Lecture Notes on Added Mass

2.016 Hydrodynamics Reading #6 2.016 Hydrodynamics Prof. A.H. Techet Added Mass For the case of unsteady motion of bodies underwater or unsteady flow around objects, we must consider the additional effect (force) resulting from the fluid acting on the structure when formulating the system equation of motion. This added effect is added mass. Most floating structures can be modeled, for small motions and linear behavior, by a system equation with the basic form similar to a typical mass-spring-dashpot system described by the following equation: mx + bx + kx = f (t ) (6.1) where m is the system mass, b is the linear damping coefficient, k is the spring coefficient, f(t) is the force acting on the mass, and x is the displacement of the mass. The natural frequency ω of the system is simply ω= k . m (6.2) In a physical sense, this added mass is the weight added to a system due to the fact that an accelerating or decelerating body (ie. unsteady motion: dU dt ≠ 0 ) must move some volume of surrounding fluid with it as it moves. The added mass force opposes the motion and can be factored into the system equation as follows: x mx + bx + kx = f (t ) − ma  (6.3) where ma is the added mass. Reordering the terms the system equation becomes: ( m + ma ) x + bx + kx = f (t ) (6.4) From here we can treat this again as a simple spring-mass-dashpot system with a new mass m′ = m + ma such that the natural frequency of the system is now ω′ = k k = m′ m + ma (6.5) It is important in ocean engineering to consider floating vessels or platforms motions in more than one direction. Added mass forces can arise in one direction due to motion in a different direction, and thus we can end up with a 6 x 6 matrix of added mass coefficients. version 3.0 updated 8/30/2005 -1- ©2005 A. Techet 2.016 Hydrodynamics Reading #6 Looking simply at a body in two-dimensions we can have linear motion in two directions and rotational motion in one direction. (Think of these coordinates as if you were looking down on a ship.) Two dimensional motion with axis (x,y) fixed on the body. 1: Surge, 2: Sway, 6: Yaw The unsteady forces on the body in the three directions are: − F1 = m11 du du1 du + m12 2 + m16 6 dt dt dt (6.6) −F2 = m21 du du1 du + m22 2 + m26 6 dt dt dt (6.7) −F6 = m61 du du1 du + m62 2 + m66 6 dt dt dt (6.8) Where F1, F2, and F6, are the surge (x-) force, sway (y-) force and yaw moments respectively. It is common practice in Ocean Engineering and Naval Architecture to write the moments for roll, pitch, and yaw as F4, F5, and F6 and the angular motions in these directions as X4, X5, and X6. This set of equations, (6.6)-(6.8), can be written in matrix form, F = [ M ]u , ⎡ m11 F = ⎢⎢ m21 ⎢⎣ m61 version 3.0 updated 8/30/2005 m12 m22 m62 -2- ⎛ du1 ⎞ ⎟ ⎜ m16 ⎤ ⎜ dt ⎟ du m26 ⎥⎥ ⎜ 2 ⎟ ⎜ dt ⎟ ⎟ m66 ⎥⎦ ⎜ ⎜ du6 ⎟ ⎜ ⎟ ⎝ dt ⎠ (6.9) ©2005 A. Techet 2.016 Hydrodynamics Reading #6 Considering all six degrees of freedom the Force Matrix is: ⎡ m11 ⎢m ⎢ 21 ⎢m F = ⎢ 31 ⎢ m41 ⎢ m51 ⎢ ⎢⎣ m61 m12 m22 m13 m23 m14 m24 m15 m25 m32 m33 m34 m35 m42 m43 m44 m45 m52 m62 m53 m63 m54 m64 m55 m65 m16 ⎤ ⎛ u1 ⎞ ⎜ ⎟ m26 ⎥⎥ ⎜ u2 ⎟ m36 ⎥ ⎜ u3 ⎟ ⎥⎜ ⎟ m46 ⎥ ⎜ u4 ⎟ m56 ⎥ ⎜ u5 ⎟ ⎥⎜ ⎟ m66 ⎦⎥ ⎝⎜ u6 ⎠⎟ (6.10) We will often abbreviate how we write the Force matrix given in (6.10) using tensor notation. The force vector is written as F = Fi , where i = 1N , 2, 3 , 4N , 5, 6 , Linear Forces (6.11) Moments the acceleration vector as ui = [u1 , u2 , u3 , u4 , u5 , u6 ] , (6.12) and the added mass matrix [ma] as mij where i, j = 1, 2, 3, 4, 5, 6 . (6.13) A good way to think of the added mass components, mij , is to think of each term as mass associated with a force on the body in the i th direction due to a unit acceleration in the j th direction. For symmetric geometries the added mass tensor simplifies significantly. For example, figure 2 shows added mass values for a circle, ellipse, and square. In the case of the circle and square, movement in the 1 and 2 directions yields similar geometry and identical added mass coefficients ( m11 = m22 ). version 3.0 updated 8/30/2005 -3- ©2005 A. Techet 2.016 Hydrodynamics Circle m11 = m22 = ρ d 2 m66 = 0 Reading #6 Ellipse m11 = ρπ b 2 Square m11 = m22 = 1.51πρ a 2 m22 = ρπ a 2 m66 = 0.234πρ a 4 m66 = ρ ⎛⎜⎝ a 2 − b 2 ⎞⎟⎠ 2 Two dimensional added mass coefficients for a circle, ellipse, and square in 1: Surge, 2: Sway, 6: Yaw Using these coefficients and those tabulated in Newman’s Marine Hydrodynamics on p.145 we can determine the added mass forces quite simply. In three-dimensions, for a sphere (by symmetry): m11 = m22 = m33 = 1 ρ∀ = mA 2 (6.14) ALL OTHER mij TERMS ARE ZERO ( i ≠ j ). version 3.0 updated 8/30/2005 -4- ©2005 A. Techet 2.016 Hydrodynamics Reading #6 General 6 DOF forces and moments on a Rigid body moving in a fluid: Velocities: G Translation Velocity : U (t ) = (U1 , U 2 ,U 3 ) (6.15) G Rotational Velocity : Ω(t ) = (Ω1 , Ω 2 , Ω3 ) ≡ (U 4 ,U 5 ,U 6 ) (6.16) All rotation is taken with respect to Origin of the coordinate system (often placed at the center of gravity of the object for simplicity!). Forces: (force in the jth direction). ( i = 1, 2, 3, 4, 5, 6 and j, k , l = 1, 2, 3) Fj = −U i mij − ε jkl U i Ω k mli (6.17) Moments: ( i = 1, 2, 3, 4, 5, 6 and j, k , l = 1, 2, 3) M j = −U i m j +3, i − ε jkl U i Ω k ml +3, i − ε jkl U kU i mli (6.18) Einstein’s summation notation applies! The alternating tensor ε jkl is simply ε jkl if any j , k , l are equal ⎧ 0; ⎪ = ⎨ 1; if j , k , l are in cyclic order ⎪−1; if j , k , l are in anti-cyclic order ⎩ (6.19) The full form of the force in the x-direction (F1) is summed over all values of i: F = − U 1 m11 − U 2 m21 − U 3 m31 − U 4 m41 − U 5 m51 − U 6 m61 N1 N      j=1 i=1 i=2 i=4 i=3 i=5 i=6 − ε1kl U1 Ω k ml1 − ε1kl U 2 Ω k ml 2 − ε1kl U 3 Ω k ml 3 − ε1kl U 4 Ω k ml 4     i=1 i=3 i=2 (6.20) i=4 − ε1kl U 5 Ω k ml 5 − ε1kl U 6 Ω k ml 6   i=5 i=6 for k , l = 1, 2, 3 . version 3.0 updated 8/30/2005 -5- ©2005 A. Techet 2.016 Hydrodynamics Reading #6 Next we can choose the index k to cycle through. It is helpful to note that the only terms where k plays a role, contain ε jkl . Following the definition for ε jkl given in (6.19) and since j = 1, all terms will be zero for k = 1. Therefore k can only take the value of 2 or 3: F = − U 1 m11 − U 2 m21 − U 3 m31 − U 4 m41 − U 5 m51 − U 6 m61 N1 N      j =1 i =1 i =2 i =3 i =4 i =5 i =6 − ε12l U1 Ω 2 ml1 − ε12l U 2 Ω 2 ml 2 − ε12l U 3 Ω 2 ml 3 − ε12l U 4 Ω 2 ml 4 − ε12l U 5 Ω 2 ml 5 − ε12l U 6 Ω 2 ml 6       i =6 i =2 i =3 i=4 i =5 i =1  k =2 − ε13l U1 Ω3 ml1 − ε13l U 2 Ω3 ml 2 − ε13l U 3 Ω3 ml 3 − ε13l U 4 Ω3 ml 4 − ε13l U 5 Ω3 ml5 − ε13l U 6 Ω3ml 6       i =1 i =2 i =3 i=4 i =5 i =6  k =3 (6.21) Finally we cycle through the index l. Again it is helpful to note that the only terms where l plays a role, contain ε jkl . Following the definition for ε jkl given in (6.19) and since j = 1, and k = 2 or 3, then all terms will be zero for l = 1 and some zero for the case l = 2 and others zero when l =3. Like before l can only take the value of 2 or 3 such that l ≠ k ≠ j : F = − U 1 m11 − U 2 m21 − U 3 m31 − U 4 m41 − U 5 m51 − U 6 m61 N1 N      j =1 i =1 i =2 i =3 i=4 i =5 i =6 − ε123 U1 Ω 2 m31 − ε123 U 2 Ω 2 m32 − ε123 U 3 Ω 2 m33 − ε123 U 4 Ω 2 m34 − ε123 U 5 Ω 2 m35 − ε123 U 6 Ω 2 m36       i =6 i =1 i=4 i =5 i=2 i =3  k = 2; l =3 − ε132 U1 Ω3 m21 − ε132 U 2 Ω3 m22 − ε132 U 3 Ω3 m23 − ε132 U 4 Ω3 m24 − ε132 U 5 Ω3 m25 − ε132 U 6 Ω3m26       i=2 i =3 i =6 i =1 i=4 i =5  k = 3; l = 2 (6.22) On the second row of the equation above, the indices of the alternating tensor, ε jkl , are in cyclic order jkl = 123 ( ε123 = +1). In the third row, the indices are in anti (or reverse) cyclic order: ε132 = −1 where jkl = 132. More than likely you will never have to write out all six force equations with all the terms as the velocity and acceleration of the body will be zero in certain directions. However for a full seakeeping analysis of a ship then one day you just might need to be able to determine all the forces! version 3.0 updated 8/30/2005 -6- ©2005 A. Techet 2.016 Hydrodynamics Reading #6 Typical Example: For a body moving in the fluid with velocity K V = (1, 0,1, 0, 0,1) = (U1 , 0,U 3 , 0, 0,U 6 ) = (U1 , 0,U 3 , 0, 0, Ω3 ) (6.23) and acceleration K a = (1, 0, 0, 0, 0,1) = (U1 , 0, 0, 0, 0,U 6 ) (6.24) we can find the force on the body in the X-direction. The force in the x-direction is F1 so j=1. First substitute “1” for every instance of j in equation (6.17) to get: Fj=1 = F1 = −U i mi1 − ε1kl U i Ω k mli (6.25) Next we need to “cycle” through the possible values for i (i = 1,2,3,4,5,6). Looking at equation (6.25), it is clear that the only “ith” accelerations that will matter are the non-zero ones from (6.24), thus U1 and U 6 , and the only “ith” velocities to consider are for i = 1,3, and 6 [eqn (6.23)]. F1 = − U 1 m11 − U 6 m61 − ε1kl U1 Ω k ml1 − ε1kl U 3 Ω k ml 3 − ε1kl U 6 Ω k ml 6 N     i =1 i =6 i =1 i =3 (6.26) i =6 Now look at the k-index: ( k ≠ j ∴k = 2, 3 ) However Ω 2 = 0 and Ω3 ≠ 0 thus for k = 2 all associated terms will be zero, so we only have to deal with k = 3 . Since j = 1 and k = 3 the only value left for l, that could result in non-zero terms, is 2. F1 = − U 1 m11 − U 6 m61 N  i=1 i=6 − ε132 U1 Ω3 m21 − ε132 U 3 Ω3 m23 − ε132 U 6 Ω3 m26    i=1 i=3 i=6  (6.27) k =3; l =2 If the body in question was a simple, symmetrical sphere we could reduce this even further. Using the added mass values from (6.14) and trusting that the off-diagonal added mass terms are zero (just for the sphere), the force in the x-direction on a sphere, given (6.23) and (6.24), is F1 = − U 1 m11 N (6.28) i=1 version 3.0 updated 8/30/2005 -7- ©2005 A. Techet 2.016 Hydrodynamics Reading #6 Determining 3D Added Mass Using Slender Body Theory To formulate the added mass of a system such as a ship or submarine that can be modeled as a slender body, we first need the two-dimensional sectional added mass coefficients. We will consider a slender body to have a characteristic length in one direction that is considerably longer than its length in the other two directions. For these slender bodies we can use known 2D coefficients to find the unknown 3D added mass coefficient for the body. The added mass force acting on the body due to unsteady motion is Fj = −U i mij − ε jkl U i Ω k mli (6.29) where mij is the added mass in the ith direction due to a unit acceleration in the jth direction and i,j = 1:6. The added mass tensor, mij , is symmetric! To find the 3D added mass coefficients consider simply the body geometry, ignoring for now the actual motions of the vessel. To start, orient the 1-axis along the long axis of the slender body as shown in figure 1. The 3D added mass coefficients will be found by summing (or integrating) the added mass coefficients of the 2D cross-sectional slices along the body. Slender body oriented with the long axis in the 1-direction. version 3.0 updated 8/30/2005 -8- ©2005 A. Techet 2.016 Hydrodynamics Reading #6 The sectional added mass coefficients are tabulated for simple geometries. In general, with the slender body aligned lengthwise along the 1-axis, the 2D cross-sectional slice is aligned with the 2-3 plane, some distance x from the origin (figure 1). This 2D slice is shown in figure 2. To find the 3D coefficients we need to know the 2D coefficient of each section (strip) along the length of the vessel. For a uniform diameter cylinder this is quite simple, but for ships with complex geometry there is a bit more work involved. 2D cross-sectional slice of slender body. The 2D coefficients will be written as aij whereas the 3D coefficients are written as mij . From here on we will follow the basic formulations used in the handbook: Principles of Naval Architecture Vol III., (1989) Soc. Naval Arch. and Marine Engineers, p. 56. 1 2 3 4 5 6 1 2 m22 = ∫ a22 dx m23 = − ∫ a23 dx m24 = ∫ a24 dx L 3 L m26 = ∫ x a22 dx L m33 = ∫ a33 dx L m35 = − ∫ x a33 dx L L m44 = ∫ a44 dx 4 m46 = ∫ x a24 dx L L m55 = ∫ x 2 a33 dx 5 L m66 = ∫ x 2 a22 dx 6 L version 3.0 updated 8/30/2005 -9- ©2005 A. Techet