. 2023 Apr 3;9(4):e14807. doi: 10.1016/j.heliyon.2023.e14807

Table 4.

Energy calculation expressions for inclined solar still with baffles [39,43].

Convective mode is used to transmit heat from the glass cover to the water.	$Q_{c, w - g} = h_{c, w - g} (T_{w} - T_{g})$ (1) $h_{c, w - g} = 0.884 {[(T_{w} - T_{g}) + {\frac{(p_{w} - p_{g}) T_{w}}{268.90 \times 10^{3} - p_{w}}}]}^{(1 / 3)}$ (2)
Heat transfer through radiative mode with glass cover and water	$Q_{r, w - g} = h_{r, w - g} A_{w} (T_{w} - T_{g})$ (3) $Q_{r, w - g} = ε_{e f f e c t i v e} σ A_{w} (T_{w}^{4} - T_{g}^{4})$ (4) $h_{r, w - g} = ε_{e f f e c t i v e} σ (T_{w}^{2} + T_{g}^{2}) (T_{w} + T_{g})$ (5)
Evaporation enables the glass cover's water can lose heat.	$Q_{e, w - g} = (16.237 x 10^{- 3}) h_{c, w - g} A_{w} (p_{w} - p_{g})$ (6) $Q_{e, w - g} = h_{e, w - g} A_{w} (T_{w} - T_{g})$ (1) $h_{e, w - g} = 16.237 x 10^{- 3} h_{c, w - g} [\frac{p_{w} - p_{g}}{T_{w} - T_{g}}]$ (7) $m_{e, w - g} = (\frac{Q_{e, w - g}}{h_{f g}}) \times 3600$ (8)
Heat exchange between the ISSB's glass cover and outside air	$Q_{g - a} = Q_{c, g - a} + Q_{r, g - a}$ (9) $Q_{r, g - a} = h_{r, g - a} A_{g} (T_{g} - T_{a})$ (10) $Q_{r, g - a} = ε_{g} σ A_{g} (T_{g}^{4} - T_{s k y}^{4})$ (11) $h_{r, g - a} = \frac{ε_{g} σ [T_{g}^{4} - T_{s k y}^{4}]}{T_{g} - T_{a}}$ (12) $Q_{c, g - a} = h_{c, g - a} A_{g} (T_{g} - T_{a})$ (13) $h_{c, g - a} = 5.7 + 3.8 V_{a}$ (for velocity of wind<5 m/s) (14)
Balance of energy in the ISSB's still basin	$(α_{b}) A_{b} I = Q_{c, b - w} + Q_{l o s s}$ (15) $Q_{c, b - w} = h_{c, b - w} A_{b} (T_{b} - T_{w})$ (16) $Q_{l o s s} = U_{b} A_{b} (T_{b} - T_{a})$ (17) $T_{b} = \frac{(α_{b}) I + h_{c, b - w} T_{w} + U_{b} T_{a}}{h_{c, b - w} + U_{b}}$ (18)
Energy balance of the glass surface	$α_{g} I = h_{3} (T_{g} - T_{a}) - h_{2} (T_{w} - T_{g})$ (19) $h_{2} = h_{r, w - g} + h_{e, w - g} + h_{c, w - g}$ (20) $h_{3} = h_{r, g - a} + h_{c, g - a}$ (21) $T_{g} = [\frac{α_{g} I + h_{2} T_{w} + h_{3} T_{a}}{h_{2} + h_{3}}]$ (22)
Energy balance of water outlet temperature	$I τ_{g} (1 - α_{w}) n x y = m_{f w} C_{p w} (T_{o u t} - T_{i n}) (x d y - d x d y) - n x y Q_{b a f f l e}$ (23) $T_{o u t} = n x y [\frac{Q_{b a f f l e} + I τ_{g} (1 - α_{w})}{m_{f} C_{p w} (x d y - d x d y}] + T_{i n}$ (24) $T_{W} = T_{i n} + [\frac{n x y [I τ_{g} (1 - α_{w}) + Q_{b a f f l e}]}{2 m_{f w} C_{p w} (x d y - d x d y)}]$ (25)