j1bsen - hazard analysis

7/29/2019 j1bsen - Hazard Analysis

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The School of Chemical and Environmental Engineering

Hazard analysis


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Hazard analysis

Based on hazard identification

Determines likelihood of accidents to occurDone by one-two people

Time and effort consuming

Quantitative and qualitative techniques



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Quantitative hazard analysis

FrequencyBased on previous accident

experience or modeling

Provides quantifiablelikelihood of accident

Two processes in parallel

ConsequencesEffects of likely accident

Impact on personnel,

equipment, structures,

environment

Based on modeling,

accident experience or

judgment



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Quantitative hazard analysis

Various risk representations are possible

Risk to life classified in two groups:

Individual risk

Group risk

Frequencies and consequences of each event

combine to give a measure of overall risk



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Equipment Reliability

P(t) : probability that a component working at

t=0 has a life equal or less than t

R(t)= 1-P(t) : reliability of a component

Failure density function :

f(t)dt : probability of failure between t and t+dt

Probability of failure between t1 and t2:

dt

tdPtf

dttfttP t2

1121

,



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Equipment reliability

Hazard rate, H

Test interval, T

Demand rate, D

Failure rate, m

Fractional Dead Time, fdt (= Probability!)

PPT

Tfdt u



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Assume constant (average) failure rate: tetR mt

etPm1)(

tetf mm )(

t dttftR

0

exp m



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Therefore, probability of failure between t1

and t2:

Mean time to failure (MTF):

2121

21 ,tt

t

t

t eedtettP mm

0

1

mdtttfMTF



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Tu dttPT0

T dttPTfdt0

1

TeTPtPfdt mm 11

1)(

FormT small: and ttP m TPfdt m2

1



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Equipment reliability

Protective system that never fails: H=0

No protective system: H=D

Otherwise, hazard occurs during dead time:

H = D x fdt if D and fdt smallH = m(1-e-DT/2)



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Equipment reliablity

H = D x fdt is linearisation

H = m(1-e-DT/2

)If DT/2 is small: H = 0.5mDT

If DT/2 is large : H = m



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Occurrence estimation

Example: relief valve (failure remains hidden)

Test interval, T : 1 year

Fail-danger fault occurrence, m : 0.01/year

Demand rate : 1/year

Fractional dead time fdt = mT

Hazard rate H = 1 x 0.01 x x 1 = 0.005/year



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Hazard probability determination

Two common methods:

Fault tree analysis (FTA)

Event tree analysis (ETA)



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Fault tree analysis (FTA)

Sequence of events leading to a hazardous

incident

Start at fault and trace back all causes

AND gate: coincident inputs for system failure

OR gate: system failure occurs by failure of

any input

Originates in aerospace industry



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Fault tree analysis (FTA) Steps in a fault tree analysis

1. Define the top event PRECISELY (e.g. temperature in

reactor too high). explosion in process too vague2. Define the existing event. What are the present

conditions when top event occurs

3. Define unallowed events (e.g. natural disasters, wiring

failure,)

4. Define physical bounds of the process

5. Define equipment configuration (normal operation state,

valves open or closed,)



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Fault tree analysis (FTA) Steps in a fault tree analysis

6. Define the level of resolution (consider piece of

equipment or its components)7. Draw fault tree

8. Use AND and OR gates to define relation of events

9. Continue developing tree until all branches areterminated by basic, undeveloped or external events

10.Calculate probability of top event



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OR

Tank empty

Line broken

Fault tree analysisExample: car fails to start

Car fails

to start

No petrol

No spark

Engine not

turning

Other

Pump failed

Line blocked

Flat battery

Faulty distributor

Dirty plugs

Starter failed

Other

Battery too old

OR

OR

OR



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Logic rulesGate Input Pairing Output calculation Output Units

OR PA or PB P = PA + PB - PA xPB Probability

mA ormB m = mA + mB Frequency 1/time

PA

ormB Not permitted

AND PA and PB P = PA x PB Probability

mA and mB Convert mB to PB

PA and mB f = mA x PB Frequency 1/time



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Two protective systems in parallel

PA and PB fractional dead times

Demand rate A = DA = DDemand rate B = DB = PAD

Fractional dead time of system = 4/3PAPB

Hazard rate of system =4/3DPAPB

A

B

Demand rate D



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Occurrence estimationFor n redundant protective systems:r = n-m+1

With mnumber ofsystems that need

to work for protection

!! !1 rnr nr TPnnm



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Two protective systems in series

PA and PB fractional dead times

System is dead if A or B fails

Fractional dead time of system =

PA + PB - PAPB

Demand rate DA B



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Fault tree analysisExample: reactor overpressure

Overpressuring

of reactorAND

Failure of alarm

indicator

Failure of emergency

shutdown

Pressure switch 1 failure

P= 0.13

Pressure indicator

light failure P=0.04

Pressure switch 2 failure

P=0.13

Solenoid valve failure

P=0.34

OR

OR



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Fault tree analysisAdvantages

Begins with top event (event of interest)

Enormous insight into path to incidentCan be computerised

Disadvantages

Can become enormous

Uncertainty and subjective structure

Failure probabilities not always known



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Event tree analysisStart at an initiating event and look at the

consequence

Time sequence of event propagation is embedded

Each event depends on the precursor event

Outcomes of precursor event are either binary

(Yes/No) or multiple

Commonly used to gauge effect of release of

material



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Event tree analysis: Example



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Event tree analysis:

Technique description:

Identify the initiating event (can come from outcome

of FTA)Identify the safety function/hazard factor and

determine outcomes

Construct the event treeClassify the outcome

Estimate probability of an event tree branch

Quantify the outcomes



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Risk of a process

Risk of a complete

process assessed by:

Quantitative Risk

Analysis (QRA)

Layer Of Protection

Analysis (LOPA)

Chemical Process Safety, D.A. Crowl and J.F. Louvar, 2001



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Quantitative Risk Analysis

Major steps

1.Define potential event sequences and incidents

2.Evaluate incident consequences (toxic cloud

dispersion modelling, fire & explosion modelling)

3.Estimate incident frequencies (ETA, FTA)

4.Estimate incident impacts on people, environment

and property

5.Estimate risk by combining impact and frequencies

(graph are commonly used)



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Layer Of Protection Analysis

Semi-quantitative tool for analysing and

assessing risk

Simplified method to characterize

consequence and estimate risk

Frequency of undesired consequences arereduced by added layers of protection

Determination whether sufficient layers of

protection exist against certain scenarios



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Method:

Consequences and effects are approximated

by categories

Frequencies of consequences and

effectiveness of protection layers areapproximated



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Major steps

Identify a single consequence

Identify accident scenario and associated single causeIdentify initiating event for the scenario and estimate initiating

event frequency

Identify protection layers (PL) and their failure probability on

demandCalculate frequency of consequence

Estimate risk from consequence frequency and magnitude

Evaluate acceptability of risk



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Frequency of a consequence endpoint:

with fiI : initiating event i frequency

Pij : probability of failure of the jth PL after initiating

event I

fic : consequence frequency for initiating event i

i

jij

I

i

C

i Pff1



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Qualitative Risk Analysis


Use of experience

and past data to

express likelihoodand severity

Can be made asdetailed as is required


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Human errors



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Human errors



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Acceptable risk

If property loss and possible number of fatalities can

be predicted, quantitative assessment can be made:

Quantitative risk assessment = Incident frequency

x loss per incident

Loss per incident can be compared to cost of safetyequipment

One measure of risk is Fatal Accident Frequency Rate

(FAFR): # of fatal accidents per 108 working hours = #

of fatal accidents per 1000 (wo)men in their work life


Th S h l f Ch i l d E i t l E i i


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Acceptable risk ?

If FAFR > 0.4: priority elimination (general rule for

employees in process industry)

Hazard generally acceptable if FAFR


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Non-occupational risks

Risk/(person year)

Cancer 280 x 10-5

Road accidents (UK) 10 x 10-5

All accidents (UK) 30 x 10-5

Smoking (20cig/day) 500 x 10-5

Drinking (1btl wine/day) 75 x 10-5

All risks man 20yr 100 x 10-5

Lightning 10-7

Nuclear release 10-7

Meteorite 10-11


Th S h l f Ch i l d E i t l E i i


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ALARP

As Low As Reasonably Practicable

Comparison of level of risk to cost of lowering the risklevel


The School of Chemical and En ironmental Engineering


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ALARP




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ALARP

FAFR Risk/(person year)

Max tolerable limit

Employees

PublicPublic (nuclear)

50 10-3

10-4

10-5

Legal limit ionizing

radiation

75 1.5 x 10-3

Max tol. risk ionizing

radiationEmployees 20-25 5x10-3

Broadly acceptable risk

(public + employees)

0.05 10-6

Negligible risk (public +

employees)

0.005 (?) 10-7 (?)




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Hazard analysis


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