Introduction

Transformer is a key equipment in electrical system transmission, it plays a very important role in stable operation of power system. If a failure occurs in transformer, it will bring huge losses to power enterprises and national economy. A lot of research data show that the main cause of transformer failure is the deterioration of insulation performance1. This is because that transformer is subjected to the interaction of thermal, electrical and mechanical stresses in operation conditions, and its insulation system is gradually aging and unable to meet the requirements of safe and reliable operation, which resulting in a failure at last1. The service life of oil-paper insulated transformer is determined by the life of insulating material which is mainly composed of insulating oil and solid insulation. Among them, the performance decline of transformer insulation oil is the main reason of transformer breakdown. On the other hand, it is found that voltage is the main factor which affecting the insulation performance of transformer oil. Therefore, at present, a lot of research work has been done on transformer insulation aging and insulation life prediction. Based on a single ageing test at higher temperature and the activation energy of the oxidation reaction obtained by non-iso-thermal Differential Scanning Calorimetry measurements. Dumitran et al.2 proposed a mineral oil lifetime estimation method. To prolong the service life of transformers, Chatterjee et al.3 developed predictive models for transformer health. Due to the uncertainty and complexity between transformer remaining life and its influence factors such as load and ambient temperature, AbU-Elanien and Salama4 applied a Monte Carlo method to estimate the thermal life of transformer insulation oil. Other commonly used methods for analyzing transformer insulation aging include neural network method5,6,7,8, fuzzy analysis method9,10,11,12, grey clustering method13, genetic algorithm14,15, deep learning16,17, support vector machine18,19 and hybrid methods20,21,22,23,24, etc.

However, due to the limitation of time and funds, it is impossible and unnecessary to carry out long-term life test of insulating materials according to actual operating conditions in engineering practice. In this case, an ideal solution is to use accelerated life testing (ALT) method. In other words, by an ALT of insulating materials under high stress level, fault data of transformer can be obtained in a short time, and the corresponding life model can be established, so as to extrapolate the life characteristics under normal operational conditions. ALT method can also provide a theoretical basis for transformer overhaul and maintenance plan, and then simultaneously determine the best time for transformer insulation oil overhaul and replacement25,26,27. In literature, the relationship between insulation oil voltage and lifetime is often described with inverse power law (IPL) model28,29, and failure times of insulation oil generally follow Weibull distribution30,31,32. Therefore, a Weibull IPL life model is applied in this paper to analyze the reliability and reliable life of transformer insulation oil.

Therefore, the novelty and main contributions of this work can be summarized as follows:

  1. 1.

    Numerical method of model parameter estimation of IPL-Weibull distribution for insulating oil lifetime is proposed, the variances of model parameters and reliability indices, which including mean lifetime, reliability, reliable lifetime and failure rate, are derived.

  2. 2.

    The reliability of transformer insulating oil used at usual operational conditions are predicted by the proposed method, and the point and interval estimations of reliability indices are evaluated. The feasibility and correctness of the proposed method are validated by real lifetime data of transformer insulating oil in literature.

The paper is organized as follows. In “Methodology”, the details on lifetime data modelling for transformer insulating oil using IPL-Weibull distribution, model parameter estimation, as well as the point and interval estimation for reliability indices are given. In “Case study”, some information about ALT for a real case in literature are described. Results and comparison with the existing method are presented with details in “Results and discussion”. Conclusions are drawn in the final and concluding section.

Methodology

Reliability modelling for transformer insulating oil with accelerated life testing

Some basic assumptions of ALT for insulating oil are given as follows:

  1. 1.

    The lifetime of insulating oil follows the Weibull distribution under usual use stress level and accelerated stress level;

  2. 2.

    The failure mechanism of insulating oil is the same at different stress levels;

  3. 3.

    The acceleration model of insulating oil is the same at each stress level.

Suppose that failure time t of insulating oil follows a two-parameter Weibull distribution, then the cumulative distribution function (CDF) F(t) and probability density function (pdf) f(t) of failure time t of insulating oil are respectively given by

$$F\left( t \right) = 1 - \exp \left[ { - \left( {{t \mathord{\left/ {\vphantom {t \eta }} \right. \kern-\nulldelimiterspace} \eta }} \right)^{\beta } } \right]$$
(1)

and

$$f\left( t \right) = \beta \eta^{ - \beta } t^{\beta - 1} \exp \left[ { - \left( {{t \mathord{\left/ {\vphantom {t \eta }} \right. \kern-\nulldelimiterspace} \eta }} \right)^{\beta } } \right]$$
(2)

where η is scale parameter and β is shape parameter, η > 0, β > 0.

The IPL model can be used for insulating oil where voltage is the main stress. Therefore, its characteristic life is

$$\eta \left( U \right) = {1 \mathord{\left/ {\vphantom {1 {\left( {KU^{N} } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {KU^{N} } \right)}}$$
(3)

where K > 0, N > 0 are model parameters, and U > 0 is voltage stress.

Therefore, from Eqs. (1) and (3), the reliability R of insulating oil at given time t can be given by

$$R\left( t \right) = 1 - F\left( t \right) = \exp \left[ { - \left( {KU^{N} t} \right)^{\beta } } \right]$$
(4)

Numerical method of model parameter estimation

Assume that n stress levels of insulating oil in ALT are U1 < U2 < … < Ui < … < Un (i = 1, 2, …, n), its failure time at the ith stress level is ti1 < ti2 < … < tij < … < \(t_{{im_{i} }}\)(j = 1, 2, …, mi), then substituting Eq. (3) into Eq. (2), the responding pdf of all failure time of insulating oil in ALT is

$$f\left( {t_{i,j} } \right) = \beta \left( {KU_{i}^{N} } \right)^{\beta } t_{i,j}^{\beta - 1} \exp \left[ { - \left( {KU_{{_{i} }}^{N} t_{i,j} } \right)^{\beta } } \right]$$
(5)

According Eq. (5), the likelihood function of failure time ti,j of insulating oil is

$$L\left( {t_{i,j} ;\beta ,N,K} \right) = \prod\limits_{i = 1}^{n} {\left( {\prod\limits_{j = 1}^{{m_{i} }} {f\left( {t_{i,j} } \right)} } \right)} = \prod\limits_{i = 1}^{n} {\prod\limits_{j = 1}^{{m_{i} }} {\left[ {\beta \left( {KU_{i}^{N} } \right)^{\beta } t_{i,j}^{\beta - 1} \exp \left( { - \left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } } \right)} \right]} }$$
(6)

Thus, the corresponding log-likelihood function is

$$\Lambda = \ln L\left( {t_{i,j} ;\beta ,N,K} \right) = \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\ln \left[ {\beta \left( {KU_{i}^{N} } \right)^{\beta } t_{i,j}^{\beta - 1} } \right]} } - \sum\limits_{i = 1}^{n} {\sum\limits_{k = 1}^{{m_{i} }} {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } } }$$
(7)

Taking the first partial derivatives of Eq. (7) with respect to the model parameters β, K and N, and setting them equal to zero, respectively. The maximum likelihood estimates (MLE) of model parameters satisfy Eqs. (8)–(10):

$$\frac{\partial \Lambda }{{\partial \beta }} = \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\frac{1}{\beta } + \ln (KU_{i}^{N} t_{i,j} )} \right]} } - \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } \ln (KU_{i}^{N} t_{i,j} )} \right] = 0} }$$
(8)
$$\frac{\partial \Lambda }{{\partial N}} = \beta \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left\{ {\ln U_{i} \left[ {1 - \left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } } \right]} \right\}} = 0}$$
(9)
$$\frac{\partial \Lambda }{{\partial K}} = \frac{\beta }{K}\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {1 - \left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } } \right] = 0} }$$
(10)

However, Eqs. (8)-(10) have no closed-form solutions, and a numerical method, which consists of three steps, is needed.

  1. Step 1:

    Estimate the first model parameter β using the least square method. Twice taking the logarithmic of Eq. (1) with some mathematical transformations, the following Eq. (11) can be obtained as

    $$\ln \left[ { - \ln \left( {1 - F_{i} \left( {t_{i,j} } \right)} \right)} \right] = \beta_{i} \left( {\ln t_{i,j} - \ln \eta_{i} } \right)$$
    (11)

    where Fi (ti, j) can be given by midpoint estimate or median rank estimate, and the estimators are

    $$\hat{F}_{i} (t_{i,j} ) = {{\left( {j - 0.5} \right)} \mathord{\left/ {\vphantom {{\left( {j - 0.5} \right)} {m_{i} }}} \right. \kern-\nulldelimiterspace} {m_{i} }},j = 1,2,...,m_{i}$$
    (12)

    and

    $$\hat{F}_{i} (t_{i,j} ) = {{\left( {j - 0.3} \right)} \mathord{\left/ {\vphantom {{\left( {j - 0.3} \right)} {\left( {m_{i} + 0.4} \right)}}} \right. \kern-\nulldelimiterspace} {\left( {m_{i} + 0.4} \right)}},j = 1,2,...,m_{i}$$
    (13)

Therefore, the least square method33 can be used to obtain an estimate of the parameter βi by combining Eq. (11) with Eqs. (12) or (13). ALT does not change failure mechanism, and thus the estimators of βi (i = 1, 2, …, n) should be equal each other. However, there exit some errors and uncertainty in test, the estimators of βi are not exactly the same. Therefore, its weighted estimated value can be given by

$$\hat{\beta } = {{\sum\limits_{i = 1}^{n} {m_{i} } \hat{\beta }_{i} } \mathord{\left/ {\vphantom {{\sum\limits_{i = 1}^{n} {m_{i} } \hat{\beta }_{i} } {\sum\limits_{i = 1}^{n} {m_{i} } }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{i = 1}^{n} {m_{i} } }}$$
(14)
  1. Step 2:

    Estimate the second model parameter K with an iterative method. After some mathematical calculations, Eq. (9) and Eq. (10) can be changed as

    $$\sum\limits_{i = 1}^{n} {m_{i} \ln U_{i} } = K^{\beta } \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left( {U_{i}^{N} t_{i,j} } \right)^{\beta } \ln U_{i} } }$$
    (15)
    $$\sum\limits_{i = 1}^{n} {m_{i} } = K^{\beta } \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left( {U_{i}^{N} t_{i,j} } \right)^{\beta } } }$$
    (16)

Divided the left and right sides of Eq. (16) by the corresponding two sides of Eq. (15), one can get

$$\frac{{\sum\limits_{i = 1}^{n} {m_{i} } }}{{\sum\limits_{i = 1}^{n} {\left( {m_{i} \ln U_{i} } \right)} }} = \frac{{\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left( {U_{i}^{N} t_{i,j} } \right)^{{\hat{\beta }}} } } }}{{\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {U_{i}^{N} t_{i,j} } \right)^{{\hat{\beta }}} \ln U_{i} } \right]} } }}$$
(17)

Equation (17) contains only parameter N, and its estimated value \(\hat{N}\) can be solved by an iterative method. One iterative expression for the second parameter N would be

$$\hat{N}^{{\left( {l + 1} \right)}} = \frac{1}{{\ln U_{1} }}\left\{ {\ln \left[ {D - \sum\limits_{j = 2}^{{m_{i} }} {\left( {U_{1}^{{\hat{N}^{\left( l \right)} }} t_{1,j} } \right)^{{\hat{\beta }}} } - \sum\limits_{i = 2}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left( {U_{i}^{{\hat{N}^{\left( l \right)} }} t_{i,j} } \right)^{{\hat{\beta }}} } } } \right]^{{{1 \mathord{\left/ {\vphantom {1 {\hat{\beta }}}} \right. \kern-\nulldelimiterspace} {\hat{\beta }}}}} - \ln t_{1,1} } \right\}$$
(18)

where \(\hat{N}^{{\left( {l + 1} \right)}}\) and \(\hat{N}^{\left( l \right)}\) are the (l + 1)th and l th iteration values of parameter N, and

$$D = {{\left( {\sum\limits_{i = 1}^{n} {m_{i} } } \right) \times \left( {\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {U_{i}^{{\hat{N}^{\left( l \right)} }} t_{i,j} } \right)^{{\hat{\beta }}} \ln U_{i} } \right]} } } \right)} \mathord{\left/ {\vphantom {{\left( {\sum\limits_{i = 1}^{n} {m_{i} } } \right) \times \left( {\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {U_{i}^{{\hat{N}^{\left( l \right)} }} t_{i,j} } \right)^{{\hat{\beta }}} \ln U_{i} } \right]} } } \right)} {\sum\limits_{i = 1}^{n} {\left( {m_{i} \ln U_{i} } \right)} }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{i = 1}^{n} {\left( {m_{i} \ln U_{i} } \right)} }}$$
  1. Step 3:

    Estimate the last model parameter N by the iterative method. Finally, substituting the estimated value \(\hat{N}\) into Eq. (8), the estimated value \(\hat{K}\) of the last parameter K can be obtain, and its (r + 1) th iteration value is

    $$\hat{K}_{{^{1} }}^{{^{{\left( {r + 1} \right)}} }} = \left( {\frac{{\sum\limits_{i = 1}^{n} {\left( {{{m_{i} } \mathord{\left/ {\vphantom {{m_{i} } {\hat{\beta }}}} \right. \kern-\nulldelimiterspace} {\hat{\beta }}}} \right)} + \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\ln \left( {\hat{K}_{{^{1} }}^{{^{\left( r \right)} }} U_{i}^{{\hat{N}}} t_{i,j} } \right)} } }}{{\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {U_{i}^{{\hat{N}}} t_{i,j} } \right)^{{\hat{\beta }}} \ln \left( {\hat{K}_{{^{1} }}^{{^{\left( r \right)} }} U_{i}^{{\hat{N}}} t_{i,j} } \right)} \right]} } }}} \right)^{{{1 \mathord{\left/ {\vphantom {1 {\hat{\beta }}}} \right. \kern-\nulldelimiterspace} {\hat{\beta }}}}}$$
    (19)

    where \(\hat{K}_{1}^{\left( r \right)}\) is the r th iteration value of parameter K. When the estimated value \(\hat{K}\) of parameter K is obtained by using Eq. (9) or Eq. (10), its value is

    $$\hat{K}_{2} = \left( {{{\sum\limits_{i = 1}^{n} {\left( {m_{i} \ln U_{i} } \right)} } \mathord{\left/ {\vphantom {{\sum\limits_{i = 1}^{n} {\left( {m_{i} \ln U_{i} } \right)} } {\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {U_{i}^{{\hat{N}}} t_{i,j} } \right)^{{\hat{\beta }}} \ln U_{i} } \right]} } }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {U_{i}^{{\hat{N}}} t_{i,j} } \right)^{{\hat{\beta }}} \ln U_{i} } \right]} } }}} \right)^{{{1 \mathord{\left/ {\vphantom {1 {\hat{\beta }}}} \right. \kern-\nulldelimiterspace} {\hat{\beta }}}}}$$
    (20)

    or

    $$\hat{K}_{3} = \left( {{{\sum\limits_{i = 1}^{n} {m_{i} } } \mathord{\left/ {\vphantom {{\sum\limits_{i = 1}^{n} {m_{i} } } {\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left( {U_{i}^{{\hat{N}}} t_{i,j} } \right)^{{\hat{\beta }}} } } }}} \right. \kern-\nulldelimiterspace} {\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left( {U_{i}^{{\hat{N}}} t_{i,j} } \right)^{{\hat{\beta }}} } } }}} \right)^{{{1 \mathord{\left/ {\vphantom {1 {\hat{\beta }}}} \right. \kern-\nulldelimiterspace} {\hat{\beta }}}}}$$
    (21)

In general, at this case, \(\hat{K}_{1} \ne \hat{K}_{2} = \hat{K}_{3}\), the estimated value \(\hat{K}\) can be selected according to the maximum log-likelihood value of Eq. (7), and given by

$$\hat{K} = \left\{ \begin{gathered} \hat{K}_{1} {, }\ln L\left( {\hat{\beta },\hat{N},\hat{K}_{1} } \right) \ge \ln L\left( {\hat{\beta },\hat{N},\hat{K}_{2} } \right) \hfill \\ \hat{K}_{2} {, }\ln L\left( {\hat{\beta },\hat{N},\hat{K}_{1} } \right) < \ln L\left( {\hat{\beta },\hat{N},\hat{K}_{2} } \right) \hfill \\ \end{gathered} \right.$$
(22)

Thus, all estimated values of three model parameters can be obtained.

Interval estimates for model parameters and reliability indices

To estimate the confidence bounds of model parameters and reliability indicators, their variances are needed. The variance–covariance matrix of model parameters can be given by Fisher information matrix method as follows34:

$$\left[ {\begin{array}{*{20}c} {{\text{Var}}(\hat{\beta })} & {{\text{Cov}}(\hat{\beta },\hat{N})} & {{\text{Cov}}(\hat{\beta },\hat{K})} \\ {{\text{Cov}}(\hat{N},\hat{\beta })} & {{\text{Var}}(\hat{N})} & {{\text{Cov}}(\hat{N},\hat{K})} \\ {{\text{Cov}}(\hat{K},\hat{\beta })} & {{\text{Cov}}(\hat{K},\hat{N})} & {{\text{Var}}(\hat{K})} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\Delta_{11} } & {\Delta_{12} } & {\Delta_{13} } \\ {\Delta_{21} } & {\Delta_{22} } & {\Delta_{23} } \\ {\Delta_{31} } & {\Delta_{32} } & {\Delta_{33} } \\ \end{array} } \right]_{{\beta = \hat{\beta }, \, N = \hat{N}, \, K = \hat{K}}}^{ - 1}$$
(23)

where

$$\begin{gathered} \Delta_{11} = - \frac{{\partial^{2} \Lambda }}{{\partial \beta^{2} }} = \sum\limits_{i = 1}^{n} {\frac{{m_{i} }}{{\beta^{2} }} + } \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } \ln^{2} \left( {KU_{i}^{N} t_{i,j} } \right)} \right]} } \hfill \\ \Delta_{22} = - \frac{{\partial^{2} \Lambda }}{{\partial N^{2} }} = \beta^{2} \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } \ln^{2} U_{i} } \right]} } \hfill \\ \Delta_{33} = - \frac{{\partial^{2} \Lambda }}{{\partial K^{2} }} = \frac{\beta }{{K^{2} }}\left[ {\sum\limits_{i = 1}^{n} {m_{i} + \left( {\beta - 1} \right)\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } } } } } \right] \hfill \\ \end{gathered}$$
$$\begin{gathered} \Delta_{12} = \Delta_{21} = - \frac{{\partial^{2} \Lambda }}{\partial \beta \partial N} = \sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\ln U_{i} \left\{ {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } \left[ {\beta \ln (KU_{i}^{N} t_{i,j} ) + 1} \right]} \right\}} } - \sum\limits_{i = 1}^{n} {m_{i} \ln U_{i} } \hfill \\ \Delta_{13} = \Delta_{31} = - \frac{{\partial^{2} \Lambda }}{\partial \beta \partial K} = \frac{1}{K}\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left\{ {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } \left[ {\beta \ln (KU_{i}^{N} t_{i,j} ) + 1} \right]} \right\}} } - \frac{1}{K}\sum\limits_{i = 1}^{n} {m_{i} } \hfill \\ \Delta_{23} = \Delta_{32} = - \frac{{\partial^{2} \Lambda }}{\partial N\partial K} = \frac{{\beta^{2} }}{K}\sum\limits_{i = 1}^{n} {\sum\limits_{j = 1}^{{m_{i} }} {\left[ {\left( {KU_{i}^{N} t_{i,j} } \right)^{\beta } \ln U_{i} } \right]} } \hfill \\ \end{gathered}$$
(24)

In general, the MLE of model parameters are asymptotically a normal distribution. Therefore, the interval estimation IEθ of the model parameter θ = (β, N, K) with a confidence of 100(1-α)% can be constructed as

$${\text{IE}}_{{\varvec{\theta}}} = \hat{\user2{\theta }} \pm z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} = \left[ {\hat{\user2{\theta }} - z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} , \, \hat{\user2{\theta }} + z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } \right] = \left[ {{\varvec{\theta}}_{L} ,{\varvec{\theta}}_{U} } \right]$$
(25)

where \(\hat{\user2{\theta }}\) is the estimated value of model parameter,\({\text{Var}}\left( {\hat{\user2{\theta }}} \right)\) is the variance of model parameter, Zα/2 is the quantile of standard normal distribution, L and U are the lower bound and upper bound, respectively.

When model parameters are non-negative, they are more close to follow a lognormal distribution34. Therefore,

$${{\left( {\ln \hat{\user2{\theta }} - \ln {\varvec{\theta}}} \right)} \mathord{\left/ {\vphantom {{\left( {\ln \hat{\user2{\theta }} - \ln {\varvec{\theta}}} \right)} {\sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} }}} \right. \kern-\nulldelimiterspace} {\sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} }} \sim N\left( {0,1} \right)$$

Thus, the log-transformed interval estimation IElnθ of model parameter θ can be given as

$$\begin{aligned} {\text{IE}}_{{\ln {\varvec{\theta}}}} & = \ln \left( {\hat{\user2{\theta }}} \right) \pm z_{\alpha /2} \sqrt {{\text{Var}}\left( {\ln \left( {\hat{\user2{\theta }}} \right)} \right)} = \ln \left( {\hat{\user2{\theta }}} \right) \pm z_{\alpha /2} \sqrt {\left( {\frac{1}{{\hat{\user2{\theta }}}}} \right)^{2} {\text{Var}}\left( {\hat{\user2{\theta }}} \right)} \\ & = \ln \left( {\hat{\user2{\theta }}} \right) \pm {{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } \mathord{\left/ {\vphantom {{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } {\hat{\user2{\theta }}}}} \right. \kern-\nulldelimiterspace} {\hat{\user2{\theta }}}} \\ \end{aligned}$$
(26)

So, from Eq. (26), the interval estimation IEθ of model parameter θ is

$$\begin{aligned} {\text{IE}}_{{\varvec{\theta}}} & = \hat{\user2{\theta }}\exp \left( { \pm {{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } \mathord{\left/ {\vphantom {{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } {\hat{\user2{\theta }}}}} \right. \kern-\nulldelimiterspace} {\hat{\user2{\theta }}}}} \right) \\ & = \left[ {\hat{\user2{\theta }}\exp \left( { - {{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } \mathord{\left/ {\vphantom {{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } {\hat{\user2{\theta }}}}} \right. \kern-\nulldelimiterspace} {\hat{\user2{\theta }}}}} \right), \, \hat{\user2{\theta }}\exp \left( {{{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } \mathord{\left/ {\vphantom {{z_{\alpha /2} \sqrt {{\text{Var}}\left( {\hat{\user2{\theta }}} \right)} } {\hat{\user2{\theta }}}}} \right. \kern-\nulldelimiterspace} {\hat{\user2{\theta }}}}} \right)} \right] \\ \end{aligned}$$
(27)

Commonly used reliability indices include mean time-to-failure tMTTF, the reliability R with a given time t, the reliable life tR for a given reliability and failure rate function.

Based on Eqs. (2) and (4), tMTTF can be obtained as follows:

$$t_{{{\text{MTTF}}}} = E\left( t \right) = \int_{0}^{\infty } t f\left( t \right){\text{d}}t = \int_{0}^{\infty } {R\left( t \right)} {\text{d}}t = \frac{1}{{KU^{N} }}\Gamma \left( {1 + \left( {{1 \mathord{\left/ {\vphantom {1 \beta }} \right. \kern-\nulldelimiterspace} \beta }} \right)} \right)$$
(28)

and the variance of tMTTF is

$$\begin{aligned} {\text{Var}}\left( {t_{{{\text{MTTF}}}} } \right) & = \int_{0}^{\infty } {\left[ {t - E\left( t \right)} \right]} f\left( t \right){\text{d}}t = \int_{0}^{\infty } {t^{2} } f\left( t \right){\text{d}}t - \left[ {E\left( t \right)} \right]^{2} \\ & = \frac{1}{{K^{2} U^{2N} }}\left[ {\Gamma \left( {1 + {2 \mathord{\left/ {\vphantom {2 \beta }} \right. \kern-\nulldelimiterspace} \beta }} \right) - \Gamma^{2} \left( {1 + {1 \mathord{\left/ {\vphantom {1 \beta }} \right. \kern-\nulldelimiterspace} \beta }} \right)} \right] \\ \end{aligned}$$
(29)

where Г(·) is gamma function, and according to Eq. (4), the reliable life tR for a specified reliability can be given by

$$t\left( R \right) = {{\left( { - \ln R} \right)^{{{1 \mathord{\left/ {\vphantom {1 \beta }} \right. \kern-\nulldelimiterspace} \beta }}} } \mathord{\left/ {\vphantom {{\left( { - \ln R} \right)^{{{1 \mathord{\left/ {\vphantom {1 \beta }} \right. \kern-\nulldelimiterspace} \beta }}} } {\left( {KU^{N} } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {KU^{N} } \right)}}$$
(30)

According to Eqs. (2) and (4), the failure rate function of insulating oil is got by

$$\lambda \left( t \right) = {{f\left( t \right)} \mathord{\left/ {\vphantom {{f\left( t \right)} {R\left( t \right)}}} \right. \kern-\nulldelimiterspace} {R\left( t \right)}} = \beta \left( {KU^{N} } \right)^{\beta } t^{\beta - 1}$$
(31)

To obtain the confidence bound of reliability, reliable life and failure rate of insulating oil, the equivalent change method is applied. In Eq. (4), set \(u = \ln \left[ {\left( {KU^{N} t} \right)^{\beta } } \right] = \beta \left( {\ln K + N\ln U + \ln t} \right)\), then

$$\begin{aligned} {\text{Var}}\left( {\hat{u}} \right) & = \left( {\frac{\partial u}{{\partial \beta }}} \right)^{2} {\text{Var}}\left( {\hat{\beta }} \right) + \left( {\frac{\partial u}{{\partial N}}} \right)^{2} {\text{Var}}\left( {\hat{N}} \right) + \left( {\frac{\partial u}{{\partial K}}} \right)^{2} {\text{Var}}\left( {\hat{K}} \right) \\ & \;\;\;{ + }2\left( {\frac{\partial u}{{\partial \beta }}} \right)\left( {\frac{\partial u}{{\partial N}}} \right){\text{Cov}}\left( {\hat{\beta },\hat{N}} \right) + 2\left( {\frac{\partial u}{{\partial \beta }}} \right)\left( {\frac{\partial u}{{\partial K}}} \right){\text{Cov}}\left( {\hat{\beta },\hat{K}} \right) + 2\left( {\frac{\partial u}{{\partial N}}} \right)\left( {\frac{\partial u}{{\partial K}}} \right){\text{Cov}}\left( {\hat{N},\hat{K}} \right) \\ & = \left( {{u \mathord{\left/ {\vphantom {u \beta }} \right. \kern-\nulldelimiterspace} \beta }} \right)^{2} {\text{Var}}\left( {\hat{\beta }} \right) + \left( {\beta \ln U} \right)^{2} {\text{Var}}\left( {\hat{N}} \right) + \left( {{\beta \mathord{\left/ {\vphantom {\beta K}} \right. \kern-\nulldelimiterspace} K}} \right)^{2} {\text{Var}}\left( {\hat{K}} \right) \\ & \;\;\;{ + }\left( {2u\ln U} \right){\text{Cov}}\left( {\hat{\beta },\hat{N}} \right) + \left( {{{2u} \mathord{\left/ {\vphantom {{2u} K}} \right. \kern-\nulldelimiterspace} K}} \right){\text{Cov}}\left( {\hat{\beta },\hat{K}} \right) + \left( {{{\left( {2\beta^{2} \ln U} \right)} \mathord{\left/ {\vphantom {{\left( {2\beta^{2} \ln U} \right)} {\hat{K}}}} \right. \kern-\nulldelimiterspace} {\hat{K}}}} \right){\text{Cov}}\left( {\hat{N},\hat{K}} \right) \\ \end{aligned}$$
(32)

So, after obtaining the lower uL and upper bound uU of u with Eqs. (25), (27) and (32), the interval estimation of reliability R can be given by

$${\text{IE}}_{R} = \left[ {\exp \left( { - \exp u_{U} } \right), \, \exp \left( { - \exp u_{L} } \right)} \right]$$
(33)

Similarly, in Eq. (30), set \(v = \ln t = \frac{1}{\beta }\ln \left( { - \ln R} \right) - \ln K - N\ln U\), then

$$\begin{aligned} {\text{Var}}\left( {\hat{v}} \right) & = \left[ {{{\ln \left( { - \ln R} \right)} \mathord{\left/ {\vphantom {{\ln \left( { - \ln R} \right)} {\hat{\beta }^{2} }}} \right. \kern-\nulldelimiterspace} {\hat{\beta }^{2} }}} \right]^{2} {\text{Var}}\left( {\hat{\beta }} \right) + \left( {\ln^{2} U} \right){\text{Var}}\left( {\hat{N}} \right) + \left( {{1 \mathord{\left/ {\vphantom {1 {\hat{K}}}} \right. \kern-\nulldelimiterspace} {\hat{K}}}} \right)^{2} {\text{Var}}\left( {\hat{K}} \right) \\ & \;\;\;{ + }\left[ {{{2\ln U\ln \left( { - \ln R} \right)} \mathord{\left/ {\vphantom {{2\ln U\ln \left( { - \ln R} \right)} {\hat{\beta }^{2} }}} \right. \kern-\nulldelimiterspace} {\hat{\beta }^{2} }}} \right]{\text{Cov}}\left( {\hat{\beta },\hat{N}} \right) + \left[ {{{2\ln \left( { - \ln R} \right)} \mathord{\left/ {\vphantom {{2\ln \left( { - \ln R} \right)} {\left( {\hat{K}\hat{\beta }^{2} } \right)}}} \right. \kern-\nulldelimiterspace} {\left( {\hat{K}\hat{\beta }^{2} } \right)}}} \right]{\text{Cov}}\left( {\hat{\beta },\hat{K}} \right) \\ & \;\;\;{ + }\left( {{{2\ln U} \mathord{\left/ {\vphantom {{2\ln U} {\hat{K}}}} \right. \kern-\nulldelimiterspace} {\hat{K}}}} \right){\text{Cov}}\left( {\hat{N},\hat{K}} \right) \\ \end{aligned}$$
(34)

The confidence interval [vL, vP] of v can also be estimated using Eqs. (25), (27) and (34). Finally, the interval estimate of reliable life is given as

$${\text{IE}}_{t} = \left[ {\exp v_{L} , \, \exp v_{U} } \right]$$
(35)

Finally, in Eq. (31), set \(w = \ln \lambda = \ln \beta + \beta \left( {\ln K + N\ln U} \right) + \left( {\beta - 1} \right)\ln t\), then

$$\begin{aligned} {\text{Var}}\left( {\hat{w}} \right) & = \left[ {\left( {{1 \mathord{\left/ {\vphantom {1 {\hat{\beta }}}} \right. \kern-\nulldelimiterspace} {\hat{\beta }}}} \right) + \ln \left( {\hat{K}U^{{\hat{N}}} t} \right)} \right]^{2} {\text{Var}}\left( {\hat{\beta }} \right) + \left( {\hat{\beta }\ln U} \right)^{2} {\text{Var}}\left( {\hat{N}} \right) + \left( {{{\hat{\beta }} \mathord{\left/ {\vphantom {{\hat{\beta }} {\hat{K}}}} \right. \kern-\nulldelimiterspace} {\hat{K}}}} \right)^{2} {\text{Var}}\left( {\hat{K}} \right) \\ & \;\;\;{ + }2\ln U\left[ {1 + \hat{\beta }\ln \left( {\hat{K}U^{{\hat{N}}} t} \right)} \right]{\text{Cov}}\left( {\hat{\beta },\hat{N}} \right) + \frac{2}{{\hat{K}}}\left[ {1 + \hat{\beta }\ln \left( {\hat{K}U^{{\hat{N}}} t} \right)} \right]{\text{Cov}}\left( {\hat{\beta },\hat{K}} \right) \\ & \;\;\;{ + }\left( {{{2\hat{\beta }^{2} \ln U} \mathord{\left/ {\vphantom {{2\hat{\beta }^{2} \ln U} {\hat{K}}}} \right. \kern-\nulldelimiterspace} {\hat{K}}}} \right){\text{Cov}}\left( {\hat{N},\hat{K}} \right) \\ \end{aligned}$$
(36)

The interval estimate of failure rate can also be given as

$${\text{IE}}_{\lambda } = \left[ {\exp w_{L} , \, \exp w_{U} } \right]$$
(37)

Case study

Using constant ALT, Ref.35 gives 76 failure-to-time data of a transformer insulation oil under n = 7 high voltage stress levels including U1 = 26, U2 = 28, U3 = 30, U4 = 32, U5 = 34, U6 = 36 and U7 = 38 kV, the numbers of failure time data for each stress level are m1 = 3, m2 = 5, m3 = 11, m4 = 15, m5 = 19, m6 = 15 and m7 = 8, respectively, the usual use voltage is 20 kV. The testing data are shown in Table 1.

Table 1 Constant ALT data for a transformer insulating oil with different voltages.
Full size table

Results and discussion

When the least square method is used to estimate shape parameter β of accelerated life model, the midpoint estimate and median rank estimate can be used to fit the reliability of insulating oil. The estimates \(\hat{\beta }\) of shape parameter β are shown in Table 2. It can be seen that the different estimate methods give different results. At the same time, even the same method is used, the estimated results for shape parameter β in the different stress level i are also different.

Table 2 The least square estimate for shape parameter.
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The failure probability of failure time t is an approximate straight line after logarithmic transformation. Figure 1 is the failure probability diagram of insulating oil at different high voltages under median rank estimate.

Figure 1
figure 1

Probability plot under different voltages for insulating oil. Figure created using Matlab R2014a (https://www.mathworks.com).

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It can be seen from Fig. 1 that the slope of failure probability line at two stress levels of 26 kV and 32 kV is small, and the slope of the other five voltages are basically parallel. The main reason is that there are only 3 failure data obtained at the lowest stress level of 26 kV. The first failure datum 5.79 is particularly far less than the other two data 1579.52 and 2323.70, so the estimators have a certain deviation.

Akaike information criterion (AIC) and Bayesian information criterion (BIC)36 are most widely used in selecting the optimal model, the values of AIC and BIC are given by

$${\text{AIC}} = - 2\max \ln L + 2p,{\text{ BIC}} = - 2\max \ln L + p\ln q$$
(38)

where p is the number of estimated parameters, q is the number of all lifetime, and maxlnL is the maximized log-likelihood. Table 3 shows the comparison results of model parameter estimation using MLE with the midpoint estimate and median rank estimate.

Table 3 Results comparison for model parameters with the different methods.
Full size table

From Table 3, it is found that the MLE with median rank is superior to MLE with midpoint estimate, the former has a larger log-likelihood value and smaller AIC and BIC values. Compared with the result of Ref.35, the relative errors of log-likelihood estimation, AIC and BIC are all within 0.022%, indicating that the method proposed in this paper has a high estimation accuracy. Therefore, the estimators of model parameter are given as follows: \(\hat{\beta }\) = 0.8017, \(\hat{N}\) = 17.7318 and \(\hat{K}\) = 6.70E-29, respectively.

The variance–covariance matrix of model parameters β, N, K is given by:

$$\left[ {\begin{array}{*{20}c} {{\text{Var}}(\hat{\beta })} & {{\text{Cov}}(\hat{\beta },\hat{N})} & {{\text{Cov}}(\hat{\beta },\hat{K})} \\ {{\text{Cov}}(\hat{N},\hat{\beta })} & {{\text{Var}}(\hat{N})} & {{\text{Cov}}(\hat{N},\hat{K})} \\ {{\text{Cov}}(\hat{K},\hat{\beta })} & {{\text{Cov}}(\hat{K},\hat{N})} & {{\text{Var}}(\hat{K})} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {4.8882 \times 10^{ - 3} } & { - 5.1966 \times 10^{ - 4} } & { - 1.1305 \times 10^{ - 31} } \\ { - 5.1966 \times 10^{ - 4} } & {9.0867 \times 10^{ - 1} } & { - 2.1224 \times 10^{ - 28} } \\ { - 1.1305 \times 10^{ - 31} } & { - 2.1224 \times 10^{ - 28} } & {4.9679 \times 10^{ - 56} } \\ \end{array} } \right]$$

The normal operating voltage of insulating oil is U0 = 20 kV, therefore, using Eq. (28), the mean lifetime under normal use circumstances is 1.4406E + 05 min, the corresponding variance is Var(tMTBF) = 3.2824E + 08, based on Eqs. (25), (27) and (29), the interval estimator of mean lifetime is [1.2247E + 04, 4.9916E + 05]. The other point estimation and 95% confidence interval estimation of reliability indices and model parameters for transformer insulation oil under normal operating circumstances are shown in Table 4. Note that the values of reliability and failure rate are all given with mean lifetime 1.4406E + 05, reliable life is calculated by 90% reliability. Therefore, u = 9.8895E−02, v = 8.9476, w = − 12.0001, the corresponding variances are Var(u) = 1.5772E−01, Var(v) = 3.2893E−01 and Var(w) = 1.6033E−01, respectively.

Table 4 Reliability analysis results for insulating oil.
Full size table

It can be seen from Table 4, the insulating oil has a higher mean lifetime 1.4406E + 05, but its reliability at mean lifetime is very low, only 33.16%, and the reliable life of insulating oil with 90% reliability is 7689.68 min, only 5.34% of mean lifetime. At this case, if the average life is taken as the criterion of maintenance and replacement only, it will bring some potential safety hazards. Therefore, the replacement time should be determined according to the reliability and mean lifetime of the transformer during maintenance simultaneously.

Figure 2 is the plot of mean time-to-failure vs. voltage, two sides confidence bounds are also given. It can be found that as the increasing of voltage, the MMTF decreases rapidly. The MMTFs are 1374.43 min and 369.34 min when voltages are 26 kV and 28 kV, especially when voltage is greater than 30 kV, the value of MMTF not exceeds 200 min. At 38 kV, it is 1.64 kV only.

Figure 2
figure 2

Interval estimation of MTTF. Figure created using Matlab R2014a (https://www.mathworks.com).

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The interval estimations of reliability, reliable life and failure rate of insulating oil under normal use conditions are shown in Figs. 3, 4, 5.

Figure 3
figure 3

Interval estimation of reliability. Figure created using Matlab R2014a (https://www.mathworks.com).

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Figure 4
figure 4

Interval estimation of reliable life. Figure created using Matlab R2014a (https://www.mathworks.com).

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Figure 5
figure 5

Interval estimation of failure rate. Figure created using Matlab R2014a (https://www.mathworks.com).

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Conclusions

Based on the theory of constant ALT, a probabilistic method for reliability assessment of transformer insulating oil is proposed using two-parameter Weibull IPL life model. Numerical procedure of model parameter estimation is given. The reliability indices such as the MMTF, reliability, reliable life and failure rate of insulating oil under normal service conditions were predicted by using the proposed method, and the point estimation and interval estimation were calculated. Compared with the existing method, the method proposed in this paper has a high accuracy, the relative errors of log-likelihood estimation, AIC and BIC do not exceed 0.022%. The results of reliability analyses show that even the insulating oil has a higher mean lifetime, the reliable life is not always high. Therefore, mean lifetime and reliable life should be considered simultaneously in replacement and maintenance of insulating oil.