next up previous
Next: Exercises Up: Axisymmetric Incompressible Inviscid Flow Previous: Flow Around a Submerged

Flow Around a Submerged Prolate Spheroid

Consider the conformal map

$\displaystyle z+{\rm i}\,\varpi = c\,\cosh(\xi+{\rm i}\,\eta),$ (7.142)

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

$\displaystyle z$ $\displaystyle = c\,\cosh\xi\,\cos\eta,$ (7.143)
$\displaystyle \varpi$ $\displaystyle = c\,\sinh\xi\,\sin\eta.$ (7.144)

Let

$\displaystyle a$ $\displaystyle =c\,\cosh\xi_0,$ (7.145)
$\displaystyle b$ $\displaystyle =c\,\sinh\xi_0.$ (7.146)

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

$\displaystyle \left(\frac{\varpi}{b}\right)^2+\left(\frac{z}{a}\right)^2=1.$ (7.147)

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

$\displaystyle \psi(\xi\rightarrow\infty,\eta)$ $\displaystyle = 0,$ (7.148)
$\displaystyle \psi(\xi_0,\eta)$ $\displaystyle = -\frac{1}{2}\,V\,c^{\,2}\,\sinh^2\xi_0\,\sin^2\eta,$ (7.149)

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

$\displaystyle \frac{d}{d\xi}\left(\frac{1}{\sinh\xi}\,\frac{dF}{d\xi}\right) - \frac{2}{\sinh\xi}\,F=0.$ (7.150)

The solution that satisfied the constraint (7.148) is

$\displaystyle F(\xi)= -\frac{1}{2}\,B\left(\cosh\xi + \sinh^2\xi\,\ln\left[\tanh(\xi/2)\right]\right).$ (7.151)

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 $ \tanh(\xi_0/2)=[(1-e)/(1+e)]^{1/2}$ . The constraint (7.149) yields

$\displaystyle B = \frac{V\,a^{\,2}\,e^{\,2}}{e\,(1-e^{\,2})^{\,-1}+(1/2)\ln\left[(1-e)/(1+e)\right]}.$ (7.152)

Hence,

$\displaystyle \psi(\xi,\eta)= -\frac{1}{2}\,V\,b^{\,2}\left\{\frac{\cosh\xi + \...
...-1}+e^{\,-2}\,(1-e^{\,2})\,(1/2)\ln\left[(1-e)/(1+e)\right]}\right\}\sin^2\eta.$ (7.153)

Finally, from Equation (7.84),

$\displaystyle \frac{\partial\phi}{\partial\eta} = -\frac{1}{c\,\sinh\xi\,\sin\eta}\,\frac{\partial\psi}{\partial\xi},$ (7.154)

which can be integrated to give

$\displaystyle \phi(\xi,\eta) =-V\,a\left\{\frac{1 +\cosh\xi\,\ln\left[\tanh(\xi...
...ght]} {(1-e^{\,2})^{\,-1}+(1/2\,e)\ln\left[(1-e)/(1+e)\right]}\right\}\cos\eta.$ (7.155)

It is easily demonstrated that

$\displaystyle \psi(\xi_0,\eta)= -\frac{1}{2}\,V\,b^{\,2}\,\sin^2\eta,$ (7.156)

and

$\displaystyle \phi(\xi_0,\eta)= -V\,a\left\{\frac{1 +(1/2\,e)\ln\left[(1-e)/(1+...
...ght]} {(1-e^{\,2})^{\,-1}+(1/2\,e)\ln\left[(1-e)/(1+e)\right]}\right\}\cos\eta.$ (7.157)

Thus,

$\displaystyle K$ $\displaystyle = -\pi\,\rho\left.\int_{\eta=0}^{\eta=\pi} \phi\,d\psi\right\vert _{\xi=\xi_0}$    
  $\displaystyle = -\frac{2}{3}\,\pi\,\rho\,a\,b^{\,2}\,V^{\,2}\left\{\frac{1 +(1/...
...(1+e)\right]} {(1-e^{\,2})^{\,-1}+(1/2\,e)\ln\left[(1-e)/(1+e)\right]}\right\}.$ (7.158)

It follows that the added mass of the spheroid is

$\displaystyle m_{\rm added} = \frac{4}{3}\,\pi\,\rho\,a\,b^{\,2}\,G(e),$ (7.159)

where

$\displaystyle G(e)=-\left\{\frac{1 +(1/2\,e)\ln\left[(1-e)/(1+e)\right]} {(1-e^{\,2})^{\,-1}+(1/2\,e)\ln\left[(1-e)/(1+e)\right]}\right\}.$ (7.160)

is a monotonic function that takes the value $ 1/2$ when $ e=0$ , and asymptotes to $ (1-e)\,\ln[1/(1-e)]$ as $ e\rightarrow 1$ .


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