Download - Mining Equipment Maintenance

8/13/2019 Mining Equipment Maintenance

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Mining EquipmentMaintenance

Fundamental Concepts


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Old Underground Coal Mines


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Modern Longwall Mine


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The effect of Mechanisation on Productivity

0

5000

10000

15000

20000

25000

1916

1920

1925

1930

1935

1940

1945

1950

1952

1955

1960

1969

1972

1975

1980

1985

1988

1995

2003

Tonnes/Man-Year

Produced in QueenslandCoal Mines


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Maintenance Costs

30-50% of the operating costs Annual bill is about $10 billion

Equally significant is the cost of lostproduction when the machine is down

Every 1% improvement in equipment

availability or productivity improves the

company profits by up to 3.5%


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Objective Function forMaximising Company profits

The Annual Production for a given

investment is a suitable objective function


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Total Annual Production

Is this total annual production ?Tonnes/Hour x Total Hours

More realistic prediction is

Tonnes/Hour x (TotalHours - Downtime) Probably the following is more illuminating

Tonnes/Hour x (TotalHours -

-Scheduled Downtime

- Breakdown maintenance)


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Total Hours of PossibleProduction


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Subtract PlannedMaintenance


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Subtract BreakdownsThis is the time available

for production. Obviously,you want to maximise thistime. That is why the ratioof this time to the totaltime is a very importantKPI.


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Availability

TH - PM - BMA= TH

TH Total HoursPM Planned Maintenance

BM Breakdown Maintenance


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Availability

!Operating TimeTH - PM - BM

A= TH Total Time

Is also expressed in terms of MTBF and MTTR as

MTBFA=MTBF+MTTR


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Availability

!Operating TimeTH - PM - BM

A= TH Total Time

Is also expressed in terms of MTBF and MTTR as

MTBFA=MTBF+MTTR

This is the textbookdefinition of AVAILABILITY


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AvailabilityOperating Time = N x MTBF

Downtime = N x MTTR

Total Time = N x (MTBF+MTTR)

Operating Time MTBFA=

Total Time (MTBF

N x

N )x +MTTR

!


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Maximise the Return to the Company

The aim for the company is to maximise the return from its

productive assets.

Let us keep it simple and express this as maximising theproduction through the year.

To achieve this,

(a) The machine availability must be high

(b) The machine must be producing at a high rate when it is

operating

The solution is a compromise between the two.

We will see this in an example.


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Optimum Bucket SizeThe recommended suspended load for your dragline is 150

tonnes. A competitor put a larger bucket on a similar unitand they are running it at a suspended load of 170 tonnes.

Your Manager wants you to install an even bigger bucket,

taking your suspended load up to 210 tonnes. You know

that the safe static load for this dragline is 250 tonnes.

-Does a larger bucket necessarily mean higher production?

-Why not increase it to 250 tonnes if this is the safe load?

-How would you determine the optimum bucket size?


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Question TH - PM - BMA= THPresent Production Po = 9000 t/h.

Total Mine Operating Time in a Year = 8640h

12-hour planned maintenance shift every month.Breakdown maintenance downtime is expected tovary with the production rate as

3336o

PBM wR RhereP

! " !P is the production rate with the bigger bucket.

Find R that maximises the annual production.


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Solution

L

ostpr

oduction

due

to

Breakd

ownm

aintenanc

e

Optimum Point


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Dependence on MTBF


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Maintenance

Maintenance must be considered in thecontext of asset utilisation

Mining is an asset-rich industry

Optimum utilisation of these assets is the

only way a company will stay competitive

This is a task for both production and

maintenance engineers.


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Some Basic Concepts


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FailureLoss of abilityof an item to

perform itsrequiredfunction


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Failure

Broken teethon shearerdowndrive gear


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Failure

Cost associated with

this failure:

Lost production whenthe machine was down

Replacement gear

Maintenance labour


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Failure Rate, MTBF, MTTRMTTR MTBF MTTR MTBF

0 100 200 Hours

R

epair

Repair

Repair

Failure


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Failure Rate, MTBF, MTTRMTTR MTBF MTTR MTBF

0 100 200 Hours

R

epair

Repair

Repair

MTBF = Mean Time Between Failure (100 h)

Failure Rate = Number of failures per unit time (0.01 h-1)

MTTR = Mean Time To Repair (20 h)


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Historical RecordsFailures occur randomly. The repair time is also not constant.How do we find MTBF and MTTR?

If we treat failure as a random event, then we can use the

well-established tools of probability and statistics to modelthe uptime, downtime and availability for our equipment.


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Random Failure?

All nature is but art, unknown to thee

All chance, direction thou canst not see

All discord, harmony not understood

Alexander Pope, Essay on Man


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Poisson ProcessPoisson distribution is commonly used in forecasting to

represent the number of occurrences of a specific event ina given continuous interval.

Ships arriving at a dock on a given day

Traffic accidents on the SE freeway in a month

Mad cow disease breakouts in the world in one year

Typos per page in a long report typed by Hal Gurgenci

Cable shovel failures in one day of operation


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Poisson Distribution

# $( ) !

xte tp x

x

% %&

!

This is the probability distribution of the Poisson random

variable X representing the number of outcomes occurring in

a given time interval t. The parameter % is the averagenumber of outcomes per unit time.


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ReliabilityAssume failure events follow a Poisson distribution.What is the probability of having NO FAILURES in a

given time interval t?

This can be found by substituting x=0 in the Poisson

distribution function:

# $0

(0)0!

t

te t

p e

%

%%& &! !

This is referred to as the survival probability or the

reliability


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Reliability( ) tt e %&!

Reliability is the probability that a product will

operate throughout a specified period without failure

when maintained in accordance with the manufacturer's

instructions; and

when not subjected to the environmental or operational

stresses beyond limits stipulated by the manufacturer


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The value of e

Two centuries ago, a Polish Statistician, Ladislaus

Bortkiewicz, investigated the Prussian army fatalities

caused by horse kicks. According to army reports, the rate

was about one fatality every 1.64 years. Ladislaus

collected the reports for one year. These were 200 reportsand 109 recorded no deaths at all.

Can you estimate the value of

eusing the above data?


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ExampleA major piece of equipment fails twice a day on average.

Consider its reliability over a period of month.

What is the probability of failure at any time during that

month?

R li bilit


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Reliability

2t

e

&

!

Z I


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Zoom In

2t

e&

!


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Failure ProbabilityIf the reliability, R(t), is the probability to survive through

time t, then the probability of failing through that period is

1 R(t)

or

( ) 1 ( ) 1

t

F t R t e

%&

! & ! &Let us plot this


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Failure ProbabilityThis chart gives the probability of

the failure from 0 to time t. In

other words, it is the Cumulative

Distribution Function for failures.

How would we find the probability

density function or p.d.f. This is

sometimes useful.

We differentiate the cumulative

distribution function.


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Failure Distribution FunctionCumulative

Distribution Function ( ) 1 ( ) 1 tF t R t e %&! & ! &

The time-derivative of the c.d.f. gives the Probability

Distribution Function or p.d.f.

( ) tdF

f t e

dt

%% &! !

This form of p.d.f. is called the Exponential Distribution.

It represents the case when the hazard rate or failure rate, ,

is constant over time.


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Failure p.d.f.

The probability of failure through aunit time interval is given by

1

( )t

t

p f t dt'

!

(


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Let us zoom into the p.d.f.The probability of failure through a unit time interval is

given by the area under the curve

1 1 1( ) ( ) 1 ( ) ( )

2 2

t

tp f t dt f t f t f t

'! ) ' " ) ' )(


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Hazard RateThe hazard rate is the conditional probability of failure in a

small time interval (t, t+dt). It is conditional on there being

no failure until t:

( )( )

( )

f th t

t

!

For exponential failure distribution, the hazard rate is

constant:

( )( )( )

t

t

f t eh tR t e

%

%

% %&

&! ! !


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Constant Hazard Rate

The exponentialdistribution corresponds to

a constant failure rate a.k.a.

constant hazard rate


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Is % always constant ?

( ) t

f t e %

% &

!

Failure p.d.f.The constant % is the failure rate(%= 1/MTBF)

So far, we treated % as constant

This is called the exponentialdistribution

Is this always true?

Let us first give an example

Reliability function

( ) tR t e %&!


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Uniformly Increasing RateAssume the failure

rate is increasing by

the formula 0.1twhere t is measured

in days

It starts from zero and

at the end of the

month, the hazard rate

is 3 failures per day.

How do we generate the reliability function for this

component?

Reliability with varying


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Reliability with varyinghazard rate

dN Nmtdt ! &

In a sample of N, dN will fail in a time interval dt

dN mtdtN

! &2

ln ln2

tN m C! & '

m=0.1

2 2/ 2 / 2mt mt oN Ce N e

& &! ! Where N is the value of N at t=0

Then the reliability function is 2/ 2mt

o

NeN

&!

This corresponds to a Weibull distribution


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Weibull Reliability

t

R e

*

+, -&. /

0 1!R The probability of surviving through time t* Shape factor

+ Scale factor


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Weibull Distribution Curves

( ) 1 ( ) 1

t

F t R t e

*

+

, -&. /

0 1! & ! &c.d.f.

1

( )

tdF

f t t edt

*

+*

*

*

+

, -&. /&

0 1! !p.d.f.

1( )( )( )

f th t tR t

***+

&! !Hazard Rate

W ib ll s ith diff nt


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Weibull curves with different

hazard functions

t

R e

*

+, -&. /0 1!

1( )h t t***+

&!


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Bathtub CurveInfant

Mortality Useful Working LifeFallingApart

Fai

lur e

Rat e, %

Life in operation, hours

Optimum Point to Replace


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Actual Failure Patterns

These curves are the failurepatterns observed on

aircraft components in a

study completed in 1978 byNowlan and Heap.

This shows that only 4% of

the components go througha bathtub curve.

4%

2%

5%

7%

82%(68% of this with infant mortality)


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RELIABILITY OF SYSTEMS

Series Systems

Parallel Systems (Redundancy)


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Series SystemsA series system is a chain of components. When one of these parts fails,

the entire system fails.

Series Systems BA C

B CR R R R!


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Parallel SystemsThe failure for a parallel system means the failure of each

individual component. The system failure probability is then the

product of individual failure probabilities (1 R).

A

B

C

1 (1 )(1 )(1 )A B CR R R R! & & & &

Most mining machinery

systems are series

systems. In other words,the failure of one

component fails the

entire system. The

redundancy in miningcan be provided by

having multiple systems,

eg spare trucks or

shovels.


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Managing Reliability Optimum utilisation of its capital investment in

equipment is essential for company profits

Equipment reliability plays a major role in this

Therefore, managing reliability is a core business

for a mining company This is a task for both production and maintenance

engineers. In the rest of this presentation, we will

talk about the maintenance function.


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Maintenance Function Preventive Maintenance Prevent failures by performing a set of maintenance

tasks at periodic intervals Service

Inspection

Replacement

Corrective Maintenance Repair after a failure to bring the machine back to an

operating state

Which one delivers higher overall systemavailability?


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Corrective Maintenance We assume that corrective maintenance

brings the system to as new state.

Then it has no effect on system reliability

Its impact on system availability is measure

by the Mean Time To Repair (MTTR)


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Mean Time To Repair Fault Identification What caused the failure? What needs to be repaired?

Set-up time Find and bring the right person to the job

Actual repair

Logistic delays Waiting for the spare part

Restart time Time spent to bring the system back to normal

operation after the fault is repaired


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How to minimise MTTR Identify the failed components quickly.

This is achieved by experienced operators,on-line fault detection tools

For frequent failures have the repair crewwith the right skills on standby

Ditto for the frequently failing spare parts

Design the equipment and the operatingprocedure to minimise re-starting time


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PM Trade-Offs The cost of failure

MTTR

The cost of the repair and the replacement

The cash cost of the planned maintenanceaction (salaries, consumables, etc)

The opportunity cost (lost production)

Preventive Maintenance


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Preventive Maintenance

Issues Service

Effect on System Reliability

Inspection

P-F time (between potential and actual systemfailure)

Replacement

Failure distribution curves

Effect on System Reliability


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Would inspections help?

Failure

P-F Interval

The time when we can recognise

Potential Failure

Time

Scheduled inspections help when


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Scheduled inspections help when

Potential failure condition is clearly defined

The P-F interval is consistent

It is practical to inspect at intervals less than

the P-F interval

The P-F interval is long enough to

implement corrective maintenance action

Scheduled replacements help when


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Scheduled replacements help when

The component breakdown has costly

consequences (eg chain of failures, distance

from the workshop, etc)

The dominant failure mode is age-relatedwith the hazard rate consistently increasing

above an acceptable value at around the setreplacement period

Weibull curves with different


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ff

hazard functions

t

R e

*

+, -&. /0 1!

1( )h t t***+&!


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Periodic replacements - 10.5*! Decreasing Hazard Rate

Scheduled replacementincreases failure

probability


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Periodic replacements - 21*!

Constant Hazard Rate

Scheduled replacementhas no effect on failure

probability


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Periodic replacements - 32*!

Increasing Hazard Rate

Scheduled replacementdecreases failure

probability


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Reliability ResearchA significant part of the academic and research community

has been continuing to develop increasingly complex

mathematical models of the engineering systems and the

expected modes of failure under various loading

assumptions.

While the intellectual rigour in these studies and the amount

of effort that go into them cannot be ignored, the

applications to real manufacturing and mining processes

have been limited primarily for the lack of data needed to

support these models


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Industry Data

Analysis and Representation Tools


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Pareto Analysis Pareto Principle : Significant few and

Insignificant many In any application, a large part of the

failures are due to a small number of causes A Pareto plot helps to identify the most

significant causes

The benefit is incurred only by attending thesignificant issues


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Pareto Chart

Pareto Analysis for LongwallFace Equipment Failures

Shearer Drive Shaft


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Scatter Plot Diagrams A scatter plot is a logarithmic plot of

MTTR against the number of failures N.

Since the total downtime associated with

each failure is NxMTTR, constantdowntime curves appear as lines on

logarithmic axes.log log

DowntimeTTR MTTR C N

N! 2 ! &


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Longwall Scatter Plot

Lines of

Constant

Downtime


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Reliability Analysis Pareto Analysis and Scatter Plots are good tools to

identify the reliability sinks in the equipment

The next step is to calculate the failure probability

distribution curves for all critical components.

The MTTR statistics may also be required if

MTTR is not reasonably constant for each item

This step requires high quality data


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Features of High Quality Data Large enough set to have at least 4-5 failure

events for each target failure mode Cover a sufficiently long time period to

eliminate local effects Uniform operating conditions over this

period

Accuracy free of collectors bias


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Failure History Example

Suppose that we have the failure log for acomponent as 180, 216, 930, 990, 1300 and

1850 hours.

Estimate the MTBF assuming an exponential

probability distributions.


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Failure Data (example)Failure times

180h

216h

930h

990h1300h

1850h

36

714

36

714

60

310

550

60

310

550

36 714 60 310 550334

5MTBF

' ' ' '! !


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Curve FittingThe failure log was 180, 216, 930, 990, 1300 and 1850 hours.The TBF array , TBF = {36, 714, 60, 310, 550}.

100%80%60%40%20%Fi

7145503106036TBF

54321i

This is cumulative probability distribution data. We can then compare it

against exponential or Weibull distributions. For example, use the

following form to try an exponential distribution:

1ln 1 ( )

1 ( )

tt F t e

F t

%% &3 4

! 5 & !6 7&8 9

Estimate the Failure


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Distribution

TBF, h

Rough

Estimate

Median

Estimate

36 20% 10%

60 40% 30%310 60% 50%

550 80% 70%714 100% 90%

Exponential Fit334

t

t% &&


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( ) 1 1F t e e! & ! &


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Longwall Equipment

Failure Probability Distribution

Functions for some Critical Items

AFC Blockage/Overload (2002 only)


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1ln

1 ( )F t

3 46 7&8 9

Exponential

AFC Blockage/Overload (2002 only)


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Weibull

AFC Chain Failures


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Exponential

1ln

1 ( )F t

3 46 7&8 9

AFC Chain Failures


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Weibull

BSL Chain Failures

3 4


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Exponential

1ln

1 ( )F t

3 46 7&8 9

BSL Chain Failures


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Weibull

Shearer Cable Failures

3 4


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Exponential

1ln

1 ( )F t

3 46 7&8 9

Shearer Cable Failures


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Weibull

Wh t i t b d ?


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What is to be done? Increase overall availability Minimise the time spent on PM

Decrease number of breakdowns More effective PM

Condition monitoring with long enough P-F time

Engineering changes

Design changes

Changes to the operating procedure

Decrease MTTR

Do all this while maximising the profit Learn lessons for next equipment purchase

The Reliability Function?


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Failed =6Failed=4Failed =2Failed =0

t=3t=1t=0 t=2

t=4 t=5 t=6 t=7

Failed=8 Failed=10 Failed =12 Failed=14

Is this best represented by a Weibull or an exponential distrubution?