G-S two iterations, lines 0.026+j0.11

Bus 1 is the slack bus, bus 2 a PV (generator) bus and bus 3 a PQ (load) bus. Lines 1-2, 2-3 and 3-1 each have $z = 0.026 + j0.110$ pu (no shunt admittance given). Gauss-Seidel is used with no acceleration factor and no Q-limits (none are given).

Bus data

Initial values: $V_2^{(0)} = 1.03\angle 0^\circ$, $V_3^{(0)} = 1.0\angle 0^\circ$.

Y-bus

$$\begin{aligned} y_{12} &= \frac{1}{0.0260 + j0.1100} = 2.0351 - j8.6099\ \text{pu} \\ y_{23} &= \frac{1}{0.0260 + j0.1100} = 2.0351 - j8.6099\ \text{pu} \\ y_{13} &= \frac{1}{0.0260 + j0.1100} = 2.0351 - j8.6099\ \text{pu} \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} 4.0701 - j17.2198 & -2.0351 + j8.6099 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & 4.0701 - j17.2198 & -2.0351 + j8.6099 \\ -2.0351 + j8.6099 & -2.0351 + j8.6099 & 4.0701 - j17.2198 \end{bmatrix}\ \text{pu}$$

Gauss-Seidel iterations

For the PV bus, $Q_2$ is calculated first, then $V_2$ is found and its magnitude reset to 1.03 pu, keeping the new angle. Bus 3 uses the newest $V_2$.

Iteration 1

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0611 - j0.2583 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2660\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.2660}{1.0300} = 0.9709 - j0.2583 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -4.1312 + j17.4781 \\ V_2^{(1)} &= \frac{(0.9709 - j0.2583) - (-4.1312 + j17.4781)}{4.0701 - j17.2198} = 1.0418 + j0.0500 \\ \delta_2 &= 2.7499^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 2.7499^\circ = 1.0288 + j0.0494\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.6153 + j17.6256 \\ V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-4.6153 + j17.6256)}{4.0701 - j17.2198} = 0.9863 - j0.0348 \\ &= 0.9869\angle -2.0203^\circ\ \text{pu} \end{aligned}$$

Iteration 2

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 1.2346 - j0.0838 \\ Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.1473\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.1473}{1.0288 - j0.0494} = 0.9766 - j0.0962 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -3.8038 + j17.4310 \\ V_2^{(2)} &= \frac{(0.9766 - j0.0962) - (-3.8038 + j17.4310)}{4.0701 - j17.2198} = 1.0261 + j0.0351 \\ \delta_2 &= 1.9572^\circ \Rightarrow V_2^{(2)} = 1.0300\angle 1.9572^\circ = 1.0294 + j0.0352\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{0.9863 + j0.0348} = -1.1973 + j0.5492 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -4.4939 + j17.6596 \\ V_3^{(2)} &= \frac{(-1.1973 + j0.5492) - (-4.4939 + j17.6596)}{4.0701 - j17.2198} = 0.9839 - j0.0411 \\ &= 0.9848\angle -2.3932^\circ\ \text{pu} \end{aligned}$$

Unknown variables after the 2nd iteration

Slack bus power (with the 2nd-iteration voltages)

$$\begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(0.1462 - j0.3897) \\ &= 0.1505 - j0.4014 \\ \Rightarrow P_1 &= 0.1505\ \text{pu},\quad Q_1 = 0.4014\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 0.1505 + 1.00 = 1.1505\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.4014 + 0.50 = 0.9014\ \text{pu} \end{aligned}$$

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

$$\begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0263\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.1114\ \text{pu} \end{aligned}$$

Note: $Q_{G2} = 0.1473$ pu is the value used in the 2nd iteration (found from the 1st-iteration voltages). Recomputing $Q_2$ from the final voltages gives 0.2011 pu (see the table above). The two differ because Gauss-Seidel has not yet converged.

Answer: after two iterations $V_2 = 1.0300\angle 1.9572^\circ$ pu, $V_3 = 0.9848\angle -2.3932^\circ$ pu, $Q_{G2} = 0.1473$ pu, $P_{G1} = 1.1505$ pu, $Q_{G1} = 0.9014$ pu, and real power loss $\approx 0.0263$ pu.