**3. Solution Procedure**

Here, we used the following dimensionless transformations:

$$\begin{aligned} \zeta &= \sqrt{\frac{s}{\upsilon\_f}} r, \ p = \rho\_f \text{s}^2 \text{x}^2 P(\zeta), \ T = T\_\infty (1 + (\theta\_w - 1)\theta), \\ \mathbb{C} = \mathbb{C}\_\infty &+ \frac{j\_w}{\mathcal{D}\_\mathbb{R}} \sqrt{\frac{\upsilon\_f}{s}} h(\zeta), \ u = \text{sx} f'(\zeta), \ v = -\frac{\mathbb{R}}{r + \mathbb{R}} \sqrt{\text{s}\upsilon\_f} f(\zeta) \end{aligned} \tag{11}$$

Here, prime denotes the derivative *w*, *r*, *T*, ζ and θ*<sup>w</sup>* = *T*f/*T*∞. The above transformation Equation (11) satisfies Equation (1) identically and Equations (2)–(6) are given by the following:

$$P' = \left(1 - \phi + \phi \frac{\rho\_s}{\rho\_f}\right) \frac{f'^2}{\zeta + K\_1} \tag{12}$$

$$\begin{split} \frac{1}{\left(1-\phi+\phi\frac{\varrho\_{2}}{\varrho\_{f}}\right)^{2}} \frac{2\underline{K\_{1}}}{\zeta+K\_{1}} P &= \frac{1}{\left(1-\phi\right)^{2\mathfrak{F}/10}\left(1-\phi+\phi\frac{\varrho\_{2}}{\varrho\_{f}}\right)} \left(f''' - \frac{f'}{\left(\underline{K\_{1}}+\zeta\right)^{2}} + \frac{f''}{\zeta+K\_{1}}\right) - \frac{\underline{K\_{1}}}{\zeta+K\_{1}} f'^{2} \\ &+ \frac{\underline{K\_{1}}}{\underline{K\_{1}}+\zeta} ff'' + \frac{\underline{K\_{1}}}{\left(\underline{K\_{1}}+\zeta\right)^{2}} f'f - Mf' \end{split} \tag{13}$$

$$\begin{split} \frac{1}{\text{Pr}} \Big( \frac{k\_{nf}}{k\_f} + R\_d (1 + \left( \Theta\_{\text{w}} - 1 \right) \boldsymbol{\theta} \big)^3 \Big) \Big( \boldsymbol{\theta}'' + \frac{1}{\zeta + K\_1} \boldsymbol{\theta}' \big) + \Big( 1 - \boldsymbol{\phi} + \Phi \frac{\left( \rho \mathcal{C}\_{\text{p}} \right)\_p}{\left( \rho \mathcal{C}\_{\text{p}} \right)\_f} \Big) \Big( \frac{K\_1}{\zeta + K\_1} f \boldsymbol{\theta}' \big) \\ \boldsymbol{\lambda}\_1 \boldsymbol{\theta} + 3 \boldsymbol{R}\_d (1 + \left( \Theta\_{\text{w}} - 1 \right) \boldsymbol{\theta} \big)^2 \boldsymbol{\theta}' = 0 \end{split} \tag{14}$$

$$h'' + \frac{1}{\zeta + K\_1}h' + S\_{\mathfrak{c}} \left(\frac{K\_1}{\zeta + K\_1}\right)fh' = 0\tag{15}$$

.

and

$$\begin{array}{ll} f(\zeta) = 0, \ f'(\zeta) = 1, \ \theta'(\zeta) = (1 - \theta(\zeta)) \text{Bi}, \ h'(\zeta) = -1, \ \text{as } \zeta = 0\\ f'(\zeta) \to 0, \ f''(\zeta) \to 0, \ \theta(\zeta) \to 0, \ h(\zeta) \to 0, \ \text{as } \zeta \to \infty \end{array} \tag{16}$$

$$\text{Here, } K\_1 = R\sqrt{\frac{s}{v\_f}}, \text{Bi} = \frac{h^\*\sqrt{\frac{v\_f}{s}}}{k\_f}, S\_c = \frac{v\_f}{D\_B}, R\_d = \frac{16\sigma^\* T\_\infty^{-3}}{3\text{k}\text{k}^\*}, \lambda\_1 = \frac{Q\_0}{s(\rho C\_p)\_f}, \text{and } \text{Pr} = \frac{v\_f}{\alpha\_f}$$

*Coatings* **2018**, *8*, 430

Eliminating pressure term from Equations (12) and (13) by differentiating Equation (13) with respect to ζ and then putting in Equation (12), we get the following:

$$\begin{split} f^{\text{iv}} + \frac{2f^{\prime\prime\prime}}{\zeta + K\_1} - \frac{f^{\prime\prime}}{\left(\zeta + K\_1\right)^2} + \frac{f^{\prime}}{\left(\zeta + K\_1\right)^3} + \left(1 - \Phi\right)^{25/10} \Big(1 - \Phi + \Phi \frac{p\_t}{\rho\_f}\Big) \Big\{\frac{K\_1}{\left(\zeta + K\_1\right)} \left(f^{\prime 2} - f f^{\prime \prime}\right) - \frac{K\_1}{\zeta + K\_1} \left(f^{\prime} f^{\prime \prime} - f f^{\prime \prime}\right)\Big) \\ - \frac{K\_1}{\left(\zeta + K\_1\right)^3} f f^{\prime \prime}\Big) - \left(1 - \Phi\right)^{25/10} M(\frac{f^{\prime}}{\zeta + K\_1} + f^{\prime \prime}) = 0 \end{split} \tag{17}$$

with

$$f(0) = 0, \ f'(0) = 1, \ f'(\infty) \to 0, \ f''(\infty) \to 0 \tag{18}$$

The surface drag force (*Cf*), Sherwood number (Sh*x*), and Nu*<sup>x</sup>* (Nusselt number) along *x*-direction are defined as follows:

$$\mathbf{C}\_{f} = \frac{\mathbf{\tau}\_{rx}}{\frac{1}{2}\rho u\_w^2}, \text{ Nu}\_x = \frac{\mathbf{x}q\_w}{k\_f(T\_f - T\_\infty)}, \text{ Sh}\_x = -\frac{\mathbf{x}}{(\mathbb{C} - \mathbb{C}\_\infty)}\frac{\partial \mathbb{C}}{\partial r} \tag{19}$$

where

$$
\pi\_{rx} = \mu\_{nf} (\frac{\partial u}{\partial r} - \frac{u}{r + R})\_{r=0}, \\
q\_w = (q\_r)\_w - (\frac{\partial T}{\partial r})\_{r=0} \tag{20}
$$

After putting Equations (11) and (20), Equation (19) becomes the following:

$$\begin{split} \frac{1}{2} \mathbb{C}\_{f} (\text{Re}\_{\text{x}})^{\frac{1}{2}} &= f''(0) - \frac{f'(0)}{K\_{1}} \, \bigvee\_{} \text{Nu}\_{\text{x}} (\text{Re}\_{\text{x}})^{-\frac{1}{2}} = -\left[\frac{k\_{af}}{K\_{f}} + \text{R}\_{d} (1 + (\theta\_{\text{w}} - 1)\theta(0))^{3}\right] \mathfrak{G}'(0), \\ &\text{Sh}\_{\text{x}} (\text{Re}\_{\text{x}})^{-\frac{1}{2}} = \frac{1}{h(0)} \end{split} \tag{21}$$

Here, Re*<sup>x</sup>* = *sx*<sup>2</sup> *vf* .
