Reliability 2013 1

EM 105 (B) Process Reliability Engineering Module I Definition of reliability – key key elements; failure analysis – analysis – failure failure density – failure failure rate – rate – probability of failure - bathtub curve - Basic reliability equations – equations – Reliability Reliability in terms of – failure rate – failure density - relation between reliability, failure density and hazard rate - Mean time to failure failure (MTTF) – (MTTF) – Integral Integral equation of MTTF in terms of reliability Module II Hazard models – models – constant constant hazard model – model – linearly linearly increasing hazard model – model – expressions expressions for reliability, reliability, failure density, density, and probability p robability of failure of these models – models – problems problems Module III System reliability – reliability – components components connected in series – series – components components connected in parallel – parallel – mixed configuration – reliability block diagrams (RBD) – distinction between physical configuration and logical configuration co nfiguration – – problems problems Module IV Reliability improvement methods – Redundancy – unit redundancies – element redundancies – redundancies – simplification simplification of design – design – parts parts derating – derating – operating operating environment; Cost of reliability – reliability – factors factors to be considered co nsidered for optimizing the reliability cost References

1. L.S.Srinath, “Reliability Engineering”, Affiliated East-West East -West Press Ltd., 1985 2. E. Balaguruswamy, “Reliability Engineering”, Tata McGraw Hill Publishing Co., 1984 3. Charles E. Ebling, “Reliability & Maintainability Engg.”, Tata McGraw Hill Publishing Co., 1997 4. Alessandro Birolini “Reliability Engineering Theory and Practice”, Practice”, Springer, 2007. 5. Lewis, E., “Introduction “Introduction to Reliability Engineering” Engineering”, John Wiley & Sons, 1995

2

EM 105 (B) Process Reliability Engineering Definition of reliability Reliability Reliability is the probability probability that an item will perform a required function for a specified period of time under specified operating conditions. This definition has four key elements: 1) The quantification of reliability in terms of probability. 2) A statement defining the required function – function – as as the function is defined in detail, it becomes more clear which product failures failures impair the success of the t he mission and a nd which do not. 3) A statement specifying the period of time – deterioration of materials and parts with time is natural and consequently the performance level of the unit will also go down with time. If the time t ime period is not specified, probability is a meaningless number for time oriented products. 4) A statement defining the operating condition and this could be with regard to temperature, humidity, shock, vibration, and so on. Failure data analysis Reliability is defined as the probability of a device giving satisfactory performance for a specified period under specified operating conditions. When a unit or system does not perform satisfactorily, satisfactorily, it is said to have failed. The pattern of failure can be o btained from life test results. That is, by testing a fairly large number of models until failure occurs, and observing the failure rate characteristics as a function of time. The first step, therefore, is to link reliability with experimental or field – failure data. These data will also provide a basis for formulating or constructing mathematically a failure model for general analysis. The following data is related with a series of tests conducted under certain stipulated conditions on 1000 electronic components. The total duration of the tests is 19 hours. The number of components that fail during each hourly interval is noted and the results obtained are tabulated below.

Reliability Engineering Engineering Lecture notes on Process Reliability Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

3

Time (t)

No. of failures ( f)

0

Cumulative failures (F) 0

No. of survivors (S) 1000

130 1

3

0.016 0.012

996

1.200

4

4

0.004 0.004

1000

1.110

16

12

19

0.056 0.04

984

0.714

56

40

18

0.486 0.118

0.062 944

17

0.194

118

62 16

0.283

194 0.076

882

0.103 0.258

0.064

76 15

0.286

258

64 806

0.103

286 0.028

742

14

0.317 0.031

28 13

0.101

317

31 714

0.100 0.351

0.034 683

12

0.388

351

34 11

0.100

388 0.037

649

0.101 0.429

0.041

37 10

0.475

429

41 612

0.101

475 0.046

571

9

0.526 0.051

46 8

0.101

526

51 525

0.101 0.582

0.056 474

7

0.644

582

56 6

0.100

644 0.062

418

0.100 0.712

0.068

62 5

0.787

712

68 356

0.100

787 0.075

288

4

0.870 0.083

75

Reliability (R)

0.139

870

83 213

Failure rate (Z)

1 0.130

130

2

Failure density ( f d )

2.000

0

0 Total =1

Mean = 0.376

Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

4 The failure density, failure rate, reliability and probability of failure can be defined as follows. Failure density ( f d ) This is the ratio of number of failures during a given unit interval of time to the total number of items at the very beginning of the test (also called as initial population). For the given data the failure density associated with the first unit interval is 130 f d 1 0.13 1000 83 Similarly, f d 2 0.083 and so on. 1000 Failure rate (Z) This is the ratio of number of failures during a particular unit time interval to the average population during that interval. The average population during an interval is the average of populations at the beginning and at the end of the interval. For the given data, failure rate during the first unit interval is 130 Z (1) 0.139 1000 870 2 Similarly, Z (2)

83 870 787

0.100 and so on.

2 Reliability (R) This is the ratio of survivors at any given time to the total initial population. “Probability of survival” is another name for reliability. For the given data, reliability corresponding to first hour is 870 R(1) 0.870 1000 787 R(2) 0.787 and so on. 1000 Probability of failure Probability of failure can also be termed as unreliability factor. Since survival and failure are complementary events, Probability of failure = 1- Probability of success (or reliability). For the given data, probability of failure corresponding to first hour = 1=R(1) = 1-0.787 = 0.213 Similarly, probability of failure corresponding to fifth hour = 1=R(5) = 1-0.582 = 0.418


5

0.14 0.12 y 0.1 t i s n e 0.08 d e r 0.06 u l i a F 0.04

0.02 0

0

2

4

6

8

0 1

2 1

1 4

6 1

8 1

Time interval

2.5

2 e t a r e r u l i a F

1.5

1

0.5

0

0

2

4

6

8

0 1

2 1

4 1

6 1

8 1

Time interval

1.2 1 0.8 y t i l i b a 0.6 i l e R 0.4 0.2 0

0

2

4

6

8

0 1

2 1

1 4

6 1

8 1

Time

Variation of failure density, failure rate, and reliability with time


6 Conclusions (i) Let f d 1 = failure density associated with the first unit time interval f d 2 = failure density associated with the first unit time interval

. . f d l = failure density associated with the last unit time interval Then, f d 1

f d 2

f d l



1

(ii) Let n 1 number of failed components during the first unit interval n2

nt

number of failed components during the second unit interval . . number of failed components associated with the tth unit interval

N = Total initial population th

th

Then, Reliability for the t hour is the number of survivors till the t hour divided by the initial population. N (n1 n2  nt ) nt n1 n2 That is, R(t ) 1  N N N N That is, R (t )

1

f d 1

f d 2



f d t

th (iii) Probability of failure for t hour = f d 1 f d 2 

f d t

th

(iv) Failure rate or hazard rate associated with the t hour, Z (t ) =

Z (t )

Z (t )

NR(t 1) NR(t ) NR(t 1) NR(t ) 2

2 R(t

t R(t

t ) R(t ) t ) R(t )

2

R(t 1) R(t ) R(t 1) R(t )

(for unit time interval)

(for a time interval of

t )

Problem Following table gives the results of tests conducted under severe co nditions on 1000 safety valves. Obtain the failure density and hazard rates for various t ime intervals. Time 0-4 4-8 8-12 12-16 16-20 20-24 interval No. of 267 59 36 24 23 11 failures Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

7 Solution Time

No. of failures

Failure density

Hazard rate

0 267

0.0668

0.00770

59

0.0150

0.00210

36

0.0090

0.0137

24

0.0060

0.0096

23

0.0058

0.0095

11

0.0028

0.0047

4 8 12 16 20 24 Sample calculation for time interval 0 – 4 is given below. 267

f d

4 1000

0.0668 267

Z 4

1000

(1000

0.0770

267 )

2

Probability density function Consider the life testing of N components. The total test last for T hours, at the end of which all specimens will have failed. Let n1

Number of components that failed during the 1st unit interval

n2 . . nl

Number of components that failed during the 2 unit interval

Then,

f d 1

n1

n2

n

Number of components that failed during the last unit interval 

nl

1

N

f d 2



f d l

1

where f d 1 , f d 2  etc. are the failure densities associated with the respective intervals. When the total population is large and the time interval is very small, the variation of failure density with time will be a smooth curve a nd the summation can be represented by integrals. Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

8

T

Therefore, with reference to the above figure,

f d ( )d

1

0

(where is a dummy variable) That is, the probability that a specimen will fail in T hours is 1 (that is, a certainty). In probability theory, the function f d ( ) is known as the probability density function. In order to make this function more general, the upper limit T is replaced by that no specimen in any test will last for an infinite number of hours. f d ( )d

This means

1

0

Reliability and Probability of failure in terms of failure density th th Reliability for the t hour is the number of survivors till the t hour divided by the initial population. N (n1 n2  nt ) nt n1 n2 That is, R(t ) 1  N N N N t

That is, R(t )

1

f d 1

f d 2



f d t

1

f d

d (for very small time intervals)

0 t

Probability of failure 1 R(t )

f d

d

0

Reliability in terms of hazard rate The hazard rate Z (t ) can be expressed as 2 R(t t ) R(t ) Z (t ) t R(t t ) R(t ) Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

9 When t is very small and tends to zero, the value of R(t ) approaches that of R(t t ) and we get R(t t ) R(t ) Z (t ) Lt . t 0 tR(t ) 1 dR(t ) R(t ) d (t )

Z (t )

d dt

Since

dR(t )

Lt .

t d (t ) d ln R(t ) Since dt

ln R(t )

For example,

d dx

0

R(t

t ) R(t )

R(t ) 1 dR(t ) R(t ) d (t )

1

ln( 2 x)

2 x

2

t

Integrating,

Z ( )d

ln R(t )

C

0

where C is a constant and At t

0, Reliability 1

is a dummy variable.

C 0

t

Z ( )d

ln R(t )

0 t

R(t )

exp(

Z ( )d 0

Failure density in terms of failure rate and reliability

f d (t )

When

1 N R(t t

t ) N R(t ) N

t is very small and tends to zero, f d (t )

Also we have, Z (t )

1 dR(t )

Z (t ) R(t )

R(t ) d (t ) Combining above two equations we get, f d (t ) Z (t ) R(t )

Lt .

R(t

t 0

t ) R(t ) t

dR(t ) dt

dR(t ) dt

Mean Time To Fail (MTTF) and Mean T ime Between Failures (MTBF) MTTF is the mean time to first failure and is used in case of components that are not repaired when they fail, but are replaced by new components. On the other hand, MTBF Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

10 is the mean time between two successive component failures and is used with repairable equipment or systems. Note: 1)

When the number of samples tested is small, it is possible to note the time to failure of each sample and the mean failure rate is given by the formula

Z (T ) where,

1 N (0) N (T ) N (0)

T

Z (T ) is the mean failure rate for T hours N (T) is the population remaining at time T N (0) is the population at T = 0

2) As the number of specimens tested becomes large it is tedious to record the time to failure of each specimen. Instead, the number which fail during specific intervals of time are recorded. Here mean time to failure for N specimens will be

where,

1

n1 t 2n2 t 3n3 t N t is the time interval

MTTF



l .nl t

n1 is the number of specimens that failed during the 1 st interval n2 is the number of specimens that failed during the 2 nd interval

. . n l is the number of specimens that failed during the last interval That is, in general, MTTF

1

l

N K 1

n K K t

Problem In the life testing of 10 specimens of a mini-mixer, the time to failure of each specimen is recorded as given in the following table. Calculate the mean failure rate for 900 hours and the mean time to failure for all ten specimens. Specimen Number 1 2 3 4 5 6 7 8 9 10

Time to Failure (hours) 805 810 815 820 825 832 842 856 875 900


11 Solution The mean failure rate can be calculated using the expression:

Z (T )

1 N (0) N (T )

T N (0) Here, Z (900) is to be calculated and N (0)=10 ; N (900)=0 1 900 0 1 Z (900 ) 1.11 10 3 ( per hour ) 900 900 900 Mean Time to Fail (MTTF) 1 ( MTTF ) 805 810 815 820 825 832 842 856 875 900 10

838 hours

Problem Ten transformers were tested for 500 hours each within the prescribed operating conditions, and one transformer failed exactly at the end of the 500 hours exposure. What is the failure rate for this type of transformer? Solution The mean failure rate can be calculated using the expression: 1 N (0) N (T ) Z (T ) T N (0) Z (500 )

1 500

10

9

10

1 5000

0.0002 failure / hour

Mean Time to Fail (MTTF) in integral form 1 l MTTF n K K t N K 1 Also, by definition failure density f d can be expressed as f d

n K N t

, where n K is the

th

number of failures associated with the K time interval n K f d K t N Hence the expression for MTTF can be written as l

MTTF

f d K ( tK ) t K 1

Further K t is the elapsed time t and therefore the expression for the MTTF becomes l

MTTF

tf d t K 1 l

For very small time intervals, MTTF

tf d dt 0


12 In the above equation, upper limit l is the number of hours after which there are no survivors. It is customary to replace this by infinity since all components will have failed at the end of an infinite test period. MTTF

tf d dt 0 t

Also we have, R(t )

1

fd ( )d 0

Differentiating with respect to time, dR(t )

That is, f d (t ) MTTF

t 0

dt

dR(t ) dt

dR(t )

f d (t )

dt

and substituting this in the equation for MTTF tdR(t )

dt 0

Integrating by parts, MTTF

tR(t )

R(t )dt

0



udv

uv

0

R(t )dt

MTTF

{Since R(0)

1 and R( )

0}

0


vdu

13

Hazard Models The data obtained from failure tests can be analyzed to obtain reliability, failure density, hazard rate and other necessary information. Obviously, the behavioural characteristics exhibited by one class of components differ from those exhibited by another class of components. In order to compare different behavioural characteristics and also to draw general conclusions from behavioural patterns of similar components, a mathematical model representing the failure characteristics of the components becomes necessary. The procedure involves assuming a function for hazard rate and thereby o btaining reliability and failure density by using this failure rate function. The assumed function for the hazard rate will be the hazard model. Some of the common hazard models are discussed below. Constant hazard model Here the failure rate is assumed to remain constant with t ime. That is, Z (t ) , a constant. t

R(t )

exp

t

Z ( )d

exp

0

exp

d

exp

t

0

That is, for a constant hazard model, Reliability, R (t ) Probability of failure, F (t )

t 0

1 R(t )

Failure density, f d (t ) Z (t ) R (t )

e

1 e

e

t

t

t

The variation of failure rate, reliability, probability of failure, and failure density with respect to time for a constant hazard model is shown in the following figure.

Variation of failure rate, reliability, probability of failure, and failure density for a constant hazard model


14 It can be seen that, for a constant hazard model the mean time to failure is the reciprocal of failure rate. That is, MTTF

R(t )dt

e

0

t

e

dt

t

1

0

1

e0

e

1

0 1

0

The constant hazard model is also known as exponential reliability case. Linearly increasing hazard model Here the hazard rate is assumed to increase linearly with time. That is, Z (t ) Kt , where K is a constant t

R(t )

exp

t

Z ( )d

exp

0

exp

K d 0

K

2

2

t

Kt 2

exp

2

0 Kt 2

That is, for a linearly increasing hazard model, Reliability, R(t )

e

2

Kt 2

Probability of failure, F (t )

1 R(t )

1 e

2

Kt 2

Failure density, f d (t ) Z (t ) R(t )

Kt e

2

The variation of failure rate, reliability, probability of failure, and failure density with respect to time for a linearly increasing hazard model is shown in the following figure.

Variation of failure rate, reliability, probability of failure, and failure density for a linearly increasing hazard model

It can be seen from the failure density curve that the curve has a slope equal to K at time t 0 . Also the value of f d (t ) reaches a maximum of

K e

at time t

1 K

, and tends

to zero as t becomes larger. Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

15

The Weibul Model 1 This model is expressed as Z (t ) Kt m , m Here K and m are parameters and if these are chosen appropriately, a variety of failurerate situations can be covered, including both the constant hazard and linearly increasing hazard conditions. If m If m

0 ; Z (t ) K - Constant hazard model 1 ; Z (t ) Kt - Linearly increasing model t

R(t )

exp

t

Z ( )d

exp

0

K

m

exp

d

m 1

0

Kt m

That is, in case of Weibul model, Reliability, R(t ) Kt m

Probability of failure, F (t )

1 R(t )

1 e Kt m

Failure density, f d (t ) Z (t ) R (t )

m

Kt e

m 1

K

e

t

exp 0

Kt m

1

m 1

1

m 1

1

m 1 1

m 1

Following figure shows the variation of reliability in case of Weibul model for various values of K and m

Variation of Reliability in case of Weibul model


16 Bath Tub Curve Component failure rate as a function of age follows a curve that is concave upward, as shown in the above figure. Because of its shape, this curve is also referred to as “bath tub curve”. This curve exhibits three distinct zones. The first is the short initial period called variously the early failure, infant mortality, or the burn in period. The decreasing but greater failure rate early in life of the system is due to one or more of several potential causes. The causes include inadequate testing or screening of components during selection or acceptance, damage to components during production, assembly, or testing, and choice of components which have too great a failure variability. It shall be a specific goal of the supplier to ensure that the early failure period is rigorously controlled and covered by a suitable warranty.

Bath tub curve

The failures in the second zone are termed service failures. During this period, the failure or hazard rate is constant and it represents the effective life of the product. The failures in the third zone are the wear-out failures. The incidence of failure in this zone is high since most of the components will have exceeded their service life, and consequently would have deteriorated. Hence, they are appropriately called wear-out failures. Note: Failure (death) rates for human beings are different by sex, race, nationality, and other factors but all failure rate for humans appear to exhibit this distinctive “bath tub curve”. The failure rate for infants is extremely high for the first few months, drops sharply, and remains fairly constant for many years and then slowly climbs as the person ages.

System reliability A system or a complex product is an assembly of a number of parts or components. The components may be connected in series or in parallel, or it may be a mixed system, where the components are connected in series as well as in parallel. Series configuration If the components of an assembly are connected in series the failure of any component causes the failure of the assembly or system. Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

17 Let us consider a system consisting of n units which are connected in series as shown in the following figure.

Reliability block diagram of a system having n components connected in series

Let the successful operation of by X 1 , X 2 , , X n and

these individual units their respective

be

represented probabilities

by P ( X 1 ), P ( X 2 ),, P ( X n ) . For the successful operation of the system, it is necessary that all n units function satisfactorily. Hence, the probability of the simultaneous successful operation of all units is P ( X 1 andX 2 and  and . X n ) . Therefore according to multiplication rule, R(t )

P ( X 1 andX 2 and  and . X n )

R(t )

P ( X 1 ) P ( X 2 X 1 ) P ( X 3 X 1 andX 2 )



P ( X n X 1 andX 2  andX n 1 )

In this expression, P ( X 2 X 1 ) represents the probability of the successful operation of unit 2 under the condition that unit 1 operates successfully. Similarly, P ( X n / X 1 andX 2 and  X n 1 ) represents the probability of the successful operation of unit n under the condition that all the remaining units 1,2,3,…………,n-1 are working successfully.If the successful operation of each unit is independent of the successful operation of the remaining units, then events X 1 , X 2 , , X n are independent and the above equation becomes R(t )

P ( X 1 ) P ( X 2 )  P ( X n )

That is, R(t ) R1 R2 R3  Rn , where R1 R2 R3  Rn are component reliabilities. Parallel configuration Several systems exist in which successful operation depends on the satisfactory functioning of any one of their n sub-systems or elements. They are said to be connected in parallel. Let us consider a system consisting of n units which are connected in parallel as shown in the following figure.


18

Reliability block diagram of a system having n components connected in parallel

Let X 1 , X 2 , , X n represent the successful operation of units 1,2,……….,n respectively. Similarly, let X 1 , X 2 , , X n represent the unsuccessful operation. If P ( X 1 ) is the probability of successful operation of unit 1, then P ( X 1 ) is the probability of its failure. Further, P ( X 1 )

1 P ( X 1 )

For the complete failure of the system, all n units have to fail simultaneously. If F (t ) is the probability of failure of the system, then F (t )

P ( X 1 and X 2 and  and . X n )

F (t )

P ( X 1 ) P ( X 2 X 1 ) P ( X 3 X 1 and X 2 )



P ( X n X 1 and X 2  and X n 1 )

In this expression, P ( X 3 / X 1 and X 2 ) represents the probability of failure of unit 3 under the condition that units 1 and 2 have failed. The other terms can also be interpreted in the same manner. If the unit failures are independent of each other, then F (t )

R(t )

P ( X 1 ) P ( X 2 )  P ( X n )

1

That is, R(t )

1 P X 1 1

(1 P X 2

1 R (t )



1 P X n

1 R1 (1 R2 )  (1 R n )

where R1 R2 R3  Rn are component reliabilities


19

Mixed configuration If a system is having a mixed configuration, then it will have components connected in parallel as well as in series and the following figure indicates a system having components in series and parallel.

Reliability block diagram of a system having n components connected in parallel

If R1 , R2 , R3 , R4 , R5 and R 6 are the respective component reliabilities of the components in the above configuration, then system reliability can be expressed as R(t )

R1 R2 1 (1 R3 )(1 R4 )(1 R5 ) R6

Problem A certain type of electronic component has a uniform failure rate of 0.00001 per hour. What is the reliability for a specified period of service of 10,000 hou rs? Solution 0.000001 per hour t 10000 hours R(1000 ) e (0.00001)10000

e

0.1

0.90483 (or ) 90.483%

Problem Given a MTTF of 5000 hours and a uniform failure rate. What is the probability that the system failure occurs within 200 hours? Solution 1

(per hour) 5000 t = 200 hours 1 5000

200

R(200) e 0.96079 (or ) 96.079% F (200) 1 0.96079 0.03921 (or ) 3.921% Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

20

Problem The following reliability requirements have been set on the subsystems of a communication system. Subsystem Receiver Control System Power Supply Antenna

Reliability (for a 4 – hour period) 0.970 0.989 0.995 0.996

What is the expected reliability of the overall system? Solution Assuming that all the four components are necessary for the successful operation of the system, R(t ) R1 R2 R3 R4

R(t )

(0.970)(0.989)(0.995)(0.996)

0.950 (or ) 95%

Problem An element has a probability of successful operation of 60% over a given period of time. If 4 such components are connected in parallel estimate the improvement factor Solution Reliability when 4 components (each having a reliability of 0.6) are connected in parallel can be expressed as: R(t ) 1 (1 0.6)(1 0.6)(1 0.6)(1 0.6) 0.9744 0.9744 If x is the improvement factor, then 0.6 x 0.9744 x 1.624 0.6 Problem For the Reliability Block Diagram (RBD) shown in the following figure calculate the system reliability. The respective component reliability values are also indicated in the figure.

Reliability block diagram


21 Solution R(t ) (0.95 )(0.96 ) 1 (1 0.95 )(1 0.94 ) (0.90 ) 0.8183 (or ) 81.83% Problem The MTBF of equipment is 500 hours. What is the failure rate expressed in a) Failures / hour 6 b) Failures / 10 hours c) % failures / 1000 hours Is MTBF a guaranteed failure free period? Solution Failure rate,

1

1

MTBF

500

0.002 failures / hour

0.002 106

{Answer (a)}

6 2000 failures / 10 hours {Answer (b)}

0.002 1000 2 failures / 1000 hours = 200 % failures / 1000 hours

{Answer (c)}

MTBF cannot be regarded as a guaranteed failure free period as it is only a mean value of operating times between failures. Reliability Increasing Techniques One way of achieving high reliabilities is by introducing redundant parts. For example we may have two parts in parallel such that the system operates if at least one part operates. Here the probability that the system fails is equal to the probability both parts fail. If the failures are assumed to be independent, then the system reliability will be R(t ) = 1- (1 R1)(1- R2), where R1 and R2 are the reliability of the two parts respectively. If the reliability of each part is 0.95 at t ime t, then the reliability of the system is R(t ) = 1-(1-0.95)(1-0.95) = 0.9975 By adding a redundant part we have increased the reliability of system at time t from 0.95 to 0.9975 We have been assuming that both parts are operating whenever the system is on and the failure of one part does not affect the operation of other part. This is some times called hot standby and is not always practical. We may need to provide a cold stand by where the second part is switched into service when the first one fails. Then we must also take into account the reliability of the switch. If we assume we have, as before, two components with reliability 0.95 at time t and a switching device with reliability 0.98 at time t we have the system reliability at time t as R(t ) = 0.95 + (0.05)(0.98)(0.95) = 0.9966


22 The above equation is just the probability the first part is operating plus the probability the first part fails times the probability the switch operates times the probability the second part operates. It is to be noted that there is a point of diminishing returns in using redundancy configurations; an increase in the level of parallel redundancy employed increases size, weight, cost and volume of the equipment and often requires complicated failure-sensing devices whose reliability need to be co nsidered.

As this figure illustrates, as the amount of dollars invested in reliability increases, Operation and Support (OS) costs decrease. When the reliability is combined with OS, the result is total cost. The objective is to reach the lowest point in the total cost curve at which the benefits of reliability (expressed as total operating cost) are optimized with the cost of obtaining that level of reliability. It is usually necessary to perform trade- off calculations to determine the advisability of parallel redundancy versus improvement of the reliability of the bas ic subsystem by other means. Methods for improvement could include the following considerations: 1) Review the users’ needs to see if the function of the unreliable part is really necessary to the user. If not eliminate these parts from the design. That is, decrease the number of component parts for a system and believe in the vital few. For a system that contains items connected in series, assuming independence of their individual failures, the reliability of the system is the product of the reliabilities of the individual items. If a product has 5 parts, each with reliability 0.9, then system reliability is 0.9

5

0.59 whereas, if there would have been only

three parts, the reliability of the system would be 0.9

3

0.73

2) Review the selection of any parts that are relatively new and unproven. Use standard parts whose reliability has been proved. (However, be sure that the conditions of previous use are applicable to the new product.) Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin. (for 1 st year M.Tech. Mechanical Engineering batch)

23 3) Use derating to assure that stresses applied to the parts are lower than the stresses the parts can normally withstand. 4) Use “robust” design methods that enable a product to handle unexpected environments. 5) Control the operating environment to provide conditions that yield lower failure rates Common examples are (a) potting electronic components to protect them against climate and shock, and (b) use of cooling equipments to keep down ambient temperatures. 6) Specify replacement schedules to remove and replace low-reliability parts before they reach the wear-out stage. 7) Prescribe screening tests to detect infant-mortality failures and to eliminate substandard components. The tests include “burn in”, acce lerated life tests etc. 8) Conduct research and development to attain an improvement in the basic reliability of those components which contribute most of the unreliability.


Reliability 2013 1

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