CN103762560B - The non-same famous prime minister's cross-line earthing reactance distance protecting method of double-circuit line - Google Patents

Abstract

Translated fromChinese

本发明公开了一种双回线路非同名相跨线接地电抗距离保护方法，它首先测量双回线路I回线路保护安装处的故障相电压、故障相电流和零序电流，计算双回线路II回线路的零序电流，利用I回线路零序电流和II回线路零序电流之和的相角估算I回线路接地故障点电压相角，然后判断I回线路保护整定范围处的电压领先I回线路接地故障点电压的相角落在[‑165°，15°]范围内是否成立，若成立，则保护装置发出动作跳闸信号。本发明方法只用到单端单回线路电气量，动作性能不受电力系统运行方式的影响，消除了线间零序互感、过渡电阻和负荷电流对同杆并架双回线路非同名相跨线接地电抗距离继电器动作性能的影响，保护范围稳定可靠。

The invention discloses a non-identical phase cross-line grounding reactance distance protection method for a double-circuit line. It first measures the fault phase voltage, fault phase current and zero-sequence current at the installation place of the double-circuit line I circuit protection, and calculates the double-circuit line II For the zero-sequence current of the circuit, use the phase angle of the sum of the zero-sequence current of the I circuit and the zero-sequence current of the II circuit to estimate the phase angle of the ground fault point voltage of the I circuit, and then judge that the voltage at the protection setting range of the I circuit is ahead of I Whether the phase angle of the voltage at the ground fault point of the return line is established within the range of [‑165°, 15°], if established, the protection device will send out an action trip signal. The method of the present invention only uses the electrical quantity of the single-ended single-circuit line, and the action performance is not affected by the operation mode of the power system, eliminating the inter-line zero-sequence mutual inductance, transition resistance and load current on the double-circuit line with the same pole and the same name. The influence of line grounding reactance distance relay action performance, the protection range is stable and reliable.

Description

Translated fromChinese

双回线路非同名相跨线接地电抗距离保护方法Protection method of non-identical phase cross-line grounding reactance distance for double-circuit lines

技术领域technical field

本发明涉及一种力系统继电保护技术领域，具体地说是涉及一种双回线路非同名相跨线接地电抗距离保护方法。The invention relates to the technical field of power system relay protection, in particular to a distance protection method for non-identical phase-cross-line grounding reactance of a double-circuit line.

背景技术Background technique

同杆并架双回线路具有占地面积少、造价成本低，连接电网运行稳定可靠，已成为电力系统一种常见输电线路连接方式。同杆并架双回线路线间存在零序互感，零序互感对零序补偿系数产生影响，进而产生附加阻抗，因零序互感引起的附加阻抗会导致保护装置测量到的故障阻抗大于实际故障阻抗，造成同杆并架双回线路保护区内靠近保护整定范围处发生接地故障时，保护出现误动作，对电网安全稳定运行不利。The double-circuit line paralleled on the same pole has a small footprint, low cost, and stable and reliable operation when connected to the grid. It has become a common transmission line connection method in the power system. There is zero-sequence mutual inductance between double-circuit lines on the same pole, and the zero-sequence mutual inductance affects the zero-sequence compensation coefficient, thereby generating additional impedance. The additional impedance caused by the zero-sequence mutual inductance will cause the fault impedance measured by the protection device to be greater than the actual fault When a ground fault occurs near the protection setting range in the double-circuit line protection area of the same pole, the protection will malfunction, which is not good for the safe and stable operation of the power grid.

同杆并架双回线路即使发生经杆塔直接接地故障，在土壤电阻率较低的地区过渡电阻也在10Ω附近；在电阻率较高的地方过渡电阻可达30Ω，或甚至更高。过渡电阻不为零使得保护装置计算得到的故障阻抗除包含反应真实故障距离的故障阻抗分量外，还包含了因过渡电阻而产生的附加阻抗。过渡电阻产生的附加阻抗呈阻感性或呈阻容性容易造成接地阻抗距离保护拒动或稳态超越。继电保护误动或拒动，会给电力系统安全运行带来重大的损失，甚至有可能会威胁到电力系统的稳定性。Even if a direct grounding fault occurs on the double-circuit lines paralleled on the same pole, the transition resistance is around 10Ω in areas with low soil resistivity; the transition resistance can reach 30Ω or even higher in areas with high resistivity. The non-zero transition resistance makes the fault impedance calculated by the protection device not only include the fault impedance component reflecting the real fault distance, but also include the additional impedance caused by the transition resistance. The additional impedance generated by the transition resistance is resistive-inductive or resistive-capacitive, which may easily cause the grounding impedance distance protection to refuse to operate or exceed the steady state. Misoperation or refusal of relay protection will bring significant losses to the safe operation of the power system, and may even threaten the stability of the power system.

发明内容Contents of the invention

本发明提供了一种双回线路非同名相跨线接地电抗距离保护方法，其克服了背景技术中所述的现有技术的不足。The invention provides a non-identical phase cross-line grounding reactance distance protection method for a double-circuit line, which overcomes the shortcomings of the prior art described in the background art.

本发明解决其技术问题的所采用的技术方案是：The adopted technical scheme that the present invention solves its technical problem is:

双回线路非同名相跨线接地电抗距离保护方法，其特征在于，包括如下依序步骤：The method for protecting a double-circuit line with non-identical phase-cross-line grounding reactance distance is characterized in that it includes the following sequential steps:

步骤1，保护装置测量同杆并架双回线路I回线路保护安装处的故障相电压故障相电流和零序电流其中：φ为I回线路A相、I回线路B相、I回线路C相；Step 1, the protection device measures the fault phase voltage at the installation place of the I-circuit line protection of the parallel double-circuit line on the same pole fault phase current and zero sequence current Among them: φ is the A phase of the I circuit line, the B phase of the I circuit line, and the C phase of the I circuit line;

步骤2，保护装置计算同杆并架双回线路II回线路的零序电流相角α=r₁+r₂-π-β；Step 2, the protection device calculates the zero-sequence current phase angle α=r₁ +r₂ -π-β of the line II of the double-circuit line paralleled on the same pole;

其中：$r_1={\sin }^{-1}\left(\frac{a_3{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),r_2={\sin }^{-1}\left(\frac{a_1{b}_2-a_2{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),a_1=\mathrm{Re}\left(\frac{{\stackrel{\·}{U}}_{\mathrm{I\φ}}}{Z_{I1}}\right),b_1=\mathrm{Im}\left(\frac{{\stackrel{\·}{U}}_{\mathrm{I\φ}}}{Z_{I1}}\right),$$a_2=\mathrm{Re}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),b_2=\mathrm{Im}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),a_3=b_3=\left|\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right|,\mathrm{\β}=\mathrm{Arg}\left(\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right);$Z_m为同杆并架双回线路I回线路与同杆并架双回线路II回线路之间的零序互感；Z_I0为同杆并架双回线路I回线路的零序阻抗；Z_I1为同杆并架双回线路I回线路的正序阻抗；φ=I回线路A相、I回线路B相、I回线路C相；in:$r_1={\sin }^{-1}\left(\frac{a_3{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),r_2={\sin }^{-1}\left(\frac{a_1{b}_2-a_2{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),a_1=\mathrm{Re}\left(\frac{{\stackrel{\&Center\; Dot;}{u}}_{\mathrm{I\φ}}}{Z_{I1}}\right),b_1=\mathrm{Im}\left(\frac{{\stackrel{\·}{u}}_{\mathrm{I\φ}}}{Z_{I1}}\right),$$a_2=\mathrm{Re}({\stackrel{\&Center\; Dot;}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\&Center\; Dot;}{I}}_{I0}),b_2=\mathrm{Im}({\stackrel{\&Center\; Dot;}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),a_3=b_3=\left|\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right|,\mathrm{\β}=\mathrm{Arg}\left(\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right);$ Z_m is the zero-sequence mutual inductance between the I-circuit line of the parallel double-circuit line on the same pole and the II-circuit line of the parallel double-circuit line on the same pole; Z_I0 is the zero-sequence impedance of the I-circuit line of the parallel double-circuit line on the same pole; Z_I1 is the positive sequence impedance of the I-circuit line of the parallel double-circuit line on the same pole; φ=I-circuit line A phase, I-circuit line B phase, and I-circuit line C phase;

步骤3，保护装置计算同杆并架双回线路II回线路的零序电流${\stackrel{\·}{I}}_{\mathrm{II}0}={\stackrel{\·}{I}}_{I0}(\cos \mathrm{\α}+j\sin \mathrm{\α});$Step 3, the protective device calculates the zero-sequence current of the double-circuit line II on the same pole${\stackrel{\·}{I}}_{\mathrm{II}0}={\stackrel{\&Center\; Dot;}{I}}_{I0}(\cos \mathrm{\α}+j\sin \mathrm{\α});$

步骤4，保护装置计算${\stackrel{\·}{U}}_{\mathrm{I\φ}}-Z_{I1}\frac{l_{\mathrm{set}}}{l}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}+\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{\mathrm{II}0})$领先的相角落在[-165°，15°]范围内是否成立，若成立，则保护装置发出动作跳闸信号；其中：l_set为同杆并架双回线路I回线路保护整定范围，l为同杆并架双回线路I回线路长度。Step 4, protection device calculation${\stackrel{\·}{u}}_{\mathrm{I\φ}}-Z_{I1}\frac{l_{\mathrm{set}}}{l}({\stackrel{\&Center\; Dot;}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\&Center\; Dot;}{I}}_{I0}+\frac{Z_m}{3Z_{I1}}{\stackrel{\&Center\; Dot;}{I}}_{\mathrm{II}0})$ take the lead Whether the phase angle of the phase angle is established within the range of [-165°, 15°], if established, the protection device will send out an action trip signal; where: l_set is the setting range of the I-circuit line protection of the double-circuit line on the same pole, and l is the same The length of the I-circuit line of the pole-parallel double-circuit line.

本技术方案与背景技术相比，它具有如下优点：Compared with the background technology, this technical solution has the following advantages:

本发明方法首先测量同杆并架双回线路I回线路保护安装处的故障相电压、故障相电流和零序电流，计算同杆并架双回线路II回线路的零序电流相角，计算同杆并架双回线路II回线路的零序电流，利用同杆并架双回线路I回线路零序电流和同杆并架双回线路II回线路零序电流之和的相角估算I回线路接地故障点电压相角，计算I回线路保护整定范围处的电压，然后判断I回线路保护整定范围处的电压领先I回线路接地故障点电压的相角落在[-165°，15°]范围内是否成立，若成立，则保护装置发出动作跳闸信号。本发明方法只用到单端单回线路电气量，不需要引入另一回线路电气量，动作性能不受电力系统运行方式的影响，在电力系统运行方式发生较大改变时具有很强的适应能力。本发明方法计及线间零序互感和接地故障点电压的影响，消除了线间零序互感、过渡电阻和负荷电流对同杆并架双回线路非同名相跨线接地电抗距离保护动作性能的影响，保护范围稳定可靠。The method of the present invention firstly measures the fault phase voltage, fault phase current and zero-sequence current at the protection installation place of the I-circuit line of the parallel double-circuit line on the same pole, calculates the phase angle of the zero-sequence current of the II-circuit line of the double-circuit line parallel on the same pole, and calculates The zero-sequence current of the II circuit of the parallel double-circuit line on the same pole is estimated by the phase angle of the sum of the zero-sequence current of the I circuit of the parallel double-circuit line on the same pole and the zero-sequence current of the II circuit of the parallel double-circuit line on the same pole I The phase angle of the voltage at the ground fault point of the return line, calculate the voltage at the setting range of the I return line protection, and then judge that the voltage at the setting range of the I return line protection is ahead of the phase angle of the voltage at the ground fault point of the I return line at [-165°, 15° ] within the range, if it is established, the protection device will send out an action trip signal. The method of the present invention only uses the electrical quantity of the single-ended single-circuit line, and does not need to introduce the electrical quantity of another circuit. The operating performance is not affected by the operation mode of the power system, and it has strong adaptability when the operation mode of the power system changes greatly. ability. The method of the invention takes into account the influence of the zero-sequence mutual inductance between lines and the voltage of the grounding fault point, and eliminates the influence of the zero-sequence mutual inductance, transition resistance and load current on the double-circuit line with the same pole and the double-circuit line with the same name. The impact of the protection range is stable and reliable.

附图说明Description of drawings

下面结合附图和实施例对本发明作进一步说明。The present invention will be further described below in conjunction with drawings and embodiments.

图1为应用本发明的同杆并架双回线路输电系统示意图。Fig. 1 is a schematic diagram of a double-circuit line power transmission system on the same pole parallel to the rack applying the present invention.

具体实施方式detailed description

请查阅图1，图1中PT为电压互感器2、CT为电流互感器3。保护装置1对输电线路保护安装处的电压互感器2所获得的电压波形和电流互感器3所获得的电流波形分别进行采样得到电压、电流瞬时值。Please refer to FIG. 1 . In FIG. 1 , PT is the voltage transformer 2 and CT is the current transformer 3 . The protection device 1 samples the voltage waveform obtained by the voltage transformer 2 and the current waveform obtained by the current transformer 3 respectively at the place where the transmission line protection is installed to obtain instantaneous values of voltage and current.

保护装置1对采样得到的电压、电流瞬时值利用傅里叶算法计算同杆并架双回线路I回线路保护安装处的故障相电压故障相电流和零序电流其中，φ=I回线路A相、I回线路B相、I回线路C相。The protection device 1 uses the Fourier algorithm to calculate the instantaneous value of the voltage and current obtained by sampling, and calculates the fault phase voltage at the installation place of the I-circuit line protection of the parallel double-circuit line on the same pole fault phase current and zero sequence current Among them, φ=phase A of the I-circuit line, phase B of the I-circuit line, and phase C of the I-circuit line.

进而，保护装置1计算同杆并架双回线路II回线路的零序电流相角α=r₁+r₂-π-β；Furthermore, the protection device 1 calculates the zero-sequence current phase angle α=r₁ +r₂ -π-β of the line II of the double-circuit line paralleled on the same pole;

其中：$r_1={\sin }^{-1}\left(\frac{a_3{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),r_2={\sin }^{-1}\left(\frac{a_1{b}_2-a_2{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),a_1=\mathrm{Re}\left(\frac{{\stackrel{\·}{U}}_{\mathrm{I\φ}}}{Z_{I1}}\right),b_1=\mathrm{Im}\left(\frac{{\stackrel{\·}{U}}_{\mathrm{I\φ}}}{Z_{I1}}\right),$$a_2=\mathrm{Re}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),b_2=\mathrm{Im}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),a_3=b_3=\left|\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right|,\mathrm{\β}=\mathrm{Arg}\left(\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right);$Z_m为同杆并架双回线路I回线路与同杆并架双回线路II回线路之间的零序互感；Z_I0为同杆并架双回线路I回线路的零序阻抗；Z_I1为同杆并架双回线路I回线路的正序阻抗；φ=I回线路A相、I回线路B相、I回线路C相；in:$r_1={\sin }^{-1}\left(\frac{a_3{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),r_2={\sin }^{-1}\left(\frac{a_1{b}_2-a_2{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),a_1=\mathrm{Re}\left(\frac{{\stackrel{\·}{u}}_{\mathrm{I\φ}}}{Z_{I1}}\right),b_1=\mathrm{Im}\left(\frac{{\stackrel{\&Center\; Dot;}{u}}_{\mathrm{I\φ}}}{Z_{I1}}\right),$$a_2=\mathrm{Re}({\stackrel{\&Center\; Dot;}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),b_2=\mathrm{Im}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\&Center\; Dot;}{I}}_{I0}),a_3=b_3=\left|\frac{Z_m}{3Z_{I1}}{\stackrel{\&Center\; Dot;}{I}}_{I0}\right|,\mathrm{\β}=\mathrm{Arg}\left(\frac{Z_m}{3Z_{I1}}{\stackrel{\&Center\; Dot;}{I}}_{I0}\right);$ Z_m is the zero-sequence mutual inductance between the I-circuit line of the parallel double-circuit line on the same pole and the II-circuit line of the parallel double-circuit line on the same pole; Z_I0 is the zero-sequence impedance of the I-circuit line of the parallel double-circuit line on the same pole; Z_I1 is the positive sequence impedance of the I-circuit line of the parallel double-circuit line on the same pole; φ=I-circuit line A phase, I-circuit line B phase, and I-circuit line C phase;

进而，保护装置1计算同杆并架双回线路II回线路的零序电流${\stackrel{\·}{I}}_{\mathrm{II}0}={\stackrel{\·}{I}}_{I0}(\cos \mathrm{\α}+j\sin \mathrm{\α});$Furthermore, the protection device 1 calculates the zero-sequence current of the line II of the parallel double-circuit line on the same pole${\stackrel{\&Center\; Dot;}{I}}_{\mathrm{II}0}={\stackrel{\&Center\; Dot;}{I}}_{I0}(\cos \mathrm{\α}+j\sin \mathrm{\α});$

最后，保护装置1计算${\stackrel{\·}{U}}_{\mathrm{I\φ}}-Z_{I1}\frac{l_{\mathrm{set}}}{l}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}+\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{\mathrm{II}0})$领先的相角落在[-165°，15°]范围内是否成立，若成立，则保护装置发出动作跳闸信号；其中：l_set为同杆并架双回线路I回线路保护整定范围，l为同杆并架双回线路I回线路长度。Finally, protector 1 calculates${\stackrel{\·}{u}}_{\mathrm{I\φ}}-Z_{I1}\frac{l_{\mathrm{set}}}{l}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}+\frac{Z_m}{3Z_{I1}}{\stackrel{\&Center\; Dot;}{I}}_{\mathrm{II}0})$ take the lead Whether the phase angle of the phase angle is established within the range of [-165°, 15°], if established, the protection device will send out an action trip signal; where: l_set is the setting range of the I-circuit line protection of the double-circuit line on the same pole, and l is the same The length of the I-circuit line of the pole-parallel double-circuit line.

本发明方法首先测量同杆并架双回线路I回线路保护安装处的故障相电压、故障相电流和零序电流，计算同杆并架双回线路II回线路的零序电流相角，计算同杆并架双回线路II回线路的零序电流，利用同杆并架双回线路I回线路零序电流和同杆并架双回线路II回线路零序电流之和的相角估算I回线路接地故障点电压相角，然后判断I回线路保护整定范围处电压领先I回线路接地故障点电压的相角落在[-165°15°]范围内是否成立，若成立，则保护装置发出动作跳闸信号。本发明方法只用到单端单回线路电气量，不需要引入另一回线路电气量，动作性能不受电力系统运行方式的影响，在电力系统运行方式发生较大改变时具有很强的适应能力。本发明方法计及线间零序互感和接地故障点电压的影响，消除了线间零序互感、过渡电阻和负荷电流对同杆并架双回线路非同名相跨线接地电抗距离继电器动作性能的影响，保护范围稳定可靠。The method of the present invention firstly measures the fault phase voltage, fault phase current and zero-sequence current at the protection installation place of the I-circuit line of the parallel double-circuit line on the same pole, calculates the phase angle of the zero-sequence current of the II-circuit line of the double-circuit line parallel on the same pole, and calculates The zero-sequence current of the double-circuit line II of the double-circuit line paralleled on the same pole is estimated by using the phase angle of the sum of the zero-sequence current of the I line of the parallel double-circuit line on the same pole and the zero-sequence current of the second line of the double-circuit line paralleled on the same pole I The voltage phase angle of the ground fault point of the return line, and then judge whether the phase angle at which the voltage at the protection setting range of the I return line is ahead of the voltage of the ground fault point of the I return line is within the range of [-165°15°]. If it is established, the protection device will send out Action trip signal. The method of the present invention only uses the electrical quantity of the single-ended single-circuit line, and does not need to introduce the electrical quantity of another circuit. The operation performance is not affected by the operation mode of the power system, and it has strong adaptability when the operation mode of the power system changes greatly. ability. The method of the invention takes into account the influence of the zero-sequence mutual inductance between lines and the voltage of the grounding fault point, and eliminates the influence of the zero-sequence mutual inductance, transition resistance and load current on the same-pole parallel double-circuit line and the non-identical phase cross-line grounding reactance distance relay action performance The impact of the protection range is stable and reliable.

以上所述，仅为本发明较佳实施例而已，故不能依此限定本发明实施的范围，即依本发明专利范围及说明书内容所作的等效变化与修饰，皆应仍属本发明涵盖的范围内。The above is only a preferred embodiment of the present invention, so the scope of the present invention cannot be limited accordingly, that is, the equivalent changes and modifications made according to the patent scope of the present invention and the content of the specification should still be covered by the present invention within range.

Claims (1)

1. the non-same famous prime minister's cross-line earthing reactance distance protecting method of double-circuit line, it is characterised in that include following sequential steps:

Step 1, protector measuring double-circuit lines on the same pole road I returns the faulted phase voltage of route protection installation placeFault phaseElectric currentAnd zero-sequence currentWherein: φ is I loop line road A phase, I loop line road B phase, I loop line road C phase；

Step 2, protection device calculates the zero-sequence current phase angle α=r on II loop line road, double-circuit lines on the same pole road₁+r₂-π-β；

Wherein:$r_1={\sin }^{-1}\left(\frac{a_3{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),r_2={\sin }^{-1}\left(\frac{a_1{b}_2-a_2{b}_1}{\sqrt{{\left(a_3{b}_1\right)}^2+{\left(a_1{b}_3\right)}^2}}\right),a_1=\mathrm{Re}\left(\frac{{\stackrel{\·}{U}}_{\mathrm{I\φ}}}{Z_{I1}}\right),b_1=\mathrm{Im}\left(\frac{{\stackrel{\·}{U}}_{\mathrm{I\φ}}}{Z_{I1}}\right),$$a_2=\mathrm{Re}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),b_2=\mathrm{Im}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}),a_3=b_3=\left|\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right|,\mathrm{\β}=\mathrm{Arg}\left(\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{I0}\right);$Z_mForZero-sequence mutual inductance between I loop line road, double-circuit lines on the same pole road and II loop line road, double-circuit lines on the same pole road；Z_I0For parallel lines on same towerThe zero sequence impedance on double-circuit line I loop line road；Z_I1Positive sequence impedance for I loop line road, double-circuit lines on the same pole road；φ=I loop line road APhase, I loop line road B phase, I loop line road C phase；

Step 3, protection device calculates the zero-sequence current on II loop line road, double-circuit lines on the same pole road${\stackrel{\·}{I}}_{\mathrm{II}0}={\stackrel{\·}{I}}_{I0}(\cos \mathrm{\α}+j\sin \mathrm{\α});$

Step 4, protection device calculates${\stackrel{\·}{U}}_{\mathrm{I\φ}}-Z_{I1}\frac{l_{\mathrm{set}}}{l}({\stackrel{\·}{I}}_{\mathrm{I\φ}}+\frac{Z_{I0}-Z_{I1}}{Z_{I1}}{\stackrel{\·}{I}}_{I0}+\frac{Z_m}{3Z_{I1}}{\stackrel{\·}{I}}_{\mathrm{II}0})$LeadingPhase corner [-165 °, 15 °] in the range of whether set up, if setting up, then protection device sends action trip signal；Wherein: l_setFor parallel lines on same towerDouble-circuit line I loop line road protection seting scope, l is that double-circuit lines on the same pole road I returns line length.