A Theoretical Analysis of the Interaction Between Pores and Inclusions During the Continuous Casting of Steel

Abstract

A mathematical model is derived to predict the trajectories of pores and inclusions that are nucleated in the interdendritic region during the continuous casting of steel. Using basic fluid mechanics and heat transfer, scaling analysis, and asymptotic methods, the model accounts for the possible lateral drift of the pores as a result of the dependence of the surface tension on temperature and sulfur concentration. Moreover, the soluto–thermocapillary drift of such pores prior to final solidification, coupled to the fact that any inclusions present can only have a vertical trajectory, can help interpret recent experimental observations of pore-inclusion clusters in solidified steel castings.

Appendices

Appendix A: Thermocapillary Drift

Fig. 13

Schematic of geometry for a drifting gas bubble

Here, the formulae for the velocity of the bubble given in Young et al.[35] and Kuznetsov et al.[17] are re-derived, and it is verified that diffusion dominates convection when parameters for the casting of steel are used; in fact, it turns out that Eq. [9] in Reference 35 and Eq. [7] in Reference 17 do not agree with each other. The steady-state axisymmetric governing equations for conservation of mass, momentum, and heat in spherical polar $\left( r,\theta ,\phi \right)$ coordinates, as shown in Figure 13, and in the frame of reference in which the bubble is fixed at the origin are, for $j=g,l,$

$$\begin{aligned} \frac{1}{r^{2}}\frac{\partial }{\partial r}\left( r^{2}u_{r}^{\left( j\right) }\right) +\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left( u_{\theta }^{\left( j\right) }\sin \theta \right) =0, \end{aligned}$$

(A.1)

$$\begin{aligned} 0=-\frac{\partial p^{\left( j\right) }}{\partial r}+\frac{1}{r^{2}} \frac{\partial }{\partial r}\left( r^{2}\tau_{rr}^{\left( j\right) }\right) +\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left( \tau_{r\theta }^{\left( j\right) }\sin \theta \right) -\frac{\left( \tau_{\theta \theta }^{\left( j\right) }+\tau_{\phi \phi }^{\left( j\right) }\right) }{r}, \end{aligned}$$

(A.2)

$$\begin{aligned} 0=-\frac{1}{r}\frac{\partial p^{\left( j\right) }}{\partial \theta }+\frac{1}{r^{2}}\frac{\partial }{\partial r}\left( r^{2}\tau_{r\theta }^{\left( j\right) }\right) +\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left( \tau_{\theta \theta }^{\left( j\right) }\sin \theta \right) +\frac{\left( \tau_{r\theta }^{\left( j\right) }-\tau_{\phi \phi }^{\left( j\right) }\cot \theta \right) }{r}, \end{aligned}$$

(A.3)

$$\begin{aligned} \rho ^{\left( j\right) }c_{p}^{\left( j\right) }\left( u_{r}^{\left( j\right) }\frac{\partial T^{\left( j\right) }}{\partial r}+\frac{u_{\theta }^{\left( j\right) }}{r}\frac{\partial T^{\left( j\right) }}{\partial \theta }\right) = \end{aligned}$$

$$\begin{aligned} k^{\left( j\right) }\left\{ \frac{1}{r^{2}}\frac{\partial }{\partial r}\left( r^{2}\frac{\partial T^{\left( j\right) }}{\partial r}\right) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \frac{\partial T^{\left( j\right) }}{\partial \theta }\right) \right\} , \end{aligned}$$

(A.4)

where $u_{r}^{\left( j\right) }$ and $u_{\theta }^{\left( j\right) }$ are the radial and angular velocity components, $p^{\left( j\right) }$ is the pressure, $c_{p}^{\left( j\right) }$ is the specific heat capacity of the fluid of phase j, and $k^{\left( j\right) }$ is its thermal conductivity. In Eqs. [A.2] and [A.3], $\tau_{rr}^{\left( j\right) } ,\tau_{r\theta }^{\left( j\right) },\tau_{\theta \theta }^{\left( j\right), }$ and $\tau_{\phi \phi }^{\left( j\right) }$ denote the components of the stress tensor, which are given by

$$\begin{aligned} \tau_{rr}^{\left( j\right) }=2\mu ^{\left( j\right) }\frac{\partial u_{r}^{\left( j\right) }}{\partial r},\quad \tau_{\theta \theta }^{\left( j\right) }=\frac{2\mu ^{\left( j\right) }}{r}\left( \frac{\partial u_{\theta }^{\left( j\right) }}{\partial \theta }+u_{r}^{\left( j\right) }\right) , \end{aligned}$$

(A.5)

$$\begin{aligned} \tau_{\phi \phi }^{\left( j\right) }=\frac{2\mu ^{\left( j\right) }}{r}\left( u_{r}^{\left( j\right) }+u_{\theta }^{\left( j\right) } \cot \theta \right) ,\quad \tau_{r\theta }^{\left( j\right) }=\mu ^{\left( j\right) }\left( r\frac{\partial }{\partial r}\left( \frac{u_{\theta }^{\left( j\right) }}{r}\right) +\frac{1}{r}\frac{\partial u_{r}^{\left( j\right) }}{\partial \theta }\right) . \end{aligned}$$

(A.6)

The boundary conditions are as follows. At $r=a,$

$$\begin{aligned} u_{r}^{\left( g\right) }&=u_{r}^{\left( l\right) }=0,\end{aligned}$$

(A.7)

$$\begin{aligned} u_{\theta }^{\left( g\right) }&=u_{\theta }^{\left( l\right) }, \end{aligned}$$

(A.8)

$$\begin{aligned} \mu ^{\left( g\right) }\left( r\frac{\partial }{\partial r}\left( \frac{u_{\theta }^{\left( g\right) }}{r}\right) +\frac{1}{r}\frac{\partial u_{r}^{\left( g\right) }}{\partial \theta }\right) -\mu ^{\left( l\right) }\left( r\frac{\partial }{\partial r}\left( \frac{u_{\theta }^{\left( l\right) }}{r}\right) +\frac{1}{r}\frac{\partial u_{r}^{\left( l\right) } }{\partial \theta }\right) =\frac{1}{r}\frac{\partial \gamma }{\partial \theta }, \end{aligned}$$

(A.9)

$$\begin{aligned} p^{\left( g\right) }-2\mu ^{\left( g\right) }\frac{\partial u_{r}^{\left( g\right) }}{\partial r}-p^{\left( l\right) }+2\mu ^{\left( l\right) } \frac{\partial u_{r}^{\left( l\right) }}{\partial r}=\frac{2\gamma }{r}, \end{aligned}$$

(A.10)

$$\begin{aligned} T^{\left( g\right) }&=T^{\left( l\right) },\end{aligned}$$

(A.11)

$$\begin{aligned} k^{\left( g\right) }\frac{\partial T^{\left( g\right) }}{\partial r}&=k^{\left( l\right) }\frac{\partial T^{\left( l\right) }}{\partial r}, \end{aligned}$$

(A.12)

where $\gamma$ is the surface tension. As $r\rightarrow \infty ,$

$$\begin{aligned} u_{r}^{\left( l\right) }\rightarrow -V\cos \theta ,\quad u_{\theta }^{\left( l\right) }\rightarrow V\sin \theta ,\quad T^{\left( l\right) }\rightarrow T^{0}+T_{1}r\cos \theta , \end{aligned}$$

(A.13)

where V is the thermocapillary drift speed. If $\gamma =\gamma \left( T\right) ,$ linearizing about $T=T^{0}$ gives

$$\begin{aligned} \gamma =\gamma ^{0}+\gamma_{T}\left( T-T^{0}\right) , \end{aligned}$$

(A.14)

where $\gamma ^{0}=\gamma \left( T^{0}\right) ,$ and $\gamma_{T}:=\left( d\gamma /dT\right)_{T=T_{0}}$ is a constant.

Nondimensionalizing with

$$\begin{aligned} R=\frac{r}{a},\quad {U}_{R}^{\left( j\right) }=\frac{{u}_{r}^{\left( j\right) }}{V},\quad {U}_{\theta }^{\left( j\right) }=\frac{{u}_{\theta }^{\left( j\right) }}{V},\quad P^{\left( j\right) }=\frac{p^{\left( j\right) }}{\mu ^{\left( j\right) }V/a},\quad \Theta ^{\left( j\right) }=\frac{T^{\left( j\right) }-T_{0}}{aT_{1}}, \end{aligned}$$

(A.15)

Eqs. [A.1] through [A.4] become, on dropping the tildes and using the forms given in Eqs. [A.5] and [A.6] for the stress components in Eqs. [A.2] and [A.3],

$$\begin{aligned} \frac{1}{R^{2}}\frac{\partial }{\partial R}\left( R^{2}U_{R}^{\left( j\right) }\right) +\frac{1}{R\sin \theta }\frac{\partial }{\partial \theta }\left( U_{\theta }^{\left( j\right) }\sin \theta \right) =0, \end{aligned}$$

(A.16)

$$\begin{aligned} 0=-\frac{\partial P^{\left( j\right) }}{\partial R}+\frac{2}{R^{2}} \frac{\partial }{\partial R}\left( R^{2}\frac{\partial U_{R}^{\left( j\right) }}{\partial R}\right) + \frac{1}{R\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \left\{ R\frac{\partial }{\partial R}\left( \frac{U_{\theta }^{\left( j\right) }}{R}\right) +\frac{1}{R}\frac{\partial U_{R}^{\left( j\right) }}{\partial \theta }\right\} \right) -\frac{2}{R^{2}}\left\{ 2U_{R}^{\left( j\right) }+\frac{1}{\sin \theta }\frac{\partial }{\partial \theta }\left( U_{\theta }^{\left( j\right) }\sin \theta \right) \right\} , \end{aligned}$$

(A.17)

$$\begin{aligned}0&=-\frac{1}{R}\frac{\partial P^{\left( j\right) }}{\partial \theta }+\frac{1}{R^{2}}\frac{\partial }{\partial R}\left( R^{2}\left\{ R\frac{\partial }{\partial R}\left( \frac{U_{\theta }^{\left( j\right) }}{R}\right) +\frac{1}{R}\frac{\partial U_{R}^{\left( j\right) }}{\partial \theta }\right\} \right)\\ &\quad+\,\frac{2}{R\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \left\{ \frac{1}{R}\frac{\partial U_{\theta }^{\left( j\right) } }{\partial \theta }+\frac{U_{R}^{\left( j\right) }}{R}\right\} \right) \\&\quad +\,\frac{1}{R}\left\{ R\frac{\partial }{\partial R}\left( \frac{U_{\theta }^{\left( j\right) }}{R}\right) +\frac{1}{R}\frac{\partial U_{R}^{\left( j\right) }}{\partial \theta }-\frac{2\cot \theta }{R}\left( U_{R}^{\left( j\right) }+U_{\theta }^{\left( j\right) }\cot \theta \right) \right\} , \end{aligned}$$

(A.18)

$$\begin{aligned} Pe^{\left( j\right) }\left( U_{R}^{\left( j\right) }\frac{\partial \Theta ^{\left( j\right) }}{\partial R}+\frac{U_{\theta }^{\left( j\right) }}{R}\frac{\partial \Theta ^{\left( j\right) }}{\partial \theta }\right) =\frac{1}{R^{2}}\frac{\partial }{\partial R}\left( R^{2}\frac{\partial \Theta ^{\left( j\right) }}{\partial R}\right) +\frac{1}{R^{2}\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \frac{\partial \Theta ^{\left( j\right) }}{\partial \theta }\right) , \end{aligned}$$

(A.19)

where

$$\begin{aligned} Pe^{\left( l\right) }=Pe,\quad Pe^{\left( g\right) }=\alpha Pe, \end{aligned}$$

(A.20)

with Pe denoting the Péclet number and being given by $Pe=Va/\alpha_{l}$. As for the boundary conditions: at $R=1,$

$$\begin{aligned} U_{R}^{\left( g\right) }&=0,\quad U_{R}^{\left( l\right) }=0,\end{aligned}$$

(A.21)

$$\begin{aligned} U_{\theta }^{\left( g\right) }&=U_{\theta }^{\left( l\right) }, \end{aligned}$$

(A.22)

$$\begin{aligned} \mu \left( \frac{\partial U_{\theta }^{\left( g\right) }}{\partial R}-U_{\theta }^{\left( g\right) }\right) -\left( \frac{\partial U_{\theta }^{\left( l\right) }}{\partial R}-U_{\theta }^{\left( l\right) }\right) ={\mathcal{C}}_{1}\frac{\partial \Theta }{\partial \theta }, \end{aligned}$$

(A.23)

$$\begin{aligned} \mu \left( P^{\left( g\right) }-2\frac{\partial U_{R}^{\left( g\right) } }{\partial R}\right) -\left( P^{\left( l\right) }-2\frac{\partial U_{R}^{\left( l\right) }}{\partial R}\right) =2\left( {\mathcal{C}}_{0}+{\mathcal{C}}_{1}\Theta \right) , \end{aligned}$$

(A.24)

$$\begin{aligned} \Theta ^{\left( g\right) }&=\Theta ^{\left( l\right) },\end{aligned}$$

(A.25)

$$\begin{aligned} k\frac{\partial \Theta ^{\left( g\right) }}{\partial R}&=\frac{\partial \Theta ^{\left( l\right) }}{\partial R}, \end{aligned}$$

(A.26)

where

$$\begin{aligned} \mu =\mu ^{\left( g\right) }/\mu ^{\left( l\right) },\quad k=k^{\left( g\right) }/k^{\left( l\right) },\quad {\mathcal{C}}_{0}=\frac{\gamma ^{0}}{V\mu ^{\left( l\right) }},\quad {\mathcal{C}}_{1}=\frac{a\gamma_{T}T_{1}}{V\mu ^{\left( l\right) }}; \end{aligned}$$

(A.27)

as $R\rightarrow \infty ,$

$$\begin{aligned} U_{R}^{\left( l\right) }\rightarrow -\cos \theta ,\quad U_{\theta }^{\left( l\right) }\rightarrow \sin \theta ,\quad \Theta ^{\left( l\right) }\rightarrow R\cos \theta . \end{aligned}$$

(A.28)

Introducing streamfunctions $\psi ^{\left( j\right) }$ such that

$$\begin{aligned} U_{R}^{\left( j\right) }=-\frac{1}{R^{2}\sin \theta }\frac{\partial \psi ^{\left( j\right) }}{\partial \theta },\quad U_{\theta }^{\left( j\right) }=\frac{1}{R\sin \theta }\frac{\partial \psi ^{\left( j\right) }}{\partial R}, \end{aligned}$$

(A.29)

it is known that these streamfunctions will satisfy

$$\begin{aligned} \left( \frac{\partial ^{2}}{\partial R^{2}}+\frac{\sin \theta }{R^{2}} \frac{\partial }{\partial \theta }\left( \frac{1}{\sin \theta }\right) \frac{\partial }{\partial \theta }\right) ^{2}\psi ^{\left( j\right) }=0, \end{aligned}$$

(A.30)

which comes from algebraic manipulation of Eqs. [A.17], [A.18], and [A.29]; the general solutions will be

$$\begin{aligned} \psi ^{\left( j\right) }=\left( A^{\left( j\right) }R^{4}+B^{\left( j\right) }R^{2}+C^{\left( j\right) }R+\frac{D^{\left( j\right) }}{R}\right) \sin ^{2}\theta , \end{aligned}$$

(A.31)

where $A^{\left( j\right) },B^{\left( j\right) },C^{\left( j\right) },D^{\left( j\right) }$ are all constants to be determined. Thus,

$$\begin{aligned} U_{R}^{\left( j\right) }&=-\frac{2}{R^{2}}\left( A^{\left( j\right) }R^{4}+B^{\left( j\right) }R^{2}+C^{\left( j\right) }R+\frac{D^{\left( j\right) }}{R}\right) \cos \theta ,\end{aligned}$$

(A.32)

$$\begin{aligned} U_{\theta }^{\left( j\right) }&=\frac{1}{R}\left( 4A^{\left( j\right) }R^{3}+2B^{\left( j\right) }R+C^{\left( j\right) }-\frac{D^{\left( j\right) }}{R^{2}}\right) \sin \theta . \end{aligned}$$

(A.33)

Also, if $Pe\ll 1,$ Eq. [A.19] reduces to

$$\begin{aligned} \frac{1}{R^{2}}\frac{\partial }{\partial R}\left( R^{2}\frac{\partial \Theta ^{\left( j\right) }}{\partial R}\right) +\frac{1}{R^{2}\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \frac{\partial \Theta ^{\left( j\right) }}{\partial \theta }\right) =0; \end{aligned}$$

(A.34)

the computations indicate that, for the mushy region, $3\times 10^{-5} \le \left| Pe\right| \le 0.2,$ showing that this simplification is valid. The relevant solution to equation [A.34] is

$$\begin{aligned} \Theta ^{\left( j\right) }=\left( E^{\left( j\right) }R+\frac{F^{\left( j\right) }}{R^{2}}\right) \cos \theta , \end{aligned}$$

(A.35)

with $E^{\left( j\right) }$ and $F^{\left( j\right) }$ also constants to be determined.

First, Eq. [A.28c] requires that $E^{\left( l\right) }=1,$ whereas the fact that $\Theta ^{\left( g\right) }$ is finite at $R=0$ requires that $F^{\left( g\right) }=0;$ Eqs. [A.25] and [A.26] then lead to

$$\begin{aligned} F^{\left( l\right) }=\frac{1-k}{2+k},\quad E^{\left( g\right) }=\frac{3}{2+k}. \end{aligned}$$

(A.36)

Note from Table II that $k\ll 1,$ giving $F^{\left( l\right) }\approx 1/2,E^{\left( g\right) }\approx 3/2,$ so that

$$\begin{aligned} \Theta ^{\left( g\right) }=\frac{3}{2}R\cos \theta ,\quad \Theta ^{\left( l\right) }=\left( R+\frac{1}{2R^{2}}\right) \cos \theta . \end{aligned}$$

(A.37)

Next, Eqs. [A.28a] and [A.28b] require $A^{\left( l\right) }=0,\quad B^{\left( l\right) }=1/2,$ whereas the fact that $U_{R}^{\left( g\right) }$ and $U_{\theta }^{\left( g\right) }$ are finite at $R=0$ requires that $C^{\left( g\right) }=0,\quad D^{\left( g\right) }=0.$ There are four constants left to be determined—$C^{\left( l\right) },D^{\left( l\right) },A^{\left( g\right) },B^{\left( g\right) }$—with four Eqs. [A.21a], [A.21b], [A.22], and [A.23]. Since $\mu \ll 1,$ these conditions give

$$\begin{aligned} A^{\left( g\right) }=\frac{1}{4}-\frac{{\mathcal{C}}_{{1}}}{4},\quad B^{\left( g\right) }=-\frac{1}{4}+\frac{{\mathcal{C}}_{{1}}}{4},\quad C^{\left( l\right) }=-\frac{1}{2}-\frac{{\mathcal{C}}_{{1}}}{4},\quad D^{\left( l\right) }=\frac{{\mathcal{C}}_{{1}}}{4}. \end{aligned}$$

(A.38)

Now,

$$\begin{aligned} U_{R}^{\left( g\right) }=-2A^{\left( g\right) }\left( R^{2}-1\right) \cos \theta ,\quad U_{\theta }^{\left( g\right) }=2A^{\left( g\right) }\left( 2R^{2}-1\right) \sin \theta , \end{aligned}$$

(A.39)

$$\begin{aligned} U_{R}^{\left( l\right) }&=-\left( 1-\frac{\left( 2D^{\left( l\right) }+1\right) }{R}+\frac{2D^{\left( l\right) }}{R^{3}}\right) \cos \theta ,\end{aligned}$$

(A.40)

$$\begin{aligned} U_{\theta }^{\left( l\right) }&=\left( 1-\frac{\left( 2D^{\left( l\right) }+1\right) }{2R}-\frac{D^{\left( l\right) }}{R^{3}}\right) \sin \theta , \end{aligned}$$

(A.41)

$$\begin{aligned} P^{\left( g\right) }=-20A^{\left( g\right) }R\cos \theta +P_{\infty }^{\left( g\right) },\quad P^{\left( l\right) }=\frac{\left( 2D^{\left( l\right) }+1\right) \cos \theta }{R^{2}}+P_{\infty }^{\left( l\right) } \end{aligned}$$

(A.42)

and so

$$\begin{aligned} \frac{\partial U_{R}^{\left( g\right) }}{\partial R}&=-4A^{\left( g\right) }R\cos \theta ,\quad \frac{\partial U_{R}^{\left( l\right) } }{\partial R}=-\left( \frac{\left( 2D^{\left( l\right) }+1\right) }{R^{2}}-\frac{8D^{\left( l\right) }}{R^{4}}\right) \cos \theta ,\\ \frac{\partial U_{\theta }^{\left( g\right) }}{\partial R}&=8A^{\left( g\right) }R\sin \theta ,\quad \frac{\partial U_{\theta }^{\left( l\right) } }{\partial R}=\left( \frac{1}{R^{2}}\left( \frac{1}{2}+D^{\left( l\right) }\right) +\frac{3D^{\left( l\right) }}{R^{4}}\right) \sin \theta . \end{aligned}$$

Still ${\mathcal{C}}_{1}$ needs to be determined. There should be no net force parallel to the flow,[1] implying that

$$\begin{aligned} \int_{0}^{\pi }\left( \sigma_{R\theta }^{\left( l\right) }\sin ^{2} \theta -\sigma_{RR}^{\left( l\right) }\sin \theta \cos \theta \right)_{R=1}\rm {d}\theta =0. \end{aligned}$$

(A.43)

Now,

$$\begin{aligned} \tau_{RR}^{\left( l\right) }=2\frac{\partial U_{R}^{\left( l\right) } }{\partial R},\quad \tau_{R\theta }^{\left( l\right) }=R\left( \frac{U_{\theta }^{\left( l\right) }}{R}\right) +\frac{1}{R}\frac{\partial U_{R}^{\left( l\right) }}{\partial \theta }, \end{aligned}$$

so that

$$\begin{aligned} \left( \tau_{RR}^{\left( l\right) }\right)_{R=1}=-2\left( 1-6D^{\left( l\right) }\right) \cos \theta ,\quad \left( \tau_{R\theta }^{\left( l\right) }\right)_{R=1}=6D^{\left( l\right) }\sin \theta ; \end{aligned}$$

also,

$$\begin{aligned} \left( P^{\left( l\right) }\right)_{R=1}=\left( 2D^{\left( l\right) }+1\right) \cos \theta +P_{\infty }^{\left( l\right) }. \end{aligned}$$

Performing all the integrals in Eq. [A.43] leads to $C^{\left( l\right) }=0,$ giving ${\mathcal{C}}_{1}=-2,$ and hence

$$\begin{aligned} V=-\frac{\gamma_{T}aT_{1}}{2\mu ^{\left( l\right) }}. \end{aligned}$$

(A.44)

It is often overlooked, but nevertheless deserves to be noted, that the solutions obtained above made no use of the normal stress condition [A.24]; indeed, the left-hand side of [A.24] is

$$\begin{aligned} \mu (P_{\infty }^{(g)}-9\cos \theta )-(P_{\infty }^{(l)}+8\cos \theta ), \end{aligned}$$

whereas the right-hand side is

$$\begin{aligned} 2({\mathcal{C}}_{0}-3\cos \theta ). \end{aligned}$$

This difference is explained by the fact that the original Eqs. [A.1] through [A.13] should be formulated so that the location of the gas–liquid interface is not known, but should be determined as part of the problem; hence the bubble would, in general, not be spherical. This issue has been considered by Levan,[19] and requires us to determine how nonspherical it might become. In the present case, where $\mu \ll 1,$ one requires

$$\begin{aligned} -(P_{\infty }^{(l)}+8\cos \theta )\approx 2({\mathcal{C}}_{0}-3\cos \theta ), \end{aligned}$$

which implies that $P_{\infty }^{(l)}=-2{\mathcal{C}}_{0}$; thence, the bubble will be spherical if ${\mathcal{C}}_{0}\gg 1,$ i.e., if the $\cos \theta$ terms are negligible compared with the leading-order terms. Thus, it is apparent that the spherical approximation is valid provided that $\left| {\mathcal{C}}_{0}\right| \gg 1.$ From the computations, for the mushy region, $8.3\times 10^{4}\le \left| {\mathcal{C}}_{0}\right| \le 1.8\times 10^{8},$ as required.

Appendix B: Solutocapillary Drift

For solutocapillary drift, the formulation is slightly different to that in Appendix A, since the sulfur concentration is only defined for the liquid, i.e., $r\ge a;$ however, the governing equations for $u_{r}^{\left( l\right) },u_{\theta }^{\left( l\right) },p^{\left( l\right) },p_{r}^{\left( g\right) },u_{\theta }^{\left( g\right) },p^{\left( g\right) }$ remain unchanged. Consequently, [A.4] is replaced by

$$\begin{aligned} \rho ^{\left( j\right) }\left( u_{r}^{\left( l\right) }\frac{\partial c}{\partial r}+\frac{u_{\theta }^{\left( j\right) }}{r}\frac{\partial c}{\partial \theta }\right) =D\left\{ \frac{1}{r^{2}}\frac{\partial }{\partial r}\left( r^{2}\frac{\partial c}{\partial r}\right) +\frac{1}{r^{2}\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \frac{\partial c}{\partial \theta }\right) \right\} . \end{aligned}$$

(B.1)

The boundary conditions for c are

$$\begin{aligned} \frac{\partial c}{\partial r}&=0\quad {\text {at }}\quad r=a,\end{aligned}$$

(B.2)

$$\begin{aligned} c&\rightarrow c_{0}+c_{1}r\cos \theta \quad \text {as }\quad \it{r} \rightarrow \infty ; \end{aligned}$$

(B.3)

thus, [B.2] replaces [A.25] and [A.26]. The remaining boundary conditions are as before, except that here $\gamma =\gamma \left( c\right)$ and linearizes about $c=c^{0}$ so that

$$\begin{aligned} \gamma =\gamma ^{0}+\gamma_{c}\left( c-c^{0}\right) , \end{aligned}$$

(B.4)

where $\gamma ^{0}=\gamma \left( c^{0}\right) ,$ and $\gamma_{c}:=\left( d\gamma /dc\right)_{c=c^{0}}$ is a constant.

Nondimensionalizing with

$$\begin{aligned} C=\frac{c-c_{0}}{ac_{1}}, \end{aligned}$$

(B.5)

Eq. [B.1] gives, for $R>1,$

$$\begin{aligned} Le\left( U_{R}^{\left( l\right) }\frac{\partial C}{\partial R} +\frac{U_{\theta }^{\left( l\right) }}{R}\frac{\partial C}{\partial \theta }\right) =\frac{1}{R^{2}}\frac{\partial }{\partial R}\left( R^{2} \frac{\partial C}{\partial R}\right) +\frac{1}{R^{2}\sin \theta }\frac{\partial }{\partial \theta }\left( \sin \theta \frac{\partial C}{\partial \theta }\right) , \end{aligned}$$

(B.6)

subject to

$$\begin{aligned} C&\rightarrow R\cos \theta \quad \text {as }\, \it{R}\rightarrow \infty ,\end{aligned}$$

(B.7)

$$\begin{aligned} \frac{\partial C}{\partial R}&=0\quad {\text {at }}\,R=1, \end{aligned}$$

(B.8)

where Le denotes the Lewis number and is given by $Le=aV/D$. If $Le\ll 1,$ then

$$\begin{aligned} C=\left( R+\frac{1}{2R^{2}}\right) \cos \theta \end{aligned}$$

(B.9)

and

$$\begin{aligned} V=-\frac{\gamma_{c}ac_{1}}{2\mu ^{\left( l\right) }}. \end{aligned}$$

(B.10)

Eq. [B.9] looks identical to, and Eq. [B.10] very similar to, the second equation in [A.37] and Eq. [A.44], respectively; however, this similarity is only a coincidence that arises because $k\ll 1.$ From the computations, for the mushy region, $10^{-3}\le \left| Le\right| \le 10,$ meaning that C and V will not be as given by [B.9] and [B.10], respectively, for all of the mushy region. Resolving this inconsistency requires a more sophisticated approach to the governing equations, whereby the left-hand side of [B.6] is not neglected. This approach is beyond the scope here, but see, for example References 2 through 4 and 29.

Appendix C: Separation of Time Scales

The model clearly builds on the possible separation of time and length scales. While the latter is feasible, the former has yet to be justified. The velocities determined in Appendices A and B are the terminal values. Instead of [A.43], one has

$$\begin{aligned} \frac{4}{3}\pi a^{3}\rho ^{\left( g\right) }\frac{\rm{{d}}^{2}\it{y}_{b} }{\rm{{d}}\it{t}^{2}}=a^{2}\int_{0}^{\pi }\left( \sigma_{r\theta }^{\left( l\right) }\sin ^{2}\theta -\sigma_{rr}^{\left( l\right) }\sin \theta \cos \theta \right)_{r=a}{\rm{{d}}}\theta , \end{aligned}$$

and the left-hand side can be neglected if

$$\begin{aligned} \frac{a^{2}\rho ^{\left( g\right) }\left[ y_{b}\right] }{\left[ t\right] ^{2}}\ll \mu ^{\left( l\right) }V. \end{aligned}$$

Here, $\left[ y_{b}\right]$ is a characteristic length scale for the drift motion of the bubble and $\left[ t\right]$ is a characteristic time scale for this motion; it is appropriate to take $\left[ y_{b}\right] \sim W,$ $\left[ t\right] \sim L/V_{\rm cast},$ since the bubble clearly cannot move further than half-width of the casting, and it clearly cannot move through the mush for longer than the time taken for molten steel to pass from $z=0$ to $z=L.$ Then,

$$\begin{aligned} \frac{a^{2}\rho ^{\left( g\right) }WV_{\rm cast}^{2}}{L^{2}}\ll \mu ^{\left( l\right) }V. \end{aligned}$$

In the least favorable case,

$$\begin{aligned} \mu ^{\left( l\right) }V=\frac{1}{2}a\min \left( \gamma_{c}c_{1},\gamma_{T}T_{1}\right) , \end{aligned}$$

and thus

$$\begin{aligned} \frac{2a\rho ^{\left( g\right) }WV_{\rm cast}^{2}}{L^{2}}\ll \min \left( \gamma_{c}c_{1},\gamma_{T}T_{1}\right) . \end{aligned}$$

This inequality will be satisfied, since the left-hand side is around $10^{-14}$ kg m$^{-1}$s$^{-2},$ whereas

$$\begin{aligned} 24\,\text { kg\, m}^{-1}\,\text {s}^{-2}\ll \min \left( \gamma_{\it{c}}c_{1},\gamma_{T} \it{T}_{1}\right) \ll 1.6\times 10^{6}\,\text { kg \, m}^{-1}\,\text {s}^{-2}. \end{aligned}$$

Appendix D: Material Balance

To calculate the size of the particles TiN and Al$_{2}$O$_{3}$ based on material balance:

$$\begin{aligned} N \times \frac{4 \ \pi }{3} \times r^{3} = (x_{ss}-x_{eq}) \times V_{M} \end{aligned}$$

(D.1)

where N is the number of particles, r cm is the particle radius, $x_{ss},\ x_{eq}$ mol are the supersaturated and equilibrium concentrations, respectively, and $V_{M}$ cm$^{3}$/mol is the molar volume.

Supersaturated and equilibrium concentrations have been calculated as weight percentage. To change the supersaturated value to the number of moles, the following procedure is adopted:

According to random walk in one dimension, melt in between dendrite arms is uniform in concentration at constant solid fractions. Al$_{2}$O$_{3}$ and TiN precipitate at  92 pct solid. The width of the liquid at this solid fraction ranges from $5 \times 10^{-3}$ cm to $5 \times 10^{-4}$ cm based on DAS which changes between 90 and 600 $\mu$m. The number of moles at supersaturated and equilibrium concentrations can only be calculated at a certain defined volume. Hence a sphere with a diameter equal to DAS is considered since random walk is instantaneous at constant solid fractions. The volume fraction of Fe, Cr, and Ni in the sphere is calculated according to

$$\begin{aligned} {\rm{Fe}}_{\rm{Volume}\ {\rm{fraction}}} = \frac{\frac{\omega_{\rm{Fe}}}{\rho_{\rm{Fe}}}}{\frac{\omega_{\rm{Fe}}}{\rho_{\rm{Fe}}}+\frac{\omega_{\rm{Cr}}}{\rho_{\rm{Cr}}}+\frac{\omega_{\rm Ni}}{\rho_{\rm Ni}}}, \end{aligned}$$

(D.2)

where $\omega$ represents the weight fraction of the elements. The number of moles in the sphere is

$$\begin{aligned} {\rm{Total}}\ {\rm{number}} \ {\rm{of}}\ {\rm{moles}} = \frac{V_{\rm{sphere}}}{V_{M({\rm{Fe}})}+V_{M({\rm Ni})}+V_{M({\rm{Cr}})}}. \end{aligned}$$

(D.3)

The concentration of the material inside the sphere is assumed to be composed of Fe, Ni, and Cr since other alloying elements exist in dilute amounts. Mole fraction of Al, O, Ti, and N is calculated as

$$\begin{aligned} O_{\rm{mole}\ {\rm fraction}} = \frac{\frac{\omega_{\rm{O}}^{\rm{SS}}}{\rho_{\rm{O}} \times V_{M(\rm O)}} }{\frac{\omega_{\rm{Fe}}}{\rho_{\rm{Fe}} \times V_{M(\rm Fe)}}+\frac{\omega_{\rm Ni}}{\rho_{\rm Ni} \times V_{M(\rm Ni)}}+\frac{\omega_{\rm{Cr}}}{\rho_{\rm{Cr}} \times V_{M(\rm Cr)}}}, \end{aligned}$$

(D.4)

where SS denotes the supersaturated amount of the element, $\rho_{\rm{O}}$ is the oxygen density (g/cm$^{3}$), and $V_{\rm{O}}$ is the oxygen molar volume (cm$^{3}$/mol). The number of moles for Al, O, Ti, and N is calculated as

$$\begin{aligned} \rm{Oxygen}\ {\rm{moles}} &= \rm{Mole}\ {\rm{fraction}}\ \rm {oxygen} \\&\quad \times \,{\rm{Total}}\ \rm{number} \ \rm{of}\ {\text{moles}}. \end{aligned}$$

(D.5)

The formation of Al$_{2}$O$_{3}$ is controlled by concentration of aluminum since the fraction of oxygen over aluminum in the sphere is 1.09, while molar fraction of oxygen to aluminum in Al$_{2}$O$_{3}$ is 1.5. The formation of TiN is controlled by concentration of titanium since fraction of nitrogen to titanium in TiN is one while 1.2 in the sphere. The number of moles from Eq. [D.5] for aluminum and titanium is used in Eq. [D.1] as $x_{ss}-x_{eq}$. The radius of one particle of TiN which forms at  92 pct solid where DAS is 90 $\mu$m is $r=2.8 \times 10^{-5}$ cm and where DAS is 600 $\mu$m is $r=1.8 \times 10^{-4}$ cm. The radius of one particle of Al$_{2}$O$_{3}$ which forms at  92 pct solid where DAS is 90 $\mu$m is $r=3.4 \times 10^{-5}$ cm and where DAS is 600 $\mu$m is $r=2.2 \times 10^{-4}$ cm.