**2. Problem Formulation**

Consider an axisymmetric unsteady MHD water base nanofluid flow between continuously stretchable disks with hall current effect amid non-conducting rotating disks at *z* = 0 and *z* = *h*. The disks rotate at constant angular velocities Ω<sup>1</sup> and Ω<sup>2</sup> about its axis. Magnetic field *B*<sup>0</sup> that is uniformly distributed is applied in the normal direction of the disks (Figure 1). Furthermore, the stretching rates of the disks are *a*<sup>1</sup> and *a*2. Temperature *T*<sup>2</sup> = *T*<sup>0</sup> + *Br* <sup>1</sup>−*ct* refers to the temperature of upper disk while the disk's temperature at *z* = *h* is *T*<sup>1</sup> = *T*<sup>0</sup> + *Ar* <sup>1</sup>−*ct* in a thermally stratified medium.

**Figure 1.** Schematic picture of the fluid flow.

For isothermal cubic autocatalysis, a model for homogeneous and heterogeneous reactions with reactants as chemical species are *A*∗ and *B*∗ and was proposed by Merkin and Chaudary [41] and is given by:

$$A^\* + 2B^\* \to 3B^\*,\\
rate = \mathcal{K}\_\mathbb{C} = ab^2.\tag{1}$$

$$A^\* \to B^\*, \text{ rate} = K\_\circ a\_\prime \tag{2}$$

The continuity equation is

$$
\nabla \vec{V} = 0,\tag{3}
$$

The momentum equations are

$$\frac{\partial \mu}{\partial t} + (\vec{V}.\nabla)\mu = \frac{-1}{\rho\_{nf}}p\_r^\* + \frac{\mu\_{nf}}{\rho\_{nf}}(\nabla^2\mu) - \frac{\sigma\_{nf}}{\rho\_{nf}}\frac{B\_O}{1+m^2}(\mu - mv),\tag{4}$$

$$\frac{\partial \upsilon}{\partial t} + (\overrightarrow{V}.\nabla)\upsilon = \frac{\mu\_{nf}}{\rho\_{nf}}(\nabla^2 \upsilon) - \frac{\sigma\_{nf}}{\rho\_{nf}}\frac{B\_O^{-2}}{1+m^2}(\upsilon+mu),\tag{5}$$

$$\frac{\partial w}{\partial t} + (\vec{V} . \nabla) w = \frac{-1}{\rho\_{nf}} p\_z^\* + \frac{\mu\_{nf}}{\rho\_{nf}} + \nabla^2 w\_\prime \tag{6}$$

The relevant energy equation is

$$(\rho \mathbb{C}\_{\mathcal{P}})\_{nf} (\overrightarrow{V}.\nabla) T = -\nabla.\overrightarrow{q}\_{\prime} \tag{7}$$

where *T* represents the temperature, *Cp* the specific heat and <sup>→</sup> *q* the heat flux. Heat flux in perspective of Cattaneo–Christov expression is satisfied.

$$
\overrightarrow{\overline{q}} + \varepsilon\_1 (\frac{\overrightarrow{\partial \overline{q}}}{\partial t} + \overrightarrow{V}.(\overrightarrow{\nabla \overline{q}}) - \overrightarrow{\overline{q}}.(\overrightarrow{\nabla V}) + (\overrightarrow{\nabla \cdot V})\overrightarrow{q} \,) - k \nabla T. \tag{8}
$$

Here, <sup>1</sup> is the thermal relaxation time and *k* is the thermal conductivity. Utilizing the incompressibility condition, we arrive at

$$
\overrightarrow{\ddot{q}} + \epsilon\_1 (\overrightarrow{\frac{\partial \ddot{q}}{\partial t}} + \overrightarrow{V}.(\nabla \overrightarrow{\dot{q}}) - \overrightarrow{\dot{q}}.\overrightarrow{\nabla V}) - k\nabla T.\tag{9}
$$

Eliminating <sup>→</sup> *q* from Equations (9) and (7), we get

$$\begin{aligned} T\_{l} + \mathfrak{u}T\_{l} + \mathfrak{w}T\_{z} + \mathfrak{e}\_{1}(T\_{ll} + \mathfrak{u}\_{l}T\_{l} + 2\mathfrak{u}T\_{lr} + 2\mathfrak{w}T\_{lz} + \mathfrak{w}\_{l}T\_{z} + \mathfrak{u}\mathfrak{u}\_{l}T\_{l} + \mathfrak{w}\mathfrak{w}\_{z}T\_{z} \\ + \mathfrak{u}\mathfrak{w}\_{l}T\_{l} + \mathfrak{w}T\_{l}T\_{z} + 2\mathfrak{u}\mathfrak{w}T\_{lz} + \mathfrak{u}^{2}\mathfrak{w}\_{rr} + \mathfrak{w}^{2}T\_{zz} = \frac{\mathsf{K}\_{\mathfrak{u}f}}{(\mathfrak{o}\mathbb{C}\_{\mathfrak{p}})\_{\mathfrak{n}f}} \Big(\nabla^{2}T\Big). \end{aligned} \tag{10}$$

As <sup>→</sup> *V* = (*u*, *v*, *w*) is the velocity vector, we obtain the following governing equations after applying the boundary layer theory:

$$
u\_r + \frac{u}{r} + w\_z = 0,\tag{11}$$

$$u\_{l} + uu\_{l} + wu\_{z} - \frac{v^{2}}{r} = \frac{-1}{\rho\_{nf}} p\_{r}^{\*} + \frac{\mu\_{nf}}{\rho\_{nf}} (u\_{l\tau} + \frac{1}{r}u\_{l} - \frac{u}{r^{2}} + u\_{zz}) - \frac{\sigma\_{nf}}{\rho\_{nf}} \frac{B\_{O}}{1 + m^{2}} (u - mv)\_{r} \tag{12}$$

$$uv\_l + uv\_{l'} + uv\_z + \frac{uv}{r} = \frac{\mu\_{nf}}{\rho\_{nf}}(v\_{l'l'} + \frac{1}{r}v\_{l'} - \frac{v}{r^2} + v\_{zz}) - \frac{\sigma\_{nf}}{\rho\_{nf}}\frac{B\_O}{1+m^2}(v+mu),\tag{13}$$

$$
\mu w\_l + \mu w\_{l'} + \mu w\_{z} - \frac{v^2}{r} = \frac{-1}{\rho\_{nf}} p\_z^\* + \frac{\mu\_{nf}}{\rho\_{nf}} \left( w\_{rr} + \frac{1}{r} w\_{r} + w\_{zz} \right) \tag{14}
$$

$$\begin{aligned} T\_I + uT\_I + wT\_z + \varepsilon\_1 (T\_{fl} + u\_lT\_r + 2uT\_{lr} + 2wT\_{lz} + w\_lT\_z + u\mu\_r T\_r + ww\_z T\_z \\ + uw\_l T\_I + wT\_{l'}T\_z + 2uwT\_{lz} + u^2w\_{l'} + w^2T\_{zz} = \frac{K\_{nf}}{\left(\rho \overline{\mathbb{C}\_p}\right)\_{nf}} \left(T\_{l7} + \frac{1}{r}T\_{r} + T\_{zz}\right), \end{aligned} \tag{15}$$

$$a\_t + ua\_r + va\_z = D\_A(a\_{rr} + \frac{1}{r}a\_r + a\_{zz}) - K\_c ab^2,\tag{16}$$

$$ub\_t + ub\_r + wb\_z = D\_B \left( b\_{rr} + \frac{1}{r} b\_r + b\_{zz} \right) + K\_c ab^2. \tag{17}$$

The associated boundary conditions are

$$\begin{array}{l} u = \frac{r\Omega\_1}{1 - \text{cl}}, \; v = \frac{r\Omega\_1}{1 - \text{cl}}, w = 0, T = T\_1(r) = T\_0 + \frac{Ar}{1 - \text{cl}}, \\\ D\_A \frac{\partial v}{\partial z} = K\_s a, D\_B \frac{\partial b}{\partial z} = -K\_s a, \text{at } z = 0, \end{array} \tag{18}$$

$$u = \frac{m\_2}{1 - ct}, \ v = \frac{r\Omega\_2}{1 - ct}, w = 0, T = T\_2(r) = T\_0 + \frac{Br}{1 - ct}, \ a \to a\_0, b \to 0, \ z = h. \tag{19}$$

Here, *T*<sup>0</sup> is the reference temperature. *A* and *B* are the dimensional constant with dimension [<sup>T</sup> · <sup>L</sup>−1].

Thermo-physical properties of CNTS are represented in mathematical form as follows:

$$A = \frac{\mu\_{nf}}{\mu\_f} = \frac{1}{\left(1 - \phi\right)^{2.5}}\tag{20}$$

$$B = \frac{\rho\_{nf}}{\rho\_f} = (1 - \phi) + \frac{\rho\_{CNT}}{\rho\_f} \phi\_\prime \tag{21}$$

$$\mathbb{C} = \frac{\left(\rho \mathbb{C}\_p\right)\_{nf}}{\left(\rho \mathbb{C}\_p\right)\_f} = \left(1 - \phi\right) + \frac{\left(\rho \mathbb{C}\_p\right)\_{\text{CNT}}}{\left(\rho \mathbb{C}\_p\right)\_f} \phi\_\prime \tag{22}$$

$$D = \frac{k\_{nf}}{k\_f} = \frac{(1 - \phi) + 2\phi \frac{k\_{CNT}}{k\_{CNT} - k\_f} \ln \frac{k\_{CNT} + k\_f}{2k\_f}}{(1 - \phi) + 2\phi \frac{k\_f}{k\_{CNT} - k\_f} \ln \frac{k\_{CNT} + k\_f}{2k\_f}},\tag{23}$$

$$\frac{\sigma\_{nf}}{\sigma\_{f}} = 1 + \frac{3\phi \left(\frac{\sigma\_{CNT}}{\sigma\_{f}} - 1\right)}{\left(\frac{\sigma\_{CNT}}{\sigma\_{f}} + 2\right) - \left(\frac{\sigma\_{CNT}}{\sigma\_{f}} - 1\right)}.\tag{24}$$

Table 2 represents the thermos-physical characteristics of CNTs and H2O.

**Table 2.** Thermo-physical properties of water and carbon nanotubes.


Following transformation are used to convert the above nonlinear partial differential equations to dimensionless ordinary differential equations.

$$\begin{array}{l} \mu = \frac{r\Omega\_1}{1 - ct} f'(\eta), \; v = \frac{r\Omega\_1}{1 - ct} g(\eta), \; w = \frac{2t\Omega\_1}{\sqrt{1 - ct}} f(\eta), \; \theta = \frac{T - T\_2}{T\_1 - T\_o}, \\\ p^\* = \frac{\rho \Omega\_1 v}{(1 - ct)^2} \{P(\eta) + \frac{r^2}{2h^2} \varepsilon\}, \; \eta = \frac{z}{h\sqrt{1 - ct}}, \; a = c\_0 \overline{\eta}, b = c\_0 \overline{l}. \end{array} \tag{25}$$

Equation (11) is satisfied automatically, Equations (12) to (17) are transformed into the following form:

$$A\_1(f' + \frac{\eta}{2}f'') + \text{Re}(f'^2 - 2ff' - g^2) + \varepsilon - \frac{\sigma\_{\text{nf}}}{\sigma\_f} \frac{M \text{Re}(f' - mg)}{B(1 + m^2)} = \frac{A}{B}f'''',\tag{26}$$

$$\frac{B}{A}\text{Re}[\left(\mathbf{g} + \frac{1}{2}\,\eta\mathbf{g'}\right)\mathbf{A}\_1 + 2(f'\,\mathbf{g} - f\mathbf{g'})] - \frac{\sigma\_{nf}}{\sigma\_f}\frac{MR\mathbf{e}(\mathbf{g} + mf')}{A(1+m^2)} = \mathbf{g''},\tag{27}$$

$$\frac{\partial p^\*}{\partial z} = (A\_1(f + \eta f') - 4ff')B(1 - ct)Rc - 2\frac{(1 - ct)}{A}f'',\tag{28}$$

$$\begin{array}{l} A\_{1} \big( \mathfrak{s} + \theta + \frac{1}{2} \eta \theta' \big) + (\mathfrak{s} + \theta) f' - 2f \theta' + \chi \big( \mathfrak{s} + \theta + \frac{7}{8} \eta \theta' \big) + \\\ f' \big( f' + \frac{1}{2} \eta f'' \big) (\mathfrak{s} + \theta) + 2f' \big( \mathfrak{s} + \theta + \frac{1}{2} \eta \theta' \big) - 4f \big( \frac{1}{2} \eta \theta'' + \frac{3}{2} \theta' \big) + \\\ (f + \eta f') \theta' + f'^2 (\mathfrak{s} + \theta) - \frac{4}{A\_{1}} f' \theta' - 2f f'' (\mathfrak{s} + \theta) + \frac{4}{A\_{1}} f^{2} \theta'' - \frac{4}{A\_{1}} f f' \theta' \\\ = \frac{\mathsf{D}}{\mathsf{C}} \big( \frac{1}{\mathrm{PrR} \mathfrak{c}} \theta'' + \frac{1}{\mathrm{Pr}} (\mathfrak{s} + \theta) \big), \end{array} \tag{29}$$

$$\frac{1}{2}\eta\widetilde{\boldsymbol{\varphi}}^{\prime} - \frac{2}{A\_1}f\widetilde{\boldsymbol{\varphi}}^{\prime} - \frac{1}{Sc}\widetilde{\boldsymbol{\varphi}}^{\prime\prime} + k1\widetilde{\boldsymbol{\varphi}}^2 = 0,\tag{30}$$

$$\frac{1}{2}\frac{1}{2}\overline{\eta l'} - \frac{2}{A\_1}\overline{f l'} - \frac{\delta}{Sc}\overline{l''} - k1\overline{q}l^2 = 0,\tag{31}$$

with transformed boundary conditions

$$\begin{aligned} f(0) &= 0, f(1) = 0, f'(0) = \gamma\_1, f'(1) = \gamma\_2, g(0) = 1, \\ g(1) &= \Omega, \theta(0) = 1 - s, \theta(1) = 0, P(0) = 0, \end{aligned} \tag{32}$$

where

$$\begin{split} M &= \frac{\frac{\rho\_f B\_0^2 (1 - \text{cl}t)}{\rho\_f}, A\_1 = \frac{\epsilon}{\Omega\_1}, \gamma\_1 = \frac{a\_1}{\Omega\_1}, \gamma\_2 = \frac{a\_2}{\Omega\_2}, \text{Sc} = \frac{\mu\_c^2 \epsilon}{\Omega\_A}, Pr = \frac{\nu\_f (\rho \mathbb{C}\_p)\_f}{k\_f}, \Omega = \frac{\Omega\_2}{\Omega\_1}, \\\ k1 &= \frac{\mathbb{K} \cdot \text{cl}^2 (1 - \text{cl}t)}{\varepsilon}, k2 = \frac{\mathbb{K} \cdot \text{cl} (1 - \text{cl}t)^{1/2}}{D\_A} \delta = \frac{D\_\mathbb{R}}{\mathbb{D}\_A}, \gamma = \frac{\varepsilon \text{cl}\_1}{1 - \text{cl}t}, D = \frac{k\_{nf}}{k\_f}, B = \frac{(\rho \mathbb{C}\_p)\_{nf}}{\left(\rho \mathbb{C}\_p\right)\_f}. \end{split} \tag{33}$$

By assuming the chemical species alike, we take diffusion coefficient of both species equal, so that δ = 1. And thus we have1*l*(η) + ϕˇ(η) = 1, we get from Equations (30) and (31)

$$
\frac{1}{Sc}\overleftarrow{\rho}^{\prime\prime} - \frac{1}{2}\eta\overleftarrow{\rho}^{\prime} + \frac{2}{A\_1}f\overleftarrow{\rho}^{\prime} - \mathbf{k}1(1-\overleftarrow{\rho})^2\overleftarrow{\varphi} = 0,\tag{34}
$$

$$
\widetilde{\varphi}'(0) = \mathcal{K}2\widetilde{\varphi}'(0), \ \widetilde{q}\nu(1) \to 1,\tag{35}
$$

Differentiating Equation (26), we get

$$A\_1 \{ \frac{3}{2} f'' + \frac{\eta}{2} f'''' \} + \text{Re}(2f f'''' - 2 \text{gg}') - \frac{\sigma\_{\text{nf}}}{\sigma\_f} \frac{\text{MRe}(f'' - \text{mg}')}{B(1 + m^2)} = \frac{A}{B} f'''' \} \tag{36}$$
