209 lines
6.0 KiB
TeX
209 lines
6.0 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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线路功率(不考虑接地导纳)
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\begin{equation}
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\begin{aligned}
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\dot{I}_{12}&=(V_1e^{j \theta_1} - V_2e^{j \theta_2})
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(G_{ij}+jB_{ij})
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\end{aligned}
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\end{equation}
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共轭后
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\begin{equation}
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\begin{aligned}
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\dot{I}^*_{12}&=(V_1e^{-j \theta_1} - V_2e^{-j \theta_2})
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(G_{ij}-jB_{ij})
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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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\dot{S}_{12}&=V_1e^{j \theta_1}
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\dot{I}^*_{12} \\
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&=V_1e^{j \theta_1}
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(V_1e^{-j \theta_1} - V_2e^{-j \theta_2})
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(G_{ij}-jB_{ij}) \\
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&=(V_1e^{j \theta_1}V_1e^{-j \theta_1}
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-V_1e^{j \theta_1}V_2e^{-j \theta_2}
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)
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(G_{ij}-jB_{ij}) \\
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&=[V_1^2-V_1V_2e^{j (\theta_1 - \theta_2) }
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]
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(G_{ij}-jB_{ij}) \\
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&=\{V_1^2-V_1V_2[cos(\theta_1 - \theta_2)+jsin (\theta_1 - \theta_2) ]\}
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(G_{ij}-jB_{ij}) \\
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&=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij} \\
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&+[V_1^2-V_1V_2cos(\theta_1 - \theta_2)](-jB_{ij}) \\
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&-jV_1V_2sin (\theta_1 - \theta_2)G_{ij} \\
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&-jV_1V_2sin (\theta_1 - \theta_2)(-jB_{ij}) \\
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&=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\
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&-j[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-jV_1V_2sin (\theta_1 - \theta_2)G_{ij}
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\end{aligned}
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\end{equation}
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有功传输功率
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\begin{equation}
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\begin{aligned}
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P_{ij}&=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\
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&=V_1^2G_{ij}-V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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无功传输功率
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\begin{equation}
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\begin{aligned}
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Q_{ij}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \theta_2)G_{ij} \\
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&=-V_1^2B_{ij}-V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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线路有功功率Jacobi
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\begin{equation}
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\begin{aligned}
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\frac{\partial P_{ij}}{\partial V_1}=
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2V_1G_{ij}-V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial P_{12}}{\partial V_2}=
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-V_1[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial P_{12}}{\partial \theta_1}&=
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-V_1V_2[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\
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&=V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial P_{12}}{\partial \theta_2}&=
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-V_1V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\
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\end{aligned}
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\end{equation}
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线路无功功率Jacobi
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\begin{equation}
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\begin{aligned}
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\frac{\partial Q_{12}}{\partial V_1}&=
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-2V_1B_{12}-V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial Q_{12}}{\partial V_2}&=
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-V_1[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial Q_{12}}{\partial \theta_1}&=
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-V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial Q_{12}}{\partial \theta_2}&=
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-V_1V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\
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&=V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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变压器支路功率
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\newline
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变压器模型为
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\newline
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\begin{center}
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----k:1----z----高压侧 \\
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or \\
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i----k:1----z----j \\
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\end{center}
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变压器支路功率
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\newline
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变压器有功输送功率
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\begin{equation}
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\begin{aligned}
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P_{ij}&=[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]G_{ij}-V_1V_2sin (\theta_1 - \theta_2)B_{ij} \\
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&=\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}]
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\end{aligned}
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\end{equation}
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变压器无功输送功率
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\begin{equation}
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\begin{aligned}
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Q_{ij}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \theta_2)G_{ij} \\
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&=-\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}]
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\end{aligned}
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\end{equation}
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变压器输送有功功率Jacobi
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\begin{equation}
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\begin{aligned}
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\frac{\partial P_{ij}}{\partial V_1}=
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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}]
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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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\frac{\partial P_{12}}{\partial V_2}=
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-\frac{V_1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial P_{12}}{\partial \theta_1}
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&=\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial P_{12}}{\partial \theta_2}&=
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-\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\
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\end{aligned}
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\end{equation}
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变压器输送无功功率Jacobi
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\begin{equation}
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\begin{aligned}
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\frac{\partial Q_{12}}{\partial V_1}&=
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-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}]
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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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\frac{\partial Q_{12}}{\partial V_2}&=
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-\frac{V_1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial Q_{12}}{\partial \theta_1}&=
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-\frac{V_1}{k}V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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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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\frac{\partial Q_{12}}{\partial \theta_2}
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&=\frac{V_1}{k}V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
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\end{aligned}
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\end{equation}
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ps.已检验过线路的公式。
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\end{document}
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