先进的PFH计算安全完整性系统与高诊断

P. Holub, J. Börcsök
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引用次数: 4

摘要

根据IEC 61508对“按需故障概率”(PFH)的计算表明,对于使用相同初始值的1001和1002系统,正如预期的那样,1002系统比1001系统具有更好的安全完整性。这意味着,一个1002系统比一个1001系统有更低的故障概率。比较在高需求或连续运行模式下运行的相同系统,在某些情况下,特别是对于具有高诊断的系统,使用IEC 61508的PFH公式,1001系统的“每小时故障概率”(PFH)值低于1002系统。根据IEC 61508的公式,也有可能系统在高模式或连续模式下运行的时间越长,PFH值显然越好。这两种结果在工业现实中都是不可复制的。因此,本文将首先分析在使用PFH系统时发生危险事件的必要条件。修改后的PFH计算只考虑那些确实发生危险事件的情况。在第二步中,本文提出了一种新的方法,利用合理的参数来计算PFH系统的PFH值,考虑到高诊断率和极低故障率。本文所提出的方法也可以用先进的马尔可夫模型进行正确的数学验证。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Advanced PFH calculations for safety integrity systems with high diagnostic
The calculation of "Probability of Failure on Demand" (PFH) according to IEC 61508 shows for a 1oo1 and a 1oo2 system using the same initial values that - as expected - a 1oo2 system is a better safety integrity system than a 1oo1 system. This means, that a 1oo2 system has a lower probability of failure than a 1oo1 system. Comparing the same systems operating in a high demand or continuous mode of operation, there will be some cases, especially for systems with high diagnostics, where a 1oo1 system has a lower ¿Probability of Failure per Hour¿ (PFH) value than a 1oo2 system, using the PFH formulas according to IEC 61508. Using the equation according to IEC 61508, it is also possible that the PFH value is apparently better the longer the system runs either in the high mode or continuous mode of operation. Both results are irreproducible in industrial reality. Therefore, this paper will analyze first what conditions are necessary to get a hazard event when using a PFH system. Only those cases are considered in the modified PFH calculation, in which indeed a hazard event occurs. In a second step, it presents a new approach with the help of plausible arguments in order to calculate the PFH value of a PFH system considering a high diagnostic and a very low failure rate. The presented approach can also be correctly verified with mathematics using the advanced Markov model also described in this paper.
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