13.021 - Marine Hydrodynamics, Fall 2003

Lecture 13

Copyright © 2003 MIT - Department of Ocean Engineering, All rights reserved.

13.021 - Marine Hydrodynamics
Lecture 13

Some Properties of Added-Mass Coefficients

1.

$m_{ij}=\rho \cdot$[function of geometry only]
F, M = [linear function of $m_{ij}$] $\times$ [function of $\underset{\mbox{\tiny{not of motion history}}}{\mbox{\underline {instantaneous}}}$ $U,\dot {U},\Omega$]

2.

Relationship to momentum of fluid:

Linear momentum $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$ }}\over {L}}$ in fluid:

$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightha... ...at}}\ \infty}{\underbrace{\infty}} } {\rho \phi \hat {n}dS}$$

$$L_x \left( {t = T} \right) = \int\!\!\!\int\limits_B {\rho \Phi n_x dS}$$

Force on fluid by body = $-F(t) = -(-m_{A}\dot {U}) = m_{A}\dot {U}$.

$$\int\limits_0^T {dt\left[ { - F\left( t \right)} \right]} = \... ... {t = 0} \right) = \int\!\!\!\int\limits_B {\rho \Phi n_x dS}$$

Therefore, $m_{A}$ = total fluid momentum for body moving at $U = 1$ (regardless of how we get there from rest) = fluid momentum per unit velocity of body.
K.B.C. $\frac{\partial \phi }{\partial n} = \nabla \phi \cdot \hat {n} = \left( {u,0,0} \right) \cdot \hat {n} = Un_x ,\mbox{ }\frac{\partial \Phi }{\partial n} = n_x$ for $U = 1$.

$$\mbox{ }m{ }_A = \rho \int\!\!\!\int\limits_B {\Phi \frac{\partial \Phi }{\partial n}} dS$$

For general 6 DOF:

$$\underset{\begin{array}{l}\scriptstyle{ j - }\mbox{\tiny{forc... ...d momentum due to\ } \\ i \mbox{\ body motion} \end{array}$$

3.

Symmetry of added mass matrix $m_{ij }= m_{ji}$.

$$\begin{align}m_{ji} & = \rho \int\!\!\!\int\limits_B {\Phi _i } \left( {\frac{\p... ...bla \Phi _j + \Phi _i \nabla ^2\Phi _j } \right)d\upsilon } \notag \end{align}$$

Therefore,

$$m_{ji} = \rho \mathop{\int\!\!\!\int\!\!\!\int}\limits_{\kern... ...V} {\nabla \Phi _i \cdot \nabla \Phi _j d\upsilon } = m_{ij}$$

4.

Relationship to kinetic energy of fluid. In general, for body motion $U_{i}\{(U_{1}, U_{2}, \ldots , U_{6})\}$.

$$\phi = \underset{\scriptstyle{\sum}\mbox{\tiny{\ notation}}}{... ...ce{U_i \Phi _i}} ; \Phi _i = \mbox{\ potential for\ } U_i = 1$$

$$\begin{align}K.E. & = \frac{1}{ 2}\rho \mathop{\int\!\!\!\int\!\!\!\int}\limits... ...a \Phi _j d\upsilon } = \textstyle{1 \over 2}m_{ij} U_i U_j \notag \end{align}$$

K.E. depends only on $m_{ij}$ and instantaneous $U_{i}$.

5.

Use of symmetry to simplify $m_{ij}$. From 36 $\underset{\mbox{\tiny{symmetry}}}{\to}$ 21 $\to$ ? Choose coordinate system so that some $m_{ij} = 0$ by symmetry. e.g. example 1: port-starboard symmetry (sym w.r.t. $x_{3})$

$$\begin{array}{l} m_{ij} = \left[ {{\begin{array}{*{20}c} ... ...\end{array} \hspace{0.5in} \mbox{12 independent coefficients}$$

example 2: rotational (axi) sym. about $x_{1}$

$$m_{ij} = \left[ {{\begin{array}{*{20}c} {m_{11} } \hfill & ... ...6} = m_{35}, \mbox{ so 4 different coefficients} \end{array}$$

Exercise: How about 3 planes of symmetry (e.g. a cuboid); a cube; a sphere?? Work out detail!