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Seakeeping: Complex numbers Compute: (3 4i ) (2 2i ) (3 4i ) (2 2i ) (3 4i ) (2 2i ) 3 4i 2 2i Euler's formula brings trigonometry and exponential function together: eix cos x i sin x where i 1 . Show for one example that Re Aˆ eix A sin( x ) , where Â is the complex amplitude with real part Re() and imaginary part Im() and absolute value A = Â and the phase shift is tan Re( Aˆ ) / Im( Aˆ ) . Take as an example Â = 3+4i. Solution We use elementary mathematical operations and get: (3 4i ) (2 2i ) 5 2i (3 4i ) (2 2i ) 1 6i (3 4i ) (2 2i ) 6 8i 6i 8 14 2i 3 4i 3 4i 2 2i 2 14i 0.25 1.75i 2 2i 2 2i 2 2i 8 Compute the phase shift: arctan Re( Aˆ ) Re( 3 4i ) 3 arctan arctan 36.87° Im( 3 4i ) 4 Im( Aˆ ) The (real valued) amplitude is: A Aˆ Re( Aˆ )2 Im( Aˆ ) 32 42 5 Re Aˆ eix Re(3 4i )(cos x i sin x) 3 cos x 4 sin x and A sin( x ) 5 sin( x 36.87) 5(sin x cos 36.87 sin 36.87 cos x) 4 sin x 3 cos x The two expressions are equal. Seakeeping: Celerity of wave A surf board travels with Froude number Fn = 0.5 on a deep-water wave of 4 m length. How long must the surf board be to ‘ride’ on the wave, i.e. to have the same speed as the wave? Solution The speed of the surf board is: V Fn gL The celerity of the wave is: c g 2 Both velocities must be the same for surf riding: V c Fn gL g 2 L 2 F 2 n 4 2.5 m 2 0.52 Seakeeping: Velocity and pressure in elementary wave Consider a regular deep-water wave of wave length =100 m and wave height 5 m. a) What is the orbital velocity in 10 m depth? b) What is the amplitude of the pressure fluctuation (absolute and in percent of the hydrostatic pressure)? The pressure fluctuation is (linearised): p f t Solution a) The amplitude of the wave is half the wave height: h = 2.5 m. 2 0.0628 m1 100 The frequency is: g k 9.81 0.0628 0.785 s1 The wave number is: k 2 The orbital velocity is the maximum of the x-velocity (for deep waves). The x-velocity is: v x x Re h e kz e i (t kx ) Thus the amplitude is: v x h e kz 0.785 2.5 e 0.062810 1.05 m/s b) The pressure fluctuation (=pressure without hydrostatic part) is given by: p f t g Re h e kz e i (t kx ) The amplitude of the pressure fluctuation is then: p f gh e kz 1025 9.81 2.5 e 0.062810 13.4 kPa The hydrostatic pressure is p stat gz 1025 9.81 10 100.6 kPa Thus the pressure fluctuation is 13% of the hydrostatic pressure. Seakeeping: Velocity and acceleration in shallow-water wave The potential of a regular wave on shallow water is given by: Re ich sinh( kH ) cosh( k ( z H ))ei (t kx ) wave length = 100 m wave amplitude h = 3 m water depth H = 30 m Determine velocity and acceleration field at a depth of z = 20 m below the water surface! The following parameters are given: Solution The velocity is derived by differentiation of the shallow-water potential: h h vx x Re cosh( k ( z H ))ei (t kx ) cosh( k ( z H )) cos(t kx) sinh( kH ) sinh( kH ) ih h vz z Re sinh( k ( z H ))ei (t kx ) sinh( k ( z H )) sin( t kx) sinh( kH ) sinh( kH ) This derivation used the relation c = /k. The individual values are: 2 2 k 0.06283 m1 100 gk tanh( kH) 9.81 0.06283 tanh( 0.06283 30) 0.767 s1 This yields: vx 0.861 cos(0.767t 0.06283x) vz 0.480 sin( 0.767t 0.06283x) The accelerations are obtained by differentiating the velocities with respect to time: ax xt 0.661 sin( 0.767t 0.06283x) az zt 0.368 cos(0.767t 0.06283x) The velocities have a phase shift of 90° and different amplitudes. The particles will thus trace an ellipse. Seakeeping: Wave breaking Wave breaking occurs theoretically when particle velocity in x-direction is larger than the celerity c of the wave. In practice, wave breaking occurs for wave steepness h/ exceeding 1/14. What is the theoretical and practical limit for h concerning wave breaking in a deep-water wave of = 100 m? Solution Practically the limit is h = /14 = 100/14 = 7.143 m. 2 0.0628 m1 100 The maximum particle velocity in x-direction is: vx, max h ekh The wave number is: k The celerity is: c 2 k The condition c = vx,max yields: 1 e kh kh The above expression is solved iteratively for kh, using kh e kh . The iteration starts with the practical limit: k0h = 7.143 0.0628 = 0.4488, then yields: 1 0.4488 11 0.5667 2 0.6384 12 0.5674 3 0.5281 13 0.5670 kh 0.5671 h 4 0.5897 14 0.5672 5 0.5545 15 0.5672 0.5671 9.026 m 0.0628 6 0.5744 16 0.5671 7 0.5631 17 0.5671 8 0.5695 18 ... 9 0.5658 10 0.5679 Seakeeping: Wave with two buoys The position of two buoys is given by their (x,y) coordinates in [m] as sketched below. The buoys are excited by a regular wave of = 62.8 m, amplitude h = 1 m, and angle = 30° to the x-axis. a) What is the maximum vertical relative motion between the two buoys if they follow exactly the waves? b) What is the largest wave length to achieve maximum vertical relative motion of twice the wave amplitude? Solution a) The wave number of this wave is: k 2 2 0.1 m1 62.8 Transform coordinates in local ξ-system. Points on ξ = const. have same ζ-values for regular waves. The two buoys have then coordinates: ξ1 = 0 ξ2 = x2 cos 30° y2 sin 30° = 30 cos 30° 10 sin 30° = 21 m The relative difference between the wave elevations is: 1 2 Re h ei (t k ) Re h ei (t k 1 2) Reh e 1 e it ik 2 ) The amplitude (= maximum) of this relative motion is given by: h e it 1 e ik 2 ) h 1 cos( k 2 ) 2 sin( k 2 ) 2 1 1 cos(0.1 21) sin( 0.1 21) 2 2 = 1.73 m b) The maximum difference between wave elevations is twice the wave amplitude if the buoys are spaced by an odd multiple of half the wave length. The longest wave is obtained for a spacing of half the wave length: 2 21 m = 42 m All other wave lengths fulfilling the criterion are even shorter. Seakeeping: Encounter frequency A ship travels with 28.28 kn in deep sea in regular sea waves. The ship travels east, the waves come from southwest. The wave length is estimated to be between 100 m and 200 m. The encounter period Te is measured at 31.42 s. a) What is the length of the seaway? b) Two days before, a storm started in an area 1500 km southwest of the ship's position. Can the waves have their origin in this storm area? Solution The speed is V = 28.28 0.5144 m/s = 14.55 m/s. The encounter period yields the encounter frequency: The value of is: eV g e 2 2 0.2 s1 Te 31.42 0.2 14.55 0.2966 9.81 There are three possible frequencies which could excite this encounter frequency: g 9.81 1 1 4 cos 1 1 4 0.2966 cos 45 1.123 Hz 2V cos 2 14.55 cos 45 g 9.81 2 1 1 4 cos 1 1 4 0.2966 cos 45 0.6684 Hz 2V cos 2 14.55 cos 45 g 9.81 3 1 1 4 cos 1 1 4 0.2966 cos 45 0.2854 Hz 2V cos 2 14.55 cos 45 1 The corresponding wave lengths are: 2 9.81 49 m 1.1232 2g 2 9.81 2 2 138 m 2 0.66842 2g 2 9.81 3 2 756 m 3 0.28542 1 2g 2 1 Only 2 = 138 m fits the observed bandwidth of wave lengths. Wave groups travel with group velocity. For deep ocean water, this is cgr 1 g 1 9.81 138 7.34 m/s 2 2 2 2 At this speed, waves can travel within two days: s = 48 h 60 min/h 60 s/min 7.34 m/s = 1268 km < 1500 km The waves can thus not originate from the storm area. Seakeeping: Undamped free heave oscillation The differential equation of a free undamped heave oscillation is: m m33 z cz 0 m is the displacement of the ship, m33 the added mass for heave motion. c is the restoring force coefficient. The (circular) natural frequency is subsequently: z c m m33 Consider the ship with the following particulars: L = 125 m, B = 17 m, T = 7 m, CB = 0.75, Cwp = 0.8, m33 = 0.9m. a) Determine the (circular) natural frequency and natural period of heave! b) Initial conditions are: At t = 0 we have z = 0 and z =0.5 m/s. What is the maximal load a winch of 4000 kg mass exerts on its foundation? Solution a) The natural frequency follows from: z c m m33 g Cwp L B CB L B T (1 m33 / m) g Cwp CBT (1 m33 / m) 9.81 0.8 0.89 Hz 0.75 7 (1 0.9) 2 2 Tz 7.08 s z 0.89 b) The winch exerts its maximal load when gravity g and amplitude of heave oscillation are superposed. We write the heave motion as: z Re Kˆ ei z t Kˆ K1 iK 2 is determined by the initial conditions: z Re Kˆ ei z 0 Re Kˆ K 0 t 0 zt 0 Re i Kˆ e K 1 i z 0 z 2 z 0.5 K 2 Then the amplitude of the heave acceleration is given by: z z2 K2 0.892 0.56 0.45 m/s2 The maximal load of the winch is then: mwinch ( g z) 4000 (9.81 0.45) 41 kN 0.5 z 0.5 0.56 m 0.89 Seakeeping: Power requirements for a wave maker A wave maker is to be designed for a towing tank of width B = 4 m and depth H = 2.5 m. The wave maker shall be designed for a wave of 5 m length and 0.2 m amplitude. a) What is the power requirement for the motor for the wave maker if we assume 30% total efficiency between motor and wave? (For the considered wave length, the depth can be regarded as ‘deep’. The power requirement of the wave is (energy/meter [in direction of wave propagation]) * group velocity, as the energy in a wave is transported with group velocity.) b) After switching the wave maker off, for a long time there is still a wave motion with period 40 s observed in the tank. How long is the tank if the motion is due to the lowest natural frequency of the tank? Solution a) The group velocity is: 1 1 g 1 9.81 5 cgr c 1.397 m/s 2 2 2 2 2 The average energy per area is: E 1 1 gh 2 1000 9.81 0.22 196.2 N/m 2 2 Thus the power of the wave is: P E B cgr 196.2 4 1.397 1096 W The power of the motor then needs to be: Pm P / 0.3 3.65 kW b) The wave length is now long compared to the depth. Thus we have to use finite water depth expressions: k tanh( kH ) With 2 g 2 0.157 Hz, this yields: T k tanh( kH ) 0.157 2 0.002515 m1 9.81 This equation has to be solved iteratively. As the convergence is slow, it is useful to start with a good estimate. For small x, tanh(x) x. This would yield k 0.002515 / 2.5 0.03172 . We solve the problem using Newton's iteration: f (k ) f ' (k ) With f k tanh( 2.5k ) 0.002515 and f ' tanh( 2.5k ) k / cosh 2 (2.5k ) , we obtain within 7 kk iterations: k 0.03175 2 2 198 m. k 0.03175 The lowest natural frequency is at /2 = L, i.e. the tank has a length of 99 m. Seakeeping: Wave maker for required wave length and amplitude A wave maker consists of a cylinder which moves up and down at a wall. The wave maker has a cross section of half a Lewis section with Cm = 0.8. The draft of the section is T = 1 m, the half-width is B/2 = 0.6 m. Determine the amplitude and frequency of motion to create waves of 5 m length and 0.2 m amplitude. Note: The ratio between wave amplitude and body motion amplitude is Az . Solution The frequency corresponding to = 5 m is: 2g 2 9.81 3.51 Hz 5 Important parameters for Lewis section curves: 2 B 3.512 0.6 0.754 2g 9.81 B 0.6 H 0.6 2T 1 The curves for Lewis sections give Az = 0.52. Thus: h 0.2 u3 0.38 Az 0.52 m Seakeeping: Wave maker with Lewis section moves floating body A wave maker consists of a cylinder which moves up and down at a wall. The wave maker has a cross section of half a Lewis section with Cm = 0.8. The draft of the section is T = 1 m, its half-width B/2 = 0.6 m. The wave maker performs heave motions with 0.6 m amplitude and = 3.51 Hz. The basin is 10 m wide. The mass of the wave maker itself is negligibly small. Use linear theory in your solution approach. a) What is the amplitude of the force needed to move the wave maker? b) What is the average power supply if an efficiency of 70% is assumed? c) In the middle of the basin is a cylinder of Cm = 0.8, B = 1.2 m, T = 1 m, L = 5 m, arranged with axis parallel to the wave maker axis. The cylinder has all six degrees of freedom suppressed. What is the exciting force on the cylinder? d) If the cylinder in the middle of the basin is free to heave, what is its heave amplitude? Assume deep water. Solution a) The basic equation for heave motion is: m m 2 33 in33 c33 u3 f (The force is negative as now the body excites the water and not the water the body.) All quantities are in this equation per length, i.e. for the final computation they need to be multiplied by the tank width. The individual quantities are: 2 B 3.512 0.6 0.754 2g 9.81 B 0.6 H 0.6 2T 1 The curves for Lewis sections give Cz = 0.6, Az 0.52 . Thus: B2 1.22 0.6 1000 339 kg/m 8 8 g 2 1000 9.812 n33 Az2 3 0.522 601.8 kg/ms 3.513 c33 gB 1000 9.81 1.2 11770 N/m2 m33 Cz The mass of the wave maker is negligible: m = 0 Thus: 3.51 (0 339) i 3.51 601.8 11770 0.6 f 2 f (4556 1267i) N/m This force is for waves radiated in both directions, i.e. so far we assumed symmetry. The total force amplitude is then given by: | F | 1 1 | f | BTank 45562 1267 2 10 23.6 kN 2 2 b) The power needed to drive the wave maker is given by: P E BTank cgr gh2 c 1 BTank kN 2 2 The wave amplitude is: h | u3 | Az 0.6 0.52 0.312 m The celerity is: c Thus: g 9.81 2.79 m/s 3.51 P 1000 9.81 0.3122 2.79 1 10 9.5 kW 2 2 0.7 Now we check the result, using a different approach. The power needed is due to the damping force times the velocity. The mass force and the restoring force are phase shifted by 90° to the velocity and thus do not contribute to the power. Again we take into account that we have only half a cross section: 1 Im( f ) 1 1 1267 1 P BTank ( zˆ) 10 (3.51 0.6) 9.5 kW 2 2 2 2 0.7 c) We read from the diagrams for Lewis sections: Re fˆe 0.25 gBh Im fˆe 0.25 gBh f e, r 0.25 gBh 0.25 1000 9.81 1.2 0.312 918.2 N/m f e,i 918.2 N/m This then gives the absolute value for the three-dimensional force: Fe f e2,r f e2,i L 918.22 918.22 5 6.5 kN d) The basic equation for heave motion is: m m 2 33 in33 c33 u3 fe The mass (per length) follows from Archimedes: m Cm B T 1000 0.8 1.2 1.0 960 kg/m As before, we have then: 3.51 960 339 i 3.51 601.8 11770 u 2 4233.8 2112.3i u3 (918.2 918.2i) u3 0.0872 0.2602 ) 0.274 m 3 (918.2 918.2i ) u3 (0.087 0.260i) m Seakeeping: Cylinder with Lewis cross section in regular waves A cylinder of Lewis cross section (L = 10 m, B = 1 m, T = 0.4 m, Cm = 0.8) floats parallel to the wave crests of regular waves = 5 m, h = 0.25 m) which excite heave motions. Assume that the form of the free surface is not changed by the cylinder. What is the amplitude of relative motion between cylinder and free surface predicted by linear theory? Solution Fundamental equation for heave motion: m m 2 33 in33 c33 u3 fe The individual quantities needed are: 2 9.81 3.51 Hz 5 2 B 3.512 1 0.628 2g 2 9.81 B 1 H 1.25 2T 2 0.4 2g Then the curves for Lewis sections yield: Cz = 0.62, Az 0.68 , f er* 0.39 , f ei* 0.38 . This in turn gives: B2 12 0.62 1000 243 kg/m 8 8 m Cm B T 1000 0.8 1.0 0.4 320 kg/m m33 Cz n33 Az2 2 g 2 2 1000 9.81 0 . 68 1029 kg/ms 3 3.513 c33 gB 1000 9.81 1.0 9810 N/m2 f e gB f er* f ei*i h 1000 9.81 1.0 (0.39 0.38i ) 0.25 956.5 932.0i N/m We assume = 1000 kg/m3 here for convenience. As appears in all terms, the final result is not influenced by the choice of . This yields the heave motion amplitude: 3.51 243 320 i 3.51 1029 9810 u 2 3 u3 956.5 932.0i 956.5 932.0i 0.287 0.0366i m 2874 3612i The amplitude of relative motion is determined from: r u3 h 0.287 0.0366i 0.25 0.037 0.0366i 0.0372 0.03662 = 0.052 m Seakeeping: Lewis section in forced heave An infinite cylinder of width B = 2 m, draft T = 1 m and with Lewis cross section of coefficient Cm = 0.8 floats in equilibrium in fresh water. Then a harmonic force per length with period Te = 3.14 s and amplitude fe=1000 N/m is applied. What motion results after a long time (when the initial start-up has decayed)? What wave length and what wave amplitude are generated, assuming a ‘deep’ basin? Solution The motion has only one degree of freedom, namely heave. The basic equation is then: m m 2 33 in33 c33 u3 fe u3 is the complex amplitude of heave motion. The individual quantities needed are: 2 2 2 Hz Te 3.14 2 B 22 2 0.408 2g 2 9.81 B 2 H 1 2T 2 1 Then the curves for Lewis sections yield: Cz = 0.7, Az 0.51 . This in turn gives: B2 22 m33 Cz 0.7 1000 1100 kg/m 8 8 m Cm B T 1000 0.8 2.0 1.0 1600 kg/m n33 Az2 2 g 2 2 1000 9.81 0 . 51 3129 kg/ms 3 23 c33 gB 1000 9.81 2.0 19620 N/m2 This yields the heave motion amplitude: 2 1600 1100 i 2 3129 19620 u 2 3 u3 1000 0.0754 0.0535i m 8820 6258i The amplitude is: u3 0.07542 0.05352 0.0925 m The wave length follows from: 2 g 2 2g 2 15.4 m The wave amplitude follows from Az : h Az u3 0.51 0.0925 0.047 m 1000 Seakeeping: Power generator A cylinder for power generation is moored in regular waves such that the cylinder axis is parallel to the wave crests. The cylinder dimensions are L = 50 m, B = 10 m, T = 5 m, the cross section is a Lewis section with Cm = 0.8. The waves have = 50 m and h = 1 m. The mooring adds a restoring force coefficient c = 2.5106 N/m and a damping (from the generator) of d = 1.5106 kg/s. The mooring remains at all times under tension as the buoyancy of the cylinder is much larger than its weight due to its mass of 106 kg. The sea water has a density of 1025 kg/m3. What is the amplitude of heave motions for the cylinder? What would be the amplitude without generator? Solution The basic equation of motion is: m m 2 33 L i (d n33L) (c c33L) u3 Lf e The individual quantities needed are: 2g 2 9.81 1.11 Hz 50 2 B 1.112 10 0.628 2g 2 9.81 B 10 H 1 2T 2 5 Then the curves for Lewis sections yield: Cz = 0.61, Az 0.65 , f er* 0.39 , f ei* 0.36 . This in turn gives: B2 102 50 0.61 1025 1.2275 106 kg 8 8 g 2 1025 9.812 L n33 L Az2 3 50 0.652 1.5235 106 kg/s 3 1.11 L c33 L gB 50 1025 9.81 10 5.0275 106 N/m L m33 L Cz L f e L gB f er* f ei*i h 50 1025 9.81 10 (0.39 0.38i ) 1 (1.961 1.81i ) 106 N Then the basic equation yields: 1.11 1 1.2275 i1.11 (1.5 1.5235) (2.5 5.0275) u 2 3 u3 1.961 1.81i 1.961 1.81i 0.453 0.061i m 4.783 3.356i The amplitude is: u3 0.4532 0.0612 0.46 m The basic equation modifies for the case of no generator (i.e. d = 0): m m L i (n L) (c c L) u Lf 1.11 1 1.2275 i1.11 1.5235 (2.5 5.0275) u 2 33 33 33 3 e 2 3 u3 1.961 1.81i 0.483 0.208i m 4.783 1.691i u3 0.4832 0.2082 0.53 m 1.961 1.81i Seakeeping: Catamaran in waves A raft consists of two cylinders with D = 1 m diameter and L = 10 m length. The two cylinders have a distance of 2e = 3 m from centre to centre. The raft has no speed and is located in regular waves coming directly from abeam. The waves have = 3 m and wave amplitude h = 0.5 m. Assume the two cylinders to be hydrodynamically independent, i.e. waves created by one cylinder are not reflected at the other cylinder. The raft has a draft of 0.5 m, centre of gravity in the centre of the connecting plate, radius of moment of inertia for rolling is kx = 1 m. The density of the water is 1000 kg/m3. The Cm = /4 is close enough to 0.8 to use the Lewis section curves for this Cm. Neglect sway motion. What is the maximum roll angle? Hint: The centre of gravity of a semi-circle is 4r/(3) from the flat baseline, where r is the radius. Solution The mass of the raft is: D 2 12 10 2 7854 kg 8 The mass moment of inertia is: m k x2 7854 12 7854 kgm2 2 2 k 0.6667 m1 Wave number: 3 2g 2 9.81 Wave frequency: 2.56 s1 3 8 L 2 1000 The metacentric height is: 4 D3 L / 12 DLe 2 2 GM KB BM KG 1 D T D 2 L / 4 3 4 13 10 / 12 1 10 1.52 2 1 1 5.23 m 0.5 12 10 / 4 3 B 1 .0 2 B 2.562 1.0 1 Parameters for Lewis curves: and 0.333 2T 2 0.5 2g 2 9.81 Then the curves for Lewis sections yield: Cz = 0.8, Az 0.4 , f er* 0.6 , f ei* 0.25 2 ( m44 ) in44 c44 u4 f e 4 The basic equation of roll motion is: The hydrodynamic added mass m44 and damping n44 (moments) are derived from the added mass for heave (forces) times lever e: m44u4 2 m33 u3 e The factor 2 is due to the two cylinders. The term in parentheses is the vertical force. With u3 e u4 , we get for the whole 3-d raft: B 2 1.02 2 m44 2 Cz L e 2 1000 0.8 10 1.52 14137 kgm2 8 8 2 2 g 1000 9.81 10 1.52 41300 kgm2/s n44 2 Az2 3 L e2 2 0.42 3 2 . 56 The restoring force constant for roll motion is: c44 mg GM 7854 9.81 5.23 403 kNm The exciting moment has to consider the phase shift in the wave between the two hulls. The force on the left floater is then: fˆl fer* if ei* gBh L eike The force on the right floater is: fˆr fer* if ei* gBh L eike The phase shift is thus in the last term. Combine the two forces with lever e to get: f e 4 e f er* if ei* gBh L e ike eike 1.5 0.6 0.25i 1000 9.811.0 0.5 10 2i sin( 0.6667 1.5) = (30956 74293 i) kgm2/s2 Then our basic equation becomes: 2.56 (7854 14137) i 2.56 41300 40300 u (30956 74293i) 258.9 105.7i u4 (30.96 74.29i) u4 0.002 0.288i 2 The (real) roll amplitude is then u4 0.0022 0.2882 0.288 16.5 The assumption of linearity is thus no longer justifiable. 4 Seakeeping: Strip method roll A circular cylinder (length L = 20 m, D = 1 m, radius of moment of inertia kxx = 0.4 m, mass m = 8000 kg) floats in water of density = 1019 kg/m3. The centre of gravity is 0.1 m under the centre of the circle. The cylinder is excited by regular waves with wave crests parallel to the cylinder axis. The waves have amplitude h = 0.15 and length = 3.14 m. The following hydrodynamic properties are known: a) hydrodynamic mass for sway: 0.4 m b) hydrodynamic damping for sway given by ratio of amplitudes Ay 0.3 c) real part of horizontal exciting force: 0.8 orbital velocity of wave times damping coefficient as given in b) d) imaginary part of horizontal exciting force: 0.6 wave steepness times cylinder weight The coordinate system should have its origin in the centre of the circle. Then the resultant force of the hydrodynamic forces passes through the origin. Determine the complex amplitude of roll motion! Solution We use a coordinate system with z pointing down, y pointing right in direction of wave propagation. x is then the cylinder axis. The displacement of the fully submerged cylinder is: D 2 4 L 12 4 20 1019 16000 kg As the cylinder mass is 8000 kg, the draft of the cylinder is T = D/2 = 0.5 m. The frequency is: 2g 2 9.81 4.43 Hz 3.14 The orbital velocity involves differentiation with respect to time (factor ), the steepness with respect to y (factor k). The hydrodynamic quantities ‘hidden’ in a) to d) are then: m22 0.4 m 0.4 8000 3200 kg 2 g 2 2 1019 9.81 L 0 . 3 20 2040 kg/s 3 4.433 Re Fe 0.8 h n22 0.8 4.43 0.15 2040 1084 N Im Fe 0.6 k h g m 0.6 2 h m 0.6 4.432 0.15 8000 14130 N n22 Ay2 We can use the formulae for strip method and simplify them. As the waves come just from abeam, no ‘symmetric’ motions (heave, surge, pitch) are excited (with linear theory). Also, as the cylinder has always the same cross section, no yaw motion is excited. The fundamental equation of motions is: (M A) iN S u Eh 2 Sway motion does not excite roll motion, neither does roll motion excite sway for a circular cylinder with origin as chosen. There is no exciting roll moment either as in potential theory only normal pressures exist and for the circular cylinder all normal vectors pass through the origin, i.e. excite no moment. Similarly, there is no radiation moment, i.e. no damping and added mass for roll. Sway has no restoring term. The metacentre of a circular cylinder is the centre of the circle. Thus GM = 0.1 m. Thus we can write for the two degrees of freedom: mzg m 1 M m z mzg z g 0 m 3200 0 A 22 m 0 0 0 0 z g 8000 800 k xx2 800 1280 0 n 2040 0 N 22 m 0 0 0 0 0 0 0 0 S m 0 mgGM 0 7848 Thus: 219800 15700 9037i 0 0 0 u2 1080 14130i 15700 25120 0 0 0 7848 u4 0 219800 9037i 15700 u2 1080 14130i 15700 17272 u4 0 The solution of this system of equations is: u2 0.002 0.069i u4 0.002 0.063i Thus the roll amplitude is u4 0.0022 0.0632 = 0.063 = 3.6°. Seakeeping: Pontoon with crane in waves Consider a pontoon with a heavy-lift derrick as sketched. The pontoon has L = 100 m, B = 20 m, D = 10 m, mp = 107 kg. The load at the derrick has mass ml = 106 kg. The height of the derrick over deck is 10 m. This is where the load can be considered to be concentrated in one point. The longitudinal position of the derrick is 20 m before amidships. A force F = 106 N acts on the forward corner. We assume homogeneous mass distribution in the pontoon. a) Consider the pontoon ‘in air’ (without hydrodynamic masses) and determine the acceleration vector u ! b) Consider the pontoon statically in water and determine u! Solution All numbers are given in standard units. We use the coordinate system as in the book with z pointing down. Origin is at K. 0 0 0 0 F 106 a) The general 6-component force vector is: F 7 F 10 m 10 F 50m 5 107 0 0 The mass matrix needs the following expressions: m mp ml 107 106 1.1 107 m zg mp zg , p ml zg ,l 107 (5) 106 (20) 7 107 m xg mp xg , p ml zg ,l 107 0 106 20 2 107 The moments of inertia are computed for pontoon and load: xx , p ( z 2 y 2 ) d m p L 3 B3 m B 2 107 2 202 B D3 D p D 2 10 4 3 4 3 4 = 0.667 109 xx,l ml zg2,l 106 204 0.4 109 Thus xx xx, p xx,l 1.067 109 . Correspondingly we get: xz , p xz d m 0 xz ,l ml xg ,l zg ,l 106 (20) 20 0.4 109 Thus xz xz , p xz ,l 0.4 109 . yy , p ( x 2 z 2 ) d m = 8.667 10 9 p B 3 L3 m L2 107 2 1002 L D3 D p D 2 10 4 3 4 3 4 yy ,l ml xg2,l zg2,l 106 (202 202 ) 0.8 109 yy yy , p yy ,l 9.467 109 Thus zz, p ( x 2 y 2 ) d m = 8.667 109 p D 12 LB 3 BL3 B 12 mp 2 L2 107 202 1002 12 yy ,l ml xg2,l yg2,l 106 (202 02 ) 0.4 109 zz zz, p zz,l 9.067 109 Thus 0 0 0 70 0 11 0 11 0 70 0 20 0 0 11 0 20 0 This yields the mass matrix: M 106 70 0 1067 0 400 0 70 0 20 0 9467 0 20 0 400 0 9067 0 , where F and u are generalised 6-component Now we can use the fundamental equation F M u vectors. Since we solve manually, it is advisable to decouple the 66 system of equations into 33 systems for symmetric and anti-symmetric degrees of freedom: 11u1 70u1 11u3 20u3 11u2 70u4 70u2 1067u4 20u2 400u4 70u5 0 u1 20u5 1 u3 9467u5 50 u5 0.0367 m / s 2 0.1014 m / s 2 0.00577 s 2 20u6 0 u2 400u6 10 u4 9067u6 0 u6 b) Basic equation of motion is: 0.103 m / s 2 0.0163 s 2 0.000492 s 2 (M A) i N S u F 2 e e e Now we consider the quasi-static case, i.e. e = 0. F is then the ‘exciting’ force. Then: S u F The hydrostatic data are: Aw L B 100 20 2000 m2, xw = 0, T GM = KB + BM KG = m 1.1 107 3 5.5 m, Aw 10 2000 T B2 z g 2.75 + 3.06 6.36 = 2.45 m, 2 12T T L2 z g 2.75 151.51 6.36 147.9 m 2 12T 0 0 0 0 0 0 0 0 0 0 0 0 19.62 0 0 S 106 This gives: 0 264.4 0 0 0 0 0 0 0 15960 0 0 0 0 0 GM L 0 0 0 0 0 0 It is impossible to make any statement on u1, u2, and u6. The remaining degrees of freedom form a simple ‘system’ of equations: 19.62u3 1 u3 0.05 m 264.4u4 10 u4 0.038 15960u5 50 u5 0.0031 Seakeeping: Probability of exceeding operational limit A ship sails in a natural seaway of approximately Te = 6 s encounter period between ship and waves. The bridge of the ship is located forward near the bow. During one 1 hour, 6 times a downward acceleration occurred exceeding gravity acceleration g. For downward accelerations exceeding 1.5g, severe injuries for the crew have to be expected. What is the probability for this happening if the ship continues sailing with the same speed in the same seaway for 12 hours? Solution The number of amplitudes encountered per hour (=3600 s) is: n 3600 3600 600 Te 6 The actual number of amplitudes for acceleration exceeding g in one hour was 6. Thus the probability that one amplitude exceeding g occurs is W(1g) = 600/6 = 0.01. On the other hand, we can write this probability formally: W (1g ) e(1g ) 2 /(2 m0 ) 0.01 m0 0.109 g 2 The probability for an amplitude exceeding 1.5g is correspondingly: W (1.5g ) e(1.5 g ) /(2 m0 ) 3.29 105 The number of amplitudes encountered in 12 h is 12 n 12 600 7200. Thus the probability for one 2 e1.5 2 /(20.109) amplitude exceeding 1.5g in 12 h is: W 7200 3.29 105 = 0.24 = 24 % Seakeeping: Roll motion of cargo ship in waves A new cargo ship type features relatively large beam and extreme sectional flare at the ship ends to allow many container on deck. The ship has the following main data: LWL = 94m, B = 16 m, D = 8.40 m, T = 5 m, radius of inertia (including added mass) k' = 0.4 B = 6.4 m. The service speed is V = 11.8 kn. Investigate whether (respectively when) there is a danger of capsizing due to resonance in wave induced rolling for possible metacentric heights (0.30 m < GM < 1.00 m). Consider the following scenarios: a) waves from abeam ( < 150 m) b) following waves ( < 150 m) If there is a danger of capsizing, specify the conditions which are most critical. Assume deep water. Solution a) waves from abeam Resonance appears for e = n. For elementary waves on deep water this yields: 2g 2 k ' 2 2 6.4 2 257.4 GM GM GM Thus for 0.30 m < GM < 1.00 m only wave lengths 257 m < < 858 m lead to resonance. For the g GM k '2 given range below 150 m, there is no danger of capsizing. b) following waves Resonance appears for n 1 n with n = 1, 2, 3, ... e 2 For following waves (angle of encounter = 0°) we have: e With 2g / we get: e 2g 2 g V 2V The first resonance lies at n = ½ e . The natural frequency of roll is n Thus we get for the first resonance: Solving for GM yields: GM g GM k' g GM 1 2g V k' 2 k ' 2 2 3 k ' 2 V 2 k ' 2 V 2 2 g g2 3 Take all quantities in standard units (m and s), to get with V = 11.8 kn = 6.07 m/s: GM 6.4 2 2 3 6.4 2 6.07 2 6.4 2 6.07 2 64.34 625.11 1518.3 2 9.81 9.81 2 2 3 3 This is evaluated for various wave lengths in the given interval: 30 40 50 60 80 100 120 150 [m] GM [m] 0.03 0.09 0.13 0.15 0.17 0.17 0.17 0.16 The case GM > 0.30 m is never reached. Thus there is no danger of capsizing for n = 1. The second resonance lies at n = e. Due to the quadratic relation, we have thus four times the values for GM: 30 40 50 60 80 100 120 150 [m] GM [m] 0.11 0.35 0.50 0.60 0.67 0.68 0.66 0.62 Thus for most wave lengths, there is a danger of resonance for smaller GM values. The third resonance lies at n = 1.5 e. 30 40 50 60 80 100 120 150 [m] GM [m] 0.11 0.35 0.50 0.60 0.67 0.68 0.66 0.62 The third resonance is also reached, but only for wave lengths much shorter than the ship length. Thus the expected fluctuation in righting levers will be smaller as for second resonance. Correspondingly the probability of capsizing will be lower. Higher resonances will shift resonance to even shorter waves which will be uncritical. Seakeeping: Roll motion of submarine in waves Consider a surfaced submarine with speed V = 6 kn. Neglect the sail and model the submarine hull as a circular cross section. Neglect damping. a) What is the expected roll amplitude in elementary waves from abeam? b) Evaluate the danger of parametric roll excitation in elementary following waves! Ship data: L = 70.00 m, D = 8.00 m (diameter), T = 7.00 m, KG = 3.60 m, k' = 3.00 m (radius of the transverse mass moment of inertia incl. added mass) Wave data (deep water): = 70 m (wave length); h = 1.75 m (wave amplitude) Solution: a) waves from abeam e Wave circular frequency: 2g 2 9.81 0.93835 s1 70 For a circular cylinder, the metacentre is always in the centre of the circle: D 8.00 KG 3.60 0.40 m 2 2 g GM 9.81 0.40 The natural frequency for roll is: n 0.6603 s1 2 k' 3.002 0.93835 1.4211 the roll excitation lies in the super-critical region. The roll amplitude For e n 0.6603 for undamped roll is thus (with k = 2/): 1 1 2 | || uˆ4 | k h 1.75 0.15406 8.83 2 2 1 (e / n ) 1 1.4211 70 GM KM KG b) following waves Encounter frequency e is then: 2 9.81 2 6 0.5144 0.6613 s1 g 70 70 Although the roll natural frequency n is close to encounter frequency e, there is no parametric e 2 V 2g 2V excitation, because for a circular cylinder there is no fluctuation of righting levers. The righting lever is always h = GM sin , with GM = 0.40 m = const. Seakeeping: Cruise ship in waves from abeam A small cruise ship is designed. As the destination island has no deep-water port, the passengers will disembark via small boats. To allow normal disembarkation, the metacentric height should be always sufficiently high to have maximum roll amplitude = 6° in regular waves from abeam up to wave steepness max = 4.5° up to wave length of max = 60 m. Assume a linear restoring moment. For the small roll angles considered, damping can be neglected. The radius of inertia including added mass is k' = 4 m. What is the resulting requirement for the metacentric height? Solution The requirement 6° for max = 4.5° means for the amplification function: Vratio max 6 1.33 4.5 The equation for V yields then limiting ratios of frequency: (e / n ) 2 0.25 1 Vratio 1.33 (e / n ) 2 1.75 1 (e / n ) 2 Thus the frequency interval to be avoided is 0.25 (e / n ) 2 1.75 . With the relations for a deep water regular wave: e 2g and for the natural frequency of roll n g GM k '2 we get for the interval to be avoided: 0.25 2 k '2 1.75 GM For various wave lengths, we can then determine the interval to be avoided: 20 m 40 m 60 m interval to avoid 2.87 m GM 20.11 m 1.44 m GM 10.05 m 0.96 m GM 6.70 m One can quickly see that short wave lengths pose no problem. The limiting value is GM 0.96 m. This requirement will hardly be possible for a passenger ship due to damaged stability requirements.