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Next: Flow Around a Submerged Up: Axisymmetric Incompressible Inviscid Flow Previous: Conformal Maps

Flow Around a Submerged Oblate Spheroid

Consider the conformal map

$\displaystyle z+{\rm i}\,\varpi = c\,\sinh(\xi+{\rm i}\,\eta),$ (7.115)

where $ c$ is real and positive. It follows that

$\displaystyle z$ $\displaystyle = c\,\sinh\xi\,\cos\eta,$ (7.116)
$\displaystyle \varpi$ $\displaystyle = c\,\cosh\xi\,\sin\eta.$ (7.117)

Let

$\displaystyle a$ $\displaystyle =c\,\cosh\xi_0,$ (7.118)
$\displaystyle b$ $\displaystyle =c\,\sinh\xi_0.$ (7.119)

Thus, in the meridian plane, the curve $ \xi=\xi_0$ corresponds to the ellipse

$\displaystyle \left(\frac{\varpi}{a}\right)^2+\left(\frac{z}{b}\right)^2=1.$ (7.120)

We conclude that the surface $ \xi=\xi_0$ is an oblate spheroid (i.e., the three-dimensional surface obtained by rotating an ellipse about a minor axis) of major radius $ a$ and minor radius $ b$ . The constraints (7.87) and (7.90) yield

$\displaystyle \psi(\xi\rightarrow\infty,\eta)$ $\displaystyle = 0,$ (7.121)
$\displaystyle \psi(\xi_0,\eta)$ $\displaystyle = -\frac{1}{2}\,V\,c^{\,2}\,\cosh^2\xi_0\,\sin^2\eta,$ (7.122)

respectively. Setting $ \psi(\xi,\eta)= F(\xi)\,\sin^2\eta$ , and substituting into the governing equation, (7.86), we obtain

$\displaystyle \frac{d}{d\xi}\left(\frac{1}{\cosh\xi}\,\frac{dF}{d\xi}\right) - \frac{2}{\cosh\xi}\,F=0,$ (7.123)

which can be rearranged to give

$\displaystyle \frac{d}{d\xi}\left(\cosh\xi\,\frac{dF}{d\xi}-2\,\sinh\xi\,F\right)=0.$ (7.124)

On integration, we get

$\displaystyle \cosh\xi\,\frac{dF}{d\xi}-2\,\sinh\xi\,F = B,$ (7.125)

which can be rearranged to give

$\displaystyle \frac{d}{d\xi}\left(\frac{F}{\cosh^2\xi}\right)= \frac{B}{\cosh^3\xi}.$ (7.126)

It follows that

$\displaystyle F(\xi)$ $\displaystyle =-B\,\cosh^2\xi \int_{\xi}^\infty \frac{d\xi'}{\cosh^3\xi'}$    
  $\displaystyle =-\frac{1}{2}\, B\,\cosh^2\xi\left[ \frac{\sinh\xi'}{\cosh^2\xi'}-\tan^{-1}\left(\frac{1}{\sinh\xi'}\right)\right]^{\infty}_{\xi}$    
  $\displaystyle = \frac{1}{2}\,B\left[\sinh\xi -\cosh^2\xi\,\tan^{-1}\left(\frac{1}{\sinh\xi}\right)\right],$ (7.127)

where use has been made of the constraint (7.121). Let $ e=(1-b^{\,2}/a^{\,2})^{1/2}$ be the eccentricity of the spheroid. Thus, $ \cosh\xi_0=1/e$ , $ \sinh\xi_0=(1-e^{\,2})^{1/2}/e$ , $ c=e\,a$ , and $ \tan^{-1}(1/\sinh\xi_0)=\sin^{-1}e$ . The constraint (7.122) yields

$\displaystyle -\frac{1}{2}\,V\,a^{\,2} = F(\xi_0)= \frac{1}{2}\,B\left[\frac{(1-e^{\,2})^{1/2}}{e}-\frac{\sin^{-1}e}{e^{\,2}}\right],$ (7.128)

or

$\displaystyle B = - \left[\frac{V\,a^{\,2}\,e^{\,2}}{e\,(1-e^{\,2})^{1/2}-\sin^{-1}e}\right].$ (7.129)

Hence,

$\displaystyle \psi(\xi,\eta) =-\frac{1}{2}\,V\,a^{\,2}\,e^{\,2}\left[\frac{\sin...
...\xi\,\tan^{-1}(1/\sinh\xi)} {e\,(1-e^{\,2})^{1/2}-\sin^{-1}e}\right]\sin^2\eta.$ (7.130)

Finally, from Equation (7.84),

$\displaystyle \frac{\partial\phi}{\partial\eta} = -\frac{1}{c\,\cosh\xi\,\sin\eta}\,\frac{\partial\psi}{\partial\xi},$ (7.131)

which can be integrated to give

$\displaystyle \phi(\xi,\eta) =-V\,a\,e\left[\frac{1 -\sinh\xi\,\tan^{-1}(1/\sinh\xi)} {e\,(1-e^{\,2})^{1/2}-\sin^{-1}e}\right]\cos\eta.$ (7.132)

It is easily demonstrated that

$\displaystyle \psi(\xi_0,\eta)= -\frac{1}{2}\,V\,a^{\,2}\,\sin^2\eta,$ (7.133)

and

$\displaystyle \phi(\xi_0,\eta) = V\,a\left[\frac{e-(1-e^{\,2})^{1/2}\,\sin^{-1} e}{\sin^{-1} e-e\,(1-e^{\,2})^{1/2}}\right]\cos\eta.$ (7.134)

Thus,

$\displaystyle K = -\pi\,\rho\left.\int_{\eta=0}^{\eta=\pi} \phi\,d\psi\right\ve...
...rac{e-(1-e^{\,2})^{1/2}\,\sin^{-1} e}{\sin^{-1} e-e\,(1-e^{\,2})^{1/2}}\right].$ (7.135)

It follows that the added mass of the spheroid is

$\displaystyle m_{\rm added} = \frac{4}{3}\,\pi\,\rho\,a^{\,3}\,G(e),$ (7.136)

where

$\displaystyle G(e)= \frac{e-(1-e^{\,2})^{1/2}\,\sin^{-1} e}{\sin^{-1} e-e\,(1-e^{\,2})^{1/2}}$ (7.137)

is a monotonic function that varies between $ 1/2$ when $ e=0$ and $ 2/\pi$ when $ e=1$ .

Figure: Contours of the Stokes stream function for the case of a disk of radius $ a$ , lying in the $ x$ -$ y$ plane, and placed in a uniform, incompressible, irrotational flow directed parallel to the $ z$ -axis.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter07/disk.eps}}
\end{figure}

In the limit $ e\rightarrow 1$ , our spheroid asymptotes to a thin disk of radius $ a$ moving through the fluid in the direction perpendicular to its plane. Expressions (7.130), (7.132), and (7.136) yield

$\displaystyle \psi(\xi,\eta)$ $\displaystyle = -\frac{1}{2}\,V\,a^{\,2}\,\sin^2\eta,$ (7.138)
$\displaystyle \phi(\xi,\eta)$ $\displaystyle = \frac{2}{\pi}\,V\,a\,\cos\eta,$ (7.139)

and

$\displaystyle m_{\rm added} = \frac{8}{3}\,\rho\,a^{\,3},$ (7.140)

respectively. Here, $ z=\sinh\xi\,\cos\eta$ and $ \varpi=\cosh\xi\,\sin\eta$ . It follows that, in the instantaneous rest frame of the disk,

$\displaystyle \psi(\varpi,z)=- \frac{1}{4}\,V\left[a^{\,2}+z^{\,2}+\varpi^{\,2}...
...rpi^{\,2})^2-4\,a^{\,2}\,\varpi^{\,2}}\,\right] + \frac{1}{2}\,V\,\varpi^{\,2}.$ (7.141)

This flow pattern, which corresponds to that of a thin disk placed in a uniform flow perpendicular to its plane, is visualized in Figure 7.6.


next up previous
Next: Flow Around a Submerged Up: Axisymmetric Incompressible Inviscid Flow Previous: Conformal Maps
Richard Fitzpatrick 2016-03-31