超高压交流同塔四回输电线路零序参数精确测量方法

摘要

本发明公开了一种新型超高压同塔四回输电线路零序参数精确测量方法。同时测量四回输电线路首末两端的零序电压和零序电流，利用全球卫星定位系统（GPS）的同步授时功能，实现对零序电压和零序电流的同步采样；利用基于分布参数的超高压同塔四回输电线路模型，得到相对应的微分方程组，再通过本发明给出的测量方式与计算方法得到超高压同塔四回输电线路的零序电阻、零序电感、零序电容参数。本发明方法基于分布参数模型，因此特别适合于长距离交（直）流输电线路的参数测量，能极大提高测量精度，可满足实际工程测量的需要。

主权项

一种超高压交流同塔四回输电线路零序参数精确测量方法，其特征在于，定义同塔四回输电线路包括线路a、线路b、线路c和线路d，其中，规则一：线路a和线路b的零序自阻抗相同，均为Z₁₁，线路c和线路d的零序自阻抗相同，均为Z₂₂，线路a与线路b之间的零序互阻抗参数为Z_ab；线路a和线路c之间的零序互阻抗参数与线路b和线路d之间的零序互阻抗参数相同，均为Z_ac；线路a和线路d之间的零序互阻抗参数与线路b和线路c之间的零序互阻抗参数相同，均为Z_ad；线路c和线路d之间的零序互阻抗参数为Z_cd；规则二：线路a和线路b的零序自电纳相同，均为Y₁₁，线路c和线路d的零序自电纳相同，均为Y₂₂，线路a与线路b之间的零序互电纳参数为Y_ab；线路a和线路c之间的零序互电纳参数与线路b和线路d之间的零序互电纳参数相同，均为Y_ac；线路a和线路d之间的零序互电纳参数与线路b和线路c之间的零序互电纳参数相同，均为Y_ad；线路c和线路d之间的零序互电纳参数为Y_cd；测量步骤包括：步骤1，停电测量同塔四回输电线路，将四回线路首末端三相分别短接；线路测量包括：测量方式一：线路a首端加压，末端接地；线路b首端悬空，末端接地；线路c首端悬空，末端接地；线路d首端悬空，末端接地；测量方式二：线路a首端悬空，末端接地；线路b首端悬空，末端接地；线路c首端加压，末端接地；线路d首端悬空，末端接地；步骤2，利用全球卫星定位系统的同步授时功能，同步测量线路a、线路b、线路c和线路d首端和末端的零序电压数据和零序电流数据；步骤3，对步骤2所得每个独立测量方式下得到的零序电压测量数据和零序电流测量数据，采用傅立叶算法得到该独立测量方式下首端和末端的零序基波电压相量和零序基波电流相量，再利用这些相量数据将同塔四回输电线路的零序参数求解出来；所需求解的参数包括单位长度的零序自阻抗参数Z₁₁、Z₂₂，零序互阻抗参数Z_ab、Z_ac、Z_ad、Z_cd，零序自电纳参数Y₁₁、Y₂₂，零序互电纳参数Y_ab、Y_ac、Y_ad、Y_cd，然后根据以下两式求解出所有的零序参数：$\left\{\begin{array}{c}{Z}_{11}=R_a+j\omega L_a=R_b+j\omega L_b\\ Z_{22}=R_c+{\mathrm{j\omega L}}_c=R_d+{\mathrm{j\omega L}}_d\\ Z_{ab}=R_{ab}+{\mathrm{j\omega L}}_{ab}\\ Z_{ac}=R_{ac}+{\mathrm{j\omega L}}_{ac}=R_{bd}+{\mathrm{j\omega L}}_{bd}\\ Z_{ad}=R_{ad}+{\mathrm{j\omega L}}_{ad}=R_{bc}+{\mathrm{j\omega L}}_{bc}\\ Z_{cd}=R_{cd}+{\mathrm{j\omega L}}_{cd}\end{array}\right.$  式一$\left\{\begin{array}{c}{Y}_{11}=j\omega C_a=j\omega C_b\\ Y_{22}={\mathrm{j\omega C}}_c={\mathrm{j\omega C}}_d\\ Y_{ab}={\mathrm{j\omega C}}_{ab}\\ Y_{ac}={\mathrm{j\omega C}}_{ac}={\mathrm{j\omega C}}_{bd}\\ Y_{ad}={\mathrm{j\omega C}}_{ad}={\mathrm{j\omega C}}_{bc}\\ Y_{cd}={\mathrm{j\omega C}}_{cd}\end{array}\right.$  式二其中，因为线路的泄漏电流很小，所以不考虑电导参数；同塔四回线路定义为平行架设且长度同为l，定义a、b、c、d四回输电线路首端的零序基波电压相量分别为末端的零序基波电压相量分别为首端的零序基波电流相量分别为末端的零序基波电流相量分别为零序参数求解过程如下：步骤3.1，由零序参数Z₁₁、Z₂₂、Z_ab、Z_ac、Z_ad、Z_cd和Y₁₁、Y₂₂、Y_ab、Y_ac、Y_ad、Y_cd得到输电线路首末端电压电流的关系：其中A_aa,A_ab,A_ba,A_bb，B_aa,B_ab,B_ba,B_bb,A′_aa,A′_ab,A′_ba,A′_bb，B′_aa,B′_ab,B′_ba,B′_bb是和线路参数有关的中间变量；$\left[\begin{array}{c}{\stackrel{·}{U}}_{a1}+{\stackrel{·}{U}}_{b1}\\ {\stackrel{·}{U}}_{c1}+{\stackrel{·}{U}}_{d1}\\ {\stackrel{·}{I}}_{a1}+{\stackrel{·}{I}}_{b1}\\ {\stackrel{·}{I}}_{c1}+{\stackrel{·}{I}}_{d1}\end{array}\right]=\left[\begin{array}{cccc}{A}_{aa}& A_{ab}& B_{aa}& B_{ab}\\ A_{ba}& A_{bb}& B_{ba}& B_{bb}\\ C_{aa}& C_{ab}& A_{aa}& A_{ba}\\ C_{ba}& C_{bb}& A_{ab}& A_{bb}\end{array}\right]\left[\begin{array}{c}{\stackrel{·}{U}}_{a2}+{\stackrel{·}{U}}_{b2}\\ {\stackrel{·}{U}}_{c2}+{\stackrel{·}{U}}_{d2}\\ {\stackrel{·}{I}}_{a2}+{\stackrel{·}{I}}_{b2}\\ {\stackrel{·}{I}}_{c2}+{\stackrel{·}{I}}_{d2}\end{array}\right]$  式三$\left[\begin{array}{c}{\stackrel{·}{U}}_{a1}-{\stackrel{·}{U}}_{b1}\\ {\stackrel{·}{U}}_{c1}-{\stackrel{·}{U}}_{d1}\\ {\stackrel{·}{I}}_{a1}-{\stackrel{·}{I}}_{b1}\\ {\stackrel{·}{I}}_{c1}-{\stackrel{·}{I}}_{d1}\end{array}\right]=\left[\begin{array}{cccc}{A}_{aa}^{\prime }& A_{ab}^{\prime }& B_{aa}^{\prime }& B_{ab}^{\prime }\\ A_{ba}^{\prime }& A_{bb}^{\prime }& B_{ba}^{\prime }& B_{bb}^{\prime }\\ C_{aa}^{\prime }& C_{ab}^{\prime }& A_{aa}^{\prime }& A_{ba}^{\prime }\\ C_{ba}^{\prime }& C_{bb}^{\prime }& A_{ab}^{\prime }& A_{bb}^{\prime }\end{array}\right]\left[\begin{array}{c}{\stackrel{·}{U}}_{a2}-{\stackrel{·}{U}}_{b2}\\ {\stackrel{·}{U}}_{c2}-{\stackrel{·}{U}}_{d2}\\ {\stackrel{·}{I}}_{a2}-{\stackrel{·}{I}}_{b2}\\ {\stackrel{·}{I}}_{c2}-{\stackrel{·}{I}}_{d2}\end{array}\right]$  式四步骤3.2，通过步骤1中的两种测量方式计算中间变量A_aa,A_ab,A_ba,A_bb，B_aa,B_ab,B_ba,B_bb,A′_aa,A′_ab,A′_ba,A′_bb，B′_aa,B′_ab,B′_ba,B′_bb：$\left[\begin{array}{cc}{{\stackrel{·}{I}}_{a1}}^1+{{\stackrel{·}{I}}_{b1}}^1& {{\stackrel{·}{I}}_{a1}}^2+{{\stackrel{·}{I}}_{b1}}^2\\ {{\stackrel{·}{I}}_{c1}}^1+{{\stackrel{·}{I}}_{d1}}^1& {{\stackrel{·}{I}}_{c1}}^2+{{\stackrel{·}{I}}_{d1}}^2\end{array}\right]=\left[\begin{array}{cc}{A}_{aa}& A_{ba}\\ A_{ab}& A_{bb}\end{array}\right]\left[\begin{array}{cc}{{\stackrel{·}{I}}_{a2}}^1+{{\stackrel{·}{I}}_{b2}}^1& {{\stackrel{·}{I}}_{a2}}^2+{{\stackrel{·}{I}}_{b2}}^2\\ {{\stackrel{·}{I}}_{c2}}^1+{{\stackrel{·}{I}}_{d2}}^1& {{\stackrel{·}{I}}_{c2}}^2+{{\stackrel{·}{I}}_{d2}}^2\end{array}\right]$  式五$\left[\begin{array}{cc}{{\stackrel{·}{U}}_{a1}}^1+{{\stackrel{·}{U}}_{b1}}^1& {{\stackrel{·}{U}}_{a1}}^2+{{\stackrel{·}{U}}_{b1}}^2\\ {{\stackrel{·}{U}}_{c1}}^1+{{\stackrel{·}{U}}_{d1}}^1& {{\stackrel{·}{U}}_{c1}}^2+{{\stackrel{·}{U}}_{d1}}^2\end{array}\right]=\left[\begin{array}{cc}{B}_{aa}& B_{ab}\\ B_{ba}& B_{bb}\end{array}\right]\left[\begin{array}{cc}{{\stackrel{·}{I}}_{a2}}^1+{{\stackrel{·}{I}}_{b2}}^1& {{\stackrel{·}{I}}_{a2}}^2+{{\stackrel{·}{I}}_{b2}}^2\\ {{\stackrel{·}{I}}_{c2}}^1+{{\stackrel{·}{I}}_{d2}}^1& {{\stackrel{·}{I}}_{c2}}^2+{{\stackrel{·}{I}}_{d2}}^2\end{array}\right]$  式六$\left[\begin{array}{cc}{{\stackrel{·}{I}}_{a1}}^1-{{\stackrel{·}{I}}_{b1}}^1& {{\stackrel{·}{I}}_{a1}}^2-{{\stackrel{·}{I}}_{b1}}^2\\ {{\stackrel{·}{I}}_{c1}}^1-{{\stackrel{·}{I}}_{d1}}^1& {{\stackrel{·}{I}}_{c1}}^2-{{\stackrel{·}{I}}_{d1}}^2\end{array}\right]=\left[\begin{array}{cc}{A}_{aa}^{\prime }& A_{ba}^{\prime }\\ A_{ab}^{\prime }& A_{bb}^{\prime }\end{array}\right]\left[\begin{array}{cc}{{\stackrel{·}{I}}_{a2}}^1-{{\stackrel{·}{I}}_{b2}}^1& {{\stackrel{·}{I}}_{a2}}^2-{{\stackrel{·}{I}}_{b2}}^2\\ {{\stackrel{·}{I}}_{c2}}^1-{{\stackrel{·}{I}}_{d2}}^1& {{\stackrel{·}{I}}_{c2}}^2-{{\stackrel{·}{I}}_{d2}}^2\end{array}\right]$  式七$\left[\begin{array}{cc}{{\stackrel{·}{U}}_{a1}}^1+{{\stackrel{·}{U}}_{b1}}^1& {{\stackrel{·}{U}}_{a1}}^2+{{\stackrel{·}{U}}_{b1}}^2\\ {{\stackrel{·}{U}}_{c1}}^1+{{\stackrel{·}{U}}_{d1}}^1& {{\stackrel{·}{U}}_{c1}}^2+{{\stackrel{·}{U}}_{d1}}^2\end{array}\right]=\left[\begin{array}{cc}{B}_{aa}^{\prime }& B_{ab}^{\prime }\\ B_{ba}^{\prime }& B_{bb}^{\prime }\end{array}\right]\left[\begin{array}{cc}{{\stackrel{·}{I}}_{a2}}^1-{{\stackrel{·}{I}}_{b2}}^1& {{\stackrel{·}{I}}_{a2}}^2-{{\stackrel{·}{I}}_{b2}}^2\\ {{\stackrel{·}{I}}_{c2}}^1-{{\stackrel{·}{I}}_{d2}}^1& {{\stackrel{·}{I}}_{c2}}^2-{{\stackrel{·}{I}}_{d2}}^2\end{array}\right]$  式八上述各式中，带上标“1”或“2”的电压相量或电流相量，表示该相量为线路接线方式1或接线方式2下的测量数据经傅里叶算法计算得到的零序基波电压相量或零序基波电流相量；步骤3.3，根据以下公式计算特征根p₁,p₂,p′₁,p′₂；$\left\{\begin{array}{c}ch{p}_{1}l+ch{p}_{2}l=A_{aa}+A_{bb}\\ {\mathrm{chp}}_{1}l-{\mathrm{chp}}_{2}l=-\sqrt{{(A_{aa}-A_{bb})}^2+4A_{ab}{A}_{ba}}\end{array}\right.$  式九$\left\{\begin{array}{c}ch{p}_1^{\prime }l+ch{p}_2^{\prime }l=A_{aa}^{\prime }+A_{bb}^{\prime }\\ {\mathrm{chp}}_1^{\prime }l-{\mathrm{chp}}_2^{\prime }l=-\sqrt{{(A_{aa}^{\prime }-A_{bb}^{\prime })}^2+4A_{ab}^{\prime }{A}_{ba}^{\prime }}\end{array}\right.$  式十然后通过p₁,p₂,p′₁,p′₂求解A₁,A₂,B₁,B₂以及A′₁,A′₂,B′₁,B′₂；$\left\{\begin{array}{c}{A}_1=\frac{{\mathrm{chp}}_{1}x}{{p_1}^2-{p_2}^2}\\ A_2=\frac{{\mathrm{chp}}_{2}x}{{p_1}^2-{p_2}^2}\\ B_1=\frac{{\mathrm{shp}}_{1}x}{p_1({p_1}^2-{p_2}^2)}\\ B_{2=}\frac{{\mathrm{shp}}_{2}x}{p_2({p_1}^2-{p_2}^2)}\end{array}\right.$  式十一$\left\{\begin{array}{c}{A}_1^{\prime }=\frac{{\mathrm{chp}}_1^{\prime }x}{p_1^{\prime 2}-p_2^{\prime 2}}\\ A_2^{\prime }=\frac{{\mathrm{chp}}_2^{\prime }x}{p_1^{\prime 2}-p_2^{\prime 2}}\\ B_1^{\prime }=\frac{{\mathrm{shp}}_1^{\prime }x}{p_1^{\prime }(p_1^{\prime 2}-p_2^{\prime 2})}\\ B_2^{\prime }=\frac{{\mathrm{shp}}_2^{\prime }x}{p_2^{\prime }(p_1^{\prime 2}-p_2^{\prime 2})}\end{array}\right.$  式十二步骤3.4，将A_aa,A_ab,A_ba,A_bb,A′_aa,A′_ab,A′_ba,A′_bb代入下式计算矩阵T₁和T₃；$\left[\begin{array}{c}{A}_{aa}\\ A_{ab}\\ A_{ba}\\ A_{bb}\end{array}\right]=A_1\left[\begin{array}{c}{p_1}^2-T_{14}\\ T_{12}\\ T_{13}\\ {p_1}^2-T_{11}\end{array}\right]-A_2\left[\begin{array}{c}{p_2}^2-T_{14}\\ T_{12}\\ T_{13}\\ {p_2}^2-T_{11}\end{array}\right]$  式十三$\left[\begin{array}{c}{A}_{aa}^{\prime }\\ A_{ab}^{\prime }\\ A_{ba}^{\prime }\\ A_{bb}^{\prime }\end{array}\right]=A_1^{\prime }\left[\begin{array}{c}{p}_1^{\prime 2}-T_{34}\\ T_{32}\\ T_{33}\\ p_1^{\prime 2}-T_{31}\end{array}\right]-A_2^{\prime }\left[\begin{array}{c}{p}_2^{\prime 2}-T_{34}\\ T_{32}\\ T_{33}\\ p_2^{\prime 2}-T_{31}\end{array}\right]$  式十四步骤3.5，代入B_aa,B_ab,B_ba,B_bb,B′_aa,B′_ab,B′_ba,B′_bb和矩阵T₁、T₃计算阻抗矩阵Z₁、Z₂；$\left[\begin{array}{c}{B}_{aa}\\ B_{ab}\\ B_{ba}\\ B_{bb}\end{array}\right]=B_1\left[\begin{array}{c}\left({p_1}^2-T_{14}\right)\left(Z_{11}+Z_{ab}\right)+T_{12}\left(Z_{ac}+Z_{ad}\right)\\ \left({p_1}^2-T_{14}\right)\left(Z_{ac}+Z_{ad}\right)+T_{12}\left(Z_{22}+Z_{cd}\right)\\ \left({p_1}^2-T_{11}\right)\left(Z_{ac}+Z_{ad}\right)+T_{13}\left(Z_{11}+Z_{ab}\right)\\ \left({p_1}^2-T_{11}\right)\left(Z_{22}+Z_{cd}\right)+T_{13}\left(Z_{ac}+Z_{ad}\right)\end{array}\right]-B_2\left[\begin{array}{c}\left({p_2}^2-T_{14}\right)\left(Z_{11}+Z_{ab}\right)+T_{12}\left(Z_{ac}+Z_{ad}\right)\\ \left({p_2}^2-T_{14}\right)\left(Z_{ac}+Z_{ad}\right)+T_{12}\left(Z_{22}+Z_{cd}\right)\\ \left({p_2}^2-T_{11}\right)\left(Z_{ac}+Z_{ad}\right)+T_{13}\left(Z_{11}+Z_{ab}\right)\\ \left({p_2}^2-T_{11}\right)\left(Z_{22}+Z_{cd}\right)+T_{13}\left(Z_{ac}+Z_{ad}\right)\end{array}\right]$  式十五$\left[\begin{array}{c}{B}_{aa}^{\prime }\\ B_{ab}^{\prime }\\ B_{ba}^{\prime }\\ B_{bb}^{\prime }\end{array}\right]=B_1^{\prime }\left[\begin{array}{c}\left(p_1^{\prime 2}-T_{34}\right)\left(Z_{11}-Z_{ab}\right)+T_{32}\left(Z_{ac}-Z_{ad}\right)\\ \left(p_1^{\prime 2}-T_{34}\right)\left(Z_{ac}-Z_{ad}\right)+T_{32}\left(Z_{22}-Z_{cd}\right)\\ \left(p_1^{\prime 2}-T_{31}\right)\left(Z_{ac}-Z_{ad}\right)+T_{33}\left(Z_{11}-Z_{ab}\right)\\ \left(p_1^{\prime 2}-T_{31}\right)\left(Z_{22}-Z_{cd}\right)+T_{33}\left(Z_{ac}-Z_{ad}\right)\end{array}\right]-B_2^{\prime }\left[\begin{array}{c}\left(p_2^{\prime 2}-T_{34}\right)\left(Z_{11}-Z_{ab}\right)+T_{32}\left(Z_{ac}-Z_{ad}\right)\\ \left(p_2^{\prime 2}-T_{34}\right)\left(Z_{ac}-Z_{ad}\right)+T_{32}\left(Z_{22}-Z_{cd}\right)\\ \left(p_2^{\prime 2}-T_{31}\right)\left(Z_{ac}-Z_{ad}\right)+T_{33}\left(Z_{11}-Z_{ab}\right)\\ \left(p_2^{\prime 2}-T_{31}\right)\left(Z_{22}-Z_{cd}\right)+T_{33}\left(Z_{ac}-Z_{ad}\right)\end{array}\right]$  式十六$Z_1=\left[\begin{array}{cc}{Z}_{11}+Z_{ab}& Z_{ac}+Z_{ad}\\ Z_{ac}+Z_{ad}& Z_{22}+Z_{cd}\end{array}\right]$$Z_2=\left[\begin{array}{cc}{Z}_{11}-Z_{ab}& Z_{ac}-Z_{ad}\\ Z_{ac}-Z_{ad}& Z_{22}-Z_{cd}\end{array}\right]$步骤3.6将矩阵T₁、T₃、Z₁、Z₂代入下式，计算零序导纳矩阵；$Y=Z_1^{-1}{T}_1$  式十七$Y_2=Z_2^{-1}{T}_3$  式十八$Y=\left[\begin{array}{cc}{Y}_1-Y_{ab}& -Y_{ac}-Y_{ad}\\ -Y_{ac}-Y_{ad}& Y_{22}-Y_{cd}\end{array}\right]$$Y_2=\left[\begin{array}{cc}{Y}_1+Y_{ab}& -Y_{ac}+Y_{ad}\\ -Y_{ac}+Y_{ad}& Y_{22}+Y_{cd}\end{array}\right]$步骤3.7，计算零序阻抗和零序导纳参数；$\frac{1}{2}(Z_1+Z_2)=\left[\begin{array}{cc}{Z}_{11}& Z_{ac}\\ Z_{ac}& Z_{22}\end{array}\right]$  式十九$\frac{1}{2}(Z_1-Z_2)=\left[\begin{array}{cc}{Z}_{ab}& Z_{ad}\\ Z_{ad}& Z_{cd}\end{array}\right]$  式二十$\frac{1}{2}(Y_1+Y_2)=\left[\begin{array}{cc}{Y}_{11}& -Y_{ac}\\ -Y_{ac}& Y_{22}\end{array}\right]$  式二十一$-\frac{1}{2}(Y_1-Y_2)=\left[\begin{array}{cc}{Y}_{ab}& Y_{ad}\\ Y_{ad}& Y_{cd}\end{array}\right]$  式二十二最后，由Z₁₁、Z₂₂、Z_ab、Z_ac、Z_ad、Z_cd及Y₁₁、Y₂₂、Y_ab、Y_ac、Y_ad、Y_cd得到对应的同塔四回输电线路零序电阻、零序电感、零序电容参数；其中，符号sh(·)表示双曲正弦函数，符号ch(·)表示双曲余弦函数，符号arch(·)表示反双曲余弦函数，ω＝2πf，f为电力系统频率50Hz，l表示同塔四回线路长度。