G-S first iteration with line charging

Bus 1 is the slack bus, bus 2 a PV bus and bus 3 a PQ bus. The line shunt admittance j0.04 pu is taken as the total line-charging admittance of each line, so $j0.02$ pu is placed at each end (nominal-π model). Gauss-Seidel is used with no acceleration factor.

Net injections: $P_2 = 1.5 - 0.5 = 1.0$ pu; $P_3 = -1.2$ pu, $Q_3 = -0.5$ pu. Initial values: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$.

i) Y-bus

$$\begin{aligned} y_{series} &= \frac{1}{0.026 + j0.110} = 2.0351 - j8.6099\ \text{pu} \\ Y_{ii} &= 2y_{series} + 2(j0.02) = 4.0701 - j17.1798 \\ Y_{ij} &= -y_{series} = -2.0351 + j8.6099 \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} 4.0701 - j17.1798 & -2.0351 + j8.6099 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & 4.0701 - j17.1798 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & -2.0351 + j8.6099 & 4.0701 - j17.1798 \end{bmatrix}\ \text{pu}$$

ii) Generator reactive power at bus 2 (first iteration)

$$\begin{aligned} \textstyle\sum_k Y_{2k}V_k^{(0)} &= 0.0611 - j0.2171 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2236\ \text{pu} \\ Q_{G2} &= Q_2 + Q_{L2} = 0.2236 + 0 = 0.2236\ \text{pu} \end{aligned}$$

No Q-limits are given, so bus 2 remains a PV bus.

iii) Voltages at buses 2 and 3 (first iteration)

Iteration 1

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0611 - j0.2171 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2236\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.2236}{1.0300} = 0.9709 - j0.2171 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -4.1312 + j17.4781 \\ V_2^{(1)} &= \frac{(0.9709 - j0.2171) - (-4.1312 + j17.4781)}{4.0701 - j17.1798} = 1.0419 + j0.0501 \\ \delta_2 &= 2.7554^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 2.7554^\circ = 1.0288 + j0.0495\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.6161 + j17.6254 \\ V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-4.6161 + j17.6254)}{4.0701 - j17.1798} = 0.9885 - j0.0353 \\ &= 0.9891\angle -2.0473^\circ\ \text{pu} \end{aligned}$$

So $V_2^{(1)} = 1.0300\angle 2.7554^\circ$ pu and $V_3^{(1)} = 0.9891\angle -2.0473^\circ$ pu.

iv) Total network real power loss (after first iteration)

Slack bus power with the first-iteration voltages:

$$\begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(-0.0351 - j0.3556) \\ &= -0.0362 - j0.3662 \\ \Rightarrow P_1 &= -0.0362\ \text{pu},\quad Q_1 = 0.3662\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = -0.0362 + 1.00 = 0.9638\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.3662 + 0.50 = 0.8662\ \text{pu} \end{aligned}$$

Total loss = sum of the calculated net injections at all buses (equal to the sum of all line losses):

$$\begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0290\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = -0.0013\ \text{pu} \end{aligned}$$

Answer: $Q_{G2} = 0.2236$ pu, $V_2 = 1.0300\angle 2.7554^\circ$ pu, $V_3 = 0.9891\angle -2.0473^\circ$ pu, total real power loss $= 0.0290$ pu (about 2.90 MW on a 100 MVA base).