SYSTEMS RELIABILTY 1. SYSTEMS are basically built of different components and /or subsystems. For...
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Transcript of SYSTEMS RELIABILTY 1. SYSTEMS are basically built of different components and /or subsystems. For...
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 failedThe SYSTEM is considered Working ONLY IF ALL components are working
A B C
If A, B and C are Events of having components A, B and C working, then System RELIABILITY is given by:
CBAPRS
Assuming that the state of any of the components is independent of the others
CBAS
S
RRRRCPBPAPR
)()()(
Generally, if there N components are connected in SERIES, the System Reliability will be
N
KKS RR
1
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
N
KK
N
KKS RFR
11
111
CBACBAS
CBAS
RRRFFFRFFFF
11111
Then
A special case of having TWO Identical components
2
22
2
)21(111
RRR
RRRR
S
S
M Out Of N (MOON)
2/4
A
B
C
D
DCBADCBA
DCBADCBADCBA
DCBACBAADCBA
DCBADCBADCBAS
RRFFRFRFFRRFRFFRFRFRFFRRRRRFRRFRRFRRFRRRRRRRR
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 AlternativesOR 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 IDENTICAL2234 64 FRFRRRS
And Generally Having M Out Of N
N
MK
KNKNKS FRCR
6!2 !2
!4 4!1!3!4
)!(!! 4
243
CC
mnmnC n
m
STANDBY SYSTEMSc
Main Unit
STANDBY Unit
STANDBY Unit)( MXPRS
M = Number of Standby Components
`X is a Discrete Random Variable distributed according to POISSON’s Distribution
...!3!2
1!
32 ttteXteR t
MX
oX
Xt
S
1...
!3!3
!2!211
0
MRdtMTTF
Systems in SERIES OF PARALLELA
B
C
D
Consider first the PARALLEL and then the SERIES,, we get
DCBAS FFFFR 11
Generally,
N
G
M
kKS
M
kKG
N
GGS
G
G
G
G
G
G
FR
FRRR
1 1
11
1
1
N is the number of PARALLEL Groups connected in SERIESMG 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
Cut Sets
1,2 3,4 1,5,3
2,4,5
1 2 3 4 1 5 3 2 4 5 F .05 .1 .15 .2 .05 .15 .25 .1 .2 .25
F cut sets
.005 .03
.00188 .005
R cut sets
0.995 0.97 0.99812 0.995
R system
0.9582
MINIMUM CUTSETS METHOD
Systems in PARALLEL OF SERIES A
B
C
DConsider first the SERIES and then the PARALLEL,, we get
DBCAS RRRRR 111
Generally
N
G
M
KKS
N
G
M
KKGGS
G
G
G
G
G
G
RR
RFFR
1 1
1 1
11
11
N is the number of PARALLEL BranchesMG 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 ReliabilityRS is the system Reliability N=6 the total number of components
K =2 The number of consecutive components If Failed, the system fails
2
1
0
1 )1(,2,
N
J
JNJJNJS RRCNRR
3
0
67 )1(6,2,J
JJJJS RRCRR
334
3425
2
561
6
)1()1(
)1(6,2,
RRCRRC
RRCRRRS
RELIABILITY & MTTF FOR
SYSTEMS WITH COMPONENTS HAVING
CONSTANT FAILURE RATE
CONSTANT FAILURE RATE (CFR)
Given a System composed of N components connected in SeriesThe Reliability of the K th component is given by
tK
KeR The System Reliability:
N
KK
K
tN
K
tS eeR 1
1
The Mean Time To Failure MTTF of the SYSTEM
N
KK
S
t
SS
MTTF
dtedtRMTTF
N
KK
1
0
0
1
1
As the number of components in series increases, MTTF of the system DECREASES
SERIES
CONSTANT FAILURE RATE (CFR) PARALLELGiven a System composed of N components connected in ParallelThe Reliability of the K th component is given by t
KKeR
The System Reliability:
N
K
tN
KKS
KeRR11
1111
The Mean Time To Failure MTTF of the SYSTEM
dtedtRMTTFN
K
tSS
K
0 10
11
Example: Take =2 and the components are identical with the same failure rate λ
tttS eeeR 22
211
5.1212
20
2
0
dteedtRMTTF ttSS
Two Identical ComponentsWith Failure Rate λ =0.01 (MTTF=100 hrs)
R=0.9
Configuration IN SERIES IN PARALLEL
MTTFS
1 / 2λ 1.5 / λ
50 hrs 150 hrs
RSR2
0.81
2R - R2
0.99
0 30 60 90 120
150
180
220
280
340
400
460
520
00.0020.0040.0060.008
0.010.012
Time
Two components each with λ=0.01 are put in Parallel
λ=0.01
λ=0.01
00.20.40.60.8
1
RS
R
ttS
t
eeR
eR
22
t
t
S
ttS
t
eeh
eetf
hetf
212
12)(
)(
Each component ofThe FOUR:Λ=0.02, R=0.95Find System Hazard RateAnd Reliability
9905.0)95.0(1122 SR
222 ttS eeR
2211 tS eR
121144
2
0
4
0
3
0
2
0
22
0
dtedtedte
eedtRMTTF
ttt
ttS
995.0)95.01(122 SR
432
11
0
4
0
2
0
22
0
dtedte
dtedtRMTTF
tt
tS
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 0.002 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 0. 15 failing short. Valve 5 has probability of 0.05 of failing openand a probability of 0.2 failing short . Find system reliability.
4
5
11
2
3
N
m m1
11
N
Km m11
N
m m1
11
N
Km m11