Linear AC Power Flow¶

Following the formulation presented in Linearized AC Load Flow Applied to Analysis in Electric Power Systems [1], we obtain a way to solve circuits in one shot (without iterations) and with quite positive results for a linear approximation.

$\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{bmatrix} \times \begin{bmatrix} \Delta \theta\\ \Delta |V|\\ \end{bmatrix} = \begin{bmatrix} Rhs_1\\ Rhs_2\\ \end{bmatrix}$

Where:

$A_{11} = -Im\left(Y_{series}[pqpv, pqpv]\right)$
$A_{12} = Re\left(Y_{bus}[pqpv, pq]\right)$
$A_{21} = -Im\left(Y_{series}[pq, pqpv]\right)$
$A_{22} = -Re\left(Y_{bus}[pq, pq]\right)$
$Rhs_1 = P[pqpv]$
$Rhs_2 = Q[pq]$

Here, $Y_{bus}$ is the normal circuit admittance matrix and $Y_{series}$ is the admittance matrix formed with only series elements of the $\pi$ model, this is neglecting all the shunt admittances.

Solving the vector $[\Delta \theta + 0, \Delta |V| + 1]$ we get $\theta$ for the pq and pv nodes and $|V|$ for the pq nodes.

For equivalence with the paper:

$-B' = -Im(Y_{series}[pqpv, pqpv])$
$G = Re(Y_{bus}[pqpv, pq])$
$-G' = -Im(Y_{series}[pq, pqpv])$
$-B = -Re(Y_{bus}[pq, pq])$

[1]	Rossoni, P. / Moreti da Rosa, W. / Antonio Belati, E., Linearized AC Load Flow Applied to Analysis in Electric Power Systems, IEEE Latin America Transactions, 14, 9; 4048-4053, 2016