\documentclass[10pt,a4paper,final]{report} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{fontspec}%使用xetex \setmainfont[BoldFont=黑体]{宋体} % 使用系统默认字体 \XeTeXlinebreaklocale "zh" % 针对中文进行断行 \XeTeXlinebreakskip = 0pt plus 1pt minus 0.1pt % 给予TeX断行一定自由度 \linespread{1.5} % 1.5倍行距 \begin{document} 线路功率(不考虑接地导纳) \begin{equation} \begin{aligned} \dot{I}_{12}&=(V_1e^{j \theta_1} - V_2e^{j \theta_2}) (G_{ij}+jB_{ij}) \end{aligned} \end{equation} 共轭后 \begin{equation} \begin{aligned} \dot{I}^*_{12}&=(V_1e^{-j \theta_1} - V_2e^{-j \theta_2}) (G_{ij}-jB_{ij}) \end{aligned} \end{equation} \begin{equation} \begin{aligned} \dot{S}_{12}&=V_1e^{j \theta_1} \dot{I}^*_{12} \\ &=V_1e^{j \theta_1} (V_1e^{-j \theta_1} - V_2e^{-j \theta_2}) (G_{ij}-jB_{ij}) \\ &=(V_1e^{j \theta_1}V_1e^{-j \theta_1} -V_1e^{j \theta_1}V_2e^{-j \theta_2} ) (G_{ij}-jB_{ij}) \\ &=[V_1^2-V_1V_2e^{j (\theta_1 - \theta_2) } ] (G_{ij}-jB_{ij}) \\ &=\{V_1^2-V_1V_2[cos(\theta_1 - \theta_2)+jsin (\theta_1 - \theta_2) ]\} (G_{ij}-jB_{ij}) \\ &=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij} \\ &+[V_1^2-V_1V_2cos(\theta_1 - \theta_2)](-jB_{ij}) \\ &-jV_1V_2sin (\theta_1 - \theta_2)G_{ij} \\ &-jV_1V_2sin (\theta_1 - \theta_2)(-jB_{ij}) \\ &=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\ &-j[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-jV_1V_2sin (\theta_1 - \theta_2)G_{ij} \end{aligned} \end{equation} 有功传输功率 \begin{equation} \begin{aligned} P_{ij}&=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\ &=V_1^2G_{ij}-V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 无功传输功率 \begin{equation} \begin{aligned} Q_{ij}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \theta_2)G_{ij} \\ &=-V_1^2B_{ij}-V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 线路有功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial P_{ij}}{\partial V_1}= 2V_1G_{ij}-V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial V_2}= -V_1[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_1}&= -V_1V_2[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\ &=V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_2}&= -V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\ \end{aligned} \end{equation} 线路无功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_1}&= -2V_1B_{12}-V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_2}&= -V_1[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_1}&= -V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_2}&= -V_1V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\ &=V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 变压器支路功率 \newline 变压器模型为 \newline \begin{center} ----k:1----z----高压侧 \\ or \\ i----k:1----z----j \\ \end{center} 变压器支路功率 \newline 变压器有功输送功率 \begin{equation} \begin{aligned} P_{ij}&=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\ &=\frac{V_1^2}{k^2} G_{ij}-\frac{V_1}{k} V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 变压器无功输送功率 \begin{equation} \begin{aligned} Q_{ij}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \theta_2)G_{ij} \\ &=-\frac{V_1^2}{k^2}B_{ij}-\frac{V_1}{k} V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} 变压器输送有功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial P_{ij}}{\partial V_1}= 2\frac{V_1}{k^2}G_{ij}-\frac{V_2}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial V_2}= -\frac{V_1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_1} &=\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial P_{12}}{\partial \theta_2}&= -\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\ \end{aligned} \end{equation} 变压器输送无功功率Jacobi \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_1}&= -2\frac{V_1}{k^2}B_{12}-\frac{V_2}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial V_2}&= -\frac{V_1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_1}&= -\frac{V_1}{k}V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} \begin{equation} \begin{aligned} \frac{\partial Q_{12}}{\partial \theta_2} &=\frac{V_1}{k}V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \end{aligned} \end{equation} ps.已检验过线路的公式。 \end{document}