INTERNAL TRANSFORMER COMPOSITE-DEFECT FUZZY DIAGNOSTIC METHOD BASED ON GAS DISSOLVED IN OIL

Abstract

A transformer internal composite defect fuzzy diagnosis method based on gas dissolved in oil, comprising: a step of acquiring monitoring data of volume concentrations of five types of monitored feature gas; a step of determining ratio codes; a step of modifying a three-ratio method; a step of fuzzifying a boundary range; a step of calculating probabilities of the ratio codes; a step of calculating a probability of occurrence of each defect fault; and finally obtaining a fault type of a transformer. The method has the beneficial effects that: the method is simple and easy to achieve, and particularly suitable for being applied to an on-line transformer state monitoring system; based on a concept of fuzzy logic, diagnosis of composite defects of the transformer under a complicated state and evaluation of the degree of severity can be achieved, and the problem of sudden change caused by criterion boundary absolutisation can be effectively avoided; and multi-feature information such as an attention value and a ratio of the gas dissolved in the oil are merged and analysed, thereby effectively improving the diagnosis reliability.

Claims

1. A fuzzy diagnosis method for transformer internal composite defect based on gas dissolved in oil, comprising the following steps: (I) acquiring monitoring data of volumetric concentrations of five types of monitored characteristic gases, i.e., hydrogen, methane, ethane, ethylene and acetylene; calculating the sum of the volumetric concentrations of methane, ethane, ethylene and acetylene (i.e., volumetric concentration of total hydrocarbons) from the monitoring data; judging whether the monitoring data of the five types of characteristic gases or the volumetric concentration of total hydrocarbons exceeds an alert value, which is selected as per the Chinese Standard GB/T7252-2001; if the monitoring data or the volumetric concentration of total hydrocarbons exceeds the alert value, further diagnosis is required; in that case, going to step (II); otherwise judging that the transformer has no defect or fault, if the monitoring data and the volumetric concentration of total hydrocarbons are normal; (II) determining ratio codes: first, the ratios are set as follows: $r_1=\frac{c_1\left(C_2.Math.H_2\right)}{c_2\left(C_2.Math.H_4\right)},.Math.r_2=\frac{c_3\left({\mathrm{CH}}_4\right)}{c_4\left(H_2\right)},.Math.r_3=\frac{c_2\left(C_2.Math.H_4\right)}{c_5\left(C_2.Math.H_6\right)},.Math.r_4=\frac{c_1\left(C_2.Math.H_2\right)}{c_5\left(C_2.Math.H_6\right)}\times \frac{c_2\left(C_2.Math.H_4\right)}{c_1\left(C_2.Math.H_2\right)}=r_3\times \frac{c_2\left(C_2.Math.H_4\right)}{c_1\left(C_2.Math.H_2\right)},$ wherein, c.sub.1(C.sub.2H.sub.2), c.sub.2(C.sub.2H.sub.4), c.sub.3(CH.sub.4), c.sub.4(H.sub.2) and c.sub.5(C.sub.2H.sub.6) respectively represent the volumetric concentration of five types of characteristic gases (acetylene, ethylene, methane, hydrogen and ethane), in unit of μL/L; then, the ratio codes are determined according to the following rules: if r.sub.1<0.1, the ratio code of r.sub.1 is 0; if 0.15_r.sub.1<1, the ratio code of r.sub.1 is 1; if 5.1r.sub.1<3, the ratio code of r.sub.1 is 1; if r.sub.1 the ratio code of r.sub.1 is 2; if r.sub.2<0.1, the ratio code of r.sub.2 is 1; if 0.1≦r.sub.2<1, the ratio code of r.sub.2 is 0; if 15-r.sub.2<3, the ratio code of r.sub.2 is 2; if the ratio code of r.sub.2 is 2; if r.sub.3<0.1, the ratio code of r.sub.3 is 0; if 0.15r.sub.3<1, the ratio code of r.sub.3 is 0; if 1.5r.sub.3<3, the ratio code of r.sub.3 is 1; if r.sub.3≧3, the ratio code of r.sub.3 is 2; if r.sub.4≦1.5, the ratio code of r.sub.4 is 0; if r.sub.4>1.5, the ratio code of r.sub.4 is 1; (III) Correcting the method for determining the types of transformer defects or faults on the basis of three ratios as specified in the Chinese Standard GB/T7252-2001: based on the types of transformer defects or faults corresponding to the three ratio codes specified in the Chinese Standard GB/T 7252-2001, a ratio code 011 corresponding to the type of partial discharge defect or fault is added; a fourth ratio r.sub.4 is added on the basis of the three ratio codes; for the type of defect or fault with ratio code 101 diagnosed with the three-ratio method, if the transformer is judged as having a spark discharge defect or fault; if r.sub.4>1.5, the transformer is judged as having an arc discharge defect or fault; thus, obtaining a method for judging the types of transformer defects or faults according to ratio codes as follows: if the ratio code of r.sub.1 is 0, the ratio code of r.sub.2 is 1, the ratio code of r.sub.3 is 0, 1 or 2, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is partial discharge; if the ratio code of r.sub.1 is 0, the ratio code of r.sub.2 is 0, the ratio code of r.sub.3 is 1, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is low-temperature overheat lower than 300□; if the ratio code of r.sub.1 is 0, the ratio code of r.sub.2 is 2, the ratio code of r.sub.3 is 0, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is low-temperature overheat lower than 300□; if the ratio code of r.sub.1 is 0, the ratio code of r.sub.2 is 2, the ratio code of r.sub.3 is 1, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is 300-700° C. moderate-temperature overheat; if the ratio code of r.sub.1 is 0, the ratio code of r.sub.2 is 0 or 2, the ratio code of r.sub.3 is 2, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is high-temperature overheat higher than 700□; if the ratio code of r.sub.1 is 2, the ratio code of r.sub.2 is 0, 1 or 2, the ratio code of r.sub.3 is 0, 1 or 2, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is spark discharge; if the ratio code of r.sub.1 is 1, the ratio code of r.sub.2 is 0, the ratio code of r.sub.3 is 1, and the ratio code of r.sub.4 is 0, the type of transformer defect or fault is spark discharge; if the ratio code of r.sub.1 is 1, the ratio code of r.sub.2 is 0, the ratio code of r.sub.3 is 1, and the ratio code of r.sub.4 is 1, the type of transformer defect or fault is arc discharge; if the ratio code of r.sub.1 is 1, the ratio code of r.sub.2 is 0, 1 or 2, the ratio code of r.sub.3 is 0 or 2, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is arc discharge; if the ratio code of r.sub.1 is 1, the ratio code of r.sub.2 is 1 or 2, the ratio code of r.sub.3 is 1, and the ratio code of r.sub.4 is 0 or 1, the type of transformer defect or fault is arc discharge; (IV)blurring the boundary ranges of the ratios r.sub.1, r.sub.2, r.sub.3 and r.sub.4 with a semi-Cauchy rising/falling function, and representing the rising edges and falling edges of the boundaries with the semi-Cauchy rising/falling function as follows: ${\mu }_a\left(r\right)=\{\begin{array}{cc}1,& r\le A\\ \frac{1}{1+{\left(\frac{A-r}{a}\right)}^2},& \mathrm{others}\end{array}.Math..Math.{\mu }_a\left(r\right)=\{\begin{array}{cc}1,& r\ge A\\ \frac{1}{1+{\left(\frac{A-r}{a}\right)}^2},& \mathrm{others}\end{array}$ wherein, μ.sub.d(r) is a falling edge function; μ.sub.a(r) is a rising edge function; A is a boundary parameter; a is a distribution parameter; the values of A and a are as follows: the rising edge boundary parameter of r.sub.1 is 0.08, and the corresponding distribution parameter is 0.01, the falling edge boundary parameter of r.sub.1 is 3.1, and the corresponding distribution parameter is 0.1; the rising edge boundary parameter of r.sub.2 is 0.06, and the corresponding distribution parameter is 0.02; the falling edge boundary parameter of r.sub.2 is 0.6, and the corresponding distribution parameter is 0.2; the rising edge boundary parameter of r.sub.3 is 0.8, and the corresponding distribution parameter is 0.1; the falling edge boundary parameter of r.sub.3 is 3.6, and the corresponding distribution parameter is 0.3; the boundary parameter of r.sub.4 is 1.43, and the corresponding distribution parameter is 0.1; (V) obtaining the probabilities of the cases that the ratio codes of the ratios r.sub.1, r.sub.2 and r.sub.3 are 0, 1 and 2 respectively and the probabilities of the cases that the ratio code of r.sub.4 is 0 or 1 respectively, with the semi-Cauchy rising/falling function; the expressions are as follows: Probability f-code0(r.sub.1) of the case that the ratio code of r.sub.1 is 0: $f-\mathrm{code0}.Math..Math.\left(r_1\right)=\{\begin{array}{cc}1& \left(r_1\le 0.08\right)\\ \frac{1}{1+{\left(\frac{0.08-r_1}{0.01}\right)}^2}& \left(r_1>0.08\right)\end{array}$ Probability f-code1(r.sub.1) of the case that the ratio code of r.sub.1 is 1: $f-\mathrm{code}.Math..Math.1.Math.\left(r_1\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.08-r_1}{0.01}\right)}^2}& \left(r_1\le 0.08\right)\\ 1& \left(0.08\le r_1\le 3.1\right)\\ \frac{1}{1+{\left(\frac{3.1-r_1}{0.1}\right)}^2}& \left(r_1>3.1\right)\end{array}$ probability f-code2(r.sub.1) of the case that the ratio code of r.sub.1 is 2: $f-\mathrm{code}.Math..Math.2.Math..Math.\left(r_1\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{3.1-r_1}{0.1}\right)}^2}& \left(r_1<3.1\right)\\ 1& \left(r_1\ge 3.1\right)\end{array}$ probability f-code0(r.sub.2) of the case that the ratio code of r.sub.2 is 0: $f-\mathrm{code}.Math..Math.0.Math.\left(r_2\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.06-r_2}{0.02}\right)}^2}& \left(r_2<0.06\right)\\ 1& \left(0.06\le r_2\le 0.6\right)\\ \frac{1}{1+{\left(\frac{0.06-r_2}{0.02}\right)}^2}& \left(r_2>0.6\right)\end{array}$ probability f-code1(r.sub.2) of the case that the ratio code of r.sub.2 is 1: $f-\mathrm{code}.Math..Math.1.Math..Math.\left(r.Math..Math.2\right)=\{\begin{array}{cc}1& \left(r_2\le 0.06\right)\\ 1+{\left(\frac{0.06-r_2}{0.02}\right)}^2& \left(r_2>0.06\right)\end{array}$ probability f-code2(r.sub.2) of the case that the ratio code of r.sub.2 is 2: $f-\mathrm{code}.Math..Math.2.Math..Math.\left(r_2\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.6-r_2}{0.2}\right)}^2}& \left(r_2<0.6\right)\\ 1& \left(r_2\ge 0.6\right)\end{array}$ probability f-code0(r.sub.3) of the case that the ratio code of r.sub.3 is 0: $f-\mathrm{code}.Math..Math.0.Math..Math.\left(r_3\right)=\{\begin{array}{cc}1& \left(r_3\le 0.8\right)\\ \frac{1}{1+{\left(\frac{0.8-r_3}{0.1}\right)}^2}& \left(r_3>0.8\right)\end{array}$ probability f-code1(r.sub.3) of the case that the ratio code of r.sub.3 is 1: $f-\mathrm{code}.Math..Math.1.Math.\left(r_3\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.6-r_3}{0.1}\right)}^2}& \left(r_3<0.8\right)\\ 1& \left(0.8\le r_3\le 3.6\right)\\ \frac{1}{1+{\left(\frac{3.6-r_3}{0.3}\right)}^2}& \left(r_3>3.6\right)\end{array}$ probability f-code2(r.sub.3) of the case that the ratio code of r.sub.3 is 2: $f-\mathrm{code}.Math..Math.2.Math..Math.\left(r_3\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{3.6-r_3}{0.3}\right)}^2}& \left(r_3>3.6\right)\\ 1& \left(0.8\le r_3\le 3.6\right)\end{array}$ probability f-code0(r.sub.4) of the case that the ratio code of r.sub.4 is 0: $f-\mathrm{code}.Math..Math.0.Math..Math.\left(r_4\right)=\{\begin{array}{cc}1& \left(r_4\le 1.43\right)\\ \frac{1}{1+{\left(\frac{1.43-r_4}{0.1}\right)}^2}& \left(r_4>1.43\right)\end{array}$ Probability f-code1(r.sub.4) of the case that the ratio code of r.sub.4 is 1: $f-\mathrm{code}.Math..Math.1.Math..Math.\left(r_4\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{1.43-r_4}{0.1}\right)}^2}& \left(r_4<1.43\right)\\ 1& \left(r_4\ge 1.43\right)\end{array}$ (VI)representing the probabilities of ratio codes with maximum value logic and minimum value logic, and thereby obtaining a fuzzy multi-value form of the diagnostic result of the types of transformer defects or faults; the probabilities of the types of transformer defects or faults are as follows:
f(partial discharge)=min[f-code0(r.sub.1), f-code1(r.sub.2)];
f(low-temperature overheat)=max{min[f-code0(r.sub.1), f-code0(r.sub.2), f-code1(r.sub.3)], min[f-code0(r.sub.1),
f-code2(r.sub.2), f-code0(r3)]};
f(moderate-temperature overheat)=min[f-code0(r.sub.1), f-code2(r.sub.2), f-code1(r3)];
f(high-temperature overheat)=max{min[f-code0(r.sub.1), f-code0(r.sub.2), f-code2(r.sub.3)], min[f-code0(r.sub.1), f-code2(r2), f-code2(r.sub.3)]};
f(spark discharge)=max{f-code2(r.sub.1), min[f-code 1(r.sub.1), f-code0(r.sub.2), f-code1(r.sub.3), f-code0(r.sub.4)]};
f(arc discharge)=max{min[f-code 1(r.sub.1), f-code0(r.sub.2), f-code1(r.sub.3), f-code1(r.sub.4)], min[f-code1(r.sub.1), f-code0(r.sub.3)], min[f-code 1(r.sub.1), f-code2(r.sub.3)], min [f-code1(r.sub.1), f-code1(r.sub.2), f-code1(r.sub.3)], min[f-code 1(r.sub.1), f-code2(r.sub.2), f-code 1(r.sub.3)]}.

Description

DESCRIPTION OF THE DRAWINGS

[0052] FIG. 1 is a diagnostic flow chart according to the present invention;

[0053] FIG. 2 shows the fuzzy boundary when the ratio code of the ratio r.sub.3 is 2.

EMBODIMENTS

[0054] Hereunder the present invention will be further described in an example, with reference to FIGS. 1-2.

[0055] The implementation steps of this example are as follows:

[0056] (I) Acquiring monitoring data of volumetric concentrations of five types of monitored characteristic gases, i.e., hydrogen, methane, ethane, ethylene and acetylene; calculating the sum of the volumetric concentrations of methane, ethane, ethylene and acetylene (i.e., volumetric concentration of total hydrocarbons) from the monitoring data; judging whether the monitoring data of the five types of characteristic gases or the volumetric concentration of total hydrocarbons exceeds an alert value; the alert value is selected as per the Chinese Standard GB/T7252-2001 “Guide to the Analysis and Diagnosis of Gases Dissolved in Transformer Oil”; if the monitoring data or the volumetric concentration of total hydrocarbons exceeds the alert value, further diagnosis is required; in that case, going to step (II); otherwise judging that the transformer has no defect or fault, if the monitoring data and the volumetric concentration of total hydrocarbons are normal;

[0057] (II) Determining ratio codes:

[0058] First, the ratios are set as follows:

[00014]$r_1=\frac{c_1\left(C_2.Math.H_2\right)}{c_2\left(C_2.Math.H_4\right)},.Math.r_2=\frac{c_3\left({\mathrm{CH}}_4\right)}{c_4\left(H_2\right)},.Math.r_3=\frac{c_2\left(C_2.Math.H_4\right)}{c_5\left(C_2.Math.H_6\right)},.Math.r_4=\frac{c_2\left(C_2.Math.H_4\right)}{c_5\left(C_2.Math.H_6\right)}\frac{c_2\left(C_2.Math.H_4\right)}{c_1\left(C_2.Math.H_2\right)}=r_3\frac{c_2\left(C_2.Math.H_4\right)}{c_1\left(C_2.Math.H_2\right)}.Math.i$

[0059] Where, c.sub.1(C.sub.2H.sub.2), c.sub.2(C.sub.2H.sub.4), c.sub.3(CH.sub.4), c.sub.4(H.sub.2), and c.sub.5(C.sub.2H.sub.6) respectively represent the volumetric concentration of five types of characteristic gases (acetylene, ethylene, methane, hydrogen and ethane), in unit of μL/L;

[0060] Then, the ratio codes are determined according to the rules shown in Table 1:

TABLE-US-00001 TABLE 1 Rules for Determining Ratio Codes Ratio Range Ratio Code Ratio Range Ratio Code r.sub.1, r.sub.2 or r.sub.3 r.sub.1 r.sub.2 r.sub.3 r.sub.4 r.sub.4 <0.1 0 1 0 ≦1.5 0 ≧0.1~<1 1 0 0   ≧1~<3 1 2 1 >1.5 1 ≧3 2 2 2

[0061] wherein, the ratio codes of r.sub.1, r.sub.2 and r.sub.3 are obtained according to the ratio code rules specified in the Chinese Standard GB/T 7252-2001 “Guide to the Analysis and Diagnosis of Gases Dissolved in Transformer Oil”; a ratio r.sub.4 and a r.sub.4 ratio code rule are added on the basis of the ratio code rules specified in the Chinese Standard GB/T 7252-2001 “Guide to the Analysis and Diagnosis of Gases Dissolved in Transformer Oil”;

[0062] (III) Correcting the three-ratio method for judging the types of transformer defects or faults specified in the Chinese Standard GB/T 7252-2001 “Guide to the Analysis and Diagnosis of Gases Dissolved in Transformer Oil” on the basis of analysis of 728 typical real fault cases of the State Grid Corporation of China; thus, rules for judging the types of transformer defects or faults based on ratio codes are obtained, as shown in Table 2.

[0063] A fourth ratio r.sub.4 is added on the basis of three ratio codes; for the type of defect or fault with ratio code 101 diagnosed with the three-ratio method, if r.sub.4≦1.5, the transformer is judged as having a spark discharge defect or fault; if r.sub.4>1.5, the transformer is judged as having an arc discharge defect or fault;

[0064] Based on the codes for transformer faults and defects specified in the Chinese Standard GB/T 7252-2001 “Guide to the Analysis and Diagnosis of Gases Dissolved in Transformer Oil”, a ratio code 011 corresponding to partial discharge fault or defect is added.

TABLE-US-00002 TABLE 2 Method for Judging the Types of Transformer Defects or Faults according to Ratio Codes Combinations of Ratio Codes r.sub.1 r.sub.2 r.sub.3 r.sub.4 Type of Defect or Fault 0 1 0, 1, 2 0, 1 Local discharge 0 1 0, 1 Low-temperature overheat lower 2 0 0, 1 than 300□ 2 1 0, 1 300-700° C. moderate-temperature overheat 0, 2 2 0.1 High-temperature overheat higher than 700□ 2 0, 1, 2 0, 1, 2 0, 1 Spark discharge 1 0 1 0 1 Arc discharge 0, 1, 2 0, 2 0, 1 1, 2 1

[0065] (IV) Blurring the boundaries of the codes in Table 1 with a semi-Cauchy rising/falling function, and representing the rising edges and falling edges of the boundaries with the semi-Cauchy rising/falling function, in order to change the either-or absolutized boundary judgment; then, obtaining the probabilities of the cases that the ratio codes of the ratios r.sub.1, r.sub.2 and r.sub.3 are 0, 1 and 2 respectively (represented by f-code0(r.sub.i), f-code1(r.sub.i), f-code2(r.sub.i) respectively) and the probabilities of the cases that the ratio code of r4 is 0 or 1 respectively, with the semi-Cauchy rising/falling function. For example, the probability of the case that the ratio code of the ratio r.sub.3 is 2 is represented by f-code2(r.sub.3), and the fuzzy boundary is represented by the semi-Cauchy rising edge function, as shown in FIG. 2.

[0066] The boundary ranges of the ratios r.sub.1, r.sub.2, r.sub.3 and r.sub.4 are blurred with a semi-Cauchy rising/falling function, and the rising edges and falling edges of the boundaries are represented with the semi-Cauchy rising/falling function as follows:

[00015]${\mu }_d\left(r\right)=\{\begin{array}{cc}1& ,r\le A\\ \frac{1}{1+{\left(\frac{A-r}{a}\right)}^2}& ,\mathrm{others}\end{array}.Math..Math.{\mu }_a\left(r\right)=\{\begin{array}{cc}1& ,r\ge A\\ \frac{1}{1+{\left(\frac{A-r}{a}\right)}^2}& ,\mathrm{others}\end{array}$

[0067] wherein, μ.sub.d(r) is a falling edge function; μ.sub.a(r) is a rising edge function; A is a boundary parameter; a is a distribution parameter; the values of A and a are optimal values obtained through verification of the data of 728 typical real fault cases of the State Grid Coporation of China, as shown in Table 3.

TABLE-US-00003 TABLE 3 Boundary Parameter A and Distribution Parameter a A.sub.1(r.sub.1) A.sub.2(r.sub.1) A.sub.1(r.sub.2) A.sub.2(r.sub.2) A.sub.1(r.sub.3) A.sub.2(r.sub.3) A(r.sub.4) 0.08 3.1 0.06 0.6 0.8 3.6 1.43  a.sub.1(r.sub.1)  a.sub.2(r.sub.1)  a.sub.1(r.sub.2)  a.sub.2(r.sub.2)  a.sub.1(r.sub.3) a.sub.2 (r.sub.3)  a(r.sub.4) 0.01 0.1 0.02 0.2 0.1 0.3 0.1 

[0068] In the Table 3:

[0069] The rising edge boundary parameter of r.sub.1, A.sub.1(r.sub.1), is 0.08, and the corresponding distribution parameter a.sub.1(r.sub.1) is 0.01;

[0070] The falling edge boundary parameter of r.sub.1, A.sub.2(r.sub.1), is 3.1, and the corresponding distribution parameter a.sub.2(r.sub.1) is 0.1;

[0071] The rising edge boundary parameter of r.sub.2, A.sub.1(r.sub.2), is 0.06, and the corresponding distribution parameter a.sub.1(r.sub.2) is 0.02;

[0072] The falling edge boundary parameter of r.sub.2, A.sub.2(r.sub.2), is 0.6, and the corresponding distribution parameter a.sub.2(r.sub.2) is 0.2;

[0073] The rising edge boundary parameter of r.sub.3, A.sub.1(r.sub.3), is 0.8, and the corresponding distribution parameter a.sub.1(r.sub.3) is 0.1;

[0074] The falling edge boundary parameter of r.sub.3, A.sub.2(r.sub.3), is 3.6, and the corresponding distribution parameter a.sub.2(r.sub.3) is 0.3;

[0075] The boundary parameter of r.sub.4, A(r.sub.4), is 1.43, and the corresponding distribution parameter a(r.sub.4) is 0.1;

[0076] (V) Obtaining the probabilities of the cases that the ratio codes of the ratios r.sub.1, r.sub.2 and r.sub.3 are 0, 1 and 2 respectively and the probabilities of the cases that the ratio code of r.sub.4 is 0 or 1 respectively, with the semi-Cauchy rising/falling function; the expressions are as follows:

[0077] Probability f-code0(r.sub.1) of the case that the ratio code of r.sub.1 is 0:

[00016]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.0.Math.\left(r_1\right)=\{\begin{array}{cc}1& \left(r_1\le 0.08\right)\\ \frac{1}{1+{\left(\frac{0.08-r_1}{0.01}\right)}^2}& \left(r_1>0.08\right)\end{array}& \left(\mathrm{expression}.Math..Math.1\right)\end{array}$

[0078] Probability f-code1(r.sub.1) of the case that the ratio code of r.sub.1 is 1:

[00017]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.1.Math.\left(r_1\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.08-r_1}{0.01}\right)}^2}& \left(r_1\le 0.08\right)\\ 1& \left(0.08\le r_1\le 3.1\right)\\ \frac{1}{1+{\left(\frac{3.1-r_1}{0.1}\right)}^2}& \left(r_1>3.1\right)\end{array}& \left(\mathrm{expression}.Math..Math.2\right)\end{array}$

[0079] Probability f-code2(r.sub.1) of the case that the ratio code of r.sub.1 is 2:

[00018]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.2.Math.\left(r_1\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{3.1-r_1}{0.1}\right)}^2}& \left(r_1<3.1\right)\\ 1& \left(r_1\ge 3.1\right)\end{array}& \left(\mathrm{expression}.Math..Math.3\right)\end{array}$

[0080] Probability f-code0(r.sub.2) of the case that the ratio code of r.sub.2 is 0:

[00019]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.0.Math.\left(r.Math..Math.2\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.06-r_2}{0.02}\right)}^2}& \left(r_2<0.06\right)\\ 1& \left(0.06\le r_2\le 0.6\right)\\ \frac{1}{1+{\left(\frac{0.6-r_2}{0.2}\right)}^2}& \left(r_2>0.6\right)\end{array}& \left(\mathrm{expression}.Math..Math.4\right)\end{array}$

[0081] Probability f-code1(r.sub.2) of the case that the ratio code of r.sub.2 is 1:

[00020]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.1.Math.\left(r.Math..Math.2\right)=\{\begin{array}{cc}1& \left(r_2\le 0.06\right)\\ \frac{1}{1+{\left(\frac{0.06-r_2}{0.02}\right)}^2}& \left(r_2>0.06\right)\end{array}& \left(\mathrm{expression}.Math..Math.5\right)\end{array}$

[0082] Probability f:code2(r.sub.2) of the case that the ratio code of r.sub.2 is 2:

[00021]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.2.Math.\left(r_2\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.6-r_2}{0.2}\right)}^2}& \left(r_2<0.6\right)\\ 1& \left(r_2\ge 0.6\right)\end{array}& \left(\mathrm{expression}.Math..Math.6\right)\end{array}$

[0083] Probability f-code0(r.sub.3) of the case that the ratio code of r.sub.3 is 0:

[00022]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.0.Math.\left(r_3\right)=\{\begin{array}{cc}1& \left(r_3\le 0.8\right)\\ \frac{1}{1+{\left(\frac{0.8-r_3}{0.1}\right)}^2}& \left(r_3>0.8\right)\end{array}& \left(\mathrm{expression}.Math..Math.7\right)\end{array}$

[0084] Probability f-code1 (r.sub.3) of the case that the ratio code of r.sub.3 is 1:

[00023]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.1.Math.\left(r_3\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{0.6-r_3}{0.1}\right)}^2}& \left(r_3<0.8\right)\\ 1& \left(0.8\le r_3\le 3.6\right)\\ \frac{1}{1+{\left(\frac{3.6-r_3}{0.3}\right)}^2}& \left(r_3>3.6\right)\end{array}& \left(\mathrm{expression}.Math..Math.8\right)\end{array}$

[0085] Probability f-code2(r.sub.3) of the case that the ratio code of r.sub.3 is 2:

[00024]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.2.Math.\left(r_3\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{3.6-r_3}{0.3}\right)}^2}& \left(r_3>3.6\right)\\ 1& \left(0.8\le r_3\le 3.6\right)\end{array}& \left(\mathrm{expression}.Math..Math.9\right)\end{array}$

[0086] Probability f-code0(r.sub.4) of the case that the ratio code of r.sub.4 is 0:

[00025]$\begin{array}{cc}f.Math.\text{-}.Math.\mathrm{code}.Math..Math.0.Math.\left(r_4\right)=\{\begin{array}{cc}1& \left(r_4\le 1.43\right)\\ \frac{1}{1+{\left(\frac{1.43-r_4}{0.1}\right)}^2}& \left(r_4>1.43\right)\end{array}& \left(\mathrm{expression}.Math..Math.10\right)\end{array}$

[0087] Probability f-code1(r.sub.4) of the case that the ratio code of r.sub.4 is 1:

[00026]$\begin{array}{cc}f-\mathrm{code}.Math..Math.1.Math.\left(r_4\right)=\{\begin{array}{cc}\frac{1}{1+{\left(\frac{1.43-r_4}{0.1}\right)}^2}& \left(r_4<1.43\right)\\ 1& \left(r_4\ge 1.43\right)\end{array}& \left(\mathrm{expression}.Math..Math.11\right)\end{array}$

[0088] (VI) Replacing the 0 logic and 1 logic in the ratio code judgement rule with minimum value logic and maximum value logic respectively, carrying out defect and fault diagnosis according to the correspondence between the ratio codes and the types of transformer defects or faults, and representing the result of diagnosis in a fuzzy multi-value form; the result is represented in the form of probability, the result of diagnosis is the probability of occurrence of defect, i.e., severity; the sum of the probabilities of all kinds of faults is 1; the probabilities of the ratio codes are represented by maximum value logic and minimum value logic, and the probabilities of the faults are:

f(partial discharge)=min[f-code0(r.sub.1), f-code1(r.sub.2)];   (expression 12)

f(low-temperature overheat)=max{min[f-code0(r.sub.1), f-code0(r.sub.2), f-code1(r.sub.3)], min[f-code0(r.sub.1), f-code2(r.sub.2), f-code0(r3)]};   (expression 13)

f(moderate-temperature overheat)=min[f-code0(r.sub.1), f-code2(r.sub.2), f-code1(r3)];

f(high-temperature overheat)=max{min[f-code0(r.sub.1), f-code0(r.sub.2), f-code2(r.sub.3)], min[f-code0(r.sub.1), f-code2(r2), f-code2(r.sub.3)]};   (expression 14)

f(spark discharge)=max{f-code2(r.sub.1), min[f-code 1(r.sub.1), f-code0(r.sub.2), f-code1(r.sub.3), f-code0(r.sub.4)]};   (expression 15)

f(arc discharge)=max{min[f-code 1(r.sub.1), f-code0(r.sub.2), f-code1(r.sub.3), f-code1(r.sub.4)], min[f-code1(r.sub.1), f-code0(r.sub.3)], min[f-code 1(r.sub.1), f-code2(r.sub.3)], min [f-code1(r.sub.1), f-code1(r.sub.2), f-code1(r.sub.3)], min[f-code 1(r.sub.1), f-code2(r.sub.2), f-code 1(r.sub.3)]}.   (expression 16)

EXAMPLE 1

[0089] The oil chromatogram test data of a transformer (volumetric concentrations of five characteristic gases and total hydrocarbons, in unit of μL/L) is listed in Table 4.

TABLE-US-00004 TABLE 4 Oil Chromatogram Test Data of a Transformer c.sub.z(total Test Date c.sub.4(H.sub.2) c.sub.3(CH.sub.4) c.sub.5(C.sub.2H.sub.6) c.sub.2(C.sub.2H.sub.4) c.sub.1(C.sub.2H.sub.2) hydrocarbons) Apr. 26, 2012 31.33 10.52 1.98 4.01 6.09 22.60

[0090] As can be seen from Table 4, the volumetric concentration of acetylene exceeds the alert value, thus the transformer is abnormal. [0091] 1. Calculating the four ratios respectively:

[00027]$r_1=\frac{c_1\left(C_2.Math.H_2\right)}{c_2\left(C_2.Math.H_4\right)}=1.52,.Math.r_2=\frac{c_3\left({\mathrm{CH}}_4\right)}{c_4\left(H_2\right)}=2.03,.Math.r_3=\frac{c_2\left(C_2.Math.H_4\right)}{c_5\left(C_2.Math.H_6\right)}=2.03,.Math.r_4=\frac{c_1\left(C_2.Math.H_4\right)}{c_5\left(C_2.Math.H_6\right)}\times \frac{c_2\left(C_2.Math.H_4\right)}{c_1\left(C_2.Math.H_2\right)}=1.33;$ [0092] 2. Calculating with the expressions 1 to 11, to obtain the probabilities of the ratio codes of the four ratios:

[0093] f-code0(r.sub.1)=0; f-code1(r.sub.1)=1; f-code2(0=0.004;

[0094] f-code0(r.sub.2)=1; f-code1(r.sub.2)=0.0051; f-code2(r.sub.2)=0.37;

[0095] f-code0(r.sub.3)=0.00657; f-code l(r.sub.3)=1; f-code2(r.sub.3)=0.035;

[0096] f-code0(r.sub.4)=-1; f-code2(r.sub.4)=0.5; [0097] 3. Calculating with the expression 12 to 16, to obtain the probabilities of the faults:

[0098] f(partial discharge)=0%;

[0099] f(low-temperature overheat)=0%;

[0100] f(moderate-temperature overheat)=0%;

[0101] f(high-temperature overheat)=0%;

[0102] f(spark discharge)=66.7%;

[0103] f(arc discharge)=33.3%; [0104] 4. Diagnosing the transformer faults

[0105] It is judged from the above probabilities of faults, the transformer has spark discharge fault and arc discharge fault.

[0106] While the above embodiments are only preferred embodiments of the present invention, the feasible embodiments of the present invention are not exhausted. Those having ordinary skills in the art may make obvious modifications without departing from the principle and spirit of the present invention; however, all of such modifications shall be deemed as falling into the scope of protection of the present invention as defined by the claims.