G-S load flow with given Ybus, losses

Bus 1 is the slack bus; buses 2 and 3 are load (PQ) buses. Gauss-Seidel is used with the given $Y_{bus}$, flat start $V_2^{(0)} = V_3^{(0)} = 1\angle 0^\circ$ pu, and no acceleration factor.

Data

$Y_{ii} = -j12$ pu and $Y_{ij} = j6$ pu. Net injections ($P_i = P_{Gi} - P_{Li}$):

$$\begin{aligned} P_2 - jQ_2 &= -0.7 + j0.5 \\ P_3 - jQ_3 &= -0.6 + j0.3 \end{aligned}$$

Gauss-Seidel iterations

Iteration 1

$$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-0.7000 + j0.5000}{1.0000} = -0.7000 + j0.5000 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= j12.3000 \\ V_2^{(1)} &= \frac{(-0.7000 + j0.5000) - (j12.3000)}{-j12.0000} = 0.9833 - j0.0583 \\ &= 0.9851\angle -3.3949^\circ\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{1.0000} = -0.6000 + j0.3000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= 0.3500 + j12.2000 \\ V_3^{(1)} &= \frac{(-0.6000 + j0.3000) - (0.3500 + j12.2000)}{-j12.0000} = 0.9917 - j0.0792 \\ &= 0.9948\angle -4.5644^\circ\ \text{pu} \end{aligned}$$

Iteration 2

$$\begin{aligned} V_2^{(2)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-0.7000 + j0.5000}{0.9833 + j0.0583} = -0.6793 + j0.5488 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= 0.4750 + j12.2500 \\ V_2^{(2)} &= \frac{(-0.6793 + j0.5488) - (0.4750 + j12.2500)}{-j12.0000} = 0.9751 - j0.0962 \\ &= 0.9798\angle -5.6339^\circ\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_3^{(2)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{-0.6000 + j0.3000}{0.9917 + j0.0792} = -0.5772 + j0.3486 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= 0.5772 + j12.1506 \\ V_3^{(2)} &= \frac{(-0.5772 + j0.3486) - (0.5772 + j12.1506)}{-j12.0000} = 0.9835 - j0.0962 \\ &= 0.9882\angle -5.5864^\circ\ \text{pu} \end{aligned}$$

Results after the 2nd iteration

Slack bus power

$$\begin{aligned} P_1 - jQ_1 &= V_1^{*}\textstyle\sum_k Y_{1k}V_k = (1.0500)(1.1543 - j0.8484) \\ &= 1.2121 - j0.8908 \\ \Rightarrow P_1 &= 1.2121\ \text{pu},\quad Q_1 = 0.8908\ \text{pu} \\ P_{G1} &= P_1 + P_{L1} = 1.2121 + 2.00 = 3.2121\ \text{pu} \\ Q_{G1} &= Q_1 + Q_{L1} = 0.8908 + 0.60 = 1.4908\ \text{pu} \end{aligned}$$

Network losses

Line flows, using $y_{ij} = -Y_{ij} = -j6$ pu and $S_{ij} = V_i[(V_i - V_j)y_{ij}]^{*}$:

The network is purely reactive (no resistance), so the real power loss is zero:

$$\begin{aligned} P_{loss} &= P_1 + P_2 + P_3 = 1.2121 + (-0.6108) + (-0.6012) = 0.0000\ \text{pu} \\ Q_{loss} &= Q_1 + Q_2 + Q_3 = 0.8908 + (-0.4318) + (-0.2873) = 0.1717\ \text{pu} \end{aligned}$$

(Here $P_2$, $P_3$, $Q_2$, $Q_3$ are recalculated with the 2nd-iteration voltages.)

Answer: $V_2 = 0.9798\angle -5.6339^\circ$ pu, $V_3 = 0.9882\angle -5.5864^\circ$ pu, $P_{G1} = 3.2121$ pu, $Q_{G1} = 1.4908$ pu, real power loss $= 0.0000$ pu, reactive power loss $= 0.1717$ pu.