Thermal conductivity of fluid as described by \cite{dada2016} in following way:
\(k=k_\infty\left(1+a\theta\right)\) here \(a=\frac{K_w-K_\infty}{K_\infty}\), Where \(a\) presents thermal conductivity parameter, \(b, c, T_r\) are means constants. Where, numerical calculations for liquid are obtained when \(b>0\) and \(b<0\) for gases.

Solution Procedures

Following similarity transformations of   \cite{radiation2015} ,we convert the partial differential equation with associated boundary conditions are in ordinary differential equations:
\[u=cxf',\ v=-\sqrt{c}\nu_\infty f\left(\eta\right),\ \eta\left(y\right)=-\sqrt\frac{{c}}{\nu_\infty}y, \ \theta(\eta)=\frac{T-T_\infty}{T_w-T_\infty}, \ \ \ \ \ \ \ \ \ \ \ \ \xi(x)=\sqrt\frac{c}{\nu_\infty}x \]The continuity equation satisfied after defining the velocity component as follows  \(u=\frac{\delta \psi}{\delta y}\) and \(v=\frac{\delta \psi}{\delta x}\)    and the remaining momentum and energy equations are in following form:
\[f'''-\frac{\theta'}{\theta-\theta_r}f''-\frac{\theta-\theta'}{\theta_r}ff'+\frac{\theta-\theta'}{\theta_r}f'^2+\frac{\theta-\theta'}{\theta_r}Mf'+\frac{\theta-\theta'}{\theta_r}\left(\frac{2\beta}{(\eta+\alpha)^4}-\lambda_1\right)\theta=0\]
\[\left(1+a\theta\right)\theta''+a\theta'^2+\Pr f\theta'-2\beta\lambda(\epsilon+\theta)\frac{f}{(\eta+\alpha)^3}=0\]