**2. Example 1: regular perturbation method to thermal analysis of convective-radiative fin with end cooling and thermal contact resistance**

Consider a convective-radiative fin of temperature-dependent thermal conductivity *k(T),* length L and thickness δ, exposed on both faces to a convective environment at temperature *T*<sup>∞</sup> and a heat transfer co-efficient *h* subjected to magnetic field shown in **Figure 1**. The dimension *x* pertains to the length coordinate which has its origin at the tip of the fin and has a positive orientation from the fin tip to the fin base. In order to analyze the problem, the following assumptions are made. The following assumptions were made in the development of the model


Applying thermal energy balance on the fin and using the above model assumptions, the following nonlinear thermal model is developed

$$\frac{d}{d\mathbf{x}}\left[\left[1+\lambda(T-T\_{a})\right]\frac{dT}{d\mathbf{x}}\right]-\frac{h}{k\_{d}\delta}(T-T\_{a})-\frac{\sigma\varepsilon}{k\_{d}\delta}\left(T^{4}-T\_{a}^{4}\right)-\frac{\sigma B\_{o}^{2}u^{2}}{k\_{d}A\_{cr}}(T-T\_{a})=\mathbf{0} \tag{1}$$

The boundary conditions are

$$\infty = 0, \quad -k(T)\frac{\partial T}{\partial \mathbf{x}} = h\_{\epsilon}(T - T\_{a}) + \sigma \in \left(T^{4} - T\_{a}^{4}\right) \tag{2}$$

$$\infty = L, \quad -k(T)\frac{\partial T}{\partial \mathbf{x}} = h\_t(T\_b - T) + \sigma \in \left(T^4 - T\_a^4\right) \tag{3}$$

**Figure 1.**

*(a) Schematic of the convective-radiative longitudinal straight fin with magnetic field. (b) Schematic of the longitudinal straight fin geometry showing thermal contact resistance and boundary conditions.*

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Considering a case when a small temperature difference exists within the material during the heat flow. This actually necessitated the use of temperature-invariant physical and thermal properties of the fin. Also, it has been established that under such scenario, the term *T4* can be expressed as a linear function of temperature. Therefore, we have

$$T^4 = T\_a^4 + 4T\_a^3(T - T\_a) + 6T\_a^2(T - T\_a)^2 + \dots \cong 4T\_a^3T - 3T\_a^4 \tag{4}$$

On substituting Eq. (4) into Eq. (1), one arrives arrived at

$$\frac{d}{d\mathbf{x}}\left[\left[1+\lambda(T-T\_{\infty})\right]\frac{dT}{d\mathbf{x}}\right]-\frac{h}{k\_{a}\delta}(T-T\_{d})-\frac{4\sigma\varepsilon T\_{d}^{3}}{k\_{a}\delta}(T-T\_{d})-\frac{\sigma B\_{o}^{2}u^{2}}{k\_{a}A\_{c\tau}}(T-T\_{d})=0.\tag{5}$$

The boundary conditions

$$\infty = 0, \quad -k(T)\frac{\partial T}{\partial \mathbf{x}} = h\_{\epsilon}(T - T\_{a}) + 4\sigma \in T\_{a}^{3}(T - T\_{a}) \tag{6}$$

$$\infty = L, \quad -k(T)\frac{\partial T}{\partial \mathbf{x}} = h\_c(T\_b - T) + 4\sigma \in T\_a^3(T - T\_a) \tag{7}$$

On introducing the following dimensionless parameters in Eq. (8) into Eq. (5),

$$\begin{split} X &= \frac{x}{L}, \; \theta = \frac{T - T\_a}{T\_b - T\_a}, \; Ra = \frac{gk\beta(T\_b - T\_a)b}{a\nu k\_r}, \; N = \frac{4\sigma\_d b \, T\_a^3}{k\_a}, \; Ha = \frac{\sigma B\_0^2 u^2}{k\_a A\_{cr}}, \\ Bi\_\varepsilon &= \frac{h\_\varepsilon b}{k\_a}, \; Bi\_\varepsilon = \frac{h\_\varepsilon b}{k\_a}, \; \; M^2 = \frac{hb^2}{k\_a \delta}, \; \; \varepsilon = \lambda (T\_b - T\_a) Bi\_{\varepsilon \, eff} = \frac{(h\_\varepsilon + \sigma \varepsilon)b}{k\_a}, \\ Bi\_{\varepsilon \, eff} &= \frac{(h\_\varepsilon + \sigma \varepsilon)b}{k\_a} \end{split} \tag{8}$$

The dimensionless form of the governing Eq. (5) is arrived at as

$$\frac{d}{dX}\left[ (\mathbf{1} + \epsilon \theta) \frac{d\theta}{dX} \right] - M^2 \theta - Nr \theta - Ha \theta = 0 \tag{9}$$

On expanding Eq. (9), one has

$$\frac{d^2\theta}{dX^2} + \epsilon\theta \frac{d^2\theta}{dX^2} + \varepsilon \left(\frac{d\theta}{dX}\right)^2 - M^2\theta - Nr\theta - Ha\theta = 0\tag{10}$$

The boundary conditions are

$$X = \mathbf{0}, \quad (\mathbf{1} + \epsilon \theta) \frac{d\theta}{dX} = -Bi\_{\epsilon, \epsilon \varnothing} \theta \tag{11}$$

$$X = \mathbf{1}, \quad (\mathbf{1} + \epsilon \theta) \frac{d\theta}{dX} = -Bi\_{c, \theta \overline{f}} (\mathbf{1} - \theta) \tag{12}$$
