13.021 Marine Hydrodynamics, Fall 2003

Lecture 4

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

13.021 - Marine Hydrodynamics
Lecture 4

Introduction

Governing Equations so far:

Knowns	Number of Equations		Number of Unknowns
$\rho$	Continuity(conservation of mass)	1	$v_{i}$	3
$F_{i}$	Euler (conservation of momentum)	3	$\tau_{ij}$	6 3 of 9 eliminated by symmetry
		4		9

Therefore, some constitutive relationships are needed to relate $v_{i}$ to $\tau$$_{ij}$.

1.7 Newtonian Fluid

1. Consider a fluid at rest ($v_{i}$ $\equiv$ 0). Then according to Pascal's Law:

   $$\tau_{ij} = -p_{s}\delta_{ij} \hspace{1.0in} \underset{ }{\tau} =\left[ \begin{array}{ccc} -p_{s} & 0 & 0 \ 0 & -p_{s} & 0 \ 0 & 0 & -p_{s} \end{array} \right]$$

   
   
   where $p_{s}$ is the hydrostatic pressure and $\delta_{ij}$ is the Kroenecker delta function, equal to $1$ if $i = j$ and 0 if $i \ne j$.

2. Consider a fluid in motion. The fluid stress is defined as:

   $$\tau _{ij} =-p\delta _{ij} +\hat{\tau}_{ij}$$

   
   
   where $p$ is the thermodynamic pressure and $\hat{\tau}_{ij}$ is the dynamic stress, which is related to the velocities empirically.

Experiments with a wide class of fluids, "Newtonian" fluids, obtain that:

$$\hat{\tau}_{ij} \approx \underset{\overbrace{\frac{\partial }{\partial t} \left( \frac{\... ...rac{\partial u_{k} }{\partial x_{m} }}}{\mbox{\underline {velocity gradient}}}$$

$$\hat{\tau}_{ij} \approx \underset{\underset{\underset{\underset{... ...\partial u_{k} }{\partial x_{m} } \hspace{5mm}\hspace{5mm}i, j, k, m = 1, 2, 3$$

Note that the shear stress is proportional to the rate of strain.

For isotropic fluids, this reduces to:

$$\hat{\tau } = \mu\left(\frac{\partial u_{i}}{\partial x_{j}}+\fr... ...cdot\vec{v}}{\underbrace{\left(\frac{\partial u_{l}}{\partial x_{l}}\right)}},$$

where the fluid properties are:

• $\mu$ - (coefficient of) dynamic viscosity.
• $\lambda$ - bulk elasticity, `second' coefficient of viscosity

For incompressible flow, $\frac{\partial u_{l} }{\partial x_{l} } =0$. Therefore, for an incompressible, isotropic, Newtonian fluid the viscous stress $\hat{\tau}_{ij}$ is given as

$$\hat{\tau }_{ij} = \mu\left(\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}}\right)$$

1.8 Navier-Stokes equations

Equations	Number of Equations	Unknowns	Number of Unknowns
Euler	3	$u_{i}$	3
continuity	1	$p$	1
constitutive (Newtonian)	6 (symmetry)	$\tau_{ij}$	6
	10		10
	$\underset{\mbox{closure}}{\underbrace{\hspace{2.35in}}}$

Substitute the equation for the stress tensor

$$\tau_{ij} = -p\delta _{ij} +\mu \left( \frac{\partial u_{i} }{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i} } \right)$$

for a Newtonian fluid into Euler's equation:

$$\rho \frac{Du_{i} }{Dt} =F_{i} +\frac{\partial \tau _{ij} }{\partial x_{j} }$$

where

$$\frac{\partial \tau _{ij} }{\partial x_{j} } =-\frac{\partial p}... ...al u_{i} }{\partial x_{j} } +\frac{\partial u_{j} }{\partial x_{i} } \right)}}$$

and $\frac{\partial u_{j}}{\partial x_{j}} = 0$ due to continuity. Finally,

$$\frac{Du_{i} }{Dt} = \frac{\partial u_{i} }{\partial t} +u_{j} \... ...rac{\partial ^{2} u_{i}}{\partial x_{j}\partial x_{j} } +\frac{1}{\rho } F_{i}$$ Tensor form

$$\frac{D\vec{v} }{Dt} = \frac{\partial \vec{v} }{\partial t} +\ve... ...{\rho } \nabla p +\nu \nabla^{2}\vec{v} +\frac{1}{\rho } \vec{F} \hspace{30pt}$$ Vector form

where $\nu \equiv \frac{\mu }{\rho }$ denoted as the Kinematic viscosity [ $L^{2} /T$ ].

• Navier-Stokes equations for incompressible, Newtonian fluids

  	Number of Equations		Number of Unknowns
  continuity	1	$p$	1
  Navier-Stokes	3 (symmetry)	$u_{i}$	3
  	4		4

1.9 Boundary Conditions

1. Kinematic Boundary Conditions: Specifies kinematics (position, velocity, ...) On a solid boundary, velocity of the fluid = velocity of the body. i.e. velocity continuity.

   $$\vec{v} = \vec{u}$$

   
   
   where $\vec{v}$ is the fluid velocity at the body and $\vec{u}$ is the body surface velocity
   • $\vec{v}\cdot\hat{n} = \vec{u}\cdot\hat{n}$ no flux - continuous flow
   • $\vec{v}\cdot\hat{t} = \vec{u}\cdot\hat{t}$ no slip - finite shear stress

   figure

   

2. Dynamic Boundary Conditions: Specifies dynamics ( pressure, sheer stress, ... )
   

   Stress continuity:

   $$p = p'+p_{\mbox{interface}}$$

   $$\tau _{ij} = \tau _{ij}'+\tau _{ij \mbox{ interface}}$$

   
   
   The most common example of interfacial stress is surface tension.

   figure

   

Surface Tension

• Notation: $\Sigma$ [Tension force / Length] $\equiv$ [Surface energy / Area].

• Surface tension is due to the inter molecular forces attraction forces in the fluid.

• At the interface of two fluids, surface tension implies in a pressure jump across the interface. $\Sigma$ gives rise to $\Delta$p across an interface.

• For a water/air interface: $\Sigma$ = 0.07 N/m. This is a function of temperature, impurities etc...

• 2D Example:

  $\underset{\approx 1}{\underbrace{\cos \frac{d\theta }{2} }} \cdot \Delta p\cd... ... }{\underbrace{\sin \frac{d\theta }{2} }} \approx 2\Sigma \frac{d\theta }{2}$

  $\therefore \Delta p=\frac{\Sigma }{R}$

  Higher curvature implies in higher pressure jump at the interface.

  figure

  

• 3D Example: Compound curvature

  $$\Delta p=\left( \frac{1}{R_{1} } +\frac{1}{R_{2} } \right) \Sigma$$

  
  
  where R$_{1}$ and R$_{2}$ are the principle radii of curvature.

1.10 Body Forces - Gravity

• Conservative force:

  $$\vec{F} = -\nabla\varphi \mbox{ for some \ensuremath{\varphi}} ,$$

  
  
  where $\varphi$ is the force potential.

  $$\oint \vec{F}\cdot d\vec{x} = 0 \int_{1}^{2}\vec{F}\cdot d\vec{x} = -\int_{1}^{2}\nabla\varphi\cdot d\vec{x} = \varphi(\vec{x}_{1})-\varphi(\vec{x}_{2})$$

  
  

• A special case of a conservative force is gravity.

  $$\vec{F} = -\rho g \hat{k},$$

  
  
  with Gravitational potential:

  $$\varphi =\rho gz \rightarrow \vec{F} = -\nabla\varphi=\nabla(-\rho gz) = -\rho g \hat{k},$$

  
  
  where $-\rho g z$ is the hydrostatic pressure $p_{s} = -\rho gz = -\varphi$.
  

  Navier-Stokes equation:

  $$\rho \frac{D\vec{v} }{Dt} = -\nabla p+\vec{F}+\rho\nu\nabla^{2}\vec{v}$$

  $$=-\nabla p-\nabla\rho gz+\rho \nu \nabla ^{2} \vec{v}$$

  $$=-\nabla \left( p+\rho gz\right) +\rho \nu \nabla ^{2} \vec{v},$$

  
  
  but $p - p_{s} = p_{d}$ and p $_{s}= -\ensuremath{\rho}gz$, where $p$ is the total pressure and $p_{d}$ is the dynamic pressure. Therefore,

  $$\rho \frac{D\vec{v} }{ Dt} = -\nabla p_{d}+\rho\nu\nabla^{2}\vec{v}$$

  
  

• Presence of gravity body force is equivalent to replacing total pressure $p$ by dynamic pressure $p_{d}$ in the Navier-Stokes(N-S) equation.

• Solve the N-S equation with $p_{d}$, then calculate $p = p_{d} + p_{s} = p_{d} - \rho gz$.