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CFR in series

19 January, 2016 - 17:08
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Let us consider n different elements, each with its own constant failure rate λi and reliability Ri=e−λi∙t, arranged in series and let us evaluate the overall reliability RS. From Table 4.3 we have:

  
R_{S}=\prod_{i=1}^{n}R_{i}=\prod_{i=1}^{n}e^{-\lambda_{i} \cdot t}=e^{-\sum_{i=1}^{n}\lambda _{i}\cdot t}    

Since the reliability of the overall system will take the form of the type RS=e−λs∙t, we can conclude that:

  
R_{S}=e^{-\sum_{i=1}^{n}\lambda _{i}\cdot t}=e^{-\lambda _{s}\cdot t}\rightarrow \lambda _{s}=\sum_{i=1}^{n}\lambda _{i}    

In a system of CFR elements arranged in series, then, the failure rate of the system is equal to the sum of failure rates of the components. The MTTF can thus be calculated using the simple relation:

  
MTTF=\frac{1}{\lambda _{s}}=\frac{1}{\sum_{i=1}^{n}\lambda _{i}}   

For example, let me show the following example. A system consists of a pump and a filter, used to separate two parts of a mixture: the concentrate and the squeezing. Knowing that the failure rate of the pump is constant and is λP=1,5∙10−4 failures per hour and that the failure rate of the filter is also CFR and is λF=3∙10−5, let’s try to assess the failure rate of the system, the MTTF and the reliability after one year of continuous operation.

To begin, we compare the physical arrangement with the reliability one, as represented in the following figure:

media/image14.png
Figure 4.14 Physical and reliability modeling of a pump and a filter producing orange juice 
 

As can be seen, it is a simple series, for which we can write:

  
\lambda _{s}=\sum_{i=1}^{n}\lambda _{i}=\lambda _{P}+\lambda _{F}=1.8\cdot 10^{-4}\left [ \frac{failures}{h} \right ]     

MTTF is the reciprocal of the failure rate and can be written:

  
MTTF=\frac{1}{\lambda _{s}}=\frac{1}{1.8\cdot 10^{-4}}=5,555\left [ h \right ]   

As a year of continuous operation is 24⋅365=8,760 hours, the reliability after one year is:

R_{S}=e^{-\lambda _{s}\cdot t}=e^{-1.8\cdot 10^{-4}\cdot 8760}=0.2066

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