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## Reliability Engineering - Part 1

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**Product Probability Law of Series Components**If a system comprises a large number of components, the system reliability may be rather low, even though the individual components have high reliabilities, e.g. V-1 missile in WW II.**Transition of Component States**Normal state continues Component fails Failed state continues N F Component is repaired**Definitions of Reliability**• The probabilitythat an item will adequately perform its specified purpose for a specified period of time under specified environmental conditions. • The ability of an item to perform an required function, under given environmental and operational conditions and for a stated period of time. (ISO 8402)**Definition of Quality**The totality of features and characteristics of a product or service that bear on its ability to satisfy or implies needs (ISO 8402). Quality denotes the conformity of the product to its specification as manufactured, while reliability denotes its ability to continue to comply with its specification over its useful life. Reliability is therefore an extension of quality into the time domain.**REPAIR -TO-FAILURE PROCESS**MORTALITY DATA t=age in years ; L(t) =number of living at age t t L(t) t L(t) t L(t) t L(t) 0 1,023,102 15 962,270 50 810,900 85 78,221 1 1,000,000 20 951,483 55 754,191 90 21,577 2 994,230 25 939,197 60 677,771 95 3,011 3 990,114 30 924,609 65 577,822 99 125 4 986,767 35 906,554 70 454,548 5 983,817 40 883,342 75 315,982 10 971,804 45 852,554 80 181,765 After Bompas-Smith. J.H. Mechanical Survival : The Use of Reliability Data, McGraw-Hill Book Company, New York , 1971.**HUMAN RELIABILITY**t L(t), Number Living at Age in Years Age t R(t)=L(t)/N F(t)=1-R(t) repair= birth failure = death Meaning of R(t): (1) Prob. Of Survival (0.86) of an individual of an individual to age t (40) (2) Proportion of a population that is expected to Survive to a given age t. 0 1,023,102 1. 0. 1 1,000,000 0.9774 0.0226 2 994,230 0.9718 0.0282 3 986,767 0.9645 0.0322 4 983,817 0.9616 0.0355 5 983,817 0.9616 0.0384 10 971,804 0.9499 0.0501 15 962,270 0.9405 0.0595 20 951,483 0.9300 0.0700 25 939,197 0.9180 0.0820 30 924,609 0.9037 0.0963 40 883,342 0.8634 0.1139 45 852,554 0.8333 0.1667 50 810,900 0.7926 0.2074 55 754,191 0.7372 0.2628 60 677,771 0.6625 0.3375 65 577,882 0.5648 0.4352 70 454,548 0.4443 0.5557 75 315,982 0.3088 0.6912 80 181,765 0.1777 0.8223 85 78,221 0.0765 0.9235 90 21,577 0.0211 0.9789 95 3,011 0.0029 0.9971 99 125 0.0001 0.9999 100 0 0. 1.**1.0**P 0.9 Survival distribution 0.8 0.7 0.6 0.5 Probability of Survival R(t) and Death F(t) 0.4 0.3 Failure distribution 0.2 0.1 0 10 20 30 40 50 60 70 80 90 100**Time to Failure**The time elapsing from when the unit is put into operation until it fails for the first time, i.e. a random variable T. It may also be measured by indirect time concepts: • The number of times a switch is operated • The number of kilometers driven by a car • The number of rotations of a bearing • etc.**State Variable**The state of the unit at time t can be described by the state variable**Reliability, R(t)**= probability of survival to (inclusive) age t = the number of surviving at t divided by the total sample Unreliability, F(t) = probability of death to age t (t is not included) =the total number of death before age t divided by the total population**Reliability - R(t)**• The probability that the component experiences no failure during the the time interval (0,t] or, equivalently, the probability that the unit survives the time interval (0, t] and is still functioning at time t. • Example: exponential distribution**Unreliability - F(t)**• The probability that the component experiences the first failure during (0,t]. • Example: exponential distribution**FALURE DENSITY FUNCTION f(t)**No. of Failures (death) Age in Years 0.02260 0.00564 0.00402 0.00327 0.00288 0.00235 0.00186 0.00211 0.00240 0.00285 0.00353 0.00454 0.00602 0.00814 0.01110 0.01500 0.01950 0.02410 0.02710 0.02620 0.02020 0.01110 0.00363 0.00071 000012 0 1 2 3 4 5 10 15 20 25 30 35 40 45 50 60 65 70 75 80 85 90 95 99 100 23,102 5,770 4,116 3,347 2,950 12,013 9,543 10,787 12,286 14,588 18,055 23,212 30,788 41,654 56,709 99,889 123,334 138,566 134,217 103,554 56,634 18,566 2,886 125 0 0.00540 0.00454 0.00284 0.00330 0.00287 0.00192 0.00198 0.00224 0.00259 0.00364 0.00393 0.00436 0.00637 0.00962 0.01367 0.01800 0.02200 0.02490 0.02610 0.02460 0.01950 0.00970 0.00210 - - -**140**120 100 Number of Deaths (thousands) 80 60 40 20 Age in Years (t) 20 40 60 80 100**0.14**0.12 0.10 Failure Density f (t) 0.8 0.6 0.4 0.2 Age in Years (t) 20 40 60 80 100**Failure Density - f(t)**(exponential distribution)**CALCULATION OF FAILURE RATE r(t)**Age in Years No. of Failures (death) Age in Years No. of Failures (death) r(t)= r(t)= 0 1 2 3 4 5 10 15 20 25 30 35 23,102 5,770 4,116 3,347 2,950 12,013 9,534 10,787 12,286 14,588 18,055 23,212 0.02260 0.00570 0.00414 0.00338 0.00299 0.00244 0.00196 0.00224 0.00258 0.00311 0.00391 0.0512 40 45 50 55 60 65 70 75 80 85 90 95 99 30,788 41,654 56,709 76,420 99,889 123,334 138,566 134,217 103,554 56,634 18,566 2,886 125 0.00697 0.00977 0.01400 0.02030 0.02950 0.04270 0.06100 0.08500 0.11400 0.14480 0.17200 0.24000 1.20000**Random failures**Early failures Wearout failures 0.2 0.15 Failure Rate r(t) 0.1 0.05 20 40 60 80 100 Failure rate r(t) versus t.**Failure Rate,**(faults/time) Period of Approximately Constant failure rate Infant Mortality Old Age Time Figure 11-2 A typical “bathtub” failure rate curve for process hardware. The failure rate is approximately constant over the mid-life of the component.**Comments on Bathtub Curve**• Often units are tested before they are distributed to the users. Thus, much of the infant mortality will be removed before the units are delivered for use. • For the majority of mechanical units, the failure rate will usually show a slightly increasing tendency in the useful life period.**Failure Rate - r(t)**• The probability that the component fails per unit time at time t, given that the component has survived to time t. • Example: The component with a constant failure rate is considered as good as new, if it is functioning.**As Good As New?**This implies that the probability that a unit will be functioning at time t+x, given that it is functioning at time t, is equal to the probability that a new unit has a time to failure longer than x. Hence the remaining life of a unit, functioning at time t, is independent of t. The exponential distribution has no “memory.”**Interpretation of Failure Rate**In actuarial statistics, the failure rate is called the force of mortality (FOM). The failure rate or FOM is a function of the life distribution of a single unit and an indication of the “proneness of failure” of the unit after time t has elapsed.**Failure-Rate Experiment**Split the time interval (0, t) into disjoint intervals of equal length dt. Then put n identical units into operation at time t=0. When a unit fails, record the time and leave that unit out. For each interval, determine • The number of units n(i) that fail in interval i. • The functioning times of the individual units in interval i. If a unit has failed before interval i, its functioning time is zero.**Mean Time to Failure - MTTF**Variance of Time to Failure**Failure Rate**Failure Density Unreliability Reliability 1 1 Area = 1 1 - F (t) f (t) F (t) R (t) 0 0 t (a) t (b) t (c) t (d) Figure 11-1 Typical plots of (a) the failure rate (b) the failure density f (t), (c) the unreliability F(t), and (d) the reliability R (t).**TABLE 11-1: FAILURE RATE DATA FOR VARIOUS SELECTED PROCESS**COMPONENTS1 Instrument Fault/year Controller 0.29 Control valve 0.60 Flow measurement (fluids) 1.14 Flow measurement (solids) 3.75 Flow switch 1.12 Gas - liquid chromatograph 30.6 Hand valve 0.13 Indicator lamp 0.044 Level measurement (liquids) 1.70 Level measurement (solids) 6.86 Oxygen analyzer 5.65 pH meter 5.88 Pressure measurement 1.41 Pressure relief valve 0.022 Pressure switch 0.14 Solenoid valve 0.42 Stepper motor 0.044 Strip chart recorder 0.22 Thermocouple temperature measurement 0.52 Thermometer temperature measurement 0.027 Valve positioner 0.44 1Selected from Frank P. Lees, Loss Prevention in the Process Industries (London: Butterworths, 1986), p. 343.**Example**Consider two independent components with failure ratesλ1andλ2, respectively. Determine the probability that component 1 fails before component 2. Similarly,**A System with n Components in Parallel**• Unreliability • Reliability**A System with n Components in Series**• Reliability • Unreliability**Upper Bound of Unreliability for Systems with n Components**in Series**Assumptions of Homogeneous Poisson Process(HPP)**Suppose we are studying the occurrence of a certain event A inthe course of a given time period. Let us assume • A canoccur at any time in the interval. The probability of A occurring in the interval is independent of t and may be written as where is a positive constant. • The probability of more than one event A in this interval is , which is a function with property • Let (t11,t12], (t21,t22],…be any sequence of disjoint intervals in the time period in question. Then the events “A occurs in (tj1,tj2],” j= 1, 2, …, are independent. The process is said to have intensityλ**Probability of No Event Occurring in (0, t]**Let N(t) denotes the number of times the event A occurs during the period (0, t]. Let**Exponential Distribution**Let T1 denotes the time point when A occursfor the first time. T1 is a random variable and Thus, thee waiting time T between consecutive occurrences in a HPP is exponentially distributed.**Poisson Distribution**Since This distribution is called the Poisson distribution with parameter and random variable n Notice that, since the expected number of occurrences of event A per unit time (t=1) is , expresses the intensity of the process.**Example 1**Suppose that exactly one event (failure) of a HPP with intensityλis known to have occurred in the interval (0, t0]. Determine the distribution of the time T1 at which this event occurred.**Example 1**In other words, the time at which the first failure occurs in uniformly distributed over (0,t0]. The expected time is thus**Example 2**Suppose that the failure of a system are occurring in accordance with a HPP. Some failures develop into a consequence C, and others do not. The probability of this development is p and is assumed to be constant for each failure. The failure consequences are further assumed to be independent of each other. Determine the distribution of the consequences.**Example 2**Thus, M(t) is also a HPP with intensity (pλ) . The mean number of C consequences in (0,t] is (pλt).**Gamma Distribution**Consider a unit that is exposed to a series of shocks which occur a HPP with intensity λ. The time intervals T1, T2, T3, …, between consecutive shocks are then independent and exponentially distributed with parameter λ. Assume that the unit fails exactly at the kth shock, and not earlier. The time to failure of the unit is then gamma distributed (k, λ).**Gamma Distribution**Consider a homogeneous Poisson distribution, The waiting time until the kth occurrence of event A in a HPP with intensity λis gamma distributed. The gamma distribution (1, λ) is an exponential distribution with parameter λ.**Weibull Distribution**For majority of mechanical units, the failure rates are slightly increasing in the useful life period (not constant). A distribution often used when r(t) is monotonic is the Weibull distribution. The time to failure T of a unit is said to be Weibull distributed with scale parameterλand shape parameterα.**Weibull Distribution**• α=1: The Weibull distribution reduces to exponential distribution. • α>1: The failure rate is increasing. • α<1: The failure rate is decreasing.