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Circular section

For a circular section of diameter $b$, $\mathcal S$=1 and the domain is given by

\begin{displaymath}
\Omega\raisebox{1ex}{$\circ$}=\{\mathbf{X}:X^2+Z^2<1/4\}.
\end{displaymath} (7.60)

The forced flow solution is (Stokes 1845; Lamb 1932, p. 585; Bird et al. 1960, p. 46):
\begin{displaymath}
\frac{v_f\!\!\raisebox{1ex}{$\circ$}}{-\,\mathrm{d}p/\,\mathrm{d}Y} = \frac{1-4(X^2+Z^2)}{16}.
\end{displaymath} (7.61)

A solution to equation (7.20) can be obtained in closed form by transforming the independent variables $(X,Z)$ to $(X,R)$, where

\begin{displaymath}
R = (X^2+Z^2)^{1/2}
\end{displaymath} (7.62)

whence it becomes apparent that (7.20) admits solutions of the form $v_n\!\!\raisebox{1ex}{$\circ$}=Xf(R)$ provided
\begin{displaymath}
f''(R)-\frac{3}{R}f'(R)+1 = 0.
\end{displaymath} (7.63)

For $f(1/2)=0$ and $f$ bounded as $R\rightarrow 0$, the solution of this is:
$\displaystyle f(R)$ $\textstyle =$ $\displaystyle \frac{1-4R^2}{32}$ (7.64)
$\displaystyle \mbox{whence}\qquad v_n\!\!\raisebox{1ex}{$\circ$}$ $\textstyle =$ $\displaystyle \frac{X[1-4(X^2+Z^2)]}{32}.$ (7.65)

This solution was reported by Ostroumov (1958) for the analogous single fluid heat transfer problem. It is also similar to the fully developed horizontal flow in a long axially heated cylinder (Klosse & Ullersma 1973; Bejan & Tien 1978).


next up previous contents
Next: Elliptic section Up: Sections other than rectangular Previous: Sections other than rectangular   Contents
Geordie McBain 2001-01-27