Ansition Point 0.0.0.0.XPS Specimen #3 XPS Specimen #0.0 0 1 10log10 (day)Figure five. Acceleration test result. Figure five. Acceleration test outcome.e (Rt/R0)1.1 1.0 0.9 Transition PointT0.0 0.001log10 (day) log10 (day)10100Appl. Sci. 2021, 11,Figure five. Acceleration test result. Figure five. Acceleration test outcome.ten of1.1 1.1 1.0 1.0 0.9 0.9 0.8 0.eight 0.7 0.7 0.6 0.6 0.five 0.5 0.four 0.4 0.three 0.3 0.two 0.two 0.1 0.1 0.0 0.Relative Thermal Resistance /R ) Relative Thermal Resistance (R(Rt/R0) SB-612111 medchemexpress tTransition Point Transition PointXPS Specimen #3 XPS Specimen #3 XPS Specimen #4 XPS Specimen #0.1 0.log10 (day) log10 (day)110100Figure six. Conversion of relative thermal resistance. Figure 6. Conversion of relative thermal resistance. Figure six. Conversion of relative thermal resistance.1.1 1.1 1.0 1.0 0.9 0.9 0.8 0.eight 0.7 0.7 0.6 0.6 0.5 0.five 0.four 0.four 0.3 0.3 0.two 0.two 0.1 0.1 0.0 0.0 1Relative Thermal Resistance (Rt/R Relative Thermal Resistance (Rt/R0) 0)ten to 50 mm Thickness scaling conversion 10 to 50 mm Thickness scaling conversionTransition Point Transition PointXPS Specimen #3 XPS Specimen #3 XPS Specimen #4 XPS Specimen #10log10 (day) log10 (day)1001,000 1,10,000 ten,Figure 7. Relative thermal resistance on applying the scaling factor Clevidipine-d7 References calculated by Equation (13). Figure 7. Relative thermal resistance on applying the scaling element calculated by Equation (13). Figure 7. Relative thermal resistance on applying the scaling aspect calculated by Equation (13).The thermal resistance from the 50 mm thick extruded insulation material just after 25 years, which consists of 9125 days, was calculated using Equation (11) as follows: t10mm specimen (t10 ) = 912510=(14)This means that the time for a 10 mm sliced specimen is shortened down to 365 days, which corresponds to 9125 days for any complete 50 mm thick slab. As observed in Figure 7, the thermal resistances of sample 1 and 2 immediately after 365 days, corresponding to 25 years, have been calculated as about 0.309 m2 /W and 0.288 m2 /W, respectively. On converting these results into thermal resistances for a thickness of 50 mm, the thermal resistances of samples 1 and two had been calculated as 1.542 m2 /W and 1.442 m2 /W, respectively. Similarly, we calculated the 25-year-average thermal resistance working with Equations (7), (11) and (13). 9125 t50mmslab,typical (t50,av ) = = 2886 (15) 10 t10mmspecimen,typical (t10,av )= t50,av = 288610=(16)Appl. Sci. 2021, 11,11 ofThe thermal resistances of specimens three and 4 just after 115 days were calculated to roughly 0.341 m2 /W and 0.300 m2 /W, respectively. On converting these outcomes into thermal resistances for any thickness of 50 mm, the thermal resistances of samples 1 and 2 have been calculated as 1.704 m2 /W and 1.499 m2 /W, respectively. Consequently, the 25-year-average thermal resistance of 50 mm thick insulation material becomes equivalent with thermal resistance values of 50 mm thick slab and 10 mm sliced specimen soon after 9125 days, 2886 days, and 115 days, respectively. In Table four, the thermal resistances of specimens 3 and four were summarized. Initial worth, the worth right after 25 years, plus the value in 25 years typical were obtained from data in Figure five and calculation by Equations (15) and (16). The change prices in the thermal resistance by initial value had been calculated by dividing degradation worth from initial value by the worth right after 25 years. It varied from 37 to 41 for the thermal resistance soon after 25 years and from 30 to 38 for the 25-year-average thermal resistance, which meant that thermal insu.
