This document discusses risk assessment methods for chemical processes including incident identification, consequence analysis, reliability analysis, fault trees, and event trees. It provides examples of calculating reliability, MTBF, availability, and unavailability for systems. Fault trees and event trees are presented as tools to analyze risks from initiating events and safety functions. Layer of protection analysis is introduced as a method to assess risks and the effectiveness of added safety layers in processes.

1. RISK ASSESSMENT
Chapter 11: Chemical Process Safety by Daniel A.
Crowl
MNIT Jaipur
Presented by:
Mayank Mehta Hardik K. Sharma

2. RISK ASSESSMENT
 Incident Identification
 How an accident occur?
 Probability analysis
 Consequence Analysis
 Analyse expected damage
(life, environment, capital)
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3. EXAMPLE 11.1
The water flow to a chemical reactor cooling coil is controlled by the
system shown in Figure. The flow is measured by a differential
pressure (DP) device, the controller decides on an appropriate
control strategy, and the control valve manipulates the flow of
coolant.
Determine the overall failure rate, the unreliability, the reliability, and the
MTBF for this system. Assume a 1-yr period of operation.
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4. MEANING OF TERMS
 Reliability
The probability that the component will not fail during
the time interval (0, t) is given by a Poisson
distribution R = exp(-μt)
 Unreliability:
The complement of the reliability is called the failure
probability (or sometimes the unreliability), P= 1-
R
 MTBF: mean time between failures
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5. SOLUTION
 These process components are related in series. Thus, if
any one of the components fails, the entire system fails.
Component Reliability
R = exp(-μt)
Failure
Probability
P = 1 - R
Faliure rate
(faults/yr) p
Control valve 0.55 0.45 0.60
Controller 0.75 0.25 0.29
DP Cell 0.24 0.76 1.41
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6.  The overall reliability for components in series:
R = R1.R2.R3 = 0.55* 0.75*0*24 = 0.10
 The failure probability:
 Overall failure rate:
0.10 = exp(-μ)
μ = -ln(0.10) = 2.30 failures/yr
 MTBF:
P = 1 - R = 1 - 0.10 = 0.901yr
MTBF = 1/μ = 0.43 yr.
This system is expected to fail, on average, once every 0.43 yr
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7. EXAMPLE 11.2
A diagram of the safety systems in a certain chemical reactor
is shown in Figure. This reactor contains a high-pressure
alarm to alert the operator in th event of dangerous reactor
pressures. It consists of a pressure switch within the reactor
connected to an alarm light indicator.
For additional safety an automatic high-pressure reactor
shutdown system is installed. This system is activated at a
pressure somewhat higher than the alarm system and
consists of a pressure switch connected to a
solenoid valve in the reactor feed line.
The automatic system stops the flow of reactant in the event
of dangerous pressures. Compute the overall failure rate, the
failure probability, the reliability, and the MTBF for a high-
pressure condition. Assume a I-yr period of operation.
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9. SOLUTION:
A dangerous high-pressure reactor situation occurs only when both the alarm system and
the shutdown system fail. These two components are in parallel. For the alarm system the
components are in series:
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10. For the shutdown system the components are also in series:
The two systems are combined using Equation 11-6:
For the alarm system alone a failure is expected once every 5.5 yr. Similarly, for a
reactor with a highpressure shutdown system alone, a failure is expected once every
1.80 yr. However, with both systems in parallel the MTBF is significantly improved and
a combined failure is expected every 13.7 yr.
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11. REVEALED AND UNREVEALED FAILURES
 The equipment to fail without the operator being
aware of the situation. This is called an unrevealed
failure.
 Failures that are immediately obvious are called
revealed failures.
 A flat tire on car = revealed failure
a flat spare tire = unrevealed failure
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12. τr = (1/n)* ∑ τri
τo = (1/n)* ∑ τro
n is the number of times the failure
τri is the period for repair for a particular
failure
O=operational
MTBF = 1/ μ = τr + τ0
τi= inspection interval
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13.  availability and unavailability
A + U = 1
A= availability , U = unavailability
A = τ0 / (τr + τ0) = μ * τ0
U = τr / (τr + τ0) = μ * τr
U= 0.5* μ * τi
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14. EXAMPLE 11.3
 Compute the availability and the unavailability for
both the alarm and the shutdown systems of
Example 11-2. Assume that a maintenance
inspection occurs once every month and that the
repair time is negligible.
 Solution:
Both systems: unrevealed failures
For the alarm system the failure rate is μ = 0.18
faultslyr.
The inspection period is 1/12 = 0.083 yr. 14
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15.  U= 0.5* μ * τi
= 0.5*0.18*0.083
= 0.0075
A= 1- U = 0.992
The alarm system is available 99.2% of the time.
 For the shutdown system, μ = 0.55 faultslyr
 U= 0.5* μ * τi
= 0.5*0.55*0.083
= 0.023
A= 1- U = 0.977
The shutdown system is available 97.7% of the time. 15
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16. FAULT TREES
Wide application in
Nuclear Plants and
Aerospace industries.
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17. EXAMPLE: FAULT TREE FOR EX.2
Alarm and Shutdown System
 Top event: Damage to reactor as a result of over
pressuring.
 Existing event: High process pressure.
 Unallowed events: Failure of mixer, electrical
failures, wiring failures, tornadoes, hurricanes,
electrical storms.
 Equipment configuration: Solenoid valve open,
reactor feed flowing.
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18. FAULT TREE
A= AND gate
B,C= OR gate
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19. MINIMAL CUT SETS FOR THE FAULT TREE
The different unique sets of events
leading to the top event are the minimal
cut sets
A
A B C
A B 1 C
2 C
A B 1 C 3
2 C
1 4
A B 1 C 3
2 C 3
1 4
2 4
1, 3
2, 3
1, 4
2, 4
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20. EVENT TREES
When an accident occurs in a plant, various safety systems come into
play to prevent the accident from propagating.
The event tree approach includes the effects of an event initiation
followed by the impact of the safety systems.
The typical steps in an event tree analysis are :-
1. Identify an initiating event of interest,
2. Identify the safety functions designed to deal with the initiating event,
3. Construct the event tree, and
4. Describe the resulting accident event sequences
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21. Reactor with high-temperature alarm and temperature controller System:-
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22. EVENT TREE COMPONENTS :
INITIATING EVENT- loss-of-coolant
SAFETY FUNCTIONS:
• The first safety function is the high-temperature alarm.
• The second safety function is the operator noticing the high reactor temperature
during normal inspection.
• The third safety function is the operator re-establishing the coolant flow
by correcting the problem in time.
• The final safety function is invoked by the operator performing an emergency
shutdown of the reactor.
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23. Fig. : Event Tree for a Loss-of-coolent accident for the Reactor
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24. LAYER OF PROTECTION ANALYSIS
 LOPA is a semi-quantitative tool for analyzing and
assessing risk.
 This method includes simplified methods to
characterize the consequences and estimate the
frequencies.
 Various layers of protection are added to a process,
for example, to lower the frequency of the
undesired consequences.
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RISK ASSESSMENT
Chapter 11: Chemical Process Safety by Daniel A. Crowl
MNIT Jaipur
Presented by:
Mayank Mehta Hardik K. Sharma