## 3 Heat Load Estimation for an Electric Bus
### Solar Heat Loads
#### 3.4.2 Heat Loss through the Walls of the Bus--\(\dot{Q}_{wall}\)  
The walls of the bus are made from three materials, that is (from inside to outside): laminated wood (MDF), polyurethane foam, and a sheet of steel. Their thickness and thermal conductivities are presented in Table 3.  
As for the simulation, in Figure 4a, the model of the three layers of the wall is presented, with a temperature of about \(-8.1\)\({}^{\circ}\)C on the exterior wall of the bus, with main thermal loss from the steel and the foam in the wall. The interior side of the vehicle is on the left side of the image (Figure 4a). Vehicle speed is incorporated into the model as the velocity of the fluid (the exterior air). For the triple layered wall of the bus (Figure 4b), the overall heat transfer coefficient is calculated as follows:  
\[\frac{1}{U_{wall}}=\frac{1}{h_{i}}+\left(\frac{L_{1}}{k_{2}}+\frac{L_{1}}{k_{2} }+\frac{L_{1}}{k_{3}}\right)+\frac{1}{h_{o}}+r_{cwall},\ \ \ [\text{K}/\text{m}^{2}\text{W}], \tag{15}\]  
The thermal contact resistance for the wall is the sum of the thermal resistance for the two interfaces \(\text{I}_{\text{f1}}\) and \(\text{I}_{\text{Z}}\) (Figure 4b). For the first interface thermal resistance estimation (between the layer of wood, L(1), and foam, L(2)), we assumed that the heat transferred through the surface area of the wall is constant, and we can write the following:  
\[\frac{\dot{Q}}{A}=\frac{\Delta T}{\frac{L1}{\text{K}}}, \tag{16}\]  
where \(\Delta T=T_{i}-T_{1}\) [\({}^{\circ}\)C] is the difference of temperature along the layer, L(1). Figure 4: (**a**) Heat transfer through the wall. (**b**) Schema of the three layers of the wall of the bus. \begin{table}
\begin{tabular}{c c c} \hline \hline
**Material** & **Thickness [m]** & **Thermal Conductivity [W/mK]** \\ \hline Iron & 0.0008 & 76.2 \\ Polyurethane foam & 0.01 & 0.04 \\ Laminated wood & 0.004 & 0.107 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Thickness and thermal conductivities for the materials of the wall of the bus. For the interface zone (between the layers L(1) and L(2)):  
\[\frac{\dot{Q}}{A}=\frac{\Delta T_{interface}}{r_{c12}}, \tag{17}\]  
\(\Delta T_{interface}=T_{1}-T_{2}\) [\({}^{\circ}\)C], with \(T_{1}\) and \(T_{2}\) indicating the temperatures on the interface of the layers L(1) and L(2), between wood and foam layers respectively (Figure 4b). The interface temperature \(T_{1}\) and \(T_{2}\) are estimated from the Comsol simulations (Figure 4a), resulting \(T_{1}=16.27\) \({}^{\circ}\)C and \(T_{2}=16.12\) \({}^{\circ}\)C.  
With these values, and using the Equations (16) and (17), the first interface shows \(r_{c12}=0.002\) [K/m\({}^{2}\) W]. Similarly, but considering \(T_{3}=-8\) \({}^{\circ}\)C and \(T_{4}=-8.14\) \({}^{\circ}\)C, the second interface is calculated, between the second and the third layer, as follows: \(r_{c12}=8\)\(\cdot\)\(10^{-6}\) [K/m\({}^{2}\) W], for the wall of the bus, is \(r_{cwall}=\)0.00208 [K/m\({}^{2}\) W]. For the speed of 40 km/h, with 20 \({}^{\circ}\)C inside and \(-10\) \({}^{\circ}\)C outside temperatures, and for an area of the wall of the bus of \(A_{wall}=27\) m\({}^{2}\), the U value is determined as \(U_{wall}=2.283\) W/m\({}^{2}\) K and the heat loss through the wall is \(\dot{Q}_{wall}=1849\) W.  
4.3 Heat Loss through the Roof--\(\dot{Q}_{roof}\)--And the Floor of the Bus--\(\dot{Q}_{floor}\)  
The roof of the bus is composed of a layer of iron sheet, a layer of polyurethane foam (as isolation), and a volume area filled with various equipment (air conditioning, pipes, power supply equipment, and power conductors from the power collecting system). These equipment elements are made mostly from metals and cover about 50% of the roof area. Due to this complex structure of the roof, in order to estimate the U value for the roof, including this complex area (composed of metals materials and air), we consider this area as composed of two parallel layers of metals and air.