Load flow equations from Ybus

Load flow equations relate the power injected at each bus to the bus voltages through the bus admittance matrix. They follow from the nodal equation $I_{bus} = Y_{bus}V_{bus}$.

Step 1: Current injection at bus i

For an $n$-bus network,

$$\begin{bmatrix} I_1 \\ I_2 \\ \vdots \\ I_n \end{bmatrix} = \begin{bmatrix} Y_{11} & Y_{12} & \cdots & Y_{1n} \\ Y_{21} & Y_{22} & \cdots & Y_{2n} \\ \vdots & & & \vdots \\ Y_{n1} & Y_{n2} & \cdots & Y_{nn} \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \\ \vdots \\ V_n \end{bmatrix}$$

so the current injected at bus $i$ is

$$I_i = \sum_{k=1}^{n} Y_{ik}V_k$$

Step 2: Complex power injection

The net complex power injected at bus $i$ is $S_i = P_i + jQ_i = V_iI_i^{*}$. Taking the conjugate,

$$\begin{aligned} P_i - jQ_i &= V_i^{*}I_i \\ &= V_i^{*}\sum_{k=1}^{n} Y_{ik}V_k \end{aligned}$$

Here $P_i = P_{Gi} - P_{Li}$ and $Q_i = Q_{Gi} - Q_{Li}$.

Step 3: Polar form (used in N-R)

Put $V_i = |V_i|\angle\delta_i$ and $Y_{ik} = |Y_{ik}|\angle\theta_{ik}$:

$$P_i - jQ_i = \sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\angle(\theta_{ik} - \delta_i + \delta_k)$$

Equating real and imaginary parts:

$$\begin{aligned} P_i &= \sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\cos(\theta_{ik} - \delta_i + \delta_k) \\ Q_i &= -\sum_{k=1}^{n}|V_i||V_k||Y_{ik}|\sin(\theta_{ik} - \delta_i + \delta_k) \end{aligned}$$

These are the static load flow equations: $2n$ real, non-linear equations in $|V|$ and $\delta$, solved by iteration.

Step 4: Form used in Gauss-Seidel

Separating the $k = i$ term in Step 2:

$$V_i = \frac{1}{Y_{ii}}\left[\frac{P_i - jQ_i}{V_i^{*}} - \sum_{k\ne i} Y_{ik}V_k\right]$$

At each bus two of the four quantities ($P$, $Q$, $|V|$, $\delta$) are given and the other two are found by solving these equations.