A transformer fault location method

Abstract

The application discloses a transformer fault position determination method, wherein through setting parameters such as short-circuit voltage percentage among each winding of a transformer device and three-side rated voltage obtained by a system micro-service, high, medium and low three-side winding impedance is calculated. Then, according to the three-side winding impedance, three-side short-circuit current value ranges corresponding to six potential transformer fault positions are calculated; at the same time, three-side actual short-circuit current values at the fault moment are obtained according to obtained fault recording data. According to the comparison result of the short-circuit current ranges corresponding to the six potential transformer fault positions and the actual short-circuit current values, the transformer fault position is determined. Through the transformer device setting parameters and the recording data, the transformer fault position can be quickly positioned by the application, and the maintenance efficiency and safety are improved.

Description

Technical Field

[0001] This invention relates to a method for determining the location of transformer faults, and more particularly to a method for determining the location of transformer faults based on the variation law of short-circuit current. Background Technology

[0002] 220kV transformers are crucial equipment in high-voltage substations, undertaking the vital tasks of power transmission and voltage level transformation, and are one of the core components of the power grid. A transformer failure can lead to anything from a complete substation blackout, affecting power transmission and load supply in some areas, to a more severe disruption of the grid's stability, resulting in a wider-ranging power outage. Therefore, ensuring the safe and reliable operation of transformers is of paramount importance.

[0003] If a fault occurs during transformer operation, the transformer protection device will issue a trip command, disconnecting the circuit breakers on all sides of the transformer and isolating it from the power grid to prevent damage caused by prolonged exposure to short-circuit current. After the transformer trips and stops operating, maintenance personnel need to determine the fault location as quickly as possible, identifying whether the fault is on the high-voltage, medium-voltage, or low-voltage side, and whether the fault is inside the transformer windings or on the external leads. This provides maintenance personnel with accurate and reliable fault diagnosis information, facilitating rapid fault location and troubleshooting, shortening transformer downtime, and improving the safety and stability of the power grid and the reliability of power supply to users.

[0004] When a transformer fails, its protection device trips to isolate the fault. This device provides both primary and backup protection. Primary protection typically uses differential current protection, reliably distinguishing between internal and external transformer faults. Backup protection generally employs impedance and overcurrent protection principles, providing backup protection for the primary protection. When the primary protection fails to operate or its sensitivity is insufficient, the backup protection takes over to isolate the fault. While transformer protection provides fast and reliable protection, it has limitations. It can only isolate faults but cannot pinpoint their exact location, failing to provide maintenance personnel with a basis for troubleshooting. This significantly impacts the efficiency of transformer fault repair and has become a major factor hindering the safe and stable operation of the power grid and the improvement of power supply reliability. Summary of the Invention

[0005] The technical problem to be solved by the present invention is to provide a method for determining the location of transformer faults. The method uses the fault short-circuit current to construct a fault location determination algorithm, which can determine the specific location of the fault based on the change law of the short-circuit current on the high and medium voltage sides when a transformer fault occurs, thereby assisting maintenance personnel in quickly diagnosing the fault.

[0006] To address the aforementioned technical problems, one aspect of the present invention provides a method for determining the location of a transformer fault, comprising the following steps:

[0007] Step S10: Obtain transformer fault equipment parameters, the parameters including at least the inter-winding short-circuit voltage percentage, the three-sided rated voltage, and the three-sided rated capacity; the inter-winding short-circuit voltage percentage includes: high-medium winding short-circuit voltage percentage, high-low winding short-circuit voltage percentage, and medium-low winding short-circuit voltage percentage.

[0008] Step S11: Based on the transformer parameters and the percentage of short-circuit voltage on the three windings of the transformer, obtain the impedance values ​​of the high, medium and low voltage windings.

[0009] Step S12: Perform per-unit value reduction on the impedance values ​​of the three-sided windings;

[0010] Step S13: Pre-determine multiple potential transformer fault locations, and calculate the short-circuit current on each side corresponding to each potential transformer fault location based on the impedance values ​​of the three windings after per-unit value reduction.

[0011] Step S14: Calculate the short-circuit current at the time of the fault by analyzing the fault recording data, and use the Fourier algorithm to calculate the high-voltage side current value I. H and the medium voltage side current value I M ;

[0012] Step S15: Based on the calculated short-circuit current on each side corresponding to each potential transformer fault location, and the high-voltage side current value I... H Medium voltage side current value I M By comparing the fault locations, the final location of the transformer fault is determined.

[0013] More specifically, step S11 further includes:

[0014] Step S110: Calculate the winding impedances of the high, medium, and low voltage sides of the transformer according to the following formula:

[0015] Percentage of short-circuit voltage on the high-voltage side winding: U k1 % = (U k(1-2) %+U k(3-1) %-U k(2-3) %) / 2

[0016] Percentage of short-circuit voltage on medium-voltage side winding: U k2 % = (U k(1-2) %+U k(2-3) %-U k(3-1) %) / 2

[0017] Low-voltage side winding short-circuit voltage percentage: U k3 % = (U k(2-3) %+U k(3-1) %-U k(1-2) %) / 2

[0018] In the formula: U k1 % represents the percentage of short-circuit voltage on the high-voltage side, U k(1-2) % represents the percentage of short-circuit voltage between high-strength windings, U k(3-1) % represents the percentage of short-circuit voltage between high and low windings, U k(2-3) % represents the percentage of short-circuit voltage between medium and low windings;

[0019] Step S111: Based on the transformer parameters and the percentage of short-circuit voltage on the three windings of the transformer, calculate the impedance values ​​of the high, medium, and low voltage windings using the following formula:

[0020] High-voltage side winding impedance: X H =U k1 *U N1 *U N1 / S N1

[0021] Medium voltage side winding impedance: X M =U k2 *U N2 *U N2 / S N2

[0022] High-voltage side winding impedance: X L =U k3 *U N3 *U N3 / S N3

[0023] In the formula: U ki (i = 1, 2, ... 3) represents the short-circuit voltage of the high, medium, and low voltage windings, U Ni (i = 1, 2, ..., 3) represents the rated voltage of the high, medium, and low voltage sides, S Ni (i = 1, 2, ... 3) represents the rated capacity of the high, medium, and low sides.

[0024] More specifically, step S13 further includes:

[0025] When the potential transformer fault location is at f1 on the high-voltage side lead, the three-sided short-circuit current value is calculated according to the following formula:

[0026] High-voltage side short-circuit current value

[0027] In the formula: E H X is the high-voltage side power supply. H This represents the impedance of the high-voltage side system.

[0028] Medium voltage side short circuit current value

[0029] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. II This is the impedance of the medium-voltage side winding.

[0030] More specifically, step S13 further includes:

[0031] When the potential transformer fault location is inside the high-voltage winding (f2), the three-sided short-circuit current value is calculated according to the following formula:

[0032] The short-circuit current on the high-voltage side is:

[0033] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This refers to the impedance of the high-voltage side winding.

[0034] The short-circuit current measured on the medium voltage side is:

[0035] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II This is the impedance of the medium-voltage side winding.

[0036] More specifically, step S13 further includes:

[0037] When the potential transformer fault location is at f3 on the medium-voltage side lead, the three-sided short-circuit current value is calculated according to the following formula:

[0038] The short-circuit current on the high-voltage side is:

[0039] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. II The impedance of the medium-voltage side winding;

[0040] The short-circuit current value on the medium voltage side is:

[0041] In the formula: E M X is the high-voltage side power supply. M This represents the impedance of the high-voltage side system.

[0042] More specifically, step S13 further includes:

[0043] When the potential transformer fault location is f4 inside the medium-voltage winding, the three-sided short-circuit current value is calculated according to the following formula:

[0044] The short-circuit current value on the high-voltage side is:

[0045] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This refers to the impedance of the high-voltage side winding.

[0046] Medium voltage side short circuit current value I 4M for:

[0047] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II This is the impedance of the medium-voltage side winding.

[0048] More specifically, step S13 further includes:

[0049] When the potential transformer fault location is at f5 on the low-voltage side lead, the three-sided short-circuit current value is calculated according to the following formula:

[0050] High-voltage side short-circuit current I 5H for:

[0051] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. III The impedance of the medium-voltage side winding;

[0052] Medium voltage side short circuit current value I 5M for:

[0053] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II X is the impedance of the medium-voltage side winding. III This is the impedance of the medium-voltage side winding.

[0054] More specifically, step S13 further includes:

[0055] When the potential transformer fault location is inside the low-voltage winding at f6, the three-sided short-circuit current value is calculated according to the following formula:

[0056] The low-voltage side short-circuit current is:

[0057] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This refers to the impedance of the high-voltage side winding.

[0058] Medium voltage side short circuit current value

[0059] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II This is the impedance of the medium-voltage side winding.

[0060] More specifically, step S15 further includes:

[0061] If the data calculated in step S13 satisfies the following formula, then it is determined that there is a fault in the high-voltage side lead:

[0062]

[0063] In the formula, I H I M These are the high-voltage side current value and the medium-voltage side current value calculated in step S14, respectively. 1H I 1M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs on the high-voltage side lead, calculated in step S13.

[0064] If the data calculated in step S13 satisfies the following formula, then it is determined that there is an internal fault in the high-voltage side winding:

[0065]

[0066] In the formula: I 1H I 1M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs on the high-voltage side lead, calculated in step S13, respectively. 2H I 2M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when an internal fault occurs in the high-voltage side winding, as calculated in step S13.

[0067] If the data calculated in step S13 satisfies the following formula, then it is determined that there is a fault at the medium-voltage side lead:

[0068]

[0069] In the formula: I 3H I 3M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the medium-voltage side lead obtained in step S13;

[0070] If the data calculated in step S13 satisfies the following formula, then it is determined that there is an internal fault in the medium-voltage side winding:

[0071]

[0072] In the formula: I 3H I 3M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the medium-voltage side lead obtained in step S13, respectively. 4H I 4M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when an internal fault occurs in the medium-voltage side winding, as calculated in step S13.

[0073] If the data calculated in step S13 satisfies the following formula, then it is determined that there is a fault at the low-voltage side lead:

[0074]

[0075] In the formula: I 5H I 5M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the low-voltage side lead obtained in step S13;

[0076] If the data calculated in step S13 satisfies the following formula, it is determined that there is a low-voltage side winding fault.

[0077]

[0078] In the formula: I 5H I 5M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the low-voltage side lead, calculated in step S13, respectively. 6H I 6M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when the low-voltage side winding is faulty, calculated in step S13.

[0079] More specifically, it further includes:

[0080] Based on the criteria, the specific location of the fault is determined, and a transformer fault location display interface is generated, along with a transformer fault report.

[0081] Implementing this invention has the following beneficial effects:

[0082] This invention provides a method for determining the location of transformer faults. It uses the difference in the magnitude of the short-circuit current felt on the high-voltage side and the medium-voltage side to determine the location of the transformer fault. It uses the variation pattern of the short-circuit current on the high, medium, and low voltage sides to determine the specific location of the fault, providing maintenance personnel with a basis for fault diagnosis and facilitating the rapid determination of the transformer fault location.

[0083] The present invention has a simple implementation scheme. It only requires obtaining the magnitude of the short-circuit current on the high-voltage side and the medium-voltage side to determine the location of the transformer fault. The scheme does not require determining the power direction and is not affected by the PT disconnection. Attached Figure Description

[0084] To more clearly illustrate the technical solutions in the embodiments of the present invention or the prior art, the drawings used in the description of the embodiments or the prior art will be briefly introduced below. Obviously, the drawings described below are only some embodiments of the present invention. For those skilled in the art, obtaining other drawings based on these drawings without creative effort still falls within the scope of the present invention.

[0085] Figure 1 This is a schematic diagram of the main flow of an embodiment of a transformer fault location determination method provided by the present invention;

[0086] Figure 2 This is a diagram of the microservice interface for the setting value involved in the present invention;

[0087] Figure 3 This is a schematic diagram of the 220kV three-winding transformer structure involved in the present invention;

[0088] Figure 4 A more detailed flowchart of the transformer fault location process of this invention;

[0089] Figure 5 This is a schematic diagram illustrating a fault on the high-voltage side of a three-winding circuit as described in the present invention. Detailed Implementation

[0090] The technical solutions of the embodiments of the present invention will be clearly and completely described below with reference to the accompanying drawings. Obviously, the described embodiments are only some embodiments of the present invention, and not all embodiments. Based on the embodiments of the present invention, all other embodiments obtained by those skilled in the art without creative effort are within the scope of protection of the present invention.

[0091] Figure 1 This diagram illustrates the main flow of an embodiment of a transformer fault location determination method provided by the present invention, and is combined with... Figures 2 to 5 As shown, in this embodiment, the method includes the following steps:

[0092] Step S10: Obtain transformer fault equipment parameters. The parameters include at least the inter-winding short-circuit voltage percentage, the three-sided rated voltage, and the three-sided rated capacity. The inter-winding short-circuit voltage percentage includes: high-medium winding short-circuit voltage percentage, high-low winding short-circuit voltage percentage, and medium-low winding short-circuit voltage percentage. It can be understood that the transformer equipment parameters refer to the relevant data of the faulty transformer itself, which is included at the factory and will not be changed.

[0093] In practical applications, the required data is determined based on functional requirements. Transformer fault equipment parameters are accessed through the setting system microservice, and inter-system interaction is achieved via a RESTful interface, enabling data sharing between systems. The setting value microservice interface can be found in [reference needed]. Figure 2 As shown. The method for obtaining tuning parameters based on the microservice of the tuning system in step S10 is as follows:

[0094] Data analysis determines the relevant data that needs to be accessed based on functional requirements, mainly including the short-circuit voltage percentage Uk% (including the short-circuit voltage percentage between the high-frequency windings Uk%). k(1-2) %, percentage of short-circuit voltage between high and low windings U k(3-1) %, percentage of short-circuit voltage between medium and low windings U k(2-3) %), three-sided rated voltage U N and three-sided rated capacity S N wait.

[0095] The standardized definition of the relevant data model is as follows:

[0096] Table 1 Standardized Definition of Model

[0097]

[0098]

[0099] The push notification method is as follows:

[0100] The push method for the line-based forced delivery auxiliary decision-making service interface is described as follows:

[0101] HTTP method: GET

[0102] Request: http: / / ipAddress:port / dmp / fault / list / {faultId}

[0103] The input parameters (URL parameters) are shown in Table 2 below:

[0104] Table 2 Input Parameters

[0105] Parameter name Required type describe access_token yes String access_token obtained through API gateway faultId yes String Fault ID

[0106] The output parameters are shown in Table 3 below:

[0107] Table 3 Output Parameters

[0108] Parameter name type describe code Int 200: Success; other values ​​indicate failure. See the error code table for details. data JSONArray JSON object format model data message String Return Information Description

[0109] Step S11: Calculate the impedance values ​​of the high, medium, and low voltage windings by combining the transformer parameters and the percentage of short-circuit voltage on the three windings. A schematic diagram of the three-winding transformer structure can be found by referring to... Figure 3 As shown.

[0110] More specifically, step S11 further includes:

[0111] Step S110: Since the short-circuit voltage percentage given in the transformer parameters is the short-circuit voltage between pairs of windings, not for a single winding, it is first necessary to calculate the short-circuit voltage percentage parameter for a single winding. Based on the transformer parameters, the winding impedances of the high, medium, and low voltage sides of the transformer are calculated according to the following formula:

[0112] Percentage of short-circuit voltage on the high-voltage side winding: U k1 % = (U k(1-2) %+U k(3-1) %-U k(2-3) %) / 2

[0113] Percentage of short-circuit voltage on medium-voltage side winding: U k2 % = (U k(1-2) %+U k(2-3) %-U k(3-1) %) / 2

[0114] Low-voltage side winding short-circuit voltage percentage: U k3 % = (U k(2-3) %+U k(3-1) %-U k(1-2) %) / 2

[0115] In the formula: U k1 % represents the percentage of short-circuit voltage on the high-voltage side, U k(1-2) % represents the percentage of short-circuit voltage between high-strength windings, U k(3-1) % represents the percentage of short-circuit voltage between high and low windings, U k(2-3) % represents the percentage of short-circuit voltage between medium and low windings;

[0116] Step S111: Based on the transformer parameters and the percentage of short-circuit voltage on the three windings of the transformer, calculate the impedance values ​​of the high, medium, and low voltage windings using the following formula:

[0117] High-voltage side winding impedance: X H =U k1 *U N1 *UN1 / S N1

[0118] Medium voltage side winding impedance: X M =U k2 *U N2 *U N2 / S N2

[0119] High-voltage side winding impedance: X L =U k3 *U N3 *U N3 / S N3

[0120] In the formula: U ki (i = 1, 2, ... 3) represents the short-circuit voltage of the high, medium, and low voltage windings, U Ni (i = 1, 2, ..., 3) represents the rated voltage of the high, medium, and low voltage sides, S Ni (i = 1, 2, ... 3) represents the rated capacity of the high, medium, and low sides.

[0121] Step S12: Perform per-unit value reduction on the impedance values ​​of the three-sided windings;

[0122] Because transformers have turns ratios, the impedance values ​​of different windings need to be normalized when calculating short-circuit current. A per-unit calculation method is used to normalize the impedances of each side to overcome the differences in voltage levels across the transformer. Since a complete and mature solution already exists for this process, it will not be elaborated further.

[0123] Step S13: Pre-determine multiple potential transformer fault locations, and calculate the short-circuit current on each side corresponding to each potential transformer fault location based on the impedance values ​​of the three windings after per-unit value reduction.

[0124] Understandably, based on transformer principles, the calculation methods for short-circuit currents differ depending on the location of the fault. Therefore, during the analysis, the transformer fault locations are subdivided into the following six locations, and the short-circuit currents on each side are calculated for each type of fault. The locations can be referenced... Figure 3 The content marked in the table, specifically, is used to distinguish internal faults f2, f4 and f6. Therefore, it is assumed that the fault location is located at most 95% of the position of the end of each winding, and not at the end of the winding (if the fault is located at the very end of each winding, then f2, f4 and f6 are the same fault point). The transformer fault location subdivision table is shown below.

[0125] Table 4 Input Parameters

[0126]

[0127]

[0128] Therefore, more specifically, step S13 further includes:

[0129] When the potential transformer fault location is at f1 on the high-voltage side lead, according to transformer principles, the high-voltage side short-circuit current is the largest when the high-voltage side lead is faulty. The short-circuit current values ​​on all three sides are calculated using the following formula:

[0130] High-voltage side short-circuit current value

[0131] In the formula: E H X is the high-voltage side power supply. H This represents the impedance of the high-voltage side system.

[0132] Medium voltage side short circuit current value

[0133] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. II This is the impedance of the medium-voltage side winding.

[0134] Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0135] More specifically, step S13 further includes:

[0136] When the potential transformer fault location is inside the high-voltage winding (f2), according to transformer principles, the short-circuit current measured on the high-voltage side will decrease when there is a fault inside the high-voltage winding. When a short-circuit fault occurs at the end of the high-voltage winding, the short-circuit current values ​​on all three sides are calculated using the following formula:

[0137] The short-circuit current on the high-voltage side is:

[0138] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This refers to the impedance of the high-voltage side winding.

[0139] The short-circuit current measured on the medium voltage side is:

[0140] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II This represents the impedance of the medium-voltage side winding. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0141] More specifically, step S13 further includes:

[0142] When the potential transformer fault location is at f3 on the medium-voltage side lead, the three-sided short-circuit current value is calculated according to the following formula:

[0143] The short-circuit current on the high-voltage side is:

[0144] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. II The impedance of the medium-voltage side winding;

[0145] The short-circuit current value on the medium voltage side is:

[0146] In the formula: E M X is the high-voltage side power supply. M This represents the system impedance on the high-voltage side. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0147] More specifically, step S13 further includes:

[0148] When the potential transformer fault location is at f4 inside the medium-voltage winding, i.e., when a short-circuit fault occurs at the end of the medium-voltage winding, the three-sided short-circuit current values ​​are calculated according to the following formula:

[0149] The short-circuit current value on the high-voltage side is:

[0150] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This refers to the impedance of the high-voltage side winding.

[0151] Medium voltage side short circuit current value I 4M for:

[0152] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II This represents the impedance of the medium-voltage side winding. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0153] More specifically, step S13 further includes:

[0154] When the potential transformer fault location is at f5 on the low-voltage side lead, the three-sided short-circuit current value is calculated according to the following formula:

[0155] High-voltage side short-circuit current I 5H for:

[0156] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. III The impedance of the medium-voltage side winding;

[0157] Medium voltage side short circuit current value I 5M for:

[0158] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II X is the impedance of the medium-voltage side winding. III This represents the impedance of the medium-voltage side winding. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0159] More specifically, step S13 further includes:

[0160] When the potential transformer fault location is inside the low-voltage winding at f6, the three-sided short-circuit current value is calculated according to the following formula:

[0161] The low-voltage side short-circuit current is:

[0162] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This refers to the impedance of the high-voltage side winding.

[0163] Medium voltage side short circuit current value

[0164] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II This represents the impedance of the medium-voltage side winding. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0165] Step S14: Use the Fourier algorithm to calculate the high-voltage side current value I at the time of the transformer fault recording data. H and the medium voltage side current value I M It can be understood that fault recording data refers to the voltage and current data recorded by the transformer fault recorder before and after a transformer fault, generating a fault recording data file. This mainly includes analog channel data such as the voltage and current data of the high, medium, and low voltage sides (A, B, C), as well as various digital channel data for opening, closing, and blocking. This invention primarily uses the voltage and current data of the high, medium, and low voltage sides (A, B, C) from the fault recording data.

[0166] Step S15, based on the high-voltage side current value I at the actual fault moment of the transformer described in step S14. H Medium voltage side current value I M The fault location is determined by comparing the short-circuit current range on each side corresponding to each potential transformer fault location calculated in step S13, and the final transformer fault location is determined.

[0167] More specifically, step S15 further includes:

[0168] According to transformer principles, when a fault occurs at the high-voltage side lead, considering certain error factors and reliability requirements, the current threshold for a high-voltage side fault is set to 0.95 times the short-circuit current at the high-voltage side bus outlet. When a fault occurs at the high-voltage side lead, because the fault point is far from the measurement point on the medium-voltage side, the corresponding impedance is larger, resulting in a larger error. Therefore, the current threshold for a medium-voltage side fault is set to 0.85 times the short-circuit current measured on the medium-voltage side. Therefore, if the data calculated in step S13 simultaneously satisfies the following two equations, a high-voltage side lead fault is determined to exist:

[0169]

[0170] In the formula, I H I M These are the high-voltage side current value and the medium-voltage side current value obtained from the measured calculation during the transformer fault in step S14, respectively. 1H I 1M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs on the high-voltage side lead, calculated in step S13.

[0171] According to transformer principles, when there is an internal fault in the high-voltage winding, the high-voltage current is less than I. 1H But greater than I 2H The medium-voltage side current is greater than I. 1M But less than I 2M Because the impedance of the high-voltage side winding is relatively large, any fault in the high-voltage side winding will lead to a significant decrease in the short-circuit current. Therefore, the upper limit of the threshold value can be taken as 0.95I. 1H To ensure reliability, the lower limit of the threshold value is taken as the short-circuit current at the end of the high-voltage winding during a fault, and no reliability coefficient is required. The same treatment applies to the medium-voltage side. Therefore, if the data calculated in step S13 satisfies the following formula, it is determined that there is an internal fault in the high-voltage winding:

[0172]

[0173] In the formula: I 1H I 1M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs on the high-voltage side lead, calculated in step S13, respectively. 2HI 2M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when an internal fault occurs in the high-voltage side winding, as calculated in step S13.

[0174] According to transformer principles, when a fault occurs at the medium-voltage side lead, considering certain error factors and reliability requirements, the current threshold for a medium-voltage side fault is set to 0.95 times the short-circuit current at the bus outlet on that side. Since the fault point is far from the measurement point on the high-voltage side, the current threshold for a high-voltage side fault is set to 0.85 times the measured short-circuit current on that side. Therefore, if the data calculated in step S13 satisfies the following formula, a fault is determined to exist at the medium-voltage side lead:

[0175]

[0176] In the formula: I 3H I 3M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the medium-voltage side lead obtained in step S13;

[0177] According to transformer principles, when there is an internal fault in the medium-voltage winding, the medium-voltage side current is less than I. 3M But greater than I 4M The high-voltage side current is less than I 4H But greater than I 3H Among them, the impedance of the high-voltage side winding is relatively large while that of the medium-voltage side winding is relatively small. For the high-voltage side, the short-circuit current caused by a fault in the medium-voltage side winding is not significantly lower than the short-circuit current caused by a fault at the end of the high-voltage side winding. Therefore, the upper limit of the threshold value of the high-voltage side short-circuit current can be directly taken as I. 4H The lower limit value is directly taken as I. 3H For the medium-voltage side, because the impedance of the winding on this side is relatively small, the lower limit of the short-circuit current can be taken as 0.95I to ensure reliability. 4M The upper limit can be directly taken as I. 3M Without adding a reliability coefficient, if the data calculated in step S13 satisfies the following formula, it is determined that there is an internal fault in the medium-voltage side winding:

[0178]

[0179] In the formula: I 3H I 3M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the medium-voltage side lead obtained in step S13, respectively. 4H I 4M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when an internal fault occurs in the medium-voltage side winding, as calculated in step S13.

[0180] According to transformer principles, when a fault occurs at the low-voltage side lead, although the impedance value of the low-voltage side winding is not large, the impedance value becomes very significant when referred to the high-voltage and medium-voltage sides. Considering certain error factors and reliability requirements, the threshold values ​​for the short-circuit current on both the high-voltage and medium-voltage sides are taken as a coefficient of 0.85. Therefore, if the data calculated in step S13 satisfies the following formula, it is determined that a fault exists at the low-voltage side lead:

[0181]

[0182] In the formula: I 5H I 5M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the low-voltage side lead obtained in step S13;

[0183] According to transformer principles, when a fault occurs in the low-voltage winding, the medium-voltage current is less than I. 6M But greater than I 5M The high-voltage side current is less than I 6H But greater than I 5H This is because the impedance of the low-voltage side winding increases significantly when referred to the high-voltage and medium-voltage sides. Considering the impact of error factors and reliability requirements, for the high-voltage and medium-voltage sides, the lower limit is taken as 1.05 times the short-circuit current at the low-voltage side bus outlet, and the upper limit is taken as 0.95 times the short-circuit current at the end of the low-voltage side winding when there is a fault. Therefore, if the data calculated in step S13 satisfies the following formula, it is determined that there is a fault in the low-voltage side winding.

[0184]

[0185] In the formula: I 5H I 5M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the low-voltage side lead, calculated in step S13, respectively. 6H I 6M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when the low-voltage side winding is faulty, calculated in step S13.

[0186] More specifically, it further includes:

[0187] Based on the criteria, the specific location of the fault is determined, and a transformer fault location display interface is generated. At the same time, a transformer fault report is generated to provide maintenance personnel with a basis for fault diagnosis and assist them in quickly determining the location of the transformer fault.

[0188] To better understand the present invention, the steps described above are explained below with reference to a specific example.

[0189] First, in step S10, the setting parameters are accessed based on the microservice platform. The short-circuit setting parameters of the 220kV three-winding transformer are obtained, and the following data parameters of the transformer fault equipment in the setting system are accessed through the microservice platform: short-circuit voltage percentage Uk% (including the short-circuit voltage percentage between the high and medium windings Uk%). k(1-2) %, percentage of short-circuit voltage between high and low windings U k(3-1) %, percentage of short-circuit voltage between medium and low windings U k(2-3) %), rated voltage U on three sides of the transformer N and the rated capacity S of the three sides of the transformer N The specific input and output parameters are as follows:

[0190] Request input parameters (URL parameters):

[0191] Table 5 Input Parameters

[0192] Parameter name Required type describe access_token yes String access_token obtained through API gateway faultId yes String Fault ID (aeaog1jj4jgp3aae)

[0193] Output parameters

[0194] Table 6 Input Parameters

[0195]

[0196]

[0197] In step S11, the percentage of short-circuit voltage on all three sides is first calculated.

[0198] Since the short-circuit voltage percentage given in the transformer parameters is the short-circuit voltage between pairs of windings, not for a single winding, it is necessary to first calculate the short-circuit voltage percentage parameter for a single winding. According to the transformer parameters, the calculation formulas for the winding impedances of the high, medium, and low voltage sides of the transformer are as follows.

[0199] Percentage of short-circuit voltage on the high-voltage side winding:

[0200] U k1 % = (U k(1-2) %+U k(3-1) %-U k(2-3) %) / 2 = (14.22% + 30.68% - 22.02%) / 2 = 11.44%

[0201] Percentage of short-circuit voltage on medium-voltage side winding:

[0202] U k2 % = (U k(1-2) %+U k(2-3) %-U k(3-1) %) / 2=(14.22%+22.02%-30.68%) / 2=2.78%

[0203] Low-voltage side winding short-circuit voltage percentage:

[0204] U k3 % = (U k(2-3) %+U k(3-1) %-U k(1-2) %) / 2 = (30.68% + 22.02% - 14.22%) / 2 = 19.24%

[0205] In the formula: U k1 % represents the percentage of short-circuit voltage on the high-voltage side, U k(1-2) % represents the percentage of short-circuit voltage between high-strength windings, U k(3-1) % represents the percentage of short-circuit voltage between high and low windings, U k(2-3) % represents the percentage of short-circuit voltage between medium and low windings.

[0206] Then, the impedance values ​​of the three-sided windings are calculated.

[0207] Based on the transformer parameters and the percentage of short-circuit voltage on the three windings, the impedance values ​​of the high, medium, and low voltage windings are calculated using the following formulas.

[0208] High-voltage side winding impedance: X H =U k1 *U N1 *U N1 / S N1 =U k1 *220*220 / 100 / 240=23.07(Ω)

[0209] Medium voltage side winding impedance: X M =U k2 *U N2 *U N2 / S N2 =U k2 *115*115 / 100 / 240=1.53(Ω)

[0210] High-voltage side winding impedance: X L =U k3 *U N3 *U N3 / S N3 =uk3*10.5*10.5 / 100 / 80=0.27(Ω)

[0211] In the formula: U ki (i = 1, 2, ..., 3) represent the short-circuit voltages of the high, medium, and low voltage windings, respectively, U Ni (i = 1, 2, ..., 3) represent the rated voltages of the high, medium, and low voltage sides, respectively, and S... N i (i = 1, 2, ... 3) represent the rated capacities of the high, medium, and low sides, respectively.

[0212] In step S12, impedance per-unit value reduction is performed.

[0213] Because transformers have turns ratios, the impedance values ​​of different windings need to be normalized when calculating short-circuit current. A per-unit calculation method is used to normalize the impedances of each side to overcome the differences in voltage levels across the transformer. Since a complete and mature solution already exists for this process, it will not be elaborated further.

[0214] In step S13, the short-circuit current on each side is calculated.

[0215] Based on the principle of transformers, the calculation methods for short-circuit current on each side under different fault locations are performed. In the calculation process, the system impedance of 220kV is set to 1.6Ω and the system impedance of 110kV is set to 2.1Ω.

[0216] Step S13.1: Calculation of short-circuit current on each side when a fault occurs at f1 (high voltage side lead).

[0217] According to transformer principles, the short-circuit current on the high-voltage side is the largest when there is a fault in the high-voltage side lead. The short-circuit currents on all three sides are calculated as follows:

[0218] High-voltage side short-circuit current value

[0219] In the formula: E H X is the high-voltage side power supply. H This represents the impedance of the high-voltage side system.

[0220] Medium voltage side short circuit current value

[0221] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. II This is the impedance of the medium-voltage side winding.

[0222] Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0223] Step S13.2: Calculation of short-circuit current on each side when a fault occurs at f2 (inside the high-voltage side winding).

[0224] According to transformer principles, the short-circuit current measured on the high-voltage side will decrease when there is an internal fault in the high-voltage winding. When a short-circuit fault occurs at the end of the high-voltage winding, the high-voltage side short-circuit current is:

[0225]

[0226] In the formula: E H X is the high-voltage side power supply.H X is the impedance of the high-voltage side system. I This refers to the impedance of the high-voltage side winding.

[0227] The short-circuit current value measured on the medium-voltage side is denoted as I. 2M Its value is:

[0228]

[0229] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II The impedance of the medium-voltage side winding;

[0230] Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0231] Step S13.3: Calculation of short-circuit current on each side when a fault occurs at f3 (medium voltage side lead).

[0232] The high-voltage side short-circuit current is denoted as:

[0233]

[0234] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. II This is the impedance of the medium-voltage side winding.

[0235] The short-circuit current measured on the medium voltage side is:

[0236] In the formula: E M X is the high-voltage side power supply. M This is the system impedance on the high-voltage side. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0237] Step S13.4: Calculation of short-circuit current on each side when a fault occurs at f4 (inside the medium-voltage winding).

[0238] When a short-circuit fault occurs at the end of the medium-voltage winding, the short-circuit current on the high-voltage side is: In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This represents the impedance of the high-voltage side winding.

[0239] Medium voltage side short circuit current value I 4M for: In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. IIThis is the impedance of the medium-voltage side winding.

[0240] Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0241] Step S13.5: Calculation of short-circuit current on each side when a fault occurs at f5 (low-voltage side lead).

[0242] When a fault occurs at the low-voltage side lead, the short-circuit current I measured on the high-voltage side... 5H for:

[0243]

[0244] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I X is the impedance of the high-voltage side winding. III This is the impedance of the medium-voltage side winding.

[0245] Medium voltage side short circuit current value I 5M for:

[0246]

[0247] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II X is the impedance of the medium-voltage side winding. III This represents the impedance of the medium-voltage side winding. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0248] Step S13.6: Calculation of short-circuit current on each side when a fault occurs at f6 (inside the low-voltage side winding).

[0249] When a short-circuit fault occurs at the end of the low-voltage side winding, the short-circuit current is:

[0250]

[0251] In the formula: E H X is the high-voltage side power supply. H X is the impedance of the high-voltage side system. I This represents the impedance of the high-voltage side winding.

[0252] Short-circuit current measured on the medium voltage side

[0253] In the formula: E M X is the high-voltage side power supply. M X is the impedance of the high-voltage side system. II This represents the impedance of the medium-voltage side winding. Since there is no power supply on the low-voltage side, no short-circuit current will be measured.

[0254] In summary, the short-circuit current conditions corresponding to the six locations are summarized in the table below:

[0255] Table 7 Input Parameters

[0256] Serial Number Fault location High-voltage side current (kA) Medium voltage side current (kA) 1 High voltage side lead 79.39 6.65 2 High voltage side winding Greater than 5.15 Less than 17.5 3 Medium voltage side lead 4.13 30.24 4 Medium voltage side winding Less than 5.15 Greater than 17.5 5 Low voltage side lead 0.89 1.91 6 low-voltage side winding Less than 5.15 Less than 17.5

[0257] In step S14, the fault recording data is used to calculate the short-circuit current at the time of the fault, and the effective values ​​of the high-voltage and medium-voltage side currents are calculated using the Fourier algorithm, namely I0 and I0. H =83.62kA, I M =6.14kA.

[0258] Step S15: Determine the short circuit location.

[0259] The fault location is determined based on the magnitude of the short-circuit current at different locations, with the specific criteria as follows:

[0260] Step S15.1: Determine the fault at the high-voltage side lead.

[0261] As mentioned above, the short-circuit current I at the high-voltage side lead is... 1H The short-circuit current on the medium-voltage side is 79.39 kA. 1M The current is 6.65kA. If the measured current values ​​on the high-voltage side and medium-voltage side meet the following criteria, then the fault is at the high-voltage side lead:

[0262]

[0263] In the formula, I H I M These are the measured current values ​​on the high-voltage side and the medium-voltage side, respectively. 1H I 1M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side calculated in step S13.1, respectively.

[0264] If both expressions in the criterion are satisfied at the same time, it is judged as a high-voltage side lead fault.

[0265] Since the measured effective values ​​of the high and medium voltage side currents are respectively I H =83.62kA, I M =6.14kA. Substituting this into the above formula, if the condition is met, then it can be determined that there is a fault in the high-voltage side lead wire.

[0266] Step S15.2: Determine the internal fault of the high-voltage side winding.

[0267] High-voltage side short-circuit current I during internal fault of high-voltage side winding 2H The short-circuit current on the medium-voltage side is approximately 5.15 kA. 2MThe current is approximately 17.5kA. If the measured current values ​​on the high-voltage side and the medium-voltage side meet the following criteria, then it indicates an internal fault in the high-voltage side winding:

[0268]

[0269] In the formula: I H I M These are the measured current values ​​on the high-voltage side and the medium-voltage side, respectively. 1H I 1M The short-circuit current values ​​on the high-voltage side and medium-voltage side calculated in step S13.1 are respectively, I 2H I 2M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side calculated in step S13.2, respectively.

[0270] Since the measured effective values ​​of the high and medium voltage side currents are respectively I H =83.62kA, I M =6.14kA. Substituting this into the above formula, it is determined that the condition is not met, thus it can be concluded that this type of fault does not exist.

[0271] Step S15.3: Fault criterion at the medium-voltage side lead wire.

[0272] When a fault occurs at the medium-voltage side lead, the high-voltage side short-circuit current I 3H The short-circuit current on the medium-voltage side is approximately 4.1 kA. 3M The current is approximately 30kA. If the measured current values ​​on the high-voltage side and medium-voltage side meet the following criteria, then the fault is at the medium-voltage side lead:

[0273]

[0274] In the formula: I H I M These are the measured current values ​​on the high-voltage side and the medium-voltage side, respectively. 3H I 3M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side calculated in step S13.3, respectively.

[0275] Since the measured effective values ​​of the high and medium voltage side currents are respectively I H =83.62kA, I M =6.14kA. Substituting this into the above formula, it is determined that the condition is not met, thus it can be concluded that this type of fault does not exist.

[0276] Step S15.4: Determine the internal fault criteria of the medium-voltage side winding.

[0277] High-voltage side short-circuit current I when there is an internal fault in the medium-voltage winding 4H The short-circuit current on the medium-voltage side is approximately 5.15 kA. 4MThe current is approximately 17.5kA. If the measured current values ​​on the high-voltage side and medium-voltage side meet the following criteria, then it indicates an internal fault in the medium-voltage side winding:

[0278]

[0279] In the formula: I H I M These are the measured current values ​​on the high-voltage side and the medium-voltage side, respectively. 3H I 3M The short-circuit current values ​​on the high-voltage side and medium-voltage side calculated in step S13.3 are I. 4H I 4M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side calculated in step S13.4, respectively.

[0280] Since the measured effective values ​​of the high and medium voltage side currents are respectively I H =83.62kA, I M =6.14kA. Substituting this into the above formula, it is determined that the condition is not met, thus it can be concluded that this type of fault does not exist.

[0281] Step S15.5: Determine the fault criteria at the low-voltage side lead.

[0282] When a fault occurs at the low-voltage side lead, the high-voltage side short-circuit current I 5H The short-circuit current I on the medium-voltage side is approximately 0.89 kA. 5M The current is approximately 1.91 kA. If the measured current values ​​on the high-voltage and medium-voltage sides meet the following criteria, then the fault is at the low-voltage side lead:

[0283]

[0284] In the formula: I H I M These are the measured current values ​​on the high-voltage side and the medium-voltage side, respectively. 5H I 5M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side calculated in step S13.5, respectively.

[0285] Since the measured effective values ​​of the high and medium voltage side currents are respectively I H =83.62kA, I M =6.14kA. Substituting this into the above formula, it is determined that the condition is not met, thus it can be concluded that this type of fault does not exist.

[0286] Step S15.6: Determine the fault criteria for the low-voltage side winding.

[0287] High-voltage side short-circuit current I during low-voltage side winding fault 6H The short-circuit current on the medium-voltage side is approximately 5.15 kA. 6MThe current is approximately 17.5kA. If the measured current values ​​on the high-voltage and medium-voltage sides meet the following criteria, then the low-voltage winding is faulty:

[0288]

[0289] In the formula: I H I M These are the measured current values ​​on the high-voltage side and the medium-voltage side, respectively. 5H I 5M The short-circuit current values ​​on the high-voltage side and the medium-voltage side calculated in step S13.5 are respectively, I 6H I 6M These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side calculated in step S13.6, respectively.

[0290] Since the measured effective values ​​of the high and medium voltage side currents are respectively I H =83.62kA, I M =6.14kA. Substituting this into the above formula, it is determined that the condition is not met, thus it can be concluded that this type of fault does not exist.

[0291] Then, based on the criteria, the specific location of the fault is determined to be a fault in the high-voltage side lead, generating a transformer fault location display interface. For example... Figure 5 The diagram shows a schematic interface for a three-winding high-voltage side fault. A report is generated to provide maintenance personnel with troubleshooting information and assist them in quickly locating the transformer fault.

[0292] Implementing this invention has the following beneficial effects:

[0293] This invention provides a method for determining the location of transformer faults, which uses the difference in the magnitude of the short-circuit current sensed on the high-voltage side and the medium-voltage side to determine the location of transformer faults.

[0294] By analyzing the variation patterns of short-circuit currents on the high, medium, and low voltage sides, the specific location of the fault can be determined, providing maintenance personnel with a basis for troubleshooting and facilitating the rapid identification of transformer fault locations.

[0295] The present invention has a simple implementation scheme. It only requires obtaining the magnitude of the short-circuit current on the high-voltage side and the medium-voltage side to determine the location of the transformer fault. The scheme does not require determining the power direction and is not affected by the PT disconnection.

[0296] The above description is merely a preferred embodiment of the present invention and should not be construed as limiting the scope of the invention. Therefore, any equivalent variations made in accordance with the claims of the present invention are still within the scope of the present invention.

Claims

1. A method for determining the location of a transformer fault, characterized in that, Includes the following steps: Step S10: Obtain transformer fault equipment parameters, the parameters including at least the percentage of short-circuit voltage between windings, the rated voltage on all three sides, and the rated capacity on all three sides; The percentage of short-circuit voltage between windings includes: percentage of short-circuit voltage between high-medium windings, percentage of short-circuit voltage between high-low windings, and percentage of short-circuit voltage between medium-low windings; Step S11: Calculate the impedance of the three windings of the transformer: Based on the transformer parameters and the percentage of short-circuit voltage of the three windings of the transformer, obtain the impedance values ​​of the high, medium and low voltage windings. Step S12: Perform per-unit value reduction on the impedance values ​​of the three-sided windings; Step S13: Pre-determine multiple potential transformer fault locations, and calculate the short-circuit current on each side corresponding to each potential transformer fault location based on the impedance values ​​of the three windings after per-unit value reduction. Step S14: Based on the fault recording data, use the Fourier algorithm to calculate the actual short-circuit current values ​​on the high-voltage side and medium-voltage side of the transformer at the time of the fault. Step S15: Based on the short-circuit currents on each side corresponding to each potential transformer fault location calculated, and the actual current values ​​on the high-voltage side and medium-voltage side of the transformer at the time of the fault, the fault location is determined and the final transformer fault location is determined. Step S15 further includes: If the data calculated in step S13 satisfies the following formula, then it is determined that there is a fault in the high-voltage side lead: In the formula, , These are the high-voltage side current value and the medium-voltage side current value calculated in step S14, respectively. , The short-circuit current values ​​of the high-voltage side and medium-voltage side when the high-voltage side lead is faulty, calculated in step S13; If the data calculated in step S13 satisfies the following formula, then it is determined that there is an internal fault in the high-voltage side winding: ； In the formula: , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs on the high-voltage side lead, calculated in step S13. , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when an internal fault occurs in the high-voltage side winding, as calculated in step S13. If the data calculated in step S13 satisfies the following formula, then it is determined that there is a fault at the medium-voltage side lead: ； In the formula: , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the medium-voltage side lead obtained in step S13; If the data calculated in step S13 satisfies the following formula, then it is determined that there is an internal fault in the medium-voltage side winding: In the formula: , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the medium-voltage side lead obtained in step S13. , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when an internal fault occurs in the medium-voltage side winding, as calculated in step S13. If the data calculated in step S13 satisfies the following formula, then it is determined that there is a fault at the low-voltage side lead: In the formula: , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the low-voltage side lead obtained in step S13; If the data calculated in step S13 satisfies the following formula, it is determined that there is a low-voltage side winding fault. ； In the formula: , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when a fault occurs at the low-voltage side lead, calculated in step S13. , These are the short-circuit current values ​​on the high-voltage side and the medium-voltage side when the low-voltage side winding is faulty, calculated in step S13.

2. The method as described in claim 1, characterized in that, Step S11 further includes: Step S110: Calculate the winding impedances of the high, medium, and low voltage sides of the transformer according to the following formula: Percentage of short-circuit voltage on the high-voltage side winding: U k1 %= (U k(1-2) %+ U k(3-1) % -U k(2-3) %) / 2 Percentage of short-circuit voltage on medium-voltage side winding: U k2 %=(U k(1-2) %+ U k(2-3) % -U k(3-1) %) / 2 Low-voltage side winding short-circuit voltage percentage: U k3 %=(U k(2-3) %+ U k(3-1) % -U k(1-2) %) / 2 In the formula: U k1 % represents the percentage of short-circuit voltage on the high-voltage side, U k(1-2) % represents the percentage of short-circuit voltage between high-strength windings, U k(3-1) % represents the percentage of short-circuit voltage between high and low windings, U k(2-3) % represents the percentage of short-circuit voltage between medium and low windings; Step S111: Based on the transformer parameters and the percentage of short-circuit voltage on the three windings of the transformer, calculate the impedance values ​​of the high, medium, and low voltage windings using the following formula: High-voltage side winding impedance: X H = U k1 *U N1 *U N1 / S N1 Medium voltage side winding impedance: X M = U k2 *U N2 *U N2 / S N2 High-voltage side winding impedance: X L = U k3 *U N3 *U N3 / S N3 In the formula: U ki (i=1,2,…3) represents the short-circuit voltage of the high, medium, and low voltage windings, U Ni (i=1,2,…3) represents the rated voltages of the high, medium, and low voltage sides, and S. Ni (i=1,2,…3) represents the rated capacity of the high, medium, and low sides.

3. The method as described in claim 2, characterized in that, Step S13 further includes: When the potential transformer fault location is at the high-voltage side lead... When the short-circuit current is applied, calculate the three-sided short-circuit current value according to the following formula: High-voltage side short-circuit current value ; In the formula: For high-voltage side power supply, The impedance of the high-voltage side system; Medium voltage side short circuit current value ; In the formula: For high-voltage side power supply, For the high-voltage side system impedance, For the high-voltage side winding impedance, This is the impedance of the medium-voltage side winding.

4. The method as described in claim 2, characterized in that, Step S13 further includes: When the potential transformer fault location is inside the high-voltage side winding When the short-circuit current is applied, calculate the three-sided short-circuit current value according to the following formula: The short-circuit current on the high-voltage side is: In the formula: For high-voltage side power supply, For the high-voltage side system impedance, This refers to the impedance of the high-voltage side winding. The short-circuit current measured on the medium voltage side is: ; In the formula: For high-voltage side power supply, For the high-voltage side system impedance, This is the impedance of the medium-voltage side winding.

5. The method as described in claim 2, characterized in that, Step S13 further includes: When the potential transformer fault location is at the medium-voltage side lead... When the short-circuit current is applied, calculate the three-sided short-circuit current value according to the following formula: The short-circuit current on the high-voltage side is: ; In the formula: For high-voltage side power supply, For the high-voltage side system impedance, For the high-voltage side winding impedance, The impedance of the medium-voltage side winding; The short-circuit current value on the medium voltage side is: ; In the formula: For high-voltage side power supply, This represents the impedance of the high-voltage side system.

6. The method as described in claim 2, characterized in that, Step S13 further includes: When the potential transformer fault location is inside the medium-voltage side winding When the short-circuit current is applied, calculate the three-sided short-circuit current value according to the following formula: The short-circuit current value on the high-voltage side is: In the formula: For high-voltage side power supply, For the high-voltage side system impedance, This refers to the impedance of the high-voltage side winding. Medium voltage side short circuit current value for: In the formula: For high-voltage side power supply, For the high-voltage side system impedance, This is the impedance of the medium-voltage side winding.

7. The method as described in claim 2, characterized in that, Step S13 further includes: When the potential transformer fault location is at the low-voltage side lead... When the short-circuit current is applied, calculate the three-sided short-circuit current value according to the following formula: High-voltage side short-circuit current for: ; In the formula: For high-voltage side power supply, For the high-voltage side system impedance, For the high-voltage side winding impedance, The impedance of the medium-voltage side winding; Medium voltage side short circuit current value for: ; In the formula: For high-voltage side power supply, For the high-voltage side system impedance, For the impedance of the medium voltage side winding, This is the impedance of the medium-voltage side winding.

8. The method as described in claim 2, characterized in that, Step S13 further includes: When the potential transformer fault location is inside the low-voltage side winding When the short-circuit current is applied, calculate the three-sided short-circuit current value according to the following formula: The low-voltage side short-circuit current is: ; In the formula: For high-voltage side power supply, For the high-voltage side system impedance, This refers to the impedance of the high-voltage side winding. Medium voltage side short circuit current value ; In the formula: For high-voltage side power supply, For the high-voltage side system impedance, This is the impedance of the medium-voltage side winding.

9. The method as described in claim 8, characterized in that, Further includes: Based on the criteria, the specific location of the fault is determined, and a transformer fault location display interface is generated, along with a transformer fault report.

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