CN101752835A

CN101752835A - Transformer compound current differential protection method and compound current differential relay

Info

Publication number: CN101752835A
Application number: CN 201010034085
Authority: CN
Inventors: 柳焕章; 屠黎明; 张德泉; 肖远清; 陈学道; 聂娟红; 李锋; 黄少锋; 冯勇; 尹梁方
Original assignee: Beijing Sifang Automation Co Ltd; Central China Grid Co Ltd
Current assignee: Beijing Sifang Automation Co Ltd; Central China Grid Co Ltd
Priority date: 2010-01-13
Filing date: 2010-01-13
Publication date: 2010-06-23
Anticipated expiration: 2030-01-13
Also published as: CN101752835B

Abstract

The invention discloses a composite current differential protection method and a composite current differential relay for steady-state quantities and fault components of transformer internal faults. Since the conventional steady-state quantity differential protection relay is not sensitive enough when the transformer body inter-turn fault occurs when the transformer is heavily loaded or when the external fault is converted into an intra-region inter-turn fault (simultaneously exists in the external region), the present invention proposes a method based on Composite current differential protection method and differential relay for steady-state quantity and fault component. The relay uses the steady-state quantity to calculate the differential current and braking current to form a steady-state quantity double K differential relay, and uses the fault component (phase current sudden change , negative sequence current , zero sequence current ) to calculate the differential current and braking current constitute another fault component double K differential relay. The two differential relays are synthesized into a compound current differential relay of steady-state quantity and fault component. The test results show that the differential protection method and the relay based on this method have good sensitivity, safety and rapidity.

Description

Transformer Composite Current Differential Protection Method and Composite Current Differential Relay

技术领域technical field

本发明属于继电保护技术中变压器保护领域，特别涉及一种变压器差动保护方法和差动继电器。The invention belongs to the field of transformer protection in relay protection technology, and in particular relates to a transformer differential protection method and a differential relay.

背景技术Background technique

目前，变压器稳态量差动保护继电器在变压器重载发生变压器本体匝间故障时或区外故障转换为区内匝间故障时(区外区内同时存在)，由于负荷电流或区外穿越性电流较大且大部分为制动电流，而匝间故障时故障电流相对较小主要为差动电流，导致稳态量差动保护继电器灵敏度不够无法动作。At present, when the transformer steady-state quantity differential protection relay is overloaded and the inter-turn fault of the transformer body occurs, or when the external fault is converted into an inter-internal fault (existing in the external area at the same time), due to the load current or the external through-circuit The current is large and most of it is the braking current, while the fault current is relatively small during the turn-to-turn fault, which is mainly the differential current, resulting in insufficient sensitivity of the steady-state differential protection relay to operate.

如果采用减小比率制动系数K的方法，则在区外故障时由于变压器各侧CT的传变特性不一致、CT的暂态误差等因数可能会导致差动保护的误动，牺牲了差动保护的安全性。因此必须有条件的调整比率制动系数K，使之在区内故障时减小，在其他情况下增大，这样就在不影响差动继电器的安全性的前提下提高了差动继电器的灵敏度。If the method of reducing the ratio braking coefficient K is adopted, the differential protection may malfunction due to factors such as the inconsistency of the transmission characteristics of the CTs on each side of the transformer and the transient error of the CTs during an external fault, sacrificing the differential protection. Protected security. Therefore, the proportional braking coefficient K must be adjusted conditionally so that it decreases when there is a fault in the zone and increases in other cases, so that the sensitivity of the differential relay is improved without affecting the safety of the differential relay. .

发明内容Contents of the invention

为解决现有技术中变压器稳态量差动保护继电器在变压器重载发生变压器本体匝间故障时或区外故障转换为区内匝间故障时(区外区内同时存在)灵敏度不够的问题，同时必须兼顾继电器的安全性，本发明提出了一种基于稳态量与故障分量的变压器复合电流差动保护方法和复合电流差动继电器。In order to solve the problem of insufficient sensitivity of the transformer steady-state quantity differential protection relay in the prior art when the inter-turn fault of the transformer body occurs when the transformer is heavy-loaded or when the external fault is converted into an intra-region inter-turn fault (simultaneously exists outside the region), At the same time, the safety of the relay must be taken into consideration. The present invention proposes a transformer compound current differential protection method and a compound current differential relay based on steady-state quantities and fault components.

本发明公开了一种基于稳态量与故障分量的变压器复合电流差动保护方法，所述方法包括以下步骤：The invention discloses a transformer compound current differential protection method based on steady-state quantity and fault component. The method includes the following steps:

一种基于稳态量与故障分量的变压器复合电流差动保护方法，其特征在于，所述方法包括以下步骤：A transformer composite current differential protection method based on steady-state quantities and fault components, characterized in that the method comprises the following steps:

(1)通过变压器各侧电流互感器测量并计算变压器各侧稳态量电流和故障分量电流，其中所述故障分量电流为突变量电流、负序电流或零序电流中的任一种；(1) Measure and calculate the steady-state quantity current and the fault component current on each side of the transformer through the current transformers on each side of the transformer, wherein the fault component current is any one of sudden change current, negative-sequence current or zero-sequence current;

(2)将所述变压器各侧稳态量电流绝对值最大的一侧作为稳态量差动保护的一端，将其余侧的稳态量电流和等效定为所述稳态量差动保护的另一端，上述稳态量差动保护的两端电流分别为

稳态量差动电流记为

其中

(2) The side with the largest absolute value of the steady-state quantity current on each side of the transformer is used as one end of the steady-state quantity differential protection, and the steady-state quantity current and the equivalent of the other sides are determined as the steady-state quantity differential protection The other end of the above steady-state quantity differential protection, the currents at both ends are respectively

The steady-state differential current is recorded as

in

(3)将所述变压器各侧故障分量电流绝对值最大的一侧作为故障分量差动保护的一端，将其余侧的故障分量电流和等效定为所述故障分量差动保护的另一端，上述故障分量差动保护的两端电流分别为

故障分量差动电流记为

其中

(3) The side with the largest absolute value of the fault component current on each side of the transformer is used as one end of the fault component differential protection, and the sum of the fault component currents on the remaining sides is equivalently defined as the other end of the fault component differential protection, The currents at both ends of the above fault component differential protection are respectively

The fault component differential current is denoted as

in

(4)当同时满足以下变压器复合电流差动保护动作方程以及稳态量差动保护附加条件和故障分量差动保护附加条件时，变压器复合电流差动保护动作：(4) When the following transformer composite current differential protection action equations and additional conditions for steady-state variable differential protection and additional conditions for fault component differential protection are satisfied at the same time, the transformer composite current differential protection operates:

K，k，k₁为制动系数，K, k, k ₁ is the braking coefficient,

K取值范围为0.5～2之间；K ranges from 0.5 to 2;

k取值范围为0.1～1之间；The value range of k is between 0.1 and 1;

k₁取值范围为0.1～1之间；The value range of k ₁ is between 0.1 and 1;

I_set为稳态量差动保护设定的阈值；I _set is the threshold set by the steady-state quantity differential protection;

FI_set分别为突变量、负序及零序附加条件设定的阈值。FI _set is the threshold value set for the mutation amount, negative sequence and zero sequence additional conditions respectively.

根据上述差动保护方法，本发明还进一步公开了一种基于稳态量与故障分量的变压器复合电流差动继电器，所述复合电流差动继电器采用稳态量计算差流和制动电流构成一个稳态量双K差动继电器，采用故障分量电流计算差流和制动电流构成另一个故障分量附加条件的差动继电器，其中故障分量电流为突变量电流、负序电流或零序电流中的任一种；当复合电流差动继电器满足复合电流差动保护动作方程以及变压器稳态量差动保护附加条件和变压器故障分量差动保护附加条件时，该继电器动作。According to the above differential protection method, the present invention further discloses a transformer composite current differential relay based on steady-state quantity and fault component. The composite current differential relay uses steady-state quantity to calculate differential current and brake current to form a Steady-state double K differential relay, using fault component current to calculate differential current and brake current to form another differential relay with additional conditions for fault component, wherein the fault component current is one of sudden change current, negative sequence current or zero sequence current Any one; when the composite current differential relay satisfies the composite current differential protection action equation, the additional conditions for the differential protection of the transformer steady-state quantity and the additional conditions for the differential protection of the transformer fault component, the relay operates.

该复合电流差动继电器的实现方法，进一步优选但不局限于以下方式来实现：The implementation method of the composite current differential relay is further preferably but not limited to the following way:

(1)将多侧差动转换为两侧差动(1) Convert multi-side differential to two-side differential

采用由多侧差动转换为两侧差动的方法。The method of converting from multi-side differential to two-side differential is adopted.

稳态量差动：Steady state differential:

找到多侧中的稳态量电流绝对值最大侧，且将该侧定为一端，将其余侧的稳态量电流和等效定为另一端。分别命名M和∑-M，常作为电流的下标。Find the side with the largest absolute value of the steady-state current among the multiple sides, and set this side as one end, and set the sum of the steady-state currents of the other sides as the other end. Named M and ∑-M respectively, often used as the subscript of current.

——稳态量差动电流

——Steady-state differential current

——各侧中稳态量电流最大侧电流 ——The steady-state current of each side and the maximum side current

——其余侧等效电流

——Equivalent current of other side

故障分量包括相电流突变量

负序电流

零序电流

记为

The fault component includes the phase current mutation

negative sequence current

Zero sequence current

recorded as

找到多侧中的故障分量电流绝对值最大的一侧，下面以突变量电流为例将突变量电流绝对值最大侧定为一侧，将其余侧的突变量电流和等效定为另一侧。分别命名M和∑-M，常作为电流的下标。Find the side with the largest absolute value of the fault component current among the multiple sides. Taking the sudden change current as an example, set the side with the largest absolute value of the sudden change current as one side, and set the sudden change current and equivalent of the other sides as the other side . Named M and ∑-M respectively, often used as the subscript of current.

——突变量差动电流 ——Sudden change differential current

——各侧中突变量电流最大侧电流

——The maximum side current of the sudden change amount current on each side

——其余侧等效电流

——Equivalent current of other side

(2)稳态量与故障分量复合电流差动继电器动作方程(2) Action equation of steady-state quantity and fault component composite current differential relay

双K值比率差动方程的通用表达式：将两侧电流中的大电流

乘小制动系数k；小电流

乘大制动系数K。分别取不同制动系数有利于改善动作特性。如下式：The general expression of the double K-value ratio differential equation: the large current in the current on both sides

Multiply the small braking coefficient k; small current

Multiply the large braking coefficient K. Taking different braking coefficients is beneficial to improve the action characteristics. as follows:

$| | {\overset{\cdot \cdot}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; | | k k {\overset{\cdot &Center Dot;}{I I}}_{M m} - - K K {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} | | = = | | k k / / K K {\overset{\cdot &Center Dot;}{I I}}_{M m} - - {\overset{\cdot \cdot}{I I}}_{Σ Σ - - M m} | |$

稳态量分相电流差动如下式所述：The steady-state quantity split-phase current differential is described as follows:

$| | {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; K K | | k k {\overset{\cdot &Center Dot;}{I I}}_{M m} - - (({\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} - - {k k}_{11} {\overset{\cdot &Center Dot;}{I I}}_{11 Σ Σ - - M m})) | | - - - - - - ((11))$

其中，

为

的电流正序分量；in,

for

positive sequence component of the current;

K，k，k₁为制动系数，K, k, k ₁ is the braking coefficient,

K取值范围为0.5～2之间；K ranges from 0.5 to 2;

k取值范围为0.1～1之间；The value range of k is between 0.1 and 1;

FI_set分别为突变量、负序及零序附加条件设定的阈值。。故障分量电流差动如下式所述：FI _set is the threshold value set for the mutation amount, negative sequence and zero sequence additional conditions respectively. . The fault component current differential is described as follows:

$| | F f {\overset{\cdot \cdot}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; K K | | kF f {\overset{\cdot &Center Dot;}{I I}}_{M m} - - F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} | | - - - - - - ((22))$

将式(1)×式(2)得到稳态量与故障分量的复合电流差动继电器。令K＝1，得到下式：Formula (1) × formula (2) to get the composite current differential relay of steady-state quantity and fault component. Let K=1, get the following formula:

$| | F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | | {| | \overset{\cdot \cdot}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; | | kF f {\overset{\cdot \cdot}{I I}}_{M m} - - F f {\overset{\cdot \cdot}{I I}}_{Σ Σ - - M m} | | | | k k {\overset{\cdot \cdot}{I I}}_{M m} - - (({\overset{\cdot \cdot}{I I}}_{Σ Σ - - M m} - - {k k}_{11} {\overset{\cdot &Center Dot;}{I I}}_{11 Σ Σ - - M m})) | |$

不等式两边除以

得到下式：Divide both sides of the inequality by

Get the following formula:

$| | {\overset{\cdot \cdot}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; \frac{| | kF f {\overset{\cdot \cdot}{I I}}_{M m} - - F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} | |}{| | F f {\overset{\cdot \cdot}{I I}}_{Σ Σ} | |} | | k k {\overset{\cdot &Center Dot;}{I I}}_{M m} - - (({\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} - - {k k}_{11} {\overset{\cdot &Center Dot;}{I I}}_{11 Σ Σ - - M m})) | |$

令

当K＜0.5时，强制性K＝0.5，得到：make

When K<0.5, mandatory K=0.5, get:

$| | {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; K K | | k k {\overset{\cdot &Center Dot;}{I I}}_{M m} - - (({\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} - - {k k}_{11} {\overset{\cdot &Center Dot;}{I I}}_{11 Σ Σ - - M m})) | | - - - - - - ((33))$

复合电流差动继电器附加动作条件：Additional operating conditions for composite current differential relays:

与现有技术相比，本发明具有以下优点：Compared with the prior art, the present invention has the following advantages:

在变压器重载发生变压器本体匝间故障时或区外故障转换为区内匝间故障时(区外区内同时存在)，由于负荷电流或区外穿越性电流较大且为制动电流，而匝间故障时故障电流相对较小主要为差动电流，导致稳态量差动保护继电器灵敏度不够而保护无法动作。故障分量电流差动与负荷无关，动作性能优良。但是，故障分量电流差动也有自身的问题：如，零序、负序是三相运算的结果，不能选相。当发生区内外同时性故障，若外部故障强于内部故障，可能拒动。又如，相电流突变量，一般突变量差动不能长期投入，计算突变量时，减数为当前电流，被减数为前一周或前两周电流，由于被减数的复杂性，前一周或前两周可能无故障、可能已经存在故障、可能发生振荡等。为了长期投入突变量差动，必须附加苛刻的限制条件

和

这两个条件同时满足几乎就是区内故障了。更何况，故障分量差动只是有限度地改变稳态量分相电流差动的制动系数。两类差动有机结合，相互补充、相互制约。总之，故障分量差动帮助稳态量差动提高灵敏度；稳态量差动帮助故障分量差动提高安全性。When the inter-turn fault of the transformer body occurs under heavy load of the transformer or when the fault outside the zone is transformed into an inter-turn fault inside the zone (existing in the zone outside the zone at the same time), due to the large load current or the through-current outside the zone and it is a braking current, and The fault current is relatively small during the turn-to-turn fault, which is mainly the differential current, which leads to the insufficient sensitivity of the steady-state differential protection relay and the protection cannot operate. The current differential of the fault component has nothing to do with the load, and the action performance is excellent. However, the fault component current differential also has its own problems: for example, zero-sequence and negative-sequence are the results of three-phase operations, and phase selection cannot be performed. When simultaneous faults inside and outside the zone occur, if the external fault is stronger than the internal fault, it may refuse to move. Another example is the sudden change of phase current. Generally, the differential of the sudden change cannot be input for a long time. When calculating the sudden change, the subtrahend is the current current, and the minuend is the current of the previous week or two weeks. Due to the complexity of the subtrahend, the previous week Or there may be no faults in the previous two weeks, there may be faults, oscillations may occur, etc. In order to invest in the mutation difference for a long time, strict restrictions must be attached

and

Satisfying these two conditions at the same time is almost an intra-area failure. What's more, the fault component differential only changes the braking coefficient of the steady-state quantity split-phase current differential to a limited extent. The two types of differentials are organically combined, complement each other and restrict each other. In a word, the differential of the fault component helps the differential of the steady-state quantity to improve the sensitivity; the differential of the steady-state quantity helps the differential of the fault component to improve the safety.

附图说明Description of drawings

图1为典型的变电站变压器的系统图及差动保护范围说明；Figure 1 is a system diagram of a typical substation transformer and a description of the range of differential protection;

图2为变压器典型的差动继电器动作曲线；Figure 2 is a typical differential relay action curve of a transformer;

图3为本发明提出的变压器差动继电器动作曲线；Fig. 3 is the action curve of the transformer differential relay proposed by the present invention;

图4为本发明功能实现的流程图。Fig. 4 is a flow chart of the function realization of the present invention.

具体实施方式Detailed ways

为使本发明的上述目的、特征和优点能够更加明显易懂，下面结合附图对本发明的具体实施方式做详细的说明。参见图1，典型的变电站变压器的系统图及差动保护范围说明。In order to make the above objects, features and advantages of the present invention more comprehensible, specific implementations of the present invention will be described in detail below in conjunction with the accompanying drawings. See Figure 1, the system diagram of a typical substation transformer and the description of the differential protection range.

图1中曲线所包括的范围为差动保护继电器保护的范围，称为区内故障，采用变压器各侧电流构成。曲线外的故障称为区外故障。图1中虚线方框为变压器本体内的绕组故障称为匝间故障。The range covered by the curve in Figure 1 is the range protected by the differential protection relay, which is called the fault in the area, and is formed by the current on each side of the transformer. Faults outside the curve are called out-of-area faults. The dotted box in Figure 1 is the winding fault in the transformer body, which is called inter-turn fault.

参见图2，该图为变压器典型的差动继电器动作曲线。其动作方程为：See Figure 2, which is a typical differential relay action curve for a transformer. Its action equation is:

$| | {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; K K | | {\overset{\cdot &Center Dot;}{I I}}_{M m} - - {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} | |$

附加条件：Additional conditions:

$| | {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; {I I}_{set set}$

图2中纵坐标为差动电流，横坐标

为制动电流，I_set为最小动作电流，K为斜率为一固定的整定值。比率差动保护曲线是由差动电流和制动电流构成的动作曲线。曲线上方为动作区。The ordinate in Figure 2 is the differential current, the abscissa

I set is the braking current, I _set is the minimum operating current, and K is the slope with a fixed setting value. The ratio differential protection curve is an action curve composed of differential current and braking current. Above the curve is the action zone.

本发明的一种基于稳态量与故障分量的变压器复合电流差动保护方法，如图4所示，其具体实现方法如下：A transformer compound current differential protection method based on steady-state quantity and fault component of the present invention, as shown in Figure 4, its specific implementation method is as follows:

(1)通过变压器各侧电流互感器测量并计算变压器各侧稳态量电流和突变量电流，并以高压侧的电流为基准，将其他侧电流归算到高压侧；(1) Measure and calculate the steady-state current and sudden change current on each side of the transformer through the current transformers on each side of the transformer, and use the current on the high-voltage side as a reference to attribute the current on other sides to the high-voltage side;

以高压侧的电流为基准，其他各侧向高压侧归算。计算各侧的平衡系数为：Based on the current of the high-voltage side, other sides are reduced to the high-voltage side. Calculate the balance factor for each side as:

计算变压器各侧一次额定电流：

Calculate the primary rated current of each side of the transformer:

式中：S_e为变压器最大额定容量；U_1e为变压器各侧额定电压(应以运行的实际电压为准)。In the formula: S _e is the maximum rated capacity of the transformer; U _1e is the rated voltage of each side of the transformer (the actual operating voltage should prevail).

计算变压器各侧二次额定电流：

Calculate the secondary rated current of each side of the transformer:

式中：I_1e为变压器各侧一次额定电流；n_LH为变压器各侧CT变比。以高压侧为基准，计算变压器中、低压侧平衡系数：In the formula: I _1e is the primary rated current of each side of the transformer; n _LH is the CT transformation ratio of each side of the transformer. Based on the high-voltage side, calculate the balance coefficient of the medium and low-voltage sides of the transformer:

${K K}_{ph pH . . M m} = = \frac{{I I}_{22 e e . . H h}}{{I I}_{22 e e . . M m}} = = \frac{{I I}_{11 e e . . H h} / / {n no}_{LH LH . . H h}}{{I I}_{11 e e . . M m} / / {n no}_{LH LH . . H h}} = = \frac{{S S}_{e e} / / \sqrt{33} {U u}_{11 e e . . H h}}{{S S}_{e e} / / \sqrt{33} {U u}_{11 e e . . M m}} \cdot &Center Dot; \frac{{n no}_{LH LH . . M m}}{{n no}_{LH LH . . H h}} = = \frac{{U u}_{11 e e . . M m}}{{U u}_{11 e e . . H h}} \cdot &Center Dot; \frac{{n no}_{LH LH . . M m}}{{n no}_{LH LH . . H h}}$

${K K}_{ph pH . . L L} = = \frac{{U u}_{11 e e . . L L}}{{U u}_{11 e e . . H h}} \cdot \cdot \frac{{n no}_{LH LH . . L L}}{{n no}_{LH LH . . H h}}$

将其他侧各相电流与相应的平衡系数相乘，即得幅值补偿后的各相电流。Multiply the current of each phase on the other side with the corresponding balance coefficient to obtain the current of each phase after amplitude compensation.

(2)各侧电流相位补偿(2) Current phase compensation on each side

变压器各侧CT二次电流相位由软件自校正，以在Y侧进行校正相位(以11点接线变压器为例)。其校正方法如下：The secondary current phase of the CT on each side of the transformer is self-corrected by the software to correct the phase on the Y side (take the 11-point connection transformer as an example). Its correction method is as follows:

Y0侧： $\begin{matrix} {\overset{\cdot}{I}}_{A}^{'} = ({\overset{\cdot}{I}}_{A} - {\overset{\cdot}{I}}_{B}) / \sqrt{3} \\ {\overset{\cdot}{I}}_{B}^{'} = ({\overset{\cdot}{I}}_{B} - {\overset{\cdot}{I}}_{C}) / \sqrt{3} \\ {\overset{\cdot}{I}}_{C}^{'} = ({\overset{\cdot}{I}}_{C} - {\overset{\cdot}{I}}_{A}) / \sqrt{3} \end{matrix}\}$ Y0 side: $\begin{matrix} {\overset{&Center Dot;}{I}}_{A}^{'} = ({\overset{&Center Dot;}{I}}_{A} - {\overset{&Center Dot;}{I}}_{B}) / \sqrt{3} \\ {\overset{&Center Dot;}{I}}_{B}^{'} = ({\overset{&Center Dot;}{I}}_{B} - {\overset{\cdot}{I}}_{C}) / \sqrt{3} \\ {\overset{\cdot}{I}}_{C}^{'} = ({\overset{\cdot}{I}}_{C} - {\overset{&Center Dot;}{I}}_{A}) / \sqrt{3} \end{matrix}\}$

式中：

为Y侧CT二次电流；

为Y侧校正后的各相电流。其它接线方式可以类推。In the formula:

is the Y-side CT secondary current;

is the corrected current of each phase on the Y side. Other wiring methods can be analogized.

差动电流与制动电流的相关计算，都是在电流相位校正和平衡补偿后的基础上进行。The correlation calculation of differential current and braking current is carried out on the basis of current phase correction and balance compensation.

(3)差动电流计算(3) Differential current calculation

差动电流的计算方法如下：The calculation method of differential current is as follows:

${\overset{\cdot \cdot}{I I}}_{Σ Σ} = = {Σ Σ}_{i i = = 11}^{N N} {\overset{\cdot \cdot}{I I}}_{i i}$

式中：

为差动电流；

为所有侧相电流之和。In the formula:

is the differential current;

is the sum of all side phase currents.

(4)多侧差动转换为两侧差动的方法。(4) The method of converting multi-side differential into two-side differential.

1)稳态量差动：1) Steady-state differential:

找到多侧中的稳态量电流绝对值最大侧，且将该侧定为一侧，将其余侧的稳态量电流和等效定为另一侧。分别命名M和∑-M，常作为电流的下标。Find the side with the largest absolute value of the steady-state current among the multiple sides, and set this side as one side, and set the sum of the steady-state currents of the other sides as the other side. Named M and ∑-M respectively, often used as the subscript of current.

——稳态量差动电流

——Steady-state differential current

——各侧中稳态量电流最大侧电流

——The steady-state current of each side and the maximum side current

——其余侧等效电流

——Equivalent current of other side

2)故障分量差动(包括相电流突变量

负序电流

零序电流

)下面以相电流突变量

为例：2) Fault component differential (including phase current sudden change

negative sequence current

Zero sequence current

) as the phase current mutation

For example:

找到多侧中的突变量电流绝对值最大侧，且将该侧定为一侧，将其余侧的突变量电流和等效定为另一侧。分别命名M和∑-M，常作为电流的下标。Find the side with the largest absolute value of the sudden change current among the multiple sides, and set this side as one side, and set the sudden change current sum of the other sides as the other side. Named M and ∑-M respectively, often used as the subscript of current.

——突变量差动电流

——Sudden change differential current

——各侧中突变量电流最大侧电流

——The maximum side current of the sudden change amount current on each side

——其余侧等效电流

——Equivalent current of other side

(5)稳态量与故障分量复合电流差动继电器动作方程双K值比率差动方程的通用表达式：将两侧电流中的大电流

乘小制动系数k；小电流

乘大制动系数K。分别取不同制动系数有利于改善动作特性。如下式：(5) The general expression of the double K value ratio differential equation of the steady-state quantity and fault component composite current differential relay action equation: the large current in the current on both sides

Multiply the small braking coefficient k; small current

$| | {\overset{\cdot \cdot}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; K K | | k k {\overset{\cdot \cdot}{I I}}_{M m} - - (({\overset{\cdot \cdot}{I I}}_{Σ Σ - - M m} - - {k k}_{11} {\overset{\cdot \cdot}{I I}}_{11 Σ Σ - - M m})) | | - - - - - - ((11))$

其中，

为

的电流正序分量；in,

for

positive sequence component of the current;

K，k，k₁为制动系数，K, k, k ₁ is the braking coefficient,

K取值范围为0.5～2之间；K ranges from 0.5 to 2;

k取值范围为0.1～1之间；The value range of k is between 0.1 and 1;

故障分量电流差动如下式所述：The fault component current differential is described as follows:

$| | F f {\overset{\cdot \cdot}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; K K | | kF f {\overset{\cdot \cdot}{I I}}_{M m} - - F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} | | - - - - - - ((22))$

$| | F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | | {| | \overset{\cdot \cdot}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; | | kF f {\overset{\cdot &Center Dot;}{I I}}_{M m} - - F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} | | | | k k {\overset{\cdot &Center Dot;}{I I}}_{M m} - - (({\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} - - {k k}_{11} {\overset{\cdot &Center Dot;}{I I}}_{11 Σ Σ - - M m})) | |$

不等式两边除以

得到下式：Divide both sides of the inequality by

Get the following formula:

$| | {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | | &GreaterEqual; &Greater Equal; \frac{| | kF f {\overset{\cdot &Center Dot;}{I I}}_{M m} - - F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} | |}{| | F f {\overset{\cdot &Center Dot;}{I I}}_{Σ Σ} | |} | | k k {\overset{\cdot &Center Dot;}{I I}}_{M m} - - (({\overset{\cdot &Center Dot;}{I I}}_{Σ Σ - - M m} - - {k k}_{11} {\overset{\cdot &Center Dot;}{I I}}_{11 Σ Σ - - M m})) | |$

令

当K＜0.5时，强制性K＝0.5，得到：make

When K<0.5, mandatory K=0.5, get:

其动作曲线如图3。Its action curve is shown in Figure 3.

从图3和图2的对比可以看出，常规典型的比率制动曲线是一条固定的曲线，对于任何情况都是一陈不变的。本发明提出的继电器则是一簇曲线，其通过故障的严重程度来调整曲线斜率。在变压器重载发生变压器本体匝间故障时或区外故障转换为区内匝间故障时(区外区内同时存在)，由于负荷电流或区外穿越性电流较大且为制动电流，而匝间故障时故障电流相对较小，导致稳态量差动保护继电器灵敏度不够无法动作。故障分量电流差动与负荷无关，动作性能优良。但是，故障分量电流差动也有自身的问题：如，零序、负序是三相运算的结果，不能选相。当发生区内外同时性故障，若外部故障强于内部故障，可能拒动。又如，相电流突变量，一般突变量差动不能长期投入，计算突变量时，被减数为当前电流，减数为前一周或前两周电流，由于减数的复杂性，前一周或前两周可能无故障、可能已经存在故障、可能发生振荡等。为了长期投入突变量差动，必须附加苛刻的限制条件

和

这两个条件同时满足几乎就是区内故障了。更何况，突变量差动只是有限度地改变稳态量分相电流差动的制动系数。两类差动有机结合，相互补充、相互制约。总之，故障分量差动帮助稳态量差动提高灵敏度；稳态量差动帮助故障分量差动提高安全性。It can be seen from the comparison of Fig. 3 and Fig. 2 that the conventional typical ratio braking curve is a fixed curve, which is unchanged for any situation. The relay proposed by the present invention is a cluster of curves, and the slope of the curve is adjusted according to the severity of the fault. When the inter-turn fault of the transformer body occurs under heavy load of the transformer or when the fault outside the zone is transformed into an inter-turn fault inside the zone (existing in the zone outside the zone at the same time), due to the large load current or the through-current outside the zone and it is a braking current, and The fault current is relatively small when there is a turn-to-turn fault, so the sensitivity of the steady-state differential protection relay is not enough to operate. The current differential of the fault component has nothing to do with the load, and the action performance is excellent. However, the fault component current differential also has its own problems: for example, zero-sequence and negative-sequence are the results of three-phase operations, and phase selection cannot be performed. When simultaneous faults inside and outside the zone occur, if the external fault is stronger than the internal fault, it may refuse to move. Another example is the phase current mutation amount. The general mutation amount differential cannot be input for a long time. When calculating the mutation amount, the subtrahend is the current current, and the subtrahend is the current of the previous week or two weeks. Due to the complexity of the subtrahend, the previous week or There may have been no faults in the previous two weeks, there may have been faults, oscillations may have occurred, etc. In order to invest in the mutation difference for a long time, strict restrictions must be attached

and

Satisfying these two conditions at the same time is almost an intra-area failure. What's more, the abrupt change differential only changes the braking coefficient of the steady-state separate-phase current differential to a limited extent. The two types of differentials are organically combined, complement each other and restrict each other. In a word, the differential of the fault component helps the differential of the steady-state quantity to improve the sensitivity; the differential of the steady-state quantity helps the differential of the fault component to improve the safety.

本发明还公开了一种基于稳态量与故障分量的变压器复合电流差动继电器，所述复合电流差动继电器采用稳态量计算差流和制动电流构成一个稳态量双K差动继电器，采用故障分量电流计算差流和制动电流构成另一个故障分量双K差动继电器；当复合电流差动继电器满足以下动作方程时，该继电器动作：The invention also discloses a transformer compound current differential relay based on steady state quantity and fault component. The compound current differential relay adopts steady state quantity to calculate differential current and brake current to form a steady state quantity double K differential relay , using the fault component current to calculate the differential current and braking current to form another fault component double K differential relay; when the composite current differential relay satisfies the following action equation, the relay operates:

K，k，k₁为制动系数，K, k, k ₁ is the braking coefficient,

K取值范围为0.5～2之间；K ranges from 0.5 to 2;

k取值范围为0.1～1之间；The value range of k is between 0.1 and 1;

Claims

1. A transformer compound current differential protection method based on steady-state quantity and fault component, is characterized in that, described method comprises the following steps:

(1) Measure and calculate the steady-state quantity current and the fault component current on each side of the transformer through the current transformers on each side of the transformer, wherein the fault component current is any one of sudden change current, negative-sequence current or zero-sequence current;

The steady-state differential current is recorded as

in

The fault component differential current is denoted as

in

(4) When the following transformer composite current differential protection action equations and additional conditions for steady-state variable differential protection and additional conditions for fault component differential protection are satisfied at the same time, the transformer composite current differential protection operates:

in,

K, k, k ₁ is the braking coefficient,

The range of K value is between 0.5 and 2;

The value range of k is between 0.1 and 1;

The value range of k ₁ is between 0.1 and 1;

I _set is the threshold set by the steady-state quantity differential protection;

FI _set are respectively the thresholds set by the fault component differential protection.

2. A transformer composite current differential relay based on steady-state quantities and fault components, characterized in that the composite current differential relay uses steady-state quantities to calculate differential current and brake current to form a steady-state quantity double K differential Relays, and differential relays that use the fault component current to calculate the differential current and brake current to form another additional condition for the fault component, wherein the fault component current is any one of the sudden change current, negative sequence current or zero sequence current; when combined When the current differential relay satisfies the compound current differential protection action equation, the additional conditions for the differential protection of the transformer steady-state quantity and the additional conditions for the differential protection of the transformer fault component, the relay operates. 3. The transformer composite current differential relay according to claim 2, characterized in that the multi-side differential of the transformer is converted into two-side differential, namely:

The side with the largest absolute value of the steady-state quantity current on each side of the transformer is used as one end of the steady-state quantity double-K differential relay, and the steady-state quantity current sum of the other sides is equivalently determined as the steady-state quantity double-K differential relay At the other end of the relay, the currents at both ends of the above-mentioned steady-state quantity double K differential relay are recorded as

The steady-state differential current is recorded as

in

The side with the largest absolute value of the fault component current on each side of the transformer is used as one end of the differential relay with additional conditions for the fault component, and the fault component current at the other sides is equivalent to the differential relay with the additional conditions for the fault component At the other end, the currents at both ends of the above-mentioned fault component differential protection are denoted as

The fault component differential current is denoted as

in

3. The transformer composite current differential relay according to claim 3, characterized in that, the composite current differential protection action equation and the additional conditions for the differential protection of the transformer steady-state quantity and the additional conditions for the differential protection of the transformer fault component are preferably as follows :

| {\overset{&Center Dot;}{I}}_{Σ} | &Greater Equal; K | k {\overset{&Center Dot;}{I}}_{m} - ({\overset{&Center Dot;}{I}}_{Σ - m} - k_{1} {\overset{&Center Dot;}{I}}_{1 Σ - m}) |

Transformer composite current differential protection action equation;

| {\overset{&Center Dot;}{I}}_{Σ} | &Greater Equal; I_{set}

Additional conditions for transformer steady-state differential protection;

| f {\overset{&Center Dot;}{I}}_{Σ} | &Greater Equal; {FI}_{set}

Additional conditions for transformer differential protection;

in, for

positive sequence component of the current;

K, k, k ₁ is the braking coefficient,

K ranges from 0.5 to 2;

The value range of k is between 0.1 and 1;

The value range of k ₁ is between 0.1 and 1;

FI _set is the threshold value set for the mutation amount, negative sequence and zero sequence additional conditions respectively.