Optimizing Gas-Filled Interspaces for a Computational Analysis of the Performance of Double-Glazed Photovoltaic Thermal Hybrid

In this work, we have chosen to apply our approach on the PV-T arrangement shown below Figure 1. To find the temperatures and pass the heat transfer coefficients, an iterative approach is used. In addition to being quantitatively investigated, the impacts of Six gases on useful energy, overall heat coefficients, and exit temperature are contrasted. The "$d=d_1=d_2$" gaps (which range from 2 to 25 mm) are taken to be equal and optimized in order to minimize heat loss and, as a result, maximize useful energy.

- The electrical analogy is used to establish heat transfer coefficients.

- Conditions of quasi-steady-state govern the system's operation.

- Flow direction is the sole determinant of air temperature.

- The flow of air is perpendicular to the cross-sectional plane.

- Radiant energy cannot be absorbed by the fluid.

- Energy radiation is absorbed by glass surfaces.

111.png

Figure 1. Cross section of the Photovoltaic-Thermal hybrid panel with a rectangular duct

2.1 Energy balance equations

The temperature of each collector component is presented in Figure 1. $\mathrm{T}_{\mathrm{g} 2}$ is the temperature of the top surface of glass 2. $\mathrm{T}_{\mathrm{g} 1}$ The other temperatures are of inner surface of glass 1 and the gas between them. placing the absorbent plate $T_p$ beneath the inner glass cover and separating them with a gas layer, between the $T_p$ absorber and the upper side of the bottom plate $\mathrm{T}_{\mathrm{Bp}}$, the fluid (air) to be heated circulates $T_f(i)$ in the PVT collector, which leads to an increase in the heat gain of this fluid. the energy balance equation is applied for each element as follows [5, 24, 48]:

2.1.1 The glass cover 2

$\mathrm{k}_{\mathrm{eq}, \mathrm{g} 1-\mathrm{g} 2} \cdot\left(\mathrm{T}_{\mathrm{g} 1}-\mathrm{T}_{\mathrm{g} 2}\right)+\mathrm{S}_2=\left(\mathrm{h}_{\mathrm{rg} 2-\mathrm{a}}+\mathrm{h}_w\right) \cdot\left(\mathrm{T}_{\mathrm{g} 1}-\mathrm{T}_{\mathrm{g} 2}\right)$                 (1)

$\mathrm{k}_{\mathrm{eq}, \mathrm{g} 1-\mathrm{g} 2}=$ standing for the double-glass equivalent transfer coefficient $\left(\frac{\mathrm{W}}{\left(\mathrm{K} \cdot m^2\right)}\right)$.

$\mathrm{k}_{\mathrm{eq}, \mathrm{g} 1-\mathrm{g} 2}=\left[\left(e_{\mathrm{g} 1} / K_{\mathrm{g} 1}\right)+\left(e_{\mathrm{g} 2} / K_{\mathrm{g} 2}\right)+\left(h_{\mathrm{cg} 1-\mathrm{g} 2}+h_{\mathrm{rg} 1-\mathrm{g} 2}\right)\right]^{-1}$

$\mathrm{S}_2=$ The energy that glass cover 2 absorbs $\left(\frac{\mathrm{W}}{\left(\mathrm{m}^2\right)}\right)$, which is stated as:

$S_2=\alpha_{g 2} \cdot S$

2.1.2 The glass cover 1

$\begin{array}{r}\left(h_{r, p-g 1}+h_{c, p-g 1}\right) \cdot\left(T_p-T_{g 1}\right)+S_1 =U_{g 1-a} \cdot\left(T_{g 1}-T_a\right)\end{array}$        (2)

$\mathrm{S}_1=$ The energy that glass cover 1 absorbs $\left(\frac{\mathrm{W}}{\left(m^2\right)}\right)$, which is stated as:

$\mathrm{U}_{\mathrm{g} 1-\mathrm{a}}=$ The heat transfer coefficient $\left(\frac{\mathrm{W}}{\left(\mathrm{K} \cdot \mathrm{m}^2\right)}\right)$ between glass cover 1 and the ambient.

$S_1=\alpha_{g 1} \cdot \tau_{g 2} \cdot S$

$\begin{gathered}\mathrm{U}_{\mathrm{g} 1-\mathrm{a}}=\left[\left(h_{\mathrm{cg} 2-\mathrm{a}}+h_{\mathrm{rg} 2-\mathrm{a}}\right)^{-1}+\left(e_{\mathrm{g} 1} / K_{\mathrm{g} 1}\right)+\left(e_{\mathrm{g} 2} / K_{\mathrm{g} 2}\right)\right.  \left.+\left(h_{\mathrm{cg} 2-\mathrm{g} 1}+h_{\mathrm{rg} 2-\mathrm{g} 1}\right)^{-1}\right]^{-1}\end{gathered}$

2.1.3 The absorber (PV module)

This is the sum of all solar energy that reaches the absorber plate minus the amount of energy that is transformed into electrical energy, denoted as $S_p$ :

$\begin{gathered}S_p=h_{c p-f} \cdot\left(T_p-T_f\right)+\left(h_{r, p-g 1}+h_{c, p-g 1}\right) \cdot\left(T_p-T_{g 1}\right) \\ +h_{r, p-B p} \cdot\left(T_p-T_{B p}\right)\end{gathered}$                 (3)

Or,

$\begin{gathered}S_p=h_{c p-f} \cdot\left(T_p-T_f\right)+U_T \cdot\left(T_p-T_a\right)+h_{r, p-B p} \\ \cdot\left(T_p-T_{B p}\right)\end{gathered}$             (4)

$\mathrm{U}_{\mathrm{T}}=$ the top heat loss coefficient $\left(\frac{\mathrm{W}}{\left(\mathrm{K} \cdot \mathrm{m}^2\right)}\right)$ between the absorber and the ambient.

$\begin{aligned} & \mathrm{U}_{\mathrm{T}}=\left[\left(h_{\mathrm{cg} 2-\mathrm{a}}+h_{\mathrm{rg} 2-\mathrm{a}}\right)^{-1}+\left(e_{\mathrm{g} 1} / K_{\mathrm{g} 1}\right)+\left(e_{\mathrm{g} 2} / K_{\mathrm{g} 2}\right)\right.  &\left.+\left(h_{\mathrm{cg} 2-\mathrm{g} 1}+h_{\mathrm{rg} 2-\mathrm{g} 1}\right)^{-1}+\left(h_{\mathrm{cp}-\mathrm{g} 1}+h_{\mathrm{rp}-\mathrm{g} 1}\right)^{-1}\right]^{-1}\end{aligned}$

where,

$\mathrm{S}_{\mathrm{p}}=\tau_{g 1} \cdot \tau_{g 2} \cdot \tau_{p o} S\left[\alpha_p(1-F)+\alpha_{p v} \cdot F\left(1-\eta_{p v}\right)\right]$     (5)

$\eta_{\mathrm{PV}}$ this is how the PV module's electrical efficiency is determined [49-52]:

$\eta_{\mathrm{Pv}}=\eta_{r f}\left(1-\beta_{r f}\left(T_p-T_{r f}\right)\right)$            (6)

where, $\eta_{\mathrm{rf}}$ is the reference electrical efficiency at standard conditions $\left(S=1000 \mathrm{~W} / \mathrm{m}^2\right.$ and $\left.\mathrm{T}_{\mathrm{rf}}=298.15 \mathrm{~K}\right)$ [53]. The temperature coefficient is assumed as $\beta_{\mathrm{rf}}=0.0041 \mathrm{~K}^{-1}$ for crystalline silicon modules [54]. 

Eqs. (5) and (6) can be combined to give the equation:

$\begin{aligned} S_p=\tau_{g 1} \cdot \tau_{g 2} \cdot \tau_{p o} S[ & \alpha_p(1-F) \\ & \left.+\alpha_{p v} \cdot F\left(1-\eta_{r f}\left(1-\beta_{r f}\left(T_p-T_{r f}\right)\right)\right)\right]\end{aligned}$              (7)

2.1.4 A working fluid's steady-state energy equilibrium equation [55, 56]

$\frac{\dot{m} . \mathrm{C}_f}{W} \frac{\mathrm{d} T_f}{\mathrm{dx}}=h_{c B p-f}\left(T_{B p}-T_f\right)+h_{c p-f}\left(T_p-T_f\right)$            (8)

The coefficients of heat loss for the absorber plate in relation to both air and the bottom plate.

$U_{p-f}=\left[\frac{1}{h_{c p-f}}+\frac{e_p}{K_p}+\frac{A_e e_i}{A_c K_{i n s}}\right]^{-1}$

$U_{p-B p}=\left[\frac{1}{h_{r p-B p}}+\frac{e_p}{K_p}\right]^{-1}$

2.1.5 The bottom plate

$\begin{aligned} & h_{r, p-B p} \cdot\left(T_p-T_{B p}\right) =h_{c B p-f} \cdot\left(T_{B p}-T_f\right)+U_b  \cdot\left(T_{B p}-T_a\right)\end{aligned}$              (9)

$\mathrm{U}_{\mathrm{b}}=\left[\frac{1}{\mathrm{~h}_{\mathrm{w}}}+\frac{\mathrm{e}_{\mathrm{ins}}}{\mathrm{K}_{\mathrm{ins}}}+\frac{\mathrm{A}_{\mathrm{e}} \mathrm{e}_{\mathrm{i}}}{\mathrm{A}_{\mathrm{c}} \mathrm{K}_{\mathrm{ins}}}\right]^{-1}$

Finally, Eqs. (3), (8) and (9) can be written $h_{c p-f}$ and $h_{r p-B p}$ are substituted by $U_{p-f}$ and $U_{p-B p}$, respectively.

$\begin{aligned} & U_{p-f} \cdot\left(T_p-T_f\right)+\left(h_{r, p-g 1}+h_{c, p-g 1}\right) \cdot\left(T_p-T_{g 1}\right)+U_{p-B p} \cdot\left(T_p-T_{B p}\right)=S_p\end{aligned}$                  (10)

$\begin{aligned} & U_{p-B p} \cdot\left(T_p-T_{B p}\right) \\ & \quad=h_{c B p-f} \cdot\left(T_{B p}-T_f\right)+U_b \quad \cdot\left(T_{B p}-T_a\right)\end{aligned}$                  (11)

$\frac{\dot{m}_f . C_f}{W} \frac{d T_f}{d x}=h_{c B p-f}\left(T_{B p}-T_f\right)+U_{p-f}\left(T_p-T_f\right)$         (12)

By combining and arranging Eqs. (10) and (12):

$\left(T_p-T_f\right)=S_p \cdot x_1-y_1\left(T_f-T_a\right)$                (13)

$\left(T_{B p}-T_f\right)=\mathrm{S}_{\mathrm{p}} \cdot \mathrm{x}_2-y_2\left(T_f-T_a\right)$                    (14)

The coefficients are $\mathrm{x}_1, \mathrm{x}_2$ and $\mathrm{y}_1, \mathrm{y}_2$ :

$\begin{aligned} & \mathrm{X}_1  =\frac{\left(U_{p-B p}+U_b+h_{c B p-f}\right)}{\left[\left(U_{p-f}+U_T+U_{p-B p}\right)\left(U_{p-B p}+U_b+h_{c B p-f}\right)-U_{p-B p}{ }^2\right]}\end{aligned}$               (15)

$\mathrm{y}_1=\frac{U_T\left(U_p-B p+U_b+h_{c ~ B p-f}\right)+U_{p-B p} U_b}{\left[\left(U_{p-f}+U_T+U_{p-B p}\right)\left(U_{p-B p}+U_b+h_{C B p-f}\right)-U_{p-B p}{ }^2\right]}$                (16)

$\begin{aligned} & \mathrm{x}_2  =\frac{U_{p-B p}}{\left[\left(U_{p-f}+U_T+U_{p-B p}\right)\left(U_{p-B p}+U_b+h_{c B p-f}\right)-U_{p-B p}{ }^2\right]} \\ & \end{aligned}$                    (17)

$\mathrm{y}_2=\frac{w}{\left[\left(\mathrm{U}_{\mathrm{p}-\mathrm{f}}+\mathrm{U}_{\mathrm{T}}+\mathrm{U}_{\mathrm{p}-\mathrm{Bp}}\right)\left(\mathrm{U}_{\mathrm{p}-\mathrm{Bp}}+\mathrm{U}_{\mathrm{b}}+\mathrm{h}_{\mathrm{cBp-f}}\right)-\mathrm{U}_{\mathrm{p}-\mathrm{Bp}}{ }^2\right]}$          (18)

$\begin{aligned} w=\left(U_{p-f}\right. & \left.+U_T+U_{p-B p}\right)\left(U_{p-B p}+U_b+h_{c B p-f}\right) +U_T U_{p-B p} -\left(U_{p-f}+U_T+U_{p-B p}\right)\left(U_{p-B p}+h_{c B p-f}\right)\end{aligned}$

Substituting Eqs. (13), (14) with Eq. (12) can be written

$\frac{\dot{m_f} \cdot C_f}{W} \frac{d T_f}{d x}=\mathrm{S}_{\mathrm{p}}\left(\mathrm{x}_2 \cdot h_{c B p-f}-\mathrm{x}_1 \cdot U_{p-f}\right)+\left(y_2 \cdot h_{c B p-f}-y_1 U_{p-f}\right)\left(T_f-T_a\right)$           (19)

If:

$\mathrm{A}=\left(\mathrm{x}_2 \cdot h_{c B p-f}-\mathrm{x}_1 \cdot U_{p-f}\right)$                  (20)

$\mathrm{B}=\left(y_2 \cdot h_{c B p-f}-y_1 U_{p-f}\right)$                   (21)

(19) becomes:

$\frac{\dot{m_f} \cdot C_f}{W} \frac{d T_f}{d x}=\frac{\mathrm{d} \emptyset_u}{\mathrm{dA}_c}=A \cdot \mathrm{S}_{\mathrm{p}}-\mathrm{B}\left(T_f-T_a\right)$                 (22)

$\begin{aligned} \frac{\mathrm{d} \emptyset_u}{\mathrm{dA}_c}=A \cdot\left(\mathrm{S}_{\mathrm{p}}-\frac{\mathrm{B}}{A}\right. & \left.\left(T_f-T_a\right)\right) =F^{\prime} \cdot\left(\mathrm{S}_{\mathrm{p}}-\mathrm{U}_L\left(T_f-T_a\right)\right)\end{aligned}$         (23)

where, $F^{\prime}=A$ and $\mathrm{U}_L=\frac{\mathrm{B}}{A}$.

Ultimately, we get the following after solving this final equation and applying the boundary conditions:

This is the useful heat energy of the considered FPPV-T [57, 58]:

$\emptyset_u=\mathrm{A}_c \mathrm{~F}_R\left(\mathrm{S}_{\mathrm{p}}-\mathrm{U}_L\left(T_{f i}-T_a\right)\right)$             (24)

$\mathrm{F}_R=\frac{\dot{m}_f \cdot C_f}{\mathrm{A}_c \cdot \mathrm{U}_L}\left(1-\exp \left(-\frac{F^{\prime} \cdot \mathrm{A}_c \cdot \mathrm{U}_L}{\dot{m}_f \cdot C_f}\right)\right)$            (25)

By discretizing the working fluid equation using the Crank-Nicolson technique, we get (12) as a solution.

$\begin{aligned} \frac{T_f(i+1)-T_f(i)}{\Delta x} & =\frac{W}{\dot{m}_f \cdot C_f}\left(h_{c B p-f} T_{B p}+U_{p-f} T_p\right) -\frac{1}{2} \frac{W}{\dot{m}_f \cdot C_f}\left(h_{c B p-f}\right.  \left.+U_{p-f}\right)\left(T_f(i+1)+T_f(i)\right)\end{aligned}$               (26)

Eqs. (1)-(2)-(10) and (11) and (26) can be expressed as a 5 $\times 5$ matrix $([a][T]=[b])$, as follows:

$\left[\begin{array}{ccccc}a_1 & a_2 & 0 & 0 & 0 \\ 0 & a_3 & a_4 & 0 & 0 \\ 0 & a_5 & a_6 & a_7 & a_8 \\ 0 & 0 & a_9 & a_{10} & a_{11} \\ 0 & 0 & a_{12} & a_{13} & a_{14}\end{array}\right]\left[\begin{array}{c}T_{g 2} \\ T_{g 1} \\ T_p \\ T_{B p} \\ T_f(i+1)\end{array}\right]=\left[\begin{array}{l}b_1 \\ b_2 \\ b_3 \\ b_4 \\ b_5\end{array}\right]$                   (27)

where:

$a_1=\left(k_{e q, g 1-g 2}+h_{r g 2-a}+h_w\right)$

$a_2=-k_{e q, g 1-g 2}$

$a_3=\left(h_{r, p-g 1}+h_{c, p-g 1}+U_{g 1-a}\right)$

$a_4=-\left(h_{r, p-g 1}+h_{c, p-g 1}\right)$

$a_5=-\left(h_{r, p-g 1}+h_{c, p-g 1}\right)$

$a_6=\left(h_{r, p-g 1}+h_{c, p-g 1}+h_{c, p-f}+U_{p-B p}\right)$

$a_7=-U_{p-B p}$

$a_8=-h_{c, p-f}$

$a_9=U_{p-B p}$

$a_{10}=-\left(U_{p-B p}+U_p+h_{c, B p-f}\right)$

$a_{11}=h_{c, B p-f}$

$a_{12}=\frac{\Delta x \cdot W}{\dot{m}_f \cdot C_f} U_{p-f}$

$a_{13}=\frac{\Delta x \cdot W}{\dot{m}_f \cdot C_f} h_{c B p-f}$

$a_{14}=-\left(1+\frac{\Delta x \cdot W}{2 \cdot \dot{m}_f \cdot C_f}\left(h_{c B p-f}+U_{p-f}\right)\right)$

$b_1=S_2+\left(h_{r g 2-a}+h_w\right) \cdot T_a$

$b_2=S_1+U_{g 1-a} \cdot T_a$

$b_3=S_p$

$b_4=-U_p \cdot T_a$

$b_5=\left(\frac{\Delta x \cdot W}{2 \cdot \dot{m}_f \cdot C_f}\left(h_{c B p-f}+U_{p-f}\right)-1\right) \cdot T_f(i)$

2.2 Natural convection and radiative coefficients related to the glass cover g2

In Eq. (28) and Eq. (29) we applied the following formulas $\left(h_w=h_{c g 2-a}\right)$ for the natural convection that happens when the wind moves across glass cover 2 to the atmosphere [55, 59, 60].

$\mathrm{hw}=3.0 \cdot \mathrm{V}_w+2.8 \quad \mathrm{~V}_w \leq 5 \mathrm{~m} / \mathrm{s}$          (28)

$\mathrm{hw}=6.15 \cdot \mathrm{V}_w^{0.8} \quad \mathrm{~V}_w>5 \mathrm{~m} / \mathrm{s}$       (29)

The coefficient of radiative heat loss to sky:

$\mathrm{h}_{\mathrm{rg} 2-\mathrm{a}}=\sigma \cdot \varepsilon_{\mathrm{g} 2} \cdot \frac{\left(\mathrm{T}_{\mathrm{g} 2}^4-\mathrm{T}_{\mathrm{s}}^4\right)}{\left(\mathrm{Tg}_2-\mathrm{T}_{\mathrm{a}}\right)}$               (30)

$\mathrm{T}_{\mathrm{s}}=$ Sky temperature $(\mathrm{K})$. The empirical relation is used to calculate it [57, 61]:

$\begin{aligned} T_s=\mathrm{T}_a(0.711+ & 0.0056 T_{d p}+0.000073 T_{d p}^2  +0.013 \cos (15 t))^{0.25}\end{aligned}$        (31)

$0^{\circ} \mathrm{C}<\mathrm{T}_{\mathrm{dp}}<93^{\circ} \mathrm{C}$

$\begin{aligned} & \mathrm{T}_{\mathrm{dp}}=6.54+14.526 \ln \left(\mathrm{p}_{\mathrm{w}}\right)+0.7389 \ln \left(\mathrm{p}_{\mathrm{w}}\right)^2 + 0.09486 \ln \left(\mathrm{p}_{\mathrm{w}}\right)^3 \\ &+0.4569\left(\mathrm{p}_{\mathrm{w}}\right)^{0.1984}\end{aligned}$         (32)

$T_{d p}=$ the dew-point temperature $(\mathrm{K})$ can be calculated directly by one of the following equations [62]:

$T_{d p}<0^{\circ} \mathrm{C}$

$T_{d p}=6.09+12.60 \ln \left(p_w\right)+0.4959 \ln \left(p_w\right)^2$                  (33)

$\mathrm{RH}=100 . \phi$ And: $\phi=p_w / p_{s w}$

$\mathrm{p}_{\mathrm{sw}}=$ the atmospheric pressure at which vaporization occurs (in pascal) [63]:

for $T>0{ }^{\circ} \mathrm{C}$

$\begin{aligned} \mathrm{p}_{\mathrm{sw}}(\mathrm{T})=0.61121 \cdot \exp \left(\left(18.678 -\frac{\mathrm{T}}{234.5}\right)\left(\frac{\mathrm{T}}{257.14+\mathrm{T}}\right)\right)\end{aligned}$                   (34)

for $T<0{ }^{\circ} \mathrm{C}$

$\begin{aligned} \mathrm{p}_{\mathrm{sw}}(\mathrm{T})=0.61115 \cdot \exp \left(\left(23.036 -\frac{\mathrm{T}}{233.7}\right)\left(\frac{\mathrm{T}}{279.82+\mathrm{T}}\right)\right)\end{aligned}$                  (35)

2.3 Natural convection coefficients in the double-glazing and between the inner glass cover and the absorber

The coefficients of radiative heat transfer between the two layers of glass, the selective absorber and the inner glass cover, the absorber and the bottom plate are, respectively.

$h_{r g 2-g 1}=\sigma \cdot \frac{\left(T_{g 1}+T_{g 2}\right)\left(T_{g 1}{ }^2+T_{g 2}{ }^2\right)}{\left(\frac{1}{\varepsilon_{g 1}}+\frac{1}{\varepsilon_{g 2}}-1\right)}$              (36)

$\mathrm{h}_{r g 1-p}=\sigma \cdot \frac{\left(\mathrm{T}_{g 1}+\mathrm{T}_p\right)\left(\mathrm{T}_{\mathrm{g} 1}{ }^2+\mathrm{T}_{\mathrm{p}}{ }^2\right)}{\left(\frac{1}{\varepsilon_{g 1}}+\frac{1}{\varepsilon_p}-1\right)}$          (37)

$\mathbf{h}_{r p-B p}=\sigma \cdot \frac{\left(\mathbf{T}_p+\mathbf{T}_{B p}\right)\left(\mathbf{T}_{\mathrm{p}}{ }^2+\mathbf{T}_{\mathrm{Bp}}{ }^2\right)}{\left(\frac{1}{\varepsilon_p}+\frac{1}{\varepsilon_{B p}}-\mathbf{1}\right)}$            (38)

2.4 Natural convection coefficients in the double-glazing and between the inner glass cover and the absorber

The air in the gap between the plates is intended to be replaced by gases such as Argon, Krypton, Xenon, Carbon monoxide and Sulfur dioxide. The temperature that manifests at a distance x from the PV-T entry determines their thermo-physical characteristics.

$\rho_{g a z}$ the density of every gas is calculated by the formula:

$\rho_{g a z}=\left(P_a \cdot M_m\right) /\left(Z \cdot R \cdot T_m\right)$                 (39)

2.4.1 Thermo-physical

The following polynomials can be used to compute thermal and transport properties [43, 64, 65].

The Air

$\begin{aligned} K_{\text {air }}=\left(0.0965 \cdot T_m\right. & -9.960 \cdot 10^{-6} \cdot T_m^2-9 \cdot 310 \cdot 10^{-8} \cdot T_m^3 \\ & \left.+8.882 \cdot 10^{-11} \cdot T_m^4\right) \cdot 10^{-3}\end{aligned}$               (40）

$\rho_{\text {air }}=\left(0.02897 * P_a\right) /\left(0.9997 * 8.314 * T_m\right)$             （41）

$\begin{aligned} & C_{p \text { air }}=1047.63657-0.372589265 \cdot T_m +9.4530421 \cdot 10^{-4} \cdot T_m^2 -6.02409443 \cdot 10^{-7} \cdot T_m^3  +1.2858961 \cdot 10^{-10} \cdot T_m^4\end{aligned}$               (42)

$\begin{gathered}\mu_{\text {air }}=7.72488 \cdot 10^{-8} \cdot T_m-5.95238 \cdot 10^{-11} \cdot T_m^2 \\ +2.71368 \cdot 10^{-14} \cdot T_m^3\end{gathered}$            (43)

The Argon gas

$\begin{aligned} & K_{A r}  =\left(0.0606 T_m\right. +3.151 \cdot 10^{-5} \cdot T_m^2-1 \cdot 525 \cdot 10^{-7} \cdot T_m^3  \left.+1.223 \cdot 10^{-10} \cdot T_m^4\right) \cdot 10^{-3}\end{aligned}$                 (44)

$\rho_{A r}=P_a /\left(0.99937 * 209.17 * T_m\right)$                 (45)

$C_{p A r}=521.55 \mathrm{~J} /(\mathrm{Kg} \cdot \mathrm{K})$         (46)

$\begin{aligned} & \mu_{A r}=7 \cdot 91722 \cdot 10^{-8} \cdot T_m+2.93448 \cdot 10^{-11} \cdot T_m^2  -1.73227 \cdot 10^{-13} \cdot T_m^3 +1.41721 \cdot 10^{-16} \cdot T_m^4  \end{aligned}$                 (47)

The Krypton gas

$\begin{aligned} & K_{K r}  =\left(0.327 T_m+1.632 \cdot 10^{-6} \cdot T_m^2-2 \cdot 060 \cdot 10^{-8} \cdot T_m^3\right. \\ & \left.+1.042 \cdot 10^{-11} \cdot T_m^4\right) \cdot 10^{-3}\end{aligned}$                    (48)

$\rho_{K r}=P_a /\left(0.99793 * 100.18 * T_m\right)$             (49)

$C_{p K r}=249.2 \mathrm{~J} /(\mathrm{Kg} . \mathrm{K})$            (50)

$\begin{aligned} & \mu_{K r}=8.56676 \cdot 10^{-8} \cdot T_m+1.91155 \cdot 10^{-11} \cdot T_m^2 \\ & -1.05643 \cdot 10^{-13} \cdot T_m^3 \\ & +1.09675 .10^{-16} \cdot T_m^4 \\ & \end{aligned}$             (51)

The Xenon gas

$\begin{aligned} & K_{X e}=\left(0.0209 T_m-1.629 \cdot 10^{-5} \cdot T_m^2\right. \\ &+3.703 \cdot 10^{-8} \cdot T_m^3 \\ &\left.-3.322 \cdot 10^{-11} \cdot T_m^4\right) \cdot 10^{-3}\end{aligned}$            (52)

$\rho_{X e}=P_a /\left(0.99471 * 64.645 * T_m\right)$             (53)

$C_{p X e}=160.09 \mathrm{~J} /(\mathrm{Kg} . \mathrm{K})$              (54)

$\begin{gathered}\mu_{X e}=7.33337 \cdot 10^{-8} \cdot T_m+3.93839 \cdot 10^{-11} \cdot T_m^2  -1.05562 \cdot 10^{-13} \cdot T_m^3 +6.29110 \cdot 10^{-17} \cdot T_m^4\end{gathered}$              (55)

The Carbon Monoxide gas

$\begin{aligned} K_{C O 2}=(0.0872 & T_m+1.659 \cdot 10^{-6} \cdot T_m^2 \\ & -6.481 \cdot 10^{-8} \cdot T_m^3 \\ & \left.+5.244 \cdot 10^{-11} \cdot T_m^4\right) \cdot 10^{-3}\end{aligned}$           (56)

$\rho_{\mathrm{CO2}}=P_a /\left(0.99964 * 298.4 * T_m\right)$       (57)

$C_{p \text { CO2 }}=1042.1 \mathrm{~J} /(\mathrm{Kg} . \mathrm{K})$            (58)

$\begin{aligned} & \mu_{C 02}=7 \cdot 11110 \cdot 10^{-8} \cdot T_m-2.80674 \cdot 10^{-11} \cdot T_m^2 \\ &-5.36367 \cdot 10^{-14} \cdot T_m^3 \\ &+6.27741 \cdot 10^{-17} \cdot T_m^4\end{aligned}$             (59)

The Sulfur dioxide

$\begin{aligned} & K_{N e}=\left(0.0475 T_m-1.622 \cdot 10^{-4} \cdot T_m^2\right. \\ &+4.816 \cdot 10^{-7} \cdot T_m^3 \\ &\left.-3 \cdot 747 \cdot 10^{-10} \cdot T_m^4\right) \cdot 10^{-3}\end{aligned}$          (60)

$\rho_{N e}=P_a /\left(0.98285 * 143.74 * T_m\right)$          (61)

$C_{p N e}=656.2 \mathrm{~J} /(\mathrm{Kg} \cdot \mathrm{K})$             (62)

$\begin{aligned} & \mu_{N e}=5.28546 \cdot 10^{-8} \cdot T_m-1.02879 \cdot 10^{-10} \cdot T_m^2 \\ & +3.28719 .10^{-13} \cdot T_m^3 \\ & -3.21789 .10^{-16} \cdot T_m^4 \\ & \end{aligned}$            (63)

The formula determines the average temperature $T_m$ at local point (i) for the two glass covers and the inner glass cover with the absorber (PV module).

$T_m(i)=\frac{T_{g 1}(i)+T_{g 2}(i)}{2}$          (64)

$T_m(i)=\frac{T_{g 1}(i)+T_p(i)}{2}$             (65)

2.4.2 Natural convection in the filled-gas space and coefficients calculation

By calculating the Nusselt number and Rayleigh number provided by Eismann, we can assess the convective heat loss across the gap between the both panes of glass and the inside glass-cover and the absorber (PV module) which vary from 2 mm to 25 mm.

$R a_{g 1-g 2}^{\prime}=\frac{g \cdot \beta^{\prime} \cdot d^3 \cdot\left(T_{g 1}-T_{g 2}\right)}{v_{g a z} \cdot \alpha_{D g a s}} \cos (\beta)=R a \cos (\beta)$                (66)

And,

$\begin{gathered}R a_{g 1-p}^{\prime}=\frac{g \cdot \beta^{\prime} \cdot d^3 \cdot\left(T_p-T_{g 1}\right)}{v_{g a z} \cdot \alpha_{D \operatorname{gas}}} \cos (\beta) \\ =R \cos (\beta)\end{gathered}$                (67)

$\beta^{\prime}=1 / T_m, \alpha_{D_{\text {gas }}}={ }^K g a z /\left(C_{p_{\text {gas }} .} \rho_{\text {gaz }}\right), v_{g a z}={ }^{\mu g a z} / \rho_{\text {gaz }}$

The Hollands correlation [17] is used by a majority of researchers to anticipate the heat transfer coefficient in situations of natural convection. However, Eismann [19] has created an enhanced and more precise equation, which is being utilized in the current study. The formulation he derived is presented below:

$N u=N u_{\text {cond }}+N u_{\text {CONV } I}+N u_{\text {CONV II }}$             (68)

$\mathrm{Nu}_{\text {cond }}=1$          (69)

$\begin{aligned} N u_{\text {CONV } I}=1.44(1 & \left.-\frac{1708}{R a^{\prime}+1708 R c}\right)(1 \\ & \left.-\frac{1708(\sin (1.8 \beta))^{1.6}}{R a^{\prime}+1708 R c}\right)\end{aligned}$             (70)

$\mathrm{Nu}_{\text {CONV II }}=\left(\left(\frac{R a^{\prime}+5830 R c}{5830}\right)^{0.39}-1\right)(1+C \cdot R c)$               (71)

C = 0.29.

$\mathrm{Rc}=0$ is first iteration's starting value was taken into account. Upon completion of the initial cycle, the value is initialized to:

$\mathrm{Rc}=\exp \left(-\frac{A_C \cdot F^{\prime} \cdot U_L}{\dot{m}_f \cdot C_p}\right)$ 

Following are the estimated values of the heat transfer coefficients between the two panes of glass and between the inside of the glass and absorber:

$h_{c g 1-g 2}=\frac{N u_{g 1-g 2}+K_{g a z}}{d}$              (72)

$h_{\mathrm{c} g 1-\mathrm{p}}=\frac{\mathrm{Nu}_{\mathrm{g} 1-\mathrm{p}}+\mathrm{K}_{\mathrm{gaz}}}{d}$               (73)

2.5 Forced convection heat transfer coefficients

By analyzing the flow regime, one may determine the forced convection heat transfer coefficients between the plate bottom and airflow, and between the selective absorber (PV module) and airflow.

2.5.1 Laminar flow in an air duct causes forced convection. is modeled using the following correlations [66]

$\begin{aligned} \mathrm{Nu}=8.235[1- & 2.0421\left(\frac{\mathrm{H}}{W}\right)+3.0853\left(\frac{\mathrm{H}}{W}\right)^2 \\ & -2.4765\left(\frac{\mathrm{H}}{W}\right)^3+1.0578\left(\frac{\mathrm{H}}{W}\right)^4 \\ & \left.-0.1861\left(\frac{\mathrm{H}}{W}\right)^5\right]\end{aligned}$                (74)

The following equation is to approximate the friction factor:

$\begin{aligned} & \mathcal{F}=24\left[1-1.3553\left(\frac{\mathrm{H}}{W}\right)+1.9467\left(\frac{\mathrm{H}}{W}\right)^2\right. \\ &-1.7012\left(\frac{\mathrm{H}}{W}\right)^3+0.9564\left(\frac{\mathrm{H}}{W}\right)^4 \\ &\left.-0.2537\left(\frac{\mathrm{H}}{W}\right)^5\right]\end{aligned}$            (75)

2.5.2 Turbulent flow in an air duct causes forced convection. is modeled using the following correlations [67, 68]:

$\mathrm{Nu}=\frac{\left(\frac{\mathcal{F}}{8}\right)(R e-1000) \cdot \operatorname{Pr}}{1.07+12.7\left(\frac{\mathcal{F}}{8}\right)^{0.5} \operatorname{Pr}^{2 / 3}-1}$            (76)

For, $0.5<P r \leq 2000$ and $4000<R e \leq 5 \times 10^6$.

Considering the relative roughness $\left(r / D_h\right)$, the friction factor is given by:

$\begin{aligned} & \mathcal{F}=\left[-2 . \log _{10}\left(\frac{2 r}{7.54 D_h}\right.\right. \\ &\left.\left.-\frac{5.02}{R e} \log _{10}\left(\frac{2 r}{7.54 D_h}+\frac{13}{R e}\right)\right)\right]^{-2}\end{aligned}$                   (77)

For, $4000<R e \leq 10^8$ and $2.10^{-8}<\left(r / D_h\right) \leq 0.1$.

Here are the heat transfer coefficients in the region of the airflow and the absorber (PV module) and, the region of the bottom plate and the flowing air, respectively.

$\mathrm{h}_{c p-f}=\frac{N u_{p-f} \cdot K_{a i r}}{D_h}$             (78)

$\mathrm{h}_{c B p-f}=\frac{N u_{B p-f} \cdot K_{\text {air }}}{D_h}$                 (79)

2.6 Energy study

The PV/T air collectors are evaluated thermo-hydraulic and electrically based on a number of parameters, including effective thermal efficiency, pressure loss, power consumption by fans, and power generation by electrical means. The ratio of the total incident solar radiation to the heat benefit less the equivalent fan power is known as the effective thermal efficiency $\left(\eta_{t h}\right)$, and it can be expressed as follows:

$\eta_{t h}=\frac{\emptyset_u-P_f}{\left(S . A_c\right)}$                (80)

$\emptyset_u$ the useful heat collected is given by Eq. (24).

The equivalent electrical efficiency of PV panel $\left(\eta_{E P V}\right)$ i estimated as:

$\eta_{\mathrm{EPv}}=\frac{\eta_{\mathrm{Pv}}}{C_f}$              (81)

$\eta_{\mathrm{Pv}}$ the electrical efficiency of the PV module is calculated in Eq. (6).

$C_f$ is the conversion factor of the thermal power plant (in the range 0.29-0.4 [50-52, 69]), and assumed equal to 0.36 .

The total combined FPPV-T collector (hybrid) efficiency $\left(\eta_C\right)$ is obtained as follows [51, 69] :

$\eta_{\mathrm{C}}=\eta_{t h}+\eta_{\mathrm{EPv}}$                 (82)

2.7 Exergy study

Energy analysis is helpful in figuring out how well PV/T collectors work, how energy degrades during thermal and electrical conversion processes, and how to design and run PV/Ts to use energy as efficiently as possible.

The exergy analysis is conducted by considering the total exergy input, exergy outflow, and exergy destroyed from the system, in accordance with the principles of the second law of thermodynamics. The exergy efficiency of the FPPV-T air heater can be determined by calculating the ratio of the desired output exergy, known as product exergy, to the exergy inflow. Exergy is a measure of the energy's quality. The optimal spacing between the FPPV-T plates can be determined using an alternative approach that considers both the maximum exergy and the exergy efficiency. This method is used to verify the gaps identified by graphical analysis of maximum energy efficiency and usable energies.

The application of the exergy balance equation to the FPPV-T results in [58, 70-72]:

$\sum \dot{E}_{\text {int }}-\sum \dot{E}_{\text {out }}=\sum \dot{E}_{\text {des }}$                  (83)

Or,

$\dot{E}_{\text {des }}=\dot{E}_{\text {sun }}-\left(\dot{E}_{\text {air out }}-\dot{E}_{\text {air int }}\right)-\dot{E}_{p v}-\dot{E}_{\text {fan }}$                 (84)

The following represents the input exergy that solar radiation supplies which reaches the PV/T collector surface [32, 73]:

$\dot{E}_{\text {sun }}=S_p A_c\left[1-\frac{4}{3}\left(\frac{T_a}{T_{\text {sun }}}\right)+\frac{1}{3}\left(\frac{T_a}{T_{\text {sun }}}\right)^4\right]$                  (85)

The solar surface temperature is regarded as a significant contributor to exergy $T_{\text {sun }}=5770 \mathrm{~K}$.

The fan's exergy is quantified by:

$\dot{E}_{\text {fan }}=\frac{T_a}{T_{i n t}} P_{\text {fan }}$               (86)

And, $P_{\text {fan }}=\frac{m_f \cdot \Delta p}{\rho_{\text {air } \cdot} \eta_{f a n} \cdot \eta_{m o t}}$.

The thermal exergy:

$\dot{E}_{\text {th }}=\dot{E}_{\text {air out }}-\dot{E}_{\text {air int }}$               (87)

Or,

$\begin{aligned} \dot{E}_{t h} & =\dot{m}_f \cdot\left[\left(h_{\text {air out }}-h_{\text {air int }}\right)\right. \\ & \left.-T_a\left(S_{\text {air out }}-S_{\text {air int }}\right)\right]\end{aligned}$          (88)

The variables h and S represent the enthalpy and entropy of air, respectively, as well as the enthalpy and entropy differential between the inlet and output air masses.

$\begin{array}{r}\dot{E}_{\text {th }}=\emptyset_u-\dot{m}_f \cdot T_a\left(\mathrm{C}_p \ln \left(\frac{T_{\text {out }}}{T_{\text {int }}}\right)\right. \\ \left.-\mathrm{R}_{\text {air }} \ln \left(\frac{p_{\text {out }}}{p_{\text {int }}}\right)\right)\end{array}$                (89)

The electrical exergy [74]:

$\dot{E}_{p v}=\eta_{\mathrm{Pv}} \mathrm{A}_c \mathrm{~S}$                (90)

The photovoltaic thermal (PV-T) exergy:

$\dot{E}_{P V-T}=\dot{E}_{t h}+\dot{E}_{p v}$            (91)

According to Eq. (84), exergy destruction refers to the portion of solar radiation exergy that is not utilized by the system, in addition to the electrical and thermal exergy production. Therefore, it is represented as:

$\begin{aligned} & \dot{E}_{\text {des }}=\mathrm{S}_p \mathrm{~A}_c\left\lceil 1-\frac{4}{3}\left(\frac{T_a}{T_{\text {sun }}}\right)+\frac{1}{3}\left(\frac{T_a}{T_{\text {sun }}}\right)^4\right]-\dot{E}_{P V-T} \\ & -\left(\frac{T_a}{T_{i n t}} \frac{\dot{m}_f \cdot \Delta p}{\rho_{\text {air }} \cdot \eta_{f a n} \cdot \eta_{\text {mot }}}\right) \\ & \end{aligned}$                  (92)

The exergy efficiency of the flat plate photovoltaic-thermal (FPPV-T) can be determined for the specific gaseous systems being investigated [71, 73-75].

$\eta_{\text {ex }}=\frac{\dot{E}_{\text {out }}}{\dot{E}_{\text {int }}}=\frac{\dot{E}_{P V-T}+\left(\frac{T_a}{T_{\text {int }}} \frac{\dot{m}_f \cdot \Delta p}{\rho_{\text {air }} \cdot \eta_{\text {fan }} \cdot \eta_{\text {mot }}}\right)}{\mathrm{S}_p \mathrm{~A}_c\left[1-\frac{4}{3}\left(\frac{T_a}{T_{\text {sun }}}\right)+\frac{1}{3}\left(\frac{T_a}{T_{\text {sun }}}\right)^4\right]}$               (93)

The procedure of choosing the filling gas is influenced by both the gap (2 mm and 25 mm) and the physical properties of the gas. For this reason, the use of the Eismann correlation [19] replaced the Holland connection because it is thought to be more accurate. It is fresh in this theoretical study, and. because it is impossible to report on performance improvements when a specific gas is separated from its ideal gaps in the photovoltaic thermal hybrid panel. Additionally, the energy analysis employed to compute heat loss through double glazing and absorber (PV model) and ascertain the natural convection characteristics in photovoltaic thermal (PV-T) panels is novel.

112.png

Figure 2. Averaged heat transfer coefficient between inner glass and absorber (PV model)

The confined gases' coefficients of heat transfer (Air, CO, and Argon) between the absorber and the inner glass cover can reach up to ( $13 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}$ ) when the gas-filled region is less than $(10 \mathrm{~mm})$. however, when the gas-filled region is leas than $(6 \mathrm{~mm})$, the heat transfer coefficients of the other gases (sulfur dioxide, Krypton, and Xenon) are up to $\left(5 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}\right.$ ) (Figure 2). On the one band, there is a slight decrease in the heat transfer coefficients of trapped gases in double glazing. The curves for Air and Carbon oride are very gimilar. It can range from to $\left(2.9945 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}\right)$ and $\left(2.8895 \mathrm{~W} \cdot \mathrm{m}^{-2}, \mathrm{~K}^{-1}\right)$ at $(10 \mathrm{~mm})$ and $\left(2.8921 \mathrm{~W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-1}\right) \quad, \quad\left(2.7871 \mathrm{~W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-1}\right)$ at $(25 \mathrm{~mm})$ respectively suggesting that a gap greater than $(10 \mathrm{~mm})$ has no impact. Additionally, the Argon is a gas with a lower heat transfer coefficient than air and $C O$ It can range from to $\left(2.1341 \mathrm{~W} \cdot \mathrm{m}^{-2}, \mathrm{~K}^{-1}\right)$ at $(10 \mathrm{~mm})$ and $\left(2.059 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}\right)$ at $(25 \mathrm{~mm})$. but the sulfur dioxide car range from to $\left(2,3444 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}\right)$ at $(6 \mathrm{~mm})$ and $\left(2.0611 \mathrm{~W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-1}\right)$ at $(25 \mathrm{~mm})$. This is a result of doubleglazed windows improved argon-based heat isolation capabilities. making it appropriate for (DV-T) system the sulfur dioxide stays better than Argon whose value is somewhat $\left(1.4162 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}\right)$ at $(6 \mathrm{~mm})$ and they're approaching equal value between $\left(2.1941 \mathrm{~W} \cdot \mathrm{m}^{-2}, \mathrm{~K}^{-1}\right)$ at $(16 \mathrm{~mm})$ and $\left(2.062 \mathrm{~W} \cdot \mathrm{m}^{-2}, \mathrm{~K}^{-1}\right)$ at $(25 \mathrm{~mm})$ and can also be enclosed in the gaps to minimize heat loss via the double glass and absorber The Xenon curve indicates that it has a value half that of air. It can range from $\left(1.4163 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}\right)$ for $(6 \mathrm{~mm})$ to $\left(1.2347 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}\right)$ for $(25 \mathrm{~mm})$, 50 leas gas would be required to achieve better insulation in comparison to other gases (Air, Argon, $\mathrm{Krypton}, \mathrm{SO}_2, \mathrm{CO}$ ) (Figure 3), as reported in the research of Antonanzas et al. [34] When producing domestic hot water (DHW) or hot water suitable for space beating, the same result in our research, Overall, Argon was identified as the most suitable filling gas considering economic and environmental factors (Figure 4).

The air's overall beat loss coefficients are, $2.7569 \mathrm{~W} /$ $\left(\mathrm{m}^2, \mathrm{~K}\right)$ to $2.7395 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}$. It is not comparable to $\mathrm{CO}$. the latter ranges between $2.7104 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}$ to $2.6912 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}$ for the game spaces from $(10 \mathrm{~mm})$, to $(25 \mathrm{~mm})$, in contrast, Xenon's ranges which are from 1.9889 to $1.9168 \mathrm{~W}, \mathrm{~m}^{-2}, \mathrm{~K}^{-1}$ (from 5 , to $25 \mathrm{~mm}$ ). In other words, at $(5 \mathrm{~mm})$, the two gases exhibit a difference of $1.5 \mathrm{~W} . \mathrm{m}^{-2}, \mathrm{~K}^{-1}$, but $\mathrm{Krypton}$ and Sulfur dioxide has values of 2.2577 and $2.4672 \mathrm{~W}^{-2}, \mathrm{~K}^{-1}$ (Figure 5).

113.png

Figure 3. Averaged heat transfer coefficient inside double-glass

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Figure 4. Convective coefficients for different distances between covers and gases [34]

115.png

Figure 5. Overall heat loss coefficient of photovoltaic thermal hybrid

When thermal insulation is achieved through fenestration with vertical double-glazing, a U-value is frequently utilized. The fenestration of Chapter 15 displays graphical results from ASHRAE 2017 [62] that exhibit the similar trend as Figure 5. Our mathematical model is validated by the fact that the ideal widths of the gaps filled by Air, Argon and Krypton are practically equal.

Because Xenon produces a lower UL value and the heat transfer coefficients, at $5 \mathrm{~mm}$ and $25 \mathrm{~mm}$, the useful heat energy of PV-T increases to 5.2924 and $5.3698 \mathrm{KW}$, respectively. However, the useful heat energy by Air is 3.977 KW and $4.5793 \mathrm{KW}$ (From 5 to $25 \mathrm{~mm}$ ), while CO gives $4.027-4.621 \mathrm{KW}$ (From 5 to $25 \mathrm{~mm}$ ). They have almost the same values. When these three gases are compared, it can be shown that the energy obtained differs by approximately $1.3-0.72 \mathrm{KW}$ (From 5 to $10 \mathrm{~mm}$ ), more useful heat energy is produced by the remaining tested noble gases than by air (Figure 6).

116.png

Figure 6. Useful heat energy of PV-T according to the gas utilized

The effect of Gap between for each gas the two plates on th thermal efficiency Where Xenon gas is less valuable ranges between 0.1764 to 0.179 (From 5 to $25 \mathrm{~mm}$ ) However, the thermal efficiency by air is 0.1512 at $10 \mathrm{~mm}$ so the Xenon gas offers an advantage of $2 \%$. Still, as th influence of the gap between each gas increases, the therma efficiency rises as well. Ranging from between 0.15 to 0.1 (From 5 to $25 \mathrm{~mm}$ ) For all used gases the stability 0 electrical efficiency in terms of the gap between each of th two plates gas 0.1141 , is evidence of its correlation with th absorber's average temperature $\left(\overline{\mathrm{T}_p}=\overline{\mathrm{T}_{p v}}\right)$ (Figure 7).

117.png

Figure 7. The thermal and electrical efficiency of PV-T

The total combined efficiency is the combination the thermal and equivalent electrical efficiency Eq. (82), for all used gases, the range is 0.47 to 0.50 (From 5 to 25 mm). Nevertheless, total combined efficiency rises in tandem with the influence of the spacing between each gas. The PV-T's overall combined efficiency increases with Xenon noble gas to 0.4935 at 5 mm and to 0.496 at 25 mm, followed closely by Krypton at the same spacing with a value of 0.4843-0.4901. Argon at 25 mm displayed a total efficiency of 0.4816, still showcases competitive performance. on the other hand, gases like sulfur dioxide (SO₂), carbon monoxide (CO), and air exhibit lower total efficiencies, hence the Xenon gas offers a 3% improvement, followed closely by Krypton a 2% and Argen a 1%. Further, these findings underscore the potential for significant total efficiency gains through careful gas selection and spacing optimization. Additionally, the study identifies the suitability of greenhouse dryers for high moisture crops and natural convection for low moisture crops, providing valuable insights for agricultural applications (Figure 8).

The plotted curves in Figure 8 illustrate that the highest efficiency for Argon is attained when there's a 9 mm gas gap between the absorber and double-glazing. This outcome mirrors the findings reported by Vestlund et al. [76]. In their 2012 study, even though their investigation was limited to a solar water heater utilizing double windows. the researchers’ findings echoed the consensus that this particular gas filling distance consistently leads to optimal efficiency consistently, highlighting its relevance across diverse research scopes within the realm of solar heating systems (Figure 9).

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Figure 8. The total combined efficiency of PV-T

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Figure 9. The efficiency of an argon-filled solar collector at Tw = Ta = 25℃ as a function of the distance between the absorber plate and cover glazing with pressure temperatures [76]

For the specific PV-T utilized in our simulation, we observe that the output temperatures for Xenon grow to $302.6634 \mathrm{~K}$ at $5 \mathrm{~mm}$, compared to $301.6299 \mathrm{~K}$ at $10 \mathrm{~mm}$ for air. The difference in output temperatures obtained is about $1.02 \mathrm{~K}$.

120.png

Figure 10. PV-T output temperature delivered depending on the filling gas in the space

The most desirable width of the gas-filled space, which yields the overall heat loss factor and lowest local heat transfer coefficient this is, the most useful heat energy and the total combined efficiency—can be found using the simulation results (Figure 1-Figure 10).

The originality emphasised at the outset stems from employing exergy analysis, encompassing destruction exergy, exergy efficiency, and exergy output, in correlation with gas selection and the gaps between double-glazing and the absorber. Notably, the exergy investigation conducted in this study brings to light that Xenon stands out as a more effective option in reducing exergy destruction compared to other gases examined. This particular insight underscores the pivotal role of gas selection in mitigating losses within the system, as evidenced by the comprehensive exergy study conducted within this research endeavor (Figure 11).

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Figure 11. Exergy destruction in FPPV-T for different enclosed gases

The obtained exergy efficiency values for different gasfilled spaces are as follows: Xenon at $5 \mathrm{~mm}$ resulted in an efficiency of 0.5296 , followed closely by Krypton at the same spacing with a value of 0.5264 . Argon at $10 \mathrm{~mm}$ displayed an efficiency of 0.5252 , while $\mathrm{SO}_2$ at $5 \mathrm{~mm}$ showed a value of 0.5240 . The efficiency dropped slightly with $\mathrm{CO}$ at $10 \mathrm{~mm}$, which yielded 0.5215 , and air at $10 \mathrm{~mm}$ had an efficiency of 0.5210 . These values signify the varying levels of exergy efficiency achieved across different gas types and spacings. Xenon and Krypton at smaller spacings notably showcased higher efficiency compared to Argon, $\mathrm{SO}_2, \mathrm{CO}$, and air at larger spacings (Figure 12).

Furthermore, the exergy analysis applied to the investigated PV-T system resulted in substantial improvements akin to those observed in the comprehensive combined efficiency analysis. Specifically, Xenon exhibited a significant exergy output of $3.8229 \mathrm{KW}$ at a $5 \mathrm{~mm}$ interspace, surpassing other gases. Krypton, Argon, and $\mathrm{SO}_2$ yielded slightly lower exergy outputs of $3.7994 \mathrm{KW}, 3.7912 \mathrm{KW}$, and $3.7845 \mathrm{KW}$, respectively, with interspaces of $5 \mathrm{~mm}, 10 \mathrm{~mm}$, and $10 \mathrm{~mm}$. Conversely, commonly used insulating gases such as air and $\mathrm{CO}$ within double-glazing showed comparatively lower exergy outputs of $3.7610 \mathrm{KW}$ and $3.7643 \mathrm{KW}$, respectively, at the same $10 \mathrm{~mm}$ spacing. These findings underscore the varying exergy outputs across different gases and spacings within the PV-T system, highlighting Xenon's superior exergy output compared to other gases evaluated (Figure 13).

The outcomes derived from this study were focused on identifying the most suitable spaces filled with specific insulating gases, employing two distinct methodologies. Firstly, a novel natural convection correlation was employed to assess the heat transfer coefficients and the overall heat loss coefficient. Secondly, both conventional and exerge methods were utilized to determine the optimal spaces between the Floating Plate Photovoltaic-Thermal (FPPV-T) plates. These combined approaches aimed to ascertain the ideal gas spacings that would optimize the system's thermal performance and minimize heat loss, utilizing advanced convection correlations and diverse analytical techniques.

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Figure 12. Exergy efficiency of the FPPV-T for different enclosed gases

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Figure 13. Exergy output by FPPV-T for different enclosed gases