# Creeping flow around a bubble

Posted in math on Friday, June 27 2014

When fluid flows around a gas bubble, circulation within the bubble dissipates energy away from the interface and the interfacial shear stress is reduced. In this problem (4B.3 Transport Phenomena) I look at what happens when that shear stress is negligible.

Flow is coming up along the positive z-axis, and is creeping, incompressible, steady-state, Newtonian flow. Furthermore I am going to assume it is axisymmetric about z (i.e. it is independent of φ in spherical coordinates).

The bubble has radius R and is centered in the coordinate system.

By assuming the flow is axisymmetric I can use a stream function ψ such that (in spherical coordinates) (see Table 4.2-1 Transport Phenomena (2nd Ed)): $$v_{\theta} = \frac{1}{r \sin \theta} \frac{\partial \psi}{\partial r}$$

$$v_{r} = \frac{-1}{r^2 \sin \theta} \frac{\partial \psi}{\partial \theta}$$

This pair of velocities satisfies the continuity equation with symmetry about the z axis.

To eliminate the pressure field and other such considerations, I will seek to solve the equation of change for the vorticity: $$\frac{D}{Dt} \left[ \nabla \times \mathbf{v} \right] = \nu \nabla^2 \left[\nabla \times \mathbf{v} \right]$$

Creeping, incompressible, steady-state, flow eliminates the L.H.S, leaving: $$\nu \nabla^2 \left[\nabla \times \mathbf{v} \right] = 0$$

In terms of the stream function: $$\nu \left( \frac{\partial^2}{\partial r^2} + \frac{\sin \theta}{r^2} \frac{\partial}{\partial \theta} \left( \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \right) \right) \left( \frac{\partial^2}{\partial r^2} + \frac{\sin \theta}{r^2} \frac{\partial}{\partial \theta} \left( \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \right) \right) \psi = 0$$

This is the equation of change.

### Boundary Condition 1

The first boundary condition is that far away from the bubble the flow should be undisturbed, that is: $$\lim \limits_{r \to \infty} v_z = v_{\infty}$$

Since: $$v_z = v_r \cos \theta - v_\theta \sin \theta$$

This implies that in the limit $r \to \infty$: $$v_{\theta} = - v_{\infty} \sin \theta$$

$$v_r = v_{\infty} \cos \theta$$

Plugging in to the equations defining the stream function, first for the vθ: $$v_{\theta} = - v_{\infty} \sin \theta = \frac{1}{r \sin \theta} \frac{\partial \psi}{\partial r}$$

$$\frac{\partial \psi}{\partial r} = - r v_{\infty} \sin^2 \theta$$

$$\psi(r,\theta) = - \frac{1}{2} v_{\infty} r^2 \sin^2 \theta + fun(\theta)$$

then for the vr: $$v_r = v_{\infty} \cos \theta = \frac{-1}{r^2 \sin \theta} \frac{\partial \psi}{\partial \theta}$$

$$\frac{\partial \psi}{\partial \theta} = - v_{\infty} r^2 \sin \theta \cos \theta$$

$$\psi(r,\theta) = - \frac{1}{2} v_{\infty} r^2 \sin^2 \theta + fun(r)$$

So obviously, $$\lim \limits_{r \to \infty} \psi = - \frac{1}{2} v_{\infty} r^2 \sin^2 \theta$$

That is boundary condition 1.

### Boundary Condition 2

The second boundary condition is that special property of bubbles, that the interfacial shear stress is negligible.

$$\mbox{at r = R, }\tau_w = \tau_{r \theta} = 0$$

$$\tau_{r \theta} = -\mu \left[ r \frac{\partial}{\partial r} \left( \frac{v_\theta}{r} \right) + \frac{1}{r} \frac{\partial v_r}{\partial r} \right] = 0$$

Substituting in for vr and vθ:

$$\frac{-2}{r^2 \sin \theta} \frac{\partial \psi}{\partial r} + \frac{1}{r \sin \theta} \frac{\partial^2 \psi}{\partial r^2} + \frac{\cos \theta}{r^3 \sin^2 \theta} \frac{\partial \psi}{\partial \theta} - \frac{1}{r^3 \sin \theta} \frac{\partial^2 \psi}{\partial \theta^2} = 0$$

Now, taking a glance back at the first boundary condition, we might suppose that: $$\psi (r,\theta) = f(r) \sin^2 \theta$$

Plugging that into the above expression yields, at r=R: $$\frac{1}{r} \frac{d^2}{d r^2} f - \frac{2}{r^2} \frac{d}{d r} f + \frac{2}{r^3} f = 0$$

That is boundary condition 2.

### Boundary Condition 3

The third boundary condition is simply a matter of inspection. Since there is no shear at the interface, momentum cannot exit at the wall, which then implies (by conservation of momentum) $\mbox{at r=R, } v_r = 0$

### Solving the Equation of Change

In the discussion of the second boundary condition a separation of variables was proposed for the stream function $\psi = f(r) \sin^2 \theta$

Substituting that into the equation of change leads, rather tediously, to...

$$\frac{d^4 f}{dr^4} - \frac{4}{r^2} \frac{d^2 f}{dr^2} + \frac{8}{r^3} \frac{d f}{dr} - \frac{8}{r^4} f = 0$$

This is a Cauchy-Euler equation and it has solutions of the form: $$f(r) = C_4 r^4 + C_2 r^2 + C_1 r + \frac{C_0}{r}$$

Using BC1 we see that C4 = 0 and $$C_2 = - \frac{1}{2} v_{\infty}$$

thus

$$f(r) = - \frac{1}{2} v_{\infty} r^2 + C_1 r + \frac{C_0}{r}$$

Using BC2 we find $$C_0 = 0$$

Using BC3 we find $$C_1 = \frac{v_{\infty} R}{2}$$

Thus $$f(r) = - \frac{1}{2} v_{\infty} r^2 + \frac{1}{2} v_{\infty} R r = - \frac{v_{\infty} r^2}{2} \left[ 1 - \frac{R}{r} \right]$$

Finally: $$\psi = - \frac{v_\infty r^2 \sin^2 \theta}{2} \left[ 1 - \frac{R}{r} \right]$$

$$v_r = v_\infty \cos \theta \left[ 1 - \frac{R}{r} \right]$$

$$v_{\theta} = - v_\infty \sin \theta \left[ 1 - \frac{R}{2r} \right]$$

The figure at the start of this post is the velocity field along the plane y=0 when R=1 and $v_\infty= 1$. I generated this graphic using sage, and the following snippet of code:

x,z = var('x z')

def v(x,z):
r = sqrt(x**2 + z**2)
theta = arctan(x/z)
phi = arctan(0/x)

vr = cos(theta)*(1-(1/r))
vth = -sin(theta)*(1-0.5*(1/r))

factor = heaviside(r-1)
vx = factor*(vr*sin(theta)*cos(phi) + vth*cos(theta)*cos(phi))
vz = factor*(vr*cos(theta) - vth*sin(theta))
return (vx,vz)

p = plot\_vector\_field(v(x,z), (x,-2,2), (z,-2,2))
p += circle((0,0),1, fill=True)
p.axes\_labels(['x axis', 'z axis'])
p.show()