The present analysis represents models of the governing equations of the inclined single solar basin still by formulating both the temperature of the water flow and that of the absorber as functions of two independent parameters: the space coordinate along the axial distance for the brine water flow in the still as the first independent parameter and, the time as the second. Furthermore, the solar still was divided into axial elements with the assumption of linear variation of the temperature of the water flow, the cover, and the absorber along the element. In this way, only the time was kept as a single independent variable for each axial element, and the lumped model was employed for each element, which led to a coupled system of differential equations. Lumped models are frequently applied for analysis purposes in solar systems; Edenburn (1976), Garld and Kuehn (1989) and Ammari and Nimir (2003).
The resulting system of differential equations was, therefore, solved simultaneously and computationally using Butcher’s Runge–Kutta method of the fifth order, Chapra and Canale (1990). The temperature of each of the water flow, the glass cover, and the absorber elements as functions of the axial direction and time were predicted. In addition, the performance of the solar still was evaluated by computing the system efficiency over the period of a day. The inclined single solar basin still was computationally divided into a number of elements along the axial fluid flow direction. Each element was fragmented into three parts: the absorber plate, the brine water flowing over the absorber, and the glass cover as shown in Fig. 2. The energy balance equation was then applied simultaneously to each of the three parts of the elements. The solar absorption and energy transfer mechanisms are also shown in Fig. 2. The inclined single solar basin still was fully simulated as in the experimental setup. The simulation takes into account the position of the system to the south and inclined at an angle of 20°.
Governing equations
The energy conservation equation is applied simultaneously to each of the three parts of an element, glass cover, absorber plate and saline water, taking into consideration the unsteadystate thermal performance of each part (Ammari and Nimir 2003). Three separate control volumes are considered in the analysis of an element of the solar still: the glass cover, the absorber plate and part of the insulation, and the saline water as shown in Fig. 2. However, the following assumptions as reported by Shanmugan (2014) are considered for the energy balance equations of glass cover, absorber plate, and saline water:

There is no vapor leakage in solar still.

It is an air tight basin and hence no heat loss.

There is no temperature gradient across the basin water and glass cover of solar still.

Water level inside the basin maintained at constant level.

The governing heat transfer coefficients in the still are temperaturedependent.
Considering an element of length Δx and width W, the energy balance equation for the control volume of the glass cover is
$$ q_{I,g} + q_{c,w  g} + q_{r,w  g} + q_{evap} = q_{c,g  a} + q_{r,g  a} + \frac{{\partial e_{g} }}{\partial t}, $$
(1)
where
$$ q_{I,g} = I(t)A(1  \beta_{g} )\alpha_{g} :{\text{ heat transfer by radiation from ambient air to glass cover}} $$
(2a)
$$ q_{c,w  g} = Ah_{c,w  g} ((T_{w} (x,t)  T_{g} (x,t)):{\text{ heat transfer by convection from water to glass cover}} $$
(2b)
$$ q_{r,w  g} = Ah_{r,w  g} \left( {(T_{w} (x,t)  T_{g} (x,t)} \right):{\text{ heat transfer by radiation from water to glass cover}} $$
(2c)
$$ q_{\text{evap}} = Ah_{\text{evap}} \left( {(T_{w} (x,t)  T_{g} (x,t)} \right):{\text{ heat transfer of evaporation}} $$
(2d)
$$ q_{c,g  a} = Ah_{c,g  a} \left( {(T_{g} (x,t)  T_{a} (x,t)} \right):{\text{ heat transfer by convection from glass cover to ambient air}} $$
(2e)
$$ q_{r,g  a} = Ah_{r,g  a} \left( {(T_{g} (x,t)  T_{\text{sky}} (x,t)} \right):{\text{heat transfer by radiation from glass cover to ambient air}} $$
(2f)
$$ \frac{{\partial e_{g} }}{\partial t} = \rho_{g} {\text{Cp}}_{g} V_{g} \left( {\frac{{\partial T_{g} \left( {x,t} \right)}}{\partial t}} \right){: }{\text{rate of energy change of glass cover}} $$
(2g)
By substitution of Eqs. (2a–2g) into Eq. (1) and rearrangement of parameters, the following differential equation is obtained for the temperature of the glass cover element:
$$ \frac{{\partial T_{g} \left( {x,t} \right)}}{\partial t} = \frac{A}{{\rho_{g} Cp_{g} V_{g} }}\left[ \begin{aligned} I\left( t \right)\,\left( {1  \beta_{g} } \right)\alpha_{g} + \left( {h_{c,w  g} + h_{r,w  g} + h_{evap} } \right)\,\left( {T_{w} \left( {x,t} \right)  T_{g} \left( {x,t} \right)} \right) \hfill \\  h_{c,g  a} \left( {T_{g} \left( {x,t} \right)  T_{a} \left( t \right)} \right)  h_{r,g  a} \left( {T_{g} \left( {x,t} \right)  T_{sky} \left( t \right)} \right) \hfill \\ \end{aligned} \right] $$
(3)
The energy balance equation for the absorber plate and part of the insulation is
$$ q_{I,p} = q_{c,p  w} + q_{{p,{\text{loss}}}} + \frac{{\partial e_{p} }}{\partial t}, $$
(4)
where
$$ q_{I,p} = I(t)A\,\tau_{g} \tau_{w} \,\alpha_{p} :{\text{ heat transfer by radiation to absorber plate}} $$
(5a)
$$ q_{c,p  w} = Ah_{c,p  w} \left( {(T_{p} (x,t)  T_{w} (x,t)} \right):{\text{ heat transfer by convection from absorber plate to water}} $$
(5b)
$$ q_{{p,\,{\text{loss}}}} = q_{{p,{\text{bottom}}}} + q_{{p,\,{\text{side}}}} :{\text{ heat loss from the absorber plate}} $$
(5c)
$$ q_{{p,\,{\text{bottom}}}} = {{A\left[ {T_{p} (x,t)  T_{a} (x,t)} \right]} \mathord{\left/ {\vphantom {{A\left[ {T_{p} (x,t)  T_{a} (x,t)} \right]} {R_{{b,\,{\text{total}}}} }}} \right. \kern0pt} {R_{{b,\,{\text{total}}}} }}:{\text{ heat transfer from the bottom of the absorber to air}} $$
(5d)
$$ R_{{b,\,{\text{total}}}} = R_{{b,\,{\text{insulation}}}} + R_{{b,\,{\text{casing}}}} + R_{{b,\,{\text{convection}}}} $$
(5e)
$$ q_{{p,{\text{side}}}} = {{A_{s} \left[ {T_{p} (x,t)  T_{a} (x,t)} \right]} \mathord{\left/ {\vphantom {{A_{s} \left[ {T_{p} (x,t)  T_{a} (x,t)} \right]} {R_{{s,\,{\text{total}}}} }}} \right. \kern0pt} {R_{{s,\,{\text{total}}}} }}:{\text{ heat transfer from the side of the absorber to air}} $$
(5f)
$$ R_{{s,\,{\text{total}}}} = R_{{s,\,{\text{insulation}}}} + R_{{s,\,{\text{casing}}}} + R_{{s,\,{\text{convection}}}} :{\text{ total thermal resistance}} $$
(5g)
$$ \frac{{\partial e_{p} }}{\partial t} = \left( {\rho_{p} Cp_{p} V_{p} + r\rho_{i} Cp_{i} V_{i} } \right)\left( {\frac{{\partial T_{p} \left( {x,t} \right)}}{\partial t}} \right):{\text{ rate of energy change of the absorber plate}} $$
(5h)
And, r represents the percentage of the insulation layer attached to the absorber plate assumed to have the same temperature as the plate.
Substituting Eqs. (5a–5h) into Eq. (4) and rearranging the parameters yields the following differential equation for the temperature of the absorber plate, pipe and insulation part of the element,
$$ \frac{{\partial T_{p} \left( {x,t} \right)}}{\partial t} = \frac{1}{{\rho_{p} Cp_{p} V_{p} + r\rho_{i} Cp_{i} V_{i} }}\left[ \begin{aligned} AI\left( t \right)\,\alpha_{p} \tau_{g}  Ah_{c,p  w} \left( {T_{p} \left( {x,t} \right)  T_{w} \left( {x,t} \right)} \right) \hfill \\  \left( {\frac{A}{{R_{{b,{\text{total}}}} }} + \frac{{A_{s} }}{{R_{{s,{\text{total}}}} }}} \right)\left( {T_{p} \left( {x,t} \right)  T_{a} \left( t \right)} \right) \hfill \\ \end{aligned} \right] $$
(6)
The energy balance equation for the part of the element
of brine water film flowing over the absorber plate is
$$ q_{I,w} + q_{c,p  w} + q_{in} = q_{c,w  g} + q_{r,w  g} + q_{\text{evap}} + q_{\text{out}} + \frac{{\partial e_{w} }}{\partial t}, $$
(7)
where
$$ q_{I,w} = I(t)A\,\tau_{g} \,\alpha_{w} :{\text{ heat transfer by radiation to water film}} $$
(8a)
$$ q_{c,p  w} = Ah_{c,p  w} \left( {(T_{p} (x,t)  T_{w} (x,t)} \right):{\text{ heat transfer by convection from absorber to water film}} $$
(8b)
$$ q_{\text{in}} = \dot{m}_{w} C_{pw} T_{w,i} (x,t):{\text{ heat in the water film element of tilted solar still}}. $$
(8c)
$$ q_{c,w  g} = Ah_{r,w  g} \left( {(T_{w} (x,t)  T_{g} (x,t)} \right):{\text{ heat transfer by radiation from water to glass cover}} $$
(8d)
$$ q_{r,w  g} = Ah_{r,w  g} \left( {(T_{w} (x,t)  T_{g} (x,t)} \right):{\text{ heat transfer by radiation from water to glass cover}} $$
(8e)
$$ q_{\text{evap}} = Ah_{\text{evap}} \left( {(T_{w} (x,t)  T_{g} (x,t)} \right):{\text{ heat transfer of evaporation}} $$
(8f)
$$ q_{\text{out}} = (\dot{m}_{w}  \dot{m}_{c} )C_{pw} T_{w,o} (x,t):{\text{ heat out from the water film element of the tilted solar still}}. $$
(8g)
$$ \frac{{\partial e_{w} }}{\partial t} = \rho_{w} Cp_{w} V_{w} \left( {\frac{{\partial T_{w} \left( {x,t} \right)}}{\partial t}} \right):{\text{ rate of energy change of water film}} $$
(8h)
The implemented lumped model assumes linear temperature distribution across the fluid element, that is,
$$ \begin{aligned} & T_{w} \left( {x,t} \right) = \frac{1}{2}\left[ {T_{w,\,i} \left( {x,t} \right) + T_{w,\,o} \left( {x,t} \right)} \right], \\ &\quad {\text{and}}\quad \frac{{\partial T_{w} \left( {x,t} \right)}}{\partial t} \cong \frac{{\partial T_{w,o} \left( {x,t} \right)}}{\partial t} \\ \end{aligned} $$
These two assumptions are assumed viable for the brine water element, which is too short in length, regardless of the length of the absorber.
Therefore, the differential equation for the temperature of the brine water element becomes
$$ \frac{{\partial T_{w,0} \left( {x,t} \right)}}{\partial t} = \frac{1}{{\rho_{w} Cp_{w} V_{w} }}\left[ \begin{aligned} I\left( t \right)\,A\,\left( {1  \rho_{g}  \cdots } \right) + Ah_{c,p  w} \left( {T_{p} \left( {x,t} \right)  \frac{1}{2}\left( {T_{w,i} \left( {x,t} \right) + T_{w,0} \left( {x,t} \right)} \right)} \right) \hfill \\  A\,\left( {h_{c,w  g} + h_{r,w  g} + h_{evap} } \right)\left( {\frac{1}{2}\left( {T_{w,i} \left( {x,t} \right) + T_{w,0} \left( {x,t} \right)} \right)  T_{g} \left( {x,t} \right)} \right) \hfill \\ + \dot{m}_{w} Cp_{w} \left( {T_{w,i} \left( {x,t} \right)  T_{w,0} \left( {x,t} \right)} \right)  \dot{m}_{c} Cp_{w} T_{w,0} \left( {x,t} \right) \hfill \\ \end{aligned} \right] $$
(9)
Note that
$$ \begin{aligned} A &= W\Delta x \\ V_{w} &= \delta W\Delta x, \end{aligned} $$
where δ is the thickness of the brine water film layer over the absorber plate, which was assumed based on the experimental observations, and it was on average around 0.1 mm.
The condensation rate \( \dot{m}_{c} \) is obtained by
$$ \dot{m}_{c} = \frac{{q_{\text{evap}} }}{{h_{fg} }} = \frac{{Ah_{\text{evap}} \left( {T_{w} \left( {x,t} \right)  T_{g} \left( {x,t} \right)} \right)}}{{h_{fg} }} $$
(10)
The performance of a solar basin still could be evaluated by calculating its efficiency, η(t), as
$$ \eta \left( t \right) = \frac{{q_{\text{evap}} }}{I\left( t \right)A} $$
(11)
The correlations employed to estimate the convection heat transfer coefficients, h
_{
c,w−g
}, h
_{
c,g−a
}, and h
_{
c,p−w
}, the radiation heat transfer coefficients, h
_{
r,w−g
} and h
_{
r,g−a
}, and the evaporating heat transfer coefficient h
_{evap} are given in “Appendix”. In the simulation process, the axial direction was for 0 ≤ x ≤ L, where L is the solar still total length, and the time was 12 h from sunrise to the sunset, 6 a.m. ≤ t ≤ 6 p.m.