2.1.7. The Viscosity of Silica Glass

Following Voyce et al. [22], we choose the correlation by Urbain et al. [44] for the viscosity in the temperature range 1400 ◦C ≤ T ≤ 2500 ◦C and the one by Hetherington et al. [45] for the temperature range 1000 ◦C ≤ *T* ≤ 1400 ◦C. Urbain et al. [44] utilized the rotating cup technique to measure the viscosity in the interval between 10−<sup>1</sup> and 10<sup>5</sup> Poise and the isothermal penetration method to measure the viscosity in the interval between 108 and 10<sup>13</sup> Poise. Because the isothermal penetration method does not allow a glass metastable equilibrium to be reached, we opt for the correlation provided by Hetherington et al. [45]. They used a fibre elongation technique to measure the viscosity of the silica in this temperature range, checking that the silica was in an equilibrium condition [46]. Therefore, we utilize the following correlation:

$$\mu = 5.8 \cdot 10^{-8} \exp\left(\frac{515400}{8.3145T}\right) \tag{34}$$

for the temperature range 1400 ◦C ≤ *T* ≤ 2500 ◦C. Instead, we use

$$
\mu = 3.8 \cdot 10^{-14} \exp\left(\frac{712000}{8.3145T}\right) \tag{35}
$$

for the temperature range 1000 ◦C<T ≤ 1400 ◦C. In (34) and (35), the viscosity *μ* is in *Pa* · *s* and the temperature *T* is in *K*.

#### **3. Final Asymptotic Equations**

*Non-Rotating Capillary*

We proceed with a regular parameter expansion of the unknowns

$$\begin{aligned} \overline{w} &= \overline{w}\_0(\overline{t}, \overline{z}) + \epsilon^2 \overline{w}\_1(\overline{t}, \overline{z}, \overline{r}) \\ \overline{u} &= \overline{u}\_0(\overline{t}, \overline{z}, \overline{r}) + \epsilon^2 \overline{u}\_1(\overline{t}, \overline{z}, \overline{r}) \\ \overline{p} &= \overline{p}\_a + \epsilon^2 \overline{P}(\overline{t}, \overline{z}, \overline{r}) \\ \overline{T} &= \overline{T}\_0(\overline{t}, \overline{z}, \overline{r}) + \epsilon^2 \overline{T}\_1(\overline{t}, \overline{z}, \overline{r}) \end{aligned}$$

and utilize it in the previously derived non-dimensional mass conservation (Equation (19)), momentum (Equations (20) and (21)), and energy equations (Equation (22)) and boundary conditions (Equations (23) and (24)). By retaining the terms of the expansion at most up to order 2, we can obtain the equations of Fitt et al. [19]:

$$\frac{\partial(\overline{h}\_1^2 \overline{w}\_0)}{\partial \overline{z}} = \frac{\overline{p}\_0 \overline{h}\_2^2 \overline{h}\_1^2 - \overline{\gamma} \overline{h}\_2 \overline{h}\_1 (\overline{h}\_2 + \overline{h}\_1)}{\mu (\overline{h}\_2^2 - \overline{h}\_1^2)}\tag{36}$$

$$\frac{\partial(\overline{h}\_2^2 \overline{w}\_0)}{\partial \overline{z}} = \frac{\overline{p}\_0 \overline{h}\_2^2 \overline{h}\_1^2 - \overline{\gamma} \overline{h}\_2 \overline{h}\_1 (\overline{h}\_2 + \overline{h}\_1)}{\mu (\overline{h}\_2^2 - \overline{h}\_1^2)}\tag{37}$$

$$\frac{\partial}{\partial \overline{z}} \left( 3\overline{\mu} \frac{\partial \overline{w}\_0}{\partial \overline{z}} (\overline{h}\_2^2 - \overline{h}\_1^2) + \overline{\gamma} (\overline{h}\_2 + \overline{h}\_1) \right) = 0 \tag{38}$$

These are evolution equations for the inner and outer fibre surface and an axial momentum equation; for more details see [19]. In Equation (38), we have neglected the inertia terms due to the small values that the Reynolds number *Re* typically assumes; as a matter of fact, *Re* ≈ <sup>10</sup>−<sup>8</sup> [29]. With regard to the energy equation, we consider the case at the leading order, where *Pe* = *<sup>P</sup>*, <sup>2</sup> and *Br* <sup>=</sup> *Br* <sup>6</sup> 2 :

$$\begin{split} \overline{P} \left( \frac{\partial \overline{T}}{\partial \overline{t}} + \overline{u} \frac{\partial \overline{T}}{\partial \overline{r}} + \overline{w} \frac{\partial \overline{T}}{\partial \overline{z}} \right) &= \frac{1}{\overline{r}} \frac{\partial}{\partial \overline{r}} \left( \overline{r} \frac{\partial \overline{T}}{\partial \overline{r}} \right) + \frac{1}{\overline{r}} \frac{\partial}{\partial \overline{r}} \left( \gamma\_R \overline{r} \frac{\partial \overline{T}^4}{\partial \overline{r}} \right) \\ + 2 \overline{Br} \overline{\mu} \left[ \left( \frac{\partial \overline{u}}{\partial \overline{r}} \right)^2 + \left( \frac{\overline{u}}{\overline{r}} \right)^2 + \left( \frac{\partial \overline{w}}{\partial \overline{z}} \right)^2 \right] &+ \overline{Br} \overline{\mu} \left[ \frac{\partial \epsilon \overline{u}}{\partial z} + \frac{\partial \overline{w}}{\partial \epsilon \overline{r}} \right]^2 \end{split} \tag{39}$$

This case corresponds to the general situation, where the convective heat transport balances the heat transfer across the fibre and the viscous dissipation [24]. Following Taroni et al. [24], we change variables to a more convenient coordinate frame that suits the geometry of the capillary:

$$\overline{\mathfrak{x}}(\overline{r},\overline{z}) = \frac{\overline{r} - \overline{h}\_1(\overline{z})}{\overline{h}\_2(\overline{z}) - \overline{h}\_1(\overline{z})} \to \overline{r} = \overline{\mathfrak{x}}(\overline{h}\_2 - \overline{h}\_1) + \overline{h}\_1. \tag{40}$$

The continuity in Equation (19) at the leading order becomes

$$\overline{u}\_0 = -\frac{\partial \overline{u}\_0}{\partial \overline{\zeta}} \frac{[\overline{\mathfrak{X}}(\overline{h}\_2 - \overline{h}\_1) + \overline{h}\_1]}{2} + \frac{\overline{A}(\overline{\zeta})}{[\overline{\mathfrak{X}}(\overline{h}\_2 - \overline{h}\_1) + \overline{h}\_1]} \tag{41}$$

where *A* is

$$\overline{A} = \frac{\overline{p}\_0 \overline{h}\_2^2 \overline{h}\_1^2 - \overline{\gamma} \overline{h}\_2 \overline{h}\_1 (\overline{h}\_2 + \overline{h}\_1)}{2\overline{\mu} (\overline{h}\_2^2 - \overline{h}\_1^2)}\tag{42}$$

and the axial momentum equation after being integrated once with respect to *z* assumes the form

$$\frac{\partial \overline{w}\_0}{\partial \overline{\zeta}} = \left( \frac{\overline{F} - \overline{\gamma} (\overline{h}\_2 + \overline{h}\_1)}{3 \overline{\mu} (\overline{h}\_{20}^2 - \overline{h}\_{10}^2) \overline{W}\_0} \right) \overline{w}\_0 \tag{43}$$

where the constant *F* arising from the integration physically represents the tension needed to pull the fibre. In addition, the following identity

$$
\overline{h}\_2^2 - \overline{h}\_1^2 = \frac{(\overline{h}\_{20}^2 - \overline{h}\_{10}^2)\overline{W}\_0}{\overline{w}\_0} \tag{44}
$$

has been used, and represents an equation of mass conservation. We now change the variables in Equation (39) according to Equation (40) and utilize the asymptotic expansions previously defined to obtain

$$\begin{split} \frac{\partial \overline{T}}{\partial \overline{\zeta}} &= \overline{\Psi} \frac{\partial \overline{T}\_0}{\partial \overline{\chi}} + \overline{\Omega} \left( \frac{\partial^2 \overline{T}\_0}{\partial \overline{\chi}^2} \right) + \overline{\Lambda} \left( \frac{\partial \overline{T}\_0}{\partial \overline{\chi}} \right)^2 \\ &+ \frac{\overline{Br} \overline{\mu}}{\overline{w}\_0} \left[ 3 \left( \frac{\partial \overline{w}\_0}{\partial \overline{\zeta}} \right)^2 + \left( \frac{2 \overline{A}}{\overline{\Theta}^2} \right)^2 \right] \end{split} \tag{45}$$

where

$$\begin{aligned} \overline{\Pi} &= \left[ \overline{\boldsymbol{x}} (\overline{\boldsymbol{h}}\_2 - \overline{\boldsymbol{h}}\_1) + \overline{\boldsymbol{h}}\_1 \right] (\overline{\boldsymbol{h}}\_2 - \overline{\boldsymbol{h}}\_1) \\ \overline{\Theta} &= \left[ \overline{\boldsymbol{x}} (\overline{\boldsymbol{h}}\_2 - \overline{\boldsymbol{h}}\_1) + \overline{\boldsymbol{h}}\_1 \right] \\ \overline{\Lambda} &= \frac{12 \gamma\_R \overline{T}\_0^2}{\overline{\varpi}\_0 (\overline{\boldsymbol{h}}\_2 - \overline{\boldsymbol{h}}\_1)^2} \\ \overline{\Omega} &= \frac{1 + 4 \gamma\_R \overline{T}\_0^3}{\overline{\varpi}\_0 (\overline{\boldsymbol{h}}\_2 - \overline{\boldsymbol{h}}\_1)^2} \\ \overline{\Psi} &= \frac{1}{\overline{\varpi}\_0} \left[ \frac{1 + 4 \gamma\_R \overline{T}\_0^3}{\overline{\Pi}} - \frac{\overline{A}}{\overline{\Pi}} + \bar{P} \frac{\partial \overline{\varpi}\_0}{\partial \overline{\zeta}} \frac{\overline{\Theta}}{2(\overline{\boldsymbol{h}}\_2 - \overline{\boldsymbol{h}}\_1)} \right] \end{aligned}$$

The thermal boundary conditions in Equation (25) at *r* = *h*<sup>2</sup> and at *r* = *h*<sup>1</sup> become

$$\frac{\partial \overline{T}\_0}{\partial \overline{\mathbf{x}}}\bigg|\_{\overline{\mathbf{X}}=1} = \left[ -\frac{\epsilon \alpha (\overline{T}\_0^4 - \overline{T}\_f^4) + \epsilon \beta (\overline{T}\_0 - \overline{T}\_a)}{1 + 4\gamma\_R \overline{T}\_0^3} \right] (\overline{h}\_2 - \overline{h}\_1) \tag{46a}$$

$$\left. \frac{\partial \overline{T}\_0}{\partial \overline{\boldsymbol{\pi}}} \right|\_{\overline{\boldsymbol{\pi}} = 0} = -\frac{\boldsymbol{\epsilon} \beta (\overline{T}\_0 - \overline{T}\_a)}{1 + 4 \gamma\_R \overline{T}\_0^3} (\overline{h}\_2 - \overline{h}\_1) \tag{46b}$$

 
