\def\ptl{\partial}
\def\uv{{\bf u}}
\def\vv{{\bf v}}
\def\wv{{\bf w}}
\def\nv{{\bf n}}
\def\del{\nabla}
\nopagenumbers
Fluid Dynamics I \hfil PROBLEM SET 4 \hfil Due October 13, 2003
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1. (Reading: Batchelor 543-545). Consider axisymmetric motion,
with velocity $\uv = (u_r,u_{\theta},u_z)$, of a fluid of constant
density. The equation of continuity in cylindrical polar coordinates is
$${\ptl u_z\over\ptl z}+{1\over r}{\ptl r u_r\over\ptl r} = 0.$$
(a) Show that this equations is satisfied if
$$u_z={1\over r}{\ptl \psi\over\ptl r}, u_r =-{1\over r}
{\ptl \psi\over\ptl z},$$
for some function $\psi$, and verify that
$$\omega_{\theta} = -{1\over r}L(\psi ), L= {\ptl^2\over\ptl z^2}
+{\ptl^2\over\ptl r^2}-{1\over r}{\ptl\over\ptl r}.$$
(b) Assume the flow is {\it steady}. i.e. $\ptl \uv / \ptl t = 0$.
Show that if
$$\uv\cdot\del Q= u_r{\ptl Q\over \ptl r}+u_z{\ptl Q\over\ptl z }=0 $$
Then $Q$ is a function of $\psi$ alone, $Q=F(\psi )$.
(c) Applied to Bernoulli's expression
$$\uv\cdot \del ({p\over \rho}+{1\over 2}q^2 ) = 0$$
show that $${p\over \rho} +{1\over 2} (u_r^2+u_{\theta}^2+u_z^2)=H(\psi )$$
for some function $H$.
Applied to the equation for $u_{\theta}$, show that
$r u_\theta = C(\psi )$ for some function $C$. Show
that this is a special case of Kelvin's theorem.
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2. Continuing problem 1, we showed in class that
$$(u_r{\ptl \over \ptl r}+u_z{\ptl \over\ptl z })(\omega_\theta / r) =
{2\over r^2} u_\theta {\ptl u_\theta\over \ptl z } .$$
From this show (using results of problem 1) that $\psi$ satisfies
$$L(\psi) = -CC_\psi+r^2 f(\psi)$$
for some functions $C,f$. Finally, show from the momentum equation
that $f=H_\psi$ where $H$ is defined in problem
1.
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3. (Reading: Milne-Thomson p. 89, Batchelor p. 384) (a) Prove Kelvin's minimum energy theorem:
In a simply-connected domain V let $\uv = \del \phi, \del^2 \phi = 0$,
with $\ptl \phi /\ptl n = f$ on the boundary $S$ of $V$. (This $\uv$
is unique in a simply-connected domain). If $\vv$ is any
differentiable vector field satisfying $\del\cdot\vv = 0$ in $V$ and
$\vv\cdot \nv = f$ on $S$, then
$$\int_V |\vv |^2 dV \geq \int_V |\uv |^2 dV .$$
(Hint: Let $\vv = \uv+\wv$ , and apply the divergence theorem to the cross
term.)
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\end