8. Single Fatality risk tolerance*
0.01% per year
Compared to…
Multiple Fatality risk tolerance*
0.001% per year
*Generalized risk tolerance in an industrial environment
9. Let’s try to make these
number resonate with
you with some relativity
15. Step 1: Identify a single
consequence to analyze*
*this is commonly done during the HAZOP to screen out high risk
scenarios for LOPA
16. In our example, the re-
boiler condensate pot
can overpressure
leading to vessel
rupture and resulting in
a single fatality
17. D-101 Re-boiler Condensate Pot
PSV Size ½
111 SET @ 700kPag
To atmosphere at safe
Steam location
Drawing Ref. D-101
LT
3/4” 2” 253
3”
HLL=2550 mm
PG
253
NLL=1650 mm
LLL=250 mm LC
2”
3” 253
3/4”
6”
LY LV
253 253
Condensate
6” 6” 6”
Drawing Ref.
6”
18. Step 2: Define the
tolerable frequency for
the consequence
19. Multiple
• 0.00001/year or 0.001%/year
Fatality
Single
• 0.0001/year or 0.01%/year
Fatality
Hospitalized • 0.001/year or 0.1%/ year
Injury
20. Step 3: Assess the
probability of the
initiating events*
*this could be identified during the HAZOP as causes
21. The level control valve
fails in the closed
position leading to
overpressure
22. D-101 Re-boiler Condensate Pot
PSV Size ½
111 SET @ 700kPag
To atmosphere at safe
location
Steam
Drawing Ref. D-101
LT
3/4” 2” 253
3”
HLL=2550 mm
PG
253
NLL=1650 mm
LLL=250 mm LC
2”
3” 253
3/4”
6”
LY LV
253 253
Condensate
6” 6” 6”
Drawing Ref.
6”
23. Let’s say this control loop has
a 0.1 probability of failure in a
year
24. Step 4: Identify
independent protection
layers* and assign a risk
reduction factor
*this could be identified during the HAZOP as safeguards
25. Important
The protection layer must
be independent from the
initiating event and
independent for other
safeguards used for this
consequence
26. D-101 Re-boiler Condensate Pot
PSV Size ½
111 SET @ 700kPag
To atmosphere at safe
location
Steam
Drawing Ref. D-101
LT
3/4” 2” 253
3”
HLL=2550 mm
PG
253
NLL=1650 mm
LLL=250 mm LC
2”
3” 253
3/4”
6”
LY LV
253 253
Condensate
6” 6” 6”
Drawing Ref.
6”
27. Let’s say the pressure safety
valve will reduce the
likelihood of tank rupture by
100
or you can say…
Risk Reduction of 100
you can also say…
the Probability of Failure on
Demand of 0.01
28. Step 5: Calculate the
expected frequency of the
consequence scenario
29. Expected frequency =
initiating event frequency x
probability of failure of
safeguards
Expected frequency = 0.1
valve failure per year x 0.01
probability of safety valve
failure
30. D-101 Re-boiler Condensate Pot
PSV Size ½
111 SET @ 700kPag
To atmosphere at safe
location
Steam
Drawing Ref. D-101
LT
3/4” 2” 253
3”
HLL=2550 mm
PG
253
NLL=1650 mm
LLL=250 mm LC
2”
3” 253
3/4”
6”
LY LV
253 253
Condensate
6” 6” 6”
Drawing Ref.
6”
31. Given a person will be around the
vessel when ruptured…
Our expected frequency of a
fatality in this scenario is 0.001
per year
Or
0.1% chance of a fatality per year
32. D-101 Re-boiler Condensate Pot
PSV Size ½
111 SET @ 700kPag
To atmosphere at safe
location
Steam
Drawing Ref. D-101
LT
3/4” 2” 253
3”
HLL=2550 mm
PG
253
NLL=1650 mm
LLL=250 mm LC
2”
3” 253
3/4”
6”
LY LV
0.1% chance of
253 253
Condensate
rupture leading to a
6” 6” 6”
Drawing Ref.
fatality 6”
33. Step 6: Decide if risk is
acceptable based on the
tolerable frequency
34. Expected Tolerable
frequency frequency
of a single of a single
fatality = fatality =
0.001/year 0.0001/year
35. That’s 10 times more
likely than the maximum
frequency your company can
accept for a single fatality
36. Step 7: Determine the
additional safeguards to
reduce the risk to meet
the tolerable frequency
37. Let’s add a high pressure
shutdown to the inlet as
a safeguard
38. D-101 Re-boiler Condensate Pot
XV PSV Size ½
253 111 SET @ 700kPag
To atmosphere at safe
location
Steam
Drawing Ref. D-101
LT
2” 253
3”
HLL=2550 mm
HH
PT
253
NLL=1650 mm
LLL=250 mm LC
2”
3” 253
6”
LY LV
253 253
Condensate
6” 6” 6”
Drawing Ref.
6”
39. This safeguard consist of a
pressure sensor, logic
solver (independent from
the level control) and a
valve as a final element
40. This safeguard is a safety
instrumented function (SIF)
XV
253
PT
253
41. Since we need to reduce the
risk by a factor of 10…
The probability of failure on
demand of the safety
instrumented function must
be less than 0.1
42. Or you can say the safety
instrumented function
must meet the
requirements of safety
integrity level 1
44. Expected frequency with the
new safeguard
= 0.1 probability of valve
failure per year
x 0.01 probability of safety
valve failure
x 0.1 probability of the safety
instrumented function failure
=0.0001/year
45. New Tolerable
expected frequency
frequency of a single
of a single fatality =
fatality = 0.0001/yea
0.0001/yea r
r