A. Appendix I

The applied solution is based on the implicit scheme as described by Patankar in 1980 [23]. Figure 3 shows the control volumes and the nodal points upon which the numerical solution was based; in the graph, only a very rough discretization was considered with 6 � 6 nodal points just for illustrative purposes. Normally, for nodal points in the interior of the geometry under investigation, the implicit scheme works on five adjacent points; in this way, for every internal point P, the four adjacent nodal points of interest are nominated as E (east), W (west), N (north), and S (south). The numerical discretization equation for the heat transfer equation at P is then given by Eq. (14)

$$
\sigma\_P T\_P = \sigma\_E T\_E + \sigma\_W T\_W + \sigma\_N T\_N + \sigma\_S T\_S + b\_P \tag{14}
$$

where the temperature coefficients and the constant term are given by Eq. (15):

$$\begin{aligned} a\_E &= \frac{k\_\epsilon \Delta \mathbf{y}}{(\delta \mathbf{x})\_\epsilon}, \quad a\_W = \frac{k\_w \Delta \mathbf{y}}{(\delta \mathbf{x})\_w}, \quad a\_N = \frac{k\_n \Delta \mathbf{x}}{(\delta \mathbf{y})\_n}, \quad a\_S = \frac{k\_s \Delta \mathbf{x}}{(\delta \mathbf{y})\_s}, \\\ a\_P &^0 = \frac{\rho c \Delta \mathbf{x} \Delta \mathbf{y}}{\Delta t}, \quad b\_P = \mathbf{S}\_C \Delta \mathbf{x} \, \Delta \mathbf{y} + a\_P \, ^0 T\_P \, ^0 \mathbf{y}, \\\ a\_P &= a\_E + a\_W + a\_N + a\_S + a\_P ^0 - \mathbf{S}\_P \, \Delta \mathbf{x} \, \Delta \mathbf{y} \end{aligned} \tag{15}$$

The described formulation is proven to satisfy the energy balance within the control volume that is included by the sides "n," "e," "w," and "s." Extra care is required in the derivation of the discretization equations at the borders of the studied geometry. For example, Eq. (16) describes the formulation for the temperature of a typical point on the outer area of the octagonal billet that is cooled by the water-cooled copper mold; the heat transfer is driven by a lower water temperature (Tf), by a rate given by a heat transfer coefficient h:

$$\begin{aligned} a\_C T\_C &= a\_G T\_G + a\_D T\_D + a\_F T\_F + b\_C\\ a\_G &= \frac{k\_w \Delta \mathfrak{y}}{\left(\delta \mathfrak{x}\right)\_w}, \ a\_D = \frac{k\_n \left(\Delta \mathfrak{x}/2\right)}{\left(\delta \mathfrak{y}\right)\_n}, \ a\_F = \frac{k\_s (\Delta \mathfrak{x}/2)}{\left(\delta \mathfrak{y}\right)\_s} \end{aligned}$$

$$a\_C^{-0} = \frac{\rho c \,\Delta vol}{\Delta t}, \ b\_C = S\_C \,\Delta vol + a\_C^{-0} \, T\_C{}^0 + h \, T\_f \, \Delta \mathfrak{y} \tag{16}$$

$$a\_C = a\_G + a\_D + a\_F + a\_C{}^0 - S\_P \, \Delta vol + h \, \Delta \mathfrak{y}$$

$$\Delta vol = (\Delta \mathfrak{x} \, \Delta \mathfrak{y}/2)$$

Similarly, another very interesting point of analysis on the derivation of the discretization equation is at a diagonal point A; the formulation is presented by Eq. (17); in this case, there is no heat transfer (due to symmetry) in the area above the diagonal (adiabatic formulation):

A Numerical Solution Model for the Heat Transfer in Octagonal Billets DOI: http://dx.doi.org/10.5772/intechopen.84305

$$\begin{aligned} a\_A T\_A &= a\_B T\_B + a\_N T\_N + b\_A \\ a\_B &= \frac{k\_\epsilon \left(\Delta \mathfrak{y}/2\right)}{\left(\delta \mathfrak{x}\right)\_\epsilon}, \quad a\_N = \frac{k\_\epsilon \left(\Delta \mathfrak{x}/2\right)}{\left(\delta \mathfrak{y}\right)\_\epsilon}, \\ a\_A \,^0 &= \frac{\rho c \,\Delta \nu o l}{\Delta t}, \quad b\_A = \mathrm{S}\_C \,\Delta \nu o l + a\_A \,^0 T\_A \,^0 \\ a\_A &= a\_B + a\_N + a\_A \,^0 - \mathrm{S}\_P \,\Delta \nu o l, \quad \Delta \nu o l = \left(\Delta \mathfrak{x} \Delta \mathfrak{y}/4\right) \end{aligned} \tag{17}$$
