Thermal transport of combined electroosmotically and pressure driven nanofluid flow in soft nanochannels

Abstract

In this paper, the heat transfer characteristics of the nanofluid through a parallel plate soft nanochannel are investigated under the fully developed condition. The flow is actuated by the combined effects of pressure gradient and implied electric field. Based on the ion partitioning effect and the Debye–Hückel linearization, the analytical solutions for electrokinetic flow in such nanochannel are obtained. Meanwhile, the uniform wall heat flux is utilized in the analysis, and the influences of viscous dissipation and the Joule heating are taken into account. The results for pertinent dimensionless parameters are presented graphically and discussed in brief. The relevant result reveals that the ion partitioning effect can greatly impact the electrostatic potential, velocity and temperature distribution. Furthermore, this ion partitioning effect can exert an influence on heat transfer of the nanofluid. The present study also indicates the possibility of alteration in the nanofluid heat transfer by the use of nanoparticle volume.

Appendix

The coefficients in Eqs. (41) and (42) are presented as follows:

$$ \begin{aligned} f_{1} (\bar{y}) & = \frac{1}{{(\beta_{11} + \beta_{21} )}}\left[ {( - 1 + S + \overline{Br} (\beta_{12} + \beta_{22} ) + \alpha^{2} \overline{Br} \beta_{23} } \right] \\ & \quad \left( {\frac{1}{24}\eta \bar{y}^{4} + u_{r} \eta A_{1} \bar{\lambda }^{2} \cos \,h\left( {\frac{{\bar{y}}}{{\bar{\lambda }}}} \right) + \frac{1}{2}B_{2} \bar{y}^{2} } \right) \\ & \quad - \frac{1}{2}S\bar{y}^{2} - \overline{Br} \left[ {\frac{1}{12}\eta^{2} \bar{y}^{4} + \frac{1}{2}\left(\frac{1}{{\bar{\lambda }}}u_{r} \eta A_{1} \right)^{2} } \right. \\ & \quad \left( {\frac{{\bar{\lambda }^{2} }}{4}\cos \,h\left( {\frac{{2\bar{y}}}{{\bar{\lambda }}}} \right) - \frac{1}{2}\bar{y}^{2} } \right) \\ & \quad \left. { - 2u_{r} \eta^{2} A_{1} \left( {\bar{\lambda }\bar{y}\sin \,h\left( {\frac{{\bar{y}}}{{\bar{\lambda }}}} \right) - 2\bar{\lambda }^{2} \cos \,h\left( {\frac{{\bar{y}}}{{\bar{\lambda }}}} \right)} \right)} \right] + K_{1} \bar{y} \\ \end{aligned} $$

$$ \begin{aligned} f_{2} (\bar{y}) & = \frac{1}{{(\beta_{11} + \beta_{21} )}}[( - 1 + S + \overline{Br} (\beta_{12} + \beta_{22} ) + \alpha^{2} \overline{Br} \beta_{23} ] \\ & \quad \left( {\frac{{D_{1} }}{{a^{2} }}\cos \,h(a\bar{y}) + \frac{{D_{2} }}{a}\sin \,h(a\bar{y}) + \frac{{C_{1} }}{{b^{2} }}\cos \,h(b\bar{y})} \right. \\ & \quad \left. { + \frac{{C_{2} }}{{b^{2} }}\sin \,h(b\bar{y}) + \frac{{C_{3} }}{2}\bar{y}^{2} } \right) - \frac{1}{2}S\bar{y}^{2} \\ & \quad - \overline{Br} \left[ {\frac{1}{2}(D_{1} a)^{2} \left( {\frac{1}{{4a^{2} }}\cos \,h(2a\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) + \frac{1}{4}C_{1} C_{2} \sin \,h(2b\bar{y})} \right. \\ & \quad + \frac{1}{2}(D_{2} a)^{2} \left( {\frac{1}{{4a^{2} }}\cos \,h(2a\bar{y}) + \frac{1}{2}\bar{y}^{2} } \right) + \frac{1}{4}D_{1} D_{2} \sin \,h(2a\bar{y}) \\ & \quad + \frac{1}{2}(C_{1} b)^{2} \left( {\frac{1}{{4b^{2} }}\cos \,h(2b\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) + \frac{1}{2}(C_{1} b)^{2} \left( {\frac{1}{{4b^{2} }}\cos \,h(2b\bar{y}) + \frac{1}{2}\bar{y}^{2} } \right) \\ & \quad + \frac{{(D_{1} C_{1} + D_{2} C_{2} )ab}}{{(a + b)^{2} }}\cos \,h((a + b)\bar{y}) + \frac{{(D_{2} C_{2} - D_{1} C_{1} )ab}}{{(a - b)^{2} }}\cos \,h((a - b)\bar{y}) \\ & \quad + \frac{{(D_{1} C_{2} + D_{2} C_{1} )ab}}{{(a + b)^{2} }}\cos \,h((a + b)\bar{y}) + \frac{1}{2}D_{2}^{2} \left( {\frac{1}{{4a^{2} }}\cos \,h(2a\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) \\ & \quad + \frac{{D_{1} D_{2} }}{{4a^{2} }}\sin \,h(2a\bar{y}) + \frac{1}{2}C_{2}^{2} \left( {\frac{1}{{4b^{2} }}\cos \,h(2b\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) \\ & \quad + \frac{1}{2}C_{2}^{2} \left( {\frac{1}{{4b^{2} }}\cos \,h(2b\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) + \frac{{C_{1} C_{2} }}{{4b^{2} }}\sin \,h(2b\bar{y}) \\ & \quad + \frac{1}{2}C_{3}^{2} \bar{y}^{2} + \frac{{(D_{1} C_{2} - D_{2} C_{1} )ab}}{{(a + b)^{2} }}\sin \,h((a - b)\bar{y}) \\ & \quad - \overline{Br} \alpha^{2} \left( {\frac{1}{2}D_{1}^{2} \left( {\frac{1}{{4a^{2} }}\cosh (2a\bar{y}) + \frac{1}{2}\bar{y}^{2} } \right)} \right. \\ & \quad + \frac{1}{2}D_{2}^{2} \left( {\frac{1}{{4a^{2} }}\cos \,h(2a\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) \\ & \quad + \frac{{D_{1} D_{2} }}{{4a^{2} }}\sin \,h(2a\bar{y}) + \frac{1}{2}C_{2}^{2} \left( {\frac{1}{{4b^{2} }}\cos \,h(2b\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) \\ & \quad + \frac{{C_{1} C_{2} }}{{4b^{2} }}\sin \,h(2b\bar{y}) + \frac{1}{2}C_{3}^{2} \bar{y}^{2} + \frac{1}{2}C_{2}^{2} \left( {\frac{1}{{4b^{2} }}\cos \,h(2b\bar{y}) - \frac{1}{2}\bar{y}^{2} } \right) \\ & \quad + \frac{{(D_{1} C_{1} + D_{2} C_{2} )}}{{(a + b)^{2} }}\cos \,h((a + b)\bar{y}) + \frac{{(D_{1} C_{1} - D_{2} C_{2} )}}{{(a - b)^{2} }}\cos \,h((a - b)\bar{y}) \\ & \quad + \frac{{(D_{1} C_{2} + D_{2} C_{1} )}}{{(a + b)^{2} }}\sin \,h((a + b)\bar{y}) + \frac{{(D_{2} C_{1} - D_{1} C_{2} )}}{{(a + b)^{2} }}\sin \,h((a - b)\bar{y}) \\ & \quad \left. { + \frac{{2C_{3} D_{1} }}{{a^{2} }}\cos \,h(a\bar{y}) + \frac{{2C_{3} D_{2} }}{{a^{2} }}\sin \,h(a\bar{y})} \right) + K_{3} \bar{y} \\ \end{aligned} $$

where $K_{1} = 0$, $K_{2} = f_{2} (\bar{d} - 1) - f_{2} ( - 1) - f_{1} (\bar{d} - 1)$, $K_{3} = \frac{{{\text{d}}f_{1} (\bar{y})}}{{{\text{d}}\bar{y}}}\left| {_{{\bar{y} = - 1 + \bar{d}}} } \right. - g(\bar{y})\left| {_{{\bar{y} = - 1 + \bar{d}}} } \right.$, $K_{4} = - f_{2} ( - 1)$, and $g(\bar{y}) = \frac{{{\text{d}}f_{2} (\bar{y})}}{{{\text{d}}\bar{y}}} - K_{3}$.