13.021 Marine Hydrodynamics, Fall 2003

Lecture 3

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

13.021 - Marine Hydrodynamics
Lecture 3

1.2 - Stress Tensor

Stress Tensor $\tau_{ij:}$. The stress (force per unit area) at a point in a fluid needs nine components to be completely specified, since each component of the stress must be defined not only by the direction in which it acts but also the orientation of the surface upon which it is acting. The first index specifies the direction in which the stress component acts, and the second identifies the orientation of the surface upon which it is acting. Therefore, the $i^{th}$ component of the force acting on a surface whose outward normal points in the $j^{th}$ direction is $\tau_{ij}$.

Figure 1: Shear stresses on an infinitesimal cube whose surface are parallel to the coordinate system.

Figure: Consider an infinitesimal body at rest with a surface PQR that is not perpendicular to any of the Cartesian axis. The unit normal vector to that surface is $\hat{n} = n_{1}\hat{x}_{1}+n_{2}\hat{x}_{2}+n_{3}\hat{x}_{3}$. The area of the surface $= A_{0}$, and the area of each surface perpendicular to $X_{i}$ is $A_{i} = A_{0}n_{i}$, for $i=1,2,3$.

Newton's law: $\sum_{all 4 faces}F_{i}$ = (volume force) $_{\mathit{i}}$ for i = 1, 2, 3

If $\delta$ is the typical dimension of the body	: surface forces $\sim \delta^{2}$
	: volume forces $\sim \delta^{3}$

An example of surface forces is the shear force and an example of volumetric forces is the gravity force. At equilibrium, the surface forces and volumetric forces are in balance. As the body gets smaller, the mass of the body goes to zero, which makes the volumetric forces equal to zero and leaving the sum of the surface forces equal zero. So, as $\delta \rightarrow 0, \sum_{all 4 faces}F_{i} = 0 \mbox{ for } i = 1, 2, 3$ and $\therefore \tau_{i}A_{0} = \tau_{i1}A_{1} + \tau_{i2}A_{2} + \tau_{i3}A_{3} = \tau_{ij}A_{j}$. But the area of each surface $\perp$ to $X_{i}$ is $A_{i} = A_{0}n_{i}$. Therefore $\tau_{i}A_{0} = \tau_{ij}A_{j} = \tau_{ij}(A_{0}n_{j})$, where $\tau_{ij}A_{j}$ is the $\sum$ notation (represents the sum of all components). Thus $\tau_{i} = \tau_{ij}n_{j}$ for i = 1, 2, 3, where $\tau_{i}$ is the component of stress in the $i^{th}$ direction on a surface with a normal $\vec{n}$ . We call $\tau$ $_{\mathit{i}}$ the stress vector and we call $\tau_{ij}$ the stress matrix or tensor.

Example: Pascal's Law for hydrostatics

In a static fluid, the stress vector cannot be different for different directions of the surface normal since there is no preferred direction in the fluid. Therefore, at any point in the fluid, the stress vector must have the same direction as the normal vector $\vec{n}$ and the same magnitude for all directions of $\vec{n}$ .

$$\hspace{5mm}\tau _{ij} =\overset{\mbox{no summation}}{\overbrace... ...ray}{ccc}-p_{1} & 0 & 0 0 & -p_{2} & 0 0 & 0 & -p_{3}\end{array} \right]$$

where $p_{i}$ is the pressure acting perpendicular to the $i^{th}$ surface. If $p_{o}$ is the pressure acting perpendicular to the surface PQR, then $\tau_{i} = -n_{i}p_{0}$ , but $\tau_{i} = \tau_{ij}n_{j} = -(p_{i})\delta_{ij}n_{j} = -(p_{i})(n_{i})$. Therefore $p_{o} = p_{i}$ , i = 1, 2, 3 and $\vec{n}$ is arbitrary.

Symmetry of the Stress Tensor

To prove the symmetry of the stress tensor we follow the steps:

Figure 3: Material element under tangential stress.

1. The $\sum$ of surface forces $=$ body forces $+$ mass$\times$ acceleration. Assume no symmetry. Balance of the forces in the $i^{th}$ direction gives:

   $$(\delta)(\tau_{ij})_{TOP}-(\delta)(\tau_{ij})_{BOTTOM} = O(\delta^{2}),$$

   
   
   since surface forces are $\sim \delta^{2}$, where the $O(\delta^{2})$ terms include the body forces per unit depth. Then, as $\delta \rightarrow 0, (\tau_{ij})_{TOP} = (\tau_{ij})_{BOTTOM}$.

2. The $\sum$ of surface torque $=$ body moment $+$ angular acceleration. Assume no symmetry. The balance moment with respect to $o$ gives:

   $$(\tau_{ji}\delta)\delta - (\tau_{ij}\delta)\delta = O(\delta^{3}),$$

   
   
   since the body moment is proportional to $\delta^{3}$. As $\delta \rightarrow 0$ , $\tau_{ij} = \tau_{ji}$.

1.3 Mass and Momentum Conservation

Consider a material volume $\vartheta _{m}$ and recall that a material volume is a fixed mass of material. A material volume always encloses the same fluid particles despite a change in size, position, volume or surface area over time.

1.3.1 Mass Conservation

The mass inside the material volume is:

$$M(\vartheta _{m} )=\iiint\limits_{\vartheta _{m(t)} } \rho d\vartheta$$

Figure 4: Material volume $\vartheta _{m}(t)$ with surface $S_{m}(t)$.

Therefore the time rate of increase of mass inside the material volume is:

$\frac{d}{dt} M(\vartheta _{m} )=\frac{d}{dt} \iiint\limits_{\vartheta _{_{m} } (t)} \rho d\vartheta =0,$

which implies conservation of mass for the material volume $\vartheta _{m}$.

1.3.2 Momentum Conservation

The velocity of fluid inside the material volume in the $i^{th}$ direction is denoted as $u_{i}$. Linear momentum of the material volume in the $i^{th}$ direction is

$$\iiint\limits_{\vartheta _{m(t)} } \rho u_{i} d\vartheta$$

Newton's law of motion: The time rate of change of momentum of the fluid in the material control volume must equal the sum of all the forces acting on the fluid in that volume. Thus:

$$\frac{d}{dt} ( )_{i} = ( )_{i} + ( )_{i}$$

$$\frac{d}{dt} \iiint\limits_{_{\vartheta _{m} (t)} } \rho u_{i} d\vartheta = \iiint\limits_{\vartheta _{m} (t)} F_{i} d\vartheta + \iint\limits_{S_{m} (t)} \underset{_{\tau _{i} } }{\underbrace{\tau _{ij} n_{j} }} dS$$

Divergence Theorems:	For vectors: $\iiint\limits_{\vartheta} \underset{\frac{\partial v _{j} }{\partial x_{j} }}... ...{-12pt}\iint\limits_{S} \underset{v_{j}n_{j}}{\underbrace{ \vec{v}.\hat{n}}}dS$
	For tensors: $\iiint\limits_{\vartheta } \frac{\partial \tau _{ij} }{\partial x_{j} } d\vartheta =\bigcirc\hspace{-12pt}\iint\limits_{S} \tau _{ij} n_{j} dS$

Thus using divergence theorems:

$\frac{d}{dt} \iiint\limits_{\vartheta _{m(t)} } \rho u_{i} d\vartheta =\iiint... ...left( F_{i} +\frac{\partial \tau _{ij} }{\partial x_{j} } \right) d\vartheta$,

which gives the conservation of the momentum for the material volume $\vartheta _{m}$.

1.4 Kinematic Transport Theorems

Consider a flow through some moving control volume $\vartheta (t)$ during a small time interval $\Delta$t. Let $f\left( \vec{x},t \right)$ be any (Eulerian) fluid property per unit volume of fluid (e.g. mass, momentum, etc.). Consider the integral

$$I(t)=\iiint\limits_{\vartheta(t)} f\left( \vec{x},t \right) d\vartheta$$

According with the definition of the derivative, we can write

$$\begin{split} \frac{d}{dt} I(t) & = \lim _{\Delta t\rightarr... ...s_{\vartheta (t)} f(\vec{x},t) d\vartheta \right\} \end{split}$$

Figure 5: Control volume $\vartheta$ and its bounding surface $S$ at instants $t$ and $t+\Delta t$.

Next, we consider the steps

1. Taylor series expansion of f about t.

   $$f(\vec{x},t+\Delta t) = f(\vec{x},t)+\Delta t\frac{\partial f}{\partial t}(\vec{x},t)+O((\Delta t)^{2})$$

   
   

2. $\iiint\limits_{\vartheta (t+\Delta t)} d\vartheta =\iiint\limits_{\vartheta (t)} d\vartheta +\iiint\limits_{\Delta \vartheta } d\vartheta$ where $\iiint\limits_{\Delta \vartheta } d\vartheta = \iint\limits_{S(t)}\left[U_{n}(\vec{x},t)\Delta t\right]dS$ and $U_{n}(\vec{x},t)$ is the normal velocity of $S(t)$.

   Figure 6: Element of the surface $S$ at instants $t$ and $t+\Delta t$.

   

So we have

$$\frac{d}{dt} I(t)=\lim _{\Delta t\rightarrow 0} \frac{1}{\Delta ... ...dSU_{n} f-\iiint\limits_{\vartheta (t)} d\vartheta f+O(\Delta t)^{2} \right\}$$

Kinematic Transport Theorem (KTT) $\sim$ Leibnitz rule in 3D

$\frac{d}{dt} \iiint\limits_{\vartheta(t)} f(\vec{x},t)d\vartheta = \iiint\lim... ...)}{\partial t}d\vartheta + \iint\limits_{S(t)} f(\vec{x},t)U_{n}(\vec{x},t)dS$

If the control volume is a material volume: $\vartheta(t)=\vartheta _{m} (t)$ and U$_{n}$ $=\vec{v}\cdot\hat{n}$, where $\vec{v}$ is the fluid particle velocity. Then the Kinematic Transport theorem (KTT) assume the form

$$\frac{d}{dt} \iiint\limits_{\vartheta_{m}(t)} f(\vec{x},t)d\vart... ...tiny {Einstein Notation}})} {\underbrace{f(\vec{x},t)(\vec{v}\cdot\hat{n})}}dS$$

Using the divergence theorem:

$$\iiint\limits_{\vartheta} \underset{\frac{\partial}{\partial x_{... ...t\limits_{S}\underset{\alpha_{i}n_{i}}{\underbrace{\vec{\alpha}\cdot\hat{n}}}dS$$ (1)

1$^{st}$ Kinematic Transport Theorem (KTT)

$\frac{d}{dt} \iiint\limits_{\vartheta_{m}(t)} f\left( \vec{x},t\right)d\varth... ...rtial x_{i}}(fv_{i})}{\underbrace{ \nabla\cdot(f\vec{v})}}\right]d\vartheta,$

where f is a fluid property per unit volume.

1.5 Continuity Equation

Let the fluid property per unit volume be mass per unit volume ( f $= \rho$)

$$0\underset{\underset{\underset{\mbox{\footnotesize {of mass}}} {... ...ho }{\partial t} +\nabla \cdot \left( \rho \vec{v} \right) \right] d\vartheta$$

since $\vartheta _{m}$ is arbitrary, so the integrand $\equiv 0$ everywhere. Therefore, the differential form of conservation of mass i.e. Continuity equation follows:

$$\begin{split} & \frac{\partial \rho }{\partial t} +\nabla \c... ...\right.}}+\left.\rho\nabla\cdot\vec{v} \right] = 0 \end{split}$$

Therefore,

$\frac{D\rho }{Dt} +\rho \nabla \cdot \vec{v} = 0$

In general, $\rho = \rho(p,T,\ldots)$. We consider the special case of incompressible flow (Note, the density of the entire flow is not constant when we have more than one fluid, like water and oil, as illustrated in the picture above).

Figure 7: Interface of two fluids (oil-water)

Therefore, for an incompressible flow:

$\frac{D\rho }{Dt} =0$

Then $\underset{\mbox{\tiny {rate of volume dilatation}}}{\underbrace{\nabla \cdot \vec{v} \mbox{ or } \frac{\partial v_{i} }{\partial x_{i} } =0}}$, which is the Continuity equation for incompressible fluid.

1.6 Euler's Equation (differential form of conservation of momentum)

2$^{nd}$ Kinematic Transport Theorem ( = 1$^{st}$ KTT + continuity equation). If $G$ = fluid property per unit mass, then $\rho$$G$ = fluid property per unit volume

$$\begin{split} \frac{d}{dt} \iiint\limits_{\vartheta_{m}(t)} ... ... t} +\vec{v} cdot\nabla G\right)}}\right]d\vartheta,\end{split}$$

and the 2$^{nd}$ Kinematic Transport Theorem (KTT) follows:

$\frac{d}{dt} \iiint\limits_{\vartheta _{m} } \rho Gd\vartheta =\iiint\limits_{\vartheta _{m} } \rho \frac{DG}{Dt} d\vartheta$

Application:

We consider $G$ as the $i^{th}$ momentum per unit mass ( $v_{i}$). Then,

$$\iiint\limits_{\vartheta _{m} (t)} \left( F_{i} +\frac{\partial \... ...parrow}}{=}\iiint\limits_{\vartheta_{m}(t)} \rho \frac{Dv_{i} }{Dt} d\vartheta$$

But $\vartheta _{m}(t)$ is an arbitrary material volume, therefore the integral identity gives Euler's equation

$\rho \frac{Dv_{i} }{Dt} \equiv \rho\left(\frac{\partial v_{i}}{\partial t}+\u... ...cdot\nabla v_{i}}}\right) = F_{i} +\frac{\partial \tau_{ij} }{\partial x_{j} }$,

and its Vector Tensor Form

$\rho \frac{D\vec{v}}{Dt} \equiv \rho \left( \frac{\partial\vec{v}}{\partial t} +\vec{v}\cdot\nabla\vec{v} \right) = \vec{F}+\nabla\cdot\underset{ }{\tau}$