Fractional modeling and synchronization of ferrofluid on free convection flow with magnetolysis

Abstract

The dispersion of ferromagnetism in free convection flow can lead the magnetization process in reduction due to misalignments of the magnetic domains. In this context, an intensive viscoelastic model is subjected to the magnetization process through non-integer-order differentiation based on singular kernel. The geometry of the problem is tackled for vertical tunnel on the basis of width d saturated by porous medium in which oscillating pressure gradient is invoked. The non-fractional governing equations have been treated for dimensionality of homogeneity. The fractionalized solutions for dimensionless velocity, temperature and Nusselt number have been investigated by employing the techniques of Laplace transforms with its inversion. The magnetized mathematical model has been disseminated for the sake of physical parameters subject the variants in fractional parameter. Finally, our results have been emphasized for rheological parameter so-called Peclet number, Reynolds number, magnetic parameter, Grashof number, Prandtl number on the variants of fractional time Caputo parameter.

Appendix

$$\begin{aligned} F(a_{0},s,b_{0},c_{0})= & {} \frac{\sinh [a_{0}\sqrt{s+b_{0}}]}{s\sinh [c_{0}\sqrt{s+b_{0}}]}\nonumber \\= & {} \sum _{k=0}^{\infty }\Bigg [\frac{1}{q} e^{-[(2k+1)c_{0}-a_{0}]\sqrt{s+b_{0}}}-\frac{1}{q}e^{-[(2k+1)c_{0}+a_{0}]\sqrt{s+b_{0}}}\Bigg ],\nonumber \\ f(a_{0},t,b_{0},c_{0})= & {} L^{-1}\left\{ F(a_{0},s,b_{0},c_{0})\right\} \nonumber \\= & {} \displaystyle \sum _{k=0}^{\infty }[\psi _{k}(a_{0},t,b_{0},c_{0})- \psi _{k}(-a_{0},t,b_{0},c_{0})], \end{aligned}$$

(A1)

$$\begin{aligned} \displaystyle \psi _{k}(a_{0},t,b_{0},c_{0})= & {} \frac{1}{2}\Big [e^{-[(2k+1)c_{0}-a_{0}]\sqrt{b_{0}}}erfc\left( \frac{(2k+1)c_{0}-a_{0}}{2\sqrt{t}}-\sqrt{b_{0}t}\right) \nonumber \\&+e^{[(2k+1)c_{0}-a_{0}]\sqrt{b_{0}}}\nonumber \\&\quad \times erfc\Big (\frac{(2k+1)c_{0}-a_{0}}{2\sqrt{t}} +\sqrt{b_{0}t}\Big )\Big ],\nonumber \\ H(a_{0},s,b_{0},c_{0})= & {} \displaystyle \frac{\sinh [a_{0}\sqrt{s^{\alpha }+b_{0}}]}{s^{\alpha }\sinh [c_{0}\sqrt{s^{\alpha }+b_{0}}]}= F(a_{0},s^{\alpha },b_{0},c_{0}),\nonumber \\ h(a_{0},t,b_{0},c_{0})= & {} {\left\{ \begin{array}{ll} \displaystyle \int _0^\infty {t^{-1}}f(a_{0},x,b_{0},c_{0})\phi (0,-\alpha ,-xt^{-\alpha })\hbox {d}x,\, 0<\alpha <1\\ f(a_{0},t,b_{0},c_{0}),\quad \alpha =1, \end{array}\right. } \nonumber \\= & {} L^{-1}\left\{ H(a_{0},s,b_{0},c_{0})\right\} , \end{aligned}$$

(A2)

$$\begin{aligned} U(s)= & {} \frac{1}{s}e^{-a_{0}{s^{\sigma }}},a_{0}\geqslant 0,\quad 0<\sigma <1,\nonumber \\ u(t)= & {} L^{-1}\left\{ U(s)\right\} (t)=\phi (1,-\sigma ,-a_{0}t^{-\sigma }),\nonumber \\ u(0)= & {} \mathop {\hbox {lim}}\limits _{t\rightarrow 0^{+}}u(t)=\mathop {\hbox {lim}}\limits _{s\rightarrow \infty }sU(s)= \mathop {\hbox {lim}}\limits _{s\rightarrow \infty }s\frac{1}{s}e^{-a_{0}s^{\sigma }}=0,\nonumber \\ L\left\{ u^{\prime }(t)\right\}= & {} sU(s)-U(0)=sU(s)=e^{-a_{0}s^{\sigma }}, \nonumber \\ u^{\prime }(t)= & {} t^{-1}\phi (0,-\sigma ,-a_{0}t^{-\sigma }). \end{aligned}$$

(A3)