CN105223449A - 一种非对称输电线路参数在线测量方法 - Google Patents

Abstract

本发明提供了一种非对称输电线路参数在线测量方法，其包括以下步骤：步骤一，在被测线路的两端安装同步测量装置，通过该同步测量装置同步测量被测线路两端的电压和电流信号，并进行m(m≥2)次不同的测量；步骤二，对步骤一获得的m组被测线路两端的电压和电流信号进行傅里叶变换，计算m组被测线路两端的工频电压相量和电流相量，包括第1组至第m组被测线路的首端电压、电流相量；末端的电压相量、电流相量；步骤三，计算被测线路A、B、C相的自导纳Y_a、Y_b、Y_c及各相的互导纳Y_ab、Y_bc、Y_ac；步骤四，计算被测线路A、B、C相的自阻抗Z_a、Z_b、Z_c，各相的互阻抗Z_ab、Z_bc。本发明可以计算出每相的自阻抗、自导纳参数以及各相之间的互阻抗、互导纳参数，正序、零序阻抗和导纳参数，以及各序间的序耦合阻抗、导纳参数。

Description

一种非对称输电线路参数在线测量方法

技术领域

本发明涉及一种非对称输电线路参数在线测量方法。

背景技术

输电线路参数的在线测量法能够反映实际工作情况下线路参数随运行方式、导线温度等条件的变化，且不会影响系统的正常运行和潮流的优化分布，是输电线路参数测量的未来发展方向。目前输电线路参数在线测量方法的研究主要有两个方向：一是针对多回路输电线路零序互感参数的测量，提出了增量法、积分法和微分法等方法。然而，这种方法只能测量输电线路的零序阻抗参数，无法测量线路的导纳参数及正序参数。此外，电力系统在正常运行时，线路中的零序分量很小，难以满足测量要求。二是基于传输线分布参数理论，通过均匀传输线的特性阻抗和传播常数求解输电线路的分布参数。该方法计算单回三相对称线路的分布参数比较准确，但对于非对称输电线路，由于需要求解的输电线路参数个数多于一次测量所得到的电气方程个数，因而无法计算非对称输电线路的分布参数。

中国专利申请201510176601.2公开了一种三相非对称输电线路阻抗参数在线测量方法，包括：步骤一，在受测三相非对称输电线路的两端安装同步测量装置，并在n个时刻通过该同步测量装置同步测量受测三相非对称输电线路两端的三相电压和电流信号；步骤二，对步骤一获得的n组三相电压和电流信号进行快速傅里叶变换，以计算得到n组受测三相非对称输电线路两端的三相电压工频相量和受测三相非对称输电线路首端的三相电流工频相量；步骤三，计算出每一相的自阻抗参数和各相之间的互阻抗参数；步骤四，计算出正序、负序、零序阻抗参数以及各序间的耦合阻抗参数。该技术方案的不足在于，仅仅实现三相非对称输电线路阻抗参数的在线测量，而对于重要性不亚于阻抗参数的导纳参数，则没有涉及。

发明内容

针对现有技术的不足，本发明的目的是提供一种非对称输电线路参数在线测量方法。

为了实现上述目的，本发明提供了一种非对称输电线路参数在线测量方法，其适用于长度小于150km、电压等级在500kV以下的输电线路参数在线测量，其包括以下步骤：

步骤一，在被测线路的两端安装同步测量装置，通过该同步测量装置同步测量被测线路两端的电压和电流信号，并进行m(m≥2)次不同的测量；

步骤二，对步骤一获得的m组被测线路两端的电压和电流信号进行傅里叶变换，计算m组被测线路两端的工频电压相量和电流相量，包括第1组至第m组被测线路的首端电压、电流相量 末端的电压相量、电流相量

步骤三，按以下公式计算被测线路A、B、C相的自导纳Y_a、Y_b、Y_c及各相的互导纳Y_ab、Y_bc、Y_ac：

$Y^{\′}={\left({{\stackrel{\·}{U}}_p}^T{\stackrel{\·}{U}}_p\right)}^{-1}\left({{\stackrel{\·}{U}}_p}^T\stackrel{\·}{I}\right)$

步骤四，按以下公式计算被测线路A、B、C相的自阻抗Z_a、Z_b、Z_c，各相的互阻抗Z_ab、Z_bc，Z_ac：

$Z^{\′}={\left({\stackrel{\·}{I}}^{\′T}{\stackrel{\·}{I}}^{\′}\right)}^{-1}\left({\stackrel{\·}{I}}^{\′T}\stackrel{\·}{U}\right)$

本发明的输电线路参数计算方法中，先求导纳参数、再求阻抗参数。

根据本发明另一具体实施方式，步骤三中，

$\stackrel{\·}{I}={\left[\begin{array}{ccccccc}{\stackrel{\·}{I}}_a^{\left(1\right)}& {\stackrel{\·}{I}}_b^{\left(1\right)}& {\stackrel{\·}{I}}_c^{\left(1\right)}& ...& {\stackrel{\·}{I}}_a^{\left(m\right)}& {\stackrel{\·}{I}}_b^{\left(m\right)}& {\stackrel{\·}{I}}_c^{\left(m\right)}\end{array}\right]}^T$

Y′＝[Y_aY_bY_cY_abY_bcY_ac]^T

${\stackrel{\·}{U}}_p=\left[\begin{array}{cccccc}{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{U}}_{ap}^{\left(m\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}\end{array}\right]$

${\stackrel{\·}{I}}_a^{\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{I}}_{a2}^{\left(i\right)},{\stackrel{\·}{I}}_b^{\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{I}}_{b2}^{\left(i\right)},{\stackrel{\·}{I}}_c^{\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{I}}_{c2}^{\left(i\right)},{\stackrel{\·}{U}}_{ap}^{\left(i\right)}=\left({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a2}^{\left(i\right)}\right)/2,$ ${\stackrel{\·}{U}}_{bp}^{\left(i\right)}=({\stackrel{\·}{U}}_{b1}^{\left(i\right)}+{\stackrel{\·}{U}}_{b2}^{\left(i\right)})/2,{\stackrel{\·}{U}}_{cp}^{\left(i\right)}=({\stackrel{\·}{U}}_{c1}^{\left(i\right)}+{\stackrel{\·}{U}}_{c2}^{\left(i\right)})/2,$ i＝1,2,…,m，m≥2

根据本发明另一具体实施方式，步骤四中，

$\stackrel{\·}{U}={\left[\begin{array}{ccccccc}{\stackrel{\·}{U}}_a^{\left(1\right)}& {\stackrel{\·}{U}}_b^{\left(1\right)}& {\stackrel{\·}{U}}_c^{\left(1\right)}& ...& {\stackrel{\·}{U}}_a^{\left(m\right)}& {\stackrel{\·}{U}}_b^{\left(m\right)}& {\stackrel{\·}{U}}_c^{\left(m\right)}\end{array}\right]}^T$

Z′＝[Z_aZ_bZ_cZ_acZ_bcZ_ac]^T

${\stackrel{\·}{I}}^{\′}=\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_a^{\′\left(1\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(1\right)}& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& {\stackrel{\·}{I}}_a^{\′\left(1\right)}\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{I}}_a^{\′\left(m\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(m\right)}& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& {\stackrel{\·}{I}}_a^{\′\left(m\right)}\end{array}\right]$

${\stackrel{\·}{U}}_a^{\left(i\right)}={\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a2}^{\left(i\right)},{\stackrel{\·}{U}}_b^{\left(i\right)}={\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b2}^{\left(i\right)},{\stackrel{\·}{U}}_c^{\left(i\right)}={\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c2}^{\left(i\right)}$

${\stackrel{\·}{I}}_a^{\′\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)}{Y}_a/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{ac}/2$

${\stackrel{\·}{I}}_b^{\′\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)}{Y}_b/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{bc}/2$

${\stackrel{\·}{I}}_c^{\′\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)}{Y}_c/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ac}/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{bc}/2$

i＝1,2，…，m

根据本发明另一具体实施方式，进一步包括如下步骤：

步骤五，按以下公式计算被测线路每回的正序阻抗Z₁、负序阻抗Z₂、零序阻抗Z₀、正序与负序间的序耦合阻抗Z₁₂、零序与正序间的序耦合阻抗Z₀₁、零序与负序间的序耦合阻抗Z₀₂：

$Z_{012}=A^{-1}{Z}_{abc}A=\left[\begin{array}{ccc}{Z}_0& Z_{01}& Z_{02}\\ Z_{01}& Z_1& Z_{12}\\ Z_{02}& Z_{12}& Z_2\end{array}\right]$

根据本发明另一具体实施方式，步骤五中，

$Z_{abc}=\left[\begin{array}{ccc}{Z}_a& Z_{ab}& Z_{ac}\\ Z_{ab}& Z_b& Z_{bc}\\ Z_{ac}& Z_{bc}& Z_c\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{\α}}^2& \mathrm{\α}\\ 1& \mathrm{\α}& {\mathrm{\α}}^2\end{array}\right]$

根据本发明另一具体实施方式，进一步包括如下步骤：

步骤六，按以下公式计算被测线路每回的正序导纳Y₁、负序导纳Y₂、零序导纳Y₀、正序与负序间的序耦合导纳Y₁₂、零序与正序间的序耦合导纳Y₀₁、零序与负序间的序耦合导纳Y₀₂：

$Y_{012}=A^{-1}{Y}_{abc}A=\left[\begin{array}{ccc}{Y}_0& Y_{01}& Y_{02}\\ Y_{01}& Y_1& Y_{12}\\ Y_{02}& Y_{12}& Y_2\end{array}\right]$

根据本发明另一具体实施方式，步骤六中，

$Y_{abc}=\left[\begin{array}{ccc}{Y}_a& Y_{ab}& Y_{ac}\\ Y_{ab}& Y_{bb}& Y_{bc}\\ Y_{ac}& Y_{bc}& Y_{cc}\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{\α}}^2& \mathrm{\α}\\ 1& \mathrm{\α}& {\mathrm{\α}}^2\end{array}\right]$

根据本发明另一具体实施方式，同步测量装置的采集精度在16位二进制数以上、采样频率在50kHz以上。

与现有技术相比，本发明具有以下有益效果：

本发明仅需在被测线路的两端安装同步测量装置，在测得m(m≥2)组线路两端的电压和电流信号之后，由以上计算公式不仅可以计算出被测线路每相的自阻抗、自导纳参数以及各相之间的互阻抗、互导纳参数，还可以通过矩阵变换计算出被测线路的正序、零序阻抗和导纳参数，以及各序间的序耦合阻抗、导纳参数；另外，经仿真验证，本发明测得的三相非对称输电线路参数具有很高的精度。

附图说明

图1为实施例1中，三相非对称输电线路模型；

图2为实施例1中，输电线路参数测试模型。

具体实施方式

实施例1

本实施例的三相非对称输电线路参数在线测量方法，适用于长度小于150km、电压等级在500kV及以下的输电线路，包括以下步骤：

步骤一，在被测的同塔多回输电线路两端安装同步测量装置，并在m个不同的时刻(在m个不同时刻的线路电流差异要足够大，以保证所建立的方程线性独立)通过该同步测量装置同步测量被测线路两端的电压和电流信号，以获得m组不同测量时刻的测量数据，其中，m≥2。

上述的同步测量装置可选用现有技术中基于GPS同步时钟实现线路两端电流、电压信号同步测量的测量装置，其采集精度应在16位二进制数以上，采样频率应在50kHz以上。

步骤二，对步骤一获得的m组被测线路两端的电压和电流信号进行傅里叶变换，计算m组被测线路两端的工频电压相量和电流相量，包括第1组至第m组被测线路的首端电压、电流相量 末端的电压相量、电流相量

步骤三，按以下公式(1)计算被测线路A、B、C相的自导纳Y_a、Y_b、Y_c及各相的互导纳Y_ab、Y_bc、Y_ac。

$Y^{\′}={\left({{\stackrel{\·}{U}}_p}^T{\stackrel{\·}{U}}_p\right)}^{-1}\left({{\stackrel{\·}{U}}_p}^T\stackrel{\·}{I}\right)---\left(1\right)$

式中 $\stackrel{\·}{I}={\left[\begin{array}{ccccccc}{\stackrel{\·}{I}}_a^{\left(1\right)}& {\stackrel{\·}{I}}_b^{\left(1\right)}& {\stackrel{\·}{I}}_c^{\left(1\right)}& ...& {\stackrel{\·}{I}}_a^{\left(m\right)}& {\stackrel{\·}{I}}_b^{\left(m\right)}& {\stackrel{\·}{I}}_c^{\left(m\right)}\end{array}\right]}^T$

Y′＝[Y_aY_bY_cY_abY_bcY_ac]^T

${\stackrel{\·}{U}}_p=\left[\begin{array}{cccccc}{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{U}}_{ap}^{\left(m\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}\end{array}\right]$

${\stackrel{\·}{I}}_a^{\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{I}}_{a2}^{\left(i\right)},{\stackrel{\·}{I}}_b^{\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{I}}_{b2}^{\left(i\right)},{\stackrel{\·}{I}}_c^{\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{I}}_{c2}^{\left(i\right)},{\stackrel{\·}{U}}_{ap}^{\left(i\right)}=({\stackrel{\·}{U}}_{a1}^{\left(i\right)}+{\stackrel{\·}{U}}_{a2}^{\left(i\right)})/2,{\stackrel{\·}{U}}_{bp}^{\left(i\right)}=({\stackrel{\·}{U}}_{b1}^{\left(i\right)}+{\stackrel{\·}{U}}_{b2}^{\left(i\right)})/2,$ ${\stackrel{\·}{U}}_{cp}^{\left(i\right)}=({\stackrel{\·}{U}}_{c1}^{\left(i\right)}+{\stackrel{\·}{U}}_{c2}^{\left(i\right)})/2,$ i＝1,2,…,m

步骤四，按以下公式(2)计算被测线路A、B、C相的自阻抗Z_a、Z_b、Z_c，各相的互阻抗Z_ab、Z_bc，Z_ac。

$Z^{\′}={\left({\stackrel{\·}{I}}^{\′T}{\stackrel{\·}{I}}^{\′}\right)}^{-1}\left({\stackrel{\·}{I}}^{\′T}\stackrel{\·}{U}\right)---\left(2\right)$

式中

$\stackrel{\·}{U}={\left[\begin{array}{ccccccc}{\stackrel{\·}{U}}_a^{\left(1\right)}& {\stackrel{\·}{U}}_b^{\left(1\right)}& {\stackrel{\·}{U}}_c^{\left(1\right)}& ...& {\stackrel{\·}{U}}_a^{\left(m\right)}& {\stackrel{\·}{U}}_b^{\left(m\right)}& {\stackrel{\·}{U}}_c^{\left(m\right)}\end{array}\right]}^T$

Z′＝[Z_aZ_bZ_cZ_acZ_bcZ_ac]^T

${\stackrel{\·}{I}}^{\′}=\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_a^{\′\left(1\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(1\right)}& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& {\stackrel{\·}{I}}_a^{\′\left(1\right)}\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{I}}_a^{\′\left(m\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(m\right)}& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& {\stackrel{\·}{I}}_a^{\′\left(m\right)}\end{array}\right]$

${\stackrel{\·}{U}}_a^{\left(i\right)}={\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a2}^{\left(i\right)},{\stackrel{\·}{U}}_b^{\left(i\right)}={\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b2}^{\left(i\right)},{\stackrel{\·}{U}}_c^{\left(i\right)}={\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c2}^{\left(i\right)}$

${\stackrel{\·}{I}}_a^{\′\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)}{Y}_a/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{ac}/2$

${\stackrel{\·}{I}}_b^{\′\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)}{Y}_b/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{bc}/2$

${\stackrel{\·}{I}}_c^{\′\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)}{Y}_c/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ac}/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{bc}/2$

i＝1,2，…，m

步骤五，按公式(3)计算被测线路的正序阻抗Z₁、负序阻抗Z₂、零序阻抗Z₀、正序与负序间的序耦合阻抗Z₁₂、零序与正序间的序耦合阻抗Z₀₁、零序与负序间的序耦合阻抗Z₀₂。

步骤六，按公式(4)计算被测线路的正序导纳Y₁、负序导纳Y₂、零序阻抗Y₀、正序与负序的耦合导纳Y₁₂、零序与正序的耦合导纳Y₀₁、零序与负序的耦合导纳Y₀₂。

输电线路的序阻抗参数为：

$Z_{012}=A^{-1}{Z}_{abc}A=\left[\begin{array}{ccc}{Z}_0& Z_{01}& Z_{02}\\ Z_{01}& Z_1& Z_{12}\\ Z_{02}& Z_{12}& Z_2\end{array}\right]---\left(3\right)$

输电线路的序导纳参数为：

$Y_{012}=A^{-1}{Y}_{abc}A=\left[\begin{array}{ccc}{Y}_0& Y_{01}& Y_{02}\\ Y_{01}& Y_1& Y_{12}\\ Y_{02}& Y_{12}& Y_2\end{array}\right]---\left(4\right)$

式中

$Z_{abc}=\left[\begin{array}{ccc}{Z}_a& Z_{ab}& Z_{ac}\\ Z_{ab}& Z_b& Z_{bc}\\ Z_{ac}& Z_{bc}& Z_c\end{array}\right],Y_{abc}=\left[\begin{array}{ccc}{Y}_a& Y_{ab}& Y_{ac}\\ Y_{ab}& Y_{bb}& Y_{bc}\\ Y_{ac}& Y_{bc}& Y_{cc}\end{array}\right]$

上述公式(1)和公式(2)的推导过程如下：

I建立三相非对称输电线路模型

对于长度小于150km，电压等级在500kV及以下的输电线路，输电线路可以用集中参数π型模型来近似替代，建立三相非对称输电线路模型如图1所示。

图1中，Z_a、Z_b、Z_c分别为每相自阻抗，Z_ab、Z_ac、Z_bc分别为两相之间的互阻抗，Y_a、Y_b、Y_c分别为每相对地导纳，Y_ab、Y_ac、Y_bc分别为两相之间的互导纳。

列写图1所示电路的方程：

$\left\{\begin{array}{c}{\stackrel{\·}{U}}_{a1}={\stackrel{\·}{I}}_a^{\′}{Z}_a+{\stackrel{\·}{I}}_b^{\′}{Z}_{ab}+{\stackrel{\·}{I}}_c^{\′}{Z}_{ac}+{\stackrel{\·}{U}}_{a2}\\ {\stackrel{\·}{U}}_{b1}={\stackrel{\·}{I}}_a^{\′}{Z}_{ab}+{\stackrel{\·}{I}}_b^{\′}{Z}_b+{\stackrel{\·}{I}}_c^{\′}{Z}_{bc}+{\stackrel{\·}{U}}_{b2}\\ {\stackrel{\·}{U}}_{c1}={\stackrel{\·}{I}}_a^{\′}{Z}_{ac}+{\stackrel{\·}{I}}_b^{\′}{Z}_{bc}+{\stackrel{\·}{I}}_c^{\′}{Z}_c+{\stackrel{\·}{U}}_{c2}\\ {\stackrel{\·}{I}}_{a1}-{\stackrel{\·}{I}}_{a2}=\left({\stackrel{\·}{U}}_{a1}+{\stackrel{\·}{U}}_{a2}\right)\frac{Y_a}{2}+\left({\stackrel{\·}{U}}_{a1}+{\stackrel{\·}{U}}_{a2}-{\stackrel{\·}{U}}_{b1}-{\stackrel{\·}{U}}_{b2}\right)\frac{Y_{ab}}{2}+\left({\stackrel{\·}{U}}_{a1}+{\stackrel{\·}{U}}_{a2}-{\stackrel{\·}{U}}_{c1}-{\stackrel{\·}{U}}_{c2}\right)\frac{Y_{ac}}{2}\\ {\stackrel{\·}{I}}_{b1}-{\stackrel{\·}{I}}_{b2}=\left({\stackrel{\·}{U}}_{b1}+{\stackrel{\·}{U}}_{b2}\right)\frac{Y_b}{2}+\left({\stackrel{\·}{U}}_{b1}+{\stackrel{\·}{U}}_{b2}-{\stackrel{\·}{U}}_{a1}-{\stackrel{\·}{U}}_{a2}\right)\frac{Y_{ab}}{2}+\left({\stackrel{\·}{U}}_{b1}+{\stackrel{\·}{U}}_{b2}-{\stackrel{\·}{U}}_{c1}-{\stackrel{\·}{U}}_{c2}\right)\frac{Y_{bc}}{2}\\ {\stackrel{\·}{I}}_{c1}-{\stackrel{\·}{I}}_{c2}=\left({\stackrel{\·}{U}}_{c1}+{\stackrel{\·}{U}}_{c2}\right)\frac{Y_c}{2}+\left({\stackrel{\·}{U}}_{c1}+{\stackrel{\·}{U}}_{c2}-{\stackrel{\·}{U}}_{b1}-{\stackrel{\·}{U}}_{b2}\right)\frac{Y_{bc}}{2}+\left({\stackrel{\·}{U}}_{c1}+{\stackrel{\·}{U}}_{c2}-{\stackrel{\·}{U}}_{a1}-{\stackrel{\·}{U}}_{a2}\right)\frac{Y_{ac}}{2}\end{array}\right.---\left(5\right)$

式中 ${\stackrel{\·}{I}}_a^{\′}={\stackrel{\·}{I}}_{a1}-{\stackrel{\·}{U}}_{a1}{Y}_a/2-({\stackrel{\·}{U}}_{a1}-{\stackrel{\·}{U}}_{b1})Y_{\mathrm{ab}}/2-({\stackrel{\·}{U}}_{a1}-{\stackrel{\·}{U}}_{c1})Y_{\mathrm{ac}}/2$

${\stackrel{\·}{I}}_b^{\′}={\stackrel{\·}{I}}_{b1}-{\stackrel{\·}{U}}_{b1}{Y}_b/2-({\stackrel{\·}{U}}_{b1}-{\stackrel{\·}{U}}_{a1})Y_{\mathrm{ab}}/2-({\stackrel{\·}{U}}_{b1}-{\stackrel{\·}{U}}_{c1})Y_{\mathrm{bc}}/2$

${\stackrel{\·}{I}}_c^{\′}={\stackrel{\·}{I}}_{c1}-{\stackrel{\·}{U}}_{c1}{Y}_c/2-({\stackrel{\·}{U}}_{c1}-{\stackrel{\·}{U}}_{a1})Y_{\mathrm{ac}}/2-({\stackrel{\·}{U}}_{c1}-{\stackrel{\·}{U}}_{b1})Y_{\mathrm{bc}}/2$

可见，(5)式中共有12个未知数，但仅有6个独立的方程，显然是一个无解的欠定方程。但(5)式中的后3个方程仅与导纳及线路两端的电压和电流有关，而前3个方程不仅与阻抗、还与导纳及线路两端的电压、电流有关。为此，本发明提出了先求导纳参数、再求阻抗参数的输电线路参数计算方法。

II导纳参数的求解

取(5)式的后3个方程，即

$\left\{\begin{array}{c}{\stackrel{\·}{I}}_a=Y_a{\stackrel{\·}{U}}_{ap}+Y_{ab}\left({\stackrel{\·}{U}}_{ap}-{\stackrel{\·}{U}}_{bp}\right)+Y_{ac}\left({\stackrel{\·}{U}}_{ap}-{\stackrel{\·}{U}}_{cp}\right)\\ {\stackrel{\·}{I}}_b=Y_b{\stackrel{\·}{U}}_{bp}+Y_{ab}\left({\stackrel{\·}{U}}_{bp}-{\stackrel{\·}{U}}_{ap}\right)+Y_{bc}\left({\stackrel{\·}{U}}_{bp}-{\stackrel{\·}{U}}_{cp}\right)\\ {\stackrel{\·}{I}}_c=Y_c{\stackrel{\·}{U}}_{cp}+Y_{ac}\left({\stackrel{\·}{U}}_{cp}-{\stackrel{\·}{U}}_{ap}\right)+Y_{bc}\left({\stackrel{\·}{U}}_{bp}-{\stackrel{\·}{U}}_{cp}\right)\end{array}\right.---\left(6\right)$

式中， ${\stackrel{\·}{I}}_a={\stackrel{\·}{I}}_{a1}-{\stackrel{\·}{I}}_{a2},{\stackrel{\·}{I}}_b={\stackrel{\·}{I}}_{b1}-{\stackrel{\·}{I}}_{b2},{\stackrel{\·}{I}}_c={\stackrel{\·}{I}}_{c1}-{\stackrel{\·}{I}}_{c2},{\stackrel{\·}{U}}_{\mathrm{ap}}=({\stackrel{\·}{U}}_{a1}+{\stackrel{\·}{U}}_{a2})/2,{\stackrel{\·}{U}}_{\mathrm{bp}}=({\stackrel{\·}{U}}_{b1}+{\stackrel{\·}{U}}_{b2})/2,$ ${\stackrel{\·}{U}}_{\mathrm{cp}}=({\stackrel{\·}{U}}_{c1}+{\stackrel{\·}{U}}_{c2})/2$

显然(6)式共有6个未知数，3个独立的方程，也是一个欠定方程，无法求得线路的导纳参数。为此，本发明提出了对线路进行多次测量，形成超定方程求解线路导纳参数的方法。在保证任意2次测量量之间具有一定差别(保证所建立的方程独立)的情况下，建立超定方程：

$\left\{\begin{array}{c}{\stackrel{\·}{I}}_a^{\left(1\right)}=Y_a{\stackrel{\·}{U}}_{ap}^{\left(1\right)}+Y_{ab}\left({\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}\right)+Y_{ac}\left({\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}\right)\\ {\stackrel{\·}{I}}_b^{\left(1\right)}=Y_b{\stackrel{\·}{U}}_{bp}^{\left(1\right)}+Y_{ab}\left({\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}\right)+Y_{bc}\left({\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}\right)\\ {\stackrel{\·}{I}}_c^{\left(1\right)}=Y_c{\stackrel{\·}{U}}_{cp}^{\left(1\right)}+Y_{ac}\left({\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}\right)+Y_{bc}\left({\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}\right)\\ \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{I}}_a^{\left(m\right)}=Y_a{\stackrel{\·}{U}}_{ap}^{\left(m\right)}+Y_{ab}\left({\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}\right)+Y_{ac}\left({\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}\right)\\ {\stackrel{\·}{I}}_b^{\left(m\right)}=Y_b{\stackrel{\·}{U}}_{bp}^{\left(m\right)}+Y_{ab}\left({\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}\right)+Y_{bc}\left({\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}\right)\\ {\stackrel{\·}{I}}_c^{\left(m\right)}=Y_c{\stackrel{\·}{U}}_{cp}^{\left(m\right)}+Y_{ac}\left({\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}\right)+Y_{bc}\left({\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}\right)\end{array}\right.---\left(7\right)$

式中， ${\stackrel{\·}{I}}_a^{\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{I}}_{a2}^{\left(i\right)},{\stackrel{\·}{I}}_b^{\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{I}}_{b2}^{\left(i\right)},{\stackrel{\·}{I}}_c^{\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{I}}_{c2}^{\left(i\right)},{\stackrel{\·}{U}}_{ap}^{\left(i\right)}=({\stackrel{\·}{U}}_{a1}^{\left(i\right)}+{\stackrel{\·}{U}}_{a2}^{\left(i\right)})/2,$ ${\stackrel{\·}{U}}_{bp}^{\left(i\right)}=({\stackrel{\·}{U}}_{b1}^{\left(i\right)}+{\stackrel{\·}{U}}_{b2}^{\left(i\right)})/2,{\stackrel{\·}{U}}_{cp}^{\left(i\right)}=({\stackrel{\·}{U}}_{c1}^{\left(i\right)}+{\stackrel{\·}{U}}_{c2}^{\left(i\right)})/2$

分别为线路两端电压、电流相量的第i(i＝1,2，…，m)次测量值。

将(7)式写成矩阵形式：

$\stackrel{\·}{I}={\stackrel{\·}{U}}_p{Y}^{\′}---\left(8\right)$

式中 $\stackrel{\·}{I}={\left[\begin{array}{ccccccc}{\stackrel{\·}{I}}_a^{\left(1\right)}& {\stackrel{\·}{I}}_b^{\left(1\right)}& {\stackrel{\·}{I}}_c^{\left(1\right)}& ...& {\stackrel{\·}{I}}_a^{\left(m\right)}& {\stackrel{\·}{I}}_b^{\left(m\right)}& {\stackrel{\·}{I}}_c^{\left(m\right)}\end{array}\right]}^T$

Y′＝[Y_aY_bY_cY_abY_bcY_ac]^T

${\stackrel{\·}{U}}_p=\left[\begin{array}{cccccc}{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{U}}_{ap}^{\left(m\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}\end{array}\right]$

应用复频域内的最小二乘法，得(8)式的解为：

$Y^{\′}={\left({{\stackrel{\·}{U}}_p}^T{\stackrel{\·}{U}}_p\right)}^{-1}\left({{\stackrel{\·}{U}}_p}^T\stackrel{\·}{I}\right)---\left(9\right)$

III阻抗参数的求解

在求得线路的导纳参数之后，建立求解线路阻抗参数的超定方程：

$\left\{\begin{array}{c}{\stackrel{\·}{U}}_a^{\left(1\right)}=Z_a{\stackrel{\·}{I}}_a^{\′\left(1\right)}+Z_{ab}{\stackrel{\·}{I}}_b^{\′\left(1\right)}+Z_{ac}{\stackrel{\·}{I}}_c^{\′\left(1\right)}\\ {\stackrel{\·}{U}}_b^{\left(1\right)}=Z_{ab}{\stackrel{\·}{I}}_a^{\′\left(1\right)}+Z_b{\stackrel{\·}{I}}_b^{\′\left(1\right)}+Z_{bc}{\stackrel{\·}{I}}_c^{\′\left(1\right)}\\ {\stackrel{\·}{U}}_c^{\left(1\right)}=Z_{ac}{\stackrel{\·}{I}}_a^{\′\left(1\right)}+Z_{bc}{\stackrel{\·}{I}}_b^{\′\left(1\right)}+Z_c{\stackrel{\·}{I}}_c^{\′\left(1\right)}\\ \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{U}}_a^{\left(m\right)}=Z_a{\stackrel{\·}{I}}_a^{\′\left(m\right)}+Z_{ab}{\stackrel{\·}{I}}_b^{\′\left(m\right)}+Z_{ac}{\stackrel{\·}{I}}_c^{\′\left(m\right)}\\ {\stackrel{\·}{U}}_b^{\left(m\right)}=Z_{ab}{\stackrel{\·}{I}}_a^{\′\left(m\right)}+Z_b{\stackrel{\·}{I}}_b^{\′\left(m\right)}+Z_{bc}{\stackrel{\·}{I}}_c^{\′\left(m\right)}\\ {\stackrel{\·}{U}}_c^{\left(m\right)}=Z_{ac}{\stackrel{\·}{I}}_a^{\′\left(m\right)}+Z_{bc}{\stackrel{\·}{I}}_b^{\′\left(m\right)}+Z_c{\stackrel{\·}{I}}_c^{\′\left(m\right)}\end{array}\right.---\left(10\right)$

式中

${\stackrel{\·}{U}}_a^{\left(i\right)}={\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a2}^{\left(i\right)},{\stackrel{\·}{U}}_b^{\left(i\right)}={\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b2}^{\left(i\right)},{\stackrel{\·}{U}}_c^{\left(i\right)}={\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c2}^{\left(i\right)}$

${\stackrel{\·}{I}}_a^{\′\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)}{Y}_a/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{ac}/2$

${\stackrel{\·}{I}}_b^{\′\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)}{Y}_b/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{bc}/2$

${\stackrel{\·}{I}}_c^{\′\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)}{Y}_c/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ac}/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{bc}/2$

i＝1,2，…，m

同理，将(10)式写成矩阵形式

$\stackrel{\·}{U}={\stackrel{\·}{I}}^{\′}{Z}^{\′}---\left(11\right)$

式中

$\stackrel{\·}{U}={\left[\begin{array}{ccccccc}{\stackrel{\·}{U}}_a^{\left(1\right)}& {\stackrel{\·}{U}}_b^{\left(1\right)}& {\stackrel{\·}{U}}_c^{\left(1\right)}& ...& {\stackrel{\·}{U}}_a^{\left(m\right)}& {\stackrel{\·}{U}}_b^{\left(m\right)}& {\stackrel{\·}{U}}_c^{\left(m\right)}\end{array}\right]}^T$

Z′＝[Z_aZ_bZ_cZ_acZ_bcZ_ac]^T

${\stackrel{\·}{I}}^{\′}=\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_a^{\′\left(1\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(1\right)}& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& {\stackrel{\·}{I}}_a^{\′\left(1\right)}\\ \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}& \begin{array}{c}.\\ .\\ .\end{array}\\ {\stackrel{\·}{I}}_a^{\′\left(m\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(m\right)}& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& {\stackrel{\·}{I}}_a^{\′\left(m\right)}\end{array}\right]$

输电线路的阻抗参数为：

$Z^{\′}={\left({\stackrel{\·}{I}}^{\′T}{\stackrel{\·}{I}}^{\′}\right)}^{-1}\left({\stackrel{\·}{I}}^{\′T}\stackrel{\·}{U}\right)---\left(12\right)$

IV序参数的求解

取 $Z_{abc}=\left[\begin{array}{ccc}{Z}_a& Z_{ab}& Z_{ac}\\ Z_{ab}& Z_b& Z_{bc}\\ Z_{ac}& Z_{bc}& Z_c\end{array}\right],Y_{abc}=\left[\begin{array}{ccc}{Y}_a& Y_{ab}& Y_{ac}\\ Y_{ab}& Y_{bb}& Y_{bc}\\ Y_{ac}& Y_{bc}& Y_{cc}\end{array}\right]$

输电线路的序阻抗及导纳参数分别为：

$Z_{012}=A^{-1}{Z}_{abc}A=\left[\begin{array}{ccc}{Z}_0& Z_{01}& Z_{02}\\ Z_{01}& Z_1& Z_{12}\\ Z_{02}& Z_{12}& Z_2\end{array}\right]---\left(13\right)$

$Y_{012}=A^{-1}{Y}_{abc}A=\left[\begin{array}{ccc}{Y}_0& Y_{01}& Y_{02}\\ Y_{01}& Y_1& Y_{12}\\ Y_{02}& Y_{12}& Y_2\end{array}\right]---\left(14\right)$

式中，Z₀、Z₁、Z₂、Y₀、Y₁、Y₂分别为输电线路的零序、正序和负序的阻抗与导纳，Z₀₁、Z₁₂、Z₀₂、Y₀₁、Y₁₂、Y₀₂分别为线路的零序、正序与负序之间的序耦合阻抗与导纳。

本实施例中，利用Matlab仿真软件搭建的输电线路参数测试模型如图2所示，线路电压等级为220kV，长度为16km。通过10次测量，计算出的输电线路参数如表1所示。

表1输电线路参数的计算结果及其相对误差

由表1可见，仅仅进行了10次测量，本实施例方法的计算结果与给定的线路参数值基本一致，线路零序阻抗模及阻抗角误差分别为‐0.011％、0.0026％，正序阻抗模及阻抗角误差分别为‐0.0051％、0.0003％，正序导纳误差为0.064％，零序导纳的误差为‐0.119％。

以上是对本发明做的示例性描述，凡在不脱离本发明核心的情况下做出的简单变形或修改均落入本发明的保护范围。

Claims (8)

1.一种非对称输电线路参数在线测量方法，其适用于长度小于150km、电压等级在500kV以下的输电线路参数在线测量，其包括以下步骤：

步骤一，在被测线路的两端安装同步测量装置，通过该同步测量装置同步测量被测线路两端的电压和电流信号，并进行m(m≥2)次不同的测量；

步骤二，对步骤一获得的m组被测线路两端的电压和电流信号进行傅里叶变换，计算m组被测线路两端的工频电压相量和电流相量，包括第1组至第m组被测线路的首端电压、电流相量 末端的电压相量、电流相量

步骤三，按以下公式计算被测线路A、B、C相的自导纳Y_a、Y_b、Y_c及各相的互导纳Y_ab、Y_bc、Y_ac：

$Y^{\′}={\left({{\stackrel{\·}{U}}_p}^T{\stackrel{\·}{U}}_p\right)}^{-1}\left({{\stackrel{\·}{U}}_p}^T\stackrel{\·}{I}\right);$

步骤四，按以下公式计算被测线路A、B、C相的自阻抗Z_a、Z_b、Z_c，各相的互阻抗Z_ab、Z_bc，Z_ac：

$Z^{\′}={\left({\stackrel{\·}{I}}^T{\stackrel{\·}{I}}^{\′}\right)}^{-1}\left({\stackrel{\·}{I}}^{\′T}\stackrel{\·}{U}\right).$

2.如权利要求1所述的在线测量方法，其特征在于，步骤三中， $\stackrel{\·}{I}={\left[\begin{array}{ccccccc}{\stackrel{\·}{I}}_a^{\left(1\right)}& {\stackrel{\·}{I}}_b^{\left(1\right)}& {\stackrel{\·}{I}}_c^{\left(1\right)}& ...& {\stackrel{\·}{I}}_a^{\left(m\right)}& {\stackrel{\·}{I}}_b^{\left(m\right)}& {\stackrel{\·}{I}}_c^{\left(m\right)}\end{array}\right]}^T$

Y′＝[Y_aY_bY_cY_abY_bcY_ac]^T

${\stackrel{\·}{U}}_p=\left[\begin{array}{cccccc}{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}& {\stackrel{\·}{U}}_{bp}^{\left(1\right)}-{\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{bp}^{\left(1\right)}& {\stackrel{\·}{U}}_{cp}^{\left(1\right)}-{\stackrel{\·}{U}}_{ap}^{\left(1\right)}\\ \·& \·& \·& \·& \·& \·\\ \·& \·& \·& \·& \·& \·\\ \·& \·& \·& \·& \·& \·\\ {\stackrel{\·}{U}}_{ap}^{\left(m\right)}& 0& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{ap}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}\\ 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}& {\stackrel{\·}{U}}_{bp}^{\left(m\right)}-{\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}& 0& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{bp}^{\left(m\right)}& {\stackrel{\·}{U}}_{cp}^{\left(m\right)}-{\stackrel{\·}{U}}_{ap}^{\left(m\right)}\end{array}\right]$

${\stackrel{\·}{I}}_a^{\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{I}}_{a2}^{\left(i\right)},{\stackrel{\·}{I}}_b^{\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{I}}_{b2}^{\left(i\right)},{\stackrel{\·}{I}}_c^{\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{I}}_{c2}^{\left(i\right)},{\stackrel{\·}{U}}_{ap}^{\left(i\right)}=({\stackrel{\·}{U}}_{a1}^{\left(i\right)}+{\stackrel{\·}{U}}_{a2}^{\left(i\right)})/2,$ ${\stackrel{\·}{U}}_{bp}^{\left(i\right)}=({\stackrel{\·}{U}}_{b1}^{\left(i\right)}+{\stackrel{\·}{U}}_{b2}^{\left(i\right)})/2,{\stackrel{\·}{U}}_{cp}^{\left(i\right)}=({\stackrel{\·}{U}}_{c1}^{\left(i\right)}+{\stackrel{\·}{U}}_{c2}^{\left(i\right)})/2,$ i＝1,2,…,m，m≥2。

3.如权利要求1所述的在线测量方法，其特征在于，步骤四中，

$\stackrel{\·}{U}={\left[\begin{array}{ccccccc}{\stackrel{\·}{U}}_a^{\left(1\right)}& {\stackrel{\·}{U}}_b^{\left(1\right)}& {\stackrel{\·}{U}}_c^{\left(1\right)}& ...& {\stackrel{\·}{U}}_a^{\left(m\right)}& {\stackrel{\·}{U}}_b^{\left(m\right)}& {\stackrel{\·}{U}}_c^{\left(m\right)}\end{array}\right]}^T$

Z′＝[Z_aZ_bZ_cZ_acZ_bcZ_ac]^T

${\stackrel{\·}{I}}^{\′}=\left[\begin{array}{cccccc}{\stackrel{\·}{I}}_a^{\′\left(1\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(1\right)}& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(1\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(1\right)}& {\stackrel{\·}{I}}_a^{\′\left(1\right)}\\ \·& \·& \·& \·& \·& \·\\ \·& \·& \·& \·& \·& \·\\ \·& \·& \·& \·& \·& \·\\ {\stackrel{\·}{I}}_a^{\′\left(m\right)}& 0& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}\\ 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_a^{\′\left(m\right)}& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0\\ 0& 0& {\stackrel{\·}{I}}_c^{\′\left(m\right)}& 0& {\stackrel{\·}{I}}_b^{\′\left(m\right)}& {\stackrel{\·}{I}}_a^{\′\left(m\right)}\end{array}\right]$

${\stackrel{\·}{U}}_a^{\left(i\right)}={\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a2}^{\left(i\right)},{\stackrel{\·}{U}}_b^{\left(i\right)}={\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b2}^{\left(i\right)},{\stackrel{\·}{U}}_c^{\left(i\right)}={\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c2}^{\left(i\right)}$

${\stackrel{\·}{I}}_a^{\′\left(i\right)}={\stackrel{\·}{I}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)}{Y}_a/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{a1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{ac}/2$

${\stackrel{\·}{I}}_b^{\′\left(i\right)}={\stackrel{\·}{I}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)}{Y}_b/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ab}/2-({\stackrel{\·}{U}}_{b1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)})Y_{bc}/2$

${\stackrel{\·}{I}}_c^{\′\left(i\right)}={\stackrel{\·}{I}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{c1}^{\left(i\right)}{Y}_c/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{a1}^{\left(i\right)})Y_{ac}/2-({\stackrel{\·}{U}}_{c1}^{\left(i\right)}-{\stackrel{\·}{U}}_{b1}^{\left(i\right)})Y_{bc}/2$

i＝1,2，…，m。

4.如权利要求1所述的在线测量方法，其特征在于，进一步包括如下步骤：

步骤五，按以下公式计算被测线路每回的正序阻抗Z₁、负序阻抗Z₂、零序阻抗Z₀、正序与负序间的序耦合阻抗Z₁₂、零序与正序间的序耦合阻抗Z₀₁、零序与负序间的序耦合阻抗Z₀₂：

$Z_{012}=A^{-1}{Z}_{abc}A=\left[\begin{array}{ccc}{Z}_0& Z_{01}& Z_{02}\\ Z_{01}& Z_1& Z_{12}\\ Z_{02}& Z_{12}& Z_2\end{array}\right].$

5.如权利要求4所述的在线测量方法，其特征在于，步骤五中，

$Z_{abc}=\left[\begin{array}{ccc}{Z}_a& Z_{ab}& Z_{ac}\\ Z_{ab}& Z_b& Z_{bc}\\ Z_{ac}& Z_{bc}& Z_c\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{\α}}^2& \mathrm{\α}\\ 1& \mathrm{\α}& {\mathrm{\α}}^2\end{array}\right]$

6.如权利要求1所述的在线测量方法，其特征在于，进一步包括如下步骤：

步骤六，按以下公式计算被测线路每回的正序导纳Y₁、负序导纳Y₂、零序导纳Y₀、正序与负序间的序耦合导纳Y₁₂、零序与正序间的序耦合导纳Y₀₁、零序与负序间的序耦合导纳Y₀₂：

$Y_{012}=A^{-1}{Y}_{abc}A=\left[\begin{array}{ccc}{Y}_0& Y_{01}& Y_{02}\\ Y_{01}& Y_1& Y_{12}\\ Y_{02}& Y_{12}& Y_2\end{array}\right].$

7.如权利要求6所述的在线测量方法，其特征在于，步骤六中，

$Y_{abc}=\left[\begin{array}{ccc}{Y}_a& Y_{ab}& Y_{ac}\\ Y_{ab}& Y_{bb}& Y_{bc}\\ Y_{ac}& Y_{bc}& Y_{cc}\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& {\mathrm{\α}}^2& \mathrm{\α}\\ 1& \mathrm{\α}& {\mathrm{\α}}^2\end{array}\right]$

8.如权利要求1-7之一所述的在线测量方法，其特征在于，所述同步测量装置的采集精度在16位二进制数以上、采样频率在50kHz以上。