 Download Download Presentation T73S04 (R5V2/3) Session 38B Probabilistic Assessments Rick Bradford

# T73S04 (R5V2/3) Session 38B Probabilistic Assessments Rick Bradford

Download Presentation ## T73S04 (R5V2/3) Session 38B Probabilistic Assessments Rick Bradford

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1. T73S04 (R5V2/3) Session 38BProbabilistic AssessmentsRick Bradford Full procedure here http://rickbradford.co.uk/ProbabilisticProcedureForR5V2_3.pdf

2. Exclusion Clause • I am quite ignorant about statistics • Probabilistic assessment is not statistical analysis • My spin: probabilistic assessment is deterministic assessment done many times with suitable probability weighting of the inputs – “Monte Carlo”

3. Introduction & Purpose • Why probabilistic R5V2/3? • If deterministic R5V2/3 gives a lemon • If components have already failed • If you want to know about lifetime / reality • Trouble with ‘bounding’ data is, • It’s not bounding • It’s arbitrary • Most of the information is not used

4. What have we done? (mostly 316H) • 2008-2010: HPB/HNB Bifs – R5V4/5 E/REP/BBAB/0009/AGR/08, Rev.001 • 2011: HYA/HAR Bifs – Creep Rupture (BIFLIFE) E/REP/BBAB/0019/HAR/11 and E/REP/BBAB/0021/HAR/11 • 2012: HYA/HAR Bifs – R5V2/3 (BIFINIT) E/REP/BBAB/0023/AGR/12 and E/REP/BBAB/0025/HAR/12 • 2013: HPB/HNB Bifs – R5V2/3 with carburisation E/REP/BBAB/0027/AGR/13 • 2013-14 Various DNB FPU probabilistics (R6) • 2015 in progress Spine Welds 12.3 R5V4/5

5. Psychology Change • Best estimate rather than conservative • Including best estimate of error / scatter • The conservatism comes at the end… • ..in what “failure” (initiation) probability is regarded as acceptable… • …and this may depend upon the application (safety case v lifetime)

6. So what is acceptable? • Will vary – consult Customer • For “frequent” plant might be ~0.2 failures per reactor year (e.g., boiler tubes) • For HI/IOGF plant might be 10-7pry to 10-5 pry (maybe) • The assessment would be the same, but the latter is more dependent on the uncertain tails of the distributions

7. What’s the Downside? • MINOR: Probabilistic assessment is more work than deterministic • MAJOR: Verification • The only way of doing a meaningful verification of a Monte Carlo assessment is to do an independent Monte Carlo assessment! • Or is it…..? • Learning Points: Can be counterintuitive • Acceptance by others • Brainteaser

8. Limitation re R5V2/3 Applications • Only the crack initiation part of R5V2/3 addressed • Not the “precursor” assessments • Primary stress limits • Stress range limit • Shakedown • Cyclically enhanced creep • Complete job will need to address these separately

9. Computing Platform • So far we’ve used Excel • Latin Hypercube add-ons available • RiskAmp / “@Risk” Being developed • Most coding in VBA essential • Minimise output to spreadsheet during execution • Matlab might be a natural platform • I expect Latin Hypercube add-ons would also be available – but not checked • Develop facility within R-CODE/DFA - No

10. Run Times • Efficient coding crucial • Typically 50,000 – 750,000 trials • (Trial = assessment of whole life of one component with just one set of randomly sampled variables) • Have achieved run times of 0.15 to 0.33 seconds per trial on standard PCs (~260 load cycles) • Hence 2 hours to 3 days per run

11. Methodology • We shall assume Monte Carlo • Monte Carlo is just deterministic assessment done many times • So the core of the probabilistic code is the deterministic assessment

12. Hysteresis Cycle Construction • R5V2/3 Appendix A7 • Always sketch what the generic cycle will look like for your application • Helpful to write down the intended algorithm in full as algebra • Recall that the R5 hysteresis cycle construction is all driven by the elastically calculated stresses • Example – see http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf • Remember that the dwell stress cannot be less than the rupture reference stress • R5V2/3 Appendix A7 • Always sketch what the generic cycle will look like for your application • Helpful to write down the intended algorithm in full as algebra • Recall that the R5 hysteresis cycle construction is all driven by the elastically calculated stresses • Example – see http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf • Remember that the dwell stress cannot be less than the rupture reference stress

13. Non-Closed Cycles • Actual plant cycle sequences may not produce cycles whch are closed even in principle • e.g., cold – operating – hot standby • My approach described in detail here, http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf • e.g., symmetrisation: find reverse stress datum for left and right hand half-cycles separately and use the average.

14. Primary Reset Issue: 316H • Is creep strain reset to zero at the start of each dwell – so as to regenerate the initial fast primary creep rate? • Existing advice is unchanged for deterministic assessments… • Reset primary creep above 550oC • Do not reset primary creep at or below 550oC… • …use continuous hardening instead (creep strain accumulates over cycles)

15. Primary Reset Issue: 316H • For probabilistic assessments I advise the use of primary reset at all temperatures (for justification see E/REP/BBAB/0022/AGR/12) • But with two alleviations, • Application of the zeta factor, z • Only reset primary creep if the previous unload caused significant reverse plasticity • “significant” plasticity in this context has been taken as >0.01% plastic strain, though 0.05% may be OK

16. The zeta factor

17. Probabilistics • Is it all just normal distributions? • No • Also Log-normal, also… • All sorts of weird & wonderful pdfs • Or just use random sampling of a histogram…

18. Normal and Log-Normal PDFs • Normal pdf • Log-normal is the same with z replaced by ln(z) • Integration measure is then d(ln(z))=dz/z

19. Go tohttp://rickbradford.co.uk/PeterHoltNotesOnPDFs.pdf

20. Non-Standard Distribution:Elastic Follow-Up

21. Non-Standard Distribution: Overhang

22. Non-Standard Distribution Thermal Transient Factor wrt Reference Trip

23. How Many Distributed Variables • Generally – lots! (20 to 40) • If a quantity is significantly uncertain… • …and you have even a very rough estimate of its uncertainty… • …then include it as a distributed variable. • The Latin Hypercube can handle it

24. Where are the pdfs? • “But what if no one has given me a pdf for this variable”, I hear you cry. • Ask yourself, “Is it better to use an arbitrary single figure – or is it better to guestimate a mean and an error?” • If you have a mean and an error then any vaguely reasonable pdf is better than assuming a single deterministic value

25. How is Probabilistics Done? • (Monte Carlo) probabilistics is just deterministic assessment done many times • This means random sampling (i.e. each distributed variable is randomly sampled and these values used in a trial calculation) • But how are the many results weighted?

26. Options for Sampling: (1)Exhaustive(Numerical Integration) • Suppose we want +/-3 standard deviations sampled at 0.25 sd intervals • That’s 25 values, each of different probability. • Say of 20 distributed variables • That’s 2520 ~ 1028 combinations • Not feasible – by a massive factor

27. Options for Sampling: (2)Unstructured Combination • Each trial has a different probability • Range of probabilities is enormous • Out of 50,000 trials you will find that one or two have far greater probability than all the others • So most trials are irrelevant • Hence grossly inefficient / results vary wildly from run to run (no convergence)

28. Options for Sampling: (3)Random but Equal Probability • Arrange for all trials to have the same probability • Split all the pdfs into “bins” of equal area (= equal probability) – say P • Then every random sample has the same probability, PN, N = number of variables

29. Equal Area “Bins” Illustrated for 10 Bins (More Likely to Use 10,000 Bins)

30. Bins v Sampling Range • 10 bins = +/- 1.75 standard deviations (not adequate) • 300 bins = +/- 3 standard deviations (may be adequate) • 10,000 bins = +/- 4 standard deviations (easily adequate for “frequent”; not sure for “HI/IOGF/IOF”)

31. Optimum Trial Sampling Strategy • Have now chosen the bins for each variable • Bins are of equal probability • So we want to sample all bins for all variables with equal likelihood • How can we ensure that all bins of all variables are sampled in the smallest number of trials? • (Albeit not in all combinations)

32. Answer: Latin Hypercube • N-dimensional cube • N = number of distributed variables • Each side divided into B bins • Hence BN cells • Each cell defines a particular randomly sampled value for every variable • i.e., each cell defines a trial • All trials are equally probable

33. Latin Hypercube • A Latin Hypercube consists of B cells chosen from the possible BN cells such that no cell shares a row, column, rank,… with any other cell. • For N = 2 and B = 8 an example of a Latin Hypercube is a chess board containing 8 rooks none of which are en prise. • Any Latin Hypercube defines B trials which sample all B bins of every one of the N variables.

34. Example – The ‘Latin Square’ • N=2 Variables and B=4 Samples per Variable 1 2 3 4 • B cells are randomly occupied such that each row and column contains only one occupied cell. • The occupied cells then define the B trial combinations. 1 2 3 4

35. Generation of the Latin Square • A simple way to generate the square/hypercube 3 4 2 1 • Assign the variable samples in random order to each row and column. • Occupy the diagonal to specify the trial combinations. • These combinations are identical to the ones on the previous slide. 3 1 4 2

36. Homework • Given N and B, how many different Latin hypercubes are there?

37. Range of Components • Modelling just one item – or a family of items? • Note that distributed variables do not just cover uncertainties but can also cover item to item differences, • Temperature • Load • Geometry • Metal losses

38. Plant History • A decision is required early on… • Model on the basis of just a few idealised load cycles… • …or use the plant history to model the actual load cycles that have occurred • Can either random sample to achieve this • Or can simply model every major cycle in sequence if you have the history (reactor and boiler cycles) • Reality is that all cycles are different

39. Cycle Interaction • Even if load cycles are idealised, if one or more parameters are randomly sampled every cycle will be different • Hence a cycle interaction algorithm is obligatory • And since all load cycles differ, the hysteresis cycles will not be closed, even in principle • This takes us beyond what R5 caters for • Hence need to make up a procedure

40. Unapproved Cycle Interaction • “Symmetrisation” of the hysteresis cycles has no basis when they are not repeated • Suggested methodology is, • http://rickbradford.co.uk/HysteresisCycleConstructionMethodologyForInteractingCycles.pdf • This leads to “symmetrisation on average” a = 0 symmetrises every cycle a = 0.93 is believed reasonable

41. Multiple Assessment Locations • In general you will need to assess several locations to cover just one component • E.g., a weld location, a stress-raiser location, and perhaps a second parent location • Crack initiation conceded when any one location cracks • So need to assess all locations in parallel at the same time

42. Correlations Between Locations • Are the material property distributions the same for all locations? • Even if they are the same distributions, is sampling to be done just once to cover all locations? (Perfect correlation) • Or are the properties obtained by sampling separately for each location (uncorrelated) • Ditto for the load distributions

43. Time Dependent Distributions • Most distributed variables will be time independent • Hence sampled once at start of life, then constant through life • But some may involve sampling repeatedly during service life • E.g., transient loads are generally different cycle by cycle

44. Time Dependent Distributions • Cycle-to-cycle variations in cyclic loading may be addressed… • Deterministically, from plant data • Probabilistically but as time independent (sampled just once) – not really right • Probabilistically sampled independently on every cycle • Latter case can be handled outwith the Latin Hypercube but must be on the basis of equal probabilities • Combination of the above for different aspects of the cyclic loading

45. Imposing Correlations • Correlations can be extremely important to the result • Proprietary software will include facilities for correlating variables • Input the correlation coefficient • If writing your own code, here’s how correlation may be imposed…

46. Imposing Correlations • For correlation between variables x and y, invent a new variable v, sampled independently of x, and then set y to, • This will generate the desired correlation Cxy between x and y. • For mutual correlations between three or more variables, see Appendix A of http://rickbradford.co.uk/ProbabilisticProcedureForR5V2_3.pdf

47. Results • If n cases of crack initiation occur in a run of N trials, then your estimate of the crack initiation probability is P = n/N • If n relates to the period from now to the end of life, then so does P • If n relates to a specified year, then so does P • If n relates to a specified individual component, then so does P • If the N trials randomly sample all the components in the reactor, then P is the reactor-average cracking probability per component

48. Results • If there are C components per reactor, the probability of cracking per reactor is then CP where P is the reactor-average cracking probability per component provided that CP << 1 • More precisely, for any CP, the probability of one or more cracks per reactor is, 1 – (1 – P)C

49. How Many Trials Do You Need? • It will depend on the application • The smaller the probability of cracking, the larger the number of trials needed to calculate it by Monte Carlo • You cannot justify a cracking probability of 10-4 per component in 1000 trials of single components. You would need ~100,000 single-component trials

50. How Many Trials Do You Need? • Suppose you want to justify a cracking probability of 0.1 per reactor year for a reactor consisting of 1000 components. • Suppose you run for 10 future years, then you need to justify 1 crack per 1000 components, i.e., 10-3 • So 10,000 single-component trials would be appropriate