**Appendix A**

For *α* = 0, the governing equations reduce to

$$f^{\prime\prime} + \frac{1}{2}\eta f^{\prime} = 0,\tag{A1}$$

$$
\theta'' + \frac{\text{Pr}}{2}\eta \theta' + N\_b \theta' \phi' + N\_l \theta'^2 = 0,\tag{A2}
$$

$$
\rho^{\prime\prime} + \frac{Le}{2} \eta \phi^{\prime} + \frac{N\_l}{N\_b} \theta^{\prime\prime} = 0. \tag{A3}
$$

We derive the analytical expressions for the skin friction, Nusselt number and Sherwood number subject to the boundary conditions in Equation (A1). The exact solution of Equation (A1) is:

$$f'(\eta) = 1 - \text{erf}\left(\frac{1}{2}\eta\right) \tag{A4}$$

It is noted that the magnitude of thermophoretic and Brownian diffusion parameters for nanoparticles is very small [1,9], therefore we consider *Nb* and *Nt* of *O*(*ε*), *ε* → 0 . We expand *θ* and *φ* in small parameter *ε* and write

$$\begin{cases} \theta = \theta\_0 + \varepsilon \theta\_1 + \dots \\ \phi = \phi\_0 + \varepsilon \phi\_1 + \dots \end{cases} \tag{A5}$$

By substituting the expressions in Equation (A5) in Equations (A2) and (A3), the leading order boundary value problem is given by

$$
\theta\_o'' + \frac{\text{Pr}}{2} \eta \theta\_o' = 0,\tag{A6}
$$

$$
\rho \phi\_o'' + \frac{L\mathcal{e}}{2} \eta \phi\_o' + \frac{\tau}{\mathcal{\beta}} \theta'' = 0,\tag{A7}
$$

where *β* and *τ* are constants of *O*(1) such that *Nb* = *βε* and *Nt* = *τε*. The solution of above boundary value problem can be written as

$$\begin{split} \theta\_o'(y) &= \frac{\sqrt{\text{Pr}}}{\sqrt{\pi}} e^{-\frac{1}{4}\text{Pr}\eta^2}, \\ \phi\_o'(y) &= -\frac{\text{Pr}\{\left(\sqrt{\text{Pr}}e^{-\frac{1}{4}\text{Pr}\eta^2} - \sqrt{\text{Tr}}e^{-\frac{1}{4}\text{Lip}^2}\right) + \beta(Lc - \text{Pr})\sqrt{\text{Tr}}c^{-\frac{1}{4}L\eta^2}}{\sqrt{\pi}\beta(Lc - \text{Pr})}. \end{split} \tag{A8}$$

The first order system can be written as

$$\frac{\partial^2 \theta\_1}{\partial y^2} + \frac{1}{2} \text{Pr}\eta \frac{\partial \theta\_1}{\partial y} + \beta \frac{\partial \theta\_o}{\partial y} \frac{\partial \phi\_0}{\partial y} + \tau \left(\frac{\partial \theta\_o}{\partial y}\right)^2 = 0 \tag{A9}$$

$$\frac{\partial^2 \phi\_1}{\partial y^2} + \frac{1}{2} L e \eta \frac{\partial \phi\_1}{\partial y} + \frac{\tau}{\beta} \frac{\partial^2 \theta\_1}{\partial y^2} = 0 \tag{A10}$$

with the boundary conditions

$$\begin{cases} \theta\_1 = 0, \,\phi\_1 = 0 \text{ at } \eta = 0\\ \theta\_1 = 0, \,\phi\_1 = 0 \text{ as } \eta \to \infty \end{cases} \tag{A11}$$

For the above boundary value problem, the exact solution is given by

$$\theta\_1'(\eta) = e^{-\frac{1}{4}\text{Pr}\eta^2} \left( 1 - \frac{\sqrt{\text{Pr}}}{\sqrt{\pi}} \beta \text{erf}\left(\frac{1}{2}\sqrt{L\epsilon\eta}\right) - \frac{\sqrt{\text{Pr}}}{\sqrt{\pi}} \frac{\text{r}}{L\epsilon - \text{Pr}} \left( \text{Le}\,\text{erf}\left(\frac{1}{2}\sqrt{\text{Pr}}\eta\right) - \text{Pref}\left(\frac{1}{2}\sqrt{L\epsilon\eta}\right) \right) \right) \tag{A12}$$

$$\begin{array}{l} \Phi\_{1}^{\prime}(\eta) = \frac{\mathsf{r}}{\mathsf{Pr}\,\overline{\mathsf{m}}\,(L\mathsf{e}-\mathsf{Pr})}e^{-\frac{1}{4}L\eta^{2}}\Big(\mathrm{Pr}\sqrt{\pi}e^{-\frac{1}{4}(\mathrm{Pr}-L\mathrm{e})\eta^{2}} + \frac{\mathsf{r}\mathrm{Pr}\,L\mathrm{e}\,\mathrm{erf}\left(\frac{1}{2}\sqrt{2\mathrm{Pr}-L\eta}\right)}{\sqrt{2\mathrm{Pr}-L\mathrm{e}}} -\\\sqrt{\mathrm{Pr}^{3}}\Big(\frac{(L\mathsf{e}-\mathsf{Pr})\beta-\mathrm{Pr}\,\mathrm{r}}{L\mathrm{e}-\mathrm{Pr}}\bigg)\Big(\mathrm{erf}\Big(\frac{1}{2}\sqrt{L}\mathrm{e}\eta\Big)e^{-\frac{1}{4}(\mathrm{Pr}-L\mathrm{e})\eta^{2}} - \frac{\sqrt{L}\mathsf{e}}{\sqrt{\mathrm{Pr}}}\mathrm{erf}\Big(\frac{1}{2}\sqrt{\mathrm{Pr}}\eta\Big)\Big) + ((L\mathsf{e}-\mathrm{Pr})\beta-\mathrm{r}\mathrm{Pr}) \\\sqrt{L}\mathrm{e}\,\mathrm{erf}\Big(\frac{1}{2}\sqrt{\mathrm{Pr}}\eta\Big) - \mathrm{r}\frac{\sqrt{L\mathrm{e}^{3}\mathrm{Pr}}}{L\mathrm{e}-\mathrm{Pr}}\bigg(\sqrt{L}\mathrm{e}\,\mathrm{erf}\Big(\frac{1}{2}\sqrt{\mathrm{Pr}}\eta\Big)e^{-\frac{1}{4}(\mathrm{Pr}-L\mathrm{e})\eta^{2}} - \sqrt{\mathrm{Pr}}\mathrm{erf}\Big(\frac{1}{2}\sqrt{L\mathrm{e}}\ \eta\Big)\Big), \end{array} \tag{A13}$$

where erf is the error function and erfi is the imaginary error function.

### **References**


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