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Reliability of systems 1 Reliability of Systems Serial Configuration In this configuration all components of system needs to function for the system to function. 1 2 3 N System reliability (Rs) Rs(t) = R1X R2X R3 …. Rn Min(R1, R2,, R3, …, Rn) In such case system reliability can never be greater than the smallest component reliability Constant Failure Rate System Where s = i Rs (t ) Ri (t ) exp( i t ) exp i t exp( s t ) i 1,n Weibull system t i t i Rs (t ) exp exp i 1,n i i 1,n i t dR (t ) 1 (t ) s i dt Rs (t ) i 1,n i i i 1 It is evident from above expression of (t) that system does not exhibit Weibull type failure though every component has Weibull failure distribution Reliability of systems 1 2 Parallel configuration If components in parallel or redundant, all units fail for the system to fail. 2 3 Rs (t ) 1 i 1, n [1 Ri (t )] N Rs(t) Max( R1, R2 , R3 ,...Rn ) In parallel configuration system reliability is always greater than maximum reliability of the components Constant failure rate (CFR) system Rs (t ) 1 (1 e it ) i 1,n Weibull system t i Rs (t ) 1 1 exp i 1,n i Combined Series and Parallel Systems A complex system contains both series and parallel components. Such systems need to be broken down in series and parallel subsystems. Finally the reliability of the system may be obtained based on the relationship among the subsystems. Reliability of systems 3 Low Level Redundancy Each component comprising the system may have one or more parallel components. Let each component has reliability R A B Rlow [1 (1 R ) 2 ]2 (2 R R 2 ) 2 High Level redundancy A B The entire system may be placed in parallel with one or more identical system Rhigh [1 (1 R ) ] 2 R R 2 2 2 A 4 B A B Comparison of High and Low level of redundancy By comparing the two reliability, it may be said that the reliability of low level redundancy is greater than the reliability of the high-level redundancy. It may be seen below: (Rlow – Rhigh) = (2R – R2)2 – (2R2 – R4) = 2R2(R2-2R +1) = 2R2(R-1)2 0 Reliability of systems 4 k-out-of-n Redundancy K-out-of-n components to function for the system to function. If k=1, complete redundancy occurs, and if k=n, the n components are, in effect, in series. The reliability may be obtained from the binomial probability distribution If R is reliability of each independent trial, than the probability of k or more successes among the n components may be represented as: n Rs R x (1 R ) n x x k x n n! x x!(n x)! n MTTF Rs (t )dt 0 Complex Configuration In many cases the component configuration may be such that the system reliability cannot be simply decomposed into series and parallel relationships. Such networks may be analyzed either by Decomposition or Enumeration. Decomposition In this case system is broken down in subsystem and reliability of each system is determined separately A A E B D B I D C A C B D C II Reliability of systems 5 The above shown network is broken down in two sub network, one in which component E is failed (II) and one in which component E is function with reliability RE (I). The total reliability of the system may be computed as: Rs = RE.RI + (1-RE).RII Enumeration All possible combination of success and failures of each component and the resulting success and failures of the system are identified. For each possible combination of component success or failures, the probability of the intersection of these events is computed (considering mutually exclusive events). Sum of success probabilities or one minus sum of failure probabilities is the system reliability Reliability of systems 6 System Structure function The system structure function is defined as: 1 system operates ( X 1, X 2 , X 3 ,... X n ) 0 system has failed For k-out-of-n system, the system structure function would be: 1 if X i k i 1, N ( X 1, X 2 , X 3 ,... X n ) 0 if X i k i 1, N Reliability for series system Pr{( X 1, X 2 , X 3 ,... X n ) 1} Pr{ X 1 1, X 2 1, X 3 1,..., X n 1} R1.R2 R3 ...Rn Reliability for parallel system Pr{ ( X 1, X 2 , X 3 ,... X n ) 1} Pr{max( X 1 , X 2 , X 3 ,..., X n ) 1} 1 Pr{allX i 0} 1 (1 R1 )(1 R2 )(1 R3 )...(1 Rn ) Reliability for k-out-of-n system Pr{( X 1, X 2 , X 3 ,... X n ) 1} Pr{ X i k} In case R1= R2= R3 … = Rn, the reliability can be determined using binomial probability distribution i 1 n Reliability of systems 7 Cutsets and minimum cutsets: A cutset is a set of components whose failure will result in a system failure. A minimal cutsets is one in which all the components must fail in order for the system to fail. Pathways and minimum pathways: A pathway is a set of components whose functioning ensures that the system function. A minimal pathways are the one in which all components within the set must function for the system to function. System bounds Lower bounds RL 1 (1 Rk ) i 1 kS c Where c is the number of minimal cutsets and S is the representation of those components composing ith minimal cutest. High bounds p RH 1 1 Rk i 1 kT Where p is the number of minimal pathways and T is the representation of those components composing ith minimal pathway. Common-Mode Failure Several components may fail by common mode failure such as same power source, external load, vibrations etc. A common-mode failure can be depicted in series with those components sharing the failure mode. In order to represent system in series, it must be possible to separate independent failures from common-mode failures. Reliability of systems 8 Three-state devices Many components may have three states such as open, short failure, and an operating stage. Examples includes, electric circuits, flow valves. The interesting problem with respect to system comprising these components is that redundancy may not always increase the system reliability. Two assumptions are made in analysis: Failures modes are mutually exclusive All components composing the system are independent Problem statement: A system comprising of two switches. Series structure For the system to fail short, both switches to fail short, for system to fail open, at least one switch must fail open. qoi = the probability that component I fails open qsi = the probability that component I fails short R = 1-[(qs1. qs2)+( qo1 + qo2 - qo1.qo2)] = (1- qo1) (1- qo2) - qs1. qs2 In other words: R (1 qoi ) q si i 1 i 1 n n Parallel structure For the system to fail short, one or both switches to fail short, for system to fail open, both switch must fail open. R = 1-[(qo1. qo2)+( qs1 + qs2 – qs1.qs2)] = (1- qs1) (1- qs2) – qo1. qo2 In other words: R (1 q si ) qoi i 1 i 1 n n Reliability of systems 9 Low-level Redundancy Block diagram of m three state devices in a low-level redundant configuration 1 1 1 2 2 2 m m m N components 1 2 m The generalized expression for series redundant system having at least one unopened path in each redundant string minus at least one short in each path RL (1 qoi ) [1 (1 q si ) n ] n i 1 i 1 m m High-level Redundancy Block diagram of m three state devices in a high-level redundant configuration 1 1 1 2 2 2 m m m N components 1 2 m The generalized expression for series redundant system having no path completely shorted minus at least one open in each path is as: m m RH 1 qsi 1 (1 qoi ) i 1 i 1 n n

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posted: | 1/13/2010 |

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