- Marine Hydrodynamics. Lecture 14. F, M = [linear function of m ij ] [function of instantaneous U, U, Ω] not of motion history.

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1 Marine Hydrodynamics, Spring 2005 ecture Marine Hydrodynamics ecture Some Properties of Added-Mass Coefficients 1. m ij = ρ [function of geometry only] F, M = [linear function of m ij ] [function of instantaneous U, U, Ω] not of motion history 2. Relationship to fluid momentum. F(t) where we define Φ to denote the velocity potential that corresponds to unit velocity U = 1. In this case the velocity potential φ for an arbitrary velocity U is φ = UΦ. The linear momentum in the fluid is given by = ρ vdv = ρ φdv = + ρφˆ nds V V Green s B theorem x (t = T ) = ρuφn x ds = U ρφn x ds }{{} φ 0 at B B The force exerted on the fluid from the body is F (t) = ( m A U )= m A U. 1

2 T T Newton s aw T dt [ F (t)] = m A Udt = m A U] }{{ 0 = x (t = T ) x (t = 0)= U ρφn x ds } 0 0 m A U B Therefore, m A = total fluid momentum for a body moving at U = 1 (regardless of how we get there from rest) = fluid momentum per unit velocity of body. K.B.C. φ = φ nˆ =(U, 0, 0) nˆ = Un x, φ = Un x UΦ = Un x n n n n Φ m A = ρ Φ ds n B Φ = n x For general 6 DOF: Φ j m ji = ρ Φ i n j ds = ρ Φ i ds = j fluid momentum due to }{{}}{{} n i body motion j force/moment B potential due to body B i direction of motion moving with U i =1 3. Symmetry of added mass matrix m ij = m ji. ( ) Φ j m ji = ρ Φ i ds = ρ Φ i ( Φ j nˆ)ds = ρ (Φ i Φ j ) dv n B = ρ Φ i Φ j +Φ i 2 Φ }{{} j dv V =0 B Divergence Theorem V Therefore, m ji = ρ V Φ i Φ j dv = m ij 2

3 4. Relationship to the kinetic energy of the fluid. For a general 6 DoF body motion U i =(U 1,U 2,...,U 6 ), φ = U }{{} i Φ i ;Φ i = potential for U i =1 notation 1 1 K.E. = ρ φ φdv = 2 ρ U i Φ i U j Φ j dv 2 V V 1 = ρu i U j Φ i Φ j dv = 1m 2 2 iju i U j K.E. depends only on m ij and instantaneous U i. V 5. Symmetry simplifies m ij. From 36 21?. Choose such coordinate system symmetry that some m ij = 0 by symmetry. Example 1 Port-starboard symmetry. m11 m m 16 F x m m 26 F y m 33 m 34 m 0 F m 35 z ij = m 44 m 45 0 M x m 55 0 M y m 66 M z 12 independent coefficients U 1 U 2 U 3 Ω 1 Ω 2 Ω 3 3

4 Example 2 Rotational or axi-symmetry with respect to x 1 axis. m ij = m m m 35 m 22 0 m m 55 0 m 55 where m 22 = m 33,m 55 = m 66 and m 26 = m 35, so 4 different coefficients Exercise How about 3 planes of symmetry (e.g. a cuboid); a cube; a sphere?? Work out the details. 4

5 3.21 Slender Body Approximation Definitions (a) Slender Body = a body whose characteristic length in the longitudinal direction is considerably larger than the body s characteristic length in the other two directions. (b) m ij = the 3D added mass coefficient in the i th direction due to a unit acceleration in the j th direction. The subscripts i, j run from 1 to 6. (c) M kl = the 2D added mass coefficient in the k th direction due to a unit acceleration in the l th direction. The subscripts k, l take the values 2,3 and 4. x 2 x 5 x 2 O x x 1 x x 3 4 x 4 x 6 x 3 Goal To estimate the added mass coefficients m ij fora3d slender body. Idea Estimate m ij of a slender 3D body using the 2D sectional added mass coefficients (strip-wise M kl ). In particular, for simple shapes like long cylinders, we will use known 2D coefficients to find unknown 3D coefficients. m ij = [M kl (x) contributions] 3D 2D Discussion If the 1-axis is the longitudinal axis of the slender body, then the 3D added mass coefficients m ij are calculated by summing the added mass coefficients of all the thin slices which are perpendicular to the 1-axis, M kl. This means that forces in 1-direction cannot be obtained by slender body theory. 5

6 Procedure In order to calculate the 3D added mass coefficients m ij we need to: 1. Determine the 2D acceleration of each crossection for a unit acceleration in the i th direction, 2. Multiply the 2D acceleration by the appropriate 2D added mass coefficient to get the force on that section in the j th direction, and 3. Integrate these forces over the length of the body. Examples Sway force due to sway acceleration Assume a unit sway acceleration u 3 = 1 and all other u j,u j = 0, with j = 1, 2, 4, 5, 6. It then follows from the expressions for the generalized forces and moments (ecture 12, JNN 4.13) that the sway force on the body is given by f 3 = m 33 u 3 = m 33 m 33 = f 3 = F 3 (x)dx A unit 3 acceleration in 3D results to a unit acceleration in the 3 direction of each 2D slice (U 3 = u 3 = 1). The hydrodynamic force on each slice is then given by F 3 (x) = M 33 (x)u 3 = M 33 (x) Putting everything together, we obtain m 33 = M 33 (x)dx = M 33 (x)dx Sway force due to yaw acceleration Assume a unit yaw acceleration u 5 = 1 and all other u j,u j = 0, with j = 1, 2, 3, 4, 6. It then follows from the expressions for the generalized forces and moments that the sway force on the body is given by f 3 = m 35 u 5 = m 35 m 35 = f 3 = F 3 (x)dx For each 2D slice, a distance x from the origin, a unit 5 acceleration in 3D, results to a unit acceleration in the -3 direction times the moment arm x (U 3 = xu 5 = x). The hydrodynamic force on each slice is then given by F 3 (x) = M 33 (x)u 3 = xm 33 (x) 6

7 Putting everything together, we obtain m 35 = xm 33 (x)dx Yaw moment due to yaw acceleration Assume a unit yaw acceleration u 5 = 1 and all other u j,u j = 0, with j = 1, 2, 3, 4, 6. It then follows from the expressions for the generalized forces and moments that the yaw force on the body is given by f 5 = m 55 u 5 = m 55 m 55 = f 5 = F 5 (x)dx For each 2D slice, a distance x from the origin, a unit 5 acceleration in 3D, results to a unit acceleration in the -3 direction times the moment arm x (U 3 = xu 5 = x). The hydrodynamic force on each slice is then given by F 3 (x) = M 33 (x)u 3 = xm 33 (x) However, each force F 3 (x) produces a negative moment at the origin about the 5 axis F 5 (x) = xf 3 (x) Putting everything together, we obtain m 55 = x 2 M 33 (x)dx In the same manner we can estimate the remaining added mass coefficients m ij - noting that added mass coefficients related to the 1-axis cannot be obtained by slender body theory. 7

8 In summary, the 3D added mass coefficients are shown in the following table. The empty boxes may be filled in by symmetry. m 22 = M 22 dx m 23 = M 23 dx m 24 = M 24 dx m 25 = xm 23 dx m 26 = xm 22 dx m 33 = M 33 dx m 34 = M 34 dx m 35 = xm 33 dx m 36 = xm 32 dx m 44 = M 44 dx m 45 = xm 34 dx m 46 = xm 24 dx m 55 = x 2 M 33 dx m 56 = x 2 M 32 dx m 66 = x 2 M 22 dx 8

9 3.22 Buoyancy Effects Due to Accelerating Flow Example Force on a stationary sphere in a fluid that is accelerated against it. Then, 3 a φ (r, θ, t) = U (t) r + cos θ }{{} 2r 2 dipole for sphere φ = U 3a cos θ t r=a 2 ( ) 3 φ r=a = 0, U sin θ, φ 2 = U 2 sin 2 θ 2 r=a 8 π [ ] F x =( ρ) ( 2πr ) 2 dθ sin θ ( cos θ) U 3a cos θ + 9 U 2 sin 2 θ 2 8 π π = U 3πρa 3 dθ sin θ cos 2 θ +ρu 2 9π a 2 dθ cos θ sin 3 θ }{{}}{{} ( ) F x = U ρ 2πa 3 }{{} 0 4 πa 3 ρ + 2 πa 3 ρ 3 3 =ρ =2/3 =0 9

10 Part of F x is due to the pressure gradient which must be present to cause the fluid to accelerate: U U U U 1 p x-momentum, noting U = U (t) : + U + v + w = (ignore gravity) t }{{} x y z ρ x }{{}}{{} dp = ρu for uniform (1D) accelerated flow dx Force on the body due to the pressure field F = pnds ˆ = pdv ; F x = B V B V B p dv = ρ U x Buoyancy force due to pressure gradient = ρ U 10

11 Analogue: Buoyancy force due to hydrostatic pressure gradient. Gravitational acceleration g U = fluid acceleration. p s = ρgy Summary: Total force on a fixed sphere in an accelerated flow ( ) F x = U ρ + m (1) = U 3 2 ρ = 3 Um }{{} (1) }{{} p s = ρgĵ F s = ρg ĵ Buoyancy added mass 1 ρ 2 In general, for any body in an accelerated flow: F x = F buoyany + Um (1), Archimedes principle where m (1) is the added mass in still water (from now on, m) F x Added mass coefficient = Um for body acceleration with U m c m = ρ in the presence of accelerated flow C m =1+c m 11

12 Appendix A: More examples on symmetry of added mass tensor Symmetry with respect to Y (= X-Z plane symmetry) 12 non-zero, independent coefficients Symmetry with respect to X and Y (= Y-Z and X-Z plane symmetry) 7 non-zero, independent coefficients 12

13 Axisymmetric with respect to X-axis 4 non-zero, independent coefficients Axisymmetric with respect to X axis and X (= Y-Z plane symmetry) 3 non-zero, independent coefficients 13