CHEN_SENG 430 - HW 8 Sol (1)
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Jan 9, 2024
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CHEN/SENG 430 Fall 2023
HW 8 Solution
1.
The hazard of exposure to radioactive chemicals is mitigated with 3 independent barriers. If 1 barrier works,
the exposure is prevented. The probability of each barrier to fail is 0.001 and the consequence of hazard
exposure is 3000 cancer-deaths per year. Develop an event tree showing all branches and outcome. What is
the probability of exposure. What is the risk due to the hazard?
Solution:
Probability of exposure=0.001×0.001×0.001=1×10
-9
; Risk=P×C= 1×10
-9
×3000 cancer-deaths = 3×10
-6
deaths
2.
Consider the event tree and the fault trees below:
Recall that a risk profile curve (Farmer’s curve) is a plot of frequency vs. consequence (injuries in this case).
Solution:
3.
Given the fault tree gates shown below and the set of failure probabilities.
a)
Determine an expression for the probability of the top event in terms of the component failure
probabilities.
b)
Determine the minimal cut sets.
c)
Compute a value for the failure probability of the top event. Use both the expression of part (a) and
the fault tree itself for each of the fault trees given.
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Figure: Fault Tree Gates
Table: Component failure probability data
Solution:
i)
a) AND gate.
P(T) = P(1) * P(2)
b) Minimal cut set: (
1, 2)
c) Eqn from a): P(T) = 0.1 * 0.2 = 0.02
Fault tree: P(T) = P(1 and 2) = 0.1 * 0.2 =
0.02
ii)
a) OR gate.
P(T) = P(1) + P(2) - [P(1)*P(2)]
b) Minimal cut sets:
(1)
(2)
c) Eqn from a): P(T) = 0.1 + 0.2 - (0.02) = 0.28
Fault tree: P(T) = P(1 or 2) = 0.1 + 0.2 - (0.1*0.2) =
0.28
iii)
a) First gate is OR, gate 2 OR comp 1. Gate 2 is AND gate.
P(T) = P(1) + P(G2) - [P(1)*P(G2)] where P(G2) = P(2) * P(3)
→
P(T) = P(1) + [P(2)*P(3)] - [P(1)*(P(2)*P(3))]
b) Minimal cut sets:
(1)
(2,3)
c) Eqn from a): P(T) = 0.1 + (0.2 * 0.3) - (0.1 * 0.2 * 0.3) = 0.154
Fault tree: P(G2) = 0.2 * 0.3 = 0.06 → P(T)= 0.1+0.06
-(0.1*0.06) =
0.154
iv)
a) Gate 1 is AND gate, Gate 2 AND comp 1. Gate 2 is an OR gate, comp 2 or 3.
P(T) = P(1) * P(G2) and P(G2) = P(2) + P(3) - [P(2) * P(3)]
→
P(T) = P(1) * [P(2) + P(3) - (P(2)*P(3))]
b) Minimal cut sets:
(1,2)
(1,3)
c) Eqn from a): P(T) = 0.1 * (0.2 + 0.3 - (0.2*0.3)) =
0.044
Fault tree: P(1AND2) = 0.1 * 0.2 = 0.02
P(1AND3) = 0.1 * 0.3 = 0.03
P(T) = 0.02 + 0.03 - (0.02*0.03) =
0.0494
v)
a) Gate 1 is AND gate, Gate 2 AND Gate 3. Gate 2 is OR gate, comp 1 or 2. Gate 3 is OR gate, comp 3 or 4.
P(T) = P(G2) * P(G3)
P(G2) = P(1) + P(2) - [P(1)*P(2)]
P(G3) = P(3) + P(4) - [P(3)*P(4)]
P(T) = [P(1) + P(2) - [P(1)*P(2)]] * [P(3) + P(4) - [P(3)*P(4)]]
b) Minimal cut sets:
(1,3)
(2,3)
(1,4)
(2,4)
c) Eqn from a): P(T) = [0.1+0.2-0.02] * [0.3+0.4-0.12] =
0.1624
Fault tree: P(1AND3) = 0.1*0.3 = 0.03
P(1AND4) = 0.1*0.4 = 0.04
P(2AND3) = 0.2*0.3 = 0.06
P(2AND4) = 0.2*0.4 = 0.08
P(T) = 0.03 + 0.04 + 0.06 + 0.08 - (0.03*0.04*0.06*0.08) =
0.20999
4.
The storage tank system in your plant is used to store process feedstock. The current design for this system is
shown below in Figure 1.
Figure: Level Control System with Alarm
To prevent overfilling, your plant’s storage tank is equipped with a high-level alarm and a high-level shutdown
system, which is connected to a solenoid valve that stops the flow of input stock. Overfilling of storage tanks
is a common problem, especially in the process industries. Develop a fault tree for the top event of “storage
tank overflows”. Use failure data used at your plant and listed in the table below to estimate the failure
probability of the top event and the expected number of occurrences per year.
Failure rate data for selected system components
(Lees, F.P.,
Loss Prevention in the Process Industries,
Elsevier, 2004)
Equipment
Failures/yr
P
yr
Code
Letter
Controller
0.29
0.25
Control valve
0.60
0.45
FIC
A
Flow measurement (fluids)
1.14
0.68
Flow switch
1.12
0.67
High level switch (liquids)
1.70
0.82
LIC
F
Indicator lamp
0.044
0.04
AI
D
Level measurement (liquids)
1.70
0.82
LI
B
Solenoid valve
0.42
0.34
SV
E
Strip chart recorder
0.22
0.20
CR
C
For this system, the fault tree components are:
Control System
Letter
Protection System
Letter
FIC, flow control valve
A
AI, alarm indicator (not shown)
D
LI, level measurement
B
SV, solenoid valve
E
CR, strip chart recorder
C
LIC, level controller (switch)
F
a)
Determine the minimal cut sets.
b)
Identify the two most likely failure modes.
c)
Make suggestions for development of your system design to reduce residual risk.
Solution:
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a)
Determine the minimal cut sets.
The minimal cut sets based on the logic of the presented fault tree are as follows:
Min Cut
sets
Cut set
freq
A
E
0.004
A
F
0.016
B
E
0.011
B
F
0.045
C
D
E
0.000007
C
D
F
0.0003
total
0.076
b)
Identify the two most likely failure modes.
The most likely failure modes are due to the two-component cutsets, B F and A F, which cut both success
paths and result in the mishap of storage tank overflow.
c)
Make suggestions for development of your system design to reduce residual risk.
This design should be developed to include an additional control valve, solenoid valve, and level switch
to reduce the dependence of mishap rate on the A (control valve), E (solenoid valve), and F (level switch)
and thereby to reduce significantly the residual risk that must be subsequently managed.