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\begin{document}
{\noindent\bf MAE224A: Environmental Fluid Mechanics}
\hfill Fall Quarter 2010
\bigskip

\hfill \textrm{http://maecourses.ucsd.edu/~sllewell/MAE224A\_2010/}
\bigskip

\begin{center}
\bf\LARGE Solution I
\end{center}

\paragraph{1.} Consider a tensor of arbitrary order, $F(\xb,t)$,
integrated over the region $V(t)$, which may be either a fixed or
material volume. The time derivative of this tensor is written as
\[
\frac{\dd}{\dd t}\int_{V(t)}F \,\dd V,
\]
where $\dd/\dd t$ emphasizes the fact that $F$ remains solely a
function of time after integration in space. Exploring the 1D case,
with boundaries at $x = a(t)$ and $x = b(t)$, one can use Leibniz's
theorem, which shows how to differentiate an integral whose integrand
$F$ as well as the limits of integration are functions of the variable
with respect to which the integration occurs. The time derivative of
the integral becomes
\[
\frac{\dd}{\dd t}\int_{a(t)}^{b(t)}F(x,t) \,\dd x =
\int_{a}^{b}\frac{\partial F }{\partial t} \,\dd x+\left ( \frac{\dd b}{\dd
    t} F(b,t)-\frac{\dd a}{\dd t} F(a,t) \right ).
\]
The last two terms on the right-hand side represent the gain of $F$ at
the outer boundary moving at a rate $\dd b/\dd t$, and the loss of $F$
at the inner boundary at a rate $\dd a/\dd t$, respectively. These may
be combined to a more general integral, representing the outward flux
through an area element $\dd \Ab$. Rewriting the above equation, and
generalizing Leibniz's theorem,
\[
\frac{\dd}{\dd t}\int_{V(t)}F(\xb,t) \,\dd V =
\int_{V(t)}\frac{\partial F }{\partial t} \,\dd V+\int_{A(t)}F\,\dd
\Ab \cdot \ub_A,
\]
where $\ub_A$ is the velocity of the boundary and $A(t)$ is the
surface of $V(t)$.

For a material volume, $\Vb(t)$, the surfaces move with the fluid, and
consequently $\ub_A = \ub$. The above equation may be re-written as
the Reynolds transport theorem, following a material volume:  
\[
\frac{\DD}{\DD t}\int_{\Vb}F(\xb,t) \,\dd \Vb =
\int_{\Vb}\frac{\partial F }{\partial t} d\Vb+\int_{A}F \,\dd \Ab
\cdot \ub_A.
\]
Another form of the transport theorem may be obtained by manipulating
the above via Gauss's theorem, and defining a new function $f$, where
$F = \rho f$. Transforming the surface integral into a volume
integral, and using index notation for simplicity,
\[
\frac{\DD}{\DD t}\int_{\Vb}\rho f\,\dd \Vb = \int_{\Vb}\left
  [\frac{\partial F }{\partial t} + \frac{\partial }{\partial
    x_j}(\rho f u_j)  \right ] \,\dd\Vb.
\]
By virtue of the product rule, the RHS becomes
\[
= \int_{\Vb}\left [\rho \frac{\partial f }{\partial t} +f
  \frac{\partial \rho }{\partial t}+f\frac{\partial }{\partial
    x_j}(\rho u_j)+ \rho u_j \frac{\partial f}{\partial x_j} \right ]
\,\dd\Vb.
\]
The continuity equation gives
\[
\frac{\partial \rho}{\partial t}+\frac{\partial }{\partial x_j}(\rho u_j) = 0
\]
so that
\[
\frac{\DD}{\DD t}\int_{\Vb}\rho f \,\dd \Vb =  \int_{\Vb}\left [\rho
  \frac{\partial f }{\partial t} + \rho u_j \frac{\partial f}{\partial
    x_j} \right ] d\Vb =  \int_{\Vb}\rho \, \frac{\DD f }{\DD t} \,\dd
\Vb.
\]

\smallskip

SOURCE: Kundu et al., $\textbf{Fluid Mechanics}$, ed. 4, pgs. 82--84.



\paragraph{2.} Upon nondimensionalizing the momentum equations with a
characteristic length scale, $L$, and characteristic velocity, $U$,
the non-dimensional Reynolds number is obtained:
\[
Re = \frac{UL}{\nu},
\]
where $\nu$ is the kinematic viscosity of the fluid. This quantity is
a ratio of inertia forces to viscous forces; if small, inertia forces
are negligible, and viscous and pressure forces are in approximate
balance in steady state (e.g. creeping flow around a sphere). When
$Re$ is large, inertia forces dominate, and the viscous term in the
momentum equations may be omitted as a first approximation, outside of
the boundary layer.

The Rossby and Ekman numbers are obtained by nondimensionalizing the
rotating momentum equations. The Rossby number is given by
\[
Ro = \frac{U^2/L}{fU} = \frac{U}{fL}
\]
and represents the ratio of the nonlinear advection terms to the
Coriolis force. When $Ro$ is large, the earth's rotation is
negligible, as opposed to the nonlinear acceleration terms. When $Ro$
is small, for example when looking at very long length scales, the
Coriolis force may not be ignored, but the neglect of nonlinear terms
is justified for the flow.

The Ekman number is defined as
\[
E = \frac{\nu}{fL^2}
\]
and is a ratio of the viscous force to the Coriolis force. Likewise,
when the flow in question is over a long characteristic length scale,
and consequently $E$ is very small, the viscous forces are negligible
in comparison to the force created by earth's rotation. 

\medskip

(i) For mantle convection, the following approximate values were found:
\[
\nu \approx  3 \times 10^{17} \;\mathrm{m^2/s}, \; \; \;
 U \approx 1.5 \times 10^{-9} \;\mathrm{m/s},   \; \; \;
 D \approx 2.9 \times 10^6  \;\mathrm{m}   
\]
The Coriolis frequency is defined as $f = 2\Omega \sin \theta$, where
$\Omega$ is the earth's rate of rotation, equal to $0.73 \times
10^{-4} \; \mathrm{s^{-1}}$, and $\theta$ is the latitude. Here, the
latitude is assumed to be $90 ^{\circ}$N, giving $f = 1.46 \times
10^{-4} \; \mathrm{s^{-1}}$. Therefore, the three dimensionless
parameters are:
\[
Re = 1.45 \times 10^{-20},   \; \; \;
Ro = 3.54 \times 10^{-12},   \; \; \; 
E = 2.44 \times 10^8  
\]
These values make sense intuitively, since mantle convection is a slow
creeping motion dominated by viscous effects.

\smallskip

SOURCE: Lowrie, William. $\textbf{Fundamentals of Geophysics}$, pg. 328. 

\medskip
(ii) For the Great Red Spot of Jupiter, the following approximate
values were found:
\[
\nu \approx  1 \times 10^{6} \;\mathrm{m^2/s},   \; \; \;
U \approx 2 \;\mathrm{m/s},   \; \; \;
D \approx 1.3 \times 10^7  \;\mathrm{m},   \; \; \;
f \approx 1.3 \times 10^{-4} \;\mathrm{s^{-1}}    
\]
These correspond to a Reynolds, Rossby, and Ekman number of:
\[
Re = 26,   \; \; \; 
Ro = 1.18 \times 10^{-3},   \; \; \; 
E = 4.55 \times 10^{-5}  
\]
In this case, it is expected that viscous forces do not dominate the
flow, but rather the Coriolis terms. The result for $Re$ seems
small. These may not be the best values to take.

\smallskip

SOURCE: Stone, P.H. et al., \textit{Concerning the existence of Taylor
  columns in atmospheres}, Journal of Royal Meter. So., 1968,
pgs. 578-579.

\medskip

(iii) The outflow from the Point Loma wastewater treatment plant has
the following approximate values:
\[
\nu \approx 1 \times 10^{-6} \; \mathrm{m^2/s},   \; \; \;
U \approx 0.731 \; \mathrm{m/s},   \; \; \; 
D = 3.6576 \; \mathrm{m}
\]
Located at a latitude of  $33 ^{\circ}$N, the Coriolis frequency is $f
= 7.95 \times 10^{-5} \; \mathrm{s^{-1}}$. These values give the
following nondimensional parameters:
\[
Re = 2.67 \times 10^{6},   \; \; \; 
Ro = 2.51 \times 10^{3},   \; \; \; 
E = 9.40 \times 10^{-4}  
\]

In this case, $Re$ is large, and one would therefore expect a
turbulent outflow from the pipe. Furthermore, the Rossby number is
large, and consequently the nonlinear acceleration terms cannot be
ignored, as is expected for turbulent flow.

\smallskip

SOURCE: City of San Diego Facilities website, Point Loma Wastewater
Treatment Plant.

{\tt http://www.sandiego.gov/mwwd/facilities/ptloma.shtml}


\paragraph{3.} The vertical momentum equation for a static fluid of
density $\rho$ is
\[
0 = -\frac{\dd p}{\dd z} - \rho g.
\]
This equation may be readily integrated in $z$ to obtain
\[
P = - \rho gz,
\]
where the atmospheric pressure, $P_0$, has been taken to be 0 at
$z=0$. Now consider a body with boundary $S$ that is at least
partially submerged in the fluid. The fluid exerts a force $d\Fb =
-P\, \mathbf{\hat{n}} \,dS$ on an area element $dS$ on the body, where
$\mathbf{\hat{n}}$ is the unit outward vector normal to the
surface. The net forces in the $\mathbf{\hat{x}}$ and
$\mathbf{\hat{y}}$ directions are 0 by symmetry, so that the total
force in the z-direction is:
\[
F_z=-\int_{S} P \,\dd S.
\]
Substituting for P,
\[
F_z=\rho g \int_{S} z\,\dd S.
\]
By virtue of the divergence theorem, the force becomes
\[
F_z=\rho g \int_{V} \frac{\partial z}{\partial z} \,\dd V = \rho g
\int_{V} \,\dd V = \rho g V = mg.
\]
Therefore, the force acting on the solid object immersed in the fluid
is equal to the weight of the displaced fluid. Note that the
hydrostatic equation is used on the boundary to obtain $F_z$. The
transformation from surface to volume integral is a mathematical
procedure and the physical contents of the interior of the volume are
irrelevant.

\smallskip

SOURCE: Physics Wiki, Archimedes' Principle


\paragraph{4.} The fluid in the tank is in solid-body rotation, and
therefore the inviscid Euler equations in the radial and vertical
directions may be used in cylindrical coordinates, and simplifies to
\[
-\rho \frac{u_\theta^2}{r}=-\frac{\partial p}{\partial r}, \qquad 0 =
-\frac{\partial p}{\partial z}-\rho g.
\]
The pressure differences between two neighboring points must be:
\[
\dd p = \frac{\partial p}{\partial r} \,\dd r + \frac{\partial
  p}{\partial z} \,\dd z = \rho \,\Omega^2r \,\dd r - \rho g \,\dd z,
 \]
where $\Omega r$ has been substituted for $u_\theta$. This may be
integrated between arbitrary points 1 and 2 so that
\[
P_2 - P_1 =\frac{1}{2} \rho \,\Omega^2(r_2^2-r_1^2)-\rho g(z_2-z_1)
\]

Surfaces of constant pressure, and thus an outline of the surface, occur at
\[
z_2-z_1 = \frac{1}{2} \,\frac{\Omega^2}{g}(r_2^2-r_1^2),
\]
which are paraboloids of revolution. If at height $z_1$ = 0 (taken at
the bottom of the tank), the radius is $r_1$ = 0, since all the fluid
is pushed up against the side walls, then the above equation becomes
\[
z_2 = \frac{1}{2} \,\frac{\Omega^2}{g}r_2^2.
\]
Working in 2D, the area under the parabola must equal the area of
fluid in the tank, which is 2HL. Therefore,
\[
2\int_{0}^{L} \frac{1}{2} \,\frac{\Omega^2}{g} x^2 \,\dd x = 2HL
\]
Integrating the LHS,
\[
 \frac{1}{6} \,\frac{\Omega^2}{g}\,L^3 = HL
\]
Solving for $\Omega_c$,
\[
{\Omega_c} = \sqrt{\frac{6gH}{L^2}}
\]

If $\Omega$ is increased above this value, the minimum point on the
parabola of the free surface (i.e. line of constant pressure) will
have to drop below z = 0, so that at z = 0, the bottom of the tank is
dry.
\end{document}
