A Sequential Inspection Model based on Risk Quantitative Constraint and Component Importance

Bai Senyang,Cheng Zhijun,Zhao Qian,Jia Xiang,Yao Hang

Table 1 The calculation results of sequential inspection interval (${{r}^{*}}=0.903$)

The serial number ${{t}_{ik}}/h$ ${{x}_{ik}}/({}^\circ /h)$ ${{a}_{ik}}$ ${{D}_{ik}}$ ${{Q}_{ik}}$ ${{({{\sigma }^{2}})}_{ik}}$ $\Delta {{\hat{t}}_{ik}}/h$ Remarks The initial state (${{t}_{0}}=0$) 0.00000 0.00000 0.04426 0.00017 0.00737 0.00053 4.71120 The degradation data exceeds the failure threshold 0.6 $({}^\circ /h)$ at the sixth inspection, which needs to be replaced. ${{t}_{01}}$ 4.7112 0.19203 0.04419 0.00016 0.00028 0.00083 4.71120 ${{t}_{02}}$ 9.4224 0.35605 0.04273 0.00012 0.00015 0.00059 3.80590 ${{t}_{03}}$ 13.2283 0.27352 0.04106 0.00008 0.00100 0.00434 3.38730 ${{t}_{04}}$ 16.6156 0.31776 0.04068 0.00007 0.00068 0.00356 3.01020 ${{t}_{05}}$ 19.6258 0.30006 0.04012 0.00007 0.00050 0.00285 3.58900 ${{t}_{06}}$ 23.2148 1.84797 —— —— —— —— —— Replace as new (${{t}_{1}}$) 23.2148 0.00000 0.04012 0.00007 0.00050 0.00285 8.75680 The degradation data exceeds the failure threshold at the fifth inspection, which needs to be replaced. ${{t}_{11}}$ 31.9716 0.28173 0.03950 0.00006 0.00024 0.00104 5.26090 ${{t}_{12}}$ 37.2325 0.19451 0.03773 0.00005 0.00101 0.00657 4.05290 ${{t}_{13}}$ 41.2854 0.31754 0.03721 0.00005 0.00068 0.00546 2.60630 ${{t}_{14}}$ 43.8917 0.49501 0.03660 0.00005 0.00052 0.00515 0.54880 ${{t}_{15}}$ 44.4405 0.62656 —— —— —— —— —— Replace as new (${{t}_{2}}$) 44.4405 0.00000 0.03660 0.00005 0.00052 0.00515 7.92460 The degradation data exceeds the failure threshold at the fourth inspection, which needs to be replaced. ${{t}_{21}}$ 52.3651 0.20505 0.03616 0.00005 0.00032 0.00122 6.86210 ${{t}_{22}}$ 59.2272 0.20363 0.03453 0.00004 0.00031 0.00273 5.85850 ${{t}_{23}}$ 65.0857 0.49155 0.03359 0.00004 0.00025 0.00410 0.69760 ${{t}_{24}}$ 65.7833 0.66387 —— —— —— —— … … ... … … … ...