ETAP - reliability assesment

© 1996-2009 Operation Technology, Inc. – Workshop Notes: Reliability Assessment
Reliability Assessment

Power System Reliability
Analysis
• Reliability
– The probability of a system performing its
function adequately for the period of time and
operation conditions intended
• Adequacy
– Sufficient facilities within the system to satisfy
customer demand
• Security
– Ability of the system to respond favorably to
disturbances arising within that system

Analysis
– Used in system planning and operation
– Reliability Assessment for:
• Generation station and generation capacity
• Composite generation and transmission system
• Distribution system
• Substation and switching stations
• Protection system

Analysis
• Various Indices to Measure Reliability
– Customer Interruption Frequency
– Customer Interruption Duration
– Customer Curtailment – Power/Energy Not
Served
– Reliability Worth Study
– Minimize Total Cost: Reliability Cost and
Consumer Interruption Cost

Distribution System Reliability
Analysis
• Concerned with availability and quality of power
supply at each customer’s service entrance –
Adequacy Analysis
• Statistics show that failures in distribution systems
contribute as much as 90% towards the
unavailability of supply to a load as compared with
other parts of electric power systems

Reliability Analysis Using ETAP
• Assesses distribution system reliability level for
radial and looped systems with a very efficient
algorithm
• Considers single and double contingencies
• Assesses reliability level for system and each load
point based on component failure model and
system configuration
• Performs sensitivity analysis to identify the optimal
location to make greatest improvement on system
reliability at minimum cost

Single & Double Contingencies
• Consider a system with two or more parallel
branches.
• In case of Single Contingency Analysis failure of
only one branch at a time is considered.
• In case of Double Contingency Analysis
simultaneous failure of two branches at a time is
also considered in addition to failure of one branch
at a time.

Single & Double Contingency
Example
Single Contingency:
• Failure of T5 or T6 at
a time is considered.
Double Contingency
• Simultaneous failure
of T5 & T6 and failure
of T5 or T6 at a time is
considered.

Component Model
A - Active Failure Rate (No of Failures/Year)
• Causes the operation of the protection devices around
the failed component, i.e. a short-circuit fault
• Failed component itself (and those components that are
directly connected to this failed component) restores to
service after repair or replacement
P- Passive Failure Rate (No of Failures/Year)
• Does not cause the operation of protection around the
failed component, i.e. an open circuit fault
• Failed component itself restores to service after repair
or replacement

Component Model
• Mean Time To Repair in hours (MTTR)
Time required to repair a component outage
and/or restore the system to its normal operating
state
• Mean Repair Rate (No of repairs per year) ( )
= 8760/MTTR
• Mean Time To Failure (years) (MTTF):
MTTF = 1.0/( A+ P)

Component Model
• Mean Time Between Failure (Year) (MTBF)
MTBF = MTTF + MTTR/8760
• Forced Outage Rate (Unavailability) (FOR)
FOR = MTTR/(MTBF*8760)
• Switching Time
– Time in hours for isolating a fault occurred at the
component
– Assume that CB/Fuse trip a fault instantaneously
• Time for replacing a failed element by a spare one,
in hours rP

Single-Component Concepts
• Two-State Model
– A two-state up/down representation is used for
the operation/repair cycle of a component (such
as lines, cables, transformers, breakers, fuses,
switches, loads and busbars)
DOWN
= ( A+ P) Up
Down
MTTF
……..
……..
MTTF
MTTR MTTR
UP

Model for Components in
Series/Parallel
Two Components in Series
sys
rr
sys
rrrr
sysr 2211
)
22
)(
11
(
2211
1, r1
Component 1 Component 2
21sys
2, r2

Model for Components in
Series/Parallel
Two Components in Parallel
1, r1
Component 1
Component 2
2, r2
)
21
(
21
2211
1
)
21
(
21 rr
rr
rr
sys
21
21
rr
rr
sysr

System Modeling
• Fault Current Interruption
– Only overcurrent protection devices (CB and
fuse) can interrupt fault current
– Fault current interruption is assumed to be
instantaneous
– Assumed to have no effect on components with
multiple source connection and isolated from
fault by CB/fuses

System Modeling
• Fault Isolation
– All switching devices can isolate faults. CBs and
fuses isolate fault instantaneously
– Switches isolate fault at switching time of the
faulted component
– Switching time for a load is equal to that of the
closest component

System Modeling
Normally Open Tie: Open tie PD can be closed
(switching time) to provide back up power
– Two terminal buses should be energized
– Can have several PDs connected in series and
with one or more open

Library for Reliability Analysis
• Component Reliability
– Data for each type of component - transformer,
bus, line, etc.
• Active Failure Rate
• Passive Failure Rate
• Repair Time
• Switching Time
• Replacement Time
• …
– Typical data from IEEE Standard

• Sector Customer Interruption Cost
– Standard Industrial Classification (SIC) is used
to divide customers into seven categories of
large user, industrial, commercial, agriculture,
residential, government & institutions and office
& buildings.
– Sector Customer Damage Functions (SCDF) are
interruption costs for several discrete outage
durations.

– A log-log interpolation of the cost data is used
where the interruption duration lies between two
separate times.
– If fault duration is outside the range, a linear
extrapolation with the same slope as that
between the two largest durations are used to
calculate the interruption cost.

Indices
Nej jei ,
Average Failure Rate at Load Point i, i(f/yr)
• e,j - The average failure rate of element j (or element
combination j, such as double contingency).
• Ne - The total number of the elements whose faults will
interrupt load point i.
Annual Outage Duration at Load Point i, Ui(hr/yr)
Nej ij
r
jei
U
,
• rij --Failure duration at load point i due to a failed element j.

Indices
Average Outage Duration at Load Point, ri(hr)
ii
U
i
r /
Expected Energy Not Supplied Index at Load Point, EENSi
(MWhr/yr)
iU
i
PiEENS Pi - the average load of load point i.
Expected Interruption Cost Index at Load Point, ECOSTi
(k$/yr)
Nej jeij
rf
i
P
i
ECOST
,
)(
The EENS and ECOST for a bus are calculated based on loads that
are directly connected to that bus due to the outage of that bus.
Where f(rij) is the SCDF.

Indices
Interrupted Energy Assessment Rate Index at Load Point,
IEARi ($/kWhr)
System Average Interruption Frequency Index, SAIFI
(f/customer.yr)
System Average Interruption Duration Index, SAIDI
(hr/customer.yr)
iEENS
iECOST
iIEAR
i
N
i
N
iSAIFI
servedcustomerofnumberTotal
onsinterrupticustomerofnumberTotal
Where N is the number of customers at load point i
i
N
i
N
i
U
SAIDI
durationsoninterrupticustomerofSum

Indices
Customer Average Interruption Duration Index,
CAIDI(hr/customer interruption)
Average Service Availability Index, ASAI(pu)
ii
N
i
N
i
U
CAIDI
onsinterrupticustomerofnumberTotal
sdurationoninterrupticustomerofSum
8760
8760
demandedhoursCustomer
serviceavailableofhoursCustomer
i
N
i
U
i
N
i
N
ASAI
Where 8760 is the number of hours in a calendar year

Indices
Average Service Unavailability Index, ASUI(pu)
System Expected Energy Not Supplied Index, EENS (MWhr/yr)
ASAIASUI 1
EENS = Total energy not supplied by the system = EENSi
System Expected Interruption Cost Index, ECOST(k$/yr)
ECOST = ECOSTi

Indices
Average Energy Not Supplied Index, AENS
(MWhr/customer.yr)
System Interrupted Energy Assessment Rate Index,
IEAR($/kWhr)
i
N
i
EENS
AENS
systemby thesuppliednotenergyTotal
EENS
ECOSTIEAR

Data:
Active failure rate for breakers: a = 0.003 failure/year
Passive failure rate for breakers: p = 0.002
failure/year
Failure rate for Bus, Utility: = 0.001 failure/year
MTTR for breakers: 30 hours
MTTR for buses, utility: 2 hours

Reliability Indices at LP1:
Failure rate for the main bus:
• CB1 fails actively OR passively
• CB2 and CB3 fail actively
• Utility fails
• Main bus itself fails
yearfailure
BusMainUtilityCBaCBaCBpaLP
/013.0
001.0001.0003.0003.0005.0
3211
The main bus would be de-energized if:

Annual unavailability for the main bus:
yearhour
MTTR
MTTRMTTR
MTTRMTTRU
BusMainBusMain
UtilityUtilityCBaCB
CBaCBCBpaCBLP
/334.0
001.02001.02003.030003.030005.030
33
22111
Time to replace the main bus:
hours
U
rLP 692.25
013.0
334.0
1

Results for the Single Contingency case

Results for the Double Contingency case

Calculations for the Double Contingency case:
For simplicity for hand calculations:
• Failure rates for the breakers connecting the transformers to the buses are
taken to be zero
• Failure rates of the two transformers are taken to be 1; MTTR = 200 hr.
Therefore the failure rate at Bus 2 due to double contingency:
yearfailures
rr
rr
double /0436681.0
8760
20012001
1
8760
)200200(11
8760
1
8760
)(
2211
2121

Calculations for the Double Contingency case:
Failure rate for the single contingency case:
Therefore total failure rate at Bus 2 :
yearfailures
gledoubleBus
/0546681.0011.00436681.0
sin2
yearfailures
A
CBBusMainBus
P
CB
A
CBUgle
/011.0
62111sin

ETAP - reliability assesment

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