Power Systems Reliability

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POWER SYSTEMS RELIABILITY


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Activity 1: Whats Your Buying Pattern?

(10 minutes)


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Subject : Probability and Statistics

Target Audience : H.S. Level (Grade 12)

General Objective: To teach the students the

concept of reliability as anapplication of probability.


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What is reliability?

Reliability measures the probability that a

system or part of a system will work.

So, if I say that the reliability of my walkman is

0.98 (or 98%), I mean that my walkman is

working 98% of the time and failing 2% of the

time.

Power Systems

Reliability


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What is a system?

A system is a collection of components, subsystem

and/or assemblies arranged to a specific design

in order to achieve desired functions with acceptable

performance and reliability.

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Reliability


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Models of a System

A) Series Structures

1 2

For this system to work, both components 1 and

2 must work.

1 2 . . . n

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Reliability


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A Personal Computer

Power

Supply

MotherboardProcessor

Hard

Drive

A Simple Series system

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B) Parallel Structures

1

2

1

2.

.

.

n

I n a parallel system, the system wi l l work as long

as at least one component works.

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C) Combination of Series and Parallel Structures

12

3

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Reliability


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Finding A Systems Reliability

1 2 n. . .

A) A Series System

For a pure series system, the system reliability is

equal to the product of the reliabilities of itsconstituent components. Or:

ns

RRRR ...21

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Example:Three components are connected in series and make

up a system. Component 1 has a reliability of 0.95,

component 2 has a reliability of 0.98 and

component 3 has a reliability of 0.97 for a missionof 100 hours. Find the overall reliability of the

system for a 100-hr mission.

Solution:

= (0.95)(0.98)(0.97)

= 0.90 = 90%

Rs = (R1)(R2)(R3)

Power Systems

Reliability


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Activity 2A:Computing Series SystemsReliability

0.8

1)0.9 Rs = (0.8)(0.9)

= 0.72

2)

0.7 0.6Rs = (0.7)(0.6)

= 0.42


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Activity 2B: Effect of the Number ofComponents in a Series System

S

y

s

t

e

m

R

e

l

ia

b

i

l

i

t

y

1.00

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.20

0.10

0.00

1 2 3 4 56 7 8 9 10

N u m b e r o f C o m p o n e n t s

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B) A Parallel System

1

2

n

.

.

.

For a pure parallel system,

the overall system reliability

is equal to the product of thecomponent unreliabilities.

Thus, the reliability of theparallel system is given by:

Rs = 1[(1R1)(1-R2)(1-Rn)]

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Example: Three components are connected in paralleland make up a system. Component 1 has a reliability of

0.95, component 2 has a reliability of 0.98 and component

3 has a reliability of 0.97 for a mission of 100 hours.

Find the overall reliability of the system for a 100-hrmission.

Solution:

Rs = 1[(1R1)(1R2)(1R3)]= 1[(10.95)(1-0.98)(1-0.97)]

= 0.99997

= 99.997 %

Power Systems

Reliability

R1 = 0.95

R2 = 0.98

R3 = 0.97


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Activity 3A:Computing Parallel Systems Reliability

0.9

1)

0.8

Rs = 1[(1- 0.9)(1- 0.8)]= 0.98

2)0.7

0.8

Rs = 1[(1- 0.7)(1- 0.8)]

= 0.94


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Activity 3B:Effect of the Number of Componentsin a Parallel System

S

y

s

t

e

m

R

e

l

ia

b

i

l

i

t

y

1.00

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.20

0.10

0.00

1 2 3 4 56 7 8 9 10

N u m b e r o f C o m p o n e n t s

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C) Combination of Series and Parallel StructuresExample:Consider a system with three components. Units 1 and 2

are connected in series and Unit 3 is connected in parallel

with the first two, as shown in the figure below. Findthe reliability of the system.

Solution:

R1 = 0.99 R2= 0.98

R3 = 0.97

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First, the reliability of the segment consisting of Units 1and 2 is calculated:

R1,2 = (R1)(R2)

= (0.99)(0.98)

= 0.97

The reliability of the overall system is then calculated

by treating Units 1 and 2 as one with a reliability of

0.97 connected in parallel with Unit 3.

Therefore:

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Reliability


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R1,2= 0.97

R3 = 0.97

Rs = 1[(10.97)(10.97)]

= 0.9991

= 99.91%

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Reliability


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Activity 4: Combination of Series and Parallel

Structure (10 minutes)

R1=0.7

1) R2=0.6

R3=0.5

This is equivalent to:

R1=0.7 R2,3=0.8

Rs = (0.7)(0.8) = 0.56


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Activity 4: Combination of Series and Parallel

(contd) Structure (10 minutes)

R1=0.72) R2=0.9

R3=0.5

This is equivalent to:

R1,2=0.63

Rs = 1- [(1 - 0.63)(10.5)]

= 0.815 = 81.5%R3=0.5


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Power Systems

Reliability

The Bathtub CurveReliability specialists often describe the lifetime of a population of

products using a graphical representation called the bathtub curve.

It characteristically describes the life of many products, as well as

humans.F

A

I

L

U

R

E

R

A

T

E

Time (hours, miles, cycles, etc)

EARLY LIFE

(burn-in or

break-in or

infant mortality

period)

(failure rate

decreases with time)

USEFUL LIFE

(or normal life)

(failure rate approx. const)

WEAROUT LIFE

(failure rate

increases with time)


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Activity 5: Real-life Applications of theBathtub Curve

(10 minutes)


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Power Systems Reliability