Q at PV bus and initial Jacobian

Bus 1 is the slack bus ($1\angle 0^\circ$), bus 2 a PV bus ($P_2 = 4.0$ pu, $|V_2| = 1.05$ pu) and bus 3 a PQ bus ($P_3 = -5$, $Q_3 = -4$ pu). Initial guesses: $V_2^{0} = 1.05\angle 0^\circ$, $V_3^{0} = 1\angle 0^\circ$.

Y-bus

With $y_{12} = -j40$, $y_{13} = -j20$, $y_{23} = -j20$:

$$Y_{bus} = \begin{bmatrix} -j60.0000 & j40.0000 & j20.0000 \\ j40.0000 & -j60.0000 & j20.0000 \\ j20.0000 & j20.0000 & -j40.0000 \end{bmatrix} \Rightarrow |Y_{11}| = |Y_{22}| = 60,\ |Y_{33}| = 40\ (\theta = -90^\circ),\ |Y_{12}| = 40,\ |Y_{13}| = |Y_{23}| = 20\ (\theta = 90^\circ)$$

i) Reactive power at bus 2

$$\begin{aligned} Q_2 &= -\sum_{j=1}^{3}|V_2||V_j||Y_{2j}|\sin(\theta_{2j} - \delta_2 + \delta_j) \\ &= -\left[(1.05)(1)(40)\sin 90^\circ + (1.05)^2(60)\sin(-90^\circ) + (1.05)(1)(20)\sin 90^\circ\right] \\ &= -\left[42 - 66.15 + 21\right] = 3.1500\ \text{pu} \end{aligned}$$

Since $Q_{L2} = 0$, $Q_{G2} = 3.1500$ pu.

ii) Jacobian for the first iteration

Unknowns $\delta_2$, $\delta_3$, $|V_3|$; equations $\Delta P_2$, $\Delta P_3$, $\Delta Q_3$.

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, $j \ne i$):

$$\begin{aligned} \frac{\partial P_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial P_i}{\partial |V_i|} &= 2|V_i||Y_{ii}|\cos\theta_{ii} + \sum_{j\ne i} |V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial P_i}{\partial |V_j|} &= |V_i||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_i} &= \sum_{j\ne i} |V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial \delta_j} &= -|V_i||V_j||Y_{ij}|\cos\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_i|} &= -2|V_i||Y_{ii}|\sin\theta_{ii} - \sum_{j\ne i} |V_j||Y_{ij}|\sin\phi_{ij} \\ \frac{\partial Q_i}{\partial |V_j|} &= -|V_i||Y_{ij}|\sin\phi_{ij} \end{aligned}$$ $$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0500)(1.0000)(40.0000)\sin(90.00^\circ) + (1.0500)(1.0000)(20.0000)\sin(90.00^\circ) = 63.0000 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0500)(1.0000)(20.0000)\sin(90.00^\circ) = -21.0000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0500)(20.0000)\sin(90.00^\circ) = -21.0000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(20.0000)\sin(90.00^\circ) + (1.0000)(1.0500)(20.0000)\sin(90.00^\circ) = 41.0000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(40.0000)\cos(-90.00^\circ) + (1.0000)(20.0000)\cos(90.00^\circ) + (1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(20.0000)\cos(90.00^\circ) + (1.0000)(1.0500)(20.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(40.0000)\sin(-90.00^\circ) - (1.0000)(20.0000)\sin(90.00^\circ) - (1.0500)(20.0000)\sin(90.00^\circ) = 39.0000 \end{aligned}$$ $$[J^{0}] = \begin{bmatrix} \frac{\partial P_2}{\partial \delta_2} & \frac{\partial P_2}{\partial \delta_3} & \frac{\partial P_2}{\partial |V_3|} \\ \frac{\partial P_3}{\partial \delta_2} & \frac{\partial P_3}{\partial \delta_3} & \frac{\partial P_3}{\partial |V_3|} \\ \frac{\partial Q_3}{\partial \delta_2} & \frac{\partial Q_3}{\partial \delta_3} & \frac{\partial Q_3}{\partial |V_3|} \end{bmatrix} = \begin{bmatrix} 63 & -21 & 0 \\ -21 & 41 & 0 \\ 0 & 0 & 39 \end{bmatrix}$$

The off-diagonal blocks are zero at the flat start because the network has no resistance and all angles are zero.

For use in the first iteration the mismatch vector is $[\Delta P_2,\ \Delta P_3,\ \Delta Q_3]^T = [4.00,\ -5.00,\ -3.00]^T$ (since $P_2^0 = P_3^0 = 0$ and $Q_3^0 = -1.0000$).

Answer: $Q_2 = 3.1500$ pu; $J^{0} = \begin{bmatrix} 63 & -21 & 0 \\ -21 & 41 & 0 \\ 0 & 0 & 39 \end{bmatrix}$.