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Designing for System Reliability

Designing for System Reliability. Dave Loucks, P.E. Eaton Corporation. To Facility. From Main. To UPS. To UPS. From Emergency Bus. From Distribution Bus. What Reliability Is Seen At The Load?. For example, if power flows to load as below:

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Designing for System Reliability

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  1. Designing for System Reliability Dave Loucks, P.E. Eaton Corporation

  2. To Facility

  3. From Main To UPS To UPS

  4. From Emergency Bus From Distribution Bus

  5. What Reliability Is Seen At The Load? • For example, if power flows to load as below: • Assume outage duration exceeds battery capacity Utility UPS Breaker Load

  6. Series Components • For example, if power flows to load as below: • Assume outage duration exceeds battery capacity Utility UPS Breaker Load 99.9% 99.99% 99.99%

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

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

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

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

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

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

  13. Parallel Components • What if power flows to load like this: • Solve each path independently UPSa UPSa 99.99% 99.99% 99.99 % 99.99 % UPSb UPSb 99.99% 99.99% PF(a or b) = 0.01 % x 0.01% = 0.000001 % R(t) = 1 - PF(a or b) = 99.999999%

  14. Parallel Components • Multiply the two Probabilities of Failure, PF(a) and PF(b) and subtract from 1 UPSa or UPSb Utility Static ATS Load 99.9% 99.99% 99.89 % 99.999999% PF(total) = PF(u) + PF(a or b) + PF(s) = 0.1% + 0.000001% + 0.01% = 0.110001%

  15. Parallel Components • Multiply the two Probabilities of Failure, PF(a) and PF(b) and subtract from 1 UPSa or UPSb Static ATS Load 99.99% 99.99 % 99.999999% PF(total) = PF(a or b) + PF(s) = 0.000001% + 0.01% = 0.010001%

  16. Summary Table Comments?

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

  18. Availability • Increase Mean Time Between Failures (MTBF) • Decrease Mean Time To Repair (MTTR)

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

  20. 95% Availability from Different MTBF/MTTR combinations MTBF 1000 hrs 800 hrs 600 hrs 400 hrs 10.5 21.1 31.6 42.1 52.6 200 hrs

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

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

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

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

  25. Combine Reliabilities:Parallel Sources Source 1 Source 2 Source 1 or Source 2 99.9% 99.9% PF1 x PF2 = PF both PFboth = 0.11% x 0.11% PFboth = 0.0121% R(t) = 1 – PF both = 100% - 0.0121% = 99.99% 99.99% 99.99% 52 52 K K PF1 = 0.0121% R(t) = 99.99% 99.999% 99.999% 99.99% 99.99% 99.999% 99.999% 99.99% 99.99% What is the reliability at this point? What is the reliability at this point? Design 1

  26. Combine Reliabilities:Parallel Sources + Tx + Sec. Bkr. Source 1 Source 2 Source 1 or Source 2 99.9% 99.9% Rsource x Rtx x Rmb = R(t) = .9999 x .99999 x .9999 = .9998 99.99% 99.99% 52 52 K K 99.999% 99.99% R(t) = 99.98% 99.999% 99.999% 99.99% 99.99% What is the reliability at this point? What is the reliability at this point? Design 1

  27. Combine Reliabilities:… + bus and feeder breaker Source 1 Source 2 Source 1 or Source 2 99.9% 99.9% Rsource x Rtx x Rmb = R(t) = .9998 x .99999 x .9999 = .99969 99.99% 99.99% 52 52 K K 99.999% 99.99% 99.999% 99.99% What is the reliability at this point? R(t) = 99.969% Design 1

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

  29. Secondary Selective Source 1 Source 2 Source 1 Source 2 99.9% 99.9% R(t) = Rs x Rmb x Rtx x Rsb = .999 x .9999 x .99999 x .9999 = .99879 R(t) = Rs x Rmb x Rtx x Rsb = .999 x .9999 x .99999 x .9999 = .99879 99.99% 99.99% 52 52 99.999% 99.999% 99.99% 99.99% 99.999% 99.999% R(t) = 99.88% 99.999% 99.999% 99.88% 99.99% 99.99% 99.99% 99.99% What is the reliability at this point? Design 2

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

  31. Secondary Selective Source 1 Source 2 Source 1 Source 2 99.9% 99.9% R(t) = Rs x Rmb x Rtx x Rsb = .999 x .9999 x .99999 x .9999 = .99879 R(t) = Rs x Rmb x Rtx x Rsb = .999 x .9999 x .99999 x .9999 = .99879 99.99% 99.99% 52 52 99.999% 99.999% 99.99% 99.99% 99.999% 99.999% Rs1 = 99.879% Rbus2 = 99.999% Rbus1 = 99.999% 99.99% Rtie = 99.99% Rpath1 = Rs1 x Rbus1 x Rtie x Rbus2 x Rfdr Rpath1 = .99879 x .99999 x .9999 x .99999 x .9999 Rpath1 = .99857 PF1 = 1 – Rpath1 = 1-.99857 = 0.00143 = 0.143% Rfdr = 99.99% Design 2

  32. Secondary Selective Source 1 Source 2 Source 1 Source 2 99.9% 99.9% R(t) = Rs x Rmb x Rtx x Rsb = .999 x .9999 x .99999 x .9999 = .99879 R(t) = Rs x Rmb x Rtx x Rsb = .999 x .9999 x .99999 x .9999 = .99879 99.99% 99.99% 52 52 99.999% 99.999% 99.99% 99.99% 99.999% 99.999% Rs2 = 99.879% Rbus2 = 99.999% 99.99% Rpath2 = Rs2 x Rbus2 x Rfdr Rpath2 = .99879 x .99999 x .9999 Rpath2 = ..99868 PF2 = 1 – Rpath2 = 1-.99868 = 0.00132 = 0.132% Rfdr = 99.99% Design 2

  33. Secondary Selective Source 1 Source 2 Source 1 Source 2 99.9% 99.9% R(t) = 1 – (PF1 x PF2) = 100% - (0.143% x 0.132%) = 100% - (0.0189%) = 99.981% 99.99% 99.99% 52 52 99.999% 99.999% 99.99% 99.99% 99.999% 99.999% 99.99% R(t) = 99.981% Design 2

  34. Comparison Comments?

  35. Value Analysis • 99.97% x 8760 = failure once every 8757 hours • 99.98% x 8760 = failure once every 8758 hours • Assuming 1 hour repair time, we will see two, 1-hour outages after 8758 hours • Meaning 1/8758 hours (0.411 seconds) expected outage per year • As with UPS example, what is 0.411 seconds worth? • What is cost differential of higher reliability solution?

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

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

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

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

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

  41. Uniform Series Present Value Equations

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

  43. Reliability Tools • Eaton Spreadsheet Tools • IEEE PCIC Reliability Calculator • Commercially Available Tools • Financial Tools (web calculators)

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

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

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

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

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

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

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