Transcript Improving Electrical System Reliability

Maintenance
Strategies with Limited
Resources
© 2005 Eaton Corporation. All rights reserved.
Goal

We will learn if you have reliable electrical power
systems

Why that is important

What to do if you don’t

Learn a financial method to rank alternative
methods of improving reliability
Agenda

Where We Are Today

What Can Be Done?

IEEE 493 Process (Identify Candidates)

Predicting Failures

Economic Analysis / Prioritization

Solutions Available
Where We Are Today
Where We Are Today

Fact # 1:
The aggregate economic loss of electrical power
disruptions has climbed to more than $100 billion per
year or more than 1% of U.S. Gross Domestic
Product!

Recent events have demonstrated the fragility of our aging power
grid. With transmission networks operating close to their stability
limits, minor faults can cause cascading outages. Capacity
limitations in several regions can lead to economic losses that
cascade through the economy, causing loses for not only
residential, but also commercial and industrial institutions.
Where We Are Today

Fact # 2:
The recent power outages have been in the works for
the last several years!

U.S utilities have always taken pride in their uptime and system
performance. Over the past several years, due to industry-wide
deregulation, market pressures for rate reductions, business
restructuring and downsizing, overall investment in infrastructure
has not been at traditional levels. The decoupling of transmission,
distribution and generation has caused disruption in traditional
business models and industry workings (i.e. the vertically
integrated utility no longer exists in deregulated markets)
Where We Are Today

Fact # 3:
De-regulation has contributed to loss of stability.

When the Federal Energy Regulatory Commission dictated that
the electrical transmission network was to be opened to the free
market, it allowed anyone to transmit power over any
transmission line. This allowed generators outside a customer’s
service area to bid on a distant customer’s power requirements
and be guaranteed access to that customer over the transmission
system. As a result, the owners of the transmission lines lost
some of their ability to maintain stability since these lines now
carry power generated outside their control. As demand has
continued to increase at an average 2% per year and because
few will accept new transmission lines in their backyard, grid
stability continues to degrade over time.
Where We Are Today

Fact # 4:
The 2000 dotcom implosion and the resultant
relaxation of electrical demand has temporarily
relaxed the demands placed on electrical
transmission system, but the problem remains.

The recent retraction of the economy has reduced electrical
demand temporarily. However little or no new transmission was
constructed during that economic downturn. Electrical demand is
again returning to historic levels. We can only expect the
problem to return and even become worse as the economy
expands to more robust levels.
Conclusion

This problem will worsen before it improves

Universities must take action themselves
Agenda

Where We Are Today

What Can Be Done?

IEEE 493 Process (Identify Candidates)

Predicting Failures

Economic Analysis / Prioritization

Solutions Available
What Can Be Done?
A. Protect Yourself From External Problems
1. Install Local Backup Power
•
Natural Gas, LPG, Diesel from 1 kW to over 10 MW
2. Install Voltage Correction
•
•
•
SAG Correction Equipment
UPS
Capacitors
3. Engineering Review
•
•
•
•
Audits
Site Supervision
Equipment Commissioning
Turnkey Installation
What Can Be Done?


Many university outages are not caused by loss of utility
power, but rather by internal problems
Protect yourself from internal problems

Equipment failure, accelerated by:
•
•
•
•
•

Dust, dirt, moisture, rodents, etc.
Thermal cycling, vibration induced loosening, etc.
Obstruction of ventilation, etc.
Operator error
Reduction in funding for preventative maintenance
But when resources are tight, where should they be
spent to give maximum uptime?
What Can Be Done?
B. Protect Yourself From Internal Problems
1. Bring in an expert
2.
•
Reliability Study
•
 Site Survey (e.g. evaluation of on-site generation, UPS,
calculation of reliability of existing system, etc.)
 Thermography
Coordination Study
•
Harmonics Study
Install Predictive Diagnostic equipment
•
Early warning of pending failure in MV Equipment
Agenda

Where We Are Today

What Can Be Done?

IEEE 493 Process (Identify Candidates)

Predicting Failures

Economic Analysis / Prioritization

Solutions Available
IEEE 493 Process
1. Establish Current Condition of Facility
2. Determine Likelihood of Serious Problem Based on
this Condition
3. Sort to Find Equipment Most at Risk to Cause Problems
4. Identify the Predictive Techniques that Gives
Early Warning of Problems at that Equipment
What Is “Current Condition?”

The ‘Quiet Crisis’
Term created by Paul Hubbel,
Deputy Director, Facilities and Services,
Marine Corps. Government Executive Magazine, Sept 2002.
When he was asked “why isn’t preventative
maintenance adhered to more closely in government
facilities?”
“We call it the ‘quiet crisis’ because a lot of maintenance
problems take time to occur and are not noticed until
damage occurs”.
Current Condition?


This leader says many follow the “fix when broken” but
not before mentality?
What happens if the
power goes out at your
facility for an extended
period of time?
For example, we would expect that
facilities (such as universities & hospitals)
won’t have power outage problems!
06-AUG-03 A power failure forced New York University Medical
Center to shut its emergency room and turn away visitors and
some patients yesterday, as the hospital staff struggled with limited
use of the air-conditioning, computers and other equipment. Chris
Olert, a spokesman for Consolidated Edison, said, "We know that it
was a combined failure of some of our equipment and some of the
hospital's equipment, but we don't know exactly what triggered
it.“ He said a cable feeding electricity to the medical center was
damaged and had to be bypassed. When the power failed, the
hospital's backup generators automatically turned on, but they
could not carry the entire load, so hospital officials shut down some functions to preserve
electricity for the most crucial ones. "No critical services were affected," said Lynn Odell, a
hospital spokeswoman. But hospital employees interviewed outside the building and family
members of some patients said things were seriously disrupted for a time.
One worker told of a darkened pharmacy with dormant computers, where pharmacists using
flashlights filled out paperwork by hand and responded to orders by telephone rather than
computer. Another spoke of a stiflingly hot surgery department where some medicines
spoiled in a nonworking refrigerator. (New York Times 8-6-2003)
Fairview Southdale Hospital, MN
21-SEP-02 …in the case of Fairview Southdale Hospital's
power failure, which lasted an hour, a power line wound up
carrying more than its normal load for four days, which melted a fuse. That set off a chain of
events that wound up overloading the hospital's emergency generators, causing them
to fail, too. Mark Enger, the hospital's president, said the power failure could have been
life-threatening. The power dropped but didn't go out entirely. It dropped enough, however,
to trigger the hospital's three emergency generators. But because of the way the system is
wired, the generators' cooling fans failed to work, the generators overheated and the
hospital lost all power at about 11 a.m. Enger said there were eight surgeries going
when the power failed. Four of the surgeons were able to finish their operations, while the
other four finished quickly and re-scheduled the surgeries for the next day. Partial power
was restored by noon, and full power was restored the next day, Enger said. Xcel is
adamant that its maintenance practices are not posing widespread service problems. But
one NSP worker said the hospital's power failure is but a symptom of the electrical grid's ill
health. "Their work force is stretched so thin, they're putting people in jeopardy," he said..
(St. Paul MN, Pioneer Press 8-6-2003)
Rhode Island Hospital
On April 16, 2002, Rhode Island Hospital and Women &
Infants Hospital lost power for an afternoon, leaving
various parts of the 33-building campus dark for varying
amounts of time. At Women & Infants, an emergency
generator immediately turned on, emergency lights came
on, and no essential services were disrupted, according to
a Providence Journal article. However, at Rhode Island Hospital, the backup electric
system did not work, prompting Providence Mayor Vincent A. Cianci Jr., to tell the
Providence Journal, “A hospital of this magnitude and this size should not have these
problems.” Surprisingly, power outages do happen with alarming frequency to big hospitals.
In fact, they’ve happened at Rhode Island Hospital campus before. In September 1999,
a blackout plunged the entire campus into darkness. The backup systems failed once
again and this time a patient died after his respirator failed. Then, in January 2000,
another power failure forced the hospital to rely on backup generators for nearly two hours
and shut down nonessential equipment and lights. A faulty ceramic insulator at a
substation on the hospital campus caused the failure. Then a damaged coil prevented
some of the backup power from flowing back into one of the hospital buildings. (EC&M
Magazine 8-1-2002)
Current Condition?





Why would there be such problems in critical tested
systems?
Budget Cuts / Management Redirection of Maintenance
Funds?
This results in “Crisis Mode Operation” or “Fix What’s
Broken and Skip the Rest” mentality
But if you operate this way, how do you guess what will
break next and where money should be targeted?
Is there an analytical way of targeting scarce
resources?
IEEE 493 Step 1
1. Establish Current Condition of Facility
2. Determine Likelihood of Serious Problem Based on
this Condition
3. Sort to Find Equipment Most at Risk to Cause Problems
4. Identify the Predictive Techniques that Gives
Early Warning of Problems at that Equipment
IEEE 493 Step 1

What is the likelihood of a loss of MV
power at a particular point in the
university?

How many sources of supply

Are there one or more single-points-of-failure?

What is the likelihood of those points failing?

Use algebra to combine probabilities
IEEE 493 Step

For example, if power flows from utility like this:
Utility
Switch
Breaker
Load
IEEE 493 Step

For example, if power flows from utility like this:
Utility
Switch
Breaker
99.9%
99.99%
99.99%
Load
IEEE 493 Step

For example, if power flows from utility as below:
Utility
99.9%
(8.7 hr/yr)

Switch
x
99.99%
x
(0.87 hr/yr)
Load
Breaker
99.99%
(0.87 hr/yr)
=
99.88%
(10.5 hr/yr)
Overall reliability is poorer than any component
reliability
Finding “Downtime Per Year”

Downtime / Year

Made up of two components
•
How often failures occur
•
 Measured as Mean-Time-Between-Failures
 MTBF
 e.g. 80000 hours
How long a failure knocks you out
 Measured as Mean-Time-To-Repair
 MTTR
 e.g. 8 hours
Finding “Downtime Per Year”
Multiply together to determine downtimeper-year


How often : e.g.. 1 failure every 80000 hours

How long : e.g.. 8 hours each failure
8
8 hrs
length
of
outages
(MTTR)
hours
0
1
2
3
4
5
6
7
frequency of outage
(MTBF) years
8
9 80008 hrs
80000 hrs
Finding “Downtime Per Year”
Multiply together to determine downtimeper-year


How often : e.g.. 1 failure every 80000 hours

How long : e.g.. 8 hours each failure
8
8 hrs
length
of
outages
(MTTR)
hours
How thick is this lower line, if we
take the same probability and
spread it over 80000 hours?
0
1
2
3
4
5
6
7
frequency of outage
(MTBF) years
8
9
80000 hrs
Finding “Downtime Per Year”

Multiply together to determine downtimeper-year


How often : e.g.. 1 failure every 80000 hours
How long : e.g.. 8 hours each failure
hr
1

MTTR*
y ear
MTBF
hr
Failure
*
Failure y ear
8 hr
*
Failure

hr
y ear
Failure
hr
 0.876
1 y ear
y ear
80000 hr *
8760 hr
IEEE 493 Step 2
1. Establish Current Condition of Facility
2. Determine Likelihood of Serious Problem Based on
this Condition
3. Sort to Find Equipment Most at Risk to Cause Problems
4. Identify the Predictive Techniques that Gives
Early Warning of Problems at that Equipment
IEEE 493

Likelihood of failure of electrical equipment


Low
Consequences of failure of electrical equipment

High
Substation Transformer
Substation Transformer
Consequences of Failure


After certain failures (such as that transformer
failure), power cannot be restored for many
hours
Because of the large amount of energy within
electrical equipment, that equipment can fail
explosively

This may cause mechanical damage which limits
ability to quickly repair and restore equipment to
service
Equipment Destruction
Forensic Analysis of
Equipment Destruction
Equipment Destruction
Worker Injury
Equipment Destruction
Equipment Destruction
Equipment Destruction
Consequences

This equipment isn’t going to be repaired in a few
hours, or even a few days.
IEEE 493

Likelihood (Failures / yr)


Low
Consequences (Hrs / failure)

High
IEEE 493-1997 (Gold Book)
Analysis
IEEE Std 493-1997, Table 7-1
Product
Failures/yr Hours/Failure
Prot. Relays
.0002
5
Low
likelihood
High
consequences
Note: could be higher, but protective relays are usually
installed in draw-out cases, so the delay in repairing is
due to:
1) diagnosing the problem
2) locating a spare
3) installing a spare
4) reprogramming setpoints to match previous unit
5) restarting system
IEEE 493-1997 (Gold Book)
Analysis
IEEE Std 493-1997, Table 7-1
Failures/yr * Hours/Failure = Hours/Yr
Product
Failures/yr Hours/Failure Hours/Yr
f(x) %
Prot. Relays
.0002
99.999989
5
.001

hr 
 8760  
yr 

f (x) 
8760

hr 
 8760  0.001 
yr 

f (x) 
 0.99999989  99.999989
8760
IEEE 493-1997 (Gold Book)
Analysis
IEEE Std 493-1997, Table 7-1
Failures/yr * Hours/Failure = Hours/Yr
Product
Failures/yr Hours/Failure Hours/Yr
f(x) %
Prot. Relays
.0002
5
.001
99.999989
LV Swgr Bkrs
.0027
4
.0108
99.99988
MV Swgr Bkrs
.0036
2.1
.0076
99.99991
83.1*
.2992
99.9966
LV Cable (1000 ft)
.00141
10.5
.0148
99.99983
MV Cable (1000 ft)
.00613
26.5
.1624
99.9981
Disc. Switches .0061
3.6
.022
99.9997
Transformer
.003
342
1.026
99.988
LV Swgr Bus
.0024
24
.0576
99.9993
MV Swgr Bus
.0102***
26.8
.2733
99.9969
* when no on-site spare is available
** below ground *** 3 connected to 3 breakers
IEEE 493 Step 1
SW1
CBL1


TX1
CBL2
Example:
Likelihood of failure at an
ICU =

BKR1
52
51
RLY1
BUS1

BKR2
52
51
BKR3
52
51
BKR4
52
51
BKR5
52
51
RLY2
RLY3
RLY4
RLY5
CBL3
CBL4
CBL5
CBL6
ICU
f(SW1) x f(CBL1) x f(TX1) x
f(CBL2) + f(BKR1) x f(RLY1) x
f(BUS1) x f(BKR5) x f(RLY5) x
f(CBL6)
f(…) means probability of failure
of that component
ICU Failure Scenario
f(Utility) + f(SW1) + f(CBL1)+ f(TX1)+ f(CBL2) + f(BKR1)+ f(RLY1)+
f(BUS1) + f(BKR5) + f(RLY5) + f(CBL6)
SW1

f(Utility) = 8.76 hrs/yr (99.9% )
CBL1
TX1

f(SW1) = .022 hrs/yr (99.9997%
CBL2

f(CBL1300ft) = 300/1000 * 0.1624 = 0.049 hrs/yr (99.9994%)
BKR1 52
51 RLY1

f(TX1)= 1.026 hrs/yr (99.988%)

f(CBL2100ft) = 100/1000 * 0.1624 = 0.0162 hrs/yr (99.9998%) BUS1
BKR5
51
52

f(BKR1), (BKR5) = .2992 hrs/yr (99.9966%)
RLY5
CBL6

f(BUS1) = .2733 hrs/yr (99.9969%)

f(RLY1), (RLY5) = .001 hrs/yr (99.999989%)

f(CBL6300ft) = 300/1000 * 0.1624 = 0.049 hrs/yr (99.9994%)

Total = 99.9%Utility x 99.9997%SW1 x 99.9994%CBL1 x 99.988%TX1 x
99.9998%CBL2 x 99.9966%BKR1 x 99.999989%RLY1 x 99.9969%BUS1 x
99.9966%BKR5 x 99.999989%RLY5 x 99.9994%CBL6
= 99.88% (10.8 hrs/yr)
ICU Failure Scenario




Obviously, this is completely unacceptable
reliability, but our example isn’t exactly accurate
An ICU will always be connected to a critical
backup bus fed from a generator
It may or may not have redundant transformers
How do we recalculate with a generator and
transfer switch in the system?
Utility Only Reliability
99.9%
UTILITY
99.9997%
99.9994%
SW1
CBL1
99.988%
TX1
99.9998%
CBL2
52
99.9966%
51
99.999989%
RLY1
BKR1
BUS1
99.9969%
BKR5
52
51
RLY5
CBL6
ICU
Reliability at this point
= f(UTILITY) x f(SW1) x
f(CBL1) x f(TX1) x f(CBL2) x
f(BKR1) x f(RLY1) x
f(BUS1)
= 99.88%
Adding Generation
GEN
GEN1
f(GEN1) =
99.9%
f(GEN1) x f(CBLG11000ft)
= 99.9% x 99.9998%
= 99.8998%
Critical Bus
CBLG11000ft
f(CBLG11000ft) =
99.9998%
Combining Busses
99.88%
Normal Bus
SWN1
99.8998%
Critical Bus
SWG1
Combining Busses
99.88%
Normal Bus
f(SWN1) = 99.9997% SWN1
99.8998%
Critical Bus
SWG1
f(Normal Bus x f(SWN1)
= 99.88% x 99.9997%
= 99.8797%
What is the reliability at this point?
f(SWG1) = 99.9997%
f(Critical Bus x f(SWN1)
= 99.8998% x 99.9997%
= 99.8995%
Revisit Normal-Bus Reliability
Reliability 
TotalTime  UnavailableTime
TotalTime
AvailableTime
UnavailableTime
Available
Not Running
Running
Unavailable
TotalTime
“Reliability”
Reliability 
8760 hrs  x hrs
 0.9987  99.87%
8760 hrs
x  8760  8760 * 99.87%  11.4 hrs
AvailableTime =
8760-11.4 =
8748.6 hrs =
UnavailableTime =
11.4 hrs =
Available
Running
8757.3 hrs/yr
Unavailable
TotalTime = 8760 hours = 1 year
Not Running
11.4 hrs/yr
Transferring to a less than
reliable source?
8760 hrs
Source 1
(Utility)
99.9% chance will switch
to a source that is 99.9%
reliable within 10 seconds
0.1% chance it won’t switch
and outage last 11.4 hours.
11.4 hrs
(utility outage)
= Unavailable Time
= 0.999*(10)
+ (0.001*11.4*60*60)
= 51.03 sec
= 0.0142 hr
Source 2
(Generator)
Total Time = S1A+S1U+S2A+S2U
= 8760 + 0.0142 + 11.4 + 0.00889
= 8771.423 hr
Source 1
(Utility)
99.9% chance that will
switch in 1 second
to a source that is 99.87%
reliable.
0.1% chance of not switching
and resulting in an outage that
(according to the Gold Book)
averages 8.6 hours.
= Unavailable Time
= (0.999*1 sec) +
(0.001*8.6*60*60)
= 32 sec
= 0.00889 hr
Combined Reliability
TotalTim e UnavailableTim e
Reliabilit y 
TotalTim e
8771.423 hrs  0.0142 hr  0.00889 hr 
Reliabilit y 
8771.423 hrs
8771.423 - 0.02309
Reliabilit y 
8771.423
8771.4
Reliabilit y 
 99.9997%
8771.423
Combining Busses
99.88%
Normal Bus
f(SWN1) = 99.9997% SWN1
f(Normal Bus x f(SWN1)
= 99.88% x 99.9997%
= 99.8797%
99.8998%
Critical Bus
SWG1
f(SWG1) = 99.9997%
f(Critical Bus x f(SWN1)
= 99.8998% x 99.9997%
= 99.8995%
99.9997%
at this point
Combining Busses
99.88%
99.8998%
Normal Bus
Critical Bus
SWN1
SWG1
XFR1
99.9997%
at this point
BKR5
52
51
RLY5
CBL6
ICU
99.9957%
= f(FDR5)
= 99.9960%
Reliability at ICU
= f(ICU)
= f(XFR1)*f(FDR5)
=.999997*.99996
= .999957
= 99.9957%
or
(0.38 hr/yr)
or
(23 min/yr)
(outages per year)
But…


Remember that these are failure calculations
based on “proper” maintenance
Also does not include normal age-induced
degradation
Equipment Failure Timing


Initial failures (installation problems, infant
mortality of installed components).
Degradation over time (temperature, corrosion,
dirt, surge)
Likelihood
Of
Failure
2.33 hrs/yr
(average)
Initial
Failures
Degradation
Failures
Area under hatch
marks represents
the total likelihood
of a failure
Time
Equipment Failure Timing

Poor maintenance reduces equipment life since
failures due to degradation come prematurely
soon. IEEE says add 10% to likelihood of
downtime.
Likelihood
Of
Failure
Initial
Failures
Likelihood of failure is
higher because
postponed maintenance
increases problems due
to corrosion,
misalignment, etc, that
would be picked up in a
PM program
Early
Degradation
Failures
2.59 hrs/yr
(average)
Time
Results

Good Maintenance* = 11.4 hrs/year downtime

Fair Maintenance = 11.4 hrs + 10% =12.5 hr/yr

12.5 – 11.4 = 1.1 hr/yr less downtime

1 hour per year more downtime
Is that worth spending any time fixing?
… but this is only a simple example
* Utility source to transfer switch
Real Systems Are Much
Larger


17 MV breakers
14 MV loop feed
switches



31 MV internal
bus runs




Typical Large MV System
3 switching
elements
42 total
(17+14)
4000’ MV cable
15 MV
transformers
3 standby
generators
What is the likelihood of a power failure at this
location?
Just looking at a portion of the equipment…
42 MV disconnect switches (42 * .022 = 0.924 hrs/yr)
4000’ MV cable (4000/1000 * 0.1624 = 0.649 hrs/yr)
15 MV transformers (15.39 hrs/yr)
30000’ LV cable (30000/1000 * 0.0148 = 0.444 hrs/yr)
31 MV bus run with 17 MV breakers
(31(0.2733) + 17(.2992)= 8.47 + 5.08 = 13.55 hrs/yr)
17 protective relays (17*.001 = 0.017)
Total = 0.924 + 0.649 + 15.39 + 0.444 + 13.55 + 0.017







= 31 hrs/yr (total duration of outages somewhere in this facility)
(Assuming a 1 hr/per failure means you would expect an electrical problem
31 times per year or 1 every week and a half!)
What is the likelihood of a power failure at
this location?
Just looking at a portion of the equipment…
42 MV disconnect switches (42 * .022 = 0.924 hrs/yr)
4000’ MV cable (4000/1000 * 0.1624 = 0.649 hrs/yr)
15 MV transformers (15.39 hrs/yr)
30000’ LV cable (30000/1000 * 0.0148 = 0.444 hrs/yr)
31 MV bus run with 17 MV breakers
(31(0.2733) + 17(.2992)= 8.47 + 5.08 = 13.55 hrs/yr)
17 protective relays (17*.001 = 0.017)
Total = 0.924 + 0.649 + 15.39 + 0.444 + 13.55 + 0.017







= 31 hrs/yr (total duration of outages somewhere in this facility)
(Assuming a 1 hr/per failure means you would expect an electrical problem
31 times per year or 1 every week and a half!)
Step 3
1. Establish Current Condition of Facility
2. Determine Likelihood of Serious Problem Based on
this Condition
3. Sort to Find Equipment Most at Risk to Cause Problems
4. Identify the Predictive Techniques that Gives
Early Warning of Problems at that Equipment
MV
Breakers
Prot.
Relays
MV Bus
LV Cable
MV Xfrm
MV
Cable
16
14
12
Expected Hours of
Outages per year 10
8
6
4
2
0
MV Sw
MV Transformers Win!
(Lose?)
Step 4
1. Establish Current Condition of Facility
2. Determine Likelihood of Serious Problem Based on
this Condition
3. Sort to Find Equipment Most at Risk to Cause Problems
4. Identify the Predictive Techniques that Gives
Early Warning of Problems at that Equipment
Now What?



We now know how to figure “how many minutes
of outage will occur each year” for each device.
But how do we reduce that value?
We can recognize that failures can be predicted
if we recognize the early warning signs


The so-called “Predictive Indicator”
Once we know that, we can identify the likely
cause and fix the problem before it is serious.
Agenda

Where We Are Today

What Can Be Done?

IEEE 493 Process (Identify Candidates)

Predicting Failures

Economic Analysis / Prioritization

Solutions Available
Predicting Failures
Equipment Failure
Leads to …
Initiating Causes
Hints at …
Predictive Indicator
Failure Contributing Causes
(Source:IEEE 493-1997)
Combined Analysis of Switchgear Bus and Circuit Breaker Failure
Contributing Causes (%)
Switchgear Bus Failure Contributing Cause Insulated Bare Bus Breakers
(%) Percentage
Bus
Thermocycling
6.6%
12.5%
Mechanical Structure Failure
3.0%
8.0%
Mechanical Damage From Foreign Source
6.6%
Shorting by Tools or Metal Objects
15.0%
Shorting by Snakes, Birds, Rodents, etc.
3.0%
Malfunction of Protective Relays
10.0%
4.0%
Improper Setting of Protective Device
4.0%
Above Normal Ambient Temperature
3.0%
Exposure to Chemical or Solvents
3.0%
15.0%
Exposure to Moisture
30.0%
15.0%
Exposure to Dust or Other Contaminants
10.0%
19.0%
Exposure to Non-Electrical Fire or Burning
6.6%
Obstruction of Ventilation
8.0%
Normal Deterioration from Age
10.0%
4.0%
11.0%
Severe Weather Condition
3.0%
4.0%
Testing Error
4.0%
Lubricant Loss, or Deficiency
18.0%
Lack of Preventive Maintenance
18.0%
Other - Breaker Related
40.5%
Totals
94.8% 100.0%
100.0%
Totals
19.1%
11.0%
6.6%
15.0%
3.0%
14.0%
4.0%
3.0%
18.0%
45.0%
29.0%
6.6%
8.0%
25.0%
7.0%
4.0%
18.0%
18.0%
Normalized
to 100%
7.5%
4.3%
2.6%
5.9%
1.2%
5.5%
1.6%
1.2%
7.1%
17.7%
11.4%
2.6%
3.1%
9.8%
2.8%
1.6%
7.1%
7.1%
254.3%
100.0%
Contributing
Cause
Initiating
Switchgear Bus & Breaker
Most Probable Initiating Cause for Failure
Failure Contributing Cause (%) Contributor
Thermocycling
Loose connections, load current, internal
temperature, ambient, cubicle heaters, etc.
Mechanical Structure Failure
Fatigue, vibration, electrical loose components
Mechanical Damage From
Foreign Source
Shorting by Tools or Metal
Objects
Shorting by Snakes, Birds,
Rodents, etc.
Malfunction of Protective Relays
Improper Setting of Protective
Device
Above Normal Ambient
Temperature
Exposure to Chemical or
Solvents
Exposure to Moisture
Exposure to Dust or Other
Contaminants
Exposure to Non-Electrical Fire
or Burning
Obstruction of Ventilation
Normal Deterioration from Age
%
7.5%
4.3%
Accidental action during maintenance / Enclosure
Openings
Accidental action during maintenance / Enclosure
Openings
Enclosure Openings
2.6%
Relay failure
Improper relay settings
5.5%
1.6%
Ambient Temperature
1.2%
Corona or Surface Tracking / Enclosure Openings
7.1%
Corona or Surface Tracking / Enclosure Openings
/ Cubicle Heater Circuit Failure
Corona or Surface Tracking
17.7%
5.9%
1.2%
11.4%
External activity
2.6%
Clogged door or other filters
Normal deterioration: corona or surface tracking of
the insulation; contacts, interrupters, springs,
mechanisms, etc.
Severe Weather Condition
External activity
Testing Error
External activity
Lubricant Loss, or Deficiency
Overheating of the equipment and lubrication,
aged lubricants or loss-of lubricants
Lack of Preventive Maintenance External activity
3.1%
9.8%
2.8%
1.6%
7.1%
7.1%
Initiating Causes
Predictive Indicators
Most Probable Initiating Cause for Failure
Contributor
Loose connections, load current, internal
temperature, ambient, cubicle heaters, etc.
Fatigue, vibration, electrical loose
components
Accidental action during maintenance /
Enclosure Openings
Accidental action during maintenance /
Enclosure Openings
Enclosure Openings
Relay failure
Improper relay settings
Ambient Temperature
Corona or Surface Tracking / Enclosure
Openings
Corona or Surface Tracking / Enclosure
Openings / Heater Circuit Failure
Corona or Surface Tracking
External activity
Clogged door or other filters
Normal deterioration: corona or surface
tracking of the insulation; contacts,
interrupters, springs, mechanisms, etc.
External activity
External activity
Overheating of equipment and lubrication
age or loss-of lubricants
External activity
Available Solutions to address Initiating
Causes
On-Line Thermal Model Analyzer &
Thermography for Hot Spots
Thermography for Hot Spots and Future
Vibro-acoustics of electrical equipment
Safety during maintenance & Visual
Inspections
Safety during maintenance & Visual
Inspections
Visual Inspections
Periodic Relay Testing
Periodic Power System Study
On-Line Thermal Model Analyzer
Partial Discharge Detection & Visual
Inspection
Partial Discharge Detection & Visual Inspection & On-Line Thermal Model Analyzer
Partial Discharge Detection (External visual
inspection can not detect internal bus)
On-Line Thermal Model Analyzer &
Inspection of External area
On-Line Thermal Model Analyzer &
Thermography for Hot Spots
Partial Discharge Detection and
Thermography for Hot Spots
%
7.5%
4.3%
2.6%
5.9%
1.2%
5.5%
1.6%
1.2%
7.1%
17.7%
11.4%
2.6%
3.1%
9.8%
None
Safety during maintenance & Improved
preventive maintenance
Future vibro-acoustics of electrical equipment
2.8%
1.6%
Improve preventive maintenance
7.1%
7.1%
Available Predictive Tools
Available Solutions to address
Initiating Causes
On-Line Thermal-Model Analyzer
Totals
Normalized % of Total On-Line Predictive Diagnostic - Monitoring
to the new Failure
Capabilities Available
100%
Causes
Addressed
32.1%
18.1%
Thermography for Hot Spots
24.7%
Future vibro-acoustics of electrical
11.4%
equipment
Safety during maintenance
10.1%
Visual Inspections (Switchgear
37.1%
Enclosure and Surrounding Area)
Periodic Relay Testing
5.5%
Periodic Power System Study
1.6%
Partial Discharge Detection
46.0%
Improve preventive maintenance
8.7%
Totals 177.2%
13.9%
6.4%
5.7%
20.9%
3.1%
0.9%
26.0%
4.9%
100.0%
15.6% Technology available for
continuous monitoring
12.0% Yes - Periodic
5.6% Not fully commercially available
12.0%
4.9% NA
18.1% Periodic - Plant Personnel / Safety and
Operating Procedures
2.7% Periodic Relay Testing
0.8% Periodic Power System Study
22.4% Yes - Periodic
22.4%
4.2% NA
86.3% Total Causes address by CBM:
50.1%
• Top 4 in order of importance are:
- Partial Discharge Diagnostics (22.4%)
- Visual Inspection (18.1%)
- On-Line Thermal Analyzer (15.6%)
- Thermographic Inspections (12.0%)
15.6%
CBM – Condition Based Maintenance
What If We Implemented
One Predictive Solution?

Partial Discharge – 22.4% of failures detected


Caveat: Only works on medium voltage (>1000 volts)
Our example:

15.39 hrs/yr from transformer failure
•

8.47 hrs/yr from MV bus failure
•

22.4% reduction  11.94 hrs/yr
22.4% reduction  6.57 hrs/yr
5.08 hrs/yr from MV breaker failure
•
22.4% reduction  3.94 hrs/yr
Reduction In Outages

Transformer Failure (was 15.39 hrs/yr, now 11.94 hrs/yr)


MV bus failure (was 8.47 hrs/yr, now 6.57 hrs/yr)


Saving 1.9 hrs/yr
MV breaker failure (was 5.08 hrs/yr, now 3.94 hrs/yr)


Saving 3.45 hrs/yr
Saving 1.1 hr/yr
Total Savings from PD
6.45 hrs/yr



Failures/yr
1 hr/failure = 6 fewer failures?
6 hr/failure = 1 fewer failure?
So what is the correct
answer?
1000
100
10
1
0.1
0.01
0.01
0.1
1
10
Hrs/failure
100
1000
Determining Number of Failures
Start by capturing the hrs/failure and hours/yr for each device
Category
Failures/yr
Hours/Failure
Hours/Yr
Prot. Relays
.0002
5
.001
LV Swgr Bkrs
.0027
4
.0108
MV Swgr Bkrs
.0036
2.1 / 83.1*
.0076/.2992
LV Cable (1000 ft)
.00141
10.5
.0148
MV Cable (1000 ft)
.00613
26.5
.1624
Disc. Switches
.0061
3.6
.022
Transformer
.003
342
1.026
LV Swgr Bus
.0024
24
.0576
MV Swgr Bus
.0102***
26.8
.2733
* when no on-site spare is available
** below ground *** 3 connected to 3 breakers
Average Outage
Next, total the number of devices and determine the total hours of
failure for each device and for all devices
Device
MV Breaker
MV Transformer
MV Bus
Quantity Hrs/Failure Combined
17
83.1
1412.7
15
342
5130
31
26.8
830.8
63
7373.5
117.04 hours/failure
(weighted average)
Divide the total time by the number of devices.
(7373.5 hrs)
(63)
Result = Average hours per failure = 7373.5/63 = 117.04 hrs/failure
Compute Likely Failure
Rate
100
10
0.055 failures/
year
1
0.1
0.01
0.1

Total Savings from PD
6.45 hrs/yr
1
10
100
0.001

1 hr/failure = 6.45 fewer failures per year

6.45 hr/failure = 1 fewer failure per year

117.4 hr/failure = ? fewer failures per year
117 hrs/
failure
•
6.45/117 = .0.055
•
Answer: ? = 0.055 failures/yr or 1 fewer failure every 18 years
1000
Agenda

Where We Are Today

What Can Be Done?

IEEE 493 Process (Identify Candidates)

Predicting Failures

Economic Analysis / Prioritization

Solutions Available
How Much Does It Cost?



We know that if we install PD sensors on all this
equipment, statistically it will result in 1 less outage every
eighteen years.
Each PD sensor costs ~ $5K installed in quantity
We have 23 items to be monitored


15 transformers
17 breakers with 31 bus runs
•
•


Assume 1 PD sensor can monitor 1 vertical 2-high structure
17/2 = 8.5, round up to 9. 15+9 = 23
$7000 * 23 = $161000
Does saving an outage once every 18 years justify
spending $161000?
Cost Savings Through
Reduced Outage (Detected by PD)
Total Exposure = Median Outage duration * % Related to Insulation * Downtime Cost
Your Costs May Vary…
Obviously, the more critical the application, the more important the solution
At $1000 / hour of downtime costs (lab, stadium, imaging)

Loss of one of the small power transformers would require
•
•

Cost of a 1000 kVA indoor dry, MV power transformer
•
•
•



342 hours to repair (remember fire video) and
cost: $342000 of downtime ($24000 / day)
Assume $18/kVA or $18000
Assume labor $50/hr, 3 man-days labor
Total cost = (1000 * $18) + ($50 * 3 * 8) = $18000 + $1200
Total cost = $19200
Downtime = $342000
Material = $19200
Total Loss = Downtime + Material = $361200
Compute Payback



Our cost is $161000
Our savings is $361200 once every 18 years or $20000
per year
If injury results, you can add $100000 - $1000000 (if selfinsured)

Assume we expect a 10% return on invested capital

Assume 10 year project life

Assume 2.5% inflation rate
Compute Equivalent
Payback





Cost = $161K, Savings = $20K/yr, N=10 years,
inflation = 2.5%, capital cost = 10%
Is this a good investment?
First Pass Analysis
Simple Payback = $161K/$20K = 8 years
Total Cash to bottom line =
= (Annual Savings * Project Life) - Installed Cost
= ($20000*10) - $161K = $200000 - $161000
= $39000 positive cash flow (life of project)
0% cost of capital (simple payback) attractive return,
but does this cover cost of capital (10%) considering
the reduction in value of money over time (2.5%
inflation)?
Compounded IRR
Calculator
 Cost 
1 
log10 
 a    1
Savings 
a 

n
log10 a
Cost
Savings
a
i
g
n
installed cost of equipment
annual savings
(1+g)/(1+i)
interest rate
annual inflation rate
duration (payback period in years)
Compounded IRR
Calculator
 Cost 
1 
log10 
 a    1
Savings 
a 

n
log10 a
Cost
Savings
a
i
g
n
$161,000
$20,000
(1+g)/(1+i) = (1+0.025)/(1+0.1) = 0.932
10%
2.5%
10
Run The Numbers…
Cost
Savings
a
i
g
n
$161,000
$20,000
(1+g)/(1+i) = (1+0.025)/(1+0.1) = 0.932
10%
2.5%
10
C 
 161000 
1 
1  
log10   a    1 log10 
 0.932 
  1
S
a 
20000 
0.932  


n

log10 a
log10 0.932
Compute Payback
 161000 
1  
log10 
 0.932 
  1
20000 
0.932   log10 8.05 0.141  1

n

log10 0.932
 0.0306
log10 8.05 0.141  1 log10  0.135
n

 0.0306
 0.0306
invalid
n
 invalid
 0.0306
What Does ‘invalid’ Mean?

Based on a cost of $161000, an annual savings
of $20000, a required rate of return of 10%, and
inflation rate of 2.5%…



Means that this does not provide any payback over
any time period
Using the cost-of-money numbers listed here
…this project is not financially viable.
Note: assuming an 5.75% cost of money vs 10%


N turns out to be 9.69 (<10)
In that case, the project is financially viable
Great, I’ve Found Problems,
Now what?



You can certainly replace with new
or…
If you catch it before it fails catastrophically, you
can rebuild
Many old electrical devices can be rebuilt to like
new condition
Agenda

Where We Are Today

What Can Be Done?

IEEE 493 Process (Identify Candidates)

Predicting Failures

Economic Analysis / Prioritization

Solutions Available
LV Refurbished
Power Breakers

LV Equipment Retrofit / “Roll-In” Replacements
- (W) - C-H
- ITE - GE
- AC - FPE
- Siem - R-S
510- Upgraded Trip
810-KW-Comm-O/C
610- Display
910-Harmonics
LV Rack-In Replacement
With New (In Old Equipment)
Old Breaker:
• Parts no longer
available
Modern Breaker:
• New warranty
• Installed in the old
structure
Motor Control Upgrades
Breaker-to-Starter Conversions:
- circuit breaker used to start motor
- only good for 1000 or less operations
- replace breaker with starter
- now good for 1,000,000 operations
Continuous
Partial
Discharge
Monitor
MCC Bucket Retrofits
- new breaker and starter
MV Vacuum Replacement
•Vacuum replacement for Air Break in same space
•Extensive Product Availability
• ANSI Qualified Designs
• 158 Designs
• Non-Sliding Current Transfer
• SURE CLOSE - Patented (MOC Switches)
• 2-Year Warranty - Dedicated Service
• Factory Trained Commissioning Engineers
• Full Design & Production Certification
• ANSI C37.59 Conversion Standard
• ANSI C37.09 Breaker Standard
• ANSI C37.20 Switchgear Standard
• Design Test Certificate Available on Request
Can’t Buy a Spare? Class 1
Recondition Instead

Receiving & Testing

Complete Disassembly

Detailed Inspection and
Cleaning

New Parts

OEM Re-assembly

Testing

Data-Base Tracking
Spot Network Upgrade
Network
Protector
Class 1
Recondition
Network Relay
Upgrades...
Transformer Oil Processing
Other Services Available:
• Samples Obtained On-Site
• Mail-in
Sampling
Kits
• Self
Powering
Generator
• Complete Transformer Testing
• On-Site
Testing
& Analysis
- PF, PCB
& Dissolved
Gas Analysis
• Vacuum Filling & Start-up
• Reclamation & Retesting
• Samples Obtained On-Site
On-Board Testing
Dielectric Testing
Karl Fischer Moisture Test
Acid Titration Testing
For more information
www.eatonelectrical.com
1. Search “calculators”
2. Select “life extension”
Calculator will determine
rate-of-return for life
extension solution.
Greatly simplifies math.
Thank You