## 2 Materials and Methods
\begin{table}
\begin{tabular}{c l c c c} \hline \hline
**Wall Type** & **Building Material** & **Thickness (m)** & \
\begin{tabular}{c} **Heat Transfer** \\ **Coefficient (W/mK)** \\ \end{tabular} &
\begin{tabular}{c} **Conduction** \\ **Resistance (m\({}^{2}\)K/W)** \\ \end{tabular} \\ \hline \multirow{7}{*}{Stone wall} & Indoor environment & - & - & 0.130 \\  & Interior plaster (a) & 0.03 & 1.40 & 0.021 \\  & Stone wall (b) & 0.35 & 2.50 & 0.140 \\  & Exterior plaster (c) & 0.03 & 1.40 & 0.021 \\  & External environment & - & - & 0.040 \\  & Total heat transfer coefficient = (1/total & & 2.841 W/m\({}^{2}\)K \\  & conduction resistance) & & & \\ \hline \multirow{7}{*}{Solid brick} & Indoor environment & - & - & 0.130 \\  & Interior plaster (a) & 0.02 & 0.87 & 0.023 \\  & Solid brick (b) & 0.22 & 0.75 & 0.293 \\  & Exterior plaster (c) & 0.03 & 1.40 & 0.021 \\  & External environment & - & - & 0.040 \\  & Total heat transfer coefficient = (1/total & & 1.972 W/m\({}^{2}\)K \\  & conduction resistance) & & & \\ \hline \multirow{7}{*}{Perforated brick} & Indoor environment & - & - & 0.130 \\  & Interior plaster (a) & 0.02 & 0.87 & 0.023 \\  & Perforated brick (b) & 0.19 & 0.45 & 0.422 \\  & Exterior plaster (c) & 0.03 & 1.40 & 0.021 \\  & External environment & - & - & 0.040 \\  & Total heat transfer coefficient = (1/total & & 1.572 W/m\({}^{2}\)K \\  & conduction resistance) & & & \\ \hline \multirow{7}{*}{Ytong wall} & Indoor environment & - & - & 0.130 \\  & Interior plaster (a) & 0.02 & 0.87 & 0.023 \\ \cline{1-1}  & Ytong (b) & 0.15 & 0.18 & 0.833 \\ \cline{1-1}  & Exterior plaster (c) & 0.03 & 1.40 & 0.021 \\ \cline{1-1}  & External environment & - & - & 0.040 \\ \cline{1-1}  & Total heat transfer coefficient = (1/total & & 0.955 W/m\({}^{2}\)K \\ \cline{1-1}  & conduction resistance) & & \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Exterior wall components. Figure 2: Exterior wall types (see Table 1 for parameters a, b, X, and c). Thermal insulation represents a cost-effective energy-saving method that can be applied in facilities and buildings experiencing heat loss. The initial investment cost is relatively low, and the savings in energy costs can be significant, resulting in a relatively short payback period. The economic advantage of insulation applications is not only limited to fuel or energy saving but also provides gain in initial investment cost and a reduction in equipment capacities used in energy production [23]. In order to restrict the amount of energy utilized for the purpose of heating of buildings, thus enhancing energy savings and establishing the standard calculation methodology to be employed during the calculation of energy requirements, it is necessary to first determine the energy consumption of existing buildings for the purpose of heating [24]. The annual energy requirement (E\({}_{\text{h}}\)) to heat the unit area is determined by the following:  
\[\text{E}_{\text{h}}=\frac{86400.\text{U.DD}}{\eta} \tag{1}\]  
where DD is the degree-day value, U is the total heat transfer coefficient of the external wall (W/m\({}^{2}\) K), and \(\eta\) is the efficiency of the heating system. The total heat transfer coefficient U of the external wall is calculated using Equation (2), taking into account the resistances of the layers that comprise the wall. \[\text{U}=\frac{1}{\text{R}_{\text{i}}+\text{R}_{\text{w}}+\text{R}_{\text{in} }+\text{R}_{0}} \tag{2}\]  
\(\text{R}_{\text{i}}\) represents the internal surface heat convection resistance, R\({}_{0}\) is the external surface heat convection resistance, R\({}_{\text{w}}\) is the heat conduction resistance of the wall, and R\({}_{\text{in}}\) stands for the heat conduction resistance of the insulation layer.