Find |V3|, θ3 and QG2 with π lines

Bus 1 is the slack bus ($1.0\angle 0^\circ$), bus 2 a PV bus ($P_2 = 0.6661$ pu, $|V_2| = 1.04$ pu) and bus 3 a PQ bus ($P_3 = -2.5$, $Q_3 = -1.0$ pu). Each line is a nominal-π with $z_L = j0.1$ pu and $j0.02$ pu at each end. The unknowns are found from the converged load flow (Newton-Raphson, polar form, flat start $V_2^0 = 1.04\angle 0^\circ$, $V_3^0 = 1\angle 0^\circ$).

Y-bus

$$\begin{aligned} y_L &= 1/j0.1 = -j10\ \text{pu} \\ Y_{ii} &= 2(-j10) + 2(j0.02) = -j19.96 \\ Y_{ij} &= +j10 \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} -j19.9600 & j10.0000 & j10.0000 \\ j10.0000 & -j19.9600 & j10.0000 \\ j10.0000 & j10.0000 & -j19.9600 \end{bmatrix}\ \text{pu}$$

First iteration (shown in full)

With $Y_{ij} = |Y_{ij}|\angle\theta_{ij}$, $V_i = |V_i|\angle\delta_i$ and $\phi_{ij} = \theta_{ij} - \delta_i + \delta_j$:

$$\begin{aligned} P_i &= \sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ Q_i &= -\sum_{j=1}^{n} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \end{aligned}$$

Jacobian elements (polar form, $j \ne i$):

$$\begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned}$$

At the flat start the $\cos$ terms vanish ($\theta = \pm 90^\circ$), so $P_2^0 = P_3^0 = 0$, and

$$\begin{aligned} Q_2^{0} &= 0.7887, \quad Q_3^{0} = -0.4400 \\ \Delta P_2 &= 0.6661, \quad \Delta P_3 = -2.5000, \quad \Delta Q_3 = -1.0 - (-0.4400) = -0.5600 \end{aligned}$$ $$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0400)(1.0000)(10.0000)\sin(90.00^\circ) + (1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = 20.8000 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0400)(1.0000)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = -10.4000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(10.0000)\sin(90.00^\circ) + (1.0000)(1.0400)(10.0000)\sin(90.00^\circ) = 20.4000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(19.9600)\cos(-90.00^\circ) + (1.0000)(10.0000)\cos(90.00^\circ) + (1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(10.0000)\cos(90.00^\circ) + (1.0000)(1.0400)(10.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(19.9600)\sin(-90.00^\circ) - (1.0000)(10.0000)\sin(90.00^\circ) - (1.0400)(10.0000)\sin(90.00^\circ) = 19.5200 \end{aligned}$$ $$\begin{bmatrix} 0.6661 \\ -2.5000 \\ -0.5600 \end{bmatrix} = \begin{bmatrix} 20.8000 & -10.4000 & 0.0000 \\ -10.4000 & 20.4000 & 0.0000 \\ 0.0000 & 0.0000 & 19.5200 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} \Rightarrow \begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix} = \begin{bmatrix} -0.0393 \\ -0.1426 \\ -0.0287 \end{bmatrix}$$

Iterations to convergence

The same steps are repeated (Jacobian recalculated each time) until all mismatches are below $10^{-5}$ pu:

The mismatches in each row are those at the start of that iteration; the angles and $|V_3|$ are the values at its end.

Results

Converged voltages: $V_2 = 1.0400\angle -2.285^\circ$ pu, $V_3 = 0.9614\angle -8.490^\circ$ pu.

Reactive power at bus 2:

$$\begin{aligned} Q_2 &= -\sum_{j}|V_2||V_j||Y_{2j}|\sin(\theta_{2j} - \delta_2 + \delta_j) \\ &= 1.2571\ \text{pu} \end{aligned}$$

Since there is no load at bus 2, $Q_{G2} = Q_2$.

Answer: $|V_3| = 0.9614$ pu, $\theta_3 = -8.49^\circ$, $Q_{G2} = 1.2571$ pu. (Slack bus: $P_{G1} = 1.8339$ pu, $Q_{G1} = 0.0597$ pu.)