**2. Problem Formulation**

Consider a steady three-dimensional Casson nanofluid thin-film flow over a rotating disk. The rotation of the disk is due to the angular velocity (Ω) in its own plane as displayed in Figure 1. An angle, β, is made by the inclined disk with the horizontal axis. Also, *h* denotes the film thickness of the nanofluid, and *W* represents the spraying velocity. The radius of the disk is very large as compared to the liquid film thickness and hence the termination influence is unnoticed. *g* is gravitational acceleration, *T*<sup>0</sup> is the temperature at the film surface, while *T*<sup>w</sup> represents the surface temperature of the disk. Likewise, *C*<sup>0</sup> and *C*<sup>h</sup> are the concentration on the film and on the disk surfaces, respectively. Pressure is a function of the z-axis only and the ambient pressure (*P*0) at the sheet of the film is kept constant. The equations of continuity, momentum, concentration, and energy for a steady state are shown in Equations (1) to (6) [8–10]:

$$
\mu\_x + \mu\_y + \mu\_z = 0 \tag{1}
$$

$$
\rho\_{nf} \left( \mu u\_x + vu\_y + wu\_z \right) = \left( 1 + \frac{1}{\mathcal{I}} \right) \mu\_{nf} \left( u\_{xx} + u\_{yy} + u\_{zz} \right) + \overline{\mathcal{g}} \sin \beta \tag{2}
$$

$$
\mu v v\_x + v v\_y + w v\_z = \left(1 + \frac{1}{\mathcal{V}}\right) \frac{\mu\_{nf}}{\rho\_{nf}} (v\_{xx} + v\_{yy} + v\_{zz}) \tag{3}
$$

$$
\mu w\_x + \upsilon w\_y + \upsilon w\_z = \left(1 + \frac{1}{\mathcal{I}}\right) \frac{\mu\_{nf}}{\rho\_{nf}} (w\_{xx} + w\_{yy} + w\_{zz}) - \overline{\mathcal{g}} \cos\frac{\mathcal{\beta}}{\Omega'} - \frac{P\_z}{\rho\_{nf}} \tag{4}$$

$$
\mu T\_x + \nu T\_y + \nu T\_z = \frac{k\_{nf}}{\left(\rho\_{cp}\right)\_{nf}} \left(T\_{xx} + T\_{yy} + T\_{zz}\right) \tag{5}
$$

$$u\mathbf{C}\_{x} + v\mathbf{C}\_{y} + w\mathbf{C}\_{z} = D\_{\beta} \left( \mathbf{C}\_{xx} + \mathbf{C}\_{yy} + \mathbf{C}\_{zz} \right) + \left( \frac{D\_{T}}{T\_{0}} \right) \left( T\_{xx} + T\_{yy} + T\_{zz} \right) \tag{6}$$

**Figure 1.** Geometry of the problem.

In the above equations, *u*, *v*, and *w* represent the velocity components in the *x*, *y*, and *z* axis, respectively.

The boundary conditions are as follows:

$$\begin{aligned} \mu &= -\Omega y, \upsilon = \Omega x, w = 0, T = T\_{\mathfrak{w}\prime} \mathbb{C} = \mathbb{C}\_{h} & \text{at } z = 0\\ \mu\_{z} = \upsilon\_{z} = 0, w = 0, T = T\_{\mathfrak{w}\prime} \mathbb{C} = \mathbb{C}\_{0\prime} P = P\_{0} & \text{at } z = h \end{aligned} \tag{7}$$

Consider the similarity transformations of the form:

$$\begin{aligned} u &= -\Omega y \text{g}(\eta) + \Omega x f'(\eta) + \overline{\text{g}}k(\eta) \sin \frac{\beta}{\text{f}'} \\ v &= \Omega x \text{g}(\eta) + \Omega y f'(\eta) + \overline{\text{g}}s(\eta) \sin \frac{\beta}{\text{f}'} \\ w &= -2\sqrt{\Omega v\_{nf}} f(\eta), \ T = (T\_0 - T\_w)\theta(\eta) + T\_w \\ \eta \phi(\eta) &= \frac{\mathbb{C} - \mathbb{C}\_w}{\mathbb{C}\_0 - \mathbb{C}\_w}, \eta = z\sqrt{\frac{\Omega}{v\_{nf}}} \end{aligned} \tag{8}$$

The transformations introduced in Equation (8) are implemented in Equations (2) to (7). Equation (1) is proved identically and Equations (2) to (6) are obtained in the forms:

$$\left(1+\frac{1}{\mathcal{I}}\right)f'''' - f'^2 + g^2 + 2ff'' = 0\tag{9}$$

$$\left(1 + \frac{1}{\gamma'}\right)k'' + \text{gs} - kf' + 2kf = 0\tag{10}$$

$$\left(1+\frac{1}{\gamma}\right)\mathbf{g}^{\prime\prime}-2\mathbf{g}f^{\prime}+2\mathbf{g}^{\prime}f=\mathbf{0}\tag{11}$$

$$\left(1 + \frac{1}{\gamma}\right) \mathbf{s}^{\prime\prime} - k \mathbf{g} - s f^{\prime} + 2 \mathbf{s}^{\prime} f = 0 \tag{12}$$

If θ(η) and φ(η) are a function of *z* only, Equations (5) and (6) take the forms:

$$(1 - \frac{4}{3}R)\theta' + 2\text{Pr}\frac{A\_2 A\_3}{A\_1 A\_4}(I\_1 f' + j\_1)\theta = 0\tag{13}$$

$$
\phi'' + 2\mathfrak{S}\mathfrak{c}f\phi' + \frac{\mathcal{N}t}{\mathcal{N}b}\theta'' = 0\tag{14}
$$

$$\begin{array}{l} f(0) = 0, \ f'(0) = 0, \ f''(\delta) = 0, \ \phi(0) = 0, \ \phi(\delta) = 1\\ \end{array} \\ \begin{array}{l} g(0) = 0, \ g'(\delta) = 0, \ k(0) = 0, \ k'(\delta) = 0\\ \end{array} \\ \begin{array}{l} (15) = 0, \ s'(\delta) = 0, \ \theta(0) = 0, \ \theta'(\delta) = 1. \end{array} \tag{15}$$

Physical parameters and other dimensionless numbers of interest are defined as:

$$\begin{array}{l} \text{Pr} = \frac{v\_f}{\alpha\_f}, \text{ Sc} = \frac{\mu}{\rho f D}, \text{ Nb} = \frac{(\rho c)\_p D\_b(\mathbb{C}\_h)}{(\rho c)\_f \alpha} \\ \text{Nt} = \frac{(\rho c)\_p D\_T(T\_H)}{(\rho c)\_f \alpha T\_c}, \text{ S} = \frac{\mu}{\Omega} \end{array} \tag{16}$$

Here, *Pr* is the Prandtl number, *Sc* is the Schmidt number, *Nb* is the Brownian motion parameter, and *Nt* is the thermophoretic parameter.

Where the normalized thickness constant is presented as:

$$\delta = h \sqrt{\frac{\Omega}{v\_{nf}}} \tag{17}$$

The condensation velocity is defined as:

$$f(\delta) = \frac{W}{2\sqrt{\Omega\nu}} = a\tag{18}$$

The pressure can be attained by the integration of Equation (4).

For the exact solution, let *Pr* = 0 and using θ(δ) = 1, the exact solution is:

$$
\theta'(0) = \frac{1}{\delta} \tag{19}
$$

An asymptotic limit for small, δ, is defined in Equation (17). The reduction of θ'(0) for rising δ is not monotonic. So, *Nu* is defined as:

$$N\mu = \frac{k\_{\rm nf}}{k\_f} \frac{(T\_z)\_w}{(T\_0 - T\_w)} = A\_4 \delta\theta'(0) \tag{20}$$

The Sherwood number is defined as:

$$Sh = \frac{\left(\mathbb{C}\_z\right)\_w}{\mathbb{C}\_0 - \mathbb{C}\_w} = \delta\phi'(0) \tag{21}$$
