Maintenance and Reliability
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Transcript of Maintenance and Reliability
Maintenance and Reliability
BRAVEHEART
Gutierrez, KojiHizon, CarloLiterato, DioniSengson, RichardTongol, Earl
MaintenanceMaintenance of power generating of power generating plantsplants
Every year each plant is taken Every year each plant is taken off-off-line for 1-3 weeks maintenanceline for 1-3 weeks maintenance
Every three years each plant is Every three years each plant is taken off-line for 6-8 weeks taken off-line for 6-8 weeks for for complete overhaul and turbine complete overhaul and turbine inspectioninspection
Each overhaul has Each overhaul has 1,800 tasks 1,800 tasks and and requires requires 72,000 labor hours72,000 labor hours
OUC performs over OUC performs over 12,000 12,000 maintenance tasks maintenance tasks each yeareach year
Every day a plant is down Every day a plant is down costs costs OUC $110,000OUC $110,000
Unexpected outages Unexpected outages cost cost between $350,000 and between $350,000 and $600,000 per day$600,000 per day
Preventive maintenance Preventive maintenance discovered a cracked rotor discovered a cracked rotor blade which could have blade which could have destroyed a $27 million destroyed a $27 million piece of equipmentpiece of equipment
STRATEGIC IMPORTANCE OF MAINTENANCEAND RELIABILITY
Failure has far reaching effects Failure 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 of Reduced value of
investment in plant and investment in plant and equipmentequipment
ReliabilityReliability1.1. Improving individual componentsImproving individual components
2.2. Providing redundancyProviding redundancy
MaintenanceMaintenance1.1. Implementing or improving preventive Implementing or improving preventive
maintenancemaintenance
2.2. Increasing repair capability or speedIncreasing repair capability or speed
IMPORTANT TACTICS
Employee InvolvementEmployee Involvement
Information sharingSkill trainingReward systemEmployee empowerment
Maintenance and Reliability Maintenance and Reliability ProceduresProcedures
Clean and lubricateMonitor and adjustMake minor repairKeep computerized records
ResultsResults
Reduced inventoryImproved qualityImproved capacityReputation for qualityContinuous improvementReduced variability
MAINTENANCE STRATEGY
RELIABILITY
Improving individual componentsImproving individual components
RRss = R = R11 x R x R22 x R x R33 x … x R x … 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
RELIABILITY EXAMPLE
RRss
RR33
.99
RR22
.80
RR11
.90
Reliability of the process isReliability of the process is
RRss = R = R11 x R x R22 x R x R33 = .90 x .80 x .99 = .713 or 71.3%= .90 x .80 x .99 = .713 or 71.3%
PRODUCT FAILURE RATE
Basic unit 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 = 11FRFR((NN))
PRODUCT FAILURE RATE EXAMPLE
2020 air conditioning units designed for use in air conditioning units designed for use in NASA space shuttles operated for NASA space shuttles operated for 1,0001,000 hours hoursOne failed after One failed after 200 200 hours and one after hours and one after 600600 hours hours
FRFR((%%)) = (100%) = 10%= (100%) = 10%22
2020
FRFR((NN)) = = .000106 = = .000106 failure/unit hrfailure/unit hr2220,000 - 1,20020,000 - 1,200
MTBF MTBF = = 9,434 = = 9,434 hrshrs11.000106.000106
PRODUCT FAILURE RATE EXAMPLE
2020 air conditioning units designed for use in air conditioning units designed for use in NASA space shuttles operated for NASA space shuttles operated for 1,0001,000 hours hoursOne failed after One failed after 200 200 hours and one after hours and one after 600600 hours hours
FRFR((%%)) = (100%) = 10%= (100%) = 10%22
2020
FRFR((NN)) = = .000106 = = .000106 failure/unit hrfailure/unit hr2220,000 - 1,20020,000 - 1,200
MTBF MTBF = = 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
PROVIDING REDUNDANCY
Provide backup components to increase Provide backup components to increase reliabilityreliability
++ xx
Probability Probability of first of first
component component workingworking
Probability Probability of needing of needing
second second component component
Probability Probability of second of second
component component workingworking
(.8)(.8) ++ (.8)(.8) xx (1 - .8)(1 - .8)
= .8= .8 ++ .16 = .96.16 = .96
A redundant process is installed to support the A 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 has Reliability has increased increased
from from .713.713 to to .94.94
PROVIDING REDUNDANCY EXAMPLE
Two types of maintenanceTwo types of maintenance Preventive maintenance – routine Preventive maintenance – routine
inspection and servicing to keep inspection and servicing to keep facilities in good repairfacilities in good repair
Breakdown maintenance – emergency Breakdown maintenance – emergency or priority repairs on failed equipmentor priority repairs on failed equipment
MAINTENANCE
Output ReportsOutput Reports
Inventory and 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
Maintenanceand work order
schedule
Inventory of spare parts
Repair history file
COMPUTERIZED MAINTENANCE SYSTEM
MAINTENANCE COSTS
TRADITIONAL VIEW
Total Total costscosts
Breakdown Breakdown maintenance maintenance costscosts
Co
sts
Co
sts
Maintenance commitmentMaintenance commitment
Preventive Preventive maintenance maintenance costscosts
Optimal point (lowestOptimal point (lowestcost maintenance policy)cost maintenance policy)
Co
sts
Co
sts
Maintenance commitmentMaintenance commitmentOptimal point (lowestOptimal point (lowest
cost maintenance policy)cost maintenance policy)
Total Total costscosts
Full cost of Full cost of breakdownsbreakdowns
Preventive Preventive maintenance maintenance costscosts
MAINTENANCE COSTS
FULL COST VIEW
MAINTENANCE COST EXAMPLE
1.1. Compute the expected number of Compute the expected number of breakdownsbreakdowns
Number of Number of BreakdownsBreakdowns
FrequencyFrequency Number of Number 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 of Number of breakdownsbreakdowns
Expected number Expected number of breakdownsof breakdowns
Corresponding Corresponding frequencyfrequency== xx
= (0)(.1) + (1)(.4) + (2)(.3) + (3)(.2)= (0)(.1) + (1)(.4) + (2)(.3) + (3)(.2)
= 1.6= 1.6 breakdowns per month breakdowns per month
2.2. Compute the expected breakdown cost per month Compute the expected breakdown cost per month with no preventive maintenancewith no preventive maintenance
Expected Expected breakdown costbreakdown cost
Expected number Expected number of breakdownsof breakdowns
Cost per Cost per breakdownbreakdown== xx
= (1.6)($300)= (1.6)($300)
= $480= $480 per month per month
MAINTENANCE COST EXAMPLE
3.3. Compute the cost of preventive maintenanceCompute the cost of preventive maintenance
Preventive Preventive maintenance costmaintenance cost
Cost of expected Cost of expected breakdowns if service breakdowns if service contract signedcontract signed
Cost of Cost of service contractservice contract
==
++
= (1= (1 breakdown/month breakdown/month)($300) + $150)($300) + $150/month/month= $450= $450 per month per month
Hire the service firm; it is less expensive
MAINTENANCE COST EXAMPLE
1.1. Well-trained personnelWell-trained personnel
2.2. Adequate resourcesAdequate resources
3.3. Ability to establish repair plan Ability to establish repair plan and prioritiesand priorities
4.4. Ability and authority to do Ability and authority to do material planningmaterial planning
5.5. Ability to identify the cause of Ability to identify the cause of breakdownsbreakdowns
6.6. Ability to design ways to Ability to design ways to extend MTBFextend MTBF
INCREASING REPAIR CAPABILITIES
OperatorOperator Maintenance Maintenance departmentdepartment
Manufacturer’s Manufacturer’s field servicefield service
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
SimulationSimulation Computer analysis of complex situationsComputer analysis of complex situations
Model maintenance programs before they Model 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 problems Computers help users identify problems
and select course of actionand select course of action
ESTABLISHING MAINTENANCE POLICIES
THANK YOU!
koji.carlo.dioni.richard.earlBRAVEHEART