## 2 Mathematical Model
### Heating and Cooling Energy Demands on an Annual Scale  
To compute the yearly cooling and heating energy demands, in addition to the \(HDD\) and \(CDD\), some essential parameters, such as overall heat transfer coefficient (\(U\)), cooling system performance coefficient (\(COP\)), and heating system efficiency (\(\eta\)), should be determined. The wall structure considered in the analysis consists of interior and exterior plaster, hollow bricks, and insulation material, as shown in Figure 2. In this study, it is presupposed that heating relies on natural gas, while cooling is powered by electricity, and the required values are detailed in Table 1. The selection of financial parameters for this study is based on published data from the Central Bank of Turkiye [30].  
The following equation is used to establish the overall heat transfer coefficient. Here, \(R_{w}\) represents the wall's thermal resistance in the absence of insulation, \(h_{i}\) corresponds to the heat transfer coefficient on the interior side, \(h_{d}\) corresponds to the heat transfer coefficient on the exterior side, \(k\) stands for the insulation material's thermal conductivity, and \(x\) represents its thickness. The assumed heat transfer coefficients for the interior and exterior are 8.29 W/m\({}^{2}\)K and 28.35 W/m\({}^{2}\)K, respectively.  
\[U=\frac{1}{\frac{1}{h_{i}}+R_{w}+\frac{x}{k}+\frac{1}{h_{d}}} \tag{14}\]  
Suppose a wall's total thermal resistance without insulation is called (\(R_{t,w}\)). In that case, the equation of the wall's overall heat transfer coefficient with insulation can be simplified as [3]:\[U=\frac{1}{R_{t,\text{av}}+\frac{\pi}{k}} \tag{15}\]  
The yearly energy demand for heating per unit area because of heat loss through the wall may be computed using the following equation [31].  
\[q_{A,H}=86,400\ HDD\ U/\eta \tag{16}\]  
Here, \(\eta\) is the heating system's efficiency, and it is usually accepted as 0.93 for natural gas heating systems [31]. Likewise, the annual energy demand for cooling per unit area because of heat gain through the wall is found as follows [31]:  
\[q_{A,C}=86,400\ CDD\ U/COP \tag{17}\]  
Here, \(COP\) is the coefficient of cooling system performance and depends on operational factors. However, it is commonly considered to be 2.5 [24].  
\begin{table}
\begin{tabular}{c c c} \hline \hline  & **Parameters** & **Value** \\ \hline x cm & Insulation material & \(k=0.03\) W/mK \\
3 cm & Exterior plaster & \(k=0.87\) W/mK \\
2 cm & Interior plaster & \(k=0.87\) W/mK \\
20 cm & Hollow brick & \(k=0.45\) W/mK \\  & Heat transfer coefficient (Inside) & \(h_{t}=8.29\) W/m\({}^{2}\)K \\  & Heat transfer coefficient (Outside) & \(h_{o}=28.35\) W/m\({}^{2}\)K \\ \hline  & Expanded polystyrene (EPS) & \\  & _Material cost_ & \(C_{\text{its}}=120\) S/m\({}^{3}\) \\  & _Conductivity_ & \(k=0.039\) W/mK \\  & _Density_ & \(\rho=20\) kg/m\({}^{3}\) \\ \hline  & Natural gas & \\  & _Price_ & \(C_{f}=0.29\) S/m\({}^{3}\) \\  & _Lower heating value_ & \(Hu=34526.10^{3}\) J/m\({}^{3}\) \\  & _Heating system’s efficiency_ & \(\eta=93\%\) \\ \hline  & Electricity & \\  & _Price_ & \(C_{e}=0.0947\) S/kWh \\  & _Coefficient of cooling system performance_ & \(COP=2.5\) \\  & _Discount rate_ & \(g=20\%\) \\  & _Inflation rate_ & \(i=17\%\) \\  & _Lifetime_ & \(LT=20\) \\  & _Present worth factor_ & \(PWF=15.5\) \\ \hline \hline \end{tabular}
\end{table}
Table 1: Wall composition, insulation material’s properties, heating and cooling sources, and financial parameters.  
Figure 2: The wall components.