*2.2. Boundary Conditions*

• At the surface of the rotating disk

$$\begin{array}{c} \text{(i)} \ \overline{u} = 0, \\ \text{(ii)} \ \overline{v} = \overline{r} \,\overline{\Psi}, \\ \text{(iii)} \ \overline{w} = 0, \\ \text{(iv)} \ \overline{T} = \overline{T}\_a - \frac{\overline{T}\_b}{2} \overline{r}^2 \end{array} \text{; when } \overline{z} = 0 \tag{7}$$

• At the free surface of the rotating disk

$$\begin{array}{c} \mbox{(i)} \begin{array}{l} \mbox{(i)} \cdot \frac{\partial \widetilde{\boldsymbol{u}}}{\partial \widetilde{\boldsymbol{t}}} = \overline{\boldsymbol{u}} \boldsymbol{\tau} \\ \mbox{(ii)} \cdot \overline{\boldsymbol{\mathcal{T}}} + 2\overline{\boldsymbol{\mu}}\_{nf} \frac{\partial \widetilde{\boldsymbol{u}}}{\partial \widetilde{\boldsymbol{\mathcal{T}}}} = \boldsymbol{0}, \\ \mbox{(iii)} \cdot \overline{\boldsymbol{\mathcal{T}}}\_{nf} \left( \frac{\partial \widetilde{\boldsymbol{u}}}{\partial \widetilde{\boldsymbol{\mathcal{T}}}} + \frac{\partial \widetilde{\boldsymbol{u}}}{\partial \widetilde{\boldsymbol{\mathcal{T}}}} \right) = \frac{\partial \widetilde{\boldsymbol{T}}}{\partial \widetilde{\boldsymbol{\mathcal{T}}}} \frac{\partial \widetilde{\boldsymbol{\mathcal{T}}}}{\partial \widetilde{\boldsymbol{\mathcal{T}}}}, \\ \mbox{(iv)} \cdot \overline{\boldsymbol{\mu}}\_{nf} \left( \frac{\partial \widetilde{\boldsymbol{u}}}{\partial \widetilde{\boldsymbol{z}}} \right) = \frac{\partial \widetilde{\boldsymbol{T}}}{\partial \widetilde{\boldsymbol{z}}} \frac{\partial \widetilde{\boldsymbol{\mathcal{T}}}}{\partial \boldsymbol{\mathcal{T}}}, \\ \mbox{(v)} \cdot \frac{\partial \widetilde{\boldsymbol{T}}}{\partial \widetilde{\boldsymbol{z}}} + L \left( \overline{\boldsymbol{T}} - \boldsymbol{T}\_{\mathcal{S}} \right) = \boldsymbol{0}. \end{array} \tag{8}$$

where *L* denotes heat transfer coefficient and σ stands for surface tension.

By using suitable transformations [40], the governing equations can be obtained as:

$$2F + \frac{\partial W}{\partial z} = 0\tag{9}$$

$$\operatorname{Re}\mathcal{Q}\_1\left(\frac{\partial F}{\partial t} + F^2 + \mathcal{W}\frac{\partial F}{\partial z}\right) = \frac{\partial^2 F}{\partial z^2} + G^2 \tag{10}$$

$$\operatorname{Re}\mathcal{Q}\_1 \left( \frac{\partial G}{\partial t} - G \frac{\partial W}{\partial z} + W \frac{\partial G}{\partial z} \right) = \frac{\partial^2 G}{\partial z^2} \tag{11}$$

$$\operatorname{RePr\mathcal{Q}}\_2\left(\frac{\partial \Gamma}{\partial t} - \Gamma \frac{\partial \mathcal{W}}{\partial z} + \mathcal{W} \frac{\partial \Gamma}{\partial z}\right) = \frac{k\_{nf}}{k\_f} \frac{\partial^2 \Gamma}{\partial z^2} \tag{12}$$

$$\operatorname{RePr} \mathcal{Q}\_2 \left( \frac{\partial \pi}{\partial t} + W \frac{\partial \pi}{\partial z} \right) = \frac{k\_{nf}}{k\_f} \left( \frac{\partial^2 \pi}{\partial z^2} + 2\Gamma \right) \tag{13}$$

$$\begin{cases} \text{(i)}.F(z,t) = 0, \\ \text{(ii)}.G(z,t) = 0, \\ \text{(iii)}.W(z,t) = 0, \\ \text{(iv)}.\Gamma(z,t) = 0, \\ \text{(v)}.\pi(z,t) = 0, \\ \text{(v)}.\pi(z,t) = 0, \\ \text{(vi)}.H(t) = 1, \end{cases} \}; \text{ at } t = 0 \tag{14}$$

$$\begin{array}{l} \mbox{(i)}.F(z,t) = 0, \\ \mbox{(ii)}.G(z,t) = 1, \\ \mbox{(iii)}.W(z,t) = 0, \\ \mbox{(iv)}.\Gamma(z,t) = 1, \\ \mbox{(v)}.\tau(z,t) = 0, \end{array} \mbox{; at } z = 0 \tag{15}$$

here, Re is the Reynolds number and Pr denotes the Prandtl number, whereas ∅<sup>1</sup> and ∅<sup>2</sup> represent dimensionless constants.

For free surface

(i). *<sup>∂</sup><sup>F</sup> <sup>∂</sup><sup>z</sup>* = *α*(1 − *φ*) 2.5Γ, (ii). *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>z</sup>* = 0, (iii). *<sup>∂</sup>*<sup>Γ</sup> *<sup>∂</sup><sup>z</sup>* = 0, (iv). *<sup>∂</sup>*<sup>τ</sup> *<sup>∂</sup><sup>z</sup>* = 0, (v). *dH dt* = *W*. ⎫ ⎪⎪⎪⎪⎪⎬ ⎪⎪⎪⎪⎪⎭ ; at *z* = *H*(*t*) (16)
