**2. Problem Formulation**

Assume a three-dimensional nanofluid thin liquid flow through a steady rotating disk. The disk rotates on its own axis due to the angular velocity (Ω) as shown in Figure 1. With the horizontal axis the inclined disk creates an angle β. The radius of the disk is greater as it is associated to the thickness of the fluid film and therefore the last impact is unobserved. Here *T*<sup>0</sup> and *Tw* denote the temperatures of the surface film and the surface of the disk respectively. Similarly, at the film *C*0, *Ch* are the concentrations on the disk surface respectively. The equations for the steady state flow are displayed as follows [50,55,56]:

$$
u\_x + v\_y + w\_z = 0\tag{1}$$

$$
\Delta u u\_x + \nu u\_y + w u\_z = \nu u\_{zz} - \mathbf{v}' u\_{zzzz} - \frac{\sigma B\_0^2 u}{\rho} \tag{2}
$$

$$
\Delta w\_x + \nu v\_y + wv\_z = \mathbf{v}v\_{zz} - \mathbf{v'}v\_{zzzz} - \frac{\sigma B\_0^2 \upsilon}{\varrho} \tag{3}
$$

$$
\Delta w\_x + vw\_y + ww\_z = \mathbf{v}w\_{zz} - \mathbf{v'}w\_{zzzz} - \frac{\sigma B\_0^2 w}{\rho} \tag{4}
$$

$$
\mu T\_X + \sigma T\_Y + wT\_z = \frac{k\_{nf}}{\left(\rho\_{cp}\right)\_{nf}} \left(T\_{zz}\right) + \pi \left[D\_B \mathbb{C}\_z T\_z + \frac{D\_T}{T\_0} \left(T\_z\right)^2\right] \tag{5}
$$

$$
u \mathbb{C}\_x + v \mathbb{C}\_y + w \mathbb{C}\_z = D\_\beta \left( \mathbb{C}\_{zz} \right) + \left( \frac{D\_T}{T\_0} \right) (T\_{zz}) \tag{6}$$

with boundary conditions:

$$\begin{cases} \mathbf{u} = -\Omega \mathbf{y}, \mathbf{v} = \Omega \mathbf{x}, \mathbf{w} = \mathbf{0}, T = T\_{\mathbf{w}\prime} \mathbf{C} = \mathbf{C}\_{\mathbf{h}} & \text{at} \qquad \mathbf{z} = \mathbf{0} \\\mathbf{u}\_{\overline{z}} = \mathbf{v}\_{\overline{z}} = \mathbf{0}, \mathbf{w} = \mathbf{0}, T = T\_{\overline{w}\prime} \mathbf{C} = \mathbf{C}\_{\mathbf{0}} & \text{at} \qquad \mathbf{z} = \mathbf{h} \end{cases} \tag{7}$$

using the Similarity transformation [50]:

$$\begin{cases} u = -\Omega \text{yg}\left(\mathfrak{n}\right) + \Omega x f'\left(\mathfrak{n}\right) + \overline{\mathfrak{g}} k\left(\mathfrak{n}\right) \sin\frac{\beta}{\Omega^\*}\\ \nu = \Omega \text{xg}\left(\mathfrak{n}\right) + \Omega y f'\left(\mathfrak{n}\right) + \overline{\mathfrak{g}} s\left(\mathfrak{n}\right) \sin\frac{\beta}{\Omega^\*}\\ w = -2\sqrt{\Omega \nu\_{nf}} f\left(\mathfrak{n}\right), T = \left(T\_0 - T\_w\right) \Theta\left(\mathfrak{n}\right) + T\_w\\ \eta \Phi\left(\mathfrak{n}\right) = \frac{\mathbb{C} - \mathbb{C}\_w}{\mathbb{C}\_0 - \mathbb{C}\_w}, \eta = z\sqrt{\frac{\Omega}{\nu\_{nf}}} \end{cases} \tag{8}$$

we obtain:

$$\text{kg}^{\prime\prime} - 2\text{g}f^{\prime} + 2\text{g}^{\prime}f - \text{Kg}^{\text{vii}} - \text{Mg} = 0\tag{9}$$

$$f^{\prime\prime\prime} - f^{\prime2} + g^2 + 2ff^{\prime\prime} - \mathcal{K}f^{\prime\bar{n}} - \mathcal{M}f^{\prime} = 0\tag{10}$$

$$k'' + \text{gs} - kf' + 2kf - \text{K}k^{\text{ru}} - \text{M}k = 0\tag{11}$$

$$s'' - kg - sf' + 2s'f - \mathcal{K}s^{v\bar{i}} - \mathcal{M}s = 0\tag{12}$$

**Figure 1.** Geometrical description of the problem.

Now if θ(η) and φ(η) depend only on *z*, then Equations (5) and (6) are reduced to the following forms:

$$
\Theta'' + 2\text{Pr}\Theta f + \text{PrN}b\eta\phi'\theta' + \text{Pr}\Omega\text{N}t\theta'^2 = 0\tag{13}
$$

$$\left(\Phi'' + 2\text{c}f\phi' + \frac{\text{Nt}}{\text{N}b}\theta'' + \frac{\text{S}}{2}\left(\eta\phi' + \eta^2\phi''\right) = 0\tag{14}$$

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

while, Pr *M*, *Sc*,*Nt*,*K*,*S* and *Nb* are defined as below [57]:

$$\begin{aligned} \text{Pr} &= \frac{\rho\_{\text{P}} \mathbf{v}\_{nf}}{k}, \mathbf{S} \mathbf{c} = \frac{\mu}{\rho\_{f}^{\prime} D^{\prime}}, \mathbf{N} \mathbf{b} = \frac{\pi D\_{\text{R}}}{\mathbf{v}} \left( \mathbf{C}\_{0} - \mathbf{C}\_{\text{w}} \right) \\ \text{N} &= \frac{\pi D\_{\text{T}}}{\mathbf{v} \mathbf{T}\_{\text{w}}} (T\_{0} - T\_{\text{w}}), \mathbf{S} = \frac{\mathbf{g}}{\mathbf{T}\_{\text{I}}}, \mathbf{M} = \frac{\sigma \beta\_{\text{0}}^{2}}{\Omega \rho}, \mathbf{K} = \frac{\mathbf{v}^{\prime} \Omega}{\mathbf{v}\_{\text{ref}}^{2}} \end{aligned} \tag{16}$$

here, the constant of the normalized thickness is defined as:

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

where the velocity of condensation is written as:

$$f\left(\delta\right) = \frac{W}{2\sqrt{\Omega \mathbf{v}}} = \infty \tag{18}$$

Direct integration of Equation (4) gives the pressure term. Using Pr = 0 and θ(δ) = 1, we get

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

For small δ an asymptotic limit is explained by Equation (17). The decrease of θ(0) for increasing δ is not monotonic, thus *Nu* is signified as [57]:

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

Similarly, the Sherwood number is given by:

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