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.

1. Maintenance and Reliability
BRAVEHEART
Gutierrez, Koji
Hizon, Carlo
Literato, Dioni
Sengson, Richard
Tongol, Earl

2.  MaintenanceMaintenance of power 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 OF MAINTENANCE
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. Improving individual 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
Information sharing
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 componentsImproving individual 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 of the 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
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 =
11
FRFR((NN))

9. PRODUCT FAILURE RATE EXAMPLE
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 RATE EXAMPLE
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 backup components 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 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

13.  Two types of 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
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
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
Maintenance commitmentMaintenance commitment
PreventivePreventive
maintenancemaintenance
costscosts
Optimal point (lowestOptimal point (lowest
cost maintenance policy)cost maintenance policy)

16. CostsCosts
Maintenance commitmentMaintenance commitment
Optimal point (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 the expected 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 the cost 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-trained personnel
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 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

22.  SimulationSimulation
 Computer analysis 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. THANK YOU!
koji.carlo.dioni.richard.earl
BRAVEHEART