Component reliability Jrn Vatn 1 The state of

Component reliability Jørn Vatn 1

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

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

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

Evident and hidden function n 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 n 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 5

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

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

Unavailability (probability of failure on demand) 8

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

Exercise n Define all combination of component states in the example RBD n List those combination that represent a system fault state 1 2 3 System Up Up Down Etc. 10

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

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

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)) 13

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

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

k-out-of-n structures n 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 n We often write k oo n to denote a k out of n system, for example 2 oo 3 n Example 1: We have three pumps which each has a 50% of the total required capacity n Thus, it is sufficient that only two pumps is functioning to achieve sufficient pump capacity 2 oo 3 system. n Example 2: We have mounted 3 fire detectors in a process area n 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 16

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

Binary state variables n Note that the state variables are binary, and takes the values one or zero. n We note that n 1 n = 1 for n > 1 n 0 n = 0 for n > 1 n Thus we could replace xin with xi for n > 1 n 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 18

Exercise n Find the structure function for the following RBD 19

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 20

Component reliability n 21

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

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 xin with xi for n > 1. Denote the result M(x) 4. The system reliability is found by replacing the xi’s in M(x) with the corresponding pi’s, i. e. , 5. h(p) = p. S = M(x x = p) n Note 1: All the Xi’s must be independent n Note 2: Step 2 could be extremely time consuming 23

Exercise n Find the system reliability for the RBD: n When the following reliability parameters are given: Component 1 2 3 4 5 MTTF MDT 20 000 8 4 000 24 1 000 48 4 000 12 24

Solution Component MTTF 1 2 3 4 5 20 000 4 000 1 000 4 000 MDT 8 24 48 48 12 0. 99960 0. 99404 0. 95420 0. 99701 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 25