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![A redundant process is installed to support theA redundant process is installed to support the
earlier example where Rearlier example where Rss = .713= .713
RR11
0.90
0.90
RR22
0.80
0.80
RR33
0.99
= [.9 + .9(1 - .9)] x [.8 + .8(1 - .8)] x .99= [.9 + .9(1 - .9)] x [.8 + .8(1 - .8)] x .99
= [.9 + (.9)(.1)] x [.8 + (.8)(.2)] x .99= [.9 + (.9)(.1)] x [.8 + (.8)(.2)] x .99
= .99 x .96 x .99 = .94= .99 x .96 x .99 = .94
Reliability hasReliability has
increasedincreased
fromfrom .713.713 toto .94.94
PROVIDING REDUNDANCY EXAMPLE](https://image.slidesharecdn.com/report002opeman-130709001348-phpapp02/85/Maintenance-and-Reliability-12-320.jpg)
Uploaded byEarl Tongol
Maintenance and Reliability
This document discusses maintenance and reliability of power generating plants. It notes that every year each plant undergoes 1-3 weeks of maintenance and every three years they undergo 6-8 weeks of complete overhaul. Preventive maintenance is important as it discovered a cracked rotor blade that could have destroyed expensive equipment. The document discusses ways to improve reliability such as improving individual components and providing redundancy. It also discusses the importance of maintenance for operational and financial reasons and strategies for effective maintenance like employee involvement, proper procedures, and computerized record keeping.
![002 Maintenance M Overview 20 06 06[1]](https://cdn.slidesharecdn.com/ss_thumbnails/002-maintenance-m-overview-20-06-061-1221944333735887-9-thumbnail.jpg?width=640&height=640&fit=bounds)
Maintenance and Reliability
- 1. Maintenance and Reliability BRAVEHEART Gutierrez,Koji Hizon, Carlo Literato, Dioni Sengson, Richard Tongol, Earl
- 2. MaintenanceMaintenance ofpower generatingof power generating plantsplants Every year each plant is takenEvery year each plant is taken off-off- line for 1-3 weeks maintenanceline for 1-3 weeks maintenance Every three years each plant isEvery three years each plant is taken off-line for 6-8 weekstaken off-line for 6-8 weeks forfor complete overhaul and turbinecomplete overhaul and turbine inspectioninspection Each overhaul hasEach overhaul has 1,800 tasks1,800 tasks andand requiresrequires 72,000 labor hours72,000 labor hours OUC performs overOUC performs over 12,00012,000 maintenance tasksmaintenance tasks each yeareach year Every day a plant is downEvery day a plant is down costscosts OUC $110,000OUC $110,000 Unexpected outagesUnexpected outages costcost between $350,000 andbetween $350,000 and $600,000 per day$600,000 per day Preventive maintenancePreventive maintenance discovered a cracked rotordiscovered a cracked rotor blade which could haveblade which could have destroyed a $27 milliondestroyed a $27 million piece of equipmentpiece of equipment
- 3. STRATEGIC IMPORTANCE OFMAINTENANCE AND RELIABILITY Failure has far reaching effectsFailure has far reaching effects on a firm’son a firm’s OperationOperation ReputationReputation ProfitabilityProfitability Dissatisfied customersDissatisfied customers Idle employeesIdle employees Profits becoming lossesProfits becoming losses Reduced value ofReduced value of investment in plant andinvestment in plant and equipmentequipment
- 4. ReliabilityReliability 1.1. Improvingindividual componentsImproving individual components 2.2. Providing redundancyProviding redundancy MaintenanceMaintenance 1.1. Implementing or improving preventiveImplementing or improving preventive maintenancemaintenance 2.2. Increasing repair capability or speedIncreasing repair capability or speed IMPORTANT TACTICS
- 5. Employee InvolvementEmployee Involvement Informationsharing Skill training Reward system Employee empowerment Maintenance and ReliabilityMaintenance and Reliability ProceduresProcedures Clean and lubricate Monitor and adjust Make minor repair Keep computerized records ResultsResults Reduced inventory Improved quality Improved capacity Reputation for quality Continuous improvement Reduced variability MAINTENANCE STRATEGY
- 6. RELIABILITY Improving individual componentsImprovingindividual components RRss = R= R11 x Rx R22 x Rx R33 x … x Rx … x Rnn wherewhere RR11 = reliability of component 1= reliability of component 1 RR22 = reliability of component 2= reliability of component 2 and so onand so on
- 7. RELIABILITY EXAMPLE RRss RR33 .99 RR22 .80 RR11 .90 Reliability ofthe process isReliability of the process is RRss = R= R11 x Rx R22 x Rx R33 = .90 x .80 x .99 = .713 or 71.3%= .90 x .80 x .99 = .713 or 71.3%
- 8. PRODUCT FAILURE RATE Basicunit of measure for reliabilityBasic unit of measure for reliability FRFR((%%)) = x= x 100%100% Number of failuresNumber of failures Number of units testedNumber of units tested FRFR((NN)) == Number of failuresNumber of failures Number of unit-hours of operating timeNumber of unit-hours of operating time Mean time between failuresMean time between failures MTBF =MTBF = 11 FRFR((NN))
- 9. PRODUCT FAILURE RATEEXAMPLE 2020 air conditioning units designed for use inair conditioning units designed for use in NASA space shuttles operated forNASA space shuttles operated for 1,0001,000 hourshours One failed afterOne failed after 200200 hours and one afterhours and one after 600600 hourshours FRFR((%%)) = (100%) = 10%= (100%) = 10% 22 2020 FRFR((NN)) = = .000106= = .000106 failure/unit hrfailure/unit hr 22 20,000 - 1,20020,000 - 1,200 MTBFMTBF = = 9,434= = 9,434 hrshrs11 .000106.000106
- 10. PRODUCT FAILURE RATEEXAMPLE 2020 air conditioning units designed for use inair conditioning units designed for use in NASA space shuttles operated forNASA space shuttles operated for 1,0001,000 hourshours One failed afterOne failed after 200200 hours and one afterhours and one after 600600 hourshours FRFR((%%)) = (100%) = 10%= (100%) = 10% 22 2020 FRFR((NN)) = = .000106= = .000106 failure/unit hrfailure/unit hr 22 20,000 - 1,20020,000 - 1,200 MTBFMTBF = = 9,434= = 9,434 hrshrs11 .000106.000106 Failure rate per trip FR = FR(N)(24 hrs)(6 days/trip) FR = (.000106)(24)(6) FR = .153 failures per trip
- 11. PROVIDING REDUNDANCY Provide backupcomponents to increaseProvide backup components to increase reliabilityreliability ++ xx ProbabilityProbability of firstof first componentcomponent workingworking ProbabilityProbability of needingof needing secondsecond componentcomponent ProbabilityProbability of secondof second componentcomponent workingworking (.8)(.8) ++ (.8)(.8) xx (1 - .8)(1 - .8) = .8= .8 ++ .16 = .96.16 = .96
- 12. A redundant processis installed to support theA redundant process is installed to support the earlier example where Rearlier example where Rss = .713= .713 RR11 0.90 0.90 RR22 0.80 0.80 RR33 0.99 = [.9 + .9(1 - .9)] x [.8 + .8(1 - .8)] x .99= [.9 + .9(1 - .9)] x [.8 + .8(1 - .8)] x .99 = [.9 + (.9)(.1)] x [.8 + (.8)(.2)] x .99= [.9 + (.9)(.1)] x [.8 + (.8)(.2)] x .99 = .99 x .96 x .99 = .94= .99 x .96 x .99 = .94 Reliability hasReliability has increasedincreased fromfrom .713.713 toto .94.94 PROVIDING REDUNDANCY EXAMPLE
- 13. Two typesof maintenanceTwo types of maintenance Preventive maintenance – routinePreventive maintenance – routine inspection and servicing to keepinspection and servicing to keep facilities in good repairfacilities in good repair Breakdown maintenance – emergencyBreakdown maintenance – emergency or priority repairs on failed equipmentor priority repairs on failed equipment MAINTENANCE
- 14. Output ReportsOutput Reports Inventoryand purchasing reports Equipment parts list Equipment history reports Cost analysis (Actual vs. standard) Work orders – Preventive maintenance – Scheduled downtime – Emergency maintenance Data entry – Work requests – Purchase requests – Time reporting – Contract work Data FilesData Files Personnel data with skills, wages, etc. Equipment file with parts list Maintenance and work order schedule Inventory of spare parts Repair history file COMPUTERIZED MAINTENANCE SYSTEM
- 15. MAINTENANCE COSTS TRADITIONAL VIEW TotalTotal costscosts BreakdownBreakdown maintenancemaintenance costscosts CostsCosts MaintenancecommitmentMaintenance commitment PreventivePreventive maintenancemaintenance costscosts Optimal point (lowestOptimal point (lowest cost maintenance policy)cost maintenance policy)
- 16. CostsCosts Maintenance commitmentMaintenance commitment Optimalpoint (lowestOptimal point (lowest cost maintenance policy)cost maintenance policy) TotalTotal costscosts Full cost ofFull cost of breakdownsbreakdowns PreventivePreventive maintenancemaintenance costscosts MAINTENANCE COSTS FULL COST VIEW
- 17. MAINTENANCE COST EXAMPLE 1.1.Compute the expected number ofCompute the expected number of breakdownsbreakdowns Number ofNumber of BreakdownsBreakdowns FrequencyFrequency Number ofNumber of BreakdownsBreakdowns FrequencyFrequency 00 2/20 = .12/20 = .1 22 6/20 = .36/20 = .3 11 8/20 = .48/20 = .4 33 4/20 = .24/20 = .2 ∑∑ Number ofNumber of breakdownsbreakdowns Expected numberExpected number of breakdownsof breakdowns CorrespondingCorresponding frequencyfrequency== xx = (0)(.1) + (1)(.4) + (2)(.3) + (3)(.2)= (0)(.1) + (1)(.4) + (2)(.3) + (3)(.2) = 1.6= 1.6 breakdowns per monthbreakdowns per month
- 18. 2.2. Compute theexpected breakdown cost per monthCompute the expected breakdown cost per month with no preventive maintenancewith no preventive maintenance ExpectedExpected breakdown costbreakdown cost Expected numberExpected number of breakdownsof breakdowns Cost perCost per breakdownbreakdown== xx = (1.6)($300)= (1.6)($300) = $480= $480 per monthper month MAINTENANCE COST EXAMPLE
- 19. 3.3. Compute thecost of preventive maintenanceCompute the cost of preventive maintenance PreventivePreventive maintenance costmaintenance cost Cost of expectedCost of expected breakdowns if servicebreakdowns if service contract signedcontract signed Cost ofCost of service contractservice contract == ++ = (1= (1 breakdown/monthbreakdown/month)($300) + $150)($300) + $150/month/month = $450= $450 per monthper month Hire the service firm; it is less expensive MAINTENANCE COST EXAMPLE
- 20. 1.1. Well-trained personnelWell-trainedpersonnel 2.2. Adequate resourcesAdequate resources 3.3. Ability to establish repair planAbility to establish repair plan and prioritiesand priorities 4.4. Ability and authority to doAbility and authority to do material planningmaterial planning 5.5. Ability to identify the cause ofAbility to identify the cause of breakdownsbreakdowns 6.6. Ability to design ways toAbility to design ways to extend MTBFextend MTBF INCREASING REPAIR CAPABILITIES
- 21. OperatorOperator MaintenanceMaintenance departmentdepartment Manufacturer’sManufacturer’s field servicefieldservice Depot serviceDepot service (return equipment)(return equipment) Preventive maintenance costs less and is faster the more we move to the left Competence is higher as we move to the right HOW MAINTENANCE IS PERFORMED
- 22. SimulationSimulation Computeranalysis of complex situationsComputer analysis of complex situations Model maintenance programs before theyModel maintenance programs before they are implementedare implemented Physical models can also be usedPhysical models can also be used Expert systemsExpert systems Computers help users identify problemsComputers help users identify problems and select course of actionand select course of action ESTABLISHING MAINTENANCE POLICIES
- 23.