fault tree analysis part 8 - probability calculation
TRANSCRIPT
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Fault Tree Analysis
Part 8 - Probability Calculation
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RESULTS OF PROBABILITY CALCULATIONS
1) The probability of the top event.
2) The “importance” of the cut sets and primal events.
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PROBABILITY OF EVENTS CONNECTED BY AN “AND” GATE
In general, if events X and Y are probabilistically dependent, then
Pr( ) Pr( )Pr( )X Y X Y YWhere, is the probability that X occurs given that Y occurs.
If events X and Y are probabilistically independent, then
Pr( )X Y
Pr( ) Pr( )
Pr( ) Pr( )Pr( )
X Y X
X Y X Y
Usually, it is assumed that the basic events in a fault tree are independent.
Thus,
1 2 1 2Pr( ) Pr( )Pr( ) Pr( )
n nB B B B B B
and
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PROBABILITY OF EVENTS CONNECTED BY AN “OR” GATE
1 2 1 2
1 2
Pr( ) 1 Pr( )
1 [1 Pr( )][1 Pr( )] [1 Pr(B )]n n
n
B B B B B B
B B
2,n
1 2 1 2 1 2Pr( ) Pr( ) Pr( ) Pr( )Pr( )B B B B B B
3,n1 2 3 1 2 3
1 2 2 3 3 1
1 2 3
Pr( ) Pr( ) Pr( ) Pr( )
Pr( )Pr( ) Pr( )Pr( ) Pr( )Pr( )
Pr( )Pr( )Pr( )
B B B B B B
B B B B B B
B B B
Note,
1 2 1 2 1 2Pr( . . ) Pr( ) Pr( ) 2Pr( )Pr( )B EOR B B B B B
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PROBABILITY OF EVENTS CONNECTED BY A m-OUT-OF-n
VOTING GATEAssume
1 2Pr( ) Pr( ) Pr( )
nB B B Q
thenkQ Q (1 Q)
nn k
systemk m
n
k
! =
!( )!
n n
k k n k
where
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SHORT-CUT CALCULATION METHODS
Information Requiredj
j
(1) failure rate cons tan t
(2) repair rate cons tan t
(3) min imum cut sets
Approximation of Event Unavailability
When time is long compared with MTTR and , the following approximation can be made,
Where, is the MTTR of component j.
tjj
jj
jjeQ )(1
1.0j
j
jjj
j
j
j
j
j
jj
jj
tQ
1)(
)(lim
j
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Z
AND
X Y
IF X and Y are Independent
YX
YXZ
YXYXZ
)(
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AND-Gate Algorithm
N
j j
AND
ANDAND
N
j j
N
jjj
N
j j
jANDANDAND
N
j
N
ll
j
jN
j
N
jl
llj
AND
j
N
jj
N
jj
AND
Q
QQw
tQtww
1
111
1 11 1
11
11
1
)()(
)(
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Z
OR
X Y
( )
( )
z x y
x x y yz
x y
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OR-Gate Algorithm
M
ii
M
iii
OR
OROR
M
ii
OROR
i
M
ii
M
ii
OR
Q
w
1
1
1
11
)(
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COMPUTATION OF ACROSS LOGIC GATES
,
2 INPUTS 3 INPUTS n INPUTS
1 2 1 2( )
1 2
1 2
1 2
1 1 2 2
1 2
1 2 3 2 3 1 3 1 2( )
1 2 3
2 3 1 3 1 2
1 2 3
1 1 2 2 3 3
1 2 3
1 2 2 3 1 3
1 2 1
(
)n n
n
1 2
11 1 1
n
1 2 n
1 1 2 2
1 2 3
n n
n
AND
GATES
OR
GATES
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COMPUTING TOP EVENT PROBABILITY
1.Compute q (= ) for each primal Event.
2.Compute the Probability or Failure Rate for each
Cut Set (QK). Use the “AND” Equation.
3.Compute the Top Event Probability or failure rate.
Use the “OR” Equation.
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Example
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HEAT EXCHANGERHOTNITRICACID
TEMPERATURE SENSOR
TO REACTOR
1
3
2 8
AIR TO OPEN TRC SET POINT
5
6
COOLING WATER
4
7
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M 7
M 4
T 2
WATERLEAKSINTDACID
T 8EXT. FIRE AT
HWAT EXCHANGER
M 2
M 8
M 1
M 3
T 1
T 4
T 7
P 5
TEMRSENSORFAILSLOW
P 6
SETPOINT
EXT.FIREATTRC
INSTRUMENTAIR
PRESSURE
+1
-10 (HX FOULED)
+1
0 V
AL
VE
ST
UC
K
+1 -1(C
ON
TR
OL
VA
LV
E
RE
VE
RSE
D)
+1 +1
+1+1
+1
+1
-1
+1
+1
+1
+1 0
(TEM
P. S
ENSO
R S
TUC
K)
-10+1
-1 (TRC REVERSED)
0 (TRC STUCK)
0 (ON MANUAL)
-1
+1
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CUT SET IMPORTANCE
The importance of a cut set K is defined as
k
k
s
QI
Q
Where, is the probability of the top event. may be interpreted as the conditional probability that the cut set occurs given that the top event has occurred.
SQ
KIK
PRIMAL EVENT IMPORTANCEThe importance of a primal event is defined asX
1
1
1 M
X KK
S
M
X KK
i QQ
i I
or
Where, the sum is taken over all cut sets which contain primal event .X
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[ Example ] TOP
OR
OR
1 2
G2 AND
3 OR
AND 5
3 4
G3
G4
G5
OR
6 ANDG7
G2 3
G6
GATE CUT SETS
2
5
4
7
3
6
1
(1) (2)
(3 , 4)
(3 , 4) (5)
(1 , 3) (2 , 3)
(3 , 4 , 3) (3 , 5)
(6) (1 , 3) (2 , 3)
(1) (2) (3 , 4) (3 , 5) (6) (1 , 3) (2 , 3)
Hence, the minimal cut sets for this tree are : (1) , (2) , (6) , (3 , 4) and (3 , 5).
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As an example , consider the tree used in the section on cut sets.
The cut sets for this tree are (1) , (2) , (6) , (3,4) ,(3,5). The following data
are given from which we compute the unavailabilities for each event.
1yr Event ( )yr hr
1 .16 1.5E-5 (.125) 2.4E-6
2 .2 1.5E-5 (.125) 3.0E-6
3 1.4 7E-4 (6) 9.8E-4
4 30 1.1E-4 (1) 3.3E-3
5 5 1.1E-4 (1) 5.5E-4
6 .5 5.5E-5 (.5) 2.75E-5
Now, compute the probability of occurrence for each cut set and top event
probability. Cut Set K
Q (1) 2.4E-6
(2) 3.0E-6
(6) 2.75E-5
(3,4) 3.23E-6
(3,5) 5.39E-7
SQ 3.67E-5
iq
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THE COMMON–MODE FAILURES WITHIN FAULT TREES
S
1
2
3
SWITCH
PUMP 2
(STAND – BY)POWER 2
POWER 1
PUMP 1
(RUNNING)
Shared Power Source
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PUMP 2
SPEED
S
PUMP 1
SPEED PUMP 1
MECH.
FAILURE
POWER 1
FAILURE
PO
WE
R 2
FA
ILU
RE
PU
MP
2 M
EC
H.
FA
ILU
RE
+10 0
10
0
(PU
MP
1 S
PE
ED
= -
10)
-1
SW
ITC
H S
TU
CK
2M
3M
1M
-10
-10
0
1
0
1
0
-10+1
+1
0
-10
+1
+1
0
+10
PO
WE
R 1
.
FA
ILU
RE
0
0
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3M ( 10)
AND
G1
OR
P1 Mech Fail.
LocalPower 1Failure
64.97 1041.69 101/ 2.68 yrs.
3.97 hrs.
G2
Pump 1
Shut Down OR
P2MechFail.
LocalPower 2Faiture
SwitchStuck
G3 22.94 10
1/90.8 yrs.
3.95 hrs.
1/3.04 yrs.
4.64 wks.
Pump 2 NotStarted
1 2 3 4 5
41.67 10
1/3 yrs.
4 hrs.
62.3 10
1/ 25 yrs.
5 hrs.
33.8 10
1/5 yrs.
1 wk.
45.5 10
1/35 yrs.
1 wk.
22.5 10
1/10 yrs.
3 mo.
LocalPower 1Failure
2
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GATE MIN CUT SETS
G2
G3
G1
(1) , (2)
(2) , (3) , (4) , (5)
(1 , 2) , (1 , 3) , (1 ,4) , (1 ,5)
(2 , 2) , (2 , 3) , (2 ,4) , (2 , 5)
(1 , 2) , (1 , 3) , (1 , 4) , (1 , 5)
(2) , (2 , 3) , (2 , 4) , (2 , 5)
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3M ( 10)
OR
67.23 101/ 20 yrs.
1.27 hrs.
262.3 10 AND
6q 4.2 10
1/120 yrs.
4.4 hrs.
1 541.67 10 22.5 10
AND
76.4 101/ 20 yrs.
4.3 hrs.
1 341.67 10 33.8 10
AND
89.2 101/5333 yrs.
4.6 hrs.
1 441.67 10 45.5 10
COMP q-1(Yr )
1
2
3
4
5
41.67 1062.3 1033.8 1045.5 1022.5 10
1/3
1/25
1/5
1/35
1/10
4 Hr.
5 Hr.
1 Week
1 Week
3 Months
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Cut Set KQ K
(2)
(1 , 3)
(1 , 4)
(1 , 5)
6
7
8
6
2.3 10
6.4 10
9.2 10
4.2 10
1 / 25 Yr.
1 / 762 Yr.
1 / 5333 Yr.
1 / 120 Yr.
TOP Event6
TOPQ 7.23 10
TOP1/ 20Yr.
UnavailabilityImportances
Q 6 6
(2)
Q
(1,3)
Q
(1,4)
Q
(1,5)
I 2.3 10 / 7.23 10 .32
I .088
I .0012
I .58
Q
1
Q
2
Q
3
Q
4
Q
5
I .669
I .32
I .088
I .0012
I .53
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Unreliability Importances
R
(2)
R
(1,3)
R
(1,4)
R
(1,5)
1/ 25I 0.8
1/ 20I .026
I .004
I 0.167
R
1
R
2
R
3
R
4
R
5
I .197
I 0.8
I .026
I .004
I .167