Unsymmetric Two-Phase Current¶

Initial Short-Circuit Current¶

The two-phase initial short-circuit current is calculated in the same way the three-phase current is calculated, only with a source voltage of $$c \cdot \sqrt{2} \cdot V_N$$ instead of $$\frac{c \cdot V_N}{\sqrt{3}}$$:

$\begin{split}\begin{bmatrix} \underline{I}''_{k2, 1} \\ \vdots \\ \underline{I}''_{k2, m} \\ \end{bmatrix} = \begin{bmatrix} \frac{c_1 \cdot \sqrt{2} \cdot V_{N, 1}}{Z_{11} + Z_{fault}} \\ \vdots \\ \frac{c_n \cdot \sqrt{2} \cdot V_{N, n}}{Z_{nn} + Z_{fault}} \end{bmatrix}\end{split}$

Peak Short-Circuit Current¶

The peak short-circuit current is calculated as:

$\begin{split}\begin{bmatrix} i_{p2, 1} \\ \vdots \\ i_{p2, n} \\ \end{bmatrix} = \sqrt{2} \begin{bmatrix} \kappa_{1} \\ \vdots \\ \kappa_{1} \\ \end{bmatrix} \begin{bmatrix} \underline{I}''_{k2, 1} \\ \vdots \\ \underline{I}''_{k2, n} \\ \end{bmatrix}\end{split}$

where the factor $$\kappa$$ is calculated for each bus as defined here.

Thermal Short-Circuit Current¶

The equivalent

$\begin{split}\begin{bmatrix} \underline{I}_{th2, 1} \\ \vdots \\ \underline{I}_{th2, n} \\ \end{bmatrix} = \begin{bmatrix} \sqrt{m_1 + n_1} \\ \vdots \\ \sqrt{m_n + n_n} \\ \end{bmatrix} \begin{bmatrix} \underline{I}''_{k2, 1} \\ \vdots \\ \underline{I}''_{k2, n} \\ \end{bmatrix}\end{split}$

where the factors m and n are calculated for each bus as defined here.