Energy Engineering - RELIABILITY, SAFETY AND RISK ANALYSIS C

Full exam

First name and Last name: _________________________________________________________________________________________________ RELIABILITY , SAFETY AND RISK ANALYSIS EXAM: 11 /07 /2022 Note: 1. Make sure you write your first name, last name, student number or person code, and sign every exam sheet; 2. The exam consists of 2 exercises (E1, E2) and 2 open questions (Q1, Q2); 3. The exam time is 2 hours and 30 minutes for students of Part A+B , who also ha ve to solve Q1.4, and 2 hours and 15 minutes for Part C . Exercise 1 [E1 ] A pipeline (P) of a liquid natural gas production system connects an extraction platform with a liquefaction plant . The pipeline has a constant failure rate ������������. Due to the high corrosivity of the gas residues , the pipeline is periodically cleaned . The time interval between the end of the cleaning and the beginning of the next cleaning is � and the cleaning intervention lasts ������. You are required to: E1.1) Plot the time evolution of the system instantaneous unavailability; E1.2) Evaluate the system average unavailability over the maintenance cycle (you can assume ������� ≪ 1); E1.3) Find the optimal interval between cleaning interv entions ( ��������� ) that minimizes the average unavailability ; E1.4) Assume that ������� is the probability that the maintenance team fail s to correctly reconnect the pipeline after the cleaning intervention . Repeat E1.3) and comment the results . The plant manager considers the possibility of replacing the pipeline P with two pipelines P1 and P2 working in parallel and with constant failure rate ������1 and ������2, respectively. Assuming that : • P2 is cleaned after P1; • the time interval between the end of the cleaning of P1 and the beginning of the next cleaning of P2 is ������ 2; • the time interval between the end of the cleanin g of P2 and the beginning of the next cleaning of P1 is ������ 2; • Both the testing and repair of P1 and the testing and repair of P2 last ������ 2. You are required to: E1.5) repeat E 1.1) and E1.2) (you can assume that �������= 0). Exercise 2 [E2 ] The probability of failure to start on demand of a pump of a fire protection system is �. Considering past experience on similar pumps , plant experts sugg est to use the following prior distribution for �: ������′(�= 0.1)= 0.3, ������′(�= 0.3)= 0.5, ������′(�= 0.6)= 0.2 E2.1) A test on 5 pumps is performed and only one of them fail ed to activate on demand . You are required to f ind the posterior probability distribution of �; E2. 2) Assuming that the prior distribution of � is a Beta distribution with parameters �= �= 3, you are required to find the posterior probability distribution of �; E2. 3) Assume that a new test is performed on 500 pumps and 50 of them fail ed to start on demand. You are required to : E2. 3.1) find the new posterior probability distribution of �; E2. 3.2) Compare the expected value and variance of � in E2. 2) and E2. 3.1). Comment the results; E2. 4) Using the posterior distribution found in E2.3.1), f ind the probability that there are no uncontrolled fire during 1 year . Assum e that the the mean occurrence rate of a fire in the plant is once every 4 months . Solution of the integral is not required . Question 1 [Q1] Consider the network system of the Figure below: Consider the top event: “no flow from left to right”. You are required to: Q1.1) Find the minimal cut sets of the system and compute the Top Event probability; Q1.2) Provide the pseudocode for the Monte Carlo estimation of the probability of the Top Event and the uncertainty of the estimate; Q1.3) Compute the probability of the Top Event and the uncertainty of the estimate by simulating with the Monte Carlo method the system state ������ = 4 time s. Use the random numbers sampled from a uniform distribution �[0,1) reported in the Table below and comment the results . Iteration RANDOM NUMBERS 1 0,7577 0,0318 0,3171 0,7952 0,7547 2 0,7431 0,2769 0,9502 0,1869 0,2760 3 0,3922 0,0462 0,0344 0,4898 0,6797 4 0,6555 0,0971 0,4387 0,4456 0,6551 Component Failure Probability A 0.15 B 0.25 C 0.05 D 0.5 E 0.2 Q1.4 ) Provide the pseu docode for the Monte Carlo estimation of the Birnbaum’s Importance Measures of components A and C and compute them using the random numbers sampled from �[0,1) reported in the Table below. Iteration RANDOM NUMBERS 1 0,4984 0,7513 0,9593 0,8407 2 0,9597 0,2551 0,5472 0,2543 3 0,3404 0,5060 0,1386 0,8143 4 0,5853 0,6991 0,1493 0,1935 5 0,2238 0,8909 0,2575 0,9293 Question 2 [Q2] Q2.a) Define the instantaneous availability of a component; Q1. b) Consider a continuously monitored component whose failure time, �, is described by an exponential probability density function with failure rate ������ and whose repair time, �������, is described by the probability density function ������������������(������). You are requir ed to find the integro -differential equation for the component instantaneous availability. Q1 .c) Solve the integro -differential equation in the Laplace transform domain. Q1. d) Estimate the component limiting availability �∞. Table of Points Students Exercise 1 Exercise 2 (E.2.1,E2.2, E2.3) Exercise 2 (only E2.4 ) Question 1 Question 2 A+B (10CFU) 7 7 1 9 7 C (8CFU) 7.5 7.5 1 7.5 7.5 Q1.1 The minimal cut sets of the systema are: • ������1= �� ; → ������(������1)= 0.15 ∗0.25 = 0.0375 ; • ������2= �; → ������(������2)= 0.05 ; • ������1= �� ; → ������(������3)= 0.5∗0.2= 0.1; The structure function of the system is: ������(������)= 1− (1− ������1)(1− ������2)(1− ������3) The probability of the TOP event is: ������(������(������)= 1)= �[1− (1− ������1)(1− ������2)(1− ������3)]= = �[������1+ ������2+ ������3− ������1������2− ������1������3− ������2������3+ ������1������2������3]= ������(������1)+ ������(������2)+ ������(������3)− ������(������1������2)− ������(������1������3)− ������(������2������3)+ ������(������1������2������3)= = 0.0375 + 0.05 + 0.1− 0.0375 ∗0.05 − 0.0375 ∗0.1− 0.05 ∗0.1+ 0.0375 ∗0.05 ∗0.1= 0.1771 We cannot apply the rare event approximation since we have not low pro bability events . Q1.2 N→ Number of MC trials ; M→ Number of components in the system (i.e., 5) ; ������ = [�1,… ,��] → Components fai lure probability; ▪ For n=1 to N (i.e., for each MC trial) • Find the state ������ = [������1,… �������] of the system components : o Sample M random numbers ������⃗= (������1,… ,�������) from a uniform distribution ������[0,1) (i.e., ������~������[0,1)); o For j=1 to M If ������⃗(������)< �� �������= 1 (i.e., failed) ; Else �������= 0 (i.e., working) ; End if End for • Determine if �������(������) is a failed/working state for the system : o If ������ ==cut set �������= 1 (i.e., failed); Else �������= 0 (i.e., working ); End if End for ▪ Estimate the TOP event probability ��������� and its standard deviation �������������� : ��������� = ∑ ������������ ������ � ; �������������� = √��������������������������������� (������) � Q1.3 It can be noticed that, due to the small number of MC iterations, the uncertainty of the estimate (i.e., �������������� ) is significant (i.e., �������������� ��������� ~1). To reduce �������������� , the Number of MC trial s should be increased ; thus, the estimate of ��������� approaches the re al values ��������� = 0.1771 . Q1. 4 The Birnbaum’s I mportance Measures �������� of the component ������ is the is the probability that the components other than ������ of the system are in such a state that the system works if ������ work s and the system is failed if ������ is failed : ��������= Pr {������(������,�������= 1)− ������(������,�������= 0)= 1} PSEUDOCODE TO COMPUTE �������� N→ Number of MC trials ; ▪ For n=1 to N (i.e., for each MC trial) o Sample the state of the components other than ������ (see po int Q1.2 ); o If ������(������,�������= 1)− ������(������,�������= 0)== 1 ���� �������������������(�)= 1 ; Else ���� �������������������(�)= 0 ; End if End for ▪ Estimate th e Birnbaum’s Importance Measures : ��������= ∑ ���� �������������������(�) ������ � ; Iteration A B C D E PHI 1 0 1 0 0 0 0 2 0 0 0 1 0 0 3 0 1 1 1 0 1 4 0 1 0 1 0 0 p_TOP 0,25 sigma_TOP (Bessel's correction) 0,25 sigma_TOP 0,216506 STATE (0-->WORKING; 1-->FAILED) ▪ The Birnbaum’s I mportance Measures of component A ▪ The Birnbaum’s I mportance Measures of component C Iteration A B C D E PHI(A=0,B,C,D,E) A B C D E PHI(A=1,B,C,D,E) Critical path 1 0 0 0 0 0 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 0 0 0 0 0 0 3 0 0 0 1 0 0 1 0 0 1 0 0 0 4 0 0 0 1 1 1 1 0 0 1 1 1 0 5 0 1 0 1 0 0 1 1 0 1 0 1 1 Birnbaum’s 0,2 Birnbaum's Index of A STATE (0-->WORKING; 1-->FAILED) Iteration A B C D E PHI(A,B,C=0,D,E) A B C D E PHI(A,B,C=1,D,E) Critical path 1 0 0 0 0 0 0 0 0 1 0 0 1 1 2 0 0 0 0 0 0 0 0 1 0 0 1 1 3 0 0 0 1 0 0 0 0 1 1 0 1 1 4 0 0 0 1 1 1 0 0 1 1 1 1 0 5 0 0 0 1 0 0 0 0 1 1 0 1 1 Birnbaum’s 0,8 Birnbaum's Index of c STATE (0-->WORKING; 1-->FAILED) RELIABILITY SAFETY AND RISK ANALYSIS – EXAM RESULTS DATE : 1 1/07/2022 PERSON CODE EX1 EX2 (E2.1, E2.2, E2.3) EX2 (E2.4) Q1 Q2 WRITTEN SCORE AGGREGATED SCORE 10496633 5,4 6,7 0,0 0,1 6,0 18,0 22 10501104 1,9 6,2 0,0 6,9 6,3 21,3 25 10531480 5,7 6,8 0,0 5,3 5,0 22,8 23 10537555 4,0 7,5 0,1 6,9 4,3 22,8 26 10539739 3,7 7,5 0,0 6,5 7,1 24,8 25 10573054 4,6 6,2 0,0 7,4 3,9 22,1 25 10573622 4,7 6,8 0,0 7,5 7,0 26,0 27 10607418 6,9 6,2 1,0 8,0 7,0 29,1 29 10608371 5,4 7,0 0,7 8,1 7,0 28,2 27 10614798 5,3 4,7 0,2 8,0 1,8 20,0 21 10619194 4,4 5,6 0,3 7,2 2,7 20,2 23 10619483 6,2 6,8 0,6 8,4 6,8 28,8 28 10620340 5,4 7,0 0,2 7,6 6,6 26,8 27 10622554 6,8 7,0 0,0 2,1 5,3 21,3 23 10635479 6,3 7,5 0,0 6,3 7,4 27,5 28 10660185 4,8 7,3 0,2 6,9 7,5 26,7 28 10662331 5,1 7,0 0,2 8,2 7,0 27,5 28 10664903 1,3 2,3 0,0 3,0 0,0 6,6 not passed 10750754 4,5 7,5 0,0 0,6 7,5 20,1 23 10775616 3,6 6,2 0,3 2,2 6,0 18,2 22 10809212 4,5 3,3 0,2 8,0 5,0 21,0 24 10816778 5,6 6,2 0,2 1,8 6,9 20,6 24 10818073 3,0 6,1 0,0 1,5 2,5 13,1 not passed 10826261 3,3 3,5 0,0 0,9 3,4 11,1 not passed 10833308 5,1 6,3 0,2 7,5 6,6 25,7 25 10834390 3,9 7,0 0,2 7,1 6,9 25,1 27 10840925 5,4 7,0 0,2 5,3 5,0 23,0 24 10842843 5,5 5,3 0,2 7,9 7,0 25,9 25 10864298 6,4 6,2 0,3 8,6 7,0 28,5 28 Table of Points Students EX 1 EX2 (E2.1, E2.2, E2.3) EX 2 ( E2.4 ) Q1 Q22 Total Part A+B (10 CFU) 7 7 1 9 7 31 Part C (8 CFU) 7,5 7,5 1 7,5 7,5 31