## 4 Mathematical Modeling  
### Energy Analysis  
In the still, the energy obtained by saltwater equals the amount of energy lost via convection, radiation, and evaporation heat transfer processes between glass and water basins plus the energy gained by saltwater [24]. \[I(t)\alpha_{w}A_{w}+q_{con,b-w}=q_{rad,w-gl}+q_{con,w-gl}+q_{evap,w-gl}+m_{w}C_{ p,w}\frac{dT_{w}}{dt} \tag{2}\]  
where the heat transfer rates by convection, radiation, and evaporation can be calculated using the following relationships. \[q_{con,w-gl}=h_{con,w-gl}A_{W}\Big{(}T_{w}-T_{gl}\Big{)} \tag{3}\]  
\[q_{rad,w-gl}=h_{rad,w-gl}A_{W}\Big{(}T_{w}-T_{gl}\Big{)} \tag{4}\]  
\[q_{evap,w-gl}=h_{evap,w-gl}A_{W}\Big{(}T_{w}-T_{gl}\Big{)} \tag{5}\]  
The coefficients of heat transfer by convection, radiation, and evaporation between glass and basin water can be calculated using the following relations [25]:  
\[h_{con,w-g}=0.884\Bigg{(}\Big{(}T_{w}-T_{gl}\Big{)}+\frac{\Big{(}P_{w}-P_{gl} \Big{)}T_{w}}{(268900-P_{w})}\Bigg{)}^{0.333} \tag{6}\] \begin{table}
\begin{tabular}{c c c} \hline \hline
**Description** & **Quantity** & **Value** \\ \hline Solar radiation & \(I_{t}\) & \(\pm\)3.7 W/m\({}^{2}\) \\ Temperature & \(T\) & \(\pm\)0.75 \({}^{\circ}\)C \\ Water productivity & \(m_{w}\) & 0.87\% \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Measured variable uncertainties. \[h_{rad,w-gl}=\epsilon_{eff}\sigma\bigg{[}\left(T_{w}+273.15\right)^{2}-\left(T_{gl }+273.15\right)^{2}\bigg{]}\times\left[T_{w}+T_{gl}+546\right] \tag{7}\]  
\[h_{exp,w-gl}=162.76\cdot 10^{-4}\times h_{con,w-gl}\times\left(\frac{P_{w}-P_{gl }}{T_{w}-T_{gl}}\right) \tag{8}\]  
where, \(P_{gl}\), and \(P_{w}\) is the vapour pressure at glass and water temperatures respectively, can be calculated using the following relations:  
\[P_{w}=e^{\left(25.317-\frac{5144}{T_{gl}+273.15}\right)} \tag{9}\]  
The heat transfer rates by convection and radiation between ambient and glass can be calculated using the following relations:  
\[P_{gl}=e^{\left(25.317-\frac{5144}{T_{gl}+273.15}\right)} \tag{10}\]  
\[q_{con,gl-a}=h_{con,gl-a}A_{g}\Big{(}T_{gl}-T_{a}\Big{)} \tag{11}\]  
\[q_{rad,gl-sky}=h_{rad,gl-sky}A_{g}\Big{(}T_{gl}-T_{sky}\Big{)} \tag{12}\]  
The coefficients of heat transfer by convection and radiation between ambient and glass can be calculated using the following relations:  
\[h_{con,gl-sky}=5.7+2.8V;for\ V>5\ \text{m/s} \tag{13}\]  
\[h_{con,gl-sky}=2.8+3V;for\ V<5\ \text{m/s} \tag{14}\]  
\[h_{rad,gl-sky}=\epsilon_{eff}\sigma\bigg{[}\left(T_{gl}+273.15\right)^{4}- \left(T_{sky}+273.15\right)^{4}\bigg{]}/\left[T_{gl}-T_{sky}\right] \tag{15}\]  
The still's daily thermal efficiency is determined as follows [26]:  
\[h_{rad,w-gl}=\epsilon_{eff}\sigma\bigg{[}\left(T_{w}+273.15\right)^{2}-\left(T_{gl }+273.15\right)^{2}\bigg{]}\times\left[T_{w}+T_{gl}+546\right] \tag{7}\]0  
where \(h_{fg}\) is the water latent heat of vaporization in J/kg estimated by Equation (17) as follows [27]:  
\[h_{rad,w-gl}=\epsilon_{eff}\sigma\bigg{[}\left(T_{w}+273.15\right)^{2}-\left(T_{gl }+273.15\right)^{2}\bigg{]}\times\left[T_{w}+T_{gl}+546\right] \tag{7}\]1  
### Energy Analysis  
Exergy analysis of the still is performed by employing the governing equations formulated based on the second law of thermodynamics. Consequently, the exergy balance's general form can be expressed as described as follows [21]:  
\[h_{rad,w-gl}=\epsilon_{eff}\sigma\bigg{[}\left(T_{w}+273.15\right)^{2}-\left(T_{gl }+273.15\right)^{2}\bigg{]}\times\left[T_{w}+T_{gl}+546\right] \tag{7}\]2  
where \(E_{x,dest}\) refers to the consumed or destroyed exergy amount, \(E_{x,out. }\) and \(E_{x,in.