1. 1 Component reliability Jørn Vatn

2. 2 The state of a component is either “up” or “down” T 1, T 2 and T 3 are ”Uptimes” D 1 and D 2 are “Downtimes”

3. 3 Time to failure - TTF The term ‘time to failure’ (TTF) denotes the time from a unit is put into service, until it fails That is TTF is equivalent to T 1 In some situations we also use the term time to failure to denote T 2, T 3 It should however be denoted that the distribution of subsequent uptimes are not necessarily identical A range of quantities relates to TTF Distribution function, F(t) = Pr(T  t) Survivor function R(t) = 1 - F(t) = Pr(T > t) Hazard rate z(t) Mean Time To Failure (without maintenance), MTTF = MTTF WO

4. 4 Availability (A) and unavailability (U) Availability is the probability that the component is able to perform its designated function Availability is thus the probability that the component is functioning well when we consider an arbitrary point of time Availability may also be seen as the average “uptime” Unavailability is the probability that the component is not able to perform its designated function Unavailability is thus the probability that the component is not functioning well when we consider an arbitrary point of time Unavailability may also be seen as the average “downtime” A+U = 1 = 100%

5. 5 Evident and hidden function An evident function means that the a failure of the component immediately will be detected, i.e. when we go from “Up” to “Down will be known to the operator under normal operation procedures A hidden function means that we do not continuously monitor the unit, thus we will not detect a failure before the unit is demanded, or we perform a functional test

6. 6 Unavailability for evident functions The uptimes and downtimes are characterized by: Mean Time To Failure = MTTF = E(T 1 ) = E(T 2 ) = … Mean Down Time = MDT = E(D 1 ) = E(D 2 ) = … Graphically we observe that the average “downtime” is given by where = 1/MTTF = failure rate (assume constant failure rate)  = 1/MDT = repair rate (assume constant repair rate)

7. 7 Hidden functions A hidden function means that a component failure is not immediately detected, i.e. We do not know when we are going from “Up” to “Down” In this situation D 1,D 2,… represent the ”non detected” downtime + the time of repair In order to reduce the non-detected downtime, the component is tested (function test) periodically Time between testing is denoted  (test interval) If a failure occurs in a test period we will in average be down half of the interval, i.e.  /2

8. 8 Unavailability (probability of failure on demand)

9. 9 Reliability block diagram (RBD) Reliability block diagrams are valuable when we want to visualise the performance of a system comprised of several (binary) components The interpretation of the diagram is The system is functioning if it is a connection between a and b, i.e. it is a path of functioning components from a to b. The system is in a fault state (is not functioning) if it does not exist a path of functioning components between a and b

10. 10 Exercise Define all combination of component states in the example RBD List those combination that represent a system fault state 123System Up DownUp Down Etc.

11. 11 Structure function For components we have For structures (systems) we have  denotes the structure function, and depends on the x i ’s (x is a vector of all the x i ’s).

12. 12 Structure function for simple structures For a serial structure we have  (x) = x 1  x 2 ...  x n For a parallel structure (redundancy) we have  (x) = 1-(1-x 1 )(1- x 2 ) …(1- x n ) For structures composed of serial- and parallel structures we may combine the formulas above

13. 13 Example  (x) =  I  II because I and II are in serial  I = x 1  II = 1-(1-x 2 )(1- x 3 ) because 2 and 3 are in paralell   (x) = x 1 (1-(1-x 2 )(1- x 3 ))

14. 14 Pivotal decomposition Some structure are more complex than just a collection of serial and parallel structures We may then often use the rule of pivotal decomposition to establish the structure function Consider component number i in the structure The rule of pivotal decomposition now states:  (x) = x i  (x | x i =1) + (1-x i )  (x | x i =0) where  (x | x i =1) is the structure function if component i is functioning (all the time), and  (x | x i =0) is the structure function if component i is not functioning (is not there)

15. 15 Example Component 5 is causing “problems” for us If component 5 is always functioning we have  (x | x 5 =1) = [1-(1-x 1 ) (1-x 3 )] [1-(1-x 2 ) (1-x 4 )] If component 5 is never functioning we have  (x | x 5 =0) = [1-(1-x 1 x 2 ) (1-x 3 x 4 )] In total  (x) = x 5  (x | x 5 =1) + (1-x 5 )  (x | x 5 =0) = x 1 x 2 +x 3 x 4 -x 1 x 2 x 3 x 4 +x 1 x 4 x 5 -x 1 x 2 x 4 x 5 - x 1 x 3 x 4 x 5 +2x 1 x 2 x 3 x 4 x 5 +x 2 x 3 x 5 -x 1 x 2 x 3 x 5 -x 2 x 3 x 4 x 5

16. 16 k-out-of-n structures A k-out-of-n system is a system that functions if and only if at least k out of the n components in the system is functioning We often write k oo n to denote a k out of n system, for example 2 oo 3 Example 1: We have three pumps which each has a 50% of the total required capacity Thus, it is sufficient that only two pumps is functioning to achieve sufficient pump capacity  2 oo 3 system. Example 2: We have mounted 3 fire detectors in a process area To avoid shut-down due to false alarms, we vote the detectors, and require 2 fire indications to shut-down and start fire fighting  2 oo 3 system

17. 17 Structure function for k oo n Example: 2 oo 3   (x) = x 1 x 2 +x 1 x 3 +x 2 x 3 -2x 1 x 2 x 3

18. 18 Binary state variables Note that the state variables are binary, and takes the values one or zero. We note that 1 n = 1 for n > 1 0 n = 0 for n > 1 Thus we could replace x i n with x i for n > 1 This simplifies the expansion of the terms in the structure function, and is also a prerequisite for system reliability assessments to come later in the course

19. 19 Exercise Find the structure function for the following RBD

20. 20 Solution  I (x) = 1-(1-x 3 )(1-x 4 ) = x 3 +x 4 -x 3 x 4  II (x) = x 5  I (x) = x 3 x 5 +x 4 x 5 -x 3 x 4 x 5  III (x) = 1-(1-x 2 )(1-  II (x)) = x 2 +x 3 x 5 +x 4 x 5 -x 3 x 4 x 5 -x 2 x 3 x 5 -x 2 x 4 x 5 +x 2 x 3 x 4 x 5  (x) = x 1  III (x) = x 1 x 2 +x 1 x 3 x 5 +x 1 x 4 x 5 -x 1 x 3 x 4 x 5 -x 1 x 2 x 3 x 5 - x 1 x 2 x 4 x 5 +x 1 x 2 x 3 x 4 x 5

21. 21 Component reliability

22. 22 System reliability (h(p) = p S )

23. 23 1.Obtain the structure function  (x) 2.Multiply out all terms in  (x) (to get a sum of products) 3.Remove all exponents in powers of x, i.e. replace x i n with x i for n > 1. Denote the result  M (x) 4.The system reliability is found by replacing the x i ’s in  M (x) with the corresponding p i ’s, i.e., 5.h(p) = p S =  M (x  x = p) Note 1: All the X i ’s must be independent Note 2: Step 2 could be extremely time consuming

24. 24 Exercise Find the system reliability for the RBD: When the following reliability parameters are given: ComponentMTTFMDT 120 0008 24 00024 31 00048 41 00048 54 00012

25. 25 Solution h(p) = p S =  M (x  x = p) = p 1 p 2 +p 1 p 3 p 5 +p 1 p 4 p 5 -p 1 p 3 p 4 p 5 -p 1 p 2 p 3 p 5 -p 1 p 2 p 4 p 5 +p 1 p 2 p 3 p 4 p 5 = 0.99957 Component MTTFMDT 120 00080.99960 24 000240.99404 31 000480.95420 41 000480.95420 54 000120.99701