Flow Around a Submerged Prolate Spheroid

Consider the conformal map

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

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

$$z = c\,\cosh\xi\,\cos\eta,$$ (7.143)

$$\varpi = c\,\sinh\xi\,\sin\eta.$$ (7.144)

Let

$$a =c\,\cosh\xi_0,$$ (7.145)

$$b =c\,\sinh\xi_0.$$ (7.146)

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

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

$$\psi(\xi\rightarrow\infty,\eta) = 0,$$ (7.148)

$$\psi(\xi_0,\eta) = -\frac{1}{2}\,V\,c^{\,2}\,\sinh^2\xi_0\,\sin^2\eta,$$ (7.149)

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

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

The solution that satisfied the constraint (7.148) is

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

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

Hence,

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

Finally, from Equation (7.84),

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

which can be integrated to give

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

It is easily demonstrated that

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

and

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

Thus,

$$K = -\pi\,\rho\left.\int_{\eta=0}^{\eta=\pi} \phi\,d\psi\right\vert _{\xi=\xi_0}$$

$$= -\frac{2}{3}\,\pi\,\rho\,a\,b^{\,2}\,V^{\,2}\left\{\frac{1 +(1/... ...(1+e)\right]} {(1-e^{\,2})^{\,-1}+(1/2\,e)\ln\left[(1-e)/(1+e)\right]}\right\}.$$ (7.158)

It follows that the added mass of the spheroid is

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

where

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