Probabilistic Calculations

22.39 Elements of Reactor Design, Operations, and

Safety

Lectures 10-11

Fall 2006

George E. Apostolakis Massachusetts Institute of Technology

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General Formulation

X T = φ ( X 1 ,…X n ) φ ( X )

X T 1

N

( 1

1

M i )

N

M i 1

N N 1 N

N 1 N

X T M i

M i M j

... ( 1 )

M i

i 1

i 1 j i 1

i 1

X T : the TOP event indicator variable

M i : the ith minimal cut set or accident sequence

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TOP-event Probability

P X

N P M

1 N 1 P N M

T i

i

1 1

N

P X T

P M i

1

Rare-event approximation

The question is how to calculate the probability of M i

P ( M i )

P ( X i

... X i )

P A

B

k

m

P A B P B

Co nditional probability:

I n d e p e n d e n t e v e n ts :

P A

B

P A

P ( A B ) = P ( A ) P ( B )

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MinCutSet Probability

P ( M ) P ( X 1 X 2 X 3 )

P ( X 1 ) P ( X 2 X 3

/ X 1 )

P ( X 1 ) P ( X 2

/ X 1 ) P ( X 3 / X 1 X 2 )

For independent events:

P ( M )

P ( X 1 X 2 X 3 )

P ( X 1 ) P ( X 2 ) P ( X 3 )

For accident sequences, we must incl ude the initiating-event frequency per year:

fr ( M ) fr ( IE X 1 X 2 ) fr ( IE ) P ( X 1 X 2 / IE ) fr ( IE ) P ( X 1 / IE ) P ( X 2 / IE X 1 )

fr X T

CDF

N

fr

1

M i

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Example: 2-out-of-4 System

1

2

3

4

M 1 = X 1 X 2 X 3 M 2 = X 2 X 3 X 4 M 3 = X 3 X 4 X 1 M 4 = X 1 X 2 X 4

X T = 1 ( 1 M 1 ) (1 M 2 ) (1 M 3 ) (1 M 4 )

X T = (X 1 X 2 X 3 + X 2 X 3 X 4 + X 3 X 4 X 1 + X 1 X 2 X 4 ) - 3 X 1 X 2 X 3 X 4

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2- out-of-4 System (cont’d)

P( X T = 1) = P(X 1 X 2 X 3 + X 2 X 3 X 4 + X 3 X 4 X 1 + X 1 X 2 X 4 )

3P(X 1 X 2 X 3 X 4 )

Assume that the components ar e i ndependent and nominally identical with failure probability q. Then,

P( X T = 1) = 4q 3 3 q 4

Rare-event approximation: P( X T = 1) 4q 3

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Overview

We need models for:

The frequency of initiating events.

The probabili ty that a component will fail on demand.

The probabili ty that a component will run for a period of tim e given a successful start.

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Initiating Events: T he Poisson Distribution

Used typically to model the occurrence of initiating events.

Discrete Random Variable: Number of events in (0, t)

The r ate λ is assumed to be constant; t he events are

independent.

The probability of exactly k events in (0, t) is (pmf):

( t ) k

e

Pr[ k ]

t

k !

k! 1*2*…*(k-1)*k 0! = 1

m λ t

σ 2 λ t

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Example of the Poisson Distribution

A component fails due to "shocks" that occur, on the average, once every 100 hours. What is the probability of exactly one replacement in 100 hours? Of no replacement?

λ t = 10 -2 *100 = 1

Pr[1 repl.] = e - t = e -1 = 0.37 = Pr[no replacement]

Expected number of replacements: 1

Pr[ 2 repl ]

1 1 2

e

2 !

e 1

2

0 . 185

Pr[k 2] = 0.37 + 0.37 + 0.185 = 0.925

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Reliability and Availability

Reliability : Probability of successful operation over a period (0, t).

Availability : Probability the item is working at time t.

Note :

In industrial applications, the term “reliability” includes the probability that a safety system wi ll start successfully and operate for a period (0, t).

The term “unavailability” usually refers to maintenance.

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Failure while running

T: the time to failure of a component (continuous random variable).

F(t) = P[T t]: failure distribution (unreliability)

R(t) 1-F(t) = P[t T]: reliability

m: mean time to failure (MTTF)

f(t): failure density, f(t)dt = P{failure occurs between t and t+dt} = P [t T t+dt]

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The Hazard Function or Failure Rate

h t

f t

f ( t )

t

R t

1 F ( t )

F t

1 ex p h s ds .

0

The distinction between h(t) and f(t) :

f(t)dt: unconditional probability of failure in (t, t +dt), f(t)dt = P [t T t+dt]

h(t)dt: conditional probability of failure in (t, t +dt) given that the component has survived up to t.

h(t)dt = P [t T t+dt/{ t T}]

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The “Bathtub” Curve

I

II

III

h(t)

0 t t t

1 2

I I nfant Mortality

II Useful Life

III Aging (Wear-out)

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The Exponential Distribution

f(t) =

λ e λ t

λ 0 t 0 (failure density)

F ( t )

1

e λ t

R ( t )

e λ t

h(t) = λ constant (no memory; the only pdf w ith

this property) useful life on bathtub curve

F(t)

λ t for

λ t 0.1 ( another rare-event approximation)

m 1

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Example: 2-out-of-3 system

Each sensor has a MTTF equal to 2,000 hours. What is the unreliability of the system for a period of 720 hours?

Step 1: System Logic.

X T = ( X A X B + X B X C + X C X A ) - 2 X A X B X C

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Example: 2-out-of-3 system (2)

Step 2: Probabilistic Analysis.

For nominall y identical components:

P ( X T ) = 3q 2 2 q 3

q F ( t ) 1 e t

5 x 10 4 hr 1

System Unreliability:

F T ( t )

3 1

e t 2

2 1

e t 3

Rare event approximation:

F T ( t )

3 ( t ) 2

2 ( t ) 3

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A note on the calculation of the MTTF

Proof

MTTF

R ( t ) dt

0

MTTF

tf ( t ) dt

0

t (

0

dR

dt

) dt

tdR

0

0

tR

R ( t ) dt

0

R ( t ) dt

0

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A note on the calculation of the MTTF (cont.)

since

and

f ( t ) dF

dt

d ( 1 R )

dt

dR

dt

R ( t

) 0

faster

than

t

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MTTF Examples

Single exponential component:

t 1

MTTF e

1

0

dt

Series system:

MTTF dte m t

0

m

1

system

1- out-of-2 system :

2- out-of-3 system :

MTTF

dt 2 e t

0

e 2 t 3

2

MTTF R T

( t ) dt [ 1

3 ( 1 e t ) 2

2 ( 1 e t ) 3 ] dt 5

6

0 0

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MTTF Examples: 2-out-of-3 System

Using the result for F T (t) on slide 15, we get

MTTF R T ( t ) dt [ 1 3 ( 1 e t ) 2 2 ( 1 e t ) 3 ] dt

0 0

MTTF

1 1

2 3

6

5

The MTTF for a single exponential component is: 1

The 2-out-of-3 system is slightly worse.

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The Weibull f ailure model

W e ib u l l H a z a r d Ra t e Cu r v e s

0 . 0035

0. 003

0 . 0025

0. 002

0 . 0015

0. 001

0 . 0005

0

b= 0 . 8 b= 1 . 5

b= 1 . 0 b= 2 . 0

0 200 40 0 600 80 0 1000

Adj u sting the value of b, we can model any part of the bathtub curve.

h ( t )

b b t b 1

R ( t )

e t b

For b = 1 the exponential distribution.

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The Model of the World

Deterministic , e.g., a mechanistic computer code

Probabilistic (Aleatory) , e.g., R(t/ ) = exp(- t)

Both deterministic and aleatory models of the world have assumptions and parameters.

How confident are we about the validity of these assumptions and the numerical values of the parameters?

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The Epistemic (state-of-knowledge) Model

Uncertainties in assumptions are not handled routinely. If necessary, sensitivity studies are performed.

Parameter uncertainties are reflected on appropriate probability distributions.

For the failure rate: π ( λ ) d λ = Pr(the failure rate has a value in d λ about λ )

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Unconditional (predictive) probability

R ( t )

R ( t

/ )

( ) d

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Communication of Epistemic Uncertainties: The discrete case

Suppose that P( = 10 -2 ) = 0.4 and P( = 10 -3 ) = 0.6 Then, P(e -0.001t ) = 0.6 and P(e -0.01t ) = 0.4

R(t) = 0.6 e -0.001t + 0.4 e -0.001t

1.0

exp(-0.001t)

0.6

exp(-0.01t)

0.4

t

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Communication of Epistemic Uncertainties: The continuous case

Co urtesy o f US NRC.

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The lognormal distribution

It is very common to use the lognormal distribution as the epistemic distribution of failure rates.

1

(ln ) 2

2

( ) exp 2 

2 2

m exp

2

95

05

e 1 . 645

e 1 . 645

median

: 50

e

EF

95

50

95

05

50

05

Y ln

Y is normally distributed with mean μ a nd standard deviati o n σ

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Co urtesy o f US NRC.

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28

Co urtesy o f US NRC.

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SIMPLIFIED SYSTEM DIAGRAM

Co urtesy o f US NRC.

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HIGH PRESSURE INJECTION DURING LOOP 1-0F-3 TRAINS FOR SUCCESS

Co urtesy o f US NRC.

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HPIS Analysis (1-out-of-3)

In the RSS HPIS, the three pump trains have a common suction line from the RWST. The South Tex a s Pr oject design has separate suction lines for the three trains, as the fault tree shows.

Q total = Q singles + Q doubleFail’s + Q test&maint + Q CCF

Representativ e singl e failure s (single-elemen t mcs) :

Check valve SI 225 fails to open

Check valve SI-25 fails to open

RWST discharge line ruptures

Other

Q singles = 1.1x10 -3 (“point estimate”)

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HPIS: Double Failures

Representativ e doubl e failure s (double-elemen t mcs) :

RWST supply MOVs 1115B and 1115D fail to open

Born Injection Tank (BIT) inle t MOVs 1867A and 1867B fail to open

BIT discharge MOVs 1867C and 1867D fail to open

Service water pumps; cooling water pumps; BIT cooling system

other

Q(MOVs 1867C and 1867D fail to open) = P(X 1 ) P(X 2 ) where P(X i ) is a lognormal with median 1.9x10 -2 and EF = 3

Q doubleFail’s = 2.5x10 -3 (“point estimate”)

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HPIS: Other Contributions

Q test&m aint is negligible because of the 1-out-of-3 redundancy (if one train is out, double failures must occur for the system to fail).

Q CCF (MOVs 1867C and 1867D fail to open) = β P(X 1 ) =

= 0.075x1.9x10 -2 = 1.4x10 -3

Monte Carlo simulation yields (RSS):

Q total,median = 8.6x10 -3

Q total,upper = 2.7x10 -2

Q total,lower = 4.4x10 -3

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In some important cases, CDF and LERF cannot be calculated.

Decision Options

Risk- Informed Decision

Expert Panel Delibera tion

Impact on CDF and LERF

SSC

Categories Based on Importance Measures

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Fussell-Vesely Importance Measure

Pr[ M ( i ) ]

0 i i

FV i

k k R R R 0 R 0

1 R

R 0

R 0 The base-case risk metric (CDF or LERF) =

Pr[ M k ]

k

M

( i ) k

R i

The k th accident sequence containing event i

The risk metric (CDF or LERF) with the i th component up (unavailability equal to zero)

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Risk Reduction Worth (RRW)

R 0

RRW

R 0 R i

i R i

R i 1

R

F V i 0

1

R 0

1

RR W i

FV i is the fractional decrease in th e risk metric when event i is always true (component i is alw ays available; its unavailability i s set equal to zero).

This importance measure is parti c ularly useful for identifying improvements to the reliability of elements which can most reduce risk.

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F-V Ranking

Lo ss Of O f fsite Po wer Initiating Event 0.831

DIESEL GENERATOR B FAILS 0.437

DIESEL GENERATOR A FAILS 0.393

COM M ON CAUS E FAI L URE OF DI E S EL GENERATORS 0.39

OPERATOR FAILS TO RECOVER OFFSITE POW E R (SEAL LOCA) 0.388

RCP SEALS FAIL W/O COOL ING AND INJECTION 0.344

OPERATOR FAILS TO RE COVER OFFSITE POW E R

BEFORE BATTERY DEPLETION 0.306

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Risk Achievement Worth (RAW)

RAW i

R i

R 0

R +i The risk metric (CDF or LERF) with the i th component always down (its unavailability is set equal to 1)

RAW presents a measure of the “worth” of the basic event in “achieving” the present level of risk and indicates the importance of maintaining the current level of reliability for the basic event.

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RAW Ranking

Loss Of Offsite Power Initiating Event

51,940

Steam Generator Tube Rupture Initiating Event

41,200

Small Loss Of Coolant Accident Initiating Event

40,300

CONTROL ROD ASSEMBLIES FAIL TO INSERT

3,050

COMMON CAUSE FAILURE OF DIESEL GENERATORS

271

RPS BREAKERS FAIL TO OPEN

202

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Comments on Importance Measures

Importance measures are typically evaluated for individual SSCs, not groups.

T he various categories of risk significance are determined by defining threshold values for the importance measures. For example, in some applications, a SSC is in the "high" risk-significant category when FV > 0.005 and RAW > 2.0.

Importance measures are strongly affected by the scope and quality of the PRA. For example, incomplete assessments of risk contributions from low- power and shutdown operations, fires, and human performance will distort the importance measures.

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