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Conformal Maps

As we saw in Section 6.7, conformal maps are extremely useful in the theory of two-dimensional, irrotational, incompressible flows. It turns out that such maps also have applications to the theory of axisymmetric, irrotational, incompressible flows. Consider the general coordinate transformation

$\displaystyle z+ {\rm i}\,\varpi = f(\xi+{\rm i}\,\eta),$ (7.79)

where $ f$ is an analytic function. The Cauchy-Riemann relations (see Section 6.3) yield

$\displaystyle \frac{\partial z}{\partial\xi}$ $\displaystyle =\frac{\partial\varpi}{\partial\eta},$ (7.80)
$\displaystyle \frac{\partial \varpi}{\partial\xi}$ $\displaystyle =-\frac{\partial z}{\partial\eta}.$ (7.81)

It follows, from the previous two expressions, that $ \nabla\xi\cdot\nabla\eta=0$ . In other words, $ \xi$ and $ \eta$ are orthogonal coordinates in the meridian plane. [Incidentally, we are assuming that $ (\xi,\,\eta,\,\varpi)$ are a right-handed set of coordinates.] Furthermore, it can also be shown from the Cauchy-Riemann relations that

$\displaystyle h_\xi=h_\eta= \left[\left(\frac{\partial z}{\partial\xi}\right)^2...
...}\right)^2+ \left(\frac{\partial \varpi}{\partial\eta}\right)^2\right]^{\,1/2},$ (7.82)

where $ h_\xi=\vert\nabla\xi\vert^{\,-1}$ and $ h_\eta=\vert\nabla\eta\vert^{\,-1}$ . Writing the flow velocity in terms of a velocity potential, so that $ {\bf v}=-\nabla\phi$ , or, alternatively, in terms of a Stokes stream function, so that $ {\bf v} = \nabla\varphi\times\nabla\psi$ , we get

$\displaystyle v_\xi$ $\displaystyle =-\frac{1}{h_\xi}\,\frac{\partial\phi}{\partial\xi} =-\frac{1}{\varpi\,h_\eta}\,\frac{\partial\psi}{\partial \eta},$ (7.83)
$\displaystyle v_\eta$ $\displaystyle = -\frac{1}{h_\eta}\,\frac{\partial\phi}{\partial\eta} =\frac{1}{\varpi\,h_\xi}\,\frac{\partial\psi}{\partial \xi}.$ (7.84)

Of course, writing the velocity field in terms of a Stokes stream function ensures that the field is incompressible, which also implies that $ \nabla^{\,2}\phi=0$ . The additional requirement that the field be irrotational yields $ \omega_\varphi=0$ . Making use of the analysis of Appendix C, this requirement reduces to

$\displaystyle \frac{\partial(h_\eta\,v_\eta)}{\partial\xi} -\frac{\partial(h_\xi\,v_\xi)}{\partial \eta}=0,$ (7.85)


$\displaystyle \frac{\partial}{\partial\xi}\left(\frac{1}{\varpi}\,\frac{\partia...
...partial\eta}\left(\frac{1}{\varpi}\,\frac{\partial\psi}{\partial\eta}\right)=0.$ (7.86)

Figure 7.5: An axisymmetric solid body moving through an incompressible irrotational fluid.
\epsfysize =2.5in

Let $ \xi=\xi_0$ represent the surface of an axisymmetric solid body moving with velocity $ {\bf V} =V\,{\bf e}_z$ through an incompressible irrotational fluid that is at rest a long way from the body. Let the fluid occupy the region $ \xi_0<\xi<\infty$ , where $ \xi\rightarrow\infty$ far from the body. (See Figure 7.5.) Let $ \eta$ be an angular coordinate such that $ \eta=0$ on the positive $ z$ -axis, and $ \eta=\pi$ on the negative $ z$ -axis. The fact that the fluid is at rest at infinity implies that $ \psi $ asymptotes to a constant a long way from the body. Without loss of generality, we can chose this constant to be zero. Thus, one constraint on the system is that

$\displaystyle \psi(\xi\rightarrow\infty,\eta)= 0.$ (7.87)

The appropriate constraint at the surface of the body is that

$\displaystyle v_\xi = {\bf V}\cdot{\bf e}_\xi,$ (7.88)

where $ {\bf e}_\xi=\nabla\xi/\vert\nabla\xi\vert$ . However, we can write $ {\bf V} = \nabla\varphi\times\nabla\psi_0$ , where $ \psi_0(\varpi,z)= -(1/2)\,V\,\varpi^{\,2}$ . (See Section 7.6.) Hence, from Equation (7.83), the previous constraint becomes

$\displaystyle \frac{\partial\psi}{\partial\eta} = \frac{\partial\psi_0}{\partial\eta}$ (7.89)

when $ \xi=\xi_0$ . Integrating, making use of the constraint (7.87) (which implies that $ \psi=0$ on the $ z$ -axis, where $ \psi $ is constant, by symmetry), we obtain

$\displaystyle \psi(\xi_0,\eta)= -\frac{1}{2}\,V\,\varpi^{\,2}.$ (7.90)

We can also set the velocity potential, $ \phi$ , to zero at infinity, and on the $ z$ -axis.

The total kinetic energy of the fluid surrounding the moving body is

$\displaystyle K = \frac{1}{2}\,\rho\int v^{\,2}\,dV = \frac{1}{2}\,\rho\int(\nabla\phi)^2\,dV = \frac{1}{2}\,\rho\int\nabla\cdot(\phi\,\nabla\phi)\,dV,$ (7.91)

where we have made use of the fact that $ \nabla^{\,2}\phi=0$ . Here, $ \rho$ is the fluid mass density, and $ d V$ is an element of the volume obtained by rotating the area $ ABCDEFA$ , shown in Figure 7.5, about the $ z$ -axis. Making use of the divergence theorem, we obtain

$\displaystyle K=\frac{1}{2}\,\rho\oint \phi\,\nabla\phi\cdot{\bf n}\,2\pi\,\var...
...{ABC}\phi\,\frac{1}{h_\xi}\,\frac{\partial\phi}{\partial\xi}\,2\pi\,\varpi\,ds.$ (7.92)

where $ ds$ is an element of the curve $ ABCDEFA$ , and $ {\bf n}$ is an outward pointing, unit, normal vector to the area $ ABCDEFA$ . Here, we have made use of the fact that the velocity potential is zero at infinity (i.e., on $ DEF$ ), and also on the $ z$ -axis (i.e., on $ CD$ and $ FA$ ). On the curve $ ABC$ , we can write $ ds= h_\eta\,d\eta$ . Furthermore, it follows from Equations (7.82) and (7.83) that $ h_\xi=h_\eta$ , and $ \partial\phi/\partial\xi = \varpi^{\,-1}\,\partial\psi/\partial\eta$ . Thus,

$\displaystyle K= -\pi\,\rho\left.\int_{0}^\pi \phi\,\frac{\partial\psi}{\partial\eta}\,d\eta\right\vert _{\xi=\xi_0},$ (7.93)


$\displaystyle K = -\pi\,\rho\left.\int_{\eta=0}^{\eta=\pi} \phi\,d\psi\right\vert _{\xi=\xi_0}.$ (7.94)

As a simple example, consider the conformal map

$\displaystyle z+{\rm i}\,\varpi = c\,\exp\,(\xi+{\rm i}\,\eta),$ (7.95)

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

$\displaystyle z$ $\displaystyle = c\,{\rm e}^{\,\xi}\,\cos\eta,$ (7.96)
$\displaystyle \varpi$ $\displaystyle = c\,{\rm e}^{\,\xi}\,\sin\eta,$ (7.97)

which implies that

$\displaystyle z^{\,2}+\varpi^{\,2} = r^{\,2},$ (7.98)


$\displaystyle r(\xi)=c\,{\rm e}^{\,\xi}.$ (7.99)

Thus, the constant-$ \xi$ surfaces are concentric spheres of radius $ r(\xi)$ . If we set

$\displaystyle a = c\,{\rm e}^{\,\xi_0}$ (7.100)

then the problem reduces to that of a sphere, of radius $ a$ , moving through a fluid that is at rest at infinity. This problem was solved, via different methods, in Section 7.10. The constraints (7.87) and (7.90) yield

$\displaystyle \psi(\xi\rightarrow\infty,\eta)$ $\displaystyle = 0,$ (7.101)
$\displaystyle \psi(\xi_0,\eta)$ $\displaystyle = -\frac{1}{2}\,V\,c^2\,{\rm e}^{\,2\,\xi_0}\,\sin^2\eta,$ (7.102)

where use has been made of Equation (7.97). This suggests that we can write

$\displaystyle \psi(\xi,\eta)=F(\xi)\,\sin^2\eta.$ (7.103)

Substitution into the governing equation, (7.86), gives

$\displaystyle \frac{d}{d\xi}\left({\rm e}^{-\xi}\,\frac{dF}{d\xi}\right) -2\,{\rm e}^{-\xi}\,F = 0,$ (7.104)

whose most general solution is

$\displaystyle F(\xi)=A\,{\rm e}^{\,2\,\xi} + B\,{\rm e}^{-\xi}.$ (7.105)

The constraints (7.101) and (7.102) yield

$\displaystyle A$ $\displaystyle = 0,$ (7.106)
$\displaystyle B$ $\displaystyle =-\frac{1}{2}\,V\,c^{\,2}\,{\rm e}^{\,3\,\xi_0},$ (7.107)

respectively. Thus, we obtain

$\displaystyle \psi(\xi,\eta) = -\frac{1}{2}\,V\,a^{\,3}\,\frac{\sin^2\eta}{c\,{\rm e}^{\,\xi}}.$ (7.108)

Now, from Equations (7.84) and (7.97),

$\displaystyle \frac{\partial\phi}{\partial\eta} = -\frac{1}{\varpi}\,\frac{\par...
...artial\xi} = -\frac{1}{2}\,V\,a^{\,3}\,\frac{\sin\eta}{(c\,{\rm e}^{\,\xi})^2},$ (7.109)

which can be integrated to give

$\displaystyle \phi(\xi,\eta)= \frac{1}{2}\,V\,a^{\,3}\,\frac{\cos\eta}{(c\,{\rm e}^{\,\xi})^2}.$ (7.110)

Note that the previous expression is formally the same as expression (7.63), as long as we make the identifications $ V\rightarrow V_z(t)$ , $ c\,{\rm e}^{\,\xi}\rightarrow r$ , and $ \eta\rightarrow\theta$ .

On the surface of the sphere, $ \xi=\xi_0$ , we obtain

$\displaystyle \psi(\xi_0,\eta)$ $\displaystyle = -\frac{1}{2}\,V\,a^{\,2}\,\sin^2\eta,$ (7.111)
$\displaystyle \phi(\xi_0,\eta)$ $\displaystyle = \frac{1}{2}\,V\,a\,\cos\eta.$ (7.112)


$\displaystyle K= -\pi\,\rho\left.\int_{\eta=0}^{\eta=\pi} \phi\,d\psi\right\ver...
...rho\,V^{\,2}\int_{-1}^1 \mu^{\,2}\,d\mu= \frac{\pi}{3}\,a^{\,3}\,\rho\,V^{\,2}.$ (7.113)

As is clear from the analysis of Section 7.10, the sphere's added mass can be written

$\displaystyle m_{\rm added} = \frac{K}{(1/2)\,V^{\,2}} = \frac{2\pi}{3}\,\,a^{\,3}\,\rho.$ (7.114)

Hence, we arrive at the standard result that the added mass is half the displaced mass [i.e., half of $ (4\pi/3)\,a^{\,3}\,\rho$ ].

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