1. © 2005 Eaton Corporation. All rights reserved. Designing for System Reliability Dave Loucks, P.E. Eaton Corporation

3. To Facility

4. To UPS From Main

5. From Emergency BusFrom Distribution Bus

6. What Reliability Is Seen At The Load? UtilityUPSBreaker Load l For example, if power flows to load as below: n Assume outage duration exceeds battery capacity

7. Series Components UtilityUPSBreaker Load 99.9%99.99% l For example, if power flows to load as below: n Assume outage duration exceeds battery capacity

8. Series Components l For example, if power flows to load as below: n Assume outage duration exceeds battery capacity UtilityUPSBreaker Load 99.9% (8.7 hr/yr) 99.99% (0.87 hr/yr) 99.99% (0.87 hr/yr) x+x+ x+x+ ==== 99.88% (10.5 hr/yr) l Overall reliability is poorer than any component reliability

9. Series Components l For example, if power flows to load as below: n Assume outage duration exceeds battery capacity UtilityUPSBreaker Load 99.9% (8.7 hr/yr) P F * = 0.1% 99.99% (0.87 hr/yr) 0.01% 99.99% (0.87 hr/yr) 0.01% x++x++ x++x++ ====== 99.88% (10.5 hr/yr) 0.12% l P F = (1 – Reliability) = 1 – R(t) * P F = probability of failure

10. Series Components l For example, if power flows to load as below: n If outage duration less than battery capacity UPSBreaker Load 99.99% (0.87 hr/yr) 0.01% 99.99% (0.87 hr/yr) 0.01% x++x++ ====== 99.98% (1.74 hr/yr) 0.02%* P F = l Batteries Depleted  99.88% reliable l Batteries Not Depleted  99.98% reliable

11. Parallel Components l What if power flows to load like this: n Assume outage duration exceeds battery capacity Utility UPS Static ATS Load UPS

12. Parallel Components Utility UPS Static ATS Load UPS 99.9% 99.99% ?? % l What if power flows to load like this: n Assume outage duration exceeds battery capacity

13. Parallel Components Utility Load 99.9% 99.99% P F * = 0.1%P F (a or b)++=?? % UPS Static ATS UPS 0.01% l What if power flows to load like this: n Assume outage duration exceeds battery capacity

14. Parallel Components 99.99% P F (a or b) =0.01 %x=0.000001 % 99.99 % UPS a UPS b 0.01% l What if power flows to load like this: n Solve each path independently 99.99% UPS a UPS b 99.99 % R(t) = 1 - P F (a or b) = 99.999999%

15. Parallel Components Utility Load 99.9% 99.99% 99.999999% 99.89 % UPS a or UPS b Static ATS l Multiply the two Probabilities of Failure, P F (a) and P F (b) and subtract from 1 P F (total) = P F (u) + P F (a or b) + P F (s) = 0.1% + 0.000001% + 0.01% = 0.110001%

16. Parallel Components Load 99.99% 99.999999% 99.99 % UPS a or UPS b Static ATS l Multiply the two Probabilities of Failure, P F (a) and P F (b) and subtract from 1 P F (total) = P F (a or b) + P F (s) = 0.000001% + 0.01% = 0.010001%

17. Summary Table ConfigurationReliability Single UPS system (long term outage) 99.88% Single UPS system (short term outage) 99.98% Redundant UPS system (long term outage) 99.89% Redundant UPS system (short term outage) 99.99% Comments?

18. Value Analysis Is going from this: UtilityUPSBreaker Utility UPS Static ATS 99.89% (only battery) 99.99% UPS 99.88% (only battery) 99.89% to this worth it? 0.01% difference

19. Availability l Increase Mean Time Between Failures (MTBF) l Decrease Mean Time To Repair (MTTR)

20. Relationship of MTBF and MTTR to Availability 1000 hrs 800 hrs 600 hrs 400 hrs 200 hrs MTBF

21. 95% Availability from Different MTBF/MTTR combinations 1000 hrs 800 hrs 600 hrs 400 hrs 200 hrs MTBF 10.521.131.642.152.6

22. Breakeven Analysis l Total Economic Value (TEV) n Simple Return (no time value of money) TEV S = (Annual Value of Solution x Years of Life of Solution) – Cost of Solution l Assume an outage > 0.1s costs $10000/yr l Assume cost of solution is $30000 l Assume life of solution is 10 years

23. Breakeven Analysis l Since solution eliminates this potential 0.41 second outage, we “save” $10000 each year l Total Economic Value (TEV) n Simple Return (no time value of money) n TEV S = (Annual Value of Solution x Years of Life of Solution) – Cost of Solution TEV S = (($10000 x 1) x 10) – $100000 TEV S = $100000 - $30000 = $70000

24. Let’s Examine a More Complex System 52 Source 1Source 2 What is the reliability at this point? KK 52 Source 1 What is the reliability at this point? 52 Source 2 99.9% 99.99% 99.999% 99.99% 99.999% 99.9% 99.99% 99.999% 99.99% 99.9% 99.99% 99.999% 99.99% 99.999% Design 1Design 2

25. Primary Selective 52 Source 1Source 2 What is the reliability at this point? KK 99.9% 99.99% 99.999% 99.99% 99.999% Source 1Source 2 What is the reliability at this point? 99.999% 99.99% 99.999%.999 x.9999 = 99.89% 99.89% Convert to Design 1

26. Combine Reliabilities: Parallel Sources 52 Source 1Source 2 What is the reliability at this point? KK 99.9% 99.99% 99.999% 99.99% 99.999% Source 1or Source 2 What is the reliability at this point? 99.999% 99.99% 99.999% P F1 x P F2 = P F both P F both = 0.11% x 0.11% P F both = 0.0121% R(t) = 1 – P F both = 100% - 0.0121% = 99.99% P F1 = 0.0121% R(t) = 99.99% Design 1

27. Combine Reliabilities: Parallel Sources + Tx + Sec. Bkr. 52 Source 1Source 2 What is the reliability at this point? KK 99.9% 99.99% 99.999% 99.99% 99.999% Source 1or Source 2 What is the reliability at this point? 99.99% 99.999% R source x R tx x R mb = R(t) =.9999 x.99999 x.9999 =.9998 R(t) = 99.98% Design 1

28. Combine Reliabilities: … + bus and feeder breaker 52 Source 1Source 2 What is the reliability at this point? KK 99.9% 99.99% 99.999% 99.99% 99.999% Source 1or Source 2 R source x R tx x R mb = R(t) =.9998 x.99999 x.9999 =.99969 R(t) = 99.969% Design 1

29. Secondary Selective 52 Source 1 What is the reliability at this point? 52 Source 2 99.9% 99.99% 99.999% 99.99% 99.9% 99.99% 99.999% 99.99% 99.999% 99.99% Design 2

30. Secondary Selective 52 Source 1 What is the reliability at this point? 52 Source 2 99.9% 99.99% 99.999% 99.99% 99.9% 99.99% 99.999% 99.99% 99.999% Source 1Source 2 99.99% 99.999% R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R(t) = 99.88%99.88% Design 2

31. Secondary Selective 52 Source 1 What is the reliability at this point? 52 Source 2 99.9% 99.99% 99.999% 99.99% 99.9% 99.99% 99.999% 99.99% 99.999% Source 1Source 2 99.99% 99.999% R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R(t) = 99.88% Design 2

32. Secondary Selective 52 Source 1 52 Source 2 99.9% 99.99% 99.999% 99.99% 99.9% 99.99% 99.999% 99.99% 99.999% Source 1Source 2 R tie = 99.99% R fdr = 99.99% R bus1 = 99.999% R bus2 = 99.999% R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R s1 = 99.879% R path1 = R s1 x R bus1 x R tie x R bus2 x R fdr R path1 =.99879 x.99999 x.9999 x.99999 x.9999 R path1 =.99857 P F1 = 1 – R path1 = 1-.99857 = 0.00143 = 0.143% Design 2

33. Secondary Selective 52 Source 1 52 Source 2 99.9% 99.99% 99.999% 99.99% 99.9% 99.99% 99.999% 99.99% 99.999% Source 1Source 2 R fdr = 99.99% R bus2 = 99.999% R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R(t) = R s x R mb x R tx x R sb =.999 x.9999 x.99999 x.9999 =.99879 R path2 = R s2 x R bus2 x R fdr R path2 =.99879 x.99999 x.9999 R path2 =..99868 P F2 = 1 – R path2 = 1-.99868 = 0.00132 = 0.132% R s2 = 99.879% Design 2

34. Secondary Selective 52 Source 1 52 Source 2 99.9% 99.99% 99.999% 99.99% 99.9% 99.99% 99.999% 99.99% 99.999% Source 1Source 2 R(t) = 1 – (P F1 x P F2 ) = 100% - (0.143% x 0.132%) = 100% - (0.0189%) = 99.981% R(t) = 99.981% Design 2

35. Comparison MethodReliability at Load 1. Primary Selective99.969% 2. Secondary Selective99.981% Comments?

36. Value Analysis 1.99.97% x 8760 = failure once every 8757 hours 2.99.98% x 8760 = failure once every 8758 hours l Assuming 1 hour repair time, we will see two, 1- hour outages after 8758 hours n Meaning 1/8758 hours (0.411 seconds) expected outage per year l As with UPS example, what is 0.411 seconds worth? l What is cost differential of higher reliability solution?

37. Breakeven Analysis l Total Economic Value (TEV) n Simple Return (no time value of money) TEV S = (Annual Value of Solution x Years of Life of Solution) – Cost of Solution n Discounted Return (borrowed money has cost) TEV D = (NPV(annual cash flow, project life, interest rate) – Cost of Solution l Assume 0.411 sec of downtime costs $20000/yr l Assume cost of solution is $75000 l Assume life of solution is 10 years

38. Breakeven Analysis l Total Economic Value (TEV) n Simple Return (no time value of money) n TEV S = (Annual Value of Solution x Years of Life of Solution) – Cost of Solution TEV S = (($20000 x 1) x 10) – $200000 TEV S = $200000 - $75000 = $125000 Discounting cash flow at 10% cost of money n TEV D = NPV($20000/yr, 10 yrs, 10%) – $30000 TEV D = $122891 – $75000 = $47891

39. Solve for Equivalent Interest Rate l Knowing initial cost of  $75000 and annual benefit of  $20000 what is the equivalent return? Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9 Year 10 Year 0 $75000 $20000

40. Uniform Series Present Value l P – Present Value l A – Annuity payment l n – Number of periods l i – Interest per period n =

41. Uniform Series Present Value Equations l But what if PV, A and n are known and i is unknown? l Iterative calculation or l P – Present Value l A – Annuity payment l n – Number of periods l i – Interest per period

42. Uniform Series Present Value Equations niPVRS of EquationComment 105%$75000$154435i too low 1030%$75000$61831i too high 1020%$75000$83849i too low 1025%$75000$71410i too high 1023.413%$75000 correct value

43. Breakeven Analysis l Total Economic Value (TEV) n Simple Return (no time value of money) n TEV S = (Annual Value of Solution x Years of Life of Solution) – Cost of Solution TEV S = (($20000 x 1) x 10) – $200000 TEV S = $200000 - $75000 = $125000 Discounting cash flow at 10% cost of money n TEV D = NPV($20000/yr, 10 yrs, 10%) – $30000 TEV D = $122891 – $75000 = $47891 IRR = 23.413% effective return

44. Reliability Tools l Eaton Spreadsheet Tools l IEEE PCIC Reliability Calculator l Commercially Available Tools l Financial Tools (web calculators)

45. Web Based Financial Analysis l www.eatonelectrical.com l search for “calculators” l Choose “Life Extension ROI Calculator”

46. Web Based Financial Analysis l Report provides financial data l Provides Internal Rate of Return l Use this to compare with other projects competing for same funds l Evaluates effects due to taxes, depreciation l Based on IEEE Gold Book data

47. Uncertainty – Heart of Probability l Probability had origins in gambling n What are the odds that … l We defined mathematics resulted based on: n Events What are the possible outcomes? n Probability In the long run, what is the relative frequency that an event will occur? “Random” events have an underlying probability function

48. Normal Distribution of Probabilities l From absolutely certain to absolutely impossible to everything in between Absolutely Certain 100% 0% Absolutely Impossible Most likely value

49. Distribution System Reliability l How do you predict when something is going to fail? l One popular method uses exponential curve Absolutely Certain 37% of them are working 50% of them are working 69% 50%

50. Mean Time Between Failures l The ‘mean time’ is not the 50-50 point (1/2 are working, 1/2 are not), rather… l When device life (t) equals MTBF (1/ ), then: l The ‘mean time’ between failures when 37% devices are still operating

51. MTBF Review l Remember, MTBF doesn’t say that when the operating time equals the MTBF that 50% of the devices will still be operating, nor does it say that 0% of the devices will still be operating. It says 37% (e -1 ) of them will still be working. l Said another way; when present time of operation equals the mean (1/2 maximum life), the reliability is 37%

52. Exponential Probability l Assumes (1/MTBF) is constant with age l For components that are not refurbished, we know that isn’t true. n Reliability decreases with age ( gets bigger) l However, for systems made up of many parts of varying ages and varying stages of refurbishment, exponential probability math works well.

53. Reliability versus MTBF l Assume at time = 0 n Reliability equals 100% (you left it running) l At time > 0, n Reliability is less than 100%

54. Converting MTBF to Reliability l Unknown n Reliability = ? l Known n MTBF (40000 hrs) n t (8760 hrs = 1 year) UPS

55. Great, I’ve Found Problems, Now what? l You can certainly replace with new or… l If you catch it before it fails catastrophically, you can rebuild l Many old electrical devices can be rebuilt to like new condition

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