## 6 Simulation of a Realistic Wall
### The Initial and the Boundary Conditions  
In this part, the final time (the end of the analyzed time span) is \(t_{\mathrm{fin}}=10\),000 s. The time step size is measured in seconds as well. In all cases and all boundaries, zero Neumann boundary conditions are used, which prohibits the flow of conductive heat at the boundaries:  
\[\frac{\partial u}{\partial x}(x,z=0,t)=\frac{\partial u}{\partial x}(x,z=1,t) =\frac{\partial u}{\partial z}(x,z=0,t)=\frac{\partial u}{\partial z}(x,z=1,t )=0\]  
This is accomplished by setting the necessary resistances to infinity and setting the value for the matrix elements representing heat conduction through the boundary as zero. (A) In the case of the surface area simulation, the heat transfer by radiation and convection is happening in the \(y\) direction, i.e., orthogonal to the plane of Figure 9.  
A linear function of the \(x\) variable is applied as the initial condition:  
\[u(x,z,\ t=0)=300-280x\]  
For the convection heat transfer coefficient \(h_{c}\), we have taken values from the literature [2], as one can see in Table 3. Regarding radiation, the Stefan-Boltzmann constant is a universal number \(5.67\cdot 10^{-8}\frac{W}{m^{2}\cdot K^{4}}\). The surface is not an ideal black body; thus, we multiplied the Stefan-Boltzmann constant by the appropriate emissivity constant to obtain realistic values for \(\sigma^{*}\). We estimate the value of \(q^{*}\) for the heat source term, which includes the solar radiation as it is shown below. The temperature of the ambient air is taken to be \(27\ ^{\circ}\mathrm{C}\approx 300\) K.  
Due to the nonzero temperature \(u_{a}\) of the air (in Kelvin), the expression \(q\) also contains the convective heat gain. We can obtain the value of \(q\) as follows  
\[q=\frac{q^{*}}{c\rho\Delta y}+\frac{h_{c}}{c\rho\Delta y}\cdot u_{\mathrm{a}}\]  
The convective and radiative energy transfer occurs in the \(y\) direction, i.e., perpendicular to the surface. Consequently, these terms are proportional to the element's free surface area, which is \(\Delta x\Delta z\) here. This information yields the following values for the coefficients in Equations (2) and (3):  
\[K=\frac{h_{c}}{c\rho\Delta y}\,\ \ \sigma=\frac{\sigma^{*}}{c\rho\Delta y}\.\]  
We assumed that the lower-half side of the surface is in the shade, resulting in much less incoming heat there. Specifically, we have  
* the first portion of N (sunny side): \[q=\frac{1}{c\rho\Delta y}\times 800\frac{W}{m^{2}}+\frac{h_{c}}{c\rho\Delta y }\times 300\mathrm{K};\] \begin{table}
\begin{tabular}{c c c c c} \hline \hline  & \(h_{c}\) (\(\frac{W}{m^{2}\cdot K^{4}}\)) & \(\sigma^{*}\) (\(\frac{W}{m^{2}\cdot K^{4}}\times 10^{-8}\)) & \(q^{*}_{shadow}\)(\(\frac{W}{m^{2}}\)) & \(q^{*}_{sumny}\)(\(\frac{W}{m^{2}}\)) \\ \hline All elements & 4 & 4 & 300 & 800 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: The parameters of convection, radiation, and heat source in case of wall surface area [39]. the second portion of \(N\) (shaded side):  
\[q=\frac{1}{c\rho\Delta y}\times 300\frac{W}{m^{2}}+\frac{h_{c}}{c\rho\Delta y} \times 300\text{K}.\]  
(B) In the case of the simulation of the cross-sectional area with the thermal bridge, the interior components cannot absorb or lose heat via convection, radiation, and the heat source. According to Table 4, elements on the right and left sides may transfer heat in the \(x\) direction through radiation and convection. The coefficient values in our equations are obtained as follows:  
\[K=\frac{h_{c}}{c\rho\Delta x},\sigma=\frac{\sigma^{*}}{c\rho\Delta x},q=\frac{ q^{*}}{c\rho\Delta x}+\frac{h_{c}}{c\rho\cdot\Delta x}\cdot u_{\text{a}}\]  
The ambient air temperature is taken to be 20 \({}^{\circ}\)C \(\approx\) 293 K and 40 \({}^{\circ}\)C \(\approx\) 313 K on the brick and the insulation side (inside and outside of the building), respectively.