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FSE I – Pre-Class Exercise Solutions
1.
What does the Safety Integrity Level (SIL) measure?
The safety integrity level is a measure of risk reduction. The SIL that is selected during the
requirements portion of the safety life cycle is a measure of the risk reduction required to
make the process risk tolerable. During the verification stage of the safety life cycle the
amount of risk reduction that an SIS can provide is quantitatively determined.
2.
The probability of:
(Probability Multiplication)
P(A and B) = PA * PB
P(A or B) = PA + PB – (PA * PB), or 1 – (1 - PA)*(1 - PB) (Probability Addition)
(where A and B are not mutually exclusive)
If A and B are mutually exclusive
P(A or B) = PA + PB
3.
Name three different consequences that can occur as the result of a flammable material
release.
1.Flash Fire
2.Jet Fire
3.Pool Fire
4.Vapor Cloud Explosion
5.Fireball
6.Toxic release with no fire
4.
What are the three parts of an event tree?
1.Initiating Events
2.Branches or propagation steps or escalating events
3.Outcomes
5.
How are the initiating events and layers of protection logically related to the outcome
probability in a layer of protection analysis? What type of probability math is used to
relate them?
The probability of an outcome is the probability that an initiating event occurs AND all of
the protection layers fail. Probability multiplication is used to determine the outcome
probability.
6.
Where can information on what initiating events and layers of protection are involved with
a hazard be found?
The process hazards analysis (PHA) often using the HAZOP method is a systematic study
of a process that is designed to identify hazards that exist. The PHA will identify all
hazards that already have an SIS in place and all locations where an SIS is recommended.
In addition, the causes, consequences, and safeguards are listed.
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1
7.
What measure is used in LOPA to demonstrate the effectiveness of a safeguard, and how is
it calculated?
The effectiveness of safeguards is demonstrated as Probability of Failure on Demand
(PFDavg). PFDavg is a function of an items failure (λ) and test interval (TI). These
quantities are related by the following equation:
PFDavg = (λ * t) / 2
8.
Name two methods that can be used to assign SIL given that a consequence and likelihood
have been determined.
Risk Matrix
Risk Graph
Frequency Based Target
Individual Risk Based Target
9.
What standards are available to assist in design of burner management systems in your
plant’s location?
NFPA 85 and NFPA 86 in the US
AS 3814 / AG 501 and AS 1375 in Australia
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2
FSE I - Application Exercise 1
Title:
1.
Tolerable Risk
Develop a tolerable risk guideline and risk matrix for environmental risks ranging from 1
per 10 years to 1 per 100,000 year events and ranging from release inside the plant with
small consequences up to a release outside the plant with large permanent consequences?
Assume all extreme risks will be reduced and all moderate risks will be reduced where
practical.
1/10
yrs
1/100
yrs
1/1000
yrs
1/10,000
yrs
1/100,000
yrs
2.
Internal release with
small consequences
Internal release with
large consequences or
External release with
small temporary
consequences
External release with
large temporary or
small permanent
consequences
External release with
large permanent
consequences
Moderate
Moderate
Extreme
Extreme
Acceptable
Moderate
Moderate
Extreme
Acceptable
Moderate
Moderate
Moderate
Acceptable
Acceptable
Moderate
Moderate
Acceptable
Acceptable
Acceptable
Moderate
Compare your tolerance with that of the example matrix in the slides and identify the
equality points. (Where does the tolerable frequency match for different consequences?)
In the proposed answer,
Recordable injury roughly matches internal release with small consequences.
Lost time injury roughly matches internal release with large consequences or external release
with small temporary consequences.
Permanent injury roughly matches external release with large temporary or small permanent
consequences.
Many deaths roughly matches external release with large permanent consequences.
3.
Are there any significant points where the risk tolerance is inconsistent? For example does
the tolerance for external releases with large temporary consequences match that for many
human fatalities?
In the proposed answer, most items are generally consistent depending on the view of one death
vs external release with small permanent consequences. Better definition on the large and small
consequences is probably needed to make this a more useful working guide. Note that with the
same number of categories and the same risk tolerances, the matrix can be combined with the
one from the slides relatively easily by incorporating a definitions table for the four different
consequence magnitudes.
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3
FSE I - Application Exercise 2
Title:
1.
Probability
An insurance company studied 32400 persons for six months. There were 1800 accidents.
If this dangerous condition is equally likely at any moment, what is the probability of the
dangerous condition in any given year?
The probability of an event is the number of outcomes divided by the number of chances.
In this problem there is one outcome (i.e., accident) and nine chances (i.e., nine years). So,
P=
2.
outcomes
1800accidents
1
=
= = 0.11
chances 16200 person _ years 9
We toss three fair coins. What is the probability of getting three heads?
The probability of getting three heads is the ANDing of the probabilities of getting a head
on each of three individual tosses. For each individual toss the probability of heads is ½.
P1 = P2 = P3 = 0.5
Poverall = P1 * P2 * P3 = 0.5 * 0.5 * 0.5 = 0.125
3.
A system will fail if a power supply fails or a controller fails. The probability of a power
supply failure during the next year is 0.05. The probability of a controller failure in the
next year is 0.01. What is the probability of system failure?
The probability of system failure is given if the power supply OR the controller fails.
The events are logically OR’d so use probability addition. Also, the events are not
mutually exclusive (i.e., both the sight glass and transmitter can fail at the same time), so
use the form:
P (A or B) = PA + PB – PA * PB
Psystem failure = 0.05 + 0.01 – 0.05 * 0.01 = 0.0595
4.
A check valve has a probability of not stopping reverse flow of 0.015 in a one-year
interval. The probability of getting a dangerous condition in the next year is 0.004. What
is the probability of dangerous condition AND not having the check valve operate?
The occurrence of the described situation is the logical ANDing of two probabilities. Use
probability multiplication.
Poverall = 0.015 * 0.004 = 0.00006
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FSE I - Application Exercise 3
1. A fault tree is shown below. What is outcome frequency?
Freq. = 10 / year
Fa
Pb
P = 0.05
AND
Pc
P = 0.1
Outcome Frequency = Fa * Pb * Pc = 10 * 0.05 * 0.1 = 0.05 /year
2. A fault tree is shown below. What is the output probability?
P = 0.001
P = 0.002
P = 0.005
Pa
Pb
OR
Pc
Probability = 0.001 + 0.002 + 0.005 – 0.001*0.002 – 0.001*0.005 – 0.002*0.005 +
0.001*0.002*0.005 = 0.007983
OR
Approx. Probability = 0.001 + 0.002 + 0.005 = 0.008
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3. A fault tree is shown below. What is the output probability?
P = 0.004
P = 0.010
P = 0.006
Pa
Pb
OR
P = 0.01988
Pc
AND
P = 0.000159
P = 0.080
AND
P = 0.008
P = 0.100
P for the top OR gate = 1 – (1 – 0.004)*(1 – 0.010)*(1 – 0.006) = 0.01988
or
Approximate P for the top OR gate = 0.004 + 0.010 + 0.006 = 0.020
P for the bottom AND gate = 0.080*0.100 = 0.008
Total Probabilty = 0.01988*0.008 = 0.000159
Approximate Total Probabilty = 0.020*0.008 = 0.00016
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6
FSE I - Application Exercise 4
Title:
1.
Consequence Analysis Overview
Your company is estimating the risk posed by the failure of a new railroad track switching
system. Estimate the average consequence, in terms of injuries and fatalities, of a train
accident using the following data.
In 1996,
550 Fatalities
10,948 Injuries
2,443 Accidents
Data from Transportation Statistics Annual Report 1998, Bureau of Transportation
Statistics, US Department of Transportation, BTS98-S-01.
The average consequence is calculated by dividing the total consequence by the number of
opportunities.
Average Consequence = (# consequences) / (# opportunities)
Average Fatalities = 550 / 2,443 = 0.225
Average Injuries = 10,948 / 2,443 = 4.48
2.
Explain why average industry loss data may not be a valid way to estimate the consequence
for chemical accidents?
For industry average data to be valid two conditions must be satisfied. 1) There must be a
large amount of incidents from which to draw data. 2) Each of the incidents must occur
under roughly similar circumstances. Neither of these two conditions are true for chemical
accidents. Luckily, the amount of chemical accidents is fairly small. Additionally, all
chemical plants are very different. It is very unlikely that potential consequences of
different plants will be similar enough to allow statistical analysis.
3.
A high-pressure vessel containing flammable gas that is liquefied under pressure undergoes
an incident where it is expected to instantaneously rupture. What type of incident outcome
can be expected if there is a source of ignition? If there is no source of ignition?
If there is a source of ignition, a fireball will occur. If there is no source of ignition,
possible consequences include equipment damage and other economic losses.
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FSE I - Application Exercise 5
Title:
Event Tree Analysis
PROCEDURE:
1. Draw an event tree that describes that following situation: (Use the back of this sheet)
•
•
•
A toxic release can be initiated by a delivery driver pumping more material into a storage
tank than the available capacity.
The delivery driver may or may not realize there is not enough capacity for the material
that he is delivering, and then not attempt to transfer the material.
The driver may carefully monitor the level in the storage tank and stop the material
transfer before a release occurs.
INITIATING EVENT
More m aterial than
available space
BRANCH 1
Driver does not
notice lack of
available Space
BRANCH 2
Driver does not detect
high level in tank
after starting pum p
OUTCOME
TRUE
Spill
FALSE
No Event
TRUE
FALSE
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No Event
8
2. Using the following data, quantify the frequency at which toxic releases occur.
•
•
•
Based on historical data, delivery drivers are requested to deliver to storage tanks that do
not have the required capacity approximately 3 times per year.
Due to a training initiative educating the drivers on the hazards of overfilling the tank the
probability that the driver will try to fill a tank that does not have sufficient capacity is
estimated at 0.01.
The probability that the driver will not detect a high level condition after he has begun
transfer is estimated at 0.1.
INITIATING EVENT
More m aterial than
available space
3 /year
BRANCH 1
Driver does not
notice lack of
available Space
BRANCH 2
Driver does not detect
high level in tank
after starting pum p
TRUE
0.1
FALSE
0.9
TRUE
0.01
FALSE
0.99
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OUTCOME
Spill
0.003 /year
No Event
0.027 /year
No Event
2.97 /year
FSE I - Application Exercise 6
Title:
Layer of Protection Analysis
PROCEDURE:
1.
Draw a LOPA diagram that describes that following situation
Reactant A
(through manhole)
FT
01
FIC
01
To Safe
Location
PSE
05
UC
01
PSH
04
Reactant B
Reaction
Inhibitor
Cooling Water
Supply
Drain
TSH
07
Product
Solution
UC
01
PROCESS:
A pharmaceutical company has developed a new process to produce one of its drugs. The
process creates an aqueous solution that is withdrawn from the bottom of the pressurized, water
cooling jacketed, continuously stirred tank reactor. The vessel is charged by filling it with 250
kg of water and manually dumping 125 kg, or 5 bags of reactant A into the vessel. After the
vessel is charged and closed, the stirring mechanism is started and the vessel’s jacket is flooded
with cooling water. After the stirring and cooling have been established a small metered rate of
0.005 kg/min of reactant B is continuously added to the solution. Reactants A and B combine to
form the desired product. Each batch operates for three weeks, and 12 batches are operated per
year.
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HAZARDS:
The reaction of A and B is nearly instantaneous and highly exothermic. Safe operation of this
process requires that an excess amount of reactant B never be allowed into the reactor, and that
cooling water continuously be flowing through the jacket. Hazard analysis determined that the
following events could cause a “runaway” reaction and physical explosion of the vessel.
1. Failure of controller FIC-01 causing uncontrolled reactant B entry into the reaction
vessel.
2. Failure of cooling water supply causing heat and pressure to build up in the vessel.
The following layers of protection were identified as a safeguard against explosion of the vessel
due to runaway reaction.
1. A rupture disk set to relieve the pressure well below the design pressure of the vessel.
2. Operator intervention to high vessel temperature, high vessel pressure and low cooling
water flow alarms. The alarm system is independent from the control system with no
common components.
It was also noted in the hazard assessment that the rupture disk pressure relief would not be
effective in the situation where controller FIC-01 failed, because pressure can not be vented as
fast as it is generated.
2.
Quantify the LOPA Diagrams
The following frequencies and failure probabilities were determined by a process engineer after
reviewing the history of the plant.
Flow control fails open:
Cooling Water Pump Fails:
1/25 /year
1/75 /year
Rupture Disk PFD:
Operator Response to Cooling Water Loss:
Operator Response to Control Failure:
0.0956
0.1
0.1
In this case, use fraction is a layer of protection. An accident can only occur when the hazard is
present.
3 weeks/batch * 7 days/week * 12 batches/year = 252 days/year of operation.
Use fraction is 252 days / 365 days = 0.69 = 69%
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1.
IE #1
INITIATING EVENT
FIC-01 Failure
PL #1
PL #2
Operator Failure Use Fraction
OUTCOME
Explosion
No Event
IE#2
INITIATING EVENT
Pump Failure
PL #1
PL#2
Operator Failure Rupture Disk Fails Use Fraction
OUTCOME
Explosion
No Event
2.
IE #1
INITIATING EVENT
FIC-01 Failure
PL #1
PL #2
Operator Failure Use Fraction
0.69
OUTCOME
Explosion
2.76E-03
0.1
1/25 /yr
No Event
IE#2
INITIATING EVPL #1
PL#2
Pump Failure Operator Failure Rupture Disk Fails Use Fraction
0.69
0.0956
OUTCOME
Explosion
8.80E-05
0.1
1/75 /yr
No Event
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FSE I - Application Exercise 7
Title:
Quantifying Initiating Events and Layers of Protection
PROCEDURE:
Use the excerpts from “Guidelines for Process Equipment Reliability Data” to quantify the rates
and / or probabilities of the following situations.
1.
A motor driven fan fails to provide cooling air, initiating an accident.
Use data from “Guidelines for Process Equipment Reliability Data” table 3.3.4, use the mean
failure rate. Failure mode of interest is “Fails while running”.
9.09 failures / 106 hours
converting to failures per year,
9.09 failures 8760hours
*
= 0.08 failures / year
106 hours
1year
* Initiating events described in frequency
2.
A flexible hose ruptures, initiating an accident
Use data from “Guidelines for Process Equipment Reliability Data” table 3.2.5, use the mean
failure rate. Failure mode of interest is “Rupture”.
0.570 failures / 106 hours
converting to failures per year,
0.570 failures 8760hours
*
= 0.005 failures / year
106 hours
1year
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3.
A non-operated check valve, with a periodic inspection and test interval of four years,
fails to prevent an accident.
Use data from “Guidelines for Process Equipment Reliability Data” table 3.5.1.2, use the
mean failure rate. Use catastrophic, which are given per unit time, not failures per attempt.
3.18 failures / 106 hours
PFDavg = (λ * t) / 2
PFDavg = (0.00000318 * 4 * 8760) / 2 = 0.055
* Protection layers must be described by a probability. In the case of periodic inspection
and test, average probability of failure on demand, which is a function of failure rate
and test interval, is the best probability to use.
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FSE I - Application Exercise 8
Title:
Assigning Safety Integrity Levels
PROCEDURE:
An accident can occur that will cause the release of 2,000-pounds of highly toxic phosgene from
a reactor that makes polycarbonate resin. Risk analysis has shown that the probable loss of life
due to this release is 75.6 fatalities per event. The analysis also showed that the accident has an
unmitigated frequency of once per 892 years. Use all four SIL assignment methods described in
this section to select safety integrity levels.
* Individual risk target for the facility is 1.0 x 10-4/year.
SOLUTIONS:
a.
Risk Matrix
Consequence Æ Extensive
Likelihood Æ 1/892 year = 1.2 x 10-3 Æ Moderate
High
Moderate
Low
2
1
NR
3b
2
1
3a
3b
3
SIL = 3
Note b: One Level 3 Safety Instrumented Function may not provide sufficient risk
reduction at this risk level. Additional review is required (see note d).
Note d: This approach is not considered suitable for SIL 4.
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b.
Risk Graph
Consequence Æ CD
Occupancy Æ FB
* No credit taken for lack of occupancy this factor is consolidated in the PLL = 75.6
estimate
Probability of Avoidance Æ PB
* No credit taken for lack of occupancy this factor is consolidated in the PLL = 75.6
estimate
Demand Rate Æ W1
Following Risk Graph Path yields SIL = 3
c.
Frequency Based Target
Select target based on consequence
Æ Extensive, 1.0 x 10-6
RRF = (1/892) / 1.0 x 10-6 = 1121
* Selected SIFRRF must be greater than 1121, so an SIF w/ SIL = 4
d.
Individual Risk Target
Select target based on consequence
Ftarget = 1.0 x 10-4 / 75.6 = 1.32 x 10-6
RRF = (1/892) / 1.32 x 10-6 / = 849
* Selected SIFRRF must be greater than 849, so an SIF w/ SIL = 3
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16
FSE I - Application Exercise 9
Title:
Comprehensive SIL Selection Exercise
PROCEDURE:
A chemical processor has just performed an upgrade of a process heater. The upgrade was
complex enough for the Management of Change procedures to be used. During the process a
new HAZOP was performed on the process section.
Review the HAZOP study to determine if there are any new SIS requirements. If so, select a
safety integrity level. The process plant’s tolerable risk target is based on the risk integral with a
target individual risk of 1.0 x 10-4.
Process Diagram:
Vent to Safe
Location
To Users
PSV
02
Vent to Flare
PT
07
PIC
07
Wet Gas from
Reciprocating
Compressor
LT
06
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LIC
06
17
Process Description:
A “wet” hydrocarbon gas is compressed by a reciprocating compressor into a flash drum. In the
flash drum liquid and vapor separate. The liquid is withdrawn from the bottom of the flash drum
under level control and vapor is withdrawn from the top of the vessel and either compressed and
sent to downstream users or sent to flare under pressure control. The flare line has not been
sized to pass the full discharge of the wet gas compressor to flare.
HAZOP Report Output
SIF:
Consequence:
Initiating event:
Protection Layers:
•
•
•
•
•
Open vent valve upon high pressure in vessel
Overpressure and rupture of vessel
Outlet vapor compressor fails
Operator intervention
Relief Valve
Relief valve is pilot operated, tested annually.
“Wet gas” compressor is a motor driven reciprocating compressor
“Vapor withdrawal” compressor is a motor driven reciprocating compressor
Operator is well trained, but only has 15 seconds to perform a shutdown before an
accident occurs.
Consequence analysis has determined a PLL=0.15 for the overpressure and explosion of
the flash drum.
SOLUTION
Step 1 – The LOPA diagram for the overpressure consequence is as follows.
INITIATING EVENT
Outlet vapor
com pressor fails
PL #1
Operator Fails
PL #2
Relief valve fails
OUTCOME
Overpressure
No Event
Step 2 – Quantify the LOPA diagram.
INITIATING EVENT
Outlet v apor
com pressor fails
PL #1
Operator Fails
PL #2
Relief v alv e fails
0.00415
OUTCOME
Ov erpressure
8.96E-02
1
21.6 /year
No Ev ent
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18
Vapor withdrawal compressor failure – Table 3.3.2.1
2470.0 failures / 106 hours Æ 21.6 failures per year
Operator Failure – Simplified Method
Conditions for PFD=0.1 are not met – use PFD = 1.0
Relief valve fails – Table 4.3.3.1
4.15 failures / 103 demands
PFD = 0.00415
Step 3 – Select SIL (Individual Risk / Risk Integral)
Ftarget = 1.0 x 10-4 / 0.15 = 6.67 x 10-4
PFD = 6.67 x 10-4 / 0.0896 = 7.44 x 10-3
RRF = 134
SIL = 3 (or SIL 2 with a RRF suitably greater than 134)
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19
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