N-R Jacobian and first iteration (new course)

Bus 1 is the slack bus ($1\angle 0^\circ$), bus 2 a PV bus ($P_2 = 0.6$ pu, $|V_2| = 1$ pu) and bus 3 a PQ bus ($P_3 = -0.5$, $Q_3 = -0.4$ pu). Flat start: $V_2^0 = V_3^0 = 1\angle 0^\circ$. The line lengths are not needed since $Y_{bus}$ is given: $|Y_{ii}| = 12\angle -90^\circ$, $|Y_{ij}| = 6\angle 90^\circ$.

(i) Jacobian matrix

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 every $\phi_{ij} = \pm 90^\circ$, so all $\cos$ terms vanish:

$$\begin{aligned} \frac{\partial P_2}{\partial \delta_2} &= (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) + (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = 12.0000 \\ \frac{\partial P_2}{\partial \delta_3} &= -(1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = -6.0000 \\ \frac{\partial P_2}{\partial |V_3|} &= (1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial P_3}{\partial \delta_2} &= -(1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = -6.0000 \\ \frac{\partial P_3}{\partial \delta_3} &= (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) + (1.0000)(1.0000)(6.0000)\sin(90.00^\circ) = 12.0000 \\ \frac{\partial P_3}{\partial |V_3|} &= 2(1.0000)(12.0000)\cos(-90.00^\circ) + (1.0000)(6.0000)\cos(90.00^\circ) + (1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_2} &= -(1.0000)(1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial \delta_3} &= (1.0000)(1.0000)(6.0000)\cos(90.00^\circ) + (1.0000)(1.0000)(6.0000)\cos(90.00^\circ) = 0.0000 \\ \frac{\partial Q_3}{\partial |V_3|} &= -2(1.0000)(12.0000)\sin(-90.00^\circ) - (1.0000)(6.0000)\sin(90.00^\circ) - (1.0000)(6.0000)\sin(90.00^\circ) = 12.0000 \end{aligned}$$ $$[J^{0}] = \begin{bmatrix} 12 & -6 & 0 \\ -6 & 12 & 0 \\ 0 & 0 & 12 \end{bmatrix}$$

(ii) Unknowns after the first iteration

Calculated powers at the flat start: $P_2^0 = P_3^0 = 0$ and $Q_3^0 = -[6 - 12 + 6] = 0$. Mismatches:

$$\begin{aligned} \Delta P_2 &= 0.6 - 0 = 0.6 \\ \Delta P_3 &= -0.5 - 0 = -0.5 \\ \Delta Q_3 &= -0.4 - 0 = -0.4 \end{aligned}$$ $$\begin{bmatrix} 0.6 \\ -0.5 \\ -0.4 \end{bmatrix} = \begin{bmatrix} 12 & -6 & 0 \\ -6 & 12 & 0 \\ 0 & 0 & 12 \end{bmatrix}\begin{bmatrix} \Delta\delta_2 \\ \Delta\delta_3 \\ \Delta|V_3| \end{bmatrix}$$

From the first two rows: $12\Delta\delta_2 - 6\Delta\delta_3 = 0.6$ and $-6\Delta\delta_2 + 12\Delta\delta_3 = -0.5$, giving

$$\begin{aligned} \Delta\delta_2 &= \frac{0.6(12) + 6(-0.5)}{144 - 36} = \frac{4.2}{108} = 0.0389\ \text{rad} = 2.228^\circ \\ \Delta\delta_3 &= \frac{12(-0.5) + 6(0.6)}{108} = \frac{-2.4}{108} = -0.0222\ \text{rad} = -1.273^\circ \\ \Delta|V_3| &= -0.4/12 = -0.0333\ \text{pu} \end{aligned}$$

So $V_2^{1} = 1.0000\angle 2.228^\circ$ pu and $V_3^{1} = 0.9667\angle -1.273^\circ$ pu.

With these voltages ($P_i - jQ_i = V_i^{*}\sum_k Y_{ik}V_k$):

$$\begin{aligned} Q_2 &= 0.2154\ \text{pu} \\ P_1 &= -0.1044\ \text{pu}, \quad Q_1 = 0.2060\ \text{pu} \end{aligned}$$

Answer: $J^0$ as above; after one iteration $\delta_2 = 2.228^\circ$, $|V_3| = 0.9667$ pu, $\delta_3 = -1.273^\circ$, $Q_2 = 0.2154$ pu, $S_1 = -0.1044 + j0.2060$ pu.