Axisymmetric Stokes Flow Around a Solid Sphere

Consider a solid sphere of radius $a$ that is moving under gravity at the constant vertical velocity $V\,{\bf e}_z$ through a stationary fluid of density $\rho$ and viscosity $\mu$ . Here, gravitational acceleration is assumed to take the form ${\bf g} = -g\,{\bf e}_z$ . Provided that the typical Reynolds number,

$${\rm Re} = \frac{2\,\rho\,V\,a}{\mu},$$ (10.94)

is much less than unity, the flow around the sphere is an example of axisymmetric Stokes flow. Let us transform to a frame of reference in which the sphere is stationary, and centered at the origin. Adopting the standard spherical coordinates $r$ , $\theta$ , $\varphi$ , the surface of the sphere corresponds to $r=a$ , and the surrounding fluid occupies the region $r>a$ . By symmetry, the flow field outside the sphere is axisymmetric (i.e., $\partial/\partial \varphi = 0$ ), and has no toroidal component (i.e., $v_\varphi=0$ ). The physical boundary conditions at the surface of the sphere are

$$v_r(a,\theta) = 0,$$ (10.95)

$$v_\theta(a,\theta) =0:$$ (10.96)

that is, the normal and tangential fluid velocities are both zero at the surface. A long way from the sphere, we expect the fluid velocity to asymptote to ${\bf v}=-V\,{\bf e}_z$ . In other words,

$$v_r(r\rightarrow \infty,\theta) \rightarrow -V\,\cos\theta,$$ (10.97)

$$v_\theta(r\rightarrow\infty,\theta) \rightarrow V\,\sin\theta.$$ (10.98)

Let us write

$${\bf v} =\nabla\varphi\times\nabla\psi,$$ (10.99)

where $\psi(r,\theta)$ is the Stokes stream function. As we saw in the previous section, axisymmetric Stokes flow is characterized by

$${\cal L}^{\,2}(\psi)=0.$$ (10.100)

Here, the differential operator ${\cal L}$ is specified in Equation (10.88). The boundary conditions (10.95)-(10.98) reduce to

$$\left.\frac{\partial\psi}{\partial r}\right\vert _{r=a} = 0,$$ (10.101)

$$\left.\frac{\partial\psi}{\partial\theta}\right\vert _{r=a} = 0,$$ (10.102)

$$\psi(r\rightarrow\infty,\theta) \rightarrow \frac{1}{2}\,V\,r^{\,2}\,\sin^2\theta.$$ (10.103)

Equation (10.103) suggests that $\psi(r,\theta)$ can be written in the separable form

$$\psi(r,\theta) = \sin^2\theta\,f(r).$$ (10.104)

In this case,

$$v_r(r,\theta) = -\frac{2\,\cos\theta\,f(r)}{r^{\,2}},$$ (10.105)

$$v_\theta(r,\theta) = \frac{\sin\theta}{r}\,\frac{df}{dr},$$ (10.106)

and Equations (10.100)-(10.103) reduce to

$$\left(\frac{d^{\,2}}{dr^{\,2}} - \frac{2}{r^{\,2}}\right)^2 f = 0,$$ (10.107)

$$f(a)=\left.\frac{df}{dr}\right\vert _{r=a} = 0,$$ (10.108)

$$f(r\rightarrow\infty) \rightarrow \frac{1}{2}\,V\,r^{\,2}.$$ (10.109)

Figure: Contours of the stream function $\psi$ in the $x$ -$z$ plane for Stokes flow around a solid sphere.

Let us try a test solution to Equation (10.107) of the form $f(r)=\alpha\,r^{\,n}$ . We find that

$$\alpha\left[n\,(n-1)-2\right]\,[(n-2)\,(n-3)-2] = 0,$$ (10.110)

which implies that $n=-1$ , $1$ , $2$ , $4$ . Hence, the most general solution to Equation (10.107) is

$$f(r) = \frac{A}{r} + B\,r + C\,r^{\,2} + D\,r^{\,4},$$ (10.111)

where $A$ , $B$ , $C$ , $D$ are arbitrary constants. However, the boundary condition (10.109) yields $C= (1/2)\,V$ and $D=0$ , whereas the boundary condition (10.108) gives $A=(1/4)\,V\,a^{\,3}$ and $B=-(3/4)\,V\,a$ . Thus, we conclude that

$$f(r) = \frac{V\,(r-a)^2\,(2\,r+a)}{4\,r},$$ (10.112)

and the stream function becomes

$$\psi(r,\theta) = \sin^2\theta\,\frac{V\,(r-a)^2\,(2\,r+a)}{4\,r}.$$ (10.113)

(See Figure 10.4.) From Equation (10.87), the fluid vorticity is

$$\omega_\varphi(r,\theta) = \frac{{\cal L}(\psi)}{r\,\sin\theta} =... ...{dr^{\,2}}-\frac{2}{r^{\,2}}\right) f = \frac{3\,V\,a\,\sin\theta}{2\,r^{\,2}}.$$ (10.114)

(See Figure 10.5.) Moreover, from Equation (10.81),

$$\nabla P = -\mu\,\nabla\times \mbox{\boldmath$\omega$} = \mu\,\nabla\phi\times \nabla(\omega_\phi\,r\,\sin\theta).$$ (10.115)

Hence,

$$\frac{\partial P}{\partial r} = -\frac{3\,\mu\,V\,a\,\cos\theta}{r^{\,3}},$$ (10.116)

$$\frac{\partial P}{\partial \theta} = -\frac{3\,\mu\,V\,a\,\sin\theta}{2\,r^{\,2}},$$ (10.117)

which implies that the effective pressure distribution within the fluid is

$$P(r,\theta) = p_0 + \frac{3\,\mu\,V\,a\,\cos\theta}{2\,r^{\,2}},$$ (10.118)

where $p_0$ is an arbitrary constant. (See Figure 10.6.) However, $P = p + \rho\,{\mit\Psi}$ , where ${\mit\Psi} = g\,z=g\,r\,\cos\theta$ . Thus, the actual pressure distribution is

$$p(r,\theta) = p_0- \rho\,g\,r\,\cos\theta + \frac{3\,\mu\,V\,a}{2\,r^{\,2}}\,\cos\theta.$$ (10.119)

Figure: Contours of the vorticity, $\omega _\phi$ , in the $x$ -$z$ plane for Stokes flow around a solid sphere. Solid/dashed lines correspond to opposite signs of $\omega _\varphi$ .

Figure: Contours of the effective pressure, $P-p_0$ , in the $x$ -$z$ plane for Stokes flow around a solid sphere. Solid/dashed lines correspond to opposite signs of $P-p_0$ .

From Section 1.20, the radial and tangential components of the force per unit area exerted on the sphere by the fluid are

$$f_r(\theta) = \sigma_{rr}(a,\theta) = \left(-p + 2\,\mu\,\frac{\partial v_r}{\partial r}\right)_{r=a},$$ (10.120)

$$f_\theta(\theta) =\sigma_{r\theta}(a,\theta) = \mu\left(\frac{1}{r}\,\frac{\partia... ... \theta}+\frac{\partial v_\theta}{\partial r} -\frac{v_\theta}{r}\right)_{r=a}.$$ (10.121)

Now, $v_r(a,\theta)=v_\theta(a,\theta)=0$ . Moreover, because $\nabla\cdot{\bf v}=0$ , it follows from Equation (1.168) that $(\partial v_r/\partial r)_{r=a}=0$ . Finally, Equation (10.87) yields $(\partial v_\theta/\partial r)_{r=a}=\omega_\phi(a,\theta)$ . Hence,

$$f_r(\theta) = -p(a,\theta) = -p_0+ \rho\,g\,a\,\cos\theta-\frac{3\,\mu\,V}{2\,a}\,\cos\theta,$$ (10.122)

$$f_\theta(\theta) = \mu\,\omega_\varphi(a,\theta) = \frac{3\,\mu\,V}{2\,a}\,\sin\theta.$$ (10.123)

Thus, the force density at the surface of the sphere is

$${\bf f}(\theta) = -\frac{3\,\mu\,V}{2\,a}\,{\bf e}_z + \left(-p_0+\rho\,g\,a\,\cos\theta\right){\bf e}_r.$$ (10.124)

It follows that the net vertical force exerted on the sphere by the fluid is

$$F_z = \oint_S {\bf f}\cdot{\bf e}_z\,dS$$

$$= -\frac{3\,\mu\,V}{2\,a}\,4\pi\,a^{\,2} + 2\pi\,a^{\,2}\int_0^\pi\left(-p_0+\rho\,g\,a\,\cos\theta\right)\cos\theta\,\sin\theta\,d\theta,$$ (10.125)

which reduces to

$$F_z = -6\pi\,a\,\mu\,V + \frac{4\pi}{3}\,a^{\,3}\,\rho\,g.$$ (10.126)

By symmetry, the horizontal components of the net force both average to zero. We can recognize the second term on the right-hand side of the previous equation as the buoyancy force due to the weight of the fluid displaced by the sphere. (See Chapter 2.) Moreover, the first term can be interpreted as the viscous drag acting on the sphere. Note that this drag acts in the opposite direction to the relative motion of the sphere with respect to the fluid, and its magnitude is directly proportional to the relative velocity.

Vertical force balance requires that

$$F_z = M\,g,$$ (10.127)

where $M$ is the sphere's mass. In other words, in a steady state, the weight of the sphere balances the vertical force exerted by the surrounding fluid. If the sphere is composed of material of mean density $\overline{\rho}$ then $M=(4\pi/3)\,a^{\,3}\,\overline{\rho}$ . Hence, in the frame in which the fluid a large distance from the sphere is stationary, the steady vertical velocity with which the sphere moves through the fluid is

$$V = \frac{2}{9}\,\frac{a^{\,2}\,g}{\nu}\left(1-\frac{\overline{\rho}}{\rho}\right),$$ (10.128)

where $\nu=\mu/\rho$ is the fluid's kinematic viscosity. Obviously, if the sphere is more dense than the fluid (i.e., if $\overline{\rho}/\rho>1$ ) then it moves downward (i.e., $V<0$ ), and vice versa. Finally, the typical Reynolds number of the fluid flow in the vicinity of the sphere is

$${\rm Re} = \frac{2\,\rho\,V\,a}{\mu} = \frac{4}{9}\,\frac{a^{\,3}\,g}{\nu^{\,2}}\left\vert 1-\frac{\overline{\rho}}{\rho}\right\vert.$$ (10.129)

For the case of a grain of sand falling through water at $20^\circ\,{\rm C}$ , we have $\overline{\rho}/\rho\simeq 2$ and $\nu=1.0\times 10^{-6}\,{\rm m^2/s}$ (Batchelor 2000). Hence, ${\rm Re} = (a/6\times 10^{-5})^{\,3}$ , where $a$ is measured in meters. Thus, expression (10.128), which is strictly speaking only valid when ${\rm Re}\ll 1$ , but which turns out to be approximately valid for all Reynolds numbers less than unity, only holds for sand grains whose radii are less than about $60$ microns. Such grains fall through water at approximately $8\times 10^{-3}\,{\rm m/s}$ . For the case of a droplet of water falling through air at $20^\circ\,{\rm C}$ and atmospheric pressure, we have $\overline{\rho}/\rho=780$ and $\nu=1.5\times 10^{-5}\,{\rm m^2/s}$ (Batchelor 2000). Hence, ${\rm Re} = (a/4\times 10^{-5})^{\,3}$ , where $a$ is measured in meters. Thus, expression (10.128) only holds for water droplets whose radii are less than about $40$ microns. Such droplets fall through air at approximately $0.2\,{\rm m/s}$ .

At large values of $r/a$ , Equations (10.105), (10.106), and (10.112) yield

$$v_r(r,\theta) = -V\,\cos\theta + \frac{3}{2}\,V\,\cos\,\theta\,\frac{a}{r} + {\cal O}\left(\frac{a}{r}\right)^2,$$ (10.130)

$$v_\theta(r,\theta) = V\,\sin\theta - \frac{3}{4}\,V\,\sin\theta\,\frac{a}{r}+{\cal O}\left(\frac{a}{r}\right)^2.$$ (10.131)

It follows that

$$[\rho\,({\bf v}\cdot\nabla)\,{\bf v}]_r = \rho\left(v_r\,\frac{\p... ...\theta}- \frac{v_\theta^{\,2}}{r}\right) \sim \frac{\rho\,V^{\,2}\,a}{r^{\,2}},$$ (10.132)

and

$$(\mu\,\nabla^{\,2}{\bf v})_r \sim \mu\,\frac{\partial^{\,2} v_r}{\partial r^{\,2}}\sim \frac{\mu\,V\,a}{r^{\,3}}.$$ (10.133)

Hence,

$$\frac{[\rho\,({\bf v}\cdot\nabla)\,{\bf v}]_r }{(\mu\,\nabla^2{\bf v})_r} \sim \frac{\rho\,V\,r}{\mu}\sim {\rm Re}\left(\frac{r}{a}\right),$$ (10.134)

where ${\rm Re}$ is the Reynolds number of the flow in the immediate vicinity of the sphere. [See Equation (10.94).] Our analysis is based on the assumption that advective inertia is negligible with respect to viscosity. However, as is clear from the previous expression for the ratio of inertia to viscosity within the fluid, even if this ratio is much less than unity close to the sphere--in other words, if ${\rm Re}\ll 1$ --it inevitably becomes much greater than unity far from the sphere: that is, for $r\gg a/{\rm Re}$ . In other words, inertia always dominates viscosity, and our Stokes flow solution therefore breaks down, at sufficiently large $r/a$ .