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似乎公式没推错，只是牛顿法对初值敏感，带入潮流的电压和相角迭代几次就可以收敛。

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Signed-off-by: dmy <dugg@21cn.com>
This commit is contained in:
dmy
2013-09-26 22:04:35 +08:00
parent a9d49d8edd
commit 59f7810cab
14 changed files with 484 additions and 68 deletions
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71
公式/公式.tex
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@@ -221,6 +221,8 @@ ps.已检验过线路的公式。
\par
线路支路功率二阶导数
有功部分
\newline
海森矩阵对电压
\begin{equation}
\begin{aligned}
\frac{\partial^2 P_{12}}{\partial V_1^2}&=
@@ -243,7 +245,7 @@ ps.已检验过线路的公式。
\frac{\partial^2 P_{12}}{\partial V_2^2}&=0
\end{aligned}
\end{equation}
海森矩阵对相角
\begin{equation}
\begin{aligned}
\frac{\partial P_{12}}{\partial \theta_1^2}&=
@@ -268,7 +270,41 @@ ps.已检验过线路的公式。
\frac{V_1}{k} V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}
海森矩阵对电压相角
\begin{equation}
\begin{aligned}
\frac{\partial P_{12}}{\partial V_1 \partial \theta_1}&=
-\frac{V_2}{k}[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\
&=\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 P_{12}}{\partial V_1 \partial \theta_2}&=
-\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 P_{12}}{\partial V_2 \partial \theta_1}&=
-\frac{V_1}{k}[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_2)B_{ij}] \\
&=\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 P_{12}}{\partial V_2 \partial \theta_2}&=
-\frac{V_1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}
无功部分
\newline
海森矩阵对电压
\begin{equation}
\begin{aligned}
\frac{\partial^2 Q_{12}}{\partial V_1 \partial V_1}&=
@@ -288,7 +324,7 @@ ps.已检验过线路的公式。
\frac{\partial^2 Q_{12}}{\partial V_2^2}&=0
\end{aligned}
\end{equation}
海森矩阵对电压
\begin{equation}
\begin{aligned}
\frac{\partial^2 Q_{12}}{\partial \theta_1^2}&=
@@ -310,6 +346,37 @@ ps.已检验过线路的公式。
&=\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}
海森矩阵对电压相角
\begin{equation}
\begin{aligned}
\frac{\partial Q_{12}}{\partial V_1 \partial \theta_1}&=
-\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 Q_{12}}{\partial V_1 \partial \theta_2}&=
-\frac{V_2}{k}[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{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 Q_{12}}{\partial V_2 \partial \theta_1}&=
-\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 Q_{12}}{\partial V_2 \partial \theta_2}&=
-\frac{V_1}{k}[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\
&=\frac{V_1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
\end{aligned}
\end{equation}
以上公式对线路和没有计及接地支路的变压器适用，只是线路中变比$k$为1.
\end{document}

12
公式/内点法公式.tex
View File

@@ -15,7 +15,7 @@ f(x)=[z-h(x)]^T W [z-h(x)]
\end{equation}
最优条件等价为
\begin{equation}
\bigtriangledown f(x)=J^T W [z-h(x)]=0
\bigtriangledown f(x)=J^T W [h(x)-z]=0
\end{equation}
其中$J$是$h(x)$的$Jacobi$矩阵
\newline
@@ -55,6 +55,16 @@ f(x)=[z-h(x)]^T W [z-h(x)]
J ^T W J + \tilde{Q}
\end{equation}
其中
$J$具有以下形式
\begin{equation}
J=\left[ %左括号
\begin{array}{cc} %该矩阵一共3列，每一列都居中放置
\frac{\partial f_1}{ \partial x_1} & \frac{\partial f_1}{ \partial x_2}\\ %第一行元素
\frac{\partial f_2}{ \partial x_1} & \frac{\partial f_2}{ \partial x_2}\\ %第二行元素
\end{array}
\right] %右括号
\end{equation}
\begin{equation}
J ^T W J=
(\omega_1 \frac{ \partial f_1 }{\partial x_1} \frac{ \partial f_1 }{\partial x_1}