## 3 Results and Discussion
### Accelerated Test Results  
For the accelerated test, the thermal conductivity of the 10 mm sample was measured repeatedly for approximately 3200 hr, and the results were converted into thermal resistance values, as seen in Figure 5. The relative thermal resistance ratio (\(\rm{R_{t}/R_{0}}\)) was calculated from the resultant thermal resistance values, and the time axis was subjected to a logarithmic transformation to derive the results shown in Figure 6. The scaling factors of the 10 mm and 50 mm thickness were compared, respectively, based on Equation (13) to express the results shown in Figure 7. As shown in Equation (13), the scaling factor for the 10 mm thick specimen corresponding to the original 50 mm thick XPS is 0.04. By applying 0.04 to the actual time of measurement for 50 mm thick material, the time for 10 mm thick specimen was obtained and graphed in log scale without any change of measured thermal resistance values. From scaling conversion, the corresponding thermal resistance of the time for a 10 mm thick specimen represents the thermal resistance of 50 mm thick material in actual ageing time. In addition, relative thermal resistances in dimensionless values were graphed as suggested in ISO 11561 [6] to compare the trend of thermal resistance degradation as a function of time. In Figures 5-7, it can be observed that the initial thermal resistance of specimens varied within ranges from 0.4 to 2.5 m\({}^{2}\).K/W. As explained in condition details, the intrinsic nature of specimens 3 and 4 can be regarded as the same owing to the cut from the same 50 mm slab. However, unexpectable inhomogeneities might induce the initial thermal resistance difference. Figure 4: Thermal resistance of two XPS specimens. Redrawn based on previous results [15; 16]. Figure 5: Acceleration test result. The thermal resistance of the 50 mm thick extruded insulation material after 25 years, which consists of 9125 days, was calculated using Equation (11) as follows:  
\[\mathrm{t_{10 mm\ specimen}(t_{10})}=9125\cdot\left(\frac{10}{50}\right)^{2}=365 \tag{14}\]  
This means that the time for a 10 mm sliced specimen is shortened down to 365 days, which corresponds to 9125 days for a full 50 mm thick slab. As seen in Figure 7, the thermal resistances of sample 1 and 2 after 365 days, corresponding to 25 years, were calculated as approximately 0.309 m\({}^{2}\cdot\)K/W and 0.288 m\({}^{2}\cdot\)K/W, respectively. On converting these results into thermal resistances for a thickness of 50 mm, the thermal resistances of samples 1 and 2 were calculated as 1.542 m\({}^{2}\cdot\)K/W and 1.442 m\({}^{2}\cdot\)K/W, respectively. Similarly, we calculated the 25-year-average thermal resistance using Equations (7), (11) and (13). \[\mathrm{t_{50 mmslab,average}(t_{50,av})}=\frac{9125}{\sqrt{10}}=2886 \tag{15}\]  
\[\mathrm{t_{10 mmspecimen,average}(t_{10,av})}=\mathrm{t_{50,av}}\cdot\mathrm{ S}=2886\cdot\left(\frac{10}{50}\right)^{2}=115 \tag{16}\]  
Figure 6: Conversion of relative thermal resistance. Figure 7: Relative thermal resistance on applying the scaling factor calculated by Equation (13). The thermal resistances of specimens 3 and 4 after 115 days were calculated to approximately 0.341 m\({}^{2}\cdot\)K/W and 0.300 m\({}^{2}\cdot\)K/W, respectively. On converting these results into thermal resistances for a thickness of 50 mm, the thermal resistances of samples 1 and 2 were calculated as 1.704 m\({}^{2}\cdot\)K/W and 1.499 m\({}^{2}\cdot\)K/W, respectively. Consequently, the 25-year-average thermal resistance of 50 mm thick insulation material becomes similar with thermal resistance values of 50 mm thick slab and 10 mm sliced specimen after 9125 days, 2886 days, and 115 days, respectively. In Table 4, the thermal resistances of specimens 3 and 4 were summarized. Initial value, the value after 25 years, and the value in 25 years average were obtained from data in Figure 5 and calculation by Equations (15) and (16).