CN104965151A - Fault distance detecting method based on voltage fault component of fault point - Google Patents

Fault distance detecting method based on voltage fault component of fault point Download PDF

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CN104965151A
CN104965151A CN201510284181.XA CN201510284181A CN104965151A CN 104965151 A CN104965151 A CN 104965151A CN 201510284181 A CN201510284181 A CN 201510284181A CN 104965151 A CN104965151 A CN 104965151A
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voltage
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柳焕章
谢俊
李勇
孙晓彦
陈祥文
王英英
金明亮
李会新
郭崇军
陈红雨
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Beijing Sifang Automation Co Ltd
State Grid Corp of China SGCC
Central China Grid Co Ltd
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State Grid Corp of China SGCC
Central China Grid Co Ltd
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Abstract

一种基于故障点电压突变量的故障测距方法,该法先计算得到故障点电压和故障点电压突变量,再根据故障点电压和故障点电压突变量之间同相位的原理列出以故障距离为未知数的一元二次方程,对该方程进行求解,即得到故障距离。本设计不仅对于线路单相和相间故障均适用,而且实用性强、计算精度和效率均很高。

A fault location method based on the sudden change in voltage at the fault point. This method first calculates the voltage at the fault point and the sudden change in voltage at the fault point, and then lists the faults according to the principle of the same phase between the voltage at the fault point and the sudden change in voltage at the fault point. The distance is an unknown quadratic equation, and the equation is solved to obtain the fault distance. This design is not only suitable for single-phase and phase-to-phase faults of the line, but also has strong practicability, high calculation accuracy and high efficiency.

Description

一种基于故障点电压突变量的故障测距方法A Fault Location Method Based on Voltage Change at Fault Point

技术领域technical field

本发明属于电力系统继电保护技术领域,具体涉及一种输电线路短路故障单端测距方法。The invention belongs to the technical field of electric power system relay protection, and in particular relates to a single-end ranging method for a short-circuit fault of a power transmission line.

背景技术Background technique

输电线路是电力系统发电、输送电等的基本设备,在电力系统中占有非常重要的地位。输电线路故障时,若不能及时切除或误切除,则对主系统影响较大,容易造成电网解列,以及变压器越级跳闸等重大事故。高压/超高压/特高压线路保护担负输送电的重要任务,是电力系统的重要枢纽。它们造价极为昂贵,一旦因故障而遭到损坏,检修难度大、时间长,对国民经济造成的直接和间接损失十分巨大。Transmission line is the basic equipment for power generation and transmission in the power system, and occupies a very important position in the power system. When the transmission line is faulty, if it cannot be cut off in time or by mistake, it will have a great impact on the main system, and it will easily cause major accidents such as grid splitting and transformer tripping. High-voltage/ultra-high voltage/ultra-high voltage line protection undertakes the important task of power transmission and is an important hub of the power system. They are extremely expensive to manufacture, and once they are damaged due to malfunctions, they will be difficult to repair and take a long time, causing huge direct and indirect losses to the national economy.

输电线路单端故障测距方法仅利用输电线路一端电气量进行故障定位,无须通讯和数据同步设备,运行费用低且算法稳定,在输电线路中获得广泛应用。输电线路单端故障测距方法主要分为行波法和阻抗法。行波法利用故障暂态行波的传送性质进行单端故障测距,精度高,不受运行方式、过渡电阻等影响,但对采样率要求很高,需要专门的录波装置,应用成本高。阻抗法利用故障后的电压、电流量计算故障回路阻抗,根据线路长度与阻抗成正比的特性进行单端故障测距,简单可靠,但故障测距精度收到过渡电阻和负荷电流等因素影响严重,尤其当过渡电阻较大时,因过渡电阻引起的附加故障距离的影响,故障距离测量结果会严重偏离真实故障距离,甚至出现故障测距失败。The single-end fault location method of transmission lines only uses the electrical quantity at one end of the transmission line for fault location, does not require communication and data synchronization equipment, has low operating costs and stable algorithms, and has been widely used in transmission lines. Single-ended fault location methods for transmission lines are mainly divided into traveling wave method and impedance method. The traveling wave method utilizes the transmission properties of the fault transient traveling wave to perform single-ended fault location. It has high precision and is not affected by the operation mode and transition resistance. However, it requires a high sampling rate, requires a special wave recording device, and has high application costs. . The impedance method uses the voltage and current after the fault to calculate the fault loop impedance, and performs single-ended fault location according to the characteristic that the line length is proportional to the impedance. It is simple and reliable, but the fault location accuracy is seriously affected by factors such as transition resistance and load current. , especially when the transition resistance is large, due to the influence of the additional fault distance caused by the transition resistance, the measurement result of the fault distance will seriously deviate from the real fault distance, and even the failure of fault distance measurement will occur.

中国专利:申请公布号为CN104090212A,公布日为2014年10月8日的发明专利公开了一种无需故障前电压记忆的线路相间故障定位方法,该方法首先测量输电线路保护安装处的两故障相电压突变量、两故障相电流突变量和正常相电压,利用正常相电压计算输电线路正常运行时的故障相间电压,然后利用线路故障后相间短路故障点电压突变量幅值(即等于输电线路正常运行时的故障相间电压幅值)最大且由相间短路故障点向输电线路两端单调递减这一特性实现线路相间短路故障单端定位。虽然该方法无需对线路故障前电压进行记忆,继电保护硬件无需设置线路故障前电压保持回路,但其仅仅适用于线路相间故障,适用范围较窄,且用固定步长代入比较进行求解,步长大则计算精度不够,步长小则计算时间太长。Chinese patent: The application publication number is CN104090212A, and the invention patent published on October 8, 2014 discloses a line-phase fault location method without pre-fault voltage memory. The method first measures the two fault phases at the transmission line protection installation. The voltage mutation, the two fault phase current mutations and the normal phase voltage, use the normal phase voltage to calculate the fault phase-to-phase voltage when the transmission line is in normal operation, and then use the amplitude of the voltage mutation at the phase-to-phase short-circuit fault point after the line fault (that is, equal to the transmission line normal During operation, the fault phase-to-phase voltage amplitude) is the largest and monotonically decreases from the phase-to-phase short-circuit fault point to both ends of the transmission line, which realizes the single-end location of the line-phase short-circuit fault. Although this method does not need to memorize the voltage before the line fault, and the relay protection hardware does not need to set the voltage holding circuit before the line fault, it is only applicable to the phase-to-phase fault of the line, and the scope of application is narrow, and the fixed step is substituted into the comparison to solve the problem. If the step size is large, the calculation accuracy is not enough, and if the step size is small, the calculation time will be too long.

发明内容Contents of the invention

本发明的目的是克服现有技术存在的适用范围窄、计算精度和效率低的问题,提供一种适用范围广、计算精度和效率高的基于故障点电压突变量的故障测距方法。The purpose of the present invention is to overcome the problems of narrow application range, low calculation accuracy and efficiency in the prior art, and provide a fault distance measurement method based on fault point voltage mutation with wide application range and high calculation accuracy and efficiency.

为实现以上目的,本发明的技术方案如下:For realizing above object, technical scheme of the present invention is as follows:

一种基于故障点电压突变量的故障测距方法,依次包括以下步骤:A method for fault distance measurement based on the sudden change in voltage at a fault point, comprising the following steps in sequence:

步骤1:保护装置对保护装置安装处的线路互感器的电压电流波形采样得到电压电流瞬时值;Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

步骤2:若为单相故障,采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相电流保护安装处的相电压单相接地故障时的零序电流I0的相量形式;若为相间故障,则采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相间电压保护安装处的故障相间电流其中,表示线路的a、b、c三相;Step 2: If it is a single-phase fault, use the Fourier algorithm to obtain the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I 0 in the case of a single-phase ground fault; if it is a phase-to-phase fault, the Fourier algorithm is used to obtain the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the three phases a, b, and c of the line;

步骤3:根据式1求出故障点电压U':Step 3: Calculate the fault point voltage U' according to formula 1:

U′=U-IZlx    式1U'=U-IZ l x Formula 1

上式中,In the above formula,

Zl为每公里正序阻抗,x为保护安装处到短路故障点的故障距离;Z l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point;

若为单相故障,则k为零序补偿系数;If it is a single-phase fault, then k is zero sequence compensation coefficient;

若为相间故障,则 In the case of a phase-to-phase fault, the

步骤4:先取故障前N周波采样点和故障后M周波采样点,再采用步骤2所述方法求出故障前保护安装处的故障相/相间电压U1、故障前保护安装处的故障相/相间电流I1、故障后保护安装处的故障相/相间电压U2、故障后保护安装处的故障相/相间电流I2,然后采用步骤3所述方法求出故障后故障点电压U'2,故障前故障点电压U'1,最后将得到的U'1、U'2带入式2求出故障点电压突变量ΔU′:Step 4: First take the N-cycle sampling point before the fault and the M-cycle sampling point after the fault, and then use the method described in step 2 to obtain the fault phase/phase-to-phase voltage U 1 at the protection installation before the fault, and the fault phase/phase voltage at the protection installation before the fault. Phase-to-phase current I 1 , fault phase/phase-to-phase voltage U 2 at the post-fault protection installation, fault phase/phase-to-phase current I 2 at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' 2 , the fault point voltage U' 1 before the fault, and finally put the obtained U' 1 and U' 2 into Equation 2 to obtain the fault point voltage mutation value ΔU′:

ΔU'=U'2-U'1    式2ΔU'=U' 2 -U' 1 Formula 2

其中,M、M均为正整数;Among them, M and M are both positive integers;

步骤5:将步骤3得到的U'、步骤4得到的ΔU′带入式3,并结合式4得到关于保护安装处到短路故障点的故障距离x的一元二次方程,进而计算出其数值:Step 5: Put U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

Re ( ΔU ′ . C ) Im ( ΔU ′ . C ) = Re ( U ′ ) Im ( U ′ )     式3 Re ( Δ U ′ . C ) Im ( Δ U ′ . C ) = Re ( u ′ ) Im ( u ′ ) Formula 3

C=1.0∠(180°-(ArgZl+δ))    式4C=1.0∠(180°-(ArgZ l +δ)) Formula 4

上式中,In the above formula,

Arg(Zl)为线路正序阻抗角,δ为系统阻抗角大于线路正序阻抗角的角度,Im(·)为变量的虚部,Re(·)为变量的实部,C为幅值是1的向量。Arg(Z l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1s.

所述步骤4中,M为大于等于2、且小于故障持续周波数的正整数,N为1或2。In the step 4, M is a positive integer greater than or equal to 2 and less than the number of continuous fault cycles, and N is 1 or 2.

所述步骤5中,若为单相故障,则δ为-8°~0°;若为相间故障,则δ为0°~+4°。In the step 5, if it is a single-phase fault, then δ is -8° to 0°; if it is an interphase fault, then δ is 0° to +4°.

与现有技术相比,本发明的有益效果为:Compared with prior art, the beneficial effect of the present invention is:

1、本发明一种基于故障点电压突变量的故障测距方法利用故障点电压与故障点电压突变量这两个向量之间存在的相位关系,列出了以保护安装处到短路故障点的故障距离为未知数的一元二次方程,由于该方程与负荷电流和接地电阻无关,因此不受负荷电流和高电阻接地故障的影响,不仅对于线路单相和相间故障均适用,而且实用性强,计算精度和效率均很高。因此,本发明不仅适用范围广,而且计算精度和效率高。1, a kind of fault location method based on fault point voltage sudden change of the present invention utilizes the phase relation that exists between these two vectors of fault point voltage and fault point voltage sudden change, has listed to protect installation place to short-circuit fault point A quadratic equation with unknown distance to fault. Since this equation has nothing to do with load current and ground resistance, it is not affected by load current and high-resistance ground fault. The calculation accuracy and efficiency are high. Therefore, the present invention not only has a wide application range, but also has high calculation accuracy and efficiency.

附图说明Description of drawings

图1为本发明的线路接地短路故障示意图。Fig. 1 is a schematic diagram of a line grounding short circuit fault according to the present invention.

图中,为保护安装处的故障相电流保护安装处的相电压故障点电压U'、故障点电流If、保护安装处到短路故障点的故障距离x。In the figure, for the fault phase current at the protection installation Protect the phase voltage at the installation Fault point voltage U', fault point current I f , fault distance x from the protection installation to the short-circuit fault point.

具体实施方式Detailed ways

下面结合具体实施方式对本发明作进一步详细的说明。The present invention will be described in further detail below in combination with specific embodiments.

一种基于故障点电压突变量的故障测距方法,依次包括以下步骤:A method for fault distance measurement based on the sudden change in voltage at a fault point, comprising the following steps in sequence:

步骤1:保护装置对保护装置安装处的线路互感器的电压电流波形采样得到电压电流瞬时值;Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

步骤2:若为单相故障,采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相电流保护安装处的相电压单相接地故障时的零序电流I0的相量形式;若为相间故障,则采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相间电压保护安装处的故障相间电流其中,表示线路的a、b、c三相;Step 2: If it is a single-phase fault, use the Fourier algorithm to obtain the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I 0 in the case of a single-phase ground fault; if it is a phase-to-phase fault, the Fourier algorithm is used to obtain the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the three phases a, b, and c of the line;

步骤3:根据式1求出故障点电压U':Step 3: Calculate the fault point voltage U' according to formula 1:

U′=U-IZlx    式1U'=U-IZ l x Formula 1

上式中,In the above formula,

Zl为每公里正序阻抗,x为保护安装处到短路故障点的故障距离;Z l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point;

若为单相故障,则k为零序补偿系数;If it is a single-phase fault, then k is zero sequence compensation coefficient;

若为相间故障,则 In the case of a phase-to-phase fault, the

步骤4:先取故障前N周波采样点和故障后M周波采样点,再采用步骤2所述方法求出故障前保护安装处的故障相/相间电压U1、故障前保护安装处的故障相/相间电流I1、故障后保护安装处的故障相/相间电压U2、故障后保护安装处的故障相/相间电流I2,然后采用步骤3所述方法求出故障后故障点电压U'2,故障前故障点电压U'1,最后将得到的U'1、U'2带入式2求出故障点电压突变量ΔU′:Step 4: First take the N-cycle sampling point before the fault and the M-cycle sampling point after the fault, and then use the method described in step 2 to obtain the fault phase/phase-to-phase voltage U 1 at the protection installation before the fault, and the fault phase/phase voltage at the protection installation before the fault. Phase-to-phase current I 1 , fault phase/phase-to-phase voltage U 2 at the post-fault protection installation, fault phase/phase-to-phase current I 2 at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' 2 , the fault point voltage U' 1 before the fault, and finally put the obtained U' 1 and U' 2 into Equation 2 to obtain the fault point voltage mutation value ΔU′:

ΔU'=U'2-U'1    式2ΔU'=U' 2 -U' 1 Formula 2

其中,M、M均为正整数;Among them, M and M are both positive integers;

步骤5:将步骤3得到的U'、步骤4得到的ΔU′带入式3,并结合式4得到关于保护安装处到短路故障点的故障距离x的一元二次方程,进而计算出其数值:Step 5: Put U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

Re ( ΔU ′ . C ) Im ( ΔU ′ . C ) = Re ( U ′ ) Im ( U ′ )     式3 Re ( Δ U ′ . C ) Im ( Δ U ′ . C ) = Re ( u ′ ) Im ( u ′ ) Formula 3

C=1.0∠(180°-(ArgZl+δ))    式4C=1.0∠(180°-(ArgZ l +δ)) Formula 4

上式中,In the above formula,

Arg(Zl)为线路正序阻抗角,δ为系统阻抗角大于线路正序阻抗角的角度,Im(·)为变量的虚部,Re(·)为变量的实部,C为幅值是1的向量。Arg(Z l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1s.

所述步骤4中,M为大于等于2、且小于故障持续周波数的正整数,N为1或2。In the step 4, M is a positive integer greater than or equal to 2 and less than the number of continuous fault cycles, and N is 1 or 2.

所述步骤5中,若为单相故障,则δ为-8°~0°;若为相间故障,则δ为0°~+4°。In the step 5, if it is a single-phase fault, then δ is -8° to 0°; if it is an interphase fault, then δ is 0° to +4°.

本发明的原理说明如下:Principle of the present invention is described as follows:

本发明方法应用于电力系统的高压、超高压、特高压线路,依据故障点电压突变量和故障点电压同相位的原理,较好的消除负荷电流和接地电阻对接地距离保护的影响。具体计算过程如下:The method of the invention is applied to high-voltage, ultra-high-voltage, and ultra-high-voltage lines of the power system, and based on the principle that the fault point voltage mutation amount and the fault point voltage are in phase, it can better eliminate the influence of load current and grounding resistance on grounding distance protection. The specific calculation process is as follows:

由于故障点电压突变量ΔU'=-IfZΣ(If为故障点电流,ZΣ为从故障点看进去的总阻抗即系统等效阻抗),且单相故障和相间故障均属于电阻性接地,即故障点电流If与故障点电压U'同相位,因此U'超前ΔU'(180°-Arg(ZΣ))度,令Arg(ZΣ)=Arg(Zl)+δ,其中Arg(Zl)为线路正序阻抗角,δ为系统阻抗角大于线路正序阻抗角的角度;Since the voltage mutation at the fault point ΔU'=-I f Z Σ (I f is the current at the fault point, Z Σ is the total impedance seen from the fault point, that is, the equivalent impedance of the system), and both single-phase faults and phase-to-phase faults belong to resistance grounding, that is, the fault point current I f is in the same phase as the fault point voltage U', so U' is ahead of ΔU'(180°-Arg(Z Σ )) degrees, so that Arg(Z Σ )=Arg(Z l )+δ , where Arg(Z l ) is the line positive sequence impedance angle, δ is the angle at which the system impedance angle is greater than the line positive sequence impedance angle;

令C=1.0∠(180°-(ArgZl+δ)),则Arg(ΔU′.C)=Arg(U'),即ΔU′.C、U'这两个相量的角度相等,所以他们的实部除以虚部相等,可列出以下方程:Let C=1.0∠(180°-(ArgZ l +δ)), then Arg(ΔU′.C)=Arg(U’), that is, the angles of the two phasors ΔU′.C and U’ are equal, so Their real part divided by the imaginary part is equal, and the following equation can be listed:

Re ( ΔU ′ . C ) Im ( ΔU ′ . C ) = Re ( U ′ ) Im ( U ′ )     式3 Re ( Δ U ′ . C ) Im ( Δ U ′ . C ) = Re ( u ′ ) Im ( u ′ ) Formula 3

and

ΔU'=U'2-U'1=U2-I2Zix-(U1-I1Zix)=(U2-U1)-(I2-I1)Zix=ΔU-ΔIZlx,ΔU'=U' 2 -U' 1 =U 2 -I 2 Z i x-(U 1 -I 1 Z i x)=(U 2 -U 1 )-(I 2 -I 1 )Z i x= ΔU-ΔIZ l x,

U'=U-IZlxU'=U-IZ l x

其中,ΔU为故障前后电压突变量,ΔI为故障前后电流突变量;Among them, ΔU is the voltage mutation before and after the fault, and ΔI is the current mutation before and after the fault;

所以式3可变为:So formula 3 can be changed to:

ReRe (( ΔUΔ U .. CC -- xΔIZxΔIZ ll .. CC )) ImIm (( ΔUΔ U .. CC -- xΔIZxΔIZ ll .. CC )) == ReRe (( Uu -- xIZwxya ll )) ImIm (( Uu -- xIZwxya ll )) -- -- -- (( 11 ))

将(1)左右相乘,得(2):Multiply (1) left and right to get (2):

Re(ΔU.C)Im(U)-x.Re(ΔU.C)Im(IZl)-xRe(ΔIZlC)Im(U)+x2Re(ΔIZlC)Im(IZl)=Re(ΔU.C)Im(U)-x.Re(ΔU.C)Im(IZ l )-xRe(ΔIZ l C)Im(U)+x 2 Re(ΔIZ l C)Im(IZ l )=

Im(ΔU.C)Re(U)-x.Im(ΔU.C)Re(IZl)-xIm(ΔIZlC)Re(U)+x2Im(ΔIZlC)Re(IZl)Im(ΔU.C)Re(U)-x.Im(ΔU.C)Re(IZ l )-xIm(ΔIZ l C)Re(U)+x 2 Im(ΔIZ l C)Re(IZ l )

                                               (2) (2)

整理方程,得(3):After rearranging the equations, we get (3):

x2[Re(ΔIZlC)Im(IZl)-Im(ΔIZlC)Re(IZl)]+x 2 [Re(ΔIZ l C)Im(IZ l )-Im(ΔIZ l C)Re(IZ l )]+

x[Im(ΔU.C)Re(IZl)+Im(ΔIZlC)Re(U)-Re(ΔU.C)Im(IZl)-Re(ΔIZlC)Im(U)]+    (3)x[Im(ΔU.C)Re(IZ l )+Im(ΔIZ l C)Re(U)-Re(ΔU.C)Im(IZ l )-Re(ΔIZ l C)Im(U)]+ ( 3)

Re(ΔU.C)Im(U)-Im(ΔU.C)Re(U)=0Re(ΔU.C)Im(U)-Im(ΔU.C)Re(U)=0

上式即为以x为未知数的一元二次方程,该一元二次方程其中一个合理的解,即为保护安装处到短路故障点的故障距离。The above formula is a quadratic equation with x as the unknown. One of the reasonable solutions of this quadratic equation is the fault distance from the protection installation to the short-circuit fault point.

M:为躲过直流分量衰减时间,M应大于等于2。M: In order to avoid the decay time of the DC component, M should be greater than or equal to 2.

实施例1:Example 1:

以下按照本方法对某一线路B相接地故障(故障位置350公里,接地电阻100欧)进行故障测距,参见图1;In the following, according to this method, fault location is carried out on a line B-phase ground fault (fault location 350 kilometers, grounding resistance 100 ohms), see Figure 1;

本方法依次进行以下步骤:This method carries out the following steps in sequence:

步骤1:保护装置对保护装置安装处的线路互感器的电压电流波形采样得到电压电流瞬时值;Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

步骤2:采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相电流保护安装处的相电压单相接地故障时的零序电流I0的相量形式,其中,表示线路的b相;Step 2: Using the Fourier algorithm, calculate the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I 0 in case of a single-phase-to-ground fault, where, Indicates phase b of the line;

步骤3:根据式1求出故障点电压U':Step 3: Calculate the fault point voltage U' according to formula 1:

U′=U-IZlx    式1U'=U-IZ l x Formula 1

上式中,In the above formula,

k为零序补偿系数,Zl为每公里正序阻抗,x为保护安装处到短路故障点的故障距离; k is the zero-sequence compensation coefficient, Z l is the positive-sequence impedance per kilometer, and x is the fault distance from the protection installation to the short-circuit fault point;

步骤4:先取故障前1周波采样点和故障后2周波采样点,再采用步骤2所述方法求出故障前保护安装处的故障相电压U1、故障前保护安装处的故障相电流I1、故障后保护安装处的故障相电压U2、故障后保护安装处的故障相电流I2,然后采用步骤3所述方法求出故障后故障点电压U'2,故障前故障点电压U'1,最后将得到的U'1、U'2带入式2求出故障点电压突变量ΔU′;Step 4: Take the sampling point of the 1 cycle before the fault and the sampling point of the 2 cycles after the fault, and then use the method described in step 2 to obtain the fault phase voltage U 1 of the protection installation before the fault and the fault phase current I 1 of the protection installation before the fault , the fault phase voltage U 2 at the post-fault protection installation, the fault phase current I 2 at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' 2 and the pre-fault fault point voltage U' 1 , and finally put the obtained U' 1 and U' 2 into formula 2 to obtain the voltage mutation value ΔU' at the fault point;

ΔU'=U'2-U'1    式2ΔU'=U' 2 -U' 1 Formula 2

步骤5:将步骤3得到的U'、步骤4得到的ΔU′带入式3,并结合式4得到关于保护安装处到短路故障点的故障距离x的一元二次方程,进而计算出其数值:Step 5: Put the U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

Re ( ΔU ′ . C ) Im ( ΔU ′ . C ) = Re ( U ′ ) Im ( U ′ )     式3 Re ( Δ U ′ . C ) Im ( Δ U ′ . C ) = Re ( u ′ ) Im ( u ′ ) Formula 3

C=1.0∠(180°-(ArgZl+δ))    式4C=1.0∠(180°-(ArgZ l +δ)) Formula 4

上式中,In the above formula,

Arg(Zl)为线路正序阻抗角,δ为系统阻抗角大于线路正序阻抗角的角度,Im(·)为变量的虚部,Re(·)为变量的实部,C为幅值是1的向量;Arg(Z l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1;

由于系统阻抗角大小是不知道的,但是和线路阻抗角的大小差不多,且略大于线路阻抗角,而在单相故障时,有零序电流,线路零序阻抗角要小于线路正序阻抗角,可取δ为-6°。Since the system impedance angle is unknown, it is similar to the line impedance angle and slightly larger than the line impedance angle. In the case of a single-phase fault, there is zero-sequence current, and the line zero-sequence impedance angle is smaller than the line positive-sequence impedance angle. , it is desirable to take δ as -6°.

解得x1=-11,x2=348。取x=x2=348,即故障位置在距保护安装处348公里处,与故障实际位置350公里非常接近。Solving x 1 =-11, x 2 =348. Take x=x 2 =348, that is, the fault location is 348 kilometers away from the protection installation, which is very close to the actual fault location of 350 kilometers.

实施例2:Example 2:

以下按照本方法对某一线路AB相间接地故障(故障位置350公里,接地电阻50欧)进行故障测距;Carry out fault distance measurement to a line AB phase-to-ground fault (fault location 350 kilometers, grounding resistance 50 ohms) according to this method as follows;

本方法依次进行以下步骤:This method carries out the following steps in sequence:

步骤1:保护装置对保护装置安装处的线路互感器的电压电流波形采样得到电压电流瞬时值;Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current;

步骤2:采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相间电压保护安装处的故障相间电流其中,表示线路的a、b相间;Step 2: Using the Fourier algorithm, calculate the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the a and b phases of the line;

步骤3:根据式1求出故障点电压U':Step 3: Calculate the fault point voltage U' according to formula 1:

U′=U-IZlx    式1U'=U-IZ l x Formula 1

上式中,In the above formula,

Zl为每公里正序阻抗,x为保护安装处到短路故障点的故障距离; Z l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point;

步骤4:先取故障前1周波采样点和故障后2周波采样点,再采用步骤2所述方法求出故障前保护安装处的故障相间电压U1、故障前保护安装处的故障相间电流I1、故障后保护安装处的故障相间电压U2、故障后保护安装处的故障相间电流I2,然后采用步骤3所述方法求出故障后故障点电压U'2,故障前故障点电压U'1,最后将得到的U'1、U'2带入式2求出故障点电压突变量ΔU′;Step 4: First take the sampling points of the 1 cycle before the fault and the sampling points of the 2 cycles after the fault, and then use the method described in step 2 to obtain the fault phase-to-phase voltage U 1 at the protection installation before the fault and the fault phase-to-phase current I 1 at the protection installation before the fault , the fault phase-to-phase voltage U 2 at the post-fault protection installation, the fault phase-to-phase current I 2 at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' 2 and the pre-fault fault point voltage U' 1 , and finally put the obtained U' 1 and U' 2 into formula 2 to obtain the voltage mutation value ΔU' at the fault point;

ΔU'=U'2-U'1    式2ΔU'=U' 2 -U' 1 Formula 2

步骤5:将步骤3得到的U'、步骤4得到的ΔU′带入式3,并结合式4得到关于保护安装处到短路故障点的故障距离x的一元二次方程,进而计算出其数值:Step 5: Put the U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value :

Re ( ΔU ′ . C ) Im ( ΔU ′ . C ) = Re ( U ′ ) Im ( U ′ )     式3 Re ( Δ U ′ . C ) Im ( Δ U ′ . C ) = Re ( u ′ ) Im ( u ′ ) Formula 3

C=1.0∠(180°-(ArgZl+δ))    式4C=1.0∠(180°-(ArgZ l +δ)) Formula 4

上式中,In the above formula,

Arg(Zl)为线路正序阻抗角,δ为系统阻抗角大于线路正序阻抗角的角度,Im(·)为变量的虚部,Re(·)为变量的实部,C为幅值是1的向量;Arg(Z l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1;

在相间故障时,由于没有零序电流,此时可取δ为+2°。In the phase-to-phase fault, since there is no zero-sequence current, it is desirable to take δ as +2° at this time.

解得x1=-19,x2=348。取x=x2=349,即故障位置在距保护安装处349公里处,与故障实际位置350公里非常接近。Solving x 1 =-19, x 2 =348. Take x=x 2 =349, that is, the fault location is 349 kilometers away from the protection installation, which is very close to the actual fault location of 350 kilometers.

上述实施例显示本发明方法具有很高的实用性和计算精度。The above examples show that the method of the present invention has high practicability and calculation accuracy.

Claims (3)

1.一种基于故障点电压突变量的故障测距方法,其特征在于:1. A fault location method based on fault point voltage mutation amount, is characterized in that: 该方法依次包括以下步骤:The method sequentially includes the following steps: 步骤1:保护装置对保护装置安装处的线路互感器的电压电流波形采样得到电压电流瞬时值;Step 1: The protection device samples the voltage and current waveform of the line transformer where the protection device is installed to obtain the instantaneous value of voltage and current; 步骤2:若为单相故障,采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相电流保护安装处的相电压单相接地故障时的零序电流I0的相量形式;若为相间故障,则采用傅氏算法,根据采样得到电压电流瞬时值求出保护安装处的故障相间电压保护安装处的故障相间电流其中,表示线路的a、b、c三相;Step 2: If it is a single-phase fault, use the Fourier algorithm to obtain the fault phase current at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Protect the phase voltage at the installation The phasor form of the zero-sequence current I 0 in the case of a single-phase ground fault; if it is a phase-to-phase fault, the Fourier algorithm is used to obtain the fault phase-to-phase voltage at the protection installation according to the instantaneous value of the voltage and current obtained by sampling Fault phase-to-phase current at the protective installation in, Indicates the three phases a, b, and c of the line; 步骤3:根据式1求出故障点电压U':Step 3: Calculate the fault point voltage U' according to formula 1: U′=U-IZlx              式1U'=U-IZ l x Formula 1 上式中,In the above formula, Zl为每公里正序阻抗,x为保护安装处到短路故障点的故障距离;Z l is the positive sequence impedance per kilometer, x is the fault distance from the protection installation to the short-circuit fault point; 若为单相故障,则 k为零序补偿系数;If it is a single-phase fault, then k is zero sequence compensation coefficient; 若为相间故障,则 In the case of a phase-to-phase fault, the 步骤4:先取故障前N周波采样点和故障后M周波采样点,再采用步骤2所述方法求出故障前保护安装处的故障相/相间电压U1、故障前保护安装处的故障相/相间电流I1、故障后保护安装处的故障相/相间电压U2、故障后保护安装处的故障相/相间电流I2,然后采用步骤3所述方法求出故障后故障点电压U'2,故障前故障点电压U'1,最后将得到的U'1、U'2带入式2求出故障点电压突变量ΔU′:Step 4: First take the N-cycle sampling point before the fault and the M-cycle sampling point after the fault, and then use the method described in step 2 to obtain the fault phase/phase-to-phase voltage U 1 at the protection installation before the fault, and the fault phase/phase voltage at the protection installation before the fault. Phase-to-phase current I 1 , fault phase/phase-to-phase voltage U 2 at the post-fault protection installation, fault phase/phase-to-phase current I 2 at the post-fault protection installation, and then use the method described in step 3 to obtain the post-fault fault point voltage U' 2 , the fault point voltage U' 1 before the fault, and finally put the obtained U' 1 and U' 2 into Equation 2 to obtain the fault point voltage mutation value ΔU′: ΔU'=U'2-U'1             式2ΔU'=U' 2 -U' 1 Formula 2 其中,M、M均为正整数;Among them, M and M are both positive integers; 步骤5:将步骤3得到的U'、步骤4得到的ΔU′带入式3,并结合式4得到关于保护安装处到短路故障点的故障距离x的一元二次方程,进而计算出其数值:Step 5: Put the U' obtained in Step 3 and ΔU' obtained in Step 4 into Equation 3, and combine Equation 4 to obtain the quadratic equation of the fault distance x from the protection installation to the short-circuit fault point, and then calculate its value : Re ( ΔU ′ . C ) Im ( ΔU ′ . C ) = Re ( U ′ ) Im ( U ′ )               式3 Re ( Δ U ′ . C ) Im ( Δ U ′ . C ) = Re ( u ′ ) Im ( u ′ ) Formula 3 C=1.0∠(180°-(ArgZl+δ))            式4C=1.0∠(180°-(ArgZ l +δ)) Formula 4 上式中,In the above formula, Arg(Zl)为线路正序阻抗角,δ为系统阻抗角大于线路正序阻抗角的角度,Im(·)为变量的虚部,Re(·)为变量的实部,C为幅值是1的向量。Arg(Z l ) is the positive sequence impedance angle of the line, δ is the angle at which the system impedance angle is greater than the positive sequence impedance angle of the line, Im(·) is the imaginary part of the variable, Re(·) is the real part of the variable, and C is the amplitude is a vector of 1s. 2.根据权利要求1所述的一种基于故障点电压突变量的故障测距方法,其特征在于:2. a kind of fault location method based on fault point voltage mutation amount according to claim 1, is characterized in that: 所述步骤4中,M为大于等于2、且小于故障持续周波数的正整数,N为1或2。In the step 4, M is a positive integer greater than or equal to 2 and less than the number of continuous fault cycles, and N is 1 or 2. 3.根据权利要求1或2所述的一种基于故障点电压突变量的故障测距方法,其特征在于:3. according to claim 1 or 2 described a kind of fault location method based on fault point voltage mutation amount, it is characterized in that: 所述步骤5中,若为单相故障,则δ为-8°~0°;若为相间故障,则δ为0°~+4°。In the step 5, if it is a single-phase fault, then δ is -8° to 0°; if it is an interphase fault, then δ is 0° to +4°.
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