上海理工大学学报  2020, Vol. 42 Issue (4): 399-403 PDF

Modeling of failure process of an equipment and its simulation study
ZHENG Rui
School of Public Economy and Management, Shanghai University of Finance and Economics, Shanghai 200433, China
Abstract: The failure of an equipment is usually a random process, so the preventive maintenance (PM) interval of the equipment is mainly formulated by maintenance engineers according to their experiences, or by the specification of PM interval given by the manufacturing enterprise, which results into two main problems of over maintenance and insufficient maintenance. In order to optimise the preventive maintenance interval, a time delay model, in the condition of Poisson process defects and perfect inspection was established based on the maintenance data of recorded failures and inspection data. Then the failure process in case of perfect inspection was simulated by computer programs. The simulation results prove the validity of the established time delay model. The established time delay model and the simulation system demonstrate intuitively the failure process, reveal clearly the failure mechanism, and solve the difficulty of model validation in case of no failure data record. The failure process model is benefitial to make maintenance decision.
Key words: time delay     preventive maintenance     simulation     failure process

1 问题描述

 图 1 在（Ti-1,Ti）期间各次故障发生时间及PM活动观察到的缺陷次数 Fig. 1 Observed failures and number of defects over time（Ti-1,Ti）

2 完全检查下设备故障模型的建立 2.1 假定条件及有关符号

a. 检查是完全的，PM活动能检查出所有存在的缺陷；以及所有检查出的缺陷在较短时间内得到修复；

b. 缺陷的发生率为齐次泊松过程（HPP，homogeneous Poisson process）；

c. 各个缺陷的发生是相互独立的，以及缺陷发生和故障发生也是两个独立的事件。

2.2 最大似然函数

 $\begin{split} L= & \prod\limits_{{{i}}={\rm{1}}}^{{{{m}}}} {\left\{ {} \right.} p({\text{在}}{\rm{ }}{T_i}{\text{时检查出}}{n_i}{\text{个缺陷}})\! \times \\ & \left. {\prod\limits_{{{j}}={\rm{1}}}^{{{{k}}_{{{i - 1}}}}} {\left[ {p({\text{在}}{\rm{ }}{t_{(i - 1)j}}{\text{时发生故障}})} \right]}} \times p({\text{在}}({\rm{ }}{T_{i - 1}},{T_i}) \right.\\ &\left.{\text{期间各记录故障之间无故障再发生}}) \right\} \\[-10pt]\end{split}$ (1)

 $\begin{split} \log\; L &= \sum\limits_{i = 1}^m {\left\{ {} \right.} \log\; p({\text{在}}{T_i}{\text{时检查出}}{n_i}{\text{个缺陷}}) + \\ & \left.{\displaystyle\sum\limits_{j = 1}^{{k_{i - 1}}} {\left[ {\log\; p({\text{在}}{t_{(i - 1)j}}{\text{时发生故障}})} \right] }}\right.+\log \;p({\text{在}}({\rm{ }}{T_{i - 1}},\\&\left.{\rm{ }}{T_i}){\text{期间各记录故障之间无故障再发生}}) \right\} \\[-10pt]\end{split}$ (2)

a. p（在ti−1）j发生故障）的计算。

 $v(t)=\int_{{T_{i - 1}}}^t \lambda f(t - u){\rm{d}}u=\lambda F(t - {T_{i - 1}})$ (3)

 $\begin{split}& p({\text{在}}\left( {{t_{\left( {i - 1} \right)j}},{t_{\left( {i - 1} \right)j}} + \Delta t} \right){\text{区间发生一个故障}}) = \\&\quad v\left( {{t_{\left( {i - 1} \right)j}}} \right){\rm{ }}\Delta t \end{split}$ (4)

b. $p({\text{在}}({\rm{ }}{T_{i - 1}},{\rm{ }}{T_i}){\text{期间各记录故障之间无}}$ 故障再发生)的计算。

 $\begin{split}& p({{{\text{在}} (}}{t_{(i - 1)(j - 1)}},{t_{(i - 1)j}}){{{\text{期间没有故障发生}}}}) = \\&\quad\mathop {\rm{e}}\nolimits^{ - \int_{{t_{(i - 1)(j - 1)}}}^{{t_{(i - 1)j}}} {v(t){\rm{d}}t} } \\[-6pt] \end{split}$ (5)

 $\begin{split} \sum {\log } &p({\text{在}}({T_{i - 1}}{\rm{,}}{T_i}{{){\text{期间故障之间无故障再发生}})}} =\\& \sum\limits_{j = 1}^{{k_{i - 1}}} {\left( { - \int_{{t_{(i - 1)(j - 1)}}}^{{t_{(i - 1)j}}} {v(t){\rm{d}}t} } \right)} - \\ & \int_{{t_{(i - 1){k_{i - 1}}}}}^{{T_i}} {v(t){\rm{d}}t} = - \int_{{T_{i - 1}}}^{{T_i}} {v(t){\rm{d}}t} \\[-10pt]\end{split}$ (6)

c. P（在Ti时检查出ni个缺陷）。

Ti时刻检查出缺陷个数的均值ENpTi）为

 $\begin{split}E{N_p}({T_i})=&\int_{{T_{i - 1}}}^{{T_i}} {\lambda R({T_i} - u){\rm{d}}u=} \\ & \int_{{T_{i - 1}}}^{{T_i}} {\lambda [1 - F({T_i} - u){\rm{d}}u} \end{split}$ (7)

 $\begin{split} p({\text{在}}{T_i}{\text{时检查出}}{n_i}{\text{个缺陷}}) =\\ \frac{{{{(E{N_p}({T_i}))}^{{n_i}}}{{\rm{e}}^{ - E{N_p}({T_i})}}}}{{{n_i}!}} \end{split}$ (8)

 $\begin{split} \log\; L=&\sum\limits_{i=1}^m \Bigg\{ ({n_i}\log\; E{N_p}({T_i}) - E{N_p}({T_i}) - \log \;{n_i}!) +\\&\left. \sum\limits_{j=1}^{{k_{i - 1}}} {\log \;v({t_{(i - 1)j}})} - \int_{{T_{i - 1}}}^{{T_i}} {v(t){\rm{d}}t} \right\} \end{split}$ (9)

3 故障过程的仿真与结果分析

 图 2 完全检查下设备故障过程的计算机模拟仿真流程 Fig. 2 Flowchart of simulation of the failure process of the equipment in case of perfect inspection

Step 1 　产生缺陷的过程（服从泊松分布）。

a. 产生U~U（0，1）；

b. 置ui=ui−1−（1/λ）ln U

Step 2 　计算每一个缺陷的时间延迟函数。

a. 产生U~U（0，1）；

b. 置hi=−（1/α）ln U

Step 3 　统计每一周期中缺陷的个数和故障的次数。

4 结　论

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