ARTÍCULO 5º.-

Components are simulated for a specified mission time for depicting the duration of available (up) and unavailable (down) states. Up and down states will come alternatively, as these states are changing with time they are called state time diagrams. Down state can be due to unexpected failure and its recovery will depend upon the time taken for repair action. Duration of the state is random for both up and down states. It will depend upon PDF of time to failure and time to repair respectively.

Tofirst understand how component failures and simple repairs affect the system and to visualize the steps involved, the simulation procedure is explained here with the help of the two examples. Thefirst example is a repairable series system and the second example is two unit parallel system.

Example 1—Series System A typical power supply system consists of grid sup- ply, circuit beaker, transformer and bus. The success criterion is that the availability of power supply at the bus, which demands the successful functioning of all the components. So, it is simple four component series system. The reliability block diagram (Functional diagram) is shown in Fig.4.3.

In addition to success/failure logic, failure and repair Probability Density Functions (PDF) of components are the input to the simulation. The PDF of failure and repair for the components are given in Table 4.3. It is assumed that all the component failure/repair PDFs are following exponential distribution. However, the procedure is same even when component PDFs are non-exponential except that the random variants will be different as per the PDF. The simulation procedure is as follows:

Step 1: Generate a random number

Step 2: Convert this number into a value of operating time using a conversion method on the appropriate times to failure distribution (exponential in the present case)

Step 3: Generate a new random number

Table 4.2 Generation of random samples for different distributions

Distribution Random samples

Uniform (a, b) aþ ðb  aÞUi

Exponential (λ) _1

klnðUiÞ

Weibull (α, β) _{að ln U}

iÞ1=b

Normal (μ, σ) Xi¼ Xsr þ l

Xs¼ ð2 ln UiÞ1=2cosð2pUiþ1Þ

Lognormal (μ, σ) Generate Y = ln(X) as a normal variate with mean

Step 4: Convert this number into a value of repair time using conversion method on the appropriate times to repair distribution

Step 5: Repeat step 1–4 for a desired period of operating life. Step 6: Repeat steps 1–5 for each component

Step 7: Compare the sequences for each component and deduce system failure times, frequencies and other parameters.

Step 8: Repeat steps 1–7 for the desired number of simulations

Typical overview of up and down states for Class IV supply system is shown in Fig. 4.4. System failure time is sum of the component failure times if they are mutually exclusive. If there is any simultaneous failure of two or more components, failure time of component having largest value is taken for system failure time.

Reliability Evaluation with Analytical Approach

In the analytical (or algebraic analysis) approach, the system’s PDF/other reliability indices are obtained analytically from each component’s failure distribution using probability theory. In other words, the analytical approach involves the determi- nation of a mathematical expression that describes the reliability of the system in terms the reliabilities of its components.

Considering components to be independent the availability expression for power supply system (A) is given by the following expression:

A¼ A1A2A3A4 Ai¼ li liþ ki A¼ l1l2l3l4 ðl1þ k1Þðl2þ k2Þðl3þ k3Þðl4þ k4Þ Grid Supply Circuit

Breaker Transformer Bus

Fig. 4.3 Functional diagram of typical Class IV supply system (Problem 1)

Table 4.3 Failure and repair

rate of components S. no. Component Failure rate (/h) Repair rate (/h)

1 Grid supply 2e-4 2.59

2 Circuit breaker 1e-6 0.166 3 Transformer 2e-6 0.0925926 4 Bus 1e-7 0.0925926 5 Pump (1 and 2) 3.7e-5 0.0925926

Ai,λi, andμiare the availability, failure rate and repair rate of ith component (i = 1, 2, 3 and 4).

Example 2—Parallel System Typical emergency core cooling system of Nuclear Power Plant consists of a two unit injection pump active redundant system. One pump operation is sufficient for the successful operation of the system. The failure of the system occurs when both the pumps fail. The reliability block diagram (Functional diagram) is shown in Fig.4.5.

Typical overview of up and down states for emergency injection pumps branch having two pumps in parallel is shown in Fig.4.6. System failure time is the time when there is simultaneous failure of two pumps.

Considering components in the two-unit active redundant parallel pump system, to be independent, the unavailability (Q) is given by the following expression:

Grid supply Circuit Breaker Transformer Bus Class IV (System) Time Functional state Failure state

Q¼ Q1Q2 Qi¼ k i liþ ki Q¼ k1k2 ðl1þ k1Þðl2þ k2Þ

Qi,λi, and μi are the unavailability, failure rate and repair rate of ith component (i = 1 and 2).

Table4.4gives the comparison of both the approaches, analytical and simula- tion, for the two problems. In addition to the parameters such as average unavailability, expected number of failures, failure frequency, Mean time between failures and mean time to repair, simulation can give Cumulative Density Function (CDF) of random variable time between failures for the system under consideration.

Pump 1

Pump 2

Fig. 4.5 Functional block diagram of two unit pump active redundant system

Time Pump 1 Pump 2 System Functional state Failure state

Fig. 4.6 Overview of up and down states for emergency injection pumps branch

The CDF of problem 1 and problem 2 are shown in Figs.4.7and4.8respectively. Mission times of 107 and 108 h are considered for problem 1 and problem 2 respectively. Simulation results are from 104iterations in both the cases.

Table 4.4 Summary of results

Parameter Series Parallel

Analytical Simulation Analytical Simulation

Average unavailability 1.059e-4 1.059e-4 1.59e-7 1.61e-7

Avg. no. of failures 2030.78 2031.031 2.955 2.997

Failure frequency (/h) 2.031e-4 2.031e-4 2.95e-8 2.99e-8

Mean time between failure (h) 4924.21 4923.51 33.84e+6 33.36e+6

Mean time to repair (h) 0.5214 0.5214 5.39 5.37

0 0.2 0.4 0.6 0.8 1

0.00E+00 1.00E+04 2.00E+04 3.00E+04 4.00E+04 5.00E+04

Time between failure (hrs)

Cum. Prob. Fig. 4.7 CDF of series system 0 0.2 0.4 0.6 0.8 1

0.00E+00 2.00E+06 4.00E+06 6.00E+06 8.00E+06 1.00E+07

Time between failure (hrs)

Cum. Prob.

Fig. 4.8 CDF of parallel system