# Failure Trends of High-Voltage Porcelain Insulators Depending on the Constituents of the Porcelain
## 1 Introduction
Data distributed in accordance with Weibull function should approximately follow a straight line, as shown in Figures 5 and 6. Moreover, as discussed in articles, the correlation coefficient should be calculated to check the adequacy of the Weibull distribution [17, 18, 19, 20, 21]. Here, the correlation coefficient was calculated to be  
Figure 5: Cumulative probability of failure of the samples based on the mechanical load test with electric voltage (M&E) of the porcelain insulators. Figure 6: Cumulative probability of failure of the samples based on the thermal and mechanical load test (T&M) of the porcelain insulators. 0.64, 0.84, and 0.95 respectively, for the three samples A, B and C. The correlation coefficient was calculated using the built-in function in a spreadsheet. The adequacy of the Weibull distribution can be evaluated if the correlation coefficient is greater than \(2/\sqrt{n}\)[22], where \(n\) stands for number of samples. All cases reported satisfy the adequacy of a Weibull distribution in accordance with the above formula. Further, the data was sorted in ascending order and a breakdown probability was assigned. A simple approximation used to calculate the probability of failure was used:  
\[F(i,n)=\frac{i-0.44}{n+0.25}\times 100\% \tag{2}\]  
where \(i\) is the rank of the data point. For each breakdown load value, (\(t_{i}\)), a value was assigned:  
\[Y_{i}=\ln(t_{i}). \tag{3}\]  
For each probability of failure, \(F(i,n)\) was expressed as:  
\[X_{i}=\ln\left(-\ln\left(1-\frac{F(i,n)}{100}\right)\right). \tag{4}\]  
The estimates of failure, i.e., \(\alpha\) and \(\beta\) was calculated using:  
\[\alpha=\text{slope of the Weibull plot}, \tag{5}\]  
\[b=-\alpha\ln(\beta) \tag{6}\]  
where \(b\) is the y-intercept. Considering a two-parameter Weibull distribution, the failure probability was calculated using:  
\[F(t)=1-e^{-(\frac{t}{\alpha})^{\beta}} \tag{7}\]  
where \(F(t)\) is the cumulative probability of failure, \(t\) is the failure load, \(\alpha\) is the shape factor, \(\beta\) is the scale factor, and \(n\) is the number of samples  
In this study, six cases were considered to find the lifetime expectancy of the insulators. Three case study probabilities of failure were based on the mechanical load test with electric voltage (M&E) on insulators. However, another three studies dealt with the probability of failure based on the thermal and mechanical load test on the insulators. The correlation factor for all six cases were checked as stated before in this section. These cases satisfied the Weibull adequacy criterion where the correlation coefficient was greater than \(2/\sqrt{n}\). The correlation coefficient was calculated to be 0.64, 0.84, and 0.95, respectively, for the three case studies mentioned above. Consequently, \(\alpha\) and \(\beta\) values were calculated to be \(\alpha\) = 7.69 and \(\beta\) = 20,952.22, \(\alpha\) = 32.27 and \(\beta\) = 22,378.61, and \(\alpha\) = 45.66 and \(\beta\) = 22,743.4, respectively, for samples A, B, and C. The probability of failure was also calculated according to the Weibull function using Equation (7). The probability of failure was calculated to be 9%, 3%, and 0.4%, respectively, for the same sequence of samples mentioned above. Further, the probability of failure based on the thermal and mechanical load test on samples A, B, and C were calculated. The correlation factors were 0.65, 0.97, and 0.98, respectively. It was checked that for all cases, the correlation coefficient was greater than \(2/\sqrt{n}\). Consequently, the \(\alpha\) and \(\beta\) values were calculated to be \(\alpha\) = 6.96 and \(\beta\) = 21,284.8, \(\alpha\) = 46.84 and \(\beta\) = 22,859.07, and \(\alpha\) = 59.82 and \(\beta\) = 23,078.26, respectively, for samples A, B, and C. Additionally, the probability of failure was calculated to be 8%, 1.2%, and 1%, respectively, for samples A, B, and C. All six cases of load failure were re-evaluated by implementing the normal function. The calculated probability of failure was 7%, 3%, and 0.1%.