G-S with PV bus Q limits and losses

The (1,1) element printed as $4 - j2$ is a misprint: each row of this $Y_{bus}$ must sum to zero (no shunt elements), so $Y_{11} = 4 - j12$ pu is used. Line lengths are not needed because $Y_{bus}$ is given.

Bus data

Limits: $0.1 \le Q_{G2} \le 0.4$ pu, and $Q_2 = Q_{G2} - Q_{L2} = Q_{G2} - 0.5$. Start: $V_2^{(0)} = 1\angle 0^\circ$, $V_3^{(0)} = 1\angle 0^\circ$.

$$Y_{bus} = \begin{bmatrix} 4.0000 - j12.0000 & -2.0000 + j6.0000 & -2.0000 + j6.0000 \\ -2.0000 + j6.0000 & 4.0000 - j12.0000 & -2.0000 + j6.0000 \\ -2.0000 + j6.0000 & -2.0000 + j6.0000 & 4.0000 - j12.0000 \end{bmatrix}\ \text{pu}$$

Gauss-Seidel iterations

Iteration 1

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= -0.1000 + j0.3000 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = -0.3000\ \text{pu} \end{aligned}$$

$Q_{G2} = Q_2 + Q_{L2} = 0.2000$ pu lies within $0.10 \le Q_{G2} \le 0.40$, so bus 2 remains a PV bus.

$$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{0.5000 + j0.3000}{1.0000} = 0.5000 + j0.3000 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -4.1000 + j12.3000 \\ V_2^{(1)} &= \frac{(0.5000 + j0.3000) - (-4.1000 + j12.3000)}{4.0000 - j12.0000} = 1.0150 + j0.0450 \\ \delta_2 &= 2.5385^\circ \Rightarrow V_2^{(1)} = 1.0000\angle 2.5385^\circ = 0.9990 + j0.0443\ \text{(magnitude reset to }1.00\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{1.0000} = -0.6000 + j0.3000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.3638 + j12.2055 \\ V_3^{(1)} &= \frac{(-0.6000 + j0.3000) - (-4.3638 + j12.2055)}{4.0000 - j12.0000} = 0.9870 - j0.0154 \\ &= 0.9871\angle -0.8912^\circ\ \text{pu} \end{aligned}$$

Iteration 2

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.5457 + j0.4417 \\ Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = -0.4171\ \text{pu} \end{aligned}$$

$Q_{G2} = Q_2 + Q_{L2} = 0.0829$ pu violates the limit $0.10 \le Q_{G2} \le 0.40$. So $Q_{G2}$ is fixed at 0.10 pu, i.e. $Q_2 = 0.10 - 0.50 = -0.4000$ pu, and bus 2 is treated as a PQ bus in this iteration (its voltage magnitude is not reset).

$$\begin{aligned} V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{0.5000 + j0.4000}{0.9990 - j0.0443} = 0.4818 + j0.4218 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -3.9819 + j12.2528 \\ V_2^{(2)} &= \frac{(0.4818 + j0.4218) - (-3.9819 + j12.2528)}{4.0000 - j12.0000} = 0.9989 + j0.0390 \\ &= 0.9997\angle 2.2359^\circ\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{0.9870 + j0.0154} = -0.6030 + j0.3133 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.3318 + j12.2155 \\ V_3^{(2)} &= \frac{(-0.6030 + j0.3133) - (-4.3318 + j12.2155)}{4.0000 - j12.0000} = 0.9859 - j0.0179 \\ &= 0.9860\angle -1.0398^\circ\ \text{pu} \end{aligned}$$

Unknowns after the 2nd iteration

In the 2nd iteration bus 2 hit its lower Q-limit, so $Q_{G2} = 0.1$ pu and $|V_2|$ drifted slightly from 1.0 to 0.9997 pu.

Slack bus:

$$\begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0500)(0.1037 - j0.7334) \\ &= 0.1089 - j0.7701 \\ \Rightarrow P_1 &= 0.1089\ \text{pu},\quad Q_1 = 0.7701\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 0.1089 + 2.00 = 2.1089\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.7701 + 0.60 = 1.3701\ \text{pu} \end{aligned}$$

Network losses

$$\begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0239\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.0718\ \text{pu} \end{aligned}$$

Answer: $V_2 = 0.9997\angle 2.2359^\circ$ pu, $Q_{G2} = 0.1$ pu (at its lower limit), $V_3 = 0.9860\angle -1.0398^\circ$ pu, $P_{G1} = 2.1089$ pu, $Q_{G1} = 1.3701$ pu, $P_{loss} = 0.0239$ pu, $Q_{loss} = 0.0718$ pu.