*2.1. Governing Equations*

A fibre preform is slowly inserted from the top of a furnace through an opening iris and is pulled from the bottom. During drawing, the size of the preform is significantly reduced, and it achieves the required final dimensions at the exit of the furnace. The process is schematically depicted in Figure 1. To develop a suitable mathematical model for describing the fibre drawing process, we begin our study with the full three-dimensional continuity, momentum, and energy equations written in cylindrical coordinates.

**Figure 1.** Schematic diagram of the drawing process of an optical glass capillary. The furnace temperature depends on the axial position *z*, and is indicated by *Tf* , while the furnace ambient temperature is indicated by *Ta*.

2.1.1. Mass and Momentum Equations

The continuity and momentum equations in cylindrical coordinates read [41]

$$\frac{\partial \rho}{\partial t} + \frac{1}{r} \frac{(\rho ru)}{\partial r} + \frac{1}{r} \frac{\partial (\rho v)}{\partial \phi} + \frac{\partial (\rho w)}{\partial z} = 0 \tag{1}$$

$$\begin{split} \rho \left( \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial r} + \frac{v}{r} \frac{\partial u}{\partial \phi} - \frac{u^2}{r} + w \frac{\partial u}{\partial z} \right) &= f\_r - \frac{\partial p}{\partial r} + \frac{1}{r} \frac{\partial (r \mathbf{r}\_{rr})}{\partial r} \\ &+ \frac{1}{r} \frac{\partial (r \mathbf{r}\_{r\phi})}{\partial \phi} + \frac{\partial (r \mathbf{r}\_{rz})}{\partial z} - \frac{\mathbf{r}\_{\phi\phi}}{r} \end{split} \tag{2}$$

$$\begin{split} \rho \left( \frac{\partial v}{\partial t} + \mu \frac{\partial v}{\partial r} + \frac{v}{r} \frac{\partial v}{\partial \phi} + \frac{\mu v}{r} + w \frac{\partial v}{\partial z} \right) &= f\_{\phi} - \frac{1}{r} \frac{\partial p}{\partial \phi} + \frac{1}{r^{2}} \frac{\partial (r^{2} \tau\_{r\phi})}{\partial r} \\ &+ \frac{1}{r} \frac{\partial (\tau\_{\phi\phi})}{\partial \phi} + \frac{\partial (\tau\_{\phi z})}{\partial z} \end{split} \tag{3}$$

$$\begin{split} \rho \left( \frac{\partial w}{\partial t} + u \frac{\partial w}{\partial r} + \frac{v}{r} \frac{\partial w}{\partial \phi} + w \frac{\partial w}{\partial z} \right) &= f\_z - \frac{\partial p}{\partial z} + \frac{1}{r} \frac{\partial (r \tau\_{rz})}{\partial r} \\ &+ \frac{1}{r} \frac{\partial (r \tau\_{\phi z})}{\partial \phi} + \frac{\partial (r \tau\_{zz})}{\partial z} \end{split} \tag{4}$$

Herein, *ρ*, *t*, and *p* denote the density, the time, and the pressure, respectively, and *r*, *z*, and *φ* represents the radial, axial, and azimuthal coordinates. The fluid velocity is denoted by *q* = *we<sup>z</sup>* + *ue<sup>r</sup>* + *veφ*, and *ez*, *er*, and *e<sup>φ</sup>* are the unit vectors in the axial, radial, and azimuthal directions, respectively. *u*, *v*, and *w* are the radial, azimuthal, and axial components of the velocity field, and *f* = *fze<sup>z</sup>* + *fre<sup>r</sup>* + *fφe<sup>φ</sup>* = *ρg*(*e<sup>z</sup>* + *e<sup>r</sup>* + *eφ*) represents the body force. The components of the viscous stress tensor read

$$\tau\_{rr} = \mu \left( 2 \frac{\partial u}{\partial r} - \frac{2}{3} (\nabla \cdot q) \right)$$

$$\tau\_{\Phi\Phi} = \mu \left( 2 \left( \frac{1}{r} \frac{\partial v}{\partial \Phi} + \frac{u}{r} \right) - \frac{2}{3} (\nabla \cdot q) \right)$$

$$\tau\_{zz} = \mu \left( 2 \frac{\partial w}{\partial z} - \frac{2}{3} (\nabla \cdot q) \right)$$

$$\tau\_{r\Phi} = \mu \left( r \frac{\partial}{\partial r} \left( \frac{v}{r} \right) + \frac{1}{r} \frac{\partial u}{\partial \phi} \right)$$

$$\tau\_{rz} = \mu \left( \frac{\partial w}{\partial r} + \frac{\partial u}{\partial z} \right)$$

$$\tau\_{\Phi z} = \mu \left( \frac{\partial v}{\partial z} + \frac{1}{r} \frac{\partial w}{\partial \phi} \right)$$

The divergence of the velocity field may be written as

$$\nabla \cdot \boldsymbol{q} = \frac{1}{r} \frac{\partial (r\boldsymbol{u})}{\partial r} + \frac{1}{r} \frac{\partial \boldsymbol{v}}{\partial \phi} + \frac{\partial \boldsymbol{w}}{\partial z} \tag{5}$$

The left-hand side of the momentum equations denotes the temporal and convective inertial acceleration, while the right-hand side incorporates the pressure gradient, divergence of the stress tensor, and sum of the body forces.

#### 2.1.2. Two-Dimensional Mass and Momentum Equations

We assume the flow to be incompressible, and consider axis symmetry and a nonrotating capillary, say, *ρ* = *const*, *<sup>∂</sup> ∂φ* = 0, and *v* = 0. Therefore, the mass conservation equation reduces to

$$\frac{1}{r}\frac{\partial(ru)}{\partial r} + \frac{\partial w}{\partial z} = 0 \tag{6}$$

and many terms of the components of the stress tensor vanish:

$$\begin{aligned} \tau\_{rr} &= 2\mu \left( \frac{\partial \mu}{\partial r} \right) \\\\ \tau\_{\phi\phi} &= 2\mu \left( \frac{\mu}{r} \right) \\\\ \tau\_{zz} &= 2\mu \left( \frac{\partial w}{\partial z} \right) \\\\ \tau\_{r\phi} &= 0 \end{aligned}$$

$$
\tau\_{\overline{\tau}z} = \mu \left( \frac{\partial w}{\partial r} + \frac{\partial u}{\partial z} \right),
$$

$$
\tau\_{\notin \overline{\tau}} = 0
$$

Thereby, the previous momentum equations reduce to

$$\begin{split} \rho \left( \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial r} + w \frac{\partial u}{\partial z} \right) &= f\_r - \frac{\partial p}{\partial r} \\ + \frac{2}{r} \frac{\partial}{\partial r} \left( r \mu \left( \frac{\partial u}{\partial r} \right) \right) + \frac{\partial}{\partial z} \left( \mu \left( \frac{\partial w}{\partial r} + \frac{\partial u}{\partial z} \right) \right) - 2 \mu \frac{u}{r^2} \end{split} \tag{7}$$
 
$$\begin{split} \rho \left( \frac{\partial w}{\partial t} + u \frac{\partial w}{\partial r} + w \frac{\partial w}{\partial z} \right) &= f\_z - \frac{\partial p}{\partial z} + \frac{1}{r} \frac{\partial}{\partial r} \left( r \mu \left( \frac{\partial w}{\partial r} + \frac{\partial u}{\partial z} \right) \right) \\ &\qquad + 2 \frac{\partial}{\partial z} \left( \mu \frac{\partial w}{\partial z} \right) \end{split} \tag{8}$$

because the momentum equation in the azimuthal direction vanishes identically. To close the problem, boundary and initial conditions must be specified. Concerning boundary conditions, we impose initial, kinematic, and dynamic boundary conditions. The kinematic boundary conditions require that the normal components of the velocity at each interface of the capillary be continuous across the interfaces. In addition, the tangential component of the velocity must be continuous at the interfaces. At each fibre surface, they read

$$\frac{\partial h\_1}{\partial t} + w \frac{\partial h\_1}{\partial z} = \mu \tag{9a}$$

$$\frac{\partial h\_2}{\partial t} + w \frac{\partial h\_2}{\partial z} = u \tag{9b}$$

The normal boundary conditions represent a balance of forces across the surfaces of the capillary. They may be written as

$$-\mathbf{u}\_1^T \cdot \mathbf{\tau} \cdot \mathbf{u}\_1 + \frac{\gamma}{h\_1} = p\_H \tag{10a}$$

$$-\mathbf{u}\_2^T \cdot \mathbf{\tau} \cdot \mathbf{u}\_2 - \frac{\gamma}{h\_2} = p\_a \tag{10b}$$

in the normal direction, and

$$\mathbf{t}\_1^T \cdot \mathbf{\tau} \cdot \mathbf{n}\_1 = 0 \tag{10c}$$

$$\mathbf{t}\_2^T \cdot \mathbf{\tau} \cdot \mathbf{\pi}\_2 = 0 \tag{10d}$$

in the tangential direction. In Equation (10), *τ*, *t*, and *n* denote the stress tensor and the unit vectors in the tangential and normal directions, respectively. They read

$$\mathbf{r} = \begin{bmatrix} 2\mu \left(\frac{\partial w}{\partial z}\right) - p & \mu \left(\frac{\partial w}{\partial r} + \frac{\partial u}{\partial z}\right) \\ \mu \left(\frac{\partial w}{\partial r} + \frac{\partial u}{\partial z}\right) & 2\mu \left(\frac{\partial u}{\partial r}\right) - p \end{bmatrix} \tag{11a}$$

$$\mathbf{t}\_i^T = \frac{(-1)^{i+1}}{\sqrt{1 + \left(\frac{\partial l\_i}{\partial z}\right)^2}} \left(1\mathbf{e}\_z + \frac{\partial l\_i}{\partial z}\mathbf{e}\_r\right) \tag{11b}$$

$$\mathfrak{m}\_{i}^{T} = \frac{(-1)^{i+1}}{\sqrt{1 + \left(\frac{\partial l\_{i}}{\partial z}\right)^{2}}} \left(\frac{\partial l\_{i}}{\partial z} \mathfrak{e}\_{z} - \mathbf{1} \mathfrak{e}\_{r}\right) \tag{11c}$$

Moreover, *γ*, *pH*, and *pa* are the surface tension, the hole, and the ambient pressure, respectively. To close the problem, we impose the velocity at the beginning and end of the drawing process as initial conditions:

$$w(t, r, z = 0) = \mathcal{W}\_{0\prime} \qquad\qquad w(t, r, z = L) = \mathcal{W}\_1 \tag{11d}$$
