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\begin{document}


\begin{center}
{\Large{\bf Astrophysical Fluid Dynamics -- Problem Set 4}}

%{\large{\bf Due in class Wednesday Feb 21}}

\end{center}

\vspace{0.1in}
\ni
Readings: Tritton, chapter 3; Movies: Drag I and II; also readings from textbooks by Acheson and by Batchelor as pointed out in specific problems below

\vspace{0.1in}
\ni
{\large{\bf Problem 1.}}
Elementary Viscous Flows

\ni Problems taken from Acheson (Chapter 2; Problem 2.3; Problem 7.2).
These are ``classic problems'' in low Reynolds number (highly viscous)
flows. While they are not directly germane to astrophysics, 
they do provide practice with solving the Navier-Stokes equation (highly
simplified), and they arguably form part of the standard canon
of fluid mechanics which physicists of all persuasions should know.

\ni Consider an incompressible fluid of dynamic viscosity $\mu$.
For all problems below, take the Reynolds number to be $\ll 1$
and neglect gravity.

\ni (a) {\it Steady
flow between plates}: Viscous fluid flows between two stationary rigid
boundaries $y = \pm h$ under a constant pressure gradient $G = -dP/dx$.
Show that

\begin{equation}
u = \frac{G}{2\mu} (h^2 - y^2),\,\,\,\,\,v=w=0\,.
\end{equation}

\ni (b) {\it Poiseuille flow}, after the physician who studied such flows
in connection with blood vessels: Viscous fluid flows steadily down a
horizontal pipe
of circular cross-section $r=a$ under a constant pressure gradient
$G = -dp/dx$. Show that

\begin{equation}
u = \frac{G}{4\mu} (a^2-r^2)
\end{equation}

\ni (Such pipe flow is unstable at high Re, as discovered by
Osborne Reynolds, the father of the study of fluid turbulence.)

\ni (c) {\it Couette flow}: Viscous fluid occupies the space between
two infinitely long, concentric cylinders of radius $r_1$ and $r_2 > r_1$.
The cylinders rotate at angular speeds of $\Omega_1$
and $\Omega_2$, respectively. Show that in steady state, the fluid rotates
with velocity

\begin{equation}
u = u_{\theta} = Ar + \frac{B}{r}
\end{equation}

\ni where $A = (\Omega_2 r_2^2-\Omega_1 r_1^2)/(r_2^2-r_1^2)$
and $B = (\Omega_1-\Omega_2)r_1^2r_2^2/(r_2^2-r_1^2)$.

\ni (Such flow becomes unstable to the formation of Taylor
vortices if $\Omega_1 \gg \Omega_2$. Maybe we'll study this
instability later.)

\ni (d) Imagine a rigid sphere of radius $a$ immersed
in the viscous fluid, initially rotating at angular frequency $\Omega$.
Write down an order-of-magnitude expression for
the spin-down time of the sphere. For extra credit: evaluate this
spin-down time for the Earth's inner solid iron core, immersed in the outer
liquid iron core.

%\vspace{0.1in}
%\ni
%{\large{\bf Problem 1.}}
%Fountains

%\ni Tritton page 469, problem 2.

%\ni A vertical tube of diameter $d = 2$ mm has its upper end open
%to the atmosphere. At the lower end, water is maintained at a pressure
%of $P = 10^4$ N m$^{-2}$ above atmospheric.

%\ni (a) What is the longest length
%of the tube for which this pressure will induce a flow through
%the tube?

%\ni (b) What volume of water will pass through the tube per second
%if the length is half this? Assume that the flow is laminar and ignore
%the entry length (see last week's reading Tritton chapter 2).

\vspace{0.1in}
\ni
{\large{\bf Problem 2.}} Cavitation

\ni Problem taken from Blandford and Thorne, Exercises 12.11--12.12

\ni A submarine lies at depth $h \approx 500 \m$ below the water's surface.

\ni (a) The propeller radius is $L \approx 2 \m$. Estimate the
angular rotation
speed $\Omega$ for which cavitation---the creation of bubbles of water vapor
near the fast-moving propeller---occurs.
Provide a symbolic answer and a numerical answer in
revolutions per second.

\ni (b) The bubble, created with radius $R_{i}$, travels to a location where
the ambient pressure is $P_0$, much greater than the pressure where
the bubble was created. The greater pressure causes the bubble to collapse.
The rest of this problem studies this implosion.

\ni We picture the water surrounding the bubble to undergo
spherical infall with velocity $v(r,t)$, where $r$ measures the
radial distance from center of the bubble.
Show that as a result of water's incompressibility ($\rho$ = constant),

\begin{equation}
v = \frac{F(t)}{r^2}
\end{equation}

\ni where $F(t)$ is a function solely of time (and not space).

\ni (c) By integrating the momentum equation from the bubble's
surface at $r = R$ to $r = \infty$, show that the fluid velocity
$v(R)$ at the bubble's surface obeys

\begin{equation}
-\frac{1}{R} \frac{dF}{dt} + \frac{1}{2} v(R)^2 = \frac{P_0}{\rho}
\end{equation}

\ni (d) Show that

\begin{equation}
|v(R)| = \left( \frac{2P_0}{3\rho} \right)^{1/2} \left[ \left( \frac{R_i}{R} \right)^3 - 1 \right]^{1/2}
\end{equation}

\ni (e) The yield stress of a submarine hull might be
$Y \sim 10^{9} \dyne \cm^{-2}$.
Estimate the factor by which the bubble shrinks, $R_i/R$, such
that the imploding bubble would be expected to damage the hull.

\vspace{0.1in}
\ni
{\large{\bf Problem 3.}} Turning the Bend

\ni Consider a short cylindrical vat of radius $R$ containing
incompressible, viscous
fluid of density $\rho$ and kinematic viscosity $\nu$. The vat
(not the fluid) is set
spinning at angular velocity $\Omega$. The Reynolds number Re$_R$
is very small. Eventually the liquid comes into solid body rotation
with the vat (via the Ekman boundary layer, which we might study in the
future.)

%\ni (a) Estimate the timescale over which solid body rotation
%of the fluid is achieved. An approximate analytic expression suffices.

\ni (a) Calculate the run of pressure $P$ with radius $r$,
after solid body rotation sets in. Neglect atmospheric pressure
and neglect also hydrostatic pressure due to gravity (the vat
is very short). Take the pressure at the center of the vat to
be $P_0 = P(r=0)$.

\ni (b) Place the long side of your index finger underneath
a gently running tap so that water falls
only on one side of your finger. Observe how the stream tends to
hug the underside of your finger and get deflected towards the other
(unwetted) side.

\ni You can also try this with a spoon. Turn the spoon
sideways and place the bottom of the spoon into the flow
so that water curves round the spoon.
By moving the spoon in and out of the flow, feel how
the spoon seems to be {\it attracted} into the running water.
(It should feel like a kitchen magnet placed very close to the fridge.)

\ni Based on your understanding of (a),
derive an order-of-magnitude expression for the horizontal
force exerted on your finger, per unit length of wetted finger.
Express your answer in whatever variables you think are important.

\ni The effect that this problem explores is called the {\it Coanda}
effect, and is reported to be the same effect responsible for
liquids refusing to pour nicely out of containers.
If you pour beer from a pitcher that is tipped at too small an
angle from the vertical,
the beer leaves the rim only to dribble down the side
of the pitcher.

\vspace{0.1in}
\ni
{\large{\bf OPTIONAL Problem 4.}} Symmetry of Stress Tensor

\ni We introduced the stress tensor $\sigma_{ij}$ as a ``box of 9 numbers''
describing the force in the $j$-direction exerted on a surface
of constant $i$. This problem shows that the 9 numbers actually
aren't all independent; in fact the stress tensor
is symmetric: $\sigma_{ij} = \sigma_{ji}$, so that there really
are only 6 independent numbers. This result was shown by Cauchy.
The following problem is taken from textbooks by Acheson (Chapter 6; Problem 6.14)
and by Batchelor (page 11). It isn't that hard; it's essentially
an exercise in tensor calculus, and it proves a result that isn't
that obvious.\footnote{It wasn't obvious, I think, to Newton, whose hypothesis
for the viscous stress tensor $\tau_{ij} = \mu \partial u_j/\partial x_i$
is incorrectly asymmetric. By contrast, an example of a symmetric
stress tensor is
$\tau_{ij} = \mu (\partial u_j/\partial x_i + \partial u_i/\partial x_j)$.
Fluids that obey this relation are called Newtonian---giving generous credit
to Sir Issac! This symmetric form and Newton's asymmetric form
share one key property, however, which is that the stress
is linearly proportional to the velocity gradient.}

\ni Reynolds' transport theorem states that

\begin{equation}
\frac{d}{dt} \int_{V(t)} G dV = \int_{V(t)} \left( \frac{dG}{dt} + G \nabla \cdot {\mathbf u} \right) dV
\end{equation}

\ni where $V(t)$ denotes the region of space
occupied by a deforming fluid parcel, ${\mathbf u}$ is the
velocity,
$G({\mathbf x},t)$ is any scalar or vector function,
and $d/dt$ is the convective
derivative.
% The symbol $D/Dt$ denotes a simple time derivative;
% that is, the left-hand side is simply the time rate of change
% of a volume integral over the fluid parcel.

\ni (a) Argue\footnote{A complete proof is outlined in problem 6.13 of
Acheson; the solution is in the back of the book.}
that the theorem makes sense by considering $G = \rho F$,
where $\rho ({\mathbf x},t)$ is the fluid density
and $F({\mathbf x},t)$ is any function. That is, show that the above statement
implies

\begin{equation}
\frac{d}{dt} \int_{V(t)} F \rho dV = \int_{V(t)} \frac{dF}{dt} \rho dV
\end{equation}

\ni and explain why this latter statement makes sense. (Hint: what is conserved
about the fluid parcel?)

\ni (b) Conservation of angular momentum of the fluid parcel states that

\begin{equation}
\frac{d}{dt} \int_{V(t)} {\mathbf x} \times \rho {\mathbf u} dV = \int_{S(t)} {\mathbf x} \times {\mathbf t} dS + \int_{V(t)} {\mathbf x} \times \rho {\mathbf f}_{\rm body} dV
\end{equation}

\ni where $t_idS = \sigma_{ij} n_j dS$ is the force in the $i$-direction acting
on a surface having a directed area element ${\mathbf \hat{n}} dS$,
$S(t)$ denotes the surface of the fluid parcel,
and ${\mathbf f}_{\rm body}$ contains all the body forces.
The symbol $\times$ denotes the cross product: e.g.,
$({\mathbf a} \times {\mathbf b})_i = \epsilon_{ijk}a_jb_k$, where
$\epsilon_{ijk}$ is the (space- and time-independent) Levi-Civita tensor.
Clearly, the first term
on the right-hand side contains the torques due to surface stresses,
while the second term contains the torques due to body forces.

\ni Use Reynolds' transport theorem and our standard
momentum equation to show that

\begin{equation}
\int_{V(t)} \epsilon_{ijk} x_j \frac{\partial \sigma_{kl}}{\partial x_l} dV = \int_{S(t)} \epsilon_{ijk} x_j t_k dS
\end{equation}

\ni (c) Use the divergence theorem to rewrite the surface integral above as

\begin{equation}
\int_{S(t)} \epsilon_{ijk} \frac{\partial (x_j \sigma_{kl})}{\partial x_l} dV \,.
\end{equation}

\ni Thereby show that

\begin{equation}
\int_{V(t)} \epsilon_{ijk} \sigma_{kj} dV = 0
\end{equation}

\ni and that $\sigma_{kj} = \sigma_{jk}$ as a consequence.

\end{document}

