Insight into the dynamics of blood conveying alumina nanoparticles subject to Lorentz force, viscous dissipation, thermal radiation, Joule heating, and heat source

Abstract

This analysis studies the impact of the pulsating flow of Al$_2$O$_3$-blood non-Newtonian nanofluid in a channel in the presence of the magnetic field and thermal radiation. Viscous dissipation and Joule heating effects are taken into account. Blood is taken as Oldroyd-B fluid (base fluid) and Al$_2$O$_3$ as nanoparticles. The present study is important in engineering and biological models. The walls of channel are assumed to be semi-infinite in length. Assumed that the flow is fully developed and induced by a pressure gradient. Analytical solutions for flow variables are obtained using the perturbation method. The influence of different parameters on temperature and rate of heat transfer have been analysed through graphical results. The results reveal that the temperature of nanofluid accelerates by increasing viscous dissipation and heat source and frequency parameter. Further, the rate of heat transfer enhances with an increase in nanoparticle volume fraction and viscous dissipation.

Appendix

$B_1 = \frac{{A_2 }}{{A_1 }};\quad B_2 = \frac{{A_5 M^2 }}{{A_1 }};\quad B_3 = \frac{{B_2 }}{{B_1 }};\quad B_4 = - \frac{H^2}{{A_1 B_1 }};B_6 = \frac{{B_4 (e^{ - \sqrt{B_3 } } - 1)}}{{B_3 (e^{ - \sqrt{B_3 } } - e^{\sqrt{B_3 } } )}};$ $B_5 = \frac{{B_4 }}{{B_3 }} - B_6;$ $\quad B_7 = (\frac{{M^2 A_5 }}{{A_1 }} + iH^2 )\beta ^2;\quad B_8 = \frac{{B_7 }}{{B_1 }}; \quad B_9 \!=\! - \frac{{H^2\beta ^2 }}{{A_1 B_1 }};$ $B_{11} \!=\! \frac{{B_9 (e^{ - \sqrt{B_8 } } - 1)}}{{B_8 (e^{ - \sqrt{B_8 } } - e^{\sqrt{B_8 } } )}};$ $B_{10} = \frac{{B_9 }}{{B_8 }} - B_{11};\quad C_{1} = \frac{{A_4 }}{{A_3 }} + \frac{4}{3}\frac{1}{{A_3 }}Rd;$ $C_{2} = -\frac{Q \Pr }{A_3};$ $C_{3}=C_2/C_1; \qquad C_{4}=-\frac{A_2}{A_3C_1}Ec\Pr ;\qquad C_{5}=-\frac{A_5}{A_3C_1}Ec\Pr M^2;$ $C_6=C_4B_5^2B_3+C_5B_5^2;\qquad C_7=C_4B_6^2B_3+C_5B_6^2;\qquad C_8=-2C_5\frac{B_4}{B_3}B_5;$    $C_9=-2C_5\frac{B_4}{B_3}B_6;\quad C_{10}=-2C_4B_5B_6B_3+C_5\frac{B_4^2}{B_3^2}+2C_5B_5B_6;$      $C_{13}=\frac{C_6}{4B_3-C_3};\qquad C_{14}=\frac{C_7}{4B_3-C_3};$ $C_{15}=\frac{C_8}{B_3-C_3};\qquad C_{16}=\frac{C_9}{B_3-C_3};\qquad C_{17}=-\frac{C_{10}}{C_{3}};\qquad C_{12}$ $=\{1-[C_{13}(e^{-2\sqrt{B_3}}-e^{-\sqrt{C_3}})+C_{14}(e^{2\sqrt{B_3}}-e^{-\sqrt{C_3}})$ $+C_{15}(e^{-\sqrt{B_3}}-e^{-\sqrt{C_3}})+C_{16}(e^{\sqrt{B_3}}-e^{-\sqrt{C_3}})+C_{17}(1-e^{-\sqrt{C_3}})]\}/{e^{\sqrt{C_3}}-e^{-\sqrt{C_3}}};$ $C_{11}=-(C_{12}+C_{13}+C_{14}+C_{15}+C_{16}+C_{17});\quad C_{18}=iH^2\Pr -\frac{Q\Pr }{A_3};\quad C_{19}=\frac{C_{18}}{C_1};\quad C_{20}=2C_{4}B_5B_{10}\sqrt{B_3B_8};$   $C_{21}=-2C_{4}B_5B_{11}\sqrt{B_3B_8};$   $C_{22}=-2C_{4}B_6B_{10}\sqrt{B_3B_8};$   $C_{23}=2C_{4}B_6B_{11}\sqrt{B_3B_8};$   $C_{24}=2C_5B_5B_{10};\quad C_{25}=2C_5B_5B_{11};\quad C_{26}=2C_5B_6B_{10};\quad$ $C_{27}=2C_5B_6B_{11};\quad C_{28}\!=\!-2C_5B_5\frac{B_9}{B_8};\quad C_{29}\!=\!-2C_5B_6\frac{B_9}{B_8}; \quad$ $C_{30}=-2C_5B_{10}\frac{B_4}{B_3};\quad C_{31}=-2C_5B_{11}\frac{B_4}{B_3}; \quad C_{32}=2C_5\frac{B_4}{B_3}$ $\frac{B_9}{B_8}; \quad C_{35}=\frac{C_{20}+C_{24}}{(\sqrt{B_3}+\sqrt{B_8})^2-C_{19}}; \quad C_{36}=\frac{C_{21}+C_{25}}{(\sqrt{B_3}-\sqrt{B_8})^2-C_{19}}; \quad$ $C_{37}=\frac{C_{22}+C_{26}}{(\sqrt{B_3}-\sqrt{B_8})^2-C_{19}};$    $C_{38}=\frac{C_{23}+C_{27}}{(\sqrt{B_3}+\sqrt{B_8})^2-C_{19}}; C_{39}=\frac{C_{28}}{B_3-C_{19}}; C_{40}=\frac{C_{29}}{B_3-C_{19}};\quad C_{41}=\frac{C_{30}}{B_8-C_{19}};\quad C_{42}=\frac{C_{31}}{B_8-C_{19}};\quad$ $C_{43}=\frac{-C_{32}}{C_{19}};\quad C_{34}=\{-[C_{35}(e^{-(\sqrt{B_3}+\sqrt{B_8})}-e^{-\sqrt{C_{19}}})$ $+C_{36}(e^{-(\sqrt{B_3}-\sqrt{B_8})}-e^{-\sqrt{C_{19}}})+C_{37}(e^{(\sqrt{B_3}-\sqrt{B_8})}-e^{-\sqrt{C_{19}}})+C_{38}(e^{(\sqrt{B_3}+\sqrt{B_8})}-e^{-\sqrt{C_{19}}})$ $+\,C_{39}(e^{-\sqrt{B_3}}-e^{-\sqrt{C_{19}}})+C_{40}(e^{\sqrt{B_3}}-e^{-\sqrt{C_{19}}})+C_{41}(e^{-\sqrt{B_8}}-e^{-\sqrt{C_{19}}})$ $+\,C_{42}(e^{\sqrt{B_8}}-e^{-\sqrt{C_{19}}})+C_{43}(1-e^{-\sqrt{C_{19}}})]\}/{e^{\sqrt{C_{19}}}-e^{-\sqrt{C_{19}}}};$   $C_{33}=-(C_{34}+C_{35}+C_{36}+C_{37}+C_{38}+C_{39}+C_{40}+C_{41}+C_{42}+C_{43})$