ρω H + 2 1 = ρωb 2 π 2 (Kb) 2 (1 a 1) 2 (1 + a 1 + a ) 4 B ( ) 11 ρb 2 ω = 1 a 2 1 πkb as Kb 0 (1 + a 1 + a ) 2 We can easily confirm from this expression that the nondimensional value of B 11 becomes zero as you may have realized by numerical computations. If you study more, you may be able to understand other results of asymptotic values, for instance, in my private note. Good luck! 5