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Logic solver
Final elementsInput elements
http://www.puffer.com/
Example:
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LT
PT
PT
GD
GD
LT
Logic unit
GD
Logic unit
SIF
An emergency
shutdown system
A fire and gassystem
SIF
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≥ 10−5
≤ 10−4 ≥ 10000
≤ 100000
≥ 10−4
≤ 10−3 ≥ 1000
≤ 10000
≥ 10−3
≤ 10−2 ≥ 100 ≤ 1000
≥ 10−2
≤ 10−1 ≥ 10
≤ 100
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Increasing risk
EUC risk Tolerable
risk
Residual
risk
Risk reduction by
other risk reduction
measures
Risk reduction by
other technologysafety-related
systems
Risk reduction by
E/E/PEsafety-related
systems
Necessary risk reduction
Actual risk reduction
(Adapted from IEC 61508-5)
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= 1PFD
Hazardous
event
C2f d
Avoided by SIS*
Yes
No
1-PFD C1
f res
Risk reduction: How many times
that the frequency is reduced
by having the SIF
Accident
f res: Residual frequency
f d: Demand frequency
f tol: Tolerable frequency
No accident
PFD
f d . PFD = f res < f tol*Or more precisely , the SIF
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Logic solver
Voted 1oo3 (for safety)
Failure rate: λ Test interval: τFraction that are common cause failures: β
PFDIE [ (1-β)λτ]3 /4 + βλτ/2≈
PFD PFDIE + PFD LS + PFDFE≈
Example - input elements:
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Starting point
for risk reduction estimation
a
b
1
1
2
2
23
3
34
4
C = Consequence parameter
F = Exposure time parameter
P = Probability of avoiding the hazardous event
W = In the absence of the SIF under consideration
a
a
1
--- ---
---
--- = No safety requirements
a = No special safety requirements
b = A single SIF is not sufficient
1, 2, 3, 4 = Safety integrity level
W W W123
C
C
C
C
F
F
P
P
P
A
B
D
C
A
B
FF
P
P
P
A
B A
B
A
B
B
A
A
F
F P
PA
B
B
X
X6
X5
X4
X3
X2
1
Generalized arrangement(in practical implementations
the arrangement is specific to
the applications to be covered
by the risk graph)
Source: (IEC 61511, 2003)
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A
B
C
D
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= 0.1
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A
B
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A
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A B
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2
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i
f d
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W i · P F DSIF = f
< f
f
= f
PLLα
f
α
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f tol
B
B
A A
E ∗(N ) = 0.1 · 0.1 · 0.1 = 0.001
f
= IRPAAC
E ∗(N )α
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f tol
f tol
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1 2 3
P F DrequiredSIF = f
W i
i
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Inherently safer design
Control
Supervisory
Preventive
Mitigative
Barrier
Limitation
Response
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