3. Theoretical solution

There is no published work on the theoretical solution for the octagonal billet until now. However, a theoretical solution can be approximated by solving the DE of heat transfer on an equivalent circular geometry. According to Figure 1, the selected radius of that circle was Re = 156.6 mm, as given by Eq. (5):

$$R\_{\varepsilon} = \frac{1 + \cos\left(\pi/8\right)}{2} R = 0.962\ 162.84 = 156.6\ \text{mm} \tag{5}$$

This radius is actually the mean value between the circumradius and in-radius of the octagon [25]. The circumradius R being equal to 162.84 mm makes the currently studied octagonal billet equivalent in size to a 250 � 300 mm<sup>2</sup> billet produced in the Stomana plant, at Pernik, Bulgaria. Now the original problem can be converted into a heat-transfer problem of cylindrical geometry, and the DE together with the boundary conditions is formulated according to Eq. (6):

$$\begin{aligned} \frac{\partial T}{\partial t} &= a \left( \frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial r} \right) \\\\ T &= T\_0, \quad t = 0; \quad \frac{\partial T}{\partial r} = 0, \quad r = 0; \quad -k \frac{\partial T}{\partial r} = h \left( T - T\_f \right), \quad r = R\_\epsilon \end{aligned} \tag{6}$$

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

By variable transformation using Eq. (7)

$$
\Theta = \frac{T - T\_f}{T\_0 - T\_f} \tag{7}
$$

the boundary condition at r=Re becomes:

$$\frac{\partial \Theta}{\partial r} + \frac{h}{k} \Theta = 0, \ r = R\_\epsilon \tag{8}$$

In order to fulfill the boundary condition at r=Re according to Eq. (8), Eq. (9) has to be solved for β:

$$
\beta J\_1(\beta) = A \, J\_0(\beta), \,\,\, A = \frac{R\_\epsilon \, h}{k} \tag{9}
$$

where J0 and J1 are the Bessel functions of zero and first order, respectively.

This theoretical solution is presented in the book of Carslaw and Jaeger [26] and is followed here. There are an infinite number of values for β, which solve Eq. (9); nevertheless, the first six roots [26] are enough for the computations. In this way, the derived solution arrives in the form of Eq. (10):

$$\begin{aligned} \Theta &= \frac{T - T\_f}{T\_0 - T\_f} = \sum\_{n=1}^{\infty} \exp\left(-\beta\_n^{-2} \xi\right) \frac{2A \, f\_0(\beta\_n \, r/R\_\epsilon)}{\left(\beta\_n^{-2} + A^2\right) f\_0(\beta\_n)}\\ \xi &= \frac{a \, t}{R\_\epsilon}, \quad A = \frac{R\_\epsilon \, h}{k} \end{aligned} \tag{10}$$

Figure 2. Theoretical predictions of the temperature distribution in various positions in a cylindrical billet.

Eq. (10) was applied in order to compute the temperature distribution T(r,t) inside the equivalent to the octagonal geometry cylindrical billet; some results are depicted in Figure 2. In fact, the following parameters were used in these computations: thermal conductivity, k = 50 W/m/K; density, ρ = 7600 kg/m<sup>3</sup> ; heat capacity, <sup>c</sup> = 400 J/kg/K; thermal diffusivity, <sup>α</sup> <sup>=</sup> k/(ρc) = 1.645 ˜ <sup>10</sup>°5 m2 /s; heat transfer coefficient, h = 200 W/m/K; cooling water temperature,Tf = 30°C, and initial billet temperature,T0 = 500°C.
