G-S two iterations, lines 0.02+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 $z = 0.020 + j0.110$ 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

$$\begin{aligned} y_{12} &= \frac{1}{0.0200 + j0.1100} = 1.6000 - j8.8000\ \text{pu} \\ y_{23} &= \frac{1}{0.0200 + j0.1100} = 1.6000 - j8.8000\ \text{pu} \\ y_{13} &= \frac{1}{0.0200 + j0.1100} = 1.6000 - j8.8000\ \text{pu} \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} 3.2000 - j17.6000 & -1.6000 + j8.8000 & -1.6000 + j8.8000 \\ -1.6000 + j8.8000 & 3.2000 - j17.6000 & -1.6000 + j8.8000 \\ -1.6000 + j8.8000 & -1.6000 + j8.8000 & 3.2000 - j17.6000 \end{bmatrix}\ \text{pu}$$

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

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 0.0480 - j0.2640 \\ Q_2^{(1)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.2719\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.2719}{1.0300} = 0.9709 - j0.2640 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -3.2480 + j17.8640 \\ V_2^{(1)} &= \frac{(0.9709 - j0.2640) - (-3.2480 + j17.8640)}{3.2000 - j17.6000} = 1.0392 + j0.0508 \\ \delta_2 &= 2.7962^\circ \Rightarrow V_2^{(1)} = 1.0300\angle 2.7962^\circ = 1.0288 + j0.0502\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{1.0000} = -1.2000 + j0.5000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -3.7362 + j18.0368 \\ V_3^{(1)} &= \frac{(-1.2000 + j0.5000) - (-3.7362 + j18.0368)}{3.2000 - j17.6000} = 0.9899 - j0.0359 \\ &= 0.9905\angle -2.0757^\circ\ \text{pu} \end{aligned}$$

Iteration 2

$$\begin{aligned} \sum_{k=1}^{3} Y_{2k}V_k &= 1.2603 - j0.1132 \\ Q_2^{(2)} &= -\text{Im}\{V_2^{*}\textstyle\sum_k Y_{2k}V_k\} = 0.1798\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{1.0000 - j0.1798}{1.0288 - j0.0502} = 0.9782 - j0.1270 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= -2.9161 + j17.8324 \\ V_2^{(2)} &= \frac{(0.9782 - j0.1270) - (-2.9161 + j17.8324)}{3.2000 - j17.6000} = 1.0267 + j0.0346 \\ \delta_2 &= 1.9298^\circ \Rightarrow V_2^{(2)} = 1.0300\angle 1.9298^\circ = 1.0294 + j0.0347\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$ $$\begin{aligned} V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-1.2000 + j0.5000}{0.9899 + j0.0359} = -1.1924 + j0.5483 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= -3.6003 + j18.0674 \\ V_3^{(2)} &= \frac{(-1.1924 + j0.5483) - (-3.6003 + j18.0674)}{3.2000 - j17.6000} = 0.9876 - j0.0428 \\ &= 0.9886\angle -2.4788^\circ\ \text{pu} \end{aligned}$$

Unknown variables after the 2nd iteration

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

$$\begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0300)(0.1397 - j0.3651) \\ &= 0.1439 - j0.3761 \\ \Rightarrow P_1 &= 0.1439\ \text{pu},\quad Q_1 = 0.3761\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 0.1439 + 1.00 = 1.1439\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.3761 + 0.50 = 0.8761\ \text{pu} \end{aligned}$$

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

$$\begin{aligned} P_{loss} &= \textstyle\sum P_i = 0.0201\ \text{pu} \\ Q_{loss} &= \textstyle\sum Q_i = 0.1106\ \text{pu} \end{aligned}$$

Note: $Q_{G2} = 0.1798$ pu is the value used in the 2nd iteration (found from the 1st-iteration voltages). Recomputing $Q_2$ from the final voltages gives 0.2251 pu (see the table above). The two differ because Gauss-Seidel has not yet converged.

Answer: after two iterations $V_2 = 1.0300\angle 1.9298^\circ$ pu, $V_3 = 0.9886\angle -2.4788^\circ$ pu, $Q_{G2} = 0.1798$ pu, $P_{G1} = 1.1439$ pu, $Q_{G1} = 0.8761$ pu, and real power loss $\approx 0.0201$ pu.