対称座標法 一相断線故障の計算


対称座標法を用いて一相談線時の故障電圧,故障電流を求める

なお,記事中の$\dot{V}_i',\,\dot{V}_i''\,(i=a,b,c)$は故障点の端子電圧,$\dot{E}_f$は故障直前に故障点に現れていた正相の電圧,$\dot{I}_i',\,\dot{I}_i''\,(i=a,b,c)$は各相の線電流を表し,下図のような向きで定義する

断線故障 電圧と電流の定義

本記事では,下図のように,a相,b相,c相のうちa相が断線するとしている

一相断線故障

計算

対称分回路の作成

故障条件は

\begin{aligned} \dot{V}_b'&=\dot{V}_b'' \tag{1.1}\\ \end{aligned}
\begin{aligned} \dot{V}_c'&=\dot{V}_c'' \tag{1.2}\\ \end{aligned}
\begin{aligned} \dot{I}_a'&=\dot{I}_a''=0 \tag{1.3}\\ \end{aligned}
\begin{aligned} \dot{I}_b'&+\dot{I}_b''=0 \tag{1.4}\\ \end{aligned}
\begin{aligned} \dot{I}_c'&+\dot{I}_c''=0 \tag{1.5}\\ \end{aligned}

式 (1.1), (1.2)より

\begin{aligned} \left\{ \begin{aligned} \dot{V}_0'+a^2\dot{V}_1'+a\dot{V}_2'&=\dot{V}_0''+a^2\dot{V}_1''+a\dot{V}_2''\\ \dot{V}_0'+a\dot{V}_1'+a^2\dot{V}_2'&=\dot{V}_0''+a\dot{V}_1''+a^2\dot{V}_2'' \end{aligned} \right. \end{aligned}
\begin{aligned} \Leftrightarrow \left\{ \begin{aligned} (\dot{V}_0'-\dot{V}_0'')+a^2(\dot{V}_1'-\dot{V}_1'')+a(\dot{V}_2'-\dot{V}_2'')&=0\\ (\dot{V}_0'-\dot{V}_0'')+a(\dot{V}_1'-\dot{V}_1'')+a^2(\dot{V}_2'-\dot{V}_2'')&=0 \end{aligned}\tag{1.6} \right. \end{aligned}

式 (1.6) の辺々を引いて

\begin{aligned} (a^2-a)(\dot{V}_1'-\dot{V}_1'')&=(a^2-a)(\dot{V}_2'-\dot{V}_2'')\\ \dot{V}_1'-\dot{V}_1''&=\dot{V}_2'-\dot{V}_2''\tag{1.7} \end{aligned}

式 (1.7) を式 (1.6) 第一式に代入して

\begin{aligned} (\dot{V}_0'-\dot{V}_0'')+(a^2+a)(\dot{V}_1'-\dot{V}_1'')&=0\\ \dot{V}_0'-\dot{V}_0''&=\dot{V}_1'-\dot{V}_1'' \tag{1.8} \end{aligned}

式 (1.7), (1.8) より

\begin{aligned} \dot{V}_0'-\dot{V}_0''=\dot{V}_1'-\dot{V}_1''=\dot{V}_2'-\dot{V}_2'' \tag{1.9} \end{aligned}

式 (1.4), (1.5) より

\begin{aligned} \left\{ \begin{aligned} \dot{I}_0'+a^2\dot{I}_1'+a\dot{I}_2'+\dot{I}_0''+a^2\dot{I}_1''+a\dot{I}_2''&=0\\ \dot{I}_0'+a\dot{I}_1'+a^2\dot{I}_2'+\dot{I}_0''+a\dot{I}_1''+a^2\dot{I}_2''&=0 \end{aligned} \right. \end{aligned}
\begin{aligned} \Leftrightarrow \left\{ \begin{aligned} (\dot{I}_0'+\dot{I}_0'')+a^2(\dot{I}_1'+\dot{I}_1'')+a(\dot{I}_2'+\dot{I}_2'')&=0\\ (\dot{I}_0'+\dot{I}_0'')+a(\dot{I}_1'+\dot{I}_1'')+a^2(\dot{I}_2'+\dot{I}_2'')&=0 \end{aligned}\tag{1.10} \right. \end{aligned}

式 (1.10) を辺々引いて

\begin{aligned} (a^2-a)(\dot{I}_1'+\dot{I}_1'')&=(a^2-a)(\dot{I}_2'+\dot{I}_2'')\\ \dot{I}_1'+\dot{I}_1''&=\dot{I}_2'+\dot{I}_2'' \tag{1.11} \end{aligned}

式 (1.11) を式 (1.10) 第一式に代入すると

\begin{aligned} (\dot{I}_0'+\dot{I}_0'')+(a+a^2)(\dot{I}_1'+\dot{I}_1'')&=0\\ \dot{I}_0'+\dot{I}_0''&=\dot{I}_1'+\dot{I}_1'' \tag{1.12} \end{aligned}

式 (1.11), (1.12) から,

\begin{aligned} \dot{I}_0'+\dot{I}_0''=\dot{I}_1'+\dot{I}_1'' =\dot{I}_1'+\dot{I}_1'' \tag{1.13} \end{aligned}

式 (1.3) を変形して

\begin{aligned} \left\{ \begin{aligned} \dot{I}_0'+\dot{I}_1'+\dot{I}_2' &= 0\\ \dot{I}_0''+\dot{I}_1''+\dot{I}_2'' &= 0 \end{aligned} \right. \end{aligned}\tag{1.14}

式 (1.14) の辺々を足し,式 (1.13) を適用して

\begin{aligned} (\dot{I}_0'+\dot{I}_0'')+(\dot{I}_1'+\dot{I}_1'')+(\dot{I}_2'+\dot{I}_2'') &= 0\\ \dot{I}_0'+\dot{I}_0''&= 0\\ \therefore\dot{I}_0'+\dot{I}_0''=\dot{I}_1'+\dot{I}_1''=\dot{I}_2'+\dot{I}_2''&=0\tag{1.15} \end{aligned}

式 (1.9), (1.15) を満たすように対称分回路を作成すると,下図のようになる

対称座標法 一相断線故障

対称成分の導出

上の対称分回路から,対称成分を導出する

便宜上,以下のように$\dot{Z}_0,\,\dot{Z}_1,\,\dot{Z}_2$を定義する

\begin{aligned} \dot{Z}_0&= \dot{Z}_0'+ \dot{Z}_0''\tag{2.1}\\ \end{aligned}
\begin{aligned} \dot{Z}_1&= \dot{Z}_1'+ \dot{Z}_1''\tag{2.2}\\ \end{aligned}
\begin{aligned} \dot{Z}_2&= \dot{Z}_2'+ \dot{Z}_2''\tag{2.3}\\ \end{aligned}
\begin{aligned} \Delta&=\dot{Z}_0+ \dot{Z}_1+\dot{Z}_2\tag{2.4} \end{aligned}

電流$\dot{I}_1$を用いて電圧方程式を立てると,

\begin{aligned} \dot{E}_f'-\dot{Z}_1\dot{I}_1'-\dot{E}_f''-\frac{\dot{Z}_0\dot{Z}_2}{\dot{Z}_0+\dot{Z}_2}\dot{I}_1&=0\\ \end{aligned}
\begin{aligned} \therefore \dot{I}_1'=-\dot{I}_1''&= \frac{\dot{Z}_0+\dot{Z}_2}{\dot{Z}_0\dot{Z}_1+\dot{Z}_1\dot{Z}_2+\dot{Z}_2\dot{Z}_0}(\dot{E}_f'-\dot{E}_f'')\\ &= \frac{\dot{Z}_0+\dot{Z}_2}{\Delta}(\dot{E}_f'-\dot{E}_f'') \tag{2.5} \end{aligned}

さらに,

\begin{aligned} \dot{I}_2'=-\dot{I}_2''&= -\frac{\dot{Z}_0}{\dot{Z}_0+\dot{Z}_2}\dot{I}_1'\\ &= -\frac{\dot{Z}_0}{\Delta}(\dot{E}_f'-\dot{E}_f'')\tag{2.6}\\ \end{aligned}
\begin{aligned} \dot{I}_0'=-\dot{I}_0''&= -\frac{\dot{Z}_2}{\dot{Z}_0+\dot{Z}_2}\dot{I}_1'\\ &= -\frac{\dot{Z}_2}{\Delta}(\dot{E}_f'-\dot{E}_f'')\tag{2.7} \end{aligned}

電圧の対称成分は,

\begin{aligned} \dot{V}_1'&=\dot{E}_f'-\dot{Z}_1'\dot{I}_1'\\ &= \dot{E}_f'- \frac{\dot{Z}_1'(\dot{Z}_0+\dot{Z}_2)}{\Delta}(\dot{E}_f'-\dot{E}_f'')\\ &= \frac{\Delta-\dot{Z}_1'(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f'+ \frac{\dot{Z}_1'(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f''\\ &= \frac{\dot{Z}_0\dot{Z}_2+\dot{Z}_1''(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f'+ \frac{\dot{Z}_1'(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f'' \tag{2.8} \end{aligned}
\begin{aligned} \dot{V}_1''&=\dot{E}_f''-\dot{Z}_1''\dot{I}_1''\\ &= \dot{E}_f''+ \frac{\dot{Z}_1''(\dot{Z}_0+\dot{Z}_2)}{\Delta}(\dot{E}_f'-\dot{E}_f'')\\ &= \frac{\dot{Z}_1''(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f'+\frac{\Delta-\dot{Z}_1''(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f''+\\ &= \frac{\dot{Z}_1''(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f'+\frac{\dot{Z}_0\dot{Z}_2+\dot{Z}_1'(\dot{Z}_0+\dot{Z}_2)}{\Delta}\dot{E}_f'' \tag{2.9} \end{aligned}
\begin{aligned} \dot{V}_2'&=-\dot{Z}_2'\dot{I}_2'\\ &= \frac{\dot{Z}_2'\dot{Z}_0}{\Delta}(\dot{E}_f'-\dot{E}_f'')\tag{2.10}\\ \end{aligned}
\begin{aligned} \dot{V}_2''&=-\dot{Z}_2''\dot{I}_2''\\ &= -\frac{\dot{Z}_2''\dot{Z}_0}{\Delta}(\dot{E}_f'-\dot{E}_f'')\tag{2.11}\\ \end{aligned}
\begin{aligned} \dot{V}_0'&=-\dot{Z}_0'\dot{I}_0'\\ &= \frac{\dot{Z}_0'\dot{Z}_2}{\Delta}(\dot{E}_f'-\dot{E}_f'')\tag{2.11}\\ \end{aligned}
\begin{aligned} \dot{V}_0''&=-\dot{Z}_0''\dot{I}_0''\\ &= -\frac{\dot{Z}_0''\dot{Z}_2}{\Delta}(\dot{E}_f'-\dot{E}_f'')\tag{2.12}\\ \end{aligned}

以上をまとめると

\begin{aligned} \left[ \begin{array}{c} \dot{I}_0' \\ \dot{I}_1' \\ \dot{I}_2' \end{array} \right] &= - \left[ \begin{array}{c} \dot{I}_0'' \\ \dot{I}_1'' \\ \dot{I}_2'' \end{array} \right] = \frac{\dot{E}_f'-\dot{E}_f''}{\Delta} \left[ \begin{array}{c} - \dot{Z}_2 \\ \dot{Z}_0+\dot{Z}_2 \\ -\dot{Z}_0 \end{array} \right]\tag{2.13} \end{aligned}
\begin{aligned} \left[ \begin{array}{c} \dot{V}_0' \\ \dot{V}_1' \\ \dot{V}_2' \end{array} \right] &= \frac{1}{\Delta} \left[ \begin{array}{c} \dot{Z}_0'\dot{Z}_2(\dot{E}_f'-\dot{E}_f'')\\ \left(\dot{Z}_0\dot{Z}_2+\dot{Z}_1''(\dot{Z}_0+\dot{Z}_2)\right)\dot{E}_f'+ \left(\dot{Z}_1'(\dot{Z}_0+\dot{Z}_2)\right)\dot{E}_f''\\ \dot{Z}_2'\dot{Z}_0(\dot{E}_f'-\dot{E}_f'') \end{array} \right]\tag{2.14}\\ \end{aligned}
\begin{aligned} \left[ \begin{array}{c} \dot{V}_0'' \\ \dot{V}_1''\\ \dot{V}_2'' \end{array} \right] &= \frac{1}{\Delta} \left[ \begin{array}{c} - \dot{Z}_0''\dot{Z}_2(\dot{E}_f'-\dot{E}_f'')\\ \left(\dot{Z}_1''(\dot{Z}_0+\dot{Z}_2)\right)\dot{E}_f'+ \left(\dot{Z}_0\dot{Z}_2+\dot{Z}_1'(\dot{Z}_0+\dot{Z}_2)\right)\dot{E}_f''\\ -\dot{Z}_2''\dot{Z}_0(\dot{E}_f'-\dot{E}_f'') \end{array} \right]\tag{2.15} \end{aligned}

ここで,故障点の両側の対称成分を各々引くと,きれいな式になる

これは故障点の両端の電位差の対称成分を表している

\begin{aligned} \left[ \begin{array}{c} \dot{V}_0'-\dot{V}_0'' \\ \dot{V}_1'-\dot{V}_1''\\ \dot{V}_2'-\dot{V}_2'' \end{array} \right] &= \frac{\dot{Z}_0\dot{Z}_2}{\Delta}(\dot{E}_f'-\dot{E}_f'') \left[ \begin{array}{c} 1\\ 1\\ 1 \end{array} \right]\tag{2.16} \end{aligned}

各相の電圧,電流の導出

故障点両端の電圧,電流はかなり煩雑になるので,ここからは故障点における両端の電位差と電流について考える

式 (3.1), (3.2) のように$\dot{V}_a,\,\dot{V}_b,\,\dot{V}_c,\,\dot{I}_a,\,\dot{I}_b,\,\dot{I}_c$を定義する

\begin{aligned} \left[ \begin{array}{c} \dot{I}_a \\ \dot{I}_b \\ \dot{I}_c \end{array} \right] = \left[ \begin{array}{c} \dot{I}_a' \\ \dot{I}_b' \\ \dot{I}_c' \end{array} \right] &= - \left[ \begin{array}{c} \dot{I}_a'' \\ \dot{I}_b'' \\ \dot{I}_c'' \end{array} \right]\tag{3.1} \end{aligned}
\begin{aligned} \left[ \begin{array}{c} \dot{V}_a \\ \dot{V}_b \\ \dot{V}_c \end{array} \right] = \left[ \begin{array}{c} \dot{V}_a' \\ \dot{V}_b' \\ \dot{V}_c' \end{array} \right] - \left[ \begin{array}{c} \dot{V}_a'' \\ \dot{V}_b'' \\ \dot{V}_c'' \end{array} \right]\tag{3.2} \end{aligned}

$\dot{V}_a,\,\dot{V}_b,\,\dot{V}_c$の対称成分は,それぞれ$\dot{V}_0'-\dot{V}_0'',\,\dot{V}_1'-\dot{V}_1'',\,\dot{V}_2'-\dot{V}_2''$となり, $\dot{I}_a,\,\dot{I}_b,\,\dot{I}_c$の対称成分は,それぞれ$\dot{I}_0'(=-\dot{I}_0''),\,\dot{I}_1'(=-\dot{I}_1''),\,\dot{I}_2'(=-\dot{I}_2'')$となる

したがって,電圧の対称成分は,式 (2.16) から,

\begin{aligned} \left[ \begin{array}{c} \dot{V}_a \\ \dot{V}_b \\ \dot{V}_c \end{array} \right] &= \left[ \begin{array}{c} 1&1&1\\ 1&a^2&a\\ 1&a&a^2\\ \end{array} \right] \left[ \begin{array}{c} \dot{V}_0'-\dot{V}_0'' \\ \dot{V}_1'-\dot{V}_1''\\ \dot{V}_2'-\dot{V}_2'' \end{array} \right]\\ &= \frac{\dot{Z}_0\dot{Z}_2}{\Delta}(\dot{E}_f'-\dot{E}_f'') \left[ \begin{array}{c} 1&1&1\\ 1&a^2&a\\ 1&a&a^2\\ \end{array} \right] \left[ \begin{array}{c} 1\\ 1\\ 1 \end{array} \right]\\ &= \frac{3\dot{Z}_0\dot{Z}_2}{\Delta}(\dot{E}_f'-\dot{E}_f'') \left[ \begin{array}{c} 1\\ 0\\ 0 \end{array} \right]\tag{3.3} \end{aligned}

同様に,電流の対称成分は,

\begin{aligned} \left[ \begin{array}{c} \dot{I}_a \\ \dot{I}_b \\ \dot{I}_c \end{array} \right] &= \left[ \begin{array}{c} 1&1&1\\ 1&a^2&a\\ 1&a&a^2\\ \end{array} \right] \left[ \begin{array}{c} \dot{I}_0\\ \dot{I}_1\\ \dot{I}_2 \end{array} \right]\\ &= \left[ \begin{array}{c} 1&1&1\\ 1&a^2&a\\ 1&a&a^2\\ \end{array} \right] \left[ \begin{array}{c} \dot{I}_0' \\ \dot{I}_1'\\ \dot{I}_2' \end{array} \right]\\ &= \frac{\dot{E}_f'-\dot{E}_f''}{\Delta} \left[ \begin{array}{c} 1&1&1\\ 1&a^2&a\\ 1&a&a^2\\ \end{array} \right] \left[ \begin{array}{c} - \dot{Z}_2 \\ \dot{Z}_0+\dot{Z}_2 \\ -\dot{Z}_0 \end{array} \right]\\ &= \frac{\dot{E}_f'-\dot{E}_f''}{\Delta} \left[ \begin{array}{c} 0 \\ (a^2-1) \dot{Z}_0+(a^2-a)\dot{Z}_2 \\ (a-1) \dot{Z}_0+(a-a^2)\dot{Z}_2 \\ \end{array} \right]\tag{3.4} \end{aligned}

おわりに

送電系統の非対称故障計算の基本とまとめはこちら

www.charter-blog.com