13.021 Marine Hydrodynamics, Fall 2003
Lecture 4
Copyright © 2003 MIT - Department of Ocean
Engineering,
All rights reserved.
13.021 - Marine Hydrodynamics
Lecture 4
Governing Equations so far:
Knowns |
Number of Equations |
Number of Unknowns |
|
Continuity(conservation of
mass) |
1 |
![$ v_{i}$](img2.gif) |
3 |
|
Euler (conservation of
momentum) |
3 |
![$ \tau_{ij}$](img4.gif) |
6
3 of 9 eliminated by symmetry |
|
|
4 |
|
9 |
Therefore, some constitutive relationships are needed to relate
to ![$ \tau$](img5.gif)
.
- Consider a fluid at rest (
0). Then according to
Pascal's Law:
(Pascal's
Law)![$\displaystyle \hspace{1.0in} \underset{ }{\tau} =\left[
\begin{array}{ccc}
-p_{s} & 0 & 0 \
0 & -p_{s} & 0 \
0 & 0 & -p_{s}
\end{array}
\right]$](img9.gif) |
|
where
is the hydrostatic pressure and
is the
Kroenecker delta function, equal to
if
and 0 if
.
- Consider a fluid in motion. The fluid stress is defined as:
where
is the thermodynamic pressure and
is
the dynamic stress, which is related to the velocities
empirically.
Experiments with a wide class of fluids, "Newtonian"
fluids, obtain that:
linear function of the
![$\displaystyle \underset{\overbrace{\frac{\partial }{\partial t} \left(
\frac{\...
...rac{\partial u_{k} }{\partial x_{m}
}}}{\mbox{\underline {velocity gradient}}}$](img19.gif) |
|
i.e. ![$\displaystyle \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$](img20.gif) |
|
Note that the shear stress is proportional to the rate of strain.
For isotropic fluids, this reduces to:
where the fluid properties are:
- (coefficient of) dynamic viscosity.
- bulk elasticity, `second' coefficient of viscosity
For incompressible flow,
. Therefore, for an incompressible, isotropic, Newtonian fluid
the viscous stress
is given as
Equations |
Number of Equations |
Unknowns |
Number of Unknowns |
Euler |
3 |
|
3 |
continuity |
1 |
|
1 |
constitutive (Newtonian) |
6 (symmetry) |
|
6 |
|
10 |
|
10 |
|
|
Substitute the equation for the stress tensor
for a Newtonian fluid into Euler's equation:
where
and
due to continuity.
Finally,
![$\displaystyle \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}$](img33.gif) |
Tensor form |
|
![$\displaystyle \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}$](img34.gif) |
Vector form |
|
where
denoted as the
Kinematic viscosity [
].
- Navier-Stokes equations for incompressible, Newtonian fluids
|
Number of Equations |
|
Number of Unknowns |
continuity |
1 |
![$ p$](img16.gif) |
1 |
Navier-Stokes |
3 (symmetry) |
![$ u_{i}$](img27.gif) |
3 |
|
4 |
|
4 |
- Kinematic Boundary Conditions: Specifies kinematics (position, velocity,
...) On a solid boundary, velocity of the fluid = velocity of
the body. i.e. velocity continuity.
"no-slip" boundary condition |
|
where
is the fluid velocity at the body and
is
the body surface velocity
-
no flux - continuous
flow
-
no slip -
finite shear stress
figure
- Dynamic Boundary Conditions: Specifies dynamics ( pressure, sheer stress, ...
)
Stress continuity:
The most common example of interfacial stress is surface tension.
figure
- Conservative force:
where
is the force potential.
or ![$\displaystyle \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})$](img61.gif) |
|
- A special case of a conservative force is gravity.
with Gravitational potential:
where
is the hydrostatic pressure
.
Navier-Stokes equation:
but
and p
, where
is the total pressure and
is the dynamic pressure.
Therefore,
- Presence of gravity body force is equivalent to replacing
total pressure
by dynamic pressure
in the
Navier-Stokes(N-S) equation.
- Solve the N-S equation with
, then calculate
.
Karl P Burr
2003-08-27