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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