logo
  • userLoginStatus

Welcome

Our website is made possible by displaying online advertisements to our visitors.
Please disable your ad blocker to continue.

Current View

Energy Engineering - RELIABILITY, SAFETY AND RISK ANALYSIS C

Chapter 06 - Exercises collection

Divided by topic

Chapter 6 Availability and maintainability 6.1 Compressor A compressor is designed for Td = 5 years of operation. There are two significant contributions to the failure. The first is due to wear (W) of the thrust bearing and is described by a Weibull distribution �(������)= ������ ������(������ ������) ������−1 �[−(������������)������] with θ= 7.5 year and m = 2.5 . The second, which includes ali other causes ( O) is described by a constant failure rate of λO = 0.013 year -1. l. What is the reliability if no preventive maintenance is performed over the 5 -year design life? 2.If the reliability of the 5 -year design life is to be increased to at least 0.9 by periodically replacing the thrust bearing, how frequently must it be replaced? 3. Suppose that the probability of fault bearing replacement causing failure of the compressor is p = 0.02. What will the design -life reliability be with the replacement program decided in 2)? 6.2 Sequential and staggered maintenance scheme of a one -out -of-two system Consider a one -out -of two system of identica l components with constant failure rate λ. The testing and repair of each component last for τr hours. l. In the sequential maintenance scheme, the two components are tested one after the other, τ being the time between the end of the previous maintenance of the second component and the beginning of the next maintenance of the first one (in other words, every τ hours we test both components in sequence). Find the average unavailability of the system. 2. In the staggered maintenance scheme, the f irst component maintenance starts at k τ, k < l , where τ is the time interval between the end of the previous maintenance of the second component and the beginning of the next maintenance of the same second component. Find the average unavailability of the system.