Chapter 6 1. a. Section 6.1. b. Section 6.3, see also Section 6.2. c. Predictions based on most published sources of reliability data tend to underestimate the reliability that is achievable, given that high standards of design, manufacture and maintenance are applied. This is due to the fact that published data represents past experience which seldom reflects modern best practices and achievement. However, the prediction could also be an overestimate, depending on the extent to which the design, production and maintenance do reflect the highest standards. Additionally, the standard-based reliability prediction reflects the inherent reliability of the system and ignores other sources of failures such as interconnects, manufacturing defects, workmanship, etc. The engineering risks and novelty involved will also influence the accuracy and credibility of the prediction. See Sections 6.2, 6.3 and 6.4. 2. Circuit diagrams: a. Open Circuit. Only if both resistors fail open-circuit, i.e. they are in parallel. b. Short Circuit. Fails if either resistor fails short-circuit, i.e. they are in series. 3. a. For loss of flow (LF), the system will fail if either regulator fails in the closed mode (series reliability). It survives only if both survive, so the reliability is the probability of both surviving. From Eq. (2.2): R LF = R C R C = R 2 C Practical Reliability Engineering, Fifth Edition. Patrick D. T. O Connor and Andre Kleyner. 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.
46 Chapter 6 For overpressure (OP), the system will only fail if both regulators fail in the open mode. It will survive if either regulator survives. It is therefore configured as in Figure 2.2., i.e. the regulators are in a dual redundant (parallel) configuration. (Note that the reliability configuration is not necessarily thesameasthefunctional configuration. In this case, the regulators are in a series configuration in the pipework.) From Eq. (2.6): P f = [1 P(A)][1 P(B)] but P(A) = P(B) = R 0, and System Reliability (R op ) = 1 P f,so b. From Eq. (2.27), R C (1.0) = exp( 1/2) = 0.606 R OP = 1 (1 R O )(1 R O ) = 1 (1 R O ) 2 so R LF = 0.606 2 = 0.367 From Eq. (2.31), R O (1.0) = exp( (1.0/1.6) 1.8 ) = 0.651 so R OP = 1 (1 0.651) 2 = 0.878 c. Probability of survival = probability of surviving in both modes = R LF R OP = 0.367 0.878 = 0.322. 4. For loss of flow, the system will fail only if both streams fail, i.e. it survives if either stream survives. This is the situation in Figure 2.2 again, i.e. R LF = 1 (1 0.367)(1 0.367) = 0.600 (so the twin stream does increase the reliability in this mode). For overpressure, the system will fail if either stream becomes straight through i.e. its probability of survival is the probability that both streams survive. R OP = 0.878 2 = 0.771 (so the twin streaming has reduced the reliability for this mode of failure).
Chapter 6 47 5. See Figure A6.1. Fail open (overpressure) tie sets (4) cut set (1) Fail closed (loss of flow) tie sets (2) cut sets (4) Figure A6.1 Question 4 diagram. 6. Reliability of A: R(A) = exp [ 150/837] = 0.8359 Reliability of B: R(B) = exp [ (150/315) 2 ] = 0.7971 Reliability of C: z = (150 420)/133 = 2.03 SD (similar to Example 2.8) therefore, from Appendix 1 (or applying Excel =1 NORMSDIST( 2.03), R(C) = 0.9788 Therefore system reliability R(S) = 0.8359[1 (1 0.7971)(1 0.9788)] = 0.8324 7. Using formulae in Table 6.1: = 0.04 = 0.5 a. For single element R(2) = exp( 0.08) = 0.923 For two elements in standby R(2) = exp( 0.08)(1 + 0.8) = 0.997
48 Chapter 6 b. For single element For two elements in standby A(t) = A(t) = 0.5 0.54 = 0.926 0.5 2 + (0.5 0.04) 0.5 2 + (0.5 0.04) + 2 0.04 2 ) = 0.994 Assumes single repair perfect switching both of which are reasonable, and exponential distribution of ages to failure exponential distribution of repair times both of which are questionable but does it affect the result if they are not exponential? c. For single element, downtime cost For two elements, downtime cost i.e. saving in downtime cost = 6.80 per hour i.e. payback period for second element = (1 0.926) 100 = 7.40 per hour = (1 0.994) 100 = 0.6 per hour = 25000 = 3676 hours 6.8 (This is about two years of operation at 40 hours/week and 48 weeks/year.) 8. Section 6.10, Figure 6.8. 9. a. Reliability block diagram, Figure A6.2. Electrical power Stand by battery Sensor & switch Hydraulic power Controller Actuator 1 Actuator 2 Figure A6.2 Question 9. System reliability block diagram.
Chapter 6 49 b. Fault tree, Figure A6.3. Loss of actuator control OR Controller fail Hydraulic power fail No electrical power No actuator AND AND No standby power Power supply fail Actuator 1 fail Actuator 2 fail OR Standby battery fail Sensor & switch fail Figure A6.3 Question 9. Fault Tree diagram. 10. [1 (1 0.99)(1 0.995 0.995)] 0.95 0.98 [1 (1 0.99 0.99)] = 0.9124 11. Run this problem as an exercise in simulation using the methods suggested in the assignment. ( ) 1 12. a. Mean life or MTTF for Weibull distribution = Ɣ + 1 In order to calculate Ɣ-value we will use Microsoft Excel function GAMMALN in the following way: ( ) 1 Ɣ + 1 = EXP(GAMMALN(1 + 1/ ))
50 Chapter 6 Therefore = ( MTTF ) = 1 Ɣ + 1 Ɣ 83 ( 1 1.5 + 1 ) = 91.94 ( ( ) ) 25 1.5 R(25) = exp = 0.8678 91.94 For simple availability calculations, we must assume exponentially distributed lifetimes. By the terms of the question, this assumption is obviously not valid. However, an exact solution involves mathematics beyond the scope of this book. It would more likely be evaluated using simulation, as in question 8 or applying Petri nets using the methods discussed in Section 6.12. Making the constant failure rate assumption: = 1 1 = 0.120 = 83 3.5 = 0.2857 R(25) = exp ( 0.012 25) = 0.7408 a considerable error compared to 0.8678 obtained earlier under Weibull distribution assumption. A(25) = b. For two elements in active redundancy 0.2857 0.012 + 0.2857 = 0.96 R s = 1 (1 R) = 1 (1 R) 2 for two = 1 (1 0.8678) 2 = 0.9825 For availability, we have the same assumptions as for the single element. A(25) = 0.2857 2 + (2 0.012 0.2857) 0.2857 2 + (2 0.012 0.2857) + (2 0.012) 2 = 0.997 c. Covered in comments in a. and b. d. Section 6.7. 13. a. Parachute: failure is failure to open when needed. Correct deployment of the canopy and lines depends primarily on how the parachute is packed. The deployment mechanism is usually a simple device (rip cord or manual removal of a retaining pin). The overall system is simple and of mature design. Quantitative data on failures of the items would relate primarily to human errors. The analysis methods would be very unlikely to provide any useful insights.
Chapter 6 51 b. Microprocessor: there can be very many kinds of failure (incorrect operation at rated speed, incorrect math operation, memory location failures, etc.). The system is enormously complex, containing hundreds of millions of transistors, capacitors and connections, each with several potential failure modes. Also, the effects of failure modes can be dynamic, i.e. dependent upon the data and logic flow. Quantitative failure data at these levels does not exist. Therefore, these kinds of analysis cannot be applied in any practical or useful way to such a system. (The same applies to most other complex digital electronic systems. See Chapter 9.) c. Electric motor: the important failure would be no torque. This could be caused by a relatively small number of clearly-definable lower-level failures, such as open-circuit or short-circuit coils and seized bearings. Data may or may not be available. The analyses could be performed quite easily. However, unless the design is novel or critical, they would probably not be very useful. d. Static mechanical assembly: failure could be fracture under static or dynamic loading. The appropriate analyses would be load and strength analysis, possibly using finite element methods (Chapter 8). It is unlikely that useful failure data would be available. The system analysis methods in question would not provide any further enlightenment. e. Train: there could be several types of failure. These could be classified as those that cause the train to stop or be unable to start, those that reduce performance, those that inconvenience passengers, those that create hazards, etc. The train consists of a number of sub-systems, ranging from simple to very complex, and mature designs to new ones, and with many interactions. Reasonably credible data is usually available on the reliability of components and sub-systems. The analysis methods are most appropriate for this kind of system design. 14. This problem should be considered as an exercise in the application of standards-based reliability prediction methods and the specific details of the solution are at the discretion of the course instructor. The results may vary based on additional assumptions accompanying the calculations. Not all of the required parameters are listed here and therefore need to be assumed or assigned by the instructor. According to MIL-HDBK-217 failure rate for a capacitor classified as fixed, ceramic, general purpose is: p = b CV Q E [ ( ) S 3 ( ) T + 273 b = 0.0003 + 1] exp 0.3 373 Where S = Ratio of operating to rated voltage (assumed to be 0.5) and T = 60 C operating temperature. The maximum rated temperature is 100 C = 373K (shown in the denominator of the equation above). Therefore, b = 0.00412 Capacitance factor CV = 0.41C 0.11 therefore, for C = 200 pf (0.2 F) CV = 0.7343. Environmental factor E = 9.0 for G M ground mobile environment, Figure A6.4.
52 Chapter 6 Quality factor Q = 1.0 for the quality category M (assumed generic), Figure A6.4. Figure A6.4 Question 14. MIL-HDBK-217 π-factors. Therefore p = 0.02726 failures per million hours. In comparison, the calculations using Telcordia Issue 2 (Figure A6.5) produce p = 0.01207 failures per million hours, approximately half the value obtained from MIL-HDBK-217. Figure A6.5 Question 14 Telcordia, Issue 2 solution using LambdaPredict R by ReliaSoft.
Chapter 6 53 15. A Markov state-space diagram is shown in Figure A6.6. In the case of a perfect switching remove the lower left state Switch failed, standby system unavailable. Figure A6.6 Question 15. Markov chain diagram.
54 Chapter 6 A Petri net diagram is shown in Figure A6.7. In the case of a perfect switching remove the upper right state Standby system unavailable. Figure A6.7 Question 15. Petri net diagram. Please note, that unlike Markov analysis, the Petri net solution is not limited by the constant failure rate assumption. Any type of system failure distribution can be simulated using Petri nets. 16. Various methods can be applied. The list below is just a suggestion for the best approach: a. A bill of materials for an electronic system (electronics parts count). b. Mechanical drawings of the system (mechanical reliability prediction based on NSWC-06/LE10 standard). c. Bill of materials and the knowledge of the stress factors for the electronic components, such as temperature, vibration, voltage, etc. (parts stress analysis). d. Electronic schematics (electronics parts count). e. List of mechanical parts and the detailed drawings (mechanical reliability prediction based on NSWC-06/LE10 or physics of failure if material properties are also available). f. Detailed information about device geometry, material properties and the applied stresses (physics of failure methods). g. Field return and warranty data for the previous model of the system with a clear description of the differences between the models (field return based methods or fusion of field data and reliability prediction standards).
Chapter 6 55 h. Field return and warranty data for the previous model of the system with a clear description of the differences in the environmental stresses between the two models (Fusion of field data with reliability prediction standards or top-down approaches). 17. Please follow the process covered in Section 6.10. 18. Figure A6.8 Question 18. Block diagram. R BD = 1-(1-R B )(1-R D ) Parallel System Reliability R S = 1-(1-R A R BD R C )(1-R E ) Using Microsoft Excel: R A = EXP( 12*t/1000000) R B = EXP( ((t/20,000)ˆ2)) R C = 1-NORMDIST(t, 1600, 100, TRUE) R D = 1-LOGNORMDIST(t, 24, 12) R E = EXP( ((t/10000)ˆ0.8)) See the Excel solution in Figure A6.9. Figure A6.9 Question 18. Excel solution. a. R S (1200hrs) = 0.9975 b. R S (1600hrs) = 0.8949 c. Block A Useful life (Exponential distribution) Block B Wear out ( > 1) Block E Infant Mortality ( < 1)
56 Chapter 6 19. R 4 = 0.9, therefore R = 4 0.9 = 0.974 20. Please compare the reliability of three parts at the 10 years mark: R A (10yrs) = exp( 0.58 10 365 24/10 6 ) = 0.95 R B (10yrs) = 0.9 (by definition of B 10 life) R C (10yrs) = exp( (10/20) 2 ) = 0.779 R C < R B < R A ( 21. R Parallel = 1 [1 R(t)] 2 for Weibull distribution : R(t) = e t [ ( R Parallel = 1 1 e t ) ] 2 = 2e ( t ) ( ) e 2 t ) Based on Eq. (2.21): h(t) = dr(t) dt ( 1 R(t) = ( ) t 1 2 2e t ( 2 e t 22. The solution is a series system, therefore it is a product, represented by an OR gate per Figure A6.10. ) ) Figure A6.10 Question 22. Fault Tree Analysis (FTA) diagram.
Chapter 6 57 P turn signal = 1-(1 P B )(1 P C )(1 P L ) P turn signal = 1-(1 0.11)(1 0.05)(1 0.21) = 0.332 23. For a system with perfect switching: Therefore at 60 hours: R(t) = e t (1 + t) R(60hr) = e 0.006 60 (1 + 0.006 60) = 0.9488