更新电流，有功，无功计算公式。开始考虑接地支路。

Signed-off-by: facat <facat@facat.cn>

公式/公式.tex

@@ -61,11 +61,25 @@ Q_{ij}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \the
 &=-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}
+考虑接地支路后(利用Ali Abur 书上的公式)
+\newline
+有功传输功率
+\begin{equation}
+\begin{aligned}
+P_{ij}&=V_1^2(G_{ij}+G_{is})-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(B_{ij}+B_{is})-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}]
+2V_1(G_{ij}+G_{is})-V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
 \end{aligned}
 \end{equation}
 
@@ -94,7 +108,7 @@ Q_{ij}&=-[V_1^2-V_1V_2cos(\theta_1 - \theta_2)]B_{ij}-V_1V_2sin (\theta_1 - \the
 \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}]
+-2V_1(B_{12}+B_{is})-V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
 \end{aligned}
 \end{equation}
 
@@ -206,20 +220,21 @@ ps.已检验过线路的公式。
 以上公式已经可以完成状态估计，若要实现更好的收敛性，需要利用二阶导数。
 \par
 线路支路功率二阶导数
+有功部分
 \begin{equation}
 \begin{aligned}
 \frac{\partial^2 P_{12}}{\partial V_1^2}&=
-\frac{-2}{k^2}B_{12}\\
-&=\frac{-2B_{12}}{k^2}
+\frac{2}{k^2}(G_{12}+G_{1s})\\
+&=\frac{2 (G_{12}+G_{1s} ) } {k^2}
 \end{aligned}
 \end{equation}
 
 \begin{equation}
 \begin{aligned}
 \frac{\partial^2 P_{12}}{\partial V_1 \partial V_2 }&=
-\frac{-1}{k}[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}] \\
+\frac{-1}{k}[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}] \\
 &=
-\frac{-[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]}{k}
+\frac{-[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]}{k}
 \end{aligned}
 \end{equation}
 
@@ -232,26 +247,69 @@ ps.已检验过线路的公式。
 \begin{equation}
 \begin{aligned}
 \frac{\partial P_{12}}{\partial \theta_1^2}&=
-V_1V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{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}
 
 \begin{equation}
 \begin{aligned}
 \frac{\partial P_{12}}{\partial \theta_1 \partial \theta_2}&=
-V_1V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\
+\frac{V_1}{k}  V_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}]
+- \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 P_{12}}{\partial \theta_2^2}&=
--V_1V_2[-cos(\theta_1 - \theta_2)G_{ij}-sin (\theta_1 - \theta_2)B_{ij}] \\
+- \frac{V_1}{k} V_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}]
+\frac{V_1}{k} V_2[cos(\theta_1 - \theta_2)G_{ij}+sin (\theta_1 - \theta_2)B_{ij}]
+\end{aligned}
+\end{equation}
+无功部分
+\begin{equation}
+\begin{aligned}
+\frac{\partial^2 Q_{12}}{\partial V_1 \partial V_1}&=
+-2\frac{( B_{12}+B_{1s})}{k^2}
 \end{aligned}
 \end{equation}
 
+\begin{equation}
+\begin{aligned}
+\frac{\partial^2 Q_{12}}{\partial V_1 \partial V_2}&=
+-\frac{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^2 Q_{12}}{\partial V_2^2}&=0
+\end{aligned}
+\end{equation}
+
+\begin{equation}
+\begin{aligned}
+\frac{\partial^2 Q_{12}}{\partial \theta_1^2}&=
+-\frac{V_1}{k}V_2[-sin(\theta_1 - \theta_2)G_{ij}+cos (\theta_1 - \theta_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}
+
+\begin{equation}
+\begin{aligned}
+\frac{\partial^2 Q_{12}}{\partial \theta_1 \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}
+
+\begin{equation}
+\begin{aligned}
+\frac{\partial^2 Q_{12}}{\partial \theta_2^2}
+&=\frac{V_1}{k}V_2[sin(\theta_1 - \theta_2)G_{ij}-cos (\theta_1 - \theta_2)B_{ij}]
+\end{aligned}
+\end{equation}
+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}