155 lines
5.0 KiB
TeX
155 lines
5.0 KiB
TeX
\documentclass[10pt,a4paper,final]{report}
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\usepackage[utf8]{inputenc}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{fontspec}%使用xetex
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\setmainfont[BoldFont=黑体]{宋体} % 使用系统默认字体
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\XeTeXlinebreaklocale "zh" % 针对中文进行断行
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\XeTeXlinebreakskip = 0pt plus 1pt minus 0.1pt % 给予TeX断行一定自由度
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\linespread{1.5} % 1.5倍行距
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\begin{document}
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%\begin{equation}
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%\begin{aligned}
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%\frac{\partial P_{ij}}{\partial V_i}
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%&=
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%-V_{ij}(g_{ij}cos\theta_{ij}+b_{ij}sin\theta_{ij})+
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%2(g_{ij}+g_{si})V_i
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%\end{aligned}
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%\end{equation}
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线路功率（不考虑接地导纳）
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\newline
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\begin{equation}
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\begin{aligned}
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\dot{I}_{12}&=(V_1 e^{j\theta_1} - V_2 e^{j\theta_2} )
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(G+jB) \\
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&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)](G+jB) \\
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&=(V_1 cos\theta_1+ jV_1 sin\theta_1- V_2 cos \theta_2 -j V_2 sin \theta_2)(G+jB) \\
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&=(V_1 cos \theta_1 -V_2 cos \theta_2)G + j(V_1 sin \theta_1- V_2 sin \theta_2)G
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\\&+j( V_1 cos \theta_1 -V_2 cos \theta_2)B + j(jV_1 sin \theta_1-jV_2 sin \theta_2)B \\
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&=G(V_1 cos \theta_1- V_2 cos \theta_2) - B(V_1 sin \theta_1 -V_2 sin \theta_2) \\
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&+jB(V_1 cos \theta_1 - V_2 cos \theta_2) + jG(V_1 sin \theta_1 - V_2 sin \theta_2) \\
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&= V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) \\
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&+jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)
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\end{aligned}
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\end{equation}
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电流的共轭复数
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\begin{equation}
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\begin{aligned}
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\dot{I}^*_{12}&=V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) \\
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&-jV_1(Bcos \theta_1+Gsin\theta_1)+jV_2(Bcos\theta_2+Gsin\theta_2)
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\end{aligned}
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\end{equation}
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线路复功率
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\begin{equation}
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\begin{aligned}
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\dot{S}_{12}&=(V_1 e^{j\theta_1} - V_2 e^{j\theta_2} )
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\dot{I}^*_{12} \\
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&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)]\dot{I}^*_{12} \\
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&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)] \\
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&[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2)
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-jV_1(Bcos \theta_1+Gsin\theta_1)+jV_2(Bcos\theta_2+Gsin\theta_2)]\\
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&=V_1( G cos \theta_1-B sin \theta_1)
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[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2)]\\
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&-V_1( G cos \theta_1-B sin \theta_1)
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[jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)] \\
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&- V_2 (cos\theta_2+jsin\theta_2)
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[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2)]\\
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&V_2 (cos\theta_2+jsin\theta_2)
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[jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)]
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\end{aligned}
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\end{equation}
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另一种推导方式
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\begin{equation}
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\begin{aligned}
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\dot{I}_{12}=(V_1 e^{j\theta_1} - V_2 e^{j\theta_2} )
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(G+jB)
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\end{aligned}
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\end{equation}
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电流的共轭复数
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\begin{equation}
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\begin{aligned}
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\dot{I}^*_{12}=(V_1 e^{-j\theta_1} - V_2 e^{-j\theta_2} )
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(G-jB)
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\end{aligned}
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\end{equation}
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线路复功率
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\begin{equation}
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\begin{aligned}
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\dot{S}_{12}&=(V_1 e^{j\theta_1} - V_2 e^{j\theta_2} )\dot{I}^*_{12} \\
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&=(V_1 e^{j\theta_1} - V_2 e^{j\theta_2} )
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(V_1 e^{-j\theta_1} - V_2 e^{-j\theta_2} )
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(G-jB) \\
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&=(V_1 e^{j\theta_1}V_1 e^{-j\theta_1} - V_1 e^{j\theta_1}V_2 e^{-j\theta_2}
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- V_2 e^{j\theta_2}V_1 e^{-j\theta_1}+ V_2 e^{j\theta_2}V_2 e^{-j\theta_2}
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)
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(G-jB) \\
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&=(V_1^2+V_2^2-V_1V_2e^{j(\theta_1-\theta_2)} -V_1V_2e^{j(\theta_2-\theta_1)} )
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(G-jB) \\
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&=[V_1^2+V_2^2
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-V_1V_2(e^{j(\theta_1-\theta_2)}+e^{j(\theta_2-\theta_1)})
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](G-jB) \\
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&=\{V_1^2+V_2^2-V_1V_2[cos(\theta_1-\theta_2)+jsin(\theta_1-\theta_2)
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+cos(\theta_2-\theta_1)+jsin(\theta_2-\theta_1)]\}(G-jB) \\
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&=\{V_1^2+V_2^2-V_1V_2[cos(\theta_1-\theta_2)
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+cos(\theta_1-\theta_2)]\}(G-jB) \\
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&=[V_1^2+V_2^2-2V_1V_2cos(\theta_1-\theta_2)](G-jB) \\
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&=(V_1^2+V_2^2)G-j(V_1^2+V_2^2)B-j2V_1V_2cos(\theta_1-\theta_2)G-j2V_1V_2cos(\theta_1-\theta_2)(-jB) \\
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&=(V_1^2+V_2^2)G-2V_1V_2cos(\theta_1-\theta_2)B
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-j(V_1^2+V_2^2)B-j2V_1V_2cos(\theta_1-\theta_2)G
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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P_{12}=(V_1^2+V_2^2)G+2V_1V_2sin(\theta_1-\theta_2)B
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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Q_{12}=-(V_1^2+V_2^2)B+2V_1V_2sin(\theta_1-\theta_2)G
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\end{aligned}
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\end{equation}
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考虑接地支路,接地电流为
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\begin{equation}
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\label{接地电流}
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\dot{I}_{1}=V_1e^{j\theta_1}(G_d+jB_d)
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\end{equation}
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(\ref{接地电流})共轭为
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\begin{equation}
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\dot{I}_{1}^{*}=V_1e^{-j\theta_1}(G_d-jB_d)
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\end{equation}
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%Sd
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\begin{equation}
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\begin{aligned}
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\dot{S}_{1}&=V_1e^{j\theta_1}\dot{I}_{1}^{*} \\
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&=V_1e^{j\theta_1}
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V_1e^{-j\theta_1}(G_d-jB_d) \\
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&=V_1^2(G_d-jB_d)
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\end{aligned}
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\end{equation}
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因而
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\begin{equation}
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\begin{aligned}
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P_{1}&=V_1^2 G_d \\
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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Q_{1}&=-V_1^2 B_d \\
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\end{aligned}
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\end{equation}
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考虑节点支路后的注入电流
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\begin{equation}
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\begin{aligned}
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\tilde{P}_{12}&=V_1^2 G_d
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+(V_1^2+V_2^2)G+2V_1V_2sin(\theta_1-\theta_2)B
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\end{aligned}
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\end{equation}
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\begin{equation}
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\begin{aligned}
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\tilde{Q}_{12}&=-V_1^2 B_d
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-(V_1^2+V_2^2)B+2V_1V_2sin(\theta_1-\theta_2)G
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\end{aligned}
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\end{equation}
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\end{document}
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