© 2005 Eaton Corporation. All rights reserved.
Designing for System Reliability
Dave Loucks, P.E.
Eaton Corporation
ToFacility
From Emergency Bus From Distribution Bus
What Reliability Is Seen At The Load?
Utility UPS Breaker Load
For example, if power flows to load as below: Assume outage duration exceeds battery capacity
Series Components
Utility UPS Breaker Load
99.9% 99.99% 99.99%
For example, if power flows to load as below: Assume outage duration exceeds battery capacity
Series Components For example, if power flows to load as below:
Assume outage duration exceeds battery capacity
Utility UPS Breaker Load
99.9%(8.7 hr/yr)
99.99%(0.87 hr/yr)
99.99%(0.87 hr/yr)
x+
x+
==
99.88%(10.5 hr/yr)
Overall reliability is poorer than any component reliability
Series Components For example, if power flows to load as below:
Assume outage duration exceeds battery capacity
Utility UPS Breaker Load
99.9%(8.7 hr/yr)PF* = 0.1%
99.99%(0.87 hr/yr)
0.01%
99.99%(0.87 hr/yr)
0.01%
x++
x++
===
99.88%(10.5 hr/yr)
0.12%
PF = (1 – Reliability) = 1 – R(t)
* PF = probability of failure
Series Components For example, if power flows to load as below:
If outage duration less than battery capacity
UPS Breaker Load
99.99%(0.87 hr/yr)
0.01%
99.99%(0.87 hr/yr)
0.01%
x++
===
99.98%(1.74 hr/yr)
0.02%*PF =
Batteries Depleted 99.88% reliable Batteries Not Depleted 99.98% reliable
Parallel Components What if power flows to load like this:
Assume outage duration exceeds battery capacity
Utility
UPS
StaticATS
Load
UPS
Parallel Components
Utility
UPS
StaticATS
Load
UPS99.9%
99.99%
99.99%
99.99%
?? %
What if power flows to load like this: Assume outage duration exceeds battery capacity
Parallel Components
Utility Load
99.9%
99.99%
99.99%
99.99%PF* = 0.1% PF(a or b)+ + = ?? %
?? %
UPS
StaticATS
UPS
0.01%
What if power flows to load like this: Assume outage duration exceeds battery capacity
Parallel Components
99.99%
99.99%PF(a or b) = 0.01 % x = 0.000001 %
99.99 %
UPSa
UPSb
0.01%
What if power flows to load like this: Solve each path independently
99.99%
99.99%
UPSa
UPSb
99.99 %
R(t) = 1 - PF(a or b) = 99.999999%
Parallel Components
Utility Load
99.9%99.99%
99.999999%
99.89 %
UPSa
orUPSb
StaticATS
Multiply the two Probabilities of Failure, PF(a) and PF(b) and subtract from 1
PF(total) = PF(u) + PF(a or b) + PF(s)= 0.1% + 0.000001% + 0.01% = 0.110001%
Parallel Components
Load
99.99%
99.999999%
99.99 %
UPSa
orUPSb
StaticATS
Multiply the two Probabilities of Failure, PF(a) and PF(b) and subtract from 1
PF(total) = PF(a or b) + PF(s)= 0.000001% + 0.01% = 0.010001%
Summary Table
Configuration Reliability
Single UPS system (long term outage) 99.88%
Single UPS system (short term outage) 99.98%
Redundant UPS system (long term outage) 99.89%
Redundant UPS system (short term outage) 99.99%
Comments?
Value AnalysisIs going from this:
Utility UPS Breaker
Utility
UPS
StaticATS
99.89%(only battery) 99.99%
UPS
99.88%(only battery) 99.89%
to this
worth it?
0.01%difference
Availability
Increase Mean Time Between Failures (MTBF) Decrease Mean Time To Repair (MTTR)
%100
MTTRMTBF
MTBFAi
i
i
A
AMTBFMTTR
)(
100
Relationship of MTBF and MTTR to Availability
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
0 10 20 30 40 50 60 70 80 90 100
MTTR (Hours)
Av
aila
bili
ty
1000 hrs800 hrs600 hrs
400 hrs
200 hrs
MTBF
95% Availability from Different MTBF/MTTR combinations
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
0 10 20 30 40 50 60 70 80 90 100
MTTR (Hours)
Av
aila
bili
ty
1000 hrs800 hrs600 hrs
400 hrs
200 hrs
MTBF
10.5 21.1 31.6 42.1 52.6
Breakeven Analysis Total Economic Value (TEV)
Simple Return (no time value of money)• TEVS = (Annual Value of Solution x Years of Life of Solution)
– Cost of Solution
Assume an outage > 0.1s costs $10000/yr
Assume cost of solution is $30000 Assume life of solution is 10 years
Breakeven Analysis
Since solution eliminates this potential 0.41 second outage, we “save” $10000 each year
Total Economic Value (TEV) Simple Return (no time value of money) TEVS = (Annual Value of Solution x Years of Life of
Solution) – Cost of Solution
TEVS = (($10000 x 1) x 10) – $100000
TEVS = $100000 - $30000 = $70000
Let’s Examine a More Complex System
52 52
Source 1 Source 2
What is the reliabilityat this point?
K K 52
Source 1
What is the reliabilityat this point?
52
Source 299.9% 99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999%
99.9%
99.99%
99.999%
99.99%
99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999% 99.999%
Design 1 Design 2
Primary Selective
52 52
Source 1 Source 2
What is the reliabilityat this point?
K K
99.9% 99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999%
Source 1 Source 2
What is the reliabilityat this point?
99.999%
99.99%
99.99%
99.999%
.999 x .9999 = 99.89%
.999 x .9999 = 99.89%
99.89% 99.89%
Convert to
Design 1
Combine Reliabilities:Parallel Sources
52 52
Source 1 Source 2
What is the reliabilityat this point?
K K
99.9% 99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999%
Source 1 or Source 2
What is the reliabilityat this point?
99.999%
99.99%
99.99%
99.999%
PF1 x PF2 = PF both
PF both = 0.11% x 0.11%PF both = 0.0121%
R(t) = 1 – PF both
= 100% - 0.0121%= 99.99% PF1 = 0.0121%
R(t) = 99.99%
Design 1
Combine Reliabilities:Parallel Sources + Tx + Sec. Bkr.
52 52
Source 1 Source 2
What is the reliabilityat this point?
K K
99.9% 99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999%
Source 1 or Source 2
What is the reliabilityat this point?
99.99%
99.999%
Rsource x Rtx x Rmb = R(t)= .9999 x .99999 x .9999
= .9998
R(t) = 99.98%
Design 1
Combine Reliabilities:… + bus and feeder breaker
52 52
Source 1 Source 2
What is the reliabilityat this point?
K K
99.9% 99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999%
Source 1 or Source 2
Rsource x Rtx x Rmb = R(t)= .9998 x .99999 x .9999
= .99969
R(t) = 99.969%
Design 1
Secondary Selective
52
Source 1
What is the reliabilityat this point?
52
Source 299.9%
99.99%
99.999%
99.99%
99.9%
99.99%
99.999%
99.99%
99.99%
99.999% 99.999%
99.99%
Design 2
Secondary Selective
52
Source 1
What is the reliabilityat this point?
52
Source 299.9%
99.99%
99.999%
99.99%
99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999% 99.999%
Source 1 Source 2
99.99%
99.99%
99.999% 99.999%
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
R(t) = 99.88% 99.88%
Design 2
Secondary Selective
52
Source 1
What is the reliabilityat this point?
52
Source 299.9%
99.99%
99.999%
99.99%
99.9%
99.99%
99.999%
99.99%
99.99%
99.99%
99.999% 99.999%
Source 1 Source 2
99.99%
99.99%
99.999% 99.999%
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
R(t) = 99.88%
Design 2
Secondary Selective
52
Source 1
52
Source 299.9%
99.99%
99.999%
99.99%
99.9%
99.99%
99.999%
99.99%
99.99%
99.999% 99.999%
Source 1 Source 2
Rtie = 99.99%
Rfdr =99.99%
Rbus1 = 99.999%
Rbus2 = 99.999%
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
Rs1 = 99.879%
Rpath1 = Rs1 x Rbus1 x Rtie x Rbus2 x Rfdr
Rpath1 = .99879 x .99999 x .9999 x .99999 x .9999 Rpath1 = .99857PF1 = 1 – Rpath1 = 1-.99857 = 0.00143 = 0.143%
Design 2
Secondary Selective
52
Source 1
52
Source 299.9%
99.99%
99.999%
99.99%
99.9%
99.99%
99.999%
99.99%
99.99%
99.999% 99.999%
Source 1 Source 2
Rfdr =99.99%
Rbus2 = 99.999%
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
R(t) = Rs x Rmb x Rtx x Rsb
= .999 x .9999 x .99999 x .9999
= .99879
Rpath2 = Rs2 x Rbus2 x Rfdr
Rpath2 = .99879 x .99999 x .9999Rpath2 = ..99868PF2 = 1 – Rpath2 = 1-.99868 = 0.00132 = 0.132%
Rs2 = 99.879%
Design 2
Secondary Selective
52
Source 1
52
Source 299.9%
99.99%
99.999%
99.99%
99.9%
99.99%
99.999%
99.99%
99.99%
99.999% 99.999%
Source 1 Source 2
R(t) = 1 – (PF1 x PF2)= 100% - (0.143% x 0.132%)
= 100% - (0.0189%)= 99.981%
R(t) = 99.981%
Design 2
Comparison
Method Reliability at Load
1. Primary Selective 99.969%
2. Secondary Selective 99.981%
Comments?
Value Analysis
1. 99.97% x 8760 = failure once every 8757 hours
2. 99.98% x 8760 = failure once every 8758 hours Assuming 1 hour repair time, we will see two, 1-
hour outages after 8758 hours Meaning 1/8758 hours (0.411 seconds) expected outage
per year
As with UPS example, what is 0.411 seconds worth?
What is cost differential of higher reliability solution?
Breakeven Analysis
Total Economic Value (TEV) Simple Return (no time value of money)
• TEVS = (Annual Value of Solution x Years of Life of Solution) – Cost
of Solution
Discounted Return (borrowed money has cost)• TEVD = (NPV(annual cash flow, project life, interest rate) – Cost of
Solution
Assume 0.411 sec of downtime costs $20000/yr Assume cost of solution is $75000 Assume life of solution is 10 years
Breakeven Analysis
Total Economic Value (TEV) Simple Return (no time value of money) TEVS = (Annual Value of Solution x Years of Life of
Solution) – Cost of SolutionTEVS = (($20000 x 1) x 10) – $200000
TEVS = $200000 - $75000 = $125000
Discounting cash flow at 10% cost of money TEVD = NPV($20000/yr, 10 yrs, 10%) – $30000
TEVD = $122891 – $75000 = $47891
Solve for Equivalent Interest Rate
Knowing initial cost of $75000and annual benefit of $20000what is the equivalent return?
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
Year 7
Year 8
Year 9
Year 10
Year 0
$75000
$20000 $20000 $20000 $20000 $20000 $20000 $20000 $20000 $20000 $20000
Uniform Series Present Value
P – Present Value A – Annuity payment n – Number of periods
i – Interest per period
n =
Uniform Series Present Value Equations
But what if PV, A and n are known and i is unknown? Iterative calculation
n
n
ii
iAPV
1
11
11
1n
n
i
iiPVAor
P – Present Value A – Annuity payment n – Number of periods i – Interest per period
Uniform Series Present Value Equations
n
n
ii
iAPV
1
11
n i PV RS of Equation Comment
10 5% $75000 $154435 i too low
10 30% $75000 $61831 i too high
10 20% $75000 $83849 i too low
10 25% $75000 $71410 i too high
10 23.413% $75000 $75000 correct value
Breakeven Analysis
Total Economic Value (TEV) Simple Return (no time value of money) TEVS = (Annual Value of Solution x Years of Life of
Solution) – Cost of SolutionTEVS = (($20000 x 1) x 10) – $200000
TEVS = $200000 - $75000 = $125000
Discounting cash flow at 10% cost of money TEVD = NPV($20000/yr, 10 yrs, 10%) – $30000
TEVD = $122891 – $75000 = $47891
IRR = 23.413% effective return
Reliability Tools
Eaton Spreadsheet Tools IEEE PCIC Reliability Calculator Commercially Available Tools Financial Tools (web calculators)
Web Based Financial Analysis
www.eatonelectrical.com search for “calculators” Choose “Life Extension
ROI Calculator”
Web Based Financial Analysis
Report provides financial data
Provides Internal Rate of Return
Use this to compare with other projects competing for same funds
Evaluates effects due to taxes, depreciation
Based on IEEE Gold Book data
Uncertainty – Heart of Probability
Probability had origins in gambling What are the odds that …
We defined mathematics resulted based on: Events
• What are the possible outcomes?
Probability• In the long run, what is the relative frequency that an event
will occur?
• “Random” events have an underlying probability function
Normal Distribution of Probabilities
From absolutely certain to absolutely impossible to everything in between
AbsolutelyCertain
100%
0%AbsolutelyImpossible
Most likely value
Distribution System Reliability How do you predict when something is going to
fail? One popular method uses exponential curve
AbsolutelyCertain
37% of them are working
50% of them are working
69%
50%
Mean Time Between Failures
The ‘mean time’ is not the 50-50 point (1/2 are working, 1/2 are not), rather…
When device life (t) equals MTBF (1/), then:
368011
.)(
eeetRt
MTBFt
The ‘mean time’ between failures when 37% devices are still operating
MTBF Review
Remember, MTBF doesn’t say that when the operating time equals the MTBF that 50% of the devices will still be operating, nor does it say that 0% of the devices will still be operating. It says 37% (e-1) of them will still be working.
Said another way; when present time of operation equals the mean (1/2 maximum life), the reliability is 37%
Exponential Probability
Assumes (1/MTBF) is constant with age For components that are not refurbished, we
know that isn’t true. Reliability decreases with age ( gets bigger)
However, for systems made up of many parts of varying ages and varying stages of refurbishment, exponential probability math works well.
Reliability versus MTBF
Assume at time = 0 Reliability equals 100% (you left it running)
At time > 0, Reliability is less than 100%
tMTBFt eetR
1
)(
Converting MTBF to Reliability
Unknown Reliability = ?
Known MTBF (40000 hrs) t (8760 hrs = 1 year)
tMTBFt eetR
1
)(
UPS
%.
)(
.
380
21908760
40000
1
1
ee
eetRt
MTBFt
Great, I’ve Found Problems, Now what?
You can certainly replace with newor…
If you catch it before it fails catastrophically, you can rebuild
Many old electrical devices can be rebuilt to like new condition
LV Refurbished Power Breakers LV Equipment Retrofit / “Roll-In” Replacements
510- Upgraded Trip
610- Display
810-KW-Comm-O/C
910-Harmonics
- (W) - C-H
- ITE - GE
- AC - FPE
- Siem - R-S
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
MCC Bucket Retrofits- new breaker and starter
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
MV Vacuum Replacement•Vacuum replacement for Air Break in same space •Extensive Product Availability
• ANSI Qualified Designs• 158 Designs
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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...
InsulGard … “Predictive Relay”
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