R&T 2007 - Replacement Interval Calculations - Elder

8/13/2019 R&T 2007 - Replacement Interval Calculations - Elder

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Pressure Relief ValvePressure Relief Valve

Pop Test DataPop Test Data

StatisticalStatistical

Replacement Interval CalculationReplacement Interval Calculationbyby

Frederick T. ElderFrederick T. ElderIRC Research and Technology ForumIRC Research and Technology Forum

February 9, 2007February 9, 2007

(c) Frederick T. Elder(c) Frederick T. Elder


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When to Replace per IIAR 110When to Replace per IIAR 110 After a known relief, and within a reasonable time,After a known relief, and within a reasonable time,

springspring--loaded relief valves shall be replaced by newloaded relief valves shall be replaced by new

or remanufactured certified valves. If reor remanufactured certified valves. If re--seating isseating isnot complete, replacement shall be immediate.not complete, replacement shall be immediate. When a component reliability program is in place toWhen a component reliability program is in place to

verify relief valve functionality and longevity byverify relief valve functionality and longevity byhistory, testing, disassembly and inspection, andhistory, testing, disassembly and inspection, and

periodic statistical review of these activities, reliefperiodic statistical review of these activities, reliefvalves may be replaced at any interval justified byvalves may be replaced at any interval justified bythe findings of such a program. In the absence ofthe findings of such a program. In the absence ofsuch a program, each relief valve shall be replacedsuch a program, each relief valve shall be replaced

at the frequency recommended by the relief valveat the frequency recommended by the relief valvemanufacturer. In the absence of both a componentmanufacturer. In the absence of both a componentreliability program and manufacturersreliability program and manufacturersrecommendations, relief valves shall be replacedrecommendations, relief valves shall be replacedevery five years if not indicated earlier at annualevery five years if not indicated earlier at annual

inspection.inspection.


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Why Test?Why Test?

Properly assess health of NH3Properly assess health of NH3

refrigeration safety systemrefrigeration safety system

OSHA has required it in priorOSHA has required it in prior

settlement agreementssettlement agreements It may save $$$It may save $$$

It may answer a PHA questionIt may answer a PHA question


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Maintenance GuidanceMaintenance Guidance

http://www.valvehttp://www.valve--world.net/srv/ShowPage.aspx?pageID=640world.net/srv/ShowPage.aspx?pageID=640


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OutlineOutline

Background andBackground and

Advantages of WeibullAdvantages of WeibullAnalysisAnalysis

Failure CriteriaFailure Criteria

Generating Weibull PlotGenerating Weibull Plot

Weibayes AnalysisWeibayes Analysis

ExamplesExamples


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BackgroundBackground Invented byInvented by WaloddiWaloddi WeibullWeibull

in 1937in 1937 he used it forhe used it forfatigue life estimationfatigue life estimation

Dr. Robert Abernethy theDr. Robert Abernethy the

modern Weibull Analysismodern Weibull Analysis

expertexpert

Weibull Analysis first usedWeibull Analysis first usedextensively in aerospaceextensively in aerospace

applicationsapplicationsWaloddi Weibull 1887-1979


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Advantages of Weibull AnalysisAdvantages of Weibull Analysis

Main advantage:Main advantage: Small sample sizeSmall sample size

Samples may be expensiveSamples may be expensive

Reduces time/cost of testingReduces time/cost of testing

May not have many recorded failuresMay not have many recorded failures

Weibull Analysis is displayed by anWeibull Analysis is displayed by an

easy to read graphical ploteasy to read graphical plot


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Pop Test Failure CriteriaPop Test Failure Criteria

Example 250 psig valveOpens at pressures < 242.5 psig failure

Opens at pressures > 262.5 psig -- failure


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Alternate Failure CriteriaAlternate Failure Criteria

Reduce the set pressure of reliefReduce the set pressure of reliefvalves when possiblevalves when possible then expandthen expand

failure definitionfailure definition

Do not consider low pressureDo not consider low pressureopening a failure for those valvesopening a failure for those valves

where that does not create a hazardwhere that does not create a hazard


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Weibull Analysis PlotWeibull Analysis Plot

Most Weibull Analysis done from plotMost Weibull Analysis done from plot

To Plot, you need:To Plot, you need:

Failure criteriaFailure criteria

Number of failures and timesNumber of failures and times Number of suspensions and timesNumber of suspensions and times

From Plot, you get:From Plot, you get:

Predicted failure ratePredicted failure rate

Failure mechanismFailure mechanism


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Plotting DataPlotting Data

Plot scalesPlot scales X axis: Age parameter (Units of Hours in Figure)X axis: Age parameter (Units of Hours in Figure)

Y axis: Cumulative Distribution Function (CDF)Y axis: Cumulative Distribution Function (CDF)

Defines percentage of units that will fail up to an ageDefines percentage of units that will fail up to an age..


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Weibayes is used when thereWeibayes is used when there

are no or very few failures:are no or very few failures: Finding the MTTF of a unit afterFinding the MTTF of a unit after

initial testing lead to no failuresinitial testing lead to no failures

Redesigned component, severalRedesigned component, severalunits tested without failure, isunits tested without failure, is

testing sufficient?testing sufficient?

Smaller sample sizes neededSmaller sample sizes needed

with Weibayes since previouswith Weibayes since previousfailure history is knownfailure history is known


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Weibayes AnalysisWeibayes Analysis Weibayes Analysis equation,Weibayes Analysis equation,

uses anuses an assumedassumed

Can be used whenCan be used when no failuresno failures

have occurredhave occurred

Need to have back groundNeed to have back ground

failure infofailure info

Company Weibull libraryCompany Weibull library

OtherOther WeibullWeibull librarieslibraries

Where:Where:

N =N = total number oftotal number ofsuspensions and failuressuspensions and failures

r =r = number of failed unitsnumber of failed units

== assumed slopeassumed slope

t =t = time or cyclestime or cycles

/1N

i

i=1

t

r

=


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Weibayes AnalysisWeibayes Analysis Relief valve failure dataRelief valve failure data

shows typicalshows typical value ofvalue of 11 http://www.barringer1.com/http://www.barringer1.com/

wdbase.htmwdbase.htm

Weibayes can be used toWeibayes can be used todetermine replacementdetermine replacementinterval timeinterval time

Can input data intoCan input data into

Weibull program orWeibull program orcalculate by hand usingcalculate by hand usingequationequation


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Determine New Replacement IntervalDetermine New Replacement Intervalfor Test with Zero Failuresfor Test with Zero Failures

1.1. Gather suspension dataGather suspension data2.2. FindFind (as described in next slides)(as described in next slides)

3.3. Find kFind k11--value from Onevalue from One--Failure PlanFailure Plantable for your assumedtable for your assumed andand

number of samples being testednumber of samples being tested

4.4. Replacement Interval =Replacement Interval =(k(k11))


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Weibayes: FindingWeibayes: Finding With No FailuresWith No FailuresHand CalculationHand Calculation

Confidence Limit Equation for Zero Failures:Confidence Limit Equation for Zero Failures:

Use:Use:

wherewhere r=# of failuresr=# of failures

TTii=Time of each replacement=Time of each replacement

: look up this value from Chi: look up this value from Chi--squaredsquaredtable for C confidence and 2r+2 degrees of freedomtable for C confidence and 2r+2 degrees of freedom

( ){ }2 ; 2 2C f +

( )1

22 ; 2 2 0

iT C r for r

= +


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Weibayes: FindingWeibayes: Finding With No FailuresWith No FailuresWinSMITH CalculationWinSMITH Calculation

Can Select SpecificCan Select Specific

ConfidenceConfidence Enter number of units, allEnter number of units, all

as suspensionsas suspensions

Select Weibayes methodSelect Weibayes method

Choose specificChoose specific

confidence,confidence, 63.2%63.2%

confidence equivalent toconfidence equivalent to

assuming 1 failure isassuming 1 failure isimminentimminent

FindFind from Weibayes plotfrom Weibayes plot


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Weibayes Example:Weibayes Example: No FailuresNo Failures

During TestingDuring Testing

Parameters: 30 relief valves used forParameters: 30 relief valves used for5 years, 0 failures, want to increase5 years, 0 failures, want to increase

Replacement IntervalReplacement Interval

Question:Question: How many years can theHow many years can thevalves be used and have at most onevalves be used and have at most one

failure with a 90% confidence?failure with a 90% confidence?


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Data entered in WinSMITHData entered in WinSMITH

30 suspensions, 5 year time30 suspensions, 5 year time Weibayes method,Weibayes method, =1, 90% Confidence=1, 90% Confidence

=65.14


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Table of KTable of K11--values For Onevalues For One--Failure Test Plans,Failure Test Plans, =1=1

Read N=30, KRead N=30, K11=0.132=0.132

Complete table and equation to derive KComplete table and equation to derive K11--valuesvalues

included in Appendix Bincluded in Appendix B


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Replacement Interval: 65.14(0.132)= 8.6Replacement Interval: 65.14(0.132)= 8.6yearsyears

So with a 90% confidence, you can replaceSo with a 90% confidence, you can replace

the relief valves every 8.6 years and havethe relief valves every 8.6 years and haveat most one failure during that periodat most one failure during that period

Reasonable approach: 8.6 years minus 5Reasonable approach: 8.6 years minus 5

years = 3.6 years/2=1.8 years, so add 1.8years = 3.6 years/2=1.8 years, so add 1.8years to 5 year zero failure plan to haveyears to 5 year zero failure plan to have

reasonable probability of no failuresreasonable probability of no failures


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Weibayes AnalysisWeibayes AnalysisDetermine New Replacement Interval forDetermine New Replacement Interval for

One or More FailuresOne or More Failures During TestingDuring Testing

Most CommonMost Common

Typically, there will be failuresTypically, there will be failures


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Determine New Replacement Interval forDetermine New Replacement Interval forOne or More FailuresOne or More Failures During TestingDuring Testing

1.1. Gather failure and suspensionGather failure and suspensiondatadata

2.2. FindFind (as described in next slides)(as described in next slides)

3.3. Find kFind k00--value from Zerovalue from Zero--Failure Plan table for yourFailure Plan table for yourassumedassumed and number ofand number ofsamples being testedsamples being tested

4.4. Replacement Interval =Replacement Interval =(k(k00))


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Weibayes: FindingWeibayes: Finding With FailuresWith FailuresHand CalculationHand Calculation

Use Weibayes equation to findUse Weibayes equation to find

Use:Use:

to get ato get a specific confidencespecific confidence, where f=# of, where f=# offailuresfailures

: look up this value from: look up this value from

ChiChi--squared table for C confidence andsquared table for C confidence and2f+2 degrees of freedom2f+2 degrees of freedom

( ){ }

1/

2

2

; 2 2c

f

C f

=

+

( ){ }2 ; 2 2C f +


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Weibayes: FindingWeibayes: Finding With FailuresWith FailuresHand CalculationHand Calculation

ChiChi--Squared Table, C: 90% ConfidenceSquared Table, C: 90% Confidence


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Weibayes: FindingWeibayes: Finding With FailuresWith Failures

WinSMITH CalculationWinSMITH Calculation

Enter number of failures, allEnter number of failures, allwith the assumed time ofwith the assumed time of

half the usage timehalf the usage time

Enter number of suspensionsEnter number of suspensions Choose the specificChoose the specific

confidenceconfidence

FindFind from the Weibayesfrom the Weibayesplotplot


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Weibayes Example:Weibayes Example: One or MoreOne or More

FailuresFailures During TestingDuring Testing

Parameters: 30 relief valves used forParameters: 30 relief valves used for5 years, 2 failures, dont know when5 years, 2 failures, dont know when

failures occurredfailures occurred

Question:Question: How many years can theHow many years can thevalves be used and have zerovalves be used and have zero

failures with a 90% confidence?failures with a 90% confidence?


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FailuresFailures During TestingDuring Testing Data entered in WinSMITHData entered in WinSMITH

28 suspensions, 5 year time28 suspensions, 5 year time

2 failures, assumed half of 5 years, or 2.5 years2 failures, assumed half of 5 years, or 2.5 years

Weibayes method,Weibayes method, =1, 90% Confidence=1, 90% Confidence

27.26=


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FailuresFailures During TestingDuring Testing Table of KTable of K00--values For Zerovalues For Zero--Failure Test Plans,Failure Test Plans, =1=1

Read N=30, KRead N=30, K00=0.077=0.077

Complete table and equation to derive KComplete table and equation to derive K00--valuesvalues

included in Appendix Bincluded in Appendix B


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FailuresFailures During TestingDuring Testing

Replacement Interval:Replacement Interval:

27.26(0.077)= 2.1 years27.26(0.077)= 2.1 years

So with a 90% confidence, you canSo with a 90% confidence, you can

replace the relief valves every 2.1 yearsreplace the relief valves every 2.1 years

and have no failures during the intervaland have no failures during the interval


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RememberRemember

MI of pipes and vessels is also of high priorityMI of pipes and vessels is also of high priority

Relief Valves not to be placed back in serviceRelief Valves not to be placed back in service

after testingafter testing Need judgment to extend the replacement/testNeed judgment to extend the replacement/test

intervalinterval

Failed relief valve may never be neededFailed relief valve may never be needed


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Where to Buy Weibull MaterialWhere to Buy Weibull Material The New Weibull Handbook and theThe New Weibull Handbook and the

WinSMITH software packages can beWinSMITH software packages can bepurchased at:purchased at:

http://http://www.weibullnews.com/contents.hwww.weibullnews.com/contents.h

tm#Pricestm#Prices


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SourcesSources Engineering Safety Relief SystemsEngineering Safety Relief Systems, March 2006., March 2006.

by Reindl, D.T, Jekel, T.B., Available from theby Reindl, D.T, Jekel, T.B., Available from the

Industrial Refrigeration ConsortiumIndustrial Refrigeration Consortium The New Weibull HandbookThe New Weibull Handbook, Fourth Edition, 2000,, Fourth Edition, 2000,

by Robert Abernethyby Robert Abernethy

The New Weibull HandbookThe New Weibull Handbook, Fifth Edition, 2006,, Fifth Edition, 2006,by Robert Abernethyby Robert Abernethy

Fitness for Service of Pressure Relieving SystemsFitness for Service of Pressure Relieving Systems,,by W. E. Short II, presented at The 2003 ASMEby W. E. Short II, presented at The 2003 ASME

Pressure Vessels and Piping ConferencePressure Vessels and Piping Conference Reliability Testing of Relief ValvesReliability Testing of Relief Valves, by Robert E., by Robert E.

Gross, presented at The 2004 ASME PressureGross, presented at The 2004 ASME PressureVessels and Piping ConferenceVessels and Piping Conference


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SourcesSources Plant Guidelines for Technical Management ofPlant Guidelines for Technical Management of

Chemical Process SafetyChemical Process Safety

, pp 169, pp 169

--172, by Center172, by Center

for Chemical Process Safety, 1992for Chemical Process Safety, 1992

Armor SwiftArmor Swift EckrichEckrich OSHA settlementOSHA settlement

agreement of October 9, 1997agreement of October 9, 1997

IBPIBP OSHA settlement agreement of 2001OSHA settlement agreement of 2001

Code Requirements for Safety Relief SystemsCode Requirements for Safety Relief Systems,,

Todd Jekel, 2005 Research and TechnologyTodd Jekel, 2005 Research and Technology

Forum, January 20, 2005Forum, January 20, 2005


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SourcesSources Center for Chemical Process Safety (1998).Center for Chemical Process Safety (1998).

Guidelines for Pressure Relief and EffluentGuidelines for Pressure Relief and Effluent

Handling SystemsHandling Systems. (pp. 104. (pp. 104--107). Center for107). Center for

Chemical Process Safety/Chemical Process Safety/AIChEAIChE. Online version. Online version

available at: http://www.knovel.com/knovel2/available at: http://www.knovel.com/knovel2/

Toc.jsp?BookIDToc.jsp?BookID=831&VerticalID=0=831&VerticalID=0 Center for Chemical Process Safety (1989).Center for Chemical Process Safety (1989).

Process Equipment Reliability Data with DataProcess Equipment Reliability Data with Data

TablesTables. P 212. P 212


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Appendix A:Appendix A:Weibull Analysis BackgroundWeibull Analysis Background


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Advantages of Weibull AnalysisAdvantages of Weibull Analysis Weibull AnalysisWeibull Analysis

can be used for:can be used for: Failure DistributionFailure Distribution

Failure ForecastsFailure Forecasts

and Predictionsand Predictions

MaintenanceMaintenancePlanningPlanning

Effectiveness of aEffectiveness of aRedesignRedesign


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Weibull Analysis SoftwareWeibull Analysis Software WinSMITH Weibull from FultonWinSMITH Weibull from Fulton

FindingsFindings http://www.barringer1.com/wins.htmhttp://www.barringer1.com/wins.htm

Created by Wes Fulton and Dr. BobCreated by Wes Fulton and Dr. Bob

AbernetheyAbernethey


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Weibull Analysis SoftwareWeibull Analysis Software Enter age data, suspensions and failuresEnter age data, suspensions and failures

Software will:Software will: Plot DataPlot Data

CalculateCalculate EtaEta, Beta, and PVE numbers, Beta, and PVE numbers

Run a distribution analysisRun a distribution analysis

Generate a results reportGenerate a results report


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Plotting DataPlotting Data Age must be known for dataAge must be known for data

Standard Life Data: exact age of partsStandard Life Data: exact age of partsknownknown

Interval Data: Age of parts not exactlyInterval Data: Age of parts not exactly

know, so parts are groupedknow, so parts are grouped Could be from weekly, monthly, etcCould be from weekly, monthly, etc

inspectionsinspections

Age may be operating time,Age may be operating time,starts/stops, etc.starts/stops, etc.


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Plotting DataPlotting Data FailuresFailures

Establish failure modeEstablish failure mode

Every part displayingEvery part displayingthis mode constitutes athis mode constitutes a

failurefailure

SuspensionsSuspensions

Parts that failed via aParts that failed via adifferent modedifferent mode

Parts that have not yetParts that have not yet

failedfailed

Early Suspension: AgeEarly Suspension: Agebelow age of firstbelow age of first

failurefailure

Late Suspension: AgeLate Suspension: Age

above age of lastabove age of last

failurefailure


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Plotting DataPlotting Data Plot scalesPlot scales X axis: Age parameter (Units of Hours in Figure)X axis: Age parameter (Units of Hours in Figure)

Y axis: Cumulative Distribution Function (CDF)Y axis: Cumulative Distribution Function (CDF) Defines proportion of units that will fail up to an age as aDefines proportion of units that will fail up to an age as a

percentagepercentage


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Plotting DataPlotting Data

WinSMITH Data PlotWinSMITH Data Plot


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Plotting DataPlotting Data TwoTwo--parameter most widely used Weibullparameter most widely used Weibull

distributiondistribution

CDFCDF (Cumulative Distribution Function):(Cumulative Distribution Function):

F(tF(t) = 1) = 1-- ee--(t(t//))

F(tF(t) = fraction failing up) = fraction failing up--toto--time ttime t t= failure timet= failure time

= characteristic life= characteristic life

e = 2.718281, the base for natural logarithmse = 2.718281, the base for natural logarithms = slope parameter= slope parameter


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Plotting Data on Weibull PaperPlotting Data on Weibull Paper Arrange failures and suspensions in time ascendingArrange failures and suspensions in time ascending

orderorder

Set up the following table:Set up the following table:

Fill in Rank and Reverse Rank, and in Time column,Fill in Rank and Reverse Rank, and in Time column,include whether it was a Suspension or Failureinclude whether it was a Suspension or Failure

If two data points have the same time to failure, theyIf two data points have the same time to failure, theyare both presented in the column, and they will bothare both presented in the column, and they will both

get median rank valuesget median rank values


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Plotting Data on Weibull PaperPlotting Data on Weibull Paper Use equation to get Adjusted Rank (A.R.):Use equation to get Adjusted Rank (A.R.):

A.R.=[Reverse Rank X Previous Rank + (N+1)] / [Reverse Rank + 1]A.R.=[Reverse Rank X Previous Rank + (N+1)] / [Reverse Rank + 1]

UseUse BenardsBenards Median Rank formula to getMedian Rank formula to getnew Median Rank (since adjusted rank isnew Median Rank (since adjusted rank isnot an integer):not an integer):

BenardsBenards M.R.=(iM.R.=(i--0.3) X 100 / (N+0.4)0.3) X 100 / (N+0.4)

Fill out previous table, and plot:Fill out previous table, and plot: BenardsBenards M. R. on the yM. R. on the y--axisaxis

Time on xTime on x--axisaxis Draw a best fit line through the points,Draw a best fit line through the points,

make sure it is 1:1 Weibull papermake sure it is 1:1 Weibull paper

W ib ll E lW ib ll E l P iP i


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Weibull Example:Weibull Example: PreparingPreparing

Weibull Plot by HandWeibull Plot by Hand

Parameters: You are given theParameters: You are given the

following data, 8 total parts, 5following data, 8 total parts, 5failures at 49,82,96,30, and 90 hoursfailures at 49,82,96,30, and 90 hours

and, 3 suspensions at 45,10, andand, 3 suspensions at 45,10, and

100 hours100 hours Question:Question:At how many hours canAt how many hours can

you expect approximately 50% ofyou expect approximately 50% of

the parts to fail?the parts to fail?

W ib ll E lW ib ll E l P iP i


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Weibull Plot by HandWeibull Plot by Hand Set up and fill in table:Set up and fill in table:

Plot the points on 1:1 Weibull PaperPlot the points on 1:1 Weibull Paper

Draw a best fit line through the points andDraw a best fit line through the points and

draw a line across from the 50% mark anddraw a line across from the 50% mark and

down to the time axisdown to the time axis

W ib ll E lWeib ll E ample P iPreparing


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Weibull Plot by HandWeibull Plot by Hand

50% fail by 77 hours50% fail by 77 hours


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Interpreting the PlotInterpreting the Plot

PVE %: Goodness of fit indicator forPVE %: Goodness of fit indicator for

Weibull lineWeibull line 10% is acceptable, 50% is average10% is acceptable, 50% is average

N/S: Total number of data points/ NumberN/S: Total number of data points/ Number

of Suspensionsof Suspensions

Eta, Beta, PVE,and N/S


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Interpreting the PlotInterpreting the Plot EtaEta -- Characteristic life: Age at whichCharacteristic life: Age at which

63.2% of parts will fail63.2% of parts will fail Parameter most effected by suspensionsParameter most effected by suspensions


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Interpreting the PlotInterpreting the Plot

BetaBeta Slope of Weibull line: Failure ModeSlope of Weibull line: Failure Mode

Beta < 1.0 indicates infant mortalityBeta < 1.0 indicates infant mortality Beta = 1.0 indicates random failures that areBeta = 1.0 indicates random failures that are

independent of ageindependent of age

Beta > 1.0 indicates wear out failuresBeta > 1.0 indicates wear out failures


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Interpreting the PlotInterpreting the Plot Use PVE number to evaluate fit ofUse PVE number to evaluate fit of

lineline

Use Beta to evaluate failure methodUse Beta to evaluate failure method

Look for Bad Weibull characteristicsLook for Bad Weibull characteristics


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Bad WeibullBad Weibull Curved Weibull dataCurved Weibull data

Origin not at t=0, mustOrigin not at t=0, must

use threeuse three--parameterparameterWeibullWeibull

Outlying data pointsOutlying data points Look at engineeringLook at engineering

aspects of dataaspects of datarecording, test records,recording, test records,calibrations, etc.calibrations, etc.

Two different slopes ofTwo different slopes ofWeibull dataWeibull data More than one failureMore than one failure

mode represented bymode represented bydata, try to separatedata, try to separatedatadata


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Bad WeibullBad Weibull Close Serial NumbersClose Serial Numbers

Batch problemBatch problem

If PVE number is unacceptableIf PVE number is unacceptable

Look to different distributions, LogLook to different distributions, Lognormal, Threenormal, Three--parameter Weibullparameter Weibull

Careful, few data points leads to highCareful, few data points leads to highPVE numberPVE number


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Failure ForecastingFailure Forecasting Expected number ofExpected number of

failures that mayfailures that may

occur in a specificoccur in a specificperiod of timeperiod of time

Predicts:Predicts: Future failures whenFuture failures when

failed units are replacedfailed units are replaced

Future failures whenFuture failures when

failed units are notfailed units are notreplacedreplaced


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Failure ForecastingFailure Forecasting Additional input needed:Additional input needed:

Age of components in serviceAge of components in service Usage rateUsage rate

Introduction rate of new unitsIntroduction rate of new units

Failed parts replacement infoFailed parts replacement info


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Appendix B: WeibayesAppendix B: WeibayesExtrasExtras

Weibayes: FindingWeibayes: Finding With No FailuresWith No Failures


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Weibayes: FindingWeibayes: Finding With No FailuresWith No Failures

Hand CalculationHand Calculation

Assume at Least One Failure is Imminent:Assume at Least One Failure is Imminent:

Use Weibayes equation to findUse Weibayes equation to find

Assume 1 failure (r=1) since a failure isAssume 1 failure (r=1) since a failure is

imminent (yields 63% confidence)imminent (yields 63% confidence)

Use the following table to achieve differentUse the following table to achieve different

confidences:confidences:


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Zero Failure Plan Table,Zero Failure Plan Table, = 1= 1

K=[(K=[(--1/N)*ln(0.1)]1/N)*ln(0.1)](1/(1/))

Chi Squared Table for Use WithChi Squared Table for Use With


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Chi Squared Table for Use WithChi Squared Table for Use With

Weibayes Hand CalculationsWeibayes Hand Calculations

Use 0.10 column for 90% Lower Bound,Use 0.10 column for 90% Lower Bound,

0.05 for 95% Lower Bound, etc.0.05 for 95% Lower Bound, etc.


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OneOne--Failure Test Plan Table,Failure Test Plan Table, = 1= 1

(1(1

--Confidence)=(Confidence)=(

ee--(k(k))

))NN

+N(e+N(e--(k(k))

))NN--11

(1(1

--

ee--(k(k))

))

R&T 2007 - Replacement Interval Calculations - Elder

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Transcript of R&T 2007 - Replacement Interval Calculations - Elder