tutorial4 rbd fta solutions

7/27/2019 Tutorial4 RBD FTA Solutions

1/18

Fundamental Safety Engineering and Risk Management Concepts, 2012/2013H. Tan, M.J. Baker and Ian Stanley

Tutorial 4 Reliability Block Diagram and Fault Tree Analysis

1.Consider the statically determinate structural system with 7 elements shown in Figure Q1. Eachelement has a reliability ofr.

(a)Denote the event that element i fails by Fi ; write down the logical expression for the structurefailure.

(b)Draw the Reliability Block Diagram.(c)Draw the Fault Tree for the failure of the structure.(d)The reliability of each element is r. Determine the reliability of the whole structure.

[ 7r ]

Figure Q1, A statically determinant structural system.Solution:

(a)

1 2 3 4 5 6 7structure failure F F F F F F F

(b)For a determinate structural system, the whole structure fails if any of the elements fails. So this aseries network

(c)


2/18


(d)

For series network, the reliability is 7R r


3/18


2.It is desirable to increase the redundancy of the system shown in Figure Q2, consisting of threeidentical components, with reliability R=0.9, by duplicating (a) each component in parallel, or (b) thewhole system in parallel.

1) Calculate the failure probability of the new system (a), aF ;[ 0.000199 ]

2) Calculate the failure probability of the new system (b), bF ;[0.000361]

3) Prove that the failure probability of system (a) is lower than that of system (b) by giving theratio /a bF F .

[0.55]

Figure Q2

Solution:

(1)The Reliability Block Diagram of the new system (a) by duplicating each component in parallel is

The reliability of each duplicated parallel sub-system is


4/18


2

2

1 1 1

22 0.9 0.9

0.99

R R R

R R

The reliability of the top branch is 2topBranchR R R R

The reliability of the bottom branch is bottomBranchR R

Therefore the reliability of the system (a) is

2 3

2 3

1 1 1

0.99 0.99 0.99

0.999801

a topBranch bottomBranch

topBranch bottomBranch topBranch bottomBranch

R R R

R R R R

R R R

Thus the failure probability

1

1 0.999801

0.000199

a aF R

(2)

The Reliability Block Diagram of the new system (b) by duplicating the whole system component inparallel is


5/18


The reliability of each branch is 2

2 3

2 3

1 1 1

0.9 0.9 0.9

0.981

branchR R R

R R R

Therefore the reliability of the system (b) is2

2

2

2 0.981 0.981

0.999639

b branch branchR R R

Thus the failure probability

1

1 0.999639

0.000361

b bF R

(3)The failure probability of system (a) is lower than the failure rate of system (b), the ratio is

0.55a

b

F

F


6/18


3.The logical expression for the occurrence of the top event, T, of a fault tree is given by

T A B C D E F G

(a) Construct the corresponding fault tree.(b) Construct an equivalent network diagram.

Solution:

(a)The fault tree is


7/18


(b)The reliability block diagram is


8/18


4.(a) Draw the fault tree for system failure for the following network diagram and write down thecorresponding logical expression for this event.

Figure Q4

(b) If the component failures are statistically independent events with probabilities of occurrence peryear of:

21 10P a , 22 2 10P a , 23 5 10P a

21 2 3 4 10P b P b P b ,

22 10P c

21 8 10P d , 22 6 10P d

210P e ,evaluate the probability of system failure (in one year) if none of the components can be replaced orrepaired.

[ 43.11 10 ]

(c) If, instead of being statistically independent events, the conditional probability of failure ofcomponent d1 is:

1 2 0.8P d d determine the effect of this on the system failure probability.

[ 46.89 10 ]


9/18


Solution:

(a)The fault tree for system failure is

System Failure 1 2 3SS SS SS

where 1 1 2 3SS a a a is the failure of the a1-a2-a3 sub-system,2 1 2 3SS b b b is the failure of the b1-b2-b3 sub-system,

1 1 2LLSS d d

is the failure of the d1-d2 sub-system.3 1SS LLSS c e is the failure of the 1LLSS -d-e sub-system,

(b)

21 8 10P d , 22 6 10P d

2 21 1 2 1 2 8 10 6 10 0.0048P LLSS P d d P d P d

22 10P c , 210P e ,


10/18


2 2

3 1

1 1

1 1

1 1

1 1 1 1 1

1 1 0.0048 1 2 10 1 10

0.03445696

P SS P LLSS c e

P LLSS c e

P LLSS c e

P LLSS P c P e

P LLSS P c P e

2

1 10P a

, 2

2 2 10P a

, 2

3 5 10P a

2 2 2

1 1 2 3

1 1 2 3

1 1 2 3

1 1 1 1 2 1 3

1 1 10 1 2 10 1 5 10

0.07831

P SS P a a a

P a a a

P a P a P a

P a P a P a

21 2 3 4 10P b P b P b ,

32

2 1 2 3

1 1 2 3

1 1 2 3

1 1 2 3

1 1 1 1 2 1 3

1 1 4 100.115264

P SS P b b b

P b b b

P b b b

P b P b P b

P b P b P b

4

System Failure 1 2 3

1 2 3

0.07831 0.115264 0.03445696

3.11 10

P P SS SS SS

P SS P SS P SS


11/18


(c)All the other events are statistically independent, except d1 and d2 which have the conditionsprobability relationship described by

1 2 0.8P d d , 22 6 10P d

2

1 1 2

1| 2 2

0.8 6 10 0.048

P LLSS P d d

P d d P d

2 23 1 1 1 1 1

1 1 0.048 1 2 10 1 10

0.0763696

P SS P LLSS P c P e

4

System Failure 1 2 3

1 2 3

0.07831 0.115264 0.0763696

6.89 10

P P SS SS SS

P SS P SS P SS


12/18


13/18


(iii) Owing to an upgrade of the offshore platform, a new power generation configuration is required.

Figure Q5(b)

The reliability of power generation blocks (PG1, PG2, PG3 and PG4) is the same as in part (a) of thequestion (0.93). The Sub Switch Boards have reliabilities of 0.98 and the reliability of power supplyfrom the adjacent platform is 0.85.

Calculate the reliability of the Power Generation group and the overall system reliability.

[0.979, 0.979]

(iv) Whilst there is inherent redundancy in the Power Generation group, compare the results with thosefrom part (a) and explain any abnormalities.

Solution:

(i)For Figure Q7(a):

Parallel connection for Reliability of the Power Generation group, the reliability is

3

1 2 31 1 1 1 1 1 0.93 0.999657PowerGeneration train train trainR R R R

Parallel connection for transformer, the reliability is

3

transformer transformer1 transformer2 transformer31 1 1 1 1 1 0.98 0.999992R R R R

The overall system reliability is the serial connection of Power Generation group, Main Switch Board

transformer MainSwitchBoard

0.999657 0.999992 0.98

0.979656

system PowerGenerationR R R R


14/18


(ii)Assume that due to operational issues only two Power Generation Trains are available.

2

1 21 1 1 1 1 0.93 0.995100PowerGeneration train trainR R R

transformer MainSwitchBoard

0.995100 0.999992 0.98

0.9751902

system PowerGenerationR R R R

(iii)For Figure Q5(b):

2

1 2 _ _ _11 1 1 1 1 0.93 0.98 0.975198PGA PG PG Sub Switch Board R R R R

3 4 _ _ _ _ 2

2

1 1 1 1

1 1 0.93 1 0.85 0.98 0.9792797

PGB PG PG Power cable Sub Switch Board R R R R R

Therefore,

MainSwitchBoard1 1 1

1 1 0.975198 1 0.9792797 0.98

0.979496

PowerGeneration PGA PGBR R R R

3

power_distribution transformer1 transformer2 transformer31 1 1 1 1 1 0.98 0.999992R R R R

power_distribution

0.979496 0.9999920.979

system PowerGenerationR R R

(iv)In spite of the additional redundancy in Figure 7(b), the overall system reliability is governed by thereliability of the Main Switch Board as it is in series for the overall system irrespective of whether it isplaced in the Power Distribution group or the Power Distribution group.


15/18


6.You are given a system of switches connected as shown in Figure Q6. The probability per demand thata switch fails in the closed position is 10-4 and in the open position is 10-3. There are no other causes offailure.

(a)Construct a fault tree for the top event TC, the circuit is closed when it should be open.(b)Identify the minimal cut set forTC.(c)Evaluate the probability of the top event TC. [ 810 ](d)Construct a second fault tree for the top event TO, the circuit is open when it should be closed.(e)Identify the minimal cut set forTO.(f) Evaluate the probability of the top event TO. [ 64 10 ]

Figure Q6


16/18


Solution:(a) Fault tree for the top event TC, the circuit is closed when it should be open.

(b)Minimal cut sets forTC:set 1: ABset 2: CDEset 3: CDF

(c) Evaluate the probability of the top event TC.

Probability of the failure of sub-system AB is 4 4 810 10 10ABP

Probability of the failure of sub-system EF is 4 4 4 4 4 410 10 10 10 2 10 10EFP

Probability of the failure of sub-system CDEF is 4 4 4 4 4 1210 10 2 10 10 2 10 10CDEFP


17/18


Probability of the top event

8 4 128

1 1 1

1 1 10 1 2 10 10

10

CT AB CDEF P P P


18/18


(d) Fault tree for the top event TO, the circuit is open when it should be closed.

(e) Identify the minimum cut set forTO.set 1: ACset 2: ADset 3: BCset 4: BDset 5: AEFset 6: BEF

(f) Evaluate the probability of the top eventT

O.Probability of the failure of sub-system AB is 3 31 1 10 1 10ABP Probability of the failure of sub-system EF is 3 310 10

EFP

Probability of the failure of sub-system CDEF is 3 3 61 1 10 1 10 1 10CDEFP

Probability of the top event 3 3 3 3 6

6

1 1 10 1 10 1 1 10 1 10 1 10

4 10

OT AB CDEF P P P

tutorial4 rbd fta solutions

Documents