Maintenance and Renewal Cost Evaluation for Managing Assets of Electric Power Equipment and Operational Data Analysis for Failure Rate Estimation

Abstract

In recent years, “asset management” or “managing assets” technique has been expected to rationalize maintenance and operation of electric power equipment, especially for aging equipment. Some concrete support tools have been developed by considering life cycle cost for substation equipment in “Central Research Institute of Electric Power Industry, Japan,” which include failure risk evaluation. Such cost and risk evaluation are essential for comparative evaluation of different types of equipment. Failure probability is one of the most important factors for the evaluation. Because of its high reliability, electric power equipment can be expected to have a very long lifetime, therefore, durability test is not applicable, but rather relies on analysis of actual operational data. Collection, accumulation, and analysis of actual operational data are necessary for accurate evaluation. This chapter describes the evaluation method for the managing assets, and data collection and analysis to improve the accuracy of failure probability estimation.

1. Introduction

In general, power transmission and distribution equipment that can be expected to operate for more than several decades had a wide age distribution over time. The amount of capital investment is inevitably affected by society and the economy, therefore, the shift to the aging side of the age distribution is progressing as a common phenomenon in many countries in recent years. In US and European countries, the issue of aging has been discussed since the end of the 20th century [1], and the effective introduction of the so-called asset management technique for formulating maintenance and management strategies that appropriately balance risk and cost has been examined.

2. Maintenance and renewal cost evaluation for power transmission and distribution equipment

In the asset management for corporate activities, benefits as positive impacts and risks as negative impacts are generally evaluated and added for each possible activity to be considered as an evaluation index to select the optimal strategy. In the case of electric power transmission and distribution equipment, it is difficult to evaluate the contribution of individual equipment because the entire network system generates benefits, therefore in many cases, the optimal maintenance strategy is selected by minimizing the costs and statistically evaluating risks required to maintain the network size and reliability. As one simple model, CRIEPI has proposed the cumulative cost evaluation method, and some support programs have been developed [2, 3, 4, 5].

2.1 Maintenance and renewal “cost” in operation

Maintenance and renewal costs during normal operation are classified into the following four items based on their expenditure timing and characteristics of change by age.

1. Average repairing cost.

   In ordinary operations, there are some necessary repairing costs, such as oil leakage repair cost for power transformers. Generally, it can be assumed to increase with age. For example, its characteristic is assumed to be proportion to age.

2. Inspection cost.

   Generally, regulated inspection cost is needed commonly for each equipment. A periodic and a nonperiodic (which is performed at a certain age) inspection costs are considered.

3. Overhaul cost.

   Some equipment can be applied so-called “Overhaul” to realize rejuvenation as a maintenance measure. Overhaul costs depending on their effect is considered.

4. Installation cost of planned renewal.

   The installation cost of equipment can be regarded as installments over several years, by considering depreciation. The property tax should also be considered during these years.

2.2 Statistically expected failure cost as “risk”

The expense required at a failure is installation cost of renewed equipment and some so-called penalty costs. The “penalty” cost should include the lost revenue from selling electricity, the emergent recovery cost, a penalty resulting from service interruption, and so on. Since the occurrence of a failure is statistical, the expense is “statistically expected cost,” which is the product of the “cost of failure” and the “probability of failure.” This cost is not a real cash flow, but it is expressed in monetary values and can be compared and combined with maintenance and renewal costs. When some aged equipment in service is removed as a result of failure, the same number of new equipment should be installed in order to maintain its power network scale. That is, the total number of equipment does not change. From the statistical point of view, such failures are occurred every year, depending on their failure probability. This means that the age distribution changes over time, which should be considered when the cumulative cost evaluation is carried out [4].

2.3 Maintenance scenarios to be compared

In general, when the asset management techniques are utilized for maintenance and renewal planning, it is necessary to consider possible maintenance measures and scenarios, in advance. The cumulative cost evaluation should be carried out for each scenario. Therefore, this scenario setting is important for this method. As one example, time-based renewal scenarios (such as at 40 and 50 years) with and without overhaul (OH) have been considered, as shown in Figure 1. The OH is assumed to rejuvenate equipment at a certain cost. Its effect (rejuvenation years), cost, and timing are specified as parameters.

Figure 1.

Output example of CRIEPI’s support program.

3. Failure data analysis based on operation results

One of the most important items in the managing assets is the risk evaluation, and failure probability estimation for each equipment is crucial. The failure probability distribution of a product is generally obtained from endurance tests on a large number of the same products. However, it is not practical to conduct such endurance tests for electric power transmission and distribution equipment, which can generally be expected to operate for more than several decades. In order to investigate the failure probability characteristics of long-life products, statistical analysis of residual performance tests of removed products from a real field, and operation/failure results in a real field are often employed. This section describes statistical analysis methods for operational data.

3.1 Definition of failure rate

Failures that are generally considered with a failure rate for an industrial product include those that stop the operation of the product once they occur and those that are repaired repeatedly each time they occur. The former determines the service life of the product, and the rate of occurrence is usually evaluated as a function of operating hours. The latter usually focuses on the interval of occurrence of multiple failures, and the change over time of the average time or the occurrence rate in a certain operating time is evaluated. In this section, “failure rate” means the occurrence rate of the former failures at a certain age, and is expressed as a function of age.

If a large number of the same products start operating simultaneously, the percentage of those that continue to operate without failure up to a certain elapsed time t is generally called the reliability, and it is often expressed as R(t). The cumulative failure probability, which means the percentage of failures between the start of operation and t, is expressed as follows:

$$\mathrm{F}\left(\mathrm{t}\right)=1-\mathrm{R}\left(\mathrm{t}\right)$$E1

F(t) differentiated by t is often denoted as f(t).

$$\mathrm{f}\left(\mathrm{t}\right)=\frac{d}{\mathit{dt}}\mathrm{F}\left(\mathrm{t}\right)$$E2

This is the increment of F(t) at time t, i.e., the percentage of products that fail at time t, for all products. It is sometimes called the “failure probability” because it is the time derivative of the “cumulative failure probability,” but it is also called the probability density of failure because it represents the probability distribution of when the product will fail. When examining the risk of equipment in service, the probability that equipment that has been operating until age t will fail by the following year is often utilized. This is often denoted as λ(t) and is obtained as follows:

$$\lambda \left(t\right)=\frac{\mathrm{f}\left(\mathrm{t}\right)}{\mathrm{R}\left(\mathrm{t}\right)}=\frac{d\mathrm{F}\left(\mathrm{t}\right)}{\mathit{dt}}\bullet \frac{1}{\mathrm{R}\left(\mathrm{t}\right)}=-\frac{d\mathrm{R}\left(\mathrm{t}\right)}{\mathit{dt}}\bullet \frac{1}{\mathrm{R}\left(\mathrm{t}\right)}$$E3

In this chapter, this is referred to as the “failure rate,” and a method for estimating it from operation results is discussed.

3.2 Characteristics of power transmission and distribution equipment operational data for failure rate estimation

In general, to examine the failure rate characteristics of a product, an endurance test is conducted using several units of the same product. To examine aging characteristics, test samples are usually operated simultaneously and the time required for each sample to reach failure is determined. In conducting endurance tests, it is not always possible to continue the test until all samples fail due to time and cost constraints, but some statistical analysis methods can analyze data obtained by discontinuing the endurance test in the middle of the test. In the case of power transmission and distribution equipment, it is difficult to plan an endurance test in which a sufficient number of test samples are operated simultaneously, but a method to estimate the failure rate by considering the actual results of the long-term operation of a large number of facilities in a real field as a pseudo endurance test is conceivable. In this case, some considerations need to be made for the data used in the analysis.

3.2.1 Data for failure rate estimation

In order to estimate the failure rate, information on equipment that has been in operation without failure is needed in addition to the equipment that has failed. For a group of equipment that is the same type and assumed to exhibit the same failure rate aging characteristics, it is necessary to investigate the age of each failed facility as well as the age distribution of the group in operation.

3.2.2 Observation period

Failure rate estimation based on operational data for a group of equipment with age distribution starts by determining the number of failures/operation equipment at each age in order to obtain an approximation of λ(t). In doing so, it is necessary to consider the period of the survey (observation period). For example, if the past 10 years of failure history is to be investigated, the number of failures for each age can be obtained by adding up the 10 years of information for each individual failure by age of occurrence. Similarly, for the number of operations, the actual number of operations (the age distribution of equipment in operation) for each year of the observation period should be surveyed and added. For example, equipment that has continued to operate without failure at 20 years old has existed every year for the past 10 years, so they can be added up to the amount of equipment in operation at 20 years old in the aggregate. However, for example, equipment that is currently 40 years old should not be combined as “equipment that has continued to operate at 20 years old without failure” because this information is outside the observation period, though it is clear that the equipment continued to operate for 20 years without failure 20 years ago. This is because the information on equipment that has failed or has been removed outside the observation period cannot be used. Only “should have been in operation” information is likely to lead to underestimation of the failure rate. Therefore, the “observation period” is important and should be paid attention in the failure rate estimation.

3.2.3 Influence of low number of failure results

The inherent difficulty in statistical analysis of the failure rate from operational data of power transmission and distribution equipment lies in the fact that such equipment has a low failure rate and is highly reliable, and that preventive maintenance is performed to maintain high supply reliability. These suppress the occurrence of failures during operation, resulting in a decrease in the accuracy of failure rate estimation and underestimation. There is no other way to deal with these problems than to continuously accumulate appropriate data and increase the amount of data.

3.3 Procedure of statistical analysis

This section presents a computer simulation of virtual operational data and uses the results to show a specific procedure for estimating failure rates [6].

3.3.1 Simulated data

The simulating equipment is a group of 12,340 units with the aging distribution shown in Figure 2, all of which are assumed to have the same failure rate characteristics shown in Figure 3. The Weibull distribution is assumed for the failure rate characteristics, and the failure rate λ(t) and probability density of failure f(t) are expressed as follows:

Figure 2.

Assumed age distribution (12,340 units).

Figure 3.

Assumed failure characteristics.

$$\lambda \left(\mathrm{t}\right)=\frac{m}{t_s}·{\left(\frac{t}{t_s}\right)}^{m-1}$$E4

$$\mathrm{f}\left(\mathrm{t}\right)=\frac{m}{t_s}·{\left(\frac{t}{t_s}\right)}^{m-1}·exp\left\{-{\left(\frac{t}{t_s}\right)}^m\right\}$$E5

where m is the shape parameter and t_s is the scale parameter, and in Figure 2, m and t_s (years) are set to 4 and 80, respectively.

It can be simulated that after 1 year of operation of this equipment group, some equipment will fail according to the failure rate determined by each age. For each aged equipment, a random number between 0 and 1 is generated, and when the value is less than the failure rate at its age the unit is regarded to fail. In the following year, the number of failures is subtracted from the amount of equipment in each age, and the age distribution is shifted to the higher side by 1 year, and new equipment equal to the number of failures in the previous year is regarded to be installed, assuming maintenance that keeps the total amount of equipment in the group constant. If this is continued for several years, which corresponds to “observation period,” operational data can be generated as one simulation case, but its result would depend on the random number output, so different results would be obtained each time of the simulation. One simulation result whose observation period is 5 years is shown in Table 1. Table 1 also includes the combined number at each age during the observation period. The statistical analysis in the following section is performed on this combined operational data. Such simulation was performed five times.

age (year)	1st year		2nd year		3rd year		4th year		5th year		total
	operating	failed	operating	failed	operating	failed	operating	failed	operating	failed	operating	failed
1	10	0	22	0	22	0	23	0	25	0	102	0
2	10	0	10	0	22	0	22	0	23	0	87	0
3	10	0	10	0	10	0	22	0	22	0	74	0
4	20	0	10	0	10	0	10	0	22	0	72	0
5	250	0	20	0	10	0	10	0	10	0	300	0
6	350	0	250	0	20	0	10	0	10	0	640	0
7	280	0	350	0	250	0	20	0	10	0	910	0
8	200	0	280	0	350	0	250	0	20	0	1100	0
9	280	0	200	0	280	0	350	0	250	1	1360	1
10	270	0	280	0	200	0	280	0	350	0	1380	0
11	280	0	270	0	280	0	200	0	280	0	1310	0
12	300	1	280	0	270	0	280	0	200	0	1330	1
13	340	0	299	0	280	0	270	0	280	0	1469	0
14	550	0	340	0	299	0	280	0	270	0	1739	0
15	540	0	550	0	340	0	299	0	280	1	2009	1
16	580	0	540	0	550	0	340	2	299	0	2309	2
17	590	0	580	0	540	0	550	0	338	0	2598	0
18	590	1	590	1	580	0	540	0	550	0	2850	2
19	600	1	589	0	589	0	580	0	540	0	2898	1
20	600	2	599	1	589	2	589	0	580	0	2957	5
21	600	0	598	1	598	2	587	1	589	1	2972	5
22	600	0	600	0	597	0	596	1	586	0	2979	1
23	440	0	600	0	600	2	597		595	1	2832	3
24	480	3	440	0	600	0	598	1	597	4	2715	8
25	480	1	477	1	440	0	600		597	2	2594	4
26	310	0	479	0	476	2	440	1	600	1	2305	4
27	310	1	310	3	479	0	474		439	0	2012	6
28	290	0	309	2	307	0	479	1	472	1	1857	4
29	200	1	290	1	307	1	307	0	478	0	1582	3
30	210	1	199	0	289	0	306	2	307	0	1311	3
31	280	1	209	1	199	0	289	0	304	0	1281	2
32	210	2	279	0	208	0	199	0	289	2	1185	4
33	200	1	208	0	279	3	208	0	199	1	1094	5
34	180	2	199	0	208	1	276	0	208	1	1071	4
35	190	1	178	1	199	4	207	0	276		1050	9
36	150	0	189	1	177	0	195	1	207	1	918	3
37	125	0	150	1	188	0	177	0	194	1	834	2
38	100	1	125	0	149	1	188	1	177	1	739	4
39	80	1	99	1	125	0	148	1	187		639	5
40	70	0	79	2	98	0	125	2	147	1	519	5
41	50	0	70	0	77	0	98	3	123		418	3
42	40	0	50	1	70	2	77	2	95	1	332	6
43	30	0	40	1	49	1	68	1	75		262	3
44	20	1	30	0	39	2	48	0	67	1	204	4
45	15	0	19	0	30	0	37	0	48	2	149	2
46	12	0	15	0	19	0	30	1	37	0	113	1
47	7	0	12	3	15	0	19	1	29	0	82	4
48	5	0	7	0	9	0	15	0	18	0	54	0
49	3	0	5	0	7	0	9	0	15	0	39	0
50	3	0	3	0	5	0	7	0	9	0	27	0
51	0	0	3	0	3	0	5	0	7	0	18	0
52	0	0	0	0	3	0	3	1	5	0	11	1
53	0	0	0	0	0	0	3	0	2	0	5	0
54	0	0	0	0	0	0	0	0	3	0	3	0
55	0	0	0	0	0	0	0	0	0	0	0	0

Table 1.

One example of simulated operational data.

3.3.2 Hazard analysis

Failure characteristics of high voltage equipment are often expressed as a Weibull distribution. In other words, the fitting of an endurance test and operational performance data is performed assuming that the probability density distribution of failure can be expressed as a Weibull distribution function. The Hazard analysis is a statistical analysis method for this purpose. In this simulation, as described in Section 3.3.1, the true distribution is given as Weibull distributions, therefore, if hazard analysis is performed with high accuracy, the distribution is expected to be restored.

When the failure characteristic follows a Weibull distribution, the failure rate λ(t) is expressed by Eq. (4). This is integrated over time as in the following equation and is called the cumulative hazard H(t).

$$\mathrm{H}\left(\mathrm{t}\right)={\int }_0^t\lambda \left(\tau \right)\mathit{d\tau }={\left(\frac{t}{t_s}\right)}^m$$E6

Taking the natural logarithm of both sides of Eq. (6), the following equation is obtained.

$$ln\left\{\mathrm{H}\left(\mathrm{t}\right)\right\}=ln\left\{{\left(\frac{t}{t_s}\right)}^m\right\}=m\bullet ln\left(t\right)-m\bullet ln\left(t_s\right)$$E7

Plotting as y = ln{H(t)} and x = ln(t) results in a straight line with slope m and y-intercept -mln(t_s). Utilizing Eq. (7), the natural logarithm of the approximate value $\widehat{H}\left(t\right)$ of the cumulative hazard H(t) obtained from endurance tests or operational data is plotted against the natural logarithm of age t, to estimate the shape parameter m and scale parameter t_s by linear approximation. When the number of operations at age t is N(t) and the number of failures is n(t), the approximate value $\widehat{\lambda }\left(t\right)$ of the failure rate λ(t) is obtained from the following equation.

$$\widehat{\lambda }\left(\mathrm{t}\right)=\frac{\mathrm{n}\left(\mathrm{t}\right)}{\mathrm{N}\left(\mathrm{t}\right)}$$E8

Using $\widehat{\lambda }\left(t\right)$, $\widehat{H}\left(t\right)$ is obtained as follows:

$$\widehat{\mathrm{H}}\left(\mathrm{t}\right)=\sum _{\tau =1}^{\mathrm{t}}\widehat{\lambda }\left(\tau \right)=\widehat{\mathrm{H}}\left(\mathrm{t}-1\right)+\widehat{\lambda }\left(\mathrm{t}\right)$$E9

Table 2 summarizes the results of these calculations using the combined data in Table 1. Table 2 also includes the natural logarithm of t and $\widehat{H}\left(t\right)$ for graph plotting. In the hazard plot that shows the relationship between ln(t) and $ln\left(\widehat{H}\left(t\right)\right)$, plotting is carried out only when $\widehat{H}\left(t\right)$ is changed, that is, when $\widehat{\lambda }\left(t\right)\ne 0$. The hazard plots created from ln(t) and $\widehat{H}\left(t\right)$ in Table 2 are shown in Figure 4. Figure 4 includes linear approximation, from which the shape parameter m = 3.57 and scale parameter t_s = 82.2 years are obtained from the slope and the intercept.

age t (year)	total		approx. Failure rate $\widehat{\lambda }$(t)	approx. Cum. hazard $\widehat{\mathrm{H}}$(t)	ln (t)	ln ($\widehat{\mathrm{H}}$(t))
	operating	failed
1	102	0	0/102 = 0.0000	0.0000	—	—
2	87	0	0/87 = 0.0000	0.0000	—	—
3	74	0	0/74 = 0.0000	0.0000	—	—
4	72	0	0/72 = 0.0000	0.0000	—	—
5	300	0	0/300 = 0.0000	0.0000	—	—
6	640	0	0/640 = 0.0000	0.0000	—	—
7	910	0	0/910 = 0.0000	0.0000	—	—
8	1100	0	0/1100 = 0.0000	0.0000	—	—
9	1360	1	1/1360 = 0.0007	0.0007	2.197225	−7.21524
10	1380	0	0/1380 = 0.0000	0.0007	—	—
11	1310	0	0/1310 = 0.0000	0.0007	—	—
12	1330	1	1/1330 = 0.0008	0.0015	2.484907	−6.51088
13	1469	0	0/1469 = 0.0000	0.0015	—	—
14	1739	0	0/1739 = 0.0000	0.0015	—	—
15	2009	1	1/2009 = 0.0005	0.0020	2.70805	−6.22217
16	2309	2	2/2309 = 0.0009	0.0029	2.772589	−5.86005
17	2598	0	0/2598 = 0.0000	0.0029	—	—
18	2850	2	2/2850 = 0.0007	0.0036	2.890372	−5.64
19	2898	1	1/2898 = 0.0003	0.0039	2.944439	−5.54731
20	2957	5	5/2957 = 0.0017	0.0056	2.995732	−5.18698
21	2972	5	5/2972 = 0.0017	0.0073	3.044522	−4.92383
22	2979	1	1/2979 = 0.0003	0.0076	3.091042	−48,787
23	2832	3	3/2832 = 0.0011	0.0087	3.135494	−4.74832
24	2715	8	8/2715 = 0.0029	0.0116	3.178054	−4.45565
25	2594	4	4/2594 = 0.0015	0.0132	3.218876	−4.33097
26	2305	4	4/2305 = 0.0017	0.0149	3.258097	−4.20705
27	2012	6	6/2012 = 0.0030	0.0179	3.295837	−4.0245
28	1857	4	4/1857 = 0.0022	0.0200	3.332205	−3.91071
29	1582	3	3/1582 = 0.0019	0.0219	3.367296	−3.82024
30	1311	3	3/1311 = 0.0023	0.0242	3.401197	−3.72095
31	1281	2	2/1281 = 0.0016	0.0258	3.433987	−3.65846
32	1185	4	4/1185 = 0.0034	0.0291	3.465736	−3.53538
33	1094	5	5/1094 = 0.0046	0.0337	3.496508	−3.38972
34	1071	4	4/1071 = 0.0037	0.0375	3.526361	−3.28467
35	1050	9	9/1050 = 0.0086	0.0460	3.555348	−3.07858
36	918	3	3/918 = 0.0033	0.0493	3.583519	−3.00999
37	834	2	2/834 = 0.0024	0.0517	3.610918	−2.96248
38	739	4	4/739 = 0.0054	0.0571	3.637586	−2.8629
39	639	5	5/639 = 0.0078	0.0649	3.663562	−2.73448
40	519	5	5/519 = 0.0096	0.0746	3.688879	−2.59613
41	418	3	3/418 = 0.0072	0.0817	3.713572	−2.50423
42	332	6	6/332 = 0.0181	0.0998	3.73767	−2.30448
43	262	3	3/262 = 0.0115	0.1113	3.7612	−2.19587
44	204	4	4/204 = 0.0196	0.1309	3.78419	−2.03356
45	149	2	2/149 = 0.0134	0.1443	3.806662	−1.93591
46	113	1	1/113 = 0.0088	0.1531	3.828641	−1.87639
47	82	4	4/82 = 0.0488	0.2019	3.850148	−1.59987
48	54	0	0/54 = 0.0000	0.2019	—	—
49	39	0	0/39 = 0.0000	0.2019	—	—
50	27	0	0/27 = 0.0000	0.2019	—	—
51	18	0	0/18 = 0.0000	0.2019	—	—
52	11	1	1/11 = 0.0909	0.2928	3.951244	−1.22816
53	5	0	0/5 = 0.0000	0.2928	—	—
54	3	0	0/3 = 0.0000	0.2928	—	—
55	0	0	0/0 = −	^—	^—	—

Table 2.

One example of processing data for hazard plotting.

Figure 4.

One example of hazard plots.

The Weibull distribution obtained by such procedure is a “sample mean” obtained from the average characteristics of a “sample” of equipment operational data and is expected to be different each time the sample is taken, in this case, each time the operational data is simulated. On the other hand, the true failure rate characteristic, the “population mean,” is determined first in this discussion, therefore they can be compared. The failure rates and probability density distributions of failures obtained from the results of five simulations, including the data in Table 1, are shown in Figure 5. Figure 5 also shows the true failure rate characteristics. Some of the five estimates (sample mean) have failure rates that are close to the true values, while others are higher or lower. Only one of the sample means, which is expected to vary, can be observed in reality, and there is no way to know the deviation from the true value. The only fundamental solution to the low estimation accuracy due to the low failure rate described in Section 3.2.3 is to increase the amount of data by investigating and accumulating the actual operation of equipment over a long period of time.

Figure 5.

Results of hazard analysis for 5 simulated data. (a) failure rate (b) probability density of failure.

3.4 Failure data and preventive renewal data

As mentioned in section 3.2.3, another issue in estimating failure rates from actual operational results is that failures do not occur as actual results because preventive renewals are performed in actual maintenance. This section discusses the addition of renewal data to the failure rate estimation.

Reference [7] introduces the failure rate and renewal rate of transformers and points out that the failure rate does not increase over time but the renewal rate does, and that the failure rate characteristic should be what is shifted renewal rate characteristic to the right (toward the high aging side) if no preventive maintenance is performed since the failure would have occurred years later if no renewal was performed. In risk evaluation for the examination of maintenance and renewal plans, the use of failure rate characteristics based on the operational data without consideration of the preventive renewal effect is clearly an underestimate. Reference [8] analyzed operational performance data for transformers and shunt reactors and estimated failure rates assuming that equipment that was replaced before failure was the one that would fail 5 or 10 years later. In order to conduct such a study, it is necessary to investigate and accumulate the actual field data of not only operations and failures, but renewals with reasons.

An example of a survey of equipment operational data, including renewal data, is the questionnaire survey [9] conducted by “Investigating R&D Committee for Asset Management for Electric Power Equipment Based on Insulation Diagnosis” of IEE Japan. This is a survey of failure and renewal data in 10 years conducted on approximately 200 plant manufacturers and other companies. Reference [6] has tried to utilize the results of this survey to estimate failure rate characteristics by combining failure and renewal data. Although how to combine them should be examined according to the reasons for each renewal and the characteristics of the target equipment, the reasons for renewal were not investigated in the survey, therefore, the analysis was conducted by assuming that the failure should have occurred at +5 years after the year of each renewal [6]. The results for CV cables from 6 kV class to 60 kV class are shown in Figures 6 and 7. For example, the failure rate at 30 years old is about 26 times higher than without taking the renewal results into account. Considering that approximately 90% of the group is 6 kV class cables, and that diagnostic methods such as DC leakage current measurement would have been applied to this class cables, it can be assumed that most of the reasons for renewal are due to some kind of trouble, and it is highly appropriate to add up them when the failure rate is estimated.

Figure 6.

Hazard plots of CV cables.

Figure 7.

Results of failure characteristics estimation of CV cable. (a) failure rate (b) Cumulative failure probability.

4. Conclusion

In recent years, the asset management or managing assets for power transmission and distribution equipment has been actively examined. In order to optimize maintenance strategies, both risk and cost associated with operating the equipment should be considered, and both evaluated and compared in monetary values. CRIEPI is investigating managing assets support tools evaluating the cumulative cost in operation including statistically expected risk.

It is important to obtain failure rate characteristics of equipment for risk assessment. As one method, this chapter has presented a method for statistical analysis of actual equipment operation data in the field, as well as the necessary data and considerations for this method. For power transmission and distribution equipment with high reliability and low failure rates, it is necessary to accumulate actual data over a long period of time in order to accurately estimate failure rates. Among them, the information on equipment renewal, which has not necessarily been sufficiently investigated and analyzed in the past, is particularly important in the situation that the preventive maintenance is generally adopted, and must be investigated and accumulated together with the reasons for renewal.