SYSTEMS RELIABILTY 1

SYSTEMS are basically built of different components and /or subsystems. For each component, there is an assigned role in the system performance. Take the motorcar as an illustrating example. The car is a system composed of: Prime mover Transmission subsystem Car body Braking subsystem The transmission subsystem is composed of: Clutch, Transmission shaft, Differential gear, wheel axes and wheels. All these components are connected in SERIES The Braking system is composed mainly of two Alternative subsystems: Hydraulic leg actuated Brakes Mechanical hand operated brakes. The TWO subsystem are working in PARALLEL Electric power generating systems may consist of N generating sets, For purposes of Reliability Increase, Out of them M generating sets (M<N)are sufficient to provide the necessary power GENERALLY,

SYSTEMS may be built in one of the following configurations : SERIES PARALLEL (With HOT REDUNDANCY) Parallel Of Series Series Of Parallel

M Out Of N (MOON ) K Consecutive Out N: Failed 2/4 2 Consecutive Out 8: Failed System Fails System Survives

SYSTEMS IN SERIES The SYSTEM is considered failed if ONE of the components failed The SYSTEM is considered Working ONLY IF ALL components are working ABC If A, B and C are Events of having components A, B and C working, then System RELIABILITY is given by: Assuming that the state of any of the components is independent of the others Generally, if there N components are connected in SERIES, the System Reliability will be

SYSTEMS IN PARALLEL The SYSTEM is considered failed if ALL components failed A B C Generally, if there N components are connected in PARALLEL, the System Reliability will be Then A special case of having TWO Identical components

M Out Of N (MOON ) 2/4 A B C D The System operates if ALL the four components operate A B C D =1 Alternative OR THREE OUT OF FOUR operate A B C D’ ABD C’ ACD B’ BCD A’ =4 Alternatives OR TWO OUT OF FOUR operate AB C’D’ AC B’D’ AD B’C’ BC A’D’ BD A’ C’ CD A’B’ =6 Alternatives The Important Special Case, when ALL components are IDENTICALAnd Generally Having M Out Of N

STANDBY SYSTEMS c Main Unit STANDBY Unit M = Number of Standby Components `X is a Discrete Random Variable distributed according to POISSON’s Distribution

Systems in SERIES OF PARALLEL A B C D Consider first the PARALLEL and then the SERIES,, we get Generally, N is the number of PARALLEL Groups connected in SERIES M G is the number of components in Parallel in the G th Group

For COMPLEX SYSTEMS, The RELIABILITY and MTTF Cannot be obtained in a CLOSED FORM as obtained for the Previous configurations. Therefore, Special Methods will be applied as will be shown later

Complex Systems

3 Cut Sets 1,2 3,4 1,5,3 2,4,5 F F cut sets R cut sets R system MINIMUM CUTSETS METHOD

Systems in PARALLEL OF SERIESA B C D Consider first the SERIES and then the PARALLEL,, we get Generall y N is the number of PARALLEL Branches M G is the number of components in series in the G th branch

K Consecutive Out N: Failed 2 Consecutive Out 6: Failed R is the component Reliability R S is the system Reliability N=6 the total number of components K =2 The number of consecutive components If Failed, the system fails

RELIABILITY & MTTF FOR SYSTEMS WITH COMPONENTS HAVING CONSTANT FAILURE RATE

CONSTANT FAILURE RATE (CFR) Given a System composed of N components connected in Series The Reliability of the K th component is given by The System Reliability: The Mean Time To Failure MTTF of the SYSTEM As the number of components in series increases, MTTF of the system DECREASES SERIES

CONSTANT FAILURE RATE (CFR) PARALLEL Given a System composed of N components connected in Parallel The Reliability of the K th component is given by The System Reliability: The Mean Time To Failure MTTF of the SYSTEM Example: Take =2 and the components are identical with the same failure rate λ

Two Identical Components With Failure Rate λ =0.01 (MTTF=100 hrs) R=0.9 ConfigurationIN SERIESIN PARALLEL MTTF S 1 / 2λ1.5 / λ 50 hrs150 hrs RSRS R R - R

Two components each with λ=0.01 are put in Parallel λ=0.01 RSRS R

Each component of The FOUR: Λ=0.02, R=0.95 Find System Hazard Rate And Reliability

1) Unit reliability R = 10 / (10 + t) t in years. How many units in parallel are required to achieve a reliability of 0.98 in 5 years? If there is an additional common mode failure rate of as a result of environmental factors. How many units in this case? 2) A natural gas distribution network contains FIVE shut-off valves. Valves 1 – 4 have probability of 0.02 of failing open and a probability of failing short. Valve 5 has probability of 0.05 of failing open and a probability of 0.2 failing short. Find system reliability