Maintenance Strategies with Limited Resources 2005 Eaton

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Goal l We will learn if you have reliable electrical power systems l Why that is important l What to do if you don’t l Learn a financial method to rank alternative methods of improving reliability

Agenda l Where We Are Today l What Can Be Done? l IEEE 493 Process (Identify Candidates) l Predicting Failures l Economic Analysis / Prioritization l Solutions Available

Where We Are Today

Where We Are Today l Fact # 1: The aggregate economic loss of electrical power disruptions has climbed to more than $100 billion per year or more than 1% of U. S. Gross Domestic Product! n Recent events have demonstrated the fragility of our aging power grid. With transmission networks operating close to their stability limits, minor faults can cause cascading outages. Capacity limitations in several regions can lead to economic losses that cascade through the economy, causing loses for not only residential, but also commercial and industrial institutions.

Where We Are Today l Fact # 2: The recent power outages have been in the works for the last several years! n U. S utilities have always taken pride in their uptime and system performance. Over the past several years, due to industry-wide deregulation, market pressures for rate reductions, business restructuring and downsizing, overall investment in infrastructure has not been at traditional levels. The decoupling of transmission, distribution and generation has caused disruption in traditional business models and industry workings (i. e. the vertically integrated utility no longer exists in deregulated markets)

Where We Are Today l Fact # 3: De-regulation has contributed to loss of stability. n When the Federal Energy Regulatory Commission dictated that the electrical transmission network was to be opened to the free market, it allowed anyone to transmit power over any transmission line. This allowed generators outside a customer’s service area to bid on a distant customer’s power requirements and be guaranteed access to that customer over the transmission system. As a result, the owners of the transmission lines lost some of their ability to maintain stability since these lines now carry power generated outside their control. As demand has continued to increase at an average 2% per year and because few will accept new transmission lines in their backyard, grid stability continues to degrade over time.

Where We Are Today l Fact # 4: The 2000 dotcom implosion and the resultant relaxation of electrical demand has temporarily relaxed the demands placed on electrical transmission system, but the problem remains. n The recent retraction of the economy has reduced electrical demand temporarily. However little or no new transmission was constructed during that economic downturn. Electrical demand is again returning to historic levels. We can only expect the problem to return and even become worse as the economy expands to more robust levels.

Conclusion l This problem will worsen before it improves l Universities must take action themselves

Agenda l Where We Are Today l What Can Be Done? l IEEE 493 Process (Identify Candidates) l Predicting Failures l Economic Analysis / Prioritization l Solutions Available

What Can Be Done? A. Protect Yourself From External Problems 1. Install Local Backup Power • Natural Gas, LPG, Diesel from 1 k. W to over 10 MW 2. Install Voltage Correction SAG Correction Equipment • UPS • Capacitors • 3. Engineering Review • • Audits Site Supervision Equipment Commissioning Turnkey Installation

What Can Be Done? l l Many university outages are not caused by loss of utility power, but rather by internal problems Protect yourself from internal problems Equipment failure, accelerated by: n l Dust, dirt, moisture, rodents, etc. • Thermal cycling, vibration induced loosening, etc. • Obstruction of ventilation, etc. • Operator error • Reduction in funding for preventative maintenance • But when resources are tight, where should they be spent to give maximum uptime?

What Can Be Done? B. Protect Yourself From Internal Problems 1. Bring in an expert Reliability Study - Site Survey (e. g. evaluation of on-site generation, UPS, calculation of reliability of existing system, etc. ) - Thermography • • • 2. Coordination Study Harmonics Study Install Predictive Diagnostic equipment • Early warning of pending failure in MV Equipment

Agenda l Where We Are Today l What Can Be Done? l IEEE 493 Process (Identify Candidates) l Predicting Failures l Economic Analysis / Prioritization l Solutions Available

IEEE 493 Process 1. Establish Current Condition of Facility 2. Determine Likelihood of Serious Problem Based on this Condition 3. Sort to Find Equipment Most at Risk to Cause Problems 4. Identify the Predictive Techniques that Gives Early Warning of Problems at that Equipment

What Is “Current Condition? ” l The ‘Quiet Crisis’ Term created by Paul Hubbel, Deputy Director, Facilities and Services, Marine Corps. Government Executive Magazine, Sept 2002. When he was asked “why isn’t preventative maintenance adhered to more closely in government facilities? ” “We call it the ‘quiet crisis’ because a lot of maintenance problems take time to occur and are noticed until damage occurs”.

Current Condition? l l This leader says many follow the “fix when broken” but not before mentality? What happens if the power goes out at your facility for an extended period of time? For example, we would expect that facilities (such as universities & hospitals) won’t have power outage problems!

06 -AUG-03 A power failure forced New York University Medical Center to shut its emergency room and turn away visitors and some patients yesterday, as the hospital staff struggled with limited use of the air-conditioning, computers and other equipment. Chris Olert, a spokesman for Consolidated Edison, said, "We know that it was a combined failure of some of our equipment and some of the hospital's equipment, but we don't know exactly what triggered it. “ He said a cable feeding electricity to the medical center was damaged and had to be bypassed. When the power failed, the hospital's backup generators automatically turned on, but they could not carry the entire load, so hospital officials shut down some functions to preserve electricity for the most crucial ones. "No critical services were affected, " said Lynn Odell, a hospital spokeswoman. But hospital employees interviewed outside the building and family members of some patients said things were seriously disrupted for a time. One worker told of a darkened pharmacy with dormant computers, where pharmacists using flashlights filled out paperwork by hand responded to orders by telephone rather than computer. Another spoke of a stiflingly hot surgery department where some medicines spoiled in a nonworking refrigerator. (New York Times 8 -6 -2003)

Fairview Southdale Hospital, MN 21 -SEP-02 …in the case of Fairview Southdale Hospital's power failure, which lasted an hour, a power line wound up carrying more than its normal load for four days, which melted a fuse. That set off a chain of events that wound up overloading the hospital's emergency generators, causing them to fail, too. Mark Enger, the hospital's president, said the power failure could have been life -threatening. The power dropped but didn't go out entirely. It dropped enough, however, to trigger the hospital's three emergency generators. But because of the way the system is wired, the generators' cooling fans failed to work, the generators overheated and the hospital lost all power at about 11 a. m. Enger said there were eight surgeries going when the power failed. Four of the surgeons were able to finish their operations, while the other four finished quickly and re-scheduled the surgeries for the next day. Partial power was restored by noon, and full power was restored the next day, Enger said. Xcel is adamant that its maintenance practices are not posing widespread service problems. But one NSP worker said the hospital's power failure is but a symptom of the electrical grid's ill health. "Their work force is stretched so thin, they're putting people in jeopardy, " he said. . (St. Paul MN, Pioneer Press 8 -6 -2003)

Rhode Island Hospital On April 16, 2002, Rhode Island Hospital and Women & Infants Hospital lost power for an afternoon, leaving various parts of the 33 -building campus dark for varying amounts of time. At Women & Infants, an emergency generator immediately turned on, emergency lights came on, and no essential services were disrupted, according to a Providence Journal article. However, at Rhode Island Hospital, the backup electric system did not work, prompting Providence Mayor Vincent A. Cianci Jr. , to tell the Providence Journal, “A hospital of this magnitude and this size should not have these problems. ” Surprisingly, power outages do happen with alarming frequency to big hospitals. In fact, they’ve happened at Rhode Island Hospital campus before. In September 1999, a blackout plunged the entire campus into darkness. The backup systems failed once again and this time a patient died after his respirator failed. Then, in January 2000, another power failure forced the hospital to rely on backup generators for nearly two hours and shut down nonessential equipment and lights. A faulty ceramic insulator at a substation on the hospital campus caused the failure. Then a damaged coil prevented some of the backup power from flowing back into one of the hospital buildings. (EC&M Magazine 8 -1 -2002)

Current Condition? l l l Why would there be such problems in critical tested systems? Budget Cuts / Management Redirection of Maintenance Funds? This results in “Crisis Mode Operation” or “Fix What’s Broken and Skip the Rest” mentality But if you operate this way, how do you guess what will break next and where money should be targeted? Is there an analytical way of targeting scarce resources?

IEEE 493 Step 1 1. Establish Current Condition of Facility 2. Determine Likelihood of Serious Problem Based on this Condition 3. Sort to Find Equipment Most at Risk to Cause Problems 4. Identify the Predictive Techniques that Gives Early Warning of Problems at that Equipment

IEEE 493 Step 1 l What is the likelihood of a loss of MV power at a particular point in the university? n How many sources of supply n Are there one or more single-points-of-failure? n What is the likelihood of those points failing? n Use algebra to combine probabilities

IEEE 493 Step l For example, if power flows from utility like this: Utility Switch Breaker Load

IEEE 493 Step l For example, if power flows from utility like this: Utility Switch 99. 9% 99. 99% Breaker 99. 99% Load

IEEE 493 Step l For example, if power flows from utility as below: Utility Switch 99. 9% x 99. 99% x (8. 7 hr/yr) (0. 87 hr/yr) l Load Breaker 99. 99% (0. 87 hr/yr) = 99. 88% (10. 5 hr/yr) Overall reliability is poorer than any component reliability

Finding “Downtime Per Year” l Downtime / Year Made up of two components n How often failures occur - Measured as Mean-Time-Between-Failures - MTBF - e. g. 80000 hours • • How long a failure knocks you out - Measured as Mean-Time-To-Repair - MTTR - e. g. 8 hours

Finding “Downtime Per Year” Multiply together to determine downtimeper-year l How often : e. g. . 1 failure every 80000 hours n How long : e. g. . 8 hours each failure n 8 8 hrs length of outages (MTTR) hours 0 1 2 3 4 5 6 7 frequency of outage (MTBF) years 8 9 80008 hrs 80000 hrs

Finding “Downtime Per Year” Multiply together to determine downtimeper-year l How often : e. g. . 1 failure every 80000 hours n How long : e. g. . 8 hours each failure n 8 8 hrs length of outages (MTTR) hours How thick is this lower line, if we take the same probability and spread it over 80000 hours? 0 1 2 3 4 5 6 7 frequency of outage (MTBF) years 8 9 80000 hrs

Finding “Downtime Per Year” l Multiply together to determine downtimeper-year How often : e. g. . 1 failure every 80000 hours n How long : e. g. . 8 hours each failure n

IEEE 493 Step 2 1. Establish Current Condition of Facility 2. Determine Likelihood of Serious Problem Based on this Condition 3. Sort to Find Equipment Most at Risk to Cause Problems 4. Identify the Predictive Techniques that Gives Early Warning of Problems at that Equipment

IEEE 493 l Likelihood of failure of electrical equipment n l Low Consequences of failure of electrical equipment n High

Substation Transformer

Consequences of Failure l l After certain failures (such as that transformer failure), power cannot be restored for many hours Because of the large amount of energy within electrical equipment, that equipment can fail explosively n This may cause mechanical damage which limits ability to quickly repair and restore equipment to service

Equipment Destruction

Forensic Analysis of Equipment Destruction

Equipment Destruction

Worker Injury

Equipment Destruction

Consequences l This equipment isn’t going to be repaired in a few hours, or even a few days.

IEEE 493 l Likelihood (Failures / yr) n l Low Consequences (Hrs / failure) n High

IEEE 493 -1997 (Gold Book) Analysis IEEE Std 493 -1997, Table 7 -1 Product Failures/yr Hours/Failure Prot. Relays . 0002 5 Low likelihood High consequences Note: could be higher, but protective relays are usually installed in draw-out cases, so the delay in repairing is due to: 1) diagnosing the problem 2) locating a spare 3) installing a spare 4) reprogramming setpoints to match previous unit 5) restarting system

IEEE 493 -1997 (Gold Book) Analysis IEEE Std 493 -1997, Table 7 -1 Failures/yr * Hours/Failure = Hours/Yr Product Failures/yr Hours/Failure Hours/Yr f(x) % Prot. Relays . 0002 99. 999989 5 . 001

IEEE 493 -1997 (Gold Book) Analysis IEEE Std 493 -1997, Table 7 -1 Failures/yr * Hours/Failure = Hours/Yr Product Failures/yr Hours/Failure Hours/Yr f(x) % Prot. Relays . 0002 5 . 001 99. 999989 LV Swgr Bkrs . 0027 4 . 0108 99. 99988 MV Swgr Bkrs . 0036 2. 1 . 0076 99. 99991 83. 1* . 2992 99. 9966 LV Cable (1000 ft) . 00141 10. 5 . 0148 99. 99983 MV Cable (1000 ft) . 00613 26. 5 . 1624 99. 9981 Disc. Switches . 0061 3. 6 . 022 99. 9997 Transformer . 003 342 1. 026 99. 988 LV Swgr Bus . 0024 24 . 0576 99. 9993 MV Swgr Bus . 0102*** 26. 8 . 2733 99. 9969 * when no on-site spare is available ** below ground *** 3 connected to 3 breakers

IEEE 493 Step 1 SW 1 CBL 1 l l TX 1 CBL 2 BKR 1 Example: Likelihood of failure at an ICU = n RLY 1 51 52 BUS 1 n BKR 2 52 51 BKR 3 52 51 BKR 4 52 51 RLY 2 RLY 3 CBL 4 52 RLY 4 CBL 3 BKR 5 CBL 5 51 RLY 5 CBL 6 ICU f(SW 1) x f(CBL 1) x f(TX 1) x f(CBL 2) + f(BKR 1) x f(RLY 1) x f(BUS 1) x f(BKR 5) x f(RLY 5) x f(CBL 6) f(…) means probability of failure of that component

ICU Failure Scenario f(Utility) + f(SW 1) + f(CBL 1)+ f(TX 1)+ f(CBL 2) + f(BKR 1)+ f(RLY 1)+ f(BUS 1) + f(BKR 5) + f(RLY 5) + f(CBL 6) SW 1 l f(Utility) = 8. 76 hrs/yr (99. 9% ) CBL 1 TX 1 l f(SW 1) =. 022 hrs/yr (99. 9997% CBL 2 l f(CBL 1300 ft) = 300/1000 * 0. 1624 = 0. 049 hrs/yr (99. 9994%) BKR 1 52 RLY 1 l f(TX 1)= 1. 026 hrs/yr (99. 988%) l f(CBL 2 ) = 100/1000 * 0. 1624 = 0. 0162 hrs/yr (99. 9998%) BUS 1 100 ft BKR 5 52 l f(BKR 1), (BKR 5) =. 2992 hrs/yr (99. 9966%) RLY 5 CBL 6 l f(BUS 1) =. 2733 hrs/yr (99. 9969%) l f(RLY 1), (RLY 5) =. 001 hrs/yr (99. 999989%) l f(CBL 6300 ft) = 300/1000 * 0. 1624 = 0. 049 hrs/yr (99. 9994%) l Total = 99. 9%Utility x 99. 9997%SW 1 x 99. 9994%CBL 1 x 99. 988%TX 1 x 99. 9998%CBL 2 x 99. 9966%BKR 1 x 99. 999989%RLY 1 x 99. 9969%BUS 1 x 99. 9966%BKR 5 x 99. 999989%RLY 5 x 99. 9994%CBL 6 = 99. 88% (10. 8 hrs/yr) 51 51

ICU Failure Scenario l l Obviously, this is completely unacceptable reliability, but our example isn’t exactly accurate An ICU will always be connected to a critical backup bus fed from a generator or may not have redundant transformers It may l l How do we recalculate with a generator and transfer switch in the system?

Utility Only Reliability 99. 9% UTILITY 99. 9997% 99. 9994% SW 1 CBL 1 99. 988% TX 1 99. 9998% CBL 2 99. 9966% 52 51 99. 999989% RLY 1 BKR 1 BUS 1 99. 9969% BKR 5 52 51 RLY 5 CBL 6 ICU Reliability at this point = f(UTILITY) x f(SW 1) x f(CBL 1) x f(TX 1) x f(CBL 2) x f(BKR 1) x f(RLY 1) x f(BUS 1) = 99. 88%

Adding Generation GEN 1 f(GEN 1) = 99. 9% f(GEN 1) x f(CBLG 11000 ft) = 99. 9% x 99. 9998% = 99. 8998% Critical Bus CBLG 11000 ft f(CBLG 11000 ft) = 99. 9998%

Combining Busses 99. 88% Normal Bus SWN 1 99. 8998% Critical Bus SWG 1

Combining Busses 99. 88% Normal Bus f(SWN 1) = 99. 9997% SWN 1 99. 8998% Critical Bus SWG 1 f(SWG 1) = 99. 9997% f(Normal Bus x f(SWN 1) = 99. 88% x 99. 9997% = 99. 8797% What is the reliability at this point? f(Critical Bus x f(SWN 1) = 99. 8998% x 99. 9997% = 99. 8995%

Revisit Normal-Bus Reliability Available. Time Unavailable. Time Available Not Running Unavailable Total. Time

“Reliability” Available. Time = 8760 -11. 4 = 8748. 6 hrs = Unavailable. Time = 11. 4 hrs = Available Running 8757. 3 hrs/yr Unavailable Total. Time = 8760 hours = 1 year Not Running 11. 4 hrs/yr

Transferring to a less than reliable source? 8760 hrs Source 1 (Utility) = Unavailable Time = 0. 999*(10) + (0. 001*11. 4*60*60) = 51. 03 sec = 0. 0142 hr 99. 9% chance will switch to a source that is 99. 9% reliable within 10 seconds 0. 1% chance it won’t switch and outage last 11. 4 hours. 11. 4 hrs (utility outage) Source 2 (Generator) Total Time = S 1 A+S 1 U+S 2 A+S 2 U = 8760 + 0. 0142 + 11. 4 + 0. 00889 = 8771. 423 hr Source 1 (Utility) 99. 9% chance that will switch in 1 second to a source that is 99. 87% reliable. 0. 1% chance of not switching and resulting in an outage that (according to the Gold Book) averages 8. 6 hours. = Unavailable Time = (0. 999*1 sec) + (0. 001*8. 6*60*60) = 32 sec = 0. 00889 hr

Combined Reliability

Combining Busses 99. 88% Normal Bus f(SWN 1) = 99. 9997% SWN 1 99. 8998% Critical Bus SWG 1 f(SWG 1) = 99. 9997% f(Normal Bus x f(SWN 1) = 99. 88% x 99. 9997% = 99. 8797% 99. 9997% at this point f(Critical Bus x f(SWN 1) = 99. 8998% x 99. 9997% = 99. 8995%

Combining Busses 99. 88% 99. 8998% Normal Bus Critical Bus SWN 1 SWG 1 XFR 1 99. 9997% at this point BKR 5 52 51 RLY 5 CBL 6 ICU 99. 9957% = f(FDR 5) = 99. 9960% Reliability at ICU = f(ICU) = f(XFR 1)*f(FDR 5) =. 999997*. 99996 =. 999957 = 99. 9957% or (0. 38 hr/yr) or (23 min/yr) (outages per year)

But… l l Remember that these are failure calculations based on “proper” maintenance Also does not include normal age-induced degradation

Equipment Failure Timing l l Initial failures (installation problems, infant mortality of installed components). Degradation over time (temperature, corrosion, dirt, surge) Likelihood Of Failure 2. 33 hrs/yr (average) Initial Failures Degradation Failures Area under hatch marks represents the total likelihood of a failure Time

Equipment Failure Timing l Poor maintenance reduces equipment life since failures due to degradation come prematurely soon. IEEE says add 10% to likelihood of downtime. Likelihood Of Failure 2. 59 hrs/yr (average) Initial Failures Likelihood of failure is higher because postponed maintenance increases problems due to corrosion, misalignment, etc, that would be picked up in a PM program Early Degradation Failures Time

Results l Good Maintenance* = 11. 4 hrs/year downtime l Fair Maintenance = 11. 4 hrs + 10% =12. 5 hr/yr l 12. 5 – 11. 4 = 1. 1 hr/yr less downtime l 1 hour per year more downtime Is that worth spending any time fixing? … but this is only a simple example * Utility source to transfer switch

Real Systems Are Much Larger l l 17 MV breakers 14 MV loop feed switches n n l 31 MV internal bus runs n l l l Typical Large MV System 3 switching elements 42 total (17+14) 4000’ MV cable 15 MV transformers 3 standby generators

What is the likelihood of a power failure at this location? Just looking at a portion of the equipment… l 42 MV disconnect switches (42 *. 022 = 0. 924 hrs/yr) l 4000’ MV cable (4000/1000 * 0. 1624 = 0. 649 hrs/yr) l 15 MV transformers (15. 39 hrs/yr) l 30000’ LV cable (30000/1000 * 0. 0148 = 0. 444 hrs/yr) l 31 MV bus run with 17 MV breakers (31(0. 2733) + 17(. 2992)= 8. 47 + 5. 08 = 13. 55 hrs/yr) l 17 protective relays (17*. 001 = 0. 017) l Total = 0. 924 + 0. 649 + 15. 39 + 0. 444 + 13. 55 + 0. 017 = 31 hrs/yr (total duration of outages somewhere in this facility) (Assuming a 1 hr/per failure means you would expect an electrical problem 31 times per year or 1 every week and a half!)

Step 3 1. Establish Current Condition of Facility 2. Determine Likelihood of Serious Problem Based on this Condition 3. Sort to Find Equipment Most at Risk to Cause Problems 4. Identify the Predictive Techniques that Gives Early Warning of Problems at that Equipment

MV Transformers Win! (Lose? ) Expected Hours of Outages per year

Step 4 1. Establish Current Condition of Facility 2. Determine Likelihood of Serious Problem Based on this Condition 3. Sort to Find Equipment Most at Risk to Cause Problems 4. Identify the Predictive Techniques that Gives Early Warning of Problems at that Equipment

Now What? l l l We now know how to figure “how many minutes of outage will occur each year” for each device. But how do we reduce that value? n l We can recognize that failures can be predicted if we recognize the early warning signs The so-called “Predictive Indicator” Once we know that, we can identify the likely cause and fix the problem before it is serious.

Agenda l Where We Are Today l What Can Be Done? l IEEE 493 Process (Identify Candidates) l Predicting Failures l Economic Analysis / Prioritization l Solutions Available

Predicting Failures Equipment Failure Leads to … Initiating Causes Hints at … Predictive Indicator

Failure Contributing Causes (Source: IEEE 493 -1997)

Contributing Cause Initiating

Initiating Causes Predictive Indicators

Available Predictive Tools • Top 4 in order of importance are: - Partial Discharge Diagnostics (22. 4%) - Visual Inspection (18. 1%) - On-Line Thermal Analyzer (15. 6%) - Thermographic Inspections (12. 0%) CBM – Condition Based Maintenance

What If We Implemented One Predictive Solution? l Partial Discharge – 22. 4% of failures detected Caveat: Only works on medium voltage (>1000 volts) n l Our example: n n 15. 39 hrs/yr from transformer failure • 8. 47 hrs/yr from MV bus failure • n 22. 4% reduction 11. 94 hrs/yr 22. 4% reduction 6. 57 hrs/yr 5. 08 hrs/yr from MV breaker failure • 22. 4% reduction 3. 94 hrs/yr

Reduction In Outages l Transformer Failure (was 15. 39 hrs/yr, now 11. 94 hrs/yr) Saving 3. 45 hrs/yr n l MV bus failure (was 8. 47 hrs/yr, now 6. 57 hrs/yr) Saving 1. 9 hrs/yr n l MV breaker failure (was 5. 08 hrs/yr, now 3. 94 hrs/yr) Saving 1. 1 hr/yr Total Savings from PD 6. 45 hrs/yr n n n Failures/yr 1 hr/failure = 6 fewer failures? 6 hr/failure = 1 fewer failure? So what is the correct answer? Hrs/failure

Determining Number of Failures Start by capturing the hrs/failure and hours/yr for each device Category Failures/yr Hours/Failure Hours/Yr Prot. Relays . 0002 5 . 001 LV Swgr Bkrs . 0027 4 . 0108 . 0036 2. 1 / 83. 1* . 0076/. 2992 LV Cable (1000 ft). 00141 10. 5 . 0148 MV Cable (1000 ft) . 00613 26. 5 . 1624 Disc. Switches . 0061 3. 6 . 022 Transformer . 003 342 1. 026 LV Swgr Bus . 0024 24 . 0576 MV Swgr Bus . 0102*** 26. 8 . 2733 MV Swgr Bkrs * when no on-site spare is available ** below ground *** 3 connected to 3 breakers

Average Outage Next, total the number of devices and determine the total hours of failure for each device and for all devices Divide the total time by the number of devices. (7373. 5 hrs) (63) Result = Average hours per failure = 7373. 5/63 = 117. 04 hrs/failure

Compute Likely Failure Rate 0. 055 failures/ year l Total Savings from PD 6. 45 hrs/yr n 1 hr/failure = 6. 45 fewer failures per year n 6. 45 hr/failure = 1 fewer failure per year n 117 hrs/ failure 117. 4 hr/failure = ? fewer failures per year • 6. 45/117 =. 0. 055 • Answer: ? = 0. 055 failures/yr or 1 fewer failure every 18 years

Agenda l Where We Are Today l What Can Be Done? l IEEE 493 Process (Identify Candidates) l Predicting Failures l Economic Analysis / Prioritization l Solutions Available

How Much Does It Cost? l l l We know that if we install PD sensors on all this equipment, statistically it will result in 1 less outage every eighteen years. Each PD sensor costs ~ $5 K installed in quantity We have 23 items to be monitored n n 15 transformers 17 breakers with 31 bus runs • • l l Assume 1 PD sensor can monitor 1 vertical 2 -high structure 17/2 = 8. 5, round up to 9. 15+9 = 23 $7000 * 23 = $161000 Does saving an outage once every 18 years justify spending $161000?

Cost Savings Through Reduced Outage (Detected by PD) Total Exposure = Median Outage duration * % Related to Insulation * Downtime Cost

Your Costs May Vary… Obviously, the more critical the application, the more important the solution At $1000 / hour of downtime costs (lab, stadium, imaging) Loss of one of the small power transformers would require n n • • Cost of a 1000 k. VA indoor dry, MV power transformer • • • n n n 342 hours to repair (remember fire video) and cost: $342000 of downtime ($24000 / day) Assume $18/k. VA or $18000 Assume labor $50/hr, 3 man-days labor Total cost = (1000 * $18) + ($50 * 3 * 8) = $18000 + $1200 Total cost = $19200 Downtime = $342000 Material = $19200 Total Loss = Downtime + Material = $361200

Compute Payback l l l Our cost is $161000 Our savings is $361200 once every 18 years or $20000 per year If injury results, you can add $100000 - $1000000 (if selfinsured) l Assume we expect a 10% return on invested capital l Assume 10 year project life l Assume 2. 5% inflation rate

Compute Equivalent Payback l l l Cost = $161 K, Savings = $20 K/yr, N=10 years, inflation = 2. 5%, capital cost = 10% Is this a good investment? First Pass Analysis Simple Payback = $161 K/$20 K = 8 years Total Cash to bottom line = = (Annual Savings * Project Life) - Installed Cost = ($20000*10) - $161 K = $200000 - $161000 = $39000 positive cash flow (life of project) 0% cost of capital (simple payback) attractive return, but does this cover cost of capital (10%) considering the reduction in value of money over time (2. 5% inflation)?

Compounded IRR Calculator Cost installed cost of equipment Savings annual savings a (1+g)/(1+i) i interest rate g annual inflation rate n duration (payback period in years)

Compounded IRR Calculator Cost $161, 000 Savings $20, 000 a (1+g)/(1+i) = (1+0. 025)/(1+0. 1) = 0. 932 i 10% g 2. 5% n 10

Run The Numbers… Cost Savings a i g n $161, 000 $20, 000 (1+g)/(1+i) = (1+0. 025)/(1+0. 1) = 0. 932 10% 2. 5% 10

Compute Payback

What Does ‘invalid’ Mean? l Based on a cost of $161000, an annual savings of $20000, a required rate of return of 10%, and inflation rate of 2. 5%… n l l Means that this does not provide any payback over any time period Using the cost-of-money numbers listed here …this project is not financially viable. Note: assuming an 5. 75% cost of money vs 10% n n N turns out to be 9. 69 (<10) In that case, the project is financially viable

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

Agenda l Where We Are Today l What Can Be Done? l IEEE 493 Process (Identify Candidates) l Predicting Failures l Economic Analysis / Prioritization l Solutions Available

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LV Rack-In Replacement With New (In Old Equipment) Old Breaker: • Parts no longer available Modern Breaker: • New warranty • Installed in the old structure

Motor Control Upgrades Breaker-to-Starter Conversions: - circuit breaker used to start motor - only good for 1000 or less operations - replace breaker with starter - now good for 1, 000 operations Continuous Partial Discharge Monitor MCC Bucket Retrofits - new breaker and starter

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Can’t Buy a Spare? Class 1 Recondition Instead l Receiving & Testing l Complete Disassembly l Detailed Inspection and Cleaning l New Parts l OEM Re-assembly l Testing l Data-Base Tracking

Spot Network Upgrade Network Protector Class 1 Recondition Network Relay Upgrades. . .

Transformer Oil Processing Other Services Available: • Samples Obtained On-Site • Mail-in Sampling Kits • Self Powering Generator • Complete Transformer Testing • On-Site. PCB & Dissolved Gas Analysis - PF, Testing & Analysis • Vacuum Filling & Start-up • Reclamation & Retesting • Samples Obtained On-Site On-Board Testing Dielectric Testing Karl Fischer Moisture Test Acid Titration Testing

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